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In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?
- a)0.50 kW
- b)0.667 kW
- c)1.5 kW
- d)0.0341 kW
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In a heat engine operating in a cycle between a source temperature of...
T1 = 606 + 273 = 879 K
T2 = 20 + 273 = 293 K
n = T1 - T,/T1
= 879 - 293 / 879 = W/Q1 (since W = 1 kW)
0.666 = 1/Q1
Q1 = 1.5 kW
Amount of heat rejected
Q2 = Q1 - W = 1.5 - 1.0 = 0.5 kW
Most Upvoted Answer
In a heat engine operating in a cycle between a source temperature of...
In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, the least rate of heat rejection per kW net output of the engine can be calculated using the Carnot efficiency formula.
Carnot Efficiency (η) = 1 - (Ts/Th)
Where:
η = Carnot Efficiency
Ts = Sink Temperature
Th = Source Temperature
Given:
Sink Temperature (Ts) = 20°C
Source Temperature (Th) = 606°C
Calculating the Carnot Efficiency:
η = 1 - (20/606)
= 1 - 0.033
= 0.967
The Carnot Efficiency represents the fraction of heat energy that can be converted into useful work in the ideal case. Since the Carnot Efficiency is always less than 1, there will always be some heat energy that needs to be rejected to the sink.
To calculate the least rate of heat rejection per kW net output of the engine, we need to divide the heat energy that needs to be rejected by the net output power of the engine.
Let's assume the net output power of the engine is 1 kW.
Least Rate of Heat Rejection = (1 - η) / (Net Output Power)
Substituting the values:
Least Rate of Heat Rejection = (1 - 0.967) / 1
= 0.033 / 1
= 0.033 kW
Therefore, the least rate of heat rejection per kW net output of the engine is 0.033 kW.
However, the given answer options do not match this result. Option A states the correct answer is 0.50 kW, which is incorrect. It seems there may be an error in the answer options provided.
Carnot Efficiency (η) = 1 - (Ts/Th)
Where:
η = Carnot Efficiency
Ts = Sink Temperature
Th = Source Temperature
Given:
Sink Temperature (Ts) = 20°C
Source Temperature (Th) = 606°C
Calculating the Carnot Efficiency:
η = 1 - (20/606)
= 1 - 0.033
= 0.967
The Carnot Efficiency represents the fraction of heat energy that can be converted into useful work in the ideal case. Since the Carnot Efficiency is always less than 1, there will always be some heat energy that needs to be rejected to the sink.
To calculate the least rate of heat rejection per kW net output of the engine, we need to divide the heat energy that needs to be rejected by the net output power of the engine.
Let's assume the net output power of the engine is 1 kW.
Least Rate of Heat Rejection = (1 - η) / (Net Output Power)
Substituting the values:
Least Rate of Heat Rejection = (1 - 0.967) / 1
= 0.033 / 1
= 0.033 kW
Therefore, the least rate of heat rejection per kW net output of the engine is 0.033 kW.
However, the given answer options do not match this result. Option A states the correct answer is 0.50 kW, which is incorrect. It seems there may be an error in the answer options provided.
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In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer?
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In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer?.
In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a heat engine operating in a cycle between a source temperature of 606°C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?a)0.50 kWb)0.667 kWc)1.5 kWd)0.0341 kWCorrect answer is option 'A'. Can you explain this answer?.
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