# Efficiency, Stirling heat engine

dave84

## Homework Statement

We have a Stirling heat engine. I'm calculating the efficiency $\eta$.

## Homework Equations

$Q_{12} = \frac{m R T_2}{M} \ln(\frac{V_2}{V_1}) > 0$

$Q_{23} = m c_v (T_1 - T_2) < 0$

$Q_{34} = \frac{m R T_1}{M} \ln(\frac{V_1}{V_2}) < 0$

$Q_{41} = m c_v (T_2 - T_1) > 0$

$\kappa = \frac{c_p}{c_v}$

$\frac{R}{c_v M} = \kappa - 1$

## The Attempt at a Solution

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

Can anyone confirm this? I'm sorry if this is too trivial.

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Homework Helper

## Homework Statement

We have a Stirling heat engine. I'm calculating the efficiency $\eta$.

## Homework Equations

$Q_{12} = \frac{m R T_2}{M} \ln(\frac{V_2}{V_1}) > 0$

$Q_{23} = m c_v (T_1 - T_2) < 0$

$Q_{34} = \frac{m R T_1}{M} \ln(\frac{V_1}{V_2}) < 0$

$Q_{41} = m c_v (T_2 - T_1) > 0$

$\kappa = \frac{c_p}{c_v}$

$\frac{R}{c_v M} = \kappa - 1$

## The Attempt at a Solution

My result is $\eta = \frac{(\kappa-1)T_1 \ln(\frac{V_1}{V_2})+(\kappa-1)T_2 \ln(\frac{V_2}{V_1})}{(T_2 - T_1) + (\kappa - 1)T_2 \ln(\frac{V_2}{V_1})}$.

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

dave84
So $\eta = \frac{|A|}{Q_{in}} =\frac{|Q_{12}+Q_{34}|}{Q_{41} + Q_{12}}$, where $A$ is total work done and $Q$ is the input heat. The final result is simplified with $\kappa$.

$Q_{in} = Q_{41} + Q_{12}$
$Q_{out} = Q_{23} + Q_{34} = Q_{in} - A$

$Q_{out}$ is the amount of heat that is released from the heat engine.

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Homework Helper
So $\eta = \frac{|A|}{Q_{in}} =\frac{Q_{12}+Q_{34}}{Q_{41} + Q_{12}}$, where $A$ is total work done and $Q$ is the input heat. The final result is simplified with $\kappa$.

$Q_{in} = Q_{41} + Q_{12}$
$Q_{out} = Q_{23} + Q_{34} = Q_{in} - A$

$Q_{out}$ 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 $\eta = \frac{|W|}{Q_{in}}$.
You can rewrite this as:

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

Can you show us what this reduces to?

AM

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