Sizing a Gas Receiver

The requirement for the use of atmospheric pressure is important. It is also wise to remember that Pa will DRAMATICALLY change the length of time to empty a cylinder. Should the atmoshere surrounding the cylinder be much higher than the internal pressure of the cylinder (as in the bottom of the ocean) it will take a long time to empty owing to the surrounding pressure. Converse, on a mountain where the atmospheric pressure is lower than the pressure inside of the cylinder, it will take a lot less time to empty the cylinder. Thus, Pa needs to be considered when doing the calculation and the mere assumption that Pa can be assumed to be P2 needs to be approched with caution. By the way, what happens if you only empty 50% of the cylinder? Pa is then not equal to P2 and your theory goes out of the window.

Always approach a problem from the basis of first principles and you will be able to reason it out.

Hope this helps

Rod

Hello Rod

I don't think that's right. The question is based on a given, constant, value for C, and in that case, the external pressure makes no difference. The external pressure does of course affect whether any particular flow C can be achieved in the first place. Clearly the internal pressure must be higher than external to get any flow.

The answer to Rathinavelu's question is No!

P2 and Pa are different. P2 is the final pressure. Pa is the standard pressure at which C is evaluated. C is (or should be) flow at specified conditions of temperature and pressure. Common std temps are 0, 15 and 20°C. Std pressure is usually 1 atmosphere, but if you took it as 1 bar the difference is small.

Ignoring the gas density (kg per std m³), which cancels, need the Pa on the bottom to get kg on the top and kg/s on the bottom, to give answer in seconds.

Cheers..........Codey

Thank u !

Actually the reason for my question is, i want to size the gas reciever by assuming that the upstream of that reciever is upset but the gas is continuously withdrawn from the reciever at the normal flow rate.

For this reciever storage pressure is to be high enough to maintain pressure as well as flow. when i was calculating, i just think that Pa stands for minimum pressure(assuming that air reciever depressurised to atmosphere).

now i got the answer, thanks a lot for valuable answers of both guests.

Again it's me!

Actually, "C" flow rate is in SCFM, then why should the formula contain Pa to represent standard pressure again ?

Still it is confusing. The answer may be right( Pa = atm pressure) but why it is required is not cleared yet to me.

Thanks

Rathinavelu

Hello Rathinavelu

If gas density = ρ lb/SCF, mass of gas leaving the vessel to give pressure fall from P1 to P2 = V.ρ.(P1-P2) lb. Rate of outflow = C.ρ.Pa lb/min (Pa being standard pressure). So time taken = V.ρ.(P1-P2)/(C.ρ.Pa) = V.(P1-P2)/(C.Pa)

Cheers......Codey