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A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:
- a)5
- b)3.5
- c)6
- d)Not possible to find with this data
Correct answer is option 'A'. Can you explain this answer?
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A simple saturated refrigeration cycle has the following state points....
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A simple saturated refrigeration cycle has the following state points....
Given Data:
Enthalpy after compression = 425 kJ/kg
Enthalpy after throttling = 125 kJ/kg
Enthalpy before compression = 375 kJ/kg
Solution:
To find the coefficient of performance (COP) of the refrigeration cycle, we need to determine the heat absorbed (Qin) and the work done (W).
Step 1: Determine the heat absorbed (Qin):
The heat absorbed by the refrigerant during the refrigeration cycle can be calculated using the enthalpy values.
Qin = Enthalpy after throttling - Enthalpy before compression
= 125 kJ/kg - 375 kJ/kg
= -250 kJ/kg
Step 2: Determine the work done (W):
The work done in the refrigeration cycle can be calculated using the enthalpy values.
W = Enthalpy after compression - Enthalpy before compression
= 425 kJ/kg - 375 kJ/kg
= 50 kJ/kg
Step 3: Calculate the COP:
The coefficient of performance (COP) of the refrigeration cycle is given by the ratio of the heat absorbed to the work done.
COP = Qin / W
= -250 kJ/kg / 50 kJ/kg
= -5
Since the COP should be a positive value, we take the absolute value of the calculated COP.
COP = | -5 |
= 5
Therefore, the correct answer is option 'A' - 5.
Enthalpy after compression = 425 kJ/kg
Enthalpy after throttling = 125 kJ/kg
Enthalpy before compression = 375 kJ/kg
Solution:
To find the coefficient of performance (COP) of the refrigeration cycle, we need to determine the heat absorbed (Qin) and the work done (W).
Step 1: Determine the heat absorbed (Qin):
The heat absorbed by the refrigerant during the refrigeration cycle can be calculated using the enthalpy values.
Qin = Enthalpy after throttling - Enthalpy before compression
= 125 kJ/kg - 375 kJ/kg
= -250 kJ/kg
Step 2: Determine the work done (W):
The work done in the refrigeration cycle can be calculated using the enthalpy values.
W = Enthalpy after compression - Enthalpy before compression
= 425 kJ/kg - 375 kJ/kg
= 50 kJ/kg
Step 3: Calculate the COP:
The coefficient of performance (COP) of the refrigeration cycle is given by the ratio of the heat absorbed to the work done.
COP = Qin / W
= -250 kJ/kg / 50 kJ/kg
= -5
Since the COP should be a positive value, we take the absolute value of the calculated COP.
COP = | -5 |
= 5
Therefore, the correct answer is option 'A' - 5.
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A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer?
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