SOLUTION MANUAL
for
SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Transfer Analysis Analysis
3rd Edition Separations) (Formerly published as Equilibrium-Staged Separations
by Phillip C. Wankat
rd
SPE 3 Edition Solution Manual Chapter 1 New Problems Problems and new new solutions are listed listed as new immediately immediately after the solution number. number. These These new problems in chapter chapter 1 are: are: 1A3, 1A4, 1B2-1B4, 1D1.
A2.
Answers are in the text.
A3.
New problem 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 r everse osmosis or by distillation).
B2.
rd exchange), or a filter, or a New problem problem for 3 edition. Many homes have a water softener (ion exchange), carbon water “filter” (actually adsorption), or a r everse osmosis system. system.
B3.
New problem problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.
B4.
rd New problem problem for 3 edition. You probably used some of the following: chromatography, crystallization, crystallization, distillation, distillation, extraction, filtration and ultrafiltration.
D1.
New problem problem for 3rd edition. Basis 1kmol feed.
.4 kmole E
.4 MW
.6 kmol kmol Water ater
46 18.4 kg
.6 MW 18
10.8 kg
total
29.2 kg kg
Weight fraction ethanol = 18.4/29.2 = 0.630 Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.
17
Chapter 2 New Problems Problems and new new solutions are listed as as new immediately after the solution solution number. number. These These new problems are: 2A6, 2A6, 2A9 to to 2A16, 2C4, 2C4, 2C8, 2C9, 2C9, 2D1.g, 2.D4, 2.D4, 2D10, 2D13, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3.
2.A1. 2.A1.
Feed Feed to flash flash drum is a liq liquid uid at hig high h pre pressu ssure. re. At this this pres pressure sure its its ent enthal halpy py can be calcul calculated ated as a liquid. liquid. eg. h TF , Phigh
c p LIQ TF
Tref . When pressure pressur e is dropped the mixture is above
its bubble point point and is a two-phase mixture mixture (It “flashes”). In the flash mixture enthalpy enthalpy is unchanged but temperature changes. changes. Feed location cannot cannot be found from T F and z on the graph because equilibrium data is at a lower lower pressure on the graph used for this calculation. 2.A2.
Yes.
2.A4. !"# 6782)29028: 5'
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2.A6. New Problem. Problem. In a flash drum separating a multicomponent mixture, raising the pressure will: i. Decrease the drum diameter diameter and decrease the relative volatilities. volatilities. Answer is i. i. 2.A8.
a. K incr eases as T increases b. K decreases as P increases increases c. K stays stays same as mole fraction changes changes (T, p constant) -Assumption is no concentration effect in DePriester charts d. K decreases decreases as molecul molecular ar weight weight increases increases
2.A9.
New Problem. Problem. The answer is 0.22
2.A10.
New Problem. Problem. The answer is b.
2.A11.
New Problem. Problem. The answer is c.
18
2.A12.
New Problem. Problem. The answer is b.
2.A13.
New Problem. Problem. The answer is c.
2.A14.
New Problem. Problem. The answer is a.
2.A15.
New Problem. Problem. a. b. The answer is
The answer is 36ºC
3.5 to 3.6
Problem. The liquid is superheated when the pressure drops, and the energy comes fr om the 2.A16. New Problem. amount of superheat.
2.B1. 2.B1.
Must be Must be sure you you don don’t ’t violate violate Gibbs Gibbs phase phase rule rule for inten intensive sive variable variabless in in equi equili libri brium. um. Examples: F, TF , z, z, p F, z, Tdrum , Pdrum F, h F , z, z, p
F,z,y,Pdrum
F, TF , z, z, y
F, h F , z, z, y
F,z,x,p drum
F, TF , z, z, x
etc.
F,z,y,p drum
F, TF , z, Tdrum , p drum
F,z,x,Tdrum
F, TF , y, y, p
Drum dimensions, z, Fdrum , p drum
F, TF , y, y , Tdrum
Drum dimensions, z,y,p drum
F, TF , x, x, p
etc.
F, TF , x, x , Tdrum F, TF , y, y, x
2.B2. 2.B2.
This This is is essen essentia tially lly the same problem problem (disgui (disguised sed)) as proble problem m 2-D1c and e but but with with an exis existin ting g (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 0. 25 which corresponds corresponds to 2-D1c. lb mole If 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 With F = 25,000:
F.
Fnew Fold = 5, Dnew = 5 Dold = 4.90, and Lnew = 3 Dnew = 14.7 . Existing drum is too small. Feed
Fexisting
rate
drum
can
handle:
F
α
2
D.
Fexisting
D exist
1000
.98
2
4
2
.98
gives
16,660 lbmol/h
Alternatives Alternatives a) Do drums drums in parallel. Add a second second drum drum which can handle remaining 8340 lbmol/h. b) Bypass with liquid liquid mixing
19
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EF-1*) &
GI?#
Since x is not specified, specifi ed, use bypass. This produces less less vapor. c) Look Look at Eq. Eq. (2-6 (2-62), 2), which which becom becomes es
V MW v
D
3 K d r um 3 6 0 0
L
v
v
Bypass reduces V c1)
K drum improvements can be made with a better demister demister drum is already 0.35. Perhaps small improvements → Talk to the manufacturers. c2) ρv can be increased increased by increasing increasing pressure. pressure. Thus operate at higher pressure. pressure. Note this this will change change the equilibrium equilibrium data data and raise temperature. Thus a complete complete new calculation needs to be done. d) Try bypass bypass with with vapo vaporr mixin mixing. g. e) Other Other alte alterna rnativ tives es are pos possi sible ble.. V
2.C2.
2.C5.
zA
F
a. Start with
xi
KB
zB 1
Fz i L
KA
and let V
VK i
Fz i
xi
1
L
F
F
L
or x i L Ki
zi L F
Then y i
Kixi
yi
L F
K i
K i zi L F
From
1
xi
1
L F
K i
0 w e o b ta i n
Ki L F
1
1 zi L F
0 K i
20
zi
2.C7.
1
Ki
1
V
1
f
V F
F
21
