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

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online