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jhartis (Chemical)
15 Nov 03 10:41
In calculating downstream temperatures in a 50 psig compressed air system, I cane across an apparent puzzle.  I know the Joule Thomson coefficient for air at my conditions is roughly 0.25 °C/atm so an expansion from 3 atm -> 1 atm should cool a little less than 1°C (correct?).  I did an experiment pumping a Coke bottle to 50 psig, let the bottle cool back to ambient, and release the pressure.  I haven't found a way to measure the temperature, but I feel sure the air cools more than predicted by the JT effect.  I would guess it is cooling 5-10°C and there is often visible condensation.  The gas velocity inside is nearly zero throughout the process, so the kinetic energy change in the bottle should be negligible.  Any thoughts?
quark (Mechanical)
17 Nov 03 2:08
This is not a steady state process and can't be a JT expansion. Why can't you just use ideal gas equation (of state) to get the final temperture? What you should know is the initial and final mass of the air inside the bottle. (can a coke bottle withstand 50psig?)

Regards,

Believe it or not : Eratosthenes, a 3rd century BC true philologist, calculated circumference of earth with the help of a stick and it's shadow. The error was just 4% to the present day calcuated value.

Helpful Member!  25362 (Chemical)
17 Nov 03 3:19
The J-T data mentioned by jhartis is for air free of moisture and carbon dioxide.
Productmanager (Mechanical)
17 Nov 03 4:57
For joul Tomson effect to take place, the gas has to expand from high pressure to low pressure. The temperature of expanded gas can be found from the gas tables distributed by NIST. You take the difference between the enthropy at two state and using C calculate the change in tepmerature. Fast expansion is an adiabatic process and there is no loss of energy to the surrounding.
25362 (Chemical)
17 Nov 03 6:54
Just for the sake of exactitude the change in temperature on a isenthalpic-adiabatic drop in pressure (a free expansion) is called the Joule-Thomson coefficient.

Quark is correct in the sense that in a "bottle" experiment the upstream pressure drops as the expansion takes place and the pressure difference is not constant.

Although the cooling effect is practically independent of the pressure at high temperatures, it increases markedly at lower temperatures.

Thus a difference of 200 atm lowers the temperature of air at 0oC by 45 degrees, whilst at -90oC the drop is of 100 degrees.

Hydrogen, for example, that heats up on expansion at ordinary temperatures, when precooled with liquid air, can be further cooled and eventually liquified by its own free expansion.

1969grad (Mechanical)
17 Nov 03 8:30
The JT coefficient of 0.25 C/atm may or may not be valid as the coefficient does change with pressure.

To get a good JT coefficient check a pressure-enthalpy diagram assuming the expansion process to be a constant enthalpy process.  Plot the start gas pressure and temperature and then draw a straight vertical line to the end pressure and read the temperature at that point.
25362 (Chemical)
17 Nov 03 9:37
The J-T process is a throttling process without any appreciable change in kinetic energy of the downstream outgoing gas. Perry VI Table 3-151a gives J-T coefficients for air for such processes.

A quick (high flow rate) adiabatic expansion on the other hand, would result in T1/T2=(p1/p2)(k-1)/k.

The experiment carried out by jhartis seems to be more a quick adiabatic expansion than a throttling process where the pressure ratio is coming from a starting 4.4 down to a final 1.0 .
25362 (Chemical)
17 Nov 03 14:23
To make myself clear, even when the gas comes out through a true throttling process -with little cooling as the J-T coefficient brought by jhartis indicates- the gas left behind in the bottle undergoes cooling by expansion in a non-flow process. This expansion can be approximated by the adiabatic (reversible) expansion formula given above, although the "bottle" experiment doesn't seem to be adiabatic and the bottle would exchange heat with the surroundings.

So, the outgoing gas is cooler than expected by the J-T free expansion, because the "upstream" gas expanding inside the bottle is cooling down. This effect doesn't happen when the system (upstream) pressure is constant, and the only process playing a role is the isenthalpic expansion of the air coming out.
iainuts (Mechanical)
17 Nov 03 15:39
Expansion of a gas through a valve or orifice is isenthalpic, ie: enthalpy remains constant.

Expansion of a gas in a cylinder where pressure is released is approximated by an isentropic process, ie: entropy remains constant.  

In the case of the isenthalpic process, no work is done by the fluid.

In the case of the isentropic process, work is done, and the gas will be significantly cooler because of this.
quark (Mechanical)
18 Nov 03 4:57
Actually this process is adiabatic and lowest possible temperature of the gas left in the bottle can be checked by the equation provided by 25362. But I never perceived such lower temperatures when evacuating gas cylinders, for example. This is because we can't vent out the bottle so fast that any heat input into the bottle can be neglected.

