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Unformatted text preview: Chapter 5 Numerical Methods in Heat Conduction 529 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 onedimensional. 2 There is no heat generation in the plate. Properties The thermal conductivity is given to be k = 2.5 W/m ⋅ °C. Analysis The nodal spacing is given to be ∆ x =0.06 m. Then the number of nodes M becomes 6 1 m 06 . m 3 . 1 = + = + ∆ = 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 ) (since 2 2 1 1 2 1 1 = = + → = + ∆ + + + 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 under consideration, C 2 . 43 m 0.06 C 60 C) W/m 5 . 2 ( W/m 700 1 1 2 1 ° = → = ° ° ⋅ + → = ∆ + 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. 518 q ∆ x 1 • • • • • 2 3 4 • 5 T Chapter 5 Numerical Methods in Heat Conduction 530E A large plate lying on the ground is subjected to convection and radiation. Finite difference formulation is to be obtained, and the top and bottom surface temperatures under steady conditions are to be determined. Assumptions 1 Heat transfer through the plate is given to be steady and onedimensional. 2 There is no heat generation in the plate and the soil. 3 Thermal contact resistance at platesoil interface is negligible. Properties The thermal conductivity of the plate and the soil are given to be k plate = 7.2 Btu/h ⋅ ft ⋅ °F and k soil = 0.49 Btu/h ⋅ ft ⋅ °F. Analysis The nodal spacing is given to be ∆ x 1 =1 in. in the plate, and be ∆ x 2 =0.6 ft in the soil. Then the number of nodes becomes 11 1 ft 0.6 ft 3 in 1 in 5 1 soil plate = + + = + ∆ + ∆ = x L x L M The temperature at node 10 (bottom of thee soil) is given to be T 10 =50 ° F. Nodes 1, 2, 3, and 4 in the plate and 6, 7, 8, and 9 in the soil are interior nodes, and thus for them we can use the general finite...
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This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.
 Spring '08
 BENARD
 Heat Transfer

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