Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# 3 ki il il iv from eqs 2 64 and 2 62 v il exp il

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Unformatted text preview: a block tridiagonal matrix structure, interchange the last 3 equations with the middle group of 3 equations. The result is: 012 201 000 x1 -7 101 130 000 x2 -6 111 011 000 x3 -6 121 112 310 x4 -13 012 211 113 • x5 001 121 211 x6 -10 000 121 221 x7 -11 000 112 301 x8 -8 000 011 121 x9 -8 =− -14 The solution proceeds as in Example 10.7, with a matrix of the form: B1 C1 0 ∆X1 F1 A 2 B2 C2 • ∆X 2 = − F2 ∆X 3 F3 0 A 3 B3 B1 −1 −0.5 0.5 0.5 = 0 -1 1 B1 −1 0.5 0.5 - 0.5 B1 −1 −2.5 F1 = 0 −3.5 −0.5 C1 = 2 0 -1 - 2 1.5 1 which replaces C1 1 0 10 which replaces F1 0 and I = 0 1 0 which replaces B1 0 0 1 Exercise 10.12 (continued) Analysis: (continued) 2 2 0 0 1 0 -0.5 1 1 B2 − A 2 C1 = B2 − A 2 C1 −1 0.5 - 0.5 1 3 -1 0 0 1 0 1 = -3 1 B2 − A 2 C1 0.5 0.25 -1.5 Upon inversion, −1 C2 = which replaces C2 1.25 - 0.25 - 3.5 −7 F2 − A 2 F1 = −7 which replaces F2 −6.5 B2 − A 2 C1 −1 3.5 F2 = which replaces F2 -7 2.25 0 0 0 1 0 0 A 2 is replaced by 0 0 0 B2 is replaced by 0 1 0 0 0 0 0 0 1 -1.75 1.5 -1 0 8 -1.25 B3 − A 3 C2 = 0.75 1.25 1.5 Upon inversion, B3 − A 3 C2 -2.75 F3 − A 3 F2 = -9 which replaces F3 -2.25 0.13514 x7 F3 = -1.13514 which replaces F3 = − x8 -1.10811 B3 − A 3 C2 −1 x9 - 0.45947 - 2.67567 which replaces F2 = − x5 - 2.08110 F2 − C2 F3 = x4 x6 2.62161 F1 − C1 F2 = x1 - 3.72971 which replaces F1 = − x2 - 0.13513 x3 −1 1 = 9.25 -10 0.75 6 8.5 - 0.75 12.5 -1.25 1.25 0.75 Exercise 10.12 (continued) Analysis: (continued) Therefore, the solution to the 9 equations is: x1 = -2.62161 x2 = 3.72971 x3 = 0.13513 x4 = 0.45947 x5 = 2.67567 x6 = 2.08110 x7 = -0.13514 x8 = 1.13514 x9 = 1.10811 Subject: Given: Exercise 10.13 Matrix structure when equations are ordered by type. Mesh equations (10-58) to (10-60). Find: Whether matrix structure is block tridiagonal. Analysis: Many matrix structures are possible depending on the order of the equations and the order of the variables. Referring to Eqs. (10-61) to (10-64), try the following orders for a three component system, where j refers to the stage number: F = [ Hj , M1,j , M2,j , M3,j , E1,j , E2,j , E3,j ] X = [ υ1,j , υ2,j , 3,j , Tj , l1,j , l2,j , l3,j ] For a three-stage system, the incidence matrix is on the next page, where x denotes that the variable for that column appears in the equation for that row. It is seen in the incidence diagram that the matrix is not block tridiagonal or banded. If the order of the equations is maintained, it is not possible to permute the variable columns to get a blockbanded structure. Exercise 10.13 (continued) Analysis: (continued) Incidence Matrix ν 1,1 ν 1,2 ν 1,3 ν 2 ,1 ν 2 ,2 ν 2 ,3 ν 3,1 ν 3,2 ν 3,3 T1 T2 T3 l1,1 l1,2 l1,3 l2 ,1 l2 ,2 l2 ,3 l3,1 l3,2 l3,3 H1 x H2 x x H3 M1,1 x x x x x x M1,2 x x x x xx x x x x x x x x x x x x x x x x x x x M 3,2 x x x M 3,3 x x x x x x x E1,3 x x x x E 2,2 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x E 2,3 x x x x E1,2 x x x M 3,1 E 3,3 x x M 2,3 E 3,2 xxx x M 2,2 E 3,1 x x x M 2,1 E 2,1 x x M1,3 E1,1 xx x x x x x x...
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