Model Engine Maker

General Category => Chatterbox => Topic started by: MJM460 on May 11, 2017, 12:40:10 PM

Title: Talking Thermodynamics
Post by: MJM460 on May 11, 2017, 12:40:10 PM
I suggested starting this thread in response to a little side track on Chris's Lombard Hauler thread.  It seemed to spark some interest so here goes.  I am sure there are others with some knowledge of thermodynamics, so feel free to jump in to the discussion, especially if you disagree.  The thoughts here are offered in the spirit of sharing knowledge, rather than wasting it, and enhancing our understanding of the marvellous machines we make.

First I will try and address the questions already raised while they are fresh, then I will post again the information in Chris's thread to keep it all together.  Then we will see where it leads.

Flyboy Jim, pressure does work on the gas in a piston when the piston moves.  There are two basic cases.  In a cylinder with the inlet valve open, with adequate inlet piping etc so that the pressure remains constant.  Then work (W) = Pressure (P) times volume change (v).  Note that it does not matter whether the gas is air or steam or any other gas.  A pressure of 15 psi exerts the same force on the piston what ever gas is involved.

The second case is a closed piston, i.e. The inlet and exhaust ports are both closed.  Then when the piston moves, the volume increases, (expansion) and this results in a falling pressure.  The formula for work done is more complex and involves the inlet pressure, the pressure ratio and a factor which for an ideal gas and perfect engine turns out to involve the specific heat of the gas.  This factor is slightly different for air  and steam.  The maths is definitely not mental arithmetic, but easy enough with a scientific calculator or a spreadsheet such as excel.  However the result is only a 2% difference which is not significant compared with other factors such as friction and heat gain or loss.  We all know a steam cylinder loses heat, lagging is applied to reduce this heat loss which reduces the steam pressure faster than expansion alone. 

I mention the possibility of heat gain because when the gas expands, it's temperature falls.  Steam is still hot compared with the atmosphere, but if air starts at atmospheric temperature and then gets cooler, then the cylinder gains heat, and the pressure falls less quickly than due to expansion alone.  I have seem the exhaust pipe of a small industrial turbine ice up when the turbine was run on air instead of steam.

RonGinger, the expansion of water to steam is indeed around 1700:1 however that expansion occurs in the boiler.  We are talking about expansion of the gas in the cylinder, and only from the point where the inlet valve closes, for example around 50% of stroke, depending on the notching of the valve gear, to the point where the exhaust valve opens, ideally this occurs very close to the bottom of the stroke, and used for the example, but may be quite different in a real engine.  It turns out that the actual cut off point makes little difference.  My 2:1 is based on expansion in the cylinder from about 50% stroke to bottom dead centre. 

Jo, I cannot argue with your observation.  In my experience, when observation appears contradictory to theory, it usually means that there are more than one thing happening, and the one with the biggest influence may not be the suggested theory.  It is then important to look more closely at what is going on.  For example, the heat gain or loss mentioned above, or thermal expansion at steam temperatures might be changing clearances.  Could you please tell us more about the differences you see?  The point is often mentioned and there will be a lot of learning in that for us all.

Paul, your question is the big one that brings us to the purpose of this thread, how do they really work?  This post is already too long, I will start another.

MJM460
Title: Re: Talking Thermodynamics
Post by: Jo on May 11, 2017, 12:54:42 PM
Lost me  :noidea:, lets see if I can explain things in one sentence:

When gas expands it gets colder - which chills the container (cylinder) so when using air at room temperature to run an engine the result is that the engine will be chilled and things will get tighter/harder to turn over, if you use steam it will warm the engine and things will expand, making it easier to turn over the engine.

So if you want to wear out your steam engine fast - run it on compressed air   ::).

Jo
Title: Re: Talking Thermodynamics
Post by: Jasonb on May 11, 2017, 01:12:11 PM
Jo, won't the piston also see a similar temperature change to the cylinder and the clearances stay the same assuming similar materials for both.
Title: Re: Talking Thermodynamics
Post by: Gas_mantle on May 11, 2017, 01:19:42 PM

When gas expands it gets colder - which chills the container (cylinder) so when using air at room temperature to run an engine the result is that the engine will be chilled and things will get tighter/harder to turn over

So if you want to wear out your steam engine fast - run it on compressed air   ::).

Jo

Surely though by using air it won't be at room temperature as it heats up during compression, the subsequent heat lost after expansion in the cylinder is simply returning things back to room temperature isn't it ?

I'm no expert but I'd expect the cylinder to remain at room temperature (albeit with other factors like friction affecting the result)

Title: Re: Talking Thermodynamics
Post by: Jo on May 11, 2017, 01:53:33 PM
Ok I tried to make it too simple  :facepalm2: back to the theory....

Steam contains far more energy than air at the same pressure, this pressure is a symptom of the available heat energy in the operating gas. Were the gas is admitted into a cylinder through the whole of the stoke there would be very little difference in performance between steam or air (or even water). As soon as you cut off the supply stroke the gas expands while delivering work, so it draws on its heat content to provide the energy to do that work.

When compressed air expands the only heat it can acquire comes from a fall in its own temperature (which causes a reduction in specific volume) and from the cylinder walls. But when steam expands it can draw all the energy from within, and that reservoir of energy is considerable as it also includes the latent heat of evaporation of the water and if the spent steam is still a gas when it exits the engine this unused energy will have also warmed the engine.

So what does this mean in practise?

If you put your hand on a steam engine running on air you will notice that all of the engine will have over time cooled down, i.e. bearings, fits, covers/cylinders, piston etc. this will all have reduced clearances, in this situation lubrication is critical. If you put your hand on a Steam engine you will find it quickly heats up, this means the clearances are increasing, lubrication is still important but not as critical as running on air.

For this reason if you intend on running an engine on air rather than steam you alter the clearances (and the valve timing  ::)) to take it into account.

Jo
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 11, 2017, 02:11:25 PM
Here is a link to the H.K. Porter book "Modern Compressed Air Locomotives" it has a comparison of air and steam locomotives.

The storage tanks held high pressure air much higher than a steam locomotive. A reducing valve was used to reduce the pressure to 150 psi.

https://books.google.ca/books?id=uIslAAAAMAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

Dan
Title: Re: Talking Thermodynamics
Post by: Gas_mantle on May 11, 2017, 02:13:18 PM
Ok I tried to make it too simple  :facepalm2: back to the theory....


When compressed air expands the only heat it can acquire comes from a fall in its own temperature (which causes a reduction in specific volume) and from the cylinder walls.

Yes but the point I'm making is that by compressing air in the first place it becomes higher than room temperature,, once it expands (and therefore cools) within the cylinder it is returning back to room temperature but not lower.

I'd argue the energy you are using is obtained by first compressing the air (and heating it) then releasing it in the engine cylinder to expand back to atmospheric pressure and return to room temperature.



Title: Re: Talking Thermodynamics
Post by: steam guy willy on May 11, 2017, 02:14:19 PM
Although we are always looking for efficiency in our running costs, what about using Mercury instead of water air !!! The Americans did it apparently !!!
Title: Re: Talking Thermodynamics
Post by: Jasonb on May 11, 2017, 02:21:05 PM
Although an engine run on air will cool slightly the difference is generally far less than the heat gain of one run on steam. If we take ambient air temp as 20deg C a steam engine couled get upto over 100deg C during running depending on teh steam pressure and themp but I've yet to see one run on air get below freezing so the amount of contraction is minimal in practice and really not worth worrying about.

Also the bearings are usually quite a way from the cylinder so they will get even less of a temp change
Title: Re: Talking Thermodynamics
Post by: Jo on May 11, 2017, 02:36:53 PM
Although an engine run on air will cool slightly the difference is generally far less than the heat gain of one run on steam. If we take ambient air temp as 20deg C a steam engine couled get upto over 100deg C during running depending on teh steam pressure and themp but I've yet to see one run on air get below freezing so the amount of contraction is minimal in practice and really not worth worrying about.

Also the bearings are usually quite a way from the cylinder so they will get even less of a temp change

Hepolite recommended a difference of 0.05mm per 25mm ( 0.002" per inch ::) ) of bore difference for the gap on their rings between the two applications.

If you have ever seen someone run a double or triple expansion engine on air for a few hours you will notice that the LP cylinder covers begin to ice up. Which is why last year the guy at the GMES show was running his Triple on an electric motor  ;)

Jo
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 11, 2017, 02:52:55 PM
Jo,
The way Porter solved the problem of 2 stage air motors was to have an atmospheric reheater. The cold air from the HP cylinder was sent through a long set of tubes so the atmosphere could reheat the air before the LP cylinder.

Dan
Title: Re: Talking Thermodynamics
Post by: Jasonb on May 11, 2017, 03:13:26 PM

Hepolite recommended a difference of 0.05mm per 25mm ( 0.002" per inch ::) ) of bore difference for the gap on their rings between the two applications.Jo

So thats 0.0006" per inch on diameter ( 0.002 / 3.142). So taking what I said above for say an 80degree heat gain there is 0.0005" increase when running on steam but only 0.0001" decrease when running on air if there were a 20deg drop per 1" diameter.

I also assume that Hepolite refer to engines doing work where the air pressure will need to be as high as the steam pressure so we would be looking at 115psi steam  to get our engine upto 100deg C but as we tend to play with our engines when running on air and I can easily run mine on 5psi or less there will be far less cooling at those low air pressures. maybe 10%.

So I'm of the opinion that 0.00001" change of diameter on a 1" bore piston/cylinder combination  or 0.000005" clearance is just  not worth worrying about in the practical word.

J

PS Have you noticed that all my "steam" engines run alloy pistons so if anything the piston/cylinder clerance will go up on mine when cooled not down ;)
Title: Re: Talking Thermodynamics
Post by: paul gough on May 11, 2017, 09:53:39 PM
With this discussion we need to be very clear as to what we are addressing in making a point and responding to it. It is very easy to digress, be misinterpreted, and to confound theoretical explication by imprecise language. Also we should be aware that sometimes theory and practise don't obviously match the circumstance. As an example; for the purposes of argument, lets assume a steam engine and a compressed air engine were in fact the same, the steam engine would still be a superior mechanism, (moving a mass from A to B), due to being 'self sustaining', utilising all the stored energy within the fuel being burnt and transferred to the steam, whereas the air engine has only the tanks capacity so somewhat limited. However a coal burning locomotive might not be first choice in a gunpowder factory and the air engine superior, for obvious reasons. So we have to make it clear what we are trying to get across. From what I gather, the original purpose of the investigation seems to be aimed at revealing what is precisely  going on in the cylinder when a gas of equal qualities acts on a piston. I should like to have this verified, so I don't get confused. I am an old man with a feeble mind and have to be careful before opening my mouth in ventures such as this. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Maryak on May 12, 2017, 12:37:48 AM
Adiabatic - Air Compressor theory, heat and its associated energy are removed. With the heat removed the gas has little expansion available to do work.

Isothermal - Steam Engine theory, heat and its associated energy are retained. With the heat retained expansion is available to do work.

Running on air a reciprocating engine benefits from cylinder oil lubrication.

Running on saturated steam a reciprocating engine is happy with the water in the steam as the cylinder lubricant. Boilers do not like mineral oils.

Simplified but I hope you get my drift.

Regards
Bob
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 12, 2017, 03:16:12 AM
I see some real potential for this thread. To add to what Paul said.............hopefully the knowledgeable folks will keep in mind that a lot of us are pretty uninformed when it comes to dealing with Thermodynamics.........so it's okay to talk down to us.  :)

Jim
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 12, 2017, 01:21:42 PM
Wow!  That is an amazing response.  Thank you to everyone who has responded and all those who were interested enough to read on the topic.  Talk about letting the genie out of the bottle.  I do hope we can continue the thread and not disappoint.

First,  Flyboy Jim, I hope that I don't talk down to anybody.  It is a strength of this forum and a tribute to all its members that it is really minimal.  If a point is not clear, please say so, this will keep the thread relevant and grounded in reality.  If I have lost Jo on the first go, then I have a long way to go in my explanations and need all the help I can get.

Some clarifications on some of the posts.  When air is compressed, it heats.   After the compressor, air tends to cool in the air tank and piping after the compressor.  In addition, if there is a pressure regulator, there will be cooling due to expansion after the regulator.  I for one find the air hose is not noticeably warm when I run an engine on low pressure though the compressor discharge is quite hot.  But it is ok to consider the air as initially warmer than atmospheric, and the exhaust temperature is then above or below atmospheric depending on the amount of work done and consequent cooling effect.

Please don't compare an air and steam locomotive yet, that can come later if you want, as the differences are all about the processes involved in delivering gas at pressure to the piston face.

An adiabatic process is one that occurs without heat flow to or from external sources, it does not mean without heat!  Isothermal process means constant temperature process.  As doing work involves cooling, heat must be added from an external source throughout the process to keep the temperature constant.

These engines are all heat engines.  To understand why the air engine is a heat engine we must identify the absolute zero of temperature, the point at which molecular motion stops.  In the metric system, the scale is called Kelvin, and zero is -273.15 deg C.  In the imperial system the scale is called Rankine and the same zero is -459.67 deg F.  So even in Siberia, air is relatively warm on the absolute temperature scale.

Paul has it nailed when he notes that the original purpose of this thread was to understand what is going on inside the cylinder, the point where the heat of the gas molecules interacts with mechanical components to do work.  At least I suggest that is the starting point.  We can then work back to explore the thermodynamics of the other parts of the system if the interest continues.  Everything else is there only to turn the chemical energy in the fuel to energetic motion of molecules and to deliver those molecules to the face of the piston with as little loss as possible.

 Jo has kindly passed on her observations on running engines on air.  I am relieved that the engines get noticeably cooler.  I was concerned that when running lightly loaded, the heat due to friction might mask the effect, as there is less temperature change when less work is done.

I definitely agree that a big factor in the difference between running on air and steam comes down to thermal expansion and changes in clearances.  It is not a simple exercise to determine the change in clearances at different temperatures.  If everything is at a uniform temperature, there is no change in clearance.  However the inside of the cylinder is in close contact with steam, and approaches steam temperature, while the outside is in contact with the atmosphere, and much cooler.  We reduce the temperature difference by lagging but the cooler outside expands less and constrains the expansion of the inside through internal stresses in the metal.  The piston is also in close contact with the steam, but has a heat loss path through the piston rod.  You can see the problem, which effect has the larger influence?  I will go with the practical observation.  If the engine loosens up on steam and tightens on air then that is the answer.

Obviously our displacement lubricators do not work with air, so other provisions have to be made.

I also agree that there is a big difference in the energy content of air and steam.  However, the real issue here is whether that energy difference makes a difference to the conversion of heat to work in the engine.  Our starting point was air and steam at the same pressure.

So back to what happens in the cylinder.  No matter how loud we sing the song that "heat is work and work is heat", (was it Flanders and Swan?), it is still amazing that our little engines can turn heat into work.  It is not so long ago that no one could even imagine that the heat we feel streaming from the sun by day and from our fires by night can be harnessed to replace horses, let alone drive mighty locomotives across continents, ships across oceans and even aeroplanes through the skies.  And all this happens at the face of the piston where the energy in the random motion of tiny molecules interacts with the face of the piston.   Surely that is worth trying to understand.

I hope that clears a few issues, or at least puts them on hold for a while.  This post is more than long enough so next time I will try and get back to what happens at the piston face, and the difference between engine output on air compared with steam.  Thank you for your interest.

MJM460
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 12, 2017, 03:21:10 PM
Good info. As someone who is at the bottom of the "understanding chain" I'm on board so far. I appreciate you taking this one small step at a time.

I wasn't sure what a displacement lubricator was so looked it up: https://en.wikipedia.org/wiki/Automatic_Lubricator

It's easy to understand the idea of steam or air providing the pressure to move the piston, but I'd never thought of it in terms of heat. It's been over 50 years since my college physics class and I'm thinking I may have forgotten a few things.  :Doh:

Jim
Title: Re: Talking Thermodynamics
Post by: paul gough on May 12, 2017, 11:18:10 PM
MJM, Please take the position as Tutor for Thermodynamics 101, your explanations are adequate and soothing to the minds of people who otherwise might have been scorched by even hearing the term uttered. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: paul gough on May 13, 2017, 12:59:54 AM
Practical question/proposition: Jo has mentioned the 'chilling' effect to cylinders and an acquaintance who resorted to an electric motor to run his compound engine due to icing up. From our discussions so far it would seem that a simple expedient of introducing some heat to the air supply prior to its entry to the engine might help overcome this problem and at least theoretically enhance the power of the engine. I have in mind something simple; a length of copper tube, perhaps with a twisted strip of thin sheet metal inserted  to prevent laminar flow and enhance the surface area for the air to impinge upon. Under this section of pipe a couple of small 'tea candles' could be situated thus superheating, (in a minor way), the air supply. Of course more sophisticated arrangements could be developed from this, but for a quick solution for an exhibitor such a contrivance might suffice. Comments eagerly sought. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Steamer5 on May 13, 2017, 01:57:27 AM
Hi Paul,
 Both my current & last job did this for dropping natural gas pressure to prevent hydrate formation, (for those that don't know hydrates are nasty little blighters, basically methane gas molecules are small enough to fit inside water molecules & when they turn to ice are more stable than ice itself, they are also hard to get ride of) last job we used a steam heated exchanger to heat the gas prior to dropping the pressure, current job use an electric "superheater" to do the same thing.
So pre heating the air prior to using it in an engine I think would work quite well. Probably not require in a single cylinder but a triple where the pressure is progressively dropped would help out.
Mind you another  other option could be to use drier air, then maybe the ice would be kept to the outside.

Like others I'm finding this thread very informative.

Cheers Kerrin
Title: Re: Talking Thermodynamics
Post by: paul gough on May 13, 2017, 04:58:46 AM
Thanks for the comment Kerrin, I have only ever used the stuff that burns you to run engines and was 'speculating in isolation'. Nice to know the idea has some application but also learning that there is only a need to apply it to multi-expansion engines. I understand methane hydrates have a very nasty potential to impact on our atmosphere if the vast quantity of them in the frigid parts of the Northern hemisphere ever go through a phase change. I now have a curiosity itch, is there any role in more advanced pneumatic power applications for heating the air and to what degree, but this is an extra curricular activity, we have enough in this thread with thermodynamics and its asides. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Jasonb on May 13, 2017, 07:40:31 AM
I think the issue of running the tripple on air is that the compressed air really only works 100% on the HP cylinder and by the time it has found its way to the LP cylinder it is actually being drawn through the engine due to the large volume displaced which expands (cools) it far more than you would get with a simple single or double high. I'm assuming he was running it as a compound and not using the modified pupework that acted a bit like a simpling valve.
Title: Re: Talking Thermodynamics
Post by: Zephyrin on May 13, 2017, 10:33:08 AM
Little compressed air engine for model airplane were common before diesel or glow engines, they become ice cold in 1 min or so and frozen in 2 (for those able to run that long!), the same holds true for CO2 engine, based on the same principle...

obiously a steam engine that run on compressed air requires a mechanical lubricator, and not the condensing type of oiler, and also the appropriate oil.

I personaly spend a lot of energy (heat ?) to run my engine with steam...just because its so more appealing, more efficient, much softer and far less mechanical ratling noise, compressed air is second-best.
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 13, 2017, 01:19:17 PM
A time for humble pie.

Thank you to Paul and Jim for your kind comments. 

Preheating the air will raise the outlet temperature and potentially prevent ice formation.  The ratio of absolute temperatures will be the same for the same amount of work, so an inlet temperature of say 20 deg will result in less than 20 deg rise in the exhaust temperature.

It sounds like Kerrin has worked in a gas plant somewhere.  North or South Island? Hydrates are a real problem there, but are not likely in our models.  However we can get ice inside our models if the air is  not sufficiently dry.  Fortunately it melts when things warm up.  Ice on the outside only indicates internal temperature below zero C.

I started this thread with a view to creating a "build log" of a knowledge base of thermodynamics as it applies to understanding, designing and running our engines.  I hope that this will add depth to our enjoyment of our hobby. 

I am not claiming to be an expert, which is just as well as you will soon see.  I am hoping to learn as well as to contribute.  So first some humble pie!  I started by trying to understand the oft stated assertions that steam produces so much more work than air, and that expansion (expansive) work is so wonderful, but there is never an explanation.

Turned out to be not such a simple problem, particularly the expansive work part.

Certainly while the inlet valve is open there is agreement that steam, air or even water makes no difference, though it is better that we keep to compressible gases.  However, once the inlet valve closes, until the exhaust valve opens, the situation is quite different.  I also want to get back to the third stage of the power stroke, when the exhaust valve is open.  For air adiabatic expansion can be calculated using the ideal gas law.  For steam, this assumption is not accurate enough to be useful hence the variation of pressure with volume is not known, and the work output cannot be calculated.

Thanks to Jo for her comment on the energy content of steam that forced me to put aside the reversible process and ideal gas assumptions and to take a detailed look at the process for steam.

It has been a long day of heavy reading, detailed maths and all those abra-ca-dabras, enthalpy, entropy irreversibility, along with the first and second laws of thermodynamics.  I suspect that those who would be interested in the detail are quite capable of doing it themselves, and I don't want to turn off the others, so here is the short version.

For the expansion phase, the inlet and exhaust valves are both closed.  The cylinder pressure reduces as the piston continues to move.  Thus the work done in this part of the stroke is less than it would be if the inlet valve remained open.  However fuel consumption is efficiently reduced if the maximum power is not required.  The issue is that the way the pressure varies as the volume increases is not known for steam, so the work output cannot be easily calculated.

Fortunately the work done can be calculated using steam tables and the first and second laws.  The summary is that a given volume of steam does around four times the work done by the same volume of air.  So the old guys were right, though only for the expansion phase, I just never understood why.

There are offsetting factors that reduce the difference.  First, if our lagging is not perfect, the heat loss causes additional steam condensation hence further reducing the pressure and reducing the work output.  Second, the same processes that produce the extra work during expansion probably mean more loss during the exhaust stroke.  I am a bit cautious on that one as I have not yet done the maths.  Third, it is becoming apparent to me that the practicalities of our most common valve gears mean that the expansion phase is probably not a big part of the total work output, at least for a single cylinder engine, thus lessening the impact.

We need someone to document construction of a test stand and measuring the actual work output of an engine and comparing the results for air and steam.  Practical observation is the best way to determine issues where several complex processes occur at the same time.

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on May 13, 2017, 11:38:48 PM
 MJM you state;" The summary is that a given volume of steam does around four times the work done by the same volume of air. "

As the above quote can be taken as a general statement of outcomes. I would like to be sure, if I repeat this statement, that I cannot be challenged because of some point I am unaware of. Therefore does the statement hold in a broad range of circumstances or are there  constraints and caveats. Sorry if I appear pedantic but I really want to KNOW. I find thermodynamics and its associate, fluid dynamics, fascinating phenomena, so appreciate the lengths and effort you are putting in to get information across in an intelligible form. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 14, 2017, 03:07:57 AM
I'm finding out that in order to really understand what's going on with these engines I needed to go back and re-learn the basics. I googled the topic of "Thermodynamics for Dummies" but what I found was way too complicated for this dummy.  :Doh: Then I found a site called Physics4Kids: http://www.physics4kids.com/files/thermo_intro.html Now this is more like it!  :whoohoo: I think this will work. For a while there I was afraid I might have to google "Thermodynamics for Kindergartners"!  :facepalm2:

Jim
Title: Re: Talking Thermodynamics
Post by: Steamer5 on May 14, 2017, 04:58:10 AM
Hi MJM,
 Yes you are right current job is in a gas plant, North Island. Currently there are none in the South.
Previous job in petrochemical. Can't recall an issue with hydrates in the gas heater, but we were always on the the watch for issues. Learnt about them since & they are right little beggars given half a chance!

This thread is getting better & better! Jims link should make some fun reading!

Not totally on track, as we are discussing steam piston engines, but if we throw  steam turbines into the mix then they get some 30% of there power by taking the steam pressure down to a reasonable vacuum, makes one think what would happen if you could do that on a piston engine.....

Cheers Kerrin
Title: Re: Talking Thermodynamics
Post by: paul gough on May 14, 2017, 05:22:46 AM
Kerrin, Air Pumps (vacuum pumps) on condensers in ships and land installations with reciprocators did this to various degrees, pretty much essential with triple and quadruple expansion if you want to get something out of the low pressure cylinders and balance their output with the high and intermediate cylinders. Regards Paul.
Title: Re: Talking Thermodynamics
Post by: 10KPete on May 14, 2017, 05:49:16 AM
I'm not sure this will provide any illumination, but here is a document I've learned a lot from. The link is to my DropBox.

https://www.dropbox.com/s/sc9zcjtsbc6i28g/Edwards%20Wet%20Air%20pump%20copy.pdf?dl=0

Pete
Title: Re: Talking Thermodynamics
Post by: Steamer5 on May 14, 2017, 06:29:05 AM
Hi Paul,
 Thanks for the info! The grey cell got a jog & I seem to remember reading some thing about that in the dim distant past! Was think more along the lines of a loco.......they probably tried it somewhere....oh yeah think I read of it in Africa were water is in short supply!
We used ejectors for the big turbines, one was an extraction / condensing turbine.....it gulped up 160 tonnes / hr of 100 bar steam, after 7 blades, spat out, about 150 tonnes/hr  24 bar steam, & took the rest to vacuum.

Cheers Kerrin
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 14, 2017, 12:20:11 PM
Ideal Gas Laws

Great to see the continuing interests in this topic.  Thank you to all those who have contributed, as well as those who are just following along.

I will address some of the comments later, but first, thank you Paul for highlighting the point that there may be qualifications I should have included in my summary.  So no, better not too quote the initial result without qualification.  The result is really only valid for the conditions assumed for the calculation.  First it is for the expansion part of the stroke only, when both inlet and exhaust valves are closed.  I assumed a start pressure for each calculation of approximately 15 psi as suggested in an earlier thread.  I have worked in metric units so I have made small adjustments and used the values for the nearest value listed in my steam tables to minimise interpolation.  I have also assumed an adiabatic process, with everything up or down to  a steady temperature and no heat gain or loss to the atmosphere.

For air I assumed 17 deg C, on reflection 27 might have been a better choice.  Again to use table entries directly.  Obviously, steam at 17 deg C would be fully condensed, so I used 150 deg C.  This temperature would be reasonable if steam was generated in a boiler at 50 psi, and throttled to 15 psi then some heat loss in the delivery pipe.  Not identical conditions, but attempting to be realistic.  After expansion (2:1), steam tables indicate the steam will be condensing at about 95 deg C.

It is this proximity to condensing that is the reason for a large part of the departure from ideal gas laws, and when condensing starts the ideal gas law does not apply at all.  The heat released in condensing causes the pressure to be much higher than predicted by the ideal gas laws, and consequently, the extra work.  I must admit to never having thought of it in this detail before.  Thank you Jo for pointing me in this direction.  I look forward to learning what other surprises you have in store for us.  I have learned heaps.

You might ask what if I had assumed hotter steam so that it did not condense.  I believe as you get further away from the condensing area, the behaviour gets closer to the ideal gas law, but I expect the departure would be quite significant until the temperature gets quite high, dangerous and rather impractical for our purpose.  I have not done further calculations as they are rather tedious without a specialised computer program, even with a calculator.  I am trying to keep to the issues that potentially have practical application in our model building, and have not spent time in purely theoretical explorations.

The big deal about ideal gas laws?  I probably did not make this clear enough.  When we try to calculate the work done by an expanding fluid, the only things we really know are the starting pressure, temperature, and density of the fluid, and we carry the calculation to the end point at which we know only the volume and fluid composition.  Note we do not know what the pressure and temperature will be.  Early work on this problem (Boyle and Charles are two names that come to mind) gave rise to the ideal gas law which was found to be closely approximated by many gases.  Without this gas law, we do not know how the pressure varies throughout the expansion and cannot calculate the work done.. 

When the ideal gas law does not apply, we basically have to rely on empirical data, such as that contained in steam tables which contain the results of huge amounts of experimental work.  For air there are tables of standard integrals which account for the much smaller but real departure from ideal gas.  Both steam tables and standard integrals for air were used in my calculation, however we should keep in mind that the number only applies for the assumed start conditions. 

Let me try if this summary stands up: While the inlet valve is open enough that the pressure is essentially constant, there is no significant difference in air or steam or carbon dioxide or other gases, however, if our valve gear has both inlet and exhaust valves closed so the work is done by gas expansion, then steam will indeed produce more work than air at the same pressure.  We can finally agree with the old timers on that one.

Kerrin, I had a short assignment at Kapuni some years ago.  Beautiful country in the shadow of Mt Teranaki.  And black sand on the beaches.  Please keep the references to useful sources coming in.  We will talk about condensers and other equipment a bit later, but next time I plan to try and make progress where we started, on the cylinder.

MJM460
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 14, 2017, 02:36:17 PM
James Watt invented the steam engine and most writers believe he also invented the steam engine indicator. The second invention is not as widely known because he kept that one secret to keep ahead of the competition.

With a steam engine indicator, we have a PV curve for the full stroke of any piston engine. I have taken indicator cards on a 900mm bore slow speed diesel engine.

The reason I linked this thread to the Porter air locomotives is that it has PV diagrams for air engines. There are lots of sources of PV diagrams for steam engines but they are a lot harder to find for air engines.

I for one wish I could look at one of the historic diagrams and be able to say that was good or that was bad, but my study of the diagrams has not progressed beyond getting Peabody's book on the subject. (Cecil Peabody was the head of Marine Engineering at MIT)

Here is an article on the invention of the steam engine indicator and it has a list of sources on the last page.
http://www.farmcollector.com/steam-traction/story-steam-engine-indicator?pageid=1#PageContent1

Dan
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 14, 2017, 06:41:10 PM
Here is Figure 4: Idealized indicator diagram, from the article linked in the last post.
Illustration by Bruce E. Babcock

(http://www.7-8ths.info/gallery/6/213-140517103607.jpeg)

Dan
Title: Re: Talking Thermodynamics
Post by: derekwarner on May 15, 2017, 03:12:31 AM
I agree understanding what is happening with/within our model steam plants is essential

To achieve this an inexpensive digital laser pyrometer + a small digital [load cell] weight scale and a laser tachometer ...all for approx. $60.00 is inexpensive...otherwise all we can say is my fingers are burnt :cussing:.....& a few pressure gauges installed within the plant + your wrist watch

The specification for my gas tank is 105gm capacity
The net weight is 398gm
With the gas canister @ 26 degrees C, I can achieve a gross weight of 501gm......or 103gm of gas transferred  :ThumbsUp:

The gas discharge pressure from the tank during this example test was 40 PSI, however the gas tank temperature after the fill was ~~13 degrees C
The known volume of water 600ml after 6 minutes boiled and achieved 135 degrees C at the boiler discharge isolation valve [outer casing shell]
This same 135 degrees C indicates as 45 PSI [~~3 Bar] which is the boiler relief valve set point
[from tables, the same 45 PSI actually requires 143.7xx degrees C]

So from here, engine running times/speeds, exhaust steam temperatures from the engine.....to the de-oiler, and the resultant volume/weight of exhaust water can be measured

I am sure many of our members completed Applied Heat I, II & III in theoretical and steam practical work tests during our training in earlier days  :old: , and remembering the principals are important, however from an aspect of these engineering marvels...it is the understanding what is happening within our model steam plant is the real question

Derek

PS 1......Mixed units of Measurement .....a point of explanation is whilst my original training was under the Imperial system, my later and chosen system is SI......however to this day in Australia, the majority of model steam gauges [Miniature Steam Gauges UK] are marketed in Imperial PSI

When monitoring gas pressure, the indication of 40 PSI as a whole number in a visual report, stating that this pressure as an assumed 2.72 Bar from a gauge that has an accuracy of +/- 5% on FSD and reported as such is not appropriate   :disagree:

PS 2......steam table reference added 16/5
 
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 15, 2017, 09:46:48 AM
Indicator Diagrams

Thanks to Dan and Derek for interesting and informative posts.  I will follow up on testing along the lines of Derek's post when we look at the whole plant.  In the meantime, please collect lots of data in preparation.

Dan's post of an indicator diagram is right on cue and right on topic, so I will use that to round out this initial discussion.  For those unfamiliar with this diagram I will try and talk you through it as an indicator diagram provides a pictorial representation of what actually happens in the engine as opposed to prediction by theory, that necessary empirical data.

So we are all on the same page I will assume we are talking about the top face of the piston in a vertical engine, cylinder above the crankshaft.  The indicator diagram is a graphical representation of the variation of pressure within the engine throughout the cycle.  Like Dan, I am not able to tell whether a real diagram is good or bad, but I will try and talk us through what it tells us. 

The horizontal scale is the volume of the cylinder.  The right hand side is bottom dead centre, and the left side is top dead centre.  The part of the scale to the left of the diagram to the zero on the volume scale is the clearance volume when the piston is at top dead centre. 

The vertical axis is the pressure measurement.  The cycle progresses in the direction of the arrows on the loop on the inside of the actual diagram, clockwise.  When the inlet valve opens, the pressure rapidly rises to the steam supply pressure.  The top boundary of the diagram is the power stroke.  The pressure is roughly constant while the piston moves down with the inlet valve still open.  For this part of the cycle, work done by the gas is simply pressure times volume change, regardless of the gas used.

Note where the inlet valve closes.  After this is the expansion period.  With the inlet valve closed, our inlet pressure gauge no longer tells us anything useful about the pressure in the cylinder.  The curve then shows how the pressure might fall as the expansion proceeds.  Again the work done by the gas is the pressure times volume change, but as the pressure is changing, we have to break up the volume change into bits so small that we can consider the pressure constant during each small volume change.  We then add up all work done by the gas during each bit of volume change for the total.  This process is called integration.  Simple when it is said quickly, but we don't actually know the pressure at any point on the curve once the inlet valve is closed, so we can't do the sums.  The shape of the curve as mentioned in earlier posts, depends on the gas composition.  The indicator diagram actually measures the pressure, so solves the problem.

The indicator diagram is produced by an instrument that gets the piston position from the cross head, and has a pressure tapping into the cylinder, and prints out the graph.  This diagram was described as idealised, meaning the instrument was imaginary, and the diagram is to show the concepts.  In Dan's case, the instrument was real and actually measured the pressure throughout the stroke.  The instrument is delicate, expensive and not many of us have access to one.  For various practical reasons, they are only available for slow speed machines.  I know of ones that were made for 400 rpm machines but I do not know the current state of the technology.

The diagram shown is typical of the pressure path for an ideal gas, and the shape is also similar for steam, as enough indicator diagrams have been made on real machines for us to know the general shape.  If we had a diagram for air drawn on top of the steam diagram, we would find that the diagram for steam shows higher pressure than the air diagram throughout the stroke, so that the area under the steam curve is greater than the area under the air curve, meaning more work done by the gas.

What can we learn from the diagram?  First, we can see the work done during expansion is always less than if the inlet valve remained open.  Second, we use less steam and fuel if we have some expansion.  Third, while steam does more work during expansion than air, how big a proportion of the total work for the stroke depends on just where the valve closes, later closing, less proportional difference.  Finally, we need to look at the exhaust stroke before we can understand the lower boundary of the curve.

Looking at the curves, while the calculated end pressure for steam was about 15% higher than the end pressure for air, it is hard to conceive a reasonable pressure curve for steam that would support a large difference in work done, but no doubt that steam would produce more work than air during expansion.  It is likely that the adiabatic process assumption, no heat gain or loss, is responsible for reducing the difference. There is little doubt that the steam engine looses heat to the atmosphere.  This reduces the pressure and hence the work done by steam.  So yes, steam produces more work than air, but the size of the difference is not yet determined.

MJM460
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 15, 2017, 02:59:43 PM
Good explanation.

Looking at the diagram begs the question: If the inlet stayed open longer, keeping the pressure in the cylinder higher for longer, wouldn't there be more work (power) produced during the stroke. I'm sure it has something to do with reaching a "point of diminishing returns", but not sure where that would be.

Jim
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 15, 2017, 04:12:51 PM
Jim,
The term used for the point in the stroke that the valve gear closes the inlet valve in a steam engine is known as cut off.

This is expressed as a percent of piston stroke, so 100% cut off means that the valve stays open the full length of the stroke and is the max power for a given cylinder.

Looking at the diagram in post 32 by eye the cut off point is less that 50% of the stroke it is about 30% of the stroke which gives good economy.

Steam locomotives have a cut off of around 85% for full gear. The dotted line in figure 4 shows the effect of hooking up or notching back the valve gear. The cut off is chosen by the designer for the type of work the steam engine will be doing.

Further reading:
https://en.wikipedia.org/wiki/Cutoff_(steam_engine)

Dan
Title: Re: Talking Thermodynamics
Post by: PStechPaul on May 15, 2017, 08:48:14 PM
This is an interesting discussion, and I'm following to some small extent. I wonder how much effect there is based on the sinusoidal motion of the piston due it being connected to a flywheel, and the amount of torque applied based on angular position. There would seem to be some considerable "torque ripple" due to the changing pressure on the piston and its contribution to rotary torque of the flywheel through the length of the stroke. Also perhaps some sort of proportional inlet valve could adjust the volume of steam entering the cylinder to correspond to the position of the piston, and the volume of the chamber.
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 16, 2017, 03:08:05 AM
Jim,
The term used for the point in the stroke that the valve gear closes the inlet valve in a steam engine is known as cut off.

This is expressed as a percent of piston stroke, so 100% cut off means that the valve stays open the full length of the stroke and is the max power for a given cylinder.

Looking at the diagram in post 32 by eye the cut off point is less that 50% of the stroke it is about 30% of the stroke which gives good economy.

Steam locomotives have a cut off of around 85% for full gear. The dotted line in figure 4 shows the effect of hooking up or notching back the valve gear. The cut off is chosen by the designer for the type of work the steam engine will be doing.

Further reading:
https://en.wikipedia.org/wiki/Cutoff_(steam_engine)

Dan

Thanks Dan. I've heard the term "cutoff" before but didn't know exactly what it was. That helped a couple more pieces fall into place.

Jim
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 16, 2017, 12:46:07 PM
The Exhaust Stroke

Thank you for more great contributions, and to everyone else who is sticking with it.  Jim, there are not really diminishing returns, keeping the valve open to the end of the stroke results in the maximum work for a given size of cylinder.  This is not the same as maximum work from a given quantity of steam.  Cutting off early saves steam, so reduces fuel costs, when you don't need the maximum output.  Think of a locomotive going over a hill.  Going up, maximum effort is required.  Then, much less is required when going down the other side, so fuel can be saved by cutting off early.

Glad to have you with us, Paul.  There is a very large torque fluctuation through out each revolution, it even goes negative for a single cylinder engine.  We will be exploring that in the near future.

Thanks Dan for your help in explaining the indicator diagram, it helps me keep these posts a little shorter.

To continue our look at the cylinder, we now need to look at the exhaust stroke.  When the exhaust valve opens, the release point on the diagram, there is no more expansion, and the pressure in the cylinder goes into pushing steam into the exhaust system, where the remaining energy is essentially lost.  At least as far as this cylinder is concerned.

At the bottom dead centre (still thinking in terms of a vertical engine) another very significant change occurs, the piston starts to move upwards.  Why is this so important?  It is because in calculating work done, the two quantities, force and displacement, are vectors.  This means that these quantities have both magnitude and direction.  The cylinder constrains the piston to move only up and down, but opposite directions have opposite sign.  We unconsciously defined down as positive when we multiplied pressure times area to give force then by the piston displacement to give work done by the gas which we assumed was positive.  On the return stroke, the force on the piston is still down, but the piston is moving upwards which must then be negative.  So the work done by the gas in the exhaust stroke is negative.  Alternatively, we can consider that the piston is now doing work on the gas, in pushing the steam out of the cylinder.  This is shown on the diagram as the lower boundary of the diagram until the exhaust valve closes.  Continuing movement (with the inlet valve still closed) results in compression of the remaining gas shown by the upward curve on the left hand boundary of the diagram.  In a well timed engine, the compression achieves close to steam inlet pressure just as the inlet valve opens and the cycle repeats.

So we now see that the work done by the gas during the complete cycle is the work done during the down stroke minus the work done by the piston during the upwards or exhaust stroke.

I might appear to have laboured this point a little, but is really is where "the rubber hits the road" so to speak, in the amazing process of converting the heat from the fuel, to energy in the form of vibration of gas molecules and then to mechanical work.

Next we will look at the piston.  Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on May 16, 2017, 02:29:03 PM
Using the planimeter to calculate horsepower and stuff.......
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 16, 2017, 07:46:19 PM
I have only used a planimeter a few times in an engineering lab years ago. It is used to trace the indicator card lines in a complete loop. Knowing the spring constant on the steam engine indicator and the stroke reduction the work done can be calculated. This is known as indicated horsepower.

Peabody spends a bit of time describing the types of error that is involved with using an indicator. These include inertia of the moving parts and friction in the stroke reduction gear and a few other types of errors.

The modern way to do this is with a pressure transducer which will eliminate most of the types of error associated with a steam engine indicator.

The expansion curve for steam is very similar to a rectangular hyperbola. This is sometimes drawn between the point of cutoff and the release point as a standard reference line. It is also used in design to approximate the PV (pressure volume) curve and draw a design best case indicator card which can be used with a planimeter to calculate the indicated horsepower.

http://mathworld.wolfram.com/RectangularHyperbola.html

Dan
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 17, 2017, 12:26:35 PM
The other side of the piston.

Planimeters, those were the days.  I cannot help but marvel at the ability and skills of our fore fathers in designing and making those things.  I got to use one in my Dynamics of Machines subject, but not since.  And attended the 50 year anniversary last year.

So far I have tried to be very consistent in referring to the "work done by the gas" on the piston, specifically on the top face of a vertical engine.

If we consider the sides of the piston, the gas forces are always balanced by the equal and opposite force diametrically opposite.  There can be mechanical forces on the piston sides but these do not concern us at the moment.  However we cannot ignore what happens on the lower side.

There are two arrangements to consider, the first being a double acting cylinder as on the first steam locomotives we ever saw.  Most real working engines are also double acting for reasons that will become obvious if it is not self evident.  The lower side of the piston is in an enclosed cylinder, much as the upper side.  The main obvious difference from the upper side is the piston rod and the provision for it to exit the cylinder through the gland.  The lower face of the piston experiences gas forces, apart from the centre section which is occupied by the piston rod.  The piston rod does not escape gas forces, it experiences atmospheric pressure on its lower end.

The gas force on the lower face is directed upwards, and follows the same cycle as the the force on the upper face, but out of phase, so the lower face is in the exhaust phase while the upper face is experiencing the inlet phase.  Clearly there are two working strokes in each elevation.

When the pressure in the top of the cylinder is highest, the net force is down and the piston moves down, and vice versa.  The net work on the piston is available at the piston rod to overcome friction within the engine and to do the useful external work of the engine.

The situation in a single acting engine is slightly different.  The bottom of the cylinder is open to atmospheric pressure.  When gas is admitted to the top of the cylinder, the piston moves down for the power stroke.  Work is done by the piston on the air in the cylinder to expel it from the cylinder. 

As single acting engines normally exhaust to atmospheric pressure, the cylinder pressure during the exhaust must of necessity be greater than atmospheric pressure.  The force exerted by atmospheric pressure is less than the force from the gas above. Hence the net force on the piston is down, motion is upwards, so work by the gas above the piston is negative.  That is, the piston is doing work on the gas.

There is only one power stroke per revolution for a single acting engine.  During the exhaust stroke, the net output (and torque) is negative, and the engine slows down.  A flywheel stores energy by speeding up during the power stroke, and returns this energy as it slows down on the exhaust stroke.

I might appear to have laboured the point a bit, but without this understanding of how gas pressure converts energy to work, it can be quite difficult to understand why a Stirling engine, which appears to be single acting, under some circumstances is actually double acting.  (A topic for much later!)

Next time, how is all this relevant?

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: derekwarner on May 17, 2017, 10:46:40 PM
Well MJM460...as of this morning we see.... "Topic: Talking Thermodynamics  (Read 1143 times)"

This is always a good indicator [no pun intended] in understanding how many members are reading this thread so from the numbers I suggest you have a good repeat audience  :happyreader: ...yes viewing and reading the work

Derek
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 18, 2017, 02:01:22 PM
Yesterday was just one of those days!

Thanks Derek, yes, it is great to see so much interest.  I have always thought we were a bit too reluctant to get really technical, let alone the maths.  Our aero modeller colleagues include quite a bit of aerodynamics in their literature, so I don't know why we have to apologise before including a bit of thermodynamics.  It is my intention to show how relevant it all is to our engine making.

But yesterday?  It was not a good day.  If you found yesterday's post a bit hard to follow, please don't despair, it did not do much for me either when I read again today.

Just as in building we sometimes don't get a part up to our standard and have to consign it to the scrap bin, in trying to build a knowledge base, the same thing happens.  I assume that is what the "modify" button is for?  I will try it in the next few days and let everyone know when it is ready to read again.  I have at least given thought to how I should have approached it.

But that is not the complete reason for my heading.  Earlier in the day, I was working on the brown stuff, when my jigsaw went bang, and all the lights went out.  The machine was not even warm, and I did not see any smoke get out.  So I reset the breakers, restored the lights and tried to start and again bang and no lights.  It was only about 25 years old, but clearly a trip to the tool shop was required.  The man did not miss a beat.  What a pity you did not bring it in last Friday, he said.  It might have still been on warranty!  I was not quick enough to point out it was not broken then.  It was really disappointing.  Despite using power tools since primary school, that is the first I have ever broken.  But the new one is really nice. 

To top it all off I found that my belt sander was so old that the correct size belts do not seem to be available.  (Much older than the jigsaw, it was my Dad's.)  Looks like more tool shopping required.  What a blow!  I never saw what others see I retail therapy, then one day I happened on a really good tool store.  Now I get it.

I have resolved some internet technologies this week, so I will prepare a diagram to illustrate where we are up to, as I should now be able to communicate with the scanner.  But it has been a long day, and it is late, I will be very lucky to to get any thinking time tomorrow so it may be Saturday.

Thank you for all the interest.

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on May 18, 2017, 05:59:05 PM
All accomplished craftsmen proceed at a measured pace, and pause to consider the next step in a complex sequence. Putting together a concise but intelligible series of explanations is no mean feat. Also, a person never knows their subject completely until they have had to explain it to someone who doesn't understand it, an enlightenment that every instructor discovers not too far into their career.  Quality not quantity! Don't feel pressured to produce every day. I and others appreciate your efforts. Regards Paul Gough. 
Title: Re: Talking Thermodynamics
Post by: derekwarner on May 18, 2017, 11:34:35 PM
Many years ago, the Director General of Education in NSW was a chap by the name of Sir Harold Whyndam.....

His claim to fame in education was the introduction  :slap: of the Whyndam Scheme....I just happened to be in the first year of such Gini-Pig changes

The Scheme bought revolutionary changes ...yes changes like Applied Heat teachers being allowed to use their last years papers and continue with Imperial units of measure ....or those with an enquiring mind could apply the new SI system [no slide rules or calculators.....just Trig Tables]

I was working 3 shifts at the time...so on afternoon and night shift I attended college in daylight hours and studied Applied Heat in Imperial......however when on day shift, my Applied Heat teacher  in the evening [being a Combustion Engineer from the Steel Industry] decided SI was the way to go....after all he studied SI in Europe years earlier

The consequence of this is that I failed Applied Heat that year, and repeated the subject in SI the following year

Derek
Title: Re: Talking Thermodynamics
Post by: Flyboy Jim on May 19, 2017, 03:11:44 AM
Yesterday was just one of those days!

Thanks Derek, yes, it is great to see so much interest.  I have always thought we were a bit too reluctant to get really technical, let alone the maths.  Our aero modeller colleagues include quite a bit of aerodynamics in their literature, so I don't know why we have to apologise before including a bit of thermodynamics.  It is my intention to show how relevant it all is to our engine making.

But yesterday?  It was not a good day.  If you found yesterday's post a bit hard to follow, please don't despair, it did not do much for me either when I read again today.

Just as in building we sometimes don't get a part up to our standard and have to consign it to the scrap bin, in trying to build a knowledge base, the same thing happens.  I assume that is what the "modify" button is for?  I will try it in the next few days and let everyone know when it is ready to read again.  I have at least given thought to how I should have approached it.

But that is not the complete reason for my heading.  Earlier in the day, I was working on the brown stuff, when my jigsaw went bang, and all the lights went out.  The machine was not even warm, and I did not see any smoke get out.  So I reset the breakers, restored the lights and tried to start and again bang and no lights.  It was only about 25 years old, but clearly a trip to the tool shop was required.  The man did not miss a beat.  What a pity you did not bring it in last Friday, he said.  It might have still been on warranty!  I was not quick enough to point out it was not broken then.  It was really disappointing.  Despite using power tools since primary school, that is the first I have ever broken.  But the new one is really nice. 

To top it all off I found that my belt sander was so old that the correct size belts do not seem to be available.  (Much older than the jigsaw, it was my Dad's.)  Looks like more tool shopping required.  What a blow!  I never saw what others see I retail therapy, then one day I happened on a really good tool store.  Now I get it.

I have resolved some internet technologies this week, so I will prepare a diagram to illustrate where we are up to, as I should now be able to communicate with the scanner.  But it has been a long day, and it is late, I will be very lucky to to get any thinking time tomorrow so it may be Saturday.

Thank you for all the interest.

MJM460

Well....... speaking as someone at the bottom of the "knowledge chain", I pretty much got the drift of what you were saying, so all wasn't lost as far as I'm concerned.

Looking forward to more. As was said by another..........don't feel like you have to push yourself.

Bummer about your jigsaw and belt sander. I've got a few of those kinds of tools myself.

Jim

PS: Do you go by a name other than MJM360? I may have missed it.
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 19, 2017, 01:54:59 PM
The other side of the piston - Take 2

Thanks Paul for your encouragement.  Every day is not viable in the long term, there will be slower periods so thank you for your understanding.  But I would like to get past this one while it is fresh if I can.

Derek, sorry about your trials with the subject.  Such arbitrary changes don't make it easy.  I also studied in imperial units.   Even have my copy of 7 figure tables still.  Then, just as we went metric with all the advertising to help us convert, I went to work in Canada, near Brian's territory.  No metric conversion there.  When I came home, the conversion was essentially complete, and I had to work in metric units cold turkey.  Needless to say, I am still not alone in going out to buy 2.4 m of 4 x 2!  I now prefer to do my calculations in metric units and I hope you will see why when I get to talking about units of measurement.  But I know many forum members are more familiar with imperial units and I will try and remember to convert key numbers.  In the mean time, if the answer does not look right when using imperial units, just alternately multiply and divide by g until it does!  Or something like that I think.

Glad you got something from that last post Jim, I still hope to improve it for you.  Yes sentimentally sad about the tools, but it is an opportunity to go shopping, and the economy needs me.  I am a probably unnecessarily shy about using my name, even paranoid perhaps, but if we ever get the chance to meet, or you need a pm, I will make sure you have my name.

So here goes on the lower side of the piston, still thinking in terms of a vertical engine with the cylinder on top.  To make the explanation totally consistent and complementary with the top, I need to continue to talk in terms of the work done by the gas.  I have mentioned that we unconsciously defined down as the positive direction in the description of the top side power stroke, and we have to stay with that definition for the lower side. 

Why the significance of direction?  Remember that force and displacement quantities are vectors, so have magnitude and direction.  On the lower side of the piston, the force direction is up, so it is  negative.  On the power stroke, the displacement is also up and again negative.  Work is force times displacement, and when we multiply two negative quantities, we get a positive for the work done by the gas.  This is expected, because I would not classify work as a vector, I suggest there is no such thing as negative work.  When our calculations indicate negative work by the gas, it simply means the piston is doing work on the gas, gather than the other way around.

So we have the power strokes on the top and bottom of the piston both do work on the piston.  The work by gas on the top acts to push the piston down, while the work by the gas on the bottom pushes it up.  If our valve gear is timed correctly, the two strokes occur alternately, so we get two power strokes per revolution.  More output without making the engine a whole lot larger.

For a double acting engine, the processes follow the same sequence of admission, cut off, release and compression as the top side, just half a rev later.  Or, if you prefer it, 180 degrees out of phase.  There is one minor difference however.  The gas pressure is not applied to the whole area of the piston, as the area occupied by the piston rod has atmospheric pressure, not gas pressure.  This means the force and consequently the work done is slightly less.  Not much less for an engine with a small diameter rod in relation to the cylinder bore, essentially a low pressure engine, which describes all of our models.  However on a larger, very high pressure machine, the rod size can be very significant.

For a single acting engine, it might be harder to see how all this works.  This post is more than long enough, so I will continue that next time.


MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 20, 2017, 12:32:52 PM
The lower side of a single acting cylinder

I hope my last post made sense to everyone, and that it was enough improvement on my earlier attempt on the lower side of the piston.

I want to continue this time by applying the same reasoning to a single acting engine, such as the little oscillators made by several companies.  Many of us had one of these when we were very young, or at least saw them in our local toy shop.

The top side of the piston in a single acting engine acts the same as the top side of a double acting engine, but how these process work on the lower side of a single acting engine may seem a little more obscure.

The gas pressure under the piston on these engines is near enough to constant atmospheric pressure throughout both the power and exhaust strokes.

Now, in thermodynamics we have to think in terms of absolute units, where atmospheric pressure is close to the defined standard atmosphere of 101.325 kPa, or 14.696 psi, varying slightly as the succession of weather patterns pass over us.  Please don't get upset about the .325 kPa or 0.096 psi, it can't be read on a normal pressure gauge, or a slide rule.  Even the third significant figure is probably not significant in practical terms. 

There is no such thing as negative pressure on the absolute scale.  Pressure is caused by the impact of many molecules on the face of the piston, no pressure implies no molecules, as in outer space for example.  There is no mechanism for the molecules to pull the piston towards themselves.

On our single acting cylinder, atmospheric air pressure does work on the underside of the piston on the up stroke, and the piston does work on the air on the down stroke, just as for the gas on our double acting cylinder.  The problem is that the gas (atmospheric air) pressure does not vary significantly, so the work done by air on the up stroke is no larger than the work the piston must do on the down stroke.  If you are really being pedantic, it is a little bit smaller.  There is no excess to contribute to the engine output.  In fact we need a source of energy to do the work necessary for the exhaust stroke of the top side.  We will eventually see how the flywheel supplies this energy.  At least we do not need valve gear for the lower side of the piston of a single acting engine.

There we have it, I hope a little clearer this time.  Next time, a little consolidation and summary of the key points before we move further away from the piston faces.

Thanks for dropping in

MJM460
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on May 20, 2017, 04:07:15 PM
The lower side of a single acting cylinder

On our single acting cylinder, atmospheric air pressure does work on the underside of the piston on the up stroke, and the piston does work on the air on the down stroke,

I think the first half of that statement is only true for an engine with a vacuum in the upper section on the up stroke like a Newcomen engine or a flame gulper engine.

The bottom side of the piston is acting like a very low pressure air compressor. The upstroke is the suction stroke for the atmosphere, it is not pushing the piston the flywheel is making the piston move.

In any case this is a very very small amount of work and I do not even remember it even being mentioned in any class.

The large low speed two-stroke diesel ships I worked used the underside of the piston as an air pump at low speeds because that is where the turbocharger does not supply enough boost pressure. In this case the amount of work the underside of the piston is doing is large enough to be considered.

Dan
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 21, 2017, 02:31:48 PM
Looking at the whole piston

Hi Dan, thanks for a great introduction to this next post.  You will not be alone in your thinking, just a bit ahead of me.  I hope that what follows will clarify my meaning.

So far, I have been looking at the upper and lower face of the piston separately.  This approach is of interest because the point at which the energetic vibrations of the gas molecules act at a surface to do mechanical work is the surface of the piston.  The gas molecules colliding with the walls of the cylinder are seen as pressure, which over an area produces a force.

On the sides of the cylinder, no movement results from the force, so no work is done. 
(W=F x d, F x 0=0).  Similarly for the heads of the cylinder.  However the piston is not fixed relative to the rest of the cylinder, and it moves as a result of the collisions. The distance is no longer zero, so this is where the work is done.  This is where heat is harnessed to do mechanical work.

We can ignore the sides of the piston, as the gas forces are in equilibrium, no movement at right angles to the cylinder axis, so again no work by the gas.  (There may be mechanical forces but that is another discussion.)  Of course, a piston has both top and bottom surface, and as we have seen the force on each varies throughout each revolution, the direction of the force on the two sides is opposite, and of course the direction of movement alternates between up and down.

The piston moves in the direction of the larger force.  So we now need to look at the force balance on the piston.  This involves understanding the exact stage of the cycle on each side of the piston at each instant, so we can understand how the resulting force on the piston changes with time.

Now the job of the valve gear is to coordinate the movement of the valve with that of the piston so it all happens in an orderly manner and the engine is able to run continuously.

If we look at the double acting cylinder first, starting just as the piston approaches the top dead centre.  Here, the valve opens the gas supply to the top of the piston, and we have a downwards force, somewhere near the maximum for the cycle.

At the about the same time, the valve opens the lower side of the piston to the exhaust system.  The point of release.  The remaining pressure under the piston is rapidly depressurised into the exhaust The piston moving down overcomes the remaining pressure under the the piston, which is near the minimum for the cycle, and pushes the remaining steam out of the cylinder during the down movement.  The force on the top side is enough to not only enable the piston to push the exhaust out, but with excess force available to do external work.  Remember the absolute pressure cannot be negative, the pressure can only be less than that on the other side.  So the piston moves in response to the difference between the pressure on the top and the pressure on the bottom.

(For those familiar with the formula Horse Power = P x L x A x N/ 33000, it is important to understand that P is actually the difference in pressure between the two sides of the piston, but I will get to this in the next few posts.)

As the piston moves down, the valve closes the gas inlet, cutoff point, and expansion starts, then finally opens the exhaust port, or release point all on top of the piston.  During this same downwards movement, the valve has opened the exhaust then near the bottom dead centre, then later closes it to start compression.

I think it would be helpful to try and construct a pressure - volume diagram which shows both sides, aligned so we can see the relationship between the force on each side.  I will try and have a diagram with my next post when I will again try and show what the above means for a single acting cylinder.

Thanks everyone for following

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 22, 2017, 12:03:34 PM
Another clarification.

I don't think my last post really clarified much, and probably did not answer Dan's question.  Another part in the bin!  So bear with me while I try again to summarise the important points in only three or four short paragraphs.

Thermodynamics discussions are best based on absolute values of pressure.  Zero pressure is full vacuum, very difficult to achieve.  Atmospheric pressure is about 101.3 kPa. Our usual gauges read zero at this point.

The gas pressure on top of the cylinder is different from that on the lower side.  Each of these pressures varies through a similar series of processes but displaced by half a revolution.  By analysing the top and bottom separately we can see how the resulting force on the piston varies throughout the revolution.

For a single acting engine with atmospheric exhaust, the work on the lower side of the piston is as I described earlier, but because the pressure does not vary greatly throughout the cycle, there is no contribution to the engine output.  (This would not be the case if the exhaust system was lower than atmospheric pressure due to a condenser.  But let's leave this until later.)

I hope this helps clarify the approach I am taking. 

Still working on a diagram.  The conventional presentations do not easily show both sides of the piston at the same time.  I am working on it.

Next time I will summarise the key points so far and how these points can help us build a better engine.

Thanks for following

MJM460
Title: Re: Talking Thermodynamics
Post by: Maryak on May 23, 2017, 01:37:42 AM
  The conventional presentations do not easily show both sides of the piston at the same time.

By that do you mean simultaneously? I ask because in full size double acting engines using a Dobie McGuiness Indicator both strokes are recorded on one card then calculated to give the mean cylinder IHP.
Of course one stroke is recorded then the cocks switched to record the other stroke immediately after. This is a finicky and fickle operation which usually takes more than one attempt to get an acceptable card of the cylinder.

(http://i389.photobucket.com/albums/oo340/Maryak/MEP_0001_zpssndtylny.jpg)
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 23, 2017, 02:24:51 PM
The key message so far.

Thank you Maryak for a really great example of an indicator diagram showing top and bottom faces of a double acting cylinder.  It appears to be very realistic.  Did you by any chance take it yourself?  It certainly shows some interesting departures from the "ideal" diagrams we often see in text books, and many details which I would like to come back to later.  I had thought that presentation would not illustrate what I wanted, but now, on looking at it carefully, I think I will use it after all in due course.

My previous posts had many words and probably did not help as much as I would have liked.  They say if you are in a hole, stop digging.  I do not want to get bogged down in pure theory, as my interest is more into the practical areas where thermodynamics helps us understand our engines and how to make them better.

So, let's summarise the key messages to date, so we do not get lost in some of the inevitable side tracks of the discussion.

Our little engines are heat engines, engines which harness heat energy to do mechanical work.  The conversion interface is the face of the piston, where the random motion of molecules of gas is experienced as pressure.  The pressure on the piston causes it to move, thereby doing mechanical work.

We saw that the resulting force on the piston is the vector sum of the force on on the power stroke side of the piston and the other face which is exhaust pressure (or atmosphere for a single acting engine.)

Work is done when a force moves through a distance.  Force is pressure times area, the distance moved is the stroke.  If we rearrange this, we see that work equals pressure times swept volume.

So the potential work output of the engine depends only on the differential pressure on the piston and the swept volume.  No gas properties, no temperature, just differential pressure and the volume swept by the piston.

Obviously, the swept volume is determined by the basic size of the engine.  Clearly a larger engine should be able to do more work, no surprise there.

For a given size of engine then, the only thing we can or need do is deliver our chosen gas to the face of the piston at maximum pressure, while minimising the pressure on the exhaust side.

It might see surprising that it can be reduced to this one parameter.  Every part of an engine and its associated equipment is directed to that end.  However, the pressure at the piston varies greatly throughout the cycle, and as you may have seen in our discussion of air vs. steam, it is not always easy to predict the pressure at any point, especially during the expansion which occurs when the inlet valve closes.

Then, of course, having converted heat to mechanical work, we must then transmit that work to our chosen load with as little loss as possible.  As we all know , there are many interesting ways to loose some of our output to friction, and I hope to explore some of these as the discussion proceeds.

Next time I hope a brief look at converting the piston movement to a rotational output.

Thank for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on May 23, 2017, 02:47:59 PM
Hi Just a quick question ...When the Mallard Achieved the 126 MPH record what would be the temperature difference between the inlet steam and the exhaust steam ??.............Also how fast are the molecules moving in relation to the linear speed of the piston ??...........!
Willbert......
Title: Re: Talking Thermodynamics
Post by: Maryak on May 23, 2017, 10:08:35 PM

Thank you Maryak for a really great example of an indicator diagram showing top and bottom faces of a double acting cylinder.  It appears to be very realistic.  Did you by any chance take it yourself?


Your welcome. No it is not one taken by me, (I'm a little more clumsy with the stylus). It is from Southerns Verbal Notes and Sketches.

Regards
Bob
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 24, 2017, 12:31:22 PM
Velocity of molecules.

Hi steam boat willy, I am not sure about the temperature difference for the Mallard.  My grandson saw the Mallard in a museum and brought home an HO model only a month ago!  Temperature difference depends on the inlet temperature and degree of expansion, which would not be much on a record attempt when max power and speed are more important than efficiency.  After release, the steam is basically throttled by the exhaust valve to the exhaust system, and throttling involves a only a small temperature drop, so I would guess a relatively small temperature drop overall.

The speed of molecular motion is the obvious question when the collision of such tiny particles with the piston is responsible for so much pressure.  I am sure many others are also wondering.  My ancient text book, by Jolly, published in 1961 (it was nearly new when I bought it!), has a calculation for oxygen at 0 C based in ideal gas.  I substituted the relevant figures for steam at 150 C, and the found an RMS velocity of 765 m/s.  It is random motion with a wide range, but the book says RMS velocity is about 10% higher than the average (velocity distribution is not sinusoidal).  The energy in this vibration is proportional to the speed squared.

For an idea of piston speeds, my little oscillator piston has a stroke of 16 mm, and runs about 2000 rpm on a digital non contact tachometer.  This means the piston travels 64 m a minute or 1.06 m/s.  My unreliable memory recalls client standards for compressors to require less than 5 m/s.  As the limits are based on ring wear, I assume engines are similar.  The motion is roughly sinusoidal, to the maximum speed is 2 times the average and RMS speed is 0.707 times the maximum.  So a very large number of collisions of tiny particles at around 760 m/s with the piston which is 0 m/s at top and bottom dead centres up to a max of 2 m/s for my oscillator, or 10 m/s for my industrial machines.  Eighteen grams of steam at atmospheric conditions has around 6.03 x 10^23 molecules, (text about 5 bar pressure deleted with apologies).  Many, many random collisions, so many it feels like a constant pressure.

I hope that gives everyone an idea of the numbers involved, a worthwhile diversion from what I intended.  There is always another day for that.

Maryak, I know what you mean about the difficulty of getting such a good looking diagram from the instrument.  However, from that source, it is clearly well informed about a real indicator diagram, unlike most I have seen, which are constructed more to show an ideal engine cycle, and do not reflect a real machine.  It is a great example and I am sure I will be referring back to it.  Thank you.

I think that is enough for today, so back to rotational output next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: Maryak on May 25, 2017, 12:18:36 AM
The book I referred to, (Vol 1), contains many indicator diagrams with various results due to +tve -tve pressure, improper valve settings, ring wear etc. It may be worth a trip to a library or steam preservation society.

Regards
Bob
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 26, 2017, 12:01:23 PM
Producing rotation.

Not much progress yesterday.  Out to dinner with friends - pan fried salmon and steamed vegetables, all home cooked.  Delicious.

But I did notice a small error in my reply on the number of molecules.  Went from mass to volume too quickly, I have edited the offending bit out.  As our engines have nothing like 18 gm. in the cylinder, I made a rough calculation of how many would be in the swept volume of my little oscillator.  At atmospheric pressure it is about 10^20 molecules.  At 5 bar, it would be 5 times that.  Even on this small scale, a huge number of collisions.

The first engines were limited to pumping type applications which could directly use the reciprocating action of the piston.  The crank and conrod seems obvious now, but it was a major breakthrough at the time.  Patent squabbles led to the development of the scotch crank, and various clever geared mechanisms, we all know the story.  But today, patents have expired and the crank and conrod is practically universal, so I will stay with that.

The geometry of the mechanism means that the torque is zero at top and bottom dead centre, and reaches a maximum at about 90 degrees, which is about double the average.  Like a sine curve, but slightly modified by the con rod geometry.  And this assuming the pressure is constant throughout the stroke.  Any reduction in pressure through the stroke results in a reduction in torque from the constant pressure calculation.  The point is that the torque output is far from uniform.  Even worse on a single acting engine when the torque is negative on the exhaust stroke.  This fluctuation is a consequence of the geometry, and cannot be eliminated by pressure variations.

Now most of us have heard of Newtons laws and the formula F=ma.  (Force equals mass times acceleration) which applies to linear motion.  Less well know is the rotational equivalent, Torque equals Moment of Inertia times angular acceleration.  Both of these are now expressed more elegantly in two of the conservation laws of physics.  The fluctuating torque results in a fluctuating rotational speed which would be unsatisfactory in most applications, if indeed the mechanism could be persuaded to pass through the zero points top and bottom.

In order to even out the speed fluctuation, and get through the zeros, a flywheel is mounted on the shaft.  The shape of a flywheel is designed to provide maximum moment of inertia, and hence minimise the change of rotational speed caused by the changing torque.  So the conrod and crank combine with the flywheel to produce an acceptably smooth rotation.  What is acceptably smooth?  Well, for an example, the standards for my large industrial reciprocating compressors required the speed fluctuation to be limited to +/-7% within each revolution.

Perhaps more on flywheels next time?

Thanks for following.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 27, 2017, 01:18:24 PM
Flywheels - a little diversion into physics.

Last time I started on flywheels.  The flywheel accelerates in response to a torque, meaning it spins at an increasing angular velocity, and in so doing stores energy.  When the torque produced by the engine is less than that absorbed by the load, the flywheel returns this energy by slowing down.  This enables a single acting engine to run, and reduces the speed fluctuations of a double acting engine that otherwise result from the normal production of torque.  The physics of this is described by the law of conservation of angular momentum.

We have an intuitive feeling for momentum in linear systems, that "something" which makes a body in motion continue in a straight line unless acted on by an external force (Newtons Law).  It can be quantified as the product of mass and velocity.  It has the units kg.m/s.  (In imperial units ft. lb(mass)/s.)  Velocity is a vector, and when we multiply by mass to get momentum the resulting momentum quantity is also a vector (with the same direction as the velocity).

In spinning objects, such as spinning tops, gyroscopes, bicycle wheels, and ice skaters, the analogous quantity is angular momentum.  It is quantified by multiplying the moment of inertia by the rotational speed.

But what is moment of inertia?  First it must be defined in relation to an axis of rotation.  Then, for a point mass, it is mass times (distance from the axis)^2.  It has units of kg.m^2.  So moment of inertia is proportional to mass, but the distance term means that it is also dependant on the distribution of mass, or shape.  The squared part means that shape is much more important than mass.  We can see that the typical flywheel, which has a heavy rim plus minimal spokes to keep it centred on the axis of rotation, and a small hub to fix it on the shaft, should have a large moment of inertia relative to its mass.  A solid disk would have much more mass (and bearing load), but only slightly larger moment of inertia than a spoked flywheel.

Next, we need to know the rotational speed.  No great mystery in that, but note that the units of angular rotation are radians, not degrees, grad, or even revolutions.   A revolution of 360 degrees is two times Pi radians.  Rotations have no dimensions, but are vectors and have direction.  Hence rotational speed is measured in radians/s, and has the units just "per second".

Now we can multiply moment of inertia (kg.m^2) multiply by angular velocity (per second) to get units of angular momentum as kg.m^2/s.  This is the quantity which tends to average out speed fluctuations.  Like linear momentum it is a vector.  The direction is conventionally defined as coinciding with the axis of rotation.  If you make the typical "thumbs up" sign with your right hand, when your fingers follow the direction of rim rotation, your thumb shows the positive direction.  (This could require a "thumbs down", so don't be insulted, it could be just someone trying to work out the direction of the angular momentum of a spinning top!)

Two interesting points to close on.  Angular momentum and torque are both vectors.  When torques and angular momentum interact, we can predict the resulting vector mathematically by performing a vector product multiplication.  The result has a direction which is determined by a right hand rule.  This vector multiplication predicts all the movements we see when a gyroscope reacts to gravity, or we hold a spinning bicycle wheel by its axle and try to move the axle in various ways.

Second, linear momentum and angular momentum are quite independent quantities with independent conservation laws, and an object can have both at the same time.  (There is a third conservation law which can also be useful to us, but back to that later.)

I hope this answers many common questions about flywheels.  Next time I will try and return to getting more out of our engines.

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on May 30, 2017, 12:31:47 PM
Flywheels (2)

Thinking about my last post, I realised that it might be good to put some numbers to a real flywheel before moving on, just to give a sense of proportion.  I calculated the mass of the flywheel on my little oscillator, nothing complex.  It is machined from a 2 inch bar, cleaned up to 50 mm diameter. 15 mm in the axial direction.  I left the hub and rim full length, and machined a recess in each side to leave a web 6 mm thick.  Nothing fancy, you guys are inspiring me to at least drill some lightening holes or even curved spokes.  But this was a few years ago.

I believe the material is cast iron.  I have used cgs units for such a small flywheel to avoid having too many decimal places in the numbers, but for any other calculations I need to do it in ISO metric or mks system.  I was interested in a few key results.

The moment of inertia turned out to be 537 gm.cm^2.  This is 80% of the moment of inertia for a solid disk, and 436 gm.cm^2 (or 83% ) of this is due to the rim which was 15 mm wide and 6 mm in the radial direction.

What if I had made it of steel?  Well the density of steel is about 7% greater than cast iron, and both mass and moment of inertia increase by 7%.

Brass would make more difference and you might like the appearance or rust resistance.  Brass density is about 17% greater than cast iron, and gives about 17% higher moment of inertia.

What if we want to save some weight?  Perhaps Jim is designing his steam powered plane.  The density of aluminium is only 38% that of cast iron.  So if we use the same design, mass and moment inertia will be 38%.  If the original flywheel was just enough, that would not do.  Can we make a lighter flywheel from aluminium if we modify the design?  The clue is in the R^2 term in moment of inertia.  If we increase the diameter to just 66 mm diameter compared with the original 50 mm) we will have an aluminium flywheel with the same moment of inertia as the original cast iron, but only 60% of the mass.

Now as a practical matter, I found the calculation was quite sensitive to even 0.1 mm change in the overall diameter, and I think we would all agree that flywheel dimensions are not that critical, clearly most are "adequate" rather than bare minimum.  If we want our engine to fly it would be worth experimenting to find the minimum required moment of inertia using cheap steel discs.  Then we could make a lighter flywheel to that moment of inertia by increasing the diameter and using aluminium.

I hope that is of interest.

MJM460

Title: Re: Talking Thermodynamics
Post by: MJM460 on June 08, 2017, 12:37:58 PM
More on producing rotation

Sorry to have been a bit silent for a while.  Family and other activities always need due attention.

I have also decided that this thread needs pictures, surprising that no one has mentioned it, but I think I have overcome some technical challenges, so here goes.

Figure 1 has an overall sketch engine of the simple vertical engine I am thinking of in my terminology and arrangement.  You can see why I was the engineer, not the draughtsman. 

I have included a sketch of the piston showing the important gas pressures on the top and bottom faces.  Remember I am using absolute pressures and I am assuming for the moment an atmospheric exhaust system.  We will discuss condensing later.  Hence the exhaust pressure is a little above atmospheric pressure to push the exhaust gas out of the cylinder.  You can see the relationship between the pressure on the piston faces and the resulting force on the piston.  The exact pressure is not important, and is varying from moment to moment anyway.

Figure 1 also includes sketches showing the forces on the wrist pin, and the triangle of forces which shows the relative size of the forces.  The horizontal force is not due to gas on the sides of the piston, but a result of the angle of the conrod. 

Finally, I have shown the torque calculation.  The force is the conrod load, and the moment arm (a) which causes the torque is shown in red.  Torque varies from zero to a maximum on the power strokes, and for a single acting engine is actually negative during the exhaust stroke.

Figure 2 shows the gas pressure, force triangle and torque for three important cases.

Detail (a) is for the power producing down stroke.  Piston force is downward.  Note that the conrod is in compression.  I have shown the rotation as anticlockwise.  This is the normal trigonometry convention, and means that your calculator gives the correct result for the parts of the circle where trigonometric ratios are negative.  Only important if you put these calculations into a spreadsheet, then increase the crank angle in increments, and have the computer recalculate for each angle through a full revolution.

Detail (b) is for the upstroke of a double acting engine.  Note how the moment arm is shown, and that the piston force is now upward and the conrod is now in tension.

Detail (c) is for the upstroke of a single acting engine.  As there is no steam pressure under the piston, the flywheel provides the necessary energy to push the piston upwards, and expel the exhaust.  Note that the conrod stays in compression for the upstroke of a single acting engine, compared with load reversal for a double acting engine.  This factor is an important consideration in the lubrication of the wrist pin and crank pin journals.  If you are familiar with larger full size engines, the pistons are much heavier in relation to the gas forces, and momentum considerations modify this simple analysis to some extent.

I know I have been labouring these points a bit, but this is the process by which the pressure energy in our gas produces force on the piston, which produces mechanical work when the piston moves in the direction of the gas force.  The force on the piston loads the conrod which gives torque at the output shaft. 

This is the process by which energy in the gas is changed to mechanical work.  This is how our engines really work!

Thank for looking in, more next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 08, 2017, 12:39:36 PM
I had no luck attaching two figures, so here is the second

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 11, 2017, 02:11:48 PM
Completing the P-V diagram

I have included a idealised sketch P-V diagram, fig. 3, to illustrate the discussion.  Note I have assumed the pressure during admission to be constant, and I have rounded the corners to acknowledge the point that real valves take a finite time and rotation to open or close.

We have discussed the top boundary of the P-V diagram, shown in blue, admission and expansion of the power stroke, and also the lower boundary, the exhaust stroke, also in blue, but to describe the complete cycle so our engine can run continuously, we need to connect up the ends with the lines shown in red.

At the end of the power stroke, the cylinder pressure is still well above the exhaust pressure when the exhaust valve opens.  The remaining pressure expands into the exhaust system with that familiar chuff or beat of the steam locomotive in heavy load.  Of course when the valve gear is notched up to give earlier cut off, the beat is less distinct as expansion from an earlier cut off may be close to exhaust pressure when the exhaust valve opens.  From the thermodynamics point of view, the significance of this process, called the release, is that all the remaining energy in the steam goes out the exhaust and plays no further part in doing useful work.  When efficiency or water conservation is important, of course we can capture some of this heat in a feed water heater, and return it to the system to use in a later cycle.

At the end of the exhaust stroke, we have to get the cylinder contents back up to the inlet admission pressure.  There are two alternative processes possible, and in practice we probably get a little of both.  In the diagram, I have shown the exhaust valve closing a little before the bottom dead centre, resulting in compression of the steam remaining in the cylinder.  The energy for this compression of course comes from the flywheel, and so subtracts from the net output of the cycle.  Is there an alternative?

The alternative is to close the exhaust at bottom dead centre, and open the inlet valve with the cylinder pressure still at exhaust pressure, so that new steam expands into the cylinder and flows until the pressure equalises at the admission pressure.  We get a little more work output from the cycle, at the expense of using a bit more steam compared with the cycle in which we use compression.

So what is the difference?

The difference comes down to efficiency.  The energy in the steam which expands into the cylinder while it is still at low pressure is downgraded in terms of its ability to produce mechanical work.  It all goes into heat in the exhaust stream.  So there is extra steam consumption but no extra work output.  On the other hand, the energy which goes into compression remains mostly able to be converted back into work.  There are losses, but not total loss.  So a little less work output per cycle, but a steam saving which is greater than the work lost.

There is another more practical benefit to having some compression at the end of the exhaust cycle.  This is often described as cushioning, and is related to the need to stop then reverse the kinetic energy of the moving piston.  It may be even more directly related to softening any noise from slack in crank pins and wrist pins.  I don't know the full explanation, but the observation will have an explanation, whether I understand it or not.

Next time I will look at the more realistic P-V diagram that Maryak provided and how this differs from the idealised ones I have referred to so far.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 12, 2017, 12:59:10 PM
Introducing some reality.

Not so many comments these days, I hope I have not turned everyone off after the initial enthusiastic response.  I think we may now be beyond the tedious parts and things should move a bit quicker.

By now, it should be obvious that to get the most power out of our engine, we have to get the highest possible pressure to the face of the piston, and at the same time, the lowest possible exhaust pressure, so that we have the greatest differential pressure difference on the piston.

We have assumed an idealised picture of the p-v diagram.  But what happens in practice?

Maryak has kindly provided a p-v diagram (reply#53, page 4 of this thread) from a text book written for marine engineers, about 1916 I think.  It shows the diagram for both the top and the bottom of the piston.  As the textbook was intended to instruct, I assume it shows features, both good and bad which indicate the health of the cylinder.  So what can we see?  Let's look first at the diagram for the top, starting with admission at the top right, moving anticlockwise around the diagram.

We can see the admission process is not really constant pressure, however a drop from around 110 to 92 is probably a reasonable approximation as we will later see.  Then we appear to have cut-off and expansion, and the diagram again looks reasonable, ending with release, when the exhaust valve opens before bottom dead centre, so the cylinder pressure is right down at exhaust pressure when the return or exhaust stroke begins.  Finally we see compression to perhaps 60% of the inlet pressure.  The vertical section at top dead centre to the start of the down stroke probably implies steam flow in to raise the pressure up to the inlet pressure.  To those of you who have experience with a real indicator diagram, does that sound like a reasonable interpretation.

If we then look at the bottom trace, we have a different picture.  The admission starts at a pressure slightly higher pressure than for the top.  Then the expansion section requires some imagination to see, rather the pressure falls quicker than it did on the top diagram down to release.  Then a very flat exhaust pressure until the exhaust valve closes in time to give compression.  The exhaust valve seems to be quite good, as compression is a little better than for the top.  No doubt the student was asked to explain all the departures from the expected trace.  I leave you to think about what faults could cause the pressure trace illustrated, a bit of a detective logic puzzle.

We can see that the engineer did not use a planimeter, but a used simple arithmetic averaging process to find the indicated mean effective (differential) pressure in the cylinder and indicated power.  We can see the top has a higher mean effective pressure, 870, and the area of the diagram shows there are power losses on the underside of the piston which has an average of only 820, presumably psi.

Next time we will start to look at the issues which get in the way of achieving maximum differential pressure across the piston.

Thanks for dropping in.

MJM460
Title: Re: Talking Thermodynamics
Post by: Steamer5 on June 12, 2017, 01:24:21 PM
Hi MJM,
 Still following & enjoying the read & explanations & learning stuff!

Cheers Kerrin
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 13, 2017, 01:37:42 PM
Hi Kerrin,

Thanks for your reply, I am glad you are still finding it worthwhile.  I will assume you speak for many.

I have been doing lots of calculations today and it's getting late but I will have more tomorrow.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 14, 2017, 02:10:52 PM
Supply and exhaust pressure

It is time to ask what pressure we might have at the piston face.  It is obviously less than boiler pressure, but how much less?

On its way from the boiler to the cylinder, the steam must go through pipes, possibly a throttle valve, the steam chest, valves and the cylinder steam passages.  These items all involve pressure loss.  The losses are caused by wall friction in pipes, acceleration losses in entering the pipe end, and more losses when the steam expands into a larger chamber at the end of the pipe.  How can we get an idea of the magnitude of these losses?

If we look at pipe friction, there is a formula known as Darcy's formula which is used in industry to calculate pipe friction.  If we look at my little oscillator again, I have measured the steam flow by measuring the weight of water poured into the boiler before the run, and subtracting the amount I draw out with a syringe at the end.  I also note the time between the engine starting and stopping for the run.  It seems to be about 0.73 kg/hr.  Not much, but I am using a small Meths burner similar to that supplied by Mamod and other small engine manufacturers.  From this, Darcy's formula gives about 1.45 kPa per meter in a 5/32" tube (2.55 mm inside diameter).  As my supply pipe including a superheater is only about 500 mm, I think we can ignore this source of loss.  Alternatively compensate by running the boiler about 1 kPa higher.  Those of you who are familiar with metric units of pressure will appreciate that you would need a very accurate pressure gauge!  Clearly not worth doing a lot more maths to evaluate this more closely, but if you have any doubt, increasing the tube size to 3/16" reduces this loss to about 0.4 kPa/m.

The other losses occur each time the flow path changes cross sectional area.  As the mass flow is the same all the way along the path, changes of flow area involve changes of velocity.  Each time the velocity changes there is a loss of energy. 

To keep this post to a more reasonable length, I will look in more detail at these other losses, next time.

Thanks for dropping in

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 15, 2017, 01:47:32 PM
Losses along the flow path

Last time I looked at friction losses in the steam line, this time the losses due to velocity changes along the path from the boiler to the cylinder.

We all know about a convergent - divergent Venturi and have probably heard of Bernoulli, and the law which says the pressure energy can basically be converted to velocity energy and vice versa.  However there are some very important qualifications, perhaps not so well known.  First the reduction in area at the entrance must be smooth, not too quick, and the entry must be well rounded.  Not exactly what we see at the boiler outlet or other points in the steam path where the flow area reduces.  Similarly, the exit from a venturi must be very gentle, diverging at about 15 degrees or less with the entry and exit very well rounded to minimise any generation of turbulence.  You see these shapes if you look closely at injectors used on locomotives.  They have to be very carefully made and appear to be a bit of a black art.  Again, nothing like the pipe entrance to the steam chest.  Each of the sudden area reductions or expansions involves an energy loss due to turbulence which dissipates energy in the form of heat, and reduces the ability of the fluid to do work by reducing the pressure.  On a well insulated pipe, (you do insulate your pipes, don't you?) conservation of energy applies and no work is done, but the energy simply goes through to the exhaust as a higher temperature.  So how do we evaluate these losses?

Essentially the losses are proportional to the velocity squared.  So we need to know the velocity in our steam pipe.  In metric units, the energy due to velocity is just v^2/2, to give N.m/kg, but as we conventionally use kJ, kilojoules as the unit of energy we need to further divide by 1000 to give kJ/kg.  Pressure energy can be found by dividing the pressure in Pascals by the density to give N.m/kg.  Because a pascal is such a small unit, (atmospheric pressure is about 101000 Pa) we generally use kPa, so dividing by density then gives kJ/kg which is directly comparable with the velocity energy.  Then by a little mathematical manipulation, we can show a pressure drop for a change of velocity.

I don't propose to provide all the maths though it is simple enough, as the result is enough to identify some conclusions.  For an exhaust system without condensing, the pressure is close to atmospheric, and steam density about 1.7 kg/m^3.    We can then calculate the pressure loss for a change in velocity.

I tried to insert a table, but it did not come across, but basically for velocity of 10, 30, 100 and 200 m/s, the corresponding velocity pressure is 0.085, 0.765, 8.5 and 34 kPa

Each time our steam enters a pipe, we loose pressure equivalent to the change in velocity and again when it exits into a larger flow area such as the steam chest.  Some extra losses around the valve, and through the valve, entry to the steam passages and again on exit into the cylinder.  Say at least six times the velocity pressure.  Clearly providing the valves open quickly to at least
 area of the steam pipe, these losses would indicate a velocity of around 30 m/s might be acceptable, but losses increase very rapidly above that.

I calculated the velocity for my little oscillator, and found 15 m/s for 5/32" tube, 8 m/s for 3/16" and 5 m/s for 1/4 " tube for the steam supply and about double that for the exhaust.

 I was expecting worse,  but these figures certainly seem to support using 5/32" tube as is typical for supply to these small engines, with a size larger, say 3/16" for the exhaust.  Of course the flow is doubled for a double acting engine, and my little gas burner provides about twice the heat of my Meths burners, so larger tubes are a good idea.  Personally I use 3/16" tube for steam supply and 1/4" exhaust but I have only built small engines so far.  For larger engines it is a matter of calculating the velocity for your steam supply and exhaust, using density from the steam tables for the pressure you intend, and keep the openings in your fittings and steam passages close as practical to or larger than the steam pipe internal diameter.  Ideally aim for less than 20 m/s.

The item I have skipped through quickly is the valves, so I will look at this in more detail next time.

Thanks for dropping in

MJM460
Title: Re: Talking Thermodynamics
Post by: derekwarner on June 16, 2017, 04:52:37 AM
Back on May the 15th of May...I offered

'I agree understanding what is happening with/within our model steam plants is essential' and spoke of using an inexpensive digital laser pyrometer to help achieve this

So all the lagged lines and temperatures seemed understandable or as expected until I got to the de-oiler  :Mad: ......

This proprietary built de-oiler unit steam inlet is a single 5/32" OD copper tube.......[disregard the second 5/32" inlet installed  for the blowdown of the displacement lubricator] .....the discharge from the de-oiler again a single 5/32" OD copper tube]

I adapted further 5/32" x 0.014 wall brass tubing interconnected with K&S telescoping brass 1/8" x 0.014 wall internal joiners and a lot of bends ....all the way to the chimney connection to atmosphere

The end result [pressure drop caused by my tube work ID and unintentional internal orifice plates :Doh: ] was causing the steam discharge to condense in the de-oiler as opposed to exit the chimney as steam.....

Attempting some calculations, leads me to understand that my engine steam discharge line sizes are also too small  :facepalm:

So the first modification is to increase the discharge tube size from the de-oiler top plate to the chimney top to 1/4" OD x 0.014" wall without any internal reduction influence....and then trial & measure for the reduction in condensate to water to steam ratio

Dependent on that trial, a second possibility is to increase the actual engine steam discharge tube line sizes and a new manufactured squareish box type de-oiler with similar 1/4" full flow inlet/s

[I have previously confirmed that lagged steam tube to the engine provide a ~~ 3 degree C + gradient over non-lagged tubes] 

Thanks MJM460.......a great thought provoking thread :ThumbsUp:

Derek
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 16, 2017, 01:40:24 PM
Thanks Derek for your feedback and for telling us about your adventures.  It is sad to find after all the work to make a neat installation as you have, that it does not work the way you expected due to something no one ever talks about.  It is the sort of issue that prompted me to start this thread.  We need to talk about thermodynamics and other obscure engineering subjects to avoid some of the issues you have found.

You did not say what size engine, but what I can see in the photo suggests it may be two cylinder with cranks at 90 degrees which suggests double acting.  Unless you are into real watchmaking, I have no doubt that 1/8" is too small anywhere in the exhaust steam system.  Probably OK for a water outlet, but then surface tension can trip you up in those sizes.

Instead of building a new separator, how about taking the exhaust from that "filler" plug, is it 1/4" or perhaps 5/16"?  I assume it is actually for emptying, so a removable section of the exhaust would give you access after a day's run.

Of course the practicalities of connecting the exhaust to an engine cylinder often preclude using a reasonable size pipe.  But a fitting which screws into the block on one end can be larger at the other.  The loss from one small passage right at the engine is probably acceptable.  Then continue in 1/4" or, for a bigger engine, perhaps 5/16" to the separator.  And even bigger if your model permits it from the separator to the stack.  Again, you can get away with smaller entry to the separator especially if it is tangential.  The velocity will give a swirl which helps separate the condensate.  Perhaps a topic for another time.

I remember your point about the radiation temperature instrument, and they certainly have their uses, especially for temperature difference comparisons.  I will put a little post on temperature measuring a little later.  (It is on my list.)

I probably put the emphasis on the wrong part of the maths in my last post and glossed over the useful bit.  You can calculate the pressure loss for each change in velocity using

Pressure loss = density x velocity squared / 2000

Pressure will be in kPa when you use m/s for velocity, and kg / m^3 for density.  The two is a constant in the energy equation what ever units are chosen, while the divide by 1000 converts pressure in Pascals to kPa. 

You need to devise a test run generally involving weights to determine your approximate flow rate.  For those less familiar with steam tables, they do not appear to contain density, but instead have columns for specific volume which is 1/density.  With flow rate, and density, you can easily calculate velocity for each tube size.  Not mental arithmetic, for me anyway, but computers are magnificent for such tasks.

I know the Bernoulli equation implies that this velocity pressure is converted back to static pressure  when the velocity reduces, but in practice this only happens in a well made  venturi, and you can estimate that you loose a velocity pressure at each entrance and each exit for your tubes.

I hope this lets you get things working well without too much trial and error.  Look for a velocity less than 20 m/s.

Back to valves next time.

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 16, 2017, 02:08:15 PM
still following along .....with the help of my scientific dictionary !!...interesting info here.....in the old days engines only had two pipes inlet and exhaust. this made model engines look less unclutered with sleek lines and uninterrupted views. later on with all the thermodynamic knowledge they began to become more of a plumbers nightmare !!....A comparison can be made with an old Morriss minor A series engine and the modern car engine ....lots of room in the engine compartment then  and now there is hardly any room to get a spanner in .......
.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 16, 2017, 02:19:36 PM
Also can we talk about sound    ?  My morris Minor is very loud compared to a Rolls Royce so is some of the energy converted to sound...............?
Title: Re: Talking Thermodynamics
Post by: derekwarner on June 16, 2017, 11:24:26 PM
OK thanks MJM460....

I am awaiting the supply of 1/4" female copper long radius bends for the new exhaust trunking exiting the de-olier........[1/4" OD x 0.014" wall full flow]

Will report back on physical results with temperatures...however the final result  = volume of water consumed : to exhaust steam condensed & retained as water in the de-oiler......

Derek
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 17, 2017, 12:05:21 PM
Symmetry in valve events?

Thanks for the comments steam guy, sorry you need the dictionary, I was hoping to keep this understandable, so that the thermodynamics is accessible to all.  You will not be alone, so if you tell me some of the words you needed to look up, I will try and explain what I mean a bit more clearly.  As with all dictionaries, they tell you the word definition, but often do not help much with understanding the sentence.  And for some forum members, English is not the language of choice, and I want it to be accessible to everyone.

I know what you mean about the plumbing nightmare under the bonnet these days, I used to be able to sit on the mudguard with both feet comfortably down beside the engine while I worked on my old Holden, but not any more.  So much extra equipment crammed in, mostly more about pollution control, equipment such as air conditioning and modern engine control systems than pure thermodynamics, though I notice my Subaru keeps the engine air separate from the cooling air through the radiator, which is pure thermodynamic reasoning.

Yes, some of the energy is converted into noise, but a very small part of it.  It does not require much energy to make a lot of noise.  Most losses are directly or indirectly turned into heat.  I can talk a little about noise if you like, but if I could make your Morris sound like a Rolls, I would be a lot wealthier, and would be able to afford more machines (and castings!). It's on the list.

Derek, I am not quite with you on your aim with the "water retained to water discharged as steam".  Do you want to condense and collect more?  Or do you want collect minimum, just enough to collect the oil and send more water up the stack as steam after removing the oil?  These are very different issues to minimising the back pressure for power output.

Back to the valves.  Some of you may be feeling there must be more resistance in our inlet and outlet piping, especially as we are so often advised to have a free flowing exhaust.  The issue comes down to two areas.  For the inlet piping, it is velocity just as I have described, and then valve opening .  Especially for larger engines, it is important to have sufficiently large piping to keep the velocity low.  The valve opening requirements however are very different for inlet and exhaust.  Surprisingly so.  There is no symmetry there at all.

Most of our valve linkages whether a simple eccentric, Stevensons reversing hear, Hackworth, Joy or one of the many others (the W one used commonly on steam locomotives is too hard to spell), all produce a simple harmonic, or sinusoidal motion to the valves.  They move slowly at the extremes, full open, at each end, and fastest at middle of their stroke, which is of course about 90 degrees different from the piston.  This does mean they move fastest when the piston is moving slowly at the top and bottom dead centre, but they still take a finite time and number of degrees of rotation to open or close fully.

For the inlet valve things are nowhere near as bad as you might expect.   When the valve is opening, the piston is at top dead centre, minimum cylinder volume, and starts to move slowly down.  It turns out that the valve opening roughly matches the rate at which the cylinder volume is increasing (due to the piston moving down) so that the velocity in the valve port stays reasonable.  By reasonable, I mean the velocity required to admit enough steam to maintain the pressure in the cylinder stays within the maximum velocity I have suggested, and the valve opening does not cause any undue restriction.

This is especially true if the exhaust closing at the end of the preceding exhaust stroke provides a good compression before the inlet valve opens.  But even if it does not, the cylinder is at minimum volume and not much steam is required to get up to full pressure.  So it all works as expected.

The exhaust valve is very different!  Can you see why?  But this post is long enough already, so next time the exhaust valve operation.

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: Admiral_dk on June 17, 2017, 03:15:38 PM
As I see it - the exhaust valve opens slowly when the biggest volume of used steam must escape ..... (closing the exhaust works nicely).
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 17, 2017, 05:37:30 PM
Talking about exhaust's  I have always wondered why my exhaust valves are smaller than the inlet valves ??....is it the same with diesel engines ? and is it why they burn out every few years ........
Title: Re: Talking Thermodynamics
Post by: Zephyrin on June 17, 2017, 06:26:20 PM
IMHO, the best way to visualise the position of the steam valve during the cycle is the oval or elliptical diagram:
The oval diagram is in the CW rotation.
The steam valve is plotted (ordinate) in function of piston position in abscissa, normalised here in % of the stroke.
The green line corresponds to the opening or closing of steam exhaust.
Yellow lines are the positions of the external edge of the steam port, the crankshaft side is the upper line, and back side is the lower yellow line.
When the steam valve crosses a yellow line, steam port opens or closes. I.e. on upper yellow line, which corresponds to the crankshaft side of the cylinder, the admission opens near BDC, 0%, cut off being near 60% of the piston stroke.
One can easily appreciate the fast opening of the steam admission near dead center.
One also sees that the steam distribution on both sides of the cylinder is not symmetrical, owing to rod length.


Title: Re: Talking Thermodynamics
Post by: MJM460 on June 18, 2017, 01:49:41 PM
Exhaust Valve issues

Thank you for the oval diagram Zephyrin, such diagrams can be really useful in visualising valve event timing.  I will come back to that in a near future post.  Can I assume that what you refer to as the steam chest might be what I usually call the valve?  I usually think of the steam chest as the stationary part held to the cylinder block with studs.

Steam guy willy,  thanks for your post and pictures.  At first it might appear to contradict what I have said about exhaust valve opening, but of course, engine design also involves compromise.  Clearly the valves are about the largest that could fit the space.  The designer has to consider whether both should be the same size, or make one larger.  In internal combustion engines, the power output from a given engine is limited by the amount of air that can be delivered into the cylinder, despite losses in the air cleaner and carburettor.  Fuel is much easier to get enough.  So the designer has made the inlet valve a bit bigger to reduce the resistance to air entry.  For the exhaust, there is extra pressure available to get the exhaust gas out , so a little more resistance is perhaps less of a problem, even though bigger exhaust valves would be desirable.  They burn out because they pass very hot gases from (possibly) incomplete combustion at high velocity and including any unburnt carbon.  So very hot and abrasive.  Surprising they last as long as they do.  In contrast, a steam engine has the boiler pressure available to help get the steam into the cylinder.  The exhaust is the steam engine problem.

Thanks for your reply Admiral_dk, you have hit the nail squarely on the head.  At the end of the power stroke, the cylinder is full of steam that is at significant pressure, particularly if there was late cutoff, and the volume of the cylinder at this point is at maximum.  Of course, we can help a bit by early release, the torque due to the force on the piston is quite low within 10 to 15 degrees of top and bottom dead centre.  Can we make some estimate of how much of a problem we have?

I do not want to try and make an exact calculation, but if we assume early cutoff at 15 deg before the centre, and that we want the pressure to be essentially dissipated by 15 deg after, we have say, 30 degrees of rotation to get the pressure down.  At 2000 rpm, which is 33.3 revs per second or 0.03 seconds per revolution, or 0.0025 seconds to reduce the steam pressure to that of the exhaust system.  We then have to make an estimate of how much steam is to be discharged.  If we assume the pressure in the cylinder is 300 kPa(abs) and it expands to atmospheric pressure On release to exhaust, then 1922 mm^3 (including clearance volume) of my little oscillator would expand to 4545 mm^3.  Thus 2623mm^3 of "extra volume" has to be discharged in 0.0025 seconds.

Next we assume the valve is open to the full area of a 3/16" tube in the exhaust, 8.76mm^2, which requires a velocity of 120 m/s.  Applying v^2/2000 with an average density about 1 kg/m^3 gives a back pressure of 7 kPa.

While a pretty rough calculation, this does not seem too bad a result.  However as the valve is opening from closed for the first 15 degree, the average area might be considerably less than I have assumed.  My slide valve engines have an exhaust passage 7 mm wide.  So this must be open by 1.25 mm to give the assumed 8.76 mm^2.  As the total valve travel is only 5 mm, 1.25 mm within 15 degrees is probably optimistic.   If the area is only 4.4 mm^2, the pack pressure would be 4 times, so 28 kPa, and starting to impose serious negative torque on our engine output.  These calculations are very rough, but do serve to give an idea of the range of back pressure imposed by inadequate exhaust passage sizes.  I think enough to show that 5/32" is too small for this engine exhaust, 3/16" dubious and 1/4" better.  This compares with possibly 5/32" for steam supply, and 3/16" not really too large.

Next time, I will summarise the key points relating to valve events and the P-V diagram, and start to look at how our valve behaviour compares with the ideal.

MJM460
Title: Re: Talking Thermodynamics
Post by: Zephyrin on June 18, 2017, 03:04:44 PM
Of course, sorry, I did a confusion between valve and chest...I've corrected the post !
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 19, 2017, 02:33:28 PM
Thanks Zephyrin,  No problem, I have it almost worked out, but I wanted to check that we were talking about the same direction.

I have spent my time today cutting and colouring, Cutout Aided Design, otherwise known as CAD, trying to work out how best to describe the sequence.  I must admit to a severe case of eyes glazing over when trying to work it out, and have previously given up after setting the valve to just start opening as the crankshaft turns over top dead centre, and, on finding the engine ran ok, letting the exhaust look after itself.  The valve dimensions must have been good enough.

However for this thread, I want to match all the valve events to the P-V diagram to see how well it matches the ideal.  The oval diagram seems to be helping, I hope to have it sorted by tomorrow or perhaps Wednesday.  Have some babysitting commitments for the grandchildren tomorrow.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 19, 2017, 03:41:52 PM
Thanks for the info ...Actually the exhaust valve failed during a long journey and i was able to get there and and back on 3 cylinders..about 120 miles !! after this happened...A brief summarisation of thermodynamics might be " Everything is a compromise " when it comes down to building engines !!!! ;D ;D   :popcorn:
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 20, 2017, 02:43:23 PM
Hi steam guy willy,

I once lost a valve in of a V-8 while on a much longer trip, I was in the Smokey Mountains, and home at the time was in Canada.  No question of driving on.   Getting it fixed became a highlight of our holiday due to meeting some great people.  But I can really feel for you losing one cylinder out of a small four.

Got diverted onto thinking about controlling propane on Brian's thread today, so not quite ready to continue those slide valves, but am coming to terms with Zephyrins oval diagram, so a little progress, and l hope to be ready tomorrow.

MJM460
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on June 20, 2017, 04:55:42 PM
Have you considered Dockstader's valve gear programs? One of the outputs is the sine diagram.
http://www.billp.org/Dockstader/ValveGear.html

It would be nice to see some of the formulas you are using, a lot of pipe formulas are on the web in calculator form try Engineers Edge or Engineers Toolbox.

http://www.engineersedge.com/
http://www.engineeringtoolbox.com/

Dan
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 21, 2017, 11:36:19 AM

Real valves

Thanks for those suggestions Dan, I am glad to see that you are still following and your input always welcome.  I will put those topics on my list.  I am planning on a post on units of measurements and you might be surprised to see how much can be done with little more than the definitions.  But also a little Physics for the three conservation laws I have used and also the laws of thermodynamics. It is on the list, so I will get to it.

Casting our minds back to the P-V diagram, the key events in each cycle are the opening and closing of the inlet valve, followed by the opening and closing of the exhaust valve.  Remember that these events occur on the top of the piston, and on a double acting engine, they also occur on the lower side, but half a turn later so that the piston is first pushed down, then pushed up. 

When we set the valves on a new engine, we take off the steam chest cover, and we can see when the steam port is about to open, set this to happen at piston top dead centre, then we can see where on the crank rotation it closes.  We can also see the maximum opening at each end and so adjust the valve position on the rod so both ends open about equally.  But what of the exhaust?

Most books draw a cylinder cross section through the ports with the valve at mid travel, and show lap, and the relation of the exhaust cavity to the edges of the exhaust ports.  But unless more drawings are provided, I for one, find it difficult to keep track of the exhaust events as the valve is shifted only in my imagination.  So, does the exhaust open and close at the right points as Required for the valve events in the P-V diagram?  How much cut off do we get? And do release and compression occur as we want?

Of course with modern computer programs, we can model the valve movement and watch it on the screen as the crank shaft rotates.  But such programs are relatively recent, and quite expensive if we don't have access through our workplace.  For those of us a bit technology challenged is there a simple solution?

To solve this problem, I have resorted to Cutout Aided Design (CAD) shown in the first attachment.  The valve is the blue part that is a separate cut out that I can move back and forward over the outline of the cylinder ports to see when the ports open and close.  It is actually also useful to check the design for features such as how valve sealing surfaces and the bars between ports relate to provide clear port opening.

Many geometric constructions were developed when people only had hand calculations, and these were converted to allow computer calculations even before modern models appeared.  My text books have four variations and Zephyrin has shown us a fifth which is not in any of my books.  No doubt there are others.

Now that I have studied it carefully, I am with Zephyrin, in feeling that his oval diagram is probably the easiest to use.  Certainly the easiest I have seen so far.  It shows quite clearly the effect of lead and lap for both the inlet and exhaust events in a simple construction, at least simple with the benefit of a spreadsheet to do the repetitive calculations.

The oval diagram is produced by making a graph of valve position against piston position, which gives the oval outline.  The points around this oval correspond to the edge of the steam valve that determines valve opening, proceeding clockwise.  The same points can also be considered the edge of the exhaust cavity that determines exhaust port opening.  The steam lap is shown by the red line, and the valve starts to open when the inlet edge of the valve crosses this line at the piston dead centre.  The port finally  closes again when the oval crosses this red line on the way back to the centre position.  Similarly in the lower half of the diagram for the other end of the cylinder.

My first attempt is shown in the second attachment.  I have saved a little time by assuming a scotch crank design, but with a few more columns in my spreadsheet, and a little more trigonometry, the effect of the conrod can easily be included.

I think this item turned out way too long, so I have split it here and will continue to work through the diagram next time. 

Thanks for reading,

MJM460
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on June 21, 2017, 02:43:18 PM
Many geometric constructions were developed when people only had hand calculations, and these were converted to allow computer calculations even before modern models appeared.  My text books have four variations and Zephyrin has shown us a fifth which is not in any of my books.  No doubt there are others.

MJ,
I have about 50 textbooks on steam valve gears, could you say which ones you are using? You said four valve gear graphical constructions are in the books, off the top of my head I can think of Zeuner, Reuleaux, and Bilgram and the oval diagram.

Zeuner was the first one and I have his book but it is really a lot of Greek in the form of complex algebra. The way Zeuner made his first diagrams was very interesting. He made a cutaway steam engine and attached a drafting board spinning at crank speed. A pencil attached to the valve to gave the trace lines that we now call a Zeuner diagram.

I know how to construct a Reuleaux diagram and it would be my choice if the valve is not symmetrical at the center of the exhaust port.

My favorite valve diagram is the Bilgram, for me, it is much simpler to see what happens when the variables of the valve are changed.

If you post the valve dimensions needed, I can construct the diagrams I just mentioned.

The historical designers of steam engines used the drafting board like we use calculators and computers. I have studied the graphical methods used to set out Stephenson's valve gear and my graphical solution matched exactly the Excel program in Don Ashton's book. This was very nice because the answers are not in the back of the book and it was nice to know I got it correct.

Dan
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on June 21, 2017, 07:30:10 PM
MJ,
I failed to mention that one of Dockstader's programs is a Zeuner diagram. If you capture a screen shot of the Zeuner valve diagram of the valve you are discussing, I can show a step by step how to draw a Reuleaux and a Bilgram diagram.

Dan
Title: Re: Talking Thermodynamics
Post by: Zephyrin on June 21, 2017, 11:36:21 PM
Hi
Each of the Charlie Dockstader's programs also contains the oval diagram, which can be modified dynamically, amazing, you move a cursor to change a dimension and see the effect on the diagram. In these programs, PV diagram are simply a model, without thermodynamics in them, hence of limited use.
I spend hours (and night too) with it while drawing the plans of my little locos...
But the Dockstader's package is not fully compatible with the recent windows version.

Z.
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 22, 2017, 12:15:20 AM
Hi Dan and Zephyrin,

You have three out of the four in my text book (by Thomas Bevan), the fourth was called a rectangular diagram.  I also have some of the well known hobby books collected over many years.  But I defer to your expertise on these matters.

My motivation in writing this thread was the realisation over many years of hobby reading that the thermodynamics, that was basic to my every day work, gave me clear answers to many of the questions raised, but never answered in any of the hobby magazines or books I have read.   Or worse, answered wrongly.  The same questions keep coming.  And even simple maths is avoided like poison. 

I hoped that I could build a knowledge base of theory that would help us all understand our models better and remove some of the mystery.  I am not an academic and having people like you looking over my shoulder and joining the conversation is really helpful, just as in building an engine.  And I feel that any forum is best seen as a conversation.

My purpose in looking at valve diagrams was simply to show that the valve events on the P-V diagram actually relate to real valve events in our models, so informing what we are trying to achieve with valve setting.  Most of my machines were electric motor driven so I never had much call to go beyond the eye glazing stage of valve diagrams.  I will be more comfortable when I get back to thermodynamics and other stuff that I can understand.

You both clearly have the expertise in this area.  May I suggest that you start a separate thread (or threads) on valve gear and valve linkages plus any other areas that you think would help.  I would really like to understand those valves in the A and G beam engines that have two valve rods and presumably a two part valve.  If you could eventually get to that .......we would be really cooking.  (I will talk about the gas part.)

If we can get three or four such threads going it might even justify a separate sub board to keep the theory topics together.  After all, aero modellers would not think of trying to advance their hobby without discussing aerodynamics.  We could clearly do with some internal combustion engine topics as well.  A little theory helps in most endeavours.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 22, 2017, 12:10:08 PM
The Oval Diagram

Back to continuing to work through the oval diagram which was attached to my earlier post.

For an engine with no exhaust lap, the edge of the exhaust port is the horizontal line at the valve mid or zero position.  Exhaust lap would show as two horizontal lines, positioned above and below the centre point by the amount of exhaust lap.  We can see the exhaust valve closes a bit before piston top dead centre to give some compression, and opens before bottom dead centre for release.  All as expected. 

The angle of advance for this diagram had to be selected as 30 deg, to make the lap required match my measured valve equal to the valve displacement at piston dead centre.  This suggests my measured lap is really too big and I need to modify the valve ends.  On the other hand, for a simple mill engine, the diagram shows early cut off to give some expansion, and the exhaust valve closes to give some compression so it could be quite efficient for an engine with no reversing gear.

If you want to draw this diagram for your engine, start with the crank angles, then calculate the eccentric rotation as the crank rotation plus 90 degrees plus the advance angle.  It is also helpful to remember that the trigonometric ratios assume that an angle increases anticlockwise.  After that, simple trigonometry, a formula to increase the crank angle by some set amount for each row, and copy your formulae down until you reach one revolution, and use the graphing function of your spreadsheet.  However Dan and Zephryrin have provided links to web sites where the maths is already done, and use of these facilities is a good way to understand your valve setting.

In summary I am now satisfied that the traditional recommendation of no exhaust lap is a very good place to start, and in addition I now have a logical method to analyse the effect of changes in lap on either inlet or exhaust side.  But I still go back to my CAD method to check the width of the bars between the ports, and to make sure that the passages are clear and open.  No more glazed eyes.

So much for a slide valve, it all looks good when analysed according to our theory.  But what about my little oscillator?  I will look at that next time.

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on June 22, 2017, 07:08:13 PM
MJ,
I found the rectangular diagram, it is also known as the harmonic diagram and the sine diagram. It is very similar to the oval diagram only the valve motion and the piston motion are separate curves and not combined as in the oval diagram.

The other three diagrams Zeuner, Reuleaux, and Bilgram, are for a different purpose entirely. They are all graphical constructions used to calculate the relationship between the geometry of the common D slide valve. They all give the same answer. The reason to use one over the other is really personal preference. The slide valve variable that these three diagrams are used to calculate are: valve travel, angle of advance, steam lap, exhaust lap, lead, cutoff, release, compression, and admission.

Obviously to start a slide valve design at least 4 of the above variables have to be known. The Zeuner diagram program by Dockstader has slider inputs for valve travel, cutoff, lead, and exhaust lap. The rest of the variables are calculated and displayed as changes are made to the inputs. I find this a very cumbersome because I have a lot of Shay locomotive data
that includes the valve travel and the angle of advance. I also know the lead for Shay locomotives is 1/16". This information with a glance at the drawings for the steam lap and I can construct a Bilgram diagram. It would be simple to show how to draw a Bilgram diagram for the valve in this thread.

I was a marine engineer and I have a love of vertical steam engines with Stephenson reverse gear. The first time I saw a Shay I saw a marine engine that got lost in the woods and I had to know more. The study of valve design and Stephenson gear took me many years and quite a few times I thought I would never figure out the secrets of how to design Shay valve gear. Finally understanding Bilgram only led to a much harder challenge of understanding Stephenson reverse.

You mentioned not many textbooks cover the thermodynamics of steam engines, well steam engines were fairly well developed while thermodynamics was still in its infancy. I believe it was the need to understand the theory of steam engines that was the driving force behind the discipline of thermodynamics. The earliest book in my collection of steam engine design books that covers thermodynamics similar to what I had in college is The Steam-Engine Theory and Practice 1905 by William Ripper.

The Wright brothers flight at Kitty Hawk would not have been possible without the wind tunnel data they generated before the successful airplane design. They aerodynamic data they generated was the best in the world in 1902.
https://wright.nasa.gov/airplane/results.html

Dan
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 22, 2017, 09:03:48 PM
A new question.......does it take the same amount of coal/heat to boil water in an enclosed boiler at the top of a mountain as it does at sea level ? If the boiler is filled at sea level and then taken up a mountain, any difference ?? etc etc etc.......So can you to test a boiler at any altitude ??
Title: Re: Talking Thermodynamics
Post by: Jo on June 22, 2017, 10:01:32 PM
Willy you need to define the question better: the boiling point of the water will change as a function of altitude as does the combustion efficiency of the fuel. To provide the same volume of steam at altitude you need a bigger boiler or you would need to compress the air going into the combustion chamber.

Jo
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 23, 2017, 01:55:53 AM
Willy you need to define the question better: the boiling point of the water will change as a function of altitude as does the combustion efficiency of the fuel. To provide the same volume of steam at altitude you need a bigger boiler or you would need to compress the air going into the combustion chamber.

Jo
 
Thanks Jo , I am a complete novice when it comes to thermodynamics, but being an autodidact i like to find out about things. I do make and say completely wrong statements and then get told and informed about what is actually correct and then i learn stuff !! I am part of a few clubs and things and am always being advised as to what is actually going on ! !! However i still only know what i know. Sorry about this lengthy diatribe but i am a gemini so i have an excuse !!!! I like your work with all your models and wish i could be  more prolific with what i do......... :)
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 23, 2017, 05:26:38 AM
Hi Dan,  I thought you might be a marine engineer.  I have the greatest respect for your profession.  When at sea, the safety of the ship and the lives of all who sail on her are dependent on your ability to keep the engine running.  In the middle of a vast ocean, there is no one nearby to call on for help.  My activities were essentially land based, and steam power meant a turbine, so I never had need to come to terms with valve gear.

I suspect that you and I together are in a small (elite?) group of people who, when asked if we have read the classics, cheerfully reply, "Of course", thinking they mean classical thermodynamics.  After all, is there anything else you would read?   I did not mean that there were no text books on thermodynamics, just that it is glaringly absent from the modelling press.

So sincerely, please consider a thread about your adventures in valve gear, from the Stevenson's to the Shay.  If I have difficulty understanding them, I am sure there will be many others.  I for one will certainly be following along.

Now, Willy and Jo,  between you, you have mentioned in two (now three) short posts, at least six fundamental questions, the very questions which I suspect are behind nearly all of the confusion in our  understanding of steam engines and the boilers which fire them. 

I am very glad you asked, as I think you have collectively eloquently put your fingers on the big issues which most were afraid to mention.  I will start with the issue of boiling of water and get back to coal and combustion efficiency a bit later.

The basic issues of boiling water at elevation are dependant on whether you are talking about water in an open saucepan or kettle (even with a lid), or a closed space such as a boiler.  If you are  a mountaineer, or even live in Denver or Mexico City, you will observe that water boils at less than 100 deg C (212 deg F), and this might cause problems making a cup of tea or boiling an egg, the typical early school science examples.  To understand this we need to understand a few basics.

First, if we have both vapour (steam) and liquid water in a closed container, the pressure of the water vapour is dependant on the temperature alone.  If things are not happening too fast, they are considered to be in equilibrium, the pressure is uniquely determined by the temperature, and can be looked up in any steam tables.  It is called the vapour pressure.  Hold that thought for a moment.

If we could look at a scale where we could see the molecules, we would see the molecules in the gas moving fast in all directions.  They are a relatively long way apart, and each barely affected by the others except when they collide.   Some hit the vessel wall and bounce back into the fray.  But some hit the surface of the water.  Of these, some will lose enough energy in the collisions with water molecules that they are unable to escape, and stay in the liquid. 

In the water, molecules are also in random motion, but a lot slower and the molecules are closer together.  This close, there are attraction forces that keep the molecules in close proximity so that water stays together.  Some molecules hit the walls and bounce back.  Some reach the liquid surface, but are unable to escape the attractive forces of the closely spaced water molecules.  But  of these, a few with higher than average energy actually escape and join the gas.

If the number entering is the same as the number leaving the liquid, we call this equilibrium.  If there are more molecules leaving, this is not equilibrium, the water must be warmer than the equilibrium temperature, and extra molecules in the gas make their presence felt as extra pressure.  With more pressure, more molecules will return to the liquid and a new equilibrium is soon reached at a higher temperature.

If we heat an equilibrium mixture of water and vapour, some of the water will evaporate, but the temperature will not change until all the water has turned to steam.  The heat that goes into evaporating the water without increasing its temperature is called latent heat, and provides the energy necessary to get the molecules up to a velocity which allows them to escape the attractive forces at liquid spacing.

Boiling occurs when we put in heat fast enough that the water starts expanding into steam below the water  surface.  The huge change of volume causes the vigorous bubbling we call boiling when the suddenly large bubbles of steam quickly rise to the surface.

Now all of this is always true as I have described.  But notice I have only talked about a closed container containing only water.  No mention of anything else.  The confusing bit, the part that explains your confusion arises when, in addition to water there is something else also present, in particular if the something else is air due to the loose lid on our kettle.  The presence of air actually changes nothing but our perception, but that is the root cause of our problem.

That is enough to absorb in one sitting, don't be surprised if it takes more than one reading.  Next time I will explain what happens when our container also contains air.  And that will explain the conundrum of elevation.

Thanks for bearing with me

MJM460
Title: Re: Talking Thermodynamics
Post by: jadge on June 23, 2017, 07:49:55 AM
The book "The Steam Engine and Other Heat Engines" by J Alfred Ewing covers steam engines, including turbines, from both a pragmatic and thermodynamic viewpoint, aimed at undergraduate level. First published in 1894 I have a paperback copy of the 4th edition, published in 1926. The paperback copy wa published in 2013 by Cambridge University Press, who also published the original editions. It is what I use as a basis for understanding the thermodynamics of steam engines.

Andrew
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on June 23, 2017, 05:09:47 PM
MJ and Andrew,
The book I mentioned by Ripper is the 4th edition. The main reason I mentioned it is the first edition of the book printed in 1899 was the first text to include a temperature-entropy chart. The introduction of the temperature-entropy chart was by Macfarlane Gray who read his paper at the meeting if the Institution of Mechanical Engineers in Paris July 1889.

As to starting a thread on Stephenson valve gear, I have one on the web already, unfortunately, it is behind a privacy screen on 7-8ths.info which is a 7/8" scale model RR forum. The good news is I have not built the Shay engine I did the study for and I will cover the topic here when I get to the engine.

I was on vacation for a few weeks and did not read some of this very carefully and I was not near my library. You determined that the usual steam pipe was too small for your design. The problem is the assumption of 2000 rpm. This is a very fast speed for a double acting engine. Most double acting steam engines are much slower. Locomotives run about 300 rpm and marine engines vary from about 120 rpm to around 750 rpm. The reason a double acting engine does not make a good choice for high rpm's is the piston pushes and pulls the con rod, if the bearings are not kept tight there will be a knock as the force changes direction. A single acting engine is much more suitable for high rpm's because the con rod force is always in the same direction. To get the same power out of a single acting engine compared to a double acting engine of the same bore and stroke, the single acting engine has to turn at twice the speed.

Dan
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 24, 2017, 01:06:51 PM
Effects of air in the kettle

Hi Dan, I am really looking forward to following your Shay build, together with your valve gear analysis.  I hope it is not to early to start preparing popcorn, or at least planting the corn seed. 

You are probably right about 2000 rpm being too fast for running my little engine.  This would lead to needing a higher volume of steam, and requiring larger steam pipes.  It was a measured speed, using a non- contacting digital tachometer.  However, the engine was running unloaded, and I would expect it to run much slower under load.  It was also a short stroke oscillating engine which may affect the desirable running speed as the piston speed was quite moderate.  I agree with you about the possibility knocking at the crank pin due to rod load reversal, but the main purpose of the post was to show a rational basis for selecting pipe sizes.  I have noticed people here have a fair idea of the speed they want to run their engines and as you suggest, it is generally much slower.  I hope eventually to build a suitable dynamometer, probably based on a generator to allow me to measure power output, and I will repeat all my tests.

Back to boiling water at elevation-

Last time, I described boiling in a closed vessel with only water, partly liquid, partly vapour in the vessel.  So what happens in our kettle when we also have air at the liquid surface?   I suggested that it changed only our perception, but on reflection, it might have been better to say it changes everything.  Essentially, the pressure is now fixed at the local atmospheric pressure, and as boiling temperature is a function of pressure alone, it varies with the local atmospheric pressure.

Suppose we are boiling our kettle for a cup of tea at the top of a mountain.  We have been told there might be a problem, so we have brought a thermometer, which we put in the kettle and find it starts boiling at 95 deg C.  In this case, the water surface is in contact with the atmosphere through the spout and also the little vent hole you will notice in the lid.  Even the little gaps due to the fit of the lid help, so the pressure at the liquid surface is tied directly to the local atmospheric pressure.

 We look in our handy pocket extract of the steam tables, and find the equilibrium vapour pressure at 95 deg C is only 84.55 kPa.  This means the water vapour pressure at the water surface is only 84.55 kPa, and as the vigorous boiling tends to displace any air from the immediate surface, there is no gas mixture at the liquid surface. Any excess pressure in the kettle is eliminated by leakage to atmosphere, carrying with it any air that was initially in the kettle.  Further away from the surface, we have a mixture of air and water, and we don't know how much of each.  Well away from our kettle, we would find the air pressure measured by our barometer, had we brought an absolute pressure barometer such as a Mercury column, is only 84.55 kPa(absolute) compared with an average around 101.3 at sea level, and it matches the pressure in the steam tables that corresponds with our boiling temperature. This corresponds to very roughly 1500 metres.  If we were on Mt Everest, the boiling pressure would be nearer 50 kPa, and the temperature only a little above 80 deg C.  That might make tea making quite problematic.

The variation pressure (and hence in boiling temperature) with altitude varies in a complex manner depending on the atmospheric temperature variation and the amount of water in the air due to humidity, in addition to  elevation, and not necessary to know for our current discussion.  Google will find you some good information on this if you are interested.  I picked 95 deg as a point available in my steam tables, just for an example.  In fact even in your kitchen at sea level, it is difficult to get water to boil at exactly 100 deg C, due to the variation in atmospheric pressure as the weather patterns pass by, so if you are calibrating your thermometer, you need to know the atmospheric pressure at your location and make adjustments.  Again not necessary for our current purpose.

So I answer to the original question, in an open kettle, the water boiling point depends on the atmospheric pressure which depends on altitude, but in our closed boiler with only water present and not in contact with the atmosphere, the elevation does not affect the absolute pressure at which water boils.

One of the issues with boiling water in an open kettle is that boiling is a very vigorous action that is the very opposite of the reversible process necessary to keep the liquid and vapour in equilibrium.  The rapid motion as the vapour bubbles rise to the surface sweeps away any air, so the vapour near the surface is essentially all water vapour.  In an open vessel, the pressure is fixed to atmospheric pressure.  This very departure from equilibrium introduces an asymmetry between boiling and condensing.  When we get to discuss condensers, you will see a very different and perhaps surprising effect of even an small amount of air.

I hope that helps to answer some of the questions.  Next time I will discuss how these concepts apply to boiler testing.

Thanks for following along.

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 24, 2017, 02:26:03 PM
Thermometers....? If it is of the mercury type in a glass tube, is the 15Lb atmospheric pressure sort of squeezing the tube slightly.? at altitude is this squeezing less so the temperature would appear to read less ?? If the boiler is at altitude would this reduced pressure allow the tube to expand slightly ? Would the boudon tube in the pressure gauge also expand slightly thereby giving a different reading than at sea level ? or does everything cancel each other out ? Sorry to be so pedantic about this but as a novice it would be good to know how things actually behave !!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 24, 2017, 02:35:20 PM
In one of my electrically heated boilers there is a low water safety device that switches off the power. This is in the form of an insulated probe (PTFE) . When the water level drops below the end of the probe it switches off the 250Volts using a separate 9Volt battery circuit. I am surprised that as the boiler is full of steam, this wet steam does not complete the circuit. !! Is this going off at a completely separate tangent or is there some relevance to the topic in hand ??..............
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 24, 2017, 02:40:17 PM
I have noticed that when you buy a new barometer from a shop, they actually calibrate it for you, Here in Norwich the local shop that sells them is 8 meters above sea level !!!..........Also as we know there is a standard  Meter ,Yard Kilogram etc etc so is there a standard i Bar , 15Lbs  somewhere ??or is this being a bit silly !!!!!
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 25, 2017, 12:59:59 PM
Hi Willy, you are in good form tonight, but no, they are not silly questions, and your aim of wanting to understand exactly how things work is why I am writing this thread.  It's just that you are asking questions faster than I can answer.  After all, for both of us, this is our hobby, not our full time occupation.

You are quite right observing that the glass thermometer is subject to external pressure and the inside is closed off from the outside.  So it is squeezed a little less at altitude.  But I emphasise the LITTLE less.  One atmosphere is very small compared with the strength if the thermometer tube under external pressure, so it does not have much effect.  I have not had cause previously to look up a stress strain curve for glass, but I would hazard a guess (and you know engineers are not comfortable with guesses), that the effect on diameter is so small that you would not be able to measure it with your standard workshop tools, and I am trying to stay practical.  If you are really worried, you could use instead a thermocouple, properly calibrated at sea level of course, for your temperature measurement.  I am sure any pressure effect could then be safely ignored.

You know when we only had slide rules for calculation, we had an advantage.  You could only read at best three significant figures, unless you had a monster slide rule, but there was a real limit to the accuracy you could obtain.  Surveyors used seven figure log tables, but for most engineering purposes three significant figures are enough.  Any more is only useful so you can get an exact balance of calculated results.  We used to dismiss small errors as slide rule errors, but there were times when this hid a real error due to an incorrect calculation.

Thermodynamic calculations often involve subtracting two large numbers for a small result.  It would be good to be able to measure temperature to better than 0.1 deg C, but that requires laboratory quality instruments that are not affordable for most, and not really required.  I am confident the error in your thermometer from 1 atmosphere change in pressure would be much less than 0.1 deg C.

So you are right in your understanding of what happens, but I believe the magnitude is small enough not to matter.

Your electric boiler level probe does not use a mechanical contact which opens or closes, but a conductivity probe which measures a change in resistance (which results in a change in a small current in a circuit,) so the range of possible readings is continuous or analogue, not binary.  It then uses an amplifier to lift the currents to a useful level, perhaps to operate a relay, which opens to shut off the mains voltage.  So it depends on the difference in conductivity between steam and water.  Again I do not know, or have immediate access to the data, but you could use your volt meter to measure the resistance of your probe first in cold water, then as you heat the boiler to boiling, then proceed to low level so you measure the steam resistance.

You need to be very safety conscious when you do this test.  Just disconnect your level controller from the probe for the entire test and do not open up the circuit with the mains power connections.  You are the safety switch, so watch the level gauge and switch off as soon as you make the necessary readings.  Also work assuming there could be a fault which allows your probe to touch the mains voltage in the heater element (even though it is most unlikely).  Use a 600 V insulated meter rated for mains use and proceed as you would if everything was at 240 V.  Don't take any risks, again this is only your hobby.  If you are not sure, don't do it, just take my word for it or research the operation of your probe from other sources.  I expect you don't need any of these warnings, but neither of us know who else is reading this, and what their state of understanding or the condition of their equipment might be.

The topic is only limited by what I am prepared to try and answer, based on my 40 years in an industry where this understanding was as basic as arithmetic to an accountant.  I am trying to pass a little of it on to where it might be helpful.  It gives you a fair bit of latitude on topics.

Definitely not being silly about units.  I had intended to get to it anyway, but now you have asked, this is how it works at a practical level.

Pressure is force per unit area, so it's measurement is based on force and area.

Force is defined by Newtons equation, F equals mass times acceleration.  Usually written F= ma

It is the unit which most distinguishes Imperial and ISO metric units, and which causes the most confusion in all calculations involving force.  Metric for a few lines, then I will explain that further.

Mass is measured in kilograms and we have a standard bar of some expensive exotic alloy for that.

Acceleration is defined as change of velocity per second, velocity and acceleration require only length and time for measurement, and we have standards for those.

Now a very powerful analysis technique for these problems is called dimensional analysis.  Don't run away, it simply relies on a principal that in a rational equation, each side of the equals sign has to have the same units.  In addition, each term that is added or subtracted must have the same units.

In plain language, you cannot add apples and pears, and apples cannot equal oranges.  Sounds simple but that principle is behind the equations for force and the scaling of wave generation when ships move through water and many other complex problems.

So if we define F=Ma, we are saying the units of F are the same as the units of m times a.

Let's do it.  I will use square brackets to mean "the dimensions of", just read it that way, then

[F] = [m] times [a]. Then continue with [m] = kg,  and [a] = m/s^2. Read s^2 as seconds squared or
s raised to the power of 2.

So [F] = kg times meters divided by seconds squared.  Or more conveniently [F] = kg.m/s^2

The metric unit of force was given the name Newton and the symbol N, so 1 Newton = 1 kg.m^2.

Not so hard was it?  You only need mass length and time for these measurements.

You might say what about kilogram force?  Now, Wash your mouth out, that is not an ISO unit.  It is a hangover from the same issue as that curse of students everywhere, the pound force.  There was a time when people did not understand the difference between mass and force (has much changed?). Force was given the units of pound, the same as the unit of mass, and both combine in the term weight.  You can see the confusion.  People have tried to extract themselves from this by introducing either the small unit of force, the poundal, or the large unit of mass, the slug.  Personally I don't like either of them.  Alternatively they introduce a constant g, being the same value as the agreed standard acceleration due to gravity, but when to include it asks the student.  So metric for me.  The constant is nearly always one.  But I understand the issues of converting workshop tooling and material dimensions.

Now we have units of force, kg.m/s^2, we can move back to pressure.

Pressure is force per unit area.  Area is length squared, and we have a standard for length.  We have a standard for mass, and one for time, so we standards for all the dimensions we need to measure pressure.  There is no need for a standard bar.

I hope that answers the questions at least sufficient for the moment, it's getting late.

Happy reading

MJM460

Ps forgot the bourdon tube, oh well, next time.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 25, 2017, 01:31:59 PM
Thanks for this, getting a bit clearer.......you mention the Poundal and also the Slug  never heard of that, but, hopefully getting rid of all the slugs on my allotment will be a weight off my mind !! Between the 9 volt battery connection circuit and the 250 volt mains circuit there is a light sensitive IC so it should be quite safe exploring relative resistances with my 1960's AVO or would a modern digital instrument be better ?........also as you are in the antipodes does the water go down the plug hole anticlockwise...the correolis effect ? Or is this another urban myth ?
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 26, 2017, 08:36:48 AM
A little more on pressure measurement.

Hi Willy, I am glad this is making things a little clearer, but sounds like it needs more work.

I realised, after sleeping on it, that I had stopped a bit early on the pressure standard, so a little more to complete the trail from the mass, length and time standards to the units for pressure.

The unit of pressure, Newton per square meter, or N/m^2, was given the name Pascal.  Now this is a very small unit, so we usually use kiloPascal, or kPa, and you will realise that 1 kPa = 1000 Pa.

To help you appreciate just how small, atmospheric pressure, around 14.7 psi is 101.3 kPa, or
101300 Pa.  Mind you for a candle powered Stirling engine, the Pascal would be a very convenient unit.

I should also correct my statement about the pound force.  I found an appropriate reference and there was a standard pound force, defined as the force exerted by a one pound mass due to gravity.  It was complex to use, to eliminate the effect of the air displaced by the mass and so on.  When it was applied to Newtons law, which basically says F is proportional to mass times acceleration, it was found that the proportionality constant could not be 1, but had to have a value equal to the gravitational acceleration.  So the equation in imperial units had to become F = ma/g, and that constant, g, has plagued students ever since.  As the value of g varies at different points on the earth and with elevation, a standard value, agreed at 32.174, is used for the value of the constant.

The principals of dimensional analysis can not be avoided, so the constant also had to have units to make the equation dimensionally correct.  A little maths and you will see the units must be
 lbm.ft/(lbf.s^2).  Quite an awkward mouthful, but necessary to maintain the distinction between force and mass.  In addition, it violates two desirable features of a good standard unit.  First, in a fundamental equation such as F is proportional to m times a, the preferred constant is 1, and second if a constant is required, it should be dimensionless, meaning it has no units.  However the use of g with the applicable units was necessary to accommodate the traditional use of pound as a unit for weight.  Slugs are best fed with snail bait, or fed to birds, but not both.

You can see the beauty of the ISO system.  In Newtons equation, the constant is unity, and it is dimensionless, the equation becomes F = ma and with this equation we can easily deduce the units for force without an additional reference standard.  The resulting unit of Force, the kg.m/s^2, is given the name Newton.  When we apply this to pressure we get Newton per square meter.

I had intended to talk about your question about the bourdon tube.  We had better get M. Bourdon's name spelt correctly or we will have Marv chiding us, ever so gently of course.

You might have noticed in another thread, that one of our forum members is actually making his own pressure gauges, and started with a finite element analysis of the curved tube that forms the basis of the instrument.   Oh, to have had access to that software earlier in my career!  Even access now.  It is a thread well worth following.  I thought it was a bit too much like watch making for my eyes and fingers, but on closer reading, I find it involves more watch breaking. 

I can't do that finite element analysis, but I do have a good feeling for piping.  So I will put on my piping engineers hat, and see how we go.  Now, the Bourdon tube is very like a pipe if you recognise that it is nearly all bend and not much straight.  Also, it has a closed end, and is flattened  a bit, so that its cross section is oval rather than round, but it is still just a pipe and a pipe is something I understand.  So what do we get from this.  First a pipe is designed to hold pressure, and the strength required is determined by the difference between the inside and outside pressure and the diameter. If the inside pressure is higher, the stress-strain relationship for the material means it increases in cross sectional area.  If the higher pressure is outside, it decreases. 

There is also longitudinal stretching under pressure.  When the pipe is bent, the longitudinal dimension change has a very interesting effect.  The bend tends to open up a bit.  The cross section also tends to flatten a bit towards an oval shape.  If it has already been flattened a bit, the change in curvature will be even  greater.  It might surprise you that this happens, and can be easily measured, even on a 12" diameter pipe with 1/2 " thick steel walls, if you apply enough pressure.  Of course the little Bourdon tube is thinner and smaller and the straightening is a lot more noticeable.  The closed end is linked by levers and a gear to the needle so the movement is even more visible.  The scale is marked zero at the needle location when the tube is open to the atmosphere (same pressure inside and outside).  It can move either way from this zero point depending on whether the pressure being measured is greater or less than atmospheric, and the scale is calibrated accordingly. 

The important point in regard to your question, is that the straightening of the tube is due to the stress in the metal, and the radii of the  longitudinal bend and the cross section al radius, not dependant on volume contained in the tube.  Remember, the tube is open to the space whose pressure you are measuring, so any change in volume on results in a very small movement of fluid into or out of the tube.  Any volume effect is already reflected in the scale calibration.

Next time I hope to get back to the questions remaining from you and Jo's earlier comments.

Thank you for helping make this more of a conversation, with your questions.

MJM460

PS - your Avo meter will be fine.  I assumed an optoisolator in your circuit, but you have to disconnect your circuit to make an accurate resistance measurement.  Plug hole circulation is urban myth, Coriolis is real, but plumbing detail has more effect.  If you take notice you should also see different plug holes, sometimes CW, sometimes CCW rotation.  You also have Coriolis in the northern hemisphere!
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 27, 2017, 01:38:16 PM
Some more diverse issues

Hi Willy,  I have gone back and checked my list of your questions and still have a few from post #92 and Jo's post #93.  I will try and cover those, then hopefully I can get crack to valve events on my little oscillator.

You asked about testing at altitude.  I think it has been indirectly answered, but to be clear, testing is about the strength of the boiler to contain pressure, and not about the boiling point of water.  This means we are only interested in the difference between the inside and outside pressures.  It does not matter whether you test it up the mountain or at the depth of the ocean, so long as you measure the difference between the inside and outside pressures.  We have covered the pressure gauge so you know it measures the difference between the local atmospheric pressure and the measured pressure, so it does just what we want.  You can confidently test up the mountain.

You asked about coal and heat to boil water.  The heat released when burning coal, or any other fuel for that matter, is proportional to the mass of coal. You will find the value called calorific value expressed as kJ/kg of coal.  I haven't spoken about joule as a unit, a bit later, I will.  However, a Joule (J) is a measure of energy equivalent to a Newton meter of mechanical work.  We cannot convert all the energy onto work, or even most of it, but we can use the same units to measure it.  In imperial units, the calorific value of your fuel would be expressed in Btu/lb, a British Thermal Unit being a measure of heat energy.

The calorific value of the coal is not dependant on the pressure, so you need the same mass of coal to get the same heat when you burn it up the mountain.

Now, burning coal requires a certain mass of oxygen to burn a given mass of coal, in the same way as any other fuel.  At sea level, the density of air is well known and depends on temperature.  From density, expressed in kg/m^3, we can calculate the volume of a kg of air and hence the volume required by a mass of coal.  In practice, it will not burn very completely, unless we provide excess air to ensure that every molecule of carbon in the coal has a good chance to meet an available oxygen molecule.  A bit like you need more boys than girls at a dance to ensure that every girl has a good chance of meeting a suitable boy.  About 15% or 20% excess air might be a good place to start.  Too much means heat energy is lost up the stack in the oxygen that was not used and the accompanying nitrogen.

Up the mountain two things happen as you know.  The pressure is lower than at sea level, and so normally is the temperature.  The density depends on both, so both have to be taken into account. The end result is a lower density than sea level, so a greater volume of air is required to carry the same mass of oxygen required for our coal.  Greater volume involves more resistance to flow, so you would need a tall stack, or a very good blower to get enough draft to draw this extra air through the coal bed.  Alternatively you could use a forced draft, perhaps a fan, probably not a compressor, to push the air volume through the bed.  Of course if you are on Mt Everest, you would probably have oxygen bottles in your kit.  If you had enough extra over what you need to get back down, you could get the required mass of oxygen from a much smaller volume of air.  Assuming of course that it is enriched oxygen in your bottles, not just compressed air.  I assume it would be.

I am not a combustion specialist so I do not know how the altered volume of air in the bed would affect combustion.  It could be good or bad, I really don't know, so the combustion efficiency I cannot answer.  I hope someone else can join our conversation to help us both understand that one.  Also, I think there are factors both increasing and reducing heat transfer.  Again I am not sure which one predominates to advise on whether you would need more or less heat transfer area.

You asked about whether it matters if you fill the boiler at sea level (and, I assume tighten the filler connection) then take it up, or fill it at the top.  This is a much more interesting question than I thought when you first asked it. 

Had to attend the grandchildren's school concert tonight.  It is quite late, so this and your last question will have to wait until tomorrow.  Wonderful to see these young people developing their talents.  There will be good musicians for us to listen to well into the future.

I hope this is filling some of the gaps and we are building a firm foundation for the rest of our knowledge base.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on June 27, 2017, 04:02:58 PM
A small point of interest ..the local boiler inspector has said the the regulations state that when testing a boiler for leaks the water must be between 7 and 21 degrees centigrade. this is when the boiler is pumped full of water to check for weeps.  still following along ..interesting stuff ......
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 28, 2017, 01:13:22 PM
Our mountain top experience.

Hi Willy,  glad you are still following along.  I really appreciate your insightful questions that prevent me from glossing over details.  I don't know which regulations your inspector is talking about, or what is behind them, other than the fact that pressure will change with temperature, and so a steady temperature is necessary for constant pressure.  In fact, at least a tiny bubble of air is helpful in your boiler when testing, and almost impossible to avoid, otherwise it would be hard to keep the pressure steady through a test.  I do know that in industry, I have been involved in pressure testing in locations where, if we had to wait until the temperature was below 21 C, we would have had to wait until the next ice age.  Well, perhaps at night we would get there, at least in winter, but not for long enough for a big test, and it is very difficult to see any leak at night.  I have also been involved with tests where we would have had to wait until spring if we wanted to test above 7 C, we had to use antifreeze for the test fluid.  In fact, I had to purchase a whole train load, 7 full tanker cars, of antifreeze, and if my memory serves, we still had to get a top up.  I suspect there was a shortage of antifreeze in Canada that year.   Not so far from Brian's territory.  No more war stories, but best to comply with the inspectors requirements.

Back to your question about the difference between filling at sea level or at elevation for our mountain top operation.

Let's first think about the basic difference between the two cases.  The significant issues are the  pressure and temperature.  At sea level, let's assume the pressure is 100 kPa, we have a low pressure system passing over, and the temperature is 30 deg C.  (It is hot at the foot of the mountain).  We fill the boiler to the proper level, and seal the filler plug.  Our steam tables tell us the vapour pressure of the water at 30 C is 4.246 kPa, and of course this is absolute.  The total pressure is 100 kPa, atmospheric pressure is by definition absolute, so the air pressure in our boiler is just under 96 kPa (by subtraction).  We take it up the mountain, the water and trapped air do not know or care what the pressure is outside.  We use an absolute pressure gauge to measure the pressure inside the boiler instead of our standard bourdon tube type, and yes such things exist, so please accept it for now.  What pressure do we expect see on our gauge at the mountain top? 

We might be surprised to find our absolute pressure gauge, when we reach the mountain top, reads only a little below 89 kPa.  Of course our gauge cannot tell whether the water vapour or the air pressure or both have changed, we need our steam tables and some calculations to help with that.  We only know that reading is the total pressure in our sealed boiler.

Our thermometer tells us that the temperature has fallen to 5 deg C.  Just as well it is not below 0!  We do not want freezing to complicate things.  The steam tables tell us that at 5 deg C, the water pressure will be 0.872 kPa.  Near enough to 1 kPa, unless you are Mr. Keenan or Mr. Keyes.  The air mass in the boiler has not changed, and the volume of water has changed but by an insignificant amount.  However the fixed mass of air in the boiler has cooled down along with the water. 

We know that the pressure of a gas at constant volume is proportional to the absolute temperature.  The absolute temperature is temperature in deg C plus 273.  So at sea level, the absolute temperature was 273+30=303 K.  (That is K for Kelvin, and conventionally used without any degree sign.). Up the mountain, the absolute temperature is 273+5=278 K. 

Now we can calculate the air pressure in our boiler as 96*278/303=88 kPa.  The total pressure must therefore be 88 from air plus a touch under 1 for the water, or 89 kPa, just as read on our gauge.  It may even be a little less, as there is still plenty of space between those water molecules, and some of the air molecules which hit the water surface will not bounce back, but will stay in the water.  We find the solubility of air in water is higher at lower temperature. 

Next time we will light the fire, and heat our boiler slowly while we watch what is happening.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 29, 2017, 12:05:50 PM
Following on from my last post.

As we heat the boiler, (it is well insulated, so no miscellaneous heat losses,) we see the pressure and temperature rising.  There is no problem here, as our safety valve, properly set at sea level, looks only at the difference between the internal and external pressure, which is exactly what is required to protect the boiler from over pressure.  We can be confident that it will lift to protect the boiler if required.  (No ice remember.)  It does not matter if it is a spring type, or weight and lever.  (Please ignore the buoyancy effects of the weight in the rarified air if you have a weight type, they will be tiny.)

When it all gets up to 30 deg C, the temperature we had at sea level when we sealed the boiler, we find the pressure is back to 100 kPa, just as it was down there.  The air pressure has increased, and so, independently, has the water vapour pressure.  If we could see in sufficiently clearly, we might see some small bubbles caused by that extra air that dissolved being driven out.

Next we pause at 95 C, the temperature that our water boiled in our earlier saucepan experiment.  The absolute pressure is now about 202 kPa.  This means we have 202 - 84.55 = 117 kPa difference between inside and outside the boiler.  No problem so long as our safety valve setting is above 117 kPa.

Similar drill, at 95 C the water vapour pressure is 84.55 kPa.  The air pressure that was 96 kPa way back when we sealed the boiler at seal level and 30 C is now 96 x (273+95)/(273+30) = 116 kPa.  So the total is 84.55 + 116, say 202 and remembering that there will be even less air dissolved in the water at this temperature so our calculation might be a little low.

Is it boiling this time?  If we put a screw driver blade to the boiler and the handle against our ear, (being careful not to singe our hair,) it certainly sounds quiet.  If we could see inside sufficiently clearly, we would see some more of those small air bubbles as the air that was dissolved, even at 30 C, is driven out, but no boiling.  Something is different here from when we boiled the kettle with a vented lid.

In case you are worried, we can continue heating until the safety valve lifts without any problems,  good idea to test it while we are watching closely in case it is stuck, and we see the pressure continues to rise, due to both the air increasing in temperature, and due to the water vapour pressure increasing with water  temperature.  Boiling will only start when the safety valve lifts, or we open the stop valve to our engine.  So we have time to think about what is going on.

Remember the description of boiling from our saucepan experiment.  Boiling occurs when the vapour pressure of the water exceeds the total pressure at the liquid surface.  (That should be a bit clearer description of boiling than I used before, all this close looking has helped me clarify things in my own mind, particularly the role of the air in the boiler.)  In our vented kettle, the pressure is fixed by the vent to atmosphere, and the water vapour pressure soon exceeds that when heated.  Any vapour generated expands and results in a large volume of steam, which issues from the vent, or spout whistle, and the air is inevitably entrained in the flow, so is soon lost to the kettle.  So the pressure does not increase.  The turbulent boiling is not anything like equilibrium.

In our sealed boiler, conditions are close enough to equilibrium.  As the water evaporates to keep the vapour in equilibrium with its liquid at the surface, the air pressure, is not only still there, it is also rising with temperature.  This is where Dalton's law of partial pressures comes in.  It says that in a mixture of gasses, each acts on its own, as though the others were not there.  And this is close enough to what happens until very high pressures.  Essentially, while the air is still there, the water vapour pressure cannot exceed the pressure at the surface, and even with out the air, it can only equal, not exceed the surface pressure.  Water cannot undergo that huge expansion into vapour which causes the turbulence we call boiling, the pressure just continues to rise until something changes, preferably nothing more than the safety valve lifting, or the engine throttle opening.

As soon as some mass escapes the fixed volume of our boiler, the pressure must decrease a little.  In a fixed volume, the pressure is proportional to mass.  Some extra water will evaporate to replace the lost steam, but so long as the total pressure is above that equilibrium vapour pressure for the water temperature, no dramatic boiling happens.  Remember we are proceeding slowly here, to trying to stay close enough to equilibrium.  As soon as the pressure drops below the vapour pressure, some water will evaporate which tends to maintain the pressure, but we keep letting some out.   The remaining air is not very important, and some of it is inevitably entrained in the escaping steam, but is not replaced, so soon we are only dealing with water alone, and steam tables tell us the relationship between temperature and pressure, which is now totally water vapour pressure.

A bit of extra heat, and the vapour pressure will exceed the pressure at the surface, and boiling commences.  It is not very dramatic, as we know from every time we start one of our boilers with air and water sealed inside, the same processes happen.  The only difference is the precise starting conditions.  The water which evaporates, absorbs the latent heat from the remaining liquid, cooling it, so only a very limited mass can evaporate, limited by the heat coming in from our fire.

I do seem to use a lot of words in answering these questions.  I hope it is making sense.  But I will wait until tomorrow to talk more about equilibrium, and the other half of this double banger question.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on June 30, 2017, 12:52:12 PM
Equilibrium

The issue of equilibrium has been mentioned often, but perhaps its meaning and importance not clear.  Not all processes occur in equilibrium.  As I write this, I am also responsible for supervising the making of soup for lunch.  In the pot, in addition to uninteresting fillers like pumpkin and carrot, etc, there is some water in the bottom, placed there to help conduct the heat so things don't burn, and also some ice, from frozen stock added to help flavour, The impurities in the stock mean I don't really know it's melting temperature, but close enough for our purpose.

But food!  That should provoke some interest.  I notice that the water has begun to boil, (the lid lifts as necessary to prevent any pressure increase) so there is steam (water vapour), water and ice all in the pot, at the same time, along with some unknown quantity of air.  Now that interests all thermodynamicists, as we all know it is only possible to have water, ice and water vapour at the same time in equilibrium at a specific water vapour pressure and temperature, called the triple point, about 0.01 deg C and 0.6113 kPa, (if that number of decimal places is consistent with the term "about").  Now my thermometer tells me the temperature is hovering around 99 C, (you do monitor your soup temperature don't you?) and my steam tables tell me therefore, the vapour pressure should be close to 100 kPa, well above that triple point pressure.  Of course, there is also air, as my pot is not sealed.  This all  tells me that the system is not in equilibrium.  Equilibrium requires no turbulence and also that the temperature be uniform throughout, which it cannot be with ice, water and water vapour all at the same time, except at that one special temperature and pressure.  Any two, but not the whole three.  And equilibrium is a necessary condition if we are comparing our process with those ideal processes such as adiabatic, isothermal, and similar.  If we are nowhere near equilibrium, our actual results may be very different from our calculations.

But mmmm! That soup smells good.  To make the soup a bit quicker, I started with the stove on high (setting was 9) and watched carefully while everything heated up and the water started boiling.  Then I turned the stove down to only 2, which was enough to maintain the temperature just simmering, with a little excess heat to just evaporate a small amount of steam to cook and soften the vegetables which were not submerged in the water.  No wild boiling or drama, just a wisp of steam from the lid vent, probably not enough to reliably remove all the air, just enough to limit the steam lifting the lid to an amount I could accept.  This keeps the temperature relatively uniform, rather than burning the vegetables where they touch the pot.  Very efficient use of steam to carry a lot of heat from the bottom of the pot to where it is required for the cooking, unlike our small engines which convert only a tiny portion of the heat to the work that we require.  Of course, the rest of the heat is mostly still contained in the exhaust, and could very well be used to cook the vegetables, if you don't mind a bit of oil flavouring.  This is called combined heat and power, or sometimes cogeneration.  And of course some of the heat is lost to the atmosphere through our cladding.

So after a little thermodynamics lesson in the soup, it's  time for a break for lunch.  Next time, I will think about what happens in Willy's alternative scenario, carrying our boiler empty, and filling it at the top of the mountain from a handy stream.

I hope everyone is still with me,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 01, 2017, 01:39:48 PM
I had hoped to continue today, but my post is not quite ready, so I hope to continue tomorrow.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 02, 2017, 10:20:31 AM
We fill our boiler at the top of the mountain.

A bad day yesterday.  A beautiful page of writing, full of calculations, all carefully checked, but in the end it did not clearly answer the question.  The knowledge base equivalent of the beautifully machined part that does not fit.  Cutting a second time did not help, and it was all consigned to the bin.  So here goes with a second attempt.

I believe I have covered the first part of Willy's question, what if we fill (and seal) our boiler at sea level, then carry it up the mountain to operate, even if with a few too many calculations.  Now the other part, is it different if we carry the boiler up empty and fill it from a convenient stream at the top? 

We begin by looking at our starting conditions for this case.  It is a way back, so I will remind you that at the top of the mountain, the atmospheric pressure was only 84.55 kPa, remembering that atmospheric pressure is always expressed in absolute pressure units, whether it be measured as 14.7 psi or 101.3 kPa, or some other value as in this case.  Also  we found that water in our vented kettle boiled at 95 deg C.   More recently, we decided the air temperature at the mountain top was only 5 deg C, definitely a bit chilly unless you are in Barrie in early spring, when any plus temperature means it is time to put away the scarves and heavy jackets.

When we fill our boiler, both the water and air will be at 5 deg C, and we assume the metal boiler shell, no longer in our pack, has also cooled to 5 deg C.  The absolute pressure will be 84.55 kPa as before, then we install and tighten the plug and connect our absolute pressure gauge.  At 5 deg C, the water vapour pressure is 0.872 kPa as before, so the air pressure in our boiler is 84.55-0.87= 83.68 kPa.

Now you might have noticed that I have jumped around a bit on the number of significant figures and accuracy.  A barometer is in fact an absolute pressure gauge and is calibrated in hectopascals, (10^2 Pa, or hPa, it seems Mr. Apple does not know they exist!)  Standard atmospheric pressure is 1013 hPa, and that fourth significant figure is quite significant in determining the weather we will experience as the highs and lows of our weather pattern pass over.

You will also notice that I have not bothered to look at humidity.  Despite its "couldn't care less" attitude to air molecules, water is very inclusive about water molecules.  They are all treated equally, whether they arrived with the air as humidity, or they evaporated from the water.  The vapour pressure of water is only dependant on the total number of molecules in the space, which in turns depends on the temperature of the liquid at the surface.

Our operating conditions are dictated by the boiler design which is determined by the difference between inside and outside pressures.  Now, our standard pressure gauge measures just this very difference.  So up the mountain, we still operate to the gauge pressure, not the absolute pressure.  The same gauge pressure at the top of the mountain, where the atmospheric pressure is lower than at sea level, means the absolute pressure in the boiler is lower, and hence the boiling temperature of the water (once all the air is expelled) will be lower.

The other difference will be the starting conditions when we are up the mountain.  At sea level the air density is a little more, so there is a little more mass of air in the boiler which will add to the water vapour pressure so changing the total pressure in the boiler until it has all gone out mixed with steam production.  Also, air, water and the boiler mass will be sea level temperature of 30 deg C.

At the mountain top, the air density is less, and the temperature is less.

Because of the lower density, if we fill the boiler at the mountain top, we will trap a smaller mass of air when we insert the plug.  This will always exert a lower partial pressure contribution to the total pressure than the greater mass we trapped at sea level.  However, once we reach our operating pressure and allow a little steam to escape, the air will be mixed with the steam and soon gone.  Then both cases are the same.  Boiling commences as soon as the vapour pressure of the liquid exceeds the total pressure at the liquid surface.

The temperature effect is unchanged between the two cases, as we have already noted the boiler and its contents will be at the same temperature whether we fill at sea level or at the top. So our starting conditions at the top of the mountain are 5 deg C for both cases and slightly less air partial pressure for the case when we fill at the top.

I hope that answers that question, so next time on to the issue of heat.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 03, 2017, 07:09:30 AM
A question of heat

Hi Willy,  now for the final question to the list I made from your earlier posts, the question of heat.

 It was only one word in your original question, I almost missed it.  I am assuming the implied question was, "Is there a difference in the heat required when we operate the boiler up the mountain compared with sea level?"

I will work with the operating conditions we have already chosen, 275 kPa(g) as normal operating pressure.  This means a slightly lower absolute pressure on the mountain than at sea level, but the boiler has been designed for a difference between inside and outside pressure, not an absolute pressure.  Our standard pressure gauge provides the right measure of operating pressure, both on the plain and on the mountain.

We can determine the required heat input from the starting conditions and our operating conditions, using the law of conservation of energy and the steam tables.  Conservation of energy is the generalised form of the first law of thermodynamics.  I am only considering the boiler, no superheater for simplicity.  It will adequately illustrate the principle. 

Heating the water from our starting conditions until it reaches our operating pressure is a constant volume process in our sealed boiler.  A feature of this process is that no external work is done.

Careful application of the first law of thermodynamics reveals that for a constant volume process, the heat input required is equal to the change in internal energy.  Now the steam tables conveniently tabulate values of internal energy over the range we are interested in, so perhaps I had better briefly explain a bit more about steam tables.  You might have a copy in a thermodynamics text book, or you can purchase them as a separate booklet.  Or you can, as usual, ask Google.

Steam tables have two basic sections.  The first is just a solid block of figures that apply to the two phase range of conditions, that is when you have both liquid and vapour in equilibrium, and is the section used for boiling and condensing conditions.  Even this is divided into two parts.  One part has the temperature in the first column, and corresponding equilibrium pressure in the second.  The other part has pressure in the first column and the corresponding equilibrium temperature in the second.  Use the one that best suits your particular conditions, in particular whether you know the pressure, and want to know the temperature, or do you know the temperature.

The other part of the steam tables looks like a number of separate small tables.  This section covers the superheat range.  More about that section when we talk about boilers.

We will use the first section, called the saturated steam table, as we have both liquid and vapour in our boiler.  After pressure and temperature, the  next two columns show the specific volume of liquid and the specific volume of saturated vapour.  Specific volume is simply the reciprocal of density, and I assume the reason for the convention is because when calculations were done by hand, specific volume involved less division, so was easier to deal with.  Specific volume is the volume in cubic meters occupied by 1 kg of fluid.  You can see the specific volume of the liquid is very close to 0.001 m^3 or 1 litre per kg as we all know, while as vapour, it occupies a much larger volume and you can see the expansion ratio in these two columns.

Then we have normally have three columns for internal energy.  The first, Uf, is for water just about to boil, referred to as saturated liquid.  The third, Ug, is for saturated vapour, vapour when the last of the liquid has just evaporated, before superheat starts.  The second column, Ufg, is the difference between the two, which saves a step in many calculations, again for the days when calculations were done by hand.

At sea level, we sealed our boiler at 30 deg C which is our starting condition.  We could look up the internal energy of water at 30 deg C,  and again at our operating pressure, but this is where things get complicated!  Some of the operating pressure is due to the remaining air, so the water is not yet up to operating temperature.  It will not be until we let the air out, and we will also let out steam in the process.  Once we let some steam out, we no longer have a constant volume process, so internal energy no longer tells us what we need to know.

It all makes a precise calculation quite difficult. So I will try and answer your question a different way that I hope will tell you what you want to know.

I think that is enough for one session, so back to this question again tomorrow.

I hope it is all making sense so far.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 04, 2017, 11:48:18 AM
Continuing with the question of how much heat is required to boil water.

My first application of conservation of energy to this problem led to a further difficulties when my initial assumption of a constant volume process fell down.

Let's start again, this time looking only at the water.  Starting at 30 deg C, we gently heat our boiler until the pressure reaches our operating pressure of 275 kPa(g).  Now we let out a little steam air mixture to control the pressure at 275 kPa(g).  We no longer have a constant volume process, as our steam is expanding into our outlet pipe.  If we hold it all at our selected operating pressure, we have changed to a constant pressure process.  We can no longer rely on the internal energy to solve our problem, we have to go back to the first law of thermodynamics and apply it to a constant pressure process. 

This time our text book introduces the concept of enthalpy, one of your magic words that so far I have managed to avoid.  The first law analysis tells us that for a constant pressure process, the heat input equals the change in enthalpy!  Again, in most thermodynamics texts if you want to look at exactly how that is determined from conservation of energy.

So I am going to have to try and explain enthalpy in one easy lesson.  That will test me, but let's have a go.

Many of the thermodynamic problems associated with both gas and liquid are solved by applying those basics laws of physics, conservation of energy, and conservation of momentum.  I have referred to those previously.  We start with the three basic properties that we can measure, pressure, temperature and specific volume (or density if you prefer), then apply conservation of energy and/or conservation of momentum to find the solution.  When we do this, some combinations of the measured properties come up often.  Specifically, the combination U + p x v.  Internal energy plus pressure times specific volume.  It turns out that this combination behaves very like another property, but we have no gauge to measure it directly.  Steam can have the same value for U+pv for the same mass when evaluated at different pressure or temperature.  Some processes can proceed along a path on a pressure-temperature diagram, along which the value of U+pv is the same all the way.  Just as we can proceed along a path where the pressure, or the temperature (but not both) are the same all the way.  Hence this combination of U+pv behaves like a property, and is given the name enthalpy, and the symbol, h.  Enthalpy, like internal energy, has the units of kJ/kg and looks rather like a particular sort of energy.  It occurs so often, and is so useful that the next three columns of the steam tables list the value of enthalpy in a similar format to the columns for internal energy.  In case you are wondering, it occurs frequently in processes involving work at an external boundary, such as in our engines.  I hope that is sufficient for the moment.

So back to the question.  If we control the pressure in the boiler as the steam escapes, it becomes a constant pressure process which we can analyse using the enthalpy columns of the steam tables.  To find out how much heat is required to heat our water from the cusp of boiling until all the liquid is evaporated into steam while we let out some steam and air to hold the pressure constant, we can simply look up the change of enthalpy.  Furthermore, it turns out that we can use change of enthalpy right back to our starting point, even though the first part of our heating is really constant volume.  This conveniently avoids the issue of just how we changed from constant volume to constant pressure.

If we continue just looking at the water, we can now start at 30 deg C, and look up the enthalpy for water, hf = 125.79 kJ/kg.  We can then look at the enthalpy of the saturated vapour at our operating absolute pressure of 375 kPa, (275 kPa(g)) and find hg = 2735.6 kJ/kg.  Then the heat input required by the water is 2735.6 - 125.79 = 2609.8 kJ/kg.

Now if we go to the mountain top we do the same analysis with two differences.  First the starting temperature, and hence the enthalpy of the water, is lower at only 5 deg C, and second our absolute pressure at 275 kPa(g) is only 359.55 kPa absolute compared with 375 kPa at the foot of the mountain.

Now starting at 5 deg C, we can see the enthalpy of the saturated liquid is just 20.98 kJ/kg.   Our operating pressure of 359.55, say 360, is close enough to the saturation pressure for 140 deg C.  You can see hg for the steam released from the boiler will be 2733.9 kJ/kg.  When we insert the figures and do the subtraction, we see that we need 2733.9 - 20.98 = 2712.9 to boil our water on the mountain.

Comparing the figures, we see that we need 103.1 kJ/kg extra to boil our water at the mountain top.  That is interesting, but how did it come about?

If we look closely at the figures, we see that we needed 104.8 extra to heat the water from 5 deg C to our mountain top boiling point of 140 deg C, while the enthalpy of the steam at the lower absolute pressure on the mountain is only 1.7 less than it was at the higher absolute pressure on the plain.

My conclusion is therefore, that we will need more heat to produce our steam at the mountain top, and the reason is almost entirely due to the lower starting temperature.  The slightly lower absolute pressure at our operating gauge pressure makes a difference of only 1 part in about 2600, which is well beyond the accuracy of most of our experiments, while the extra heat to get from 5 to 30 degrees, (104.8 kJ/kg) is about 4% of the heat required to boil from 30 deg C.

A small boiler like we might use for these experiments might contain about 1 kg of copper.  This would require 42 kJ/kg to heat from 30 to 140 deg and about 52  to heat from 5 to 140, thus adding about 12 kJ/kg to our extra heat requirement, and we need to allow a little for the lagging.  The other losses will be somewhere near proportional to the heat required by the water.  On this basis the total heat required on the mountain would be not much more than 5% more than on the plain.

The differences we have found may not seem very much. They could have been made to stand out a bit better had I continued assuming you might want to climb Mt Everest.  But I am trying to stay practical, and let's face it, it's crowded up there these days, unlike when Sir Edmond Hillary did the climb.  It is unlikely that the crowd would be willing to hang around, stamping their feet to keep warm, while you carefully measure your boiler temperature!

Next time, back to my little oscillator and its exhaust port, I hope it will have something for everyone.

Thanks for hanging in there,

MJM460
Title: Re: Talking Thermodynamics
Post by: derekwarner on July 04, 2017, 01:01:02 PM
.....'Next time, back to my little oscillator and its exhaust port, I hope it will have something for everyone'

Looking forward to the exhaust lesson MJM.....this is where I am currently chasing my temperatures OK :cussing: ...however without clear resolution  :headscratch:

Derek
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 05, 2017, 01:30:14 PM
Returning to talking about engines.
 
Hi Derek, glad to have you still looking in.  In your previous posts, you were recognising that your exhaust piping was too small and were awaiting parts, which I assume have now arrived.  I assume that "still chasing temperatures" means you might be trying to condense the steam.  Or are you just trying to collect the oil?   Or are you trying to achieve vacuum conditions for your exhaust?  Please let me know which.  First, two posts on oscillating engines, then I will go on to exhaust.  I am sure that through that discussion, together we will be able to solve the problem.

 Those questions and comments by Willy and Jo, were deceptive in their depth, and quite a challenge to answer clearly.  I hope that I have done them justice.  Some of the issues will be discussed further when we come to boilers, and will also help our discussion of condensers.  They have given me the prompt to deal with some basic issues before I go too far.  Thank you both, and thank you to the others who have contributed by asking questions or commenting so far.

Back in post #90, I had been talking about valves and their part in conducting the sequence of processes that enable our engines to run in a continuous manner.  I had looked at the requirements for the valve opening to match the volume swept by the piston.  It turns out that the inlet valve opening matches quite closely the opening required, based on looking at required velocity in the port to maintain pressure as the piston goes down from top dead centre.

The real issue comes with the exhaust valve opening.  While the piston is slowing in its decent, the piston is full of steam which has done its work and must be expelled in the exhaust stroke and this is the most demanding event from the capacity point of view.  The slide valve is not too bad, with wide ports with sides parallel to the edge of the valve plate, so opening occurs quite quickly, and allows rapid exhaust of the pressure in the cylinder.  But looking so closely does show the advantage of those wide straight sided ports typical of a slide valve engine, in case you are tempted to just drill round ports.  A piston valve has a similar characteristic, with the ports wrapping around the valve.

That brings me back to my little oscillating engine.  Most of us have built, or contemplated building at least one of those, or at least seen them in action in toy engines like the Mamod range and others, excellent models which gave many of us our first hands on experience of steam.  I still have the one so much enjoyed by my brother and I.  Of course in those days, real trains were hauled by those magnificent steam locomotives, with the associated sound and billowing steam.  And we got coal specs in our eye when we put our heads out of the windows, which actually opened.

In an oscillating engine, also known as a wobbler, the valve port is formed by the overlapping portion of the drilled ports, one in the cylinder and one in the port block on the engine stand.  They are generally located so the openings do not overlap at all at top and bottom dead centre.  As the engine rotates, the cylinder port starts to overlap the port in the block on the stand to start admitting steam.  A typical design has them finally fully overlapping at the maximum cylinder movement, about 80 deg of rotation, after which they start closing again.  There is some variation, sometimes planned and sometimes due to accuracy of drilling in the required location.  There is only a momentary complete circular opening.  For most of the revolution, the opening is only the overlapping section of the drilled cylinder and steam ports.  In case the words are not clear, I have attached two sketches.  One shows the general port layout, the other, four steps in the port opening, the open section filled in red.  Once maximum opening is reached, they progressively close until bottom dead centre.  The exhaust port opens in a similar manner on the upstroke.  With a little trigonometry and some circle geometry in a spreadsheet, we can calculate the actual opening at each point of the engine revolution.

It should be obvious however, even without calculation, that just when the engine needs a large exhaust opening to quickly exhaust the steam, the port is nearly closed, and only slowly opens as the exhaust stroke progresses.  Consequently, the piston starts its return stroke for the exhaust with significant pressure in the cylinder, and the port is not fully open until the piston is over half way back to the top.

We have already seen that pressure in the cylinder on the exhaust stroke actually works against the piston movement, subtracting from the work done in the power stroke and reducing the output of the engine.  In a single acting engine, the work necessary to drive the exhaust stroke comes from the flywheel, and in the process, the engine slows during each exhaust stroke.  Despite this, the engines do run, and quite well, though they do not handle much load.  If we could allow the cylinder to exhaust freely, the engine would be considerably more capable.

Next time, I will look at two possible solutions that have been proposed at various times to overcome this problem.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 05, 2017, 02:10:20 PM
Hi, still following along........one of my favourite sayings is : the adiabatic enthalpy : principal  but i don't know if you can use the two words together !! It does however impress lesser mortals !! One question on a more practical note is ..Where is the best place to introduce the feed water into the boiler ??
Or is this a topic all in itself ? :happyreader:  also rather then a round hole for the Wobbler port would it be beneficial to have an oblong hole so as to introduce cut off  ?? this is also a whole new subject possibly !!
Title: Re: Talking Thermodynamics
Post by: Maryak on July 06, 2017, 12:14:38 AM
.Where is the best place to introduce the feed water into the boiler ??

In full size practice, feed water is introduced via internal feed pipes which evenly distribute the water along the length of the steam drum in watertube boilers. In in firetube boilers there is a distribution box fitted in the middle of the shell which acts as a form of feed heater. The idea is to reduce the effect of the difference in temperatures between the boiler water temperature and the feed water temperature.

Regards
Bob
Title: Re: Talking Thermodynamics
Post by: derekwarner on July 06, 2017, 01:34:01 AM
....."Or are you trying to achieve vacuum conditions for your exhaust?  Please let me know which"

MJM.....explanatory PM sent so as not to discolour  :toilet_claw: or cloud your thread......

Looking forward to the next instalment & reading with interest  ....Derek   :DrinkPint:
Title: Re: Talking Thermodynamics
Post by: Zephyrin on July 06, 2017, 07:51:42 AM
Quote
the port is nearly closed, and only slowly opens as the exhaust stroke progresses.  Consequently, the piston starts its return stroke for the exhaust with significant pressure in the cylinder, and the port is not fully open until the piston is over half way back to the top.
as opening and closure takes time, to get widely open port when needed, we need advance in the steam distribution, but a simple wobbler cannot cope to that.
And there no doubt that straight edges are preferable for the steam ports, as the rate of increase of the opened area is much higher than with circular holes for an identical valve displacement.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 06, 2017, 12:31:47 PM
Exhaust ports for oscillating engines

Hi Willy, Bob and Zephryn, glad to hear from you all again.  Also received a PM from Derek for me to think about. 

Now, adiabatic enthalpy!  That is a mouthful.  You probably can construct a sentence with the two words together, but without a little more context, they are perhaps not as complete or clear in their meaning as you might like.  I suspect those who nod in wise agreement did not really understand what you said, just as you implied.

To get away from the thermodynamics, a student of English might class "adiabatic" as an adverb, while "enthalpy" is a noun, so you can see the problem.

Enthalpy is a property, just like pressure and temperature (though we have to calculate its value, as we don't have a direct reading gauge), perhaps refer back to Post #113 for more detail.  It's value does not depend on the process involved in getting the fluid, in this case steam, to that condition, just final temperature, pressure and density.

On the other hand, Adiabatic is the description of a specific ideal process, meaning a process that occurs without heat transfer.  Expansion in a well insulated cylinder is a process that closely approximates adiabatic, as is throttling through a valve or nozzle.  An adiabatic process can be analysed using basic thermodynamics, and the work output can be predicted.  Unfortunately most real processes, such as the expansion of steam after admission is cut off, cannot be analysed, so we can't predict the amount of power our engine will produce.  Not very useful if your brief is to design an engine for the Queen Mary.  But we can compare the power produced by a real engine on test with an ideal adiabatic engine, and define an efficiency accordingly.  Just one of many definitions of efficiency, but more about efficiency another time.

So your favourite term needs to be used in the context of an enthalpy change.  An adiabatic process involves a change of enthalpy from which we can calculate the work produced.  A real engine involves a different enthalpy change from that which occurs in a true adiabatic engine, always less, due to thermodynamic losses classed as irreversibility.  Now you are prepared for the one who asks for further explanation, and you can safely refer to "adiabatic enthalpy change" as opposed to real engine enthalpy change.

Bob has very clearly answered your question about feed water entry.  We don't need quite such a complex arrangement in a model boiler, but the principle is still to minimise the temperature difference at the entry point.   Always below the surface to minimise splashing and carryover, and you could extend your inlet pipe a little to get the worst temperature gradient further from the shell.  You could think about other ways of reducing the temperature difference in preparation for a future topic.

Your final question, and Zephryn's comment about whether an oval or straight sided port would help the oscillating engine, very neatly introduces what I was thinking of today.

K. N. Harris, in his excellent book on engines, looked at this problem and suggested modifying the ports to have straight sides.  Many of us have a copy of this book, but I have attached a sketch to show what he described.  It always struck me as a great theoretical solution to the problem, but I did not have the practical knowledge to be able to implement it.  I made a test block and tried with fine files, and tried the traditional " chippies" cold chisel to get the required port shape, both without success, probably due to inadequate skill level in both cases.

Florian might be able to comment on whether his excellent thread on making a broach could be adapted to produce the required shape.  I suspect the real problem for broaching in this application is that the required holes are very shallow.  The cylinder port for example, the broach could only protrude one cylinder diameter before it hit the wall on the far side.  In the base port block, holes do not go right through, and I am not sure if there is space to drill and tap the back face and insert a plug after broaching.

Now, having a mill and rotary table, and having learned many new skills by following the experts on this forum, I have also come to wonder, as you have, about making straight sided oval ports.  I even have a 2.5 mm milling cutter, and could buy a 2.0 mm one, though the bigger ports are fine on my 12 mm bore model.  I suspect that this is what you mean, but again, I have attached a sketch to show what I propose.  Now that I have sketched out, it looks very like Mr Harris' design.  Should work.  As Zephryn says, still not a perfect solution, but surely better than round ports.   Later this year I may get a chance to give it a try.  But not with my present schedule.

Mr. Harris also presents another solution to the exhaust opening conundrum.  This one may be even more effective, and is certainly more interesting from the thermodynamics point of view.  It was even the subject of a patent in the early days of steam, so on to that next time.

Thanks for following,

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 06, 2017, 02:37:12 PM
Hi, more interesting stuff...i am thinking about a wobbler engine with triangular shaped ports with a 90 degrees offset wobbler part operating at the back of the port plates to achieve a cut off sequence.....!! or would this be clueless rather than corliss ?? !!!! might produce a drawing  !! Also i seem to remember an article in Model  Engineer some time ago with someone doing some experiments along these lines, the person might have been from Australia if i recall..Thanks for all this info,  it will help when i actually meet someone in person that also knows all about T.....
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 07, 2017, 07:15:04 AM
An alternative solution for oscillating engines.

Hi Willy, that second plate looks like an interesting concept to explore, it would be good to have some way of adjusting valve events.  Definitely not clueless.  You might need an eccentric to drive the plate to allow you to extend the shaft through the two fixed parts.  Also to allow timing adjustments.  I am not sure what the triangular port is about.  To me, ports are a source of irreversible loss, they all  turn good enthalpy into wasted heat in the exhaust.  Since we must have them, they should be large as practical, and always fully open or closed, with opening/closing times minimised.  And timing is the important thing.  But perhaps there is something I have not thought of.  Well worth sketching out and exploring the valve events.  There is also that very interesting design with the slide valve on the side of the cylinder, that has been the subject of some recent build logs. 

The background article in your picture is also  interesting, but I think it is not very helpful to talk about scaling nature.  We are building real engines, just small in size.  So some things might not work quite the way they do in full size, such as viscosity and surface tension.  However, mostly these are more obvious in ships for displacement and wave making, and aeroplanes  for wing lift and drag.  Almost certainly a few factors in making a miniature violin as well if we want the notes to be in the audible range.  It is more useful to think about which particular bit of nature you are trying to scale, than lumping all nature into one catchall term.

To round out my discussion of valve opening, and for oscillating engines in particular, I would like to apply some thermodynamics to the engine Mr Harris called a Uniflow engine.  It was not really an original idea, there was an early patent on the principle that I uncovered in a simple Google search, though it did not seem to have made its owners any great fortune.

The idea was to place an additional exhaust port low down in the cylinder, so that it was only uncovered when the piston was close to the bottom of its travel.  The idea was that the steam, instead of having to reverse its travel to be exhausted at the top, could continue moving down and out, thus avoiding the force on the piston necessary reduce the momentum of the steam and then reverse its direction.  He concludes by inviting his readers to try it, and see the extra output from the engine.  I would certainly encourage anyone interested to give it a try by adding the Uniflow port to their next engine.  And certainly a class of young people about to embark on engineering studies by building a small oscillating engine.  It will help introduce them to some basic thermodynamic principles of heat engines.  I hate to admit that I have not yet done it myself, but I have no doubt about the claimed increase in performance being real.

Observations usually are real, and when they depart from theory, my observation is that it is usually the wrong theory being applied, or often a different theory, also applicable, has more influence on the outcome.  The published explanation in the Uniflow patent appropriately describes momentum, but then alarm bells ring and I find myself skeptical about the explanation for the function of the port.  And I do not wish to be in any way disparaging about Mr Harris' effort, he has certainly made a great contribution to our hobby with his books on engines and boilers.

First, we have to deal with the fact that it is the momentum of the molecules being changed at the piston face which causes the force on the piston that drives the engine.  The crank mechanism means that the piston velocity is reduced to zero at bottom dead centre, so it matters not whether it then accelerates up, down or probably sideways to exit the port.  Of course the torque is minimal near the bottom of the stroke, so loss of torque due to opening the port is not much.

Then we note that the overall downward velocity of the steam is a very small nett velocity, superimposed on the quite high velocity of the molecules in otherwise random directions.  Remember way back, I looked up the typical molecular velocity of about 500 m/s, while the average piston speed is only about 5 m/s.  So the momentum due to the average downward velocity of only 1% of the momentum of the typical molecule, surely does not make that much difference.

I suspect that it is only when we appreciate the difficulty of opening the exhaust port enough to exhaust the pressure at the beginning of the exhaust stroke, that we see the more likely reason for the observed performance improvement.  It provides additional exhaust port area just when it is most needed.  In any case, on a single acting engine, drilling the port, but keeping it separate from the top exhaust, with some provision to block and thus disable it, would surely be an interesting experiment.  Not so easy on a double acting engine.  The piston length would have to be carefully proportioned for the port to work on both up and down strokes.  It would also be more difficult to keep it separate from the other exhaust ports so the engine could be reversed.  Still good food for thought, and a potentially interesting experiment.

That is not all that can be said about thermodynamics in relation to engines, however I believe that clears the decks enough to move on tomorrow to exhaust systems, which should help Derek solve his current issues.  (Derek, I have replied to your message). And I will return to some other engine topics as the need arises. 

I hope you have found this explanation of how an engine changes heat contained in the random motion of tiny molecules, to mechanical work, interesting and informative.

Thanks for dropping in,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 08, 2017, 11:26:26 AM
Exhaust systems.

Thinking still about a vertical engine, first single acting then adding any additional comments on the difference for double acting, lets now have a look at the exhaust system.

 Keep in mind that the work output of the engine is always reduced by the back pressure on the piston during the exhaust stroke.  It is obvious then, that we can increase the output of the engine by reducing the pressure in the exhaust system.  Now this approach is limited in extent, as we can only at best get to full vacuum, which gives only 101 kPa (14.7 psi) additional difference in pressure across the piston.  In practice, we are unlikely to achieve anywhere near this.  However, in most cases our models are operating at quite low pressure, so even minimal vacuum is a significant proportion.

There are four basic exhaust concepts, so let's look at each in turn.

First, there is the simple engine outlet to atmosphere.  This is seen on some of the beginner engines with a simple piston valve and no obvious way to collect the exhaust into a simple pipe.  Also on simple Mamod engines.  Sometimes, the exhaust is directed up the boiler stack.

This system is fine for simple engines and for engines we only intend to run on air.  It's simplicity is the big advantage.  The exhaust pressure is fixed at atmospheric at the pipe outlet, and the back pressure at the piston depends on how adequately the piping and port is sized.  However, the back pressure on the piston during the exhaust stroke will always be something above atmospheric.

As soon as we add a lubricator, whether using air or steam, the problems begin.  We will have a thin film of oil everywhere near the engine, even in our hair, if we run for very long.  If we are running a boat, a tiny film of oil on the water surface scatters light in a way that makes it very obvious, and that is quite rightly frowned upon.  We start thinking about adding an oil collector to our exhaust system, the second approach to exhaust system design.

The temperature at which oil boils is much higher than the exhaust temperature of a steam engine.  Even at 100 deg C, the vapour pressure of oil is very low, and can reasonably be ignored.  Most of the oil ends up as very tiny droplets, which are entrained in the exhaust, not even enough of them to make a visible fog.  Alternatively, they end up as part of an emulsion with the fine droplets of water in the exhaust which are what we actually see when the exhaust hits the air.  Even the tiny water droplets in a fog do have mass and density, enough to make it possible to separate them from the  un-condensed portion of the water vapour.  They do not settle rapidly due to the air viscosity (you can look up Stokes Law for more information on this). However, if the exhaust stream has some velocity, and we make it go though some rapid changes of direction, we can separate the droplets from our exhaust, and essentially all the oil with them.

The first picture below shows two separators I have built.  The one on the right works very well.  When the engine first starts, the cold metal condenses some of the steam, and water droplets run out the little gooseneck drain pipe pretty freely.  As the metal heats up, the condensing slows, and there is clear vapour out the top which only becomes visible some distance from the stack as the steam cools in the air and condenses to form some fog.  I collect the water from the drain in a little tin, and after a run, the tin contains a few ml. of oily water emulsion.  Not much compared with around 200 ml of water turned to steam in the boiler.  I have not been able to usefully compare the oil collected with how much is lost from the lubricator.  This would be interesting, but somewhat difficult due to small quantities.  Clearly some oil remains smeared on the cylinder walls and piston, and ends up in the drip tray under the engine, and some is clearly carried away, but the problem seems pretty well solved.

The separator on the left of the first picture is still a work in progress and I still do not consider it successful.  More on that later.  First how does a separator work?  Let's look more closely at the one on the right.  The horizontal steam inlet is not on the centreline of the vertical cylinder, but enters tangentially.  As a result, the steam entering is directed against the cylindrical wall and forced to follow round a circular path.  Centrifugal force results in heavier particles, our oily water fog, hitting the walls, and running down to be collected at the bottom.  Dry vapour, being less dense tends to stay closer to the centre.  Now, the vertical outlet extends down inside the cylinder to below the inlet.  Thus, the vapour whirling around in the cylinder has to make its way downward, then inward before it can exit vertically up the outlet stack.  More sudden changes of direction.  The heavier droplets tend to continue in straight paths due to their momentum.  The second picture shows the outlet removed, and you can see the internal extension.  No complex science to it.  Guided by the principle, I made it out of scrap tubing left over from a household plumbing job.  In industry, these cyclonic separators are carefully designed and velocities determined to remove a calculated percentage of particles above a certain size.  But the simple approach seems to work adequately for my purpose.

I will talk next time about the design on the left in these pictures.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 09, 2017, 12:06:23 PM
Another simple separator.

 Last time I included a description and pictures of two simple exhaust separators I have made.  Apologies for the typo in the last line yesterday, I have now corrected it.  I noted that the one on the left is still a work in progress, as I am not happy with its performance.  I was trying to achieve a very low profile to go under my paddle engine style joy valve engine posted in my engine showcase thread (currently on page 3 of Engine Showcase).  Very low cylinders are an advantage for stability in a boat, but make collecting exhaust oil difficult as the exhaust should slope down to the collector. 

Initially the larger diameter section of the vertical part was not included.  The inlet was near the top of one end of the horizontal cylinder.  The drain outlet was low on the other end.  Inside, both pies were extended to a point near the opposite end.  The idea was the engine exhaust has to turn 180 deg at the end of the internal pipe and return to near the inlet end before turning 90 deg to exit via the vertical stack.  Seemed worth a try, but the tube I had was really too small for my level of ability.  Despite several tries, the tubes seem to interfere on the inside and did not end up where required.  I did eventually discover the reason, but I decided a larger diameter cylinder would be a better solution.  I then tried making the separator part in the vertical section that you can see in the photos, back to the whirling design, and keeping the horizontal cylinder as a simple collection pot.  The inlet height was just adequate, but I could have made the whole thing vertical.  Both arrangements of this one still carry over oily water and splutter out the stack, so clearly not satisfactory.  The horizontal tube is only 19 mm o.d., I need to get some 25/25.4 mm tube and try again, but I will be lucky to get back to it this year.

The simple exhaust separator as described does quite a good job of removing oil and incidental condensate, but does not really do any useful condensing of the exhaust steam.  In fact the momentum changes in all those changes of direction actually increase the back pressure on the engine.

The next level of exhaust system development is to include a condenser.  A ship crossing the ocean might want to collect and reuse the water for example, or you may just wish to collect all the water from your exhaust to eliminate any possibility of escaping oil.  Unfortunately, a simple condenser will still not give you any vacuum to increase your engine output.  In addition, the collected water needs good oil removal treatment before it is suitable for the boiler.  Those with experience of full size steam plant will know the issue for achieving vacuum is air, but let's look first at simple condensing as a half way step for those with less experience.

In order to design a condenser, we need to know how much heat has to be rejected to condense the water.  Again we apply the first law of thermodynamics or conservation of energy.  A condenser is basically constant pressure, and there is a continuous flow of steam in and water out.  For this case, our text book tells us that the heat transferred is equal to the change in enthalpy, just as in the boiler.  The main difference is we must add heat to the water in the boiler, while we have to take heat away in the condenser.

Of course we must again begin with the starting conditions at the condenser inlet.  We can measure the exhaust temperature, and we know the pressure is fixed by the outlet to the atmosphere.  It is quite likely that the exhaust temperature, once we get steady running conditions, will be very close to  99 or100 deg C, depending on the atmospheric pressure.  However, we will normally have wet steam, that is steam that is partially condensing, and that is where our problems begin.

  If we have achieved some superheat in our boiler, more on that later, the engine exhaust will probably be in the range of 95% dry, meaning about 5% of the steam is condensed, but 95% of the latent heat is still to be removed.  We can look up the steam tables and work out the enthalpy for 95% dryness, but for our present purpose, assuming dry saturated steam will be close enough.  The enthalpy of dry saturated steam at 100 kPa and 99.6 deg C is 2676 kJ/kg, while saturated water is 419 kJ/kg.  So we have to reject 2258 kJ/kg, usually to cooling water.  Now this is a similar magnitude to the heat input to the boiler.  Please don't worry that the figures are not exactly the same as our earlier boiler example, the main difference is the engine performance, which we have not looked at, and that assumption of dry steam.  There is also a further difference of 292 kJ/kg, nearly 13%, depending on whether we just want to condense at 100 deg, or continue to cool it to say 30 deg C.

Let's have a quick look at the implications of that.  Think about the quantity of heat rejected at the condenser, compared with the boiler heat input.  It means that most of the heat from our fuel goes into evaporating water so it can go up the stack.  That is the prime reason for the very low overall thermal efficiency of a steam plant.  Even the best full size plants with every known heat recovery trick in the book, were only approaching 30% last time I looked.  I am not sure if power station scale plant running supercritical surpass that these days, or by how much, if they do.  If you have read the reports of the locomotive efficiency competitions run by some clubs, you will know the best 5" gauge locomotives are only around 5%.  There would seem to some opportunity and incentive to look at practical heat recovery methods.  A good starting point for next time.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 10, 2017, 10:46:42 AM
Waste heat recovery?

We noted last time that most of the heat from the fuel goes into evaporating water that goes out with the exhaust, this is the largest loss or energy wastage in any steam plant.  Close behind, as next largest loss, is the flue gas that goes to atmosphere directly.  Then we have friction, which also ends up as heat, and other relatively small losses.  The large amounts of heat in the flue gas and the engine exhaust surely provide some opportunity for recovery to help generating steam.  We will talk about the boiler later, but can we reuse any of the heat in the exhaust steam?

We know it's temperature is relatively low, around 100 deg.  A fundamental law of heat transfer is that heat will only flow from a higher temperature to a lower temperature.  So exhaust heat is only useful for heating objects up to 100 deg.  In fact, practical limitations on heat transfer area mean that we can possibly get to around 80 or 90 deg, though more likely lower.  In full size practice, steam plant can be integrated with district heating systems to directly replace burning of fuel for heating.  On a model, it's a bit over the top to heat the captains cabin.  Feed water could be pre-heated.  Fuel needs vapourising, a process that absorbs heat that otherwise comes from the fuel.  And exhaust heat can be rejected to cooling water.  In practice, we have plenty of heat in the exhaust, but it's low temperature limits the places it can be usefully recovered.

In the early days, condensing was by direct cold water injection, a process still useful in some situations. And some forum members are interested in historical engines which used direct injection condensing.  The condensed steam and injected water all end up at the same temperature.  How do we analyse this case?  It's not constant volume or constant pressure, but we can still look up enthalpy of the steam and water before and after mixing, and assume it all ends up at the same temperature.

Let's assume the steam is saturated vapour at 100 deg, hg=2676 kJ/kg and we want to add enough water to just condense it, and have it all end up at saturated liquid at 100 deg.  If our water starts at say 15 deg, then hf = 62.99 kJ/kg.  When it is all saturated water at 100 deg, hr = 419 kJ/kg.  The steam has to loose 2676-419=2257 kJ/kg, (it's listed in the tables as hfg, no need for sums this time), while each kg of water gains 419-63= 356 kJ.  Division on your calculator 2257/356=6.3 kg of water for each kg of steam.  To put this into perspective, the tables tell us dry saturated steam at 100 C occupies 1.67 m^3/kg, while 6.3 kg of water occupies only 0.0063 m^3, or just 6.3 litres, so a very small amount by volume.  If we want it all to end at a lower temperature, we can use a higher water flow.  We can look up the saturated water enthalpy at the temperature we choose, and recalculate the water flow requirement.

Apart from direct injection, condensing exhaust steam requires the use of a heat exchanger.  Now understanding a heat exchanger requires a basic understanding of heat transfer.  So let's look at that next time.

MJM460
 
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 10, 2017, 11:24:43 AM
hi, still following along... a useful unit of waste heat at 100 degrees could be cups of tea/coffee per person per hour perhaps..!!! ;D actually i am taking in this info seriously and often wonder about the ideal steam engine configuration . As we all know no energy is lost or created, it has to end up somewhere. Also when calculating E does the digging up the coal, oil,gas and transportation  come into the equation ?
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 10, 2017, 12:57:21 PM
Hi Willy,  glad to find that you are still following along.  And once again you put your finger on right on the important points and the source of much confusion.  And you continue to keep me on track, it is much appreciated, thank you.

I will give some thought to how best to answer  for tomorrow. 

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 11, 2017, 12:44:47 PM
Hi Willy,

Thanks again for the question on units, an omission I had better fix before proceeding.

I have been using kJ as the unit of energy, usually in the context of kJ/kg, or kiloJoules/kilogram.  If you are more familiar with imperial units, you might want to compare this with the BTU (British Thermal Unit.  For all practical purposes, one BTU = 1 kJ/kg.  It is not exact, the exact figure is 1.05505585262 kJ.  Now considering that the BTU depends on the pound mass, while the kJ depends on the kg mass, and that these two mass standards are provided by different relatively arbitrary size lumps of exotic alloy in a measurement laboratory somewhere, there was never going to be a convenient simple conversion factor.  Given the source standards, it is surprising that it is anywhere near so close.  You can see from this that a kJ/kg relates to a Btu/lb by a factor that reflects the ratio of kg to lb.  The factor is 1 Btu/lb = 2.326 kJ/kg.

But in terms of the new world rational standard units of cups of tea, we need some simple calculations.  Obviously you would want to use the heat required to heat the water from 15 deg C to 100 deg C, and of course we have to define the standard cup.

The attached picture shows three cups from our cupboard.  The formal garden party model has a capacity of 150 ml, or 0.15 kg of water.  The centre on 250 ml, and the coffee mug on the right 320 ml.  I am advised by an impeccable source, that the standard metric cup, as used in cooking is 250 ml.  Adopting this standard gives a very convenient conversion factor of 4 cups equals one kg.  None of that 12 significant figure nonsense.  And a good size for coffee as well.

So the direct injection condenser I talked about last time required 6.3 kg of water at 15 deg C for each kg of steam at 100 deg C.  So we multiply by 4, and we find a kg of steam will heat the water for 4 x 6.3 = 25.2 cups of tea, or more realistically 25 cups plus a small top up for the first finished.

Then the context changed, to how many cups per hour?  Once we introduce time, we are no longer just talking about energy, we are talking about rate of energy transfer, better known as power.  I will leave that until a later day, but need to expand on the difference.

In talking about kJ/kg, or cups of tea per kg of steam, we don't know or care about how long it takes, (unless we are next in line for a cup of tea of course).  A big engine could use a kilogram of steam in a few seconds, while my little oscillator will take several hours, they can both make the same number of cups, from each kg of steam, it will just be a very slow tea party with the little oscillator, not a Sarah Palin event at all.  Oops, no more politics!

Digging coal, transport of coal and crude oil and processing all require energy as your question implies.  Most crude oil produced so far comes out of the ground at high pressure, but this is changing as available oil is used, and more pumping is required these days.  I don't have the detailed information to do the calculations, but if you want to know how much of the worlds energy reserve is used to run the car, you would have to add the energy to transport the oil, process it in a refinery, transport it to the gas station, pump it out of the underground tank into your car.  In principal, the cost of all this energy input is included in the cost you pay when you fill up.  And the heat from all this activity is lost to the atmosphere from where part is radiated to the cold of outer space.  The calculation of how much is radiated is complex and is best left to the experts.

Finally, the ideal configuration for a steam plant.  I really don't think there is one ideal.  It is a matter of understanding the energy balance for your steam plant, and knowing exactly where all the heat is going.  Then you can look at applications which could usefully use the heat before it finally is exhausted to atmosphere thus replacing fuel that would otherwise be consumed.

Power stations operate at very  high pressure which gives higher efficiency.  They use feed water heating, air pre-heating, fuel pre-heating, reheat circuits, and every other possible means to raise efficiency.  Power stations in colder climates use district heating schemes to use exhaust heat before it is lost to the atmosphere.  I have worked in industrial plants where power is produced by turbines with exhaust pressure above atmospheric pressure, so the condensing temperature was high enough to drive a process heating operation which otherwise would have required more fuel.  This scheme, often called cogeneration, produces power at a the thermal efficiency of around 80%, based on the extra fuel consumed for power plus process compared with the minimum for the process heating alone.  So it is a matter of knowing the potential, and looking at the opportunities.  Unfortunately, such opportunities do not always exist, or the quantities do not match sufficiently well, or the temperatures just do not allow the necessary heat transfer.

I hope that resolves the question on units of energy, and provides a little useful background for many interesting discussions.

Back to heat transfer next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 12, 2017, 01:13:24 PM
My next step in exhaust system exploration is to add a condenser.  So far in our discussion of energy, we have only looked at steam properties.  These properties are purely a function of temperature, pressure and volume, and can be looked up in the steam tables.  Properties do not depend on previous history, or how the steam came to be at those conditions.  One place where there is difficulty with this approach is when an engine exhaust consists of wet steam.  I would prefer to leave how the exhaust steam condition can be estimated, and proceed on the basis of condensing saturated dry steam.  It turns out that this is normally not too bad an estimate, and besides, wet steam is a mixture, perhaps a fog of saturated water in droplets and dry steam.  We only have to condense the steam, and it is perhaps 90 - 95% of the total exhaust.

If we have hot steam and cool water in proximity, whether separated by perhaps the copper wall of a tube, or closely mixed like our direct injection process, heat will flow from the hot steam to the colder water until it is all at the same temperature, then no further heat will flow.  (Known as the zeroth law of thermodynamics.)  However with our guests waiting for their tea, we need to make sure that the necessary heat transfer occurs in an acceptable time.

A condenser has to be able to transfer heat at the rate necessary to condense all the exhaust steam.  So now we not only need to know the steam properties, listed in units of kJ/kg in the steam tables, we need to know how many kg/hr are being produced.

I have three miniature boilers, and the test measurements I have made are all under 0.6 kg/hr using methylated spirits fuel.  They are quite small boilers for small engines.  A slightly larger boiler for a twin cylinder engine might evaporate 1 kg/hr.  This figure is obviously easy to scale up or down for any boiler.

We have already found that to condense 1 kg of dry steam at 100 deg C involves heat transfer of 2257 kJ, and we want to do this in an hour.  In terms of the new national standard, teacups 25.2 per hour.  We now know how many guests we can invite to the party.  And in ISO Metric units,  where the time unit is seconds, we need a heat exchanger rated for 2257 kJ/hr = 0.627 KJ/s. If we remember that 1 kJ = 1000 J, and 1 J/s = 1 Watt which is the unit for power.  So the heating power of our exchanger will be 627 W, or 0.627 kW which is a bit more than half of a common 1 kW one bar electric radiator.  This conversion of units of mechanical power, J/s, to units of heat, and electrical power with the constant equal to 1 in each case, is another plus for the ISO metric system.

To design a heat exchanger for this rating requires some understanding of heat transfer, so we had better look at that next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 12, 2017, 02:11:01 PM
Hi, so much info here ....leading to so many more questions !!  Is the weight of the steam proportional to the pressure of the steam  ? Possibly yes. If you have a vertical pipe full of water the pressure at the bottom sends a jet quite far out however at the top of the pipe it will just dribble out !!!???   Also we now don't use Pounds /inches anymore but are the formulas still the same as in MKS units  ?? Looking at one of my books i am waiting for the 37th edition that might have been brought up to date !! Also iv you have an enclosed cylinder full of steam and then at the same temperature you squeeze the piston into it ...what will happen ??.....( my brain is starting to hurt now, sorry) !!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 12, 2017, 02:20:15 PM
Hi, Also in an infernal combustion engine the inlet/exhaust ports are different sizes and the cams can be set differently . so .could a steam engine have four ports  2 for the inlet cycle and two (larger)? for the exhaust cycle ? My brain is really starting to hurt now.!! Is it because i am an autodidact ??................
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on July 12, 2017, 04:36:26 PM
so .could a steam engine have four ports  2 for the inlet cycle and two (larger)? for the exhaust cycle ?

This sounds a lot like Corliss valve gear, see:
https://en.wikipedia.org/wiki/Corliss_steam_engine

Dan
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 12, 2017, 05:11:16 PM
Actually ,yes you are right ...i was thinking about a slide valve engine though !! I have never really looked at corliss steam engine dynamics ,only heard the annoying tic, click,clunk, etc etc... Ok thats dealt with !!  Ok  so steam pressure works in all directions yes? so if you convert a certain weight of water into steam and introduce it into a closed container, will the steam working in all directions cancel out the extra weight until it condenses and becomes water at the bottom of the container ??? or is weight and gravity not compatible and is that why rockets are not powered by steam engines ??  Sorry i am not trying to be flippant just curious and asking the same questions that Watts  pupils might have asked all those years ago !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 13, 2017, 01:21:43 PM
I had intended to talk about heat transfer this time but Willy has more questions of the type that prompted me to start this thread.  They are not frivolous questions, Willy,  I suspect that you have very eloquently put into words what others are also wondering.  So I will look at your questions before continuing with heat transfer.

Now after my last post, I had better add a small point before you ask.  One horse power is equal to 746 watts, so 1 kW = 1.34 HP.

So is weight proportional to pressure?  Back to basics for this one.  Weight is the force due to the action of gravity on a mass, so weight is proportional to mass, but requires a value for g, roughly 9.8 in metric units, or 32 in Imperial units unless you really need to be super accurate. 

If you know the pressure, volume and temperature, then the steam tables will tell you the specific volume, reciprocal of density, so you can work out the mass and hence the force due to gravity.  So pressure is a factor determining the force, but it is only one of three factors involved.

Now your column of water.  So far, in each of the problems involving the first law of thermodynamics, or conservation of energy, I have assumed there is no change of elevation.  In each case there should be a term for change of elevation, however the pressure change is also easily derived from density considerations.  The formula is P = density x g x height.  It is a case where the gravitational constant is required in an SI calculation.  You can check the formula is dimensionally correct by looking at the dimensions on each side of the equation.

[P] = [F] / [A] = [m x acceleration] / [area] = kg. m / s^2 / m^2 = kg / (m.s^2)

Similarly, for the [density x g x height] = (kg / m^3) x (m/s^2) x (m) = kg.m.m/(m^3.s^2)

So [density x g x h] = kg/(m.s^2) = [P]

This analysis does not prove the formula, that comes mostly from definitions of the various factors, but it shows that the dimensions of acceleration are required, which gives the clue to the student that g is required, a very powerful cheat if you like.

That might seem a bit off the track, but it shows that height and density combine to make pressure.  In your water column, the difference in the pressure at the hole near the top and the pressure at the hole near the bottom is proportional to the density and the vertical height between the holes.

For a water column, the density is 1000 kg/ m^3, while for steam, lets assume atmospheric pressure and at 100 deg C, the steam tables tell us the specific volume is 1.67 m^3/kg, so the density is about 0.6 kg/m^3.  It is clear the difference in pressure for every meter in height difference will be nearly 2000:1.  You can see why the difference in height for steam is not very significant for steam in a model.

The energy equation also should include a velocity term, like height difference, the difference in velocity at the inlet and outlet is usually very small for a model, and so omitted.  But when a more complete energy equation is used, there is also a velocity term.  If the equation is written for your column experiment, we should include both terms, and we see the pressure due to height can be directly related to the velocity of discharge at each hole.  As you note, the velocity from the lower hole is much higher than the velocity from the top hole.  With steam there would still be a difference, but the difference would be considerably less, and for practical purposes we would normally ignore it.

That is pretty heavy going, I hope I have provided at least a little clarity. It is also getting late here, so I will look at the mks system, and your steam cylinder and isothermal compression next time.

Thank you also to Dan for answering the valve question, I will add a few comments, but Dan is the expert on valve gear, so I am delighted to see his contribution. Preferably, "talking" should involve more conversation and participants, and be less like a lecture.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 13, 2017, 03:11:53 PM
Thanks for that.......also in my book..There are tables that mention the Latitude at Dublin.... Is this significant ?? Do thermodynamic tables alter at the equator  from the poles ??
Title: Re: Talking Thermodynamics
Post by: Maryak on July 13, 2017, 09:19:24 PM
Hi, Also in an infernal combustion engine the inlet/exhaust ports are different sizes and the cams can be set differently . so .could a steam engine have four ports  2 for the inlet cycle and two (larger)? for the exhaust cycle ? My brain is really starting to hurt now.!! Is it because i am an autodidact ??................

Some steam engines were fitted with double ported slide valves.


(http://i389.photobucket.com/albums/oo340/Maryak/dpsv_zps67t8yoii.jpg)
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 14, 2017, 02:22:31 PM
Continuing on Willy's questions from post #130 -

The next question was whether the mks system we may have been used to is the same as SI.  This was a very contentious issue at the time of adoption of SI metric, as it seemed to some that everyone was being made to change so no one was disadvantaged, and no one had a head start.  However this was a spurious argument, put up by people who did not understand physics.

The fundamental huge leap forward inherent in the SI metric system is the definition of the unit of force by application of Newton's first law and the associated equation, F= ma.  This step provides a clearly defined unit of force (the Newton) and removes the necessity to use the arbitrary constant equal in magnitude of the acceleration due to gravity, and instead allowing the constant to be one, and removing the confusion around force and mass.  This is a major contribution to a rational set of units.  Previously all systems defined force in terms of the effect of gravity on a mass, giving rise to the imperial unit of pound force, the mks system with its kgf, cgs with gmf, and probably others.  So mks system is a sort of inferial metric system using sensible size units for large industrial equipment (the chemical engineer's test tubes capacity measured in tons) and cgs is good for laboratory size equipment where grams and centimetres are more convenient.  The systems arose from historical understanding of mass and force.  Newton's law is now recognised as a fundamental law of physics, and now more eloquently expressed as the law of conservation of momentum, which applies to bodies moving in a straight line.  A very similar universal law which is totally analogous, expressed as the law of conservation of angular momentum, applies to torque, moment of inertia and acceleration of angular rotation.  Using this basic law of physics to define force results in a very rational set of units, and clearly defines the relationship between mass and force in a universal way, it applies equally on the moon, on Mars or anywhere else in the universe.  So please remember these legacy unit systems only for the purpose of working out conversion factors, and then continue in SI metric.

The question on steam enclosed in a cylinder with one end a piston is very interesting and the answer might not be what you would expect.  When you move the piston to reduce the volume, the steam is compressed.  If you have adiabatic compression, that is no heat flow in or out, the pressure increases, along with the temperature.  So the properties of the steam change toward superheated, and yes, you can compress steam.  The work input adds more than enough energy to avoid condensation.  However you asked what if you remove heat as you compress so the temperature does not rise.  Now for a permanent gas, such as air, isothermal compression means less work to do the compression.  However with steam, removing heat means the properties change toward condensation.  I have not yet found an example to rely on, to say what happens if you try isothermal compression on steam, but I would tackle the problem this way.  Using steam tables, if you use the superheat section, and look up the properties at a temperature like 150 deg C, starting from say 0.01 MPa, you will see that as you move to the next pressure and reduce the volume (your cylinder contains a constant mass, so reducing volume means reducing the specific volume), while the pressure increases, the saturation temperature also increases, so by about 0.48 MPa, the saturation temperature exceeds your 150 deg C, so steam will start condensing.  It seems likely that if you start at the saturation temperature, then the reduction in volume will cause some condensation immediately.  But if you do not remove all the heat added by the work you do in moving the piston, you can keep compressing and avoid condensing.

Your text book is up to date for its time, and much of the maths is still current, though the units are more difficult to deal with than SI.  The main trick to following the text is to keep track of which assumptions are current at any point in the text.  The pages you have shown talk about isothermal and adiabatic processes, and calculate the work done as the process proceeds.  It was already understood that you had to know the process to calculate the work done.  Modern terminology is that work done is a process dependent quantity, unlike enthalpy and enthalpy, which purely depend on the measured pressure, temperature, and specific volume, not the process followed to get there.  Personally, I find a modern text much easier to read, and it is definitely worth getting an SI version of the steam tables.

I am not sure about the Dixon table.  It looks like an early version of a steam table.  I note that the pressure is measured in inches of mercury, which is of course dependent on the local value of gravity.  Now the gravitational acceleration changes from the equator to the poles, and also with height above sea level.  Hence the use of the defined latitude (of Dublin in this case).  The last column looks like enthalpy with a reference temperature of 32 deg F, but I am not sure what the H and L columns are.  Modern steam tables use absolute pressure, and there is no dependence on latitude.

I hope that adds a bit more clarity, and a couple more blocks to our knowledge base.  I will check on what I have missed from your post for next time.

Thanks for following along

MJM460

Title: Re: Talking Thermodynamics
Post by: MJM460 on July 15, 2017, 10:35:31 AM
Hi Willy, a few more answers for you in the hope of dealing adequately with the outstanding issues.

While I basically leave valves to Dan and Maryak, I was going to add a couple of comments.  If I remember history correctly, the first engines were controlled by a boy on the taps, opening and closing them as required.  Obviously a real impediment to high speed, continuous operation.  Then some lever systems were developed to let the boy off, but still pretty clumsy.  Jobs sacrificed to technology even then!  So the slide valve was a huge step forward.  Then, I suspect that, just as increases in boiler pressure had to wait for pressure vessel technology to advance, I suspect that valve design had to wait for advances in machining technology, and even metallurgy.  The slide valve worked well, but there was a lot of development in the linkages which drive the valve to allow early cut off and reversing.  That valve of Maryaks would be a real trick to machine, perhaps requiring soldering some layers together.  Would be interesting to understand that one better.

Now post #133.  Some real confusion there.  I will try and make things a little clearer.  First introducing steam into a closed container.  You did not say if the container already had air in it, and you did not make any comment on heat transfer during the operation.  Now steam is normally hotter than atmospheric pressure, unless the steam itself is at quite low pressure.  So you can expect heat to flow from high temperature to low temperature, unless you make specific provisions.  You might for example insulate the vessel very well so heat transfer is prevented.  Let's start with that.  We assume no heat transfer in or out.  You introduced the steam to the container, rather than evaporated steam in the container.  So not a constant volume process, and no mention of moving pistons for external work.  So let's just assume some steam in the container, and the steam definitely has mass.  Steam will fill the entire space.  Pressure works in all directions, so pressure is experienced on the walls, but they do not move, so no work done.  As you imply, the steam "has weight", but weight is simply the result of gravity acting on a mass, there is no conflict there.  But I expect that you are alluding to the normal assumption of equal pressure in all directions, so what is weight doing?  Basically, just as we saw in your water column experiment, weight means that the steam pressure is higher at the bottom of the container than at the top.  Let's see if we can calculate the pressure difference.  Basically, that conservation of energy equation gives us the necessary clue.  We saw before that the pressure energy due the height if the fluid, in this case steam, is calculated as P= density x g x h.

If our steam is at 100 kPa (absolute), we have looked up the specific volume (tells us density) for that before.  I think the density was about 0.6 kg/m^3.  The value of g in SI units is near enough to 9.8 m/s^s, and as we are mostly interested in models, let's assume the height of our container is just 1 metre.  It is then easy to scale for other heights.  So P = 0.6 x 9.8 x 1 = 5.9, so what are the units?  Using our dimensional analysis, we see we have kg/(m.s^2) which is the dimensions of pressure in Newton/m^2.  So we have nearly 6 Newton's per square meter which has the name Pascal. A Pascal is a very small unit.  We have 1000 Pa = 1 kPa.  And standard atmospheric pressure is about 101.3 kPa.  So 6 Pa is 6 / 100000 times 1 atmosphere.  Or 6/100000 x 14.7 psi or 0.0009 psi.  You can see why it is normally considered not important.  The more so in most of our model sizes?

Condensing requires the steam to lose heat.  So with our assumption of no heat transfer, and no work done, no condensing.

Oh, and in case you were assuming some air in the container to which the steam was added.  So we can continue to assume no heat transfer, the air and the container and the air have to be at steam temperature, otherwise irreversible mixing occurs until the temperature is uniform.  Then Dalton's law of partial pressures tells us that the air and steam each fill the space as if the other was not there.  So it makes no difference to the steam, however, the total pressure in the container is the sum of the pressure due the air plus the pressure to the steam.  Each called a partial pressure in this case.  Dalton's law is close enough until quite high pressure.

And a water powered rocket?  Some one has no doubt tried it.  However rockets are about thrust to mass ratio, which of course gives acceleration.  You would need a large mass of water, plus fuel to heat it, then most of the heat goes into evaporating the steam before there is energy to accelerate the steam out of the rocket nozzles at high velocity.  And the latent heat is lost with escaping steam.  Conservation of momentum then tells us the resulting propulsive force.  Modern rockets whether solid fuel or liquid fuel usually employ chemical reactions to release much higher amounts of energy, and give much higher thrust to mass ratio, so make a more efficient rocket.  However, you might have seen those soda pop rockets where the pop bottle part full of water is pumped up with air, inverted so the neck is at the bottom, and fitted with a nozzle with a means of sudden release.  They go up in a spectacular manner, but the next continent is not in any danger.  No doubt something similar could be done with steam, but with a high probability of scolding the rocketeer, definitely not recommended.

So not silly questions, but questions which highlight the significant consequences of physics and thermodynamics.  I hope I have been able to provide at least a little clarity.  More building blocks in place.

Thanks to all who are looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: Stuart on July 15, 2017, 11:17:32 AM
By ek my brain hurts in a good way

All this has reminded me of when I worked in HVAC and because we used chillers 4 500 ton units ( with now banned R11 ) I had to do a refrigeration theory course amongst other at Manchester Uni.
All the talk of  e’s ( sorry my dyslexia is bad today) made me think how much I learned in the past but now do not use ,it’s been over 22 years since I was in work but we used the SI system for the HVAC calcs

Thank for taking the time to explain the behaviour of gases

A add on topic could be the explanation as to why it takes so much energy input/extraction to change the state of , gas ,fluid or solid into the next state

Stuart
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 16, 2017, 12:31:32 PM
Thanks for dropping in, Stuart, I'm glad the brain hurt is in a good way, and I appreciate the thanks.  You will understand my granddaughter who recently told her mother that "Grandpa really is hard work!"

Those 500 ton chillers should be able to keep a lot of beer cold.  I never had much to do with any of the freons, went straight to the hard stuff, like propane, propylene and ethylene.  They are very good refrigerants but there is an issue of flammability that makes them unsuitable for most uses.  Same principal though.  Refrigeration is great for getting to understand the principals, and always useful.

The issue of learning all this good stuff, then forgetting overtime due to not using it often enough, I think is pretty normal.  I don't think any theory will help you find and cure a leak, or install a unit in insufficient space, and when you have the pressure of jobs to complete there is not much time for contemplation.  But over the years I have come to realise that my career was a little unusual, in that I frequently needed to use the theory, all through my career.  And so many questions are asked in the model engineering community that are not really answered, so I was prompted to have a go.  Vixens comment on 4000 strong in Hugh Currin's thread struck a note though.  I know we have a few student members of the forum, and I would be delighted if I could help them in getting started too.  This is intended to be for everyone interested, not just the old hands. 

Now your question on the amount of energy required to cause a change of state, solid to liquid or liquid to gas.  This is an interesting question which leads into some of the fundamental ideas in physics, atoms molecules and the forces between these tiny building blocks of the universe.  Just what forces hold the separate atoms together whether in a fluid state such as a liquid, or in a solid state where the strength can literally be like steel?

In a gas, the individual molecules are moving rapidly in all directions and they are a big distance apart relative to their size, apart from the occasional collision.  Conservation of momentum applies to each molecule, so they travel in a straight line until a force makes them change, either collision with another molecule, or the wall of the vessel which contains them.  So gas molecules always fill the space, and they are so far apart relative to their size that they each act independently.  But there are forces between them none the less.  These are gravitational forces, not gravity in the sense of gravity causing weight on earth, though essentially the same force in action.  Basically, any two masses are attracted to wards each other by a gravitational force which is proportional to the product of their masses and inversely proportional to the distance between them.  Symbolically F is proportional to m1 x m2 / d^2.  And there is a universal gravitational constant which completes this equation.  You can look it up if you like.  Now gravity is classed by physicists as a weak force.  The gravity that holds us on earth does not feel weak, but then compared with an atom, the masses are large, and the distance is about the same magnitude as the size of the objects, and if we have a rocket with enough energy, it can escape.  If m1 is the earth, the mass is obviously large, and m2 our body mass (even without being political about it) is also very large compared with atoms.  You can see why gravity varies from place to place, the distance between centre of mass of the earth and the centre of our body mass changes of we climb a mountain, descend to the bottom of the sea, or travel towards the poles from the equator, ( the earth is not a perfect sphere but technically an oblate spheroid, sort of like a slightly squashed ball.

But in molecular dimensions, the masses are extremely small, and the distances large compared with the size of the molecules.  Also gravity acts over very large distances, just getting weaker as the distance increases.  It is the Suns gravitational force which holds Jupiter and Pluto in their orbits, not to mention Earth.  So the gravitational forces exist between molecules in a gas, but are very small, and are classed as weak forces.  The energy in the fast moving molecules allows them to very easily escape the attractive forces between them.

Now this may not seem relevant to your question, but it is necessary to have an understood starting point.  That's probably more than enough for one session.  I will have a go at moving from this starting point to understanding latent heat of liquids next time.

Thanks for dropping in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 16, 2017, 01:37:25 PM
more fascinating info and a whole new set of questions............. so in an enclosed vessel half full of air and water at sea level the water sinks to the bottom. if this vessel were taken to outer space how would the water know where to go ??  How does gravity get through the walls of the vessel ??What is the temperature of outer space and would the water be frozen ?   Just wondering really...........
Title: Re: Talking Thermodynamics
Post by: Stuart on July 16, 2017, 07:34:42 PM
Thanks MJM

Ever whished you had not asked :lolb:

Now my brain is even more tasked

Yes the chillers were pretty big all to keep some IBM  main frames cool long time ago they needed chilled water to cool them
‘Bluechip “ was at the same complex looking after bit on the computor floor

The most interesting session I did a Manchester uni was on vibration analysis of machines using fast Fournier transformation but at that time the resultant graphs had to be manually interpretated hence the course , it main use was to predict a machine failure before it failed , but in practice some times it beat you the failure   :censored:

Thanks again I will follow with interest
Title: Re: Talking Thermodynamics
Post by: Jo on July 16, 2017, 08:56:28 PM
Some where you lost me: Please send beer  :noidea: I feel the need for experimentation   :naughty:

Jo
Title: Re: Talking Thermodynamics
Post by: crueby on July 16, 2017, 10:30:56 PM
more fascinating info and a whole new set of questions............. so in an enclosed vessel half full of air and water at sea level the water sinks to the bottom. if this vessel were taken to outer space how would the water know where to go ??  How does gravity get through the walls of the vessel ??What is the temperature of outer space and would the water be frozen ?   Just wondering really...........

Well, I remember watching astronauts (different from us wingnuts) up in space playing with water - when squeezed out of the bottle into the air, the surface tension kept it in a sphere bobbling and wobbling about. So, that container would likely have a blob of water floating around in it. Depending on the surface of the inside of the tank, anyway... I think.... Jo is right, we need some beer to help the thought experiment get forgotten!!
  :cheers:
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 17, 2017, 12:12:31 PM
I had intended to continue to discussing latent heat for condensation and freezing, but not before I acknowledge today's contributions, they are all thought provoking and very welcome. 

Willy, I think Chris has answered your question about water in space.  Many thanks Chris.  To analyse that case, the energy equation would need to include a term for energy in surface tension.  Gravity is described by physicists as a field which acts through space, and difficult to detect unless you are working on those gravity wave detectors, apart from its interaction with mass.  A bit like magnetic and electric fields, though those are more obviously affected by solid materials.  When you find the complete explanation, you should check that your suit is good enough for the ceremony. 

Empty space does not have a measurable temperature.  Temperature results in matter as a result of the balance between energy in and energy out.  So analysis of the water temperature would depend on how the space ship external temperature is transferred to things inside.  The side of the craft facing the sun would be quite hot, no clouds to shield it, while the other side I expect would be cold, as it effectively radiates to absolute zero. Presumably there are heat insulating layers and clever tricks to make sure the inside surface temperature are acceptable to the astronauts.  We will eventually look at means of heat transfer either in the condenser topic or when we get to boilers.  Outside the space ship the total vacuum means all the water would evaporate and the resulting vapour would attempt to fill space.  Before long it would all be gone, unless the gravity of a nearby planet or other space body captured it.

Stuart, you are welcome, it's good to have you on board, and please don't go away, there is more.  I have set the scene, but still have not answered your question on latent heat.  That fast Fourier transformation technique (FFT) is interesting to apply to a frequency spectrum, but I suspect that there is something futile about trying to predict random events.  However the analysis can help us see them coming, if we can interpret the signs.  Basically outside my expertise, I will leave any more details to the maintenance experts, and I suspect that few if any of us have the equipment necessary to apply this to our models.

Sorry Jo, Ade has not yet included the "send beer" button on the forum.  I suspect he is waiting for Bill Gates to get his act together on incorporating a missing sub routine from Windows, so please don't overwhelm him with requests.  But I can provide a little more information for those who have not come across the tons unit in refrigeration.  Will that help?

A ton of refrigeration is the heat necessary to produce a ton of ice at 32 deg F, starting from water  at 32 deg F.  Please correct me if I am wrong, Stuart.  Obviously another infernal unit, but four times 500 ton units should be able to cool an awful lot of beer, especially if you don't actually want to freeze it.  Though I understand that you guys don't actually chill your beer, but serve it tepid.  Strange!  By the way, great to see you back fondling castings, we don't want them getting lonely, and best wishes for your upcoming post op assessment.  We all hope the reports will be good.

Now let's try and make a little progress on that latent heat question.  I described the molecules in a gas state last time.  When heat is lost from the gas, the molecules slow down.  We sense it as cooling.  As a gas cools, the molecular speeds slow, all relatively predictable and uniform. 

But as well as the weak force there are also forces classed as strong forces. (I should possibly be saying weak forces, not sure for the moment if there are other forces also classed as weak forces.  Perhaps we have a physicist reading who can chip in.)  Now the strong forces are very strong, even beside the force of gravity that holds the universe together.  But the strong forces act over a much smaller distance.  It is one of the strong forces, or perhaps several that, when the energy of the molecules gets sufficiently low, the strong attractive force captures any that come close but do not have sufficient energy to escape, and holds the atoms closer together.  This is when liquid droplets start to form.  They still have a large range of movement relative to their size, and they all still move in a random manner, though loosely connected way that we recognise as liquid, by those strong forces.  In this state, the random movement allows the boundary to be flexible and to conform to the shape of their container. Gravity over the whole mass causes liquid to rest in the bottom of a container, or a river to run down hill.
 
We observe that  heat travels from higher temperature to a lower temperature.  In molecular terms, the faster ones keep loosing more than they gain from their collisions with the slower ones while the slow ones are kept moving by the energy they gain from the faster ones. So there is no further temperature change with cooling until all the molecules join the liquid.  Then the liquid can continue to cool.  So the latent heat is the heat that must be lost by the gas molecules before they can no longer escape that close range where the strong attractive forces take hold.  Similarly, to evaporate the liquid, energy must be supplied to give the molecules in the liquid enough energy to escape the strong forces.  The ones that escape are the higher energy ones, so their escape cools the liquid.  Again, they all have to evaporate before the temperature can rise.  I hope I have explained it well enough to explain the constant temperature part of the observation.

 As the liquid molecules are cooled, they continue to get slower.  There comes a point where the free movement within the liquid becomes limited by the close proximity of the surrounding molecules, and the molecular motion becomes more of a vibration.  Molecules tend to jiggle until they drop into a matrix where they still all fit, and the constraints to movement by the surrounding molecules causes the mass to take on the properties of a solid.  But the molecules are still moving within that matrix.  They do not actually stick together.  If they get too close, they actually repel with enough force to resist the attractive force.  As solidification involves losing enough heat that the molecules drop into this regular pattern characteristic of a solid, melting involves adding enough energy for all the molecules to "pop out" or escape that matrix.

I am sorry if my terminology is a bit woolly here, but I am not a physicist, this is just a mechanical engineers understanding of the mechanism, but I hope that is sufficient to answer the satisfy curiosity for now.

Thank you to everyone who is following along, I hope not too much brain pain.

MJM460
Title: Re: Talking Thermodynamics
Post by: Stuart on July 17, 2017, 01:57:32 PM
sounds about right ,

note the chillers where at the bank about 22 years ago mainframe du dads are now better

we only chilled down to 42  deg F ( they were USA machines ) and the controls were also made by Jonhsons ) we did get them to change to a better system

the analysis was used to spot trends and to diagnose actual faults e.g. a peak at twice mains frequency would indicate a motor over load

still looking in  :)
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 18, 2017, 01:14:50 PM
Hi Stuart, clearly you would not want to actually freeze the water in your computer cooling application.  This highlights the point that a ton of refrigeration is a measure of the amount of heat removal capacity in terms of the amount if ice that could be produced, not necessarily the amount that is produced.  Obviously a historical unit from the early days when the main use of industrial refrigeration was to produce ice for food preservation.  There were no domestic refrigerators, the ice man came around with a horse drawn trailer and deliver ice to each household, closely followed by the milk man.  I am sure that I am not the only one who can remember the family ice chest.

To finish off your question on latent heat, it is interesting that as the solid cools further, there can be further phase changes which X-Ray diffraction reveals as different crystal structures.

We are perhaps getting into deeper theory than we need to understand and build an engine, but if would like to read more about it I would suggest a little book called Six Easy Pieces, a small collection from the Richard Feynman lectures in physics. It has been very helpful to me in developing my understanding.  His Nobel prize should be enough authority for us to rely on.  And a very interesting, readable and readily available little book.  If you get hooked, move on to Six Not So Easy pieces, which are, well, not so easy, but very interesting and worth having a go at.

A shorter post this time, I need to get the average down.  Tomorrow, I will get back to heat transfer and condensers.

Thanks for dropping in.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 19, 2017, 12:31:50 PM
The heat transfer equation

To design our condenser, we need to look at the factors which are relevant to heat transfer.  The primary factors are the temperature difference, heat transfer is faster with a higher temperature difference, area available for heat transfer, a direct measure of the necessary physical size of the exchanger.  It turns out that these factors are related in a deceptively simple looking equation.  The equation that applies to all heat transfer problems is written as follows:-

Q = U x A x (delta T).

In the equation, delta is the Greek capital letter usually written like an isosceles triangle with the base flat.  However I haven't conquered the technique on my iPad, so I will abbreviate delta T to dT, it means temperature difference.

Q is the heat transfer rate in Watts (or J/s)

U is called the overall heat transfer coefficient.  The units are w/(m^2.K)

A is the heat transfer area in m^2

Delta T is the temperature difference driving the heat transfer.  Technically the units are the absolute temperature units, Kelvin, but as it is normally in the context of a temperature difference, so you can also use deg C.  However radiation heat transfer calculations, for example, must use K.

Looks like a simple equation, and it would be if any of the terms were as simple as they appear.   The equation does tell us that the heat transfer rate is dependant on the heat transfer area, the the temperature difference, and a heat transfer coefficient.  Time to examine each of these in more detail.

The simplest to understand is the heat transfer area, A.  Measured as you would expect in square meters.  In model sizes we will mostly measure in mm, so we have to deal with a lot of zeros.  One square meter = 1,000,000 square mm, which can be conveniently written as 10E6.  You might read this as 10 exponential 6, or 10 to the power 6.  When dividing is required, you can instead multiply by 10E-6, ten to the power -6.  I believe excel allows this, and has engineering and scientific notation, unfortunately not Numbers for iPad.  I tend to use 10^6 (10 raised to the power of 6) for this reason.

The other wrinkle with area, particularly when tubes or pipes are involved, is that the area based on the outside diameter is larger than the area based on the inside diameter, so you need to keep track of which is the relevant area.

The temperature difference in dT is a bit more complicated.  In our condenser, the steam side of the heat exchanger is essentially constant temperature while the latent heat is transferred.  However if we are cooling with water, the water temperature is rising as is moved through the exchanger and takes up heat.  Similar considerations if the heat transfer is between two liquids except that then both fluid temperatures are changing.  It is necessary in these conditions to use a log mean temperature difference.  It involves the temperature difference on each end as well as the natural logarithm of the ratio of the inlet and outlet temperature differences.  I can write it out if anyone is really interested, but I don't think it will be useful for most readers.

The final factor, U is the really difficult one, a look at that next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: Kim on July 19, 2017, 03:37:22 PM
The temperature difference in dT is a bit more complicated.  In our condenser, the steam side of the heat exchanger is essentially constant temperature while the latent heat is transferred.  However if we are cooling with water, the water temperature is rising as is moved through the exchanger and takes up heat.  Similar considerations if the heat transfer is between two liquids except that then both fluid temperatures are changing.  It is necessary in these conditions to use a log mean temperature difference.  It involves the temperature difference on each end as well as the natural logarithm of the ratio of the inlet and outlet temperature differences.  I can write it out if anyone is really interested, but I don't think it will be useful for most readers.

So, why would we use natural Log of the ratio rather than just the average of the delta temperatures between the inlet and outlet?

I'm finding this thread quite interesting. Thanks MJM!
Kim
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 20, 2017, 01:00:33 PM
hi ,still following along but as i am more hands on than brain on and i find the all this a bit heavy going, especially as i was brought up on yards feet and inches.!! I am getting the jist of things but am more into the practical applications.........when i have to drink a hot cup of tea in a hurryi always put 3 teaspoons in it much to the amusement of the Baristas !! i then try to explain using words like  Adiabatic Enthalpy etc etc etc.....also were men like Watt, Woolfe, and Corliss really educated in thermodynamics or were they just practical engineers ??
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 20, 2017, 01:25:40 PM
The heat transfer coefficient.

Hi Kim, thanks for following along, I am glad you are finding it interesting.  Cooling problems proceed at a rate proportional to the temperature difference, so it proceeds most quickly when the temperature difference is greatest, and as the heat transfer proceeds the temperature difference reduces so the heat transfer slows.  An arithmetic average only gives the right answer when the process proceeds at an even rate.  When the maths is done in detail, and I have to admit I find it a bit too hard, it comes to integration of a 1/x term, which results in the natural log term, which properly accounts for the change in rate as the temperature difference decreases.  The classic example is the cooling of a cup of coffee.  Cooling proceeds at a rate proportional to the temperature difference between the coffee and the air.  Not totally true with evaporation at a free surface, but not a bad approximation for a good cappacinno with a thick layer of good insulating foam on the surface.  The temperature initially drops very quickly but then the temperature change becomes slower and slower.   The cooling follows an exponential function which also involves that natural log function.  Of course Willy's teaspoons absorb a bit of heat to start the process even quicker, but then with the lower temperature further cooling is slower.  His tea is always a little cooler than if he had not used the teaspoons.  Thanks for that example Willy.  No need for anything too technical, just specific heat of the metal and temperature difference.  But if you really want to get technical, stainless steel ones will work slower, but will have a better cooling effect.  But you don't want hot ones from the dishwasher.  I believe those gentlemen were mostly practical engineers who achieved remarkable results with very little theory and rather crude instruments to guide them.  The theory was only in its infancy at the time.  I will leave it to the historians to say just what order some of the significant events happened, but your historical text will give you some ideas of the status of the technology.  Sorry about the units, I am trying to translate a few key numbers, but please remind me if more would be helpful, or particular ones I miss.

Returning to the discussion on the heat transfer coefficient.

The heat transfer equation would be quite easy to use if U was a constant which we could look up in a book for our particular application.  In some situations it comes pretty close, for example conduction in a solid, where U = conductivity/ thickness.  We can look up the conductivity of the material we are interested in and insert it in the equation.  For example, the conductivity of pure aluminium is 236 W/m.K, while copper is 399.  Alloys tend to lower the conductivity, a few percent of copper in aluminium reduces the conductivity to 164 W/m.K, while brass is 111 W/m.K and bronze is only 26 W/m.K.  A good insulating material such as cork would be 0.042 W/m.K.  When we do the dimensional analysis, we see that the thickness has to be in metres. 

It is worth putting some typical thicknesses for copper and brass, for example say 1 mm ((0.001 m) wall thickness, to see the difference in heat transfer over a square meter for each degree of temperature difference.  Similarly say 2 mm (0.002 m) of cork.  These figures can be helpful in indicating differences in materials for some applications.  The high conductivity means that heat is transferred with not much temperature drop, especially across a thin metal wall. 

However to measure the metal wall temperature introduces difficulties.  If you just place a thermocouple on the metal surface, there will be a contact resistance that will affect the reading, and it is difficult to get really good contact.  In addition, the side of the thermocouple not contacting the metal is in contact with the surrounding air and the wiring to the meter.  The thermocouple will read something in between the metal temperature and the air temperature.

Sometimes an electrical analogy is used to analyse heat transfer problems.  The voltage  difference is the analogy of temperature, and the current the analogy of heat flow.  Resistance is the reciprocal of conductivity which is the analogy of heat conductivity.  Now to measure a voltage accurately, we all know we must have a meter which has very high resistance so it draws minimal current.  Not a problem with digital meters, but a real accuracy issue in the days of analogue meters.  In our temperature measurement the accuracy is reduced by the heat transfer from the metal to the thermocouple and on to the air.  If we cover the thermocouple with some insulating material, we reduce the heat flow, and the thermocouple will then be very close to the metal temperature. 

If we want to measure the steam temperature inside our pipe for example, we can put a thermocouple junction on the pipe and hold it firmly in place with some insulating material, a wrap or two of silicon tape is excellent for the purpose.  Some extra insulating material over the thermocouple wires is also a good idea.  Insulating tube can be purchased from an electronics component supplier.  Almost every reasonable digital voltmeter these days comes with a thermocouple as well as its voltage probes, and is well worth buying if you don't have one.

You might want to use an infrared non contact instrument.  These instruments are not affected by the heat loss but unfortunately are influenced by the radiation characteristic of the surface, known as emissivity.  This varies with surface colour, surface finish and material.  Ok for measuring change of temperature over time or different locations on an essentially identical surface, but not good for small differences, or especially the difference between the temperature of different materials.  Even the difference between shiny and tarnished surfaces will introduce a significant measurement error.  A thermocouple is a better option, though they are useful in some specific cases.

Binding the thermocouple to your steam pipe is not ideal, especially if you want to measure several different points by relocating one thermocouple.  In industry, a device called a thermowell is used.  This is basically a solid rod, with an external thread and screwed into a fitting incorporated into the pipe or even pressure vessel so most of the rod is inside the pressure shell. It is drilled with a blind hole so the thermocouple can be inserted from the outside without causing leakage.  The pipe can be properly insulated and the insulation left in place.  The thermocouple can be either permanently fitted for continuous measurement, or a portable unit used for occasional measurement.  I have made miniature thermowells like the one in the attached photo.  It is a short one for use in an elbow fitting in piping.  I use a long one as the filler plug in my boilers, so the thermocouple is closer to the liquid level.  You can see how it is fitted in the pictures of my engines in the gallery post.  This is probably the best way to measure temperature in the boiler, (which allows you to use steam tables to determine the pressure more accurately than a miniature pressure gauge).  Also probably the most practical way to measure superheater outlet, engine inlet and engine exhaust temperatures.  I make them out of brass not bronze for better conductivity, but probably should try one in copper, or a silver soldered fabrication with a copper blind tube or drilled rod, and bronze screwed part.

Next time a little about heat transfer coefficients in a condenser.

Thanks for dropping in.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 20, 2017, 02:57:28 PM
Hi, Another practical question i have is what is the ideal shape  (with no pecuniary limitations) are the cooling fins on air head IC engines ? Parallel, tapered outwards, tapered inwards, Perforated etc etc. Also if the exhaust pipe on a steam engine has more surface area for the same weight of metal is that better, also you mention the difference between brass and bronze as conductors of heat ? what is this as a percentage ? We are advised to use bronze in boilers so as to prevent dezincification !....more good stuff to stir up the grey matter. I have also heard that a matt black surface will absorb heat better ,But, it will also give off heat better ? That is why i have painted my house radiators black, much to the chargrin of the council inspector of my council house !!!
Title: Re: Talking Thermodynamics
Post by: Stuart on July 20, 2017, 03:14:52 PM
Only if the paint is very thin , the fillers that are in the paint could nullify the benefit.

The main thing to take up on is as MJM said temp difference matters , get the rads hot is you need to get the room warmed up .

Or install fan coil units then the water can be hotter due to the coil being enclosed so safer to people.

Or you can do the job properly and fit a optimiser controller , work out the heat loss though the building structure then control the heating/cooling from an outside temp sensor in a Stevenson’s screen

The joys of thermo dynamics  :lolb: :lolb:
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 20, 2017, 03:22:05 PM
Hi. Also when i made the electrically heated steam boilers, i made the brass holders for the cartridge heaters with a row of grooves to get quicker and more heat transfer. Would this have been correct thinking or would there have been no advantage ?? Also in the pics are the Pressure switch and the low water gauge that i mentioned in a previous post ! Thanks Stuart for info.......i shall get onto the council about that !!!
Title: Re: Talking Thermodynamics
Post by: Kim on July 21, 2017, 03:29:02 AM
Cooling problems proceed at a rate proportional to the temperature difference, so it proceeds most quickly when the temperature difference is greatest, and as the heat transfer proceeds the temperature difference reduces so the heat transfer slows.  An arithmetic average only gives the right answer when the process proceeds at an even rate.  When the maths is done in detail, and I have to admit I find it a bit too hard, it comes to integration of a 1/x term, which results in the natural log term, which properly accounts for the change in rate as the temperature difference decreases.  The classic example is the cooling of a cup of coffee.  Cooling proceeds at a rate proportional to the temperature difference between the coffee and the air.  Not totally true with evaporation at a free surface, but not a bad approximation for a good cappacinno with a thick layer of good insulating foam on the surface.  The temperature initially drops very quickly but then the temperature change becomes slower and slower.   The cooling follows an exponential function which also involves that natural log function. 

Thank you for the very clear explanation.  It makes perfect sense when you think about it that way!
Kim
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 21, 2017, 01:41:41 PM
Well, quite a few interesting questions there.  I will try and address each one in turn, though some will be more easily answered after the post I had prepared.  So I may defer some until tomorrow, particularly your question on fins, Willy.  The answer will make much more sense after I make some explanation of why we have fins.  But the steam exhaust pipe, I am wondering where you are coming from there.  The exhaust pipe is not basically intended as providing condensing area, although there is no harm in taking every bit of area we can get.  So long as the pipe is sloping down, that is, so the water droplets do not fall back into the engine, or sit around causing corrosion when the engine is not running.  Essentially the flow in the exhaust is sort of steady, I know it is a stream of pulses, but the pipe finds an average temperature.  The weight of metal and its specific heat do tend to stabilise things a bit there, but I am not sure it is any real advantage.  If anything, extra metal will condense more steam when the engine first starts until it gets to temperature, then if the pipe is insulated it makes no difference.  If we are wanting to loose heat from the pipe, less metal so it increases in temperature a bit at each pulse means slightly higher mean temperature difference, but I think it would be a very small difference, at least for practical tube thicknesses.  Perhaps overthinking the issue a little.

The difference between brass and bronze, and I will add stainless steel for your tea coolers, right on topic with that one.  For straight conduction issues, conductivity, k in W/(m.K) is the relevant property.  I have not found how to incorporate a table in a post, so please make yourself a table with these figures.  I have also included the specific heat, Cp in J/(kg.K), which I will come back to.
Copper, k= 399 W/(m.K), Cp=383J/(kg.K)
brass, k=111 W/(m.K), Cp=385 J/(kg.K)
Bronze k=26 W/(m.K), Cp=343 J/(kg.K)
Stainless steel, k=14.4 W/(m.K), Cp=461 J/(kg.K)

You can see there is a large difference between brass and bronze.  Copper has much higher conductivity again.  I feel that it is better not to talk about percentage, as that facilitates overlooking the contact resistance, I discussed yesterday, when thinking about the heat transfer.  In comparison, Bronze is not a great conductor at all, and stainless steel, (SS), is even worse.  Dezincification is a form of corrosion where the zinc content of brass is dissolved out, leaving behind a much weaker material which cannot be relied on to safely contain pressure.  There are many contradictory stories surrounding this issue, and I am not enough of a metallurgist to be able to explain the issue more clearly.  However, if you get dezincification of a boiler bush, you would have to condemn the boiler, or at least do major repairs to replace the bushes.  So the codes require us to make them bronze.  There is not so much emphasis on the plug which screws into the bush.  I guess this is because it is easily removed and inspected, but please don't quote me in a discussion with the boiler inspector.  The safe approach is to make any pressure containing component in bronze. 

Let's consider the impact of this on two quite different components.  First, my little thermowell.  When screwed into the pipe or boiler, this component is surrounded by the fluid being measured, and there is no intentional heat flow unless there is a temperature change.  With no heat flow, there is no temperature drop, so the conductivity does not matter.  When the temperature changes, some heat has to flow until the new stable temperature is reached.  Conductivity will affect the time taken to reach the new temperature, so the material has an influence on the speed of response of the instrument.  Temperature instruments are always slow to respond, but this makes it worse.  Perhaps I should use bronze after all.

Now the bush for your heater element.  Obviously the bush silver soldered into the boiler shell should be bronze.  Now, the well for your heating element is removable, so probably could be made of brass.   However I suspect it is not often inspected, but then, any leakage might be contained within your wiring conduit, perhaps!  Probably better to use bronze.  Heat transfer through the wall thickness of bronze will mean a temperature drop, so your element has to reach a higher temperature to drive the heat through the sleeve into the water.  Here you are wanting really good heat transfer, so your element within the tube does not have to get so hot to transfer enough heat.  In this application, bronze is not an ideal solution.  This leads me to wonder if we could make a bronze part to screw on to the the boiler, with a copper tube silver soldered in to contain the element, and a bronze plug silver soldered in the end.  Even better conductivity, and no dezincification.

I am assuming that your cartridge heater comes with the element insulated and sealed in a sheath which you have to introduce into a pressure tight component, much like a longer version of my thermowell with the hole drilled to accept the heater sheath, and flanged onto the boiler bush.  I addition you have machined grooves on the outside of your pressure tight sleeve.  I am going to leave my comments on these grooves until after I discuss convection, as I hope it will make more sense then.

Then, your household radiators.  First we need to clarify the terminology.  I know that "radiator" is the normal terminology, but it is probably actually a convection heater.  Especially if the heat comes from circulating hot water.  A common arrangement is a thermostat which controls the water flow so when the temperature is too high, the water flow is reduced.  I am not sure if the controlled temperature is the room temperature or the water temperature, though I am inclined to guess the water temperature.  So either a fairly flat panel, or a more complex column arrangement.  This is a typical heat transfer problem where transfer is by all three of conduction, convection and radiation.  Usually however, one of these modes predominates and the others relatively unimportant.  The term radiator suggests the main mode is radiation.  However radiation depends on difference between the fourth power of the absolute temperature of the heater surface and room objects.  I suggest for a hot water system, this is not a very important contributor, but most heat is transferred by heating the air in contact with the metal surface.  The warm air then rises and circulates the air in the room, thus distributing the heat.  This is called convection.  Conduction is only important if objects are in contact with the metal surface, not usually recommended, though sitting on it is very comforting when you first come in from the cold.

I am totally in agreement with Stuart on the paint, especially if the heater really is a radiator.  A very thin black paint can in principle help increase the emissivity of a radiating surface, but getting this right is tricky.   You would need to do some careful experimenting, as the wrong paint does not work so well.  It also, as you say, increases the absorption.  However, for primarily convection heating, while the metal surface detail is important, paint is more likely an insulating layer which means the temperature difference between the water and the room will be greater.  Surface roughness is probably helpful.

Thanks for joining in again, Stuart.  Good to have you on board.  It would be interesting to hear more about your idea for an optimiser for household climate control.  These days with Arduino microprocessors and the means of programming them so readily available a fully automated household climate control system should be an easy project.  However the logic for the optimisation needs to be developed.

I am glad the explanation was clear Kim, any time it is not, please ask again.  Building blocks in a knowledge base are only useful if everyone one can understand them.

My intent was to continue to look at heat transfer coefficients, but I think this post is long enough.  Better to continue next time.  Not sure I have completely answered all those questions yet either.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 21, 2017, 03:33:36 PM
Thanks for the info ...cool.... i hope i am not hijacking these posts !! just a thought but no questions this time.! this pic is the exhaust arrangement of a small double acting steam engine in the local museum. As you can see the exhaust pipe goes strait up ?!!!! However it was taken apart and reassembled, so they may have got it the wrong way round as it is bolted on. I did make a model of this and used the same configureation . I may have to write a comment in the visitors book !!they also got other parts in the wrong places as well !!! Also the cartridge heaters are designed to fit into reamed holes, and make contact when they heat up.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 22, 2017, 01:07:15 PM
Hi Willy, I am glad that you are finding the information interesting, and don't worry about hijacking the thread.  It is all about building a knowledge base that is helpful to model engines.  I always have the choice to defer a reply until later.  I am just as happy to look at radiation now as after convection, so I chose to look your heater questions when you asked.  Mostly, I find your questions right on topic, and they help keep me grounded in just what it would be useful to cover, so they are very helpful, thank you. 

That is a beautiful model you have made of an interesting museum model.  Just a bit more about the exhaust, I was thinking particularly in reference to an exhaust pipe connecting the engine to a condenser.  Points in piping which collect liquid cause all sorts of problems, hence my comments about making the piping self draining.  Of course, if there is no condenser, the exhaust steam still has to go somewhere.  In an industrial situation, a horizontal hot exhaust is quite dangerous, downwards not much better, and upwards is probably the best option.  Of course, weather conditions determine whether the steam mixes with air and is carried away, or condenses into a fine fog that is carried away, or condenses into drops which rain down on everything in the area.  On my models, I run the exhaust down to a separator/oil separator, where any water (mostly only on starting) is collected and drained, while the dry steam is discharged straight up.  I suspect that in the period of the model, there were no such niceties, or possibly no cylinder lubrication.  They may have just relied on condensate for lubrication.  So the vertical exhaust on the the museum example and your model is most likely quite ok.

The requirement for a reamed hole for your electric elements is interesting.  Can you ream a blind hole?  By the way an electric element is easier to deal with in experimental work.  You know the power dissipation of the element based on supply voltage and element resistance, and the element just gets hotter until all the heat is lost to the water.  So good contact, and a high conductivity sheath both help to limit the maximum temperature of the element.  Do you also use a smear of heat conducting grease to help even more?  I am not sure if there is anything available that would not cause other problems, such as mess or corrosion.

Yesterday, I mentioned that heat transferred by radiation was proportional to the difference in the fourth power of temperatures.  Mathematically Q = U x A x (T1^4 - T2^4).  The temperatures have to be absolute, K or Rankine if you must.  Radiant heat travels in straight lines like light, but not evenly in all directions.  Think of a car headlight compared with a room light with a translucent diffuser. So A and U have to be modified by a view factor which relates how much of the radiated heat is actually received by the receiving surface, also by directional factors and so on.  But if the water in your heater is say 65 C, the heat transferred by radiation will not be much compared with the heat from a red hot glowing element.  You can get a good idea of whether the main mechanism is radiation or convection by holding your hand facing the heater but say a metre away at the height of the middle of the heater and feeling the warmth.  Then hold your hand above the heater, again about a metre distant but directly above the heater palm facing downward, and feeling the warm air rising.  Which one gives you the most heat?  On the other hand if you have a fireplace, or campfire, or perhaps a one or two bar electric radiator, and hold your hand facing the fire, a metre will be too close for comfort due to radiant heat.

Now let's return to our condenser.  We usually have a metal wall, usually tube, with the condensing steam on one side and perhaps water on the other.  This is when things get difficult and we have to look again at types of heat transfer.

Most of us learnt in school science that heat is transferred by conduction, convection and radiation.  We then usually talk about conduction, possibly mention radiation as heat transfer from the sun to earth through space, and rarely go into convection in any depth.  Now this is not entirely without reason.  Convection is quite complicated.  And radiation not much different in complexity as we saw when looking at your household radiators.  My text book on heat transfer is significantly bigger and heavier going than my book on thermodynamics.  And much of this is due to study of convection.  I don't want to go too deeply into heat transfer, but it is necessary to just peek in and try and understand the main issues.

With convection we find that we have to look at three resistances to heat transfer which are effectively in series, to use an electrical analogy.  We are talking about conductivity which is the reciprocal of resistance.   We have the steam side of the tube, the metal tube wall, and the water side.  Now we can look up the thermal conductivity of steam at 100 deg C, it is about 0.025 W/m.K, and the conductivity of water at say 25 deg C, about 0.6 W/m.K.  Now this is not looking good for our condenser, steam is about as good an insulator as cork and water not much better.  But this is where convection comes in.  The copper tube is a good conductor, and will have only a small temperature drop from one side to the other. 

Now let's look at the water side of the tube, and conduction of heat from the tube to the water.  The water temperature is measured well away from the tube, and if the water is not flowing, perhaps trapped in a porous foam, the temperature will increase as you move the measuring point closer to the tube with a linear temperature gradient as heat flows by conduction, and will match the metal temperature very close to the wall.  However, if the water is free to move, everything changes.  The water against the wall expands as it get hotter, and the hot water rises through the more dense cooler water, and is replaced by cooler water.  This means that the temperature gradient near the wall is very steep, much steeper than if there was no water movement and only conduction.  It is  this high temperature difference across the layer very close to the wall which means much higher heat transfer than the conductivity would imply.  In addition, water has a high specific heat, so the moving water can carry away a lot of heat.  So the water side will carry away heat very well.  If we then use a pump to force the water past the tube the heat transfer will be further increased.  So the heat transfer on the water side is not only proportional to the temperature difference but also to the water velocity, whether it is a forced (pumped) flow, or natural convection flow.

The steam side has a quite different situation.  The steam cooling against the tube condenses and forms a film in the tube.  Of course gravity makes the film flow down and off, and you can see this is quite complicated to analyse.  Just to make it interesting, water is one of few substances which in this situation can form droplets which drop off the tube.  The resulting turbulence further increases the heat transfer.  I assume that now days with computers the maths is doable and coefficients can be calculated.  For most of my working life, approximate coefficients were assumed based on industry experience, and there are books with suggested figures for various situations.  In the end it should be clear that to calculate coefficients from theory is probably not practical for most of us.  I am sorry if this is too heavy going.

Let's just take away the summary that the convection coefficient for a condenser is determined by three individual coefficients, the steam side, the tube wall and the water side.  Usually one of these provides most of the resistance to heat transfer and so determines the overall heat transfer rate.

Next time I will talk about a few things we can learn from this basic understanding of heat transfer.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 22, 2017, 03:33:30 PM
Hi, I have had another look at the Bridewell exhaust pipe and there is a drain cock arrangement at the bottom of it, see pic.and there was a hydrostatic lubricator in the steam line. Also you get two types of basic reamers The Hand type have a longer taper and the morse taper type has a very short taper, about 3-5% and these are to be used on the lathe specifically for blind holes .....And Bronze is approx 88% copper  12%tin   and brass is 65% Cu and 35% Zn...so yes bronze is a much better conductor.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 23, 2017, 01:23:39 PM
Went for a bike ride this morning.  Disturbed a mob of kangaroos that were grazing by the path.  They all bounded away in that typical flowing motion and were gone.  Not the everyday experience even for for most of us in this country.  All too quick for any chance of a picture unfortunately.

Willy, I am not surprised that there is a drain cock at the bottom of the vertical exhaust of your prototype engine.  There is a lot of condensate to get rid of when the engine starts, and a little more when it is shutdown, the two occasions when it is good to be able to drain condensate.  If you have ever looked at a petroleum refinery or petrochemical plant and seen that mess of piping everywhere, you may be interested to know that every one of those pipes has a drain valve at every low point where liquid would collect, and a vent valve at every high point where air would be trapped.  A little known fact, I believe real rather than alternative fact, is that the equipment count with the best correlation to the total plant cost is the number of high and low point drains and vents.  Unfortunately, when you finally know the number, you know the cost anyway, so not very useful for budgeting.

Thank you for the information about reamers, I have often wondered which I should buy.  One of those puzzles that we all have to deal with as beginners in this hobby.

Now conductivity.  I am not sure whether you are expecting intuitively that more copper is the same as better conductivity, or if you are using a very different text book to mine.  Pure metals such as copper, aluminium, have relatively high conductivity.  Tin and zinc much lower. The conductivity of pure copper 399, and aluminium 236, while zinc is only 121 and tin 67, all in W/(m.K) from the same book.  However, when the elements are melted together to make an alloy, the conductivity is normally less than either of the components.  So brass is 111, which is less than either copper or zinc.  Similarly, bronze is only 26, so again less than copper or tin.  You could also look at Duralumin, an aluminium copper alloy with only 5% copper is only 164, and aluminium bronze, also a copper aluminium alloy but this time 95% copper, 5% aluminium is only 83.  I don't have the facilities to test these, and do not have the opportunity at the moment to check other sources, but I would suggest that while any one of these could be a misprint in one book, the for whole lot to be misprints, all the same direction, I suggest would be unusual.   Interestingly, Carbon in iron to make steel seems to increase conductivity, and chrome or nickel individually in iron have a very different result to the combination in stainless steel.  Definitely an area where intuition does not help much. 

In summary, I suggest brass is actually the better conductor, while bronze is only better than stainless steel.

Getting back to our topic of heat transfer and condensers, I hope you can  see why I am not proposing that we try and calculate a heat transfer coefficient for a new condenser.  However we can do it the other way, if we have a condenser, and we can measure how much steam it is condensing, and the inlet and outlet temperatures, we can calculate an approximate overall coefficient.   If we have a condenser, we can calculate an overall heat transfer coefficient from its performance, and this would be a reasonable guide to a similar condenser with more or less area.

We can also apply the knowledge to other problems such as when to add fins to our surface.  On an internal combustion engine, with air cooling, we can identify a film coefficient for inside the cylinder, conducting heat to the metal cylinder wall, then the conductivity of the metal wall, then the metal to the air surrounding the engine.  In this case the combustion gases are very hot, and provide a large temperature difference to drive heat transfer into the metal, and a smooth cylinder wall is an obvious necessity, so the main variables we can play with are the metal composition and its thickness.  When we look at the transfer coefficient from the metal to the surrounding air we find this is the lowest of the three and basically controls the overall heat transfer.  We can increase the coefficient by forcing a cooling air flow instead of relying on natural convection, and we can increase the metal surface area by adding fins.  Of course adding fins is not quite as simple as we would like.  Basically heat has to travel to the parts of the fin which provide the extra area.  So conductivity along the fin becomes more important.  In travelling to the end of the fin there is a temperature drop which reduces the temperature difference to the air, so the extremities of the fin are less significant as the fin becomes deeper.  In the end, a very deep fin has no heat transfer advantage over one of the "just right" depth.  The "just right" depth depends on the the ratio of the surface coefficient and the conduction coefficient, so again no simple answer, but cast iron fins would have a different ideal depth to aluminium ones.  But you can see some clues to the question on fin shape.  A fin that is thickest further from the cylinder wall is not only tricky to make, but has more metal where it is less useful.  On the other hand, one that is thicker nearer the cylinder is actually more effective than a constant thickness fin, but you will notice that I am summarising the answer rather than trying to include all the maths.

This logic applies generally to problems involving transfer to air.   Transfer from condensing steam to a copper tube wall is very good, conduction through a copper tube wall good while transfer to air from the metal is much lower.  So we see the extended surface approach on our car radiators for example.  Some indication of when fins might be a good idea is provided is given by industrial steam condensers.  A normal water cooled condenser for an engine or steam turbine exhaust is normally built using plain tubes, in very large numbers, I might add.  It is quite unusual to use air cooling for a condenser, partly because water is normally cooler than air and easily achieves a lower condensing temperature, and hence pressure.  However, I do have experience of one quite large air cooled steam condenser.  I can assure you it was very large and had very large fins on the outside of the tubes in addition to large fans to force a high air flow.  So air side film coefficients something between water and air cooling is the indicator for adding fins.

You might also be interested to know a little more about your use of teaspoons to cool your tea.  In addition to just absorbing heat to warm the spoon, thus cooling the tea, the handles of the spoons also act as extended surface area, in addition to the surfaces of the cup, so the spoon handles are fins on your tea cup which increase the cooling rate.  I am serious. The specific example is in my text book!  Also, I assume you sit the cup on two pencils, or more spoons, so air can circulate underneath instead of the bottom being insulated by the saucer.  I think it's time to get out the stop watches and thermometers and do some serious thermodynamics over a few cups of tea.

Next time I will have a look at your electric boiler sheaths for the heating elements.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 23, 2017, 02:36:45 PM
Hi MJM,  Wow.OMG. My intuitive expectations are so wildly incorrect i feel i need to go back to school, However as a Baby Boomer we did not really do Science or technology lessons.!! Interesting about conductivity (thermal not electrical) or there might be some correlation, I do not have the text books to hand to look up specific tables. When you have an amalgam of Tin /Lead as in solder the melting point is less than either of the two metals, so i assume this is correct with brass and bronze ? I have always wondered why cast iron melts at about 1100 degrees and steel about 2000  ? I use german or Nickel silver (copper/Nickel) quite a lot in my models so what is the value for this ?  Interesting info about the tea cup cooling, and i will now have to explain why i am now taking 5 teaspoons (M'Lud)!! instead of 3 !! Interesting about the fins on IC engines...On my 1920's Scootamotor the cylinder was steel with lots of very thin fins turned up out the thickness on the billet,.I notice that fins seem to be evenly placed around cylinders rather than longer at the back where there is less and higher temp air flow or are they made from an asthetically pleasing rather than a thermodynamically correct logic.? On Condensers ,The tubes in the boilers are steel where we want maximum heat transfer ,but, incondensers the tubes are invariably brass ,where we also want maximum heat transfer !! This is to do with cost considerations i suppose.....in some of the GWR locomotives the inner fire boxes were made of pure copper !!! .Thanks for all this info, i have always said that if you have all the correct Info about anything, you can make the
correct decisions about how you proceed using that info.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 24, 2017, 11:51:40 AM
Hi Willy,  intuition is often helpful in forming a hypothesis as a basis for further inquiry.  Like you, I am quite surprised when I come across examples which are not in agreement with intuition.  However, intuition still helped you formulate the question, and when I looked carefully at a few examples of data, together we found useful information.  Data leads us to modify the initial hypothesis to something like "when metals are melted together in an alloy, the thermal conductivity is less than either of the pure metals".  When I then looked at more complex alloys, like steel, or alloys of more than two metals, we have to modify the hypothesis a bit more to be even less positive.  Something like "often less than any of the components" or "may be less".  Not so definitive, but useful none the less.  Then on thinking further, steel may have other trace or significant quantities of other metals, but it also has carbon, which is of course not a metal.  And we know the properties of steel are very sensitive to even very small amounts of carbon.  I believe there is a fairly good correlation between thermal and electrical conductivity, but again, there are some exceptions.  You may have one in your electric heater cartridges where there are conflicting requirements for good electrical insulation and good thermal conductivity.

I don't think it is necessary to go back to school at our age, we already have a life time of knowledge and experience that schools are trying to prepare kids for.  Life is not long enough to learn everything.  We are better to concentrate of filling in the gaps in our knowledge in areas that interest us, a privilege of our age.  Of course we may choose to do this by private reading, discussions or even by taking a class.  I am also a baby boomer.  In high school we had to choose between science or humanities, and each stream only did one light subject from the other side.  The stuff we are talking about here was only introduced at tertiary level, but most of the real learning came from a lifetime of needing to apply these principles in my every day work.

The melting points of alloys vary over a wide range, depending on composition and these are usually described in a phase diagram.  You will also find there are true eutectic alloys that have a simple single melting point, and many other alloy compositions that melt over a temperature range.  Your tin lead solder example is one of these, where the eutectic alloy heats up then suddenly melts, while other (generally cheaper) alloys with less exact composition, start to soften and gradually melt as they get hotter.  The best source for these phase diagrams is a metallurgy text book, which you may be able to find in your local library, particularly if it has a good technical section.  Definitely not my strong subject, and unfortunately my text is not accessible for now.

The shape of the fins on your scooter engine will be determined by a lot of things, not simply thermodynamics.  In fact thermodynamics (of the engine performance) probably would prefer less heat loss for higher engine efficiency.  The main reason for the fins is so nothing melts, and the necessary cooling aims to do it uniformly so the cylinder stays straight.  Also fins might be shorter lower on the cylinder, as the piston limits the time that hot gases are near the bottom end of the cylinder, so there is less heat to be taken from there.  Again well beyond my knowledge.

When you refer to steel tubes in a boiler, I assume you are talking about steel boilers.  Perhaps locomotives, but ships and stationary engines also.  Again, heat transfer is desirable but even in a boiler, not the primary consideration.  The major thing is strength, particularly at temperature.  Copper and its alloys are not strong enough, particularly at high temperatures.  Copper is chosen for model boilers as it is easily worked by people such as us, and readily joined by soldering or brazing.  But copper strength at only 200 deg C is only half that at room temperature.  Better make sure you have plenty of water in the boiler, especially with a good hot coal fire.  Steel is stronger than copper at all temperatures and has useful strength to quite high temperatures.  But steel working (as opposed to machining) and joining by welding to maintain full strength even at temperature, is not for the average hobbyists.  Of course steel is subject to rusting, much to our dismay when we find it.  While copper is very much less so.  Some special stainless steels, known as duplex stainless steels, have been developed to resist corrosion as well as have high strength at temperature.  In fact there are groups working on making model boilers from duplex stainless, but you can safely assume that these people are real expert welders in their working life.  They are not average hobbyists, or even just good welders.  So many more factors than pure thermodynamics are involved in choosing the best material for an application.  It is necessary to have not only correct information, but also to the extent practical, complete information.

 Very few real life problems are single dimensional, most have many factors, and very few real problems are binary.  Despite computer logic, the answer is not often limited to yes or no, black or white.  Rather the answer is mostly maybe, or some shade of grey.  And rarely do we have really complete information.

Today's adventure in the long paddock, a horseman rode out onto the highway, and started walking in small circles in my lane.  Obviously a sign to stop.  Out came about 100 head of cattle, being driven across the road by two drovers on horse back.  They had put out signs to advise the traffic coming towards us, but I think they forgot our direction.  To complete the picture, we came up behind a truck with a rather superfluous "Wide Load" sign on the back.  The road is one lane each way, yes it is a main highway, each lane about 4 meters wide.  The truck was carrying a bulldozer with a cab about 6 meters high, and a blade at least 6 meters wide, protruding well over the centre line.  Even the pilot vehicles for wide loads coming towards us, usually quite pushy about claiming their side of the road and more, moved over for that one.  We had a very sedate peaceful drive behind him for about 50 km before he pulled off into a rest area.  Obviously mining in the area as well as cattle.

I was going to continue the heat transfer discussion by looking at Willy's electric heating elements, but I seem to have been side tracked.  Oh well, life is more than pure thermodynamics.  That will be tomorrow.

Thanks for dropping in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on July 25, 2017, 11:33:25 AM
In post 128 you state, "It is a matter of understanding the energy balance for your engine, knowing exactly where all the heat is going."

Well, this and all the discussion lately has raised a question I have never been able to answer satisfactorily. I would like to optimise the design and material choice for small gauge 1 locomotives with inside cylinders, fixed cut off with dimensions of the order of 8X12 mm, (to a little larger), that operate with methylated spirit fuel, single firing rate and throttle setting, hence mostly more or less stable cylinder conditions. I previously had two locos like this, twin cylindered Aster Lions, (Titfield Thunderbolts). One has a brass block cylinder and the other brass tubes silver soldered into end plates, neither had cylinder lagging. The latter engine seemed to be very slightly superior in run time/distance and a little less condensate out of the chimney, but I had not done any measuring unfortunately and have now donated this engine to a mate some thousands of kilometres to the South so can't do any analysis now.

For future engines of similar type and size I am wondering if moving to bronze cylinders built up from two tubes with end plates silver soldered together and insulated with Kaowool lagging would achieve any superior results than the use of a bronze block for the inside cylinders. I have some intuitive and experiential notions but am not competent enough to work out the thermodynamics of it all. Given that modern loco cylinders were relatively thin walled steel castings I am presuming models with big lumps of heat absorbing and transferring material like solid block brass fabrication or castings are potentially working against us. But to what degree?

So, I beg your indulgence in pondering this issue, and are able to address my specific enquiries by showing us some thermodynamics that might resolve the issue. Thanks for your labours to date, if it be any consolation some of us labour just as intensively trying to absorb your energy output. Regards, Paul Gough. P.S. Operating pressure in my loco is 50 to 60 psi so assume cyl. pressure to be somewhere around 50 or a bit more.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 25, 2017, 01:23:28 PM
Heat transfer example - electric boiler heater elements.

Hi Paul, I was just about to post this when I saw your reply.  More very interesting questions.  I will give some thought to those, and will definitely come back with my ideas.  I suspect it will continue to involve taking a little bite at a time and weaving these bites between bits on condensers, but I am heading towards boilers, so a very timely query.  I am sorry it is such heavy going, any suggestions on how to make it easier to follow would be welcome.  Even requests for more explanation on the hardest bits.  Then I can try to come up with some examples that will help.  Always a balance between too many words and not enough explanation.  By the way, is tropical Queensland anywhere near Rocky?

While this subject is perhaps a little out of order, I was talking about condensers after all, but it is the last outstanding part of Willy's question, and it is as good an example as any to round out the heat transfer discussion.

Let's look at the complete heat transfer path from the heating element to the water.  Electrical energy is used to heat a coil of wire. The wire has significant resistance, and ohms law gives us the current as V/R.  I believe the elements are connected to alternating current power supply from the mains, not sure whether it is direct to the mains or perhaps through a transformer.  The wire is probably formed into a coil to fit enough length into a compact space, so may have some inductance.  With AC, inductance causes a phase lag so the current is not in phase with the current, but the heating effect is only due to the resistance.  Electrical power is calculated in Watts as V x I.  Alternatively, a little more manipulation of the equation can give power directly from the voltage and resistance as V^2/R.  Looking at it this way is useful in understanding the impact of reduced voltage on the power output.  You can see that this formula has no components relating to temperature, or thermal conductivity.  But a heat balance on the element tells us that the temperature will get high enough to reject all the heat input from the electrical input.  We might think that ideally, we would put the element directly onto the water.  This would result in the lowest wire temperature, but there would be a real danger of someone getting electrocuted, particularly with mains power (220 - 240 V in UK, Europe and Australia for our US readers).  There is also quite a high probability that some of the electrical current might take an alternative path through the water, producing hydrogen and oxygen.  You don't want any sparks around that combination.  And there is no need, the wire has very high temperature tolerance and is unlikely to melt.  It is safer to surround the wire with an insulating material.  In industry, and even domestic hot water services,  a temperature measuring element is also incorporated into the bundle within the sheath.   This element is to protect the wire from excessive temperature by isolating the power supply when some high temperature limit is reached.  I don't know what specific material is used for the insulation, it may be some sort of ceramic or a magnesium compound.  Willy, do you know what it is?  As I have previously mentioned, there are conflicting requirements for its selection.  It needs to be a good electrical insulator, but ideally have good thermal conductivity.  In the end, there must be no compromise on the electrical insulation properties.  Then the whole lot is placed in a metal sheath so it can be safely sealed up.  The sheath is almost certainly kept thin, perhaps for better heat transfer, but it is probably not rated for significant external pressure.  I assume the manufacturer has some specification for the maximum temperature this sheath should be allowed to achieve.

 In order to use it in a pressure boiler, the user has to produce a second sheath designed to withstand external pressure, and also designed so it can be installed in the boiler without causing leakage.  Willy has produced a brass sheath similar to the thermowells I make for temperature measurement, which he has grooved to increase the surface area for heat transfer to the water. 

If we assume that steam is to be raised at 350 kPa(g), 450 kPa(a), it's temperature will be about 148 deg C.  The heat path from the manufacturers metal sheath to the water involves a contact thermal resistance between the sheath and Willy's boiler fitting, the conductivity of this fitting and finally the film resistance inherent in the convection transfer to the water.  I was interested to learn that the heater elements are designed to be fitted in a reamed hole.  This implies very close contact, possibly even slight interference that compresses the manufacturers sheath, even if just a little to ensure very good contact.  Certainly an air gap would be a real problem, and must be avoided to the extent possible.  I don't know if any heat conductive grease is used, similar to that used in assembling heat sinks on electronic components where there is the same problem.  Willy can choose the material from which to make his well.  As we have already noted brass has better conductivity than bronze which might also be chosen.  The better conductivity of brass means that the element will not get so hot.  Stainless steel could also be used, however stainless steel has even lower conductivity,

Finally, the question of fins.  Now adding fins adds surface area which you would think would always help.  However this is another of those cases where intuition does not always give us the right answer.  So how do we answer the question for this particular case.  The maths for determining the effectiveness of fins is complex.  You will find it in a text book on heat transfer if you want to look in more detail, perhaps in the local library.  My quick summary is that there is a dimensionless ratio called the Biot number, my book uses the symbol Bi, which tells us when fins are likely to be helpful.  The Biot number is defined by Bi = h x d/ k, where h is the convection film coefficient, and k is the conductivity of the fin material, and d is half the fin thickness (please don't ask why the half, it's in the book).  Fins are worth using when Bi is very small, meaning much less than 1, while they actually are counterproductive when Bi is much greater than 1.  As an indication, convection transfer to a gas generally gives a very small Bi due to a very low film coefficient on the air side.  Heating water is on the borderline, while boiling water has a very high film coefficient, especially if the temperature difference is enough to cause vigorous boiling, so fins are counterproductive.  The border between fins helping and not helping with brass fins is somewhere in the region between just gently heating water and vigorous steam production.  Unfortunately I do not know whether your fins are useful or not, so the issue becomes one to devise an experiment which will tell us whether fins help or not.  That sounds like a good spot to pause until tomorrow.

A bit more adventure in the long paddock today.  Came around a bend on the highway, only to find a small herd of cattle, perhaps about 20, standing on the road.  We stopped.  Beautiful healthy looking animals.  Collecting fresh steak for dinner had some appeal, but it tends to do a lot of damage to the car.  So we just tooted.  Half started moving on, the others moved back on the road.  No drovers in sight, so we waited.  Eventually they moved off and we went slowly past.  I reckon somebody left the gate open!

Tomorrow I will try and devise an experiment that will help Willy decide whether or not to continue machining his fins.  Has anyone got any suggestions?

MJM460
Title: Re: Talking Thermodynamics
Post by: Kim on July 25, 2017, 02:39:07 PM
The Biot number is defined by Bi = h x d/ k, where h is the convection film coefficient, and k is the conductivity of the fin material, and d is half the fin thickness (please don't ask why the half, it's in the book).

Intuitively, it seems that the 1/2 the fin thickness makes sense because the maximum path for the heat transfer would only be 1/2 the thickness of the fin.  The heat can conduct to the outside on either side of the fin, it doesn't have to travel all the way through the fin.

Might be wrong, but this logic makes sense to me :)
Kim
Title: Re: Talking Thermodynamics
Post by: paul gough on July 25, 2017, 02:49:00 PM
Thanks very much for offering to provide some ideas which might resolve the question at issue. Please don't feel obliged to provide explanations that labour the point, it is my responsibility to work at understanding what is presented then to seek further assistance if needed and as this is not paid work for you please engage with my enquiry at your leisure and when you think it appropriate to the discussion. My comment about 'labouring just as intensively' was not meant to be literally the case, more a bad metaphor to indicate I/we are paying attention, working at keeping up with the prodigious information flow, and attempting to assimilate it as best I/we can, I guess  something akin to diligent students.   

As a Far Nth. Queenslander,I don't regard Rockhampton as being in the tropics, just on the edge of it, a bit of a conceit I admit. I reside in the hills behind Cairns. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: jadge on July 25, 2017, 03:21:16 PM
I had a most interesting chat with steam_guy_willy a couple of years back at the Forncett Model Engineers day about his electric boiler. I was interested as I wanted a simple boiler to provide steam for injector testing.

The heating elements used are obtainable from commercial distributors such as RS in the UK. They are basically resistance heating elements running on 240VAC, at least the ones I was interested in. Although they contain a helical wound element I doubt that the inductance will have any impact at 50Hz. One could get an estimate of inductance from the Wheeler formula, but even if it was 1mH that's only about 0.3ohm at 50Hz. If the heater is 500W then the DC resistance is on the order of 115ohms when hot.

The internal insulating material is magnesium oxide; average thermal conductivity but high dielectric strength. The outer case is usually sealed stainless steel.

The heating element depends upon a good heatsink to work. If you have an element about the size of your finger dissipating 500W and relying primarily on convection in air it's going to get mighty hot, mighty quick! Hence the need for a reamed hole and thermal grease. The manufacturer of the elements sell a high temperature grease, but rather stupidly RS do not stock it. Since I need the boiler to run at 170psi (~190°C) that rules out most of the general heatsink compounds. In high power electronics any sort of air gap, or even air pockets, is death to heat transfer, and thus to the semiconductor device.

As an aside I decided not to follow the steam_guy_willy design on the electronics side. For £18 I got a PID controller, thermocouple and solid state relay from Ebay. So it should be simple to implement a closed loop control system, based on temperature, and thus indirectly steam pressure. Mind you it might be prudent to add a proper safety valve, and possibly a water level detector.

The main thing I know about thermodynamics is that, despite doing a course in it at university, I still don't really understand entropy.  :noidea:

Andrew
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 25, 2017, 03:22:24 PM
Hi MJM, These are the RS Components  Cartridge heaters that i used in this boiler. They are the 500 watt 100mm x10mm ...250volt ones. the boiler dimensions are 3" x6". The boiler takes about 7 mins to get to 2 bar (30 psi) , this is to get over the attention span problems of the youth today !!!. I do not use any sealant just push them in slowly as the air needs to come out. Interestingly when i demonstrated it at Beeleigh mill because it was outside they had a large extension cable un wound and it took about half an hour to heat up !! I think the inductance of the coil had something to do with it. This boiler was featured in the Engineering in Miniature magazine a few years ago......No Questions at the mo ,so you can catch up !
.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 25, 2017, 03:32:26 PM
Hi Just seen the above comment that overlapped mine. when i talked to RS they did not mention the high temp grease so i did not use it.!! I used the control system designed by a friend that used a 9 volt battery for the low water safety circuit to trigger the 250 volt circuit, incidentally you do need a fairlly good battery ,as although the led's light up the circuit will not function properly !!
Title: Re: Talking Thermodynamics
Post by: Admiral_dk on July 25, 2017, 09:01:53 PM
The first thing that springs to mind when considering a high temperature insulator that is a good thermal conductor, is aluminium oxide as the material is used inside certain electronic components as such. The thermal conductivity of a few um is almost the same as aluminium and it will withstand thousand volts or more (depending on layer thickness). Aluminium oxide is extremely strong / tough, but still it can be damaged and then the insulation properties are gone, so ....  :noidea:

Best wishes

Per
Title: Re: Talking Thermodynamics
Post by: jadge on July 25, 2017, 10:14:50 PM
Some power semiconductors, especially RF power devices, used beryllium oxide as an electrical insulator, but also with a very high thermal conductivity. There were dire warnings about not cutting open such devices and creating dust, as BeO is carcinogenic.

Andrew
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 26, 2017, 01:28:45 PM
Hi,  Entropy !!  Is this the elephant in the room ?? James Clark Maxwell certainly thought so........!! Found this reference in quite a good book that should be on every model engineers shelf? !!!!!!
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 26, 2017, 01:50:34 PM
Thank you everyone for the helpful comments and contributions.  I will try and do justice to each of you, however it is getting very late due to an unexpected water leakage problem we discovered this afternoon and that had to be attended to first.  Have just about everything dry, and back in place and will attend to repairing the fault tomorrow.

Kim, I believe that you are on the right track with your suggestion of the two sides of the fin meaning there will be a two somewhere, alternatively a half will have to appear in a number of other places.  I think it is just a matter of convenience which approach is used.  On the issue of conduction path, remember that heat has to flow from where the fin attaches to the cylinder along the the complete length of the fin to the tip, not just through the fin thickness.  This is why there are diminishing returns on making fins ever longer.

Paul, FNQ makes sense, thank you.  Unfortunately turning left at Rocky, but if you ever visit the Deep South, please do yell out.  I am trying to find that balance between too many words and too superficial.  So please do say something any time you need a bit more explanation.  Life does not have to be that hard.

Thanks jadge for the extra information on the heaters.  I totally agree with you that the inductance would not significantly change anything.  Also, only the resistance is responsible for the heat, and we only measure resistance with a normal multimeter.  Good information on the MgO and the stainless steel sheath thank you.  Interesting that the manufacturer does sell high temperature grease.  However, I also wonder if there is any problem inserting a close fitting sheath in a reamed hole with a coat of grease, could the air get out?  Or would it make a very effective pneumatic ram?  The PID controller is a great solution for precise pressure control.  I think Willy's system is a simple low level protection,  and he expects to use all the heat, so temperature control is not really an issue.  And I would always recommend having the safety valve.  And low level protection is also a good idea.  While pressure inferred from temperature measurement is probably more accurate than little pressure gauges, when the system is first heated there is still air in the boiler, so the pressure can be higher than you would expect based on the temperature, until the air is lost with the first steam production.  Remember the mountain top experiments?

Willy, thanks for the element data.  Looks like a good range available for many boiler sizes and capacities.  If you are using those little rectangular 9 Volt batteries, they will not last very long with an LED and a relay.  The relay should be energised to make contact in the 240 V circuit, so that in case of a fault the power is isolated.  The ones I am thinking of would not last very long as they have very little energy capacity.  I would recommend a 9 V plug pack with a reasonable current capacity.  After all you already have 240 V power for your heater.  Alternatively a 12 V motorcycle battery with the appropriate regulator to supply your 9 V would have enough capacity to last a reasonable time.  An unusually long boiler heat up time at your display venue could be due to excessive voltage drop in the long extension lead, especially if there is a poor contact somewhere along the way.  Remember the power is V^2/R, so a small under voltage does significantly impact on your heater power.  I do not advocate exposing mains power terminals to measure the voltage unless you are an electrician, but you could check each plug and even the power switch with one of those infra red temperature devices.  They are excellent in such applications.  A hot spot indicates a poor contact, and may indicate a faulty cord or switch.

Thanks also to jadge and Admiral_dk for suggestions of further materials with suitable combinations of thermal and electrical properties.  Clearly there are many options.  Obviously the carcinogen one would be well avoided, but I suspect that enclosed in a stainless steel sheath it would be well enough protected from damage in normal use.  I suspect to find a material that could be bent or deformed without consequence would be more of a challenge.  I hope most users would understand that the element should be treated with reasonable care.  Above all it would be interesting to know which insulating material the manufacturer has selected.

Now recognising the importance of a proper heat sink to the element, and the need to avoid impediments to thermal conduction, we are back to Willy's conundrum of whether to machine the grooves in the outside of his pressure boiler insert.  In most situations, we could measure all the inlet and outlet temperatures on two samples and easily deduce which is most effective.  But in this specific electric heater case, the element temperature varies to ensure rejection of all the heat, and we can't measure that temperature.  With the heat input to the boiler always equal to the element power rating, there is no expected variation in time to heat, whether the outer surfaces has fins or not.

I would suggest that the mathematics to calculate the difference is certainly too hard for me, and I am not sure if the heat transfer coefficient can be calculated with sufficient accuracy to give a clear answer.  I suggest the most definitive way to answer the question is to make two element carriers, one with grooves and one without.  And conduct an experiment.  But what experiment?

I have implied that the element resistance is constant, however resistance of a metal is temperature dependent, and does change as the temperature rises.  I seem to recall that the resistance would increase as the temperature rises.  Increased resistance means reduced current.  So in principal, if we had an ammeter in the circuit, we might expect to see a different current in each case, and the lower current implies lower temperature, hence better heat transfer.  But I am not sure how much the temperature changes, and whether the measurements would be sensitive enough.  Also I am definitely not advocating putting your multimeter in the mains circuit to measure the current.  Perhaps one of those clamp meters or a Hall effect current detector could be used if it is sensitive enough.

Another possibility is to make the two element carriers say 12 mm longer than required, assuming the boiler shell is long enough to accommodate the longer element mounting.  A thermocouple could then be tucked in after the element is inserted, and perhaps a little kaowool or similar fibre  insulation pushed into hold everything secure.  The thermocouple would then measure a temperature which might be close to the element sheath temperature, and more importantly, might vary in the same way as the element temperature.  Again the lower temperature implies better heat transfer to the water, and hence gives our answer.  I think I would find this method the most practical, and probably most likely to provide a definitive answer.  But I would have to try it to find out.  What do you think?  Is there another method that might resolve the issue?

I hope the plumbing issues will be resolved tomorrow and that at last I can not only address that condenser, but also think a bit about Paul's locomotives.

Thanks for following along

MJM460

PS Willy I saw your post just as I was about to post today's essay.  Looks like I will have to have a go at entropy sooner rather than later.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 26, 2017, 02:23:57 PM
Thanks for the latest info.......... New question  ........would the boiler produce steam quicker if the filler cap is left loose to allow the air to escape and not be compressed by the expansion of the heating up water ??or could you increase the pressure in the boiler first with a bike pump ???.... and ........if there is a device to vibrate the boiler at a certain frequency would this also speed up steam production ?? please only answer these questions at your leisure !! These cartridge heaters were never meant to be used in my model boiler i suspect, but used primarily to heat up large chunks of metal used in industry where one could have an exit hole to release the air. Also the grease would enable the item to be removed easily if it needed changing.
Title: Re: Talking Thermodynamics
Post by: paul gough on July 26, 2017, 04:10:50 PM
I am sitting here at 1am pondering the interaction of the 'dancing' molecules of steam, (a gas), interacting with the 'buzzing' somewhat constrained molecules of the piston surface, (a solid). Now, one can easily comprehend the affect of the force(s) of the steam molecules on the piston as analogous with the force(s) of a ball thrown against a wall or some such. But is our mechanical or engineering description/analogy all there is to it and sufficient??? After all most volume of the piston is in fact the space between the atoms, this leads me to think that some sort of other interaction(s) is/are occurring rather than just our day to day and somewhat 'gross' explanation. Are we in fact talking of molecular/atomic forces interacting and if so there would presumably be losses involved in any energy transfer between them, are these significant or relevant??? I don't want to side track this thread into the dense and dark regions of physics or expect any technical explanation, however I would appreciate having a laymans handle on the phenomena and if there is any connection or relevance to our discussions on thermodynamics. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 26, 2017, 08:22:03 PM
Hi, Paul, I an see what you are getting at and are some metals that may be a bit elastic or have large spaces between the molecular structure be better at transferring the energy in the steam propellant. err ,um, i think what i am saying are some (things) better than others when used as a piston in a steam engine ???
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 26, 2017, 08:24:29 PM
Hi MJM, Saw this and thought of you !!! I am a member of the Newcomen Society and they have in there magazines quite a lot of info on Australian engines...........
Title: Re: Talking Thermodynamics
Post by: paul gough on July 27, 2017, 12:42:06 AM
Hi Willy, The physical properties of metals at the molecular level and their various capacities to exchange energy is not something I would claim any understanding to, so whether an 'elastic' metal had different behaviours to an inelastic one I have not a clue, to me it would be like trying to speculate on what happens in a black hole. We can only hope there is a higher authority amongst us who can deliver enlightenment.

I also found the Newcomen society a valuable storehouse of very interesting information, I spent many weeks reading a large proportion of their articles going back to Volume 1 of the transactions, (Journal), while I was researching disk engines some time ago,  they are available to members via the web. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 27, 2017, 01:12:47 PM
Well at least the plumbing issue is resolved and fixed.  No amount of theory will predict that the tube had not been pushed far enough into a quick-lock fitting.  But something made me look very carefully into the fitting, make some measurements and stick a little bit of tape on the tube to mark the correct insertion.  I had just not used enough muscle.  I had to get that out of the way before discussing entropy.

I have avoided mentioning entropy up until now but Willy, your post last night suggests it is on your mind, so let's have a go.  The page you scanned from your book made a quite suitable introduction. 

You may remember that I described enthalpy as a property that only depended on pressure temperature pressure and specific volume, all things we can measure.  It did not depend on the previous history of how the substance came to have that particular pressure, temperature and specific volume.  That sentence is basically the definition of a property.  I noted that enthalpy is calculated from the pressure temperature and specific volume, we don't have an enthalpy meter.  But the calculation of enthalpy comes into many significant problems and thermodynamics (particularly problems involving heat and work, so obviously interesting to anyone designing and operating heat engines).  It is so useful that it is convenient to have the value tabulated in steam tables, and the tables for other fluids such as refrigerants.

Entropy is another such calculated property, given the symbol S.  No such thing as an entropy meter, it is the outcome of a calculation.  I think the main reason that entropy is so mysterious is that it is not a simple calculation involving things we can measure, but it requires a bit more complex calculation that has a surprising applicability.  The calculation involves summing the quotient of heat input divided by temperature as a substance undergoes an ideal reversible process, which goes in such small steps that the temperature can be considered constant within each step.  In symbols it is written
dS = dQ / T at each step, and leads to the result that e change of entropy with the process (S2-S1) is found by integration  of dQ/T through the process.  Now none of us want to have to do integral maths to understand our engines, but we don't have to.  The unexpected result is that this change in entropy, even though it is calculated by considering an ideal reversible process, is exactly the same, whether the process was reversible, or ideal, or was not reversible.  The reversible process allows the change to be calculated, but then the result also applies to any real irreversible process as well.  So we can leave it to a few boffins to calculate the values  and include them in our steam tables, and tables for other fluids.

Now apart from the interesting unexpected nature of this property, why do we care?  It comes down to the second law of thermodynamics.  If we first note that the first law calculates the heat exchanged in a process based purely on conservation of energy, but gives us no idea of whether the process can actually occur.  For example we can calculate the heat lost by your tea to your teaspoons when you plunge them in to the hot brew.  The first law tells is that it is the same amount as is gained by the spoons.  But we can also calculate how much heat would be gained by the coffee if it became hotter by cooling the teaspoons.  Now we know that can't happen.  The second law of thermodynamics tells us it can't happen, but does not really quantify why it can't happen.  This is where entropy comes in.  It turns out that the only processes that can happen (without heat or work input) are processes that result in an increase of entropy.  This is relatively easily extended to mean that the entropy of the universe is increasing.  Philosophers ponder if there is a limit, if so what happens when the limit is reached, or is there a mechanism somewhere in the universe that reverses the increase of entropy.  But I find all that way too esoteric.  I would rather leave such discussions to someone who cares.

However, accepting the concept of entropy, and having it tabulated in steam tables is very useful.  So let me illustrate by showing how it helps us understand a steam engine.  Let's assume we have a boiler, and the steam outlet pipe then loops back through the fire box a couple of times as a superheater, then on to the engine.  We can measure the temperature and pressure at the engine inlet. We would all like to know how much work our engine can produce.  We need to know the exhaust pressure, but we don't know the exhaust temperature.  We know the amount of work is given by the change of enthalpy, but we need to evaluate that change.  Now we remember that ideally steam expansion in an engine is an adiabatic process, meaning no heat transfer in or out.  And it turns out that an ideal or adiabatic process also means no change in entropy.  So if we look up the entropy of our steam at the engine inlet, it will have the same value at the exhaust of our adiabatic engine.  With a bit of interpolation of the steam tables, we can find the temperature, and enthalpy of the exhaust steam, and also the exhaust steam quality or dryness if the exhaust is wet steam.  Now we can easily calculate the change in enthalpy by subtraction.

Using entropy, just using the tabulated values in the steam tables, has allowed us to calculate the power output of an ideal engine.  The second law of thermodynamics tells us that any real engine will produce less power than an ideal or adiabatic one.  This leads to a definition of adiabatic efficiency, but that can come another time.

Now your other comments, letting the air out when your element is inserted.  Your explanation seems likely.  It would take some ingenuity, but it should be possible to devise a way to solder a return bend on the end of your reamed tube, and use say 3 mm tube vent back to the mounting flange. This would allow you to use the grease.  But your other suggestions to reduce the time to raise steam?  I am tempted to stick my neck out and suggest the only way to raise steam quicker is to put in more heat, either use a bigger element, or put in two or three elements.  You see your heater power rating is the amount of heat generated by the element per unit time.  A 500 Watt element means 500 Joules/second.  We have previously looked at how to calculate how much heat is required to heat water from cold, say 15 deg C to saturation temperature, and we can look up the specific heat of copper and calculate the heat required to increase the copper temperature from 15 to our operating temperature.  You have an advantage over fired boilers, you know how much heat is produced and you can insulate it well to limit loss to the atmosphere, though we should make an estimate of the heat that will be absorbed by the insulation.  So you only need to weigh the empty boiler, and the quantity of water you fill , and you can calculate the time required.  Similarly, once you start steam production, you can calculate the heat required per kg of steam, and so you can easily determine how much steam you can make with 500 watts, as no more heat is absorbed by the copper or insulation once steady temperature is reached.  All your effort to improve the heat transfer coefficient only reduces the temperature the element must reach to transfer the rated heat.  It does not affect the time to heat or the the amount of steam you can raise.  So long as you insulate the boiler well!  I will hold the air questions until I get to condensers, and try and address Pauls comments tomorrow. 

Paul, you have obviously understood well my explanation of how heat is changed to work in our engines. I will address your questions on this next time.

Oh and thanks for the notice about the Mildura conference.  Mildura is about 600 km from Melbourne, quite a solid drive.  Unfortunately I have too many other commitments in October, but it's good to see these events being held "locally".

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 28, 2017, 11:37:39 AM
I hope my discussion on entropy yesterday was enough for our purposes, as it did not allow me to continue looking at yesterday's other questions.  So today, Paul's question on the molecular interaction that produces force on the piston.  Paul, you have quite properly observed that while I described the action of the gas molecules in detail, I treated the piston more superficially as though it was a uniform solid.  The only thing I can say in defence is that there are good precedents for this approach.  However, it is not necessary to simplify, as treating the metal piston as jiggling molecules is quite valid, it just adds another layer of detail, and does not change the original conclusion.  It is all about scale of the motion and particle mass.  You will remember that gas molecules move at high velocity in all directions.  The velocity on average is enough that the molecules easily escape the close range attractive forces any time they come into collision with another.  I remember looking up the mean free path and it's very large compared with the size of the molecules, but unfortunately don't have access to the book at the moment.

On the other hand, the metal molecules have much less energy.  Obviously not still in the vapour state, or even liquid.  Molecules are still moving but the energy is so low that they are well in the grip of the short range attraction forces.  Their energy and hence velocity is so low that they have  dropped into a regular close packed pattern that can actually be identified by X-ray diffraction techniques and in a larger scale, in the crystal structure.  The molecules are still moving in this array, they cannot "clump together" as when the molecules get very close together there are repulsive forces.  But there is not much room between the jiggling molecules of a metal that gas molecules can penetrate and get lost in the metal.  Not much room, but not no room.  There are some gaps or faults in the regular structure that make grain boundaries in all but very carefully produced single crystal structures.  A few gas molecules sometimes get trapped in these gaps, particularly small molecules like hydrogen, and this can cause problems in welding the affected metals.  But solubility of gases in metals is much less than in liquids for example.  In addition most practical piston metal atoms are much heavier than gas molecules.

So what happens when a gas molecule collides with the metal surface?  First it approaches a nearly solid wall of vibrating metal atoms, that are moving in a tightly packed array.  Even if a gas molecule penetrates the first layer, I am guessing it does not often get through the second without colliding with one.  Of course the motion is all random so the gas molecule could collide with one moving towards it, or one moving away, or like billiard balls it could hit at an angle.  The metal atoms are on average all moving in one direction that we identify as the piston movement.  But at each collision, conservation of momentum applies as a basic law of physics.  It is the physical law behind Newton's law about bodies continuing to move unless acted on by an external force.  Unlike energy, there is no equivalent of energy conversion with momentum.  Conservation of momentum can be applied separately in each of three perpendicular directions, though only components in the direction of piston movement is of interest in production of work.  The law of conservation of energy still applies, but the sum of the energies of all the particles will be less after the collision, some does the work to accelerate the metal atom and some is converted to heat.  So conversation of momentum is easier to apply in this case.

It is worth noting that in addition to conservation of momentum, which should probably be called linear momentum, there is an analogous law of conservation of angular momentum.  This applies to the spin motion of a particle, again it applies to spin around three perpendicular axes.  The formula for angular momentum are very similar to those for linear momentum, just angular velocity and torque instead of velocity and force.  And moment of inertia, instead of mass.

So the gas particle bounces back into the gas space as it still has too much energy to be captured by the close range attractive forces, and the space between gas molecules is enough for it to be hardly noticed.  But what about the metal particle?  Well it bounces off on the opposite direction, back into the metal.  Much lower velocity as it is so much heavier than the gas molecule.  Also it is in a close packed array so as soon as it tries to move beyond it own little space, it collides with another molecule, which collides with another and so on right through the metal, and can be seen in movement at the other side millions or is it billions of atoms away.  But the close range attraction prevents the last particle popping out the other side.  So the solid piston acts like, well, like a solid.  The net force due to all this change of momentum on the top and bottom face of the piston is carried through to the crank pin where it produces torque as I have already described.

I have probably still glossed over a few details but I hope that is enough to illustrate that the molecular motion model is still valid when you include the molecular model of the metal piston.

I think the next question was about how the structure of the molecules affects the collisions, compared with solid billiard balls.  Of course solid billiard balls are a concept that most if not all of us are familiar with, even if some have invested more time in study of the billiard table than others.  Atoms, however, at a first level of detail, consist of a positively charged massive nucleus, surrounded by a cloud of negatively charged orbiting electrons, each with relatively little mass.  And mostly empty space between.  So what does a collision look like at this scale?  Well I am not a physicist, so I am not sure that I can give a definitive model.  However I would think that the strong close range repulsion forces, possibly something to do with the positively charged nuclei, might mean that the collision involves elastic forces strong enough to look like reaction of elastic solid bodies.  But it would need some detailed study of molecular physics to be sure.  Perhaps some holiday reading to look out for.  Of course it is even more complex when, instead of single atoms, we have molecules.  For example oxygen and nitrogen both involve molecules consisting of two atoms, while water has one oxygen and two hydrogen atoms, hence H2O.  These combinations mean we have atoms that are not at all like a spherical ball.  Their asymmetrical form would have some influence on the energy contained in the spin around each of the three perpendicular axes, and also the momentum change when they collide.  And of course the molecules implies another energy change in a chemical reaction which involves molecular chemistry.  Again well out of my field.  Personally I am happy to stick with the simpler assumptions, I think they give enough information for my purpose.  But I recognise that as a gap in my knowledge.

There is one other issue in the molecular model.  A characteristic of the random motion of molecules is that the magnitude of the velocity as well as the direction is random.  This means that there are large numbers of molecules with velocity well above and others well below the average.  But is is reasonable to discuss on the basis of an effective average.  Studying at this level of detail is however a whole new field. 

The other outstanding question was about the elastic properties of metals and how these affect our selection of a piston material.  I wanted to start this one from the point that all materials deform under stress.  In simple language they compress or stretch like a spring.  Just some deform less than others.  Also some are quite linear and steel is in this category, some less linear.  When the material is deformed an elastic material returns to original form when the force is removed.  There is usually a limit beyond which the deformation becomes plastic.  That is the material does not return full to the original shape.   Selection of material involves not only consideration of strength, but also temperature resistance, chemical resistance, wear resistance, friction and so on, and also consideration of manufacture such as machinability.  Availability and machinability are probably among the first priority for model engines.  All the commonly used piston materials have slightly different but adequate modulus of elasticity and strength, although care over specific choice of alloy is necessary if aluminium is required for higher temperature applications, due to both strength and thermal expansion.  Basically the differences in modulus of elasticity is not very important in the conversion of energy to work.  Thermal conductivity is important for heat transfer applications, and also when thermal insulation is required, but again not in pistons.

I hope that deals with some of the philosophical issues, next time back to those engine questions from Paul and perhaps Willy's air in the boiler issues.

Thanks for following,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on July 28, 2017, 01:27:21 PM
Thank you very much for this extended trip to the interface where the molecules interact. It has cleared some of the fog and settled my thinking as I am now comfortable that we have treated steam and steel (our piston) as equals, molecularly speaking. Your explanations have given me a pretty precise 'picture' of the actions of molecules on the piston from a macro and microscopic perspective. Now I need to get comfortable with all the thermal or heat energy issues. This I think will be very energetic mental exercise and hope my brain does not pass through some super critical phase and end up suffering runaway entropy!

Looking forward to seeing your thoughts on the model cylinder materials/construction. Regards, Paul Gough. P.S. I just noticed the average hit rate to this thread, (approx. 6000/90 days= 66). This would seem to me to be very satisfying to have a packed classroom every lecture!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 28, 2017, 03:09:11 PM
Hi..........I have just had an eureka moment and now realise that the study of Thermodynamics was to determine how long a steam engine engineer could have for a tea break !!!However this was before they invented stainless steel teaspoons !!!!
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 29, 2017, 12:29:48 PM
Hi Paul, I am glad the explanation on gas collisions with the metal surface made things clearer for you.  You know, brain meltdown depends not on energy, but on rate of energy movement, or power.  Energy is measured in Joules, J, while power is joules per second, J/s, also named Watts. So to avoid the brain meltdown, it is a case of pacing yourself. I think that was your words very early in the piece?  However the main thing is that we make it interesting and preferably also useful.

Now you were asking about cylinder arrangements for your little locomotives, so let's discuss the cylinder after all that talk about pistons.  In principal, the cylinder is only there to complete the enclosure around the piston so the steam is constrained and the high pressure can act on the piston.  Of course, when the pressure causes a force on the piston, there is an equal and opposite force on the head of the cylinder.  The magnitude of the force is P x A, and the direction perpendicular to the cylinder head.  Remember to use consistent units, so pressure is Newton/square metre, or Pascals, and area is in square metres.  Then force is in Newton's.  This force has not been mentioned so far as the cylinder head on a stationary engine is by definition stationary, meaning that it does not move in response to the force, and hence does no work.  Remember work = Force x distance.  If distance = 0, then work is zero.  So what happens to the force?  The force is transferred to the frame through the cylinder mounting fastenings, and is resisted by the main bearings.  And of course the big end bearing is providing an opposite force on the main bearings, so the whole lot is in equilibrium.

This is a very important point in the engine design.  The primary strength design of the engine is based on the magnitude of this force.  The cylinder head bolts must carry the load in tension, the cylinder mounting bolts must carry the load in shear, or there may be a shear key to transfer the load to the frame.  Similarly for the main bearing mounting.  On the piston side, the force is carried through the piston rod, and with some allowance for the angulation, through the con rod.  It is a major factor in the bearing design.  There must also be an allowance for inertia loads, and bending moments, when things are not in line, but the basic rod load or cylinder head load is the basic design load.  My compressor experience tells me that major manufacturers design a model series around nominal steps in rod load capacity and design the frame and motion parts from that, rather than start from scratch every time.

Thermodynamics assumes that expansion in an engine is a nominally adiabatic process, meaning no heat transfer in or out.  This assumption is used almost entirely because it makes mathematical analysis possible, not because it represents reality.  The performance of a real engine is determined on a test stand and compared with the ideal adiabatic engine.  So what is the effect of the heat transfer on engine performance?  Now heat input hardly ever happens so let's deal with that first.  I have not been able to quickly find an example with the first law equation for this example, and I am not going to try and derive it.  However basically it says that Heat input = change in internal energy plus the work done.   We don't know how much of the heat goes into internal energy and how much goes into work output.  So let's put together a few things.  The analysis of an ideal adiabatic process assumes the process proceeds through a series of very small steps.  And the first law applies to each of these steps.  The work output is pressure times area times the distance the piston moves in each tiny increment.  So the heat probably goes mostly into internal energy, but that will be reflected in the pressure for the next increment.  So the pressure does not drop as much as you would expect in that increment with the consequence it is higher for the next increment, which means some more work produced.  I don't know how much more, but directionally, heat input to the cylinder during expansion will increase the engine output.

In a real steam engine, the more likely position is that heat is lost from the cylinder.  We know the outside of the cylinder is hot, so it will loose heat to the air, and that heat comes from the steam.  Applying the same logic we can see that in this case, the heat loss reduces the engine work output.

Now to think about how we can use this information for your engines.  First, we can at least reduce the heat loss by adding insulation, or cladding to the cylinder.  This is done in many of the model builds on this forum, reflecting the fact that this was indeed full size practice.  Relatively simple step that adds to appearance and directionally improves the engine output by reducing losses. 

In the early days, when engines operated at very low pressure, some went a step further and tried adding a steam jacket around the cylinder.  This obviously adds a level of complexity, but the question becomes where should the steam come from?  You might be tempted to say what about recovering heat from the exhaust steam by using some of it in the jacket.  Now remembering that heat moves from a high temperature to a lower temperature, and as the exhaust temperature is lower than the average temperature in the cylinder, it will not actually provide any heat input.  However, it means the temperature difference driving the heat loss is the difference between the cylinder temperature and exhaust temperature, say 100 deg C, compared with say 20 deg C atmospheric temperature.  So there would be less heat loss, even though no actual heat gain.  If we use some of the engine supply steam in the jacket, we now have a small heat input to the engine.  However I feel that on balance of probability, that steam might produce more work if used in the engine.

I don't know how much extra power is produced by reducing the heat loss.  If we think about Willy's electric heater of 500 watts, similar in magnitude to my methylated spirits burners which produce about 600 watts, you engine burner might be similar.  I suspect the engine output from such a boiler might be around 2 to 5 watts.  If we made a dynamometer and did an engine test, we might expect to identify a difference of say 5 to 10%.  A full size test stand only achieves around 1% accuracy, so I doubt we would do better.  I don't know if the difference would be measurable, but race winning performance is certainly achieved by tiny differences, so it is probably important if we want to achieve the best we can.  At the very worst it will only help us avoid one source of burned fingers.

But what about the question of a solid block compared with a light weight fabrication?  Before we try and answer that, there is another temperature variation in the cylinder wall when the engine is running that we should be aware of.  At the beginning of the power stroke, hot steam from the boiler superheater is admitted to the cylinder.  This will tend to lose some heat to the cylinder, causing some loss in efficiency as we have discussed.  Then as the piston moves down, and the inlet valve closes, the steam starts expanding.  In expanding without heat input, the steam cools, so that it is close to exhaust temperature at the bottom of the stroke.  Then the exhaust steam is pushed out at the top of the cylinder cooling it.  So in each cycle, the cylinder is repeatedly heated and cooled, clearly a heat loss that leads it loss of efficiency.  It also causes a little variation in thermal expansion, which may in time lead to sealing problems with the  head gasket, just conjecture, I don't know if the movement is enough to cause such problems.  This temperature cycling occurs under the insulation, whether insulation is present or not.  It is probably the clue to the answer on whether to use a lightweight fabrication.  A more massive block provides a thermal inertia which would tend to stabilise the temperature a bit which may be helpful.  And a bit of mass generally helps with traction in a small engine, so I would probably lean towards a solid block (with insulation around it, rather than a very lightweight fabrication.  That also appeals to my skill level with soldering as I currently need to borrow a big enough burner, preferably with operator attached, for anything of significant size.  It think it would take some experimenting and very careful observation to determine if there is a real difference in performance.  Of course a solid block will absorb more heat on startup and hence produce more condensate initially, but once everything is warmed up and running it would not make any difference to condensate in the exhaust.  I suspect the choice will probably be determined by ease of manufacture and availability of material, rather than thermodynamic considerations. 

Similarly with material choices.  Compatibility between the piston and cylinder materials is probably the main consideration.  Thermal expansion needs to be considered.  I suspect an aluminium piston might seize in a cast iron cylinder, while a cast iron piston in an aluminium cylinder might leak excessively.  So it is worth thinking about whether the clearances in your engine will increase or decrease when it heats up.

Thanks for pointing out the viewing numbers.  I hope it means that there is interest in the topic, and assume that others will join in at the appropriate time if they would like to add something, or clarify something.  I am not a teacher or lecturer, and you can probably tell from some posts that I was not any loss to that profession, I am just trying to pass on some things I learned in my career in the hope that it will be helpful.

Next time Willy's questions about air in the boiler.  By the way Willy, we are all waiting for your posts of the cooling curves you get with and without the teaspoons in your tea.

Thanks to everyone for reading,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 29, 2017, 01:10:03 PM
Thanks for this latests info,.....as the cylinder block expands should the holes in the frames be slightly elongated away from the middle bolts so there is no shearing stresses on them ? or even worse could it distort the frames, especially if it is a solid block??can one work out the actual length increase for a freezing cold cylinder block, (early in the morning) to when it is working at 250 LBS/s   16 Bar later on in the day half way to Scotland from london ? All these forces are ones that we don't usually think about.!! They say that you can suspend a London bus from a 1/4" bolt!! but i am not sure. I shall endeavour to do that experiment soon...promise...I have just found some graph paper in a skip ! so that has prompted me !! Ok I have a   K.type thermocouple ? and i might use it on the Foreignhight scale as it will be more sensitive !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 30, 2017, 12:22:37 PM
Hi Willy,  you are quite right to point out that I stopped the discussion on cylinder construction a bit early, and expansion and how to live with it should have been included.  I quite deliberately say how to live with it, rather than control it because we basically have nearly no choice.  If you heat steel, it expands.  If you try and restrain this expansion, you will encounter tremendous forces as you would have to apply the force necessary to compress the steel back to the size it was before heating.  We can calculate the expansion, and the expansion due to forces, just as you suggest.  You can look up some properties of metals, and you will find a yield strength, an ultimate strength, a Young's Modulus and a coefficient of thermal expansion.  You may also find Poison's Ratio.  Let's look at each in turn.  The yield strength is the stress limit beyond which there will be some plastic definition.  The ultimate strength is the stress beyond which the material can be expected to break.  If you take a piece of steel strip, say 2 mm thigh by 25 mm wide and gently bend it with your fingers.  If you bend it just a little, it will spring back straight.  You may be able to bend it a little more and still have it spring back.  But there comes a point where it no longer springs back all the way to straight.  Without going into the stress analysis, in bending, the stress is not uniform as it might be if you just tried to stretch the steel lengthwise, but the point where the strip no longer springs back to straight, is the point where the highest stressed areas undergoes some plastic deformation, or permanent deformation.

Young's modulus is the property that relates the load to the deformation in the linear or elastic range.  This is the one you use to calculate the stretching or compression due to the pressure load, or other mechanical loads.  Nearly all mechanical design aims to keep the stresses within the electric limit.  Now there is considerable variation in the strength of even the best materials.  And it is quite difficult to apply a load so it is totally uniform.  In practice, quite generous safety factors are normally applied, so that there is minimal chance of the elastic limit being exceeded.

The coefficient of thermal expansion is the one you use to calculate how much the material expands as it is heated.  (Or contracts when it is cooled.). The figure for steel, from memory, i.e. not very reliable, is about 6.3 x 10^-6 in/in per degree Fahrenheit.  Funny tricks the memory plays.  But you will recognise that in/in is dimensionless, just a reminder really, so the units are really just per degree F.  But the expansion is measured in millionths of an inch in each inch of length of the component for each degree temperature change. My access to books and technology is limited at the moment, I don't have the right book with me, so apologies for being approximate.  The figures will vary for other materials.  I think aluminium is much more, brass and other copper alloys in between. 

Thermal expansion is interesting.  Things expand from a geometrical centre in all directions, whether the material is there or not.  So an engine cylinder expands just as much as a solid block.  The bore expands, not contracts.  So if the piston is the same material, and the same temperature, the clearance will also expand in proportion.  And this expansion, so long as the temperature is uniform, causes no stress in the material.  Well if not from the expansion, where do thermal stresses come from?  There are two causes for thermal stresses.  First if the temperature is not uniform.  If the temperature is not uniform, the expansion is different in different parts, but they are joined together.  So what ever stress is necessary to deform the parts so they are still joined, will appear.  Either that, or something will break.  If heating is slow, and particularly if things are insulated, the temperature can be kept uniform within reasonable limits.

The second source of thermal stresses is when different materials are in the same component.  If you fabricate a cylinder from say a brass tube and attached flanges made from a material with a lower coefficient of expansion, the flanges will tend to compress the ends of the cylinder to a slightly smaller diameter.  Of course with Paul's very small cylinders, the magnitude of this difference might be much smaller than reasonable construction tolerances, but it is worth knowing what is happening.  So the whole fabrication is best made from the one material.  Of course you may wonder about the silver solder.  When everything cools down, the solder may experience some local stresses.  Similar to the distortion of an improperly supported welded joint when it cools.  But even if there is a little plastic deformation on cooling, this is not a disaster, the stress is relieved by the plastic deformation.  Disaster is only when things are so severe that the joint cracks.  There are many build logs on this forum that illustrate that a well made silver soldered fabrication is normally quite sound.  So no reason to avoid fabricated components.

Willy, your second question was about how to support the cylinder on the frame to avoid expansion issues.  We are all familiar with the well known vertical engines.  The cylinder is supported on top of the cross head guide.  The cylinder expands upwards from the mounting face.  The close contact of the cylinder with the mounting flange means they are practically the same temperature, so both the cylinder and the head are expanding together.  If the cylinder and the cross head guide are both cast iron, they even expand the same amount.  The cross head guide and its mounting bolts are all stretched vertically by the rod load forces, but as we have observed they are usually made strong enough to resist the forces.

When the cylinder is mounted without the symmetry of the vertical engine, for example my mill engine which you can see in the engine gallery, the cylinder is mounted on its side on a steel base plate.  The centre line of the cylinder is well above the plate, as are the main bearings.  So the forces on the cylinder and main bearings are not in line with the base plate.  This of course results in some bending forces on the base plate.  Now you can see that the base plate is quite solid and easily resists the bending forces from this small engine.  There was a recent excellent build log by one of our members is Germany, my apologies to him that I can't remember his name for the moment, but look for the In Line Engine.  The cylinder is mounted on its centre line, and the symmetrical frame on each side properly balances all the loads.  It also has an interesting valve linkage and governor, but I believe the "In Line" designation refers mainly to the main rod load being in line with the centreline of the symmetrical frame.  It was beautiful workmanship.

Paul's locomotive had cylinders mounted I think between the locomotive frames.  Each frame is relatively flexible alone, but the cross bracing between frames is intended to stiffen the frames and the symmetrical nature of the frame and cylinder block configuration properly resists the bending loads as well as the tension.

I think that leaves the issue of expansion of the cylinder where it is attached to a frame on the side.  It is time to defer that one until tomorrow.

Looking forward to seeing those cooling curves Willy.  You look well equipped so far.  Degree F does give you smaller steps which may help give smoother curves.  Remember to control or at least record all the significant variables, ambient temperature, cup design, mass of tea.  May be better to do it with hot or boiling water instead of sacrificing a perfectly good cup of tea, though it is in a good cause.  The i Pad timer is very good for such experiments, it will record several times requiring only one touch for each time, so by a single touch say every 5 degrees, you will have an accurately recorded time to use in your graph.  You may want to make a thermowell by folding a drinking straw, and inserting the thermocouple in one side instead of placing it directly in your tea.  A K type thermocouple is a standard device for which the metals of the two wires are defined by standards, as is the voltage for each temperature.  Fortunately the calculation from voltage to temperature is included internally in most multimeters these days.  It should not need calibration, but it never hurts to check it at 32 F and 212 F.  Easier said than done though.  The boiling temperature is affected by atmospheric pressure, so not easy to achieve the 212,  the ice point is a bit easier as in principal you only need ice, and water in an open vessel, hence also water vapour to give 0.01 C = 32.018 F, near enough to 32, but do have plenty of ice.

Thanks everyone for following

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on July 30, 2017, 05:14:56 PM
Hi, I have done a preliminary graph and it is quite interesting asa there is a quick drop in temp, then it stays the same for a few mins then falls gradually 1 degree a minuet. it looks like the heat rushes into the cup and spoons then finds it has nowhere else to go so comes back and thinks about it then decides to make the best of a bad job and gives off the rest of the heat in a begrudgingly manner ??!!! Are there technical terms for some of these adjectives ?!!. What i did was to pour the coffee this time in to the cup with the the teaspoons and temp node already in it. In the cafe the temp of the water to make tea is actually 80 degrees rather than 100 c Next i will try it with equal dimensions bars of copper, brass, bronze, aluminium and SS. and see how this performs. In all my text books there is no mention at all of differential expansion problems, and looking at a huge Triple expansion engine that would be in the Titanic, one wonders how much the engine would grow !!?
Title: Re: Talking Thermodynamics
Post by: paul gough on July 30, 2017, 11:30:21 PM
Thanks again for the comprehensive reply on the cylinders, and the extension to 'expansion' prompted by Willys questions, thanks Wily.  I was particularly taken by your sentence, "Things expand from a geometrical centre in all directions, whether the material is there or not." Conjuring an image of a cylinder in my mind as I read this extraordinarily succinct and illuminating sentence led to an immediate grasp of the phenomena. A situation where previously my intellectual myopia had glossed over the origin, geometrical centre, because I was only thinking in terms of the bore getting bigger or the cylinder body growing as more or less separate things. Oh, if only all engineering explanations were so eloquent!!!

I have a fantastical question that keeps penetrating my mind and which I can't reconcile because I am confounded by internal energy and the spectre of irreversibility/reversibility etc. The Scenario: (1) A normally working cylinder and piston arrangement, but the cylinder body, including all covers, piston and rod etc. are somehow heated to and maintained at the temperature of the inlet steam, (say 150C); (2) All these elements heated to a high degree and maintained, (say 5x inlet temp).  (3) Would case (1) be somewhat equivalent to all the components being made of a perfect insulating material? Thus, are there any substantial changes to how our steam (heat) engine works, (or doesn't work), with these three situations???? Many thanks for your patience and effort in dealing with my confusions. Regards Paul Gough.   
Title: Re: Talking Thermodynamics
Post by: MJM460 on July 31, 2017, 01:13:55 PM
Willy's tea cooling experiment.

I was going to talk about thermal expansion but I would first like to look at Willy's tea cooling experimental results.  Willy, your graph is very well done and gives a lot of information about what is going on.  Let's look at three distinct phases, first that quick initial drop, then the steady temperature then the continuing cooling.  Your explanation is not bad so let's look at the technicalities.  Then try and deduce what it means.

That initial quick cooling is due to the teaspoons absorbing heat as they absorb heat from the coffee.  Remember stainless steel has poor thermal conductivity so it is not instantaneous.  We would expect the temperature drop to be greater for additional teaspoons.  But we have some  complicating factors.  First, the cup will also absorb some heat from the coffee.  We can get a good idea of how much heat the cup absorbs by looking at the curve for zero teaspoons.  There is a gap you have not told us about, the coffee starts at 80 deg C, which is 176 deg F, but your graph starts at 150 F.  Is that perhaps due to some air cooling of the coffee on the way from the machine to the cup?  Then I notice the 3 spoon experiment started at a lower temperature.

We can analyse what is happening in this way.  We have three components in the experiment, the cup, the coffee and the teaspoons.  Each brings an initial quota of heat to the experiment.  We could use absolute zero for the starting temperature, but the properties are not exactly constant over the necessary temperature range.  It is easier and a bit more accurate to use 0 , F or C depending on which we are working with.  The initial quota of heat for each is calculated as
Mass x specific heat x temperature - ref temperature, which we have nominated as zero, you can now see why.  When the coffee hits the cup and teaspoons, all three end up at the same temperature.  So we need only to calculate a temperature, t at which all three have the same temperature, but the total heat is still equal to the sum of the three initial quotas.  So you can see we need to know the initial temperature of the cup, the teaspoons and the hot water.  We don't know much about the cup, and it will end up hotter on the inside against the coffee, than the outside which is being cooled by the air.  However the zero teaspoons experiment probably tells us how much heat the cup absorbs, if we know it's initial temperature.

The initial starting point is probably not very important if we have the first temperature reading, and note that this is probably not at zero time for the complete curve.  However to keep the teaspoons and cup effect separate, it may be better to let the cup cool the coffee, then add the teaspoons as soon as the initial drop is complete, say as soon as you have two consecutive readings that show the levelling off, so you see the quick drop when the spoons are added separately from the effect of the cup.

Then we have that roughly constant temperature period.  We have to be careful in our explanation of this phase to be sure that we do not violate known laws of thermodynamics, in particular that heat cannot flow from one item at one temperature to any other place which is at a higher temperature.  Heat can only flow from hot to cold.  However, plunging the cold teaspoons into a hot cup of coffee is not equilibrium at every tiny increment through the change.  It is not a reversible process, but it also does not happen slowly is a series of tiny equilibrium steps.  It is likely that the coffee in the cup is initially over cooled in the immediate vicinity of the spoons, leaving it a bit cooler than the coffee in the more distant parts of the cup.  Then, the heat flows from the warmer coffee to the cooler until it is all at the same temperature as convection currents redistribute the heat in the cup.  The temperature measurement may also be affected by the location of the thermocouple, relative to the teaspoons.  It really requires a bit more experimentation.  The initial purpose of the teaspoons was to cool the coffee to drinking temperature, in which case we might try stirring the coffee to maximise heat transfer coefficients to the cup and air at the surface and keep the temperature more uniform.  Worth a try.  May have to try a few times at home with the kettle, to save some coffee cost, while determining the optimum experimental method.

Now the third stage.  Let's just look at the two that start at the same temperature.  It certainly looks linear, but if you continued until almost at air temperature, (which we should record), it would show a curve, as the heat loss slows due to the decreasing temperature difference.  You can see the curve in the zero teaspoons curve.  But notice that the curves with and without teaspoons cross over after a while.  Can we explain that?

Let me make a suggestion.  When the teaspoons are plunged into the coffee, the heat is not lost, some of it is just stored in the spoons.  It is not lost until it goes to the air either through the walls of the cup, or through evaporation at the surface.  But the lower temperature due to the spoons reduces the temperature difference driving the cooling, so the coffee cools slower, but the same total amount of heat has eventually to be transferred to the air.  The cup with no spoons initially cools faster so the heat is actually lost to the air.  So when the slow one gets down to the point where it is the same temperature as the one with the spoons, the two at that moment cool at the same rate, but the one with the spoons still has more heat remaining in the coffee plus spoons so it's temperature drops slower.  So the spoons not only initially cool the drink to drinking temperature more quickly, they actually keep it near that temperature for longer.  Bonus points to anyone who saw that coming, I certainly did not.  It also casts some doubt on my text book which suggested that the spoons added cooling surface like fins to help cool it faster.  Either it's a very small effect, or the conductivity of the stainless steel puts the handles in that category where fins actually are counterproductive.

I think everyone will now be waiting for the next experimental results.

Thank you Paul for your kind words, I am glad that my explanation was helpful.  I hope that others have found it equally helpful.  Many people are confused by this point, but one time I had to analyse a large vertical pressure vessel with heavy parts suspended inside from a nozzle in the middle of the top head.  It was during my work on this that the penny dropped. 

Enough for today, current intention is to continue on control of thermal expansion and the cylinder heating questions tomorrow.

Thanks to everyone following,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 01, 2017, 11:50:06 AM
Thanks for this synopsis and i shall rethink the next experiments as there are so many variables to consider I may start with boiling water at home rather than  coffee in the cafe as i was getting some funny looks !! Also trying to get the waitresses to come over strait away with the coffee was a bit difficult . ! was going to transpose the 3 teaspoon graph onto the starting point for the others but that would have made the graph a bit over loaded!! so , we shall do some more work over more controlled conditions.........
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 01, 2017, 01:16:39 PM
Thermal expansion issues.

Hi Willy, We are obviously thinking about this at about exactly the same time.  I should have mentioned yesterday that the end temperature would not be much affected by the order you do things in those first few seconds, however by carefully doing things in a suitable order, you can see the separate effects, and thus isolate the effect of the teaspoons from the other things such as air cooling during pouring, and heating the cup.  Of course I didn't think of the time for the waitress to deliver.  Boiling water at home allows you to control much more of this.  By the way, if you have some silver plated brass teaspoons, these have better conductivity than stainless steel so the handles might make more effective fins, would be interesting to see.  The specific heat of brass is 385 J/kg.C compared with 461 for stainless steel so you would need about 20% more mass of spoons, about 6 brass spoons compared with 5 SS ones of the same mass.  Though you could stay with the same number of spoons and see a smaller initial temperature drop.

Continuing our discussion of thermal expansion issues, let's look at a few examples.  Willy you mentioned the engines on the Titanic.

Now I have no idea of what size these engines are, or what temperature they operate at, so I will make a few wild guesses to use for calculating representative numbers.  Thermal expansion is fully proportional to the length being heated, or cooled, so if the real engines were double the height I have guessed, then the expansion will be double.  Perhaps someone will be able to join in with the actual engine dimensions.  Let's assume an engine 6 metres high from the crankshaft centre line to the top cylinder head, consisting of a cylinder 1 metre high and support structure 5 m high.  In the typical model marine engine the cylinder is mounted on a table at the top of columns.  A larger ship might have a more substantial support incorporating the cross head guide.  Further let's assume the engine was built at 15 degrees C and the steam inlet temperature is say 250 deg C.

Thermal expansion is determined by the length of the component, the temperature change and the coefficient of thermal expansion of the material.  If the engines were made of cast iron, the coefficient of thermal expansion is 13.5 by 10^-6/deg C (or 7.5 by 10^-6/deg F).

Let's assume the cylinder is well insulated and operates at the steam inlet temperature of 250 deg C.  As I have previously mentioned, steam cools during expansion so the average temperature would be a little lower, but then the steam inlet temperature may be a bit higher.  These assumptions contribute to inaccuracy, but the calculation is still useful.

The cylinder then expands by d = 13.5 x 10^-6 x 1 x (250-15) = 0.003 m or 3 mm.  This may not seem like much on an engine of that size, however the clearance between the piston and the cylinder head at top dead centre may not be much larger than that.

If we look at the cylinder support, the metal at the top is in close contact with the cylinder so let's assume 250 degrees C.  However at the bottom end, at crankshaft level, it is probably nearer engine room temperature, say 40 deg C.  We can assume that thermal expansion is linear so the expansion can be calculated by assuming it is all at the average temperature, (250+ 40)/2 = 145.

Now calculate the expansion d = 13.5 x 10^-6 x 5 x (145-15) = 0.0088 m or nearly 9 mm.

So we can estimate that the distance between the crankshaft and the top cylinder head is 9+3= 12 mm.  You can see it would be pretty small on a model engine.  It is important to understand that, so long as we allow that expansion to go unrestrained, there is no force or stress involved.  It would take very high forces to restrain the engine to its original dimensions.

So far I have assumed that the whole cylinder block is at one temperature, the steam inlet temperature.  However, the question was about a triple expansion engine, and the temperature for the the low pressure cylinder is quite different from the high pressure cylinder.

For the lp cylinder, assuming a condensing engine, the cylinder temperature might be nearer 40 deg C.  The temperature difference is then 40-15= 25 deg C, instead of 250-15=235 in the high pressure cylinder.  With a temperature difference of approximately 1/10, so the expansion of the cylinder would be 0.3 mm instead of 3 mm.  More importantly, the stand at the lp end has a thermal expansion very close to zero, compared with 9mm at the hp end.

Even with the necessary approximations, I think you can see the problem.  If the three cylinders are machined in one casting, there will be significant internal stresses in the casting as one end expands under the inlet steam temperature, while the other end undergoes minimal expansion at the lower operating temperature.

Remember at the start of the discussion on thermal expansion I said we cannot control this expansion, we can only learn to live with it.  I suspect that in this case, the engineer might set up the engine cold with the lp end a little high, so that when the engine is hot, both ends are at the same height.  He would then have very careful heat up procedures to avoid any problems during warm up.  Now we have a few marine engineers on the forum, perhaps they could chip in and comment on the thermal expansion issues they see, and what they actually do.  I don't have any practical experience with real ship engines so would appreciate any help that is offered.

There is one area that we can control with regard to the thermal expansion.  I mentioned earlier that materials expand in all directions away from a geometric centre.  However nothing says that geometric centre has to be stationary.  We can choose which part of the engine is fixed stationary, and let the expansion go from there.  The geometric centre moves away from that fixed point, and the rest of the material expands away from the altered position of the geometric centre.  On a ship, the crankshaft would be fixed in line with the propellor shaft, and the engine allowed to grow upwards from there.  The remaining problem to be solved is that the piping to the engine must be designed to be flexible enough to allow for the movement of the cylinder flanges.

I hope this simple approach to expansion sheds some light on the issue.  These days with finite element methods we can do a much more exact analysis of exactly what happens, however that might be considered a bit over the top for our model building purposes.

We should have a quick look at how expansion effects a horizontal mill engine before leaving this topic.  Also a few questions from earlier posts still to be tidied up, then back to condensers.

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 01, 2017, 03:21:55 PM
Having looked at Utube ,the engine are described as 4 cylinder triple expansion engines ??. Also all the cylinders are separate !! so they knew about thermal expansion. I do not know what the other cylinder did in the engine ...any ideas? Will look further into this and here a few pics off the web
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on August 01, 2017, 03:54:06 PM
Willy, it was common to use 2 cylinders for the low pressure expansion to keep the cylinders smaller. If you notice the violet line it is the exhaust and it is connected to both LP cylinders on the ends.

Dan
Title: Re: Talking Thermodynamics
Post by: Ye-Ole Steam Dude on August 01, 2017, 05:03:32 PM
Having looked at Utube ,the engine are described as 4 cylinder triple expansion engines ??. Also all the cylinders are separate !! so they knew about thermal expansion. I do not know what the other cylinder did in the engine ...any ideas? Will look further into this and here a few pics off the web

It appears looking at the picture from right to left, the fourth cylinder somehow helped “balance” out the crankshaft. Again looking right to left, cylinder #1 is at 0-degrees, cylinder #2 is in retard at 120-degrees ( assuming clockwise rotation), cylinder #3 is at 0-degrees, and cylinder #4 is in retard by 240-degrees. Just my observation and would like to find out.
Title: Re: Talking Thermodynamics
Post by: Maryak on August 01, 2017, 11:23:21 PM
Triple Expansion Steam Engines

1. Cylinder Volumes:
HP = 1.0
IP = 2.6
LP = 7.0

To keep the LP Cylinder a manageable diameter it was split into 2 cylinders at 3.5 x HP volume giving a 4 cylinder triple expansion engine normally arranged from front to back LP,HP,IP,LP.

With this arrangement balance was achieved by setting the cranks at:
Between Forward LP and HP 1660
Between Forward LP and IP 2700
Between Forward and Rear LPs 700
In addition suitable counterweights were added to the crankshaft.

With Titanic and Olympus, not shown is a turbine which utilised the LP exhaust to drive the centre propeller. For manoeuvering the exhaust was switched over from this turbine and sent directly to the condenser. This made emergency reversing/stopping difficult as until this exhaust was removed, the turbine continued to move the centre propeller ahead. Transfer was accomplished by some big hand operated valves!

HTH
Regards
Bob
Title: Re: Talking Thermodynamics
Post by: Ye-Ole Steam Dude on August 02, 2017, 01:38:16 AM
Thanks Bob for giving us this information.

Thomas
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 02, 2017, 02:15:30 AM
Hi all thanks for the info ,it all makes sense actually. I don't know why the HP cylinder looks different with no cylinder head bolts though. there is a wonderfull model made by someone in Germany on the web Utube that is really superb,
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 02, 2017, 11:41:21 AM
More on thermal expansion and triple expansion engines.

Thanks Willy for those great pictures, they make the arrangement very clear.  Thanks Dan for coming in.  From the pictures we can see that not only are the cylinders separate as you have noted, but also the supports, rather like three separate engines with a common crankshaft.  Because the lp cylinders each expand the same amount, the pipe can be attached to both as you have observed.  Thanks also to Thomas for the observations on the angles.  Maryak, you must have worked on those engines, thank you for all the detail.  The turbine makes great sense as it is so much more suitable than reciprocating engines for the very large volume of low pressure steam before it is condensed.

I think balancing is probably quite complex for these engines.  There is not only the the normal balancing of a reciprocating masses, and the inertia forces for such large pistons, also balancing the power strokes and also the steam distribution.  Perhaps you will be able to explain it to us all.  I can see that the steam exhaust for the hp cylinder must be taken in by the ip cylinder, but I quickly get lost extending it to the lp stage, particularly when the two cylinders are not in line or even at 180 degrees to each other.

In terms of the thermodynamics, this is a great example of the problem with trying to extract that expansive power of the steam, the volume gets huge.  Adding a turbine is a great way to expand a large volume of steam to the lowest possible pressure, which is of course limited by the temperature of the sea water available for condensing.  Even the course into northern waters would have contributed to better efficiency due to cooler water available for condensing.  Or would have done, if it was not for a small miscalculation of the risk associated with icebergs.  Obviously the advantage of the turbine was enough to justify the cost of the complexity, telling us something about the economics of shipping.  Pity the accountants never had to swing those valves, even once would probably justify the installation of power operators for that job.

A great introduction to my intended return to the topic of condensing, but I would like to know more about that valve timing and influence of the volume of the steam chests and crossover piping for that triple expansion engine.

Apologies for the walls of text.  I know this thread needs pictures, so just a little update on the adventures in the long paddock.  The long paddock seems to have been joined by the long rail.  The train in the picture was as best I could measure it with the car speedo, around 1800 metres long, about 1.1 miles.  It was stationary, took 70 seconds to pass it at 92 kph.  How about one of you people who like repetitive work modelling this one?  Simple enough, only four electric locomotives, two at the front, one in the middle and one at the tail end, and a few identical cars in between.  Might have to scale down the voltage for your club track though.  The signs only said high voltage.  I don't know what the locomotive power rating is, but my compressors with several megawatt size motors were 11000 V.  Perhaps the trains accept higher currents.  Interesting that each car was marked with direction of travel, I presume something to do with the unloading method, but also implies that it is turned around by a large loop, or perhaps the direction is only relevant to the fully loaded cars.

Thanks for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: Ye-Ole Steam Dude on August 02, 2017, 01:19:12 PM
Triple Expansion Steam Engines

1. Cylinder Volumes:
HP = 1.0
IP = 2.6
LP = 7.0

To keep the LP Cylinder a manageable diameter it was split into 2 cylinders at 3.5 x HP volume giving a 4 cylinder triple expansion engine normally arranged from front to back LP,HP,IP,LP.

With this arrangement balance was achieved by setting the cranks at:
Between Forward LP and HP 1660
Between Forward LP and IP 2700
Between Forward and Rear LPs 700
In addition suitable counterweights were added to the crankshaft.

With Titanic and Olympus, not shown is a turbine which utilised the LP exhaust to drive the centre propeller. For manoeuvering the exhaust was switched over from this turbine and sent directly to the condenser. This made emergency reversing/stopping difficult as until this exhaust was removed, the turbine continued to move the centre propeller ahead. Transfer was accomplished by some big hand operated valves!

HTH
Regards
Bob

G' day Maryak,

I am a bit confused with order of arrangement for each cylinder, in the attached first photo Cylinder No.-1 is not at the same timing as Cylinder No.-4. Your numbers show each LP ( 1 and 4 ) are each at 70 degrees. In the photo, No.-4 appears to be close to 160 degrees ( or so ) opposite from No.-1. It looks like No.1 and No.3 are the same. Am I looking at this incorrectly?

Thank you,
Thomas
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 03, 2017, 12:59:22 PM
More on Titanic engines

Hi Thomas, I can see what you mean about the angles, though I think I have interpreted them slightly differently from what you have described.  The penny dropped when I read Maryak's post, that the angles are only nominally about balance, but the odd angles come from the need to progressively expand the steam a bit in the hp cylinder, then transfer it to the ip cylinder and expand it a bit more, then finally transfer it to the lp cylinders where we not only expand it a bit more, but also provide as steady a flow as possible to the turbine which is a continuous steady flow machine.  There are compromises along the way, especially if the cut off varies as speed is reduced to cruising speed from maximum.  I will leave it to Maryak to add a bit more explanation, I have attached a little sketch showing my interpretation of his figures.  You will note I have drawn it with the hp cylinder at top dead centre.  If you think about the hp cylinder as the engine rotates, you can see that steam admission to cut off follows much the same sort of pulsation says a single cylinder engine as you would expect, with steam into the top of the cylinder for the first half rotation, and into the bottom of the cylinder for the second half.

Then if you look at the lp cylinders, the rear lp which I have labelled RLP, is exhausting at max volume, while the FLP so just starting to exhaust at minimum flow.  The combination gives a fairly steady flow to the turbine for the first half revolution.  Similarly the bottom end of the lp cylinders gives reasonable steady flow for the second half.  The angular difference between them seems to give quite good flow for the whole revolution, another reason apart from sheer size for having two lp cylinders.

So it remains to see if the ip cylinder properly accepts exhaust from the hp cylinder and exhausts it to the lp cylinders in turn.  Its a bit mind bending, but if you turn the drawing around, or redraw it so the ip cylinder is at top dead centre, and follow the similar thought process, I think you will find it works.  The complication is when the cut off varies, when the process might be a bit less smooth, and the volume of the steam crossover piping and steam chests become important.  I am hoping that Maryak will shed some light on that as well.

I am suspicious that the artist might have taken some liberties for artistic presentation, not expecting anyone to analyse it in this much detail, or maybe the artists angles will also work, I just don't know.

But my area is thermodynamics and I look for what thermodynamics tells us that makes it worthwhile building such a complex arrangement.  Why not just a three cylinder triple to provide more power than a simple single or twin cylinder engine?

Basically there are two ways that expansion to a low pressure (which can only be achieved by condensing) provides more power output from the engine.  The first we have covered earlier, the force on the piston is due to the difference in pressure between the top side and the bottom side of the piston.  More differential pressure gives more work output.  Even on a simple oscillating engine.  But then, if the valve gear can cut off admission, the steam trapped in the cylinder is able to do more work by continuing to expand to a lower pressure, to provide even more work output, providing this lower pressure can be exhausted to a lower pressure in the condenser.  Theory might let us expand to a very low pressure, and get even more work out, but as you can see in the Titanic engine example, the volume becomes huge.  Now the arrangement with the turbine following the low pressure stage is very clever, because a turbine is easily able to handle a much larger volume of steam, and expand it to a lower pressure in a reasonably sized machine.  With two large triple expansion engines exhausting to a turbine, the scale is clearly such that economics support installing the turbine.  Otherwise they would not have done it.

Of course, condensing also adds further complexity.  We need a heat exchanger.  We have already see that the heat rejected in condensing is close to the same as the heat added in the boiler for evaporation.  The heat transfer equation is the same, Q=U x A x dT.

Now the heat transfer coefficient for condensing is not as high as for boiling, but it is still quite high.   The temperature difference is the real problem.  Even in winter the water temperature is always above zero, and for most of us, at the club pond, it will be quite a bit higher.  Winter is building time, rather than sailing time.  As the water takes up heat, it's temperature rises.  It can't get above the steam inlet temperature, and realistically, even in full scale industrial condensers, it will at best get to a maximum 10 - 20 degrees below the steam temperature, and that requires a lot of tubes.  You can see it if you find a ships condenser picture.  We can probably anticipate an LMTD less than 50 degrees.  Compare that with the firing temperature in a boiler.  I really don't know the fire temperature, but I am guessing that the temperature difference is more than 200 degrees, remembering that the flue gasses are much lower than the fire temperature at the chimney end of the boiler.  So with lower heat transfer coefficient and much lower temperature difference, we need considerably more heat transfer area in the condenser than the boiler.  However, providing we use enough tubes, we can condense the steam.  And this is done on ships so that water can be reused, instead of using the considerably more expensive process of desalination of sea water for boiler feed.

Now while it is possible, I don't advocate trying to calculate the area needed for a condenser.  I would use a published design, scale it to my estimated engine steam consumption, preferably by testing, then make it as large as I can accommodate in the model.  Then I suspect my time would be more enjoyably spent building a new condenser, with the size adjusted as necessary to condense all the steam if it turns out to be insufficient.  There is actually no real problem in making the condenser too large.  You just get a nearer approach temperature, or even sub cool the water a bit.  So long as it is not too heavy, or takes too much space in your model, it will be ok.

By building a condenser, you have the possibility of increasing your engine power output by increasing the differential pressure on the piston or pistons.  Unfortunately there is another problem we have to deal with, and that will be the topic for next time, or when we deal with some of the interesting little side tracks on our project.

Thanks for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: Ye-Ole Steam Dude on August 03, 2017, 01:20:42 PM
Hello MJM460,

I appreciate your explanation and drawing, this design is so interesting and I wish that I knew more about the complete process. This kind of work is very impressive when I think about the time and era it was conceived, when you factor in what tools and knowledge was available to the designers.

Enjoying following this thread and thanks again,
Thomas
Title: Re: Talking Thermodynamics
Post by: Ye-Ole Steam Dude on August 03, 2017, 04:55:33 PM
I found this information and drawing on the internet and it is pretty straight forward in explaining the cylinder arrangement. Hope this will help.

Titanic’s 4-cylinder reciprocating engines were balanced on what was called the Yarrow, Schlick, and Tweedy system. The crank throws were not arranged at 90-degree intervals as one might assume. Instead, vibration was reduced by adjustment of the relative crank angles and crank sequence being used. Beginning with the HP cylinder piston at top dead center (TDC), the crank sequence and angles of the engines were: HP at TDC, then a 106° rotation for the IP to TDC, then a 100° rotation for the forward LP to TDC, then a 54° rotation for the aft LP to TDC, then a 100° rotation for the HP to return to TDC. This is the link to the website: http://www.titanicology.com/Titanica/TitanicsPrimeMover.htm
 

This crank arrangement is shown in the diagram below.

 

Title: Re: Talking Thermodynamics
Post by: MJM460 on August 04, 2017, 12:27:03 PM
Hi Thomas, I am glad that you are enjoying the thread, it is all about sharing knowledge, so it's great to have some interest.

That was a great find on the titanic engines.  Balance is not so easily achieved when pistons are different sizes, but I think the sequencing of valve events also has an influence on the angles between cranks.  It also appears that slightly different angles might be selected, depending on just what the designer is trying to achieve.  I am still fascinated that the two low pressure pistons with the angular displacement between the cranks are able to produce a steady enough flow to the turbine, instead of the normal zero flow at top and bottom dead centre, I had not thought of that.  But I will let Bob or others tell us more about triple expansion engines, while I return to the condenser topic.

I mentioned last time that there was more to a condenser than just removing heat from the steam.  The problem is air.  So first where does the air come from?  Two sources, each in fact quite small.  There is always some dissolved air in water.  In industry, this air is first boiled out in a vessel called a deaerator.  This is followed by a chemical oxygen scavenger.  With steel boilers air removal is particularly important as oxygen causes corrosion in the warm wet environment in the boiler.  But you can safely assume there will be some air in your model boiler feed water.  The second source comes in as soon as you are able to maintain some vacuum.  There will be some air in leakage.  The quantity is small, but it accumulates in a condenser so its presence soon becomes important.

If we remember Willy's mountain top experiences, we talked about some air in the boiler when it was sealed up.  The air and water vapour in the vapour space act independently, so the total pressure is the sum of the air pressure plus the water vapour pressure.  We had to heat the air, as well as heat the water.  Fortunately, in the boiler, the air is entrained in the steam production, so it does not accumulate and is soon near enough to zero.

In a condenser, air is a non-condensable and unless we make special provision, there is no path out, so it accumulates.  And in accumulating, the partial pressure of air increases, and hence the total pressure in the condenser increases.  Even though the steam condenses at quite a low pressure due to the cooling water temperature, the piston sees the total pressure during the exhaust stroke, so there is no advantage in the low water condensing temperature.

You can see where this is leading, prototype engines with condensers also have an air pump.  The job of the air pump is to remove the air down to the low pressure of the condensing steam, so the condenser truly does provide the low exhaust back pressure we are looking for.

In many of the prototype machines often selected for modelling, the air pump is some sort of diaphragm pump with sufficient displacement to deal with a volume of low pressure air, remembering that some of the low pressure steam will also enter the air pump with the air.  The pump then has to compress the air and steam mixture to above atmospheric pressure so it can be discharged to the exhaust stack.  An air pump is in fact an air compressor, much like a bike pump, and clearance volume is critical, unlike the water pump for which clearance volume is not so important. 

You will also recognise that the condensed water is also at low pressure compared with atmospheric pressure, so normally a water pump is used to remove the condensate.  There are even some quite ingenious designs where the air and water pumps are combined.  When you remember the purpose, and what the design is trying to achieve, these designs become a little easier to understand.

In summary, adding condensing to an engine, increases the power output due to the reduced back pressure on the piston.   We achieve this advantage even on simple engines without early cutoff and expansion.

We can condense at atmospheric pressure simply to recover the water, but there is no real power advantage to this over a simple atmospheric exhaust.  To achieve the advantage of low back pressure due to condensing, we need a condenser, a condensate pump and an air pump.

The low temperature difference inherent in condensing means we need a large heat transfer area, more than required in the boiler that produces the steam, usually provided by means of a large number of tubes.

I believe I have some questions from Willy and Paul which have not been addressed, so I will have a go at those next time.

Thanks for reading,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 04, 2017, 02:38:53 PM
Hi, more explanations all good ...I was wondering ,if i may enquire about the jet condensing arrangement on the Woolf compound engine i am making. The cylinders are double acting but the air pump is single acting? and i was thinking that when the exhaust steam is condensing and forming a vacuum ,is this vacuum impeding the opening of the flap valves to the air pump at various places in the beam engine cycle ? Here are a few pics and drawings of the Beeleigh Mill engine, I made the drawings showing the use of hippopotamus hide valves as used in some engines before they took the air pump apart and discovered ordinary metal to metal flap valves. Just a thought and i may be wrong , however the force produced by the Newcommen atmospheric engines was quite considerable, and could this force be calculate into horses power as they used to say.....Also the jet condenser valve is always open rather than connected to the valve events ?
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on August 04, 2017, 02:56:59 PM
MJM, I have to admit that I struggled with thermodynamics in school. You mentioned the engine book by K.N. Harris but I think his boiler book is more likely far better known. I think a good topic for this thread would to explain how to size a boiler to an engine using the boiler book as a starting point.

K.N. Harris was an engineer and had a lot of practical knowledge.

Dan
Title: Re: Talking Thermodynamics
Post by: Maryak on August 05, 2017, 01:14:09 AM
G' day Maryak,

I am a bit confused with order of arrangement for each cylinder, in the attached first photo Cylinder No.-1 is not at the same timing as Cylinder No.-4. Your numbers show each LP ( 1 and 4 ) are each at 70 degrees. In the photo, No.-4 appears to be close to 160 degrees ( or so ) opposite from No.-1. It looks like No.1 and No.3 are the same. Am I looking at this incorrectly?

Thank you,
Thomas

Hi Thomas, sorry to be late in responding below is a diagram which I hope explains the differences

(http://i389.photobucket.com/albums/oo340/Maryak/Drawing1_zpsv7jziyha.jpg)

Regards
Bob
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 05, 2017, 12:35:47 PM
Out on the plains, the brolgas are dancing

Hi Willy, Just in case I have created some confusion, I should make a small clarification on jet condensing.  I mentioned that steam industrial steam plants sometimes use a jet ejector instead of an air pump.  Much the same principle as the injectors used for boiler feed water on some locomotives.  It literally uses a steam jet and Venturi arrangement to create a low pressure so that air flows in, at which point an diverging nozzle increases the pressure enough to discharge the air to the atmosphere.  This is quite different to the jet condenser on your engine, which uses a spray of cool water directly injected into the exhaust steam in order to condense the steam.  No tubes involved.  Saves soldering in all those tubes.  Basically the latent heat in the steam heats the water by direct contact with the water spray.  The end result is dependent on the amount of water sprayed in.  You could use just enough to condense the steam and the whole mass of steam plus water ends up as saturated water at the steam condensing pressure, or you can add extra water and sub cool the whole mixture to some extent.

The actual condensing pressure depends on your air pump.  The air pump removes the mixture of air plus water vapour from the condenser space, thus lowering the pressure in that space.  It then compresses the air and vapour to a pressure high enough to discharge to the atmosphere.  In your engine, the air pump handles both the air/vapour mixture and the condensed water.  While it is handling both, it is strictly a compressor.  The difference is basically in the ratio of swept volume to clearance volume.  In a typical feed pump, and also on a dedicated condensate pump, the swept volume or displacement of the piston is relatively small, while the clearance volume, basically dependant on the valve chamber arrangement does not really matter and can be relatively large.  As soon as the piston starts moving towards the valve chamber, a water filled cylinder rapidly increases in pressure until the discharge valve opens and the liquid is discharged into the outlet pipe, in the case of a boiler feed pump, right up to boiler pressure.  Even a quite small bubble of air in the water pump causes a real problem.  In the presence of a small bubble, the bubble must be reduced in volume, or compressed, until the pressure is sufficient to open the discharge valve.  If the clearance volume is significant, the bubble is compressed in the clearance space but not enough to open the discharge valve.  When the piston starts moving down again, the bubble just expands to its original volume and there is no discharge flow.  In your pump, the displacement is quite large.  When the air pump piston moves up, it reduces the pressure so that all the water flows past the first flap (or ball) valve, and the stroke is such that a volume of air and water vapour also passes the first valve.  When the piston then starts moving down, the pressure is increased so that the valve in the piston opens and air and water flows to the top side of the piston.  In fact you have a two stage compressor, and on the next upstroke, not only does more air and water flow past the lower valve, the water on top of the piston is lifted, and the air above the piston compressed until the top valve opens and water and air are discharged into the top chamber.  The gland in the top plate prevents air leaking back past the pump rod and increasing the volume that has to be discharged.  When the piston is at its lowest point, the remaining volume is quite small so sufficient of the vapour is compressed to a high enough pressure to flow through the valve vent holes.  (Don't make them too small!). Not important that air pump is single acting while the engine is double acting.  Some variation or fluctuation in the condenser pressure will not matter.  Compressibility of the air, vapour will smoothe the pulsations considerably.

By the way, vacuum is a useful concept in conjunction with a pressure gauge that measures only the difference between the measured pressure and atmospheric pressure.  However, there is no such thing as negative pressure.  Zero pressure is the vacuum of deep outer space, or in your equipment, but only if you a have a really, I mean super really good vacuum pump.  Atmospheric pressure is about 14.7 psi or 101.3 kPa, just over 1 bar.  The pressure in your jet compressor, is unlikely to ever get to zero but easily somewhere in the range 10 to 14 psi, and at best possibly even lower.  The discharge of your pump must be something above 14.7 psi in order to discharge to atmosphere.  Remember the molecular model of gases.  Pressure comes from the change of momentum when the gas molecules hit a surface and bounce off.  There is no negative pressure.

So your valves will open when the pressure on the underside is higher than the pressure on the top side, so the force is available to open the valve.  When the higher pressure is on top, the valve leather or plate moves down onto the support place and closes off the vent holes, so the water and air cannot flow back.  To understand how they work, just look at the direction of pressure difference, and don't worry about vacuum.  When your engine starts, the condenser part will be full of air.  So long as the air pump removes more than is introduced by the feed water plus the inevitable air inleakage, the pressure will, fall to something near the saturation pressure of the condensed steam after a short run time and give you quite a good vacuum to increase your engine output.

The horses power of the engine can be calculated, but in addition to the force, it requires the stroke length and number of strokes per minute to complete the calculation.  I will explain that another time, as I have a few units not addressed so far.

Hi Dan, thanks for looking in.  I intend to go on to boilers when we get past condensing, and I will include your suggestion of looking at analysing Mr Harris' boiler capacity method as you suggest.  I have the book and will try and make some notes in the mean time.

Hi Maryak, thanks for that explanation.  I wonder if optimising the flow for that turbine might be one factor in the choice of crank angles for the Titanic, while for more typical engine arrangements, exhaust pulsations would not be so important so other factors such as balance can be given more importance.

A wild bird, voluntarily in the park stands about 5 ft high and has around 6 ft wingspan!

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 06, 2017, 12:30:05 PM
Hi Willy, I have held over some of your questions from earlier posts, so I think it is time to talk about them.

In post #174, you asked about some ways of raising steam a bit quicker.  I think the first step is to think about how much energy is required to raise steam, then where the heat goes, and then what is the effect of the procedures that you propose.

Your boiler is a great example, as you can insulate it very well, and then assume that all the energy from your electric element goes into the boiler and its contents.  So the heat goes into the copper of the boiler, the water you put in the boiler, and the air that is in the boiler when you tighten the plug.

I did some rough calculations assuming the empty boiler has a mass of 1 kg and holds 0.5 kg of water.  It depends on all the dimensions, but I guessed the mass of air at around 0.01 kg, but it's probably quite a bit less.  I also assumed the whole lot starts at 15 deg C.

To get the whole lot to 100 deg, but without making steam, remember that constant volume process at the top of the mountain?  The copper will absorb 33 kJ, the water about 178 kJ, and the air about 0.08 kJ.  You can see the water takes much more heat than the copper, while the air is absorbing a negligible proportion of the heat.  Your 500 watt heating element provides 500 J/sec, or 0.5 kJ/s, so requires about 7 min to get everything to 100 deg C.  Obviously plus or minus a bit depending on the actual mass of your boiler and water.  The element is surrounded by the water, and air is only heated via the water and top of the boiler shell, so does not really have any effect on  heat transfer.  Of course, if you leave the plug loose, air will escape as the heating progresses, but more importantly, some water vapour will escape with it.  Now, heating 1 kg of water from 15 to 100 deg C requires 356 kJ, but evaporating this 1 kg to steam requires 2676 kJ.  Remember I assumed the boiler only contained 0.5 kg.  You can see it takes much more heat to evaporate water than simply to heat it.  The escaping steam will take away with it much more heat than the energy absorbed by the tiny mass of air.  Clearly not the way to go.  Please also remember that while the water expands on heating, and so does compress the air, the volume change is tiny.  You can use the steam tables to find the difference in the specific volume of water at each temperature.  You will see the compression is negligible.  Similarly the steam does not compress the air.  The steam and air molecules occupy the space essentially independently, and the measured gauge pressure is the sum of the separate pressure of each component.  The air pressure does increase, but due to its increase in temperature, and the energy is accounted for in calculating the heat absorbed by air.

Now those calculations were based on raising the temperature to 100 deg C.  At this temperature, the water vapour pressure will be 101 kPa, or I atmosphere.  The air will be about 130 kPa, giving a total of 230 kPa(absolute) or 130 gauge pressure.  We could start releasing the steam air mixture, however the pressure will rapidly drop as the air escapes.  Not easy to see, as we expect the pressure to drop when some steam is let out to the engine.  And at 100 deg C we will then only produce steam at atmospheric pressure, not real useful unless we have a condenser.  But we should also be aware that while air and steam are both present in the boiler, the pressure gauge will show a higher pressure than you would expect from the steam tables.  If necessary, the safety valve will release a little, or if you open the steam valve, the engine will start but rapidly slow down and possibly stop.  About another 4 minutes will raise the temperature of water plus boiler to 150 deg, so somewhere in that 4 minutes, you will start getting enough steam to run your engine, especially if it is unloaded.

The experiment is not difficult to try, so it is worth checking the time to raise steam with the plug loose, and comparing the time to raise steam with it tight from the same temperature.  The time consuming bit is waiting for the boiler to cool between trials.  Possibly only practical to do one or two tests each day.  A brief break in machining time on separate days is probably more efficient than waiting for it all to cool.

You could of course compress the air in the boiler with a bike pump.  This would give an initial pressure reading that might imply you could open the steam valve.  If you have a suitable fitting, try it.  However, my guess would be that the engine will run for a few seconds while the air pressure reduces, and you then have to wait for the heat input to get the water up to temperature.  I would not recommend spending time making the fitting unless you want it for other purposes.

It would be interesting to know the actual mass of your boiler, and the quantity of water you use to fill and you normal heat up time to compare with the calculation.  But the more important question is whether you can make the boiler produce steam a bit quicker.  You can see now where the heat goes, and how much is required.  Your vibration idea, I assume is with the thought of increasing the heat transfer.  I think I may have already answered this one, the calculation did not need to know the heat transfer coefficient, only copper specific heat, mass of water and copper, and steam tables.  You heating element only provides so much energy, 500 watts, and just gets as hot as it needs to, to achieve adequate heat transfer.  I suspect the heating element insulation is the main limit, but if there is any effect of vibrating the boiler during heating, it is only to reduce the temperature of the element.

So is there anything that can be done to speed up the process?  Well, first we could reduce the heat required, by starting at say 80 degrees instead of 15 deg.  This is relatively easy if you fill the boiler from a boiling kettle instead of cold water.  You might go a step further by pouring in some boiling water, letting it sit for a few minutes to heat the copper, then tipping it out and refilling with water from the freshly boiled kettle.  However all of this takes time, so you would probably not achieve much difference in time from just filling with hot water in a single step, may even be slower.

You could increase the heating element power output by using a higher voltage.  Power output is just as sensitive to an increase in voltage as to a low voltage, so with a variable transformer, you could increase the voltage and shorten the heat up time.  I do not recommend this.  It would work, but you would run a real risk of exceeding the temperature at which your element burns out.  Better to stay within the rating.  However, it's worth making sure that your power cords have the current carrying capacity, so that you are not working on low voltage.

The next trick involves modifying the boiler, or making a new one, to use an element with a higher rating, or even two elements (which you connect in parallel) so you put in the required energy in a shorter time.  Probably the best solution, though a long power cord will need an even heavier rating.

I think that leaves only some outstanding questions from Paul that have not been addressed.  I will have a go at those next time.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 06, 2017, 01:37:59 PM
Thanks for that, so we use copper as it is easier to work with and join together with silver solder, and also it is traditional and not too prone to deterioration. Could one however line the inside with a stainless steel sheath the ends as well as the tube, or even plate it with something ,cadmium or nickel ?? Just thinking about this !! Also in my boiler there are two 500 watt elements actually. I was thinking with the vibration that this would help the release of the bubbles a bit quicker? So, more questions .......and will they ever end !! I do find all this quite fascinating actually and of course this is why with modern engines of all sorts there are so many additional bits and pieces that fill up the engine compartments and spaces around the engines.!!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 06, 2017, 04:37:46 PM
This is the video of this boiler from a few years ago    63c9KR0bqb8  
Title: Re: Talking Thermodynamics
Post by: Maryak on August 06, 2017, 11:03:49 PM
Hi Maryak, thanks for that explanation.  I wonder if optimising the flow for that turbine might be one factor in the choice of crank angles for the Titanic, while for more typical engine arrangements, exhaust pulsations would not be so important so other factors such as balance can be given more importance.

MJM460

I think that may well be the case. I suspect that at cruising speed the LP inlet and exhaust pressures would be higher than without the turbine.  Based on a total HP of 51000 and the difference in propeller sizes I estimate the turbine would be producing some 13000 HP of the total.

Regards Bob
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 07, 2017, 02:01:35 PM
I have been intending for some time to address Paul's questions from post #208, so this time I will start with those.

So what about the effect of jacketing the cylinder, and even supplying the jacket with much higher temperature steam.  To understand what the jacket heating will do its best to first understand the basic engine cycle and what happens to the cycle if we add or remove heat during the cycle.

Generally, the steam engine is analysed as four processes which are repeated every revolution.  First, adiabatic expansion, then constant volume cooling, third, adiabatic compression and finally constant volume heat input.  This is referred to as an ideal cycle, and the thing most ideal about it is that it can be easily analysed.  Now any real engine is not very much like this, but the ideal cycle does provide a useful basis against which we can compare the performance of our real engines.

Now if we look at the effect of unintended heat input or loss around the cycle, we can see that during the expansion process, we initially approach constant pressure process which requires heat input.  The heat supplied by the incoming steam far outweighs any heat input or loss through the cylinder walls, which can reasonably be ignored.  Then after cutoff, heat input or loss causes the pressure to fall slower or faster than we would expect based on adiabatic expansion.  Heat input means the pressure falls less quickly, so more work output is achieved, so this is beneficial, just difficult to analyse.  Obviously heat loss means faster pressure drop and less work output.

But let's continue around the cycle.  At the end of expansion, we release the remaining steam and associated heat into the exhaust.  It is not entirely constant volume, but continues into the exhaust stroke.  While you are exhausting steam, any heat input is increasing its pressure so is undesirable.   

