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A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthalpy 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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Verified Answer
A simple saturated refrigeration cycle has the following state points...
Refrigerating effect = h1−h4
Work input = h2−h1
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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
Understanding the refrigeration cycle:
In a simple saturated refrigeration cycle, the refrigerant undergoes four processes: compression, condensation, expansion (throttling), and evaporation. The refrigerant absorbs heat from the refrigerated space during evaporation and releases it to the surroundings during condensation.
Calculating the COP:
The coefficient of performance (COP) is the ratio of the desired effect (refrigeration) to the required work input (compressor work).
Step 1: Determine the heat absorbed during evaporation:
The heat absorbed during evaporation is given by the difference in enthalpy before and after throttling:
Heat absorbed = Enthalpy before throttling - Enthalpy after throttling = 375 kJ/kg - 125 kJ/kg = 250 kJ/kg
Step 2: Determine the work input during compression:
The work input during compression can be calculated using the first law of thermodynamics:
Work input = Enthalpy after compression - Enthalpy before compression = 425 kJ/kg - 375 kJ/kg = 50 kJ/kg
Step 3: Calculate the COP:
COP = Heat absorbed / Work input = 250 kJ/kg / 50 kJ/kg = 5
Conclusion:
The coefficient of performance (COP) of the refrigeration cycle is 5. Therefore, the correct answer is option A.
- Enthalpy after compression = 425 kJ/kg
- Enthalpy after throttling = 125 kJ/kg
- Enthalpy before compression = 375 kJ/kg
Understanding the refrigeration cycle:
In a simple saturated refrigeration cycle, the refrigerant undergoes four processes: compression, condensation, expansion (throttling), and evaporation. The refrigerant absorbs heat from the refrigerated space during evaporation and releases it to the surroundings during condensation.
Calculating the COP:
The coefficient of performance (COP) is the ratio of the desired effect (refrigeration) to the required work input (compressor work).
Step 1: Determine the heat absorbed during evaporation:
The heat absorbed during evaporation is given by the difference in enthalpy before and after throttling:
Heat absorbed = Enthalpy before throttling - Enthalpy after throttling = 375 kJ/kg - 125 kJ/kg = 250 kJ/kg
Step 2: Determine the work input during compression:
The work input during compression can be calculated using the first law of thermodynamics:
Work input = Enthalpy after compression - Enthalpy before compression = 425 kJ/kg - 375 kJ/kg = 50 kJ/kg
Step 3: Calculate the COP:
COP = Heat absorbed / Work input = 250 kJ/kg / 50 kJ/kg = 5
Conclusion:
The coefficient of performance (COP) of the refrigeration cycle is 5. Therefore, the correct answer is option A.
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A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthalpy 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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