Defination of Carnot Engine :
Carnot engine is a theoretical heat engine that sets the upper limit of efficiency for any engine operating between two thermal reservoirs. Its cycle consists of four reversible processes—two isothermal and two adiabatic—and its efficiency depends only on the temperatures of the hot and cold reservoirs. This post derives the Carnot efficiency while explicitly accounting for positive (expansion) and negative (compression) work.
Carnot cycle overview :
The working substance (ideal gas) undergoes the following four reversible processes in sequence:
Work and heat in each process (with signs) :
1. Isothermal expansion (A → B at ($T_1$))
For an ideal gas undergoing isothermal expansion, internal energy does not change, so heat absorbed equals work done by the gas.
- Heat absorbed:
$Q_{1} = n R T_{1} \ln\!\left(\frac{V_2}{V_1}\right)$
- Work done (positive):
$W_{AB} = +n R T_1 \ln\!\left(\frac{V_2}{V_1}\right)$
2. Adiabatic expansion (B → C)
No heat exchange; temperature drops from ($T_2$) to ($T_1$). For an ideal gas:
Heat exchange:
Q = 0
- Work done (positive):
$W_{BC} = +\frac{n R}{\gamma - 1}\,\left(T_1 - T_2\right)$
3. Isothermal compression (C → D at (T_{2}))
Gas is compressed at constant ($T_2$); heat is rejected to the cold reservoir. Work is done on the gas (negative).
- Heat rejected:
$Q_{C} = n R T_2 \ln\!\left(\frac{V_4}{V_3}\right)$
- Work done (negative):
$W_{CD} = -n R T_2 \ln\!\left(\frac{V_4}{V_3}\right)$
4. Adiabatic compression (D → A)
No heat exchange; temperature rises from \$(T_{C})$ back to $(T_{H})$. Work is done on the gas (negative).
- Heat exchange:
$Q = 0$
Work done (negative):
$W_{DA} = -\frac{n R}{\gamma - 1}\,\left(T_2 - T_1\right)$
Geometric relations from adiabats
For reversible adiabatic processes of an ideal gas:
- Adiabatic condition:
$T V^{\gamma-1} = \text{constant}$
Applying this to B → C and D → A gives:
- Volume ratio equality:
$\frac{V_B}{V_A} = \frac{V_C}{V_D}$
This relation ensures the logarithmic terms in isothermal steps align, which is key to simplifying the net work and efficiency.
Net work of the cycle (including negative work)
Sum the work over all four processes:</p>
- Total work:
$W{\text{net}} = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
Using the expressions above and noting that adiabatic works cancel in magnitude:
- Cancellation of adiabatic works:
$W_{BC} + W_{DA} = 0$
Therefore, the net work reduces to the difference between isothermal works:
- Net work from isothermals:
$W{\text{net}} = n R T_H \ln\!\left(\frac{V_B}{V_A}\right)\;-\;n R T_C \ln\!\left(\frac{V_D}{V_C}\right)$
Using the adiabatic volume relation \(\frac{V_B}{V_A} = \frac{V_C}{V_D}\):$
- Compact form:
$W{\text{net}} = n R \left(T_H - T_C\right)\,\ln\!\left(\frac{V_B}{V_A}\right)$
From the isothermal steps:
- Heat absorbed and rejected:
$Q_H = n R T_H \ln\!\left(\frac{V_B}{V_A}\right),\quad Q_C = n R T_C \ln\!\left(\frac{V_D}{V_C}\right)$
$<code>\Rightarrow QC = n R T_C \ln\!\left(\frac{V_B}{V_A}\right)</code>$
Thus the net work can also be written as:
- Energy balance form:
$W{\text{net}} = Q_H - Q_C$
Efficiency derivation
Thermal efficiency is the ratio of net work output to heat absorbed from the hot reservoir:
- Definition:
$<code>\eta = \frac{W{\text{net}}}{Q_H} = \frac{Q_H - Q_C}{QH} = 1 - \frac{Q_C}{Q_H}</code>$
For reversible isothermal processes of an ideal gas, the heat exchanged is proportional to the absolute temperature:
- Heat ratio: