Carnot's Cycle and Carnot's Ideal Engine

A Carnot cycle is a cyclic process consisting of four reversible processes performed in a definite sequence, namely two isothermal processes and two adiabatic processes.

Main Parts of a Carnot Engine

(i) Heat Source

The heat source is a reservoir of effectively infinite heat capacity maintained at a high temperature \(T_1\) K. Its temperature remains constant even when heat is supplied to the working substance. Its upper surface is perfectly conducting.

(ii) Mechanical Arrangement and Working Substance

A hollow cylinder is used whose walls are perfectly insulating and whose base is perfectly conducting. A frictionless piston made of insulating material is fitted inside the cylinder.

An ideal gas is used as the working substance.

(iii) Heat Sink

The heat sink is a reservoir of effectively infinite heat capacity maintained at a lower temperature \(T_2\) K. It receives the rejected heat from the working substance while maintaining a constant temperature.

(iv) Insulating Stand

The stand is perfectly insulated. When the cylinder is placed on this stand, the working substance can expand or contract adiabatically without exchange of heat with the surroundings.

Working of Carnot Engine

According to Carnot, maximum work is obtained when the working substance is taken through a sequence of reversible processes and finally returned to its initial state.

The Carnot cycle consists of the following four stages:

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression

(i) First Process: Isothermal Expansion (A → B)

The cylinder is placed on the heat source. The temperature of the ideal gas becomes equal to the temperature \(T_1\) K of the heat source.

Let the initial pressure and volume of the gas at state \(A\) be \(P_A\) and \(V_A\), respectively.

The piston is allowed to move upward slowly. The gas expands isothermally at temperature \(T_1\). During this expansion, the gas absorbs heat \(Q_1\) from the heat source and reaches state \(B\), where its pressure and volume are \(P_B\) and \(V_B\).

For an ideal gas undergoing an isothermal process, \(\Delta U=0\). Therefore, from the first law of thermodynamics,

\[ Q_1=W_1. \tag{1} \]

The work done is

\[ W_1 = \int_{V_A}^{V_B}P\,dV. \]

For one mole of an ideal gas,

\[ PV=RT_1. \]

Hence,

\[ W_1 = RT_1 \int_{V_A}^{V_B}\frac{dV}{V}. \]

Therefore,

\[ \boxed{ Q_1=W_1 = RT_1 \ln\left(\frac{V_B}{V_A}\right) } \tag{2} \]

Since \[ \ln x=2.303\log_{10}x, \] we may also write

\[ \boxed{ Q_1 = 2.303RT_1 \log_{10} \left(\frac{V_B}{V_A}\right) }. \]

(ii) Second Process: Adiabatic Expansion (B → C)

The cylinder is removed from the heat source and placed on the insulating stand. The working substance is therefore completely thermally isolated from the surroundings.

The gas is allowed to expand slowly and adiabatically until its temperature falls from \(T_1\) to the temperature \(T_2\) of the heat sink.

The gas reaches state \(C\), where its pressure and volume are \(P_C\) and \(V_C\), respectively.

For an adiabatic process,

\[ \boxed{ PV^\gamma=\text{constant} } \tag{3} \]

where

\[ \boxed{ \gamma=\frac{C_P}{C_V} }. \]

Thus,

\[ P_BV_B^\gamma=P_CV_C^\gamma=K.\tag{3.1} \]

The work done during adiabatic expansion is

\[ W_2 = \int_{V_B}^{V_C}P\,dV. \]

Since \[ P=\frac{K}{V^\gamma}, \] we obtain

\[ W_2 = K \int_{V_B}^{V_C} V^{-\gamma}\,dV. \]

Therefore,

\[ W_2 = \frac{K}{1-\gamma} \left[ V_C^{1-\gamma} - V_B^{1-\gamma} \right]. \]

\[ W_2 = \frac{1}{1-\gamma} \left[ KV_C^{1-\gamma} - KV_B^{1-\gamma} \right]. \]

from eq. (3.1),

\[ W_2 = \frac{1}{1-\gamma} \left[ {P_CV_C^\gamma} V_C^{1-\gamma} - {P_BV_B^\gamma} V_B^{1-\gamma} \right]. \]

\[ W_2 = \frac{P_CV_C-P_BV_B} {1-\gamma} \]

Hence,

\[ \boxed{ W_2 = \frac{P_BV_B-P_CV_C} {\gamma-1} } \]

For one mole of an ideal gas,

\[ P_BV_B=RT_1, \qquad P_CV_C=RT_2. \]

Therefore,

\[ \boxed{ W_2 = \frac{R(T_1-T_2)} {\gamma-1} }. \tag{4} \]

(iii) Third Process: Isothermal Compression (C → D)

The cylinder is removed from the insulating stand and placed on the heat sink. The gas is compressed slowly and isothermally at temperature \(T_2\) K.

