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# ch07 - CHAPTER 7 SOLUTIONS 7.1 First consider all the...

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7-1 CHAPTER 7 SOLUTIONS 7.1 First consider all the possible splits. They are as follows with identification numbers 1 to 20, where: A = C3, B = B1, C = NB, D = B2, and E = C5. The five components are ordered from A to E in decreasing volatility. 5 - Component splits 4 - Component splits 3 - Component splits 2 - Component splits 1 A/BCDE 5 A/BCD 11 A/BC 17 A/B 2 AB/CDE 6 AB/CD 12 AB/C 18 B/C 3 ABC/DE 7 ABC/D 13 B/CD 19 C/D 4 ABCD/E 8 B/CDE 14 BC/D 20 D/E 9 BC/DE 15 C/DE 10 BCD/E 16 CD/E The 14 sequences of 4 columns each are as follows: Starting with 1 Starting with 2 Starting with 3 Starting with 4 1 - 8 - 15 - 20 2 - (17 , 15) - 20 3 - (11 , 20) - 18 4 - 5 - 13 - 19 1 - 8 - 16 - 19 2 - (17 , 16) - 19 3 - (12 , 20) - 17 4 - 5 - 14 - 18 1 - 9 - (18 , 20) 4 - 6 - (17 , 19) 1 - 10 - 13 -19 4 - 7 - 11 - 18 1 - 10 - 14 - 18 4 - 7 - 12 - 17 Figure 7.12(b) is the sequence 1 - 9 - (18 , 20)

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7-2 7.2 To determine the number of possible sequences when two different types of separators are used, use Eq. (7.9) with the equation: N s T = T P- 1 N s where, P = number of products = 4 T = separator types = 2 Combining the two equations, the number of possible sequences is, ( ) [ ] ( ) ( ) [ ] ( ) 40 5 8 ! 3 ! 4 ! 6 2 ! 1 4 ! ! 1 4 2 2 ! 1 ! ! 1 2 3 1 4 1 = = = = = P P P P T N P T s
7-3 7.3 The feed contains 8 components, but only 7 products are to be produced because one of the products combines the two lightest components. Pertinent data for applying the heuristics are as follows, where the symbols A, B, etc. replace the usual component symbols: Product species Symbols Feed Mole fraction Adjacent Relative Volatility, α H 2 , C 1 A, B 0.38 5.0 C 2 = C 0.26 1.6 C 2 D 0.15 3.6 C 3 = E 0.10 1.1 C 3 F 0.06 2.9 nC 4 G 0.04 2.5 nC 5 + H 0.01 From this table, we see that the most difficult separations are C / D and E / F. Also, the largest product mole fraction in the feed is A,B at 0.38. Two good sequences, in terms of splits, are as follows: Case 1. Apply Heuristics 3 and 4 on page 251: AB / CDEFGH, CD / EFGH, C / D, EF / GH, E / F, G / H Case 2. Apply Heuristic 6 on page 251: AB / CDEFGH, C / DEFGH, D / EFGH, E / FGH, F / GH, G / H This is the same result that would be obtained from Heuristic 2 on page 251.

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7-4 7.4 From the given data, adjacent volatilities are: B / C at 1.39 and D / E at 1.25 The most plentiful component in the feed is E. Consider the sequences in Fig. 7.48 and the 6 heuristics on page 251. The best sequence applies Heuristic 3 to drop out E early, Heuristic 4 to leave the most difficult separation last, and Heuristic 6 to approximate equimolar distribution. The second best sequence is similar and applies Heuristic 3 to drop out E early and Heuristic 4 to leave the two most difficult separations last. The third best sequence applies Heuristic 4 to leave the two most difficult separations last. The worst sequence performs the most difficult separation first, the easiest separation last, and the second most difficult separation in the presence of plentiful D. Heuristics 4 and 6 appear to be the most important, with Heuristic 3 also important, but Heuristic 2 not very important.
7-5 7.5 Use the following symbols for the components: A = RH 3 , B = HCl, C = RH 2 Cl, D = RHCl 2 , and E = RCl 3 . This is the order of volatility, which not the order in the table.

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