Explanation Given: Thickness of plate is 5 cm . Thermal conductivity of plate is 111 W / m ⋅ K . Rate of heat generation is 2 × 10 5 W / m 3 . Temperature of environment is 25 ° C Heat transfer coefficient is 44 W / m 2 ⋅ K Concept used: Write the expression for general heat conduction equation in Cartesian coordinate system. ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 + ∂ 2 T ∂ z 2 + e ˙ g e n k = 1 α d T d t ...... (1) Here, T is temperature, k is thermal conductivity, α is thermal diffusivity, t is time and e ˙ g e n is heat generation. For steady state, d T d t = 0 For one dimensional heat conduction, ∂ T ∂ y = 0 ∂ T ∂ z = 0 Substitute 0 for d T d t , 0 for ∂ T ∂ y and 0 for ∂ T ∂ z in equation(1), ∂ 2 T ∂ x 2 + 0 + 0 + e ˙ g e n k = 1 α ⋅ 0 Simplify the above equation, ∂ 2 T ∂ x 2 + e ˙ g e n k = 0 Simplify the above equation, d ( d T d x ) = − e ˙ g e n k d x Integrate above equation, d T d x = − e ˙ g e n k x + C 1 ...... (2) Here, C 1 is a constant. Integrate above equation, T = − e ˙ g e n k x 2 2 + C 1 x + C 2 ...... (3) According to boundary condition, At x = 0 Q 1 = 0 ...... (4) Here, Q 1 is rate heat transfer from inner wall. From Fourier law of heat conduction per unit area, Q 1 = − k d T d x Substitute − k d T d x for Q 1 in equation (4). − k d T d x = 0 d T d x = 0 Substitute 0 for d T d x in equation (2). 0 = − e ˙ g e n k ⋅ 0 + C 1 C 1 = 0 Substitute 0 for C 1 in equation (3). T = − e ˙ g e n k x 2 2 + 0 ⋅ x + C 2 Simplify the above equation. T = − e ˙ g e n k x 2 2 + C 2 ...... (5) At x = L T = T 2 Here, L is thickness of plate and T 2 is outer surface temperature. Substitute L for x and T 2 for T in equation (5). T 2 = − e ˙ g e n k L 2 2 + C 2 C 2 = T 2 + e ˙ g e n k L 2 2 Substitute T 2 + e ˙ g e n k L 2 2 for C 2 in equation (5)

Heat and Mass Transfer: Fundamentals and Applications

5th Edition

ISBN: 9780073398181

Author: Yunus A. Cengel Dr., Afshin J. Ghajar

Publisher: McGraw-Hill Education

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Question

Chapter 2, Problem 83P

To determine

The location of highest and lowest temperature in the plate and their values.

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Chapter 2, Problem 82CP

Chapter 2, Problem 84P

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Heat and Mass Transfer: Fundamentals and Applications

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