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

9of the srk equation of state gave a bubble point

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 (10-63) and (10-64). Consequences of changing ordering in the above two equations. With the variable ordering v, T, l, as seen in Eq. (10-68), the non-zero variables in Analysis: the A j submatrix are pushed to the right, close to the B j submatrix, while the non-zero 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 non-zero variables in the A j submatrix are now pushed to the left, away from the B j submatrix, while the non-zero 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. (10-63) and (10-64) 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 (10-104). Given: Inside-out method procedure in Section 10.5. Find: Details of the subject method. Analysis: From Eq. (10-104), when uncorrected, Si,j = αI,j Sb,j . This product appears in the component material balance equations (10-83). Using initial estimates from steps 1-7 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 Inside-out method. Given: Error function for the simultaneous correction method. Find: Error function Analysis: Rewrite Eqs. (10-82) and (10-84) 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. (10-75), 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...
View Full Document

Ask a homework question - tutors are online