Still, I doubt the practicality of conducting this experiment in a coke bottle and original poster didn't turn up again to excuse my ignorance.

Regards,

Believe it or not : Eratosthenes, a 3rd century BC true philologist, calculated circumference of earth with the help of a stick and it's shadow. The error was just 4% to the present day calcuated value.

iainuts (Mechanical)
18 Nov 03 7:27
Adiabatic means only lack of heat transfer.  The first law is applicable here.  Change of internal energy = enthalpy in - enthalpy out plus etc ...  

For expansion through a valve, the adiabatic process is isenthalpic.

For expansion of gas in a cylinder, work is done by the gas.  Yes, this is an adiabatic process also, but easiest to calculate as isentropic assuming you have the NIST tables or other thermodynamic data as mentioned earlier.

Hope that clarifies just a little.   
jhartis (Chemical)
20 Nov 03 22:25
Thanks to all.  While some of these posts help, I'm still a little confused.  To answer Quarks question about the pressure capability of a Coke bottle, I pumped a standard 2 liter PET Coke bottle to nearly 150 psi, before I chickened out and quit pumping.

I see now that I really have 2 processes, the expansion of the gas in the bottle and the gas that blasts out though the valve.  In spite of the differences, the bottle gets almost as cold as the nozzle.  And Quark, I know the experiment isn't very practical, but it was some fun and piqued my curiosity, not a bad bargain.

I don't believe work is being done by either the expanded gas or the throttled discharge gas.  Or am I wrong and energy lost somehow by accelerating the discharge gas?  A JT expansion occurs though a nozzle.  What happens to the kinetic energy in the accelerated gas in that case?

If the eqn suggested by 25362 & Quark (T1/T2=(p1/p2)(k-1)/k) applies, the gas should get quite cold as mentioned.  Admittedly the process is not entirely adiabatic, but it only takes a few seconds to release the pressure in the bottle and the plastic is not very conductive nor is it a large heat sink, it should be nearly adiabatic.   How is this different from a JT expansion?
25362 (Chemical)
21 Nov 03 4:06
You are absolutely right. For an adiabatic expansion the temperature T2' of the remaining air inside the bottle would be T1/1.5 or about 300/1.5=200 K=-73oC !!

Air may function as an ideal gas and the process was quick and thus adiabatic, but definitely irreversible. There is indeed no shaft work done. However, because of friction effects and the energy spent in "pushing" the outcoming air molecules the efficiency of the expansion would be just, say, 30%.
Taking Cp=1.0 kJ/kg-K for air, the total "available" energy of expansion would be = Cp(T1-T2') equal to

                    1.0*(300-200)=100 kJ/kg air.

The final "actual" air temperature T2 inside the bottle after expanding it down to atmospheric pressure can be estimated as follows.

Assume 30% efficiency, the "lost" work would be 70 kJ/kg.

The entropy change related to this lost work would be
                   
                        S2-S1=70/200=0.35 kJ/kg-K

And since S2-S1=Cp ln(T2/T2')

                   0.35=1.0*ln(T2/200), thus T2=283 K=10oC.

Which is a reasonable result confirming your findings, and may serve to explain quark's own experience when emptying gas cylinders.

The J-T expansion didn't contribute much to this cooling, especially considering it was not a "steady" process been effected with a small (decaying) delta P.

 
unclesyd (Materials)
21 Nov 03 12:46
What would an Infra-Red camera shown in the space occupied by "jhartis's" bottle if it had burst?
quark (Mechanical)
30 Nov 03 0:35
Jhartis!

Are you still there? I am very sorry that I was very quick in deciding the practicality of the problem. I saw closed coke bottles (empty) bulging when exposed to Sun for a while. That is why I thought it was dangerous to pressurize them (ofcourse, I forgot junkyard wars). I did a lot of reading after my second post and didn't find a single clue to prove JT expansion for the gas which remains in the bottle. Some of the books stated it as adiabatic expansion.

One more thing you have to mind is that, 1lb of moisture requires 1000 btu/lb and if you measure the quantity condensed, you can have some estimation about the actual temperature of gas in the botle. I will have some discussions meanwhile and I will let you know if anything comes up further.

PS:I think 25362 is very close to the answer.  

Regards,

owg (Chemical)
30 Nov 03 9:59
I would like to know if the thermodynamic "daemon" available for online calculations at the link shown below, can help in solving this problem. I input P1 and T1 conditions for a real gas (air) of 3 atm and 60 deg F, and final conditions of 1 atm, and the initial enthalpy condition as calculated by the daemon for the P1, T1 conditions. The answer is shown below and results in a drop of about 1 deg F for the expanded gas.

