Author Topic: Talking Thermodynamics  (Read 118380 times)

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1185 on: July 10, 2019, 06:23:19 AM »
Hi Willy, no salt water crocs, only freshies.  Not too much of an issue if you don’t provoke them.  Yes fabulous location, but did the ominous orange glow show up on the site you looked up?

Only one small hot spot and repeater for all in the area at the moment, so I will write something on evaporation when I get to an area with normal service. 

MJM460
« Last Edit: July 11, 2019, 08:58:23 AM by MJM460 »
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #1186 on: July 11, 2019, 12:10:40 PM »
Well, after three hundred km, about half of which is corrugated surface with loose gravel, and four river crossings, back in the world of internet availability.  We are finding that moving that distance further inland gives us more of a continental climate, so much cooler at night.

Thank you Derek and Avtur for looking in and contributing.  I hope I can justify your confidence Derek.

Avtur, I have discussed steam tables previously, but when questions come up again, I put it down to my explanations being incomplete.  It is perhaps a complex subject to those who have not come across them before, while others of us just need a reminder.   I have not said much about ideal gases and real gases.  Obviously necessary for a more complete analysis of many engine problems, but the simplified description is probably enough to help people understand qualitatively what is going on.  But I would welcome a more complete explanation when you feel it would help understanding.  We have people with a wide range of levels of prior knowledge on the forum, and its good to have a place with information at all levels so everyone can find something new, or even a helpful reminder.

Hi Willy, back to your question.  I take it that the question comes from observing your test of the silver soldered cylinder by filling it with water.  If some water is lost then, the obvious question, is it evaporation? Or is it leakage. And was there a temperature change causing a change in volume of the water?

I think the simple answer is that unless you put in warm water, or filled the cavity when the fabrication was hot from soldering, it is unlikely to be evaporation.  But the question as an interesting one because it involves so many of the topics we have been discussing.

In principle, if everything is at the same temperature, there is no heat transfer.  (The so called zeroth law of thermodynamics.) Evaporation requires the latent heat to be supplied, so how does it happen.

Basically, the temperature of the water is due to the average energy of all the atoms making up the water, but individually the amount of energy in each atom varies over quite a range.  The atoms that escape the liquid surface are the higher energy ones, so they take that energy with them, and leave the liquid a little cooler, as we know when we blow on our hot coffee to speed up the process.  The temperature difference created, makes it possible for more heat to transfer from the air to the liquid.  Removing the water atoms which have escaped into the vapour space by blowing or a fan reduces the number which fall back into the liquid, so establishes a net loss of water atoms in the vicinity of the liquid surface, and so increases the amount of evaporation.

The lower limit of the temperature that can be obtained is known as the dew point, or the temperature at which the water vapour pressure in the air is equal to the equilibrium vapour pressure that we can look up in the steam tables.  The difference between the dew point temperature and the atmospheric temperature is dependent on the humidity of the air.  But I suggest that to calculate the rate at which the water will evaporate is quite difficult, and easier to determine by experiment for a given temperature, physical configuration, air temperature, humidity and air velocity.  Which does not solve your problem of whether there is a leak. 

I would suggest that rather than calculate the amount of evaporation, I would make up an exhaust fitting to accept a plastic tube, fill the cavity with water, plug the end of the tube and mark the water level with a pen.  This should eliminate the possibility of evaporation as equilibrium will soon be reached close to the liquid surface then no further evaporation.  If there is leakage, surely it would appear at one of the other ports which should not be connected?

And don’t forget the third possibility, if the temperature changes, the volume of the water will change, and hence so will the level.

MJM460
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Offline AVTUR

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Re: Talking Thermodynamics
« Reply #1187 on: July 11, 2019, 05:52:12 PM »
Willy and MJM

I have procrastinated too long.

I am a believer in calculations, the simpler the better. They may not give you the right answer but will give you a guide at what is likely. MJM has given an explanation of what is happening but, in most cases, the molecular level is acknowledged and then ignore.

For many following, sums are a great turn off but I will try to explain the steps and the science. I sure there will be questions, please ask them.

