This discussion focuses on calculating the drain time of a gas cylinder under pressure to atmospheric pressure, specifically for natural gas. The primary formula referenced is the differential equation dP/dt = -k(P - Patm), where the rate of escape is proportional to the pressure difference. It is established that for high-pressure cylinders, the concept of choked flow applies, making the mass flow rate independent of atmospheric pressure until the internal pressure drops significantly. The conversation emphasizes the importance of understanding compressible gas dynamics and the conditions under which the flow becomes choked.
Engineers, physicists, and students in mechanical engineering or fluid dynamics who are involved in gas flow calculations and pressure management in high-pressure systems.
Are you to assume that the cylinder is insulated? Kept at constant temperature?Jeirn said: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.
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-Patm.Jeirn said: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 said:This doesn't look like homework.
haruspex said: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-Patm.
##\frac{dP}{dt}=-k(P-P_{atm})##, for some constant k (which will depend on the nozzle).
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.boneh3ad said: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 said: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.