6.8 Carnot Engine and Carnot's Theorem

6.8 Carnot Engine and Carnot's Theorem

Sadi Carnot in 1824 described an ideal engine using only isothermal and adiabatic processes. He showed that a heat engine operating in an ideal reversible cycle between two heat reservoirs at different temperatures, would be the most efficient engine. A Carnot cycle using an ideal gas as the working substance is shown on PV diagram (Fig. 6.9). It consists of following four steps:

1. The gas is allowed to expand isothermally at temperature $T_{1}$ absorbing heat $Q_{1}$ from the hot reservoir. The process is represented by the curve $AB$.

2. The gas is then allowed to expand adiabatically until its temperature drops to $T_{2}$. The process is represented by the curve $BC$.

3. The gas at this stage is compressed isothermally at temperature $T_{2}$ rejecting heat $Q_{2}$ to the cold reservoir. The process is represented by the curve $CD$.

4. Finally the gas is compressed adiabatically to restore its initial state at temperature $T_{1}$. The process is represented by the curve $DA$.

Thermal and mechanical equilibrium is maintained all the time so that each process is perfectly reversible. As the working substance returns to the initial state, there is no change in its internal energy i.e., $\Delta U = 0$.

The net work done during one cycle equals to the area enclosed by the path $ABCDA$ of the PV diagram. It can also be estimated from net heat $\Delta Q$ absorbed in one cycle.

From 1st law of thermodynamics:

or $W = Q_{1} - Q_{2}$ ($\because \Delta U = 0$)

The efficiency $\eta$ of the heat engine is defined as:

Thus

The energy transfer in an isothermal expansion or compression turns out to be proportional to kelvin temperature. So $Q_{1}$ and $Q_{2}$ are proportional to kelvin temperatures $T_{1}$ and $T_{2}$ respectively and hence:

The efficiency is usually taken in percentage, in that case:

Thus, the efficiency of Carnot engine depends on the temperature of hot and cold reservoirs. It is independent of the nature of working substance. The larger the temperature difference of two reservoirs, the greater is the efficiency. But it can never be one or $100\%$ unless cold reservoir is at absolute zero temperature ($T_{2} = 0\,\mathrm{K}$).

Such reservoirs are not available and hence the maximum efficiency is always less than one. Nevertheless, the Carnot cycle establishes an upper limit on the efficiency of all heat engines. No practical heat engine can be perfectly reversible and also energy dissipation is inevitable. This fact is stated in Carnot's theorem:

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No heat engine can be more efficient than a Carnot engine operating between the same two temperatures.

The Carnot's theorem can be extended to state that:

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All Carnot's engines operating between the same two temperatures have the same efficiency, irrespective of the nature of working substance.

In most practical cases, the cold reservoir is near room temperature. So, the efficiency can only be increased by raising the temperature of hot reservoir. All real heat engines are less efficient than Carnot engine due to friction and other heat losses.

Example 6.3

The turbine in a steam power plant takes steam from a boiler at $427^{\circ}\mathrm{C}$ and exhausts into a low temperature reservoir at $77^{\circ}\mathrm{C}$. What is the maximum possible efficiency?

Solution

Maximum efficiency for any engine operating between temperatures $T_{1}$ and $T_{2}$ is:

$\eta = [(T_{1} - T_{2})/T_{1}] \times 100\%$ where $T_{1} = 427 + 273 = 700\,\mathrm{K}$

and $T_{2} = 77 + 273 = 350\,\mathrm{K}$

So $\eta = (T_{1} - T_{2})/T_{1} = (700\,\mathrm{K} - 350\,\mathrm{K})/700\,\mathrm{K} = 350\,\mathrm{K}/700\,\mathrm{K} = 0.5$ or $\eta = 0.5 \times 100 = 50\%$