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Creating a temperature differential on SF6 to evacuate a cylinder

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Aug12-14, 05:21 PM
P: 8
I have a 50KG cylinder with 1-2LB of SF6 at 15PSI. I want to transfer it into another cylinder by temperature differential.

Keeping it simple. I try to model it with two cylinders that has a valve on each cylinder. Connecting the cylinder with a 1/4" hose. Both valves are closed. The cylinder with the SF6 is at ambient temperature (25C). I cool the other cylinder to -40C (~-50C is solidifying point).. Open both valves.

How would I go about calculating the flow rate?

I try to find the pressure differential using ideal gas law

PV=nRT; P1/T1 = P2/T2, 15 PSI / 298.15K = P2/233.15K; P2 = 11.73PSI.

This is the wrong way since I know it is a compressible gas and the pressure should be much lower. Any help or direction would greatly be appreciated.

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Aug12-14, 06:46 PM
P: 12,113
Weird unit system :(.

PV=nRT takes compression into account, but you won't get a pressure difference if your valves are open. You just get a density difference, while the total amount of SF6 is constant (and known).

Your calculation would require the same amount of SF6 in both cylinders and closed valves.

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