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For simple vapour compression cycle, enthalpy at suction = 1600 kJ/kg, enthalpy at discharge from the compressor = 1800 kJ/kg, enthalpy at exit from condenser = 600 kJ/kg.
What is the COP for this refrigeration cycle?
- a)3·3
- b)5·0
- c)4
- d)4·5
Correct answer is option 'B'. Can you explain this answer?
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For simple vapour compression cycle, enthalpy at suction = 1600 kJ/kg,...
COP of refrigeration cycle
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For simple vapour compression cycle, enthalpy at suction = 1600 kJ/kg,...
Calculation of COP for Simple Vapour Compression Cycle
Given data:
- Enthalpy at suction (h1) = 1600 kJ/kg
- Enthalpy at discharge from the compressor (h2) = 1800 kJ/kg
- Enthalpy at exit from condenser (h3) = 600 kJ/kg
The Coefficient of Performance (COP) of a refrigeration cycle is defined as the ratio of the heat removed from the cold space to the work done on the refrigerant. It is given by the formula:
COP = Heat removed / Work done
For a simple vapour compression cycle, the heat removed is equal to the difference between the enthalpy at the condenser exit and the enthalpy at the suction, which is given by:
Heat removed = h1 - h3
The work done on the refrigerant is equal to the difference between the enthalpy at the discharge from the compressor and the enthalpy at the suction, which is given by:
Work done = h2 - h1
Substituting the given values, we get:
Heat removed = 1600 - 600 = 1000 kJ/kg
Work done = 1800 - 1600 = 200 kJ/kg
Therefore, the COP is given by:
COP = Heat removed / Work done = 1000 / 200 = 5
Thus, the correct option is B) 5.
Given data:
- Enthalpy at suction (h1) = 1600 kJ/kg
- Enthalpy at discharge from the compressor (h2) = 1800 kJ/kg
- Enthalpy at exit from condenser (h3) = 600 kJ/kg
The Coefficient of Performance (COP) of a refrigeration cycle is defined as the ratio of the heat removed from the cold space to the work done on the refrigerant. It is given by the formula:
COP = Heat removed / Work done
For a simple vapour compression cycle, the heat removed is equal to the difference between the enthalpy at the condenser exit and the enthalpy at the suction, which is given by:
Heat removed = h1 - h3
The work done on the refrigerant is equal to the difference between the enthalpy at the discharge from the compressor and the enthalpy at the suction, which is given by:
Work done = h2 - h1
Substituting the given values, we get:
Heat removed = 1600 - 600 = 1000 kJ/kg
Work done = 1800 - 1600 = 200 kJ/kg
Therefore, the COP is given by:
COP = Heat removed / Work done = 1000 / 200 = 5
Thus, the correct option is B) 5.
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For simple vapour compression cycle, enthalpy at suction = 1600 kJ/kg, enthalpy at discharge from the compressor = 1800 kJ/kg, enthalpy at exit from condenser = 600 kJ/kg.What is the COP for this refrigeration cycle?a)3·3 b)5·0 c)4 d)4·5Correct answer is option 'B'. Can you explain this answer?
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