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Receiver Pressure Equalizations Calculations

  1. Jun 26, 2009 #1
    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 dont know how to solve it with n, receiver 1 in expansion and receiver 2 in compression,

    Thanks, Dan
     
  2. jcsd
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