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Efficiency of a Carnot engine is 75%. If the cycle direction is reversed, COP of the reversed Carnot cycle is
- a)1.33
- b)0.75
- c)0.33
- d)1.75
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Efficiency of a Carnot engine is 75%. If the cycle direction is revers...
If the cycle direction is reversed it will be a heat pump not refrigerator. Students make a common mistake here and calculated (COP)R. We know that the definition of refrigeration is producing a temperature below atmospheric temperature. In heat engine lower temperature is atmospheric temperature. When we reverse this cycle then lower temperature will be atmospheric and higher temperature will be more than atmospheric means it will be a Heat Pump not a refrigerator.
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Efficiency of a Carnot engine is 75%. If the cycle direction is revers...
Given, efficiency of a Carnot engine = 75%
Efficiency of a Carnot Engine
The efficiency of a Carnot engine is given by the formula:
η = 1 - (T2/T1)
where, T1 is the temperature of the high-temperature reservoir and T2 is the temperature of the low-temperature reservoir.
Reversed Carnot Cycle
When the direction of the Carnot cycle is reversed, the engine becomes a refrigerator or a heat pump. The coefficient of performance (COP) of a refrigerator or a heat pump is given by the formula:
COP = QL/W
where, QL is the amount of heat removed from the low-temperature reservoir and W is the work input required.
COP of a Reversed Carnot Cycle
The COP of a reversed Carnot cycle is given by the formula:
COP = TH/(TH - TL)
where, TH is the temperature of the high-temperature reservoir and TL is the temperature of the low-temperature reservoir.
Solution
Efficiency of a Carnot engine = 75%
η = 0.75
1 - (T2/T1) = 0.75
T2/T1 = 0.25
T2 = 0.25T1
COP of a reversed Carnot cycle = TH/(TH - TL)
= T2/(TH - T2)
= (0.25T1)/(TH - 0.25T1)
= (1/4)/(1 - (1/4))
= 1.33
Therefore, the correct answer is option A (1.33).
Efficiency of a Carnot Engine
The efficiency of a Carnot engine is given by the formula:
η = 1 - (T2/T1)
where, T1 is the temperature of the high-temperature reservoir and T2 is the temperature of the low-temperature reservoir.
Reversed Carnot Cycle
When the direction of the Carnot cycle is reversed, the engine becomes a refrigerator or a heat pump. The coefficient of performance (COP) of a refrigerator or a heat pump is given by the formula:
COP = QL/W
where, QL is the amount of heat removed from the low-temperature reservoir and W is the work input required.
COP of a Reversed Carnot Cycle
The COP of a reversed Carnot cycle is given by the formula:
COP = TH/(TH - TL)
where, TH is the temperature of the high-temperature reservoir and TL is the temperature of the low-temperature reservoir.
Solution
Efficiency of a Carnot engine = 75%
η = 0.75
1 - (T2/T1) = 0.75
T2/T1 = 0.25
T2 = 0.25T1
COP of a reversed Carnot cycle = TH/(TH - TL)
= T2/(TH - T2)
= (0.25T1)/(TH - 0.25T1)
= (1/4)/(1 - (1/4))
= 1.33
Therefore, the correct answer is option A (1.33).
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Efficiency of a Carnot engine is 75%. If the cycle direction is reversed, COP of the reversed Carnot cycle is a)1.33 b)0.75 c)0.33d)1.75Correct answer is option 'A'. Can you explain this answer?
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