Chapter 12.8 - Efficiency of a Carnot Engine

In the previous section, we saw the basic processes in a Carnot engine. In this section, we will see the work done during each of those processes. For easy reference, the final PV diagram is shown below:

Fig.12.23

First segment from A to B
• In this segment, the gas is subjected to an isothermal expansion
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from B to C
• In this segment, the gas is subjected to an adiabatic expansion
• We know that, the work done by the ideal gas during the adiabatic expansion will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$ (see Eq.12.6 at end of section 12.3)

Third segment from C to D
• In this segment, the gas is subjected to an isothermal compression
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$

Fourth segment from D to A
• In this segment, the gas is subjected to an adiabatic compression
• We know that, the work done by the ideal gas during the adiabatic compression will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$

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Now we can find the efficiency of the Carnot engine. It can be written in 9 steps:
1. Net work, W = Work done by the gas - Work done on the gas
   ♦ Work done by the gas = Work done during segments one and two
   ♦ Work done on the gas = Work done during segments three and four
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
2. We know that, efficiency of a heat engine is given by:
$\mathbf\small{\rm{\eta = \frac{Net \; Work}{Heat \; Absorbed}}}$ (see Eq.12.9 in section 12.5)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\eta = 1-\frac{Heat \;Rejected}{Heat \; Absorbed}}}$ (see Eq.12.10 in section 12.5)
• It is easier to find efficiency using this equation
4. So our next task is to find the 'Heat absorbed' and the 'Heat rejected'
• It can be done in steps:
(i) We know that, heat absorption takes place only during the first segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(ii) We know that, heat rejection takes place only during the third segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat rejected = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\eta = 1-\frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}}}}$
⇒ $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right ) \left (\frac{ln \frac{V_C}{V_D}}{ln \frac{V_B}{V_A}}\right )}}$
6. We see that, the right side of the above equation involves temperatures and volumes. So our next task is to find the relations between them
• For that, we consider the two adiabatic processes in the cycle. It can be written in two steps:
(i) For the  adiabatic process BC, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(see Eq.12.7 of section 12.3)
(ii) Similarly, for the adiabatic process DA, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_D}{V_A} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(iii) The right sides in (i) and (ii) are the same. So we get:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{V_D}{V_A} \right )^{\gamma - 1}}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )=\left (\frac{V_D}{V_A} \right )}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
7. We see the above result in (5)
So the result in (5) becomes:
Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
8. Two expressions for efficiency:
   ♦ The above Eq.12.14 gives us an expression for efficiency
   ♦ Step (3) also gives us an expression for efficiency
• So we can equate the two:
$\mathbf\small{\rm{1-\frac{Heat \;Rejected}{Heat \; Absorbed} = 1- \left (\frac{T_2}{T_1} \right )}}$
• Thus we get Eq.12.15:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{T_2}{T_1} \right )}}$

⇒ $\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{Temperature \; of \; cold \; reservoir}{Temperature \; of \; hot \; reservoir} \right )}}$
9. Let us use try the results in (1) and (2) to find efficiency. it can be written in 3 steps:
(i) Using the result in (6), the result in (1) becomes:
W = $\mathbf\small{\rm{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}}$
(ii) Using the result in 4(i), we have:
Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(iii) Applying the above two results into (2), we get:
$\mathbf\small{\rm{\eta = \frac{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}{n R T_1\,\,ln \frac{V_B}{V_A}}=\frac{T_1 - T_2}{T_1}}}$
⇒ $\mathbf\small{\rm{\eta =1-\left (\frac{T_2}{T_1} \right )}}$
This is the same result as in Eq.12.14

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Let us see a solved example
Solved example 12.10
Three heat engines operate between the following temperature differences
    ♦ Engine A: T₁ = 1000 K, T₂ = 700 K
    ♦ Engine A: T₁ = 800 K, T₂ = 500 K
    ♦ Engine A: T₁ = 600 K, T₂ = 300 K
• Which is the most efficient engine?
• Which is the least efficient engine?
Solution:
1. Efficiency of a heat engine is given by Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
• So we get:
    ♦ Efficiency of Engine A = $\mathbf\small{\rm{\eta = 1- \left (\frac{700}{1000} \right )}}$ = 0.3
    ♦ Efficiency of Engine B = $\mathbf\small{\rm{\eta = 1- \left (\frac{500}{800} \right )}}$ = 0.375
    ♦ Efficiency of Engine C = $\mathbf\small{\rm{\eta = 1- \left (\frac{300}{600} \right )}}$ = 0.5
2. We can write:
    ♦ Engine C is the most efficient
    ♦ Engine A is the least efficient
3. Dependence of efficiency on temperature:
    ♦ We see that, the temperature difference is the same for all three engines
    ♦ In such a situation, the engine with the smallest T₂ has the most efficiency 

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In the next section, we will see Carnot refrigerator


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