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Efficiency of a combustion motor

  1. Jan 24, 2015 #1
    1. The problem statement, all variables and given/known data
    The efficiency of a heat machine is ##E=\frac{Q_2-Q_1}{Q_2}## where Q2 is the heat inserted because of the explosion and Q1 is the heat that leaves through the exhaust.
    It says in the book that E can be derived from the volumes ratio V1/V2:
    $$E=1-\frac{1}{\left(V_1/V_2 \right)^{\gamma-1}}$$ why?

    2. Relevant equations
    Adiabatic expansion: ##T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}##

    3. The attempt at a solution
    $$E=1-\frac{1}{\left(V_1/V_2 \right)^{\gamma-1}}=\frac{T_2-T_1}{T_2}$$
    But V2 and T2 have nothing to do with Q2, the heat from the explosion. also i don't know what's the connection between V1 or T1 and the heat lost through the exhaust.
    I have just started studying the second law, maybe the book is excluding some explanations, it's an old book for high school but the level is quite high
     
  2. jcsd
  3. Jan 24, 2015 #2

    CWatters

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    Not my field but perhaps look up Carnot Efficiency. That relates efficiency and temperature.
     
  4. Jan 24, 2015 #3

    BvU

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    Note that E is the maximum efficiency, achievable only if all process steps are carried out reversibly, i.e. entropy is conserved. Then Q2 = T2dS2 and idem 1 (with a - sign). Over all ##\Delta S = Q_2/T_2 - Q_1/T_1## and only ##Q_2 - Q_1## can be drawn off as work.
     
  5. Jan 24, 2015 #4
    You're right. But mostly actual cycles aren't considered till you study the subject in depth. And I think there is an easier way for this proof.

    If you know the the changes in temperatures (ΔT), you can use the equations-
    Q = mCPΔT (constant pressure) or Q = mCVΔT (constant volume)
    Knowing the relation between temperature and volume, you can introduce 'V' into the equation.

    I must say though, it isn't really too easy for a starter.
     
  6. Jan 26, 2015 #5
    Thanks i solved but it's long development so i won't write it here
     
  7. Jan 26, 2015 #6
    No problemo!
     
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