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Unformatted text preview: ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Solution to Assignment No: 5 Due Date: April 4, 2011 Instructor: J. Murthy 1. We are asked to reconstruct e from e = P + ( r e ) x 2 P W x with r e = E P P W using the superbee and minmod limiters. The superbee limiter is given by: ( r ) = max [ , min ( 2 r , 1 ) , min ( r , 2 )] The minmod limiter is given by: ( r ) = min ( r , 1 ) if r > ( r ) = if r The values of r e and the corresponding values of and e are given in Tables 2 and 1. Case (W,P,E) r e e e value Comment a 300,200,100 1 1 p + P W 2 150 linear extrapolation b 100,50,200-3 P 50 extremum, therefore use upwinding c 300,150,100 1 3 r e (= 1 3 ) P + E P 2 125 use downwind gradient d 300,250,100 3 1 p + P W 2 225 use upwind gradient Table 1: Problem 1 - e Values Using Minmod Limiter Case (W,P,E) r e e e value Comment a 300,200,100 1 1 p + P W 2 150 linear extrapolation b 100,50,200-3 P 50 extremum, therefore use upwinding c 300,150,100 1 3 2 r e (= 2 3 ) P + ( E P ) = E 100 use full downwind difference d 300,250,100 3 2 p + ( P W ) 200 use full upwind differnece Table 2: Problem 1 - e Values Using Superbee Limiter 2. We are given: d dx ( u ) = 2 A L sin parenleftbigg 2 L x parenrightbigg with ( ) = 100 and A = 100. 1 (a) Integrating the governing equation over the control volume and using higher-order upwinding for face values yields the following discrete equation for an interior point: a E = ....
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This note was uploaded on 12/29/2011 for the course ME 608 taught by Professor Na during the Fall '10 term at Purdue University-West Lafayette.
- Fall '10