Adiabatic cooling vs. Joule-Thomson effect


by One128
Tags: adiabatic, cooling, effect, joulethomson
One128
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#1
Oct2-06, 08:23 AM
P: 15
When a valve on a pneumatic tire (filled with compressed air) is released and the tire is allowed to deflate, one may notice that the valve stem will become quite cold during the process. A common, everyday observation. So far, so good.

The question is: What would we observe if the tire being deflated was filled with helium instead of air?

I have encountered two contradicting theories about the possible outcome of the experiment:

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A: The observed cooling occurs as a result of adiabatic expansion of the compressed air. The air, as it escapes from the tire valve, does work on the atmosphere by pushing it away against atmospheric pressure to make room for the expanded air. The amount of work done, delta W, equals P(atm) * delta V (the extra volume that the expanded air takes). We assume delta Q = 0 (adiabatic process), and so this work is done at the expense of internal energy of the expanding air which is why it cools down. (Joule-Thomson effect is not a major factor here and only manifests as a slight deviation from the 'ideal gas' result.)

This mechanism (internal energy converted to work) will apply to helium as well, therefore, when deflated, a helium-filled tire will also cool down.

B: What we observe is indeed a kind of adiabatic expansion: free (unrestrained) adiabatic expansion, where no external work is extracted. There is no receding piston that would take away kinetic energy of the expanding air molecules. The tire begins at atmospheric temperature, so the 'tire molecules' have the same average kinetic energy as 'atmospheric molecules' and will not pass energy to them upon collision. The compressed air molecules do not 'push' the atmospheric molecules away by collisions; expansion occurs because the 'empty space' on the boundary is more likely to get occupied by one of the more densely occuring 'tire molecules' first.

In free adiabatic expansion, where no work is extracted (delta W = 0) and no heat transferred (delta Q = 0), the internal energy must remain the same (delta U = 0). The observed cooling occurs due to Joule-Thomson effect which predicts cooling of air by isenthalpic expansion at room temperature (because air is not ideal gas and so temperature does not directly correspond to internal energy).

For helium however, Joule-Thomson inversion temperature is far below room temperature, therefore, when deflated, a helium-filled tire will warm up instead.

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A's rebuttal: B is wrong. Free expansion (as in Joule expansion experiment) assumes expanding against vacuum, and only predicts that temperature of (ideal) gas will be the same after the gas reaches equilibrium. In our experiment, we vent the compressed air into the atmosphere and we do work on it. This work results in (generally inobservable) adiabatic heating of the rest of the atmosphere.

So the zero energy change would only be true if we included the entire atmosphere in the system, and waited until it reaches equilibrium, i.e., the cold valve stem is warmed up again by the surrounding air. That doesn't mean that there can't be work done between parts of the system (tire+air inside vs. rest of the atmosphere).

If our system includes only the tire and the air inside, then this is not a free expansion experiment, and the internal energy change of this system doesn't have to be zero.

As for the Joule-Thomson effect: Joule-Thomson experiment assumes isenthalpic expansion. B suggests that the internal energy of the tire air remains the same. But that would mean that there is positive enthalpy change, because the volume of the compressed air has increased (H = U + P(atm) * V). This invalidates the assumptions of a Joule-Thomson experiment.

B's rebuttal: A is wrong. No work is done by the compressed air on the atmosphere. Sure, a reversible, isentropic adiabatic process that allows gas to expand against pressure does positive work on the environment during expansion and always results in cooling; we could do that by venting air from the tire into a balloon. That would genuinely push the atmosphere away and thus do some work.

In our experiment though, the process is irreversible; the molecules get hopelessly mixed up and we will never be able to sort them out again. Entropy of the system is increased. Irreversible adiabatic expansion does not have to exert work on the environment, and in this case it doesn't.

The process is isenthalpic and is a valid Joule-Thomson experiment; volume of the compressed air does indeed increase, but the atmosphere is not blocked from permeating this extra volume, therefore the change in enthalpy is not justified.

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Who is right and who is wrong? Will a helium-filled tire cool down or warm up when deflated?

I don't know what the correct answer is; both theories sound credible and I don't have any helium at hand to actually do the experiment.
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Q_Goest
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#2
Oct2-06, 12:05 PM
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If we draw a control volume around just the valve, and if we assume there is no heat transfer or work done on the CV, then we are left with dU=0 and enthalpy in = enthalpy out (an isenthalpic process).

For air, the process results in a cooling of the gas. For helium, the proces results in the warming of the gas (JT affect is often quoted here).

~

If we draw a CV around the gas inside the tire only, then there are two ways of looking at this.
1. We cold recognize the change in internal energy being equal to the enthalpy of the gas leaving:
dU = Hout

orů
2. We could draw a CV only around that portion of the gas which remains inside the tire before and after the event. In other words, we have a control mass which is expanding. If there is no heat transfer, this expansion is also reversible, and the gas undergoes an isentropic expansion:
dS = 0.

Both methods should result in the same result for the gas remaining in the tire at the end of some cycle.

Attached is a graph of pressure and temperature versus time for some arbitrary expansion of helium inside the tire. This was done using the first method using an iterative process. This graph can be compared at any point to helium undergoing an isentropic expansion, and every point will be found to match.

~

Conclusion:
- Helium expanding across a valve is an isenthalpic process in which the helium generally will increase in temperature.
- Helium expanding inside a fixed volume due to gas escaping the volume can be determined using the first law, and is found also to follow a line of constant entropy. This process results in a temperature drop in the helium.
Attached Files
File Type: pdf Graphs.pdf (12.5 KB, 88 views)
mattbanham90
mattbanham90 is offline
#3
Sep14-07, 09:02 AM
P: 2
thanks q_goest - was observing this thread. Just wondering on the nature of your iterative model.

I am looking at what happens when a pipe full of high pressure CO2 suddenly breaks. What I'm trying to do is estimate the rate of freezing.

If the pipe has diameter, D, and CO2 density, rho, and is of length L, temperature, T and pressure, P and is initially completely sealed at both ends and breaks at one end, how quickly will dry ice form at the end of the pipe?


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