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Adiabatic Cooling

by s.p.q.r
Tags: adiabatic, cooling
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s.p.q.r
#1
Jul19-07, 04:05 AM
P: 25
Hi,

I don't fully understand this thing. Ive read it on wiki, but it just doesn't get through. Perhaps im a dumbass.

I've 3 questions that hopefully will help clear it up.

1) How do you stop adiabatic cooling?
2) In a closed and insulated system where there is a compressed gas which is also expanding, will there be adiabatic heating?
3) If you cause a gas to expand through adiabtaic heating, will the temperature of the gas go up or down?

Cheers in advance.
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russ_watters
#2
Jul19-07, 05:33 AM
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The definition of adiabatic is "occurring without gain or loss of heat". So there is no such thing as adiabatic cooling or adiabatic heating. [edit] The way the OP was worded, it implies the addition or removal of heat - adiabatic cooling and heating is a temperature change without the addition or removal of heat. [/edit]

In an enclosed system, if you let a gas expand without adding heat, it will get cooler. If you compress it without adding heat, it will get warmer.
s.p.q.r
#3
Jul19-07, 06:14 AM
P: 25
Hi Russ and everybody,

Cheers for your response. I wonder if you could try to help me out with question 1 a bit more.
If i had a compressed gas which is released into a larger container, the loss of pressure would cause adiabatic cooling. The temperature of the gas would decrease. How can I stop the gas from decreasing in temperature as it leaves the original container? Something to do with insulation perhaps? Add heat to the hole from which the gas is escaping? Ive got no idea.

PS- From what I can gather there is such thing as adiabatic cooling and heating. I remember it from wiki, but im not sure if I understand correctly.

Thanks for your help.

russ_watters
#4
Jul19-07, 06:46 AM
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Adiabatic Cooling

Quote Quote by s.p.q.r View Post
If i had a compressed gas which is released into a larger container, the loss of pressure would cause adiabatic cooling. The temperature of the gas would decrease. How can I stop the gas from decreasing in temperature as it leaves the original container? Something to do with insulation perhaps?
By adding heat to it (making the process no longer adiabatic). If a process is adiabatic, it implies the container is already insulated. Remeber, the temperature of a gas is not the only thing that affects its heat content. Pressure and temperature are related.
PS- From what I can gather there is such thing as adiabatic cooling and heating. I remember it from wiki, but im not sure if I understand correctly.
I edited my first post and added a clarification. Unfortunately, the words "cooling" and "heating" are a little ambiguous and can really mean 'to lower [raise] the temperature of' or 'to remove [add] heat'. The way you worded it in the OP, I took it to mean the second....
s.p.q.r
#5
Jul19-07, 10:11 AM
P: 25
OK, I understand now, but, I can't get one thing out of my head. In the wiki article, it says this, "Adiabatic cooling occurs when the pressure of a substance is decreased, such as when it expands into a larger volume. An example of this is when the air is released from a pneumatic tire; the outlet air will be noticeably cooler than the tire."

So, if I understand what you said correctly, this adiabatic cooling of the tire's released air will not occur if I raise the temperature of the air (whilst still in the tire) through external methods? ie- isothermal methods. So when the warmed air is released from the tire it will be warm and not turn cold, even though there is a great loss of pressure. Is this correct?
Thanks for the help.
Q_Goest
#6
Jul19-07, 11:35 AM
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Hi spqr,
... adiabatic cooling of the tire's released air will not occur if I raise the temperature of the air (whilst still in the tire) through external methods? ie- isothermal methods. So when the warmed air is released from the tire it will be warm and not turn cold, even though there is a great loss of pressure. Is this correct?
The problem with using the term "adiabatic cooling" is exactly as Russ mentions, it is only a term which implies there is no heat transfer to/from the fluid with which you are interested in. It does not define what other processes the fluid may be undergoing.

