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Thermodynamics Entropy +2nd law problem

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Propane at 350 degrees Celsius and 600cm^3 / mol is expanded in a turbine. The exhaust is atmospheric. What is the lowest possible exhaust temperature? How much work is obtained? You may assume ideal gas behavior and the heat transfer to the surroundings is negligible.


    2. Relevant equations



    3. The attempt at a solution

    I haven't gotten too far but I would like someone to confirm what I have done so far is correct, I went straight to finding the work done which by using a form of Bernoulli's is W/n = [tex]\int vdP[/tex] to find the first pressure I did Ideal Gas Law P1=(350+273)*n*(8.314)/(600cm^3/mol) to get a specific Pressure(the problem didn't give any mole information or mass) Then I did the integral and found Work/mol but I am getting an extremely large number for the initial pressure (8.633*10^6 Pa). I'm a little iffy on if I can actually use a specific pressure here can anyone help please
     
  2. jcsd
  3. Mar 5, 2009 #2
    you can find propane mole mass: it's CH3-CH2-CH3, so it's 3×12 + 8×1 = 44 kg/kmol.
    i get the same pressure as you did. it's not that high pressure for a turbine, i guess. note that pressure cannot be specific, it doesn't depend on mass or mole mass.
    since heat transfer to the surroundings is negligible, you should use equations for isentropic processes. search wikipedia for isentropic or adiabatic or so, there must be equations like (p1/p2) = (T1/T2)^(sth).
    to calculate work, you don't have to start with that integral. just use the equations already derived from that integral.
     
  4. Mar 5, 2009 #3
    thank you very much I got it.
     
  5. Mar 5, 2009 #4
    although pressure cannot be specific is this calculation of pressure correcT? what I mean is will the pressure calculated here work for the rest of the problem?

    also how do I know this is an isentropic process
     
    Last edited: Mar 5, 2009
  6. Mar 22, 2009 #5
    sure it'll work, juse keep in mind results will be in units of Joule/kg or Joule/mole or m3/kg etc.

    this is isentropic because there is no heat transfer to the surroundings.
     
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