Author Topic: Talking Thermodynamics  (Read 126514 times)

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1275 on: October 29, 2019, 11:29:07 AM »
Hi Willy, the key to understand what is going on here is to understand how the air behaves, and how the water behaves, then how they each act when there are both in the space at the same time.  It is the same thermodynamics as when we were talking so long ago about filling a boiler at the top or the bottom of the mountain.

Behaviour of Water

The water is simple enough.   If you have a sealed pressure vessel which contains only water, so no air or anything else, and let it sit undisturbed for long enough to reach thermal equilibrium with the surrounding atmosphere, then the pressure in the vessel will be only due the the water and its temperature.  The pressure corresponding to any temperature is listed in the steam tables as we have discussed before.  Remember that reaching thermal equilibrium is just the technical description of everything being at the same temperature.  There are a few ways of achieving this, but I will leave that for another time.  But as you can easily see from the steam tables, higher temperature causes higher pressure, lower temperature, lower pressure.  The pressure gets very close to zero at zero degrees centigrade, and of course the reference point we all know is atmospheric pressure at 100 degrees centigrade.  This means that at all temperatures below 100, the pressure is below atmospheric, but so long as the vessel is designed for an external pressure of 101 kPa (14.7 psig) it will not collapse as that is the maximum external pressure possible in atmospheric air.  Remember that so long as the vessel is totally sealed, the pressure inside is quite independent of the pressure outside.

Behaviour of Air

If you now look at air, we all know that it air is a mixture of gases, the biggest part is nitrogen, then oxygen, then a number of other gases in quite small quantities, normally including a small amount of water.  Letís just look at dry air.  That is air that does not contain any water.  Perhaps it has been obtained by passing it through an air dryer such as provided in a wood quality air system.  Even those are not perfectly dry, but close enough for our discussion.  Despite being such a complex mixture you can assume that air behaves like a perfect gas, so the absolute pressure is proportional to the absolute temperature.  You donít have to treat each component separately to be close enough to understanding air behaviour.

So if our vessel is filled with air at atmospheric pressure and temperature and then sealed, when it is heated, the pressure will rise, or if it is cooled the pressure will fall.  Over ambient temperature range, the pressure does not vary a lot because you have to cool it to absolute zero temperature or minus 273 degrees C, and that is quite difficult to achieve.  On the other hand, to increase the pressure of two atmospheres, you have to double the absolute temperature.  If the vessel was sealed at say 20 degrees, or 293 degrees absolute (or Kelvin), so to double the temperature means 586 K or 313 C, not impossible, but careful not to burn your fingers.

Water and Air in the Same Space

So what happens when you have both air and water in the same vessel.  This is simple to achieve,  pour some water into a vessel, leaving it say half liquid water, the remainder is mostly air, and insert the plug.  Now what happens?  Basically, we can look at each component separately.  The temperature will be the same everywhere, while the pressure will be the sum of the pressure due to the air and the pressure due to the water.  First the temperature will even out if the water was not already at air temperature. 

Letís say it is 20 degrees C and standard atmospheric pressure of 101 kPa.  The liquid water will obviously be in the bottom part of the vessel, the air will still be at atmospheric pressure and temperature.  The vapour space will also contain some water vapour.  At 20 C, the vapour pressure will be 2.4 kPa.  Some of this will be water that was in the air as humidity, while the rest will come from some evaporation of the liquid.  When I look at the humidity here on my little weather station it says 25%.  So if the temperature is 20 C, the water vapour pressure is 25% of that 2.4 listed in the steam stables, or 0.6 kPa.  So some water will have to evaporate and the temperature even out to get to thermal equilibrium, and in the end the water vapour pressure will increase by 2.4 - 0.6 or 1.8 kPa.  The air temperature is unchanged so pressure unchanged to the total pressure in the vessel is increased by 1.8 kPa, and the air humidity in the vessel will be 100%.

If the vessel is now put outside in the desert, then as the temperature rises and falls, some water will evaporate during the day when the metal and internal temperature will exceed 65 degrees, (it will certainly burn your hand if you try and pick it up), and like wise, some will condense overnight when the vessel and its contents cool by radiation to a clear sky.  In these conditions I have had ice forming on the canvas of the caravan that makes folding down in the morning quite difficult.

