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| Calculating heat dissipation requirements for high pressure compressor... ? |
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| pipe2null:
Yea, this is not exactly an electronics question, but thermodynamics is part of electronics, right? ;) I'm trying to figure out the actual cost and complexity of setting up a high pressure air compressor (3000-4500psi), small high pressure holding tank, plus a couple pressure regulators downstream to step down to normal shop air 40-90psi. I have occasional use for high ~3000psi pressure, but not often enough to merit the cost of a high pressure system unless I'm also getting shop air out of it too. Yes, I am fully aware you can rent high pressure tanks, like $15/day from a local scuba shop. The other immediate advantage is the equivalent storage capacity of a small 3000psi tank compared to a normal 120psi tank: 1 gallon at 3k psi is roughly equivalent to a 25 gallon 120psi tank, if I did my math correctly... Right now, I'm stuck at trying to figure out how much heat must be dissipated by the compressor to pressurize the air, temporarily ignoring heat generated by the compressor itself. These types of compressors are water cooled and you normally use a couple large trashcans of cold water to dump heat to, or something similar. For simplicity, let's assume the high pressure holding tank is a 100 cubic foot scuba tank being pressurized to 3000psi. I'm pretty sure scuba tanks are rated by volume of ambient 1 atm air compressed into the cylinder, so 100 cu.ft of 1 atm air compressed to 3000psi. I vaguely recall from high school chemistry class the whole PV=nRT thing, so with zero heat transfer, the compressed air would be 204xAmbient degrees (Ambient temperature x 3000psi/14.696psi), so the heat energy that the needs to be removed from the system is equivalent to the temperature delta for the given volume of air... I think... But that's where I'm getting stuck, and I'm not even sure PV=nRT is the right equation to use for this calculation. Any pointers on the subject would be much appreciated. Ultimately I'm trying to evaluate how feasible closed loop water cooling might be, and if a PC watercooling radiator would be sufficient or if I would need a much, much beefier radiator. For my purposes, it is acceptable to shut off the compressor once or twice during a fill so the water cooling can catch up, but not every other second. Thoughts? |
| skylar:
A junkyard car radiator with 12V electric fan attached is probably cheaper, easier, and more effective to dump that sort of heat load than a CPU cooler ever will be. Also, most of your assumptions are on adiabatic compression which is valid for piston compressors, though a water jacket will change that slightly due to the Q!=0 of the walls. As for your temp calc, you ignored the volume change and your number is way off. My math shows your final volume to be 2.27% of initial and a final temperature of 1994 degF which is reasonable for adiabatic cycles. This illustrates why many 3kpsi systems are triple piston and run at low RPM, you don't want to hit the ignition temp of your lubricants (O2 compressors and tanks/valves even use a non-hydrocarbon grease due to the lower flash point of high O2 environment). I would suggest that for your high pressure needs you get a small electric 3kpsi system for <$500 on the bay and use a separate compressor for shop air. You will see much better service life for both systems and you won't have to worry about how to get 3000 psi air down to 120psi without freezing your regulators, etc. |
| coppercone2:
PV = NRT is correct Delta P is 3000 V = Volume of tank, 18 Liters N = Number of moles as corresponding to amount of air you compress, value taken at 77 cubic feet of air R = Constant, 8.314 J/mol T = Solve for Conversions: 77 cubic feet, 2180.4 liters 18 liters to 0.018 cubic meters 3000 PSI to 2.0684e+7 pascal STP Mole of air = 22.4 L 2180.4/22.4, 97.33 Moles of air Formula = PV=NRT 2.0684*10^7*0.018=97.33*8.314*t 372312=809.2*t t=460.09 K T final = 186.948 C Thats what I got. It assumes a perfect system. I filled tanks to 4000+ PSI. What you do is fill it slow and it comes off kinda hot, you can add extra when it cools down so long you respect pressure rating if you fill too fast. Fill it in a blast proof chamber. 1.225 grams per liter, * 2180 = 2670 grams of air. Rough because its a compressible gas. If you are interested in heat, then calculate how much energy it takes to raise 77 cubic feet of air by 160 C, that is how much will be removed from system when it cools down. You use a regulator to slowly fill it. |
| Picuino:
The maths are very simple. If you use a 1kW motor for the compressor, the power to be dissipated is 1kW. The non dissipating power will serve to heat the air inside the tank, which should be as little as possible. |
| wizard69:
I can't answer your heat questions but ca offer a bit of practical advice. If the opportunity is there to turn up the air pressure somebody will do it when they shouldn't!!! :palm: :palm: :palm: Thus blowing out your 80 PSI air lines and damaging anything else pressure sensitive. Working in the automation field I've seen this happen multiple times, sometimes a qualified engineer on the knob. It isn't just air lines either but that is getting off track. Instead keep the high pressure systems separated from your low pressure systems. If you need more storage for low pressure air put a tank on the roof of your building, , |
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