# Carnot engine efficiency

• tjackson3
I think I understand what you are trying to do now. In summary, the gas is cooled at constant pressure to (p_1,V_2); the gas is heated at constant volume to (p_2,V_2); the gas expands adiabatically back to (p_1,V_1). Assuming constant heat capacities, the thermal efficiency is 1 - \gamma\frac{(V_1/V_2)-1}{(p_2/p_1) - 1}

## Homework Statement

A possible ideal gas cycle operates as follows:

(i) From an initial state $$(p_1,V_1)$$, the gas is cooled at constant pressure to $$(p_1,V_2)$$;

(ii) the gas is heated at constant volume to $$(p_2,V_2)$$;

(iii) the gas expands adiabatically back to $$(p_1,V_1)$$.

Assuming constant heat capacities, show that the thermal efficiency is

$$1 - \gamma\frac{(V_1/V_2)-1}{(p_2/p_1) - 1}$$

where $$\gamma = c_p/c_v$$

## Homework Equations

Carnot efficiency: $$\nu = \frac{W}{Q_H} = 1 - \frac{T_l}{T_h}$$

In an adiabatic process, $$pV^{\gamma},TV^{\gamma-1},p^{1-\gamma}T^{\gamma}$$ are all constant.

## The Attempt at a Solution

I've been spinning my wheels a lot with this one, and I think the issue may be algebraic. My first thought was that I'd calculate the work done in the cycle. For part (i), it's just $$p(V_1-V_2)$$. For part (ii), it's zero, since dV = 0. For part (iii), I used the fact that since it's adiabatic, $$pV^{\gamma}$$ is constant, which I'll call k. Then the work becomes

$$W_3 = \frac{k}{\gamma-1}(V_2^{1-\gamma} - V_1^{1-\gamma})$$

You can set $$k = p_1V_1^{\gamma} = p_2V_2^{\gamma}$$, but there doesn't seem to be a preferable way. Adding these together gives you

$$W = p(V_1-V_2) + \frac{p_iV_i^{\gamma}}{\gamma-1}(V_2^{1-\gamma} - V_1^{1-\gamma})$$

Since none of these processes are isothermal, I can't figure out a meaningful expression for $$Q_H$$, so I'm stuck here. Any help is very much appreciated!

This is not a Carnot engine, so Carnot efficiency is not applicable.

Heat flow is into the system in during the isochoric part and out on the isobaric part. Calculate the heat flow in and the heat flow out. Apply the first law to determine the work done in terms of heat in and heat out. From that, determine the efficiency.

AM

Thank you very much, Andrew!