Application to Air Compression

The final step is to apply this basic thermodynamics to the system where an air compressor takes ambient air and compresses it into a system to drive the engine.  Letís assume a balmy spring day, about 20 deg C.  I suspect your climate is more humid than here so letís say 33%.  It could be quite a bit higher if itís about to rain and the temperature is dropping.  The water vapour pressure in that air will be 33% of 2.4 or 0.8kPa.

Letís assume that compressor discharges at about 500 kPa (absolute), or about 60 psig, (75 psia).  This is a pressure ratio of 5, so the water vapour pressure will increase five times to 4 kPa.  It is important to recognise that this pressure rise will not cause the water to condense,

The temperature is more complex to estimate, there will be a significant increase due to the work done by the compressor and the compressor efficiency.  I wonít divert to trying to calculate the discharge pressure, but the discharge side of a compressor is generally pretty hot to touch so I am guessing perhaps 70 to 100 deg C.  At 4 kPa, the condensing temperature is about 29 C, so you can see it is well and truly superheated.  But at the discharge of the compressor there should be an after cooler, a heat exchanger, air cooled on a small machine but could be water cooled on a larger machine.  The cooler may or may not be large enough to cool the compressed air below 29, but the additional cooling provided by the air receiver and the distribution piping almost certainly will cool it pretty close to that ambient temperature, and that is where the water will condense.  Ideally the system would include an air dryer to further reduce the water in the system and hence the amount of condensation but these are usually only provided in critical instrument air or breathing air or medical systems.  So water condenses in the distribution system where the temperature reaches 29 or below, leaving the air at 100% humidity at that temperature.  And that condensate will drain into the engine unless the piping is carefully designed so the off take comes off the top of the header, and drains are provided at low points.  And there will be more condensation when everything cools down overnight, so when you start the system you will get some water out at any drain point in the piping.  But while the engine is operating we can assume the air is cooled so some water drops out in the vessel, leaving the air at 100% humidity, or perhaps a little warmer with slightly lower humidity, but basically very humid air.

Air to Power a Steam Engine

Now to the question of what happens when you drive the engine with this air.  We have already talked about that water in the engine remaining from the last operation, so rather than revise that all again, letís just look at the engine running.

First the hp cylinder.   Near the start of each stroke, the valve position allows air to enter the cylinder and does work by pushing the piston, roughly at constant pressure, and the energy used is supplied by more air entering.  Then the valve reaches the point where it closes off the inlet.  The air in the cylinder continues to do work by pushing in the piston.  As the piston continues to move the air expands, the pressure falls, and significantly, the temperature falls due to the energy converted to work, just the opposite to the heating that occurred when it was compressed.  That is why the running engine feels cold, it really is cold.  The temperature is predicted by the equations for adiabatic expansion.  But as a result of that reduction in pressure and temperature, the air does get to the point where the water vapour that remains in the air will start to condense.  Not good for the cylinder lubrication, but I donít really know enough about lubrication to suggest an appropriate lubricant.  Some attention needs to be given to what lubrication is necessary.  I suggest the condensate is mostly swept out with the air in the exhaust stroke, it is only when the engine stops that any remaining water stays around to cause corrosion.

The engine is, I believe a compound, so the hp cylinder discharges into the lp cylinder.  The air is now cooler, and is saturated with water (100% humidity), air continues to expand as it transfers from the hp to the location cylinder due to the difference in piston diameter, so it continues to fall in pressure and cool.

Now the complication with a compound engine.  When new, designed to run on steam, it probably exhausted to a condenser at a pressure lower than atmospheric.  But you are not set up to achieve condensing the air, so you will be exhausting to atmospheric pressure unless you have a very good air pump.  If the pressure in the lp cylinder is below atmospheric pressure when the exhaust valve opens, outside air will rush into the lp cylinder when the exhaust valve opens.  This is effectively a reverse pulse to the engine which will make it run rough to say the least.  Of course simpling valves or what ever the engine is set up with for starting will provide a way around the issue by ensuring the air inside the engine is never below atmospheric.  But I donít know how your engine is set up in that area.

So expansion on the hp cylinder causes the air to cool, continued expansion in the lp cylinder causes it to cool further, hence the engine feels cold when it has been running.  And that cooling causes condensation of the water from the humidity in the atmosphere.

Oh, and the piston valves, I believe that they cannot lift as does a slide valve to release condensate, so those automatic condensate drain valves would seem to be quite essential.  I have even less knowledge of those.  At least I have built and operated model engines with slide valves.

