Instructor's Solution Manual - Separation Process Engineering: Includes Mass Transfer Analysis

478 Pages·2011·25.864 MB·English

by  Phillip C. Wankat

SOLUTION MANUAL
for
SEPARATION PROCESS ENGINEERING.
Includes Mass Transfer Analysis
rd
3 Edition
(Formerly published as Equilibrium-Staged Separations)
by
Phillip C. Wankat
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
10.8 kg
.6 kmol Water .6 MW18
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.
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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. h T c T T . When pressure is dropped the mixture is above
F,Phigh pLIQ F ref
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 T and z on the
F
graph because equilibrium data is at a lower pressure on the graph used for this calculation.
2.A2. Yes.
2.A4.
1.0
Equilibrium
z = 0.965
yw (pure water) w
Flash
.5 operating
line
2.A4
0
0 .5 x 1.0
w
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.
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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:
F,z,T ,P F,T ,z,p F,h ,z,p
drum drum F F
F,z,y,P F,T ,z,y F,h ,z,y
drum F F
F,z,x,p F,T ,z,x etc.
drum F
F,z,y,p F,T ,z,T , p
drum F drum drum
F,z,x,T F,T ,y,p
drum F
Drum dimensions, z,F , p F,T ,y,T
drum drum F drum
Drum dimensions, z,y,p F,T ,x,p
drum F
etc. F,T ,x,T
F drum
F,T ,y,x
F
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.
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 F.
With F = 25,000:
F F = 5, D = 5 D = 4.90, and L = 3 D = 14.7.
new old new old new new
Existing drum is too small.
F D 2 4 2
Feed rate drum can handle: F α D2. existing exist gives
1000 .98 .98
F 16,660 lbmol/h
existing
Alternatives
a) Do drums in parallel. Add a second drum which can handle remaining 8340 lbmol/h.
b) Bypass with liquid mixing
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y = .58,
V = .25 (16660) = 4150
16,660
25,000
L
Total
8340
x
Since x is not specified, use bypass. This produces less vapor.
c) Look at Eq. (2-62), which becomes
V MW
D v
3K 3600
drum L v v
Bypass reduces V
c1) K is already 0.35. Perhaps small improvements can be made with a better demister
drum
→ Talk to the manufacturers.
c2) ρ can be increased by increasing pressure. Thus operate at higher pressure. Note this
v
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.
V z z
2.C2. A B
F K 1 K 1
B A
Fz
2.C5. a. Start with x i and let V F L
i
L VK
i
Fz z
x i or x i
i L F L K i L L
i 1 K
F F i
K z
Then y K x i i
i i i L L
1 K
F F i
K 1 z
From y x 0 we obtain i i 0
i i L L
1 K
F F i
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z V
2.C7. i 1 f From data in Example 2-2 obtain:
V F
1 K 1
i F
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.
p
drum
F = L + V
F Lx Vy
T z
drum
x y
Solve for L & V
Or use lever arm-rule
z
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2.C9. New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,
F V L L
1 2
Fz L x L x Vy
i 1 i,L1 2 i,L2 i
Substituting in equilibrium Eqs. (2-60b) and 2-60c)
Fz L K x L x VK x
i 1 i,L1 L2 i,L2 2 i,L2 iV L2 i,L2
Solving,
Fz Fz
x i i
i,L2
L K L VK L K F V L VK
1 i,L2 2 i,V L2 1 i,L1 L2 1 i,V L2
Dividing numerator and denominator by F and collecting terms.
z
x i
i,liq2 L V
1 K 1 1 K 1
i,L1 L2 F i,V L2 F
K z
Since y K x , y i,V L2 i
i i,V L2 i,L2 i L V
1 K 1 1 K 1
i,L1 L2 F i,V L2 F
C C C C
Stoichiometric equations, x 1 , y 1 , thus, y x 0
i,L2 i i i,L2
i 1 i 1 i 1 i 1
K 1 z
C
i,V L2 i
which becomes 0 (2-62)
L V
i 1 1 K 1 1 K 1
i,L1 L2 F i,V L2 F
K z
Since x K x , we have x i,L1 L2 i
i,liq1 i,L1 L2 i,liq2 i,liq1 L V
1 K 1 1 K 1
i,L1 L2 F i,V L2 F
K 1 z
C
i,L1 L2 i
In addition, x x 0 (2-63)
i,liq1 i,liq2 L V
i 1 1 K 1 1 K 1
i,L1 L2 F i,V L2 F
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 0.4 1500 600 and L 900. Rest same as part a.
c. Plot x 0.2 on equil. Diagram and y x z 0.3. y zF V 1.2
intercept
V F z 1.2 0.25. From equil y 0.58.
d. Plot x 0.45 on equilibrium curve.
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L F V 1 V F .8
Slope 4
V V V F .2
Plot operating line, y x z at z 0.51. From mass balance F 37.5 kmol/h.
e. Find Liquid Density.
MW x MW x MW .2 32.04 .8 18.01 20.82
L m m w w
MW MW 32.04 18.01
Then, V x m x w .2 .8 22.51 ml/mol
L m w
.7914 1.00
m w
MW V 20.82 22.51 0.925 g/ml
L L L
P MW
Find Vapor Density. v (Need temperature of the drum)
v
RT
MW y MW y MW .58 32.04 .42 18.01 26.15 g/mol
v m m w w
Find Temperature of the Drum T:
From Table 3-3 find T corresponding to
y .58, x 20, T=81.7 C 354.7K
ml atm
1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10 4 g/ml
v mol K
Find Permissible velocity:
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u K
perm drum L v v
2-60
2 3 4
K exp A B nF C nF D nF E nF
drum lv lv lv lv
V
Since V F 0.25 1000 250 lbmol/h,
F
lb
W V MW 250 26.15 6537.5 lb/h
v v lbmol

