Thermodynamics HW Solutions 415

# Thermodynamics HW Solutions 415 - under consideration C 2...

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Chapter 5 Numerical Methods in Heat Conduction 5-29 A plate is subjected to specified heat flux and specified temperature on one side, and no conditions on the other. The finite difference formulation of this problem is to be obtained, and the temperature of the other side under steady conditions is to be determined. Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate. q 0 Δ x 1 0 2 3 4 5 T 0 Properties The thermal conductivity is given to be k = 2.5 W/m °C. Analysis The nodal spacing is given to be Δ x =0.06 m. Then the number of nodes M becomes 6 1 m 06 . 0 m 3 . 0 1 = + = + Δ = x L M Nodes 1, 2, 3, and 4 are interior nodes, and thus for them we can use the general finite difference relation expressed as ) 0 (since 0 2 0 2 1 1 2 1 1 = = + = + Δ + + + g T T T k g x T T T m m m m m m m , for m = 1, 2, 3, and 4 The finite difference equation for node 0 on the left surface is obtained by applying an energy balance on the half volume element about node 0 and taking the direction of all heat transfers to be towards the node
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Unformatted text preview: under consideration, C 2 . 43 m 0.06 C 60 C) W/m 5 . 2 ( W/m 700 1 1 2 1 ° = ⎯→ ⎯ = ° − ° ⋅ + ⎯→ ⎯ = Δ − + T T x T T k q & Other nodal temperatures are determined from the general interior node relation as follows: C 24 ° − = − − × = − = = ° − = − × = − = = ° = − × = − = = ° = − × = − = = 6 . 9 ) 2 . 7 ( 2 2 : 4 C 2 . 7 4 . 26 6 . 9 2 2 : 3 C 6 . 9 2 . 43 4 . 26 2 2 : 2 C 4 . 26 60 2 . 43 2 2 : 1 3 4 5 2 3 4 1 2 3 1 2 T T T m T T T m T T T m T T T m Therefore, the temperature of the other surface will be –24 ° C Discussion This problem can be solved analytically by solving the differential equation as described in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above. 5-18...
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