Carnot Cycle

Carnot’s cycle was first developed in the year in 1824 by a French physicist named Sadi Carnot. It is an ideal cycle that basically laid the foundation for the second law of thermodynamics. Carnot cycle also brought up the concept of reversibility.

In this lesson, we shall look at all the aspects of Carnot’s cycle as well as understand some key concepts.

Carnot’s Theorem

From the second law of thermodynamics, two important results are derived where the conclusions are taken together to constitute Carnot’s theorem. It may be stated in the following forms.

(a) ‘No engine can be more efficient than a perfectly reversible engine working between the same two temperatures’.

(b) ‘The efficiency of all reversible engines, working between the same two temperatures is the same, whatever the working substances’.

Carnot’s Cycle Proof

First Part

To prove the first part of the theorem, we consider two engines R and I working between the temperatures T1 and T2 where T1 > T2 (Fig.). Of these two engines, R is reversible and I is irreversible.

Suppose I is more efficient than R. In each cycle, R absorbs the quantity of heat Q1 from the source at T1 and rejects the quantity of heat Q1 from the source at T1 and rejects the quantity of heat Q2 to the sin at T2.

Similarly, in each cycle, I absorb the quantity of heat Q1’ from the source at T1 and gives up the quantity of heat Q2’ to the sink at T2.

Let the two engines do the same amount of work W in each cycle. According to the assumption I is more efficient than R.

Therefore,

,Q1Q2Q1>Q1Q2Q1\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{Q}_{1}}’-{{Q}_{2}}’}{{{Q}_{1}}’}>\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}

Or

WQ1>WQ1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{W}{{{Q}_{1}}’}>\frac{W}{{{Q}_{1}}}

Or

Q1>Q1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Q}_{1}}>{{Q}_{1}}'

And

Q1Q2=Q1Q2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Q}_{1}}’-{{Q}_{2}}’={{Q}_{1}}-{{Q}_{2}}

Or

Q2Q2=Q1Q1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Q}_{2}}-{{Q}_{2}}’={{Q}_{1}}-{{Q}_{1}}'

Now, because Q1>Q1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Q}_{1}}>{{Q}_{1}}'

Therefore,

,Q2>Q2\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Q}_{2}}>{{Q}_{2}}'

Now, if the two engines are coupled together so that I drives R backwards and suppose they use the same source and sink, the combination forms a self-acting machine in which I supplies external work W and R absorbs this amount of work in its reverse cycle. I in its cycle absorbs heat Q’ from the source and gives up heat Q2’ to the sink. R in its reverse cycle absorbs heat Q2 from the sink and gives up heat Q1 to the source.

The net result of the complete cycle of the coupled engines is given by;

Gain of heat by the source at T1 = Q1 – Q1

Loss of heat by the sink at T2 = Q2 – Q2

External work done on the system = 0

Thus, the coupled engines forming a self-acting machine unaided by any external agency transfer heat continuously from a body at low temperature to a body at a higher temperature.

This conclusion is contrary to the second law of thermodynamics, according to which in a cyclic process heat cannot be transferred from one body to another at a higher temperature by a self-acting machine. Hence our assumption is incorrect and we conclude that no engine can be more efficient than a perfectly reversible engine working between the same temperatures.

Second Part

The second part of the theorem may be proved by the same arguments as before. For this purpose, we consider two reversible engines R1 and R2 and assume that R2 is more efficient than R. Proceeding in the same way we can show that R2 cannot be more efficient than R1. Therefore, all reversible engines working between the same two temperatures have the same efficiency.

Thus, the efficiency of a perfectly reversible engine depends only on the temperatures between which the engine work and is independent of the nature of the working substance.

Carnot’s Ideal Heat Engine

In 1824, the French engineer Sadi Carnot conceived a theoretical engine which is free from all practical imperfections. Such an engine cannot be realised in practice. It has maximum efficiency and it is an ideal heat engine. Sadi Carnot’s heat engine requires the following important parts.

Carnot’s Ideal Heat Engine

1. A cylinder having perfectly non-conducting walls, a perfectly conducting base and is provided with a perfectly non-conducting piston which moves without friction in the cylinder. The cylinder contains one mole of perfect gas as the working substance.

2. Source. A reservoir maintained at a constant temperature T1 from which the engine can draw heat by perfect conduction. It has an infinite thermal capacity and any amount of heat can be drawn from it at constant temperature T1.

3. Heat insulating stand. A perfectly mom-conducting platform acts as a stand for adiabatic processes.

4. Sink. A reservoir maintained at a constant lower temperature T2 (T2 < T1) to which the heat engine can reject any amount of heat. The thermal capacity of sink is infinite so that its temperature remains constant at T2, no matter how much heat is given to it.