The gas moves from state \(C\) to state \(D\).

The work done by the gas is negative:

\[ W_3 = RT_2 \ln\left(\frac{V_D}{V_C}\right). \]

Since,

\[ \boxed{ W_3 = -RT_2 \ln\left(\frac{V_C}{V_D}\right) }. \tag{5} \]

Therefore, the heat rejected to the sink is

\[ \boxed{ Q_2 = RT_2 \ln\left(\frac{V_C}{V_D}\right) }. \tag{6} \]

(iv) Fourth Process: Adiabatic Compression (D → A)

The cylinder is again placed on the insulating stand. The gas is compressed slowly and adiabatically until its temperature rises from \(T_2\) to \(T_1\).

The gas returns from state \(D\) to the original state \(A\), completing the Carnot cycle.

The adiabatic relation is

\[ \boxed{ PV^\gamma=\text{constant} }. \tag{7} \]

The work done by the gas during this compression is negative. Its magnitude is

\[ \boxed{ W_4 = -\frac{R(T_1-T_2)} {\gamma-1} }. \tag{8} \]

Net Work Done in a Carnot Cycle

The work done by the gas during the first two processes is positive, whereas the work done by the gas during the last two processes is negative.

Therefore, the net work done in one complete Carnot cycle is

\[ W=W_1+W_2+W_3+W_4. \tag{9} \]

On substituting Eqs. (2), (4), (5), and (8),

\[ W = RT_1\ln\left(\frac{V_B}{V_A}\right) + \frac{R(T_1-T_2)}{\gamma-1} - RT_2\ln\left(\frac{V_C}{V_D}\right) - \frac{R(T_1-T_2)}{\gamma-1}. \]

The adiabatic terms cancel, giving

\[ W = RT_1\ln\left(\frac{V_B}{V_A}\right) - RT_2\ln\left(\frac{V_C}{V_D}\right). \]

Relation Between the Volumes in Carnot Cycle

For the adiabatic expansion

\[ TV^{\gamma-1} = constant \]

So, \(B\rightarrow C\),

\[ T_1V_B^{\gamma-1} = T_2V_C^{\gamma-1}. \tag{10} \]

Therefore,

\[ \frac{V_C}{V_B} = \left( \frac{T_1}{T_2} \right)^{1/(\gamma-1)}. \tag{11} \]

Similarly, for the adiabatic compression \(D\rightarrow A\),

\[ T_2V_D^{\gamma-1} = T_1V_A^{\gamma-1}. \tag{12} \]

Hence,

\[ \frac{V_D}{V_A} = \left( \frac{T_1}{T_2} \right)^{1/(\gamma-1)}. \tag{13} \]

Comparing Eqs. (11) and (13),

\[ \frac{V_C}{V_B} = \frac{V_D}{V_A}. \]

Therefore,

\[ \boxed{ \frac{V_B}{V_A} = \frac{V_C}{V_D} }. \tag{14} \]

Using Eq. (14), the net work becomes

\[ W = R(T_1-T_2) \ln\left(\frac{V_B}{V_A}\right). \tag{15} \]

Thus,

\[ \boxed{ W = R(T_1-T_2) \ln\left(\frac{V_B}{V_A}\right) }. \]

This net work is equal to the heat converted into useful work:

\[ \boxed{ W=Q_1-Q_2 }. \tag{16} \]

Efficiency of Carnot Engine

The efficiency of a heat engine is

\[ \eta=\frac{W}{Q_1}. \]

For the Carnot engine,

\[ Q_1 = RT_1 \ln\left(\frac{V_B}{V_A}\right). \]

Also,

\[ W = R(T_1-T_2) \ln\left(\frac{V_B}{V_A}\right). \]

Therefore,

\[ \eta = \frac{ R(T_1-T_2) \ln(V_B/V_A) }{ RT_1 \ln(V_B/V_A) }. \]

Cancelling the common terms,

\[ \boxed{ \eta = \frac{T_1-T_2}{T_1} } \tag{17} \]

Hence,

\[ \boxed{ \eta = 1-\frac{T_2}{T_1} }. \tag{18} \]

The percentage efficiency is therefore

\[ \boxed{ \eta(\%) = \left( 1-\frac{T_2}{T_1} \right)\times100 }. \tag{19} \]