Next the ideal cycle has adiabatic compression.  In real engines, this occurs only after the exhaust valve closes, but again heat input means that this process occurs at a higher temperature, and this requires more work input, and so again is a reduction of work output.

Finally we have near constant heat input as the steam valve opens and admits more steam.

I don't know if I have painted this clearly enough, but the summary is that during the downstroke, any heat input is an advantage, during the return stroke, heat input reduces the cycle output.  As it is not practical to switch the heating on and off each cycle, any advantage during expansion will be reduced but probably not eliminated on the return stroke.  Similarly, any heat loss is probably more significant on the expansion cycle than the exhaust so there is a net loss.

It is very hard to quantify these effects.  Certainly, the extent is limited by the small surface area available for heat transfer.  The example of full size practice which seems to be to apply lagging suggests there is a slight advantage to insulation, but not enough to justify the complexity of installing and supplying steam to the jacket.  An even higher temperature would give more temperature difference, but again the extent is limited by area, and it would probably be more effective to apply the heat in the boiler or superheater so the supply steam is hotter.  Then the heat is only supplied to the engine when it is needed.  But if a jacket is supplied with steam at about the inlet temperature, this comes close to the same thing as perfect insulation for the enclosed cylinder.  Obviously the jacket itself then needs lagging.

I hope that is enough to satisfy your curiosity for now.

Now on to Willy's question about the stainless steel cladding of the boiler.  First Willy, I have to ask why you would want to?  I know you can silver solder stainless, but I am not sure of the strength of the resulting job.  If you can join it by welding, stainless steel gives a boiler which has higher strength at high temperature than copper, but it is very susceptible to cracking if there are even tiny amounts of chlorides in the water.  These come in even good potable water, and are very hard to eliminate.  Hence the incentive to try and use duplex stainless steels.  And the thermal conductivity is less than copper, so it's main advantage is high temperature strength.  Copper is better for heat transfer.  I am not sure of the effect of plating on silver soldered joints, but I am sure I have seen model boilers that were externally plated for appearance.  I suspect it even has an advantage in lower emissivity that copper that has lost its shine, so less radiant heat loss.  However a simpler approach is insulation.

Steam bubbles detach very readily, and lead to very high heat transfer coefficients for boiling, so I suspect there is no real advantage in vibrating the boiler.  As I mentioned earlier, the element insulation is probably the main determinant of heat transfer rate, and the consequent element temperature necessary to dissipate the heat.

Having 2 by 500 watt elements means twice the heat input, therefore half the heat up time, and twice the steam production.  It is still necessary to know the mass of copper, and the mass of water to calculate the time more precisely.  It would be interesting to know how long your boiler does need to raise steam.

Thanks Maryak, I think that you are right about those lp cylinder pressures with and without the turbine.  The total pressure range tends to distribute itself over the number of stages, and the turbine is effectively an extra stage when it is on line.  But it depends on valve events and steam chest volumes.  Do you have any more information about that aspect of the triple expansion engine?

Thanks everyone for looking in,

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 07, 2017, 03:20:11 PM
Hi thanks for the comments and questions ,I shall endeavour to measure the actually size and water volume and times to steam up soon. Talking about steam jackets the engine i am making ,Woolf Compound by Wentworth and sons is a steam jacketed engine that has a bit later sister engine that is still working. The Ramm Brewery engine is steam jacketed as well but does have wooden cladding although not around the steam chest parts of the engine. In the Beeleigh engine there is no cladding and also no indication of how it may have been attached, ie no threaded bolt holes to secure it. This system of jacketing on this double acting engine means the cylinders have no provision for steam/ condensate removal !! so one assumes that the engine block is first brought up to temperature with the valves in midway closed position, then once the engine block is fully hot the condensate is drained off by a drain cock at the bottom of the cylinder block. the flywheel is then barred into the starting position and the cylinder steam valve opened. This enormous mass and area of cast iron must have been responsible for lots of conduction and radiation of heat continuously during its working cycles !! here are some pics of the two engines, The Beeleigh engine hose burnt down in 1875 so the wooden cladding may have rotted away over the last 140 years
Title: Re: Talking Thermodynamics
Post by: Maryak on August 08, 2017, 01:16:09 AM
Thanks Maryak, I think that you are right about those lp cylinder pressures with and without the turbine.  The total pressure range tends to distribute itself over the number of stages, and the turbine is effectively an extra stage when it is on line.  But it depends on valve events and steam chest volumes.  Do you have any more information about that aspect of the triple expansion engine?

MJM460

Sorry, I can't pin that down. The information I have been able to find gives conflicting results as to power. i.e. one source suggests 48000 HP with 16000 provided by each of the 3 engines another says total HP 51000. There is information regarding the speeds of the triples at 76 rpm and the turbine at 9psi inlet and 165 rpm. Some of this seems counter intuitive when comparing propeller locations diameters and speeds.

Based on my experience with air pumps in tip top condition a condenser vacuum of 26" Hg is the best I have seen in cold seawater at economical cruising speed.

To give a 9 psi LP exhaust I estimate LP inlet at 24 psi. This suggests that with the turbine out of the loop LP exhaust would now be around 20" Hg and LP inlet around 9 psi, all of this is assuming no linking in or out i.e. line in line valve settings, (which with a standard way shaft is always the case when going astern).

The condensers would now be "pulling" steam through the engines and the 4th stage expansion would be lost. I suspect CW pumps would also need to be operated at higher rpm to cope with the additional heat arriving in the condensers.
Title: Re: Talking Thermodynamics
Post by: paul gough on August 08, 2017, 01:46:33 AM
Thanks MJM for considering and running through what is going on with a 'heated cylinder'. I am pleased you did not think it an altogether flippant enquiry. I have been interested in whether attempts to keep cylinders 'warm' in a steam engine is/was of any significant value in practice, especially for models. Thus trying to understand what might go on in an extreme situation can deepen understanding. As I am primarily interested in locomotives it is, (usually), to these I address my investigations. All this line of inquiry was prompted by noting that the full size Lion locomotive, (Titfield Thunderbolt), had the top of the inside cylinders forming the base of the smokebox so the steam chest at least received some heat from the flue gases, but maybe the cylinders also got a bit warmer on a longer run?? As the Lion would, I presume, have been a coke burner like most locos, (railway not colliery lines), before 1850, there might not have been too much soot/ash build up to create an insulating layer on top to impede heat transfer. So I wondered if any noticeable benefit might have occurred beyond the obvious hopeful reduction in condensation formation due to the locos maximum boiler operating pressure of 50 p.s.i. I also notice on my drawing of Stephensons Patentee loco it had the cylinders and steam chest, except for cyl. covers, totally enclosed in the smokebox, presumably it would have achieved a greater heating effect on the cylinders than Lion. Regards Paul Gough.

 
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 08, 2017, 01:33:46 PM
Hi Willy, interesting that those engines had steam jacketing.  I suspect the boiler pressure was not high, and probably minimal if any superheat, so minimising losses was probably even more important than for a modern engine.  I suspect that the tapped opening you have arrowed is the condensate drain for the jacket.  Today it would have a steam trap on it, but I don't know if they had been invented then.  The engineer probably had to regularly drain the jacket by opening a valve briefly.  The engine could be cleared of condensate by barring it over when it had heated well, before opening the steam valve.  It is great that you have two contemporary engines to help you work out the authentic arrangement after so much time has passed since they were new.

Hi Maryak, I wonder if the 165 rpm is actually the turbine speed or is it the propeller speed?  Modern turbines would certainly be much faster for good efficiency, and use a gear to give the slower required propellor speed.  Certainly the engines would have been designed to be optimum when going forward with the turbine in operation.  Then operation in reverse without the turbine would be an off design condition, where lower efficiency would be acceptable.  I also assume you would not often if ever need full power in reverse.  When the lp cylinder is connected directly to the condenser, it would of course have higher than normal power due to the higher differential pressure,  but I suspect there would be some ripple effect through the ip and hp cylinders which might each operate under slightly modified pressures.  Usually a multi stage engine design aims to have roughly equal pressure ratio in each stage, and equal pressure ratio gives roughly equal power, so your figure of 13,000 HP for the turbine out of a total of 51,000 looks about right.  It would mean about 19,000 HP for each of the triple expansion engines.  I guess further explanation will have to wait until we have more information.

A vacuum of 26 inches of mercury, means an absolute pressure of only 13 kPa.  I assume that 9 psi is a gauge pressure, so absolute pressure of 163 kPa.  This means a pressure ratio of 12.3:1 for the turbine, enough for two extra reciprocating stages, but they would be impractically large.  However, a turbine can handle the volume of steam easily in a moderate size machine, even if with lower efficiency.  A really ingenious arrangement for good fuel efficiency.

No problems answering questions, Paul.  I think this thread probably more useful when it addresses the questions people are wondering about, rather than some subject no one is interested in.  It is interesting that heating or cooling of the cylinder is an asymmetric problem.  We know that heat loss is detrimental, but we can reduce heat loss until it is hardly significant by insulation, which has low conductivity.  We can even reduce heat loss to zero by steam jacketing, so the cylinder has no temperature difference to drive heat transfer.  I presume we would also lag the jacket to minimise the steam requirements. Interesting that insulation works by addressing the heat transfer coefficient, while jacketing addresses the temperature difference.

However, it is more difficult to drive this further by adding heat to the cylinder to increase the work out.  We can use high temperature in the jacket, though we probably don't have a steam source at higher temperature than the engine supply.  The size of the cylinder severely limits the area available for heat transfer, so it is difficult to achieve much heat input.  Compare the geometry and temperature difference with that of the boiler to get an idea of the relative magnitudes.  If we had a higher temperature steam available, we would probably be better to supply it to the engine directly, rather than try and drive it through the cylinder wall.  Even a high temperature source such as the smoke box is limited by the film coefficient of the gas to metal, so a high temperature heat source such as the smoke box probably does not achieve much more than a standard jacket.  However it has a great advantage in simplicity, no moving parts or extra connections, and in fact difficult to see how it could be avoided on a typical locomotive.

Now I have introduced power and efficiency this time, so next time. I had better define those terms and their units of measurement.

Thanks for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 09, 2017, 01:24:47 PM
For the dinosaurs amongst us

Came across some dinosaur foot prints near the road today.  Well not really near, more like 110 km off the road really, a little detour from the long paddock, but interesting to see a moment in time captured by nature, the time of a stampede.  And not found unaided of course.  The palaeontologists told us where they were, and how they solved the puzzle, and how this moment is still preserved for the future.

Very early in this thread, I looked at the SI metric system and the definitions of force and work.  I have avoided introducing other terms as long as I could, because in the explanation of how engines work, I wanted to concentrate on what work is, and it's definition and how any engine turns heat into work.  No fundamental difference between big and small.  However it was necessary to talk about power in the discussions on the engines of the Titanic, so I had better define it.

If you want to lift a mass of 1 kg to a height of 10 metres, you need a force of 1 kg x 9.8 m/s^2  = 9.8 N.  The distance is 10 metres, so the work required is 98 N.m.  Now any size engine can, in principal, do this.  My tiny single acting Mamod engine an with sufficient gearing and given enough time, can do the job, though I admit that it would have to be very good quality gearing with near perfect bearings so that friction did not require more work than the lifting task.  On the other hand a full size ships engine would not even notice if this task was added to its normal load, and would do it in the blink of an eye.  While the amount of work done is the same in each case, there is clearly a difference in the engines.  This difference is power.  So what is power?

Power is defined as the rate of doing work.  Alternatively, it is the amount of work done per unit of time.  The unit for measurement of power is the Watt, which is defined as one Newton.metre per second.  You can see a Watt is the power required to do one Newton.metre of work in one second.

The power developed by engines we are familiar with varies over quite a wide range.  The little Mamod engine I have already mentioned would be measured in thousandths of a watt, or milliwatts.  Similarly, many low temperature difference Stirling engines would be measured in milliwatts.  For most of my life in making model boats, 10 W was considered almost unattainable, certainly for mere mortals.  How things have changed with the advent of better batteries and brushless motors.  A typical sewing machine motor, or a mini-lathe might be 150 W.  My lathe motor is about 1 thousand watts, or one kilowatt, or kW.  Our cars might be a bit less than 200 kW, while formula one, I should know, but it escapes me for a moment, come in Jo, can you help us here?

The compressors used in automotive service centres for tyre inflation are around 5 kW, while the ones I am familiar with from my working career range from about 600 kW to 20 MW, 20 million watts, 20 megawatts.  I am sure that you can all think of others, larger and smaller.

Now we sometimes see an engine described as "powerful".  We need some understanding of what this might mean.  We have on this forum, an amazing thread by strictlybusiness on a 0.9 cu. in. engine developing 7.2 HP.  Amazing for a small engine, for a model aeroplane or boat, but not very impressive for a Formula one race car.  Unless you mean 7.2 extra HP, of course, when in that context it would probably be considered a major breakthrough.  None of which takes anything away from the incredible feat of getting so much power out of such a tiny engine.  Similarly, a formula one engine would be no use in a model in the usual size range.  So, is there a measurement which allows us to compare our model engines despite the differences size?

I suggest that when we say an engine is powerful, we actually imply "powerful for its size", so we need a meaningful measure of size to convey the idea with any precision.  We could use some representative length, but the commonly used measure is the mass of the engine, which in the SI metric system would be in kilograms.  If we describe an engine as producing so many watts per kilogram, we would have a measure, generally referred to as power to weight ratio, that is meaningful for a wide range of engine sizes.   Power to weight ratio has dimensions (W/kg in the metric system), so the numbers will depend on the units you are using, you need a conversion factor if you want to compare W/kg with HP/lbm for example which would be the equivalent in imperial units.  Power to weight ratio is an important consideration in comparing engines for a specific application.  For an aeroplane, for example you need a high power to weight ratio, but can afford to check and maintain it often.  A steam engine with its boiler, and condenser is at a severe disadvantage for this application.   For a national grid electric power generation, weight is just not an issue, you put much more priority on reliability and low fuel and maintenance costs.  The turbines used in a typical national grid power station were never meant to fly or move in any manner at all, but they reliably operate 24/7/365.

So to get back to Willy's question on the horses power of his engine, first we had better look a bit closer at the commonly used units for power.  I have already mentioned the metric unit of power is the Watt.  Interesting that this watt is exactly comparable with the electrical unit of Watts, calculated by multiplying volts by amps (by power factor if the voltage is AC).  I never have quite understood  how the definitions of volts and amps could come up with such an elegant result.  Or is the definition of the unit of resistance (ohm) the secret?  In imperial units, power is measured in Ft.lbf/second.  However we normally use the unit horsepower, first defined by James Watt, I believe.  The horsepower is defined as 550 ft.lbf/sec or 33,000 ft.lbf/min. *  I have heard that James Watt must have had a very large horse, as the figure is apparently a bit high for normal horses, however, it gave a basis for comparing the power of his steam engines with the contemporary technology, which at that time was horses.  Fascinating that Willy's engine name plate specifies the power in horses power, but the concept of the engine being able to match the power of a number of horses makes sense for his advertising purposes.  It obviously later became modified to horsepower. 

We can now see that to calculate the number of horses power of the engine, we need to know the pressure in the cylinder and the piston diameter so we can calculate the force in lbf exerted by the piston rod.  We then need to know the stroke length in feet, so we know how much work is done by each stroke.  Finally we need to know the number of strokes per minute, or hour that the engine completes.  So we multiply force by stroke length by strokes per minute and divide by 33000 * to get horses power.  And because electrical watts are watts in both imperial and metric systems, we all know that one horse power is equal to seven hundred and forty six watts.  (1 HP= 746 W, or 0.746 kW)

Next time I will,talk about efficiency, which surprisingly is a little more complex, as there are many definitions in common use.

I know the thread needs pictures, but unfortunately dinosaur footprints do not photograph very well. However, Mr Google is quite good at finding dinosaur stampede photographs that are way better than I was able to take.

Thanks everyone for reading

MJM460

* corrections 10/8/17 -  1 HP = 550 ft.lbm/sec
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 09, 2017, 02:24:31 PM
Hi MJM, I think they increased the Horses power rating by adding more weight to the flywheel and increasing the steam pressure ? presumably this would work. ?? On my BMW motor bike the R69 engine of 600 cc is rated at 35 horsepower., but the R60 engine of 600cc is only rated at 25 horse power !! I think the only difference is the compression ratio is higher for the R69 engine ??!! so does compression ratio come into the equation ?? also the R69 engine can increase the rpm by 1000 to actually go up to 100MPH. With marine and areo engines one can feather the propellers to give more speed/power so is this part of the equation as well ??  Sorry i am using the question key such a lot perhaps i could remove it !!!
Title: Re: Talking Thermodynamics
Post by: paul gough on August 09, 2017, 10:09:25 PM
Horsepower??? For the novice, a confounding species with too many breeds to choose from! Eg., Brake H.P, Indicated H.P., Shaft H.P., Boiler H.P., Drawbar H.P., etc, etc. Maybe a quick explanation why the need for various sorts is in order. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 10, 2017, 01:45:44 PM
A little more horsepower?

Oops, one horsepower = 550 ft.lbm/sec or 33000 ft.lb/min, and no one picked me up!  I have corrected yesterday's post, I hope no one has wasted too much time worrying about that one.  It would have been a quite puny horse.  My apologies if I have led you astray by this one.

Thanks Willy and Paul for your questions, please don't remove the question key, either of you.  Your questions prompt me on areas where people need a bit more clarification, and I am sure that you are not alone with these questions.

On the power of those two ancient engines, the increase in steam pressure does increase the force on the piston rod, hence the work done on each stroke.  Increased work per stroke at the same number of strokes per minute means more power.  So increasing the steam pressure would have increased the power output.  Depending on the load, it may even increase the number of strokes per minute.

The flywheel however does not produce any power.  It simply stores energy when the engine tends to speed up, and returns power when the engine tends to slow down.  Remember when I looked at the torque produced by the force in the piston rod?  It rises to a peak, and falls to zero twice every revolution.  The engine accelerates under the higher than average torque, and slows when the torque is below average.  Extra weight in the flywheel is a disadvantage because it increases the bearing loads and hence losses, but it is a necessary side effect of adding extra moment of inertia.  I would assume they added the weight to the rim where it has the most beneficial effect.  However, increasing steam pressure increases that peak of torque twice each revolution, while zero is still zero, so increasing the power by increasing steam pressure increases the speed fluctuation of the engine, not usually desirable.  It is also possible that the speed fluctuation before the modifications was already higher than desirable, thus increasing the necessity for increasing the flywheel inertia when the steam pressure was increased.

On your modern BMW engines, I assume twin cylinder, those are beautiful machines.  I don't know the specific details of the engine, but in principle you are quite right, increasing the compression ratio does contribute to increased power output.  It effectively increases the mean effective pressure, so increases the torque produced by the engine.  However there are many other aspects to tuning up, or hotting up an engine.  Some of the things that can be done include grinding the top of pistons so each has the same compression, and grinding the bottom of the piston skirts to make the two the same weight.  I have it on good authority that checking the pistons at top dead centre and making them within 0.001" of each other and grinding the bottom so the weights are within 0.1 gm makes a significant difference for a street machine, but not even close to good enough for the race track.  Paying this level of attention to bearing clearances, and every other aspect of the engine construction will significantly increase its power with otherwise standard components, and will increase the maximum rpm.  Modification to cam profiles, valve spring rates, port polishing and even different air cleaners can all make a difference, especially if you have access to a dynamometer to help you see the differences.  Not really necessary to do anything heroic to make quite a difference.  However things which increase torque are the most desirable, as torque gives acceleration, while higher top speed only loses your licence, unless you confine yourself to the track of course. 

Feathering an aeroplane propellor does not change the power output of the engine, rather it changes the load.  This allows the engine to run at the optimum rpm for the immediate task.  When climbing, maximum power is required and a certain pitch will allow the engine to run at the right rpm to provide maximum power.  Slowing down and descending requires less power, but purely reducing rpm is not necessarily good for fuel efficiency.  A lesser pitch will allow the engine to run at the optimum rpm with much lower fuel consumption.  Perhaps flyboy Jim (or we may have other pilots on the forum), can elaborate on this one. I hope that makes it a bit clearer.

Paul, thanks for pointing out the different ways in which power is measured.  I have simply provided the definition, 1 Newton.metre per second = 1 watt, however everyone wants to claim that their engine is more powerful than the competition.  There is kudos in having the highest power engine, size does matter, it seems.  The engine maker does not want to have his figures compromised by what is added after the engine is sold, while the end user only wants to know if there is enough power available to drive his generator, propellor or compressor, or just make his car or bike go faster.  So everyone measures the power in the way that best suits the vested interest.  Let's have a look at some of the common terms, and how they reflect different measurement locations.  Remember above all they nearly all use the same definition of one horse power = 550 ft.lb/sec, or 1 watt = 1 Newton.metre/sec.  Of course horse power is a specific  imperial unit, in SI terms, is is usual to just refer to power.

First the one exception, the boiler horsepower.  It is actually defined in terms of raising a defined quantity of steam at 0 psig and 212 deg F, from water at zero psig and 212 deg F.  It is approximately 33475 Btu/hr or 9811 watts.  This rate of energy transfer is about(?) 13.1547 times the equivalent of the mechanical horse power definition.  It is not a measure of mechanical power, but intended to be an indication of the size engine the boiler would be suitable for.  A nice lead in to our discussion on efficiency, as it implies a steam plant efficiency of about 7.6%.  So hold that idea of efficiency, but otherwise it is not really a useful concept.  Better to define a boiler capacity in terms of rate of steam production at a specified pressure and temperature, and match it to the engine specifications.

For a given engine, the indicated horsepower can be expected to be the highest.  It is calculated from the area of the indicator diagram which we looked at earlier in this thread.  It gives a measure of the power provided by the action of steam on the piston.  It does not allow for friction in the rings or bearings, or the power required by the feed water or lubrication pumps, or condensate pump,or air pump.  So a nice big figure, also useful when looking at and optimising valve events, but does not help much in assessing the size of generator the engine could drive, or the power available to drive the ship or aeroplane propellor.  Again, some of the marine engineers might like to come in and explain a little more about how they use these diagrams.

Brake horse power and shaft horse power are very similar.  I am never sure whether there is actually any difference in the numbers.  Brake horse power is the term used in the engine is tested on a brake, an artificial load used for the specific purpose of measuring the engine output.  Shaft horse power is I believe the same number, but the term is used to refer to the power available at the output shaft to drive the end users load.  In each case there is room for some ambiguity, and in specifying an engine it is preferable to separately list the power consumed by engine accessories, such as oil pumps, cooling water pumps, and radiator fans and specifically identify the available power after allowances for all necessary accessories.  There is some justification for the engine makers desire to not include all these loads.  In many industrial applications there are good reasons for driving these accessories with electric motors, thus leaving the entire engine output available to drive the intended load.  And of course, they are all measured with the engine new and clean.  Usually, power output gradually declines for ever afterwards, depending on the quality of maintenance.

Drawbar power is useful for prime mover applications and might apply to a traction engine as well as a locomotive.  A typical traction engine requires a considerable amount of power just to move itself.  A prospective purchaser who wanted it to pull logs out of the forest, or a plough or a trailer wants to know how much power is available for this purpose.  It would indicate the drawbar pull and towing speed.  The engine efficiency competitions run by some clubs for locomotives actually measure draw bar power and compare this with fuel consumption to calculate efficiency, another neat introduction to that topic.  If the engine is used as a stationary power supply, say for a saw mill, the power available would be much higher than the drawbar power.

That probably enough for now, does anyone have any other definitions of power output they have come across?

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 10, 2017, 02:05:44 PM
Hi MJM, thanks for the reply there is lots more to increasing Horse power than i thought......also i can give you another definition of power output from this little booklet i have in my possession   4 Man Power !!!!! thats a new one on me !! however it is in print so must be true !!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 10, 2017, 02:22:03 PM
Hi MJM ,just looked up Bisschof gas engines  So can be run by any boy or girl !! the beginning of girl power then !! Also engines available in  ! man and a half power !! Brilliant stuff.........
Title: Re: Talking Thermodynamics
Post by: simplyloco on August 10, 2017, 02:42:31 PM
Many many years ago, aspiring race mechanics like me were not only matching pistons in our Austin Mini's, we were balancing up the combustion chambers by grinding them out and measuring each volume by introducing water with a pipette through a tiny v cut in the edge of a piece of perspex placed over each chamber in turn. I'm not sure it went any faster but it felt good! Before that I skimmed the heads of my Ariel Arrow 2 cycle bike and fitted padding in the crankcase: now that DID go better!
John
Title: Re: Talking Thermodynamics
Post by: Dan Rowe on August 10, 2017, 03:02:44 PM
MJM,
Some nameplates have a continuous and max or peak HP rating. If you google horsepower litigation you will find the current legal wrangling. There are cases that go back to the time of James Watt. The invention of the steam engine indicator was kept secret for many years and is first mentioned in a court case about engine horsepower. You can not always take nameplate or catalog data as true information.

Dan
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 11, 2017, 01:00:25 PM
Horse power or man power

Great examples of different ways of rating engines.  It shows ingenuity to the fore in the days before standard measurements, and manufacturers tried various ways to convince potential customers that theirs was the one to buy.  But of course, there was no standard for power measurement, so all those terms could not be precisely compared, until James Watt defined the horsepower.

Interesting that boys and girls were treated as equal in that add, Willy, pity it all seems to have gone down hill from there.

Hi simply loco, great to have you on board.  I think smoother running resulting from better balancing of weight, compression ratio, and fuel mixture always results in an engine being able to run faster, as there are losses associated with rough running.  And vibration is a reasonable criteria for calling it the maximum power.  But the tuning tricks you learned with those activities have a long and honourable tradition.  And very rewarding to my way of thinking, what ever directions your career took later.  Unfortunately my time was spent in other areas, so I never got to learn some of those things. 

Hi Dan, modern engines still have a published maximum or peak rating, as well as a continuos rating.   Suppose there is always something that limits the maximum power output from an engine in an absolute way, but most engine ratings are determined by the criteria you impose.  A formula 1 race engine is expected to run at the maximum power possible for a bit over two hours then the mechanics get to restore it for next time.  Even then, I understand they will get up to tricks like disconnecting oil or water pumps for a little extra power on the road for those critical final timing laps.  They know just how far they can push it without damage.  We expect our personal cars to last much longer between servicing, and expect the servicing costs to be more moderate.  So continuous operation might be limited by some critical temperatures, perhaps exhaust, water or oil temperature.  If you exceed this limit, you will shorten the life of the engine, but you can exceed it for a short time, occasionally, with minimal effect on reliability.  And it's not totally unreasonable, even in your car you want to be able to put your foot down to get to the top of a hill, then are happy to coast down the other side while the engine cools a little.  Car engines are rated nearer peak power, which cannot be used continuously, while my industrial machines had to run at 100% load continuously.

I think all those legal problems can be traced back to someone seeking advantage.  The manufacturer exaggerates his claims, while the purchaser wants to pay the least even when it means buying a smaller engine, so in a way they collude with each other, and it often ends with tears.

I guess catalogue and nameplate rating are at best a reasonable average for similar engines, but even when you specify a test run and certified test results, the results really only apply to the specific conditions of the test.  And of course the test is at best only a few hours, and no real indication of how long it could run at those conditions.  Most of my clients in industry expected the engine on the test stand to demonstrate at least 10% more power than that required by the driven machine, which was also demonstrated on the test stand.  But they then expected the equipment to run 24/7 for three years or more.  Not so hard for a centrifugal machine where there is really only one moving rotor, and hydrodynamic bearings mean nor rubbing or wearing parts, even seals are separated by a gas film.  But quite a feat for a large reciprocating compressor.   They weren't portable by any reasonable assessment.

No new topic today, I will be entering tin can and string territory for communication for the next week or ten days on the long paddock, so I don't expect to be able to make regular posts.  I will check in when a freak of the atmosphere, together with adequate solar power enable it.  I believe it usually involves standing on the left foot with your right elbow in your left ear while you hold the phone high over your head pointing in the precise direction of the nearest tower which is beyond the horizon.  Or something like that.

So thanks for following along, keep thinking about thermodynamics of your engines and the regular discussion will return in about two or three weeks, rather than try to continue through odd unpredictable times in that period.

MJM460
Title: Re: Talking Thermodynamics
Post by: derekwarner on August 28, 2017, 07:24:12 AM
MJM...

We know you are out in the back blocks of OZ somewhere :shrug: ...& probably with little or poor e-mail conductivity  ......

However just wondering....should we send out a search party?   :LittleAngel:

Derek

PS....I am in Adelaide for a few weeks...& called into a Hotel called the Thirsty Camel.... I called into the bar...yelled out your name but no one answered so guessed you were not there :drinking-41:
Title: Re: Talking Thermodynamics
Post by: Stuart on August 28, 2017, 08:21:12 AM
The reference brings a smile to my face

I have posted this before, but I did armature and stator winding as part of my apprentiship

Ok a job came across from the fitters name plate said 5 up so I looked in my book for Royce 5hp not in there but a 25 hp was so of we go take all the details , they were the same as the 25hp so I called over my mentor to check , his reply was yes they are all the same in that frame size they are all 25hp but if it 3as ordered as a 5hp the factory just stamped the nameplate as such

So it’s a case of caveat emptor

Keep up the lessons always a good read
Title: Re: Talking Thermodynamics
Post by: Steam Haulage on August 28, 2017, 09:01:51 AM
First the one exception, the boiler horsepower.  It is actually defined in terms of raising a defined quantity of steam at 0 psig and 212 deg F, from water at zero psig and 212 deg F.  ? ?

I don't understand.

Jerry :old:
Title: Re: Talking Thermodynamics
Post by: Stuart on August 28, 2017, 11:18:11 AM
Jerry
That’s the energy required to change the state eg. From liquid to a gas (steam) without raising the temp.
It works the other way as well when the said gas is converted back to a liquid it gives up that energy.

The same goes to convert liquid to a solid you remove the energy

In simple terms that’s why your fridge gets hot at the back and cold inside , note you do not cool a fridge you just transfer the heat from inside to out side


Here is a couple of horse power to ponder BHP ( brake horse power  ) measured with a dyno on the output shaft note this is not the input power

Then we have nHP.  Notional hp it’s a steam equivalent form the days of yore

Just remembered some more

The old time RAC hp rating for cars this was not the hp (power ) but the engine size. Eg 12hp was a 1200cc engine

Then you had the hp of a stationary/portable  steam engine this how many good horses it took to move it to its place of work again back in time

But for this old person  :old: ( yes I do look like that inc stick ) 1hp = 746 Watts period

Soory to muddy the waters
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 28, 2017, 01:03:40 PM
On the outer Barcoo-

Back in real internet territory with regular capacity, so it is obviously time for a bit more thermodynamics!

Thanks everyone for the continuing interest.  Derek, I haven't seen any camels, thought you might have guessed a bit closer.  I am holding off the explanation until the story is complete, but I will get there.  Thanks Stewart for both your contributions.  You may be interested to know that modern outboard engines are rated in a similar way to your 5.  If you order the lower power you get a lower price, but I believe the only difference is the jet in the carburettor.  Welcome aboard, Jerry, Stewart has explained that the definition is only about latent heat, it does not even include the heat necessary to raise the water temperature to the boiling point.  It was a definition introduced in about 1876 and refined by 1884 to give buyers an idea of what size engine the boiler was suitable for.  A little more on that below.  Stuart, I suspect your definition of the horse power of those stationary engines is a little tongue in cheek, but it reminds us of how many tones of fire breathing monster was necessary to replace a horse, and the numbers were possibly similar.  I guess wood was cheaper and more readily available than hay.  I also prefer to confine my use of the term horse power to the precise technical definition of 746 watts, or even better 746 J/s, remembering that one J is one Newton metre, so directly related to the rate of doing work.

Last time we were talking about power, and units for its measurement.  And we introduced the term efficiency as the topic for this time.  Efficiency is a term which, in addition to many common uses has a specific technical meaning.  Well, it should have a specific meaning, but in the way of those wanting to sell products, even the technical meaning gets modified so there are several variations.

The technical definition I prefer is more specifically called the overall thermal efficiency.  This refers to the output power of the plant or machine as a fraction, or more commonly, a percentage of the energy in.  For a steam plant for example, we can measure the power output of the engine, or at least we can in full size practice with a well equipped test stand, and we can measure the fuel consumption, and use the fuel calorific value the calculate the energy content of the fuel.  Then simply divide the engine power output by the energy in the fuel, using consistent units of course.  We can then multiply by 100 if we want to express this fraction as a percentage.

Thinking back to the old definition of a boiler horse power, which was the horse power of the engine the boiler might be expected to drive.  One boiler horse power was defined as the energy input required to evaporate 34.5 lb of water at 212 F to steam, equal to 9811 watts.  Now if 9811 watts input is required for each horsepower of engine output, that is each 746 watts of engine output, we can divide 746 by 9811 and multiply by 100 to get an assumed plant efficiency of 7.6%.  Not very impressive really, but such was the technology of the time.  We must remember that in the way of the legal profession, this boiler horse power was not a guaranteed engine output, simply and expected output from some other manufacturers engine, and almost certainly defined the boiler operation and cleanliness.

What does 7.6% mean in terms of fuel consumption?  Well, we need to know the calorific value of the fuel.   Various standards and Google pages give the calorific value of coal as about 25.46 MJ/kg, or 25,460 kJ/kg.  The energy input of 9811 watts equals 9811 J/sec equals 35,319,000 J/hr or 35,319 kJ/hr.  Finally a simple division 35319/25460 yields 1.38 kg of coal per hour for each horse power.

I don't know how that compares with Sabino or modern coal fired merchant ships, but as the definition was finally agreed in 1884, I suggest it is nearer James Watts engines than the modern counterparts.  Does anyone have any figures for the performance of modern coal fired engines in terms of coal consumption and horsepower?

Next time I will look at what efficiency means for our little oscillating engine.

Now it's hard to think of a suitable picture, for this thread, but in the effort to make sure that we don't all turn into thermodynamics nerds, I am going to suggest a little culture, just to balance the technical content.  How about a little poetry?  If you went to the right school you will remember it, but a couple of verses should jog some memories.

On the outer Barcoo,
Where the churches are few,
And men of religion are scanty.
On a road never crossed,
'cept by folk who are lost,
one Michael Magee had a shanty.

Now this Mike was the dad
of a ten year old lad,
plump, healthy and stoutly conditioned
............

Thanks for looking in again,

MJM460
Title: Re: Talking Thermodynamics
Post by: Stuart on August 28, 2017, 02:37:16 PM
Yes I have suspected that the definition of the how many horses etc is suspect but I bet it was on someone’s catalog sales pitch at some time , but in the real world it would have given the farmer a idea how big that fire belching monster was

Stuart
Title: Re: Talking Thermodynamics
Post by: Maryak on August 29, 2017, 01:54:42 AM
James Watt decided that is was a load of 550lbf that a "good" horse could raise 1 ft in 1 sec. It was also acknowledged that it was deliberately set high!!

For me it is interesting how our measurement systems relate to the world  around those who defined them at a particular point in history. The only one which is universal or perhaps earthieversal is time and at the risk of inciting a riot it's one hell of a stretch to confer on it anything relating to increments of 10's.

 
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 29, 2017, 12:09:54 PM
More on efficiency

Hi Stuart, I think the industry was still talking about water pumping for mines when the term horsepower was coined as a unit of power, but when portable engines first came in, the ratings were not very high.  I suspect to set up a pair of ploughing engines required enough horses to ensure that they would continue to receive their hay for the foreseeable future.

Hi Maryak, it's good to have you still looking in, your contributions are always appreciated.  I did understand that James Watt's definition of a horse power implied a very strong horse.  I did not realise that it was deliberately so.  Quite the opposite to the normal sales hype which tends to exaggerate the engine output.  However it would tend to attract happy repeat customers who would be delighted to find that their 4 HP engine could actually do what they previously required five to do, and keep it up all day without needing a rest, even when the tubes were a bit sooted up.

I suspect that the earthieverse standard for time probably would be either a day or a year, though the day at least does vary a bit.  Twelve months makes sense in terms of dividing it into four seasons as in moderate climates or two for the tropics.  It can also be divided into three or even six, but the moon phases occurring roughly 13 times per year mucks up the logic a bit.  As you say, trying for ten months in a year, days in a month, or hours in a day does not seem to result in a nice logical system.  It does seem necessary to accommodate the one to three hundred and sixty five and a quarter ratio of orbits around the sun to rotations of the earth, which is not easy with base 10.  But we could have ten "hours" per day, each of 100 "minutes", again each of 100 "seconds" by adjusting the length of the second.  It would then involve a different strange number of vibrations of those caesium atoms.

I understand that the second as a unit of time was selected by an ancient civilisation, who used a base 60 number system and hence the 60 seconds to a minute, 60 minutes to an hour.  Base 60 has the advantage of having many factors so you can divide it by 2, 3, 4, 5 and 6, and also 10, 12, 15, 20 and 30 without needing fractions.  They did pretty well to measure seconds, though I seem to remember from physics that a pendulum with about a metre length has a period of 1 second.  But they were working all this out before the metre was defined.  Perhaps they had a unit of length that corresponded with a one second pendulum.  We can certainly measure time pretty accurately these days with the device on our wrist, but the time differences in Formula 1 and Olympic events make a second seem pretty large.  Though I wouldn't be ashamed to come second to Ian Thorpe by one hundredth of a second or so.  Yet no one can remember who came second except the one who did it.  And those atomic vibrations would still be necessary so we can measure the difference in time a signal takes to reach us from different satellites and calculate our position on the earths surface within 5 metres.

I still have to learn not to be too quick to announce the next topic.  A little while after posting, I usually remember other aspects that I had intended to cover.  I believe that in some cultures it is called a stairways moment.  So our little engine will have to wait a bit longer, at least until I am off the stairway.

Last time I referred back to the old definition of boiler horsepower which implied a thermal efficiency of 7.6%, and noted that it was not very impressive.  So what do modern engines achieve?  Last time I looked, it was only the very best large scale plants that were just nudging 50%, and that required all the subtle tricks of heat recovery, together with high pressures and temperature.  No doubt they are a little higher now, but I suspect not much.  Most are much nearer 30%.

At the other end of the scale we have the locomotive efficiency competitions reported quite regularly In Model Engineer Magazine.  To understand the reported results, and how they relate to overall thermal efficiency, we have to note that the rankings are based on drawbar power.  That is, they have a carriage towed by the engine that measures the drawbar pull and the speed over the measured run.  So we should qualify the results as drawbar efficiency.  It is not really fair to compare drawbar efficiency with efficiency based on a stationary engine output horsepower.  It obviously is measured after all the work to drive the locomotive along the track is used, including moving the mass of the boiler, and friction introduced by the bogey cars etc.  Of course it is not an unreasonable measure for a mobile engine.  It does not matter how much power is produced by the actual engine parts, if most of it is used just driving itself and its boiler along the track.  We are definitely more interested in how much load it can pull.  I don't have access to my magazine collection at the moment, but from memory, the best 5 inch gauge locomotives are around 5%, and the smaller gauges nearer 1%.  Perhaps someone has a report handy and can confirm or update these figures.  I think those figures for a 5 in gauge loco are quite impressive and it is obvious that friction leads to proportionally greater losses in smaller engines.

If only somewhere between one and five percent of the heat available from the fuel is converted into work, what happens to the rest?  Is it really lost?  What about that basic law of physics, conservation of energy?

The first law of thermodynamics says that heat energy can be converted to work, but the second basically says not only can we not get more energy out than we put in, we will always loose some in the process.  Perhaps more accurately, we will always get less work out than the energy we put in.  Conservation of energy says we do not actually lose energy, it just goes somewhere other than getting converted into work. 

That is probably enough for one session.  Next time I will look at where the energy goes.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 30, 2017, 01:56:27 PM
Conservation of energy -

Last time, I started looking at efficiency and how we cannot convert all the energy from fuel into work.  The old definition of a boiler horsepower implies an efficiency of only 7.6%.  The law of conservation of energy says the remaining heat is not lost, it just is not converted to work.  This raises the obvious question of what happens to the rest, 92.4% of the energy from fuel that is not converted to work.  Let's start looking at where it all goes.

If we start by looking at the boiler.  Fuel needs oxygen to burn, and we normally obtain the oxygen from air.  As a part of the combustion process, carbon in the fuel is oxidised to carbon dioxide.  Hydrogen is oxidised to H2O.  Any sulphur in the fuel on oxidised to sulphur dioxide, SO2.   Air is about 80% nitrogen.  In very high temperature processes, there are even oxides of nitrogen produced, but for the most part, nitrogen just goes along for the ride, absorbing heat from combustion, and carrying quite a bit of heat up the stack.  The oxidation reactions all release heat, so the combustion products are hot.  This heat crosses the heat transfer surfaces to boil the water in the boiler.  But all the heat cannot be transferred, heat is only transferred from a hotter substance to a cooler one.  So in the end, after the maximum possible heat transfer, the whole mass of combustion products is hotter than the steam we are producing, and it goes up the stack carrying all that heat with it.  Obviously a significant heat loss from our process.  Now this would be bad enough if we only used enough air to exactly supply the requirement for combustion.  Unfortunately, it is very difficult to mix the fuel and air sufficiently well to achieve this, so even very sophisticated burners require excess air, just to achieve complete combustion.  It is not efficient to allow unturned fuel to go up the stack.  But perhaps more importantly, insufficient air means the burning only partly proceeds.  Rather than getting CO2 plus unburned fuel, we find carbon burns to carbon monoxide.  While CO2 cannot sustain breathing, it is in fact produced in our bodies during breathing, and so long as there is sufficient oxygen available, it is not particularly harmful.  However, carbon monoxide is quite toxic in even small concentrations.  This is why it is necessary to have good ventilation if we have combustion indoors.  Preferably a chimney over the fire place, as there is usually a tiny amount of CO formed in any combustion, and we cannot afford to allow its concentration to accumulate.

You may suggest leaving out the nitrogen and just burning fuel in oxygen to reduce the heat lost by the nitrogen being heated on the way through.  Separating the nitrogen is not practical for two reasons.  First, it takes a huge amount of energy to separate them, so that would not help our efficiency.   Of course in a model, we could cheat a bit, and buy in some liquid oxygen, and not count the energy used "by others" to separate it from air.  That brings up the second reason.  It would be very dangerous.  Even a slight increase in the oxygen content of the atmosphere causes combustible stuff to burn much more vigorously, and makes even stuff we normally consider non combustible burn quite well.  Given a high enough oxygen concentration, even iron, steel and concrete will burn.   I have been told that if there is a fire in an oxygen plant, there is nothing to clean up afterwards, even the concrete foundations are destroyed.  Fortunately, I don't have direct experience of this.  I think we had better just accept that nitrogen in the air will reduce our efficiency.  Perhaps 20 to 30% of the energy in the fuel is lost up the stack in a typical boiler.  Even more in a hand stoked coal fired boiler, as the practicalities of stoking mean that there is usually even more excess air than found in say a gas fired boiler.

Staying with the boiler for the moment, apart from the stack gases, the other obvious source of heat loss is the loss from the furnace or boiler casing.  If we hold our hand near the boiler we can feel the radiant heat.  If we hold our hand above the boiler, we can feel the hot air rising due to the convection losses.  This is one source of heat loss we can do something about.  If we insulate the furnace casing well, we can significantly reduce these losses.  It is even easier in a marine fire tube boiler, where the combustion is completely contained within the boiler.  We can insulate the shell quite well with as much thickness as the installation will allow and really minimise these external losses.

A slightly shorter post this time, but a good place to pause.  Next time, I will follow through the steam cycle and look at where energy is lost around the engine.  There should be some items there that we can attend to, to increase our engine efficiency.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 30, 2017, 02:47:55 PM
Hi.Good to see you back.....talking about time.......Using a year would be interesting as a leap year would through out the calculations !!  Also on efficiency a couple of pics and text from the Engineer in 1881 using the superheat to heat up the exhaust from the HP to LP cylinders......
Title: Re: Talking Thermodynamics
Post by: MJM460 on August 31, 2017, 12:08:55 PM
Hi Willy,  while I like the getting away from it all, it is also good to have communication, so good to be back and glad to have you on board again.  I guess the main disadvantage of adopting a year as a time standard is that approximate quarter day that gives rise to the leap years.  Any standard quantity has to have an agreed constant value.  Perhaps the second or the caesium atom vibration is the best after all.

Interesting to see that the reheat cycle was used so long ago, and to see just how it was implemented in those days.  The reheat cycle is used in modern power stations, where the exhaust from a high pressure turbine is piped back to the boiler, and through a secondary superheater coil, before returning to the intermediate pressure turbine.  My old textbook even shows the intermediate turbine exhaust through another reheat coil in the boiler before returning to the lp turbine.  The heat input to the reheat circuit increases the energy available for the next pressure level and so increases the power station output and efficiency without requiring more fuel.

Last time I was looking at where the heat goes in a boiler, this time let's look at the losses in the engine.  We have already looked at how the work is generated during the power stroke of the engine, but where are the "losses" in the engine?  Remember, the law of conservation of energy says the energy is not lost, it just does not get converted to work.  There are basically two quite different loss mechanisms.  First there are the processes that result in heat passing through the engine without being converted to work.  Then there are processes that use the work already created within the engine before there is any excess at the output shaft.

Heat loss from the cylinder, is heat from the steam that does not contribute to the engine output.  We have talked about reducing this loss by cylinder cladding, or even jacketing.  Cladding, being totally simple and passive is probably the most useful approach.  Jacketing is more complex, and also consumes steam that just may be more effectively used inside the cylinder.

Leakage past the piston is another area that steam is consumed without being effectively converted to work.  We can work on the piston to cylinder fit, but generally piston rings seem to be the most commonly applied design to reduce blow-by without introducing too much friction.  I will leave it to those with more expertise to describe how best to make effective rings.  If, like me, you are not up to making good rings, you can do much worse than just machine the groves as though you were going to insert rings.  Sometimes a soft packing is inserted, but the ring grooves are helpful even without it.  The flow past the piston is caused by the pressure difference between the power stroke side and the exhaust side of the piston.  This pressure difference accelerates the steam flow through the gap.  When grooves are machined in the piston, the flow expands into the groove, generally slowing but with lots of turbulence and a minimum of pressure recovery normally seen in a well designed Venturi.  More energy is consumed in accelerating the flow again into the next section of the gap.  The expansion losses occur again at the next groove.  The result of all the lost energy in each groove and the energy to reaccelerate the flow means the resistance to flow past the piston is much higher than for a plain piston, so the flow is much less.  It is called a labyrinth piston.

In case you think this would not work, I can assure you that it is used in labyrinth piston compressors.  These machines are used when the machines must be completely oil free.  The piston is supported by the cross head and piston rod so that it does not touch the cylinder walls.  Helped by a vertical configuration.  Even hydrogen can be compressed to quite high pressure s with these machines, though they do have a very large number of grooves.  Search for labyrinth piston compressors, and perhaps the Sulzer site as a manufacturer of these machines.  If you can't make good piston rings, it can't hurt to include two or three ring grooves anyway.

In a similar way, rod packing leakage reduces the output of the under side of the piston.  We usually insert a soft packing to minimise this leakage.  In a similar way, if you make some packing ring grooves they reduce the loss a little, even without the packing, which can introduce a lot of friction in a small engine if too tight.

It is worth looking at all these sources of steam blow by, but in the end they are not the major component of that 92% we are looking for.

The other main loss in the engine occurs after the energy is successfully converted to work, the work is used within the engine, so not available for doing external work.

I expect others will think of more areas where energy is lost before it is converted to work, otherwise I will look at the work lost within the engine.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on August 31, 2017, 01:08:32 PM
Thanks MJM for re-awakening my brain regarding labyrinth pistons, I shall chase them further on the web. I am currently pondering methods of eliminating friction from my small twin cylinder Gauge 1 Lion loco, 3/8 D pistons and contemplating the viability of replacing the slide valves with outside admission piston valves as the load on each of the tiny slide valves at 60 psi is 2.2 kilos and on a surface area, (underside of valve), of 0.06  sq. inches. Glad the discussion is rolling on again. Trust you enjoyed the dry country down South and got to see the bright lights of Thargomindah. Regards Paul Gough. 
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 31, 2017, 09:09:22 PM
Hi is there optimum shapes for these labyrinth grooves and do they get smaller towards the middle ? There is quite a lot of info in my old books from the 1830's on, and it is surprising the amount of knowledge that was available all those years ago to engine builders .However the pecuniary considerations tended to outweigh the actual finished engines that were produced. What would be the best wood be for cylinder lagging btw and are there tables the same as for metals regarding conduction etc etc ?
Title: Re: Talking Thermodynamics
Post by: steam guy willy on August 31, 2017, 11:54:18 PM
I have just read through your posts now and was wondering how efficient is the human body ?? in Asia where the POW men ate a lot of rice and not much else ,they still managed to do a hard days work? Vegans also manage to become successful athletes without any meat etc so how does one man power compare to 1 horsepower ? so are food calories equatable to coal calories ?  just wondering !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 01, 2017, 01:58:55 PM
More on efficiency -

Hi Paul, good to have you back.  Unfortunately I missed Thargomindah, passed by a bit north of there.  Was probably closer some years ago when I did some work in SA.  Wife says we had to bypass it as the coffee might not be as good as at Gregory.  With your little valves, I think the normal force is 60 x 0.06 = 3.6 lbf or about 1.6 kgf for each valve, so 3.2 kgf for two valves, it seems quite large in proportion to such a small engine, but it is a well known disadvantage of slide valves.  But the friction force which is at right angles to the normal force is smaller.  Steel on steel has a friction coefficient of about 0.3, but a well lubricated bronze surface should be significantly less.  Search for friction coefficient for various surfaces.  Multiply the normal force by the friction coefficient to get the force necessary to move the valves.  The point will be whether you can make piston valves in such a small size which seal well enough to be more efficient than the more simple slide valve.  I would also ask if inside admission piston valve would be better balanced for forces, involve less sealing issues with the gland packing and less rod end force influence.  I never got to see a real labyrinth piston up close, but the principle is sound, so I always put a groove or three in my small pistons on the grounds that it should be better than plain pistons.  It should be possible to make a spare piston for an engine and test both ways, but I feel the difference would be masked by the inevitable difference in diameter that I would almost certainly produce.

Hi Willy, my Heat transfer text book has a table of thermal properties including conductivity for metals and a separate table for insulating materials.  Pine, fir and spruce are all listed as 0.15 W/(m.K), while oak is listed as 0.19, and cork 0.042.  This suggests that any timber will do, if you compare with figures of 80 to 110 for copper and brass.  Just use something that will take a nice finish if the wood will be visible, but use cork if it will be covered by a metal sheet, and make it as thick as aesthetics will allow.  You are quite right to notice that the human body is in another league for efficiency, but it is still subject to the same laws of thermodynamics.  Not all the energy in the food is converted into work.  Some goes out as heat in breath temperature, breath humidity, or perspiration, or conduction to air and quite a bit is used just to breath, think and circulate blood, and digest the food.  Meat is a good source of protein necessary to build muscle, (vegans must get this from other foods), and is also good for sustained energy release.  But energy is more quickly released from carbohydrate foods such as rice.  And the body is incredibly good at conserving energy particularly in time of drought, no doubt why food restriction dieting is so difficult, our body just says why, we might be hungry tomorrow?  And then closes down to conserve as much energy as possible until more food is available.  But those POWs also paid a terrible price in terms of weight loss, far beyond what was compatible with good health.  I have a conversion Ap on my iPad that says one food calorie is equivalent to 1000 calories.  It makes sense in terms of the published calorie values of foods and the energy we produce.  My heat transfer book has quite comprehensive conversion tables but does not include food calories.  It does however consistently include in brackets (thermochemical) after calories and Btu in each description of energy units in the table, which is perhaps a pointed reference to the "nonstandard" energy unit used in those other disciplines.

I don't have a reference to a standard manpower, though I know it varies in a wide range over the population.  I was in Fremantle when the Americas cup was sailed there, (for work, really) and a university professor and his team attempted to answer the question by building a machine with a similar action to the coffee grinder winches on the boats.  They carted the machine around various pubs and challenged willing guys as to who could produce the most power.  The machine successfully gathered data until the races were over and the real winch grinders joined in.  The machine had to be rebuilt several times before it was strong enough to absorb the power of those guys.  They just blew it apart at every attempt.  Similarly, cyclists are known to have a good power to weight ratio, not the power of winch grinders, but not as heavy and so were usually favoured in the early experiments on man powered flight.  But I don't know the actual figures of power produced.  Of course, I am in another league, and don't expect to be called up as a winch grinder or for man powered flight anytime soon.

Regarding the labyrinth groves, as far as I know they are simply square groves like empty piston ring grooves.  The idea is that there is no Venturi effect with the associated pressure recovery when the gas slows and expands into the groove, the velocity energy is mostly lost in turbulence, which of course heats the gas a little, increasing the volume that has to accelerate onto the next space.  The energy lost adds up with each groove, so reduces the bypass flow.  Besides those labyrinth piston compressors, the similar labyrinth configuration is used on the shaft of steam turbines, gas turbines and on centrifugal and axial compressors, however for a really effective seal modern manufacturing techniques allow dry, non-contacting mechanical seals with really low leakage, and low power loss, so the labyrinth seals tend to be more of a first stage, especially for flammable or hazardous gases.

Didn't make much progress with the losses in engines today, but a little discussion around these interesting questions is helpful, so I really appreciate the questions.  Perhaps tomorrow.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 01, 2017, 05:33:44 PM
Hope I'm not deluding myself on the figures I presented: Valve dimensions, 0.457 x 0.185 = 0.084 sq. ins. x 60 psi = 5.04 lbs.
Underside of valve (or load bearing surface): 0.084 minus area of steam passage (0.285 x 0.091 = 0.026), thus 0.084-0.026 = 0.058, rounded off to 0.06 sq. ins. Thus approx. 5lbs or 2.2kg load on 0.06 sq. ins. If I'm wrong, please help, my old brain can't see where.

Don't know that I could get inside admission type into the space available, the outside admission type seems just possible, maybe, subject to achieving satisfactory sealing. I am thinking of one of the engineered 'plastics', perhaps a graphite embedded teflon or nylon or some such for the piston heads and valves. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Admiral_dk on September 01, 2017, 08:12:08 PM
Human Power - I kind of thought I knew, but ..... ended Googling it and according to Wikipedia https://en.wikipedia.org/wiki/Human_power (https://en.wikipedia.org/wiki/Human_power)

Quote
A trained cyclist can produce about 400 watts of mechanical power for an hour or more, but adults of good average fitness average between 50 and 150 watts for an hour of vigorous exercise. A healthy well-fed laborer over the course of an 8-hour work shift can sustain an average output of about 75 watts.

Quite a range - from not very much to rather impressive.
Title: Re: Talking Thermodynamics
Post by: Maryak on September 01, 2017, 10:45:27 PM


Last time I was looking at where the heat goes in a boiler, this time let's look at the losses in the engine. 

Heat losses Marine Steam Plant

(http://i389.photobucket.com/albums/oo340/Maryak/Heatloss.jpg)

Sometimes a picture is helpful.

Regards Bob
Title: Re: Talking Thermodynamics
Post by: derekwarner on September 02, 2017, 01:02:41 AM
Whilst a little simplistic on labyrinth grooves [for model steam engine piston valves] , I posted the following on August 6/2017 on Mammod Steam Models site
____________________________________________________________

To take this further here is some light reading

https://www.google.com.au/url?sa=... FQjCNEkXGWRYQuZJVuLfNb_X_LrJiyyrw

However for some further understanding we go back to Gas Laws..

 P1 V1 = P2 V2...

 P1 = steam pressure of the system
 V1 = the annular volume in the piston valve spool
 V2 = the annular clearance volume between the valve spool and the valve body

So we have P1 & V1 doing their work  with each stroke created by the eccentric movement

Then the same P1 travels along the minute [a ~~0.001"] annular clearance path V2, then the steam literally falls into the larger volume of the groove

This pressure drop then creates a reduced value of P2 in the groove........

Since we are talking relatively low steam pressures here, the value of P2 may well approach that of atmospheric pressure [1 Bar] and hence the body of steam in V2 will partially condense and simply shuttle back & forth as a lubricant and being contained by atmospheric pressure

The reason this steam does not spray out is that the new pressure P2  is in ~~ balance with atmosphere 

Of course, a little condensate will migrate out along the valve spool .....thus maintaining the Gas Law balance

Full sized higher pressure steam turbine engines utilize similar multi Labyrinth sealing arrangements
______________________________________________________________________________________________

Derek
Title: Re: Talking Thermodynamics
Post by: paul gough on September 02, 2017, 06:18:02 AM
Derek, I keep getting an invalid URL notice if I try to go to your 'light reading' site. Could you check there is no typos etc. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: derekwarner on September 02, 2017, 07:27:09 AM
Sorry Paul........here is the full extension thread link....I don't understand why copy & paste did not function???....

As you can see, the actual link at the very end ...FQjCN is the same leadin  ........Derek

http://redirect.viglink.com/?format=go&jsonp=vglnk_150433320089612&key=190cd6dc27c66f8edf4cd0d943583f3b&libId=j72x4w9101000abj000DAb8zr06z2&loc=http%3A%2F%2Fmodelsteam.myfreeforum.org%2Fabout97187.html&v=1&out=https%3A%2F%2Fwww.google.com.au%2Furl%3Fsa%3Dt%26rct%3Dj%26q%3D%26esrc%3Ds%26source%3Dweb%26cd%3D1%26cad%3Drja%26uact%3D8%26ved%3D0ahUKEwi-pam4ocHVAhXGG5QKHfMdCBoQFggmMAA%26url%3Dhttps%253A%252F%252Fen.wikipedia.org%252Fwiki%252FLabyrinth_seal%26usg%3DAFQjCNEkXGWRYQuZJVuLfNb_X_LrJiyyrw&ref=http%3A%2F%2Fmodelsteam.myfreeforum.org%2Fforum7.php&title=The%20Unofficial%20Mamod%20and%20Other%20Steam%20Forum%20%3A%3A%20Piston%20Valves&txt=https%3A%2F%2Fwww.google.com.au%2Furl%3Fsa%3D...FQjCNEkXGWRYQuZJVuLfNb_X_LrJiyyrw
Title: Re: Talking Thermodynamics
Post by: Steam Haulage on September 02, 2017, 09:53:03 AM
Hi All,

Labyrinth seals
Please note my liberal use of ‘may’ and ‘perhaps’ in the following.

May I suggest that the concept of the labyrinth seal goes back further than we might appreciate. Perhaps even before Jerónimo de Ayanz y Beaumont, Thomas Savery, James Watt et al were conceived.

It may be that the idea goes back into the mists of time in the joinery trade of our forefathers.

I have seen many examples of grooves in timber door frames. Before the adoption on a large scale of plastic as a construction material almost every frame would have a semi-circular groove cut into the hinge, lock and top of the frame. Apprentice joiners were instructed by the journey man they worked under that this was to allow winds to enter any gap and then reverse its direction and so prevent the draught from entering the inside of the building.

When I was a lad this groove was in common use, and visible in every house in which my family and relations lived, as well as in every door-frame at the school, built in 1909, which I attended in the fifties.

Steam Haulage

Title: Re: Talking Thermodynamics
Post by: MJM460 on September 02, 2017, 02:24:52 PM
Hi Paul, now I see where you are coming from on the area and force on your slide valve.  I don't see myself as in any position to arbitrate on right and wrong, but I would tackle the problem slightly differently.  I will leave it to you to decide on whether it is helpful.  The slide valve has two sides, just like the piston.  The steam chest side is easy, steam chest pressure over the area of the valve gives your 5.1 lbf forcing the valve against the valve face.  The other side of the valve is more complex.  The pressure in the valve cavity is a bit higher than exhaust pressure, higher by the pressure required to drive the exhaust flow through the port to the exhaust system, so we can estimate the net force on the valve over the exhaust cavity area as a bit above exhaust pressure.    However the sealing faces of the valve are more complex.  I suspect that there is a thin film of steam and oil separating the faces however slightly, and the pressure in this film varies perhaps roughly linearly between the valve chest pressure on the outer perimeter, and the valve cavity pressure around its periphery.  On this basis, I would estimate the average pressure over the sealing face is about half way between steam chest pressure and valve cavity pressure, tending to lift the valve, so tending to partially balance the pressure on the steam chest side.  Then, as I mentioned yesterday, the valve rod does not have to resist the normal force, but only the friction force, so now we have to estimate a suitable friction coefficient.  I hope this helps.

Yesterday, I mentioned inside admission, because the steam pressure forces on the piston valve are better balanced in the inside admission arrangement and the rod sealing and unbalanced rod forces are determined only by exhaust pressure.  However the challenge of making piston valves with sealing rings on such a small engine would be totally beyond me.  I suspect that you would be better to stay with the simplicity of slide valves, and if necessary, increase the piston diameter slightly to provide the extra force.  Avoiding the issue, I know.

Thanks for joining in Admiral_dk, glad to have you on board.  I think you have answered yesterday's question on human power.  Now that you have put the figures in, I seem to remember being on the tread mill for a stress test after a heart attack many years ago, closely monitored by a cardiologist and all the instrumentation, and being told that the load, which was provided by a small generator was about equal to a 60 watt globe, but fortunately I only had to keep it up for about 15 minutes.  But definitely at the low end as you might expect in the circumstances.  For an 8 hour day, I suspect your 75 watts is about a reasonably good effort when it has to be kept up all week.

Hi Maryak, thanks for another of your excellent diagrams.  It is very helpful in showing all the losses in proportion.  Interesting to note that the final figure driving the ship is only 6%, but I suspect in model sizes, the mechanical losses are a bigger proportion.  It is one of those wrinkles in the maths, that if we say halved the mechanical losses, it would not make much difference to the overall efficiency, but all the reduced loss would appear as extra output power so would have a big percentage effect on the out put power.  That is what is behind my intent to look at some of the losses that we can do something about.  At the end of the day however, as your diagram shows, the major losses are in the heat contained in the exhaust, followed by the heat carried up the stack.

Hi Derek, it is good to see someone else having a go at explaining some of the many little puzzles we come up against in every design.  If we can nut out the theory, it will definitely help inform out experimentation.  As with Paul's problem, I would take a slightly different approach, and would start with a diagram showing where conditions are known, specifically the top and bottom edges of the piston.  This pressure drop forces the flow past the piston.  Each time the flow encounters a groove, it slows down, but with this geometry, no pressure recovery.  There is then a small pressure drop to supply the energy to reaccelerate the flow into the next section of the annulus, followed by a steady pressure loss towards the next flow.  The pressure cannot get below exhaust pressure, but is always a little lower when there are grooves, so restricting the leakage flow a little.  I hope that helps your thinking a little.

Hi steam haulage, good to have you aboard also.  A truly interesting little piece of history around those grooves.  I expect the journeyman learned the explanation when he was the apprentice and so on back to time immemorial.  I suspect neither he nor his apprentice or their forbears really understood the explanation.  I suspect that we can prove using the law of conservation of energy that the wind cannot be turned back with enough energy to flow back on itself, however that does not mean there is not a labyrinth effect there that means the groves might reduce the draft a bit, and that would always be welcome.  I hope that I have continued your most appropriate use of perhaps and may.

Once again, time and words have run away, so perhaps next time for looking at what we can do on our engines.

Thanks for reading along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 02, 2017, 03:09:41 PM
Hi I have now spent all day looking at doors in Medieval city of Norwich and have not found any of these grooves yet !!! In the previous posts there is mention of grid iron valves these were used as there was less movement in the valve so contributing to greater efficiency ,however this is from the 1881 book...so would this be correct. A local engine builder uses PTFE  'rings' on his piston valves and found that the diameter should be slightly less than normal rings due to the expansion of the PTFE. lots of good info coming on this site. keep it coming !! Also this was the first i have heard of gridiron valves in the last 70 years !! How would the efficiency compare with a diesel engine in a ship ?? or a steam turbine ?
Title: Re: Talking Thermodynamics
Post by: derekwarner on September 02, 2017, 04:05:21 PM
Gents......my earlier posting in Mamod Models from August the 6th, was in answer to a question asked by another member as to what the two fine grooves on a model steam engine piston valve spool by way of the gland area could be?

I simply offered a thought and basic explanation of the pressure drop that occurs when a series of labyrinth grooves are used in fluid sealing applications

As you have noted MJM, the pressure reductions contained/constrained by a series of labyrinth grooves cannot exceed 'exhaust pressure', so again considering the model steam engine application, and from my recent trials of excessive making of condensate  :mischief: one could expect exhaust steam pressure [within the system] to be at the physical exhaust point be approaching atmosphere

In this low pressure [say 2 to 3 Bar] application, the resulting steam/condensate [as liquid] trapped in the labyrinth grooves will tend to act as a lubricant and buffer against physical displacement outside of the gland area

As mentioned, on my return to NSW, I will conduct the series of tests with the larger flow capacity exhaust system.... Derek

Derek

   
Title: Re: Talking Thermodynamics
Post by: Maryak on September 02, 2017, 11:10:21 PM
Hi All,

In full size turbines the labyrinth glands on the turbines are balanced to atmosphere with gland steam.

Often LP turbine admission is at the centre, (to balance axial thrust), with steam flowing out to the ends hence the glands are not much above condenser pressure. Without sealing steam being supplied to the glands condenser pressure would tend to rise towards atmospheric pressure causing a drop in plant efficiency.

Basically all the glands in the turbine set are interconnected at their inner and outer pockets and the gland steam system is balanced to maintain minimal leakage of gland steam from the the outer pockets. The fine adjustment valve shown below is in modern plant connected to a controller which automatically maintains the system in balance over varying load/speed ranges.....stop, ahead and astern

(http://i389.photobucket.com/albums/oo340/Maryak/GSsteam_zpsj9amcicy.jpg)

Regards Bob
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 03, 2017, 12:36:20 PM
Hi Willy, I wonder if the door detailing, being passed down from master to apprentice, became a regional thing.  I am not sure just what a grid valve looks like, however if the design enabled a shorter stroke, it would reduce the work done by the valve rod.  Remember the definition of work, force times distance.  Even with the same normal force on the valve and the same friction coefficient, half the stroke means half the work dine by the rod and half the parasitic power lost to the engine output from this cause.  A possible answer to Pauls dilemma is to reduce the valve travel, though there are practical limits on such small models.  Valve rod forces have always been known to be high however.  The early joy valve locomotives were I believe known for breaking the con rod due to the side force from the valve linkage.  Or perhaps it was just poor design.  My text book has worked examples for the efficiency of steam plants with 2 and 4 MPa operating pressure, turbine drives and 10 kPa exhaust pressure, quite good vacuum.  2 MPa gave about 30%, 4 MPa gave 35%.  Interestingly, the reheat example gave less than 1% more, but is favoured in steam plants because it gives much drier steam at the low pressure turbine exhaust, so less blade erosion by the condensate droplets.  I think reciprocating machines are a bit higher efficiency than turbines, but they are limited to much lower speeds and power outputs.  My somewhat unreliable memory recalls large industrial Diesel engines could achieve about 30% when I was using them for compressor drives.  In a simple open cycle gas turbine, about a third of the energy goes out the exhaust as tonnes of hot gas, a third is consumed by the compressor at the air inlet, and a third is available as output power, very roughly.  Again this is from using them as compressor drives.  As jet aeroplane drives, there is no mechanical output, apart from that consumed by the compressor on the front end and the fuel pumps, the power inherent in that mass of gas exhausting at the back end provides the thrust. 

We do not get this performance from our steam models due to the low boiler  pressure and normally atmospheric exhaust.  I am not advocating that we try and emulate these pressures, let alone the much higher pressure used in modern industry.  A typical petrochemical plant has steam systems operating at 10 MPa, while utility scale power plants are built to supercritical pressures, that is above 22 MPa.  Steam at any pressure is dangerous, and these very high pressures are extremely so.  Fuel costs for a model are very reasonable as the running time is generally low.  We do not need to go to those extremes in the pursuit of fuel efficiency, but it is interesting to work out just what we can do to increase the power output within the limits of our operating conditions.

Hi Derek, my little Mamod engine is a single acting oscillating engine.  I am not familiar with the design of piston valves for such small engines.  I am sure the small size makes them quite different from the common designs for the larger locomotives.  I really can't picture the grooves in the gland area that you are talking about.

Hi Maryak, gland systems can get quite complex as shown by your diagram.  But I am sure that conditions in the boiler room were unpleasant enough without extra steam leakage from shaft seals.  At the end of the day however, you have reminded us that labyrinth seals still leak, just a bit less than a plain shaft or cylinder.  A well made mechanical seal of appropriate material is still the far better solution.  I am sure that perhaps carbon rings from that material used for Stirling engine power Pistons could be made for this size, but until my skills develop, I turn the grooves as they should be better than plain pistons.

Back when we were all sidetracked by the mention of labyrinths, I had talked about some of the ways heat is lost before it is turned into work, so next I think it worth talking about some of the ways we lose work output by consuming it in the engine before it gets to the output shaft.  It always seems a bit unfair when there are so many limits to how much of our energy can be converted to work, that we then lose a significant fraction of the work we do produce to friction within the engine.  The labyrinth discussion was really prompted by the desire to reduce friction between the piston and cylinder without losing too much work due to steam bypassing the piston.  Piston rings are intended to reduce the flow, but at a cost of causing friction.  Making good rings is a challenge for a beginner, for me, anyway.  Of course a packing can also be used with a bit of judgement on how much can be added without making it too tight.  Some people advocate using o-rings, and whatever works is helpful, though it was not what O rings were designed for.  My guess is that they are initially a touch tight, and some rubber wears away to leave the clearance "just right", when they are better on a small model than any poor fitting alternative.  The concept of an abradable seal is actually used in full size compressors as an alternative to machining the seal to perfectly fit each shaft, so there is even full size precedent, even if it is in shaft seals.  Cylinder lubrication helps reduce friction, and Derek has already mentioned that the moisture in wet steam helps lubrication.  That is why it becomes more critical to use the right oil if we have a lot of superheat, so dryer steam at the exhaust with less condensate.

Then, in no particular order, we can look at bearings, main bearings, big end, and small end.  The cross head guides, and all the pivots and slides of the valve mechanism, or port face and pivot on an oscillator, all involve friction, which takes some of our valuable work and turns it into heat.  I am not suggesting that this friction can be eliminated, but attention to alignment, clearances and lubrication all leave more work available at the output shaft.  In the extreme, some modern compressors have magnetic bearings where the electronic controls vary the current to electromagnets so the magnetic forces to keep the shaft cantered in the bearings despite fluid load and unbalance of a high speed machine.  Incredibly low loss, but the electronics are mind bending, and I doubt they will be practical for model engines in the near future.  I don't know if anyone is experimenting with improved bearing design or lubrication for models.

That's enough for today, I still want to summarise the effect of the exhaust system, though the detail has probably been well enough covered.

Thanks for dropping in

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 03, 2017, 01:10:49 PM
 Hi MJM, this is the pic of the gridiron valve......It does seem to have a lot of metal to metal contact, so although it has a shorter stroke does this cancel out the friction losses? also in the pic the copper expansion joints ?? thanks for the input .........
Title: Re: Talking Thermodynamics
Post by: paul gough on September 03, 2017, 04:57:27 PM
Just quickly, part of my reasoning for outside admission piston valves is to have very short exhaust passages straight out of the top of the block and into the blast pipe which is co-incident with its centre, an initial attempt in concert with reducing the piston clearances to about ten thou to reduce exhaust and compression volumes. The block cannot be enlarged nor can boiler out put be increased without an extra wick on the metho burner, whilst this has been done previously it is forcing the tiny boiler, so larger dia. cylinders with larger steam demand is out. There is no room to move, so to speak with the Lion, it is about as small as one can go in Gauge One and still have a loco that is not a pain to operate. Putting twin inside cylinders and an eccentric driven water pump off the main axle underneath was quite a task, so I am left with trying to come up with ways of improving performance within the existing dimensional constraints. The 6 1/2 lbs of force on the 3/8" piston working against the 5 lbs load on the valve, plus any other imposts does not give this little engine much in the way of reserve or very good slow running. Labrinth piston heads and piston valves are an exploration into possibilities and I had considered combining these with the use of engineered plastics, if that is the correct term, for these components if an appropriate grade can be got. I understand Aster used Rulon for their piston rings in their piston valve engines, I understand it is a teflon with powdered mica filler, but is a bit soft for a piston head. So not only is the physical design of the engine important but trying to find alternatives to traditional materials is part of the game in situations where things are approaching functional limits. Anybody who has any ideas or experience with teflon, nylon, or like derivatives for use as pistons with steam is very much encouraged to assist. I had not considered carbon and know nothing of its characteristics for use with steam, again any info. welcome. Thanks MJM for reminding me that reducing the travel of the valve is also a valuable thing to chase if achievable. Sorry about this not so quick post. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 04, 2017, 01:33:30 PM
Hi Willy, sorry that I missed the significance of the grid iron valve drawing in your earlier post.  If I understand the system, there are separate inlet and exhaust valves, so presumably two cams and linkages.  But I can see how it reduces the stroke of each valve.  Friction is interesting.  The conventional model of friction is independent of area, just depends on the total normal force.  More area means less pressure on the surface but the total friction is assumed to be the same.  With the grid plates on edge like the drawing, even the weight is hardly a big contributor.  It the steam force pressing the valve on the face is F, then the friction force to move the valve along the face is friction coefficient times F.  It does not depend on area.  As I have mentioned to Paul, the friction coefficient for dry steel on steel is generally given as 0.3, so the friction force is only 0.3 times F.  For a very smooth, well lubricated surface, the friction coefficient might be 0.1 or less.  Now friction is hard to measure, because it tends to have a stick-slip motion.  It takes a bit bigger force to start the movement, but then a smaller force is enough to keep it moving.  So the coefficient is at best an approximation, but it does allow design to proceed.  It is possible that the area has a small secondary influence, but it is usually ignored.  Sometimes the friction coefficient is quoted as both a static friction value, the value to get movement started, and dynamic value, the value once movement begins.

Hi Paul, now I can see what you are doing.  Thank you for your clear description.  Sounds like a great project.  I can only admire your watch making skills.  Will you be posting a build log or at least some pictures?  It looks like a project where, if you can use a millimetre better than others, you have made a major breakthrough.
Remember though, the valve rod does not have to work against the 5 lbs.  The rod force will be the friction coefficient times that force, and with a little oil or even condensate lubrication, together with a nicely smooth surface, the rod force can be expected to be nearer 0.5 lbs.  It may be worth rigging up a little test rig to get a rough measurement.  You can't simulate the temperature, or pressure on the sealing face of the valve, bit you will get an idea.  Regarding sealing rings, I can imagine that you have very little room.  Generally in the applications I am more familiar with, pure teflon is considered too soft, with too low a melting point.  It softens and spreads rather than take the load.  A filled teflon, is normally used for higher load bearing properties.  Graphite or other fill materials give different properties, but availability tends to override for hobby applications, so some experimenting with the available materials in a small stationary engine rather than risk damaging your amazing locomotive.  Carbon is quite fragile and so would require a two piece piston (or ring)  construction if that is practical in your design, but is used in pump and compressor seals to quite high temperatures, so may be worth a try.   At least the experiments could be carried out with much less effort in a separate single cylinder stationary engine, allowing more trials of less mainstream ideas.

The last area of loss that I wanted to list was, as Maryaks recent heat energy diagram showed (see post #242), the heat that goes straight through to the exhaust as latent heat of the exhaust stream.  We have to condense all of the exhaust totally to water before we can pump it back to boiler pressure, or simply reject it to the atmosphere, either way, the heat is not converted to work.  That diagram nominated a figure of 70% of the fuel energy.  Regardless of the accuracy of this exact figure, it shows that it is a major factor.  Can we do anything about it?  It is low temperature heat, so not easy to use elsewhere.  Perhaps as a boiler feed water heater if we have a continuous boiler feed pump.  Perhaps a little in preheating our fuel, or even the incoming air, or for avoiding the pressure loss in butane, but the majority is unavoidable loss.  The giant cooling towers attached to modern power stations attest to the limited alternative uses for this heat apart from some very specific circumstances.  I have for example worked in plants where the turbines exhaust to a low pressure steam system, around 100 kPa(g), sacrificing some work output, but using the exhaust heat for process heating purposes.  These processes tend to be relatively constant, only minimally dependant on weather conditions, but if the exhaust steam was not available, fuel would have to be burned anyway to supply the process needs.  When you do the calculations on how much extra fuel you have to burn to drive the turbine, you get the turbine power at an efficiency of around 80%.  Much higher than can be achieved by any isolated power plant, no matter how sophisticated.  But it is really only an accounting exercise, the process requirements reduce the fuel consumption attributed to the turbine power requirement, but do not actually increase the conversion of energy in the fuel to work. I am sure the process people attribute a very poor efficiency to the turbines and take credit for using the "waste" heat to reduce the fuel costs attributed to their process.

I think that covers most of the directions energy goes other than conversion to work, so I would like to look at the thermodynamic limits on how much of the fuel energy can possibly be converted to work in an ideal power plant, which leads to different measures of efficiency.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 04, 2017, 02:15:34 PM
Hi just wondering about waste heat and thermocouples ?? at least it should power a transistor radio for the fireman to listen to the latest action from Headingly, !!! is there any info about this anywhere ? It could even illuminate pressure gauges and things on dark nights on locomotives !! I can remember in the Practical Wireless Mag from the fifties a design of a thermocouple connected to a Hooka for people to listen to the radio in India,!!
Title: Re: Talking Thermodynamics
Post by: paul gough on September 04, 2017, 06:48:15 PM
Thanks for your comments MJM. Funny you should mention utilising a stationary engine for test purpose as only last week when discussing my 'headaches' with another I mentioned how I might be driven to getting a twin horizontal stationary engine so as to test two alternative materials at the same time and measure their relative performance.. This little project is likely to run into years, as there are other challenging things that please me to tackle. 'Adequate' lubrication rather than the general flood of oil that is often a messy characteristic of these little engines  and multiple orifice blast nozzle so as to eliminate back pressure, just to mention two. One of my 'madnesses' is to develop a mechanical lubricator for Gauge One size locos. Desktop size engineering is somewhat challenging and should keep my old brain from rusting up even if the body fails and manufacturing things becomes beyond me.

 Your exploration of things theoretical presents aspects for consideration that are often unknown or overlooked and I am very pleased to be able to read your words and also the responses and questions they provoke. This thread is a unique opportunity to begin to understand what is happening  and progress the ideas of miniature mechanicians. Thank you. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 05, 2017, 01:06:14 PM
A theoretical limit-

Hi Willy, that is a good idea.  It is called the Peltier, or Seebeck effect.  Both are reversible in the sense that if you pass a current through a pair of junctions in one direction, one gets warmer and the other cooler.  Reverse the current and the hot and cold sides are reversed.  Reverse the heat flow, supply the heat instead of electrical energy, and you get a current.  Small units are readily available as drink coolers/heaters in camping and car accessory shops.  Discreet units are sometimes available in electronics outlets, and certainly on line which can be heated to get a current.  They are not very efficient but we will come back to that.  Thermocouples only give about 5 mV for 100 deg C temperature difference, or is it microvolt?  Either way you need a lot of junctions.  But some combinations of semiconductors also work, and much more efficiently.  I found a reference which implied about 6% energy conversion to electricity.  Does not sound impressive, but if 70% of our burner energy goes to the condenser, and we convert 6% of that to electricity, that makes 4.2% of the fuel turned to electricity.  For a small model that is much more than the engine output, so more than doubles the output of our plant.  Worth a few experiments perhaps.  Of course we do have to carry the heat away from the cool plate, or it soon approaches the exhaust temperature and electricity stops.  Still need a river or a cooling tower, or perhaps a good heat sink and use some of the power to drive a fan so air carries away the heat.  Try looking for Peltier or Seebeck, or Peltier-Seebeck effect in your preferred search engine.  Similarly if you have one of those three way camping fridges which have a low power absorption refrigeration cycle, it may be possible to replace the heating element with your exhaust steam as a heat source, and keep the beer cool.  Not sure if it could be scaled up to full size plant.  In industry, refrigeration plants using absorption cycle are sometimes used when there is a large source of low grade heat together with the need for refrigeration.  But to be economical, they are normally huge.

Hi Paul, thank you for your kind words of encouragement they are much appreciated, as are all the questions which keep the thread rolling along.

My thinking on a small test engine was to avoid the work and inevitable damage involved in disassembling a locomotive to do many tests.  I thought a single cylinder mill engine style would enable easy manufacture and installation of the relevant test components, so speed up the learning process.  The valve gear can be a simple eccentric for piston ring and piston valve tests, or as complex as you like to test valve gear variations.  It is a test instrument, not a historical model, so optimise it for easy changing of components.

Engineers wanting to sell their engines were not really impressed with low efficiency, and wanted to claim higher figures.  Sadie Carnot was looking at how high the efficiency of an engine could be.  I think it had previously been concluded that 100% was not possible, it was determined by the maximum and minimum temperature of the cycle, but what was possible?  He came up with a theoretical ideal (even though not very practical) cycle which he was able to show would have the maximum possible efficiency, which he showed could be calculated as 1 - Tl/Th.  This is a fraction which can be multiplied by 100 to give a percentage.  The temperatures are expressed in absolute terms so K or R , depending on your preference.  Tl is the low temperature at which heat is rejected, or condenser temperature.  Th is the maximum temperature at which heat is supplied, the boiler pressure for saturated steam, or the superheater outlet temperature. 

We could calculate the efficiency of a Carnot cycle engine operating between our cycle temperature limits.  That 4 MPa boiler we looked at earlier, had a superheater outlet temperature of   400 C or 673 K.  The condenser temperature for 10 kPa would be 46 C or 319 K.

We can calculate the efficiency of a Carnot cycle between these temperatures as 1 - 319/673 =  0.53 or 53%.  Now remember the thermal efficiency of the actual cycle was 35%, If 53% is the thermodynamic limit of what can be achieved, we can perhaps say our cycle has 35/53 = 66% of the Carnot efficiency.  Sounds a lot better than 35%, doesn't it?  We could do a similar calculation for our model.  But it also raises the question, why not build our plant to operate on the Carnot cycle?

I will try and look at that next time

Thanks for your interest and encouragement

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 06, 2017, 01:07:32 PM
After yesterday's post, I must apologise to M. Carnot for spelling his name wrongly.  Nicholas Leonard Sadi Carnot lived from 1796 to 1832, and published his work around 1824.  A salient reminder of how our life expectancy has changed since then.  If a Carnot cycle engine has the highest possible efficiency for an engine working between specified temperature limits, why not make an engine working on the Carnot cycle?  The truth is that it is not a very practical cycle.  It produces very little work and it would be difficult to make one that would overcome its own friction even.  It's only real use is to quantify the maximum possible work that can be obtained from a given temperature difference against which you can compare the performance of a real engine.  But all is not totally lost, there is actually one real engine that has the same maximum efficiency as a Carnot engine working between the same temperatures.  It is the Stirling cycle, which many forum members enjoy.  It has disadvantages in terms of power for its size unless you have a pressurised engine with sophisticated seals, but it is inherently a very efficient cycle, especially if the the regenerator is included.  It's efficiency is illustrated by the engines which produce measurable power from a tea light candle in a well known competition.  Try that with a Rankine cycle! 

A Rankine cycle, while still described in terms of ideal processes, is a much more practical cycle as history has demonstrated.

There are several reasons why a Rankine, or normal steam cycle cannot match the efficiency of a Carnot cycle engine.  The boiler heat is not all transferred at a single maximum temperature.  Most is transferred at the boiler temperature then the temperature raised to the maximum which is only reached at the superheater outlet.  In fact, the heat transfer starts at the outlet of the feed water pump or feedwater heater, which further reduces the average temperature at which heat transfer takes place.  Then we force the heat transfer by a large temperature difference to achieve the required heat transfer from a small area.  This forcing produces vigorous bubbling and circulation in the boiler, which improves the heat transfer coefficient, but absorbs energy which cannot then be converted to work.  So it is nothing like an ideal reversible heat transfer process.

I don't think there is much more needs to be said about the Carnot cycle, however when it comes to engines, there are other ways to compare efficiency which I will introduce next time.

Thanks for dropping in

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 07, 2017, 12:43:20 PM
Engine efficiency -

Last time I talked about Carnot efficiency.  While the cycle is impractical in terms of designing an engine, its strength is in indicating a theoretical upper limit to the power that can be obtained from any given cycle and its maximum and minimum temperatures.  I should have noted in passing that you cannot use a temperature lower than the surrounding atmosphere, unless you cheat by ignoring the energy necessary to produce that lower temperature.   Hence the usual minimum temperature is usually around 15 - 25 deg C, although you would have an obvious advantage if your engine is in Northern Canada, Siberia or even wintering down in Antartica.  Satellites in outer space have an even bigger theoretical advantage from this point of view. Of course for steam cycle engines, the freezing point of water provides another limitation on the minimum temperatures.  However for most practical applications, the lower temperature is pretty much fixed by our location on earth, and to get higher efficiency, we must go to ever higher steam temperatures.  A quick calculation shows why we need supercritical boilers to achieve anything near 50%

However engine manufacturers wanting to sell their product don't want their performance figures reduced by cycle conditions outside their engine.  So they like to advertise their machine efficiency by comparing with what an ideal machine could do based on the same inlet and exhaust conditions.  Now a suitable process for engine comparison is the adiabatic cycle.  Remember, no heat transfer in or out during expansion.  We can calculate the efficiency of an ideal adiabatic machine based on the actual inlet and exhaust conditions.  We then conduct a performance test for the real machine to determine its power output.  I have witnessed many compressor tests where the process is similar.  They go to great lengths to reproduce the specified guarantee conditions, though in the case of compressors, there are recognised procedures for producing equivalent conditions from which the performance at the real conditions can accurately be calculated.  This is done where for example the gas being compressed is flammable and it would be difficult to conduct the test safely on the test stand.  The the efficiency of the machine is the actual test result divided by the output of that ideal adiabatic machine and multiplied by 100 to express it as a percentage.  Real steam turbines are in the region of 80%, perhaps higher, especially for very large machines, but reciprocating machines, I don't have real information.  But our models?  I suspect friction is a bigger proportion of the potential output power in a model compared with a full size engine, so our figures are certainly much less.  This is often called adiabatic efficiency.

Remember that this approach required a test of the actual machine.  To determine the adiabatic efficiency of our model, we need to do a performance test.  This requires a load which we can suitably instrument to measure speed and torque, from which we can calculate the power.  We also need to measure the steam inlet temperature and pressure, and the exhaust pressure, and preferably exhaust temperature.  I generally run my engines unloaded.  The entire load is the internal friction in the engine.  I still have to make a suitable load to conduct a power measurement on a loaded engine.  Measuring rpm is easy with a digital non-contact tachometer which is readily available at electronics stores.  It may also be a suitable project for Picaxe or Arduino microprocessors for the electronics enthusiasts among us.  I think the best way to measure the torque is to use a little digital scale, which these days are available with a resolution of 0.1 gm.  But I still have to design a suitable machine.  The issue is easily understood from the earlier discussion on the torque produced by a reciprocating engine, it fluctuates.  In a single cylinder engine, from zero to double the average, twice each revolution.  I don't really believe the text book drag brakes can really give a meaningful reading without considerable damping.  I am thinking in terms of perhaps a model ship propellor in a container suspended on bearings so we measure the torque on the container rather than balance the whole boiler plant on a scale.  A little DC motor used as a generator may be even better, though it would still have torque fluctuations.  Perhaps it can be calibrated so the current and voltage output can be measured and power calculated.  Any suggestions for a practical small brake are more than welcome.

Next time a closer look at the adiabatic power output from typical model operating conditions.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 07, 2017, 10:53:39 PM
Looks like you'll have to move from a classroom to a lecture hall soon! 9096 reads for 120 days, 75/day,(up from 66), thats a lot of bums on seats. Thermodynamics and associated goings on can't be as daunting or esoteric as might be thought. Beyond doubt now that this thread is providing succour to an enthusiastic and curious crowd. Bravo! Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 08, 2017, 01:08:44 PM
A short one today -

Hi Paul, thanks for those encouraging words.  Have we really been going 120 days?  Time really flies when you are having fun.  I do hope that everyone is enjoying the ride, and learning something as we go.  I know I am.

That may seem like a strange statement, but when you are used to having all the test results from a well instrumented test stand, and have real information from the machine designers, it is all pretty straightforward.  When you attempt to see what you can deduce about a model from some limited home instruments, it can be a different story.