State-1:  Air > Superheated Vapor;
Given:   p1= 3.0 atm;   T1= 60.0 deg-F;   Vel1= 0.0 m/s;   
z1= 0.0 m;   
Calculated:   p_r1= 0.080629975 UnitLess;   T_r1= 2.1789098 UnitLess;   Z1= 0.99847066 UnitLess; v1= 0.27215365 m^3/kg;   u1= -92.871346 kJ/kg;   h1= -10.143444 kJ/kg;   
s1= 6.507626 kJ/kg.K;   e1= -92.871346 kJ/kg;   j1= -10.143444 kJ/kg;   

State-2:  Air > Superheated Vapor; Given: p2= 1.0 atm;   h2= -10.14344 kJ/kg;   Vel2= 0.0 m/s;   
z2= 0.0 m;   
Calculated: p_r2= 0.026876656 UnitLess; T2= 59.10502 deg-F;   T_r2= 2.1751573 UnitLess; Z2= 0.9994693 UnitLess;   v2= 0.81587 m^3/kg;   u2= -92.81147 kJ/kg;   
s2= 6.8225365 kJ/kg.K;   e2= -92.81147 kJ/kg;   j2= -10.14344 kJ/kg;   


http://eng.sdsu.edu/testcenter/testhome/indexdaemons.ht...

HAZOP at www.curryhydrocarbons.ca

25362 (Chemical)
1 Dec 03 2:22
To owg, why have you assumed a constant enthalpy expansion for the air in the bottle ?
owg (Chemical)
1 Dec 03 7:41
I was running the original problem of letting down 3 atm air to 1 atm in a steady state operation. I believe that would be enthalpic. I also ran the isentropic case for air saturated with water vapour but I did not save the results. Try this link for the thermodynamic daemon. It seems to have more features than the testcenter version referenced in my previous post.

http://www.me.udel.edu/testcenter/indexdaemons.html

HAZOP at www.curryhydrocarbons.ca

iainuts (Mechanical)
1 Dec 03 8:11
Saying "Adiabatic" is very missleading.  Adiabatic only indicates an absence of heat transfer.  

Adiabatic expansion through a flow restriction is very different from adiabatic expansion of a gas through an expansion engine.  In the first case, the gas that is expanding does no work.  In the case of the gas expanding in the cylinder work is performed by the gas.

Using the first law of thermodynamics is the proper way to solve either of these problems.  In the case of a gas being released from the bottle, the change in internal energy of the gas inside the bottle is equal to the enthalpy of the gas exiting the bottle.  

In this case, you must perform the analysis iteratively since as the pressure decays, the gas gets colder, and the enthalpy of the gas venting from the bottle drops.

If done iteratively, you will eventually get the correct answer.

Alternatively, you can draw a control volume around the gas that never leaves the bottle.  At first, it occupies a very small amount of room at the bottom of the bottle, and as it expands, it eventually displaces all the other gas and fills the bottle.

In this case, one can immediately conclude the gas that has expanded has undergone an isentropic expansion, just like the piston in a cylinder.  

Note that I've done this analysis both ways and both methods give identical answers.

The result is about -190 F.  I doubt it actually gets that cold though, since the process is not adiabatic.  Heat transfer from the thermal mass of the bottle is still very significant.  I've had to do numerous problems similar to this one at cryogenic temperatures, and I've always found that the heat transfer from the vessel is a very significant factor, so "adiabatic" it is not.

25362 (Chemical)
1 Dec 03 9:24
If the bottle were well insulated and the expansion quick enough, an adiabatic process could be safely assumed, not so for reversibility, thus it wouldn't be isentropic, and the thermophysical properties of the leaving gas would differ from those of the gas left behind in the bottle. Iainuts please comment.
iainuts (Mechanical)
1 Dec 03 10:57
" … the thermophysical properties of the leaving gas would differ from those of the gas left behind in the bottle."
Yes, this is true.  Which is why if you do this using the first law, dU = Hout, then you need to do this iteravely.  The enthalpy out (Hout) constantly changes because the internal energy of the gas inside changes.  If you do this process using lots of iterations, you'll see the properties change as the pressure decays.

Adiabatic = no heat transfer:
Really, the point is that saying the process is "adiabatic" is unsufficient to define the process.  If it is reversible, it is isentropic.  If it is irreversible, it is isenthalpic.  