Good to see that you are still thinking about thermodynamics, I hope the above helps clarify some of the issues.  Thanks to all looking in.

MJM460

« Last Edit: October 30, 2019, 11:30:34 AM by MJM460 »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1276 on: October 30, 2019, 03:16:01 AM »
Hi MJM, thanks for all this info and there is so much more to just Air and Water  than is at first apparent  Things like  'Heavy  water " that is used in Nuclear power stations   I think ??..  Is there a way of actually drying air to prevent condensation ?? This engine does have a condenser and an air pump. The air pump is used to also supply water to the boiler via a pump driven by the beam. Will the condenser  actually help with the efficiency of this engine if supplied with water when run on compressed air ? the condenser is supplied with its own water from the river ? and should they take advantage of this or not have this arrangement working ? I shall pass on your info to the restoration people so they can query it with the compressor suppliers. I will need to read your last post a few more tamest really get to grips with it.   Thanks again for your posts

Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1277 on: October 30, 2019, 09:29:13 AM »
Hi Willy, the thermodynamics is not so different from what we have previously discussed.  The main difference is in how to apply our understanding of the behaviour of air and of water separately to the case where there are both air and water in the same space.  And it is such a common occurrence that it is worth trying to understand.

I am sorry that it was such a long post, too long, even by my standards.  I will go back tonight and add some headings to more clearly separate the distinct topics.

The effect of the condenser is to lower the pressure at the engine outlet, thus increasing the overall pressure ratio across the engine and increasing the potential work output without increasing fuel consumption.

Unfortunately, as soon as we add a condenser, we have to deal with air in the system.  There is usually some air which was dissolved in the boiler feed water.  Modern high pressure boilers have steam heated deaerators to reduce this dissolved air, and the boiler treatment chemicals used during boiler operation include an oxygen scavenger to collect the air that get through the deaerator.  This is all done to reduce corrosion in the boiler and associated equipment.  Without this water treatment there will be some air in the exhaust.  More importantly, as soon as you include a condenser, and achieve pressures lower than atmospheric pressure, there will be air leakage into the system.  It will leak in through seals around piston rods and valve stems and even flanged joints in piping.

Air in the condenser reduces the partial pressure of water so lowers the temperature at which the water condenses, so the condenser is less able to transfer the necessary heat.  Furthermore, until that air is removed from the condenser, the condenser pressure increases so reducing the engine output.  Thus the air pump is included to compress the air to something above atmospheric pressure so it can be vented to the atmosphere.

The condensed water also has to be pumped so it can be discharged to atmospheric pressure, or even back to the boiler providing it is not contaminated with oil from the engine.  Some air pumps are designed to achieve the air pumping and water pumping in the same device.

It is important to recognise that the low pressure in the condenser is obtained by condensing the water in the closed volume of the condenser.  Cooling water at say 15 deg C cools the exhaust steam to a point where it starts condensing.   At 15 C, water would condense at 1.7 kPa, a pretty decent vacuum, though in practice the limited heat transfer area might only cool the steam to perhaps 25 or 30 C, so the condenser pressure might be 5 or 6 kPa, still not too bad, so long as the air pump is working well.

When you run the engine on air, you will not be surprised to hear that you will not be able to condense the air.  That requires around -180 C, and that requires quite sophisticated refrigeration equipment.  Without that condensing, there will not be any pressure reduction, so I would not suggest pumping the water.  Better to keep the system dry and free of corrosion on the water side.  Of course, the air exits the lp cylinder quite cool, so river water in the condenser tubes would probably heat it a bit, but I am not convinced it would be worth the pumping power consumption.

Clearly the more important issue for running on air is corrosion due to condensed water in the air supply.  Not so much of a problem for a model run occasionally and probably well dried out between runs, even if it is only due to storage in a warm home.  But not so easy to dry that full size museum engine which probably runs quite regularly, but probably also shuts down overnight.  Procedures for long term mothballing would not be appropriate.

I suspect the answer probably lies along the way of driving the engine with that car tyre and admitting warm air for a while to dry everything out prior to shutdown, but I have no experience in that sort of long term procedures. 