I have available results from a few runs of my engines, but I have accumulated instruments as I went along, finding out what I need as I go, as we all do with our tooling.  So inevitably many of the test results are incomplete, however I do have some that allow me to deduce a surprising amount.  Grand father duties have kept me quite busy so I have been a bit slow doing the calculations.  But I will keep working at it.  I need to fill out a few extra lines in my steam table with some intermediate values.  Then with a few results from a simple model test, I will show you what can be done in the next day or so.  I can also see quite a test program as my next project.

Thanks for looking in,

MJM460

Title: Re: Talking Thermodynamics
Post by: MJM460 on September 09, 2017, 10:59:22 AM
An engine test

Hi everyone, calculations all sorted out.  On a model test, the actual work output is quite small as you would expect, but to calculate it requires subtraction of two much larger numbers.  Consequently the large ones must be as accurate as possible, and no short cuts or back of the envelope estimates allowed.  Regardless of accuracy, which is another issue for later, I now have a set of calculations using numbers from a test of my own engine and they all make sense.  So first, what did the test involve, what calculations are involved, then most importantly, what do they mean?

The engine for this test is my first slide valve engine, all to my own design including the boiler, though obviously the concept can hardly be claimed to be original.  It is shown in the attached picture, and there is another view in the engine show case from soon after I joined the forum.  It is 12 mm bore and 20 mm stroke, and double acting.

The boiler is methylated spirits fired, it is about 2 in diameter (a standard copper tube size), 200 mm long, with four water tubes underneath.  It has a sheet metal furnace casing made from tin plate cut from a large coffee tin as a trial run for my sheet metal skill development.  Obviously looks very rusty now, so not worthy of showing on this forum, which is why it is covered in the pictures.  A new furnace casing from stainless steel or brass is on the list, and it will be fitted with some insulation.  The steam outlet (3/16 in tubing), winds back into the combustion space where there are two full turns around the space before it exits the casing.  I fill the cold boiler through a funnel, and insert a plug which protrudes 25 mm into the boiler and carries a thermocouple for boiling temperature measurement.  From the temperature, I get the steam pressure from the steam tables.  With no feed pump, run times are limited, but I can calculate steam rates by weighing the water fill, and the water I extract with a syringe when it is all cooled down.  The fuel burns out before the water level gets too low.

The steam pipe connection the boiler to the engine is insulated with a silicone tape, and could be improved.  It has a displacement lubricator and a thermowell for a thermocouple on the engine inlet.  The exhaust outlet also has a thermowell so I can measure the steam inlet and exhaust temperatures.  It then has an oil/condensate separator with a drain that I have described previously, and a vertical exhaust stack.

The burner is perhaps described as a semi vapourising type, something along the lines of a Trangia camp stove burner, a bit more vigorous than wicks, but needs development to get a bit more heat.  I would like to build one of the so called silent types, but apparently everyone knows how to do that so they are never described in articles or books that I have read, and I still don't know how.  The burner tank holds about 50 ml of Meths, and I calculate the quantities by weighing before and after a run which is just over 20 minutes from light up to extinguished.

I use the kitchen digital scale (resolution 1 gm.) for weighing water and fuel, a multimeter or two with thermocouples for temperature and the most recent addition, an electronic temperature meter with no voltmeter functions, but it has two thermocouple inlets which I can read either one at a time or as a difference.  Even to 0.1 degree, if I could get the rest of the procedure sufficiently repeatable.  I also have a non-contacting infrared thermometer which is good for comparative readings around the place, and a non-contacting digital tachometer for engine rpm.  The missing item is a means of measuring torque.  However that is not required for the test I am describing.  The engine is just free running without any external load.

I will come back to accuracy another time, just using the readings as they come so far, to see how the whole setup works.  Though I have compared the thermocouple readings at the ice point and boiling point and they are looking pretty good.  Readings are generally passing the "looks reasonable" test, and some consistency checks, so I think quite adequate for illustrative purposes.  Better accuracy can come later, but a few more repeat runs to ensure the results are repeatable is probably the main requirement.

To conduct a test, I fill the boiler with a weighed quantity of water, and fill the fuel tank which I also weigh.  Tighten the plug, set up the thermocouple instruments and record the time when I light up.  The boiler temperature takes about 6 min to get to 100 C.  No cylinder drain valve on this one, so I gently turn the engine over by hand to blow out the condensate which heats the cylinder while the boiler temperature rises.  Within about a minute, the engine runs up quite nicely and shortly after the boiler settles at about 116 C which the steam tables tell me corresponds to 0.175 MPa absolute, or about 75kPa gauge.  For the metrically challenged, this is about 10.9 psig.  At this pressure the engine runs at about 1100 rpm.

Nothing unusual in any of that, except perhaps for the instruments.  But it is probably worthwhile describing it so you can see the areas where compromise and approximations are necessary.

My notes say I filled the boiler with 200 g of water and extracted 58 g after it cooled, so 142 g evaporated.  Similarly 42 g of fuel used.

From light up to engine running was 6 min 40 seconds and I extinguished the flame after a further 8 min of engine run time.

I read the boiler temperature, engine inlet and exhaust temperatures and engine rpm often enough to judge typical figures, and runs seem pretty steady.  So I have extracted the following -

Boiler temperature 116 C.
Engine inlet 138 C
Engine exhaust 104 C
RPM 1100

It doesn't look like much but you may be surprised to find out what it can tell us.  Next time I will use these figures to get a picture of the engine thermodynamic performance.

Thanks for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 10, 2017, 11:31:57 AM
Some calculations -

The test was actually done some time ago.  Now you have an idea what was involved, and we have some results, so time to set out some calculations. 

Now don't be too intimidated by that, they are little more than simple arithmetic.  A bit tedious to do by hand, but when I started, I had no computer, and in fact no calculator.  And no, I am not that ancient.  I had a slide rule for multiplication, which together with a pencil and paper is enough, but I suggest trying that way these days is wasting time that could be better used making swarf.  In fact, the calculations are quite repetitive, so the best way to do them is in a spreadsheet.  Once we work out the formula once, it is copied, then pasted into each place it is needed.  The only trick is what calculations to do, and finding meaning in the result, and that is what this post is about.  Well perhaps this one and one or two more.

We will use the first and second laws of thermodynamics, and here is where we find the usefulness of those calculated properties, enthalpy and entropy that we discussed earlier.  With the measured results, specifically temperature, we can look up the other properties in the steam tables.  The tables list properties on a " per kg" basis, or per lb if you are using imperial tables.  We then multiply those values by the steam and fuel rates of our engine.  So let's work those out first.

The fuel burned was 42 g in 14 min 40 seconds, this is an average of 2.86 g/min or 0.0477 g/s.

The heating value of the methylated spirits is 26000 kJ/kg.  This is the lower heating value as it assumes the water produced as a product of combustion is not condensed, but goes up the stack to atmosphere as steam.  As an aside, it allows for commercial meths, in this country anyway, being 5% water, and this water must also be turned into steam, and also the vaporisation of liquid meths, which only burns as a vapour.  So we can multiply the heating value by the consumption rate to get the energy input into our boiler.  Now for model purposes, kg and kJ are very big units and all the calculations are cluttered by too many zeros.  So note that 26000 kJ/kg is the same as 26000 J/g.  Then 0.0477 g/s x 26000 J/g = 1240 J/s or 1240 watts is our heat input from the fuel. 

Now, I usually assume that this average is also the uniform rate of fuel consumption, a bit rough, but I have checked by stopping when steam pressure is achieved, extinguishing the flame and reweighing the burner.  It seems about right based on that point anyway.  More complex to design an experiment to give more detail.

We can also estimate the steam production rate.  Remember that steam production started after 6 min 40 seconds.  The heat during this time goes into heating the water, the boiler shell and the furnace.  We could calculate the heat required for the water and the boiler shell separately, but it is not of major interest, but once these items get to steam temperature, they don't absorb more heat, so we can ignore them and assume the remaining heat goes into steam production, and all the steam produced is produced in that 8 minutes of engine running.

This does not all go through the engine as some is condensed in heating the piping and cylinder, but it left the boiler as steam, so it is still steam production.

We had 142 g of steam produced in 8 min, so if we assume  an even rate we have 17.75 g/min or 0.296 g/s.  Given the assumptions, I probably should have called it 0.3.  Certainly three significant figures is not justified, but it helps with making calculations consistant.

With these basic measurements that we can all do, we already have a rate of fuel consumption, boiler heat input and steam production.

The next step involves looking at steam tables to determine the enthalpy and entropy of the steam.  From these we can then work out the boiler efficiency from how much heat from the fuel ended up as energy in the form of steam, and we can calculate the work an ideal engine would produce, and even how much work our engine produced, though that requires further explanation later.

The steam tables I have do involve quite large steps, so they do not have the precise temperatures or pressures that my boiler is operating at.  Steam properties are also published in graphical form, a common one using temperature and entropy as the axes.  Another uses enthalpy and entropy, and some even use pressure and specific volume.  They are hard to use with any accuracy, but they show the general trend of the information.  A close look at any of these will show that most properties appear as curves, with the only straight lines appearing in the two phase region where liquid and vapour both exist in equilibrium.  Fortunately, when finding the value of properties between the ones tabulated in the tables, it is normally considered accurate enough to assume the properties are linear between any two consecutive listed values.  This means we can use a simple linear method to construct a table with the additional values we need.

The boiler temperature of 116 deg is actually listed in the pressure tables, (well actually 116.06, but surely close enough) where the equilibrium pressure is 0.175 MPa.  Yes the tables uses MPa instead of kPa, however it is easy to multiply the small numbers we need by 1000, so no real problem, and no interpolation needed.  We can directly see the enthalpy, (and entropy of we needed it) of both the saturated water (hf), and the saturated steam (hg) at this temperature and pressure.

The pressures are absolute, so 0.175 MPa is 175 kPa absolute, say 75 kPa(gauge).  For the metrically challenged, about 10.9 psig.  The boiler is good for 700 kPag, and in any case has a properly set safety valve which I ease with the pliers before starting, so no problems there.

So my boiler temperature tells me the steam pressure in the boiler is 75 kPag, and the air, which causes additional pressure is soon expelled as soon as I let a bit of steam out of the boiler.  Still a gauge would be nice to have.

Next we have the engine inlet temperature of 138 deg C.  This is after the superheater, (actually even after the steam pipe and lubricator) and so we have at least 22 deg C of superheat for those who thought our super heaters were little more than driers.  You will see that it is enough to be useful.

My superheated steam tables unfortunately do not include 0.175 MPa, nor do they contain 138 deg C.  As well, they are in deg C not K, as 138 deg C is 600 K, and that might have been included.  You might be luckier when you do a test.  Worth finding tables listed each way if you come across them.  This means we have to interpolate the superheat tables, first to get a table for 0.175 MPa, then again to get values for 138 deg C.  Not very easy to work to a desirable superheater outlet temperature, as it would require adjusting the length, and even then would probably not achieve the same result at another boiler load or firing rate.

This might all be a bit tedious for those that already know how to do it, but I want to bring along those that have never used a formula in a spreadsheet as well, so please bear with me.  But enough words this time and interpolation next time

Thanks for looking in patiently.

MJM460
 
Title: Re: Talking Thermodynamics
Post by: paul gough on September 11, 2017, 02:13:44 AM
A typo that might confuse a novice from paragraph on heat input from fuel. "So note that 26000 kJ/kJ is the same as 26000 J/g." Regards Paul.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 11, 2017, 04:44:17 AM
Thanks, Paul.  Well spotted. 

I read it all at least three times, but things still slip through.  Often Apple makes very puzzling corrections, but that one was surely all my own work.   Assistance in finding these things is always welcome.

It is now corrected.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 11, 2017, 11:37:35 AM
I notice the exhaust pipe is also lagged ,Is that so you can collect the condensate from the bottom of the chimney ??
Willy
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 11, 2017, 12:37:51 PM
Interpolation

Hi Willy, welcome back.  I hope the show went well.  Two reasons for the exhaust pipe insulation.  First, it reduces the risk of burning my fingers, it is still over 100 C.  Second, quite the opposite of collecting the condensate, it reduces condensation in the exhaust pipe, so the vent steam is more dry, and so less hot  droplets raining down.  The dry steam soon mixes with the air and is dissipated before it condenses.  Remember that condensation involves rejection of heat, the opposite of evaporation in the boiler.  On the other hand, the oil does drop out in the separator and seems to mostly end up in the drain tin.

Following on from yesterday, we need to do three interpolations of the steam tables to get values for the missing temperature at the missing pressure, but it's not too hard with a spreadsheet.

I start by putting the missing temperature into the nearest two pressure tables in the superheat section of the steam tables, that is 0.1 and 0.2 MPa.  So let's start with the 0.1 MPa table.  In my table the nearest temperature below 138 is 100, and the nearest one above is 150 deg C

I set up a little table like this in my spreadsheet-

T1,  v1,  u1,  h1,  s1
Tx,  vx,  ux,   hx,  sx
T2,  v2,  u2,  h2,  s2

Please read the 1, 2, and x as subscripts, and obviously T1 is 100, v1, u1, h1 and s1 are the values at 100, Tx is 138 and the x subscripts refer to the values at 138 which so far are unknown, and the 2 subscripts refer to the values at 200.  Now, I transcribe the known values listed in the tables into just this small table.

Then I construct a formula for finding ux.  (I will get back to vx shortly)

ux = u1 + (Tx- T1)/(T2- T1) x (u2 - u1)

I type this formula into the cell for ux in the spreadsheet and when it is complete, I exit the cell.

Remember in a spreadsheet, a formula starts with an "=" sign.  You select the cells you want to use with the mouse and the spreadsheet will insert the cell reference rather than the number within the cell.  Use all the signs, including brackets, but no spaces.  In Excell, Open Office, and others you exit the cell by pressing enter.  In Numbers, on an iPad with a touch screen, you touch the green tick.  And the answer appears in the cell as the number you are looking for.  Obviously it should be a bit above 1/3 of the way between the values for u1 and u2 as a rough rationality check.

If you are not familiar with the procedure it may seem a bit tedious, but now comes the powerful bit.  You can copy that formula and paste it into as many places as you wish.  But there is more, as they say on TV.  The spreadsheet shows the cell references for the cells you select, perhaps B3 etc., but it remembers them as relative positions, eg cell above, cell below, cell two to the left etc. from the formula location.  This means you can copy it to another place, providing all the cells you want are in the same relative locations.  But you can change this by using absolute references.  This is indicated in Excell by a $ sign.  You can do three things.  With one $ sign $B3 means always column B, but the same row relative to where our formula is located.  B$3 means always row 3 but the same column relative to the formula cell location.  Then with two $ signs, $B$3 means always cell B3, regardless of where on the spreadsheet the formula is located.

In this case, the temperatures are always in the same column regardless of whether we are calculating v, u, h or s, so the cells in the formula which refer to a T1, Tx or T2 need to be edited with $ signs to fix the column, for example $B3.

I can now copy my formula for ux and paste it into the cells for vx, hx, and sx, and job done.

You may have noticed that v is smaller at T2, while all the others are getting bigger.  The same formula is used, the required negative sign is already there in the term (v2-v1) as v2 is the smaller value, so no problem at all.

Now the same process to insert a 138 deg row in the 0.2 MPa table.  Then I set up a third table consisting of the two 138 deg rows just found, and with the pressures 0.1, the required 0.175 and 0.2 in the first column instead of temperatures.  With the pressures in this location, the formula remains the same, so I can copy and paste the same formula into this third table, and I now have the table I need for 138 deg C and 0.175 MPa.

I hope that if you have been reluctant to use formulas in a spreadsheet, you might now feel encouraged to try.  That first one may well have taken longer than a pencil, paper and a calculator, but in this exercise I copied that formula, then pasted it 15 times with 15 mouse clicks and job done.  All of a sudden the initial time was well spent.  (The extra four were for that 104 deg exhaust temp.). Don't hesitate to ask a question if this is not clear.

So here is what we have so far -

Boiler 116 deg, water and vapour together means 2 phase region, so use saturated steam pressure table which gives P = 0.175, vf = 0.001057, vg = 1.1593, hf = 486.99, hg = 2700.6, sf = 1.4849, sg = 7.1717

Engine inlet 138 deg superheated, pressure assume equal boiler pressure = 0.175 MPa, interpolation gives v = 1.167, h = 2745.91, s = 7.3018.

Engine exhaust 104 deg, p = 0.1 MPa ( actually atmospheric, assumed 100 kPa), v= 1.72, h = 2684.2, s = 7.380

Units, Pressure MPa, (= 1000 kPa), specific volume, v, m3/kg, enthalpy, h, kJ/kg (= J/g), entropy, s, kJ/kg.K (= J/g.K).  K is for Kelvin, the absolute temperature unit for metric units.

Well, that is the tedious part, we have the figures for both steam rates and steam properties, but what do they mean, and what can we learn from them?

Next time, I will apply the first and second laws of thermodynamics to the boiler and to the engine for some interesting and even surprising results.

All eyes open for any typos that need correcting please,

Thanks for looking in.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 12, 2017, 12:18:32 PM
Calculated Steam properties for test and Boiler performance -

Another small correction to the post about two days ago, the total fuel burning time was actually only 14 min 40 seconds, so the fuel burned was 0.0477 g/s and the heat released 1240 watts.  I have now corrected that post.  Another case of old eyes dropping a line, or something like that.  Please let me know if you see any others.

All of those figures we gathered in the last post can be a bit daunting and hard to picture, so it will be helpful to portray them graphically.  All our readings were actually temperature, so we will use that for the vertical axis.  The second law will have us looking at entropy, and it will be useful to have that as the second (horizontal) axis.  This Temperature - Entropy diagram is quite commonly used to portray a plant cycle.  Pressure, specific volume and enthalpy can then all be shown on the diagram, but better not add too many or it will get very cluttered.  You can also illustrate your cycle on pressure - volume or even enthalpy - entropy if that makes things clearer to you.

I have hand drawn the diagram attached, but tried to keep everything in the correct relative positions.  The bell shaped curve is the boundary of the two phase area where water can exist as liquid and vapour at the same time.  It is bounded by a horizontal straight line at the bottom below which water will be solid.  To the left of the curve, water is liquid, and to the right, vapour.  Above the top of the curve, called the critical temperature, which has only one pressure, the critical pressure, water changes from clearly liquid on the left, to clearly vapour on the right with no clearly distinguishable boundary between, unlike the distinct liquid, vapour two phase region that we are so familiar with.  I should point out that the peak of the bell curve is not correctly shown in proportion, it should be at 22.09 MPa and 374.14 C, the critical pressure and temperature for steam.

I have drawn the constant pressure lines at 0.175 MPa and also for 0.1 MPa to illustrate our steam plant process.  The 0.175 line should extend to the left and down to room temperature, but that part is only important before we start generating steam, the heat up period, which was about 6 min 40 seconds.  So point 1 is the saturated water at vapour pressure for 116 deg C.  Point 2 is the saturated vapour condition, steam 116 deg C just at the point superheating starts.  Point 3 is actually the engine inlet, but I am assuming close enough to the superheater outlet, as the line between is insulated.  Point 4 is the engine outlet temperature measured in the test.  We cannot easily calculate this condition, it can only come from a test.  In full size, the manufacturers have some pretty good computer programs, so they can predict the performance for guarantee purposes, but in the end, those programs contain an efficiency which comes from a large number of previous test runs of similar engines.

So the obvious question, how have I arrived at points 4 and 5?  Let's look at the thermodynamic analysis of this process.  For the boiler, the first law says Q = h3 - h1.  That is, the heat transferred into the boiler during steam production is equal to the difference in enthalpy between the superheater outlet and the saturated liquid point.  I will come back and put some figures in that shortly.  Then for the engine, where we want to determine the work output of an ideal adiabatic engine, the second law of thermodynamics says s5 = s3 for an ideal adiabatic engine.   An adiabatic engine is used as our standard for comparison because no real engine can produce more power than an ideal adiabatic engine.  I know the exhaust pressure of both the ideal engine and our model, it is open to atmosphere, so it is atmospheric pressure.  In a perfect world I would have measured it with a calibrated barometer, but I didn't, don't have one, so I am assuming it was 100 kPa (absolute) or 0.1 MPa.  Two independent properties are enough to define all the properties of steam using the steam tables. 

Note that point 5 is shown inside the two phase region.  The value of entropy from point 3 at 0.175 MPa, when applied to the 0.1 MPa line in the steam tables confirms that to be the case.  In that area, temperature and pressure are not independent, as the is only one possible pressure for each pressure where liquid and vapour are in equilibrium.  However, temperature and entropy are independent so are sufficient to completely define the steam properties.

On the engine exhaust, I only measured the temperature.  However, the pressure is the same as at point 5, at atmospheric pressure, i.e. 0.1 MPa, where the saturation temperature is known at only 99.6 C, so our measured temperature of 104 deg means our actual exhaust is superheated.  In that area, pressure and temperature are independent, so sufficient to completely define the steam at the exhaust.  You will notice on the diagram that my engine exhaust is shown with higher entropy than it had at point 3.  The second law says that no engine exhaust entropy will be less than the inlet, and only for an ideal reversible engine, will it be equal.  For all real engines, the exhaust entropy will be higher than at the engine inlet.

For the purpose of calculating engine performance, I need to calculate the enthalpy at points 4 and 5.  The first law of thermodynamics says for the ideal engine, W = h3 - h5, and for my real engine, W = h3 - h4.  That is the work output equals the difference in enthalpy between the inlet and the exhaust.  We can perhaps accept that for the ideal engine by assuming no friction, but our real engine requires further explanation.  We will get to that later.

Now we have all the information we need, and have defined our method, so let's put in some figures.

For the boiler, the heat input h3 - h1 = 2745.91 - 486.99 = 2258.9 kJ/kg ( or J/g).

My steam production rate was 0.296 g/s, so heat in the steam = 2258.9 x 0.296 = 668 J/s = 668 Watts.  I can compare that with the heat from the fuel burned, which was 1240 watts.  I can conclude that 668 J/s of that heat was transferred into the steam, the rest was lost partly up the stack, and partly through losses from the boiler casing.  That means a boiler efficiency of 54%.  I think perhaps acceptable for a simple boiler.  I was hoping for a bit more.  Perhaps I will be able to improve it with some casing insulation, and perhaps some experimenting with air flow.  In another post, I will look at the heat transfer surface area and see if I can calculate a steam production per unit of surface area to compare with what K. N. Harris suggests.  I suspect it will be lower than his as I think he assumes coal firing, which I would expect to produce a much higher temperature than my small Meths burner.

Surprising how long it takes to describe some of these things.  I hope it is sufficiently clear to illustrate the procedure, and provide enough guidance for anyone who would like to do such a test on their own engine.  Please ask if anything is unclear.  Now I think it's time to take a break, and calculate the performance of that ideal engine next time.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 12, 2017, 10:06:23 PM
Now this is really interesting stuff!!! These numbers start to give me a ball park notion of what might be going on in my tiny Lion boiler and give me more inspiration to clamber up the learning curve. Unfortunately my computer skill ,(I don't have any), doesn't include spread sheets,  so I'll have to fall back on the old laborious ways to work things out when I master the methods. I remember many many decades ago when I was involved in Herpetology and running around sticking quick acting thermometers up tiger snake cloacas trying to determine their preferred body temperature we were lent some thermocouples and a millivolt meter from the uni. Less intrusive for the animals than a glass tube and made the job a little easier. These thermocouples were hand made by the techy at the uni., I think they were silver and something and was wondering if you knew of the metals used in the ones suitable for our steam temperature ranges. It might be possible for model engineers to make them?? Looking forward to the results on the engine. What a great thing it would be if a whole bunch of people rigged up their models and took some measurements and analysed them, maybe some interesting things might result from a broad investigation on our small sizes. Especially things like the effectiveness of lagging, its minimum thickness etc. etc. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: derekwarner on September 13, 2017, 01:30:49 AM
Morning Paul......without digressing too far from the thread, I have found my $15.00 digital laser pyrometer absolutely invaluable in understanding temperatures in my steam plant...[with simple understanding of actuals]...[without these, all I knew was that things were hot :Mad:]

My boiler steam discharge valve presents as 135 degrees C, however this is an external reading spotted on the body of the valve - from the steam tables, the boiler at 3 Bar  is producing steam at ~~142 degrees C....so everything is relative in comparison]

I can trace the temperature to the lubricator body, the steam regulator, the steam inlet fitting to the engine....the exhaust and all the way back to the de-oiler] - monitoring gas temperature is interesting......including lagging of the gas line

One simple point I can confirm [before & after test] is that insulating the steam tube from the boiler to the engine, provides steam that is ~~3.?? degrees C degrees hotter that in the uninsulated state....[and this is over approx. 220 mm long run of 1/8" OD copper tube]

My interest in this area was that I was producing too much water condensate in the de-oiler and needed to understand what was happening. To this end I have increased the exhaust tubing from a combination of 1/8" and 5/32" to 1/4" x 0.014 full flow K&S brass

[The insulated tubes with the X's were the 1/8" & 5/32" exhaust tubes to and from the de-oiler which are now redundant]


Whilst we are thinking seriously about efficiencies, a $15.00 set of digital scales [the type available were 7kg max]  are essential for confirming the volume/weight of gas burnt......[my gas tank has a published volume of 105gm......I have managed to get 103gm inserted at 26 degrees C]

Apologies MJM for deflecting out of line.... :happyreader: ....Derek
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 13, 2017, 06:01:38 AM
Hi Paul, thanks for your encouraging words, and for being kind enough not to mention that it was not one of my best efforts.  Sometimes a day does not go quite right, even when building a knowledge base.  I am determined to do better on the engine performance.  But first your thermocouples.  Obtaining these is much easier these days.  The go-to combination in industry is Copper-Constantin for all except quite special temperature ranges.  I believe the junction is electrically welded, but no need to worry.  Nearly all reasonable digital multimeters have one supplied.  The temperature- voltage characteristic is well known, even laid down in international standards.  It is a bit non linear, but that is all built into the multimeter.  Probably less expensive for the whole combination than trying to buy a length of the wire in hobby quantities plus plugs to fit your meter.  I included a picture of one of mine earlier, and Willy included a picture of his which had a nice stainless steel sheathed probe.  I have to experiment to see how the probe goes with my thermowells, but it would be great for stack gas, as well as for Willy's coffee.  I even checked mine at ice point and boiling point to see if it was really correctly calibrated.  That is probably over the top, but all the ones I have tested have come up spot on, give or take the unavoidable experimental error involved.  (Actual atmospheric pressure for boiling point, plus the difficulties of getting a true equilibrium ice/ water mixture for the ice point.  It takes lots of ice, only enough water to let you put the thermocouple in.)  Tempted to include a "war story" about purchasing small quantities of thermocouple wire, but I believe this is strongly discouraged.  If you don't have s temperature scale on your meter, it is worth considering upgrading your multimeter, Christmas is not far away.  Avoid the cheapest.  If you look at the wire into the plug, some are very flimsy.  However if your multimeter has the temperature scale but no probe, Jaycar and Altronics both sell both types separately at a reasonable price.

Back to the ideal adiabatic engine performance - see if I can make it clearer this time!

Refer back to my little sketch.  You will remember that I measure the boiler inside temperature where steam is saturated and both liquid and vapour exist together, so the steam tables gave me the pressure of 0.175 MPa (absolute).  I assumed that pressure still at the engine inlet, as the pipe is short, and no throttle valve.  I measured the engine inlet temperature, 138 deg C.  It is above saturation temperature, so steam there is obviously superheated, so temperature and pressure are independent, and two independent properties are necessary and sufficient to determine all the other properties.  I interpolated the steam tables to get the remaining steam properties, particularly specific volume, enthalpy and entropy.

Paul, you will not be the only one not used to spreadsheets, so try following the procedure from post #267, one step at a time, to put in that first formula for interpolation, and let me know what point you get into trouble.  Spreadsheet formulae are so useful for so many purposes that I don't want to leave anyone behind on that one, even if you don't want to go much further.  If you have a computer with Office on it, try Excel, or you might download Open Office, (it's free).  On an iPad, or Mac, worth buying Numbers, but there are many others.  Lotus and Multiplan were two very early ones, they are all sufficiently similar until you try much more advanced functions.  I am sure that you will not regret the effort.

The engine exhaust is more tricky.  We know the pressure is atmospheric, I assumed 100 kPa, but the steam could still be either wet or superheated.  For my real engine, I measured the exhaust temperature as 104 deg C.  Now the steam tables tell me that for 100 kPa, or 0.1 MPa, the equilibrium temperature is 99.6 deg, so our exhaust is superheated.  Hence pressure and temperature are independent and hence sufficient.  Another row of interpolation of the superheat tables and I have my exhaust steam properties.

Real engine exhaust steam by interpolation -
T = 104, P = 0.1, v= 1.72, h = 2684.22, s = 7.380.


It's worth noting that if you don't have a superheater, or only an ineffective one, your exhaust steam will probably be wet steam, and we are stuck, but superheated exhaust is more informative.

Before we look at what that means, let's look at the ideal reversible adiabatic engine performance.  Again, the exhaust pressure is 0.1 MPa, but where are we on the curve?  This is where that property, entropy comes in.  The second law of thermodynamics says for an ideal adiabatic engine operating between points 3 and 5 (on my sketch), s3 = s5.  We already have s3 = 3.018 from interpolation of the steam table for the engine inlet, so s5 = 3.018.  Two independent properties, P and s, so the steam is completely defined, but how do we find those other properties?

First, look at the saturated steam table row for 0.1 MPa, and check the entropy columns.  You will see that our ideal engine exhaust entropy of 3.018 lies between the dry vapour value, sg, and the saturated liquid value, sf, which means that the steam is wet.  For we steam we can calculate a quality factor, or dryness factor.  Quality is another of the properties, but it only exists in the region between saturated liquid and dry saturated vapour.  My text book gives it the symbol x, and you can think of it as the mass fraction of the steam in the boiler which is vapour, the rest being liquid.  For our ideal adiabatic engine exhaust the dryness fraction, x is calculated as follows _

x = (3.018 - sf)/(sg - sf). We substitute the values from the steam table and then

x = (3.018 - 1.3026)/(7.3594 - 1.3026) = 0.99 or quite close to saturation.

Now we use the dryness fraction to calculate the other properties by simple linear proportion.  We really only need h5 so let's calculate it.  We look up hf= 417.46 and hg= 2675.5, then

 h5 = hf + 0.99 x (hg - hf) = 417.46 + 0.99 x (2675.5 - 417.46) = 2652.92

If you are using paper and calculator, you will notice that between hf and hg columns there is one headed hfg.  The definition is  hfg = hg - hf to save you one subtraction.  At one time, none of us had computers, or even calculators.

Now, still talking about our ideal adiabatic engine, we apply the first law of thermodynamics which tell us that the work done on the piston, W, by each kg of steam is W = h3 - h5 = 2745.91 - 2652.92 = 92.99, say 93.  The units used by the tables are kJ/kg, or J/g.

We multiply this by our steam flow rate, 0.296 g/s to get 27.53 J/s or Watts.

That is the work done by an ideal adiabatic engine with my steam conditions and flow rate.  No engine can exceed that, and any real engine will produce less!  Not very impressive, only 2.2% of the heat released by burning the fuel, but it cannot be exceeded, or even matched by any real engine.  No point beating myself about the head because I can't get 30%.

Now to the really interesting one, the exhaust of my real engine.  What can that tell me?

That will be our topic for next time.  This one has been a bit long, but where to split it?

I hope that everyone is still with me,

Thanks for following along.

MJM460

Title: Re: Talking Thermodynamics
Post by: MJM460 on September 13, 2017, 06:29:55 AM
Thanks Derek for those excellent photos of your beautiful engine and boiler.  Your post appeared while I was inserting mine.

No apologies needed, you are right on topic.  I believe I have at last a clue as to where your condensate comes from.  I am attending my granddaughters choir performance this evening, hence my early post.  I gather from the pictures that you might not have a superheater, so after the next post, I will have a go at calculating the exhaust steam properties, starting from saturated steam instead of superheated.  It might reveal something interesting.

I also have an infrared thermometer. It is very useful, but as you are aware, it has inherent errors so is best for comparative measurements.  The outside of a tube is always cooler than the steam inside unless the tube is well insulated, but then you can't use infra red.  However, the difference between ends of a tube is probably more than close enough to the difference in temperatures inside.  Interesting that the insulation makes 3 degrees difference.  I suspect that 1/8 is very small and a significant restriction in that engine, close to being a laminar flow orifice, like a domestic fridge.  You don't want to fill your piping with temperature elbows like I have done in your boat, but on the bench for test purposes, they can be very instructive.  However, replacing the plug in your boiler with a thermowell allows use of a thermocouple to check the gauge pressure reading any time, even in the boat, by just poking in the thermocouple, so I would think may always be worthwhile.

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 14, 2017, 12:51:54 AM
Thankyou Derek and MJM for the comments regarding thermocouples/multimeters and digital thermometers, things have certainly advanced in this area since the early seventies when I was getting preferred temperature ranges of reptiles and I will look around to see what is available.

MJM, I am interested to know what constitutes a good thermowell. I take it these are used where the temperature probe cannot be inserted into the medium for direct measurement through a gland nut arrangement. Obviously one wants to get the closest to the internal temperature as possible, so does one manufacture the orifice of the well to a push fit for the probe dia. and make the profile at the bottom the same profile as the end of the probe to get contact here also? Or alternatively, does one use a clearance around the probe and some proprietary filler that sets to gain the appropriate heat transfers. The design of a good thermo-well would seem to me to be an absolute necessity to avoid introducing errors and ending up with spurious results. Are there any other considerations or do's and don'ts regarding the acquisition of data and trip ups with instrumentation etc.

Derek your temps regarding insulation are interesting, however it would be helpful to know the material used and the thicknesses. From my experience in full size stationary steam facilities pipe lagging is quite thick, as a rough guide it was as large in diameter or often larger than the size of the flanges on the pipes and nearly always had sheet metal cowling around it. Now dragging from the depths of memory, (1960s), regarding steam locos, I think the asbestos lagging rope used around the steam pipes on NSWGR locos, eg. to the air pump, was about 1/2 inch, so the dia would be roughly twice that of the pipe. I wonder if lagging on models is adequate, have you done any tests on a pipe or a vessel with a set thickness then increased it by, x2,x3,x4 thickness to see what the optimal thickness of a particular insulating material is?
Title: Re: Talking Thermodynamics
Post by: derekwarner on September 14, 2017, 01:38:41 AM
Paul......

The detail of the model lagging for the 1/8" OD copper tube from the boiler steam isolation stop vavle to the engine is as follows

3.175 diameter tube.....add a 1.78 diameter section Viton o-ring [to act as a diameter guide] at each end of the spool, wrap 1.0 diameter cotton string around the spool [superglued at ends & occasionally at bends] , paste a wet premixed cellulose [Bunnings type] Polyfiller material into the string and to exceed the major diameter of the O-Ring...[this may take 3 or 4 coats of Polyfiller & sanding to achieve the oversize diameter]

After sand back to the uniform diameter, two coats of enamel primer, one coat of gloss enamel paint....so the insulation OD for this tube is ~~ 8mm diameter, or a nominal 2 mm wall thickness.

I have bent & set ~~ all tube spools to include 90 degree bends as this lagging material has very little resistance to bending and stress cracking will appear through the enamel top coat at bends if care is not taken...[this also requires the spool fittings have sufficient clearance when being dissembled so as not to require any moment of bending]

On area of concern is the non insulated fittings in the line of components [lubricator, regulator and the entry fittings to the engine itself].....my Scottish steam regulator is a 20mm bronze cube ...[I have lagged external faces of the regulator with timber planks] but it is clearly a source of heat loss...[image below]....these screwed of flanged fittings in the lagged steam line provide an excellent point for the temperature comparisons....being as such, provide totally repeatable set points 

For exhaust lines I have used the similar insulation relationship which is governed by the 1.78 section O-Rings for the correspondingly larger diameter tubes etc

I have not conducted any varying diameter lagging test comparisons, suffice to say I can place thumb & forefinger around the steam line directly below/from the boiler isolation valve [135 degrees C]....and the lagging OD is hot, but not sufficient to burn skin

Sheet metal sheathing over lagging on full sized steam applications is I suggest only to provide resistance to any form of mechanical abrasion, as wet pasted or preformed asbestos and latter synthetic insulation material is very soft

Lagging in full sized applications includes covering all flange & connection points or and inline component parts as these are potentially a source of gross heat loss

Derek



Title: Re: Talking Thermodynamics
Post by: paul gough on September 14, 2017, 07:55:13 AM
 Derek, thanks for the detailed reply. My experiences in four full size plants has seen most of the pipe flanges exposed, i.e. high temp hot water heating systems for a uni. campus, fruit juice processor and food manufacture, only the drug manufacturer lagged most of the flanging. All these were below 200 psi though and accept that these plants were not chasing efficiencies as one of their primary goals. Interestingly the uni underground distribution system of quite a few kilometres, (about eight from memory), had eight and six inch piping with no insulation, the ground was the insulation! This was sometimes helpful in winter, a leak could be found by going for a long walk following the route of the piping, a telltale wisp of water vapour or a relatively warm wet area somewhere near the spot would indicate the leak. My experience with small, other less 'organised' plants has demonstrated insulation is often a very neglected area or of no concern to the operators. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Zephyrin on September 14, 2017, 08:33:31 AM
some friends at my club use a ribbon of plaster for orthopaedic contention to hold a layer of insulating stuff around pipes, (just like on a broken leg !), a paint layer or a teflon band keeps away the moisture.
Title: Re: Talking Thermodynamics
Post by: Admiral_dk on September 14, 2017, 12:03:05 PM
Here in Denmark we have real central heating - ei. Aarhus where I work has 300,000 people living and the heating is provided from two plants for the whole city + some villages are ; one being the electricity plant and the other the waste incinerator.

"Water" leaving the incinerator plant is between 200 and 250 degree C and around 80-90 degree C when it arrives to the buildings some 10-100Km. down the line. I don't know if there is any heat exchangers along the way.

The pipes are insulated so that the outside diameter is tre times the metal pipe in the middle - local pipes are around 250mm. ~10" and the ones leaving the plants are 2-3 times that diameter.

Pipe integrity is tested continuesly with the help of two copper wires running parallel in the insulation for the whole length of the piping. An electric pulse is transmitted out through the wires and the reflection is measured. If the resulting reflection arrives back at the right time and has the correct shape, all is well and a new pulse is transmitted. Any dammage to the insulation results in an earlier reflection and the time from transmission to reception gives the exact distance to the dammage - so they know where to dig even though there might not be a visual leak at the site.

The pipes where not insulated in my childhood and that was also the reason the rusted and broke - plus no snow or ice on the surface over them - but even back then we never had the heat missing for more than a few hours and after the insulated ones I have newer experienced any breakdowns.

EDIT : I really should make a point about the fact that the heat would have been wasted / is a waste product from the primary function at the two plants + it is a nice in the winter where we are around -10 degree C.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 14, 2017, 01:40:57 PM
Hi Paul, I don't think the thermowell design is over critical, the main thing is to provide a way of inserting the element so it is surrounded by steam without compromising the pressure containment.  No problems with liquids creeping through the wires and their insulation etc.  The main issue with contact resistance is that it slows the response of the thermowell which adds to all the other issues that make temperature measurement inherently slow response.  Special quick response designs are available, but not really needed for our application.  The first approach is usually to put a few drops of glycerine in the thermowell to improve the heat transfer, but very messy in a model where we probably do not really intend a permanent installation.  Slow response alone does not affect accuracy, as in principle, there should be no continuous heat loss once the thermocouple is heated.  So the next issue as you imply, is conduction away from the fitting where it is in the atmosphere and conduction along the wires.  Some insulation around the insertion point in the piping is probably more important on our models as the insertion fitting is closer and more bulky relative to the thermowell depth.  In industry, vortex shedding induced vibration leading to breakage is more of a problem than minor heat losses.  I make my models along the lines of the industrial ones which are basically machined from a bar, and internally drilled for the thermocouple element.  Small ones usually screwed into the appropriate pipe fitting, but larger ones are flanged.  The sheathed thermocouples could be inserted through a gland, but I think the simplicity and reliability of a thermowell is a more practical solution.

You have prompted a few ideas with your insulation comment.  Very interesting to see the different approaches in different environments.  In hydrocarbon industries, insulation falls into two basic categories, hot and cold.  Cold insulation has to be right, and cover complete, otherwise ice breaks the insulation off.  Hot insulation is applied for safety reasons, to reduce the surface temperature so people do not get burned.  And of course also for heat conservation.  It is generally a compromise between what can be maintained with normal practice, and cost.  The cheapest part, and the most result in terms of reduced heat loss is the straight pipe sections, usually with special formed sections for elbows.  Much more fiddling required to cover flanges and valves etc. which are really a relatively small part of the total heat loss.  However there are generally also higher quality specifications which are applied when heat conservation is more important.  Generally we used to use magnesia, but Calcium silicate, foam glass and fibre glass are increasingly used, unfortunately.  I say unfortunately because they wick up any oil spills then become flammable.  Installations are almost universally outdoor, so metal sheeting is applied for weather protection, and also for preventing mechanical damage, which in plant sizes includes people walking on the piping.  Obviously some different considerations from those of well protected indoor installations.  Some experimentation on different thicknesses would be interesting, but might be best carried out on a test rig with a steam pipe say 1 metre long, so there is some chance of measuring the reasonably small differences.  The first layer makes the most difference, then diminishing returns.

Thanks Derek, for describing your methods in detail.  You certainly achieve a good looking result.  As you say, avoiding damaging the insulation after installation can be a challenge.  Looks very realistic for what I have seen in ships engine rooms.  My silicone tape is not so elegant, but it is flexible, and I can always add another layer.

Hi Zephyrin, I have also heard of plaster of Paris being used in this way, I guess the source of the materials depends on who you know.  But that plaster soaked gauze is a very convenient way of applying the paste and then reinforcing the set material.

Admiral _dk, thanks for that description of your district heating plants.  Don't have them here, 37 C today, and Spring has barely started.  Also, a very interesting method for detecting insulation faults.  It is a real problem for such extensive systems, and these days, even that low temperature heat is valuable.

I wanted to get back to the calculations on that exhaust steam, but it is already late, so next time.  I am back in the long paddock, having turned my head towards home the short way.  The Buildup has started, should have left for home a week ago.  Today's temperature was accompanied by quite high humidity, which persisted through the night.  Seven thousand two hundred kilometres so far, but only 4400 home from here now.  And should be able to post most days.  Please understand if I miss a day, as there are some gaps in the service.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 16, 2017, 10:35:54 AM
Real Engine output -

Hi everyone, Internet gaps appeared quicker than I anticipated, so glad I had mentioned it.  Usually available each night, but the place we found last night only had the wrong carrier.  Along the long paddock, there are occasional reflecting antenna dishes with a little pad at the focus.  If you put your phone there, you get Internet and or phone reception, but not even a meter away.  Ok for email and messages, and stuff that can be read off line, but not much help for reading an active forum.  I will try and stop next time I see one with enough warning, and get a photo.  Possibly three or fours days time, as we should be good for the next two nights.

Last time went into acknowledging very interesting and relevant comments that people had posted, so tonight back to the test run data.  You might remember that I had calculated the power output of  an ideal adiabatic engine, using the second law to determine the exhaust entropy, which defined the steam condition so allowed calculation of the enthalpy.  The first law then allowed us to calculate the output of that ideal engine as 27.5 watts.  Next, we note that we have a measured exhaust condition for the real test engine, it was superheated to 104 deg C at atmospheric pressure, so we could find the exhaust enthalpy directly by interpolating the superheated steam tables as 2684.2 kJ/kg or J/g.

Now the first law says the work produced by our real engine is h3 - h4 = 2745.91 - 2684.22 = 61.7 kg/kg (J/g).  We multiply this by our steam rate, 0.296 g/s also W = 61.7 x 0.296 = 18.3 J/s.

Not much compared with 668 watts in our steam, only 2.7 % efficiency, based on heat in the steam, even less based on fuel energy.  But it is 66% of the output of that ideal engine, which was only 27.5 watts, and we know that cannot be exceeded or even equaled by any real engine.  So, not so shabby after all.  If I was an engine manufacturer, I would try and use an efficiency defined by comparison with an ideal engine in my sales pitch!  I think both the actual numbers, and the fact that they can be calculated at all is the most interesting thing about the whole exercise of seeing what can be found out by applying basic instrumentation and the laws of thermodynamics to a simple test run. 

Of course you are wondering how we can calculate a power output when the engine was running uncoupled, surely not producing any output at all.  It is quite correct that the engine was not producing any shaft output power, but the observation helps our understanding of what these calculations really mean.  And another illustration of different definitions of power.

The power output calculated from the first law, (h3 - h4) x steam flow rate, gives the work done by the steam on the piston in the power strokes.  Remember way back, how heat is converted to work?  Portions of this work are then used inside the engine, before we get anything to the output shaft.  Some goes into pushing the exhaust on the other side of the piston out through the ports, even before there is a net differential pressure and hence net force on the piston.  Some of the force on the piston goes into overcoming the friction of the piston in the cylinder and the remainder produces the torque necessary to overcome friction in the bearings, and pins.  Some steam, it may be quite a bit in my case, may even bypass the piston by flowing between the piston and cylinder, thus producing no work.  I only have one or two labyrinth grooves, and this will not be nearly enough, especially if the piston clearance is too great.  However, wherever the work is used, it is not available at the output shaft, so I have to get more power out of the engine to have any output at all.  I suspect it would be a bit optimistic to hope to double the power produced by the steam without increasing the losses so that we had another18 Watts available to drive something, but so many of the engines on this forum do produce enough useful work to overcome more than engine friction, that I am sure I can get something out of it.  Obviously the next project has to be a brake of some kind to provide a predictable load for the engine for more tests.

Before I do more testing, as that must be on hold for the moment, is there anything else we can learn about where this power is consumed?  Now I am not a politician, I have posed the question without being sure how I will get an answer, let alone what troubles I will get myself into in the process.  But there are a couple of formulae that relate mean effective pressure and torque to the power of an engine.  I will use those to try a few calculations and see if they yield anything interesting for next time.

Thanks for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 16, 2017, 03:34:16 PM
Hi, I have just bought the two volume set of the John Farey  "A treatise on the steam engine"  from 1827 and what a treat that is !! there are 90 pages on MR Woolf alone !!!!! A quick question on his compound engines .........................as the HP cylinder is  supplied with full steam with no cut off, how much work is done by the LP cylinder that is moving anyway as it is coupled directly with the HP side ????...Answers on a very large postcard please !!!!!Also does the water level in the boiler have any significant detrimental effects on efficiency ?? Thanks for all this info from all our subscribers .
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 17, 2017, 02:17:21 PM
Hi Willy, the second question first.  Heat transfer is determined by three resistances in series, the film resistance for transfer from the hot gases to the copper, the conductivity of the copper, then the film resistance from the copper to the fluid on the inside of the boiler.  Obviously we would the first two should be unaffected by the water level, however the internal convection coefficient is very different for steam compared with boiling water.  Boiling water has a very low film resistance, or high conductivity, partly due to the specific heat of water and partly due to the agitation by steam bubbles which continually carry away the heated fluid so it is replaced by cooler water.  This effectively increases the temperature gradient, and hence gives much better heat transfer.  Transfer to steam is much less effective, partly the lower specific heat of the steam and partly because there is no vigorous mixing due to the bubbling, just normal convection.  Usually when calculating the surface area of a small boiler, only the area below the water level is counted, and heat transfer to the steam ignored.  When the water level is low, the heat transfer will be less, so losses to the stack more, and efficiency lower.  However, all of this is for a fired boiler.  For your electric boiler, this does not apply, as the element should always be fully submerged in liquid, while the boiler shell should be well insulated so there is minimal heat loss.

The second question is a little more difficult.  I believe that what happens in the compound engine is that the volume of the exhaust side of the hp cylinder is getting smaller during the exhaust stroke, while the volume of the inlet side of the lp cylinder is getting larger, but at a greater rate due to the larger diameter.  So the total volume contained in the exhaust side of the hp plus the inlet side of the lp is getting larger or expanding.  At the hp piston face, work is being done on the gas, the energy coming from the steam on the inlet side, while at the lp piston face, the steam is doing work on the lp piston.  As always, the work is the pressure times the change in volume, but calculating the actual amount of work is complex as the pressure is changing as well as the volume.  The work input by the hp piston means the process has heat input at the same time as work is being done so it cannot be compared with an ideal adiabatic process.  With the geometry and timing all known, I assume the process could be analysed, but it is beyond me.  Is that about the right sized postcard?

Regarding the additional calculations on my little engine test, I have tried to calculate an equivalent mean effective pressure, and an equivalent torque from the usual formula, without getting any sensible answers.  I think I am making a mathematical mistake somewhere, and just need some time to check it all.  Just not sure what at the moment, really puzzling.  May need to look at some other topics for and come back to that one, unless someone has noticed the problem or had more success with the numbers.  At least I have had some success with comparing a model engine with the thermodynamic model even with the simplest of test set ups.

Crossed the Tropic of Capricorn late today, back in the temperate zone, cooler days and cool nights (making sleeping easier) only four days from the tropical heat and humidity on the coast.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 17, 2017, 04:36:05 PM
Thank you Willy for asking these questions and a very great thank you to MJM for the lucid explication of heat transfer in the boiler, it caused me to have an "Of course, what a fool am I" moment. Despite knowing of the various boundary resistances and even more if you add a soot layer on the fire side and a scale layer on the water side I had completely missed grasping the differential in heat transfer between steam and water when considering the circumstances pertaining to the poor steaming of a particular design Gauge 1 loco, a 12inch gauge loco that often defied the efforts of those who fired her and the behaviour of a vertical boiler when it had lower water levels. This one element left out of my thinking has finally explained a conundrum of many decades for the vertical and the 12" loco and illuminates perfectly part of the poor performance of the G1 loco. It is one thing to 'know' a number of facts but missing one critical one or not putting them all together can leave one sorely bereft of satisfying explanations for the behaviour of things! I owe you a beer, VB I presume. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 18, 2017, 02:43:50 AM
>Hi, MJM, I will be firing up my boiler later on today BST and could you give me all the measurements and volumes and temperatures that i need to record. Will it matter how full the boiler is ,or do you just need the volume of water. The pressure gauge is not very accurate but could i slide my temp gauge probe under the lagging. any suggestions would be welcome. It will be coupled up to my engine so that will be fun . !!
Title: Re: Talking Thermodynamics
Post by: 10KPete on September 18, 2017, 04:24:00 AM
Geez, Willy, that's a beautiful engine.....

Pete
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 18, 2017, 11:00:57 AM
Hi Paul, I am glad that I could help explain those issues, at least a little bit, it is the main purpose of this thread.  No need for a beer, but if you are ever down south, let me know and we can perhaps share a meal or at least a coffee.  Remember we have the national steam museum there, and it has some amazing historical engines actually running.  With some better understanding it will perhaps be easier to work out a solution to the problem.  With the locomotive of course, like the electric boiler, the answer only applies to boilers which are fired at least partly around the outside shell.  Obviously if it is fired through a centre flue, the flue and any return tubes must always be covered.  Not sure if you have return tubes on those little boilers.  But if you have a film of calcium salts on the inside, that will also do it.  May just need a good clean out, perhaps with vinegar.  We recently passed through a place where the water had a distinct sulphurous smell.  But it quickly faded away, the locals say the sulphur disappears but H2S does that to your nostrils anyway, and the water seemed quite good.  But we noticed that all the scale, which had accumulated from all the places where the water filters up from the ground through limestone, fell off the kettle in only two days there!

Willy, I hope I am in time, we are 10 hours ahead here, or perhaps 9.5.  It is 7 pm here as I write.  I am with Pete, that is a beautiful engine, and great to see it complete after following your build.  Will you be running the new one, or just the mill engine for today?  For the boiler performance, if you can slip the thermocouple element in under the insulation, it will give quite a good idea of the steam temperature inside.  From this we can look up the pressure in the steam tables to compare with your pressure gauge.  As it will be saturated steam, not superheated, we  can also look up the properties of the steam directly, and calculate the steam flow assuming the element rating is accurate.  Of course it may not be accurate but we can roughly check it.

If you can weigh the water into the boiler, and at the end of the run, weigh the water that you drain out when it is all cool, you can determine how much steam was generated.  If you note the time that steaming starts, and again when you switch off the power, you then know how much energy was input during steam generation.  There should be some agreement between this and the output based on element rating.  Differences will arise because before your engine runs, steam will be condensed in heating your pipes and cylinder etc. so it is difficult to determine exactly when steaming starts.  There is also an error due to the heat loss from the ends and through the insulation, but keep this to a minimum, even if it involves a little extra temporary insulation.  I don't know if you know the mass of your boiler, which I think is copper (?) and is needed if you want to analyse the heat up period, but it is not worth stripping off the insulation for it.

For the engine, it is a bit more complex.  If you have a throttle valve, you cannot assume the engine steam inlet pressure is the same as the boiler pressure.  Also you need a means of measuring the temperature at the engine inlet and exhaust.  Probably some special pipe fittings for test purposes to accommodate these measurements and delete the throttle, so not for today's run.  Some sort of tachometer is also desirable.  Measurements on that engine driving the little generator will be particularly interesting.  Voltage and current are easy, but of course we don't know the efficiency of the generator, so probably better to measure speed and torque.  But by then you will have built the test stand I need, so again, not today's project.

So I suggest today concentrate on the boiler, and make a up a few fittings when you have a gap between projects for future testing.  I am looking forward to seeing the boiler test results.

Last night I drove myself to distraction trying to make sense of the formulas for torque and mean effective pressure, and not getting answers that were consistent.  I think it is a matter of some repeat test runs, to ensure I have consistent result, and check those formulas when at last I have access to my test books again to be sure I understand the assumptions inherent in the formulas.

MJM460
Title: Re: Talking Thermodynamics
Post by: Steam Haulage on September 18, 2017, 12:31:24 PM
Hi,
In post 282 you talk of 'film resistance'. Could you provide your definition and a give me a good source of background reading on this topic -please?
Steam Haulage
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 18, 2017, 03:10:32 PM
Hi, the first test is to show the efficiency of the lagging. there is a 1/16" lagging cloth bought from the model engineers propriety show and some not very close fitting mahogany planks. The temp gauge prong is close to the boiler shell after scraping away the felt and hollowing out the plank to make a tight fit. The boiler is 16 gauge copper and is not lagged at the ends. you can see the two 500 Watt elements. The amount of water put in is 600 mL about half full. I shall endeavour to measure the length and diameter etc and i shall time the element switch  on ...to the safety vale blow off and note what the pressure gauge reads, also the rise in temp from ambient to valve lift. with your tables it should be possible to get some sort of efficiency reading !! With the new engine the inlet exhaust temp may be quite close as it is a steam jacketed cylinder

 and the exhaust comes from the bottom of the casting to the condenser. As this is all one huge lump of metal and the jacket is filled with steam first and drained at the bottom until steam escapes there may be not much difference. The engine was tested under steam before it was all taken apart to be painted ,and all was well as can be seen from a previous post on the Beeleigh mill posts.However i had to take it all apart ,make all the correct bearing parts etc etc for it and lost the timing and porting lengths a bit so may
 
have to make a few adjustments before it will work again. Thanks for taking the trouble to help with this and i shall film the results as we go along and hope there is not too much "film resistance" to confound the results !!!. Incedently in Blighty when we film stuff ,we take ' Footage'
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 18, 2017, 03:19:43 PM
And am getting in a muddle with putting the text on ,,,so to complete the last bit, in Metric countries do they take "meterage" ?? just wondering !!!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 19, 2017, 01:24:24 AM
so, i have had the boiler in steam....the length is 180mm, diameter is 75mm and the thickness is 1.7mm. the overall thickness of shell insulation and wood cladding is 10mm. The temperature started at 16 degrees centigrade and the graph shows minuets and degrees .after a slow pick up the graph was in a strait line till the safety valve blew off at 118 degrees. the valve was screwed down to give a temp of 143 degrees with the pressure gauge showing 50 Lbs square inch.I have included the graph to show the readings. I connected the boiler to the new engine but it was reluctant to run properly so i will need to make some adjustments here and there.So what can we deduce from these figures ??
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 19, 2017, 11:19:22 AM
An electric boiler test -

Hi Steam Haulage, glad to have you on board.  Film coefficient is a term associated with convection heat transfer.  The introductory texts usually talk about heat transfer from a heated plate to a fluid, liquid or gas.  The primary variables are area and temperature difference and arranged as though there was a constant heat transfer factor, so Q = (coefficient) x A x (T2 - T1)
The coefficient is near enough to a constant for conduction, but not really constant, and not a simple relationship in convection.  For a fluid heated by a plate, the coefficient is given the symbol h.  The issue is that the heat transfer alters the properties of the fluid, many of which are temperature dependent.  So the heating affects the flow, and the flow addicts the practical temperature difference.  All this goes on in an area close to the plate, overlapping and very like the boundary layer in flow situations.  If you are looking for more information, it will be in any text book on heat transfer, often titled Engineering Heat Transfer, or something similar.  These books are usually quite separate from thermodynamics texts.  My book is too old to be a useful recommendation, though much newer than some of Willy's reference books.  Spend an hour in a good technical library and choose one that seems readable before you buy, they are usually pretty heavy in every sense of the word.  Or if you know any current or recent engineering students they will know which ones are current and readable.

Hi Willy, that is a great start on the testing.  I will try and summarise what I saw in it, without requiring too large a post card.

First, a little interpolation of the steam tables gives the pressure at 118 deg C as equivalent to 12.6 psig .  If the temperature reading is a bit low, 120 deg C would give about 14.3 psig, both assuming a slightly low atmospheric pressure system at 100 kPa or14.5 psi.  No superheater, so we only need the saturated steam tables, and temperature and pressure are not independent.

With the higher safety valve setting, 143 deg C gives 42.6 psig compared with the pressure gauge 50 psig.  If the temperature was actually 145 deg C, the pressure would be 45.7, say 46 psig, so I suggest your gauge reads a bit low, but not bad for such a small gauge.  The needle even looks a bit lower than 50 in the photo.  I don't know how much error there would be in your readings, but more insulation wrapped around or even just close to the tip of the probe would help reduce the error.

With the temperature, we not only know the pressure, but the steam tables tell us the specific volume, internal energy, enthalpy and entropy at both saturated liquid and dry saturated vapour condition.  (Points 1 and 2 on the sketch for my tests).  The latent heat, or enthalpy change from 1 to 2 is 2737 - 602 = 2135 kJ//kg.

The heater of 1000 Watts is 1000 J/s so takes 2135 seconds to evaporate 1 kg of water.  We don't have the actual experimental results to compare with this, but there is more we can do.

You filled with 600 ml of water or 0.6 kg.  Water at 16 deg has an enthalpy of 67 kJ/kg.  So to heat it to point 1, you need 602 - 67 = 535 kJ/kg so 535000 x 0.6 = 321000 kJ to heat to 148 deg.

Copper has a density of about 8933 kg/m3 and specific heat of about 383 J/kg.K.  From your dimensions and the density of copper, I estimate your boiler mass to be about 0.75 kg.  So to heat the copper 16 to 148 deg, 127 deg, you need 0.75 x 535 x 127 = 36000 J.  You can see this is much smaller than the heat required for water.  The insulation will need even less, and I have ignored that.  Total heat to get from 16 to 148 = 357000J.  Your heating element should do this in 357 seconds, say 6 minutes, assuming no heat loss.  Of course the boiler ends are not insulated, and the insulation is pretty thin, so I suggest you can feel some heat coming off the insulation.  It looks like your boiler took about 9 min, so this gives you an idea of the heat loss.  Very roughly about 67% efficiency, though I have ignored the lower heat loss at lower temperature.  But as a first estimate I would say perhaps only 67% heat into steam during steaming.  On this basis, 670 J/s into the steam which requires h2 - h1, 2737 - 602 = 2135 kJ/kg, so 3187 seconds per kg of steam, or 53 minutes per kg, so about 0.314 gram/sec.  It will be interesting to see what you get.

Perhaps not as high efficiency as I would have expected, but you could try some temporary insulation of the ends, perhaps a layer of felt then whatever else you have on hand.  And wrap a folded towel around the cylindrical section.  Unlike a fired boiler you can use all sorts of insulation while you carefully monitor the temperature.  Just for a test, then perhaps layers of cork on top of your insulation, then a few extra wood strips to complete the larger circle to finish it off to compliment your beautiful engine.

Now I hope you can see how the figures are used, you can see which ones are most important to improve the accuracy.  And just how much can be learned from a simple test rig.

Oh, by the way, metric countries have put all the film in museums and archives, and just shoot video.

Should have internet tomorrow, but you never know for sure 'til you try.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 19, 2017, 09:21:50 PM
Hi MJM thanks for the figures and explanation. First  16 to 148 degrees  127 degrees .....should that be132  degrees ??  The boiler is not actually 100% steam tight as there are quite a few whiffs of steam escaping here and there !! Also the wooden lagging is not very close fitting and yes it is hot to the touch. I am wondering about temperature being able to accurately define pressure ?? as, if you have an empty boiler that is being heated ,it will still show up on the thermometer. ? .also if you open the steam valve and release all the pressure, the boiler is still very hot which will show on the reading. If you do heat up an empty boiler will it show a pressure  positive or negative as if the outside expands will it register a slight vacuum ? it is quite cool in this world of exponential curves and logarithms and stuff  that the time pressure curve is an actuall straight line !! Talking about Willys books ,i found this in John Fareys 1827 tome....However i am not actually qualified enough to agree or disagree !! I suppose the reading would have been more accurate if the thermometer was inside the boiler.? The Beeleigh engine has no lagging on it and there are no places on the cylinder block that has bolt hole attachments
, I shall look at the Ramm brewery sister engine to see if that is lagged, Thanks for all this info , really interesting.
Willbert........It looks like the Rams engine is half lagged actually.
Title: Re: Talking Thermodynamics
Post by: crueby on September 19, 2017, 09:34:06 PM
Willy - I love the line under the table in the picture you just showed - "Observe this table is entirely erroneous"

 :shrug:
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 20, 2017, 02:14:34 AM
hi Chris,  yes brilliant i love it too , you won't see that in a modern book, It was probably said by Dr Dionosor Lardner the well known vicar that also disagreed with I.K.Brunnel and other engineers of that period !!!
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 20, 2017, 12:02:15 PM
Hi Willy, when I saw your post I thought that was the inevitable consequence of balancing a calculator and an iPad on my knee, while sitting under a shady tree in bright sunshine, 1000 km from nowhere in particular.  Still, such errors are easy to make and it was in the first line of my notes.  So I reread your post and found it actually said 143 deg C, not 148.  Still, I am not above that kind of mistake and I am glad that you are checking on me.  It probably will be me next time, so let's both keep checking.   Not to worry, I put an extra line of data from the tables in my little spreadsheet, copied the formula to a few more squares, and followed through the calculations.  I was surprised how little difference it made, the difference is only a small percentage of the approximately 2000 kJ/kg latent heat so makes a small percentage difference to the answers.  The biggest difference is actually in the calculated pressure.  For 148 deg C, the pressure gauge should read 51 psig.  Again in case the temperature is reading low, 150 deg C would give 54.5 psig.  However following through the remaining calculations, enthalpy of saturated water is 623, dry saturated steam is 2744 and enthalpy to make steam is 2121 kJ/kg, always seem strange it is lower at higher pressure but that is the shape of the bell curve that outlines the two phase region, and quite small differences.  So your heaters potentially evaporate about 0.47 g/ sec, 28.3 g/min, the difference only showing in the third significant figure.  It was worth doing the calculations twice just to see the sensitivity to the variables.

Heat for water is now 556 kJ/kg, so 333600 J for 600 g of water, 37917 g for the copper total 371517 J so 371 seconds or 6.2 min.  In the right direction, but clearly not the explanation for the lower efficiency.  My first guess is still insulation.  Try the temporary fix first, just to check, then think about how to do a job more worthy of your beautiful engine.  Other significant quantities where accuracy is important are the mass of water added to the empty boiler, volume is difficult to get that accurate, if you can get access to a digital kitchen scale that reads to the nearest g, with a switch to give oz.  The other big one is the actual power input of your heating element.  Not so easy unless you have a power monitor as many shops sell for checking household appliances.  But surely your kitchen needs a new accurate scale, so you can get accurate weights doesn't it?  For the cooking of course!

The other issue is that escaping steam.  To evaporate a kg of steam requires about 2000kJ, where as to heat a kg of water by one degree requires only 4.2 kJ, so a little steam escaping is carrying away a disproportionate amount of heat and not delivering it to the engine.  Besides which it can scold your fingers, so worth fixing.

Regarding your additional questions, if you have water and steam in a closed container in equilibrium, the temperature and water vapour pressure are directly related and can be read from the steam tables.  Of course, if you also have air in the boiler, the air pressure adds to the water vapour pressure.  But you said the safety valve listed at about 118 degrees before you screwed it down a bit, so much of the air will have been swept out with the steam at that point.  Of course some air could still be there, so when you first get to the new set pressure, there could still be some air, and your gauge could read high.  If the final temperature was 143, and still some air, the gauge reading of 50 could be correct, but at 148, and the gauge reading 50 it is reading a little low compared with 51 expected.  After a bit more steam escapes, the huge volume of steam compared with the initial air volume means air is soon gone.  In summary, for steam vapour and liquid in equilibrium, temperature definitely defines the water vapour pressure.  Other gases in the space, such as air, act independently in addition to the water vapour, and the gauge and your safety valve both read/respond to the total of all the partial pressures.

If you suddenly release the pressure, the system will not be in equilibrium for a short time.  The water will cool as some evaporates to make a bit more steam.  The copper will cool by giving heat to the water.  Soon the pressure is atmospheric, the water is in equilibrium at atmospheric pressure so 100 deg C, the copper is also at 100 deg C, steam production essentially stops and the thermocouple also soon reaches/responds to that temperature.  Now the whole lot is back in equilibrium, the temperature is 100 deg C and the steam tables tell us the pressure is 101.4 kPa or 14.7 psi absolute.

If you heat the empty boiler, presumably air at atmospheric pressure and plug tight, not empty as in full vacuum, you have no water, so the steam tables are not relevant.  As the air heats, it  behaves roughly like an ideal gas, and the pressure rises in proportion to the absolute temperature, which is 473 + T, so quite slowly.  However the convection heat transfer is quite poor compared with water or steam, and the air has to carry the heat to the boiler shell, so it is unlikely to be in equilibrium.  With such poor heat transfer to cool the element, it will overheat and burn out, so not a recommended experiment.  One for theory only.

Your graph has the initial curve while the temperature gradient within the element sheath and water film are established, which involves some heat storage, then with uniform heat input and near constant specific heat of the water and copper, temperature is a linear function of heat input, but yes, interesting that it's linear, compared with all the curves involved in cooling of your coffee.

Lagging of an engine reduces the heat loss, which improves the engine efficiency, but heating just further reduces this loss.  The area is really too small to make a significant heat input.  I think I have discussed this in response to one of Paul's questions.

I hope that answers the current questions, they are good questions and helpful for directing my explanations, so keep them coming.  Remember that the principle behind using temperature to determine pressure involves two assumptions.  Water as both  liquid and vapour present at the same time in equilibrium or very close to it, and no other gases in the vapour space, both of which are sufficiently accurate soon after steam flow starts and the temperature stabilises.

By the way, I checked a couple of the figures in that extract from the book, and have to agree with the editor who added the comment.  The pressures have small errors at any temperature and the volume expansion is way out.  Like you and Chris, I like the honesty.  I suspect a modern editor may have quoted the figures for historical accuracy of the text, but recognised that they did not have good quality data at that time, so don't use them for calculations.  But a fun observation.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 21, 2017, 01:36:05 PM
Hi Willy, a little post script on yesterday's post.  I hope I did not leave you with the impression that the final temperature does not make much difference.  You can see it makes a small difference to the boiler, affecting the time to raise steam and the amount of steam raised.  For these issues, the percentage change with a small temperature difference is quite small.  However, when we get to looking at the engine, we will take that approximately 2000 kJ/kg in the steam and subtract a quite similar number for the exhaust steam to get quite a small difference available for the engine to do work.  Then a small difference in the roughly 2000 for steam will make a much larger percentage difference to the potential work output.  I will look at that next time.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 21, 2017, 03:00:21 PM
Hi MJM I have now had my 'Fitters' hat on and the engine is running freely. I connected it to an air compressor working at about 40 psi as this was easier to investigate the tight spots !! i actually needed to lift the main bearings about 3/32'' as the pistons were fouling the cylinder heads !! this would be  about an inch of packing with the full size engine.!! I think it will be quite difficult to measure the engine temperatures as the exhaust disappears below the engine into the condenser/airpump block that should be filled with water but is not and is under the engine surrounded by the box. Here is a video of the engine running with air, it is a bit more spectacular with steam of course ,what with all those not quite steam tight glands and things.

Beeleigh mill model beam engine.
Anyone
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 22, 2017, 01:10:00 PM
Hi Willy,  thank you for putting up those videos of your engine running on air and steam.  It is really great to see them running after watching the care and skill you have applied to build it with historical accuracy.  It is a real work of art.

I would not worry too much about not measuring the exhaust temperature.  I don't think it is the most important part of building such an engine.  If you really want to, your thermocouple will fit in quite a small hole and on the condenser side, could have a simple o-ring gland if there was a place for a plug that would ensure it measured the steam exhaust temperature, and not the condensing water temperature.   The exhaust pressure can also be measured if you can make an unobtrusive tapping point for a plastic u-tube manometer, but I would not spoil the appearance of a beautiful engine for it.  We don't have contractural requirements to demonstrate on our models, and we can learn all we need to know by setting up any of our other engines, particularly one that is more easily fitted with the test points.  I assume the condensing is achieved by direct injection of water, as I don't remember any tubes in the condenser, which I assume is mainly the air pump and condensate pump with a water injection nozzle.

It is worth doing what you have done with the boiler, because knowing how much steam you are making, and how much heat you are loosing due to insulation does inform your efforts to improve heat up time and steam production, and gives you an idea of the limits before you just need a bigger or more elements if you need more steam.  A spare boiler plug that acts as a thermowell and accepts your thermocouple for temperature measurement is not a big deal and easily replaced with a normal plug for display purposes would give slightly more accurate measurements.  And you have learned a lot of thermodynamics in the process.

If I heard you correctly in the video, the engine actually ran with a boiler temperature of 116 deg C.  Tomorrow I may be a bit quiet due to a family visit, but in the next couple of days I will calculate the potential output of an ideal adiabatic engine at each of the three temperatures you have mentioned, just to see the difference between 116, 143 and 148 deg C on the potential performance of the engine.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 22, 2017, 03:22:35 PM
Hi, more questions ......if you want to produce more steam to run an engine continuously would it be better to have less water in the electric boiler (no need to cover the firebox with H2O) or will the higher level make no difference ? I was surprised at how long the boiler took to cool down as well and If i have a spare couple of hours i could draw up a graph for that as well !! To get rid of the air in the boiler would it make sense to leave the steam valve open until it reaches 100 degrees  (boiling) and then close it ? will it take longer to reach operating pressure to achieve this ?  Also my boiler is at 3.8672493 meters above mean sea level if this is relevant !!!! Thanks for all the additional info.....I have some really shiny thin shim plate stuff that i could put round the wooden lagged boiler, and would this be better ? it would stop draughts taking away any convected heat perhaps ? Do you have any info on Wiredrawing that one reads about btw?
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 23, 2017, 12:23:36 AM
Also will the actual mains voltage have an effect on the input wattage available at the element it is rated at 250 volts but if only 230 v will this be relevant ,and are there tables available ? I don't know the actually voltage at the element and how much may be missing in the control Cct, however i could find out ,but that will mean taking things apart !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 23, 2017, 02:44:46 PM
Hi Willy, only a short time available tonight, so last question first.  I will try and get to the others tomorrow.  With AC, ohms law still applies though you must use impedance instead of resistance to be correct.  Fortunately your heater is almost certainly a pure resistive load so the values of resistance and impedance are the same.  Again the voltage we quote for AC, whether it be 230 or 250 volts is an RMS voltage, not the peak of the alternating voltage which is significantly higher, and the RMS voltage happens to have the same effect on a heating element as a DC voltage of that value.  So you need to use two formulae.  Ohms law V= I x R bit in the form I =  V/R.  Then power in watts, W = V x I = V^2/R.  You can calculate R = 250 ^2/ W, ie from rated voltage and rated power.  Then power at 230 V = 230^2/R. Because the voltage is squared, the effect of low voltage will be marked.

You also need to be aware that the resistance normally varies with the element temperature, usually increases with temperature.  I don't know of your element is rated based on the element hot or cold, but quite possibly cold at a defined temperature of perhaps 25 deg C. Because they don't know what temperature you will run at.  You can measure the resistance of the element cold and see if this agrees with your calculated figure from the rating.  You may be able to get a close hot reading if when your boiler is operating for some time and hot, switch off, and quickly unplug and measure.  It will cool a bit before you get a reading, but will give you an idea of the change.  Better would be if you are equipped with an AC current meter that clips on around the wires and measure the current hot and cold.  But don't fool with 230 V AC unless you really have the right equipment.  You can't measure resistance with power connected.

Finally, even with a rating of 1000 watts, unless you have a test certificate you still have to regard that as approximate.  I would hope it is within 10% of that, so in the range 900 to 1100, but I don't know what accuracy standard it is made to.  In the end measurements are necessary for accurate data, and even then the instruments should be calibrated.  Outside a formal lab environment there will be a measurement error in practice.

I will try and answer the other questions tomorrow.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 24, 2017, 01:02:09 AM
hi, thanks for that the and here is some info about the Cartridge heaters. I did not use the sealing compound though , also some pics of the cartridge heaters with their housings, if they need to be taken into account. They are a pretty complex shape though to calculate the mass/volume !! Looking forward to more info ...and i hope i have not hilacked this thread too much .In the pictures are the electrical safety valve and the water level switch. this switch device is used with a 9 volt battery Cct ,to trigger the on/off mains Cct so as to be safe from electric shocks !! the insulation is PTFE.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 24, 2017, 06:01:27 AM
More on electric boilers

Hi Willy, back to your questions in post #298.  First, you are not hijacking the thread, it is to answer questions like this that I am writing the thread, and I suspect there are many others with similar questions, and I hope they are finding it helpful and will join in when, between you and I, we miss something, or it is just still not clear to them.  And I hope also if they notice something I have got wrong.  Mistakes are just too easy in some of this stuff, I can only do my best, and correct things when they are noticed.

Does it matter whether you fill the boiler or not?  When you raise steam in an electric boiler, or any boiler for that matter, you perform two different processes in sequence.  First you heat up the whole system to your steaming temperature, then you evaporate liquid to produce steam at constant temperature.  When all the water is gone, the system temperature will start increasing if the energy input continues.  However, if you do continue with a dry boiler, or even a partially covered element or firebox, you will soon find some component reaches a mechanical strength limit at the higher than design temperature, something will melt.  On your case the element will probably burn out.  In a fired boiler, most likely the soldered joint will soften and release pressure before the copper softens enough to start yielding, as will be evidenced by bulging which will not spring back when the whole mess cools down.  It all depends on which limit is reached first in a particular boiler.

To understand the answer to your question, first consider each of those two basic processes.  While you are heating up, higher water level means more water to heat, so it will take longer for the constant energy input of your heater to get it all up to temperature.  Most of the energy input is stored in the water, copper and insulation, and a portion is lost to atmosphere as we have already seen.  This lost portion is initially low while all the temperatures are low, and increases with increasing temperature.

Once it is all up to steaming temperature, further heat input does not increase the temperature, it mostly goes into evaporating water into steam, although the heat loss to atmosphere continues, now at a constant rate due to the constant temperature.  How much steam is now found by an energy balance, and how much energy to evaporate a kg (or lbm) of steam.  We get this from the steam tables and it does not depend on how much water is available at saturated condition for the temperature, containing hf kJ/kg as enthalpy.  If you have less water during the steaming process, you will run out quicker, but it makes no difference to the steam production rate.  Steam production is completely determined by the energy input, the heat loss, and hg - hf for the water at that temperature.  And steam pressure is determined completely determined by the temperature, or vice versa.  And this pressure will be the pressure in the boiler once a little steam production has carried away the air that was in the boiler when it was filled.  Obviously less air if the boiler was more full.  But you cannot fill it completely as the steam needs some space to separate from the liquid, otherwise a lot of water will be carried over with the first steam.  Experiment is probably the best way to determine the maximum fill level before water carryover becomes too much of a problem.

The minimum water level must cover the element.  It does not matter whether it is a fired boiler or electric, the heating surface will get too hot if it is not covered by liquid as the heat transfer coefficient for vapour is very low compared with water, especially vigorously boiling water.  You could consider a little lower level and use a feed pump to maintain the level.  However, the engine will do work driving the pump before there is any available to do anything else, and the water entering the boiler will be at less than steaming temperature, so requires more heat to get to that dry saturated condition of steam.  You can only partly offset this by some boiler preheating the water (using exhaust steam for example).  You can't use an electric jug because the pump will not handle near boiling water due to its valve pressure drop and acceleration losses.

Will leaving the boiler open until at 100 deg help?  Remember the mountain top experiment?
  The pressure of the water vapour at 16 deg C is only 1.8 kPa, so 2% of the total boiler pressure at atmospheric pressure, and a nice vacuum of you can condense at that temperature.  The rest of the pressure is due to air.  Heating the air to 116 deg C increases the pressure in proportion to the absolute temperatures, so (273+116)/(273+ 16) = 1.34.  So the air at approximately 100 kPa in the cold boiler increases to 134 kPa in the sealed boiler when hot.  This is an increase in gauge pressure of 34 kPa or about 5 psi.  If this, plus the water vapour vapour pressure exceeds your safety valve setting, it will lift.  The air that escapes with that steam is not replaced, while the steam is quickly replaced by more evaporation so the air is soon gone, and the air heat content is negligible compared with the steam, so no big difference in the heat required to get to that point.  If you leave the boiler open while heating, yes the air escapes, with little saving in heat requirement, but steam escapes with it, taking away the latent heat.  So it would be slower to get to steam raising.  The air will help drive your engine, so not a significant problem.  Seal the boiler cold and start heating.

Sheet metal around the boiler.  You normally have sheet metal cladding around industrial insulation.  Yes, it reduces convection from within the insulation, but conducts heat pretty well so it gains heat from the loss through the insulation, and in turn looses that heat to the air with only minimal reduction in overall loss.  It is put there, not to increase heat conservation, but to provide weather protection and protection from mechanical damage.  Note that wet insulation is a very poor insulator, as the water has higher conductivity than insulation and higher specific heat so it takes more heat to get up to the temperature profile.  So add your metal sheet, but only to hide and protect the insulation you put under it.  But wood looks even better, even if it is brown stuff.

Wire drawing - I see this term used in two ways and I don't see it as a specific technical term.  I believe it is normally used to determine the erosion of steel by high velocity wet steam in a situation such as a gate valve, which is only designed to be fully open or shut, is used for throttling.  The pressure energy becomes high velocity at the small gap of a nearly closed valve, and if the steam is wet, it will, sooner rather than later, gouge a bigger path and the valve will no longer seal properly when shut.  A gate valve is best followed by a globe valve which is made for throttling, or at least only used in throttling mode as a controlled opening rather than sudden opening.  However, I have also seen the term used to describe reducing pressure during flow.  To me that is the cause rather than the result.

A point about accuracy - your elevation, quoted to 7 decimal places of a meter implies that the last tenth of a micron is known to plus or minus less than 5 digits.  Measuring anything to this accuracy is a real challenge for even the very best micrometers, and I challenge any surveyor to measure elevation to that precision.  I thought you might have converted inches and feet with a calculator that gave you that many figures, but one inch is exactly 25.40 millimetres by definition, and dividing by 25.4 gives about 12 feet 8 1/4 inches, but not exactly, so I don't know how you arrived at the precise figure.  For its effect on the absolute pressure, the nearest 10 metres is probably more than accurate enough.  The weather bureau publishes atmospheric pressures all reduced to sea level, and I suggest that 3 metres is close enough to just look at the weather bureau latest observation, and assume it applies without correction to your location.  If you live in Denver, you would definitely have to calculate a correction.

Getting to be a long post, but I think that brings us to today's question.  I better get back to being sociable, might get back later, otherwise tomorrow.

I hope all that is helpful

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 24, 2017, 02:03:59 PM
Hi, Ok thanks for all those replies and observations, it helps out when trying to be too clever !! I have just thought about using boilers with steam engines and your word 'Vacuum' suddenly reminded me that when using a boiler one must be very careful to allow air to enter when the engine was stopped and the boiler shut done to stop the boiler imploding when it cools down !!
Title: Re: Talking Thermodynamics
Post by: paul gough on September 25, 2017, 01:08:22 AM
Re wire-drawing, MJM is no doubt correct regarding the erosion by condensate in a seat or some such as being what should be the primary meaning for the term. However it has common usage in discussing steam circuits in locomotives, certainly in publications from Great Britain. It is normal, (or should be), in a publication to use "wire drawing of the steam" when first entering into any discussion and then shorten it to wire-drawing unless the context is already known. As MJM said, it is a pressure drop across an apparatus, eg., the drop in pressure between the entry of a superheater element or set of elements due to friction and their outlet, any steam circuit, or even can be used to discuss a loss associated with constricted ports, either their size or the opening at a specific point of the valves travel over them. In very general terms it is a lack of sufficient cross section area of the passage be it a pipe, orifice plate, valve, superheater element etc. This at least is my understanding from my reading over the years and hope it helps illuminate the phenomena. Regards Paul Gough. PS, I understand that the first or top portion of an indicator diagram where it slopes slightly can indicate wire-drawing to an experienced eye, maybe our marine engineers that are on board could expand on this.
Title: Re: Talking Thermodynamics
Post by: paul gough on September 25, 2017, 08:09:29 AM
Before someone jumps on me for lack of clarity in my PS of the preceding post. I am not referring to the normal slight down slope of the upper portion of the diagram that is due to the increase in cylinder volume from piston travel. I am trying to covey there is a departure in the trace from this expected and normal line that a discerning and experienced eye can infer wire-drawing. Unfortunately I am not able to elaborate further as I have no experience in taking extensive numbers of indicator diagrams and interpreting them, it might be that marine engineers who served on steamships might know of it and shed further light on this, I don't think there would be any steam loco designers or engineers from loco testing stations still extant let alone reading this thread to provide a clear explanation. I mention it as one example of a defect that could be seen on a diagram. Perhaps some other examples of deficiencies that appear on diagrams might be discussed as well. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 25, 2017, 11:42:26 AM
Home, sweet home -

Some of you have noticed that I have been travelling recently.  Just arrived home this afternoon, and it is always good to arrive safely home after a road trip of 12240 km, average fuel consumption 11.2 l/100 km and circumnavigating roughly a half of our continent.  Hooked up our little caravan to the Subaru Outback on July 13, and went, well, outback.  I think Hugh calls it Snow birding, we call it becoming grey nomads.  People from the southern states driving north to escape the winter and find some sunshine make for crowded roads and parks, well sometimes you might see at least 10 other cars in a day.  I suppose we all go at about the same speed, but one time we had to stop for around 5 minutes for roadworks, no one came up to wait behind us.   Then they mysteriously appear, one by one at the road houses and campgrounds, for overnight stops.  But is is a great way to get a real feel for the country, and to experience blue skies and clear starry nights in a way that is not possible in the city.  And to see some wild life in its natural habitat.  We were a little later than most in turning south to put the sun on our backs for the long road home, so perhaps that is why the road seemed less busy than normal.

Thanks Paul for some clarification of wire drawing.  It is consistent with some of my reading and it would seem to be a term more commonly used in the pressure drop sense in the marine industry.  In my industry, it was a term usually muttered while looking at the cutaway in the hardened satellite of a gate valve no longer sealing shut.  Cut like a water jet.  Like you, I would like to see someone start a thread on the indicator diagrams, they are only rarely used in my industry, then only on compressors, as the drivers are all turbines or electric motors.  So I have no experience at all on taking or interpreting them.  However perhaps it is a bit outside the scope of this forum, as I suspect that none of us have available an indicator device suitable for use on our models.

Thanks Willy for those pictures of your boiler and the screen shots of the data sheets.  It seems that the elements may actually be rated for 230 V, so the calculation would then indicate the higher power output at 250 V.  I also note that the tolerance on power is +5%, -10%, but the question of whether this applies to a hot or cold element is not answered, but I would assume cold.  So probably lower output when hot.  The grease is intended as a heat transfer compound rather than a lubricant.  I think we have discussed this before.  I do notice that your boiler seems to have flat ends without stays, though I notice the bushings have not been soldered in the picture, so you may have added stays later in the construction process.  Of course you now have the boiler pressure tested and steaming, so if it is now dimensionally stable, I guess that is practical evidence of its strength.  However, I keep wondering if your pressure housings for the elements could be extended right through and fixed at the other end, just like a large diameter hollow stay, you could then use grease on the sheaths without air displacement issues. 

I am not sure what external pressure the boiler could stand.  The effect of internal pressure being below atmospheric pressure, the condition commonly referred to as vacuum, is that the shell has the high pressure on the outside, that is why I refer to external pressure.  The commonly quoted formula for shell thickness is technically referred to as the thin shell formula, and it's derivation rests on the assumption that the material is in tension to contain the higher pressure on the inside.  Under external pressure, the shell is in compression, the failure mode is the collapse you refer to, and the strength very different, usually much less.  The boiler is relatively short, so the flat ends provide considerable stiffening against collapse, but as you say, it is good to open a path for air before the boiler cools down and avoid the issue.  And if this allows some extra steam to escape, it may speed the cooling.  You have given me the boiler dimensions, so when I get everything unpacked and put away, I will calculate the strength under external pressure for those dimensions.

I think that brings us up to date on the previous questions, so time for a good nights sleep.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 26, 2017, 12:37:37 PM
Boiler design considerations -

Yesterday I mentioned the thin shell formula often quoted for boiler shells, and noted that this formula applies for internal pressure of a cylindrical shell only.  It does not apply for external pressure such as the loading on the centre fire tube of a marine boiler.  It also does not apply to flat ends, or other shapes such as locomotive fire boxes.  It is obvious really, the diameter of curvature of a flat plate is infinite, so the thin shell formula gives an infinite thickness for any pressure.  But blank flanges for piping are flat, and not infinite thickness, there must be another formula.  And of course there is.  Blank flanges are quite thick when compared with the wall thickness of the circular  pipe, and we would not want to make our flat ends so thick.  The answer is usually found in the provision of stays which support the plate at intervals that leave the unsupported section strong enough in a reasonable thickness.  The designers of boilers carefully follow the rules for spacing of the stays.  I don't intend to branch into boiler design, but it is probably worth looking at some of the considerations involved, so it is clear why building to a published design supported by the appropriate calculations is the best course for a builder not experienced in pressure vessel design.  Boilers are even more complex to design because the surfaces subject to the combustion gases require additional calculations to determine the appropriate metal temperatures and design strength.  Allowance has to be made for the reduction in strength of the material at higher temperatures.  Hence there are separate codes for boilers and for unfired pressure vessels.

When a cylindrical shell has the high pressure on the outside, such as the marine boiler fire tube already mentioned, the failure mode is quite different from when the higher pressure is on the inside.  If subject to to great a pressure, the tube collapses, or squashes.  The collapse pressure is quite sensitive to any departures from true circular form such as dents, unlike the internal pressure case.  The formula are again quite complex, and the Australian Miniature Boiler Code for example refers to the pressure vessel code for these calculations.  Needless to say, the thickness required to resist buckling under external pressure is much greater than required for internal pressure.  This is normally reflected in the specified thickness for the design.  Again, specifying the thickness is best left to the experts.  Calculations for external pressure are included in the AS/NZ, BS and ASME pressure vessel codes.

Now Willy mentioned the possibility of collapse of his boiler due to the low pressure if it is allowed to cool to atmospheric temperature.  This is a real possibility for industrial steam containing vessels, and now days the full scale pressure vessels that I am familiar with are designed for full vacuum as well as for the required internal pressure.  The relatively small diameter of Willy's boiler, combined with the fact that the shell is not fired, so the metal temperature is well controlled, means that the design check is relatively easy.  The internal design pressure is probably the overriding consideration, especially as the maximum possible vacuum is only a little over 100 kPa or 14.7 psi.  A quick first pass on the dimensions indicates it is probably quite strong enough to resist collapse, however I would still recommend a formal check by the boiler designer, as it is never wise to commit to the results of an unchecked calculation.  The best procedure is always to admit air while the vessel cools, at least until you have the formal calculation.  Collapse under vacuum used to be demonstrated in every junior science program using a square thin walled metal can.  I presume it still is, the result is quite spectacular but not particularly dangerous as the can is usually surrounded by the sink when the cold water is splashed over it to condense the steam.  And of course steam is not escaping as the pressure is reducing, dramatically in this case, but it is a lot of work to rebuild a boiler that collapses.  Again the failure mode is much different from failure under internal pressure test.