L F V 1000 250 750 lbmol/h, and W L MW 750 20.82 15,615 lb/h,
L L
W 15615 8.89 10 4
F L V 0.0744, and n F 2.598
lv lv
W 6537.5 .925
V L
.925 8.98 10 4
Then K .442, and u .442 14.19 ft/s
drum perm 8.98 10 4
V MWv 250 26.15 454 g/lb
A 2.28 ft2.
cs u 3600 14.19 3600 8.98 10 4 g/ml 28316.85 ml/ft3
perm v
Thus, D 4A 1.705 ft. Use 2 ft diameter.
cs
L ranges from 3 D 6 ft to 5 D=10 ft
Note that this design is conservative if a demister is used.
f. Plot T vs x from Table 3-3. When T 77 C, x 0.34, y 0.69. This problem is now
very similar to 3-D1c. Can calculate V/F from mass balance, Fz Lx Vy. This is
V z y 0.4 0.34
Fz F V x Vy or 0.17
F y x 0.69 0.34
g. Part g is a new problem. V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756.

2-D2. Work backwards. Starting with x , find y = 0.62 from equilibrium. From equilibrium point
2 2
V
plot op. line of slope L V 1 V F 3 7. This gives
2 F 2
2
z 0.51 x (see Figure). Then from equilibrium, y 0.78.
2 1 1
V z x 0.55 0.51
For stage 1, 1 1 0.148.
F y x 0.78 0.51
1 1

2.D3. a. z 0.4 V F 0.6 V 6.0 kmol h, L 4.0
L
y x z V F
V
Op. eq.
2
y x 2 3
3
See graph: y 0.55 x 0.18 T~82.8 C linear interpretation on Table 2-7 .
M

24
b. Product 78.0 C x 0.30, y 0.665,
Mass Bal: Fz Lx Vy F V x Vy
or 4.0 10 V 0.3 0.665V
V 2.985 and V F 0.2985
Can also calculate V/F from slope.
V
c. F 10, 0.3 V 3 & L 7
F
L z 7 z
y x x
V V F z 0.3
If y 0.8, x 0.545 @ equil
7
Then z 0.3 0.8 0.545 0.6215.
3
7
Can also draw line of slope through equil point.
3
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