21–96 Consider two rectangular surfaces perpendicular to each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is black and is maintained at 550 K. The back sides of the surfaces are insulated. The surrounding surfaces are at 290 K, and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. T2 = 550 K Ez = 1 W = 1.6 m L2 = 1.2 m T3 = 290 K ez = 0.85 L1 = 0.8 m A, T = 400 K E = 0.75

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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21-96 Consider two rectangular surfaces perpendicular to
each other with a common edge which is 1.6 m long. The
horizontal surface is 0.8 m wide and the vertical surface is
1.2 m high. The horizontal surface has an emissivity of 0.75
and is maintained at 400 K. The vertical surface is black and
is maintained at 550 K. The back sides of the surfaces are
insulated. The surrounding surfaces are at 290 K, and can be
considered to have an emissivity of 0.85. Determine the net
rate of radiation heat transfers between the two surfaces, and
between the horizontal surface and the surroundings.
T2 = 550 K
€2 = 1
W = 1.6 m
L2 = 1.2 m
T3 = 290 K
€z = 0.85
A2
4 = 0.8 m A1
T, = 400 K
Eq = 0.75
FIGURE P21–96
Transcribed Image Text:21-96 Consider two rectangular surfaces perpendicular to each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is black and is maintained at 550 K. The back sides of the surfaces are insulated. The surrounding surfaces are at 290 K, and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. T2 = 550 K €2 = 1 W = 1.6 m L2 = 1.2 m T3 = 290 K €z = 0.85 A2 4 = 0.8 m A1 T, = 400 K Eq = 0.75 FIGURE P21–96
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