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

10 13 to 10 15 p1 0 x3 0 exercise 105 subject

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Unformatted text preview: tion or disappearance of component i by chemical reaction on Stage j is: Mj M m =1 νi ,mrk ,m, j The component material balance equations, (10-1), now become: L j −1 C i =1 xi , j −1 + V j +1 C i =1 yi , j +1 + Fj C i =1 zi , F j − L j + U j C i =1 xi , j − V j + Wj C i =1 yi , j + θ j −1 V j −1 + Wj −1 yi , j −1 − θ j V j + Wj yi , j + φ j +1 L j +1 + U j +1 xi , j +1 − φ j L j + U j xi , j − M j M m =1 νi ,mrk ,m, j = 0 The equilibrium equations, (1-2) and the summation equations (10-3) and (10-4) do not change. The energy balance is as follows, where the heat of reaction does not appear because it is assumed that enthalpies are referred to the elements: L j −1hL j −1 + V j +1hV j +1 + Fj hFj + θ j +1 V j −1 + Wj −1 hV j −1 + φ j +1 L j +1 + U j +1 hL j +1 − 1 + θ j V j + Wj hV j − 1 + φ j L j + U j hL j = 0 Exercise 10.3 Subject: Revised set of MESH equations. Given: MESH Eqs. (10-1) to (10-5). Find: Revised set of MESH equations to account for liquid pumparounds as shown in Fig. 10.2. Revised M equations like Eq. (10-7). Possible partitioning to tridiagonal sets. Analysis: As an example, consider the section of equilibrium stages below, which includes a bypass pumparound and a recycle pumparound. Let Pu,v designate a total molar liquid pumparound flow rate from Stage u to Stage v. Exercise 10.3 (continued) Analysis: (continued) The revised MESH equations for Stage j are: M i , j = L j −1 xi , j −1 + V j +1 yi , j +1 + Pj + 2, j xi , j + 2 + Pj − 2, j xi , j − 2 + F j zi , j − ( L j + U j ) xi , j − (V j + W j ) yi , j = 0 Ei , j = yi , j − K i , j xi , j = 0 (1) (2) H j = L j −1hL j−1 + V j +1hV j+1 + Fj hFj + Pj − 2, j hL j−2 + Pj + 2, j hL j+2 − (L j + U j ) hL j − (V j + W j ) hV j − Q j = 0 L j = V j +1 + j m =1 ( Fm − U m − Wm ) − V1 + Pj + 2, j = 0 (5) (6) Equations (1), (2), and (6) are combined to give the following modified Mi,j equations in terms of component mole fractions, assuming that all other variables are specified or guessed: M i , j = xi , j −1 V j + j −1 m =1 Fm − U m − Wm − V1 − Pj − 2 , j + xi , j +1 Ki , j +1V j +1 + Pj + 2 , j xi , j + 2 + Pj − 2 , j xi , j − 2 + Fj zi , j − xi , j V j +1 + j m =1 Fm − U m − Wm − V1 + Pj + 2 , j + U j + V j + Wj Ki , j = 0 Compared to Eqs. (10-7) to (10-11), we see that the modified Mi,j equations contain 5, rather than 3, unknown liquid-phase mole fractions. The additional mole fractions are xi,j+2 and xi,j-2. Therefore, the set is no longer tridiagonal. We have a banded matrix with 5 bands, one additional band due to the recycle and one due to the bypass. Exercise 10.4 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: B1 C1 0 x1 D1 A2 B2 C2 • x2 = D2 = 0 A3 B3 −160 x3 200 50 - 350 D3 0 C1 200 = = −1.25 B1 −160 p2 = C2 180 = = −0.626 ....
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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