Ice Cream and Air Compressors

Let's say you've got one of these, and you're dusting off your computer:

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As you spray, you'll notice the can getting cold. You've probably noticed this in similar contexts: propane tanks getting cold, helium tanks getting cold, compressed air tanks running pneumatic tools getting cold, etc. It seems that when air escapes from an enclosure, it gets cold. Well, there's a reason.

Imagine if you will a elementary school gymnasium filled with kids and adults. The principle of the school announces free ice cream in the field outside of the gym and flings the doors open. The kids will of course freak out and run at top speed toward the desert. The adults will follow, but being afraid to look undignified they'll saunter at a much lower speed. As a result, the rate of effusion of the fast-moving kids is higher than the rate of effusion of the slow-moving adults. As a function of time the number of kids and adults in the gym will decrease, but the ratio of adults to kids will increase. The kids are leaving faster and the thus though there are fewer of both groups, there's proportionally a lot fewer kids.

So if you have a cylinder full of randomly-moving gas molecules, it stands to reason that the ones that happen to be faster-moving will escape an opening at a greater rate. I'll prove it in a future post, but it will turn out that if the average kinetic energy of an escaping particle will be 1.333... times greater than the average kinetic energy of the remaining gas. And temperature is effectively just a measure of the average kinetic energy of the molecules in question, so if the average speed of the gas molecules is going down, so is the temperature. Voila! Cold!

The situation with the compressed air can is complicated somewhat by the fact that it's not really just a gas inside. It's a fast-evaporating liquid, which means there's a phase transition happening as well. This adds some very difficult math from a formal statistical mechanics perspective, but the overall concept remains essentially similar.

I think it's pretty cool. *rimshot*

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The propane in a tank is also a liquid that evaporates as it is used. I'd long assumed that the energy of evaporation was the major cause of the cooling.

This is common in well testing (gas wells, that is). Although I must confess, I had no idea this was the cause of the cooling down.

I'm not convinced that the distribution of molecular kinetic energies is an important contributor to the cooling of the can. The mean free path of air molecules even at atmospheric pressure is tens of nanometres. That's how far they get before they hit something else. Over the course of a few collisions they completely rethermalize. Now, the length scale of features of the air can is much larger than this, we should expect the air molecules everywhere inside the can to be in local thermal equilibrium, so there should be no great effect due to the velocity distribution allowing the higher-energy molecules to escape through the nozzle first.

This analysis would suggest that if the air can were placed inside an empty plastic bag and then suddenly discharged, the bag would not end up at a lower temperature because all of the fast and slow molecules are mixed together again in the bag. However, the bag, in inflating, is doing work against the external atmospheric pressure. The total internal energy of the gas has to be lower. In the ideal gas case where there are no intermolecular potentials to worry about, that means your bag of gas will be cooler than it was when you began. This is simply cooling via adiabatic expansion of a gas.

Quite so, #1 and #3. In both cases though the underlying principle is along the same lines. Evaporation of the liquid into gas is a result of the fastest-moving liquid molecules escaping the liquid surface and leaving as gas. Adiabatic cooling would happen even if the escaping molecules weren't distributed differently in their velocity, but they are and so the process is driven harder than it otherwise would have been.

The intermolecular distance will not erase the effect (the can would still cool in a vacuum), assuming we're at high enough temperature so that the behavior of the gas can be treated classically. For instance, the momentum distribution of the escaping gas is clearly not the same as that of the gas on the inside - the escaping gas has to have a net momentum along the direction of the nozzle, despite the very short mean free path.

Back in the '70s, it was proposed that energy collected from windmills could be stored & transported pneumatically.

Apparently the problem of heating on compression and cooling during usage was insurmountable, except possibly for rare locations requiring both thermal processes at respectively appropriate times.

By Pierce R. Butler (not verified) on 16 Apr 2009 #permalink

While I would expect that, near the nozzle opening, there is a net reduction in higher-speed molecules because they get away faster, I do not believe that this deficit is efficiently transmitted throughout the can. The length-scale of the can is millions of times that of the mean free path. The air next to the nozzle (within a few hundred nanometres) may cool somewhat, but that air is not going to cool the rest of the can.

What you're really saying is that the air near the nozzle is clamped to some lower temperature by preferential loss of higher energy molecules, and then the rest of the can cools by thermal conduction in the gas (for now let's assume that the net flow of gas can be ignored, since that effect is going to make thermal conduction even more difficult). Thermal conduction in a gas is just a result of thermalization of molecules undergoing collisions. The thermal conductivity of air is not high enough to account for the observed time scale for reduction in temperature at points far from the nozzle.

