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T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab material
having conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 per
unit kelvin _________________W/mK
having conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 per
unit kelvin _________________W/mK
Correct answer is between '150,170'. Can you explain this answer?
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T here is a slab having temperat ure at left and right faces 800 K and...
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T here is a slab having temperat ure at left and right faces 800 K and...
°C. The thickness of the slab is 0.1 m. We need to find the rate of heat transfer through the slab.
To find the rate of heat transfer through the slab, we can use Fourier's law of heat conduction:
q = -k * A * (dT/dx)
Where:
q is the rate of heat transfer
k is the thermal conductivity of the material
A is the cross-sectional area of the slab
dT/dx is the temperature gradient across the slab
First, we need to calculate the temperature gradient across the slab. The temperature difference between the left and right faces of the slab is:
ΔT = T2 - T1 = 600 K - 800 K = -200 K
Next, we need to calculate the temperature gradient:
dT/dx = ΔT / x = -200 K / 0.1 m = -2000 K/m
Now, we can calculate the rate of heat transfer:
q = -k * A * (dT/dx)
Given that the thermal conductivity of the material is k = 20 W/mK, and the cross-sectional area of the slab is A = 1 m^2, we can substitute these values into the formula:
q = -20 W/mK * 1 m^2 * (-2000 K/m)
q = 40,000 W
Therefore, the rate of heat transfer through the slab is 40,000 W.
To find the rate of heat transfer through the slab, we can use Fourier's law of heat conduction:
q = -k * A * (dT/dx)
Where:
q is the rate of heat transfer
k is the thermal conductivity of the material
A is the cross-sectional area of the slab
dT/dx is the temperature gradient across the slab
First, we need to calculate the temperature gradient across the slab. The temperature difference between the left and right faces of the slab is:
ΔT = T2 - T1 = 600 K - 800 K = -200 K
Next, we need to calculate the temperature gradient:
dT/dx = ΔT / x = -200 K / 0.1 m = -2000 K/m
Now, we can calculate the rate of heat transfer:
q = -k * A * (dT/dx)
Given that the thermal conductivity of the material is k = 20 W/mK, and the cross-sectional area of the slab is A = 1 m^2, we can substitute these values into the formula:
q = -20 W/mK * 1 m^2 * (-2000 K/m)
q = 40,000 W
Therefore, the rate of heat transfer through the slab is 40,000 W.
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T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. Can you explain this answer?
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T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. 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 T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. 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 T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. Can you explain this answer?.
T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. 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 T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. 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 T here is a slab having temperat ure at left and right faces 800 K and 600 K respectively and slab materialhaving conductivit y k = 20 W/mK at 0°C. Find the mean thermal conductivity for the slab given β = 0.01 perunit kelvin _________________W/mKCorrect answer is between '150,170'. Can you explain this answer?.
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