Systems which supply dry air employ an air dryer after the accumulator.  So, after the compressor, the after cooler, followed by an accumulator vessel, then a filter/coalescer.  Then a drier.  Commonly it is a refrigerated dryer, which will give a dew point of around -20 to -30 C.  This was not considered good enough for instrument air in a large oil or petrochemical installation and we used molecular sieves, particles of a special resin which absorbs water molecules at high pressure (compressor discharge pressure), then switches over to vent to atmosphere when the water molecules escape the resin at low pressure.  So two pressure vessels, with the appropriate automatic valving to regularly swap between high pressure and low pressure operation.  These systems gave a dew point around -70 C from memory.  A small refrigerated drier might be affordable, depending on the actual engine air consumption, which determines the drier size required, but I am not sure that the larger molecular sieve types would be affordable.  Still, worth an inquiry from suitable suppliers, especially if they could be talked into contributing to the museum by putting a loan system on display.

Thatís more than enough for another day, so I wonít start on heavy water.

Thanks to everyone looking in.

MJM460

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Offline steamer

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Re: Talking Thermodynamics
« Reply #1278 on: October 30, 2019, 12:31:15 PM »
Hi Dave, good point about the condensate, probably a good reason to have more rather than less clearance on a steam engine.  My compressors didnít like any liquid either, it turned them into very dangerous beasts.

Hi Derek, it is possible that we also used that plastic cord, I am not sure, now that you mention it.  But we were looking for about 1 mm, so not as difficult as your application.  Itís amazing what can be done by experienced people to get around difficult practical problems such as that.

MJM460

Models can also benefit from a bit of superheat just to dry it enough that by the time it gets to the cylinder, its saturated steam.  Say 100F worth of superheat is quite beneficial

Dave
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1279 on: October 30, 2019, 01:53:57 PM »
Hi MJM,  Thanks for all this ..there is so much more to these systems than what is initially visible. with our models we just have one pipe from the boiler to the engine ,and one pipe to atmosphere.  On later  full size engines there are so many extra pipes and parts that sort of spoil the visual aesthetics of the complete plant. Like in a modern car ,the whole of the engine compartment is full of 'things' unlike my Morris Minor that just has a carburettor and exhaust pipe !!! The Beeleigh Engine just has the main steam pipe  4"   and the exit pipe 1'5" from the air pump to the boiler feed pump. There is no provision for oil lubrication in the system, just a small two cocked supply valve on the top of the HP cylinder. I suppose any oil floats on top of the water in the air pump.   
   Here is a video of the air/tea pump operating !!!

Willy
« Last Edit: October 30, 2019, 02:18:24 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1280 on: October 31, 2019, 11:53:58 AM »
Hi Willy, as with all things, it is worth knowing the theory to help guide what to expect.  However, it is hard to quantify, in this case, just how much water would remain in the engine after a run, especially as it was not intended to run on air.  So while provision of dryers is possible, and will reduce the water reaching the engine, it is hard to know whether it will be enough, or even if it is necessary.  Warm air while the engine is turned over by that electric motor and car tyre might be all that is required.

With that double valves connection on top of the cylinder, you can open the top one, add a squirt of oil between the valves, then close the top one, and finally open the bottom one.  This allows some oil to enter the engine, even when running.  And I guess some oil would remain in the cylinder when the engine stops.  Probably worth doing again when the engine shuts down.

It would be interesting to hear what others with more experience of running larger engines on air have found.

That water pump has to lift the water, not just the physical height from the hot well, but also against the pressure difference between the condenser and atmospheric pressure.  Of course this wonít be needed when running on air I expect.

The model needs a bigger box for the tea, and a lever so it can be more smoothly operated against the friction of that piston o-ring, but a great model to include in the display.

MJM460


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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1281 on: January 11, 2020, 12:59:32 AM »
Hi MJM, a few more questions about boilers.   Is it possible with enough heat to evaporate all of the water in a boiler to steam assuming the boiler is strong enough to contain it without bursting.?   if you have a large boiler with a very small amount of water will the temperature of the steam rise to the same pressure  as dictated by the steam tables ie, do the steam tables only refer to boilers with the correct amount of water in it.??...If the boiler has ,theoretically 100% insulation. is it possible to raise the temperature /pressure to a high amount with say only 101 degrees of heat going into it continually..?

Hoping you are well and ok ,

Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1282 on: January 11, 2020, 09:59:53 AM »
Hi Willy, good to see you back thinking about thermodynamics again.

Some quite important questions about boilers, letís see if I can make the issues understandable.

