problem4-60 - PROBLEM 4.60 KNOWN: Rectangular plate...

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PROBLEM 4.60 K NOWN: Rectangular plate subjected to uniform temperature boundaries. F IND: Temperature at the midpoint using a finite-difference method with space increment of 0.25m SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant roperties. p ANALYSIS: For the nodal network above, 12 finite-difference equations must be written. It follows that node 8 represents the midpoint of the rectangle. Since all nodes are interior nodes, Eq. 4.29 is appropriate and is written in the form m neighbors 4T T 0. −= For nodes on the symmetry adiabat, the neighboring nodes include two symmetrical nodes. Hence, for Node 4, the neighbors are T b , T 8 and 2T 3 . Because of the simplicity of the finite-difference equations, we may proceed directly to the matrices [A] and [C] – see Eq. 4.48 – and matrix inversion can be used to find the nodal temperatures T m . 4 1 0 0 1 0 0 0 0 0 0 0 1 4 1 0 0 1 0 0 0 0 0 0 0 1 4 1 0 0 1 0 0 0 0 0 0 0 2 4 0 0 0 1 0 0 0 0 1 0 0 0 4 1 0 A
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.

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