# Efficiency, Stirling heat engine

1. Sep 4, 2014

### dave84

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

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

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

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

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

3. 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.

Last edited: Sep 4, 2014
2. Sep 4, 2014

### Andrew Mason

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. Sep 4, 2014

### 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.

Last edited: Sep 4, 2014
4. Sep 4, 2014

### Andrew Mason

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

Last edited: Sep 4, 2014