13-56 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 450 K. The vertical surface is black and is maintained at 700 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 transfer between the two surfaces and between the horizontal surface and the surroundings. FIGURE P13–56 Tz= 700 K e2 = 1 W = 1.6 m T3 = 290 K E3= 0.85 Lz= 1.2 m A2 L1 = 0.8 m A1 Tj = 450 K €1 = 0.75

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter11: Heat Transfer By Radiation
Section: Chapter Questions
Problem 11.65P
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13-56 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 450 K. The vertical surface is black and is maintained at 700 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 transfer between the two surfaces and between the horizontal surface and the surroundings.
FIGURE P13–56
Tz= 700 K
e2 = 1
W = 1.6 m
T3 = 290 K
E3= 0.85
Lz= 1.2 m
A2
L1 = 0.8 m
A1
Tj = 450 K
€1 = 0.75
Transcribed Image Text:13-56 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 450 K. The vertical surface is black and is maintained at 700 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 transfer between the two surfaces and between the horizontal surface and the surroundings. FIGURE P13–56 Tz= 700 K e2 = 1 W = 1.6 m T3 = 290 K E3= 0.85 Lz= 1.2 m A2 L1 = 0.8 m A1 Tj = 450 K €1 = 0.75
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