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

10 subject solving a set of two nonlinear equations

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Unformatted text preview: subtraction in B1 B2 − A2 C1 is cancelled. There are no subtractions in p2 . The same is found for all pj . Now investigate values of qj. From Eq. (10-15), subtractions occur in the numerator and denominator: Exercise 10.6 (continued) Analysis: (continued) qj = D j − A j q j −1 B j − A j p j −1 The denominator is identical to that already examined in pj. So we only have to investigate the numerator. Using Eqs. (10-9), as above, and (10-11), q1 = − F1zi ,1 D1 = B1 − V1 + W1 Ki ,1 + L1 + U 1 which contains no subtraction problems Therefore, the numerator in Eq. (10-15) becomes: D2 − A2 q1 = − F2 zi ,2 − L1 − F1zi ,2 − V1 + W1 Ki ,1 + L1 + U 1 = F2 zi ,2 V1 + W1 Ki ,1 + L1 + U 1 + L1 F1zi ,1 − V1 + W1 Ki ,1 + L1 + U 1 Thus, no subtractions appear in this numerator. The result is obtained for all qj. Therefore, the claim of Wang and Henke is verified. Exercise 10.7 Subject: Derivation of stagewise component material balance equations in terms of component vapor flow rates rather than liquid-phase mole fractions. Given: Eq. (10-7). Find: Component flow rates form of Eq. (10-7). Whether equations can still be partitioned into C tridiagonal matrix equations. Analysis: Start with Eqs. (10-1), (10-2), and (10-6) for Lj and Lj-1: M i , j = L j −1 xi , j −1 + V j +1 yi , j +1 + Fj zi , j − L j + U j xi , j − V j + Wj yi , j = 0 (1) Ei , j = yi , j − Ki , j xi , j = 0 (2) j L j = V j +1 + Fm − U m − Wm − V1 (3) Fm − U m − Wm − V1 (4) m=1 j −1 L j −1 = V j + m=1 Substituting Eqs. (2) to (4) into (1), to eliminate variables in L and x, M =V + i, j j −V j −1 υ m i, j − 1 F − U − W −V +υ +F z m m m i, j + 1 j i, j 1V K i, j − 1 i, j − 1 j υ υ i, j i, j F − U − W −V + U −υ −W =0 j +1 m m m 1 jVK i, j jV m j i, j j + Collecting terms in the component vapor flow rates, M =V+ i, j j − V j −1 j +1 υ i, j − 1 + K i, j − 1 i, j − 1 ( Fm − U m − Wm ) − V1 V m + j ( Fm − U m − Wm ) − V1 + U j m W 1 j −1 − υ +υ +F z =0 i, j i, j + 1 j i, j VK V j i, j j Thus, we will have a set of equations that can be partitioned into C tridiagonal sets. Exercise 10.8 Subject: Computer storage locations for tridiagonal matrix equations Given: Eqs. (10-7) to (10-11) Find: Minimum storage locations needed. Analysis: For a 100-stage column, only need: 99 storage locations for pj 100 storage locations for qj 4 storage locations for current values of Aj, Bj, Cj, and Dj This is total of 203 storage locations. If Aj, Bj, Cj, and Dj expressions are substituted into the pj and qj equations, then only need 199 storage locations. Exercise 10.9 Subject: Solving a set of two nonlinear equations by the Newton-Raphson method. Given: Two nonlinear equations in x1 and x2: 2 x12 + x2 = 17 8 x1 1/ 3 1 + x2 / 2 = 4 Find: Solutions for four different starting guesses Analysis: (a) Starting guess is x1 = 2, x2 = 5 The Newton-Raphson method converges in 5 iterations: Iteration Start 1 2 3 4 5 x1 2.00...
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