Compressed Air Nozzle Cooling Ability

[scotjon]

I'm trying to find out the cooling ability that can be achieved with compressed air.

Is there any data out that for particular nozzles that lists the cooling ability of the nozzle (kW) at various pressures and flowrates?

 

[pipehead]

You won't believe it, -40F google Vortex tube

 

[scotjon]

I've seen the vortex tube, but it doesn't suit, because it also generates heat.

I'm trying to find the cooling capacity of a normal compressed air nozzle into open space.

 

[crjones]

Snow making uses the principle and the Association Governing Ski Slopes has equation cookbook calculations. Military aircraft also use the expansion as cockpit cooling source when the fighter is on the ground. Or model using adiabatic expansion.

 

[dcasto]

You can vent the hot stream outside with a vortex cooler.

The temperature drop across a nozzle of air with a starting temperature of 90 F and 150 PSIG would would be 5 degrees F, not a lot of cooling availble there.

 

[iainuts]

Hi scotjon,

I'm trying to find the cooling capacity of a normal compressed air nozzle into open space.

I'd just like to clarify a few things. I'm assuming, the model of air expansion you're looking for is from a tank at some pressure and temperature which is expanded (through a nozzle of some type) to a lower pressure, presumably at roughly atmospheric pressure. Further, you're assuming the nominal final velocity of the air in that "open space" is zero. In other words, the air comes from this tank, through a nozzle, and is blown into a box of some sort where it dissipates its velocity and perhaps slowly is vented out of the box.

Applying the first law of thermodynamics to this, you will find the change in temperature of the air is only the adiabatic expansion of the gas from one pressure to another. This is also known as an "isenthalpic" process. The fact there is no additional velocity attributed to the air once it enters this box also indicates there is no significant kinetic energy for the gas. This is true even if the air has some small velocity. The gas velocity needs to be close to sonic to see any significant amount of energy being converted to kinetic energy such that the temperature is not a simple isenthalpic expansion. The point of all that is only that the process is isenthalpic regardless of what small final velocity you assume.

Note that there is no benefit to the type of nozzle you employ. You can put the air through an orifice, or you can have a well rounded nozzle, it doesn't matter what you use, because you are assuming the final velocity is essentially zero. If you had a converging/diverging nozzle, this would still hold true because your air, even though it may be supersonic exiting the diffuser, must still throw that velocity away and it gets converted back to heat. Regardless of what nozzle you use, if you end up at essentially zero velocity inside this 'box' then the process is isenthalpic and all nozzles will give you the same final temperature.

For an isenthalpic process, you have to consider the non-ideal properties of air to find any drop in temperature. It is because of the thermal non-ideality of air that we get any decrease (or increase) in temperature whatsoever. So to perform this calculation, you need a properties database that gives you the actual properties of air, not the ideal ones.

For air starting at 70 F and the pressure in the left column below, the final temperature is in the middle column. For air starting at 0 F, and the pressure in the left column below, the final temperature is in the right column. Note this assumes the air is expanding isenthalpiclly from the pressure in the left column to 0 psig (14.7 psia).
(psig/F/F)
30 psig / 69.06 F / -1.21 F
60 / 68.12 / -2.42
90 / 67.18 / -3.62
120 / 66.25 / -4.83
150 / 65.32 / -6.03

As you can see, there's not much cooling affect.

As dcasto points out, the warm stream of a vortex tube can be vented directly outside the box. If it is vented inside the box, the vortex tube will also give you an isenthalpic process, and the above values will still apply once the warm and cold air streams mix.