EXAMPLE 1. This is a puddle of water with air above it in a totally enclosed insulated box. When this is at equilibrium the air will be holding the maximum amount of steam, as a gas, that it is able to do so. [I am not going to use the word vapour]. This is a humidity of 100% and its value is very dependent on the air temperature. Relative humidity is the ratio of the partial pressure of the steam in the air to the saturated pressure corresponding to the air temperature. Its value can be determined from wet and dry bulb thermometer measurements; however my traditional “barometer” displays it directly.
If the air is not 100% humid it will be able at absorb water from the puddle until it reaches 100%. We have 5lb of water below a volume of 1 ft3 of air at atmospheric pressure in our box both at a temperature of 61°C (334K) [this is absolute temperature and will be required shortly]. Since we have just assembled this box the relative humidity is only 40%, typical for comfortable living. This is not at equilibrium so water will evaporate from the puddle until the air has a humidity of 100%.

Step 1. We need to find out how much steam was originally in the air. Using steam tables find the saturated pressure that corresponds to 334K [since I am an aerothermal engineer I will make no excuses for using K] which is 3.0 lbf/in2 absolute (that is above a vacuum, not above atmospheric pressure). [The significance of the pressure is water will start boiling, that is cease being a liquid, at the corresponding temperature. Remember it is the sub-atmospheric part of the steam tables, the half that no one uses]. Using the perfect gas equation........

[I think the perfect gas equation needs an introduction. It is not as nasty as it sounds. First gases away from very high energies and liquefaction are perfect. The equation is a direct result of our model of molecules bouncing off each other in free space. The equation is
    p = ρ.R.T/m     
        where p = gas pressure in absolute units, ρ = density of the gas, R = universal gas constant (2777 ft2K/s2), T = gas temperature in absolute units, m = molecular weight of the gas.

{Little example, we can use this to calculate the density of air at room temperature. p is 14.7 lbf/in2 (2117 lbf/ft2), T is 288K and m is 29
    Density (ρ) = 2117 x 29/(288 x 2777) = 0.077 lb/ft3}

The gas pressure in the equation can be split to represent each constituent of the gas (known as partial pressure) by mass]

........ the partial pressure of steam in the air, pSTEAM SAT, is the saturated pressure, 3 lbf/in2.

From the relative humidity the initial partial pressure for the steam, pSTEAM
    pSTEAM = pSTEAM SAT x relative humidity = 3.0 x 0.4 = 1.2 lbf/in2

Then mass of steam in air, MSTEAM
    MSTEAM = ρSTEAM x V = pSTEAM.m.V/(R.T)
        where M = mass, V = volume

Assuming the mass of steam in air will be small the molecular weight for air will be used
    MSTEAM = 1.2 x 144 x 18 x 1/(2777 x 334) = 0.00335 lb
        where 18 is the molecular weight of water (and steam!)

And the mass of dry air
    MAIR = (14.7 – 1.2) x 144 x 29 x 1/(2777 x 334) = 0.06078 lb

Step 2. Now to find the amount of steam that can be held by the air. Going back into the gas equation, the mass of steam in the air
    MSTEAM SAT = 3.0 x 144 x 18 x 1/(2777 x 334) = 0.00838 lb

Therefore the mass of water evaporated is
    MSTEAM EVAP =  0.00838 - 0.00335 = 0.00503 lb
        which equates to 0.139 in3 of water.

Comments
1. You may have noticed that
    MSTEAM /MSTEAM SAT = 0.00335/0.00838 = 0.4 which is the relative humidity.

However the mass ratio is not the definition since the above sums contain assumptions which do not compromise the answer. For example the evaporation of the water will lower the overall temperature because of the latent heat of evaporation and the increase in steam partial pressure would increase the pressure above the water.

2. If our model was a thermally insulated box with air being blown through it the equilibrium would never be reached since the steam is removed. After time there would be either no water left or a block of ice because of the evaporative cooling. The new model would need some knowledge of the rate of heat input to get an evaporation rate.

Willy, I realise that this does not answer your question. It is only a start. I am quite happy, if anyone is following, to continue to Example 2 (Air flowing over puddle with heat added).

This has taken quite some time. I should have been riding the bike, motorcycle, today with the lads but I over slept. My second priority, which has gone by the board, was work on the bellcrank engine.

Help! How do I use subscripts and superscripts? I wrote the above in WORD and then copied it across. I lost the paragraph spacing, underlining and subscript and superscript positions.

AVTUR
There is no such thing as a stupid question.

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1188 on: July 12, 2019, 12:35:37 AM »
hi MJM , thanks for the info...i was looking at evaporation tables and it would appear that evaporation can still take place if the ambient temp is lower than the water temp !  I think some of the water may have leaked through the soft rubber 'seal' that was held in place with a clamp....