Another case where the thin wall, formula does not give all the answers is when you put a hole in the shell for inlets, outlets level glasses and so on.  We all know the normal design solution is a bush soldered into the shell.  The dimensions of the bushes are carefully specified in the model boiler codes.  These bushes are in fact designed to properly compensate for the missing part of the shell.  So long as you don't cut down on the amount of metal, you can use the standard bush designs with confidence.  Even make them a bit bigger, but no smaller.  The problem comes when you want a larger opening, perhaps for a steam turret.  The opening has to be properly reinforced and the design supported by appropriate calculations.  This even applies to tee fittings in tubing and piping, where the metal thickness is carefully controlled so the cylindrical tube or pipe can safely contain the internal pressure.  In a forged pipe fitting, the metal thickness is so carefully controlled that it is not immediately obvious that the extra thickness is there.  But it is.

So a boiler has many components which all contribute to pressure containment, and only the circular shell with internal pressure is described by that thin shell formula.  Even then, allowance has to be made for corrosion, and fabrication techniques.  Even welded joints are not considered as strong as the base metal unless they are proven by x-ray examination.  Tennessee Whiskey is well familiar with these inspections and his certifications are testament to his skill with welding equipment.  For most of us a satisfactory silver soldered joint in a copper boiler is easier to achieve, and the code requirements for the joint design are intended to allow for the expected fabrication methods, when supported by knowledgable inspection.

If this is of any interest, perhaps a little on testing tomorrow.

Thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on September 26, 2017, 02:08:19 PM
Re -thinking about efficiencies in our system as a measure of heat in, (boiler fuel) to work done, (engine output) it is known that higher pressures and superheat give better efficiencies even in models. Logically it then should be the case that pre-heating the feed water by means of exhaust steam or exhaust gases would do likewise. But to what extent??

The question arises for us modellers and particularly for rather small models is there any point in feed water heating? Is something primitive like running a feed water line through the firebox, (I'm talking metho or small gas fired boilers here), or some similar arrangement going to be practically worth it and enhance performance, eg. save fuel or reduce the firing rate. If it is a worthy theoretical proposition then we need to ask/find out what minimum temperature increase is needed to make a feed water heater a worthwhile modification. Is there any way to work this out, other than the obvious build one and try it method? Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 26, 2017, 03:10:07 PM
Hi MJM , I was talking to the club members today about the cooling of model boilers in locomotive and one chap said that as the boiler cooled it actually sucked new feedwter in through the clacks. so there was no need for a 'snifting' valve. Also where is the best place to have the inlet water valve ? Taking into account if the valve fails ....under  the water level or above it?
Willbert.
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 27, 2017, 11:10:39 AM
More on feedwater heating -

Hi Paul, consideration of feedwater heating is an area where perhaps logic lets us down.  But let's work through two, perhaps three situations.  First without a feed pump, as in my models so far, and I think Willy's electric boiler.  Preheating in this case means boiling the jug, and filling the boiler from the jug before the plug is tightened.  Now you can see what happens.  With the water at say 80 deg by the time the plug is in and the burner lit, the enthalpy of the water is 334 kJ/kg instead of 63 kJ/kg at 15 deg C.  If our chosen pressure is say 0.198 MPa (approx 2 bar, or 30 psi) the enthalpy of saturated water is 503.7 kJ/kg then that heating phase before steaming begins has to contribute (503.7-334)/(503.7-63) or close to 40 % of the heat required to heat cold water to the steaming point.  Clearly reduces startup time, and either saves fuel or extends the run time from a fixed amount of fuel, with the attendant risk of low water level.  However, once that steaming point is reached, the heat required to produce steam is entirely determined by that chosen pressure, and the heat available from the burner, the actual steam rate is not affected by the preheating.

If the steam plant has a feed pump, then some cold water is continually added to the boiler, presumably the quantity adjusted to match the steam rate so the level remains close to constant.  This means the average temperature of the water in the boiler near the entry point, is a little below the steaming temperature.  Mixing quickly brings the feedwater up to steaming temperature, but at the expense of the quantity of steam produced.  Conservation of energy and the saturated steam pressure-temperature relationship both still apply, and the energy to heat the feedwater must come from somewhere.  Now you can see that if you have a feed water heater using exhaust steam energy which is otherwise lost to the process, less energy is needed to heat the incoming water so some of the lost steam production is restored.  As the steam production in the end has to be balanced against consumption, and assuming the load and speed is the same for both cases, the fuel consumption is slightly reduced in response to the preheating.  When you are shovelling coal in a small locomotive, it is probably hard to tell, but the theory depends only on conservation of energy which is a fundamental law of physics and always applies.  With a small Meths burner, there is probably no adjustment, so a little more steam is produced and the locomotive possibly sees slightly higher pressure and goes a little faster.  If you use a radio to throttle the steam and maintain constant speed, the boiler pressure will increase a little, as will the stack temperature and a new equilibrium is reached.  Of course the differences might be masked by the other variables, however, it is still useful to understand the theory.

Now that third case, what if you have a hand pump?  First, I suggest that the hand pump will be an intermittent operation, you are unlikely to sit there steadily working the pump continuously.  Two feedwater heater designs are possible.  You could top up the feed water tank from that electric jug, useful things those, and if all else fails you can make a cup of tea.  Probably even worth insulating the tank.  If you plan on this one, it is best to have a raised tank, so you always have a positive head on the pump suction valve, otherwise you may get a vapour lock in your pump.  But that will reduce the additional heat needed to heat the makeup water.  Alternatively you could use some exhaust heat, either by a coil in the tank, again it needs to be raised, or you could even make a little heat exchanger and pump the water through this.  A little more difficult to control, probably more practical with an engine driven pump.  The water in the exchanger will get quite close to exhaust temperature while there is no flow, then the warmer water will be moved into the boiler when pump operation resumes.

In summary, using exhaust steam to preheat the feedwater does in principal reduce fuel consumption or increase steam production, however it has no effect on the engine inlet or exhaust, unless of course it imposes a significant back pressure on the exhaust.

There is a limit as to how much exhaust heat can be recovered.  You will remember that heat only flows from a high temperature to a lower temperature.  The exhaust temperature will be at about 100 deg if you do not have a superheater, or a bit above if you do.  The boiler temperature will be something above the exhaust temperature, and we have looked at a few cases earlier, say 116 - 120 degrees.  But then we actually need a temperature difference to drive the flow, so we can't get the feedwater temperature to the exhaust temperature of 100 degrees.  And the heat required to heat the water to that temperature is a very small portion of the total heat in the exhaust steam.

Remember the equation for heat transfer, Q = U x A x (T2 - T1). To get a given heat transfer, as the temperature difference reduces, the required area increases.  Practical area in most large industrial installations rarely achieve less than 10 deg approach, and in a small model feed water heater, almost certainly a very much bigger gap. 

So there you can see the effect of feedwater heating, you know what you are trying to achieve, but as U is notoriously difficult to predict, a little test rig would be desirable to see if you get a worth while extra run length with a practical heater size.

Another heat source for feedwater heating that could potentially give a higher boiler inlet temperature is the flue gas.  The main issue is to ensure that feedwater heating only gets heat that is left after the boiler and superheater have used all that is possible.  Otherwise the feed water heating is at the expense of steam production, and I suspect that would be counterproductive.  And at the end of the day, you don't want to cool the flue gas to the point where the water component starts to condense, as that nearly always leads to corrosion problems.

Willy, the feed check valve admitting water if the pressure falls below atmospheric is a good point.  Atmospheric pressure drives the water in the direction that lifts the ball and permits flow.  If you have a feed pump.  Of course the boiler could end up over full as the cool water lowers the temperature and hence vapour pressure of the water.  The only air admitted to the boiler would be the amount dissolved in the cold water.

We discussed the location for the feedwater inlet back in post #117 when Maryak provided a very interesting picture of full size practice.  It is on page 8 of 21 of the thread (based on my forum settings).  The recommendation was not too close to the shell, and normally near but below the surface.  The second part of your question, taking into account the possibility of check valve failure,  I am sure you are thinking of whether water or steam could escape from the boiler.  There are a few things to consider.  First, if you have a feed pump, you actually have three check valves in a row.  As any one will prevent back flow, all three have to be in bad shape, and any leakage should be quite small.  I suspect an issue with the feed pump might be noticed before there was too much problem, but perhaps others have some relevant experience they can contribute.  Note that when water escapes through the leaking valves it is at boiler temperature, so superheated water when the pressure is reduced.  This means a portion will flash to steam until the excess heat is absorbed, and this will tend to spit hot water around.  Definitely not desirable.

Next time I suggest looking at the potential work output from your engine, and the differences in potential work output for the three temperatures.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 27, 2017, 01:24:09 PM
Hi, May i ask about flash steam plants with regards to feed pumps.....I don't know much about them and have always wondered how you introduce water into the coil when the pressure in the coil is very high to drive the engine ? are you trying to pump fresh water into the coils above the pressure already there. and does this require a lot of energy that reduces the efficiency of the plant? Of course with an injector the water does have to be cold.! Also is there an actual temperature difference between the steam and the water level in a boiler? If the temp is the same is it steam or water ? or have i missed something here. Are there any pictures of unclad boilers that show the temperatures with the different colours as are used with buildings.? I do have a handdraulic feed pump on my boiler, but the sight glass does not work any more as i think it is scaled up.!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 27, 2017, 01:32:47 PM
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Here is a video talking about the steam plant.....This is on Utube if you can find it .
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 27, 2017, 08:10:07 PM
Hi ,MJM Here is a vid of my steam crane also electrically heated........and a quick disappearing trick at the end !!..I will be doing another steam up of the boiler with some Rockwool. does this need to be fairly loose or quite tightly packed ?? I would imagine a bit loose to allow the air to do its part. I shall also hold it in place withe some plywood. The contemporary thinking from the club boiler inspector is to have the flanged end plates showing outwards so one can see the depth of them, however this will give more area of copper exposed to the outside to conduct the heat into the air. !https://youtu.be/2EkjRslFgSM
2EkjRslFgSM
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 28, 2017, 01:15:20 PM
Hi Willy, more very interesting and relevant questions.  I hope that I can clarify a few points for you, so here goes.

Flash steam plants.  First fluids always flow from high pressure to low pressure, you need a pump or compressor to make them flow the other way, so yes, the feed pump must provide enough pressure to exceed the highest pressure in the coils, and that occurs right at the pump piston face.  The pressure is already lower on the coil side of the discharge valve and continues to drop through the coil to the engine.  The pump does indeed use power from the engine, but this is a small portion of the power developed by the engine.  Of course in our models friction is probably a greater loss in proportion to the work done than in a full size plant, but providing the packing and piston ring are not too tight it is still a reasonably small portion.  This is because the volume change of the water is very small with pressure increase, which means that work for compression is also very small.

You mentioned injectors, there is a lot of heat added to the water that enters the injector, and if it gets up to the point where it starts to vapourise too soon, the injector passages choke up and it stops working.  With a pump, the suction stroke of the pump lowers the pressure for the incoming water, and again if due to inlet temperature, the water vapourises under the lower pressure, the pump becomes vapour locked.  However there is more latitude than with an injector.

Water and steam are nominally at the same temperature.  I say nominally, because the boiler is not normally in true equilibrium while heating is going on.  The heating surface is at a higher temperature than the water, otherwise heat would not flow.  The water starts to vapourise, but it's pressure is slightly higher than at the surface due to the water depth, very small depth in a model boiler, and the expansion to steam means the density is very low so it is rapidly displaced to the surface.  If you turn off you element when the boiler is up to pressure, boiling will quickly cease, and condition will get quite close to equilibrium, especially with good insulation.  When all the water in the boiler is in equilibrium, the liquid and vapour are at the same temperature. 

Liquid and vapour are two separate phases, which under certain conditions can exist at the same time for most fluids.  Our boilers have only water in them once the air which was initially in the boiler is displaced.  The gas bottle used by many has perhaps propane, a chemical substance classed as a hydrocarbon, which is highly flammable and a useful fuel.  Propane also has a range of conditions where liquid and vapour exist at the same time, just the pressure and corresponding temperature are different from those for water.  Similarly, most things we might normally consider gases due to their nature at atmospheric temperature, such as butane, air, carbon dioxide and even natural gas all can exist as liquid at some temperature, very low in all those cases, and have a two phase region where liquid and vapour exist in equilibrium at the same time.

So liquid and vapour (or gas) are terms which refer to the phase state of a substance.  However, water is a substance whose molecules contain two hydrogen atoms and one oxygen atom chemically bonded together.  The substance water can exist as solid, liquid or vapour, and we are all familiar with all three of these phases.  There are quite a range of conditions over which any two of those phases can exist in equilibrium at the same time, however all three can only exist at the same time and in equilibrium at one specific temperature and pressure known as the triple point.

By the way, even in your cup of tea, liquid water and vapour are approximately in equilibrium, and of course there is also air at the surface.  When the water is hotter than the equilibrium temperature for the water vapour pressure in the air, some of it evaporates, and under many conditions you can see this as a steam cloud rising from the cup.  If you blow the steam away, more liquid will evaporate to replace it, and this evaporation takes heat out of the tea.  If the tea is cooler than the equilibrium temperature for the vapour pressure in the air, some of the vapour will condense as it does on any cool surface.

In the boiler, there is liquid water and water vapour at the same time, and when in equilibrium they are at the same temperature and pressure.  However liquid water has a much higher density than water vapour, so gravity and surface tension combine to place all the liquid in a continuous phase at the bottom of the boiler, while the vapour fills all the vapour space, and there is a clear visible boundary between them.  Vigorous boiling is not true equilibrium, but the equilibrium temperature is considered the best estimate of the average temperature of the whole.

I have not seen any thermal pictures showing the temperature difference in a boiler, but I am sure that they will exist.  I don't know what temperature range is normal, certainly it will depend on the energy density which determines the necessary temperature difference for heat flow.  One of my catalogues has an instrument for viewing the temperature colours, but it is around $1000 so I will not be picking one up any time soon.

Thank you for those two videos, really excellent productions.  Clearly your safety valve works well and switching off the power reduces the steam production very quickly.  I love that crane.  It just needs to be nearer the edge of the table so it can raise and lower things from the floor.  Is the control valve just a disk valve to reverse the directions?

If you are adding rock wool, it should be packed as tight as practical.  Air doing its thing means transferring heat by convection, so you want to limit the air movement.   But rockwool is quite a suitable insulating material.  For a more permanent job it needs a metal or preferably wood cladding for appearance and mechanical protection.  Not sure if it is as bad as glass fibre, but worth taking the normal precautions for loose fine fibres when you are working with it.  Saves the itchy fingers at least.

I will come back to those flanged end plates tomorrow as it is getting late.  I suspect the above will raise more questions, but the post is long enough so I will finish off and return to the topic if there is any clarification required.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 29, 2017, 01:10:34 AM
Hi thanks for all this info , and i'm glad i don't have to pay the full rate for these consultancy fees !!! . I have noticed when using propane bottles in the summer that you do get a frost line forming at the level of the liquid  when using the gas at full tilt ! I shall do another boiler test with the extra insulation. These videos were done 8 and 10 years ago so are not of the quality and ease that we can do today on my Apple computer !! .
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 29, 2017, 11:55:15 AM
On boiling propane -

Hi Willy, just to complete yesterday's question on the way the ends are placed.  The inspector has to take into considerations the issue that get presented and I am sure that after a while the most common and troublesome problems are well known.  I would expect that the inspectors talk to each other and compare notes about the design that are people are more likely to present correctly.  If I understand the issue addressed in the code, it is to verify that the silver solder has flowed right through the joint as well as that the flange is the correct length.  I guess with those 5.5 mm cameras mentioned in another thread, inspecting the penetration on the inside is a bit easier, but the measurement is not.  Has to be measured before soldering.  I don't really know.  However the issue of heat loss is easy to control as with the flange facing outwards it is pretty easy to apply as much insulation as you need to minimise the heat loss.  So I would talk to the inspector about the design, present all the prepared parts before soldering, and go along with the instructions given, and insulate the end result (after the hydrostatic and steaming tests are approved).  Your outside cladding then determines the visible appearance of the boiler.

The propane in your gas bottle behaves very like the water in your boiler, but the equilibrium pressure and temperature are just in a different range.  At atmospheric pressure, propane boils at minus 42 deg C.  At 20 degrees, it's equilibrium pressure is about 700 kPa(g), or 100 psig.  When you open your fuel valve to burn some gas, the pressure drops as has leaves the bottle.  This means the gas and liquid are no longer in equilibrium.  The vapour pressure of the liquid is then higher than the gas pressure.  Some liquid has to be boiled off to maintain the equilibrium.  Now we don't generally put a flame under the gas bottle, so we only have the heat available in the liquid and from the atmospheric air.  The temperature difference is initially zero, so no heat transfer from air and the heat comes from the liquid which rapidly cools.  Now with a large enough bottle surface area and pleasant temperatures like we enjoy here, say 20 degrees or more, and a low fuel off take, you may soon get enough heat transfer to maintain the new low pressure.  But the liquid will be cooler than it was when the gas valve was closed.  Turn up the burner a bit, and the liquid temperature soon falls below the dew point of the atmospheric air, and moisture condenses on the outside of the bottle.  If the propane pressure falls to around 350 kPa(g), the temperature will be around zero and any further drop in pressure soon causes that moisture to freeze, as you have seen.  Much more likely in your climate where you spend much more time below 20 than above.  However in Canada, and many US states a different story, and low ambient temperatures can be insufficient to maintain the gas burner pressure requirement.  The propane itself does not freeze at these temperatures.

With butane, the pressures are quite a bit lower, so in cool climates, some exhaust steam heat, or even just conduction through a common base plate with the boiler is used to maintain the pressure in the small gas bottles generally used in models.  I can give you the butane pressures if you need them.

It's not really about consulting fees, it is a pleasure and a privilege to contribute something back to the forum where I learn so much, and my machining skills have a long way to go.  It is just useful information for many aspects of our modelling hobby, not well known within the model making community, but was basic to my work throughout my career.  It is not secret information, just basic thermodynamics.  I am delighted that you are finding it interesting, and many others obviously keep coming back despite my often wordy style.

Thanks for looking in

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 30, 2017, 01:22:20 AM
hi <Thanks for that... I was looking at a refrigeration van today and when the chap opens the door a whole lot of white swirling mist appeared. Can you/ explain what i was actually seeing it looked a bit like steam but obviously not ? !! Thanks for the info about the rock wool and i shall pack it as tight as possible............Willy
Title: Re: Talking Thermodynamics
Post by: MJM460 on September 30, 2017, 12:47:06 PM
Hi Willy, that type of question is always a little tricky to answer because there are so many contradictory intuitive thoughts about the process involved.  The way I work it out is like this.  The air in the refrigerated van is cool compared with the atmospheric air outside.  Hence when the door is opened, and the cooler, more dense air flows out and is warmed when it mixes with the outside air.  In the process the warm air is cooled.  Now if it happens to be fairly high humidity, it can be cooled below the dew point temperature, the temperature at which the humidity becomes 100%, and if cooled further, excess moisture must condense.  The moisture is well mixed with the air so condenses in tiny droplets.  These appear as a mist, like a fog or a cloud.  So what you saw was a fog of moisture condensed out of the outside air by the cool air from the van.  The tiny droplets are so small that viscous effects in the air prevent them from quickly falling into a continuous liquid phase, and they remain suspended for quite a while.  Eventually the overwhelming quantity of the warmer outside air evaporates the droplets and so the fog does not spread far.

It is often said that warm air can hold more moisture than cold air, which is a reasonable enough observation.  However it can also be looked at in terms of partial pressure.  You can look at the steam tables and see that at low temperature, the equilibrium pressure is lower.  If the water vapour partial pressure in the air reaches that equilibrium pressure, it cannot increase further and any excess results in some condensation.  This point in an air mixture is described as 100% relative humidity.  If there is less moisture in the air, so the partial pressure of the water is less than the equilibrium pressure for that temperature, we use the term humidity, or relative humidity, which is defined as the percentage of the equilibrium pressure of the moisture in air.

When you look at this way, you can easily see that if you have air with a certain moisture content, and you then cool it, then the same absolute moisture content becomes closer to the equilibrium pressure at the lower temperature.  So the humidity becomes higher and eventually if you continue cooling the air, it reaches equilibrium pressure for the new air temperature, and condensation starts as you saw at the back of the van.

I hope that dispels the mystery of the fog at the van, which is the same process as the formation of  fog on a chilly morning, or of a cloud high in the sky, and of the visible steam from a kettle or from your engine exhaust.  Water vapour itself does not reflect light and is quite invisible.

Just a short post tonight, a short night as we start daylight savings, so must put the clocks forward, after a big family day today.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: Kim on September 30, 2017, 05:26:00 PM
OK, so now it's my turn to ask a question, that came out of your answer to Willy here.

What exactly IS steam?  Steam is water in gaseous form, right?  We just happened to give it a special name. In fact we gave water is so special, it got 3 names, one for each form (Ice, water, steam - where as most things only get one name and you have to specify form "Liquid Natural Gas" for example).  But my question: how is 'steam' different than water suspended in air?

At standard atmospheric pressure, the water has to be at 100C to become steam.  So, I can see that the water vapor hovering around the door of the refrigeration truck couldn't be steam, because it is clearly not 100C.   But then what makes water vapor different from steam? Is it just the amount of energy contained in the individual water molecule?  Is that the only difference?  Is Humidity NOT water in gaseous form?  What is humidity then?

Thank you for all the interesting discussion MJM.  I did take a term of Thermodynamics in school, and I'm following most of what your saying, but clearly, making that head learning mesh with the real world is challenging me.

Kim
Title: Re: Talking Thermodynamics
Post by: steam guy willy on September 30, 2017, 07:37:10 PM
Hi Kim ,good questions there and would it be possible to answer  using words of two syllables or less ?!! And to add to this ....if you had water in a sealed tank and forced air at high pressure into it what would happen ??
Title: Re: Talking Thermodynamics
Post by: paul gough on September 30, 2017, 11:14:19 PM
Taking a step back to horsepowers and specifically boiler horsepower. Re-reading 'Perfecting the American Steam Locomotive' by J Parker Lamb, in Chapter 3, 'The Physics of Steam Power, p. 43, he gives the equation; boiler horse power (maximum) = 1/6 grate area (sq ft) X boiler pressure (psi). He states this is an approximate value and an empirical formula. I presume it was used as a comparative figure by designers, maybe a guide to match boiler capacity to maximum demand of certain size cylinders and a means of estimating fuel consumption. Were there any other uses for the resultant? Was it just a designers 'rule of thumb'? Anyone have any more on this??? Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 01, 2017, 11:51:19 AM
More on humidity -

Hi Kim, always glad to have more questions.  I think you are right on the mark pointing out that water is so special.  After all it is a basic necessity for life, at least life as we know it.  Only three words for water says something about the climate where you live.  I believe that the Eskimos have around thirty names for snow.  As a sometimes skier, I can describe about six, but thirty is amazing.  I suspect that is the heart of the dilemma.  We knew about water, ice and steam long before we knew anything about thermodynamics, so we had words for the different forms we knew, long before we had the need for more specific terms, or even knowledge to explain our observations.

To get to the basics, first the term water, defines the chemical compound consisting of two hydrogen atoms and one oxygen in each molecule, what ever its form.  But the term obviously overlaps common usage for the liquid.  We know it exists in three forms, usually known as phases, solid, liquid and gas, as do many other compounds.  And we tend to use the word steam for the gas phase, but perhaps more particularly when we can see it, like near the spout of a boiling kettle.  And sometimes we use the word vapour.  But it is not so well understood that the gas phase does not reflect light and cannot be seen, like very close the the spout of the kettle, and the mist we see a small distance from the spout is in fact fine droplets of condensed liquid.  Same as a morning fog, sea fog or even thin cloud.

I had to resort to the dictionary to check some of the normal usage definitions.  It says steam is water in the form of a gas or vapour, or water changed to this form by boiling, and extensively used for mechanical power or for heating purposes.  Also the mist which forms when when gas or vapour from boiling water condenses in air.  It also defines steam point as the equilibrium temperature of liquid and vapour phases of water at 101.315 kPa which is equal to 100 degrees C.

So it seems that you are correct in associating the word steam with boiling, and in answer to your question about the difference between steam and the water suspended in air as fog etc. I suggest it is tied to the method of generation of that form in common usage, but the first clause in the dictionary definition just says water in the form of gas or vapour, followed by the word or.

Applying thermodynamics and our knowledge of how the equilibrium pressure of the liquid and gas phases of water change with temperature, I suggest that there is no physical difference.

That leads to the question of humidity.  I suggest this is our everyday description of the moisture content normally found in air.  Normally, moisture content of air, which is proportional to the partial pressure of water vapour in the air, is not at the equilibrium pressure for the prevailing atmospheric temperature, but always less.  The weather bureau records relative humidity which is the percentage of the equilibrium pressure at that temperature.  This means that the actual moisture content of the air varies with temperature, when at the same relative humidity.    As I mentioned yesterday, the steam tables tell us the equilibrium pressure for each temperature.  The relative humidity reading tells us the proportion of that equilibrium pressure which is actually present.  It is then clear that if air with a certain partial pressure of water vapour is cooled, then eventually a temperature is reached where that partial pressure does equal the equilibrium pressure, so condensation must begin if cooling continues.

I have every sympathy for your feeling of challenge in applying the basic thermodynamics you learned.  I went into a career where it was a basic tool used every day, but while I realised that the knowledge had the answer to everyday questions, some of Willy's questions have really got me thinking carefully to make sure I had an answer that was properly supported by the theory.  I still check very carefully for relevant examples in my textbook before I post on some of the more obscure ones.

Hi Willy, I think I failed on the request for words of no more than two syllables, but I hope the longer words are generally well enough known that you won't need a dictionary to read it.  I should have mentioned yesterday, that when that condensation begins, the heat released helps warm the cooler air so the condensation soon stops, hence the limited extent of the mist.  But the new question, you have a boiler for example with water as a liquid, plus some vapour, at the equilibrium pressure for the temperature once the system has settled and all the temperatures are equal, plus some air to make a total pressure of atmospheric pressure prior to sealing the boiler.  If you now use a compressor to add more air, you increase the partial pressure of the air, but not the water vapour.  If you are extremely pedantic, you have to make sure the air is cooled to atmospheric temperature before it enters the boiler, otherwise it will just need more time to be at true equilibrium.

Some of the air will dissolve in the liquid phase, a quite small amount, and the rest just adds to the total pressure in the boiler.  You didn't say how high was high.  You boiler probably is designed for something around 100 psig when cold, or 700 kPag.  At this sort of pressure, the air in the vapour space and the water vapour act near enough to independently, as described by Dalton's law of partial pressures.  If you have a suitable vessel and suitable compressor, the situation is probably about the same at 500 psig.  You could continue to 5000 psig.  I had to specify a compressor for that once, and you don't want to go that far, but somewhere around that stage, perhaps lower or even higher, the atoms in the vapour space do get sufficiently crowded to affect each other and you can no longer simply add the partial pressure.  I am not very familiar with working in that range, and I am not sure just where it starts to be important.  Then, when you release the pressure, the amount dissolved can no longer be accommodated, and the excess dissolved air bubbles out like an opened can of soft drink.

Hi Paul, I can see that boiler horse power is still fascinating you.   As you have said, the term and also the formula are just empirical guidelines, and the formula is not dimensionally consistent unless that constant, 1/6, has the right units.  It is not just a pure number.  However there is some sense to the form of the formula, as grate area is probably a good indicator of the amount of coal that can be burned in a given time, and hence energy input per unit of time. And we know that pressure is necessary for the steam to do work.  So it makes sense that if you make a graph of potential engine brake horsepower against grate area times pressure, you might get some sort of correlation.  Especially in the days before super heaters.  Not necessarily linear, though a straight line can always be drawn as an approximation.  But whether the 1/6 factor is appropriate I really don't know.   To avoid variation due to the engine efficiency, you could use the output of that ideal adiabatic engine to make the results a bit more consistent.  Probably no help to you in designing your little locomotive boilers, but a bit of interesting history.

Thanks to everyone for looking in

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on October 01, 2017, 11:30:14 PM
Thanks for your comments on the boiler horsepower issue. I have always been fascinated by the goings on in boilers, particularly loco boilers with their high steaming rates and the relatively tight space they have to occupy. I am hoping that at some point our thermo discussions might flow toward and into the 'what is happening' from fire hole door to stack top. Almost none of us are combustion or chemical engineers but some grasp of underlying principles regarding what is required for a successful boiler and the combustion that goes on in it might expand our understanding and avoid misconceptions and advance design.

 One specific question I have and have not been able to find any satisfactory answer to is; do the boundary layers scale down, with steam, air, products of combustion remain more ore less a constant dimension from full size to model dimensions and is there any significant change in this 'dimensioning' when the fluid is relatively static, eg. in a vessel, or when moving at some velocity, such as steam in a pipe that is vented or flue gases in fire tubes, (leaving aside here 'disruption' such as by steam bubbles in the water adjacent to a heating surface)? Finally, if there is dimensional variability, what parameters might impact, eg. pressure, temperature, velocity, the proportion of the vessel eg. does the boundary layer in a 5mm  copper steam pipe differ from one  50mm in dia. or a 12 mm fire tube compared to a 50mm one? Also does the material matter, by this I mean is a higher conductance metal like copper have different boundary layer formation than that adjacent to a steel tube? Hope all this is not too headache inducing! Regards, Paul Gough.






Title: Re: Talking Thermodynamics
Post by: MJM460 on October 02, 2017, 01:57:28 PM
Boundary layers and scaling -

Hi Paul, obviously you are talking about scaling of size, and not that boiler scale that grows in our boilers if the water contains various impurities.  And you allude to the fact that the gas composition and properties are the same in our models as in full size, and how that might affect our models.

Most of the research work in fluid mechanics is done on scale models and much thought goes into how to address this very problem.  We don't have many,if any at all, cases where we can find a suitable fluid to use in a model that has scale density and viscosity in particular.  Dimensional analysis is one of the techniques used to uncover various combinations of properties that are dimensionless so experimental results can apply to all sizes.  For example the length of a pipe divided by its diameter is dimensionless, so an experiment can be done with a certain size and length of pipe and the results are found to apply to another pipe diameter if the length is such that the ratio of length to diameter is the same.  One example is the development of a velocity profile in a pipe.  At the entrance to the pipe, the flow velocity is roughly uniform across the circular cross section of the pipe.  However, the fluid velocity in contact with the pipe wall tends to be zero.  In low flow situations, the flow proceeds along the pipe, viscosity means the stationary layer at the wall slows the layer immediately inside and so on until a roughly parabolic profile develops where the velocity on the centre line is about twice the average velocity calculated from the flow through the pipe and the cross sectional area.  This profile develops fully at a length where the L/D ratio is in the range of 60, though it varies with the actual velocity.  At higher flow rates, the flow becomes turbulent, the profile is a bit more of a flat topped parabola, which is fully developed at L/D in the range 25 - 40.  Unfortunately when applied to the diameters in our models, the lengths are equal or perhaps longer than our model, so we generally operate in the range of the developing profile.  Another of those dimensionless groups, given the name Reynolds number after the one who identified it, involves viscosity, density, velocity and diameter in a dimensionless combination.  Knowing those properties do form a dimensionless combination you can work it out yourself with a little trial and error.  Then, while you cannot find corresponding "scale fluids", if you run your experiment at equal Reynolds numbers of your model and full size, many results will correlate nicely.  So Reynolds number effectively replaces velocity and compensates for viscosity, diameter and density as well.  Of course with a ship model, there is wave making to consider, but there another dimensionless group called the Froude number, will allow you to calculate a speed for your model that will make the same wave pattern as the full size ship at its appropriate speed.  But fundamentally the flow in a model boiler tube of a certain l/d can be accurately compared with the flow in a full size boiler tube of the same l/d.  One of the two must be determined by experiment, then the other can be calculated.  Of course in fluid mechanics experiments, every effort is taken to stick to the issue of flow and avoid other energy transfer.  Mostly because the properties of viscosity and even density are quite temperature sensitive.  It is hard enough to understand what is happening in the boundary between the stationary wall and the bulk fluid with steady properties, without having to deal with changing viscosity and density as well.  And the reverse applies, the velocity affects the effective temperature profile.  If the velocity is high, warmed fluid is carried on quickly and replaced by cooler fluid so the temperature gradient is effectively changed, and that changes the viscosity which changes the velocity profile.  Not to hard to get a headache in that area, and probably not very useful to go much further. 

The take away points are first, the velocity in a pipe is zero at the wall and increases towards the centre to give a roughly parabolic velocity profile, perhaps a bit flattened in the centre, and that large and small sizes can be compared provided that size and velocity at the comparison points have certain dimensionless combinations of properties controlled to be equal in the sizes being compared.

The other point you alluded to is the effect of the velocity of a fluid on the heat transfer rate.  It is worth noting that velocity has a huge impact, mostly because of its effect on the temperature gradient near the wall, and this temperature gradient plus conductivity of the fluid are the main factors influencing the film coefficient which determines the overall heat transfer rate.  It is sufficiently important that there are two distinct types of convection heat transfer recognised.  First there is natural convection, where the velocity is determined only by the change in fluid density due to the heat transfer.  Second, there is forced convection, where the fluid is given additional velocity, perhaps by a fan or a pump.

You could look at the coffee cooling experiment, perhaps tea so we don't yet have to talk about the effect of froth on the surface.  If you just let the cup sit, the tea is warmer than the air, so liquid and vapour are not in equilibrium at the surface.  Some tea evaporates to cool the air and increase the vapour pressure at the surface towards equilibrium.  However, this also warms the air near the surface, which decreases in density, and warm air rises to be replaced by more cool air.  This is called evaporative cooling, and occurs in a cooling tower, a coolgardie safe, or an evaporative air conditioner.  Similarly, there is heat transfer through the sides of the cup as we know because the cup feels hot, so it does loose heat to the air, but slower than the surface as China is not a very good conductor.  Never the less, it is an example of natural convection.  Eventually the tea cools to a drinkable temperature and we normally avoid continuing the experiment until the tea is at air temperature.  That would not only be wasteful of the tea, but also of or time, as the nearer the tea gets to air temperature, the slower the heat transfer goes, proportional to temperature difference, remember?

Now if you blow on the surface of the tea, you remove that extra vapour, so the surface is no nearer equilibrium and the evaporation and evaporative cooling rate continue at a higher rate.  Similarly blowing cool air on the sides of the cup carries away the warmed air, replacing it with cooler air, so the effective temperature gradient is greater and the cooling proceeds faster, unless of course you are a politician!  That is called forced convection.  The increase in cooling rate is not linear with velocity, so the complexity continues if you are trying to analyse the situation.  But a fan would supply more air at a higher velocity than blowing, so does further increase the cooling rate, and shortens the time to cool.  Even so, as the temperature difference between the cup and the air decreases, the heat transfer rate slows.

The flow and boundary layer problems are discussed in any engineering fluid mechanics text book, they are pretty heavy reading but worth a look in a library.  The heat transfer is described in any engineering heat transfer text book, again very heavy reading, but definitely the place to go if you want to explore the issues further.

I hope that helps.  I think back to wet steam next time, but eventually we will get to the boiler, though I am limited in not being a combustion engineer, so any contributions will be welcome.  Oh, and I will talk about the effect of copper or steel properties on the heat transfer.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 02, 2017, 02:53:19 PM
Hi, interesting stuff again ....LBSC used to say that you cannot scale nature , so interesting to note that they do make models that then work when scaled up. On reading about combustion in boilers it was noted that air intake above the fire grate helped wth the compleat combustion of the gasses above the fire. also when burning wood in an open fire you never use a grate ,(this is what my mum use to say when we had open fireplaces in the house) !! I do have a book somewhere that talks about combustion produced by Charringtons a coal supplier but i don't know where that is at the moment...........
Title: Re: Talking Thermodynamics
Post by: paul gough on October 02, 2017, 05:15:31 PM
Hi Willy, Air supply above the grate goes back a very long way, talking locos here, many of the large modern engines on the U.S. had air jets along the sides of fire boxes, most I think induced by steam, some compressed air, a few things brought about their re-application. Things had reached the limits inside a conventional firebox/combustion chamber due to enormous grate sizes of the big semi-articulated locos (Mallets) firebox volumes weren't sufficient for combustion, nor was turbulence or air fuel ratio of certain types of fuel, usually coal. The history of 'overtire jets' as they are commonly called was an interesting line of research I did a few years ago and it leads back to very early locos in Britain. There were penalties for 'smoke nuisance' and all sorts of things were tried, including 'air tubes', at this time they were at the front of the firebox, just tubes secured through the outer and inner wrappers of the firebox, later more tubes in rows and different positions some with steam jets were tried out. About the time when locomotives transitioned to burning coal rather than coke it was found that larger firebox volumes and the invention of the 'blower' to induce more draught, especially when stationary, overcame the 'smoke problem' sufficiently for the extra boiler making of 'tubes' and maintenance to override any gain, but of course the Super Power era of locos in the U.S. caused things to go full circle so to speak.

Hi MJM, thanks for the detailed comments. They go some way in explaining conditions and to the improvement in steaming of a couple of boilers, loco type, that I 'played with'. The gas fired one had a boiler excessively long and the coal fired one had fire tubes too small, thus amounting to the same thing. Introducing thin strips of metal plate twisted into a low amplitude spiral the same width as the tube dia. brought about better steaming in both boilers. The gas fired one transitioned to thin stainless strips as a permanent fix but the 12" gauge coal burner was a failure in that the strips became impossible to remove to clean the tubes due to being cemented in by the soot or burnt out. We deduced that more turbulence and slower velocity of the gases contributed to better heat transfer. Seems it wasn't far wrong.

One question please: Does the friction in small steam supply pipes to an engine you spoke of previously, 1/8" Dia I think, come from friction between the 'stationary' boundary layer and the main body of fluid or from turbulence caused by the high velocities induced by the small pipe, if both, is one dominant?? I'm  not clear on this.  Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 03, 2017, 10:45:32 AM
More on scaling models -

Hi Willy,  we always had open fires when we were young, some had a grate, some not.  A good fire can be made in either.  I just checked with my wife who is a real expert fire maker, a result of her farming background, and she also has no preference for grate or no grate.  Air over the top is helpful in making sure combustion is complete, you don't want the possibility of carbon monoxide entering the room as it is very toxic, unlike carbon dioxide.  There are a few other nasties in the partially burnt fuel, depending on just what you are burning, so a good idea to introduce some excess air.   But you also need air into the seat of the fire to support combustion.  A fire needs to be compact enough to generate a high enough temperature to exceed the required temperature for combustion of wood, and needs enough air for combustion but not enough to cool the fire and extinguish it.  In burner technology, these two air sources are called primary air and secondary air.  I always thought the grate was useful in small fireplaces to help prevent burning wood rolling out, not so necessary in a larger one where it is easier to build the fire so it is quite stable.  There will be something behind your mothers advice, there usually is, but I am not sure that I understand the real reason.  I wonder if anyone else has more detail.

Hi Paul, Thanks for the interesting information on secondary air in locomotives.  I probably signed off a bit prematurely last night, not time wise of course, but there is probably some value in following the topic a little further.x

The Reynolds number is density x velocity x diameter / viscosity.  The units for density, velocity and diameter are well known, though perhaps I should emphasise the unit for diameter is metres, not mm.  The units for viscosity are Newton seconds per metre.  You can easily check that it is dimensionless.  That formula emphasises the place of density and viscosity.  Tables often list kinematic viscosity instead of absolute viscosity, but kinematic viscosity = viscosity / density, so it can easily be used in the formula as that ratio is directly in the formula.  Kinematic viscosity has the units L^2/T.  If you have a larger tube and a smaller one, and you operate both at the same Reynolds number for equal flow patterns, viscosity and density will be the same for both, so you can see there is an inverse relationship between velocity and diameter between a smaller and a larger tube.  Now  dimensional analysis tells you that you need a quantity with the dimensions length, so a representative size.  It does not tell you what that length should be.   You might be aware that our aeronautical colleagues use the wing cord in calculating Reynolds number,  not the span, while in piping, using diameter as the representative size yields better correlations than using length when the experiments are analysed.

Another relevant dimensionless number is the ratio of cross sectional area to surface area of a tube.  Cross sectional area is important to the flow rate through the tube for a given pressure drop,  while the surface area is important to heat transfer.  For problems involving heat transfer and flow, you can't get the same conditions in both tubes for both flow and heat transfer, but some investigation might lead to an optimum diameter, where larger does not have enough surface area, and smaller does not allow enough flow.  And of course you need enough pressure drop to drive the flow.  So just finding a dimensionless ratio does not mean you can use it to scale the design.  This example illustrates that our models are small boilers, not scaled down ones.  However when trying to model a prototype, we do use a constant length ratio for as much as possible so that the overall proportion and appearance is preserved.

Those spirals are an interesting technique.  I am not sure whether they just provide extra surface area to absorb heat then transfer it to the tube wall by conduction where it contacts, or whether they modify the flow pattern so improving convection heat transfer.  Or more likely it is a complex combination of both possibly plus other factors.

The definition of friction factor uses only the wall shear stress and V^2/2g.  There is another term, apparent shear stress, which varies across the radius of the pipe.  The variation is due to the effect of turbulence and momentum transfer, but the momentum transfer requires a transverse component of velocity.  Obviously must be zero at the wall, but increases as you move into the bulk flow where the flow is turbulent.  The apparent shear stress appears to be linear between the wall and the centre, so near the wall viscous effects predominate, then moving inwards, the two become equal, then momentum transfer and turbulent effects predominate.  I will be interested to see how you use this, but again there is more detail in any engineering fluid mechanics text.  The maths gets pretty heavy quickly and I am definitely less comfortable in this area.  I hope that is sufficient for now.

Still planning to return to wet steam next time,

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on October 03, 2017, 12:32:08 PM
Again, thanks for the leads into the goings on in pipes, I really hope I have not caused any of this threads devotees to run away because of my fluid dynamics enquiries. I think I have enough grasp of things now to know what to look for when I get the chance to pursue some combustion and fluid flow tomes. I'll have to take a trip down to J.C.U. library, I think they teach some of the M.Eng. undergrad program in Cairns, so maybe I'll find a few books there. For the moment I'll retreat to my cell, meditate on your discourses and take a vow of silence, lest our followers become restive at my abstruse questions. Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: Steam Haulage on October 03, 2017, 12:39:42 PM
Hi MJM,

Could you please expand the definitions which you use in your 4th Paragraph regarding viscosity and density. I am especially interested in how you measure these values and the practical units of measurement in the various forms of apparatus used. I presume your work depends on treating water and steam/air mixtures as fluids.

Jerry
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 03, 2017, 03:07:33 PM
Hi Paul interesting about the induced air into the fire box .....would this air be better hot or cold and would it help to be dry or moist ? i think i have heard somewhere that IC engines run better in the early morning when there is more moisture in the air ??   
Hi MJM I think my mum said that the wood would reflect the heat back and forth to each other to stop it burning too quickly, but keep the radiant heat from escaping,also the grate would make it burn to quickly as well , This may have been because she would fall asleep after supper in front of the fire and when she woke up after snoring a lot she could rake the fire a bit and get a nice hot blaze going !! she also had a large metal plate to put against the fire place to increase the draft . this was after she almost burnt the house down when she used the daily paper across the fireplace that caught fire by the suction going up the chimney and floating around the room.Before i leave my mum....she used to fall asleep reading a book ,but say she never did, until we took the book away and turned it upside down !!!She would then wake up and glare at the culprit and say Time for bed !!!.......
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 04, 2017, 12:11:08 AM
Hi MJM I have redone the boiler test after insulating the end and also some of the underneath. The results seem to be quite different though First of all it would not get above 126 C without the safety valve blowing off this time and i cannot screw down the safety valve any more .It may be that the probe did not go in as far this time by about 1/4" due to the end piece of wood. Also the temp/time curve is slightly non linear. and the temp difference per minute varies quite na lot from 16 degrees in the middle to 9 degrees at beginning and end !!  So something strange going on here to think about ....the clock and thermometer are the same so it must be the boiler.... the leaks seem to be a little bit worse as well.... any way...your turn now to come up with some explanation if possible please !........Thanks.....The latest graph is the red line btw
Title: Re: Talking Thermodynamics
Post by: paul gough on October 04, 2017, 12:33:23 AM
Well, so much for my vow of silence! Hi Willy I hope the following will help.

 Regarding 'Air Tubes' see attached photos for only a couple of historical examples. First photo from the loco 'Liver', L&M Rly, 1837. Three grates, bottom two for coal and top one for coke, but note the air tubes through the front of the firebox supplying supplementary secondary air to grate 'b', (middle). The two fire hole doors also had 'perforations' to allow main supply of secondary air in above the grate, as is generally normal for most locos. This is the first example of 'air tubes' I have been able to find.

Photo 2, shows D. K. Clarks "Steam Induced Air Currents" as applied to an Eastern Counties Rly. loco, 1858. He claims it, " is quite successful in preventing smoke, making a bright fire, keeping up steam, and working economically. By inclining downwards the jets of steam, the air may be thrown at any desired inclination upon the fuel and amongst the smoke."

As you can see secondary air delivered through tubes goes back to the very first decade of railways and the steam jet type to the 1850s. These are only two examples and it was just a part of the rapid evolution of firebox design for locos happening around the middle of the 19th century.

Now addressing your question; as far as locos are concerned, I doubt that there would be any significant difference in practical terms. Remember steam locos operated outside in hot dry sunny conditions in deserts and frigid foggy or sleet driven nights atop mountains, sometimes even encountering them on the same trip or shift. I doubt that even in the sophisticated 'Super Power' era of steam locos, temp. and humidity differences in secondary, or for that matter primary air, would have concerned designers. However drivers or firemen who were especially sensitive to their engines and the various conditions under which they ran them may well have 'sensed' a difference on a sunny summer day and a snowy night, maybe would have been able to quantify it very roughly in noticing less shovels of coal on a long continuous grade, but I suspect they would have been the only ones who could. Having said this I should think the modern large stationary plant people might have different views on the matter, as they can do many things not possible on locos, but I have not looked into this area.

I too 'sensed' better running from my motor bikes when I rode, (half a century ago!), considerable distances when it was cold and foggy, especially at night, but I never tried to quantify any difference, let alone accurately do so, the bikes just seemed to 'go' better. Obviously cold air is denser and a bit more oxygen but I was riding mostly at moderate elevations also, 600-1000 metres, so less oxygen too. As to what happens to H2O in the combustion chamber of a bike I know not, perhaps it breaks down to H & O and assists. I suppose the F1 car racers might know, SCO might have an insight. When I was in the Diesel Traction Section in the N.S.W. railways it was a non issue and no-one from the design office ever mentioned it either, but diesel loco operators in Sth. and Nth. America whose locos have to traverse very high altitudes as well as work in incredible temperature/humidity ranges might have investigated it. Regards, Paul Gough.



Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 04, 2017, 11:56:45 AM
Just a quick message I think the leak at the back may have wetted the rock wool and so caused a heat loss there but am not sure !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 04, 2017, 12:20:31 PM
Viscosity and a boiler test -

Hi Paul, your questions are always welcome, I don't mind a little excursion into fluid mechanics, there are plenty of fluids flowing through pipes in a refinery, but perhaps I can leave you to start a thread about historical locomotives.  Of course real life examples are always helpful in a discussion like this, so not a very clear line, more a wide grey area.

Hi steam haulage, welcome aboard.  Viscosity is that characteristic which makes oil or honey flow more slowly than water if the jar is tipped.  In a plain journal bearing, the resistance to rotation or shear force per unit area of the bearing surface is proportional to the velocity gradient across the oil film thickness, h.  Mathematically, F/(pixDxL) is proportional to V/h, where V is the journal surface velocity.  The proportionality factor turns out to be a constant for a large group of fluids which are called Newtonian Fluids.  Of course the constant does change with temperature.  The constant is given the name viscosity, or more accurately, absolute viscosity.  (Non-Newtonian fluids are those where the factor is not a constant, but varies with the shear rate.) So absolute viscosity is a direct measure of the shear force in a fluid when it flows.   In many situations, the combination viscosity/density occurs, and this term is given the name kinematic viscosity. 

Viscosity is measured by a Saybolt viscometer, which is a strictly standardised cup like device with a short vertical tube in the bottom, all immersed in a temperature controlled oil bath.  The number of seconds it takes for 60 cm^3 of the fluid to flow out through the tube is used in an empirical formula which actually gives a value for the kinematic viscosity.  From this, absolute viscosity is found by multiplying by density.  You may have heard of Saybolt seconds.  My heat transfer book has tables of fluid properties, and it lists kinematic viscosity and density, leaving you to do the multiplication if you need absolute viscosity.  I am not sure if that answers your question, but please ask if you want more or something a little different.

Hi Willy, I am glad you mentioned the newspaper trick, I have used it often in my youth.  It is not supposed to catch fire, but if it does, it is being strongly pulled up the chimney, so I am not sure how it flew into the room.  Falling asleep during a fireside read has a long and honourable tradition, as has stirring the ashes to get the fire going again the next morning.  But I suspect that your mother is right, not if you use a grate, as it will allow extra air so the fuel is more likely to completely burn out.  But the radiant heat is what you need if you are sitting across the room, instead of standing close with your back to the fire.

Now a new boiler test, so back to the real topic at hand.  Great job on that boiler end.  It can be tidied up when you are happy with the overall result.  That safety valve is a bit of a puzzle, it allowed a higher pressure last time.  Of course, there is an inconsistency with the pressure gauge readings in your picture, which suggests there is still air in the boiler, contrary to my understanding that it is lost pretty quickly once the safety valve lifts.  Is it worth continuing the run a little longer to see if the pressure settles back?  Last time I understood that the safety valve lifted early, then was screwed down to the final pressure setting, so I assumed the air was gone.  But let's look at what else your graph shows.

First the little curve at the beginning.  As with last time, I suggest this is due to having to first heat the element then the encapsulating insulation and sheath before water starts to heat.  Then it is fairly linear until you reach 100 degrees, when the water vapour pressure finally exceeds the atmospheric pressure of the air and water vapour that was in the boiler when it was sealed.  Before this point, notice that each temperature is reached just a little earlier than last time, a small effect but almost certainly the result of your end insulation, or the moisture from that leak. Unfortunately the other end is still bare, and the area of the shell which has thin insulation is much greater than the area of the end, so the effect is not yet dramatic.   Then once you reach 100 degrees, the pressure continues to rise, but the temperature is a bit slower to rise for two reasons.  First it takes more energy to evaporate the water that is evaporating at a higher rate, and of course, that is the point where your leakage starts to be evident, and of course the leakage increases as the pressure rises.  (Actually a guess, but perhaps you can confirm.)

As your electric boiler has constant energy input, the energy input is the same now at about 125 deg C as it was previously at 143 (or 148, I am a bit unsure which).  I approximated to 125 because it explicitly occurs in my steam tables so a bit easier to use.  We calculated the heat required to get the boiler and water up to steaming temperature last time,  if we look just at the water, to get from 125 to say 143, we only need 80 J/g, but to evaporate steam at 125 degrees requires 2188.5 J/g, so a leak of about 0.46 g/s absorbs all the available energy, so prevents the temperature and pressure from rising.  The method is a bit rough, but I suggest it shows that a leak has a significant impact on the maximum pressure that can be reached.  But it is a leak of the whole engine consumption.  That would be quite a leak so probably not the whole answer.  And remember we showed last time a significant heat loss and the small area of insulation has not reduced that much.  So possibly only 60% of the heat becomes steam.  Possibly still some issue with that safety valve.  Needs some careful observation and more thought.  Of course if the engine was running, there is most of the steam, and a much smaller leak is all that is necessary to prevent the boiler reaching the original higher temperature.

I suggest the next step is some gaskets or thread sealant to eliminate the leaks, and put some insulation around the cylindrical shell, and even a bit behind the sight glass and around the other fittings.  Try and add half an inch or more over the whole boiler with just a little hole so you can see the gauge.  But you can see that some simple calculations can give some valuable insight into what is going on.  Even when there are a few assumptions, there is a pointer to the direction for further experiment.

I will still keep trying for wet steam next time, and that moisture in the engine inlet.

Thanks everyone for looking in.

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 05, 2017, 01:21:21 PM
Wet Steam -

I have been trying to get to this topic for a while in preparation for talking about engine exhaust and engine power output.

I have often referred to water in a boiler, where there are normally two distinct phases.  The liquid, which has a defined volume fills the lower part of the boiler, then is bounded by a free surface, and a gas phase which occupies the remaining space.  In a closed boiler, without significant temperature gradients, the liquid phase and the gas phase are in temperature equilibrium, and the pressure of water in the gas phase is determined by the temperature, and the values are listed in the steam tables.  If the water vapour pressure is lower than equilibrium pressure, some of the liquid will evaporate until the pressure is equal to the equilibrium pressure.  In the process the liquid will loose energy to the gas, and the temperature will fall.  Of course this means the water, liquid and gas, is a little cooler than the atmosphere, so heat will flow in, and so on.  Eventually the liquid and vapour and the atmosphere will all be at the same temperature and the gas phase will be at the equilibrium pressure for that temperature.  Now for a given volume of gas phase, the mass of water is directionally proportional to the absolute pressure. 

Let's assume the boiler initially containing only dry air, has a capacity of about 1.5 litres, and we introduce 1 kg of water, which occupies about 1 litre.  About one litre of the air initially in the boiler will be displaced back to the atmosphere, and at the stage the plug is inserted and tightened.  The air contributes to the pressure in the boiler but has no effect on the water vapour which will evaporate from the liquid until that equilibrium pressure is reached.  Let's assume the temperature is 15 deg C.  At this temperature the vapour pressure is only 1.7 kPa.  And let's further just for simplicity assume that we achieved the whole process so far in a manner that means the total mass of water in the boiler including vapour and liquid is 1 kg.  Now the specific volume of dry saturated water vapour is 77.93 m^3/kg.  So the 0.5 litre vapour space contains only 0.0065 g, and the rest of the water remains in the liquid.

We can see that the total water in the boiler is 99.97% liquid and the dryness fraction or quality of total is near enough to zero.  But this is made up of a liquid phase which is 100 % liquid with enthalpy of 419.04 kJ/kg, plus a distinct vapour phase which is essentially 100% gas phase with enthalpy of 2676.1 kJ/kg.

If we now allow this dry steam to flow to an engine, and look at the exhaust, we will generally see a healthy vapour cloud, a cloud of steam which is starting to condense, so it also contains a portion in a liquid phase (as finely divided droplets) dispersed in a stream of dry steam.  This is termed wet steam.  Like the boiler contents it is part liquid phase and part gas phase.  If the temperature could be kept constant, the fine vapour droplets would be expected to eventually settle out and form a distinct liquid phase.  In order to determine how much enthalpy change occurred over the engine, we have to determine the properties of that exhaust steam.  And the enthalpy change is the energy extracted from the steam by the engine.  The enthalpy of that wet steam is somewhere between hf and hg at that pressure.

For an atmospheric exhaust, we know the pressure.  Let's assume the weather pattern has our local pressure at that standard pressure of 101.35 kPa.  We can measure the temperature, but as we did not have a superheater, it will probably be very close to 100 deg C, depending on insulation and heat losses to atmosphere.  However the steam tables tell us that temperature,  it is not independent of the pressure, so we need another independent property of the steam to help us determine the properties of the wet exhaust steam, and specifically the enthalpy.  This one has been a bit heavy going but it will be clearer where I am going next time.

Next time, I will talk about how we use the concept of an ideal adiabatic engine and that property, entropy, to work out at least a limiting condition, that provides an estimate of the maximum change of enthalpy that could have occurred.

I hope everyone is still following

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 05, 2017, 01:40:42 PM
Hi.....If after filling the boiler to correct level we suck out all the air with a vacuum pump will the steam be raised in a shorter time ? and i have been fixing the leaks using Fibre washes rather than copper. also the boiler takes about 2.5 hours to get back to ambient temp !!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 05, 2017, 11:08:27 PM
Hi just a quick question ....Woolf talks about Steam becoming rarified ?? and i have fixed the leeks and the safety valve blows at 50 Lbs /square" whilst showing a temp of 136 degrees......but back to topic !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 06, 2017, 12:25:08 PM
Hi Willy, I am glad that the fibre washers fixed the leaks, we want to keep all the steam for the engine.  From the photo, that safety valve is blowing down well, surely all the air was gone when you took the pressure.  The equilibrium pressure for 136 degrees is only 32 psig, so the gauge looks a bit high at this point.  Perhaps some of the earlier readings were taken while there was still air in the boiler, an important point to observe carefully as the boiler heats, together with just when the safety valve lifts or you turn over the engine a few revs.  Remember the steam tables use absolute pressure, so you have to subtract atmospheric pressure from the tabulated value to get the gauge pressure.

Evacuating the air in principal would help, but when you look at the numbers, not enough to be worth the effort.  The specific heat of air at constant volume, before the safety valve lifts is only 0.72 kJ/kJ.K, and the density is only 1.2 kg/m^3, so half a litre of air at atmospheric pressure when you tighten the plug has only 0.5 g.  Hence it requires about 0.35 J/deg C.  To heat it by about 130 deg to 136, takes about 47 J.  You heater puts in 1000 J/s, so you would save less than 0.05 seconds.  I don't think you could evacuate the boiler in that time.  You can rest assured that the heat goes mostly into the water, plus the copper, plus heat leakage due to limited insulation.  Once the boiler is up to temperature, the copper takes no more as it stays at constant temperature.

The time to cool down is the limit to whether you can do more than one experiment each day.  Of course you don't need to let it cool down to refill the boiler for another run, if it is below about 90, you can do it with care and once below about 60 the plug will feel hot but not normally burn your fingers.  Of course with better insulation, it will take even longer to cool.

Your questions nicely introduced the first question I wanted to address, the air content does not take much heat.  However, until discharged it does affect the total pressure read on your gauge.  Initially the air pressure was 101.3 - 1.7, the 1.7 being the water vapour pressure, say 99.6 kPa (absolute) and when this is heated to 100 deg C from 15, it's pressure increases to 99.6 x 373/288 = 128 kPa.  The total pressure in the boiler at 100 C is 128 from air plus 101.3 from the steam, so 230 kPa (abs) or 130 kPag.  About 18 psig.  If we turn the engine a few times as soon as the pressure gets up a bit, the air will be lost reasonably quickly, while the water will evaporate to replace the lost vapour.  If you keep this process in mind as you approach steaming temperature, you will see how the air loss affect the deviation between the pressure gauge and the steam table value for your temperature.

Every thermodynamics text book has the example of some liquid in a closed volume, presumably after impurities such as air are evacuated, and applying heat until the liquid is all evaporated.  If you felt we were proceeding into the fog, you were right, but the fog did not seem to clear in front of us.  But with a nights sleep the fog has cleared somewhat.  Clearly, the textbook writers are not very practical, and do not ever apply some figures to their example.  As we saw yesterday, the pressure rises with very little of the total mass evaporating, due to the huge expansion when liquid evaporates to steam in a confined volume.  When this happens accidentally in an uncontrolled manner, such as adding water to very hot oil or metal, it is called a BLEVE, boiling liquid expanding vapour explosion, and it is well known for destroying large pressure vessels.  Our experiment assumed controlled heat addition at a slow rate.  With so little water evaporating to increase the pressure, we were never getting to the point I hoped for, where the term wet steam would make sense.  We had 1 kg of water in a 0.0015 m^3 boiler, a typical practical model starting point.  To evaporate all the water per the text book experiment, we need to get to a specific volume of about 0.0015 m^3/kg.  Now when we look for that specific volume in the saturated steam table, the table runs out at 0.003, at the critical pressure, when the temperature is 374 deg C.  Above that the water is supercritical and we are not going there.  In fact a copper boiler is designed on a copper temperature around 200 deg C.and we will exceed this at about 1554 kPa, but our boilers are not even designed for anything like that.  So we are never getting to the point usually claimed, that the pressure and temperature stay constant until all the water is evaporated, then both start to rise.  In order to demonstrate this within reasonable pressure range, our 1.5 litre boiler must start with significantly less than 1 g of water, barely enough to form a droplet, and certainly not enough to cover the heating element, and we all know that is only going to end in tears.   Another approach is needed if we are to understand wet steam.  Essentially the total mass in the boiler is always very near saturated liquid, while it is in the form of 99.9% liquid and less than 0.1% vapour.

Once we open the stop valve slowly, the steam that escapes is very nearly dry saturated vapour.  Of course many will know that if the boiler is too full, and we draw off a lot of steam, the resulting vigorous boiling will mean that some liquid is carried over with the steam.  Full size boilers have separators that separate the water from the outlet steam, but not very easy in a model.  A well designed steam dome would probably do nearly as well.

The area where understanding wet steam is most useful is in looking at the engine exhaust.  We will get to that eventually, but first, lets gradually open the stop valve and look at what happens.  A good place however to stop, so next time.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 06, 2017, 03:21:53 PM
Hi MJM , thanks for this ,I have noticed that the pressure at 100 C was quite high ...18 Lbs" by your calculation ,unlike my little pressure gauge  and found this out the hard way by assuming that when water boils in an open vessel there is no pressure !! So, When i opened the filler cap just below 100C there was a violent erruption of hot steam /water that drenched me somewhat. So question ...at what temperature is the pressure at 1Lb above atmospheric ? and how much is this due to just the water expanding, as presumably the steam has not been generated at this point ? or is the steam so saturated that it does not give the appearance of steam. If i had a very accurate pressure gauge that would be quite good at seeing what is happening inside the boiler. But do you need to make an adjustment with the air being partially compressed in the Boudon tube. Yes the boiler had blown off a few times before i took the photo as i was trying to take a photo at the exact point the safety valve lifted without drenching the camera !! The humidity level in my house has now increased quite a lot too !!!And i am enjoying a monsoon /sauna experience !! If the temp probe was inside the boiler it might give a more accurate reading as well.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 07, 2017, 01:14:21 AM
Hi MJM ..i have now done the next boiler test and we now have a strait line GREEN ..time/temp  again like the first one BLACK. I think that the RED line graph was curving because the leaks were actually getting worse/better.... The readjusted valve blew off at 135 C to release the steam/air and the next time it blew off was slightly higher 138 C. the time was also shorter to achieve this result ,so more in keeping with your first thoughts of about 8 minuets !! I was talking to a friend about my sauna/monsoon event and he told me that water does expand about 4% at 100 C or something. I was thinking that the temp/pressure tables may vary with the amount of water that is actually in the boiler to start off with ?.....this is quite fun and i always wanted to be a scientist so i could bash things and squash things and even blow things up !!! always under strict control of course. My boiler is now about 8 years old so it would be interesting to look at the scale build up as the sight glass is not functioning any more !!  good to hear about more analysis with the latest test.....If there was no heat loss with 100% insulation how long would it take to steam up to this temp/pressure or is that not possible to calculate ?? i will have to get more rockwool do any further tests of course.
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 07, 2017, 12:44:29 PM
Another boiler test -

Hi Willy, that is a very encouraging test, and a great idea to plot it on the same graph as the earlier ones to aid comparison.  I will get to that shortly.  But your first post first.

When you boil an open kettle, the pressure is held at atmospheric pressure, so zero gauge pressure, by the open path to atmosphere.  When you first tighten the plug in the boiler, it is at atmospheric pressure and temperature (unless you used that electric jug), the vapour space is mostly air at the prevailing humidity.  There will be a small amount of evaporation in the boiler, depending on the difference between the moisture in the atmosphere, and the vapour pressure at that temperature.  But eventually if we assume the boiler settles out at say 15 deg C, the saturation vapour pressure is only 1.7 kPa (absolute).  If the humidity is say 50%. Then the moisture content of the air is 0.5 x 1.7 or 0.85 kPa, and if the boiler is allowed to truly settle the extra 0.85 will evaporate, cooling the water, so more heat comes in, and slowly it will get there.  But we are playing with 0.8 kPa out of a total of 101.3 and the rest, over 99%, is air.

When you start heating the sealed boiler, you can't ignore the air, it will be close to water temperature, and when you get to 100 degrees, the air is also at 100 degrees.  And it will have increased in pressure from the original, near enough to 100 kPa, in proportion to the increase in absolute temperature rise, so 100 x (273 + 100 )/ (273 + 15 ) = 129.5 kPa.  I think I rounded it to 130 kPa last time.  At the same time, the water vapour pressure has increased to 101.3 kPa at 100 deg C.  The two apply independently so the total pressure you have in the boiler, and would be reading on an accurate pressure gauge, would be 230 kPa (absolute) or 130 kPa gauge.  This is about 18 psig.  Not a good idea to release the pressure by unscrewing the plug!

But let's look at what else is going on inside the boiler.  The steam tables tell us that the specific volume at 100 deg C is 1.673 m^3/kg, so assuming we still have very close to half a litre of vapour space, the vapour space now contains about 0.8 g of water, so about 0.08 % of the water volume has evaporated.  We are probably justified in ignoring that.  Your friend is right, you can see the expansion of water in the tables by looking at the vf columns, the specific volume of the liquid.  It's about 4.4%.  I guess that means the vapour volume is about 8% less than our initial 0.5 litres.  I certainly did overlook that, and it means that the air is compressed a little more.  But I think the sauna was caused by another point.  Boiling only starts if the water vapour pressure is above your actual atmospheric pressure when you released the plug.  Your temperature meter has a degree of inaccuracy.  If you did not check the  calibration at 0 and 100, the spec sheet that came with it will tell you the limits, but with any digital instrument, you can safely assume that the last digit is still uncertain by at least +/- 1.  That means the temperature could have been 99 or 101.  Also the tables tell you the absolute pressure, your gauge reads difference between the atmosphere and the boiler.  Depending on the weather, a low pressure system for example, your absolute pressure could have been perhaps 99.5 kPa without being extreme.  The result of these two factors is that the vapour pressure of the water could have been a little bit above atmospheric pressure.  While the plug was tight, evaporation matched the temperature rise, and boiling was suppressed by the additional pressure of the air trapped in the boiler.  When you released the plug, the excess pressure was released, and if the vapour pressure was above the actual atmospheric pressure, boiling is no longer suppressed and the huge expansion of water as it turns to steam so the reason for your boiler eruption.

I hope that you weren't scolded, but please, if you want to release the pressure in the boiler after you have started heating, use the regulator, and release it through the engine, or a whistle that you provide for the purpose.

The steam tables list the volume and energy values on a unit mass basis, that is 1kg for the metric tables or 1 lbm for imperial tables.  Obviously the pressure and temperature are not affected by the mass.  When you need to use the total energy, enthalpy or entropy, you multiply the value in the tables by the mass in your system.  I hope that is a bit clearer, but please continue to ask if there is something still not clear.

Now for that new test.  What a difference sealing those leaks made.  They seem to have been confirmed as the culprit for those curves.  And now you can more easily see the difference your insulation made.  Looks like about 1 min 15 sec shorter time to heat up.

I quickly ran the calculations again for 135 degrees assuming 16 degree atmospheric temperature, (I hope you did record the starting temperature).  I assumed 600 ml of water again, which of course it was completely emptied after the last test.  And again 750 g of copper in the boiler from my estimate last time.  And still assuming the heater is 1000 watts, though that data sheet implied that any error would make it on the low side.  This is the data needed for the calculation.  I can easily correct any figures that are not correct, the beauty of using a spreadsheet.  Heat required for the water is 300.3 kJ and the 34.2 kJ for the copper.  So again ignoring the heat stored in the insulation which will be much lower than the copper, it looks like 335 seconds or 5.6 min to heat up if there were no losses.  The actual time was 7 min, so the losses were only 20% of the heat input.  Alternatively, 80% of the heat went into the copper and water.  There is room for a little improvement with more complete and thicker insulation, but you can see only about 1 1/2 min time  savings before steaming starts, and that is still assuming the heater is a full 1000 watts.  Perhaps someone could lend you one of those energy metering devices sold for checking appliance consumption so you could check it.

I continued on to calculate how much steam production you might expect.  Once steaming begins, temperature is constant so no more storage in the water or copper.  We look up hg for the steam at 135 deg and subtract the enthalpy of the water hg.  Better still, look up hfg, which is the result of the subtraction and is the enthalpy to evaporate a kg of steam from water at the equilibrium temperature.  Divide the heat input, 1 kJ/s by the enthalpy needed for evaporation of 1 kJ to get the number of kg per second evaporated.  It is pretty small but works out to about 22.1 g/min with the current heat losses, compared with about 28 g/min if there were no losses.  It will be interesting to run the engine for a significant time without uncovering the element of course, and see what it uses.  Does your temperature control cycle the element on and off?  Or does it run continually?  Looking forward to the next step.

Well, in analysing the behaviour of the boiler when the plug was removed, I covered what I had hoped for, so two in one this evening.

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 08, 2017, 10:51:08 AM
Theory and observation -

Yesterday I tried to analyse the processes which followed Willy's unscrewing the boiler plug when the temperature gauge showed 100 deg C.  I came up with two factors contributing to the ensuing eruption, i.e. the air pressure which is also in the boiler from the original fill, and a small discrepancy in the actual temperature and the display on a digital instrument.  A bit more sleeping on it, and I have some additional factors to add.

First, when that air, now at 100 deg C escapes, it carries with it a significant amount of steam.  That means the total pressure on in the vapour space is reduced, but the steam partial pressure is also reduced, so that again, the equilibrium vapour pressure of the liquid is above the partial pressure in the vapour space.  That difference also triggers boiling a small mass of water, but resulting in a huge volume of steam.   I suspect that difference in vapour pressure would result even if all the air was gone due to the engine running or even safety valve lifting when the pressure was achieved.    A lot depends on the exact sequence and timing of events.  In the absence of an engine or a whistle to control the steam release, I would leave it till a much lower temperature, then use a socket and extension to release the plug with my hands well clear.

The whole exercise does show the difficulties of applying a little bit of theory.  There is always the possibility of other theories also applying, and sometimes the theory we did not think of is more important in the particular circumstance.  So it is important not only to learn the theory, but also to proceed cautiously until the process is understood.  Steam is hot and dangerous, and experiments with banging and squashing and blowing things up always need to be carried out with caution and due consideration of a safe vantage point.

You do not need to make an adjustment for air in the bourdon tube.  In principal, the pressure reading is affected by both the height of the gauge above or below the point where the pressure is measured and also by phase changes in the connecting tube.  Even if you have a very long horizontal tube, there is no effect.  However, some steam will condense in the tube which will act like a U-tube manometer.  The leg of the manometer with the bourdon tube will have trapped air which is at the same pressure as the water compressing it.  If there is a big change in elevation between the gauge and the boiler, the density of the water and the height of water in the tube will make a difference.  But on a model, the difference is practically only an inch or so.  And a few inches of water are not significant compared with the pressure you are measuring.  If you have 10 metres of water column, it makes a difference equal to 1 atmosphere, say 14.7 psi or 101 kPa.

Just exactly when the steam is generated is worth looking at a little closer.  The first thing is that when the temperature has evened out, so all is at the same temperature, then the water vapour pressure in the vapour space is always equal to the equilibrium vapour pressure listed in the steam tables, and that vapour pressure is in addition to the the partial pressure of any other substance, such as air, in the space.

I suspect the question is really about the difference between evaporation, which we would normally expect to happen slowly, and vigorous boiling.  Slow evaporation takes place at the surface relatively quietly.  If there is no heat source, this evaporation absorbs heat from the liquid and cools it, for example when you blow the vapour away to cool your coffee.  Under this slow evaporation there is no significant departure from equilibrium.  The pressure in the liquid varies with depth due to the liquid density, and is higher below the liquid surface.  So evaporation below the surface level is suppressed.  If heat is applied at a high rate, the temperature near the heating element will be higher than the bulk temperature and next to the element, and can become higher than the equilibrium temperature of the bulk of the water.  So some of the water then flashes into vapour, undergoing that huge expansion into bubbles that rise vigorously to the surface.

It the water is in an open vessel, the air pressure adds to the vapour pressure at the surface.  This additional pressure at the surface suppresses boiling, but the evaporation still occurs so that very near the surface the equilibrium vapour pressure is achieved.  While this equilibrium is less than atmospheric pressure, that is the temperature is less than 100 deg C, there is still air, which increases the total pressure at the surface and boiling is suppressed.  However once the water is at or nearly 100, the equilibrium vapour pressure is about atmospheric pressure, and a significant temperature gradient provided by the heating element soon means the vapour pressure exceeds the local total pressure and boiling begins.  It is that unrestrained expansion of liquid to vapour that causes the vigorous boiling, and the point it starts depends on both the bulk temperature and the heat transfer rate (due to it causing a temperature gradient.)

If the water is in a sealed boiler, still with that initial air, like your boiler, then the evaporating water and the air as the liquid heats can no longer escape and the total pressure increases due to heating the air and increase of water vapour pressure with temperature, so boiling continues to be suppressed.  However, if you then release the pressure, the air and significant steam escapes, the bulk liquid temperature exceeds the equilibrium temperature for that new pressure and unrestrained evaporation quickly proceeds, the rapid huge expansion pushing water out with it.  Preferably the escape is controlled by slowly opening your regulator and the escaping steam, together with the air that remains happily drives the engine until all the air is gone then the steam continues on.

But in the unrestrained expansion, ensuing clouds of steam consist of a mixture of dry vapour and entrained water and very fine droplets which appear as a cloud or mist or even hot rain.  The evaporation absorbs heat from the liquid, cooling it until something close to equilibrium is again attained, though by that time the eruption is complete.

It's a little long winded, but I hope that in the end it does clarify things a little.

Thanks to everyone for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 08, 2017, 06:23:40 PM
Hi Thanks for all these explanations, its a lot clearer now. When i did the time/temp graph i also wrote down the differences between each minute and found them to be slightly different ...especially after 100C so is there a valid explanation of why this should be so ?? The graph did sort of make a strait line ! I will try and sort out the boiler soon but only need a handful of rockwool so will keep a look out in the local skips. The temp controls on the panel of the boiler circuitry allows a sort of stoker control of the "fuel" by pulsing the electric to the elements . This is so one can regulate the steam production with the steam consumption required by the engine. this stops the safety valve blowing off all the time !! I have sent a photo showing the graph details ....The pencil is the minuets and the red numbers follow the red line and the black numbers are for the latest green line...I am running out of questions gradually, so will await your new post,! this has been an interesting journey and i am sure there is lots more to thermodynamics than what we think we know and assume !! There engine is not connected to the boiler at the moment as i was using it to demonstrate the Beeleigh model and was wondering what we can actually learn from running the small "Persious" engine from the boiler ?
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 09, 2017, 12:57:36 PM
Hi Willy, I am glad that things are little clearer now, I have also enjoyed the journey, I hope there is more to come.  I certainly have a few more points I was going to cover, that I hope will interest you.  You have tested the weaknesses in my understanding and I have learned a lot from working out the answers.  It has really consolidated my understanding in areas that I thought I already understood.

With regard to the insulation, rockwell is usually best, but as you have no flame, fibreglass will also do.  However it is a bit prone to breaking off fine fibres which itch the skin, so best to wear gloves.  Those disposable ones will do.  And worth using one of those paper painting masks as well.  No point in breathing in small fibres if you can avoid it.  Then wrap it with a cloth or foil cladding.

On/off control for the boiler is great for controlling the boiler pressure, but makes it hard to tell how long the element is on once steady conditions are maintained.  However in answer to your question about an engine test, I suspect it is not worth the effort at this stage, as I expect it is an unloaded test.  If you are interested, I will at some stage talk about the measurements needed to measure real power output.  However I will also write a few posts about the unloaded tests I have done on mine, and will include the differences with respect to yours. 

Tabulating those differences is a very clever way of identifying the departure from a straight line on a graph that looks pretty straight, even by my favourite method of holding my eye close to the paper level and looking along the line.  But let's look at what is going on and see if we can identify what is behind that small departure from linear.  I suggest the fact of the two 15's in a row, shows your observation of a curve is real and not just a random fluctuation due to the timing of the gauge updates.  This randomness is unavoidable with a digital instrument.  It may explain the 18 and the first 15, and may be reduced by an instrument reading in 0.1 increments.  However I think the second 15 is a definitive indication of a change in slope.

I will suggest some things that are assumed to be constant in expecting a straight line.  We are assuming constant ambient air temperature, constant convection heat transfer coefficient to the air outside the boiler, constant heat input from the heating element, and constant specific heats for the water and for the copper, constant mass of copper and water and constant temperature gradient from the element to the outside air. 

You might have a feeling for whether the air temperature was changing during that 9 minutes, but I suspect not changing much.  I have tried to find more information on the specific heat of copper and while all the tables list it, usually at 20 degrees, only a constant value seems to be given.  Perhaps I don't have the right book, but in the range from ambient to the melting point of copper, 100 degrees is not much and I suspect any change in this range is not significant.  The heating element could increase in resistance as the temperature rises.  Unfortunately, not listed on the data sheet.  We would need to do some specific measurements.  If it changes, I would expect the change to be about uniform over the whole range, rather than only start at 100 deg.

The specific heat of water is easily checked from the steam tables.  Just check the change in enthalpy between 90 and 95 deg, and divide this by 5 to get delta h per degree.  Now do this for 15 and 20 degrees.  It looks like 4.194 kJ/kg.K at 20 deg and 4.208 kJ/kg.K at 95, so it does increase and in the right direction, but only by 0.003%.  Not sure that we would see it, but could be a factor.  I did not check it at an intermediate temperature to see if it was a linear change, or did indeed start nearer 90 deg.  The convection heat transfer coefficient from the wood cladding to the air is hardest to quantify.  Instinctively, should I say dangerously, I would expect that as the test proceeds, the air moves increasingly rapidly, so might transfer heat slightly better, so might not be exactly proportional to the temperature difference, but might increase the heat loss slightly with temperature.  But it would be quite difficult to quantify this, so I am reluctant to identify this as the cause.  Then we have assumed constant mass of copper and water.  I have no doubt about the copper, but a tiny wisp of steam from a recalcitrant gasket or even the safety valve means the mass of water could be decreasing, but more importantly it takes a lot of latent heat with it, so it could show in the measurements.  I would try a piece of thin paper, taped to the end of a ruler to wave around as a sensitive test for a steam leak, not your index finger!  The safety valve is certainly a factor that we would expect to be zero until the pressure starts to approach its set pressure, then even the best safety valves are likely to leak the slightest wisp, and we are clearly not looking for something big.

Perhaps the most likely cause of the slight curvature is a part from the specific heat of water, and a little from the merest wisp of a leak.  Definitely not a definitive answer.  Some suggestions to ponder, but overall, I suggest that overall the test shows that for revel of accuracy of the available instrumentation, the assumption of linear is not too bad.  And it has provided valuable information on the heat loss, the value of insulation, where the heat goes during the heat up period, and the potential rate of steam production from your boiler.  A lot of information from a relatively simple test with readily available instruments.  And lots of other leading along the way.  So thank you for all your help.

That is enough for another day.  Thanks everyone for taking interest.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 10, 2017, 12:36:31 AM
High MJM ,thanks for this explanation and yes there were a few wisps of steam actually. Would it be possible to add accurate  boiler pressure values  to this graph, and what temperature reading would the pressure be at atmospheric to open the filler cap safely ? Perhaps the wood needs to heat up gradually before it starts to radiate the heat ?as it is insulated partly from the copper ? On the side of the boiler is a neon lamp that is connected to the elements and when the potentiometer is full on this lamp is fully on. When the pot is turned down the lamp will flash on and off in time with the pulses of  the current, the lower the pot the slower the flashes, so when it is full on the elements are getting the full currant.

Title: Re: Talking Thermodynamics
Post by: MJM460 on October 10, 2017, 12:16:00 PM
Boiler pressure and air -

Hi Willy,  I am sad to say that we can't plot the pressure directly against temperature to get a really accurate figure.   The pressure predicted from temperature has to be interpreted with an understanding of the effect of the air that is in the boiler when you first seal the plug.  Earlier in this thread, I have said a couple of times that once you start steam production, the air is soon gone.  You have prompted me to actually do the detailed calculations, and I am starting to realise that the air might take more time to become negligible than I thought.  So let's look at what really happens to the air.  First it is important to realise that the air and water vapour both fill the space independently.  So while the water vapour pressure is accurately found from the steam tables, any air remaining in the boiler adds a further component to the pressure seen by the gauge.  We have already seen that at 100 deg C, the air pressure is 130 kPa and the water vapour pressure is 100 kPa, so a total of 230 kPa.  Keep in mind that these are absolute pressures, so we subtract the atmospheric pressure, say 100 kPa, to get 130 kPa gauge or 18 psig, an 18 psi difference between the gauge and the steam tables.  If we continue to say 150 deg C, the calculation gives a gauge pressure of 75 psig compared with the water vapour pressure contribution of 54.3 psig.  The error is now just over 20 psig.  Both cases so far assume all the original air still in the boiler.

Now the air and water vapour in the vapour are fairly well mixed.  I suspect the steam being generated at the surface might mean a bit lower concentration of air near the surface, but the rapid movement of the molecules means and and water are probably reasonably well mixed further from the liquid surface.  When a little vapour escapes through the engine or the safety valve, the mass of each in the escaping steam is proportional to the partial pressures.  The lost steam is quickly replaced while the heat is applied, but the air is not.  But the air lost in the next time period is proportional to the remaining amount.  After a little has escaped with the steam, the remaining partial pressure is reduced, so the proportion of air in the escaping steam is reduced.  The first half, in a certain time, then half of the remaining amount takes a similar time.  Like the temperature of your coffee in the cooling experiment, which approaches the ambient temperature ever more slowly, the air loss happens increasingly slowly and in principal, while it gets close enough for practical purposes, it does take a significant time.  As the air is lost the discrepancy between the gauge reading and the water vapour pressure obtained from the temperature reduces, but the gauge pressure should always be a little higher than the prediction.  Fortunately the safety valve works on gauge pressure, so always protects us.  Bit if we are relying on the temperature for our pressure , we need around 20 psig safety margin between the design pressure and the intended operating pressure.  If the  safety valve is set a bit closer than that, it will lift before our intended operating pressure, the escaping steam takes a portion of the air with it, thus reducing the air pressure and reducing the difference between the steam table value and the actual gauge pressure.  The engine also runs on total pressure and does not care if it is water vapour or air, and I find the engine starts to run quite soon after 100 deg C is reached, but the air at this stage means the actual gauge pressure is a bit higher than I have assumed.

A further factor in all of this is that as the steam is used, the water level reduces, so the remaining vapour space is larger.  The steam needed to fill this volume evaporates from the water, but there is no more air, so the air pressure is further reduced as the level falls, and the error due to the air content is further reduced.  So I think that up to 150 deg C, the error is a maximum of about 20 psi which should be noticeable on the gauge, but the error reduces as steam production goes on, and the reduction in error should be noticeable.  After a few runs, you should have a good idea of how long it takes to get down to an acceptable error, then you will always have a good idea whether your pressure gauge calibration is still correct.

I must admit the error is a bit greater than I was expecting, and the time for the air to effectively be gone is certainly longer than I thought.  It looks like I had better add pressure gauges to my two small boilers which currently only have temperature measurement, plus a carefully checked safety valve.

Then, what temperature can you safely remove the plug?   In industry, you can open piping or equipment only when you have proved the pressure is zero by opening a vent valve.  Even then, sometimes the vent valve is blocked by rust or dirt, and there is still pressure inside.  That is another place where Murphy interferes with the best laid plans.  On our models if I wanted to be sure, I would add a whistle, though a vent valve would also do.  Personally I normally let it cool down overnight.  In order to refill for another run as soon as possible, I suggest even with a gauge, zero is really hard to confirm.  It clearly has to be less than 100 deg C, sixty degrees is normally considered a safe temperature to touch, though if you do not have tough skin it will still feel pretty hot.  Then I would leave the regulator open and help the engine turn over while any pressure was obvious.  At 60 deg C, the water vapour pressure is about 20 kPa absolute, so needs a lot of air to get to 100 kPa absolute or close to zero gauge.  With experience and suitable precautions against getting scolded, you may find a little higher Ok.  If it is really desirable to to get started sooner, probably the safest course is to use a hand pump to refill the boiler to the appropriate level gauge indication.  Of course, once you pump in a bit of cold water, the temperature will soon fall, so you can remove the plug if necessary to check the level.

You have mentioned that your level gauge is not very reliable.  I believe small level gauges are notorious, and the vigorous boiling of water makes the level bounce around.  Also, when boiling is occurring, part of the vapour is below the water level, and sudden changes of pressure change the effective density of the liquid phase which then upsets the level gauge as it does in a large boiler.  When the boiler is cooling for refilling, the density should be more steady, but you still have to deal with surface tension issues and the meniscus which both change the apparent level.  It may help to take the gauge apart and clean the paths into the boiler in case they are salted up, but these are practical issues and I don't have a lot of experience with gauges in model sizes.  Industrial level gauges have much larger passages and different problems.  Your colleagues in the club should be able to make more useful suggestions.

That neon light sounds like a nifty way to tell if the boiler is lightly or heavily loaded.  I like that feature, but it makes it hard to determine how much time the element is on, and hence how much heat is actually being generated.  I assume it is fully on during the heat up phase, a bit of flashing as you near the safety valve setting would be an additional cause of a curve in the heat up curve, however, I suspect that that wisp of steam is a sufficient explanation of a very slight curvature.

I hope that clarifies the issues sufficiently,

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 10, 2017, 03:34:45 PM
Hi MJM, Thanks for your further info, and leaving things to cool down provides a good excuse to go to a cafe and drink coffee etc etc . I have done a test with my thermometer and putting it in a boiling kettle only shows a reading of 93 C actually so %7 out, i have yet to get some ice but that will be interesting soon.! So all these calculations in fact are quite hypothetical, and need actual laboratory conditions to get an accurate reading and the formulae should be more like  Air pressure (at x humidity, at x height above sea level ,at x wind speed  ,at x etc etc etc)  X the rest of a formula should bring in all the variables necessary but do need to be known.....

Willbert....
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 11, 2017, 12:36:57 PM
More on boiler pressure and air-

Hi Willy, I am surprised that the thermocouple is so far out, but I am not yet convinced.  To get an accurate temperature reference at boiling water temperature of 100, you need to get as close as possible to equilibrium, just like in your boiler.  So you need a lid on the kettle, just tilted enough to admit the probe, and perhaps stack a few face washers around to minimise the steam escape, and after a while enough air is gone and you will get much closer to 100.  Careful with those face washers, the steam will make them hot.  If you get 98 per 99, I would accept your meter as reading correctly.  But also check the barometric pressure.  Either your calibrated barometer, or check the weather bureau data for your area.  Standard atmospheric pressure is 101.35 kPa absolute, or 1013.5 hPa, which is the water vapour pressure at 100 deg C.  If there is a low pressure system passing over, 1000hPa will give 99.63.  You can see this in the steam tables.  It takes a bit of interpolation for other pressures.  Similarly for zero degrees, you need water liquid and solid ice in equilibrium.  So lots of ice, minimal water, insulate your jug with a towel, probe deep into the water through the slightly tilted lid, along with a stirring stick so you can keep it well mixed, and some more face washers to minimise heat gain.  It is harder in practice to exclude the air from the zero point than the boiling point as there is not enough vapour generated to displace air.  You would need a vacuum pump to get that triple point condition.  But again if you can get within 1 or 2 degrees of the liquid-solid equilibrium, I would accept the metre as more accurate than the calibration check process.