You give a good example, the release of air from a tire. If we look only at the air inside the tire, and assuming there is no heat transfer to or from the air as it expands, we can say this is an adiabatic process. If I said "the air is undergoing adiabatic cooling" to someone, then another engineer that understood what I was saying would take into account that the gas is also undergoing an isentropic expansion inside the tire and an isenthalpic expansion through the valve. They would understand this because they could understand the term "adiabatic cooling" when put into context. On the other hand, most people would not have any idea there were different processes which can all be termed as some kind of 'adiabatic cooling process'.

Isentropic process:
The air that is expanding inside the tire is also doing work by pushing air out of the tire. In this case, the first law reduces to: dU = Hout. Similarly, we can draw a control surface around some finite volume inside the tire which we might see as being an isentropic expansion in which the entropy change is zero. Either way, the process is the same, and the air will cool off because it must do work as it follows a line of constant entropy.

Isenthalpic process:
The air passing through the valve in the tire goes from a region of higher pressure to a region of lower pressure. The air in this case is not doing work, and so the first law reduces to dU = 0 or Hin = Hout. Note also there is no heat transfer again, so we might also call this process an "adiabatic cooling" process. However, the air in this case would not cool if it were a thermodynamically ideal gas. In this case, the gas cools because of how the (real) molecules of the gas must rearrange themselves as they go from a high pressure region to a low pressure one. This cooling isn't nearly as dramatic as the isentropic process, but it results in cooling of the air without heat transfer, just like the isentropic process.

In each case, the process is adiabatic, but without understanding what process a fluid is undergoing, one can't fully define the process. By adding heat to the tire to maintain a constant temperature for example, the air still cools as it expands across the valve as it undergoes the isenthalpic process, but there won't be any cooling of the air inside the tire because it no longer undergoes an isentropic process since heat is being added and it is also no longer an adiabatic one. Note that if we added heat as the gas expanded through the valve, it would no longer be an adiabatic or isenthalpic process, and the temperature of the air would have to be calculated using the first law. The Wiki article you read is a bit missleading as the portion you quote doesn't mention anything about various types of processes.
mgb_phys
#7
Jul19-07, 11:38 AM
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So, if I understand what you said correctly, this adiabatic cooling of the tire's released air will not occur if I raise the temperature of the air (whilst still in the tire) through external methods
No it will still cool the same amount - just starting from a higher temperature it will end up at a higher temperature.

The only way to have the air come out at lower pressure and the same temperture is to add the same amount of energy you are loosing.
Imagine having the gas escape through a metal pipe with a heating element on it - you could turn up the heat to exactly balance the cooling effect of the expansion. As stated above this is then non-adiabatic.
s.p.q.r
#8
Jul19-07, 12:54 PM
P: 25
Hi Q Goest. Thanks for the lengthy response, i learnt lots. You said in your bit about Isenthalpic process, "However, the air in this case would not cool if it were a thermodynamically ideal gas." Do you mean that if the tire were filled with an ideal gas like neon or xenon, then no heat would be lost when it is released? If so, why?

Thanks
mgb_phys
#9
Jul19-07, 01:35 PM
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Quote Quote by s.p.q.r View Post
Do you mean that if the tire were filled with an ideal gas like neon or xenon
No an ideal gas isn't a noble gas - it just means it obeys the simple gas laws and we don't worry about details like the physical size of the molecules or Van der-Walls forces betwen molecules.
Basically just physicist butt-covering small print.
Algis
#10
Aug19-07, 01:35 AM
P: 6
I am new to this forum, and this is a long-winded general querry, but directed at the excellent description by Q_Goest, of isentropic and isenthalpic processes as air is let out of a tire valve.

I have two questions that were crystalized by your explanation that have hounded me for a very long time.
1. The first is that conservation of energy seems to be violated.
2. The second is how can an ideal gas ‘push’ on an ideal gas, which brings more questions. I will post question 2 separately.

Consider attaching a very large, evacuated, rigid, and thermally opaque tank to the other end of the tire valve, with a large diameter but short (i.e. unobtrusive) extension pipe.