Letís first assume that you are talking about a boiler which starts off containing water which is partly in liquid form or phase and partly vapour.  We can look at other cases later.

And letís remove any complications of any air remaining in the boiler after it is filled.  We can do this by heating the boiler until steam is being raised, the pressure is not important, but then open the stop valve so some steam escapes.  This tends to initially be a mixture of steam and air, but after some time, we can assume that essentially all the air has been removed.  We can then close the stop valve, and let the boiler cool down in preparation for the experiment.

First, can all the water in a boiler be evaporated if the boiler is strong enough?  Basically, so long as the boiler is strong enough as you assumed, the answer is yes.  So we should then ask how strong does that boiler have to be?  We can see from the steam tables that the two phase section of the tables stops at 374.14 deg C at which the pressure will be 22090 kPa absolute.  For those who prefer imperial units, that is 705.44 deg F and 3204 psia.  So the boiler does have to be really strong at quite a high temperature.  Our typical copper boilers loose their strength at a much lower temperature than that so could not do it.  Special high temperature alloy steels are required, and yes there are boilers, called super critical boilers which do operate in this region and above.

That temperature and pressure is quoted quite accurately.  It is called the critical point, and is the highest pressure and temperature at which liquid and vapour can exist at the same time.  It is quite important in higher level thermodynamic calculations but not of much interest to model engineers.  And there is room for discussion about just what liquid and vapour means above this point.  There are no distinct different phases.

Now, if the boiler initially contained only vapour, the steam conditions are entirely in the superheat zone.  The increase in pressure does not cause the steam to condense, it is always all vapour.  The upper temperature is limited by the heat source, and certainly property diagrams for steam and steam tables covering superheated steam go much higher.  My superheat tables top out at 60,000 kPa and 1300 deg C.  Not an area for hobby experiments.

If the boiler starts absolutely full of liquid water, the initial heating will increase the pressure much faster, but again I suggest the liquid condition eventually reaches the supercritical zone and becomes what we would probably call vapour without a distinct phase change.  Not a process that I have my experience in, but I hope that is enough to answer the question.

The second question really only involves clarification of the details of the first.  For the purposes of these questions, there is no ďproper amountĒ of water in the boiler.  We either start with two phases or one phase.  If there is only a small amount of water in a large boiler, it would all be boiled into vapour relatively early in the experiment, and after that the vapour in the boiler would be superheated, and the pressure rises more slowly, depending on heat input and density of the steam.  Because heat transfer is not as good as for boiling a liquid, we might get hotter spots on the shell.  The temperature and pressure are now two independent properties, and the two phase section of the tables does not apply, this is where the superheat tables come in.  And with a smaller mass of steam relative to the volume of the boiler, the pressure will always be lower than if we started with a higher mass of water in the same boiler.  The pressure is always dependent on the mass of water in the boiler.  It also is affected by temperature, but temperature and pressure are independent for superheated steam.

The third question is an entirely different issue.  We are now talking about the basic laws of thermodynamics, and no longer talking about the properties of steam. 

Remember, the basic law is that heat travels from a higher temperature area or material to a lower temperature area.  So if your heat source is at 101 degrees, the boiler will never exceed 101 degrees, because once the temperature difference is zero, there is no further heat transfer, regardless of the efficiency of the insulation.



We are both safe and well thank you, and hope your winter is not treating you too badly.  It is all quite surreal here.  The Eastern parts of our state are experiencing the worst bushfires on record, I believe the area is roughly 10 times the long term average, and make no mistake, it is a large area, and the weather is not yet being helpful.  At the same time at home, we are experiencing cool weather, even had to turn the heating back on last week.  The western part of the state is a much larger area, a few smaller fires, but not abnormal, but the whole area is tinder dry from prolonged drought, and the worst months are usually January and February.  We are not out of the woods yet by a long way.  All very hard to comprehend and quite scary.

Thanks everyone for looking in, and thank you to all those thinking about those affected by the fires.

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1283 on: January 12, 2020, 01:15:28 AM »
Hi MJM , Yes of course what we have learned about heat transfer is correct ,but. was wondering if there is a time element in the equation ?  If  the heat is continually supplied where does the added heat get dissipated if the heating element is in the boiler and the insulation is at 100% efficient.? this may be a purely hypothetical question of course ,but was thinking about the energy that is being used ?? Also, was wondering about how you calculate the Horsepower of a compound steam engine

?

