How to Calculate Gas Cylinder Drain Time from Pressure to Atmospheric?

[Jeirn]

Do you know where I can find a formula to calculate how long it would take to drain a cylinder under pressure to atmospheric pressure? if I know the volume of the cylinder and the size of the opening.

 

[haruspex]

Do you know where I can find a formula to calculate how long it would take to drain a cylinder under pressure to atmospheric pressure? if I know the volume of the cylinder and the size of the opening.

Are you to assume that the cylinder is insulated? Kept at constant temperature?
In principle it will take infinitely long, so you need to set a threshold at which you will regard the pressures as effectively equal.

 

[Jeirn]

Its a cylinder of natural gas...can you assume values and give an example of how to find time it takes for gas to fully escape through a hole in the cylinder. I know it's some kind of differential equation.

 

[berkeman]

This doesn't look like homework. Should we move the thread to the ME or General Physics forum?

Update -- Moved to the ME forum.

 

[haruspex]

Its a cylinder of natural gas...can you assume values and give an example of how to find time it takes for gas to fully escape through a hole in the cylinder. I know it's some kind of differential equation.

If we assume constant temperature, PV=nRT, with V, R and T constant. The rate of escape will be proportional to the pressure difference, P-P_atm.
$\frac{dP}{dt}=-k(P-P_{atm})$, for some constant k (which will depend on the nozzle).

If the escape is rapid, the temperature will drop. This will cause the pressure to fall faster and slow the rate of loss. Let me know if you think you need that equation. It will take a bit more work.

 

[boneh3ad]

This doesn't look like homework.

It very well might be homework. I give my compressible flow classes essentially this problem as homework.

If we assume constant temperature, PV=nRT, with V, R and T constant. The rate of escape will be proportional to the pressure difference, P-P_atm.
$\frac{dP}{dt}=-k(P-P_{atm})$, for some constant k (which will depend on the nozzle).

Actually, this is inaccurate. For a high-pressure cylinder such as a gas bottle, the pressure ratio is such that the flow would be choked and the atmospheric pressure is irrelevant. The rate of mass leaving such a bottle will depend only on the current pressure inside the bottle, the temperature inside the bottle, and the size of the opening. The ambient pressure only becomes relevant once the pressure drops to a low enough value that the flow is no longer choked.

 

[haruspex]

Actually, this is inaccurate. For a high-pressure cylinder such as a gas bottle, the pressure ratio is such that the flow would be choked and the atmospheric pressure is irrelevant. The rate of mass leaving such a bottle will depend only on the current pressure inside the bottle, the temperature inside the bottle, and the size of the opening. The ambient pressure only becomes relevant once the pressure drops to a low enough value that the flow is no longer choked.

All you are saying there is that initially the atmospheric pressure is so low compared to the pressure in the bottle that it can be ignored. So at that stage the equation I wrote is more accurate than what you propose.
At first, the additional accuracy is unnecessary, but since the question asks for the time to become effectively equalised, the more accurate form is required later. And since the one equation handles the whole domain, there is no benefit in handling the initial stage separately.

 

[berkeman]

It very well might be homework. I give my compressible flow classes essentially this problem as homework.

Good point. Let's hold off on any more replies until @Jeirn returns to clarify the source of his question... Thanks.

 

[boneh3ad]

All you are saying there is that initially the atmospheric pressure is so low compared to the pressure in the bottle that it can be ignored. So at that stage the equation I wrote is more accurate than what you propose.
At first, the additional accuracy is unnecessary, but since the question asks for the time to become effectively equalised, the more accurate form is required later. And since the one equation handles the whole domain, there is no benefit in handling the initial stage separately.

Actually no, that's not what I'm saying. I'm saying that, above a certain pressure ratio, the mass flow rate literally does not depend on downstream conditions. It's a concept called choked flow and is a fundamental principle of compressible gas dynamics.

A gas cylinder is typically pressurized to 3000 psia, and atmosphere is 14.7 psia. For air (as an example), the flow will remain choked the entire time between 3000 psia and about 28 psia. I must reiterate here that in a choked flow, the mass flow rate is physically and mathematically independent of downstream conditions; it isn't just an order of magnitude analysis.

I'd be happy to show you mathematically why this is, though I fear it would be counterproductive if this thread is actually a homework problem since it would be tantamount to giving OP the answer.

 

[berkeman]

Somebody is not reading my replieeesss..

Thread locked until the OP sends me a PM with context.