Explanation Given Information: Area of aluminium plate = 10 m m × 10 m m = ( 0.01 ) 2 m 2 . Temperature of plates are, T s , 1 = 305 K , T s , 2 = 295 K Length, L = 1 m m , L = 1 μ m and L = 10 n m . Thermal accommodation coefficient, α t = 0.92 Properties for air: K = 25.72 W / m . k ( 18 ° c ) γ = 1.41 λ m = 60 × 10 − 3 × 10 − 6 m When L = 1 m = 1 × 10 − 3 m m R t m m = L K A = 1 × 10 − 3 25.72 × ( 0.01 ) 2 = 3.8880 × 10 − 9 R t m s = λ m K A [ 2 − α α ] [ 9 γ − 5 γ + 1 ] = 60 × 10 − 3 × 10 − 6 25.72 × ( 0.01 ) 2 [ 2 − 0.92 0.92 ] [ 9 × 1.41 − 5 1.41 + 1 ] = 2.33 × 10 − 13 × 1.1739 × 3.1908 = 8.7276 × 10 − 13 Hence, R t m s < < R t m m neglect the R t m s When L = 1 μ m = 1 × 10 − 6 m R t m m = L K A = 1 × 10 − 6 25.72 × ( 0.01 ) 2 = 3.88 × 10 − 12 R t m s = λ m K A [ 2 − α α ] [ γ − 5 γ + 1 ] = 60 × 10 − 3 × 10 − 6 25

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ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.173P

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Conduction heat rate for plate spacings.

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Chapter 3, Problem 3.168P

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Introduction to Heat Transfer

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Conduction heat rate for plate spacings.

Solution Summary: The author explains the conduction heat rate for plate spacings. The temperature of plates is T_s,1=305K,

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