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Thermal Efficiency of a Heat Engine based on pV-diagram

  1. Jul 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Here's a link to the pV-diagram I am using - http://session.masteringphysics.com/problemAsset/1396784/2/p16.7.jpg
    A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K, and the ideal gas constant is R = 8.314 J/(mol·K). The thermal efficiency of the engine is closest to
    Answers)
    A - 0.026
    B - 0.53
    C - 0.40
    D - 0.26<--- This is the right answer, but I'm not sure how to get to it
    E - 0.33

    2. Relevant equations

    [tex] ΔU = Q - W [/tex]
    [tex] ΔU = nC_vΔT [/tex]
    [tex] Work = ∫PdV [/tex]

    [tex] Isothermal Work = nRT*ln(V_f/V_i) [/tex]

    3. The attempt at a solution

    I'm getting really lost on differentiating the positive/negative signs and thus finding Qin and Qout. I understand that the efficiency of a thermal engine is [tex] e = W_out/Q_H = 1 - Q_c/Q_h [/tex]

    I started finding some of the areas, but I am not sure where to subtract and add them to find Q_c and Q_h.

    Isothermal Work = nRTln(.010-.16) = 37804. J
    ΔU = 0, therefore W=Q --> So since W = 37804 J, there is 37804 J of heat added to the gas here?

    Isobaric Work = p(ΔV) = p(.16-.01)

    To solve for the pressure I re-arranged PV=nRT. First I solved for the temperature at A.

    [tex] V_A/T_A = V_C/T_C [/tex]

    This gave me the temperature at A as 51.25 Kelvin.

    Using PV=nRT.. [tex] P_A = (2.0mol)(8.314J/mol-K)(51.25 K)/(.010m^3) [/tex]
    This gave me the pressure P_A = 85218.5 Pa.

    Then the work under the Isobaric Line is p(ΔV) so I plugged in the pressure of A..
    [tex] (85218.5 Pa)(.010 m^3-.16 m^3) = -12782.775 J [/tex]

    For the isobaric line I also said that,
    [tex] ΔU = nC_vΔT = (2.0mol)(37J/mol-K)(820K - 51.25 K) = 56887.5 J [/tex]

    I think I'm close to the answer, but I'm not sure which numbers to use for the efficiency.
     
    Last edited: Jul 30, 2014
  2. jcsd
  3. Aug 2, 2014 #2

    rude man

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    How about this:
    1. net work done by system W = area inside the p-V curve
    2. compute Ta, pa and pb. This yields W.
    3. compute ∫δQ for each of the three segments = Qab + Qbc + Qca.
    Efficiency = work/heat in. 'Heat in' is the sum of the positive Qij segments.
     
  4. Aug 2, 2014 #3

    Andrew Mason

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    This cycle is not reversible, by the way.

    You seem to be doing everything correctly.

    Use W = Qh - Qc. Since Qh = Qab + Qbc

    The temperature at a = TcVa/Vc = 820/16 = 51K, so ΔT = 820-51 = 769K

    Like you I get: Qab = nCvΔT = 56.9 KJ and Qbc = nRTln(16) = 37.8 KJ, so Qh = 94.7 KJ

    So the only tricky part is determining Qc = heat flow from c-a. Since it is constant pressure, you can use Cp = Cv + R and: Qc = nCpΔT and then use W = Qh-Qc.

    OR you can do as you were doing but be careful to subtract the area under ba from the area under bc.

    I get 69.7 KJ for Qc.

    AM
     
  5. Aug 3, 2014 #4

    rude man

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    Why do you say that?

    If you compute ∑ ΔS around the loop assuming reversibility you get zero as must of course be. So reversibility seems satisfied.
     
    Last edited: Aug 3, 2014
  6. Aug 3, 2014 #5

    Andrew Mason

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    The problem is with the constant volume and constant pressure heat flows (c-a and b-c). These cannot be made reversible if you are operating between two reservoirs with a finite temperature difference (which is in this case 769K). If it was reversible, the efficiency would be 1-Tc/Th = 1-51/820 = 94%

    AM
     
  7. Aug 3, 2014 #6

    rude man

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    No one said anything about two reservoirs. An infinte number of reservoirs can be postulated along any of the three segments, making all segments reversible.
     
  8. Aug 3, 2014 #7

    rude man

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    The efficiency is work out divided by heat in = W/(Qab + Qbc)
    where
    Qab = C_V(Tb - Ta)
    Qbc = nRTb ln(Vc/Vb)
    W = nRTb{ln(Vc/Vb) - (Vc - Vb)/Vc}
    and efficiency e = W/(Qab + Qbc). All calculated on the assumption of reversibility.

    Ta is per your derivation. I did not use numbers, leaving that to the OP. He/she seems to have lost interest but might still pop up later.

    It is not necessary to compute Qbc which of course is heat removed and so is negative.
     
