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Decrease in pressure due to adiabatic expansion derivation?

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]\frac{dp}{p}[/itex] =[itex]\frac{\gamma}{\gamma-1}[/itex][itex]\frac{dT}{T}[/itex] if the decrease in pressure is due to an adiabatic expansion.


    2. Relevant equations
    Poisson equations:
    Pv[itex]^{\gamma}[/itex]
    Tv[itex]^{\gamma - 1}[/itex]

    Ideal Gas Law:
    Pv=R[itex]_{d}[/itex]T, where R[itex]_{d}[/itex] is the dry air gas constant.

    Hydrostatic Equation:
    [itex]\frac{dp}{dz}[/itex] = - ρg

    3. The attempt at a solution
    I tried making those two equations equal to each other since they are equivalent (i think) and differentiating on both sides but i ended up with [itex]\frac{\gamma - 1}{\gamma}[/itex] in the final result...

    EDIT: I was looking up adiabatic atmosphere and found this. I'm wondering where the formula p[itex]^{\gamma - 1}[/itex]T[itex]^{\gamma}[/itex] = constant comes from...
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2

    ehild

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    How is γ related to Cp and Cv?

    You can consider air as ideal gas. Use the Ideal Gas Law to eliminate V.


    ehild
     
    Last edited: Oct 24, 2013
  4. Oct 24, 2013 #3
    I think [itex]\gamma[/itex] = Cp/Cv and R = Cp - Cv.
    I used an entropy equation and made it equal to 0:
    0 = Cp*ln(T2/T1) - R*ln(p2/p1) and got (R/Cp)*ln(p2/p1) = ln(T2/T1). So I solved for R/Cv = [itex]\gamma[/itex] -1 / [itex]\gamma[/itex]... and the final result was
    (P2/P1)^(([itex]\gamma[/itex] -1) / [itex]\gamma[/itex]) = T2/T1

    Am I on the right track with this?
     
  5. Oct 24, 2013 #4

    ehild

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    You are on the right track, but write the equation with the differentials.

    ehild
     
  6. Oct 24, 2013 #5
    This is where I'm stuck... could i say P2/P1 = P and T2/T1 = T, then take the ln of the equation to bring the exponent down?
     
    Last edited: Oct 25, 2013
  7. Oct 25, 2013 #6

    ehild

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    Assume that P1=Po and T1=To are fixed, and P2=Po+dP, T2=To+dT, dT and dP very small. How are dP and dT related? Have you studied differentials?


    ehild
     
  8. Oct 26, 2013 #7
    Is dt and dp directly proportional?

    And I have calculus background from a few years ago so I'm a bit rusty on some concepts... like differentials.
     
  9. Oct 26, 2013 #8

    ehild

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    Try to refresh your Calculus knowledge.

    You have the equation (P2/P1)^((γ -1) / γ) = T2/T1.

    Write P2/P1=(Po+dP)/Po=1+dP/Po and T2/T1=1+dT/To. dP/Po << and dT/To <<1. The original equation becomes
    (1+dP/Po)(γ-1)/γ=1+dT/To.
    Use the approximation (1+δ)n ≈1+nδ,valid for δ<<1.
    You can check it, what is (1+0.001)3, for example?

    What do you get?


    ehild
     
  10. Oct 29, 2013 #9
    You would get 1.003003 so its approximately 1. Curious question, why does this approximation matter to the proof or derivation of the original question?
     
  11. Oct 29, 2013 #10

    ehild

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    You get the desired formula with that approximation.


    ehild
     
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