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vapl153
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I have a question about pressure equalization. I have 2 receiver thermically insulated (adiabatic) and connected via a valve, one receiver 10 cu/ft at 100 psig and 350 F, the other receiver is 1 cu/ft at 0 psig and 60 F. using n=1.406, what would be the equalizing pressure and temperature when I open the valve ? (the valve and piping not to be considered in the calculation).
Is this ok ?
P1= 100 psig = 114.7 psia
V1= 10 cu/ft
T2 350 F = 810 abs
P2= 0 psig = 14.7 psia
V2= 1 cu/ft
T2= 60 F = 520 abs
The combining result
P3= ?
V3= V1 + V2
T3= ?
PV= P1V1 + P2V2 = 1147 + 14.7
PV= 1161.7
PV/T= P1V1/T1 + p2v2/t2 = 1.416 + 0.02826
PV/T= 1.4443
PV/T = P3V3/T3
P3 = PV / V3 = 1.4443/11 cu/ft
P3 = 105.6 psia = 90.91 psig
T3 = (PV/T) / PV = 1.4443 / 1161.7
T3 = 804.32 abs = 344.32 F
I don't know how to solve it with n, receiver 1 in expansion and receiver 2 in compression,
Thanks, Dan
Is this ok ?
P1= 100 psig = 114.7 psia
V1= 10 cu/ft
T2 350 F = 810 abs
P2= 0 psig = 14.7 psia
V2= 1 cu/ft
T2= 60 F = 520 abs
The combining result
P3= ?
V3= V1 + V2
T3= ?
PV= P1V1 + P2V2 = 1147 + 14.7
PV= 1161.7
PV/T= P1V1/T1 + p2v2/t2 = 1.416 + 0.02826
PV/T= 1.4443
PV/T = P3V3/T3
P3 = PV / V3 = 1.4443/11 cu/ft
P3 = 105.6 psia = 90.91 psig
T3 = (PV/T) / PV = 1.4443 / 1161.7
T3 = 804.32 abs = 344.32 F
I don't know how to solve it with n, receiver 1 in expansion and receiver 2 in compression,
Thanks, Dan