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Efficiency, Stirling heat engine

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    AhKqy.png

    We have a Stirling heat engine. I'm calculating the efficiency [itex]\eta[/itex].

    2. Relevant equations

    [itex]Q_{12} = \frac{m R T_2}{M} \ln(\frac{V_2}{V_1}) > 0[/itex]

    [itex]Q_{23} = m c_v (T_1 - T_2) < 0[/itex]

    [itex]Q_{34} = \frac{m R T_1}{M} \ln(\frac{V_1}{V_2}) < 0[/itex]

    [itex]Q_{41} = m c_v (T_2 - T_1) > 0[/itex]

    [itex]\kappa = \frac{c_p}{c_v}[/itex]

    [itex]\frac{R}{c_v M} = \kappa - 1[/itex]

    3. The attempt at a solution

    My result is [itex]\eta = \frac{|(\kappa-1)T_1 \ln(\frac{V_1}{V_2})+(\kappa-1)T_2 \ln(\frac{V_2}{V_1})|}{(T_2 - T_1) + (\kappa - 1)T_2 \ln(\frac{V_2}{V_1})}[/itex].

    Can anyone confirm this? I'm sorry if this is too trivial.
     
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2

    Andrew Mason

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    You will have show your reasoning to explain how you arrived at this. You could start by showing us your expression for η in terms of Q41, Q12, Q23, and Q34.

    AM
     
  4. Sep 4, 2014 #3
    So [itex]\eta = \frac{|A|}{Q_{in}} =\frac{|Q_{12}+Q_{34}|}{Q_{41} + Q_{12}} [/itex], where [itex]A[/itex] is total work done and [itex]Q[/itex] is the input heat. The final result is simplified with [itex]\kappa[/itex].

    [itex]Q_{in} = Q_{41} + Q_{12}[/itex]
    [itex]Q_{out} = Q_{23} + Q_{34} = Q_{in} - A[/itex]

    [itex]Q_{out}[/itex] is the amount of heat that is released from the heat engine.
     
    Last edited: Sep 4, 2014
  5. Sep 4, 2014 #4

    Andrew Mason

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    I am not sure how you got your equation for efficiency. Start with [itex]\eta = \frac{|W|}{Q_{in}}[/itex].
    You can rewrite this as:

    [itex]\eta = W/Q_{in} = (Q_{in}-Q_{out})/Q_{in} = 1 - Q_{out}/Q_{in} = 1 - (Q_{23} + Q_{34})/(Q_{41} + Q_{12})[/itex]

    Can you show us what this reduces to?

    AM
     
    Last edited: Sep 4, 2014
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