Efficiency, Stirling heat engine

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  • #1
dave84
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Homework Statement



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

Homework 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]

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.
 
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Answers and Replies

  • #2
Andrew Mason
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Homework Helper
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Homework Statement



AhKqy.png


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

Homework 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]

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.
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
 
  • #3
dave84
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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.
 
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  • #4
Andrew Mason
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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.
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
 
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