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Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).
Correct answer is '1.1821'. Can you explain this answer?
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Consider a one-dimensional steady heat conduction process through a so...
Rate of Heat Transfer
To determine the rate of entropy generation, we first need to calculate the rate of heat transfer through the slab. The rate of heat transfer (Q) can be calculated using the formula:
Q = k * A * (T1 - T2) / L
where:
- Q is the rate of heat transfer
- k is the thermal conductivity of the slab
- A is the surface area of the slab
- T1 and T2 are the temperatures of the higher and lower temperature sides, respectively
- L is the thickness of the slab
Given:
- k = 15 W/m.K
- A = 1 m^2 (assuming the slab has a unit width)
- T1 = 80 °C
- T2 is not given, but we can assume it to be a lower temperature (e.g., 0 °C)
- L = 0.1 m
Using these values, we can calculate the rate of heat transfer:
Q = 15 * 1 * (80 - 0) / 0.1 = 12000 W
Rate of Entropy Generation
The rate of entropy generation (S_dot_gen) can be calculated using the formula:
S_dot_gen = Q / T_avg
where:
- S_dot_gen is the rate of entropy generation
- Q is the rate of heat transfer
- T_avg is the average temperature between the higher and lower temperature sides
To calculate T_avg, we can use the formula:
T_avg = (T1 + T2) / 2
Substituting the given values:
T_avg = (80 + 0) / 2 = 40 °C
Now, we can calculate the rate of entropy generation:
S_dot_gen = 12000 / 40 = 300 W/°C
Converting from °C to K:
S_dot_gen = 300 / (40 + 273.15) = 1.1819 W/K
Rounding off to two decimal places, the rate of entropy generation per unit area during the heat transfer process is approximately 1.18 W/m^2.K.
To determine the rate of entropy generation, we first need to calculate the rate of heat transfer through the slab. The rate of heat transfer (Q) can be calculated using the formula:
Q = k * A * (T1 - T2) / L
where:
- Q is the rate of heat transfer
- k is the thermal conductivity of the slab
- A is the surface area of the slab
- T1 and T2 are the temperatures of the higher and lower temperature sides, respectively
- L is the thickness of the slab
Given:
- k = 15 W/m.K
- A = 1 m^2 (assuming the slab has a unit width)
- T1 = 80 °C
- T2 is not given, but we can assume it to be a lower temperature (e.g., 0 °C)
- L = 0.1 m
Using these values, we can calculate the rate of heat transfer:
Q = 15 * 1 * (80 - 0) / 0.1 = 12000 W
Rate of Entropy Generation
The rate of entropy generation (S_dot_gen) can be calculated using the formula:
S_dot_gen = Q / T_avg
where:
- S_dot_gen is the rate of entropy generation
- Q is the rate of heat transfer
- T_avg is the average temperature between the higher and lower temperature sides
To calculate T_avg, we can use the formula:
T_avg = (T1 + T2) / 2
Substituting the given values:
T_avg = (80 + 0) / 2 = 40 °C
Now, we can calculate the rate of entropy generation:
S_dot_gen = 12000 / 40 = 300 W/°C
Converting from °C to K:
S_dot_gen = 300 / (40 + 273.15) = 1.1819 W/K
Rounding off to two decimal places, the rate of entropy generation per unit area during the heat transfer process is approximately 1.18 W/m^2.K.
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Community Answer
Consider a one-dimensional steady heat conduction process through a so...
Thickness (t ) = 0.1 m
Temperature (T2) = 80oC = 353 k
Heat transfer rate per unit area (Q) = 4.5 kW/m2
Thermal conductivity (k ) = 15 W/mk
⇒
⇒ 30 = 80 - T1
⇒ T1 = 50oC
Now, Entropy generation
Hence, the correct answer is 1.184.
Temperature (T2) = 80oC = 353 k
Heat transfer rate per unit area (Q) = 4.5 kW/m2
Thermal conductivity (k ) = 15 W/mk
⇒
⇒ 30 = 80 - T1
⇒ T1 = 50oC
Now, Entropy generation
Hence, the correct answer is 1.184.
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Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer?
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Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer?.
Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 800C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m2. The thermal conductivity of the slab is 15 W/m.K. The rate of entropy generation per unit area during the heat transfer process is ________ W/m2.K (round off to 2 decimal places).Correct answer is '1.1821'. Can you explain this answer?.
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