Carnot Cycle and Carnot Engine

Significance of Reversibility

A reversible process is a bidirectional process: it follows exactly the same steps in either direction. This requires the process to take place in infinitesimally small steps, with only a very small difference between the state variables of two infinitesimally close states. That is possible only if the system is in thermodynamic equilibrium with its environment throughout the change.

An irreversible process, by contrast, is unidirectional; it can take place in only one direction, and it is not in thermal equilibrium with its environment.

Since conversion of work to heat is an irreversible process, a heat engine would convert the maximum possible heat into work if all irreversible processes could be avoided. Its efficiency would then be maximum. On this reasoning Sadi Carnot, the French engineer and scientist, proposed in 1824 a hypothetical ideal engine with the maximum possible efficiency.

Why Two Isothermals and Two Adiabatics

This is a standard two-mark question. In a Carnot engine there are basically two kinds of process.

• Exchange of heat. For heat exchange to be reversible it must be isothermal. This is possible if the working substance is at the temperature $T_H$ of the source while absorbing heat, and at the temperature $T_C$ of the cold reservoir while rejecting heat. If there were any finite temperature difference during heat exchange, the flow would be irreversible.
• Work done. For work done to be reversible, the process should be adiabatic — no heat may leak in or out while the gas expands or is compressed.

Hence the cycle must include two isothermal and two adiabatic processes for maximum efficiency.

The Carnot Cycle p–V Diagram

Figure 4.27 shows the cycle as a closed loop traced clockwise through four points A, B, C, D between two isotherms and two adiabats. The upper-left region is at high pressure and small volume; the cycle runs to the right and down.

• AB: isothermal expansion at the source temperature $T_H$. The gas absorbs heat $Q_H$ from the source and does work. Curve falling gently to the right.
• BC: adiabatic expansion. No heat is exchanged; the gas does work at the cost of its internal energy and cools from $T_H$ to $T_C$. Curve falling more steeply than the isotherm.
• CD: isothermal compression at the sink temperature $T_C$. Work is done on the gas and it rejects heat $Q_C$ to the sink. Curve rising gently to the left.
• DA: adiabatic compression. No heat is exchanged; work done on the gas raises its internal energy and its temperature from $T_C$ back to $T_H$, returning the system to A. Curve rising steeply.

Efficiency of the Carnot Engine

Using the expressions for the work done during the adiabatic and isothermal processes, the efficiency of a Carnot cycle or Carnot engine works out to

$$\eta = \dfrac{W}{Q_H} = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H} = \dfrac{T_H - T_C}{T_H}$$

with $T_H$ and $T_C$ in kelvin. Equivalently, for a Carnot cycle

$$\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$$

Design rule: while designing a heat engine for maximum efficiency, the source temperature $T_H$ should be as high as possible and the sink temperature $T_C$ as low as possible.

Always remember for a Carnot engine:

• The Carnot engine is a hypothetical concept.
• Every process in it must be either isothermal or adiabatic.
• The system must maintain thermodynamic equilibrium throughout the cycle so that the cycle is reversible.
• The efficiency of a Carnot engine can never be 100 per cent unless $T_C = 0$, which is not practically possible. So even the ideal heat engine cannot have 100 per cent efficiency.

Carnot Refrigerator

A refrigerator is a heat engine operated in the reverse direction. Because each step in the Carnot cycle is reversible, the entire Carnot cycle is reversible, and running it in reverse gives the Carnot refrigerator. Its coefficient of performance is

$$K = \dfrac{Q_C}{Q_H - Q_C} = \dfrac{Q_C/Q_H}{1 - Q_C/Q_H}$$

Using $Q_C/Q_H = T_C/T_H$ for a Carnot cycle,

$$K = \dfrac{T_C}{T_H - T_C}$$

The coefficient of performance of a Carnot refrigerator therefore depends only on the temperature difference between the hot and cold reservoirs.

• When the temperature difference is very small, $K$ is very large: a large quantity of heat can be moved from the lower to the higher temperature by doing very little work.
• When the temperature difference is large, $K$ is very small: only a small quantity of heat is removed even when a large amount of work is done.

The Second Law and the Carnot Cycle

Carnot's theorem, in the textbook's own words: the Carnot engine is the most efficient heat engine; also, all Carnot engines operating between the same two temperatures have the same efficiency, irrespective of the nature of the working substance.

Two statements are being made here:

• The Carnot engine is the most efficient heat engine.
• The efficiency of a Carnot engine is independent of its working substance.

It can also be shown that a Carnot refrigerator has a greater coefficient of performance than any other refrigerator working between the same two temperatures.

Standard Results

$$\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H} = \dfrac{T_H - T_C}{T_H}$$

$$\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}, \qquad Q_C = Q_H\left(\dfrac{T_C}{T_H}\right), \qquad W = Q_H - Q_C$$

$$K_{\text{Carnot}} = \dfrac{T_C}{T_H - T_C}$$

• Temperatures always in kelvin. A Carnot engine between 500 K and 350 K has $\eta = 1 - 350/500 = 0.30$, that is 30 per cent, and a Carnot refrigerator between 250 K and 300 K has $K = 250/50 = 5$.
• The Carnot efficiency is the ceiling: no engine working between the same two temperatures can beat it, whatever its working substance.