Hi AVTUR, wow that is quite a lot of info to take in ... so can "steam" actually be quite cool ?? I have a physics expert that can take me through most of this....Presumably steam tables should start at minus 273 Degrees ??  that is quite a novel concept..Thanks for the mass of info and i do need to do some serious brain work ...which at 71 will be quite taxing....

Willy

Offline AVTUR

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Re: Talking Thermodynamics
« Reply #1189 on: July 12, 2019, 09:28:11 AM »
Willy

Quick reply.

If you connect a glass flask with some warm water in it to a vacuum pump and start pumping out the air the water will begin to boil. We were shown this at school about 58 years ago.

I have to admit I did not want to call gaseous water steam and I was not going to use the term vapour since I feel it is confusing. I started using the term "water as a gas" which is a bit clumsy. Then I realised my college notes used the word steam so I did.

AVTUR
There is no such thing as a stupid question.

Offline Admiral_dk

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Re: Talking Thermodynamics
« Reply #1190 on: July 12, 2019, 11:59:54 AM »
Avtur there are two buttons above the posting window labled "sup" and "sub" next to each other. Example :

H2O and V2

Both a first for me on this fora.

I haven't used the funktion in Word since year 1999, but you should be able to save as HTML too, so the format should be keept.

Best wishes

Per

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1191 on: July 12, 2019, 01:23:33 PM »
Hi Avtur, thanks so much for coming in with that post.  I know how much time and thought it takes when it comes to writing down the basics.  I also appreciate a view from a different industry, where you clearly use slightly different units, and even a slightly different form of the ideal gas law.  It all helps remind us that our units of measurement are somewhat arbitrary, and the most important point, is that we are consistent in any calculation.  That said, I still go to the timber shop for 2.4 m of 2 x 1!

I think much of the trouble with gas vs vapour is the common usage, with “steam” from a kettle being a visible mixture of water in gas phase and fine liquid phase droplets.  However, throughout my career in the oil industry,  the main process involves continuous boiling and condensing of many different substances and water is only an anomaly because it is not a hydrocarbon, and is a utility rather than a product.

And of course as a mechanical engineer dealing with steam turbines with or without condensers, that first page of the steam tables is quite familiar to me. 

So a broader view of the world is most welcome, thank you.

Oh, and the superscript sand subscripts, I also write in a word processor and copy and paste to the forum.  The forum does not like accepting subscripts or superscripts from my program either.

There is provision in the characters you can find along with the cartoons on the edit page as Admiral dk has noted (thanks, Admiral) but I have found them very awkward to insert into my posts. I don’t know. I have tried it a few times.  About HTML.  I like using the ^ form as in volume of a cube =L^3, but this seem not commonly understood be other forum members, so not very popular.  When used with units, I often just assume the subscript is obvious, as in lb/in2.  Or ignore the subscript notation as in P1 x V1 /T2= P2 x V2/ T2, as again it is obvious.  No easy solution with the available forum editor.

Hi Willy, water can definitely evaporate if it is warmer than the atmosphere around it, and even if it is cooler, just at different rates, and depending mostly on the humidity of the air above the water.

Certainly steam can be quite cool in the right circumstances.  My steam tables go down to 0.01 deg C which is the triple point of water, I.e.the temperature at which water can exist with solid, liquid and gas all in equilibrium.  Water as a vapour can be found in the air above ice, but no liquid below that temperature at normal atmospheric air pressure.  At the triple point the equilibrium vapour pressure is only 0.6 kPa, or about 0.08 psi and is most of that even at -75 C, and I have no information below that.  You need very high pressure to have liquid below 0 deg C.

More commonly, the condensing temperature of a steam engine or turbine, is mostly limited by the temperature of the cooling water or cooling air to something in the range of 30 to 70 deg C.

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 #1192 on: August 03, 2019, 03:32:35 AM »
Hi MJM, I have a new question after thinking about steam in cylinders...I don't know if we have covered this , as it is a hypothetical question... I you have a closed insulated cylinder with a tight fitting piston that is full of steam at a certain pressure  and then force down the piston as hard as you can ,..what will happen ??  I was thinking about this when i was pumping up someones tyres !! Presumably  the temperature will rise but what else might happen depending on the speed and force that one might apply....??

Willy

Offline AVTUR

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Re: Talking Thermodynamics
« Reply #1193 on: August 03, 2019, 09:15:43 AM »
Willy

A nice text book question. If it was air the answer would be simple but since it is steam I am interested in MJM's reply.

