# Carnot cycle

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.5.

An ideal heat engine absorbs an amount ${Q_{h}}$ from a hot reservoir at temperature ${T_{h}}$, converts an amount ${W}$ of this heat to work and expels the remaining heat ${Q_{c}}$ to a cold reservoir at tempearture ${T_{c}}$. From the first and second laws of thermodynamics, we can show that the maximum efficiency of such a heat engine is

$\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (1)$

In practice, real engines achieve considerably less than this efficiency, but is it possible to construct an engine that does achieve the maximum? It is, in principle, possible, and the path that such an engine must follow on a ${PV}$ diagram is called the Carnot cycle.

The idea is to minimize the entropy generated at each stage. If the gas absorbs an amount ${Q_{h}}$ from the hot reservoir, the entropy of the reservoir decreases by ${Q_{h}/T_{h}}$ and the entropy of the gas increases by ${Q_{h}/T_{gas}}$. If ${T_{gas}, then the net entropy generated is greater than zero. To minimize this, we must therefore make ${T_{gas}}$ as close to ${T_{h}}$ as we can. We can’t make them equal, since then no heat would flow from the reservoir to the gas, so we’ll make ${T_{gas}}$ infinitesimally less than ${T_{h}}$. Of course, this also means that it will take a very long time for the gas to absorb the heat, since such a small temperature difference means a very slow rate of heat flow.

Similarly, at the cold reservoir, to avoid extra entropy we make the temperature of the gas infinitesimally greater than ${T_{c}}$. In between the hot and cold reservoirs, we would like the heat flow to be zero, since this means no entropy flows at all in these stages.

Thus the two stages where the gas is in contact with either the hot or cold reservoir take place at a constant temperature (the temperature of the respective reservoir), so they are isothermal processes. The two stages of the cycle where the gas cools off and then warms up again take place with no heat transfer (so the temperature changes are achieved by changing the pressure and volume). Thus these two stages are adiabatic processes.

On a ${PV}$ diagram, and isothermal curve obeys the relation

$\displaystyle P=\frac{NkT}{V} \ \ \ \ \ (2)$

$\displaystyle P=\frac{K}{V^{\gamma}} \ \ \ \ \ (3)$

where ${K}$ is a constant and ${\gamma=\left(f+2\right)/f}$ where ${f}$ is the number of degrees of freedom of each molecule. For a monatomic ideal gas, ${\gamma=\frac{5}{3}}$. A Carnot cycle looks like this:

[The units are arbitrary. It’s the shape of the curve that’s important.] The gas traverses the cycle in a clockwise direction, with the red ${A\rightarrow B}$ and blue ${C\rightarrow D}$ legs being the isothermals, and the green ${B\rightarrow C}$ and yellow ${D\rightarrow A}$ legs the adiabats.

To prove that the Carnot cycle does in fact achieve the maximum efficiency, we can calculate the heats ${Q_{h}}$ and ${Q_{c}}$. In the leg ${A\rightarrow B}$, the gas is in contact with the hot reservoir and undergoes an isothermal expansion. Since the temperature doesn’t change, the energy of the gas remains constant, so the heat added goes entirely into the work of expansion of the gas. That is

 $\displaystyle Q_{h}$ $\displaystyle =$ $\displaystyle \int_{V_{A}}^{V_{B}}P\;dV\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle NkT_{h}\int_{V_{A}}^{V_{B}}\frac{dV}{V}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle NkT_{h}\ln\frac{V_{B}}{V_{A}} \ \ \ \ \ (6)$

Similarly, the heat expelled to the cold reservoir along the leg ${C\rightarrow D}$ is

$\displaystyle Q_{c}=NkT_{c}\ln\frac{V_{C}}{V_{D}} \ \ \ \ \ (7)$

[Note that ${Q_{h}}$ and ${Q_{c}}$ flow in opposite directions relative to the gas, although they both have the same sign.]

From 3 we have for the ${B\rightarrow C}$ leg

$\displaystyle P_{B}V_{B}^{\gamma}=P_{C}V_{C}^{\gamma} \ \ \ \ \ (8)$

However, at point ${B}$, the adiabat joins the isothermal curve, so

$\displaystyle P_{B}=\frac{NkT_{B}}{V_{B}} \ \ \ \ \ (9)$

and similarly for point ${C}$, so

 $\displaystyle NkT_{B}V_{B}^{\gamma-1}$ $\displaystyle =$ $\displaystyle NkT_{C}V_{C}^{\gamma-1}\ \ \ \ \ (10)$ $\displaystyle \frac{T_{B}}{T_{C}}$ $\displaystyle =$ $\displaystyle \left(\frac{V_{C}}{V_{B}}\right)^{\gamma-1} \ \ \ \ \ (11)$

Applying the same logic to the ${D\rightarrow A}$ leg, and using ${T_{A}=T_{B}}$ and ${T_{C}=T_{D}}$ we get

$\displaystyle \frac{T_{A}}{T_{D}}=\frac{T_{B}}{T_{C}}=\left(\frac{V_{D}}{V_{A}}\right)^{\gamma-1} \ \ \ \ \ (12)$

Comparing the last two equations we find that

$\displaystyle \frac{V_{D}}{V_{A}}=\frac{V_{C}}{V_{B}} \ \ \ \ \ (13)$

or, rearranging:

$\displaystyle \frac{V_{B}}{V_{A}}=\frac{V_{C}}{V_{D}} \ \ \ \ \ (14)$

Inserting this into 6 and 7 and taking the ratio we get

$\displaystyle \frac{Q_{c}}{Q_{h}}=\frac{NkT_{c}\ln\frac{V_{C}}{V_{D}}}{NkT_{h}\ln\frac{V_{B}}{V_{A}}}=\frac{T_{c}}{T_{h}} \ \ \ \ \ (15)$

Thus the efficiency of a Carnot cycle is

$\displaystyle e=1-\frac{Q_{c}}{Q_{h}}=1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (16)$

QED.

## 7 thoughts on “Carnot cycle”

1. Pingback: Air conditioners | Physics pages

2. Pingback: Heat pumps | Physics pages

3. Pingback: Stirling engine | Physics pages