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Algebra problem involving work using specific heat ratio

  1. Mar 9, 2017 #1
    NO TEMPLATE BECAUSE SUBMITTED TO NON-HOMEWORK FORUM

    I'm stuck on substituting the following (where gamma is the specific heat ratio):
    W=(P1v1 - P2v2) / (gamma-1)
    P1v1^gamma = P2v2^gamma

    substituting for v2.... <= this is where I get stuck...
    W = [(P1v1) / (gamma-1)] * [(P2 / P1)^((gamma-1)/gamma) -1]

    How did it get from the bold to the final equation?
     
    Last edited by a moderator: Mar 10, 2017
  2. jcsd
  3. Mar 9, 2017 #2
    Are you saying that you haven't been able to figure out how to do the algebra?
     
  4. Mar 9, 2017 #3
    Yes
     
  5. Mar 10, 2017 #4
    OK. I'm going to move this to the Precalculus Math forum. I'm assuming that this is a homework problem. If so, in the future, please submit homework problems to one of the Homework forums, along with using the required template.
     
  6. Mar 10, 2017 #5

    Mark44

    Staff: Mentor

    In your second equation in bold, is it ##P_1v_1^{\gamma} = P_2v_2^{\gamma}## (which is what you wrote) or did you mean ##(P_1v_1)^{\gamma} = (P_2v_2)^{\gamma}##? I'm not familiar enough with your equations to be sure that what you wrote is what you meant.
     
  7. Mar 10, 2017 #6
    Can you please tell what is to be proved and what is given ? I can't figure out from our post.
     
  8. Mar 10, 2017 #7
    It is correct as he has it written.
     
  9. Mar 10, 2017 #8
    The first step is to factor out ##P_1V_1## from the term in parenthesis in the numerator.
     
  10. Mar 10, 2017 #9
    It's from Coulson & Richardson Volume 4 problem 2.1... i'm confused with the algebra:
    W = (P1*v1 - P2*v2) / (gamma-1)
    P1*v1^gamma = P2*v2^gamma

    upload_2017-3-10_17-39-51.png
     
  11. Mar 10, 2017 #10

    haruspex

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    ##\left(\frac{v_2}{v_1}\right)^{\gamma}=\frac{P_1}{P_2}##
    ##\frac{v_2}{v_1}=\left(\frac{P_2}{P_1}\right)^{-\frac 1{\gamma}}##
     
  12. Mar 13, 2017 #11
    Thanks
     
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