Carnot Theorem and Its Applications

Carnot theorem, often known as Carnot's rule, was created in 1824 by Nicolas Léonard Sadi Carnot and is a theory that describes the greatest efficiency that any heat engine may achieve. A Carnot engine's efficiency is solely determined by the temperatures of its hot and cold reservoirs.

Carnot Theorem

This theorem has the conclusion that every Carnot heat engine operating across two heat reservoirs is equally efficient, irrespective of the working substance or operation details.

History of Nicolas Léonard Sadi Carnot

Carnot's theorem was given in 1824 by Nicolas Léonard Sadi Carnot, a French Army mechanical engineer, military scientist, and physicist known as the "Father of Thermodynamics."

Nicolas Léonard Sadi Carnot (1 June 1796 – 24 August 1832)

The Carnot theorem is a consequence of the second law of thermodynamics. It was historically founded on contemporary caloric theory and predated the formation of the second law.

Carnot Theorem Statement

According to Carnot's theorem, the heat engine's efficiency has no boundaries; it can be as high as it can be dependent on the hot and cold reservoir temperatures.

\[{\eta _{\max }} = {\eta _{carnot}}=1-\dfrac {T_C}{T_H}\]\[\]

Where,

\[T_C\] - Absolute temperature of the cold source

\[T_H\] - Absolute temperature of the hot source

Carnot’s Theorem Proof

Consider a heat engine that draws heat Q₁ from a heat reservoir at T₁, delivers work W, and dumps heat Q₂ into a heat sink at T₂.

The heat engine operates in cycles( each of which includes two isothermal and two adiabatic processes), meaning it takes in heat Q₁, performs work W, dumps heat Q₂, and then returns to its original, unaltered state.

Heat Engine and Efficiency

Consider the universe's net change in entropy \[\Delta S\]\[\].

At a steady temperature T₁, the heat reservoir emits heat Q₁. As a result, the change in its entropy is:

\[\Delta {S_1} = - \frac{{{Q_1}}}{{{T_1}}}\]

At a steady temperature T₂, the heat sink receives heat Q₂. As a result, the change in its entropy is:

\[\Delta {S_2} = - \frac{{{Q_2}}}{{{T_2}}}\]

As a result, the universe's net entropy change is,

\[\begin{array}{l}\Delta S = \Delta {S_1} + \Delta {S_2}\\\Delta S = \frac{{{Q_2}}}{{{T_2}}} - \frac{{{Q_1}}}{{{T_1}}}\end{array}\]

Using the second law of thermodynamics, \[\Delta S \ge 0\], it follows that

\[\begin{array}{l}\frac{{{Q_2}}}{{{T_2}}} - \frac{{{Q_1}}}{{{T_1}}} \ge 0\\\frac{{{Q_2}}}{{{T_2}}} \ge \frac{{{Q_1}}}{{{T_1}}}\\\frac{{{Q_2}}}{{{Q_1}}} - \frac{{{T_2}}}{{{T_1}}}\\1 - \frac{{{Q_2}}}{{{Q_1}}} \le 1 - \frac{{{T_2}}}{{{T_1}}}\end{array}\]

Because the left side reflects the efficiency of the supplied heat engine \[\eta\] and the right side indicates the efficiency of a Carnot engine,

\[\eta \le {\eta _{carnot}} \to {\eta _{\max }} = {\eta _{carnot}}\]

Hence proved

Limitations of Carnot Theorem

• The Carnot Cycle is an ideal cycle, which means it does not exist and cannot be created. Hence it is only a theoretical concept.

• The Carnot Cycle is only used to examine heat engines and does not apply to other sorts of devices.

• The output obtained per cycle is quite low.

Carnot Cycle Application

• Carnot's theorem is used in engines that turn thermal energy into work.

• In a refrigeration system, the Carnot cycle is reversed. We know that a heat engine based on the Carnot engine has the maximum efficiency. Similarly, a change cycle refrigeration system has the highest coefficient of performance.

