# Efficiency of a Stirling Engine

## Homework Statement

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.

## Homework 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.

## 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:

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

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.

Redbelly98
Staff Emeritus
Homework Helper
I get a similar, but not identical, expression.

What did you get for Q1 and Q2?

I had, Q1, the heat absorbed, is:

$$R T_2 ln(V_2/V_1) + 1.5R(T_2-T_1)$$

And Q2, the heat lost, is:

$$R T_1 ln(V_2/V_1) + 1.5R(T_1-T_2)$$

Redbelly98
Staff Emeritus
Homework Helper
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?

Andrew Mason
Homework Helper
I had, Q1, the heat absorbed, is:

$$R T_2 ln(V_2/V_1) + 1.5R(T_2-T_1)$$

And Q2, the heat lost, is:

$$R T_1 ln(V_2/V_1) + 1.5R(T_1-T_2)$$
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) :

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

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) :

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

Therefore:

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

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

So plug that into:

$$\eta = W/Q_h = (Q_h-Q_c)/Q_h$$

I get:

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

For a monatomic gas:

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

AM

Last edited:
Thank you very much for all your help. Got it now.