Willy
« Last Edit: January 12, 2020, 01:20:44 AM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1284 on: January 12, 2020, 09:30:10 AM »
Hi Willy, that is an interesting follow up to the previous question.  The issue lies in the statement of the problem.  Remember that you had already fixed the temperature, at which the heat is supplied at 101 deg, (I assume C, but it does not really matter).  You had also stated perfect insulation, or near enough to, so no heat loss, and now you are adding that energy is continually supplied.  This is too many constraints.

If heat is supplied continuously, then it must go somewhere or accumulate in the form of increasing temperature.  So the question is to look at where it goes. 

Letís look at the heat source.  You might be thinking of an electric element similar to that in your boiler.  That would make it relatively easy to add lots of insulation to reduce the heat loss to near zero.  Now, if you continue to supply electrical energy to the heating element in the enclosure, the element temperature will rise as we have discussed before.  As there is no heat loss, due to the good insulation, the temperature of the element will rise until something breaks, the element or the insulation melts.  But the temperature will rise. 

To comply with your condition of heat supply at a constant 101 degrees, you would need to interrupt the energy supply to stay within that constraint.  Either the temperature will rise, or the energy supply will be interrupted, or the insulation must fail.

Of course with other heat sources the detail will be different, but the result will be the same.

You might supply the heat using a double boiler in the way that a cook makes a delicate recipe.  The boiler would need to be slightly pressurised to get 101 degrees, as an open saucepan would be closer to 99.  But if no heat flows from the boiling fluid to your insulated box, then if heat continues to be supplied, by a flame for example, the flue gas and losses to atmosphere from the outside of the pot will rise until all the heat from the fuel combustion is lost.

Similarly if you heat is supplied by a hot fluid circulating in a coil, and no heat is transferred to your insulated box, the circulating fluid temperature will not change, and so if energy is continually supplied, the heat loss will simply be somewhere else.  You have to enlarge the boundaries of the system that you examine to see just where.

Switching subjects to calculating the power output of a compound engine, again a very different subject.

If you remember back to the early days of this thread we did talk about different definitions of the  power out put of an engine, and how these might be calculated.

The theoretical power output can be calculated from the steam supply and exhaust conditions using the Carnot temperature difference equation, but more usefully we can use the steam tables to calculate the isentropic or adiabatic power available, again based on the steam supply and exhaust conditions.  The supply conditions are generally known by direct measurement.  If we have two independent properties, the steam tables tell us all the other properties, in particular specific volume (or itís reciprocal which is density), enthalpy and entropy.  Providing the steam is either dry saturated or superheated, pressure and temperature measurements are sufficient.  If it is wet steam, pressure and temperature are not independent, so another property is required.  At the engine inlet, letís assume dry saturated or superheated with reliable pressure and temperature measurements, and we can then locate these conditions in the steam tables. 

From the tables, we want to find the enthalpy and entropy in particular, and also the specific volume.

The exhaust steam condition is more interesting.  We can measure the exhaust pressure at the cylinder exhaust connection, and while we can measure the temperature,  the steam is often wet, hence pressure and temperature are not independent, so we need something else.  Fortunately we assumed an ideal adiabatic process, so we know that the process occurs at constant entropy. 

Again we have two independent properties, pressure and entropy, and with some simple (if tedious) interpolation  calculations, we can find all the other properties, in particular the enthalpy.  All these quantities represented by these properties are based on a mass of one kg (or pound mass if you are using imperial units), so apply to any size of engine.  For a particular engine, we need to know the flow rate of steam through the engine on a per second basis, perhaps by steam flow measurement if suitable instruments are available, or a mass balance based on water lost from the boiler, or perhaps estimated using cylinder swept volumes and rpm with the steam specific volume property.

 The work done by that ideal adiabatic process is just equal to the difference in enthalpy, so we just do the simple subtraction.  The units of enthalpy in the steam tables are joules per kg, so the difference between inlet and exhaust enthalpy multiplied by the steam flow in kg/s gives us power in kilojoules/sec, which we also know as Watts.

Now two very important points.  First, no real engine can achieve that ideal work output.  No real process is actually constant entropy, the entropy always increases as expressed by the second law of thermodynamics.  In practical terms this means the actual, change in enthalpy across the engine will be less than the ideal change in enthalpy.  In this calculation we can define an isentropic efficiency as actual change in enthalpy divided by the ideal change in enthalpy.