  9. Aug 4, 2014 #8

    Andrew Mason

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    To make it reversible you would need an infinite number of hot and cold reservoirs each of whose temperatures are separated by infinitesimal temperature differences making up the temperature difference from 51K to 820K so that the heat flows occur at infinitesimal temperature differences between system and surroundings from c-a and b-c. I am not sure how you would even begin to set that up.
    My point was that if one calculates Qc one does not have to determine W from the graph: it is always the case that W = |Qh|-|Qc|.

    AM
     
    Last edited: Aug 4, 2014
  10. Aug 4, 2014 #9

    rude man

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    That nonetheless is the basis of thermodynamic analysis - approximating actual processes with reversible ones. And I believe that is what was intended in the problem.
     
  11. Aug 4, 2014 #10
    Hi guys,

    Probably I'm missing something. Doesn't the problem statement say "through the reversible cycle abca."

    Chet
     
  12. Aug 4, 2014 #11

    rude man

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    You never miss a thing, Chet! :smile:

    If it weren't reversible you couldn't calculate efficiency, do you agree? I mean, integratig δQ over the segments assumes reversibilty via

    δQ = C_V dT + p dV or δQ = C_P dT + V dp, right?
     
  13. Aug 4, 2014 #12

    Andrew Mason

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    In order to determine efficiency, you only need to assume that the work done is ∫PdV. This can be achieved by non-reversible heat flow.

    AM
     
  14. Aug 4, 2014 #13
    If the process were not reversible, the temperature and pressure within the cylinder would not be uniform.

    Chet
     
  15. Aug 4, 2014 #14

    rude man

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    That's what I was thinking - the thermodynamic coordinates would not be defined during the process so δQ could not be computed since Q is not a state function. Sound right?

    .
     
  16. Aug 4, 2014 #15

    Andrew Mason

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    It depends on how fast the changes occur. They can be pretty fast in a gas and still be practically reversible. For example, sound waves are modelled as a series of reversible adiabatic compressions and expansions, and they can occur very quickly.

    AM
     
  17. Aug 5, 2014 #16
    Sure, but in the case of sound waves, the pressure and the temperature are not spatially uniform. And they are viscously dissipated as they travel. The equations being used in our problem (involving T and p) assume that T and p are spatially uniform.

    Getting back to the problem at hand, I'm having trouble understanding what you are saying. Are you saying that:
    1. It doesn't matter whether the process is reversible or irreversible; the efficiency is the same either way.
    2. The process they described is definitely irreversible and we can calculate the efficiency from the information given (assuming, say, that p is the force per unit area at the piston).
    3. The process they described is irreversible, but we can't calculate the efficiency from the information given.

    Chet
     
  18. Aug 5, 2014 #17

    Andrew Mason

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    The point is that temperature and pressure changes in a gas occur very rapidly so you do not have to have a truly reversible process in order to apply the assumptions that T and P are essentially uniform.
    No. I am just saying that in this problem you don't need to assume that it is reversible. In this problem as long as the work output = ∫PdV, which does not require a reversible process, the efficiency will be the same whether the heat flow occurs reversibly or not. All efficiency depends on is Qh and Qc. η=1-Qc/Qh

    For example, whether heat flows into the gas from a-b reversibly or not (it hard to conceive of how a constant volume heat absorption being reversible) is irrelevant. We are only interested in the amount of heat flow into the gas. Similarly, from b-c, it does not matter whether there is a finite temperature difference between the system and surroundings during that process. We are only interested in how much heat flow occurs. Similarly from c-a.

    I guess I am saying that it is difficult to understand, in any kind of reasonable model, how a-b and c-a could be made reversible AND that we don't need the assumption of reversibility in order to answer the problem.
    Definitely not.

    AM
     
  19. Aug 5, 2014 #18
    Really? I think I could do it. a-b is just heating the gas at constant volume, using a sequence of constant-temperature reservoirs. c-a is a little more difficult, but, I'm confident I could figure out how.

    Chet
     
  20. Aug 5, 2014 #19

    rude man

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    Why isn't c-a realizable the same way: by using a sequence of constant-temperature reservoirs, of course in descending temperature, so as to maintain constant pressure pb.

    E.g. an upright cylinder with piston on top of mass m and area A, within a controlled pressure ambient of around 80,000P so that the piston can maintain the lower pressure of pb = nRTb/Vb = 2 x 8.34 x 820/0.16 = 85,485P: 80,000P + mg/A = 85,485P.
     
  21. Aug 5, 2014 #20
    Well, the way I would do it is, as you suggested, use a sequence of constant temperature reservoirs, but, such that the absolute temperatures vary linearly with the cylinder volume:

    T = 51+(820-51)*(V-0.01)/(0.16-0.01)

    This would guarantee that the pressure remains constant along c-a.

    Chet
     
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