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Unformatted text preview: ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Solution to Assignment No: 4 Due Date: March 7, 2011 Instructor: J. Murthy 1. (a) The given velocity field and mass source term can be seen to satisfy: ( V ) = S (b) UDS : For the interior cell 5 shown in Fig. 1, the discrete equations are given by the following. a E = max ( , u e y ) a W = max ( , u w y ) a N = max ( , v n x ) a S = max ( , v s x ) a P = a E + a W + a N + a S + m x y b = m x y s m = 4 ( x P + y P ) = ( F e F w + F n F s ) For the nearinlet cell 2, the discrete equations are given by: a E = max ( , u e y ) a W = max ( , u w y ) a N = max ( , v n x ) a b = max ( , v b x ) a P = a E + a W + a N + a b + m x y b = a b b + m x y s m = 4 ( x P + y P ) = ( F e F w + F n F s ) For the nearoutflow cell 6, the highPeclet number approximation is used. Therefore, the discrete equations are given by: a E = . a W = max ( , u w y ) a N = max ( , v n x ) a S = max ( , v s x ) a P = a W + a N + a S + m x y b = m x y s m = 4 ( x P + y P ) = ( F e F w + F n F s ) Other discrete equations may be derived in similar fashion. The solution may be obtained by marching from cell 1 to cell 3, from cell 4 to cell 6 and cell 7 to cell 9 because of upwinding. The solution is:cell 3, from cell 4 to cell 6 and cell 7 to cell 9 because of upwinding....
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 Fall '10
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