Explanation Given Information: Thermal conductivity of solid part k s = 10 W / m ⋅ K Thermal conductivity of fluid k f = 0.1 W / m ⋅ K Thickness of plane wall L = 1 m Value of ε max = 0.25 T ( x = 0 ) = 30 0 C Heat flux is q ″ x = 100 W / m 2 Calculation: The minimum thermal conductivity is calculated as, k e f f = 1 1 − ε k s + ε k f = 1 1 − ε max ( x L ) k s + ε max ( x L ) k f = 1 1 − 0.25 ( x 1 ) 10 + 0.25 ( x 1 ) 0.1 = 1 0.1 + 2.47 x The heat flux is calculated as, q ″ x = k e f f x [ T ( 0 ) − T ( x ) ] 100 W / m 2 = 1 0.1 + 2.475 x x [ 30 0 C − T ( x ) ] T ( x ) = 30 − 10 x − 247.5 x 2 The temperature distribution for the plane wall for minimum effective thermal conductivity is shown in figure 1. Figure 1 The maximum thermal conductivity is calculated as, k e f f = ( 1 − ε ) k s + ε k f θ = [ 1 − ε max ( x L ) ] k s + ε max ( x L ) k f = [ 1 − 0.25 ( x 1 ) ] ( 10 ) + 0.25 ( x 1 ) ( 0.1 ) = 10 − 2.475 x The heat flux is calculated as, q ″ x = k e f f x [ T ( 0 ) − T ( x ) ] 100 = 10 − 2.475 x x [ 30 − T ( x ) ] T ( x ) = 30 − 100 x 10 − 2.475 x The temperature distribution for the plane wall for maximum effective thermal conductivity is shown in figure 2.

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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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Introduction to Heat Transfer

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The steady state temperature distribution using various condition.

Solution Summary: The author explains the steady state temperature distribution using various conditions. The minimum thermal conductivity is calculated as, cq′′_x=k

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