I'd agree if the expansion were fast enough, this might be adiabatic, but that generally can't be accomplished.  In cases where there is a gas at low pressure inside a vessel, the turbulence from the expansion results in a very high convective heat transfer coefficient.  The faster this expansion occurs, the closer to adiabatic it will be, but measuring the gas temperature becomes problematic because of the lag time in your measurement and the resulting heat transfer from the bottle.

In this particular case ...  the mass of the gas in the bottle is extremely small when compared to the mass of the bottle itself.  Therefore, insulating the bottle does nothing to prevent heat transfer between the warm bottle walls and cold gas.  If the bottle and gas come to an equilibrium temperature, you'd find the bottle temperature change very little, while the gas temperature changed drastically.  There just isn't enough thermal mass in the gas for the insulation to have any impact on heat transfer.  

The thermal mass of the bottle is HUGE in comparison that of the gas.


Isentropic expansion:
When you draw a control volume around a given amount of gas, you need not concern yourself with changes outside that control volume.  You may still have heat transfer across the CV boundry, you still have changes of state inside the CV, but one does not need to concern themself with *how the change in volume* (for example) affects what goes on inside the CV.  

In the case of a piston in a cyliner, draw a CV around all the gas.  As the CV expands from 150 psig down to 0 psig, we see that it does work against a piston.  If we also assume there is no heat transfer between the CV and the walls of the cylinder (ie: this is adiabatic) then this process is obviously an isentropic expansion.  Note that this expansion is also completely reversible.  Would you agree?

In the case of a CV around ONLY the gas that remains in the bottle at the end of the process, we see the identical process.  The gas expands (against a piston of air) from 150 psig to 0 psig.  Again, we'll assume the process to be adiabatic.  Note also this process is reversible.

Both situations are identical to what is inside the CV.  The gas in each case sees a constant drop in pressure and expands appropriately.  

Note that this is a very different result when compared to expansion of gas through a restriction, such as a valve.  This is because of irreversible processes that occur to a gas in the flowing state when it expands through a restriction.  In this case, the expansion is still modeled by the first law, and enthalpy in = enthalpy out.

I honestly enjoy talking about thermodynamics and hope some of this can be of use.  Let me know if you still need clarification.

Take care,
Dave.
iainuts (Mechanical)
1 Dec 03 11:52
Just a small clarification.  This may not be completely obvious from what I've writen above.
- The gas that has left the vessel has NOT undergone an isentropic expansion.
- The gas that remains in the vessel after having expanded HAS undergone an isentropic expansion. (assuming no heat transfer.)

Note that knowing this will simplify the calculation of final temperature enormously.
25362 (Chemical)
2 Dec 03 1:56
The old Shaum's Outline Series on Thermodynamics, McGraw-Hill 1972, gives just the problem in hand.

It shows that when the rate of expansion enables the walls of the container to exchange heat with the gas so as to attain equal temperatures the process isn't, of course, adiabatic.

Given the heat capacity C and mass m of the container walls,
n1,2 the number of moles of gas at the start and at the end of the expansion, T1,2 the respective gas temperatures, Cv the constant-volume heat capacity of the gas assumed unchanged during the expansion, and k=Cp/Cv, the following formula shows the gas change in temperature:

                      T2/T1=[(n2Cv+mC)/(n1Cv+mC)]k-1

and since:
                     P1Vbottle=n1RT1, and   P2Vbottle=n2RT2

One can find both T2 and n2 by trial iterations.

It also shows that when mC is sufficiently large T2=T1 !!!

Thus, for any intermediate case, i.e., either not "too quick" expansions, or not "too large" mC container values, the final temperature of the gas in the container would be somewhere in between those for an adiabatic expansion and an isothermal one.

iainuts (Mechanical)
2 Dec 03 7:32
25362,
It's nice to find canned formulas for things, it can simplify the world a bit.  But understanding the fundamentals and applying them indicates to me a much deeper understanding of the concepts and underlying science of the problem.  

Note that when gasses become non-ideal, or when there is a change of phase, the canned formula provided will fail to give an answer.  Also, it only provides answers to the specific problem it addresses.  Understanding and applying the laws of thermodynamics can broaden the number of applications one can solve.  I would suggest using first law for this, or using control volumes to simplify the model as I've pointed out.  Then add your equations for heat transfer & thermal mass separately.  

Take care,
Dave.
25362 (Chemical)
2 Dec 03 11:12
Dave, thanks for the thoughtful advice. FYI, the formulas were developed from the first law, a development that I didn't transcribe to avoid bothering readers.

I still think the assumption that the gas in the cylinder quickly expands adiabatically and reversibly (isentropically) is a rough simplification that generally doesn't correspond with reality. See examples in this thread.

