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Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.
The maximum temperature within the plate in °C is
The maximum temperature within the plate in °C is
[2013]
- a)160
- b)165
- c)200
- d)250
Correct answer is option 'B'. Can you explain this answer?
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Consider steady one-dimensional heat flow in a plate of 20 mm thicknes...
Tmax = 165º [putting x = 5 × 10–3 is equation (1)]
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Consider steady one-dimensional heat flow in a plate of 20 mm thicknes...
°C and 40°C, respectively. The thermal conductivity and heat capacity of the material are 20 W/(m•K) and 800 J/(kg•K), respectively. Determine:
a) The temperature distribution within the plate
b) The heat transfer rate through the plate
c) The temperature at the center of the plate
a) The temperature distribution within the plate can be found using the one-dimensional heat conduction equation:
q'' = -k(dT/dx)
where q'' is the heat flux (W/m2), k is the thermal conductivity (W/(m•K)), and dT/dx is the temperature gradient (K/m).
Since the heat generation is uniform, we can assume that the heat flux is constant throughout the plate:
q'' = -k(dT/dx) = constant
Therefore,
(dT/dx) = -q''/k = -80x10^6/20 = -4x10^6 K/m
Integrating this equation with respect to x and using the boundary conditions, we get:
T(x) = -2x10^6 x^2 + 2x10^6 x + 40
The temperature distribution within the plate is given by:
T(x) = -2x10^6 x^2 + 2x10^6 x + 40, 0 ≤ x ≤ 0.02 m
b) The heat transfer rate through the plate can be found using Fourier's law:
q = -kA(dT/dx)
where A is the cross-sectional area of the plate.
Since the heat flux is constant throughout the plate, we can use the average temperature difference across the plate:
q = -kAΔT/L
where ΔT is the temperature difference between the left and right faces, and L is the thickness of the plate.
Therefore,
q = -20(0.02)(160-40)/0.02 = -3200 kW
The heat transfer rate through the plate is 3200 kW.
c) The temperature at the center of the plate can be found by setting x = L/2 in the temperature distribution equation:
T(L/2) = -2x10^6 (L/2)^2 + 2x10^6 (L/2) + 40
T(L/2) = -0.5x10^6 L^2 + 1x10^6 L + 40
Substituting the values, we get:
T(0.01) = -0.5x10^6 (0.02)^2 + 1x10^6 (0.02) + 40
T(0.01) = 120°C
Therefore, the temperature at the center of the plate is 120°C.
a) The temperature distribution within the plate
b) The heat transfer rate through the plate
c) The temperature at the center of the plate
a) The temperature distribution within the plate can be found using the one-dimensional heat conduction equation:
q'' = -k(dT/dx)
where q'' is the heat flux (W/m2), k is the thermal conductivity (W/(m•K)), and dT/dx is the temperature gradient (K/m).
Since the heat generation is uniform, we can assume that the heat flux is constant throughout the plate:
q'' = -k(dT/dx) = constant
Therefore,
(dT/dx) = -q''/k = -80x10^6/20 = -4x10^6 K/m
Integrating this equation with respect to x and using the boundary conditions, we get:
T(x) = -2x10^6 x^2 + 2x10^6 x + 40
The temperature distribution within the plate is given by:
T(x) = -2x10^6 x^2 + 2x10^6 x + 40, 0 ≤ x ≤ 0.02 m
b) The heat transfer rate through the plate can be found using Fourier's law:
q = -kA(dT/dx)
where A is the cross-sectional area of the plate.
Since the heat flux is constant throughout the plate, we can use the average temperature difference across the plate:
q = -kAΔT/L
where ΔT is the temperature difference between the left and right faces, and L is the thickness of the plate.
Therefore,
q = -20(0.02)(160-40)/0.02 = -3200 kW
The heat transfer rate through the plate is 3200 kW.
c) The temperature at the center of the plate can be found by setting x = L/2 in the temperature distribution equation:
T(L/2) = -2x10^6 (L/2)^2 + 2x10^6 (L/2) + 40
T(L/2) = -0.5x10^6 L^2 + 1x10^6 L + 40
Substituting the values, we get:
T(0.01) = -0.5x10^6 (0.02)^2 + 1x10^6 (0.02) + 40
T(0.01) = 120°C
Therefore, the temperature at the center of the plate is 120°C.
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Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. Can you explain this answer?
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Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. 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 steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. 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 steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. Can you explain this answer?.
Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. 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 steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. 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 steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.The maximum temperature within the plate in °C is[2013]a)160b)165c)200d)250Correct answer is option 'B'. Can you explain this answer?.
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