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Adiabatic compressed air and energy calculations

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data
    . Consider a pump that is required to compress air in a factory. The cylinder in the pump has an inner diameter of 2.00 cm and length 60.0 cm. Air is drawn into the pump at atmospheric pressure and 18 degrees celcius and the pump adiabatically compresses the air to a pressure of 17 atmospheres.

    (a) Calculate the volume and temperature of the compressed air.

    (b) What is the increase in internal energy of the gas during the compression?

    (c) Suppose that the air in the pump is compressed once and the compressed air and the steel of the cylinder on the pump the come into thermal equilibrium. If the cylinder is 1.00 mm thick, what will be the increase in temperature of the steel after one compression?

    How should i go about solving this question? I don't fully understand thermodynamics. I would love it if you just gave me a worked example. I find it easier to understand since my English is not very good.


    2. Relevant equations
    P1V1gamma =P2v2gamma


    3. The attempt at a solution
    I calculated the volume to be 3.16 me but that's wrong according to the answer and I'm quite stuck on how to calculate energy
     
  2. jcsd
  3. Jun 10, 2017 #2

    haruspex

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    I'm trying to see your working so that I can find your mistake, but my crystal ball doesn't seem to have the necessary range.
     
  4. Jun 10, 2017 #3
    I am sorry?.
     
  5. Jun 10, 2017 #4
    Do you know how to solve this problem?
     
  6. Jun 10, 2017 #5

    haruspex

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    You wrote that you got the wrong answer for the volume. My task, as a homework helper, is to find where you went wrong. But if you do not post your working I cannot possibly find your error.
    Please post your working (as typed text, not as an image).
     
  7. Jun 10, 2017 #6
    I'm sorry. Here is my working.
    I used PiVi^gamma = PiVi^gamma.
    In this case gamma = 7/5 as air is a diatomic molecule therefore Cp/Cv =7/2R divided by 5/2R = 1.4
    Vi = Pi*(0.01)^2* 0.6 = 1.885x10(-4) m^3
    I reformulated it to be
    Vf = Vi*(Pi/Pf)^1.4
    Vf 3.57x10(-6) m^3
     
  8. Jun 10, 2017 #7

    haruspex

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    Risk of confusion between π and Pi. Use the buttons above the text area, X2 for subscript, Σ for special characters.
    There's your mistake. Try that reformulation again.
     
  9. Jun 10, 2017 #8
    I calculated the temperature using TiVi^(gamma -1) = TfVf^(gamma -1)
    Basically it's the same method as before.
    Tf = Ti*(Vi/Vf)^2/5
    Tf = (18 +273.15)(1.885x10(-4)/3.57x10(-6))^2/5
    Tf = 291.15(4.887)
    Tf = 1422.89 K
    This doesn't seem right. I think it's because of turning 18 Degrees to Kelvin?
    If i didn't it becomes 52.8 K which still doesn't sound right
     
  10. Jun 10, 2017 #9
    Vi = Pi*Vi / Pgamma
     
  11. Jun 10, 2017 #10
    Vf= Pi*Vigamma/Pfgamma
     
  12. Jun 10, 2017 #11
    I am sorry. They are all wrong you cannot swap the position of the power therefore
    Vf = square root by gamma(Pi*Vi/Pf)
    This gives me an answer 2.88x10-4
     
  13. Jun 10, 2017 #12

    haruspex

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    Still no.
    Starting with PfVfγ=PiViγ, take it one step at a time and post all the steps.
     
  14. Jun 10, 2017 #13
    Vfγ = Pi*Viγ/Pi
    Vf = square root by γ (Pi*Viγ/Pi)
     
  15. Jun 10, 2017 #14

    haruspex

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    Right, but "square root" is the wrong terminology. You could write it as Vf=Vi(Pf/Pi)1/γ, or in LaTeX, ##V_f=V_i\left(\frac{P_i}{P_f}\right)^{\frac 1{\gamma}}##.
    What numerical answer do you get now? If you still get 2.88x10-4, please post your working.
     
  16. Jun 10, 2017 #15
    Vf = 2.49 x10(-4) m^3
     
  17. Jun 10, 2017 #16

    haruspex

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    Hmmm... I get the same except 10-5, not 10-4.
     
  18. Jun 10, 2017 #17
    oh yes sorry that is a typo. I get -5 also.
     
  19. Jun 10, 2017 #18
    With this i calculated the energy to be 654 K. Which i think is right.
    I calculated this using Tf = Ti(1/17)2/5
     
  20. Jun 10, 2017 #19

    haruspex

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    I agree with the number but not the algebra. I think you would have used Tf = Ti(17)2/7.
     
  21. Jun 10, 2017 #20
    I'm sorry I do not understand how you derived it to that from:
    TiViγ-1 = TfVfγ-1
     
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