Unformatted text preview: ,11 = li ,11 + υi ,11 − li ,10 − υi ,16 = 0
M i ,16 = li ,16 + υi ,16 (1 + S16 ) − li ,15 − li ,11 − υi ,17 = 0
Note that:
Stage 4 includes flow rates for stages 3, 4, 5, and 12.
Stage 12 includes flow rates for stages 4, 12, and 13.
Stage 11 includes flow rates for stages 10, 11, and 16.
Stage 16 includes flow rates for stages 11, 15, 16, and 17.
The same stages appear in the energy balance equations. Therefore, the matrix structure has the
diverse A and C blocks as shown in the diagram below. The matrix is not block tridiagonal. Exercise 10.16 (continued)
Analysis: (continued) Exercise 10.16 (continued)
Analysis: (continued) Matrix Structure Exercise 10.17
Subject:
Given:
Find: Effect of changing variable and equation ordering.
Ordering equations (1063) and (1064).
Consequences of changing ordering in the above two equations. With the variable ordering v, T, l, as seen in Eq. (1068), the nonzero variables in
Analysis:
the A j submatrix are pushed to the right, close to the B j submatrix, while the nonzero
variables in the C j submatrix are pushed to the left, close to the B j submatrix. Thus, the nonzero variables are arranged across the block row of these three submatrices in the most compact
manner.
If the ordering is changed to l, v, T , the nonzero variables are spread out over the entire
width of the row of the three submatrices. That is, the nonzero variables in the A j submatrix
are now pushed to the left, away from the B j submatrix, while the nonzero variables in the C j
submatrix are pushed to the right, away from the B j submatrix. This is the least compact
arrangement.
If the ordering of the equations is changed from H, M, E to E, M, H, there is little
change.
Thus, the orderings given by Eqs. (1063) and (1064) appear to as efficient or more
efficient than the other orderings given above. Exercise 10.18
Subject: Detailed method for determining the scalar multiplier, Sb , in (10104). Given: Insideout method procedure in Section 10.5.
Find: Details of the subject method.
Analysis: From Eq. (10104), when uncorrected, Si,j = αI,j Sb,j .
This product appears in the component material balance equations (1083).
Using initial estimates from steps 17 of the Initialization Procedure, values of li,j are computed
from step 9.
If the sum of li,j for the Nth stage, i.e.
B= C
i =1 is not the specified B, then use, li , N Si , j = Sb α i , j Sb , j where Sb is determined by a search, e.g. a Fibonacci search to make B = specified B.
Typically, the search might be over the range of 0.75 to 1.25. Exercise 10.19
Subject: Error function for the Insideout method. Given: Error function for the simultaneous correction method.
Find: Error function
Analysis: Rewrite Eqs. (1082) and (1084) in residual form:
Ei , j = υ i , j − α i , j Sb , j li , j = 0
M i , j = Li , j −1 − RLj + α i , j ..... li , j + α i , j .... li , j +1 + f i , j = 0
H j = hL j RLj L j +.........− hF j Fj − Q j = 0
Then, following Eq. (1075), τ= N
j =1 Hj
scale factor 2 + C
i −1 (M ) + (E )
2 i, j i, j 2 ≤ε where the scale factor is a nominal heat of vaporization, some hV j − hL j Exercise 10.20 Subject:
Distillation of a light normal paraffin hydrocarbon mixture by a rigorous...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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