Explanation Given information: The length of the plate is 1.5 m , the width of the plate is 0.6 m , thickness of the plate is 0.15 m , temperature of the one side of plate is 500 K , temperature of the other side of the plate is 350 K , the constant value is 25 W / m ⋅ K , and temperature coefficient of the thermal conductivity is 8.7 × 10 − 4 K − 1 . The thermal conductivity is varying linearly with temperature. So, average thermal conductivity of the of the medium can be used. Write the expression for average thermal conductivity of the of the medium. k a v g = k 0 ( 1 + β T ) ...... (I) Here, the average temperature is T ,average thermal conductivity of the of the medium is k a v g ,the constant is k 0 and temperature coefficient of the thermal conductivity is β . Write the expression for average temperature. T = T 1 + T 2 2 ...... (II) Here, temperature of the one side of plate T 1 and temperature of the other side of the plate T 2 , Write the expression for the rate of heat conduction through the plate. Q ˙ = k a v g A ( T 1 − T 2 L ) ...... (III) Here, the area of the plate is A and the rate of heat conduction is Q ˙ . Calculation: Substitute T 1 + T 2 2 for T in Equation (I). k a v g = k 0 ( 1 + β ( T 1 + T 2 2 ) ) ...... (IV) Substitute, 25 W / m ⋅ K for k 0 , 8.7 × 10 − 4 K − 1 for β , 500 K for T 1 and 350 K for T 2 in the Equation (IV)

Heat and Mass Transfer: Fundamentals and Applications

6th Edition

ISBN: 9781260440058

Author: CENGEL, Yunus

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 2, Problem 115P

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The rate of the heat conduction through the plate.

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Chapter 2, Problem 114P

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Heat and Mass Transfer: Fundamentals and Applications

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