Solved Examples

1. What is the efficiency of a Carnot engine operating between \[{340^ \circ }C\] and \[{70^ \circ }C\]?

Ans. Given data:

\[\begin{array}{l}{T_1} = {340^ \circ }C = (340 + 273)K\\{T_1} = 613K\\{T_2} = {70^ \circ }C = (70 + 273)K\\{T_2} = 343K\end{array}\]

Carnot's engine efficiency is given by,

\[\begin{array}{l}\eta = 1 - (\frac{{{T_2}}}{{{T_1}}})\\\therefore \,\eta = 1 - (\frac{{343}}{{613}})\\ \Rightarrow \frac{{613 - 343}}{{613}} \approx 0.44\end{array}\]

Therefore, the efficiency of the Carnot engine operating between the given temperatures is about \[44\].

2. A heat engine with a 55% efficiency performs 70kJ of work every Carnot cycle. What amount of heat is rejected in each cycle?

Ans. Given data:

\[\begin{array}{l}\eta = 55\% \\ \Rightarrow \,\eta = 0.55\\w = 70J\end{array}\]

As known,

\[\eta \] = output work/Heat absorbed \[{Q_1}\]

Therefore,

\[\begin{array}{l}0.55 = \frac{{70}}{{{Q_1}}}\\ \Rightarrow {Q_1} = 127.27\end{array}\]

Work = \[{Q_1} - {Q_2}\] (Heat absorbed - Heat rejected)

\[\begin{array}{l}70 = {Q_1} - 127.27\\{Q_1} = 57.27kJ\end{array}\]

So, the amount of heat rejected in each cycle is \[57.27kJ\].

3. Consider Carnot's cycle that produces \[2kJ\] of mechanical work during each cycle and operates between \[{T_1} = 400K\] and \[{T_2} = 560K\]. What is the amount of heat transmitted to the engine by the reservoirs?

Ans. Given data:

\[{T_1} = 400K\]

\[{T_2} = 560K\]

Output Work = \[2kJ\]

Carnot's engine efficiency is given by,

\[\begin{array}{l}\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\\ \Rightarrow \eta = 1 - \frac{{400}}{{560}}\\ \Rightarrow \eta = \frac{{560 - 400}}{{560}}\\ \Rightarrow \eta = 0.28\end{array}\]

Efficiency can also be written as:

\[\eta = \dfrac {\text{output~ work}}{\text{Heat~ supplied}}\]

\[\eta = \dfrac{{2000}}{x}\]

Where ‘x’ is the total heat supplied.

Now substitute the calculated value of efficiency in the above expression.

\[\begin{array}{l}0.28 = \frac{{2000}}{x}\\ \Rightarrow 0.28x = 2000\\ \Rightarrow x = \frac{{2000}}{{0.28}}\\ \Rightarrow x = 7142.85J\end{array}\]

Important Points to Remember

• A Carnot cycle employs an ideal gas as a working fluid.

• The working substance is subjected to a cyclic operation involving two isothermal and two adiabatic operations in a Carnot cycle.

Important Formulas to Remember from Theorem

• The formula for maximum efficiency is as follows:

\[\eta {}_{\max } = 1 - \frac{{{T_2}}}{{{T_1}}}\]

Where \[\eta \] is the efficiency, \[{T_2}\] is the absolute temperature (Kelvins) of the cold reservoir, and \[{T_1}\] is the absolute temperature (Kelvins) of the hot reservoir.

• Another formula for efficiency,

\[\eta \] = $\dfrac {\text{output~ work}}{\text{Heat~ supplied}}$

Conclusion

Carnot's theorem illustrates the principle that an irreversible engine's efficiency cannot be greater than that of a reversible engine if both operate between fixed temperature heat reservoirs.

Simple heat engines are less effective and efficient when employed between two heat reservoirs, but Carnot heat engines are more efficient. Regardless of their operation, the Carnot engines are always efficient.