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Homework Help: Thermodynamics Efficiency Of Adiabatic Compressor

  1. Nov 19, 2013 #1
    1. 100kg of methane is compressed from 0.5MPa and 300K to 3.0Mpa and 500K in a closed chamber by a compressor. The compression process is assumed to be adiabatic, calculate the efficiency of this compressor. Specific heat capacity of methane under constant pressure is 35.58 kJ/k.mol.K and ideal gas constant is 8.314 kJ/k.mol.K.
    I have managed to work out the work done by the compressor, and 2 volumes but am not sure how to use them to get the efficiency of the compressor. I believe it has something to do with carnots cycle, but I am not exactly sure. Any help would be much appreciated.

    2. PV=nRT
    Carnot efficiency

    3. W=-nCvdT
    So:Cv=35.58-8.314= 27.27
    n= 100/16 = 6.25 moles
    Therefore: W=-6.25 x 27.27 x (500-300) = -34087.5 kJ
    V1= (6.25 x 8.314 x 300)/(0.5 x 106)= 0.031 m3
    V2= (6.25 x 8.314 x 500)/(3.0 x 106)= 8.67x10-3 m3

    Any help would be much appreciated.

  2. jcsd
  3. Nov 19, 2013 #2

    rude man

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    I'm thinking efficiency = delta U/W. Since Q = 0 that makes the answer kind of obvious! But I don't know what they really had in mind with "efficiency of the compressor" either.
  4. Nov 19, 2013 #3
    How much work would have been done between the initial pressure and the final pressure if the process had been adiabatic and reversible?
  5. Nov 20, 2013 #4

    rude man

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    Good point by Chestermiller!
  6. Nov 20, 2013 #5
    I see what you mean, but even if I did that how would I get to the efficiency of the compressor? I keep finding formulas for efficiency which include Q, but since its adiabatic Q=0.
  7. Nov 20, 2013 #6
    Since Q=0 then W=deltaU so dividing them would just give me 1.
  8. Nov 20, 2013 #7
    In this context, the efficiency is the least possible amount of work that would have been required to compress the gas between the initial and final pressures, divided by the actual amount of work that was required.
  9. Nov 20, 2013 #8
    Sorry but could you please explain just a bit more i'm very confused. The work that I have worked out is the amount of work done right? So what am I dividing or multiplying it by?
  10. Nov 20, 2013 #9
    May i ask way methane? I ask because im working on a project involving methane and wondering if we are working in the same field?
  11. Nov 20, 2013 #10
    Its just a question set at university.
  12. Nov 20, 2013 #11
    Do you know how to calculate the amount of work along a reversible path? You can calculate the efficiency as the reversible work divided by the actual work. Unfortuately, the amount of reversible work is path dependent. I would tend to calculate it along an adiabatic reversible path from the initial pressure to the final pressure. But, you could also go from the initial volume to the final volume adiabatically and reversibly, and then heat it reversibly at constant volume to get to the final temperature and pressure. There is certainly ambiguity about how the efficiency should be calculated.

  13. Nov 23, 2013 #12
    For conveniences sake ill give theoretical values the subscript [th] and actual values [ac]

    As E=mc(delta)T you could look at efficiency as (delta)T[th]/(delta)T[ac] (the two mc terms cancel out).

    You can work out (delta)T[th] by finding T[2][th] which is T[1]*(P[2]/P[1])^((gamma-1)/gamma), where gamma is C[p]/C[v]

    Then (delta)T[th]=T[2][th]-T[1]

    Using your values i got 78%
  14. Nov 23, 2013 #13
    btw taki i find it amazingly coincidental that i also recieved exactly the same question from my thermodynamics lecturer recently as Q1 out of a set of 4.
    Last edited: Nov 23, 2013
  15. Dec 1, 2013 #14
    We were told to work it out using enthalpy. We used deltaU=cv(T)dt and deltaH=cp(T)dt
    Doing this we got 76.6%
    Any thoughts?
  16. Dec 1, 2013 #15
    Can i ask if your question 2 was latent heat, question 3, carnot cycle and question 4 an algebraic answer to heat exchange?
  17. Dec 2, 2013 #16
    It was indeed :) q2 cooling water with ice, q3 carnot and q4 heat transfer but with no given values. btw if your q4 is the same as mine were you able to do it? As its the only one i'm stuck on.
  18. Dec 2, 2013 #17
    What exactly did you do for the whole calculation? I'm a tad confused how you used those equations to get to the answer.
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