Ok, a bit of a process so letís summarise the whole process as the procedure to calculate the work done by the steam on the piston. 

It is perhaps disappointing to realise that a real engine cannot even deliver this amount of work at the output shaft, so we can use it to drive a saw or milling machine or pump or compressor or whatever.  We still need to take into account the energy lost in overcoming friction between piston and cylinder, bearings of all the moving parts, losses in valves and ports and so on.   We can not calculate all of these with any degree of accuracy despite sophisticated computer programmes.  These all reduce the actual engine output. 

In the end we have to do a test.  To find the actual output for the real inlet and exhaust steam conditions.  Engine manufacturers have quite complex test beds on which accurate testing can be carried out, either for a specific engine, or for the manufacturers to use to predict the performance of future engines.

I said two points to summarise.  The second point is that nowhere have I mentioned whether the engine is single acting, double acting, multi cylinder, compound, condensing or otherwise.  I have only used the steam properties at the engine inlet and exhaust, and the steam tables to calculate the maximum potential power output of an ideal engine, and note that to find the efficiency of a real engine must be determined by a test run.  All those different engine configurations are just combinations of different physical components, all of which have the same aim, to produce the maximum amount of useful work from the available energy in the steam.

There is a lot more that could be said, but I hope that is enough to answer your questions, or alternatively to prompt more if I have missed the point you were considering.

Thanks to everyone looking in.

MJM460



The more I learn, the more I find that I still have to learn!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1285 on: January 13, 2020, 02:21:39 AM »
Hi MJM, Thanks for the reply...So we never actually destroy our boilers as we have safety valves and in my boiler a low water sensor and a pressure switch. these safety features prevent any cataclysmic eventualities from happening. And in most boilers there is plenty of escape routes for the heat to dissipate . And I suppose my question was purely hypertheticle., although at the beginning of steam engine and boiler experiments there were many explosions before the theory was worked out.

The usual Horsepower is worked out with the equation     PLAN
                                                                                _________
                                                                                33,000.      but this is for simple single cylinder engines rather than Compound engines. This is of course just the theoretical  calculation rather than the actual physical calculation involving pressure and temperature gauges .  So I was thinking there might be an equation for a compound engine using similar measurements. The reason I ask is because I have the castings for a compound engine with  2 1/2" and 1 3/4" cylinders. here is a pic.  thank you for all your answers and I am happy that you are OK.

Willy
                                                                             
                                                                                 

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1286 on: January 13, 2020, 05:19:41 AM »
Hi Willy, well there are still some boiler explosions, so perhaps ďhardly everĒ.  I donít know the actual numbers, but equipment fails to operate as intended and mistakes happen.  But a lot of attention and effort goes into making all those safety systems ever more reliable.  Dual safety valves, redundant switches that shut things down, usually with voting systems so critical systems are shutdown if the instruments donít agree and so on. But we are also now much better at making the equipment strong enough, with steel production techniques and welding procedures so have reduced the number of events.  However we have to remain vigilant and not relax the standards.

That formula for horse power is interesting and like all mathematical formulae it is necessary to understand how it was developed and what assumptions are implied when it is used.  You have used imperial units and that 33000 figure is a constant necessary to account for the vagaries of the units used.  I would have to check whether the dimensions of piston diameter and stroke are in inches or feet, and similarly the units of pressure.  However if we look at the formula carefully, we see pressure multiplied by piston area, which of course gives force.  The pressure is the pressure in the cylinder.  Then force is multiplied by the stroke or distance moved by the piston in one stroke.  Now force moving through distance is mechanical work, so p x L x A is the work done per stroke of the piston.  Then N gives the number of strokes per minute giving power.  As you say, it is for a single acting cylinder.  Multiply by 2 for double acting and by the number of cylinders for a multi cylinder engine.

But, and a very big but at that, the formula implies that the pressure is constant throughout the stroke.  If the pressure varies, the force on the piston varies and the work done for each unit of distance the piston moves will vary, introducing errors into the equation.  Alternatively, by calculating the pressure over each tiny increment of stroke, adding these up and dividing the total by the stroke, we can come up with a mean effective pressure.  In addition, it is not the absolute pressure in the cylinder, but it is the difference in pressure on each side of the piston.  Remember that pressure on the exhaust side of the piston gives a force in the opposite direction to the movement of the piston, and so does negative work.  That is is subtracts from the useful output of the engine.

