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Tank Pressure Change

Tank Pressure Change

Tank Pressure Change

(OP)
Is the process in a gas baldder accumulator isotropic, so that P1*V1=P2*V2?

RE: Tank Pressure Change

By "isotropic", you presumably mean "Isentropic", meaning no heat transfer across the boundary.

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

(OP)
Whoops! That's what I meant: Isothermal.  If the pressure changes a few psi over the course of a second or two, would that be sufficient enough to assume isothermal activity? I'd assume so, myself, but I want to be sure.  Note that this change would be no more than about 15% of the initial pressure at any time.

RE: Tank Pressure Change

Say you have a vessel containing 10 litres of air at a pressure of 2 bar gauge, and its pressure increases by about 15% over a couple of seconds.  Now I would say that the process is almost certainly not isothermal but, on the other hand, neither is it adiabatic.  I think I'll just sit on the fence for that one!  However, in the real world, it makes very little difference (i.e. no difference) to me as a practising engineer since the difference is so little.

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

(OP)
Hmmm... good point :)

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

For a simulated isothermal condition to take place, the change in pressure  of a few psi in the course of one or two second should not be considered isothermal because heat transfer would not have had time to keep gas at constant temperature; in thirty minutes I would say yes.

RE: Tank Pressure Change

Can you help with a variation of the same problem? I recently have been working with an air charged piston type accumulator. I am trying to understand how volume/pressure relationship is affected by differing air qualities on the air side; for instance, in the ideal case where the air charge is instrument quality and dry (-40 degree dew point), how will accumulator performance degrade if air is saturated with water vapor.
I assume my conditions to be adiabatic.
Thanks in advance for your advice.
Tom

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