Chapter 12.9 - COP of a Carnot Refrigerator

In the previous section, we saw the efficiency of the Carnot engine. In this section, we will see details about the Carnot refrigerator

• A Carnot engine can be operated in the reverse direction to work as a Carnot refrigerator
• The four segments in the PV diagram will reverse their directions. This is shown in fig.6.24 below:

Fig.12.24

The basic working can be explained in 4 steps:
1. The cycle begins at point D. The first segment is DC
    ♦ The engine is placed in contact with the cold reservoir of temperature T₂
    ♦ The engine extracts heat from the cold reservoir and expands isothermally
2. The second segment is CB
    ♦ The engine is placed inside an insulator
    ♦ The gas is compressed adiabatically
    ♦ The temperature increases to T₁
3. The third segment is BA
    ♦ The engine is placed in contact with the hot reservoir of temperature T₁
    ♦ The gas is compressed isothermally
    ♦ During this compression, the gas rejects some heat into the hot reservoir
4. The fourth segment is AD
    ♦ The engine is placed inside an insulator
    ♦ The gas is allowed to expand adiabatically
    ♦ The temperature falls to T₂
• Thus the initial point D is reached, completing one cycle

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Next we will calculate the work done in each segment:

First segment from D to C
• In this segment, the gas place in contact with T₂ and subjected to an isothermal expansion
    ♦ Volume increases from V_D to V_C
    ♦ Pressure decreases from P_D to P_C
    ♦ Temperature remains constant at T₂
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from C to B
• In this segment, the gas is subjected to an adiabatic compression
    ♦ Volume decreases from V_C to V_B
    ♦ Pressure increases from P_C to P_B
• 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 B to A
• In this segment, the gas is subjected to an isothermal compression
    ♦ Volume decreases from V_B to V_A
    ♦ Pressure increases from P_B to P_A
    ♦ Temperature remains constant at T₁
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$

Fourth segment from A to D
• In this segment, the gas is subjected to an adiabatic expansion
    ♦ Volume increases from V_A to V_D
    ♦ Pressure decreases from P_A to P_D
• 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 coefficient of performance (COP) of the Carnot refrigerator. It can be written in 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 four
   ♦ Work done on the gas = Work done during segments two and three
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( \frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}+n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}-n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$

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• An interesting comparison:
    ♦ In the previous section for the engine, we obtained the net work as:
          $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
    ♦ In this section for the refrigerator, we obtain the net work as:
          $\mathbf\small{\rm{\left (n R T_2\,\,ln \frac{V_C}{V_D}-n R T_1\,\,ln \frac{V_B}{V_A} \right)}}$
    ♦ Both are equal in magnitude but opposite in sign
◼ That means:
    ♦ In the engine, work is done by the gas
    ♦ In the refrigerator (reversed engine), the same work is done on the gas

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2. We know that, COP of a refrigerator is given by:
$\mathbf\small{\rm{\alpha = \frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Work}}}$ (see Eq.12.12 in section 12.6)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\alpha = \frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Heat}}}$ (see Eq.12.13 in section 12.6)
• It is easier to find COP 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 = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_C}{V_D}}}$
(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 = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_B}{V_A}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}-n R T_2\,\,ln \frac{V_C}{V_D}}}}$
6. In the previous section, we compared the two adiabatic segments, and proved that: $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
• So the result in (5) becomes:
$\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_C}{V_D}-n R T_2\,\,ln \frac{V_C}{V_D}}}}$
⇒ $\mathbf\small{\rm{\alpha = \frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R \,\,ln \frac{V_C}{V_D}(T_1 - T_2)}}}$
• Thus we get:
Eq.12.16: $\mathbf\small{\rm{\alpha = \frac{T_2}{T_1 - T_2}}}$
7. Two expressions for COP:
   ♦ The above Eq.12.16 gives us an expression for COP
   ♦ Step (3) also gives us an expression for COP
• So we can equate the two. We get Eq.12.17:
$\mathbf\small{\rm{\frac{Heat \; Absorbed \; from \; cold \; reservoir}{Net \; Heat}=\frac{T_2}{T_1 - T_2}}}$

Let us see a solved example
Solved example 12.11
A refrigerator is to maintain eatables kept inside at 9 ^०C. If room temperature is 36 ^०C, Calculate the coefficient of performance
Solution:
1. Given that:
   ♦ Temperature of the hot reservoir T₁ = 36 ^०C = (273 + 36) = 309 K
   ♦ Temperature of the cold reservoir T₂ = 9 ^०C = (273 + 9 ) = 282 K
2. We have Eq.12.16: $\mathbf\small{\rm{\alpha = \frac{T_2}{T_1 - T_2}}}$
• Substituting the values, we get:
$\mathbf\small{\rm{\alpha = \frac{282}{309 - 282}}}$ = 10.44

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In the next section, we will see the second law of thermodynamics and it's application to heat engines and refrigerators


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