SETUP
So, say you have a tire with say 10 moles of ideal gas at temperature T0.
The tire has a valve which acts like a throttle. The outside of the tire valve is attached to a short hose, which is connected to a totally evacuated tank.

PROCESS
Open the valve, and let 1 mole of ideal gas escape from the tire into the tank, then close the valve.
The process will occur just as you described. Isentropic cooling of the gas in the body of the tire, as it pushes the gas that is escaping through the valve.
Isenthalpic expansion of the ideal gas through the throttle valve (tire valve) into a huge vacuum chamber. The isenthalpic expansion will not involve a change of temperature.
Allow equilibrium to be reached in the tire and in the tank.

PROBLEM (assume no external heat transfer or shaft work)
The gas in the tire will have cooled to a mean temp of b*T0 where b<1.
The gas in the tank will be at T0 or less (if the escaping gas had already isenthalpically cooled).
Internal energy (for both the tire and the tank) initially was: U = Cv N k T = Cv 10 k T0
The final total internal energy for both the tire and the tank will be at most:
U = Cv k ( 9 * b*T0 + 1 * T0) = Cv (1 + 9b) k T0 < Initial total Internal energy ???

What happened to the work done by the gas in the tire?
I know that the entropy of the total system increased, and thermodynamic potential to do work changed. But in a totally isolated system, with no external transfer of work or heat, surely U total must be conserved.

MY CONJECTURE
I think that if the gas in the tire cooled, the work accelerated the gas through the valve in a nozzle-like effect, giving it a velocity and momentum boost in one direction.
In this case, the gas escaping through the valve has internal energy proportional to CvNk*Ttotal where Ttotal = T0 + T dynamic.
In this case, conservation of internal energy could still be validated.

Am I out to lunch?
Algis
#11
Aug19-07, 01:41 AM
P: 6
This is the second question pertaining to my earlier post.

An ideal gas is composed of particles of no volume. Hence, they will never collide with each other, only with the walls.
How does an ideal gas ‘push’ the downstream gas through the tire valve?

I can sort of visualize the pushing if I assume a flexible non-elastic bladder (a grocery plastic bag for ex) inside the tire separating the pusher from the pushee. That at least gives a surface for the zero volume particles to hit.
But I suspect that the bladder visualization doesn’t solve the problem.

Is the notion of PV or ‘flow’ work just a visualization for a transfer of enthalpy by way of net particle flow?

If particles in an ideal gas truly cannot push on each other, then the only particles flowing through an orfice would be those whose trajectories are within the opening.

This would mean that the velocity distribution of flow from an orfice is solely dependent on gas temperature in the tire.
The pressure in the tire would only affect the quantity of gas flow.

Also, how could a nozzle accelerate the flow of particles of an ideal gas if the particles cannot push on each other?

Where is the high school physics lab when you need it. Please help.
mgb_phys
#12
Aug19-07, 12:02 PM
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Quote Quote by Algis View Post
How does an ideal gas ‘push’ the downstream gas through the tire valve?
It doesn't, each particle is on it's own - it bounces off the walls until it happens to go through the valve.
If particles in an ideal gas truly cannot push on each other, then the only particles flowing through an orfice would be those whose trajectories are within the opening.
Correct - remember they are bouncing off the walls at high speed and so get lots of attempts at hitting the opening.

This would mean that the velocity distribution of flow from an orfice is solely dependent on gas temperature in the tire.
The pressure in the tire would only affect the quantity of gas flow.
Correct

Also, how could a nozzle accelerate the flow of particles of an ideal gas if the particles cannot push on each other?
As the volume gets smaller the particles hit the walls more often and so the rate they hit the opening increases.

Imagine particles entering into the nozzle, it isn't pressure from the gas in the main part of the tyre that pushes them in there - it is purely statistics, there iss the same chance of a particle bouncing back into the main body of the tyre as there is for a particle in the tyre bouncing into the nozzle, this provices the 'pressure' at the entrance to the nozzle.
Q_Goest
#13
Aug19-07, 07:13 PM
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Hi Algis. Thanks very much for the compliment.