AVTUR
There is no such thing as a stupid question.

Offline MJM460

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Re: Talking Thermodynamics
« Reply #1194 on: August 03, 2019, 02:47:03 PM »
Hi Willy, it’s good to have you thinking about thermodynamics again, you must have recovered from the things that have ailed you recently.

As Avtur says, it is a good text book question, but in your usual manner, you have included just a little twist so that the precise example is not in my book.  Or perhaps I am just a bit weary tonight and unable to find it.

But let’s try for a brief answer anyway.  First your definition of the problem includes the cylinder being well insulated.  I will take this to mean that there is no heat transfer in or out.  Second, you said you would push the piston as hard and quickly as you can.  This bit departs from the usual textbook definition, which would require the process to go slowly in tiny steps so that at each stage it could be reversed, usually defined as an ideal reversible process.

So let’s look at this in two steps.  If we have a well insulated cylinder, there is no heat transfer in or out, it is called an adiabatic process.  And for an adiabatic process we can deduce from the laws of thermodynamics that the process is also isentropic, meaning there is no change of entropy.

As the piston is pushed against the gas, the gas temperature will rise, as will the pressure, as it will with any gas.  You might be thinking that as you increase the pressure, that steam might condense.  But if you look at a temperature entropy diagram for steam, you will see that at the temperature and pressure increases at constant entropy, the steam moves further away from the saturation line, so gets more superheated.  It does not condense, you can compress steam.  It is not often done because it costs more to build and operate the compressor to compress the steam than it does to condense it, pump it to a higher pressure, and evaporate it again. 

Now, you said if you compress the steam as hard (and quickly?) as you can.  The trouble is that this departs from that ideal reversible process, and the change of entropy will not be zero.  This is the second law of thermodynamics.  I have to think quite a bit more about which way this goes, and I hope that Avtur will be able to set us both straight on that one.  If you just meant as hard as you can, meaning to as high a pressure as you can, it could still be slow, close enough to reversible, and the steam will get very hot, and of course, still superheated.

And of course if your insulation is not perfect, and if you loose enough heat during that compression, the steam may condense anyway if the heat loss is sufficient, but it condenses due to the heat loss, not due to the compression.

Compression of steam is fundamentally no different from compressing any other liquifiable gas, such as propane, butane or any refrigerant.  The gas gets hotter as the pressure rises, and more superheated and then needs a condenser to remove the heat to condense it.  The heat loss from a normal compressor is not enough to condense the gas during compression.

Hi Avtur, I am glad you are looking in.  I will be interested in your answer as well.  Steam is obviously not an ideal gas when it is close to the saturation line.  So instead of calculating by the ideal gas law, I would just use the steam tables for an isentropic process.  I suggest that the main difference between steam and the usual refrigerant gases is that the relevant temperatures mean there is usually heat lost from the compressor instead of gained from the atmosphere.

Thanks everyone for looking in.

MJM460

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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1195 on: August 03, 2019, 04:24:19 PM »
Hi MJM and AVTUR,  thanks for the reply./answer...  perhaps i should just do the experiment !!! A good text book question !!  a bit like a question i once saw where the calculation of the level of the water each side of a ship turning in a circle in still water was required !!! I have always been one of those annoying pupils that sit in the front row that allways has his hand up asking the teacher for a fuller explanation !! especially when i found out that there is always an exception that makes the rule  !  ie  I before E except after C  as in science !!! 

Willy

Offline AVTUR

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Re: Talking Thermodynamics
« Reply #1196 on: August 03, 2019, 05:29:55 PM »
As I have already said the last time I did any real steam calculations was 50 or so years ago. While I kept my fluid dynamics text book the thermodynamics text book was sold when I left college. If I needed one at work I would borrow my boss's, we all did.

Unlike an ideal gas this is difficult to work out intuitively. Obviously constant entropy is the way to go and then make allowances for losses. At this point I would reach for a steam chart but I do not have one. Therefore I Goggled "steam chart" and found something very different! It was not rude or naughty, just unexpected. The only steam table I have is in Tubal Cain's "Model Engineer's Handbook". It is very condensed (pun not intended) and does not included entropy. Tubal Cain, who was a lecturer at Loughborough University, wisely avoids any mention of entropy in his handbook. As he wrote "As mentioned at the beginning, this has of necessity been a 'skim over the surface': whole books have been written on the subject, one at least running to two volumes".