As for the momentum distribution in the escaping gas, it is the same as that in the can. At least it is to me, because I am an inertial observer moving in the same direction as the gas jet but at half the nozzle flow velocity.

helium tanks getting cold

Joule-Thompson inversion temperature. Expanding helium gets warmer not cooler. See if we let you partially fill a big helium balloon and then trail a few feet of aluminum foil under it to freak out up-looking folks and radar operators. Boyancy is about 1 gram/liter. (Terrorists should not participate re power lines.)

Given that this is a flow process in almost all cases I think this is not how things work. Certainly not how it happens in a refrigeration cycle.

These evaporative systems cool because of the high latent heat of evaporation.

Coolant systems based on air expansion are quite different.

just a guess, but i think while your explanation of fast molecules leaving with more average energy lowering the temp is correct, i don't think that it is the primary cause for the decrease in temperature.

i think it the the latent heat of the phase transitions that cause most of the cooling.

i think this calls for some calulatin'!

oh, and your explanation is the reason you can cool a cup of coffee by blowing on it. essentially your are blowing away the fast molecules that come off the coffee/air interface. except now i wonder, is the effect large enough or is conduction with the air more important? maybe an experiment with a cup of hot liquid and a fan is called for!

The kids will of course freak out and run at top speed toward the desert.

What... and leave the air-conditioned gym? Why there's nothing there but sand and hard chapparal. The ice cream is just a mirage.

Rob #9: when a hot drink cools in air, the heat loss through the top surface is primarily due to two effects. One is the heat lost in warming up the air, the other is the heat lost by evaporative cooling. In still air, the space immediately above the hot drink is itself hot and humid, which reduces the degree of cooling from both of these causes. By blowing across the cup you are replacing the hot, moist air with cooler, drier air, and so increasing the rate at which heat is removed from the drink.

It's like running an electric fan in a hot room. The air coming out of the fan is somewhat warmer than ambient, having been heated by its passage through the fan mechanism, but the air still feels cooler to you because it carries away the warm, humid air that surrounds your body, replacing it with cooler, drier air.

yeah, i still think in normal conditions the gas in the can will still be in equilibrium, as well the gas leaking out. therefore the molecules in the jet escaping will have a maxwell boltzmann velocity distribution, and a well defined temperature. the temp should stay the same.

i am curious where you get the factor of 4/3 in your upcoming derivation. is it from some ration of the rms, average or most probable energy of an escaping molecule that you get from integrating the m-b speed distribution over the area of the hole in the can? i suppose i should do the integral myself!


@Winter Toad: yeah, that's basically how a dilution refrigerator works. pretty cool. *rimshot* (couldn't resist...)

By Anonymous (not verified) on 16 Apr 2009 #permalink

A dilution refrigerator is a bit more complicated than that. When you get a mixture of liquid He-3 and He-4 cold enough, it separates into two liquid phases with different isotopic ratios. If you put a pump on one of these phases, you remove He-3 preferentially, and the flow of He-3 across the interface to restore chemical equilibrium results in a net decrease in temperature. One of my fellow graduate students used such an apparatus to cool her experiment down to the vicinity of 1 mK, she had by far the lowest temperatures in the low temperature lab. Most people were happy with just pumping He-4 down to the lambda point.

I would think that the Ideal Gas Law would dominate the temperature change of a can of decompressing gas.

By complex field (not verified) on 18 Apr 2009 #permalink

Actually, the liquid and vapor are in equilibrium in the can (under high pressure) and when you open the nozzle you expel some of the vapor. After you do this the liquid and vapor set up a new phase equilibrium whereby some of the liquid molecules go into the vapor state (this is entropically favored). This results in the potential energy of the system being raised and the kinetic energy being lowered (conservation) and therefore the temperature goes down in the system (the contents of the can)

Since the liquid and vapor are in equilibrium initially, before opening the nozzle, the temperatures of both phases is the same --- this is a requirement (in this case) for phase equilibrium. Therefore, it would not be possible for the 'hotter' molecules to be mostly occupying the vapor phase -- not without some 'colder' ones to insure the temperature equivalence at least -- and therefore, this cannot be the major mechanism by which the system is cooled.

If you want real Dust Off drama, take off the little pipe attached to the push nozzle. Turn the can upside down, then give it a good blast. You'll get stuff blasting out big time, freezing and melting and every which thing. You can blow through a can pretty quickly, but it does make a dramatic demo.