I believe I understand your new set up, and at least the problem you describe is understandable, though I’m afraid there’s a bit of a error in your “PROBLEM” description here:
The gas in the tank will be at T0 or less (if the escaping gas had already isenthalpically cooled).
Just a quick pointer here that you had described it correctly when you said:
The isenthalpic expansion will not involve a change of temperature.
Let’s be careful here. The isenthalpic expansion across the valve, assuming a thermodynamically ideal gas, will not experience a temperture drop (let’s use the assumptions you’ve stated in the “PROCESS” description). So you said it correctly in the PROCESS description, but then slipped up as you described your PROBLEM statement. What I believe you meant to say is along the lines that the gas going into the tank is cooler than it was when it was inside the tire, and there it collects at this lower temperature until the pressure across the valve is equal and flow stops. (that’s where the error cropped up in your description)

Try this – draw a control volume around the tank and write the first law of thermo for it. You should end up with dU = Hin. What does this imply about the temperature of the gas inside the tank? (note that m*cp*T is greater than m*cv*T which implies a continued increase in temperature of the gas in the tank as more gas is added)

Now take a look from a different perspective. Let’s say we just started to fill the tank and we still have a large pressure difference. Draw a control surface around the gas in the tank. Now add more gas and imagine the gas coming in doesn’t mix with the gas inside your control surface. (This is what we call a control mass since we are only looking at some given mass which is undergoing a change in state.) This mass inside the control surface must continue to decrease in volume as more and more gas enters the tank, right? As the volume gets smaller and smaller (it is getting squeezed into one end of the tank as more gas comes in), and assuming there is no heat transfer (process is adiabatic), what process would we say this control mass is being subjected to? Note that this process is the mirror image of the process inside the tire, because if we draw a control surface around a volume that is changing, and say there is no heat transfer, then the change in volume and pressure must be accompanied by work. For a decrease in volume such as for the control mass in the tank, work must be done on that volume. Such a process is also considered isentropic.

I’d suggest going back and rewriting the equations for internal energy for the tire and for the tank. Note that dU = Hout for the tire, while dU = Hin for the tank. Finally, note that Hout = Hin. This is all applicable only for an isolated system in which there is no heat transfer from the environment.

The hardest part of doing this when you’re in college is probably the simple fact that you don’t have a computerized database that allows you to do a numerical analysis of this process very easily. With each bit of mass that passes through the valve, the temperature inside the tire drops and the temperature inside the tank rises. If you had a database of fluid properties, it’s easy enough to do a numerical analysis on each volume and see what happens and even graph the results over time. With that, you can easily see what happens to temperature, pressure and internal energy during this process. Hope that helps.
Algis
#14
Aug19-07, 08:02 PM
P: 6
[QUOTE=Q_Goest;1405745]Hi Algis.
With each bit of mass that passes through the valve, the temperature inside the tire drops and the temperature inside the tank rises.


Thanks Q_Goest. That was the answer I was looking for, that if the gas in the tire cools below T0, the gas in the tank will be greater than T0. At least I think that is what you said. I like the explanation.

It has been a long time since I was in college, and thermodynamics was not my major.

But say the tire (with initial interior temperature of T0) is in outer space and is somehow fixed in position so it cannot move back in recoil (no work can be done on the tire).
Let's forget about the tank - there is only the tire now.
As far as the tire interior is concerned, nothing is different.
You release some of the gas in the tire into space, and then shut the valve.
The gas in the tire is now cooler than T0 - so the average kinetic energy of the particles in the tire is now lower.
Will the gas that was released, and will forever expand into space, have same kinetic energy (i.e. average kinetic energy per particle) it had in the tank before the valve was open?

If so, isn't the total kinetic energy of the particles (encompassing both the tire and all of space) now less than before?

I am thinking that the kinetic energy of the escaped particles must be higher than when they were inside the tire, which would then allow adherance to conservation of energy.