On the other hand the assumptions made by the formulas I quoted from Schaum enable one to see another possibility, i.e., that of the high capacity container walls exchanging some heat with the gas.

You rightly speak of control volumes. One may select a small control volume (CV) containing, say, that portion of the gas that will stay in the bottle after the expansion. The control volume then expands to the full volume of the bottle doing p*dV=W  (work) albeit in an unsteady (transient conditions) process.

Energy may accumulate or deplete inside the CV. More precisely, the energy crossing the boundaries of the CV as heat -dQ and work +dW must equal the change in energy of the material contained within the control volume itself
d(mU)CV plus the energy transported by the outflowing stream
[(H+u2/2g+z)]outflowing stream.
For an ideal gas a drop in internal energy, dU, results in a drop in temperature.

In the particular case when dQ=0, du=0, dz=0,

d(mU)CV-dHoutflowing stream=dW.
If dW=0 then the change in internal energy of the gas in the bottle equals the change in enthalpy dH of the outcoming gas as you rightly said.
In particular, if du (velocity changes) are negligible one can assume reversibility if the CV expansion is done slowly and adiabatically.

But in a rapid expansion dW isn't zero. It includes not only shaft work (in this case none) but also any work resulting from the quick expansion of the CV.

This is the reason for my assumption of irreversibility, namely du is not zero, even when dQ=0. Thus we have an adiabatic, irreversible expansion.

It's a pleasure exchanging views with you.

25362 (Chemical)
2 Dec 03 12:21
In short, if I may express my thoughts qualitatively on the coke bottle experiment:

-a quick expansion may be adiabatic but it is irreversible thus not isentropic, with less gas cooling than previewed by an isentropic process;
-a slow expansion may be reversible, but not adiabatic, nearing an isothermal process;
-in fact, the actual bottle expansion process may be somewhere in between both above cases.      
iainuts (Mechanical)
2 Dec 03 13:23
25362,
I guess we can agree on some things, and disagree on others.  Perhaps it's just the presentation of concepts that differs.  I realize that's one part of the disconnect anyway.
 
- First Law (reduced for this example): dU = Hout.  If we assume dQ=0, and no work, then the gas temperture calculated by the first law is identical to the gas temperature calculated assuming the gas remaining in the vessel has undergone an isentropic expansion.  The two calculations result in identical answers.  Except:  I can appreciate that if the process is exceedingly fast (say in a reciprocating steam engine during expansion for example) turbulence may cause some minor amount of irreversibility by converting kinetic energy of the gas molecules into heat.  We would need some very high velocities to have this happen, which might be possible during free expansion.  If this process is NOT reversible, then we need to account for the irreversible components of the process somehow, and those irreversible portions of the process must convert energy to heat in some way.  This can be seen in the first law as well, since kinetic and potential energy components could be added, the only problem being a practical one of trying to determine the magnitude of those bits of energy.  Not an easy problem.
- I assume your second statement, that a slow process could be isothermal, is due to heat transfer from the thermal mass of the container, and/or from the environment.  In which case I agree.
- Yes, I essentially agree with your last statement.

Interesting discussion.  

Take care,
Dave.
MechNuke (Mechanical)
19 Feb 04 19:34
I'm working on a problem with a compressed gas bottle accelerated to high speeds when the valve is knocked off. The bottle starts off at 2400 psig and depressurizes to atmospheric (this is a missile/safety type analysis).

Since the bottle empties in a second or so, the adiabatic assumption seems apt. But, when you plug into the formula:
T = To(P/Po)^((k-1)/k)
for Argon (k=1.669) with To=530R, you end up with a final gas temperature of -365F!

Does anyone have a sense of whether this is realistic? It's important because I'm ending up with an acceptable final velocity about 100 mph by allowing T to fall to -365F, but if I pick an arbitrary minimum T of 0 F, the final velocity is an unacceptable 130 mph. The velocity impact comes mostly from that fact that the speed of sound is much lower at -365, so the choked flow ejected gas is slower.
rmw (Mechanical)
19 Feb 04 21:26
As a college student, while working in an equipment dealers shop, my friend, not heeding the warning of the older workers to leave the last wrap of the acetyline torch hose hanging on the horseshoe that held the coiled up hose, went over one day, and unwrapped all the hose, and started dragging the torch and hose across the shop.  In the process he pulled the oxygen and acyteline bottles over, and the valve broke off of the oxygen bottle, which then began to accelerate out the open shop door toward a line of brand new volkswagens at the dealership next door.

