• Written By Saurav_C

Carnot Engine: Explanation, Definition, and Efficiency

Do you know that we can convert heat into work using a heat engine? But during this conversion, a lot of energy gets wasted in the surroundings. Hence, efficiency becomes less. Then we think about the reversible Carnot cycle.

The Carnot cycle was proposed by Sadi Carnot, a young French scientist, in the year $$1824.$$ It is an ideal reversible heat engine cycle.  In this article, we will discuss about Carnot engine and the thermodynamic processes in the Carnot cycle. The Carnot engine has maximum efficiency, but it is a theoretical idealization. Here, we will also learn to derive an expression for the efficiency of a Carnot engine.

Heat Engines: Conversion of Heat into Work

We know that heat flows spontaneously from a hotter to a colder body. If we intercept this flow with a heat engine, some of it can be converted into work. A machine that can convert heat into work is known as a heat engine. To be a useful device, a heat engine must operate continuously, absorb heat at a higher temperature, and reject it at a lower temperature. That is to say; a heat engine operates between two heat reservoirs, as shown in the figure below. Moreover, the processes which take place inside an engine have to operate in a cycle. Every heat engine consists of three parts:

1. A hot body called source
2. A working substance
3. A cold body called sink

In a heat engine, a working substance absorbs heat from the source of higher temperature $$\left( {{T_h}} \right),$$ converts a part of it into useful work (motion of piston) and rejects the rest to the sink (usually atmosphere) at a lower temperature $$\left( {{T_l}} \right)$$ and comes back into the initial state.

The efficiency of a heat engine can be written as:

$$\eta = \frac{{{\rm{ Work}}\,{\rm{done }}}}{{{\rm{ Heat}}\,{\rm{absorbed }}}} = \frac{W}{{{Q_H}}} = \frac{{{Q_H} – {Q_L}}}{{{Q_H}}} = 1 – \frac{{{Q_L}}}{{{Q_H}}}$$

Carnot Heat Engine

Carnot heat engine is a reversible heat engine that is used to convert heat into work. It consists of four parts : (a) A cylinder with perfectly isolating walls and a perfectly conducting base containing a perfect gas as a working substance and fitted with an insulating frictionless piston; (b) A source of infinite thermal capacity maintained at a constant higher temperature $$\left( {{T_h}} \right);$$ (c) A sink of infinite thermal capacity maintained at a constant lower temperature $$\left( {{T_l}} \right)$$ and (d) A perfectly non-conducting stand for the cylinder.

Carnot Cycle

There are two Isothermal Processes and two Adiabatic Processes present in the Carnot cycle. The theoretical heat engine that operates on this cycle is the Carnot Engine. It consists of four operations in succession as shown in the above figure; (a) isothermal expansion (heat absorption) at a higher temperature $${T_h}\,\left( b \right)$$ the adiabatic expansion between temperature $$T_h$$ and  $${T_l}\,\left( c \right)$$ isothermal compression (heat rejection) at a constant lower temperature $${T_l}$$ and (d) adiabatic compression between temperatures $${T_l}$$ and $${T_h}.$$
The working substance (gas) is to be placed in the cylinder, which expands and compresses and makes the piston move up and down w.r.t. the processes.

It consists of $$4$$ reversible processes, as shown in the figure given below. They are:

1. Process 1-2: Reversible Isothermal Expansion.
2. Process 2-3: Reversible Adiabatic Expansion
3. Process 3-4: Reversible Isothermal Compression.
4. Process 4-1: Reversible Adiabatic Compression.

The Efficiency of the Carnot Engine

To find the efficiency of the Carnot engine, let us consider all the steps of the Carnot cycle as shown in the above figure, which is given following one by one:

(a) Step $$1 \to 2:$$ Isothermal expansion of the gas taking its state from $$\left( {{P_1},{V_1},{T_1}} \right)$$ to $$\left( {{P_2},{V_2},{T_1}} \right).$$

The heat absorbed by the gas $$\left( {{Q_1}} \right)$$ from the reservoir at temperature $$\left( {{T_1}} \right)$$ is given by the equation that is also the work done $$\left( {{W_{1 \to 2}}} \right)$$ by the gas on the environment is

$${W_{1 \to 2}} = {Q_1} = \mu R{T_1}{\mathop{\rm In}\nolimits} \left( {\frac{{{V_2}}}{{{V_1}}}} \right)……..\left( 1 \right)$$

1. Step$$2 \to 3:$$ Adiabatic expansion of the gas from $$\left( {{P_2},{V_2},{T_1}} \right)$$ to $$\left( {{P_3},{V_3},{T_2}} \right)$$

