Explanation Write the expression of general heat conduction equation. ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 + ∂ 2 T ∂ z 2 + e ˙ gen = 1 α ∂ T ∂ t ...... (I) Here, the temperature is T , the thermal diffusivity is α , and the heat generation is e ˙ gen . Since the given problem is for two dimensional steady with no heat generation. So, the term ∂ 2 T ∂ z 2 and e ˙ gen will become zero. Write the expression for the heat transfer due to conduction in y direction. ( Q cond ) y = − K ∂ T ∂ y ...... (II) Here, the heat transfer due to conduction in y direction is ( Q cond ) y and the thermal conductivity is K . Write the expression for the heat transfer due to conduction in x direction. ( Q cond ) x = − K ∂ T ∂ x ...... (III) Here, the heat transfer due to conduction in x direction is ( Q cond ) x . Write the expression for the conservation of energy along y direction. − K ∂ T ∂ y = h ( T x − T ∞ ) ...... (IV) Here, the coefficient of convective heat transfer is h , the surrounding temperature is T ∞ and the temperature at x distance is T x . Write the expression for the conservation of energy along x direction. − K ∂ T ∂ x = h ( T y − T ∞ ) ...... (V) Here, the temperature at y distance is T y . Calculation: Substitute 0 for ∂ 2 T ∂ z 2 and 0 for e ˙ gen in Equation (I). ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 + 0 + 0 = 1 α ∂ T ∂ t ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 = 1 α ∂ T ∂ t Apply the boundary conditions that is, at x = 0 , ( Q cond ) y = 0 . Substitute 0 for ( Q cond ) y and 0 for x in Equation (II)

6th Edition

ISBN: 9781265879631

Author: CENGEL

Publisher: MCG CUSTOM

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Chapter 2, Problem 143P

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The differential equation and boundary conditions.

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HEAT+MASS TRANSFER (LL)-W/ACCESS

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