Robert Morein said:
Sure. Let's do the computation for the isothermal and adiabatic cases:
Work = Int F*dl = Int{PA*dl} = A Int{P*dl} = A Int{(NRT/V)*dl}
= A Int{ (NRT/(AL))*dl} = A*Int{(NRT/(AL))*dl}
= Int{(NRT/L}*dl} = Int{NRT dV/V} = Int{PV*dV} = (1/2)PV^2 (adiabatic
solution)
Actually, that is the general solution. But since it still leaves the value
of temperature unknown, it is not very useful, and not directly solvable for
the adiabatic case. How would you calculate the temperature rise from
compression with such a formula?? Without a final temperature, how would
you calculate the final volume?
For any polytropic process, pV^n = C where 'n' has a value of 1 for
isothermal processes and is equal to the ratio of specific heats for a
reversible adiabatic process. The ratio of specific heats (Cp vs Cv) is an
interesting property of the particular working fluid chosen.
To find the temperature rise for the compression of a gas under such
conditions, it is easy to
use :
T1/T2 = (p1/p2) ^((n-1)/n)
Where 'n' is equal to the ratio of specific heats for an adiabatic process.
The work then done in compressing a gas adibatically (a near approximation
for the rapid compression seen inside a compressor cylinder) is thus....
work = int{pdV} and p=CV^(-n) thus W = int{CV^(-n) dV} From state 1 to state
2, this integral is = (CV2^(1-n) - CV1^(1-n))/(1-n) (where C = p2V2^n =
p1V1^n)
Because the ratio of specific heats *is* a function of the specific
molecule, there can be some savings by choosing the proper gas. For
example, the ratio for air and other diatomic gases is about 1.4. For
helium, neon, and the other noble gases the ratio is about 1.666
Adiabatically compressing 1 ft^3 of air from 14.7 psia to 3000 psia would
change its volume to: (14.7/3000^(1/1.4) * 1ft^3 = .022394 ft^3 for a total
work of:
W = (3000*0.022394 - 14.7*1)/(1 - 1.4) *(144) = -18894 ft-lbf
Its temperature rise, if starting at 80F (540R) would be:
T2 = 540R / ((14.7/3000)^((1.4-1)/1.4)) = 2468 R
And for helium, 1 ft^3 from 14.7 psia to 3000 psia would change its volume
to:
(14.7/3000)^(1/1.666) * 1 ft^3 = 0.041121 for a total work of:
W = (3000*.041121 - 14.7*1)/(1 - 1.666)*(144) = -23473 ft-lbf
Its temperature rise, if starting at 80F (540R) would be:
T2 = 540R / ((14.7/3000)^((1.66-1)/1.66)) = 4526 R
The severe temperature rise involved is one of the reasons why high pressure
compressors perform the compression in stages and cool the gas between
stages. A typical compression of four stages (compression ratio of about
4:1 per stage) can heat the air to > 450F at each stage outlet. The gas is
then cooled by intercoolers back down to < 200F before entering the next
stage.
The lower the ratio of specific heats, the less work needed to compress the
same volume of gas and the lesser amount of work is converted to raising
the temperature of the gas and thus eventually rejected to the environment
(unless stored in a perfectly insulated container). There exists many
gasses with ratios of specific heats lower than air (hydrazine, acetylene
and propane are some). But they all complicate handling/recover
requirements.
Of course, if you can absorb the same heat from the environment during the
expansion, then this issue falls by the way as you then recover most (but
not all) of the energy previously rejected.
Any real world situation is in between the two, but the conclusion is the
same: compressing a gas is a lousy way to store energy, because the heat
created by compression is lost to the environment. Compared to other
methods
of energy storage, such as a battery, or pumped storage, it's lousy.
Unless you can trigger a phase change as another was trying to point out. A
fluid that can be changed into a liquid at near environment temperatures by
compression and rejecting heat to the environment, can also be converted
back to a gas at nearly the same temperatures by absorbing heat from the
environment. So, much of the energy is actually 'stored' in the
environment, not the working fluid. But again, if you use some gas/fluid
other than air, you have the nasty problem of containing and holding the
expanded gas until you can 'recharge' the system.
Liquifaction can greatly increase the energy density of the storage. But
since air is so very 'far away' from liquifaction at normal environmental
conditions, an awful lot of energy would be wasted just getting air to the
condition ready for the phase change. And containing and carrying around a
huge volume of another gas once expanded is not very practical either.
daestrom