I, being the brave sort that I was, immediately ran for the other door, to escape, while others in the shop ran in all kinds of other directions to get away.  Only "Curley" the shop foreman, and a bull of a man, had the presence of mind to run after the rapidly accelerating bottle, and tackle it to prevent it from trashing the line of volkswagens.

Point of my story is that I don't know when the bottle emptied, but it sure seemed like more than a second, and for that short few second period, seemed more like an eternity.

Realistic??  I was not calculating the exit gas temperature, but I was learning to leave the last wrap on the horseshoe.  I can believe, based on emperical observation that that bottle had the capability of reaching high velocities, assuming the volkswagens were not there, except for the sudden addition of 250 or so pounds of additional payload being drug across the floor in the form of Curley's body.

My story may have not answered your question, but I sure enjoyed telling one of many "Curley" stories once again.  Good lesson for a young engineering student about the realities of high pressure.
iainuts (Mechanical)
20 Feb 04 10:14
Hey rmw, you did the right thing by running away from the bottle.  It may not have been too full at the time, but a fully charged bottle can definately kill you.  Too many people have died trying to stop a run away bottle.  Tell Curly he got lucky!  

MechNuke, you did fine on the calc, but it's not likely to be adiabatic.  The thermal mass of the bottle is sufficient to warm the contents without having to get ambient heat involved.  Without thinking too hard, you could do exactly what you've done, calculate a velocity based on adiabatic conditions, then given a relatively warm condition (ie: high heat transfer rate), and base decisions on the worst case.  
25362 (Chemical)
20 Feb 04 10:29
MechNuke, the Cp/Cv value for Argon is sensitive to pressure and temperature. At the starting conditions of your exercise it is about 2, not 1.669, which corresponds to atmospheric pressure. Am I right ? Could an adiabatic expansion condense some of the argon ?
JimLo (Mechanical)
23 Dec 04 14:39
Looking at mechnukes calculations, I would agree if he modelled adiabatically the answer would be 35 K. If you take it as a van de waal gas then the temperature is a more realistic 187K which fortunately is above argons critical temperature so no liquids. As the gas will be throttled at the neck of the bottle is it not likely that the ejected gas will not cool until it is passed the vena contractor (area of shock wave) and outside the bottle. I would expect the bottle to chill marginally say 20-30 degrees enough to get a frosting. Not that you would notice it would be gone in a flash.
sailoday28 (Mechanical)
24 Dec 04 12:29
DS=0  in adiabatic tank

energy equation    dE=hodm
m=mass in tank
e= specific internal energy of fluid in tank
 E= m*e, internal energy in tank  (KE and PE neglected)
ho= specific stagnation enthalpy of outflow=  h of fluid in tank

E=H-pV

dH -pdV-Vdp=hdm  
mdh  + hdm -pdV-Vdp =hdm
Const vol tank
Therefore               mdh-Vdp=0    dh-vdp=0

But dh=Tds+vdp         
 So that ds=0
For all fluids        

Helpful Member!  denniskb (Mechanical)
28 Dec 04 23:55
Well done jhartis a very good question and some great responses but I don't think it is over just yet. (One point though is whether you are sensing the cold of the bottle or the gas inside it?)

I have had to deal with this problem before in designing gas compressor stations and we have developed an Excel solution which iteratively solves for adiabatic expansion/compression, heat transfer and JT cooling. It seems to work well and I will try to simulate your case and advise the result.

Our compressor stations run at about 15,000 kPa (2200 psi) so when pressurising or depressurising experience a compression ratio of 150:1 in a fairly short period of time. The theoretcial temperatures from adiabatic expansion/compression are frightening but are not in fact evident in operation. The hot and cold temperatures experienced are however significant in that they can lead to valve seat damage and/or cold embrittlement of steels. In my experience many designers have not dealt with these temperature changes.

I find it quite amazing that we are unable to simply open a text book and go right to a definitive solution to what appears to be such a simple problem. Yet this is a situation that happens in real life all the time and could easily lead to catastrophic failures (of the Coke bottle?). I can well imagine a clients reaction when I tell him how long I need to provide an answer to this one.

I will run a calc and post the result for consideration.

Regards
Dennis K-B

Dennis Kirk Engineering
www.ozemail.com.au/~denniskb

jhartis (Chemical)
6 Jan 05 23:19
denniskb:

You ask if I sensed the cold of the bottle or the gas.

The bottle was ambient (summer) temperature at pressure.  I let it cool for a while for the heat of compression to dissipate before releasing the pressure.  As the gas was released, the exiting gas was noticeably cool.  The bottle's temperature also quickly cooled (maybe 20°F?) and a fog of condensed water formed in the bottle.

