A refrigerator using Carnot cycle requires 1.25 kW per tonne of refrigeration to maintain a temperature of −30°C. Find: (a) COP of the Carnot refrigerator, (b) temperature at which heat is rejected; and (c) Heat rejected per tonne of refrigeration.
The Correct Answer and Explanation is :
Solution:
Given data:
- Power required ( W = 1.25 \, \text{kW} )
- Refrigeration effect = 1 tonne of refrigeration (TR) = ( 210 \, \text{kJ/min} )
- Evaporator temperature (low temperature) ( T_L = -30^\circ C = 243 \, \text{K} )
(a) COP of the Carnot Refrigerator:
The coefficient of performance (COP) for a Carnot refrigerator is given by:
[
\text{COP} = \frac{T_L}{T_H – T_L}
]
where ( T_H ) is the temperature at which heat is rejected.
The power input ( W ) relates to the refrigeration effect (( Q_L )) as:
[
\text{COP} = \frac{Q_L}{W}
]
Given ( W = 1.25 \, \text{kW} ) and ( Q_L = 210 \, \text{kJ/min} = \frac{210}{60} = 3.5 \, \text{kW} ),
[
\text{COP} = \frac{3.5}{1.25} = 2.8
]
(b) Temperature at which heat is rejected (( T_H )):
From the COP formula:
[
T_H = T_L \left(1 + \frac{1}{\text{COP}}\right)
]
Substituting:
[
T_H = 243 \left(1 + \frac{1}{2.8}\right) = 243 \times 1.357 = 329.8 \, \text{K}
]
[
T_H = 329.8 – 273 = 56.8^\circ C
]
(c) Heat rejected per tonne of refrigeration (( Q_H )):
Heat rejected is the sum of refrigeration effect and work input:
[
Q_H = Q_L + W
]
Substituting:
[
Q_H = 3.5 + 1.25 = 4.75 \, \text{kW}
]
[
Q_H = 4.75 \, \text{kW} \times 60 = 285 \, \text{kJ/min}
]
Final Answers:
(a) COP = 2.8
(b) Heat rejection temperature = 56.8°C
(c) Heat rejected = 4.75 kW (285 kJ/min)
Explanation (300 words):
The Carnot refrigerator represents an idealized refrigeration cycle operating between two temperature reservoirs. It provides the theoretical maximum performance. The COP, a measure of efficiency, depends on the temperature difference between the heat source (condenser) and the sink (evaporator).
Given that the refrigeration effect is 1 tonne, equivalent to 3.5 kW, the power input is 1.25 kW. From the relation ( \text{COP} = \frac{Q_L}{W} ), the Carnot refrigerator’s COP is calculated as 2.8, indicating that for every 1 kW of input, 2.8 kW of heat is extracted.
To find the rejection temperature (( T_H )), the relationship between the evaporator temperature (( T_L )), COP, and ( T_H ) is used. A higher rejection temperature reduces the system’s efficiency as it increases the temperature gradient. For this case, ( T_H ) is calculated as approximately ( 329.8 \, \text{K} ) (or ( 56.8^\circ \text{C} )).
Finally, the heat rejected per tonne of refrigeration (( Q_H )) is the sum of the refrigeration effect (( Q_L )) and the input work (( W )). This represents the energy expelled to the surroundings. For the given system, ( Q_H = 4.75 \, \text{kW} ) (or ( 285 \, \text{kJ/min} )).
This example underscores the Carnot cycle’s theoretical potential to optimize refrigeration while highlighting practical constraints like operational losses that prevent real systems from achieving Carnot efficiency.