Carnot Cycle

  • State and explain Carnot’s theorem.

Assume the Carnot cycle is being performed on an ideal gas.

T-s Diagram

We start the gas at some point 1, which corresponds to an initial condition \(T_h\) and \(s_h\) and perform a series of operations on the gas that move its state around a rectangle in T - s space.

Isothermal Expansion

The first process in the Carnot cycle is an isothermal expansion, which corresponds to a vertical line in T-s space. The gas is at \(T_h\) and is brought into contact with a thermal reservoir. The volume of the gas is then increased while it is in contact with the reservoir.

Isentropic Expansion

The second process in the Carnot cycle is an isentropic expansion in which the heat reservoir is removed and the gas is allowed to expand with no input of heat . The temperature falls as the gas expands, but the entropy is fixed, since no heat is transferred to the gas. This process corresponds to a horizontal line at \(\tau_h\) in \(\tau - \sigma\) space traversed to the left.

Isothermal Compression

The third process in the Carnot cycle is a compression is then compressed at temperature \(T_l\) in contact with a heat reservoir. This isothermal compression corresponds to a vertical line at \(s_l\) traversed in the negative direction.

Isentropic Compression

Finally, the gas is compressed back to its initial volume. The gas is no longer in contact with a heat reservoir, so T increases, and the process corresponds to a horizontal path traversed to the right at entropy \(s_1\).

Work Done

Since the system returns to its initial state after a cycle, the total change in energy over this cycle is 0. \[\begin{aligned} 0 = \oint dU &= \oint \tau d\sigma - \underbrace{\oint pdV}_{W \text{ by gas}}\\ W\text{ by gas} &= \oint \tau d\sigma\\ &= \text{ area of rectangle}\\ &= \left( T_{h} - T_{l} \right)\left( s_{h} - s_{l} \right)\end{aligned}\]

P-V Diagram:

Carnot Cycle p-v Diagram

We calculate the work done over a single Carnot cycle. The analysis is now performed in terms of the pressure p and the volume V of the gas.


Isothermal Expansion:

The work done by the gas in going from points 1 to 2 is,

\[\begin{aligned} W_{1 \rightarrow 2} = +\int_{V_{1}}^{V_{2}}pdV\end{aligned}\]

Note: The integral is positive because we are finding the work performed by the gas.
Since the process is carried out on an ideal gas, use the ideal gas relation to remove p from the integral,

\[\begin{aligned} W_{1 \rightarrow 2} = N\tau_{h}\int_{V_{1}}^{V_{2}}dV = N\tau_{h}\ln\left( \frac{V_{2}}{V_{1}} \right)\end{aligned}\]

Since \(\Delta\) U = 0 for an isothermal process, all of this work must come from the heat taken up by the gas. \[\begin{aligned} W_{1 \rightarrow 2} = Q_{1}\end{aligned}\]

Isentropic Expansion:

\[\begin{aligned} W_{2 \rightarrow 3} = U(\tau_{h}) - U(\tau_{l}) = \frac{3}{2}N\left( \tau_{h} - \tau_{l} \right)\end{aligned}\]

Compression Processes:

\[\begin{aligned} W_{3 \rightarrow 4} &= N\tau_{l}\log \frac{V_{4}}{V_{3}} = Q_{2}\\ W_{4 \rightarrow 1} &= U(\tau_{l}) - U(\tau_{h})\\ &= -\frac{3}{2}N\tau\log(\tau_{h} - \tau_{l})\end{aligned}\]

Total Work:

The total work done by the gas is: \[\begin{aligned} W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 4} + W_{4 \rightarrow 1} = W_{1 \rightarrow 2} + W_{3 \rightarrow 4}\end{aligned}\]

\(W_{4 \rightarrow 1}\) is equal and opposite to the work done by the gas during the isentropic expansion. Only the isothermal processes contribute to the work done by the gas, since only when heat is transferred is work done.

We would like to express the result in terms of \(V_{1}\) and \(V_{2}\) only. Using the fact that the gas undergoes an adiabatic expansion,

\[\begin{aligned} \tau V^{\frac{2}{3}} &= \text{ constant}\\ \tau_{h}V_{2}^{\frac{2}{3}} &= \tau_{l}V_{3}^{\frac{2}{3}}\\ \Rightarrow \frac{V_{3}}{V_{2}} &= \left( \frac{\tau_{h}}{\tau_{l}} \right)^{\frac{3}{2}}\end{aligned}\]

Also, from the adiabatic compression, \[\begin{aligned} \tau_{l}V_{4}^{\frac{2}{3}} &= \tau_{h}V_{1}^{\frac{2}{3}}\end{aligned}\] Combining these results, \[\begin{aligned} \frac{V_{4}}{V_{1}} &= \left( \frac{\tau_{h}}{\tau_{l}} \right)^{\frac{3}{2}} = \frac{V_{3}}{V_{2}}\\ \frac{V_{4}}{V_{3}} &= \frac{V_{1}}{V_{2}}\\ \Rightarrow \log\left( \frac{V_{4}}{V_{3}} \right) &= \log\left( \frac{V_{1}}{V_{2}} \right)\\ &= -\log\left( \frac{V_{2}}{V_{1}} \right)\end{aligned}\]

The work done by the gas can then be written, \[\begin{aligned} W \text{ by gas} &= N\tau_{h}\log \left( \frac{V_{2}}{V_{1}} \right) - N\tau_{l}\log \left( \frac{V_{2}}{V_{1}} \right)\\ &= N(\tau_{h} - \tau_{l})\log\left( \frac{V_{2}}{V_{1}} \right)\end{aligned}\]


\[\begin{aligned} \eta &= \frac{W}{Q_{1}} = \frac{N(\tau_{h} - \tau_{l})\log\left( \frac{V_{2}}{V_{1}} \right)}{N\tau_{h}\log\left( \frac{V_{2}}{V_{1}} \right)}\\ &= \frac{\tau_{h} - \tau_{l}}{\tau_{h}} = \eta_{c}\end{aligned}\]

  • We have recovered the Carnot efficiency by examining an actual cycle. Thus our model heat engine is an example of an ideal heat engine.

  • All heal heat engines operate on this same basic process, although they included many inefficiencies.

  • Doing this same analysis in the opposite direction would be analogous to the operation of a refrigerator.

Solved Examples

Solved Example:


100% efficiency of a thermal cycle cannot be achieved because of:

Correct Answer: B

Solved Example:


A Carnot refrigerator has a lower temp. of 5$^\circ$C and rejects heat to the surround at 27$^\circ$C. If the lower temp. is decreased to -13$^\circ$C, while the surrounding temp. remains same, calculate ratio of the increase of the input work for the same heat absorbed from the cold reservoir.

Correct Answer: A

Solved Example:


During which of the following process does heat rejection takes place in Carnot cycle?

Correct Answer: C

Solved Example:


Which is the incorrect statement about Carnot cycle?

Correct Answer: B

Solved Example:


Efficiency of a Carnot engine is given as 80%. If the cycle direction be reversed, what will be the value of COP of reversed Carnot cycle: (ISRO Scientist ME 2013)

Efficiency of Carnot heat engine \begin{align*} \eta &=\dfrac {T_{1}-T_{2}}{T_{1}}\\ 0.8 &=\dfrac {T_{1}-T_{2}}{T_{1}}\\ 0.8T_{1} &=T_{1}-T_{2} \end{align*} In the case of reversed heat engine, \[COP =\dfrac {T_{1}}{T_{1}-T_{2}}= \dfrac {T_{1}}{0.8T_{1}} =\dfrac {1}{0.8} =1.25\]

Correct Answer: A