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Thermodynamics compression problem

  1. Oct 18, 2007 #1

    T-7

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    1. The problem statement, all variables and given/known data

    "A compressor takes air at 300K, 1 atm. It delivers compressed air at 2 atm, using 200W of power.

    If the process is adiabatic and reversible, what is the rate that air is delivered, and what is the final temperature of the air? [k of air = 1.40]"

    3. The attempt at a solution

    Hi. I'd like to bounce off someone these expressions I derive, to see if they're sensible.

    Since it's adiabatic, I infer that

    [tex]PV^{K} = const[/tex]

    Using the ideal gas law, and rearranging a bit, I make it that

    [tex]T_{2} = \frac{P_{1}^{1/K-1}}{P_{2}^{1/K-1}}T_{1}[/tex]

    I think that's sensible (?). It comes out as 366K.

    For the last bit, I ended up deriving the expression

    [tex]W = nR\frac{T_{1}-T_{2}}{1-1/K}[/tex]

    which to me just looks too simple (it comes out as -1920J)

    I started with the usual

    [tex]W = \int^{V_{1}}_{V_{2}}p dV[/tex]

    replacing dV with

    [tex]-nRT\frac{dP}{P^{2}}[/tex]

    and the T with

    [tex]bP^{1-1/K}[/tex]

    (b is a constant)

    to obtain, in the end, the integral

    [tex]W = -nRb \int^{P_{2}}_{P_{1}} P^{-1/K} dP[/tex]

    On integrating, and then substituting for the b's in terms of T1 and P1 or T2 and P2, I found they disappeared, leaving me with

    [tex]W = nR\frac{T_{1}-T_{2}}{1-1/K}[/tex]

    But I'm not sure that's right (?). How does it look?
     
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 19, 2007 #2

    dynamicsolo

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    Homework Helper

    I concur.

    As I say in your other thread, I get

    [tex]W = nR\frac{T_{1}-T_{2}}{1 - K}[/tex] .


    So I find dn/dt = [ (dW/Dt)(1 - K) ] / [ R · (T1 - T2) ] .

    I'm pretty sure the factor can't be [ 1 - (1/K) ]: since K > 1, this would be positive when it should be negative.
     
  4. Oct 21, 2007 #3

    T-7

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    Thanks for your reply!

    My W does come out negative, as T2 > T1. But it is a somewhat larger value of W than yours, of course. At the moment, I can't see where I've gone wrong. I can derive your result, and mine. I'll keep trying, I guess!
     
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