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A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:
- a)1.42
- b)2.33
- c)2.87
- d)3.26
Correct answer is option 'B'. Can you explain this answer?
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A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain...
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A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain...
Given:
Power required by Carnot refrigerator = 1.5 kW/ton
Temperature maintained by Carnot refrigerator = 30°C
To find:
COP of Carnot refrigerator
Formula used:
COP = (Refrigeration effect)/Work input
Refrigeration effect = Qh - Ql
Qh = Heat absorbed by the refrigerator from the high-temperature reservoir
Ql = Heat rejected by the refrigerator to the low-temperature reservoir
Solution:
1 ton of refrigeration = 3.5 kW
Power required by Carnot refrigerator = 1.5 kW/ton
Therefore, power required by Carnot refrigerator = 5.25 kW
Refrigeration effect:
Qh = Heat absorbed by the refrigerator from the high-temperature reservoir
Qh = mCpΔT
m = Mass of refrigerant
Cp = Specific heat of refrigerant
ΔT = Temperature difference between high-temperature and low-temperature reservoirs
Given that the temperature maintained by Carnot refrigerator = 30°C
Temperature of high-temperature reservoir = 30+273 = 303 K
Temperature of low-temperature reservoir = 0+273 = 273 K
ΔT = 303 - 273 = 30 K
Let us assume that 1 kg of refrigerant is used in the Carnot refrigerator
Specific heat of refrigerant (Cp) is not given. Let us assume it to be 2.5 kJ/kg.K (for R-134a refrigerant)
Qh = mCpΔT = 1 x 2.5 x 30 = 75 kJ
Ql = Heat rejected by the refrigerator to the low-temperature reservoir
Ql = Qh - Work input
Work input = Power input x Time
Assuming that the Carnot refrigerator operates for 1 hour, Work input = Power input x Time
Work input = 5.25 kW x 1 hour = 5.25 kWh = 5.25 x 3600 = 18900 kJ
Ql = Qh - Work input = 75 - 18900 = -18825 kJ
(Note: Heat rejected cannot be negative. The negative sign indicates that the assumption made for Work input is incorrect. We need to use the actual value of Work input to calculate Ql.)
Actual Work input:
COP = (Refrigeration effect)/Work input
COP = Qh/(Qh - Ql)
COP = 75/(75 + 18825)
COP = 0.004
Power input = Work input/Time
Power input = 18900/3600 = 5.25 kW
(Note: The given power input and the calculated power input are the same.)
Refrigeration effect:
Qh = mCpΔT = 1 x 2.5 x 30 = 75 kJ
Ql = Qh - Work input
Work input = Power input x Time
Work input = 5.25 kW x 1 hour = 5.25 kWh = 5.25 x 3600 = 18900 kJ
Ql = Qh - Work input = 75 - 18900 = -18825 kJ
Actual COP:
COP = Qh/(
Power required by Carnot refrigerator = 1.5 kW/ton
Temperature maintained by Carnot refrigerator = 30°C
To find:
COP of Carnot refrigerator
Formula used:
COP = (Refrigeration effect)/Work input
Refrigeration effect = Qh - Ql
Qh = Heat absorbed by the refrigerator from the high-temperature reservoir
Ql = Heat rejected by the refrigerator to the low-temperature reservoir
Solution:
1 ton of refrigeration = 3.5 kW
Power required by Carnot refrigerator = 1.5 kW/ton
Therefore, power required by Carnot refrigerator = 5.25 kW
Refrigeration effect:
Qh = Heat absorbed by the refrigerator from the high-temperature reservoir
Qh = mCpΔT
m = Mass of refrigerant
Cp = Specific heat of refrigerant
ΔT = Temperature difference between high-temperature and low-temperature reservoirs
Given that the temperature maintained by Carnot refrigerator = 30°C
Temperature of high-temperature reservoir = 30+273 = 303 K
Temperature of low-temperature reservoir = 0+273 = 273 K
ΔT = 303 - 273 = 30 K
Let us assume that 1 kg of refrigerant is used in the Carnot refrigerator
Specific heat of refrigerant (Cp) is not given. Let us assume it to be 2.5 kJ/kg.K (for R-134a refrigerant)
Qh = mCpΔT = 1 x 2.5 x 30 = 75 kJ
Ql = Heat rejected by the refrigerator to the low-temperature reservoir
Ql = Qh - Work input
Work input = Power input x Time
Assuming that the Carnot refrigerator operates for 1 hour, Work input = Power input x Time
Work input = 5.25 kW x 1 hour = 5.25 kWh = 5.25 x 3600 = 18900 kJ
Ql = Qh - Work input = 75 - 18900 = -18825 kJ
(Note: Heat rejected cannot be negative. The negative sign indicates that the assumption made for Work input is incorrect. We need to use the actual value of Work input to calculate Ql.)
Actual Work input:
COP = (Refrigeration effect)/Work input
COP = Qh/(Qh - Ql)
COP = 75/(75 + 18825)
COP = 0.004
Power input = Work input/Time
Power input = 18900/3600 = 5.25 kW
(Note: The given power input and the calculated power input are the same.)
Refrigeration effect:
Qh = mCpΔT = 1 x 2.5 x 30 = 75 kJ
Ql = Qh - Work input
Work input = Power input x Time
Work input = 5.25 kW x 1 hour = 5.25 kWh = 5.25 x 3600 = 18900 kJ
Ql = Qh - Work input = 75 - 18900 = -18825 kJ
Actual COP:
COP = Qh/(
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Community Answer
A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain...
1ton of refrigeration = 3.51 kw.... so the COP will be (desired effect/work input) =(3.51/1.5)=2.34 (option B)
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A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer?.
A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer?.
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A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A Carnot refrigerator requires 1.5 kW/ton of refrigeration to maintain a region at a temperature of – 30°C. The COP of the Carnot refrigerator is:a)1.42 b)2.33 c)2.87 d)3.26Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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