Exam (elaborations) TEST BANK FOR Separation Process Engineering Includes Mass Transfer Analysis 3rd Edition By Phillip C. Wankat

Exam (elaborations) TEST BANK FOR Separation Process Engineering Includes Mass Transfer Analysis 3rd Edition By Phillip C. Wankat (Instructor's Solution Manual) SOLUTION MANUAL for SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Analysis 3rd Edition (Formerly published as Equilibrium-Staged Separations) by Phillip C. Wankat 17 SPE 3rd Edition Solution Manual Chapter 1 New Problems and new solutions are listed as new immediately after the solution number. These new problems in chapter 1 are: 1A3, 1A4, 1B2-1B4, 1D1. A2. Answers are in the text. A3. New problem for 3rd edition. Answer is d. B1. Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse osmosis or by distillation). B2. New problem for 3rd edition. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually adsorption), or a reverse osmosis system. B3. New problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems. B4. New problem for 3rd edition. You probably used some of the following: chromatography, crystallization, distillation, extraction, filtration and ultrafiltration. D1. New problem for 3rd edition. Basis 1kmol feed. .4 kmole E  .4MW  46 18.4 kg   .6 kmol Water .6 MW 18 10.8 kg total 29.2 kg     Weight fraction ethanol = 18.4/29.2 = 0.630 Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr. 18 Chapter 2 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 2A6, 2A9 to 2A16, 2C4, 2C8, 2C9, 2D1.g, 2.D4, 2D10, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3. 2.A1. Feed to flash drum is a liquid at high pressure. At this pressure its enthalpy can be calculated as a liquid. eg. F,Phigh pLIQ F ref h T c T T . When pressure is dropped the mixture is above its bubble point and is a two-phase mixture (It “flashes”). In the flash mixture enthalpy is unchanged but temperature changes. Feed location cannot be found from TF and z on the graph because equilibrium data is at a lower pressure on the graph used for this calculation. 2.A2. Yes. 2.A4. 2.A6. New Problem. In a flash drum separating a multicomponent mixture, raising the pressure will: i. Decrease the drum diameter and decrease the relative volatilities. Answer is i. 2.A8. a. K increases as T increases b. K decreases as P increases c. K stays same as mole fraction changes (T, p constant) -Assumption is no concentration effect in DePriester charts d. K decreases as molecular weight increases 2.A9. New Problem. The answer is 0.22 2.A10. New Problem. The answer is b. 2.A11. New Problem. The answer is c. 1.0 .5 0 1.0 .5 0 xw Flash operating line yw Equilibrium (pure water) 2.A4 zw = 0.965 19 2.A12. New Problem. The answer is b. 2.A13. New Problem. The answer is c. 2.A14. New Problem. The answer is a. 2.A15. New Problem. a. The answer is 3.5 to 3.6 b. The answer is 36ºC 2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from the amount of superheat. 2.B1. Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium. Examples: drum drum F, z, T , P F F, T , z, p F F, h , z, p drum F, z, y, P F F, T , z, y F F, h , z, y drum F, z, x, p F F, T , z, x etc. drum F, z, y, p F drum drum F, T , z, T , p drum F, z, x, T F F, T , y, p Drum dimensions, drum drum z, F , p F drum F, T , y, T Drum dimensions, drum z, y, p F F, T , x, p etc. F drum F, T , x, T F F, T , y, x 2.B2. This is essentially the same problem (disguised) as problem 2-D1c and e but with an existing (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 which corresponds to 2-D1c. If lb mole F 1000 , D .98 and L 2.95 ft from Problem 2-D1e hr . Since D α V and for constant V/F, V α F, we have D α F . With F = 25,000: new old new old new new F F = 5, D = 5 D = 4.90, and L = 3 D = 14.7 . Existing drum is too small. Feed rate drum can handle: F α D2. 2 2 existing exist F D 4 1000 .98 .98 gives existing F 16,660 lbmol/h Alternatives a) Do drums in parallel. Add a second drum which can handle remaining 8340 lbmol/h. b) Bypass with liquid mixing 20 Since x is not specified, use bypass. This produces less vapor. c) Look at Eq. (2-62), which becomes v drum L v v V MW D 3K 3600 Bypass reduces V c1) Kdrum is already 0.35. Perhaps small improvements can be made with a better demister → Talk to the manufacturers. c2) ρv can be increased by increasing pressure. Thus operate at higher pressure. Note this will change the equilibrium data and raise temperature. Thus a complete new calculation needs to be done. d) Try bypass with vapor mixing. e) Other alternatives are possible. 2.C2. A B B A V z z F K 1 K 1 2.C5. a. Start with i i i Fz x and let V F L L VK i i i i i i Fz z x or x L F L K L L 1 K F F Then i i i i i i K z y K x L L 1 K F F From i i i i i K 1 z y x 0 we obtain 0 L L 1 K F F V = .25 (16660) = 4150 LTotal x y = .58, 8340 16,660 25,000 21 2.C7. i i z V 1 f V 1 K 1 F F From data in Example 2-2 obtain: V/F 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 f 0 -.09 -.1 -.09 -.06 -.007 .07 .16 .3 .49 .77 2.C8. New Problem. drum p x y z drum T F = L + V z F Lx Vy Solve for L & V Or use lever arm-rule 22 2.C9. New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are, 1 2 F V L L i 1 i,L1 2 i,L2 i Fz L x L x Vy Substituting in equilibrium Eqs. (2-60b) and 2-60c) i 1 i,L1 L2 i,L2 2 i,L2 iV L2 i,L2 Fz L K x L x VK x Solving, i i i,L2 1 i,L2 2 i,V L2 1 i,L1 L2 1 i,V L2 Fz Fz x L K L VK L K F V L VK Dividing numerator and denominator by F and collecting terms. i i,liq 2 1 i,L1 L 2 i,V L 2 z x L V 1 K 1 K 1 F F Since i i,V L2 i,L2 y K x , i,V L2 i i 1 i,L1 L2 i,V L2 K z y L V 1 K 1 K 1 F F Stoichiometric equations, C C C C i,L 2 i i i,L 2 i 1 i 1 i 1 i 1 x 1 , y 1 , thus, y x 0 which becomes C i,V L 2 i i 1 1 i,L1 L 2 i,V L 2 K 1 z 0 L V 1 K 1 K 1 F F (2-62) Since i,L1 L 2 i i,liq1 i,L1 L 2 i,liq 2 i,liq1 1 i,L1 L 2 i,V L 2 K z x K x , we have x L V 1 K 1 K 1 F F In addition, C i,L1 L 2 i i,liq1 i,liq 2 i 1 1 i,L1 L 2 i,V L 2 K 1 z x x 0 L V 1 K 1 K 1 F F (2-63) 2.D1. a. V 0.4 100 40 and L F V 60 kmol/h Slope op. line L V 3 2, y x z 0.6 See graph. y 0.77 and x 0.48 b. V 0. and L 900 . Rest same as part a. c. Plot x 0.2 on equil. Diagram and int ercept y x z 0.3. y zF V 1.2 V F z 1.2 0.25 . From equil y 0.58 . d. Plot x 0.45 on equilibrium curve. 23