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?
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Gas Cooling on Expansion 2
- Thread starter jhartis
- Start date
" … 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.
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.
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.
- 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.
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.![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
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.
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.
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.
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.
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.
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.![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
-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.
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.
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.
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.
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.
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.
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.
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.
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.
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 ?
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
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
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
-
1
- #34
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
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
- Thread starter
- #35
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
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
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
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
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
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
- Thread starter
- #38
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
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
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.
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
correction "negligible change in kinetic energy"
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