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Subject:
energy projection // weight of compressed air
Category: Science > Physics Asked by: swice-ga List Price: $5.00 |
Posted:
23 Sep 2004 13:29 PDT
Expires: 23 Oct 2004 13:29 PDT Question ID: 405417 |
Can you tell me how much total energy expressed in KwH it takes to fill up a 100cf scuba tank with air and how much the air would weigh ? |
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There is no answer at this time. |
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Subject:
Re: energy projection // weight of compressed air
From: guzzi-ga on 23 Sep 2004 16:49 PDT |
How long is a piece of string -- what pressure? |
Subject:
Re: energy projection // weight of compressed air
From: neilzero-ga on 23 Sep 2004 17:59 PDT |
I did know know the mass of a cubic foot of air (at sealevel) but I can't find it. Assume one ounce = 100 ounces = 6.875 pounds for 100cf = cubic feet. The outside volume of the tank is bouyed up in air by perhaps one ounce, but that can be ignored in most practical applications. The inside volume is needed to calculate the the KWH, but a close approximation can be obtained from the specs of typical 100cf scuba tanks. 2250 psi (about 153 atmospheres) used to be standard. Sorry I don't know how to calculate the KwH. In practical applications figure about 5 times that much energy as typical scuba compressors have rather low efficiency. Neil |
Subject:
Re: energy projection // weight of compressed air
From: racecar-ga on 29 Sep 2004 10:52 PDT |
Two more pieces of information are needed: 1) what is the volume of the scuba tank? (perhaps the volume of a 100 cf scuba tank can be found online) 2) how will the tank be filled? This may not seem like it should matter, but it does--at least the final temperature matters. If you fill the tank quickly, the process will be approximately adiabatic, meaning no heat is trasferred, which means the temperature of the air in the tank right after filling will be considerably higher than room temperature. If the tank is filled more slowly, the process will be approximately isothermal, meaning the temperature doesn't change. In either case, the energy required is just the volume of the tank times the change in pressure. For the isothermal case, this gives P*(100 - V), where P is atmospheric pressure in lb/ft^2, and V is tank volume. This gives an energy in ft-lbs. For the adiabatic case, the energy required is PV*[(100/V)^1.4 - 1]. Taking P = 2120 lb/ft^2, which is approximately correct, and V = 1 ft^3, which is a total guess, we get 210,000 ft-lb for isothermal, and 1,300,000 ft-lb for adiabatic. That's about .08 kilowatt-hours and 0.5 kilowatt-hours respectively. Probably the process in reality is neither isothermal nor adiabatic, but somwhere in between. Certainly it can't be adiabatic, because the final temperature would then be more than 1600 deg Celcius. The isothermal estimate is probably closer. 100 cf of air at 20 dec Celcius weighs 7.5 lb. |
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