Sorry for avoiding the math (I did take it and can work through it), but I am trying to get a gut feel for what happens.
Algis
#15
Aug19-07, 08:13 PM
P: 6
Thanks MGB_Phys. I am glad I am not totally off on this.
Your explanation was excellent.

But a followup question.
If the particles in an ideal gas never collide with each other, how do ideal gases mix when they reach equilibrium?

Say you have a chamber with a partition, with one side containing gas at a high temp of Th, and the other side containing gas at a low temp of Tl, and you suddenly remove the partition.

The gases will mix, and at equilibrium the average temperature Tavg for the overall volume of gas will be correct according to the applicable equations.

But if the particles cannot collide with each other, the temperature of individual particles will never change, and so your distribution of particle energies (U) will consist of two Gaussian(or something like that) peaks at Th and Tl, rather than a single distribution centered at Tavg.

Is this what happens?

Thanks.
Q_Goest
#16
Aug20-07, 08:00 PM
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Will the gas that was released, and will forever expand into space, have same kinetic energy (i.e. average kinetic energy per particle) it had in the tank before the valve was open?
How the gas is released has something to do with this unfortunately, and nozzles aren’t my strongest suit. What I can tell you is that nozzles, such as converting/diverging nozzles actually convert the internal energy of the gas to kinetic energy. Note that I’ve neglected both kinetic and potential energy terms in deriving the equations above because they generally have so little impact that they can be safely neglected. However, if you’re expanding a gas through a nozzle, you can convert a large part of that internal energy to kinetic energy, such that the flow stream is near isentropic.

I don’t think that’s what you’re getting at though. From your last post and a previous one, it sounds like you want to model the gas exiting a valve which is generally modeled as an isenthalpic process, but then you also want to model the gas as if the molecules have no volume, which is a bit different yet. I don’t think the model of a hole in a tire with molecules randomly bouncing out of the tire is a legitamate one though – the molecules colide with each other and that I believe is a necessary assumption the laws of thermodynamics rely on. I would disagree that a thermodynamically ideal gas can be modeled as a collection of molecules with no volume. If the mean free path between molecules is on the order of some characteristic dimension of the container, you can no longer apply continuum mechanics. Look into the Knudsen number and also this paper here.
Algis
#17
Aug25-07, 01:01 AM
P: 6
Actually, I was not trying to tie the two problems together.

The first case, in response to your original tire example of isentropic internal expansion and isenthalpic expansion through the valve, my question was simply where did the work (done by the isentropic expanding gas in the tire) go? I think you answered that - the escaping gas has kinetic flow energy in addition to the internal kinetic energy it had initially.

The other question regarding the ideal gas is mind-boggling. I can't believe this hasn't been resolved and carved in stone a hundred years ago. I scanned the internet, and you can find papers and discussions supporting that ideal gases have zero volume particles, yet collide with each other, and far fewer but nonetheless reputable articles or comments claiming that ideal gas particles don't collide with each other, as suggested and well stated by Mgb_Phys.

I find it amazing that in all the descriptions of ideal gas, nowhere does it clearly state 'an ideal gas is presumed to be composed of zero volume particles, yet we shall assume the particles have a non-zero cross-section to allow collisions between particles',
IF indeed that is the situation.

Thanks for the Knudsen number and mean free path links - I do remember them from school now.

I am really weak in this subject, so I apologize if I'm spewing jibberish. But from what I gather, one must assume a non-zeor collision cross section diameter d to get a finite mean free path. If d is zero (zero volume) then the mean free path is infinite. Others on the web have brought this up as well.

If the mean free path were infinite, then any flow through an opening is by definition an effusion process rather than flow. Also, I strongly suspect that an ideal gas with the presumption of no interparticle collisions cannot undergo adiabatic isentropic expansion in the absence of shaft work (moving walls).
I think this because if the particles do not collide with each other, the process (of letting ideal gas escape out of a tire) is not reversible anywhere in the tire.


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