Carnot’s Cycle And Its Stages

In order to obtain a continuous supply of work, the working substance is subjected to the to a cycle of quasi-static operations known a Carnot’ cycle. The cycle when it acts as a heat engine consists of various steps which are as follows.

Isothermal Expansion

The cylinder is first placed on the source so that the gas acquires the temperatures T1 of the source. It is then allowed to undergo quasi-static expansion. As the gas expands, its temperature tends to fall. Heat passes into the cylinder through the perfectly conducting base which is in contact with the source. The gas, therefore, undergoes slow isothermal expansion at the constant temperature T1.

Isothermal Expansion

Let the working substance during isothermal expansion goes from its initial state A (P1, V1, T1) to the state B (P2, V2, T1) at constant temperature T1 along AB. In this process, the substance absorbs heat Q1 from the source at T1 and does work W1 given by

Q1=W1=V1V2PdV=RT1logeV2V1=areaABGEA{{Q}_{1}}={{W}_{1}}=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{PdV=RT1{{\log }_{e}}}\frac{{{V}_{2}}}{{{V}_{1}}}=area\,ABGEA

Adiabatic Expansion

The cylinder is now removed from the source and is placed on the insulating stand. The gas is allowed to undergo slow adiabatic expansion, performing external work at the expense of its internal energy, until its temperature falls to T2, the same as that of the sink.

This operation is represented by the adiabatic BC, starting from the state B (P2, V2, T1) to the state

C (P3, V3, T2). In this process, there is no transfer of heat, the temperature of the substance falls to T2 and it does some external work W2 given by,

W2=V2V3P.dV=KV2V3dVVγ\,{{W}_{2}}=\int_{{{V}_{2}}}^{{{V}_{3}}}{P.dV=K\int_{{{V}_{2}}}^{{{V}_{3}}}{\frac{dV}{{{V}^{\gamma }}}}}

Because, during adiabatic process, (PVγ=constant=K)(\,P{{V}^{\gamma }}=cons\tan t=K) =KV31γP2V21γ=\,\frac{KV_{3}^{1-\gamma }-{{P}_{2}}{{V}_{2}}}{1-\gamma } =P3V3P2V21λ(P2V2γ=P3V3γ=K)\,=\,\frac{{{P}_{3}}{{V}_{3}}-{{P}_{2}}{{V}_{2}}}{1-\lambda }({{P}_{2}}V_{2}^{\gamma } ={{P}_{3}}V_{3}^{\gamma }=K) =RT2RT11γ=\,\frac{RT{}_{2}-R{{T}_{1}}}{1-\gamma } =R(T1T2)γ1=AreaBCHGB=\,\frac{R({{T}_{1}}-{{T}_{2}})}{\gamma -1} = AreaBCHGB ………(4.16)

Isothermal Compression

The cylinder is now removed from the insulating stand and is placed on the sink which is at a temperature T2. The piston is now very slowly moved inwards so that the work is done on the gas. The temperature tends to increase due to heat produced by compression since the conducting base of the cylinder is in contact with the sink, the heat developed passes to the sink and the temperature of the gas remains constant at T2. Thus the gas undergoes isothermal compression at a constant temperature T2 and gives up some heat to the sink.

This operation is represented by the isothermal CD, starting from the state (P3, V3, T2) to the state

D (P4, V4, T2). In this process, the substance rejects heat Q2 to the sink at T2 and work W3 is done on the substance given by

Q2=W3=V3V4PdV=RT2logeV4V3{{Q}_{2}}={{W}_{3}}=\int_{{{V}_{3}}}^{{{V}_{4}}}{PdV=R{{T}_{2}}{{\log }_{e}}\frac{{{V}_{4}}}{{{V}_{3}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ……(4.17)

=RT2logeV3V4=areaCHFDC\,=\,-R{{T}_{2}}{{\log }_{e}}\frac{{{V}_{3}}}{{{V}_{4}}}=area\,CHFDC

(- ve sign indicates that work is done on the working substance)

Adiabatic Compression

The cylinder is now removed from the sink and again placed on the insulating stand. The piston is slowly moved inwards so that the gas is adiabatic compression is continued till the gas comes back to its original condition i.e. state A (P1, V1, T1), thus completing one full cycle.

This operation is represented by adiabatic DA, starting from D (P4, V4, T2) to the final state A (P1, V1, T1). In this process, work W4 is done on the substance and is given by

W4=V4V1P.dV\,\,\,\,\,\,\,\,\,\,{{W}_{4}}=\int_{{{V}_{4}}}^{{{V}_{1}}}{P.dV} =R(T1T2)γ1=AreaDFEAD\,\,=-\frac{R({{T}_{1}}-{{T}_{2}})}{\gamma-1}=Area\,DFEAD\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, …..(4.18)

(- ve sign indicates that work is done on the working substance. Since W2 and W4 are equal and opposite, they cancel each other.)

