Compressed Air Receiver

[steven17]

I have a 1000 gallon air receiver charged at 150 psi.
If I lose 100 gallon air due to consumption, what will be the resulting air pressure in the tank?
I want to know the pressure drop relationship between fixed receiver tank size with initial pressure and resulting pressure due to air withdrawal.

 

[zdas04]

Last I heard PV=mR_gasT

Temperature, R_gas, and volume are constant so P₁/m₁=P₂/m₂

Pretty easy if you knew if the "100 gallon" (13.37 ft³) of air you "lost" was stated at standard conditions or at actual conditions. If it is standard then the problem is trivial (just figure out how many SCF is in the 1000 gallon tank at 150 psig and the lost volume is 10%). If it is actual then you need to iterate since some of the lost volume was lost at each pressure between 150 psig and final pressure.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[steven17]

I still don't get how I can calculate the resulting pressure from the formula of P1/m1=P2/m2. It seems that the calculation is not as straightforward as I thought.

 

[zdas04]

If the lost volume is SCF, then m₂=0.9*m₁ since standard conditions are a reasonable surrogate for mass. If you can't do that substitution then people will start questioning your ability to be here.

If the lost 100 gallons was volume at actual conditions then the problem is impossibly difficult and the best you'll be able to do is to iterate until you get a close enough approximation.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[steven17]

Assuming SCF, the P2 would be P1x0.9. That looks pretty simple.
Thanks.

 

[sailoday28]

You will have to get a handle on the mass removed.
If the air withdrawal is relativey fast, you can use an isentropic process. Is slow, use isothermal.
Regards
Sailoday28

 

[zdas04]

This is a guy that struggled with algebra. I don't think he's going to warm to the difference between isentropic and isothermal. No time component to the "loss" so I'd probably assume isothermal, but the initial charge was from a (probably 2 stage) compressor so much of the loss could have come from cooling the receiver down. I'm thinking this is a sophomore physics class and we're never going to get enough information to turn this into an interesting discussion.

In real life you would never know that you had "lost 100 gallons" of air. You would have a new pressure and calculate the new mass.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"