I look forward to your analysis.

PS, don't try this at home.  When I recently repeated this experiment, the coke bottle exploded at about 120 psi and the cap nearly took my ear off.

Jerry Hartis
denniskb (Mechanical)
6 Feb 05 5:53
Jerry,

Apologies for the delay but as usual time is short.

The calculation we use in our work includes;
- an orifice plate calculation (Crane) for the nozzle
- adiabatic expansion in the bottle over a short time period
- heat transfer from ambient to bottle to gas over the same time period and an adjustment to temperature
- Joulle Thompson cooling across the orifice

Several key inputs are estimated with the most important for this case being the Overall Heat Transfer Coefficient
- typical values air, free convection 3 - 35 W/m2.°C, forced convection air 30 - 850 W/m2.°C.
- I tried a few values then settled on 100 W/m2.°C

I started with everything at 25°C

The next most important is the time to blowdown, determined by the orifice size. You did not state whether you had a valve at the bottle outlet or how long it took to blow down.
- I tried 2, 3, 5, 10, 15 mm orifices.
- Resulting blowdown times 0.10, 0.20, 1.02, 2.55, 6.12 secs
- Resulting gas in bottle temp 10.7, -0.12, -26.3, -57.7, -71.5 °C.
- Resulting bottle temp 17.9, 19.6, 21.4, 24.2, 24.6 °C

As you can see with a slower blowdown the air temp has dropped only 14.3°C and the bottle 7.1°C which is in line with what you indicated.

I still have serious doubts about the quick blowdown results as I have experienced air blowdown from vessels and never felt anything like the -70°C condition, then again they did not happen in 0.1 sec.

I would love to see this case run as an undergraduate study of first theory and then for real in the lab.

While our method is far from perfect it provides a useful tool for our design and with time we will find ways to improve it.

Regards
Dennis K-B

Dennis Kirk Engineering
www.ozemail.com.au/~denniskb
IRstuff (Aerospace)
6 Feb 05 13:14
Argon is often used in conjunction with nitrogen for JT cooling of infrared sensors.  If I remember correctly, the boiling point of Ar is arounf 90 K, while N is 77 K.  The one real example I'm familiar with used 6000 psi tank pressure.


Ar has a better heat transfer coefficient for rapid cooling.  Sometimes, if you have the luxury of two separate tanks, the Ar is run first to rapidly cool to 90 K, followed by final cooldown to 77K with N.

TTFN

jhartis (Chemical)
24 Feb 05 0:15
denniskb,

Cool, I am impressed with your work.  FYI, I used a standard presta bicycle valve on the bottle which probably has an ID around 4 mm, but it has a valve stem which reduces the area of the exit.  

I didn't time the blowdown, but it seemed to be about 2 or 3 sec, as well as I can remember.  Well in line with the predictions per your calculation.

You mention that you did a Joule-Thompson calc across the
orifice, but the temperatures you list are for the bottle and bottle contents.  How cool is the gas exiting the bottle?

Thanks,
Jerry
sailoday28 (Mechanical)
24 Feb 05 8:20
denniskb  STATES" Joule-Thompson calc across the
orificE

Only if there is negligible change across an adiabatic orifice,can the J-T coef be used.   The J-T coef is for a constant enthalpy process.

sailoday28 (Mechanical)
24 Feb 05 8:21
correction   "negligible change in kinetic energy"
25362 (Chemical)
24 Feb 05 9:29
To denniskb.

I can understand that a gas expanding through a throttling valve does it isenthalpically. However, from the inlet to the "vena contracta" it appears the expansion follows a quasi-isentropic path with much deeper cooling to such a degree, that some gases may partly condense, CO2 and ethylene, could serve as examples.

I think this is a fact that may happen even with expanding superheated steam, precipitously forming high-velocity water droplets within the valve body, or solid hydrates in the case of moist gas.

For these reasons the internals of throttling valves are constructed from materials that can withstand the erosive action of such streams, and measures are taken to abate the resulting noise.

Am I right ?
sailoday28 (Mechanical)
24 Feb 05 9:55
25362 (Chemical)states:
I can understand that a gas expanding through a throttling valve does it isenthalpically.

For an adiabatic process the energy equation,neglecting change in elevation is
enthalpy in  +  kinetic energy in = enthalpy out + kinetic energy out.

isenthalpic only if negligible change in KE
denniskb (Mechanical)
24 Feb 05 10:17
Sorry, I had the blowdown times in the wrong order.

I have made some minor improvements to the calc so the result are slightly changed.

I have added the bottle discharge temp including JT cooling which you can see is quite small due to the small pressure drop.

