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, (101), 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, (12) and the summation equations (103) and (104) 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. (101) to (105).
Find: Revised set of MESH equations to account for liquid pumparounds as shown in Fig. 10.2.
Revised M equations like Eq. (107). 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. (107) to (1011), we see that the modified Mi,j equations contain 5, rather than
3, unknown liquidphase mole fractions. The additional mole fractions are xi,j+2 and xi,j2.
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 (1013) to (1018) 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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 Spring '11
 Levicky
 The Land

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