Now before you dismiss all this as hypothetical, please bear with me while I have a go at a more precise explanation.  A nights sleep, a powerful analysis tool, enabled me to come up with a simpler and more precise calculation of the air pressure contribution to the total pressure over time.  Looking at the boiler in isolation and trying to calculate the loss of air was not very easy, so I took the easy way out and gave only the maximum error, which occurs as you first heat up, and decided that I really did not know how long it took to loose the air.  But if we expand the control volume, to use text book terminology, to include the engine, we have a situation which is much simpler to analyse.  Now it does not make any difference to the boiler whether the engine is a conventional reciprocating engine which takes many strokes per minute, or if it is a very large piston which just accepts all the steam for your total run time in one very long stroke.  Nor does it matter to the boiler what the engine does with the steam next, just as long at it does not gain or lose heat while it takes in all this steam.  Now I have assumed that your run time ends when your 600 ml of water is reduced to 300, in order to stop while the element is still covered.  So you evaporate 300 ml, or 300 grams of water, at a constant temperature of 135 deg.  Now the initial 1.7 grams of water to give saturation vapour pressure at the 15 degrees starting temperature is not very significant compared with the total of 300 grams evaporated.  I assumed an initial vapour space of 0.5 litres.  Now when 300 grams of water evaporated at 135 deg C it makes 175 litres of vapour.  But the initial charge of air at atmospheric pressure is only 0.556 grams.  While the water evaporates into the vapour space as the total volume expands into that very big cylinder, so the mass of water in the vapour increases by evaporation of the liquid, the air mass is constant and in the end, we have expanded the initial air from 0.5 litres to 175 litres.  We had initially calculated that when the temperature reaches 135 deg, the absolute air pressure is 130 kPa.  The expansion of the air volume, remember gases always fill the whole space, the ideal gas law tells us the pressure of air at the larger volume.  Because we are assuming a constant temperature before we start the expansion, the ideal gas law is basically P1 x V1 = P2 x V2.  The condition in the boiler at any time is just a known sample of that total end volume.

I basically calculated the air pressure, total pressure and gauge pressure 10 times for the volume expanding 0.5 l to 175 l.  If the run lasts about 20 minutes, that is roughly every 2 minutes.  Now if that model is clear and accepted, the resulting gauge pressures are the answer to the problem. 

You will remember that at 135 the pressure from water vapour is 31.3 psig.  With the typical small gauge on a model boiler, with scale marks at 10 or 20 psi intervals, it would be pretty hard to read that more precisely than calling it 30.  The calculation gave 31.87 psig after just the first 30 ml was evaporated.  And 31.5 after 90 ml was evaporated, so a very rapid initial fall which then only slowly trends towards the long term value.  Now if the air remains always well mixed as assumed by the assumptions, it is never all gone.  The pressure after evaporating 300 ml was 31.40, only 0.1 psi high, but in practice it was close enough after about 4 to 6 minutes.  Not just the first blast from the safety valve, but never the less, quite quickly after the engine starts to run.  The error is really only significant during that initial heat up to safety valve set pressure that the maximum error is observed.  And that is readily calculated from the initial vapour space volume. 

While the initial humidity would normally be less than 100%, after a short time in the sealed boiler  it will be made up by evaporation from the liquid, but will make less than1% difference to the pressure contribution of the air, so I think can be ignored. 

I think this is a more rigorous approach to the calculation, and I hope it makes sense.  Like a metal part, a knowledge building block sometimes takes more than one attempt.  The first analysis that comes to mind is not necessarily satisfactory, and sometimes needs some additional work. 

The clear learning that comes from the whole conversation is the influence of the initial air in the boiler.  For pressure testing a calibrated gauge is required, and I am sure that a boiler inspector would insist on a calibrated gauge for the initial safety valve setting.  However, a model gauge calibration and safety valve setting can be checked quite accurately towards the end of a run.  And used with understanding of the influence of air, the temperature gauge is quite adequate for normal monitoring of your boiler operation.

In addition we have learned that attempting to prove zero gauge pressure by subtraction of atmospheric pressure from an absolute pressure of a similar magnitude, inferred by the temperature reading is subject to errors that make it unreliable.  Better to be safe by waiting longer for cooling, or preferably, use a whistle.

I hope that is a more satisfactory ending point than yesterday, thanks for following along.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 11, 2017, 02:31:33 PM
HiMJM, thanks for this  ,more info get my head around !! I have been testing the Thermocouple in melting ice and i think the battery may need changing as it should show 32 on the farenhieght  scale and 0 on the centigrade scale !! Also an account from the 1827 Farey book about removal of the pressure suddenly from the boiler ....very colourful language !! Here are a few pics of the readings ......Also he mentions  'Sensible heat" again !
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 12, 2017, 12:41:32 PM
Hi Willy,  I think you will find your battery ok, those instruments usually have a low battery indicator, but your open glass with one or two ice blocks just won't cut it as a zero degree reference.  Make a tray of ice blocks in the freezer compartment, many little blocks are best as there is more contact with the water.  Put your glass of water, half full, in the fridge while the ice blocks freeze, (perhaps overnight?). Then fill the glass with as many ice blocks as will fit, fold a towel to the height of your glass and wrap it around, better still leave the towel in the fridge as well, mix the ice and water really well and cover the top.  After a while to allow it to get to equilibrium it will get a lot closer to zero.  I would suggest that with the open glass which may have started at room temperature, the 4 degree reading is possibly quite accurate.  Similar care is needed to get close to equilibrium temperature at the boiling point.

Sensible heat is reflected in a temperature change you can sense with a thermometer, or even your finger if the temperature is appropriate.  You know by now that freezing or melting of ice and evaporation or condensing of water both involve quite large amounts of heat.  However both processes occur at constant temperature, and you cannot sense the heat input with a temperature instrument.  You would actually have to measure the volume of the solid, or weigh it, and measure the volume or mass of liquid, and from this infer the heat change from steam table data.  But it's a wonderful article from your book.  The language makes it hard to read, and demonstrates the difficulty of describing things if you do not have the mathematical tools, quantitative data, and even those more intangible properties of enthalpy and entropy.  A bit like Pythagorus describing his theory for right angled triangles.

I think with yesterday's calculation, I can now provide your requested graph. You will remember yesterday I said that the total pressure approached the equilibrium value quite quickly, and today I did a few more calculations in that first 10% of the run! and I am really back to my original contention that the air is soon effectively gone.  If the total run is 20 minutes, the pressure will be virtually true steam pressure in about 20 seconds.  A short safety valve blast or enough steam to warm up the engine.  This should be evident by comparing the gauge readings with the temperature readings.  So you should first see a pressure around 20 psi high, which should drop back quickly once the safety valve lifts, especially then if you release a little steam to warm up the engine cylinder.  I have two nice graphs, one for the heat up period, and one for the run time.  I am having technology troubles I getting a picture within the size limit, but I should get there tomorrow.

Just a short post tonight, I need the pictures to make it all a bit clearer,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 12, 2017, 01:28:33 PM
 Hi MJM , So Sensible heat is not the obverse of silly heat !!! I have wondered about this for some time as all the old books talk about Sensible heat !! Also ice does funny things when cooling/heating as i have been led to believe . I wonder when the words Enthalpy and Adiabatic came into being ?  Like Mr Watt and Mr Farraday,  was there a Mr Enthalpy and a Mr Adiabatic ? !!!sorry just being a bit silly there  But it would make for interesting formulas IE   1 Cholmondly-Smythe divided by 1 Monatgue-Fitzpatric = 1 Dionosius-Ericson.........!!Also from your rethinking on these posts ,if you did pull a vacuum in the boiler would it behave differently ?? or should one try that out ? .Using an electrical element would be a good way of getting uniform readings from the heat input point of view....unless you tried to do it just after the World cup Football game when everybody switched on their own kettles !! Sorry about these ramblings but i have always thought out side the box !! Good to here about your continued interest with my own small contributions

Willy
Title: Re: Talking Thermodynamics
Post by: paul gough on October 12, 2017, 10:53:59 PM
Hi Willy,  I note you have included a Dionysius in your formula. Caution! If this is of the type 'Lardner' be aware that it may contain a lot of hot air, causing unstable results. When placed under scrutiny it yields surprising behaviour, including 'criminal conversation'! An introduction to its behaviour can be had here;<https://en.wikipedia.org/wiki/Dionysius_Lardner> Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 13, 2017, 01:00:06 AM
Hi Paul ,found this in one of his books.........says it all really !!! will look your link up...........

willy.
Title: Re: Talking Thermodynamics
Post by: paul gough on October 13, 2017, 08:04:48 AM
Hi Willy, glad you get a laugh from Dr. Lardner, he is something of an engineering/science Sir Humphry from Yes Minister, a consummate obfuscator, and it seems, at times an incompetent, if not fraudulent, self promoter among other more nefarious activities. When looking at historical sources and it includes engineering historical writings, one has to be as alert to mythology as much as one would when considering the practicality of Icarus's flying equipment or the veracity of the events occurrence. All sources need to be measured against other evidence from the era to determine which is credible, (for the time), and which is to be ignored/avoided. I had two of his books and threw them in the bin. Apologies to MJM for this diversion. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 13, 2017, 08:07:26 AM
Hi Willy and Paul, that Lardner character seems to have been quite controversial.  Good to hear from you again Paul.  Willy, the page you included is pretty hard to read, without the previous page anyway.  Print is clear enough, but the language is heavy going.  I am really not sure what he is getting at.  Obviously no friend of Mr Wolfe though.  Looking now at Pauls second post, that probably means I was not far off the mark!  Personally I prefer to go to the more modern books, they tend to be easier to read as well as including more up to date ideas.  Nevertheless, a bit of humour is always appreciated, especially in a thread like this.  And your thinking outside the square, seems always to find the gap in my description that I need to fill in, so I hope everyone else is finding them as helpful as I am.  I don't know where the terms enthalpy and entropy came from.  But just as well those other characters didn't intermarry and combine hyphenated names!

Found the limitation of the Numbers spreadsheet on iPad today.  I was having trouble getting the file size of my graphs down, and of course Apple is quite sure you don't need to know file size, so it takes some trickery.  Finally decided to send it all to the desk top machine, which then would not open a Numbers spreadsheet properly.  In the end, got the figures across, re constructed the graphs, printed on a better scanner and so on.  Numbers is really great for reviewing a spreadsheet, but is more limited when it comes to building one, particularly on graphs.  However, all done now, so back on subject.

Attached are the two graphs which show the results of the calculations relevant to boiler pressure and temperature readings, I hope pretty much exactly what you wanted.  The heat up one shows the relationship of the total gauge pressure, seen on the pressure gauge to the equilibrium water vapour pressure.  The lines slowly diverge due to the increasing contribution of the air pressure as the boiler heats up.  The air pressure increase is a linear function of absolute temperature.  It has a very tiny dependence on the starting temperature, due to the very low water vapour pressure at ambient temperatures, but you can see what happens.  Unfortunately, if you let some air out by lifting the safety valve or a vent early, you would change the mass of air and need a new total pressure curve a bit lower than the one I have shown.

Once you start releasing steam, whether to the atmosphere via a safety valve, or through an engine, the Run Time graph is the one that applies.  You can see that the air contribution reduces very quickly in 1 - 2 % of the total run time to near enough to zero, considering what can be read on a small gauge.  That would be less than 20 seconds of a twenty minute run.  But the error does not totally disappear, even after 300 ml of water has been evaporated.  It still contributes 0.1 psi to the gauge pressure and that is still reducing but at a very slow rate.  You would have to evaporate as much water again to half that error.  The graph is not very dependent on the steaming temperature, but is determined mainly by the initial quantity of air in the vapour space, which was assumed to be 0.5 litres.

So, while I would reiterate the point that you always need a safety valve, and a gauge is good check that the valve is working at the correct pressure, your temperature gauge is probably much easier to read accurately for general run purposes than a tiny gauge, so long as it is used with some understanding.  After only a few minutes of engine running, the gauge and the vapour pressure from your temperature reading should match.

In principal, you could reduce the mass of air in the boiler with a vacuum pump, and this would reduce the associated error during heat up, as well as that long tail on the run time graph, so long as you have a tight shut off regulator that stays shut under vacuum.  I only have a direct line to the engine, and low boiler pressure would lift the slide valve and draw in air.  The oscillating engine should be ok though.  However I am not convinced it is worth the effort.  The graphs are easier to use than I was expecting.  So here they are.

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on October 13, 2017, 08:42:32 AM
MJM, Willy, This air/vapour, temp/pressure exercise has indeed been an interesting phenomena to have explored and explained. I have only ever considered the negative 'oxygen' aspect of air in the boiler and where practicable raised steam with a relatively 'high' water level and brought it down appropriately as it heats and expands all the while having an open vent from the steam space so that over the course of steam raising there is an evacuation of air with vapour and also the vent gives a good indicator of that point when you just start to get pressure, often a gauge does not. With a loco boiler this vent is often, for convenience, via the steamway cock and blow through cock on the gauge glass or the blower in the smokebox. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 13, 2017, 12:54:32 PM
Hi Paul, your note is a perfect addition, not out of context at all.  You see it is another example of where theory, no matter how complete, is not all that must be considered for any particular case. 

I assume that you are talking about  steel boilers, where of course a lot of effort goes into feed water treatment to remove oxygen down to minute levels in order to minimise corrosion throughout the system.  Even quite basic steam plants generally have a feed water de-aerator and a chemical treatment plan that includes a chemical oxygen scavenger.  There is no point in doing this if the oxygen scavenging system is then loaded up with oxygen introduced as part of the start up procedure.  And your procedure is a good way to eliminate that initial air by sweeping it out with steam during the warm up.  In addition, in a full size steel boiler, it is normally important to heat up slowly to minimise temperature differences which introduce thermal stresses that can be quite undesirable.

If Willy followed your procedure, he could also drive the air out of the boiler during startup, and then the temperature measurement would give a more accurate boiler pressure almost from the start, as there would not be air to add to the water vapour pressure.  However, he has a copper boiler where the corrosion issue are not very significant.  I am not sure how much oxygen is involved in scale formation, but on model run times, it is generally easily handled with a descaling procedure.  His electric heater arrangement produces very little in the way of uneven expansion as the sheath is free to expand and the water then carries heat to the shell quite evenly.  In addition, you will remember that Willy wanted to raise steam more quickly with the limited heat input from his heating element.  In that case, any early steam loss absorbs a disproportionate amount of energy due to the latent heat it carries away, and so increases the time until steam is up to pressure for running the engine.  He is also dealing with the issues of trying to read pressure accurately on a tiny model pressure gauge.  In the hobby world, accurate temperature measurement is easier these days due to readily available thermocouple instruments.

So in one case corrosion is the biggest concern, thermal expansion stresses undesirable,  accurate pressure gauges readily available, startup time not really important, it is very short as a proportion of the normal operating time of a full size machine, and these factors determine the development of the startup procedure.  Though the ships captain probably doesn't see it that way.

In the other case, corrosion not really an issue, difficult to measure pressure with small gauges, desire to reduce startup time.  In this case, understanding how the thermodynamics affects pressure inferred from temperature measurement, and starting the boiler completely sealed to minimise heat loss through vented steam may become an appropriate procedure.  However, to get rid of air through steam venting when the pressure is fixed by the open vent, then to seal and raise pressure could be in the long term, the best procedure.  It allows a little more time for examining the intricate details of the engine while the boiler heats up.  Once the engine is running, the details move too fast.

I am glad you are finding it an interesting aspect to explore,

Thanks for following along

MJM460
Title: Re: Talking Thermodynamics
Post by: paul gough on October 13, 2017, 05:55:49 PM
I guess I should have prefaced my comments by indicating I was principally addressing steel boilers from larger 'model' sizes i.e. 12 inch gauge locos to full size. Though our codes generally demand steel from anything over 8 inch dia. barrels, which is smallish, and would I presume capture a lot of 71/4 gauge models. I thought the 'scavenging' of  air via venting during steam up might have been useful knowledge/technique to somebody especially if they don't have full size experience or are isolated enthusiasts, like me, who don't have the luxury of contact with like minded peers. I agree with all your commentary. A few of years ago I visited Mr. Jaycar, (electronics parts shop), when on a trip down south looking for temp. probes and a digital readout. At the time what was on offer was a too cumbersome to incorporate into my tiny G1 Lion loco. However there was a reasonably smallish probe/digital readout combo for temp. that one could reconfigure the connections so as to calibrate it for pressure, thus achieving a digital pressure readout/gauge. The instructions with the little screen had a wiring diagram showing the particular connections that achieved any numeral or combination of numerals on the display, so a bit of patience and diligence with a small soldering iron was all it would have taken to "make" an electronic pressure gauge from the temp. probe & readout unit and which would probably be suitable for a stationary set up, e.g. Willy's. However I'm not up to date with what might be available now, maybe micro sized digital pressure sensing units can be had readily?? Hope there is something useful in all my babble. Regards Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 14, 2017, 12:50:00 AM
Hi MJM ,thanks for that ,i have been looking for a temp/preasure table on the web but could not find any. Do these graphs depend on the ratio of air (compressible ) to water (non compressible) ? Interesting about the steel boiler water treatment ! I notice that the heat up graph is non linear but this may be because the time part is missing ? Would it be possible to make a 3D model including a time element?.....
Willy
Title: Re: Talking Thermodynamics
Post by: derekwarner on October 14, 2017, 02:34:01 AM
Willy.....this may help. :happyreader:.....Imperial ,Si........any combination and in either direction.....Derek


http://www.tlv.com/global/TI/calculator/steam-table-temperature.html
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 14, 2017, 04:16:58 AM
Engine Performance and testing -

Hi Paul, making measurements on your gauge 1 engines brings a whole new level of small scale to miniature instruments. If you have a suitable plug to replace with a thermowell, you could insert the thermocouple from a cheap multimeter into the thermowell to measure temperature while the engine was at rest.  Then just pull it out when you are ready to send the engine on its way.  Alternatively, you could hold the thermocouple tip against the boiler shell with an insulating pad of felt so it is not cooled by the air, and probably get close enough.  Experiments on a stationary boiler big enough to accommodate both would give you a good idea of the accuracy of this method.   I expect that you don't ride on the tender, so you don't really need to install the instrument on board.   The real size limitations are the batteries, unless you were prepared to use hearing aid batteries, and the screen, which clearly has to be large enough to be read.

Hi Willy, this link is to an open course on Thermodynamics and it includes a copy of the steam tables that can be downloaded as an Excel spreadsheet and read by most spreadsheet programs.
https://www.ohio.edu/mechanical/thermo/index.html
The course gets heavy quickly but you may find some interesting sections. 

The graphs don't use the volume of the water space, only the vapour space, and preferably the initial temperature.  Time is not relevant to the heat up graph, the non-linear part is just the water vapour pressure-temperature curve. 

Hi Derek, they tiv site is interesting, but I find a full set of tables is even more useful.  They hard to find on line though.    Hope you are now home safely from your travels.  The thread is getting close to your exhaust issue.

Willy recently asked if there was any point in an engine test on his horizontal engine.  I guess it applies equally to the Woolf Compound Mill engine.  So let's talk about what is required for an engine test, and what we can learn from a simple test.

Ideally for a complete engine test we need to measure the inlet pressure and temperature, the exhaust pressure, and if our boiler has a superheater, the exhaust pressure and temperature, more on that later.  We need to ensure the steam flow, and engine rpm, and finally we need to measure the engine output torque.  You see, we cannot predict the actual power that will be developed by a real engine, we do have to measure it.  This applies even to full size engines, though a full size engine manufacturer has access to many previous test results of similar engines, from which a pretty good estimate can be made for the engine efficiency.  However, if you as the customer want extra assurance that you will get the promised power from your new engine, you can specify, and pay for a full test on your engine after it is built.  Alternatively, you might accept the prediction and accept a lower cost simpler test just to prove that it runs ok without mechanical or steam flow problems, quite like our first runs on air.  With a known engine efficiency, or more specifically adiabatic efficiency, you can calculate the power which would be developed by an ideal engine, and from the efficiency, you can calculate the output of your real engine from the given steam conditions.

As model builders, we don't have access to many previous tests at all, let alone tests of similar engines to the one we are building, so we don't have a knowledge base of reliable efficiency data.  We are left with doing our own tests, so we need those instruments.  The temperatures are relatively easy, we have already discussed those, the more difficult ones are the engine inlet pressure, and the torque. 

Even that inlet pressure is easy if we are prepared to accept any pressure losses in our piping and regulator as part of the engine inefficiency, and we know the boiler pressure pretty well, at least if you have been following this thread.  But measuring torque requires more thought.  Most of the trouble with torque measurement is that our engines do not produce a uniform torque. For one cylinder, it goes from maximum to zero twice each revolution.  A two cylinder engine with cranks at 90 deg means four pulses per revolution, but at least the peaks are spaced so there are no zeros, but it fluctuates considerably nevertheless.  I know every dynamics text book describes a friction brake on the flywheel, and a good flywheel should keep the speed fluctuations to around 7%.  That should help, but I am not convinced.  I know I had to do a test with such a brake in my student days, and I really can't remember how much the reading fluctuated.  But we got pretty good at estimating the average or mid position of a vibrating needle, so that's what we had to do.  Now with electric integration and averaging it should be different, but there's no fun in that!  So I am thinking about a torque measurement which somehow averages the torque.  But, is there anything we can do without torque measurement?

I suggest that is a good place to stop and continue next time.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: derekwarner on October 15, 2017, 04:36:46 AM
I am sure there would be good fishing in the Ohio River where the cooling water from that  :Mad: 2.6M kW steam power station is located

Yes MJM...back in Wollongong however have not yet completed the exhaust temperature tests....[awaiting the insulation lagging to be completed over the new 1/4" OD exhaust line tubing]

Another question :headscratch: ...forward thinking...is 'to lag or not to lag' our gas tubing supply.......from the disposable canister to the refillable tank is fixed, however from the refillable tank to the burner jet is subject to considerable temperature change.....& occasional icing

I plan to install a gas isolation valve before the actual gas jet, then relocate the gas pressure gauge location to the discharge side of the  regulator. This way will then allow me to confirm the reduced gas pressure at the jet...........and also by isolating the boiler pilot pressure to the regulator, the actual gas tank pressure can be confirmed....

Derek 
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 15, 2017, 11:19:00 AM
A simple engine test -

Good to have you back Derek.  I suspect they are not allowed to discharge cooling water into the river in Ohio, those very large concrete structures, the nearest one is issuing clouds of steam, I believe are natural draft cooling towers, and the white plume is the evaporating cooling water re-condensing in the cooler atmosphere.  A blow down flow is needed on a cooling water system, just like a continuously operating boiler and that is probably being treated and has time to cool in those low circular pools before discharge.  It looks like one tower is shut down, although there are designs which reheat the air above the coils to eliminate the plume.  I am not sure what the steam discharge from the stack is about, and that plume makes it hard to see the second stack, or precisely which is discharging that wisp of bluish smoke.  Perhaps some of our colleagues can come in with some more explanations. 

With your gas piping, remember the lpg in the container, whether the disposal or your refillable tank, boils just like water with latent heat and the two phase region, but at lower temperatures.  Your burner needs the gas phase, but the "l" in lpg is short for liquified.  The container has adequate mass in liquid form but  the liquid must evaporate into vapour for your burner.  As you don't have a heater on the container, the energy for the latent heat comes from the liquid, so it cools until heat in from the atmosphere is equal to that required for evaporation.  And at the lower temperature the vapour pressure is lower, so your burner sees ever lower pressure until the heat balance is achieved.  To get more heat in from the atmosphere, it is better not to insulate the pipe or even the bottle.  The ice on the outside of the pipe means the internal gas temperature is so low that the outside surface of the pipe is below zero, and moisture from humidity in the air first condenses then freezes on the pipe.  Removing that insulation will allow more heat in from the atmosphere, higher gas temperature, though it will still be cool, and most important, more pressure to the burner.  The regulator is probably running wide open, due to the low bottle pressure, so increasing the temperature in the bottle will help get your regulator back in control.

Back to those simple engine tests - You may remember I did a simple test on my mill engine some time ago.  Next job on the list is to repeat that a few times to ensure that I have consistent measurements.  I did measure the exhaust temperature, and came up with an efficiency for my engine.  Now I could do that test because my boiler, a fired boiler, though only Meths for fuel, has a superheater coil.  And it is quite effective.  With a boiler temperature of around 116 deg C, the superheater outlet was 138 C.  At that time, I showed the steam conditions on a temperature-entropy diagram and used the second law of thermodynamics to calculate the power output of an ideal adiabatic engine.  Yes that is repetition (tautology?), but deliberate for emphasis on the 'ideal' part of adiabatic.  I have repeated the diagram below for convenience.  The boiler outlet is point 2 and the superheater outlet is point 3.  The second law says the exhaust entropy of such an engine is equal to the inlet steam entropy, so the engine expansion shows as a vertical line on that diagram, the line from 3 to 5.  Entropy and pressure are enough to define all the exhaust steam properties.  You can see on the diagram, the adiabatic exhaust, point 5, will be only just wet steam, 99% dryness factor (by calculation).  Now my measured exhaust temperature was 104 C, meaning the real engine exhaust was superheated at atmospheric pressure, point 4 on the diagram where the remaining enthalpy is more than the enthalpy of saturated steam.  This basically means my engine had extracted less power from the steam than the ideal engine, and the ratio is the real adiabatic engine efficiency.  I don't have a torque measurement to calculate shaft power, but I know how much energy was extracted from the steam, before the inevitable friction losses and so on.

Now Willy's engine is in a different situation.  The engine itself is quite similar to mine to a casual observer, it is just better made.  But his boiler is electrically heated, and has no superheater.  In fact not easy way to add a superheater.  So the engine is receiving saturated steam, steam from point 2 on the diagram.  The ideal engine process is a vertical line down to exhaust pressure as before.  You can see from the earlier diagram that without the superheater, the exhaust steam is much more wet, or has a much lower dryness factor.  It looks quite likely that even a real engine exhaust would be in that wet, two phase region at atmospheric temperature.

Now the earlier diagram is not to scale.  It shows the two phase region as a bell shaped curve, very similar to any text book you care to pick up.  And gives the form of the curve in an easy to remember picture.  I did take care to ensure that point 5 is on the correct side of the saturated vapour line.  Also the constant pressure lines, 1-2-3 and the atmospheric pressure line through 5 and 4 are also the correct shape.  However, just to be sure that I am not being mislead by the out of scale part, I have re drawn the diagram to scale, and included it as the second attachment.  I am not the first to do this, in fact they are fairly easy to find as a complete T - S diagram for steam, covered with extra lines for every other property.  But the tiny part relevant to our model is so small it is difficult to pick up values with sufficient accuracy.  So the second  picture below is drawn to scale from the properties in the steam tables, and you can see that it is no fancy drafting job. 

We have talked about several operating pressures for Willy's boiler, I am still not sure if it was 143 or 148, and there was 135 and also 118 mentioned somewhere.  So I have plotted the constant temperature-pressure line for the boiler for 148, 135 and 116 so we can see the effect on engine performance over this range.  I have kept the temperature axis to scale to cover the exhaust at 100 deg C up to 150 C to cover over the 148.  I could not complete the bell curve on the paper at a reasonable scale, the top is at 374.14 C where the pressure is over 22,000 kPa, known as the critical point.  It is quite important in thermodynamics, so has been measured quite accurately.  Above this temperature or pressure, there is no distinct liquid phase.  But we don't need it for our calculations.  Similarly I have shown a second break in the curve which is bounded at the bottom by a straight line at zero degrees, where water freezes.  This does not have to be measured, it is a defined point on the temperature curve, as is 100 deg, the boiling point of water at 101.35 kPa and a few other points.  That is why we can use zero and 100 to check the accuracy of our temperature meters.

The enthalpy and entropy for dry saturated steam are both in the steam tables for 100 degrees, but the other temperatures needed some interpolation, a spreadsheet comes in handy for that.  Now using that second law condition, the adiabatic engine exhaust entropy is equal to the inlet entropy, I drew in the constant entropy lines which are vertical on this diagram, and calculated the dryness of the exhaust in each case.  You can see from the diagram they are very much to the dry end of the line, all above 90%, but it is better to do the calculations to get good figures for comparison.

That is enough to take in for one session, don't forget about the gas question either, so I will leave you with the diagrams to look at, and continue next time.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 15, 2017, 02:42:25 PM
Hi Mom thanks for this , I was thinking about making a superheater for this boiler but it would have to be an external one retrofitted and the electric control system would be necessary to avoid burning the element out.!! I was wondering how one might measure the exhaust pressure without restricting it as it just goes strait to atmosphere or is it calculated from the temp ?. I have just bought 2  (analogue) ? thermometers (griffen and george) and notice that the scales are different lengths so perhaps they are calibrated separately ? When using a superheater....the temp goes up and also the pressure ...,,but as it is an enclosed system is the pressure also in the boiler itself as well and how does this muddle the figures for the boiler temp/pressure etc ?? perhaps this is easily explained with the correct Maths ?!!! I may rig up the engine back to the boiler and it does have a small load on it withthw dynamo and the magnet driven cycle speedo, and perhaps this may be useful in any calculations......The magnet is placed in the periphery of the cast iron fly wheel. Looks like more fun on the way............The thermometers go up to 360 C but do they need insulating as they are about 12" long so doing superheating tests will be possible !!
Title: Re: Talking Thermodynamics
Post by: Gas_mantle on October 15, 2017, 09:32:33 PM
Hi,

I've kind of followed this thread for a while now and learnt something along the way although I'd be the first to admit as a layman some of the technical info is way above my head but I'd like to ask a few questions of the experts.

As I understand it water expands 1600x when turning to steam at atmospheric pressure, if we assume atmospheric pressure is 15 psi am I right in saying that a boiler that can convert 1 cc of water to steam per min can theoretically run a 1 cc engine at 1600rpm using 15 psi (assuming we ignore heat loss, inefficiencies etc and the engine is single acting.)

If we take a small boiler like the GLR, how much water can something like that realistically convert to steam per min ?





Title: Re: Talking Thermodynamics
Post by: MJM460 on October 16, 2017, 08:27:42 AM
Superheaters and steam volumes -

Hi Willy, your superheater design concept is spot on.  It might be well worth making that central sheath for the element hollow right through and soldered into both ends so you can put the element in, with some heat transfer grease and put a thermocouple in from the other end for your over temperature protection.

 However a few points to be aware of.  Heat transfer is not nearly as good to steam as it is to boiling water.  To put numbers on it, my heat transfer text book gives figures of 30 - 300 kJ/ m^2.K for superheated steam, and 3000 to 60000 for boiling water.  The wide range in each case makes them difficult to use in design, but you can see the difference in magnitude.  So heat transfer is limited by the steam side film coefficient.  You need lots of area and this is where fins come in.  The higher the fins the better, within reason.  Of course flow is longitudinal, so you need longitudinal fins.  Plenty of build logs present ideas on how to machine these.  You also need a lower heat intensity element, that is more surface area per watt of rating.

 It is not all bad though.  If we look at say 400 kPa absolute, 143.6 deg C, a touch over 40 psig, the enthalpy of the dry steam leaving the boiler is 2738.6, while superheated steam at the same pressure and say 200 deg C enthalpy is 2860.5 kJ/kg.  Only an extra 122 kJ/kg compared with the 2133.8 kJ/kg which the boiler needed to evaporate the water at 143 deg C to dry steam at the same temperature.  You can see that the superheater element only needs to be 50 or 60 watts to nicely compliment your boiler and achieve plenty of superheat.  But it will be harder to transfer that heat to the steam, and the element will tend to run hotter.  So a low intensity element, access for heat transfer grease and a thermocouple for your temperature protection, longitudinal fins as high as possible, and plenty of insulation.  You will need a condensate drain to aid startup which will require a bit of thought so as not to overheat anything.

Those thermometers may be individually properly calibrated for a small variation in the diameter of the capillary that carries the Mercury column up to the appropriate scale mark.  I don't know much about the manufacturing process.  I believe there is an error in the readings due to heat transfer along the glass, and it can be calculated, but it's lost in the ancient mists of school science.  Alternatively it might be possible to reduce it to an acceptable level by insulating the part of the stem below your expected reading with some felt or silicone tubing.

In the superheater, the pressure does not increase.  If anything, there is a small pressure loss due to pipe friction as the steam flows along.  Assume boiler pressure and use a generous size tube between the boiler and superheater.  That line 1-2-3 on the first diagram yesterday is a constant pressure line, the pressure is no longer a straight horizontal line outside the two phase region. 

Using a bike speedo for rpm should be ok so long as you use the wheel size set in the device to calculate rpm from the wheel circumference and speed.  The Dynamo will provide a nice little load.  Unfortunately the V and I measurements do not give us engine power as we do not know the electrical efficiency of the Dynamo.  That is why we need torque measurement.  But V x I does give a sensitive indication of load changes.  Just be aware that efficiency might also be changing.  We should be able to see a slower run speed, higher steady boiler pressure and lower steam rate with the load on, to compare with the free running unloaded engine.  We should then be able to see and  compare the power output of our adiabatic engine which will be an indicator of the actual load.  The exhaust will most likely still be in the wet region so temperature measurement is not useful.  But it  should be a worthwhile run in every way.

Hi Gas Mantle, great to have you on board.  The purpose of this thread is to try and make that technical information understood and accessible to all, so all questions welcome.  The expansion as steam changes from liquid to vapour is best found by using the specific two volume columns in the steam tables.  The precise figure for standard atmosphere is 1602 times.  At 15 psig, say 30 psi absolute, or 206.8 kPa.  Compared with the steam at atmospheric pressure the steam at 15 psig is compressed and so a smaller volume.  I will round that pressure out to 200 kPa (if you look at the tables you will see why), and see that the specific volume of the liquid vf, is 0.001061 m^3/kg, while the specific volume of the dry saturated steam, vg, is 0.8857 m^3/kg.  A simple division, using a calculator gives an expansion ratio for 15 psig of 835.  So a perfect 1cc single acting engine should run 835 revs for each cc of water evaporated at 200 kPa.  Or 1 cc per minute evaporated should give 835 rpm.  I have found it hard to match the figures in practice.  Valve leakage, piston leakage and any valve overlap all increase the steam consumption, or reduce revs, while early cut off should decrease steam consumption.  I am still working on just which it is, or possible all the above on my engines.  But it is a valid starting point. 

The fundamental limit to how much steam a boiler can raise is the amount of energy in the fuel you  burn.  Weigh your fuel container (with burner attached is ok) on the most accurate digital scale you have access to, before and after a timed run and calculate the mass of fuel burned.  Doesn't matter that boiler first heats up, then generates steam, as the burner fuel consumption should be pretty constant with time what ever.  Then look up the fuel calorific value, I can probably find a figure if you tell me what fuel you use, and multiply the two to get the energy from the fuel.  Then it depends on the boiler how much of this is turned to steam.  The gas from the stack is hot, so that tells you that energy is being lost, and it can only be cooled to a temperature something above the steam temperature in the very best boilers, so some loss from that source is inevitable.  There is also radiation from the boiler casing.  So the next experiment is to see how much steam you are getting. 

You will have seen from the calculations on Willy's boiler that it is helpful if you can weigh the bare, empty boiler, otherwise calculate it from the density of copper and the boiler dimensions and thickness.  Weigh a suitable amount of water into the boiler, light up and run for a suitable time, and carefully time from light up to first steam, and on to flame out.  Finally carefully drain and collect all the remaining water from the boiler.  The water evaporated is obtained by subtraction, and the time for that evaporation is from the first steam to the end. 

So you can now work out how much heat to raise steam, and how much heat was used to raise a known quantity of steam.  From that you have your boiler efficiency, which you can use with reasonable confidence for future runs of that boiler.  If the boiler has too little heat transfer area, the stack gases will be hotter and more heat lost.  If it has plenty of area, the stack gas will not be as hot, and if necessary you can give the burner more pressure or use a larger burner.  I don't know the particular boiler you mentioned, but testing the burner and calculating the heat transfer area will give you a good idea of how it would perform compared with other similar boilers.  I hope that helps, but don't hesitate to ask for any clarification you require.  Others will probably be wondering the same thing.

I think that is enough for a session, so I will try and return to those t-s diagrams and what we can learn from them next time.

Thanks for reading,

MJM460
Title: Re: Talking Thermodynamics
Post by: Gas_mantle on October 16, 2017, 12:40:31 PM
Thanks  :)

I seem to remember reading somewhere that small copper boilers like the sort of thing hobby engineers use should be able to evaporate 1cu" of water per min for every 400sq" of surface area in contact will the water.

Does that sound about right as a very rough guide? I can't remember where I read it and may have got the figures wrong.
Title: Re: Talking Thermodynamics
Post by: paul gough on October 16, 2017, 12:53:38 PM
Re lagging and heat losses. Reading in, 'The Efficient Use of Steam', by Oliver Lyle, 1948, the author concludes for steel pipes 3" dia or below with internal temps. of 300 degrees F., (which would work out for us at about 50 p.s.i.g.), a lagging thickness, (85% magnesia or asbestos), of 1" is sufficient, his chart shows approx. lagging surface temp. of 107 F. in still air 70 F. A lot of our model boilers are made of copper so I assume heat loss potential higher than a steel pipe. Author quotes heat loss for a bare 6" steel pipe in still air with internal temp. of 300 F. as 646 Btu/sq. ft./hr. and 98 Btu loss for 1" thick lagging. A common dia. for 5" and 71/4 gauge locos.

Question; (1) Are the insulating properties of our much used Kaowool sheet insulation comparable to the above materials?

               (2) 1" thick insulation on a 3" boiler is pretty thick by most builders standards. Are we underestimating the thickness we should
                     apply to our boilers when there is room to do so?
                                                                                            Regards, Paul Gough.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 17, 2017, 12:00:35 AM
Hi MJM ,thanks for the suggestions about the superheater design. I was going to use a 500 Watt element i have in stock but if i did use this what would happen ? in a Locomotive the superheater goes back into the firebox where the fire is at its hottest !! so do we not need to do this ?
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 17, 2017, 11:21:02 AM
Boiler potential, insulation and superheater design -

Hi Gas Mantle, that rule of thumb for boiler capacity has a very similar form to that in the K. N. Harris book on boilers.  It is not without basis, but has some big implied assumptions that really need to be stated whenever it is quoted.  Dan was asking about it some time ago and I have been waiting for an opportunity to get back to it.  For a start, look at the dimensions of the terms.  A valid equation has first to have equal dimensions on each side of the = sign.  You cannot add or multiply apples and oranges.  So that constant, 400, has dimensions to make that side of the equation dimensionally the same as the other side.  Undesirable, but no real problem, it just means that the correct units must be used, cu.in/min on one side and square inches on the other side, or a bit of work to calculate the equivalent constant for use in other systems.

   If we go back to the basic heat transfer equation (it was discussed back with condensing,) Q=U x A x delta T.  The simple form does hide some complexity.  First Q has units of Watts, but so long as consistent units are used both sides, you can equally well use Btu/hr of you prefer, and you use consistent units for the other quantities.  The delta T is log mean temperature difference, and should strictly be written delta T lm with the lm as subscripts meaning "log mean to base e".  For SI, the units are deg C, and it is a temperature difference so you can use C or K (or F or R).  A is straightforward, m^2 for SI, (or ft^2).  U is more complex, but it's units are W/m^2.hr.C, (Btu/ft^2.hr.F) and unfortunately it is not a simple constant, nor is it easy to predict a value.  Whole books on that subject are heavier in every way than thermodynamics books.  They get there in the end, but I don't intend to go there. 

If we compare the two formulae, we can see the similarities.  A cubic inch of water has a fixed mass, particularly of the temperature is specified, and that mass takes a known amount of energy to evaporate it, again "known" means using the steam tables and requires the pressure or temperature to be known.  So cubic inches can be regarded as a proxy for heat input to the water.  Now in most model boilers, neither the delta T, nor the heat transfer coefficient is known, so it is not unreasonable to combine the two into one factor.  But you can't easily determine what this factor will be.  For example, for a coal fired locomotive the temperature in the firebox will be pretty high.  A very different temperature would apply for a gas burner, or a methylated spirits burner, so any figure determined by a boiler test would really only apply to the same method of firing.  Then the log mean temperature difference, LMTD, varies with the arrangement of area and gas flow through the boiler, so would strictly only apply to similar boilers.   But more than that, both the LMTD and the maximum possible heat transfer are limited by the burner size.  Conservation of energy is a fundamental law of physics and it means in this case you cannot transfer more heat into the water than is released in the firebox.  Basically if you have a small fire in a big boiler, the stack gases will be cool to only a tiny bit above the steam temperature quite quickly, and the rest of the boiler area just gets a tiny bit closer.  On the other hand, if you put a huge burner into a small boiler, you will definitely get more steam than from the small burner, but the flue gases will arrive at the stack before they have lost all the heat that can be transferred at the steam temperature if the area was bigger, and so will go up the stack at higher temperature, wasting some the energy in the fuel.  You can see the problems, however, I would guess that formula may have been arrived at based on coal fired locomotive boilers.  Similar successful boilers might well be consistent enough in proportions of fire box and heat transfer area to come close to following this simple formula.  I don't have a better idea, so as long as the formula is used as a rule of thumb, with understanding of its limitations, it can serve as a starting point for design.  It could be useful for scaling up an existing boiler which you could test, to a larger scale similar boiler with a similar arrangement.  If you were starting a new locomotive design, you could run a few calculations in a spreadsheet for some of the past entries in the various locomotive efficiency competitions, to get an idea of the applicability of the formula.  Then add data from any tests you are able to do on your own models.

Hi Paul, generally industry places a high value on the immediate cost of materials and labour costs to apply insulation, and undervalues the long term cost of the energy losses which are usually incurred "by others", so insulation thicknesses are generally less than desirable for heat conservation.  You can also see it in domestic fridges, where the sales hype is the illogical assertion of more internal volume for storage of your food within the smallest possible outside dimensions.  The salesman does not place any value of the insulation hidden between the outside shell and the liner.  And the average plastic Esky has no insulation in the roughly 1/2 inch gap between the outside shell and liner.  And simply, yes, we put too little insulation on our models for heat conservation purposes.  But note that qualification, in our models we are trying to optimise size of the boiler, heat transfer area and appearance, and the fuel cost is not high for the typical number of operating hours, while appearance is paramount.  So we have valid reasons for using minimal thickness.  It is usually simpler to achieve performance by pouring in more heat.  But whenever we can apply some insulation, it reduces heat up times and gives more steam for a given fuel burning rate, and reduces the incidence of burnt fingers, and if it is practical, more is better.

K values for some suitable insulating materials in Watts/m.K - cork 0.04, glass fibre 0.035, kapok 0.035, plaster 0.814, polystyrene 0.157, softwoods 0.15, oak 0.19, wool 0.038.  I also have 50% magnesia 2.68.  Asbestos is 0.113, so not only dangerous, but not the best for insulation value.  It was used because it has fire resistance at really high temperatures.  All compared with copper 83 and steel 43.  Steel will have a higher temperature gradient than copper, but with insulation restricting the heat flow, both temperature gradients will be small enough to be insignificant.

OK, Willy, what will happen when you use a 1000 watt element in your superheater?  It's easy to apply a few numbers and this will require us to look at the superheat section of the steam tables. This section looks complex, but it is actually a quite simple separate table for each pressure.  Each little table starts at the saturation temperature, steam is not superheated below that, then jumps into a sensible series and has a single column for each of specific volume, internal energy, enthalpy and entropy.  So let's look at the steam from your boiler at 400 kPa abs, about 42 psig, that we used last time as the superheater input.  You will remember the enthalpy of that dry saturated steam leaving the boiler was 2738.6 KJ/kg.  And if you look back at the saturated steam section of the tables, you will see that your 1000 Watt element put 2133 KJ/kg into each kg of steam to evaporate it from water at that temperature.  Now remember your electric heater delivers its rated power output by increasing its own temperature until the heat transfer is enough.  In your superheater, your 1000 watt element will just get hotter until there is 1000 watts of heat transferred.  So you will add 2133 kJ/kg to the steam from your boiler.  The superheater outlet enthalpy will be 2738.6 + 2133 = 4871 kJ/kg.  Now we look up that enthalpy in the little superheat table for 400 kPa, and it is there.  (You can see that enthalpy is turning out to be a very useful concept).  Look across to the temperature column (it's on the left side of the page, but applies right across that line on the page).  Superheated steam with an enthalpy of 4871 KJ/kg has a temperature in the range of 1000 to 1100 deg C, you need to interpolate to get a precise figure.  I am assuming the heat loss is no greater than that from the boiler, so I assume some good insulation.  So what would that temperature mean?  Well the code design for copper boilers generally does not give a strength for temperatures above 200 or 250 deg C, so your boiler inspector would probably require a steel boiler, among other things if he knew the predicted discharge temperature.  Yes, the superheater is classed as a pressure vessel and requires testing and approval.  (Strictly, piping also).  If the copper temperature goes above 200, the strength continues to reduce, until copper melts at about 1350 deg C, so I would expect it to be pretty soft at over 1000 C and the shell would probably bulge at some point.  Not like a balloon, but a sizeable bulge, then as the boiler continues to maintain the pressure, it would split!  However silver solder melts at a somewhat lower temperature, so the ends or bushings might blow out first.  Of course the heater element has to be a bit above the steam temperature, so it might burn out before we get to this temperature.  It's a bit of a race to see which fails first.  But there are no winners in that race, whichever outcome, it will end in tears.  Better hope it's the element, 'cos steam that hot steam is very dangerous.  Just as well this is a hypothetical experiment, it's not a wise one to carry out with real equipment, however small.  Better stick with the recommendation last time of 50 to 60 watts, then the outlet temperature will only get to about 200 deg, so long as you maintain the steam flow.  I hope that makes the design process a bit clearer.  So you still need a high temperature cutout, and it should operate if there is high temperature or if the boiler power is cut.  That is an inclusive OR, either one or both should cut the superheater power.

Of course you may ask about using a controller to reduce the operating time of the heater.  It's possible with PWM of the power supply or other fancy digital controls.   The issues are reliability and heat transfer element intensity.  It is very hard to adequately cool the element when your fluid is dry steam instead of boiling water.   And all the while, you are a single failure of one of the internal electronic components away from 1000 degrees.  This is not intended to be a damper on the idea.  It is quite practical to build an electrically heated superheater as you suggest, they are used in industry.  You just have to understand the heat transfer and thermodynamics to develop a safe and realistic design.  If the elements of the lower rating are not available you could use say 100 Watt with a controller, or a transformer to reduce the voltage.

The last point, you are correct in that if the heating was from say a heat transfer oil which changes temperature as it looses heat, you would arrange the flows to be countercurrent, however the element is nominally a constant temperature device.  Think one of those one bar electric radiators, it's the same temperature all the way along as the heat is produced evenly in each metre of wire.  In the superheater, the steam temperature rises as the steam flows along the element, so the element temperature has to change in response to maintain the heat transfer all the way along.  If you connect to the other end, the hotter end of the element will always the outlet end.

I hope you find that interesting and helpful in you consideration of an electric superheater.

Thanks to everyone for looking in,

MJM460


Title: Re: Talking Thermodynamics
Post by: MJM460 on October 18, 2017, 11:42:30 AM
Back to Adiabatic engine discussion -

Those little diversions have given us an extra interesting steam condition to add to our exploration of the performance of an adiabatic engine.  You might remember that a few posts back, (post #363)
that I drew the relevant part of the T - S diagram to scale and showed three of the boiler temperatures we had been looking at for Willy's boiler.  (The second diagram)  The horizontal lines labelled 118, 135 and 143 (all deg C), show the boiler process for evaporation of water at those temperatures, while the one labelled 100 (deg C) is the condensing process for steam at standard atmospheric pressure.  The one labelled 15 shows condensation (or boiling) at the atmospheric temperature of 15 deg C.

There is a table to the right of the diagram which shows the saturation pressure in kPa (abs) for those temperatures and the entropy at the wet and dry ends of the lines, sf and sg.

 If we supply steam at one of those conditions to an adiabatic engine, then the second law of thermodynamics says that for this ideal engine, the exhaust entropy will be equal to the inlet entropy.  (Any real engine exhaust will have higher entropy.)

This means that the ideal engine expansion is that vertical line from the dry saturated steam leaving the boiler to the exhaust steam line.  The lines for the three conditions are very close together, so I have only drawn two.  You can see they all arrive at a wet steam condition.  In the wet steam or two phase area, temperature and pressure are not independent, but we now have the entropy at the exhaust condition, and this is enough to completely define the steam properties.  You see even entropy eventually comes in useful.  It is a property of the diagram that in the wet steam, the mass fractions of liquid and vapour are proportional to the values between the liquid and vapour lines for example, sf and sg, and as we now know those values, that fraction can be calculated.  Then the same fraction applies to all other properties, in particular, enthalpy.  So we can now calculate the enthalpy at the exhaust steam condition.

I have calculated those values and summarised the results in the table at the bottom of the diagram.  And the last column is the difference in the engine steam inlet enthalpy and the exhaust enthalpy, which is the work extracted from the steam by the adiabatic engine.   These are all on a per kg basis, but if you multiply by the steam rate, you have the ideal engine power output.    Remember, these are for an ideal engine, and the output of any real engine will be much less, but they are a maximum limit, and show the difference in potential output for different steam conditions.  You can see that the answer from subtracting two large quantities, the enthalpy of the supply steam and enthalpy of the exhaust steam which are a similar magnitude, is only about 10 % of either of those quantities, so it is necessary to try and carry the calculations through with the same accuracy as the tables to minimise the accumulation of errors.

You can see that as you would expect that the engine provides more output from the higher pressure steam, but you might be surprised at how much more compared with the extra heat input to the steam.  For example the ideal engine output is nearly doubled between 118 and 135 deg C inlet temperatures for about 1% extra energy input.  You can see why industry aims for ever higher temperatures and pressures.  I'm not suggesting that the model engine output is doubled by this higher inlet temperature, but there is no doubt that the potential power output is significantly increased.

It is also interesting to note that at higher temperatures, higher up that bell curve, the energy output to evaporate the steam, hfg, reduces, the increased energy contained in the higher pressure steam is all put in during the boiler heat up phase of our simple boiler, and once up to temperature the boiler can produce a small amount more steam, perhaps further increasing the engine output for our burner capacity.   Of course with continuous feedwater injection, the heat up and evaporation process happen at the same time, so for a given burner or electric element size, there is overall less steam produced.  Might be worth fiddling a bit more with the calculations to try and extract more understanding.

Now Willy has proposed a superheater.  To save a little effort in interpolation, I worked the example  at 143 deg C where the boiler pressure would be 400 kPa.  A pity it is not on the diagram, but the lines were getting to close together.  However a close look at the figures on the diagram suggests the 400 kPa steam inlet, would give exhaust steam about 0.925 dryness fraction and an engine output around 220.

When we add the superheater, the 50 watt one of course, to give 200 degrees C the steam enthalpy is 2860.5, and the entropy, from the superheat table, is 7.1706, compared with the 6.8959* (*note this figure has been corrected after the original post) of saturated steam, but still less than saturated steam at the exhaust temperature of 100 (7.3594). From this we can calculate exhaust dryness = (7.1706-1.3026)/(7.3594-1.3026) = 0.9688, clearly much more dry with the superheater, but still in the wet region.  We can then use this figure to calculate the exhaust enthalpy = 0.9688 x (2675.5-417.46)+417.46 = 2605

Finally steam inlet enthalpy - exhaust enthalpy = 2860.0 - 2605 = 255.5 KJ/kg.  This is a lot more than from the saturated steam but only little higher than if the boiler was operated at 148 deg,  possibly not enough to be worth while.  Perhaps we should make the superheater a little bigger to give say 250 deg C.  This would give us dry steam exhaust but more importantly  would show a real benefit  for the superheater compared with the boiler a little higher pressure.  I repeated the calculations for 250 deg C superheater outlet temperature, it requires 105 watt element, and increases the adiabatic engine output to 280 KJ/kg.  This looks like it would be more worthwhile.  The adiabatic engine exhaust would be very close to 104 deg C.  But worth looking at the boiler temperature required to achieve this output without superheat.

By the way, if a 100 - 110 W element is not available, remember the power is proportional to V^2.  So if you use a 240 - 110 V step down transformer, a 500 watt element rated at 240 V will produce about 105 watts at 110 V.

Another interesting point is that the specific volume of the superheated steam is higher, the superheater increases the volume of the steam.  If the engine can't run faster to use the extra volume, the system pressure right back to the boiler will rise unless the heat input is cut back.  So for a free running engine, whether loaded or not there will be a new balance of steam pressure and flow to use all the energy put in.  Remember the superheater cannot produce more pressure than the boiler.

You can see there is value in doing some calculations before you start cutting metal in order to reduce the amount of error in trial and error.  We still have to look at what all this means for a real engine.

Thanks for stopping by,

MJM460

Title: Re: Talking Thermodynamics
Post by: Gas_mantle on October 18, 2017, 07:57:49 PM
Thanks MJM,

As a general principle do small model boilers generate more steam when coal fired as opposed to gas ?
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 19, 2017, 02:39:54 AM
Hi MJM thanks for this new info.....I did hesitate when trying to make a superheater  as i thought there might be problems with this design !! I was wondering if the superheater pipe was inside  the boiler heater sheath would this work ,a bit like a locomotive superheater that is actually in the fire box.As this sheath is at the same temp as the water it is boiling it is not at a much higher temp??
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 19, 2017, 11:10:02 AM
Superheaters and real engines -

Hi Gas Mantle, I don't have direct experience with coal fired models, but I am sure there are many others on this forum who will be able to answer your question.  However a few observations can be made that might be relevant to understanding the issues.  First, with coal, you need a critical mass of hot coals to sustain combustion, just putting a match to coal will burn a little coal while it is there, but the fire does not continue when the match is removed.  You need a bit more concentrated heat  to make a coal fire which is self sustaining.  In addition you need an adequate draft to draw enough the air through the coal bed, and of course space for a big mass of coal and lots of draft are both in short supply in a small model.  You can do quite a bit with the coal particle size, but I suspect there is a minimum practical size for a coal fire, and I don't know how much coal the smallest practical fire could burn in a given time, or at what rate coal could be burned in a small model.  With gas, the particle size is molecular, and so long as you have a steady supply and the appropriate air/fuel ratio, the tiniest flame will continue.  At the other end of the scale, even in a small model, the available gas pressure means you can burn a relatively large mass of fuel with ease, and the gas pressure provides energy to achieve the required draft.   On the other hand, a coal fire has a very high transfer rate for radiation heat transfer.  The predominance of energy in that red and infra red region gives very high radiation coefficients.  The energy distribution in a clean gas flame is biased more towards the blue and beyond, which results in a lower radiant transfer coefficient.  You can feel this difference by holding your hand an appropriate distance from the flame.  So a coal fire can transfer much of its heat by radiation, where the high temperature helps more, leaving less to transfer by convection.  A gas fire does not transfer as much by radiation, so generally needs more area for convection heat transfer.  Some people insert a wire coil in the gas path to collect heat by convection from the hot flame.  The wire in turn glows red, and radiates more efficiently to the tube surface.  I don't know how well this works in practice, but it has a sound basis.

So the question is more complex than it might seem, and the answer really relies on experiment.  The theory should help understanding of the experimental results, but ultimately you need to talk with people who have done the experiments.  If you have a suitable small boiler, experimenting with each fuel, including coal size, bed depth, blower arrangements and different gas burners to see which will evaporate the most water in comparable arrangement would be the best way.  I seem to remember that Florian's Cochrane boiler is set up to allow both fuels, but it is probably not a small as you are thinking.  Comparison between different boiler arrangements is more difficult.  If the fuel rate releases the same amount of energy for two cases, they should be able to produce the same amount of steam, but coal will probably give better results with a generous firebox for radiation transfer, while gas will probably need more area in the flues for convection.  I will be interested to find out what you are able to learn.

Hi Willy, I am having trouble imagining how you would put the superheater tube in the sheath with your electric element.  Remember, you element just gets as hot as it needs to in order to loose all the heat generated, and relies on good cooling of the sheath to limit the temperature rise so the element does not melt before it rejects enough heat.  For a boiler, you slide the sheath into a close fitting tube which is part of the pressure containment, preferably with some heat transfer grease to aid transfer to keep the element cool.  You can't make it loose enough to accommodate a superheater tube, as the element cooling would not be adequate in a sheath with enough excess space.  I suppose you could use a solid rod of brass or copper, and drill two or even three holes length wise, one a close fit on your element and the other one or two fitted at each end to connect to the steam pipe.  I am not sure if you would get a high enough temperature in the steam tube to be worthwhile.  Heat only flows from hot to cold, and you need enough temperature gradient to transfer enough heat.  Remember the example with superheat to only 200 degrees compared with a higher boiler pressure.  It would be difficult to control the heat distribution to achieve adequate superheating, and at the same time adequately cool the heating element.  Your separate device concept is not only easier to analyse, it is more flexible in operating conditions, and I suspect it would be much more successful.  But an interesting idea to ponder if the specification sheet for the element allows a high enough temperature.

In a fired boiler, you are correct that the conductivity of copper means the copper is quite close to the water temperature, but still higher, as heat is transferring from copper to the water, but the flue gas is at much higher temperature to transfer all this heat via convection to the copper.  I suspect though that while the superheater tube passes through the flue tube, most superheating might occur in the part exposed to radiant heat in the firebox.  This could also be why many writers question the effectiveness of these superheaters.  You can see the arrangement is quite different from the one in my small boiler, where the steam tube is run into the fire box, two turns around the firebox, then straight out through the furnace wall from where it is insulated until it gets to the lubricator and engine inlet.

Last time, I looked at the performance of an adiabatic engine based on a few steam temperatures from Willy's boiler test, and I suspect typical of what many of us achieve.  The exhaust steam was generally in the wet region, however even the wettest exhaust steam was over 90% dry steam.  So, even in an ideal engine less than 10% of the steam arrives in the exhaust as liquid.  I don't think this is enough to explain the troubles Derek is having with condensate, but let's look at a what happens in a real engine.

Unfortunately, after being so useful in our consideration of the adiabatic engine, the second law of thermodynamics goes all wishy-washy for a real engine.  It just says the entropy of the exhaust will be greater than the entropy of the inlet steam.  No clue at all about how much more.  This means we do not know enough about the exhaust steam to do the calculations we did for the adiabatic engine.  Following on from the lack of information about entropy, the implication of exhaust entropy being higher for a real engine than an adiabatic one is that the exhaust enthalpy of a real engine will be higher than for the ideal engine, meaning that less enthalpy is converted to work in the real engine, but we don't know how much less.     In fact, it is necessary to do an engine test in order to find out how much power any real engine develops.

The simple engine test I did on my engine gave an exhaust temperature of 104 deg C.  Now this is above the saturation temperature for atmospheric exhaust pressure which means the exhaust is superheated.  Only just, but we would not expect any condensate in the engine.  Now the constant pressure line in the superheated steam region is no longer a constant temperature, and the temperature and pressure are now enough with a little interpolation to determine all the properties of steam including the enthalpy, so we can calculate the work extracted from the steam.  When the calculations are carried out, my engine extracted about 66% of the enthalpy extracted by the ideal engine.  This figure is defined as the adiabatic efficiency of the engine, and you can see why an engine manufacturer might prefer to quote this instead of the thermal efficiency.  But at least it tends to be independent of the steam conditions over a reasonable range, so there is some justification.

You will have noticed that I have been very careful not to call the work extracted from the steam the engine output.  We would hope that it is indicative of the shaft power output, but it is worth looking in a bit more detail at just what this figure represents.  A great topic for another time, but there is a little more to learn about exhaust steam first.

Thanks for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 19, 2017, 06:12:55 PM
Hi MJM, first, i was going to incorporate it into the heater sheath if i made a new boiler...and if it only requires 50 watts could i make it so the saturated steam pipe only goes in and out of the sheath for  1/10 of the length. this would require some cunning metal work !! Secondly. would it be possible to calculate the temperatures in the different places on the drawing when the boiler is at full steam production  A to G ??
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 20, 2017, 01:08:16 PM
Superheater design continued -

Hi Willy, so I was not too far out on your intention for incorporating the superheater in the boiler, but first the temperature questions.  Easy ones first.

H, atmospheric temperature, I am assuming 15 or 16 deg, i.e. your summer, not mine!

B, C and D, water and steam in the boiler, reasonable to assume all three are equal, and equal to the saturation temperature for the pressure, especially once the air is mostly gone.  The heat transfer within the boiler is largely facilitated by the mass transfer associated with vigorous boiling, which does not look much like the benign situation with a clear liquid interface like you have drawn.  The huge volume expansion when a little bit of water flashes into vapour and the associated huge density difference mean it is very turbulent, vapour quickly rises and is replaced by more water.  This is why the boiling heat transfer coefficient is so high, and the temperature is reasonably uniform throughout.

E, steam in the outlet pipe to the engine, same as B, C and D above, providing you have some insulation wrapped around it to minimise heat loss before the engine.

Calculation of F, the outside temperature of copper is a fairly straightforward variation of a problem illustrated in most heat transfer texts  In reality, it is easier because earlier we did calculate the total heat loss from your boiler from your test run.  If we proportion this heat loss to the area of the ends and the cylindrical part of the shell or put extra insulation on the ends and ignore the ends, we can use the standard formula to calculate the corresponding outside temperature.  The formula is a bit more complicated than for conduction through a flat plate because as the heat travels outward the area is increasing due to the increased radius.  It is getting late here and I have an early morning coming, so I will give you the formula, and you can easily work it out.

The basic formula is q = 2 x Pi x k x L x (T1-T2) / ln (R2/R1)

Here, q is the heat loss in Watts which in your case we already know, k is the conductivity of copper, use 399 W/m.K, which is actually for pure copper, and small amounts of alloy reduce it markedly, worth trying how much difference it would make if it was only 200, and brass is only 111.  T1 is your steam temperature, T2 is the outside temperature you are looking for.  L is the length of the boiler in m, R1 and R2 are the inside and outside radii of the shell, also on m.  The function ln means log to base e, or natural logarithm, and is one of the standard functions provided in any spreadsheet program or scientific calculator.

So you can manipulate the formula to find T2, using a calculator, or just use trial and error in a spreadsheet to find the temperature that gives the right answer.  That is the joy of a spreadsheet, but that formula is not difficult to manipulate to find T2 directly.  You might also know that if R1=R2, R2/R1=1, and ln(1) = 0, so T2-T1 =0 as you would expect, and it is still very small when the difference between R1 and R2 is small, as in a thin boiler shell.  But that ln term starts making a difference for a thick shell.

Having found T2 you can apply the same formula to the insulation, using k = about 0.035, to find the temperature difference across the insulation as q is the same for both insulation and shell.  This figure will be a little smaller than the difference between the copper temperature and the atmospheric temperature as there is a film resistance between the insulation surface and the air, and a contact resistance between the copper and the insulation.  There are standard formulae which enable us to calculate these film resistances now your test run has given us a heat loss based on the heat up phase, (as I have mentioned before they are hard to predict purely on theory).   You have already done the experimental determination, we just have to find our way through the calculations, which I will have a go at tomorrow.

Still have to tackle the temperatures A and B at the heater, and then look closer at your superheater design.  I need to get a couple of sketches into small files for that, so it won't happen tonight.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 20, 2017, 05:11:01 PM
Hi MJM , thanks for this ....ok .....a thermodynamic conundrum.....If the element is rated at 500 watts and it is 100 mm long does this workout at 5 watts per mm  ?  the total heat given off is  X degrees but at each mm is the heat given off only X ÷ 500 degrees ?  the heat given off is uniform from every square mm so perhaps not ?? Am i being a bit silly here or is this a valid question......and it  relates to only getting a 50 watts input for the superheater input from the existing 100 mm long element. !!
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 21, 2017, 10:32:04 AM
Hi Willy, not silly, but a little bit of wholly thinking there.  Remember to be consistent with the units.  You can add apples and oranges to make fruit salad, but you can't multiply them, and you never get pears.  If you stick to consistent units, the issues will be easier to understand.  The total heat from the element is 500 watts, the name given to Joules/sec.  Degrees is not part of it.  Then heat transfer comes in, a bit like flow of energy, and the flow requires a temperature difference to drive it, just like flow of electrical energy requires volts.  You can even use current as an analogy for the energy flow.  Strictly, physicists don't like to call it flow, but it's a bit like flow for most of us.  Your heater generates an amount of energy, or power, which you can calculate as V x I, or V^2/R, by conversion of electrical energy and it will be uniform along the length of the wire, just as the resistance is uniform along the length of the wire.  We don't know how the wire is arranged inside the insulation in the sheath, but reasonable to assume it is distributed uniformly, probably in a coil, possibly even on a ceramic former, we just don't know.  If you are picky, the spec sheet you posted a while back says the is a short non heating bit on each end within the 100 mm, so probably only about 85 mm, but as some heat will flow axially to the ends of the sheath 100 mm is near enough for our purposes.

Now all the heat generated by the electrical energy must escape.  If the heat flow out of the element is less than that generated, the extra heat will be stored in the form of higher temperature of the element.  So the element temperature goes up, and eventually the temperature difference between the wire and the water is enough to drive heat transfer equal the heat generated, and we at last get a stable temperature for the wire.  The temperatures are determined, not by the heater energy input, but by the heat balance, and the law of conservation of energy.

Now the heat must flow through the insulation within the sheath, the spec sheet says it is magnesia, which has a conductivity of about 2.68 W/m.K, and we don't know the thickness, then through the manufacturers sheath which I think is stainless steel, conductivity 14.4 W/m.K, again thickness unknown, then through the contact resistance between the stainless steel and your brass component, on through your pressure containing sheath, brass, conductivity 111 W/m.K, you know the thickness of that, and finally through the boiling liquid film into the water, which is very well stirred up, so a uniform temperature except for a very thin film very close to the the surface of your sheath.  All those thermal resistances need a temperature difference to drive the heat transfer.

We can actually make an estimate of the film coefficient using the general equations and your boiler test results.  I can go through it if you wish, but for the moment I will just tell you that from your boiler test, it is in the range of 5300 W/m^2.K, (which is very modest for boiling water, for which it can range from 3,000 to 60,000) based on the outside area of your pressure containing part, and this means the outside surface temperature of the brass will be in the range 10 to 100 deg hotter than the water, I suspect more likely less than 50.  The temperature gradient through the brass will depend on the thickness, but it is of the order of 3 degrees.  The temperature of the wire looks like it will be 30 - 200 degrees above the stainless steel sheath temperature depending on the thickness of the magnesia, possibly only 0.5 to 1 mm thick in your 10 mm diameter heater.

When you look at all those components, clearly your high conductivity brass is the least of the problem, and with the high film coefficient to the water, no fins are necessary.  Not sure about that contact resistance, but the spec sheet says you need H7 fit, and a design that would accommodate that heat transfer grease would be a good idea.  Clearly the highest resistance is the magnesia insulation, which by the way, allows enough current leakage to trip an earth leakage relay, so you need a good earth so that your equipment cannot build up a voltage, and I don't know what you have to do about the earth leakage protection.

The whole arrangement so far is all symmetrical and radially uniform around the centre line of the element, and it is easily analysed with simple one dimensional techniques only slightly modified.  You can make a simple electrical analogy with temperatures and thermal resistance analogous to voltages and electrical resistance.  With constant temperature water and constant heat generation along the length, it is all very uniform. 

 When you add your superheater tube, all that changes.  If you make a cross section through your pressure containment, you now have three specified temperatures, the element, which must be the highest temperature if heat is to flow the right way, and the superheater must be well above the water temperature as we saw in the engine performance discussion.  You now have three paths instead of just one, you have heater to superheater, heater to water, and superheater to the water.  The geometry of these paths is quite irregular, even though the high conductivity brass is the main heat conductor.  It can be solved, but requires finite element techniques.  Even then, it is not simple as it is a three dimensional problem.  As the steam temperature rises in the superheater, the temperature distribution has to change, so that the heater is still the highest temperature, and so that enough heat is transferred into the superheater despite much lower film coefficients.  This means there is no constant temperature along the length of the element.  Unfortunately I don't have a suitable programme.  I know we have some students on the forum and I hope they are reading.  They may be able to help.  These days students seem to use finite element like we used a slide rule, and it is good they spend their time that way, it is a very powerful tool once you know how to drive it, and enables solving of problems that can never be tackled with a slide rule, a calculator, or even a spreadsheet.

Generally, I would expect the placement of the superheater would be quite critical, it has to be close enough to the heater to get enough heat and force the heater temperature high enough, while leaving an adequate heat path direct to the water so the element cools enough.  In addition,  the rising superheater temperature would mean there are longitudinal heat flow as well as radial so not at all easy to analyse.  Any required adjustment after a test run has to be changed by changing the geometry, so not an easy exercise, but is your only available variable to control the heat flows.  Your separate superheater behaviour is much more predictable, and more controllable as you can adjust the superheater energy input independently of the boiler energy input.  I know I am a bit conservative, but in the oil industry they do not like the crash and burn approach, and even on model scale, that hot steam is very dangerous if you do not totally contain it.

I hope that helps to clarify a few more issues, and perhaps gives a slightly clearer picture of what happens inside your boiler, and how the element transfers energy to the water.

Thanks for reading,

MJM460

Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 21, 2017, 02:08:57 PM
Hi MJM ,once again thanks and i was a bit confused with the "Degrees is not part of it" ! . so i wondering how much heat(temperature) you can get from a 500Watt heater given enough seconds to do it ? In my boiler if there was no safety valve ,and the boiler was strong enough, what is the maximum temp/ pressure that one could attain ? and would a time/temp graph of this show a continuous line to infinity ? Also i can see now that an entirely separate superheater would be required as in a Loco firebox, I am now feeling a bleet sheepish with my wooly thinking !!!,And this is why we need people like you to help with our knowledge and prevent unnecessary contraptions from being built !! there have been quite a few expensive constructions in the past that have been failures, as depicted in The Engineer over the previous centuries .
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 22, 2017, 12:19:54 PM
Heat and temperature -

Hi Willy, I hope my terminology did not offend you.  And the "wooly" term refers as much to my explanation as anything else as it was obviously not sufficiently clear.  So thank you for you kind words.  We are very fortunate to have access to information and tools that Watt and his colleagues could hardy dream of.  We can calculate a likely result and narrow down to those experiments with a reasonable chance of success.  They mostly had only trial and plenty of errors, and really amazing intuition. 

I was also trying to emphasise that all those words about dimensional analysis are not just esoteric waffle, dimensional analysis is a very practical approach to every day problems.  If you check the dimensions of each part of your problem, and they are the same, then you are likely to have all the elements of the problem and that is a good start.  Especially when working in SI units, when there are very few constants which have units associated them.  Your question, as usual, puts your finger right on the area that is still not adequately explained.  So let's try again.

The electrical power input from your heating elements can be calculated from the electrical formula, P= V x I, and the answer is in watts, or J/sec.  The big advantage of SI units is that mechanical power is also measured in J/sec, and they really are equal.  You only need talk about J/s, you don't have to qualify it as to whether is is mechanical or electrical joules.  For historical reasons, the unit J/s is given the name Watt.  Nothing in your boiler is able to provide any "push back" to change the electrical energy flow, no matter what temperature it reaches.  That is why I said temperature has nothing to do with it.

Now the fundamental law of physics comes into the picture, conservation of energy.  Energy does not disappear, it can always be accounted for, even if sometimes it is hard to know where to look.  But in your boiler with the electric heating element, we know the energy input, and the total energy output, or heat losses, plus any energy stored by increase in temperature of the materials in the system is always equal to the input.  When the heat losses equals the heat input, there is no more storage, so we have a steady temperature.  Now you will remember that I divided your operation into two phases, first a heat up process then steam production.  In the heat up phase, there is no steam production, there is some heat loss to the air from the shell which you can detect by holding your hand close, and all the rest of the heat input is stored in the water, the water vapour in equilibrium with the liquid, the copper shell, and even a little in the insulation.  And we did a calculation of how much energy was required for the storage.  We can calculate this using the mass of the materials, specific heat of copper, and the steam tables for the water.  Probably should have allowed a little for the heat stored in the element which also got hotter.  Then I assumed that the difference between the heat required for the observed heating of the materials and the input from the element was lost to the atmosphere.  It seemed like a lot and I suggested some insulation which would reduce the losses.  It did not reduce it as much as I expected, and I mentioned that we only know the rated output of the element, but did not have facilities to measure it, a potential source of error.  We will get back to that.

Once the boiler is up to your selected pressure, and you open the steam regulator, it runs at essentially constant temperature, so the losses continue at a constant rate.  We should have measured the steam consumption to check the losses at this temperature, because they are not constant during the heat up period, but vary with the temperature.  Now we have constant temperature, no more storage, so there are losses to atmosphere, and heat carried out with the steam.  We can measure the steam production, the tables tell us the energy carried out in the steam, and calculate the heat losses by subtraction from the energy input.

The point is that the energy equation is a complete explanation of the energy flow, and while the balance between storage, atmospheric losses and steam production changes with temperature, you only use the temperature as a measure of the amount of energy stored, and the properties of the steam, but nothing to do with the total amount of energy input.  And all because of your electric heating element which is ideal for demonstrating these things for that reason.

Now if we look at the heat transfer, the process which describes the flow of heat we do need temperature.  The equation is basically q = U x A x temperature difference.

Now let's look at the dimensions, q is J/s.  U has the dimensions of W/m^2.K, and of course W is the same as J/s.  A is the area in square metres, or m^2, and temperature difference, whether in C or K, has the dimensions of temperature or K.

The right hand side of the equation then has the units of J/(s. m^2 .K) times m^2 times K.  You can see the metres and temperature units "cancel" out, leaving only J/s, the same as the left hand side of the equation.  So we are off to a good start in understanding heat transfer, as our units are dimensionally consistent.

So on to look at your question about the ultimate temperature and/ or pressure.  I am glad that you are aware that the boiler could bust, this must remain a purely hypothetical thought experiment.  I am pretty confident that the boiler will fail before the heating element, not a good result.  You definitely need that safety valve and a high temperature switch, both properly set.

First you insulate the boiler, right over the top, with say 3 inches of good insulation such as rock wool tightly packed, or any one of several others with a similar conductivity, around 0.04 W/m.K, so the heat loss is minimal.  And then we start the heat up process.  Initially nearly all the heat goes into the water, the copper and a little into the insulation.  And as the temperature rises, the heat loss to the atmosphere increases.  We don't set a limit when we open the regulator, we want to see how high it goes, at least before bedtime!

Now that law of conservation of energy comes in.  The energy input is constant.  If there are no losses, all the energy goes into storage, and so the temperature just rises, as the energy is stored as sensible heat (higher temperature) plus all the ways energy is stored in steam.  It is not possible to eliminate all losses, some heat will be lost to the atmosphere, and as the temperature rises this heat loss will increase, it will increase until it equals the heat input, or until something breaks.

Now I did a few calculations with increasing insulation thickness from practically none and calculated the temperature difference required to dissipate the full 1000 J/s of the heater input.  It takes very little insulation to get over 1000 deg C, way above any safe temperature for your silver soldered copper boiler.  Something will definitely break, and no safe way of doing the experiment.

So in summary, energy input is not related to temperature in your electric boiler.  Temperature is a property of steam that we can measure, and if we have two independent such properties, the steam tables can tell us all the properties of the steam.  Temperature is also the measure of sensible heat stored in the copper, and other materials as energy is stored in them.  But temperature is not a quantity that has to be defined to determine energy input.

A fired boiler has a fundamental difference which sets it apart from your electric heated boiler.  The energy input comes from the mass of fuel burned and the fuel properties, but to release the chemical energy contained in the fuel, you have to burn it which requires oxygen, which usually only comes in a mixture with four times as much nitrogen which absorbs heat but makes no contribution to the amount of heat, and the combustion products have to be allowed to escape for the fire to continue.  They take a considerable amount of heat with them, and can if necessary take it all.  And it always takes away all the heat that is not transferred into the boiler water and copper.  It is never negligible.

I hope that makes it a little clearer.  But I noticed something interesting in the process of formulating an answer.  I will send you a pm after I have posted this, please let me know if you don't receive it.

Thanks to all for looking in,

MJM460
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 23, 2017, 12:18:30 PM
Unfortunately I have not much to add today.  I spent the available time reviewing where we are up to, and doing some calculations in preparation for the next step.

I hope to have something useful to discuss tomorrow.

Apologies to all the regular readers.

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 24, 2017, 02:27:54 AM
Hi MJM i have checked out the currant going into the boiler and it looks like just over 4 amps  "Virtual" ?  so this may help with the calculations !
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 24, 2017, 12:01:40 PM
On Electric heating elements -

Hi Willy, that is a beautiful meter, and I am guessing it is at least as good as your Avometer.  With the zero clearly correct when power is off, I would say it reads pretty close to 4.0 amps.  I am guessing that "virtual amps" is the terminology of the time for what we now call RMS amps.  That stands for root mean square, and without diverging too deeply into the maths, it gives us the effective average of a sine wave.  For a pure resistive load with AC voltage we can apply ohms law that we are familiar with for DC, if we use RMS volts and amps, not the maximum voltage of the sine wave.  So it could be described as the effective volts and current that give us the same heating effect as a DC voltage of that value.  Perhaps "virtually the same as as a DC current of that value"? 

So let's look at what all that can tell us about the electric heater.  It is rated at 240 volts and 500 watts.  If we use the definition of power, power = volts times amps, we can say rated amps = power divided by volts, or 500/240 = 2.08 amps. With two elements in parallel, a total of 4.16 amps.  I am guessing you would see that 0.16 easily on that meter scale.  We can also use ohms law to calculate the expected resistance of the element, based on 240 volts rating with rated current of 2.08 amps, R = V/ I = 240/2.08 = 115 Ohms.


On the other hand, with the measured current of 4 amps, and assuming the voltage to be the rated voltage, 240 V, we get power = V x I = 240 x 4 = 960 watts, or two times 480 watts, compared with the specification 500 watts.  This time, using ohms law, we find R = V/R = 240 / 2 = 120 ohms for each element.

Already, we can see an inconsistency in the data.  The manufacturer has to be reasonably careful about the published ratings, so I would expect that data to be accurate, with the understanding that it is probably assuming all at a standard temperature of 20 deg C.  These ratings are at the specified voltage of 240 V mentioned on the data sheet.  The discrepancy in calculated resistance values would be expected if the actual voltage was less than the rated voltage.  Now I want an alternative to measuring the voltage, as mains voltage is too dangerous for unqualified people to mess with, meaning most of us, including me.  However, with an element removed from the boiler, and no power connected, we can safely measure the resistance of the element.  Now the actual resistance should be the same in both cases.  We have assumed the voltage is the same as the rated voltage, 240 V.  Now many sources I have seen mention 220 V as the normal European voltage, and there is some tolerance allowed to the electricity authorities, so it is reasonable to rate equipment for 240 V, so long as it can also operate properly at the minimum allowable supply and at the expected normal voltage.  We can assume the resistance is the same, whether connected to 240 V or passing 4 amps, so ohms law tells us the voltage must be different between the two cases.  If we measure the resistance of an element, we can calculate the voltage at 4 amps, then with voltage and current we can calculate the actual power output at our actual voltage.

Now if we do a little algebra on Ohms law, V= I x R, and on the definition of electrical power, P = V x I, we can derive P = I^2 x R, or P = V^2 / R.  Note P = power in Watts or J/s, I is in amps and V in Volts.

That second form is particularly interesting as it tells us that the power is proportional to voltage squared, which means a small difference in voltage has a big effect on power.  If the actual voltage is nearer 220 V, it would make a significant difference to the heater output.  If the resistance is 115 ohms, with 220 V, the power output would be only 420 watts each element.

There is a little more we need to understand about a resistive heater.  I have mentioned a few times that resistance increases with temperature, but I didn't have the relevant data for the wire nominated on the heater spec sheet, chrome-nickel resistive wire.  Well, I have found it.  That hyper physics site from gsu is a really great resource.  Should have thought of it earlier as I use it often.  The paid hyper physics Ap on the Ap store is money well spent, but it often turns up in response to a normal web search.  So I assumed the resistance of the element might increase by say 6 ohms when up to temperature for generating steam.  It turns out that the increase in temperature to cause this change from the specified 20 degrees would result in an element temperature of 158 degrees.  Now this is a little above our steam temperature, but probably not enough to give the required heat flow.  Hence it is likely that 6 ohms rise is a low estimate.  If we look at that voltage form of the power equation, P = V^2/R, we can see that as the increase in temperature causes the resistance to increase, the power output decreases.  This tells us that in our particular application, the temperature increase of the wire in the element and any amount of reduced voltage compared with the rated 240 V, both cause the power output of the element to be below the rated power, and that will affect our heat balance.

We need to return now to that heat balance to see how this all fits together, so that will be the topic for another post.  I hope this is all making sense so far.

Thanks for reading,

MJM460
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 25, 2017, 03:02:20 AM
Hi MJM ,i have tested another element not connected up and it reads 112.9 ohms so your calculation is pretty close !!Thanks for all the latest info and the theoretical and the actual figures depend on so many other factors that we  are not usually aware of .!! and the temp when i took the reading was 19 C....
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 25, 2017, 12:16:27 PM
Electric heaters continued -

Hi Willy, measured resistances provide hugely useful additional data for out investigation, thank you.  I think that makes two you have measured now, one was 109, and this one say 113.  This gives us some idea of the variation, and it is probably within reasonable manufacturing tolerance. 

Last time, I used the measured current, and assuming the rated voltage, calculated the resistance.  But now with that measured resistance we can manipulate ohms law again to  V = I x R.  Let's use our measured value of current, divided by two for each of two parallel elements, so 2 amps each element.  In addition let's use the average of the two resistances you have measured, or 111 ohms.  We put both of those into ohms law, so V = I x R = 2 x 111 = 222 Volts.

Now with better estimate of the voltage, we can use the power definition P = V x I = 222 x 2 = 444 watts.  Now this is quite a bit below the rated value of 500 watts for each element, and that will increase our heat up time.  The data sheet is still correct in specifying 500 watts, because it also specifies the voltage as 240.  When we use the element, we have to use the data sheet specifications and in addition apply the actual voltage we have in our location and equipment in order to understand the heat we will get from the element.

We also need to be aware of the fact that we are unlikely to have just picked out two elements one above the average and the other exactly the same amount below the average of all the elements that have been made.  In the vague shadows of my memory there were lectures on statistics, and the average called an estimate of the average based on a sample of two, and there was a method of calculating a standard deviation and with it, the likely spread.  But this thread is about thermodynamics, not statistics, so I won't go there, apart from mentioning that the difference between those two measured values gives us an idea of the variation between different elements.  To get any more accurate values, we have to measure the actual ones we intend to use.

Then there is that question of the increase in resistance with temperature.  I finally found the value of temperature coefficients resistance and the formula to calculate the resistance at different temperatures.  For Nichrome resistor wire the coefficient is 0.0004 per deg C. 

If we measure the resistance R0 at any temperature T0, then the resistance R at another temperature, T, is found using the following formula:

R = R0 x (1 + 0.0004 x (T - T0))

If we apply this to your element with R0 measured at 19, and we assume the heater wire starts at 17, and in order to transfer its output to the water it gets say 50 degrees above the water, we can calculate the resistance at each of your measured temperature points up to 135 deg C where the element would be 135 + 50 = 185.  Over this temperature range the element which measured our assumed average for the elements of 111 at 19 degrees, ranges from approx.110.9 at 17 to 117 at 185 degrees, and the power output of the element ranges from 440 watts at the start to 412 watts when it is at 185 , generating steam at 135 degrees with a constant voltage of 220 V.  Now this only 82% of the output we all assumed based on the data sheet.  It still requires that estimate of the temperature difference between the wire and the water, but it is enough to give you the idea.

Now I used a spreadsheet for all those calculations, which, because of the repetitive nature, saved quite a bit of time.  I was able to develop the formula in one cell or row, then simply copy and paste the formulae and the computer did the rest.  Actually the iPad did it.  Just a matter of learning how to use those absolute and relative references.  Then, as extra data was discovered, a simple substitution, and the spreadsheet programme updated everything in an instant.

I think that just about covers the performance of electric heaters, but if I have missed anything please ask, and I will try and cover it before returning to the calculation of those temperatures you asked for a few days back.

Thanks for following along,

MJM460
Title: Re: Talking Thermodynamics
Post by: Admiral_dk on October 25, 2017, 08:09:36 PM
Just as a little info - EU standardized the mains voltage to 230V +10/-6% one phase and 400V +10/-6% three phase more than 20 year ago (instead of the 220V / 240V in different member countries).

In practicality I actually measure between 227V and 231V with a very precise True-RMS Fluke meter at any given day over the last 22 years - and I often do as part of fault analyse on all the gear I've repaired over the years .....

Hmm - just checked WiKi to confirm the date 1995 - and discovered that there are places in the UK (a few) that still uses some very old power stations that supply 250V (still within the +10% limit) and some more that supply 240V .... Here in Denmark we adjusted our power stations from 220V to 230 within a few months after the new (back then) regulation was activated.
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 25, 2017, 11:58:53 PM
Hi all, i have just checked my mains voltage and it is 240.3 volts.......!Thanks for more info on the boiler heating front ...Due to to all the disparate parameters that can occur with all the info it is quite a formidable task to actually come up with  exact figures. ! I suppose that is why when Rolls Royce is asked about power and performance , the reply is "adequate" !!
Title: Re: Talking Thermodynamics
Post by: crueby on October 26, 2017, 12:07:55 AM
Hi all, i have just checked my mains voltage and it is 240.3 volts.......!Thanks for more info on the boiler heating front ...Due to to all the disparate parameters that can occur with all the info it is quite a formidable task to actually come up with  exact figures. ! I suppose that is why when Rolls Royce is asked about power and performance , the reply is "adequate" !!

When the Rolls Royce test driver comes back from the track all wide-eyed and quietly saying 'wow' over and over, do they translate that to 'adequate' for the brochure?!
Title: Re: Talking Thermodynamics
Post by: steam guy willy on October 26, 2017, 02:14:55 AM
Hi Chris , possibly ...It is the british stiff upper lip and "reserve ' that made us what we are/were.!  Also MJM does the specific gravity and the hard/softness of the water have any bearing on the steam ability in the boiler ??  Should i assemble for the next boiler test Hydrometer,Barometer,Ammeter, Voltmeter, temperature gauge ,pressure gauge , measuring jug etc etc etc
Title: Re: Talking Thermodynamics
Post by: Stuart on October 26, 2017, 11:32:46 AM
Steam Guy Willy

That’s why the rpm indicator in a modern roller is not as you would think.

At the bottom it says 100%power aviaiable and at what would be red line it says 0% avialable

Yes it’s calibrate  100% to 0%

But by ekk does it look odd

And no I do not have a roller but a humble BMW
Title: Re: Talking Thermodynamics
Post by: MJM460 on October 26, 2017, 01:22:38 PM
Measurement accuracy and insulation-

Hi Admiral dk, thanks the more definitive answer on what the supply voltage should be.  It looks like you have a pretty solid power supply.  So 230 V +10%/-6% means max 253, min 218.  And presumably that is defined at the substation, so in the home could we see small voltage drops depending on power drawn in the street and elsewhere in the house? 

Hi Willy,  well the measured voltage does not match any of the calculated values.  This suggests that the resistance of the heaters in the boiler may be a little different from the one you measured.  It shows the difficulty of getting really accurate data.  However it is more about understanding the sources of error and their relative importance than pinning down the exact numbers.

With power = V^2/R, you can see for example that 230 volts instead of 240, would make a 9% difference to the output of our heater.  Similarly, a 2 ohm change in resistance makes about a 2% difference to the power.

  Salt dissolved in water does change its freezing point as many will know, and I expect it also changes the boiling point a little.  But more importantly, it is likely to cause foaming and carryover, and it stays in the liquid phase when water evaporates, and eventually builds up to cause scale, so distilled water is preferred if the domestic supply is not too good.  The rest of the data required is about accuracy, I believe digital temperature instruments are pretty good, pressure gauges not so good and not easy to check if you don't have access to a dead weight tester, barometer desirable, and you can check it against weather bureau observations or just use the weather bureau observations for your area, a jug is usually not a very accurate instrument for measuring water, better to use a digital scale and the specific volume from the steam tables.  Similarly for fuel quantities.  Time is now very easy to measure sufficiently accurately, unlike the problems our predecessors had with measuring time.  But basically you only really need to measure the quantities that appear in the calculations.  Probably the most important is that the readings have to be at the appropriate time.  The voltage, current and temperature at the actual time of the test are probably most important and a good voltmeter and ammeter permanently connected inside the control system would be best.  These along with carefully weighed water fill and a once off careful measure of the boiler mass.  But we can get enough understanding  of the thermodynamics by doing the possible, and leaving the search for perfection to the laboratories that develop the steam tables and other data we rely on.  We can just keep an eye on the effect of errors in our data.

Hi Chris, glad to have you along.  I don't think there is a "awe" factor.  They just cruise quietly around the track, shave an "adequate" margin of the lap record, the passenger in the back asks when the test is going to start, because he will need to hang on to his champagne, the driver blames that D*** clock, how was he expected to concentrate o