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ee231_HW7

# ee231_HW7 - EE 231_B1 Damir Iraliyev 1246540 Homework#7...

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EE 231_B1 Damir Iraliyev 1246540 Homework#7: Question11.3 The following system of equation is designed to determine concentrations (the c’s in g/ m 3 ) in a series of coupled reactors as function of the amount of mass input to each reactor (the right-hand sides in g/day): - - = 15c1 3c2 c3 3800 - + - = 3c1 18c2 6c3 1200 - - + = 4c1 c2 12c3 2350 (a) Determine the matrix inverse (b) Use the inverse to determine the solution (c) Determine how much the rate of mass input to reactor 3 must be increased to induce a 10 g/m 3 rise in the concentration of reactor 1. (d) How much will the concentration in reactor 3 be reduced if the rate of mass input to reactors 1 and 2 is reduced by 500 and 250 g/day, respectively? Solution: (a) We write system of equations in Matrix form: ( )( )( ) - - - - - - = ={ } 1 2 3 15 3 1 318 6 4 112c1c2c3 380012002350Ac b We use LU decomposition to find the inverse matrix, [A] -1 : =- =- , : - -( ) m21 315 and m31 415 the we compute 1m21 2 and 1m31 3 And we get: - - . - . - . . 15 3 1017 4 6 20 1 811 7333 =- . . =, : - : m32 1 817 4 then we compute 2m32 3 = - - . - . . U 15 3 1017 4 6 20011 092 = - - - . L 100 31510 415 0 10341 Thus, = : - - - LY I 100 31510 415 . = 0 10341y11y12y13y21y23y23y31y32y33 100010001 : = . . . : = - = Therefore Y 100315100 287350 10341 Therefore UX Yand A 1 X - - . - 15 3 1017 4 . . = . . = 6 20011 0920x11x12x13x21x23x23x31x32x33 100315100 287350 10341 > == - . . . . . . . . . C A 10 07250 01280 01240 02070 0608 0 0321 0 0259 0 00930 0902 (b) - . . . . . . . A 10 07250 01280 01240 02070 0608 0 0321 0 0259 . . 0 00930 0902 , So = → = - * Ac b c A 1 b (c) = . . . . . . . c 0 07250 01280 01240 02070 0608 0 0321 0 0259 . . * = . . 0 00930 0902 380012002350 320 2073227 2021 . 321 5026

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(d) - . + - . + 4320 2073 10 227 2021 12 . 321 5026 = + 2350 λmass (e) =- . λmass 48 0001gday (f) = - - b2 3800 5001200 = 2502350 33009502350 = . . . . . c2 0 07250 01280 01240 02070 0608 . . . . * 0 0321 0 0259 0 00930 0902 33009502350 (g) = . . c2 280 7427201 6408 . 306 2176 (h) Therefore, c 3 = . / 306 2176 g cm3 (i) (j) Question 11.9 (k) (l) Besides the Hilbert matrix, there are other matrices that are inherently ill- conditioned.
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ee231_HW7 - EE 231_B1 Damir Iraliyev 1246540 Homework#7...

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