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

10 15 becomes d2 a2 q1 f2 zi 2 l1 f1zi 2 v1 w1

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Unformatted text preview: B2 − A2 p1 −350 − 50( −125) q2 = D2 − A2 q1 −50 − 50(0) . = = 01739 B2 − A2 p1 −350 − 50( −1.25) q1 = D1 0 = =0 B1 −160 p3 = 0 q3 = . D3 − A3q2 0 − 150(01739) = = 01917 . B3 − A3 p2 −230 − 150( −0.626) Using Eqs. (10-17) and (10-18): x3 = q3 = 0.1917 x2 = q2 - p2x3 = 0.1739 - (-0.626)(0.1917) = 0.2939 x1 = q1 - p1x2 = 0 - (-1.25)(0.2939) = 0.3674 x1 0 180 • x2 = −50 150 - 230 Applying the algorithm Eqs. (10-13) to (10-15): p1 = 0 x3 0 Exercise 10.5 Subject: Application of tridiagonal matrix algorithm. Given: Thomas algorithm equations (10-13) to (10-18) and a 3x3 matrix equation below. Find: Solution to the matrix equation. Analysis: The matrix equation is: −6 B1 C1 0 0 0 x1 D1 3 A2 B2 C2 0 0 x2 D2 0 A3 B3 C3 0 • x3 = D3 = 0 1.5 - 7.5 0 0 A4 B4 C4 x4 D4 0 0 4.5 - 7.5 0 0 0 A5 B5 x5 D5 0 0 0 3 - 4.5 0 0 0 x1 0 3 0 0 x2 0 3 0 4.5 Applying the algorithm Eqs. (10-13) to (10-15): p1 = C1 3 = = −0.5 B1 −6 p2 = C2 3 = = −1.0 B2 − A2 p1 −4.5 − 3( −0.5) p3 = C3 3 = = −0.5 B3 − A3 p2 −7.5 − 15( −1.0) . q3 = 100 − 15(0) D3 − A3q2 . = = −16.67 B3 − A3 p2 −7.5 − 15( −10) . p4 = 3 C4 = = −0.57143 B4 − A4 p3 −7.5 − 4.5( −0.5) q4 = D4 − A4 q3 0 − 4.5( −16.67) = = −14.28571 B4 − A4 p3 −7.5 − 4.5( −0.5) q5 = D5 − A5q4 0 − 4.5( −14.28571) = = −33.333 B5 − A5 p4 −4.5 − 4.5( −0.57143) q1 = D1 0 = =0 B1 −6 q2 = 0 Using Eqs. (10-17) and (10-18): x5 = q5 = -33.333 x4 = q4 - p4x5 = -14.28571 - (-0.57143)(-33.333) = -33.333 x3 = q3 - p3x4 = -16.6667 - (-0.5)(-33.333) = -33.333 x2 = q2 - p2x3 = 0.0 - (-1.0)(-33.333) = -33.333 x1 = q1 – p1x2 = 0.0 - (-0.5)(-33.333) = -16.667 • x3 = 100 3 x4 0 - 4.5 x5 0 Exercise 10.6 Subject: Solution of tridiagonal matrix equation for liquid-phase mole fractions to avoid subtraction of nearly equal quantities as claimed by Wang and Henke. Given: Equations (10-1) to (10-18) Prove: Claim of Wang and Henke. Analysis: First investigate the values of pj. From Eq. (10-14), a subtraction occurs in the denominator: Cj pj = B j − A j p j −1 Starting at the top of the column, p1 = C1/B1, therefore, the denominator of Eq. (10-14) becomes: C B2 − A2 p1 = B2 − A2 1 Clearing the fraction, we obtain B1 B2 − A2 C1 B1 Examine this difference by substituting for each of the four quantities. From Eq. (10-8), A2 = V2 + F1 − W1 − U1 − V1 However, by material balance around Stage 1 in Fig. 10.3, V2 + F1 − W1 − U 1 − V1 = L1 Therefore, A2 = L1 From Eq. (10-9), B1 = −[V2 + F1 − W1 − U 1 − V1 + U 1 + V1 + W1 Ki ,1 ] However, by material balance around Stage 1 in Fig. 10.3, V2 + F1 − W1 − U 1 − V1 = L1 Therefore, B1 = − V1 + W1 Ki ,1 + L1 + U 1 Similarly, B2 = − V2 + W2 Ki ,2 + L2 + U 2 Thus, it is clear that B1 B2 = a summation of all positive quantities, that is no subtractions. From Eq. (10-10), C1 = V2 Ki ,2 and A2 C1 = L1V2 Ki ,2 When B1B2 is expanded, it contains a term -L1V2Ki,2 . Thus, this potential...
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