- Orifice sizes      2,     3,     5,     10,    15 mm.
- Blowdown times     6.12,  2.55,  0.92,  0.22,  0.09 secs
- Gas in bottle temp 10.7, -0.12, -28.6, -61.5, -71.5 °C.
- Bottle temp        17.9,  19.6,  21.9,  24.0,  24.6 °C
- Gas outlet temp    10.5, -0.25, -28.7, -61.6, -71.6 °C

For 25362, yes the low temp can easily lead to condensation and since the velocity through the orifice is high (sonic) erosion can easily occur. Also I believe the JT cooling effect includes the reheating as the high velocity gas through the vena contracta slows down again to the downstream velocity so local temperatures at the vena contracta may be lower than indicated.

I have a pdf file of the above blowdown cases but am not sure how to make this easily available to readers.

Dennis Kirk Engineering
www.ozemail.com.au/~denniskb

sailoday28 (Mechanical)
24 Feb 05 11:32
denniskb (Mechanical)
You have stated that sonic velocity is reached.  This represents a signilficant change in kinetic energy.  I repeat again that this change does not represent an isenthalpic process.  Only then is J-T coef valid.
abeltio (Mechanical)
24 Feb 05 12:16
jarthis,
what if you use a bottle of pepsi?

saludos.
a.

25362 (Chemical)
24 Feb 05 14:37

To sailoday28. You are right in regard to the kinetic energy across the valve restriction itself. I think, however, that no real throttling process is exactly isenthalpic.
 
We speak here of two consecutive fast processes: an isentropic expansion, with an enthalpy reduction, from P1 to about 0.5 P1, followed by an enthalpy recovery after a short distance downstream, where the kinetic energy has already been reduced to a level negligibly different from that upstream, at the final pressure.

The first step has a cooling effect with the gas crossing the vapor line into the two-phase region. As denniskb says, after a certain distance the gas becomes again superheated at a lower pressure and even a lower temperature, and at practically the same original enthalpy.

The overall result being an isenthalpic throttling process.

From an example from literature, CO2 at 20 deg F and 300 psig is expanded through a throttling valve to atmospheric pressure.

The gas undergoes two temperature changes: first, to ~ -35 deg F at ~ 145 psia; then, after a short distance, the downstream gas reaches ~ -60 deg F and atmospheric pressure.

The duration of the whole process is so short that it is insufficient to cause formation of droplets.

The primary result is a pressure drop in the gas. In the absence of heat transfer, without an appreciable change in elevation or kinetic energy, and no shaft work produced, it reduces to ΔH = 0, or H2 = H1. Checking the data from the article, I found a very small overall drop (~1%) in enthalpy, which enables us to call it an isenthalpic process.

Any comments ?
sailoday28 (Mechanical)
24 Feb 05 15:04
If the adiabatic process just up and downstream of the orifice is bounded by piping, the pressure drop will result in a change in velocity up and downsteam of the orifice.  With a known pressure drop and flow, and the STAGNATION ENTHALPY remaining constant, the conservation of mass and energy (stag. enth) should be solved.  If change in velocity (and therefore KE) is negligible, the overall process may be considered isenthalpic.

It is further stated
"The duration of the whole process is so short that it is insufficient to cause formation of droplets."

I don't see what the short distance/time has to do with the process.--- Unless one is considering metastable equilibrium.  Having the thermdynamic properties in the metastable state can be tricky.
I do know for ex that the ASME stm tables used to (probably still do) have steam properties in the metastable state.
Regards

 
denniskb (Mechanical)
27 Feb 05 9:01
If anyone is still monitoring this thread I have added a new section to my website and put a pdf copy of the calculation outputs under "miscellaneous files".

Look under "design calculations" to find it.

Dennis Kirk Engineering
www.ozemail.com.au/~denniskb

jhartis (Chemical)
1 Mar 05 0:06
25362, et al

I'm struggling with much of this, but you say that for constant kinetic energy, delta H = 0.  But there is a large velocity change, as denniskb mentions, the velocity is likely sonic at the restriction and perhaps again at the discharge on the expansion to atmosphere (for cases with higher pressure in the bottle and assuming there is an internal restriction or orifice at the entry to the valve stem).

This has always been an obstacle to my understanding of JT:  Does the constant enthalpy necessary for the application of JT refer to no heat transfer from the surroundings or no change in enthalpy due to expansion (change in velocity & KE)?  If you don't have a change in kinetic energy, is it only really applicable to a piston expansion where the velocity change isn't very significant?  Am I mistaken that the change in kinetic energy for a gas acceleration comes from a change in enthalpy?

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