In a real engine, there are pressure losses in the piping, valves and cylinder passages so to measure the pressure, an indicator was invented which actually measures the pressure in the cylinder, and plots it on a diagram.  A clever device called a planimeter is used to calculate that mean effective pressure using that diagram after an engine test.  The mean effective pressure might be close to the steam chest pressure for a slow running engine with generous valve and passage sizes, providing the valves stay open for most of the stroke.  However, if you have the valves cut off early so you have some steam expansion, the pressure starts reducing for the remainder of the stroke and is nowhere near constant.

Now your compound engine is specifically arranged to take advantage of the work done by expanding the steam instead of just exhausting high pressure steam at the end of the stroke.

So you could use the formula, but you would have to estimate the pressure on each side of the piston through out the stroke.  For the high pressure cylinder this is like a normal double acting engine, though the exhaust pressure will be higher, roughly equal to the lp cylinder inlet.  Then estimate the lp cylinder pressure throughout its stroke.  And of course the exhaust side of the lp piston will experience the final exhaust or preferably condenser pressure.

Then add the results for each side of each piston for each cylinder and multiply by the number of strokes per minute.  A simple enough concept, but working out the valve timing so that the actual cylinder indicator diagram can be estimate does complicate the use of a deceptively simple looking equation.  I think I would go for the adiabatic calculation, and when you have found the engine hidden in that casting set, next project has to be a test stand with a proper brake to measure the power output.  Then you can use the brake output power to calculate the ďbrake mean effective pressureĒ which you can use as a basis for estimating what happens in future engines.

Itís a great looking set of castings by the way.  I notice that Jo quickly took interest.  But no pressure to complete the current fabulous project before you start the next one!

Thank you to all looking in.

MJM460

« Last Edit: January 13, 2020, 11:19:03 AM by MJM460 »
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1287 on: January 14, 2020, 01:12:36 AM »
Hi MJM , thanks for the latest info..I have grown up with traction Engines and they were always sold with a power rating of  between 4 and 20 NHP. this being Nominal HP. Actually they were capable of much higher actual HP . I have included a pic and citation for these powerful machines.!!! I was also talking to a friend with a Uni degree in Physics about Thermo D..and he was saying that heat and temperature were actually different, but didn't get an explanation !!

Willy,   

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1288 on: January 14, 2020, 11:44:01 AM »
Hi Willy, there were many definitions of horsepower in the early days of steam and as you say, most of them had little relationship to actual engine output.  The NHP one is surprising because it generally gives a figure lower than the actual engine output.  Most of the stake holders had an interest in claiming the highest figure they could justify in order to sell more engines.  On the other hand perhaps they wanted to minimise the figure if that meant less tax had to be paid.  While the tax man almost certainly wanted something easy to measure like bore and stroke, rather than something that required sophisticated testing.  Much the same as the tonnage of ships.

Brake horsepower, the power as measured by a brake, is a more useful measure as it involves testing of the engine, or at least a very similar one.  However all the wildly optimistic claims are still made today.  Quoted brake horse power is usually still qualified as for a new and clean engine, so may not apply after a number of hours of operation.  Also it may or may not include power to run the water pump or cooling fan, alternator, and so on.  So it is still necessary to carefully specify the power requirement of the driven load and require a margin for measurement tolerances and performance reduction over time.

I have to agree with your friend about temperature and heat being different.

Heat is a form of energy that can flow in response to a difference in temperature and can be converted to other forms, in particular in our specific hobby, it can be converted to mechanical work, though it is easier to convert mechanical work to heat. 

Temperature is a property of a substance that we can measure with a thermometer.  Generally if heat is added to an object or a fluid, the temperature will rise, but we cannot determine the amount of heat from temperature alone.  Different materials will experience a different temperature change for the same amount of heat.  We have to know a property of the substance called the specific heat.  This varies greatly for different materials, and is usually determined by experiment.  On the other hand, when water is at its boiling point, you can add or subtract heat and the temperature will not change at all, so temperature tells us nothing about the quantity of heat that is being added.

It is quite difficult to define precisely the difference between heat and temperature because temperature change is closely associated related to heat gain or loss in all our experience.  Most attempts seem to end up going in circles, and my explanation is probably no different.

I donít know if that helps at all.

MJM460

The more I learn, the more I find that I still have to learn!