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Efficiency of a Stirling Engine

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Derive an expression for the effiency of the cycle (of a Stirling Engine) working between two heat baths at temperatures T1 and T2 with volumes in the ratio V2/V1. Assume the working substance is a monatomic ideal gas.

    2. Relevant equations

    Work done=heat absorbed from going from V1 to V2 at constant temperature and is given by:

    RT ln(V2/V1)

    Heat absorbed in going from T1 to T2 is:

    (3/2)R(T2-T1)

    efficiency=1-Q2/Q1 where Q2 is the heat lost per cycle and Q1 is the heat gained per cycle.

    3. The attempt at a solution

    I have an expression but I have no idea if it was correct. I tried to google what the correct efficiency should be but I couldn't find it. My expression is:

    [tex]\eta = \frac{(T_2-T_1) (ln(V_2/V_1)+3)}{T_2 ln(V_2/V_1) +1.5(T_2-T_1)}[/tex]

    I don't think it's right, it looks a bit messy. If someone could tell me if it's correct or not that would be a great help. It would be an even greater help if someone could tell me where I have gone wrong or give me the correct expression so I can work out how to derive it.

    Thanks.
     
  2. jcsd
  3. May 14, 2009 #2

    Redbelly98

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    I get a similar, but not identical, expression.

    What did you get for Q1 and Q2?
     
  4. May 15, 2009 #3
    I had, Q1, the heat absorbed, is:

    [tex] R T_2 ln(V_2/V_1) + 1.5R(T_2-T_1) [/tex]

    And Q2, the heat lost, is:

    [tex] R T_1 ln(V_2/V_1) + 1.5R(T_1-T_2) [/tex]
     
  5. May 15, 2009 #4

    Redbelly98

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    I got, for the final term in Q2, 1.5R(T2 - T1)

    But we should clarify something: are you, like me, using T2 for the hot temperature and T1 for the cold temperature?
     
  6. May 15, 2009 #5

    Andrew Mason

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    The heat lost at constant temperature in the compression part of the cycle (V2 to V1) is equal to the work done on the gas (per mole) (dU = 0) :

    [tex]\int_{V_2}^{V_1} PdV = RT_1\int_{V_2}^{V_1}dV/V = RT_1\ln{(V_1/V_2)}[/tex]

    The heat gained at constant temperature in the expansion part of the cycle (V1 to V2) is equal to the work done by the gas (per mole) (dU = 0) :

    [tex]\int_{V_1}^{V_2} PdV = RT_2\int_{V_1}^{V_2}dV/V = RT_2\ln{(V_2/V_1)} = - RT_2\ln{(V_1/V_2)}[/tex]

    Therefore:

    [tex]|Q_h| = C_v(T_2-T_1) + RT_2\ln{(V_2/V_1)}[/tex] and

    [tex]|Q_c| = C_v(T_2-T_1) + RT_1\ln{(V_2/V_1)}[/tex]

    So plug that into:

    [tex]\eta = W/Q_h = (Q_h-Q_c)/Q_h[/tex]

    I get:

    [tex]\eta = \frac{R\ln{(V_2/V_1)}(T_2-T_1)}{C_v(T_2-T_1) + RT_2\ln{(V_2/V_1)}}[/tex]

    For a monatomic gas:

    [tex]\eta = \frac{2\ln{(V_2/V_1)}(T_2-T_1)}{3(T_2-T_1) + 2T_2\ln{(V_2/V_1)}}[/tex]

    AM
     
    Last edited: May 15, 2009
  7. May 15, 2009 #6
    Thank you very much for all your help. Got it now.
     
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