Work Done By The Engine Per Cycle

During the above cycle, the working substance absorbs an amount of heat Q1 from the source and rejects Q2 to the sink.

Hence, the net amount of heat absorbed by the gas per cycle

= Q1 – Q2

The net amount of work done by the engine per cycle

= W1 + W2 + W3 + W4

= W1 + W3 because W2=W1){{W}_{2}}=-{{W}_{1}})

From the graph, the net work done per cycle = area ABGEA + area BCHGB – area CHFDC – area DFEAD

= area ABCDA

Thus, the area enclosed by the Carnot’s cycle consisting of two isothermals and two adiabatic gives the net amount of work done per cycle.

In the cyclic process,

Net heat absorbed = Net work done per cycle.

Q1 – Q2 = W1 + W3

RT1logeV2V1RT2logeV3V4RT_{1}\log e\frac{V_2}{V_1} – RT_{2}\log e\frac{V_3}{V_4} ……….(4.19)

Since the points, A and D lie on the same adiabatic Da

T1V1γ1=T2V4γ1{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{4}^{\gamma -1} T2T1=(V1V4)γ1\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{4}}} \right)}^{\gamma -1}} ……….(4.20)

Similarly, point B and C lie on the same adiabatic BC

T1V2γ1=T2V3γ1{{T}_{1}}V_{2}^{\gamma -1}={{T}_{2}}V_{3}^{\gamma -1}

Therefore, T2T1=(V2V3)γ1\,\,\,\,\,\,\,\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{2}}}{{{V}_{3}}} \right)}^{\gamma -1}} ……..(4.21)

From equations (4.20) and (4.21),

(V1V4)γ1=(V2V3)γ1{{\left( \frac{{{V}_{1}}}{{{V}_{4}}} \right)}^{\gamma -1}}={{\left(\frac{{{V}_{2}}}{{{V}_{3}}} \right)}^{\gamma -1}}

Or

V1V4=V2V3\frac{{{V}_{1}}}{{{V}_{4}}}=\frac{{{V}_{2}}}{{{V}_{3}}}

Or

V2V1=V3V4\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{V}_{3}}}{{{V}_{4}}}

Substituting in equation (4.19), we get,

Net work done = Q1Q2=RT1logeV2V1RT2logeV2V1{{Q}_{1}}-{{Q}_{2}}=R{{T}_{1}}{{\log }_{e}}\frac{{{V}_{2}}}{{{V}_{1}}}-R{{T}_{2}}{{\log }_{e}}\frac{{{V}_{2}}}{{{V}_{1}}} W=(Q1Q2)=R(T1T2)logeV2V1W=({{Q}_{1}}-{{Q}_{2}})=R({{T}_{1}}-{{T}_{2}}){{\log}_{e}}\frac{{{V}_{2}}}{{{V}_{1}}}\,\,\,\,\,\,\,\,\,\,\,\, ……….(4.22)

Efficiency

The efficiency of the heat engine is the rate of quantity of heat converted into work (Useful output) per cycle to the total amount of heat absorbed per cycle.

Efficiency,η=usefuloutputInput=WQ1Efficiency,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\eta=\frac{useful\,output}{Input}=\frac{W}{{{Q}_{1}}} =(Q1Q2)Q1=\,\frac{({{Q}_{1}}-{{Q}_{2}})}{{{Q}_{1}}} R(T1T2)logeV2V1RT1logeV2V1\frac{R({{T}_{1}}-{{T}_{2}}){{\log }_{e}}\frac{{{V}_{2}}}{{{V}_{1}}}}{R{{T}_{1}}\log e\frac{{{V}_{2}}}{{{V}_{1}}}} η=T1T2T1\,\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}} orη=1T2T1or\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}

From the equation, we conclude that efficiency depends only upon the temperature of the source and sink and is always less than unity. The efficiency is independent of the nature of the working substance.

From equation, η=1T2T1,\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}},we get η=1,ifT2=oKi.e\eta =1,if\,T2=oK\,i.e the temperature of the sink is at absolute zero degrees. In practice, it is never possible to reach absolute zero and hence 100% conversion of heat energy into mechanical work is not possible.

Again, the efficiency is minimum or zero then T1 = T2 i.e. the temperature of the source is equal to the temperature of the sink, then η = 0 i.e. the engine does not work.

The Carnot’s heat engine is perfectly reversible. It can be operated in the reverse direction also. Then it works like a refrigerator. The heat Q2 is taken from the sink and external work is done on the working substance and heat Q2 is rejected to the source at a higher temperature (principle of a refrigerator).

Moreover, in the Carnot’s heat engine, the process of isothermal and adiabatic expansions and compressions are carried out very-very slowly i.e. quasi-static. This is an ideal case. Any practical engine can not satisfy these conditions. Therefore, all practical engines have an efficiency less than the Carnot’s engine.