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Homework Help: Propane Tank problem

  1. Apr 22, 2005 #1
    I've been having some trouble with the following Question.
    { A rigid tank contains 90% by volume propane at -40 F. The tank is then heated to 80 F. A pressure relief valve in the tank is set at 125 psi. Will any propane be vented. If yes, how much? Tables for propane at http://www.elyenergy.com/pdf/CO37.pdf }

    Please let me know if I'm doing this completely wrong.

    I was told that 90% by volume meant that 90% of the propane was liquid.

    I got v1 by using [tex]v_1=.9(.02763)+.1(3.13)[/tex]

    I was thinking that there were 3 states.

    The first process would be const. volume, up to 125 psi

    State 1 -- isochoric (v = const.) --> State 2

    state 1
    T1 =-40 F
    v1=.337 ft^3/lbm
    P1 = 16.2 psi
    state 2
    T2 = 70.54 F
    v2 = .337 ft^3/lbm
    P2 = 125 psi

    then once it reaches 125 psi, I thought that it would be an isobaric, const. pressure, process until it reached 80 F.
    State 2 -- isobaric --> State 3

    State 3
    T3 = 80 F
    v3 = ??
    P3 = 125 psi

    I was thinking if I could find v3, I could use it to find the amount (lb_mass) of propane that was vented. But I'm having some trouble with finding v3, not really sure how to solve for it. I'm pretty sure its in the super heated region though.
    Any help would be great, thanks.
    Last edited: Apr 22, 2005
  2. jcsd
  3. Apr 23, 2005 #2


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    Hi nineeyes, now I see why you wanted the specific volume over in the mech eng forum.

    Why multiply percentage volume by specific volume? The units will be ft^6/lbm. That doesn't make sense.

    What you said about state 1 and 2:
    State 1 -- isochoric (v = const.) --> State 2
    is true assuming you don't loose any product because of the relief valve lifting. In fact it will be true to state 3, so long as the relief valve doesn't lift, but we don't know if it will lift or not. Note also that the 125 psi set pressure for a relief valve is typically refering to guage pressure, not absolute pressure.

    You can simplify the problem quite a bit because you are essentially given enough information to determine the specific volume in the beginning, and at the end. The problem is figuring out what these specific volumes are. Also, specific volume is just the inverse of density, and I'd prefer to use density.

    The density (or specific volume) will stay constant unless the relief valve lifts, so let's look at what the density is at state 1, and the density at state 3. The difference can be multiplied by the volume of the tank to give you how much vented:

    State 1: .9 (ft3) * 36.19 (lbm/ft3) + .1 (ft3) * .163 (lbm/ft3) = 32.587 (lbm/ft3)

    State 3, superheated density at 125 psig (139.7 psia) and 80 F = 1.293 (lbm/ft3)

    Mass vented is (32.587 - 1.293) * Volume

    You're right that you need the state of the fluid at 80 F and 125 psig, and that really is difficult without some database.
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