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Q: energy projection // weight of compressed air ( No Answer,   3 Comments )
Question  
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 ?
Answer  
There is no answer at this time.

Comments  
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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