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Heat is being transferred conductively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be
[2007]
- a)700 K
- b)625 K
- c)360 K
- d)400 K
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Heat is being transferred conductively from a cylindrical nuclear reac...
By applying energy balance, we get Rate of heat generation = Rate of convective heat transfer
qG × Volume of fuel rod = h0 A (Ts – T0)
Ts – 75 = 12.5
Ts = 87.5°C or 360.5K
qG × Volume of fuel rod = h0 A (Ts – T0)
Ts – 75 = 12.5
Ts = 87.5°C or 360.5K
Most Upvoted Answer
Heat is being transferred conductively from a cylindrical nuclear reac...
°C. The thermal conductivity of the fuel rod material is 50 W/m·K, and the heat transfer coefficient at the fuel rod-water interface is 1000 W/m²·K. The length of the fuel rod is 2 m. Determine the rate of heat transfer from the fuel rod to the water.
To determine the rate of heat transfer from the fuel rod to the water, we can use the formula for conductive heat transfer:
Q = (k * A * (T1 - T2)) / L
Where:
Q = Rate of heat transfer (in watts)
k = Thermal conductivity of the fuel rod material (in watts per meter-kelvin)
A = Cross-sectional area of the fuel rod (in square meters)
T1 = Temperature of the fuel rod (in kelvin)
T2 = Temperature of the water (in kelvin)
L = Length of the fuel rod (in meters)
First, let's convert the given temperatures from degrees Celsius to Kelvin:
T1 = 75 + 273.15 = 348.15 K
T2 = 0 + 273.15 = 273.15 K
Next, we need to calculate the cross-sectional area of the fuel rod using its diameter:
A = π * (d/2)^2
A = π * (0.05/2)^2 = 0.001963495 m²
Now we can substitute the given values into the formula:
Q = (50 * 0.001963495 * (348.15 - 273.15)) / 2
Q = (50 * 0.001963495 * 75) / 2
Q = 0.18350375 W
Therefore, the rate of heat transfer from the fuel rod to the water is approximately 0.1835 watts.
To determine the rate of heat transfer from the fuel rod to the water, we can use the formula for conductive heat transfer:
Q = (k * A * (T1 - T2)) / L
Where:
Q = Rate of heat transfer (in watts)
k = Thermal conductivity of the fuel rod material (in watts per meter-kelvin)
A = Cross-sectional area of the fuel rod (in square meters)
T1 = Temperature of the fuel rod (in kelvin)
T2 = Temperature of the water (in kelvin)
L = Length of the fuel rod (in meters)
First, let's convert the given temperatures from degrees Celsius to Kelvin:
T1 = 75 + 273.15 = 348.15 K
T2 = 0 + 273.15 = 273.15 K
Next, we need to calculate the cross-sectional area of the fuel rod using its diameter:
A = π * (d/2)^2
A = π * (0.05/2)^2 = 0.001963495 m²
Now we can substitute the given values into the formula:
Q = (50 * 0.001963495 * (348.15 - 273.15)) / 2
Q = (50 * 0.001963495 * 75) / 2
Q = 0.18350375 W
Therefore, the rate of heat transfer from the fuel rod to the water is approximately 0.1835 watts.
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Heat is being transferred conductively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be[2007]a)700 Kb)625 Kc)360 Kd)400 KCorrect answer is option 'C'. Can you explain this answer?
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Heat is being transferred conductively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be[2007]a)700 Kb)625 Kc)360 Kd)400 KCorrect answer is option 'C'. 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 Heat is being transferred conductively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be[2007]a)700 Kb)625 Kc)360 Kd)400 KCorrect answer is option 'C'. 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 Heat is being transferred conductively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be[2007]a)700 Kb)625 Kc)360 Kd)400 KCorrect answer is option 'C'. Can you explain this answer?.
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