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Air compressor quesion

  1. Jul 17, 2017 #1
    1. The problem statement, all variables and given/known data

    Air is to be compressed from a pressure of 1 bar and a temperature of
    20°C to a pressure of 15 bar in a two stage compressor with intercooling.
    After intercooling the air temperature returns to 20°C and the polytropic
    index of compression is 1.28. If the input power per stage of compression
    is 2 kW, calculate:

    The volumetric flow rate of free air (at 1 bar and 20°C) in m3 s–1.

    2. Relevant equations
    PV = mRspecificT
    V = (mRspecificT)/P

    3. The attempt at a solution

    V = (1 x 287 x (20+273))/1x105

    V = 84091/1*105

    V = 0.841

    This attempt is using all the information I have. But I don't think this is the correct solution as the units don't balance to leave m3 s-1 (I think).

    Any direction?
     
  2. jcsd
  3. Jul 17, 2017 #2

    scottdave

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    In the formula that you use V is volume (meters3). This is what you calculated. Somehow, you need to incorporate the power (2 kW) which is a rate of energy per time. Also, you should use the fact that you are compressing the gas to a higher pressure (which takes energy).
     
  4. Jul 18, 2017 #3
    Thank you Scott for this input.

    I have found an equation in my theory notes which I missed completely when trying to solve this.

    W = n/(n-1) p1V1 [(p2/p1)(n-1/n) - 1]

    Rearranging for V1 gives me an answer of 5.416 x 10-3.

    Subbing in SI base units leaves me with m3 s-1.

    This look better?
     
    Last edited: Jul 18, 2017
  5. Jul 18, 2017 #4

    scottdave

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    What value did you use for W in this formula? Is p1 1bar and p2 is 15bar?
     
  6. Jul 18, 2017 #5
    W = 2000 J s-1, p1 = 1x105Pa & p2 = 15x105Pa

    Subbing into: W = n/(n-1) x p1V1 x [(p2/p1)(n-1/n) - 1]

    2000 = (1.28/0.28) x 105 x V1 x [(15/1)(0.28/1.28) - 1]

    2000 = 4.57 x 105 x V1 x 0.808

    V1 = 2000 / (4.57 x 105 x 0.808)

    V1 = 2000 / 369.26x103

    V1 = 5.416x10-3 m3 s-1


    In the ball park?
     
  7. Jul 18, 2017 #6

    scottdave

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    Oh OK. I had misread your n-1/n as n - (1/n) rather than (n-1)/n
     
  8. Jul 18, 2017 #7
    What you have done so far is mostly incorrect. Do you know what a two stage compressor is.? If so, please define it for us so that we know you understand.
     
  9. Jul 18, 2017 #8
    Hi Chester,

    My understanding of a two stage compressor is to take in air from the atmosphere and outputs this air at a pressure substantially higher.

    This is achieved by intercooling between stages which causes a contraction in volume causing the pressure to rise sharply in the second stage.
     
  10. Jul 18, 2017 #9
    Ah ok, anything you can add about my direction with this question?
     
  11. Jul 18, 2017 #10
    This is not correct. A two stage compressor is actually two compressors in series, with the output from the first compressor first cooled (inter-cooling) and then fed as the input of the second compressor. Were you aware of this?
     
  12. Jul 18, 2017 #11
    I wasn't.

    So from the information given we taking in air at P1 and T1 which will come out the first compressor as P2 and T2 and pass into a inter-cooler? The inter-cooler will bring the temperature down so the air passes into the second compressor as P2 and T1?

    Am I getting it in terms of what is happening to the air?
     
  13. Jul 18, 2017 #12
    Yes.
     
  14. Jul 18, 2017 #13
    So following on from that.

    The air leaves the second compressor as P3 and T3?

    So what we know about the air as it enters the second compressor is Pin = 2kW, T1= 20°C and p2 = 15 bar?
     
  15. Jul 18, 2017 #14
    No. P2 is not 15 bars. P3 is 15 bars. P2 is unknown, and the mass flow rate is unknown. But we do know that the amount of work done in each of the compressors is 2 kW. What does that tell you about the compression ratio in each compressor?
     
  16. Jul 18, 2017 #15
    In first compressor it is (p2/p1) and in the second it is (p3/p2)?
     
  17. Jul 18, 2017 #16
    How are these related, given that the inlet temperatures and the amounts of power are the same?
     
  18. Jul 18, 2017 #17
    (p2/p1) = (p3/p2)?
     
  19. Jul 18, 2017 #18
    Why don't you tentatively assume that is correct and see how it plays out.
     
  20. Jul 18, 2017 #19
    Working out for p2:

    (p2/p1) = (p3/p2)

    p22 = p1 x p3

    p2 = √p1 x p3

    Getting there?
     
  21. Jul 18, 2017 #20
    Yes. How about a number?
     
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