I am sorry but I think I have copped-out. I will try to find a steam chart.

AVTUR
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #1197 on: August 04, 2019, 10:02:44 AM »
Hi Avtur,

There are lots of short cut versions of the steam tables on the web, and I have found good ones hard to find.  This link will lead you to some good steam tables in Excel format so very useful.

https://www.ohio.edu/mechanical/thermo/property_tables/index.html

I think you might have to copy and paste that, it does not look like I succeeded in pasting a link unfortunately.  It is provided as part of a standalone course in Thermodynamics by Ohio State University.

My original thermodynamics text is very old, and I have seen it on the shelves of engineers who were older when Iwas young, and it is in Imperial Units.  These days, I refer to a modern one using SI units, much easier to use.  Willy has given me cause to refer to it often in the last year or so.  It was quite instructive, and quite enjoyable.   Nothing like a curly question to test whether you really understand the topic.  He gave me cause to re-read quite a few sections, over and over until I understood it.  But during working years my go to source was some of the various industry data books that had sections with those practical formulae that tend to be used often when dealing with compressors, turbines and hydrocarbon processing.  In the early 80’s, I got hold of an SI version of one that I have nearly worn out over many years.

MJM460

Edit - Now that I have posted it, it does look like I may have succeeded in posting a link.  Now how did I do that?  I won’t test fate by trying again!



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Offline AVTUR

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Re: Talking Thermodynamics
« Reply #1198 on: August 05, 2019, 12:36:39 PM »
MJM

Many thanks for the link. I have downloaded the steam charts and will use them now.

Willy & all

I do not know if MJM has written about steam charts. Please excuse this if he has.

I have attached the two charts. They appear to be in the public domain. If not I will let a moderator tell me off. The charts present the same data in two different forms. The first is the conventional one which is met at college. The second has an emphasis on entropy, using it for the x-axis. This is the one we are going to use.

First, a quick explanation of the chart:
The x-axis is entropy: this is a measure of lost energy according to the second law of thermodynamics. Its actual value is not important but the change in value is very important.
The y-axis is enthalpy: this is the internal energy of the steam. Again change in value is important.
The blue lines are pressure in absolute units. Red lines are temperature (I think they should be in K but I am being pedantic). The green lines are the percentage dryness of the steam (labelled quality), 0% is water and 100% is dry steam (a gas). The black line is important, above this water is a true gas – superheated steam.

Now for Willy’s question. I marked up a chart to show what happens.

We have a well lagged cylinder of steam which we are going to compress. Since no energy will enter or leave the system other that the movement of the piston, it is adiabatic. Also we are going to say that the system is reversible, we can get out what we put in (so much for the Second Law).

I have taken a starting point, Point 1, for convenience on the 0.1 MPa (atmospheric pressure) at an entropy of 7 kJ/kg.K. The chart shows that the enthalpy is 2550 kJ/kg, dryness is 94% and temperature is 100°C.

We now compress the steam to 1 MPa. There is no change in entropy since the process is reversible so we get to Point 2. The steam is superheated and at a temperature of 260°C. Again from the chart, water boils at 180°C at 1 MPa so we have 80°C of superheat. The enthalpy is now 2950 kJ/kg so we have used 400 kJ/kg to compress the steam. If we release the piston we will get all of it back and return to Point 1. In practice if we do this process quickly there will be little time for heat loss. The slower we do it the more heat will be lost.

Apologies, but I am going to take this further. We cannot get away from the Second Law. The adiabatic assumption is reasonable. However reversibility is not, there will be inefficiencies. If we repeat the about with efficiencies of 80% (next chart) we have to use more energy to reach 1 MPa. We need 500 kJ/kg so Point 2 slides up the pressure line and, hopefully to no one’s surprise, entropy has increased. The steam temperature is now 300°C. If we release the piston and allow the pressure to return to 0.1 MPa. the reversible energy released will be 450 kJ/kg (I have not drawn this line on the chart). The actual usable energy will be 360 kJ/kg and we have moved to Point 3, conveniently, dry steam. Again entropy has increased and we have lost140 kJ/kg in the process. Since the system is adiabatic where has this energy gone? It is lost to a myriad of things like friction of the piston against the cylinder (and you won’t get it back).

I hope that the above makes sense!

AVTUR
There is no such thing as a stupid question.

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #1199 on: August 05, 2019, 11:44:10 PM »
Hi AVTUR  thanks for this  a bit more clearer now....

Willy