And, the work done by the gas is given by the equation is:

$${W_{2 \to 3}} = \frac{{\mu R\left( {{T_1} – {T_2}} \right)}}{{\gamma – 1}}……….\left( 2 \right)$$

1. Step $$3 \to 4:$$ Isothermal compression of the gas from $$\left( {{P_3},{V_3},{T_2}} \right)$$ to $$\left( {{P_4},{V_4},{T_2}} \right)$$

Heat released $$\left( {{Q_2}} \right)$$ by the gas to the reservoir at temperature $$\left( {{T_2}} \right)$$ is given by the equation below, this is also the work done $$\left( {{W_{3 \to 4}}} \right)$$ on the gas by the environment.

$${W_{3 \to 4}} = {Q_2} = \mu R{T_2}{\mathop{\rm In}\nolimits} \left( {\frac{{{V_3}}}{{{V_4}}}} \right)………\left( 3 \right)$$

1. Step $$4 \to 1:$$ Adiabatic compression of the gas from $$\left( {{P_4},{V_4},{T_2}} \right)$$ to $$\left( {{P_1},{V_1},{T_1}} \right)$$

And the work done on the gas is given by the equation is:

$${W_{4 \to 1}} = \mu R\left( {\frac{{{T_1} – {T_2}}}{{\gamma – 1}}} \right)……….\left( 4 \right)$$

Now, from adding equation $$(1)$$ to $$(4),$$ the total work done by the gas in one complete cycle is given as:

$$W = {W_{1 \to 2}} + {W_{2 \to 3}} + {W_{3 \to 4}} + {W_{4 \to 1}}$$

$$W = \mu R{T_1}\,{\rm{In}}\left( {\frac{{{V_2}}}{{{V_1}}}} \right) – \mu R{T_2}{\mathop{\rm In}\nolimits} \left( {\frac{{{V_3}}}{{{V_4}}}} \right)…….\left( 5 \right)$$

And, the efficiency $$(η)$$ of the Carnot engine is:

$$\eta = \frac{W}{{{Q_1}}} = 1 – \frac{{{Q_2}}}{{{Q_1}}}$$

$$= 1 – \left( {\frac{{{T_2}}}{{{T_1}}}} \right)\frac{{\ln \left( {\frac{{{V_3}}}{{{V_4}}}} \right)}}{{\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)}}……….\left( 6 \right)$$

Now, In step $$2 \to 3,$$ since it is an adiabatic process, hence we can write,

$${T_1}V_2^{\gamma – 1} = {T_2}V_3^{\gamma – 1}$$

i.e.

$$\frac{{{V_2}}}{{{V_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^{\frac{1}{{(\gamma – 1)}}}}………..\left( 7 \right)$$

Similarly, in step $$4 \to 1,$$ it is also an adiabatic process, so we can write,

$${T_2}V_4^{\gamma – 1} = {T_1}V_1^{\gamma – 1}$$

i.e.

$$\frac{{{V_2}}}{{{V_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^{\frac{1}{{(\gamma – 1)}}}}…….\left( 8 \right)$$

Now from the above two equations $$(7)$$ and $$(8),$$ we have:

$$\frac{{{V_3}}}{{{V_4}}} = \frac{{{V_2}}}{{{V_1}}}……..\left( 9 \right)$$

Using Equation $$(6)$$ and $$(9),$$ we get the efficiency,

$$\eta = 1 – \frac{{{T_2}}}{{{T_1}}}…………\left( {10} \right)$$

Carnot Engine– Solved Examples

Q.1. The amounts of heat involved when an ideal gas is taken through a cyclic thermodynamic process through four steps are given as given as $${Q_1} = 5960\;{\rm{J}},{Q_2} = – 5585\;{\rm{J}},{Q_3} = – 2980\;{\rm{J}}$$ and $${Q_4} = 3645\;{\rm{J}}$$ respectively. And the corresponding quantities of work involved are given as $${W_1} = 2200\;{\rm{J}},{W_2} = – 825\;{\rm{J}},{W_3} = – 1100\;{\rm{J}}$$ and $${W_4},$$ respectively. Then
(a) Find the value of $${W_4}.$$
(b) What is the efficiency of the cycle?
Ans:  (a) In a cyclic process, $$\Delta U = 0$$
Therefore,
$${Q_{net}} = {W_{net}}$$
Or,
$${Q_1} + {Q_2} + {Q_3} + {Q_4} = {W_1} + {W_2} + {W_3} + {W_4}$$
Hence,
$${W_4} = \left( {{Q_1} + {Q_2} + {Q_3} + {Q_4}} \right) – \left( {{W_1} + {W_2} + {W_3}} \right)$$
$$= \{ (5960 – 5585 – 2980 + 3645) – (2200 – 825 – 1100)\}$$
Or,
$${W_4} = 765\;{\rm{J}}$$
(b)  Efficiency
$$\eta = \frac{{{\rm{ Total}}\,{\rm{work}}\,{\rm{done}}\,{\rm{in}}\,{\rm{the}}\,{\rm{cycle }}}}{{{\rm{ Heat}}\,{\rm{absorbed}}\,{\rm{(positive}}\,{\rm{heat }})}} \times 100$$
$$\eta = \frac{{{W_1} + {W_2} + {W_3} + {W_4}}}{{{Q_1} + {Q_4}}} \times 100$$
$$\eta = \frac{{(2200 – 825 – 1100 + 765)}}{{(5960 + 3645)}} \times 100$$

