Tank Pressure Change
Tank Pressure Change
(OP)
Is the process in a gas baldder accumulator isotropic, so that P1*V1=P2*V2?
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RE: Tank Pressure Change
The perfect gas equation you quoted is actually for Isothermal, meaning no temperature change. Let me demonstrate:
(P1*V1/T1) = (P2*V2/T2); the full perfect gas equation.
If temperature is constant (i.e. Isothermal), then T1 = T2 so both T1 and T2 can be removed from the equation to leave:
P1*V1 = P2*V2 (which is what you've got).
As long as the changes are happening slowly, leaving plenty of time for heat transfer (which I would suggest is the case), then you have Isothermal processes, in which case you can indeed use your equation.
Hope this helps.
Brian
RE: Tank Pressure Change
RE: Tank Pressure Change
To be more general, instead of using P1*V1 = P2*V2 (as for your isothermal case), lets assume the correct relationship is P1*(V1^n) = P2*(V2^n). If the process is isothermal, then n = 1; if it is adiabatic (a more practical / likely version of the very ideal isentropic), then n = 1.4.
Do the sums yourself; they should agree with mine thus:
Isothermal case: pressure increases by about 15%, volume decreases to about 87%.
Adiabatic case: pressure increases by about 15%, volume decreases to about 90% (the volume has not decreased so much because no heat transfer means the air's temperature has increased under compression, so it exerts a higher pressure at each volume than it otherwise would have done).
For my money, there's very little difference; what do you think?
Brian
RE: Tank Pressure Change
I suppose I'd have to determine exactly how critical estimating the volume will be. My guess is that it's not extremely vital in my case. Running my simulation for either case, I haven't noticed very much difference between the two cases, so that seems more or less in line with those results.
Thanks for the help.
RE: Tank Pressure Change
RE: Tank Pressure Change
I assume my conditions to be adiabatic.
Thanks in advance for your advice.
Tom