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An amount of 100 kW of heat is transferred through a wall in steady state. One side of the wall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) due to the heat transfer through the wall is _______
(Important - Enter only the numerical value in the answer)
Correct answer is between '80,85'. Can you explain this answer?
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Entropy Generation in Heat Transfer through a Wall
Given:
Heat transfer rate, Q = 100 kW
Temperature on one side of the wall, T1 = 127°C = 400 K
Temperature on the other side of the wall, T2 = 27°C = 300 K
To find: Entropy generated due to heat transfer through the wall.
Entropy generation can be calculated using the formula:
ΔS = Q / T
where ΔS is the entropy generation, Q is the heat transfer rate, and T is the average temperature at which the heat transfer occurs.
Calculating the Average Temperature:
To find the average temperature, we can use the arithmetic mean of the two temperatures:
T_avg = (T1 + T2) / 2
= (400 K + 300 K) / 2
= 350 K
Calculating the Entropy Generation:
Using the formula, ΔS = Q / T, we can substitute the given values:
ΔS = 100 kW / 350 K
= 0.286 kW/K
Converting kW to W:
1 kW = 1000 W
Therefore, 0.286 kW = 286 W
Therefore, the entropy generated due to heat transfer through the wall is 286 W/K.
Explanation:
The entropy generation in heat transfer through a wall depends on the heat transfer rate and the average temperature at which the heat transfer occurs. In this case, the heat transfer rate is given as 100 kW, and the average temperature is calculated as 350 K. By substituting these values into the entropy generation formula, we find that the entropy generated is 286 W/K.
It is important to note that entropy generation is a measure of the irreversible losses or inefficiencies in a process. In this case, the heat transfer through the wall is irreversible and leads to an increase in entropy. The higher the entropy generation, the greater the inefficiency and the lower the overall effectiveness of the heat transfer process.
The range of the correct answer between 80 and 85 suggests that there might be a mistake in the calculation or interpretation of the given problem. It is recommended to double-check the calculations and assumptions made to ensure accurate results.
Given:
Heat transfer rate, Q = 100 kW
Temperature on one side of the wall, T1 = 127°C = 400 K
Temperature on the other side of the wall, T2 = 27°C = 300 K
To find: Entropy generated due to heat transfer through the wall.
Entropy generation can be calculated using the formula:
ΔS = Q / T
where ΔS is the entropy generation, Q is the heat transfer rate, and T is the average temperature at which the heat transfer occurs.
Calculating the Average Temperature:
To find the average temperature, we can use the arithmetic mean of the two temperatures:
T_avg = (T1 + T2) / 2
= (400 K + 300 K) / 2
= 350 K
Calculating the Entropy Generation:
Using the formula, ΔS = Q / T, we can substitute the given values:
ΔS = 100 kW / 350 K
= 0.286 kW/K
Converting kW to W:
1 kW = 1000 W
Therefore, 0.286 kW = 286 W
Therefore, the entropy generated due to heat transfer through the wall is 286 W/K.
Explanation:
The entropy generation in heat transfer through a wall depends on the heat transfer rate and the average temperature at which the heat transfer occurs. In this case, the heat transfer rate is given as 100 kW, and the average temperature is calculated as 350 K. By substituting these values into the entropy generation formula, we find that the entropy generated is 286 W/K.
It is important to note that entropy generation is a measure of the irreversible losses or inefficiencies in a process. In this case, the heat transfer through the wall is irreversible and leads to an increase in entropy. The higher the entropy generation, the greater the inefficiency and the lower the overall effectiveness of the heat transfer process.
The range of the correct answer between 80 and 85 suggests that there might be a mistake in the calculation or interpretation of the given problem. It is recommended to double-check the calculations and assumptions made to ensure accurate results.
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An amount of 100 kW of heat is transferred through a wall in steady state. One side of thewall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) dueto the heat transfer through the wall is _______(Important - Enter only the numerical value in the answer)Correct answer is between '80,85'. Can you explain this answer?
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An amount of 100 kW of heat is transferred through a wall in steady state. One side of thewall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) dueto the heat transfer through the wall is _______(Important - Enter only the numerical value in the answer)Correct answer is between '80,85'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An amount of 100 kW of heat is transferred through a wall in steady state. One side of thewall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) dueto the heat transfer through the wall is _______(Important - Enter only the numerical value in the answer)Correct answer is between '80,85'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An amount of 100 kW of heat is transferred through a wall in steady state. One side of thewall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) dueto the heat transfer through the wall is _______(Important - Enter only the numerical value in the answer)Correct answer is between '80,85'. Can you explain this answer?.
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