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Engine thermal efficiency and Volume ratios

  1. Nov 1, 2013 #1
    The question:

    A perfect gas undergoes the following cyclic processes:
    State 1 to 2 cooling at constant pressure.
    State 2 to 3 heating at constant volume.
    State 3 to 1 adiabatic expansion.

    Deduce an expression for the thermal efficiency of the cycle in terms
    of r the volume compression ratio (r=V1/V2) and γ (where γ = ratio of specific heats Cp/Cv)

    η = 1 - γ(r-1)/(rγ-1)

    My attempt at the solution:
    First I tried sketching the cycle

    Bare with me as I present you the silly symbol art.


    2<---------':.1 v

    I'd like to work in specific terms

    As it is a perfect gas
    P1v1= RT1
    P1v2= RT2
    P2v2= RT3

    Heats from 1->2, 2->3, 3->1
    Q3->1 = 0 ; adiabatic.

    Also, polytropic relations
    v2/v1 = (P1/P2)1/n
    as r = v1/v2⇔ r = (P2/P1)1/n
    ∴ rn = P2/P1 and
    1/rn = P1/P2

    Substituting Ideal Gas expressions in terms of Tn

    Thermal efficiency
    This is what I am unsure off. I begin assuming quite a few things. First I assume that heat in the cooling process is the equivalent of heat escaping to a cold reservoir; coincidentally, heat from the pressurization is the heat INPUT from the hot reservoir. As such:

    η = [Q2->3 - Q1->2]/Q2->3

    η = 1 - Q1->2/Q2->3

    ∴ η = 1 - Cp(P1v2-P1v1)/Cv(P2v2-P1v2)

    η = 1 - γ((P1v2-P1v1)/(P2v2-P1v2))

    From here:

    P1v2/P2v2-P1v2 = P1/(P2-P1) = 1/rn-1

    P1v1/P2v2-P1v2 = r/(rn-1)

    My Result

    η = γ(1-r)/(rn-1) =/= η = 1 - γ(r-1)/(rγ-1)

    Can I assume n=γ in this situation? only 1 process is adiabatic.
    Last edited: Nov 1, 2013
  2. jcsd
  3. Nov 1, 2013 #2

    Andrew Mason

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    Apply the adiabatic condition from 3→1.

  4. Nov 1, 2013 #3
    Can you be more specific? Apply where?

    If it's adiabatic there is no heat - I do not understand how the process can be related to thermal efficiency.
  5. Nov 1, 2013 #4

    Andrew Mason

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    What you need is the relationship between P1 and P2 in terms of V1 and V2. That is determined by the adiabatic condition PVγ = K.

  6. Nov 2, 2013 #5
    Indeed, the ratios match up and allow you to express them with n=γ. Thermodynamics is always like that - an answer under your nose at all times.

    But another issue is that the numerator does not match up.
    The expected form is r-1 when I get 1-r despite getting the same denominator.
  7. Nov 3, 2013 #6

    Andrew Mason

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    You are using the absolute values of Qh and Qc so you have to make sure that the original equation reflects that. For example, Qc = |Cp(T2-T1)| = Cp(T1-T2)

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