Q.2. How much work does a Carnot engine perform when it takes $${\rm{1000}}\,{\rm{kcal}}$$ of heat from a reservoir at $${\rm{82}}{{\rm{7}}^{\rm{o}}}{\rm{C}}$$ and exhausts it to a sink at $${\rm{2}}{{\rm{7}}^{\rm{o}}}{\rm{C}}$$? And also find the efficiency of the engine?
Ans: Given, $${Q_1} = {10^6}\;{\rm{Cal}}$$
$${T_1} = (827 + 273) = 1100\;{\rm{K}}$$
And,
$${T_2} = (27 + 273) = 300\;{\rm{K}}$$
as,
$$\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}} \Rightarrow {Q_2} = \frac{{{T_2}}}{{{T_1}}} \times {Q_1} = \left( {\frac{{300}}{{1100}}} \right) \times \left( {{{10}^6}} \right)$$
$$= 2.72 \times {10^5}\;{\rm{cal}}$$
$$W = {Q_1} – {Q_2} = 7.28 \times {10^5}\;{\rm{cal}}$$
The efficiency of the cycle,
$$\eta = \left( {1 – \frac{{{T_2}}}{{{T_1}}}} \right) \times 100$$
$$\eta = \left( {1 – \frac{{300}}{{1100}}} \right) \times 100$$
$$\eta = 72.72\%$$

Summary

A machine that can convert heat into work is known as a heat engine. The efficiency of the heat engine is given as:

$$\eta = \frac{{{\rm{ Work}}\,{\rm{done }}}}{{{\rm{ Heat}}\,{\rm{absorbed }}}} = 1 – \frac{{{Q_1}}}{{{Q_2}}}$$

Carnot cycle is an ideal reversible heat engine cycle. It is a hypothetical cycle, and it cannot be achieved in practical situations. The Carnot cycle consists of four reversible processes, and these are (a) isothermal expansion, (b) adiabatic expansion, (c) isothermal compression, and (d) adiabatic compression. The  Efficiency of a Carnot engine can be given as:$$\eta = 1 – \frac{{{T_2}}}{{{T_1}}}$$

Frequently Asked Questions (FAQs) on Carnot Engine

Q.1. What are the applications of the Carnot engine?
Ans: As the Carnot engine is not possible, most heat engines are tried to be made near to Carnot engine. These Heat engines are used in thermal machines, heat pumps to generate heating, refrigerators for cooling, steam turbines are used in ships, combustion engines are used in combustion vehicles, and reaction turbines are used in an airplane.

Q.2. Is there any heat engine be more efficient than the Carnot engine?
Ans: No heat engine can be more efficient than a Carnot engine. As Carnot theorem states that any heat engine that is operating between a given constant temperature source and given temperature sink none has a higher efficiency than a reversible engine.

Q.3. What are the assumptions of the Carnot engine?
Ans: Isothermal expansions and compressions are considered quasi-equilibrium. There should be no heat losses in pipes and the other components. The cycles do not have any friction in the motion of the piston in the cylinder, and the working material should be a perfect gas.

Q.4. Why is the Carnot cycle not $$100\%$$ efficient?
Ans: For any cycle to be $$100\%$$ efficient, the engine should not lose any heat to the sink. This will be only possible when the sink will be at zero kelvin. It is impossible to reach $${\rm{0}}\,{\rm{K}}{\rm{.}}$$ Hence complete $$100\%$$ heat conversion cannot be achieved. Hence $$100\%$$ efficient engines are not possible theoretically also.

Q.5. What are the processes involved in the Carnot cycle?
Ans: It consists of $$4$$ reversible processes. They are given as follows:
1. Process $$1-2:$$ Reversible Isothermal Expansion.
2. Process $$2-3:$$ Reversible Adiabatic Expansion
3. Process $$3-4:$$ Reversible Isothermal Compression.
4. Process $$4-1:$$ Reversible Adiabatic Compression.

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