Explanation Given: The distance between parallel black disks ( L ) is 0.25 m . The diameter of lower disk ( D 1 ) is 0.2 m . The diameter of upper disk ( D 2 ) is 0.4 m . The heat of lower disk ( Q ) is 100 W . The temperature of lower disk ( T 1 ) is 500 K . The temperature of surrounding ( T s u r r ) is 300 K . Calculation: Calculate the ratio of diameter of lower disk ( D 1 ) and distance between them ( L ) using the relation. R 1 = ( D 1 2 ) L = ( 0.2 m 2 ) 0.25 = 0.4 Calculate the ratio of diameter of upper disk ( D 2 ) and distance between them ( L ) using the relation. R 2 = ( D 2 2 ) L = ( 0.4 m 2 ) 0.25 = 0.8 Refer Table 21-3 “View factor expressions for some common geometries of finite size (3-D)”. Calculate the ( s ) using the relation. s = 1 + 1 + R 2 2 R 1 2 = 1 + 1 + ( 0.8 ) 2 ( 0.4 ) 2 = 11.25 Obtain the view factor ( F 12 ) F 12 = 1 2 [ s − ( s 2 − 4 ( D 1 D 2 ) 2 ) 1 / 2 ] = 1 2 [ 11.25 − ( ( 11.25 ) 2 − 4 ( 0.4 0.2 ) 2 ) 1 / 2 ] = 0.3676 Calculate the view factor ( F 1 s u r r ) using the relation. F 1 s u r r + F 12 = 1 F 12 = 1 − F 1 s u r r F 12 = 1 − 0

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Chapter 21, Problem 91P

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The temperature of the upper disk.

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Chapter 21, Problem 90P

Chapter 21, Problem 92P

Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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