Phillip C Wankat - Instructor's Solution Manual - Separation Process Engineering_ Includes Mass Transfer Analysis-Prentice Hall (2011)

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SOLUTION MANUAL 
for 
SEPARATION PROCESS ENGINEERING. 
Includes Mass Transfer Analysis 
3rd Edition 
(Formerly published as Equilibrium-Staged Separations) 
by 
Phillip C. Wankat 
17 
SPE 3
rd
 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. 
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. 
high LIQF,P p 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 drumF,z,T ,P FF,T , z, p FF, h , z, p
drumF, z, y, P FF,T , z, y FF, h , z, y
drumF, z, x, p FF,T , z, x etc. 
drumF, z, y, p F drum drumF,T ,z,T , p
drumF, z, x,T FF,T , y, p 
Drum dimensions, drum drumz,F , p F drumF,T , y,T
Drum dimensions, drumz, y, p FF,T , x, p
etc. F drumF,T , x,T
FF,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 newF 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 
existingF 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
z zV
F K 1 K 1
2.C5. a. Start with ii
i
Fz
x and let V F L
L VK
i i
i i
i
i
Fz z
x or x
L LL F L K 1 K
F F
Then i ii i i
i
K z
y K x
L L
1 K
F F
From i ii 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 F1 K 1
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.
drump
x y 
z 
drumT
F = L + V 
zF 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 2F V L L
i 1 i,L1 2 i,L2 iFz 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,L2Fz 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 L2
z
x
L V
1 K 1 K 1
F F
Since i i,V L2 i,L2y 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,L2 i i i,L2
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 L2 ii,liq1 i,L1 L2 i,liq 2 i,liq1
1
i,L1 L2 i,V L2
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 0.4 1500 600 and L 900 . Rest same as part a. 
c. Plot x 0.2 on equil. Diagram and int ercepty 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 
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.
 L m m w wMW x MW x MW .2 32.04 .8 18.01 20.82
Then, wmL m w
m w
MWMW 32.04 18.01
V x x .2 .8 22.51 ml/mol
.7914 1.00
LL LMW V 20.82 22.51 0.925 g/ml 
Find Vapor Density. vv
P MW
RT
 (Need temperature of the drum) 
v m wm w
MW y MW y MW .58 32.04 .42 18.01 26.15 g/mol 
Find Temperature of the Drum T: 
From Table 3-3 find T corresponding to 
y .58, x 20, T=81.7 C 354.7K
4
v
ml atm
1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10 g/ml
mol K
Find Permissible velocity: 
24 
 
 
perm drum L v v
2 3 4
drum lv lv lv lv
u K
 K exp A B nF C nF D nF E nF
 2-60 
 Since 
V
V F 0.25 1000 250 lbmol/h,
F
 
 vv
lb
W V MW 250 26.15 6537.5 lb / h
lbmol
 
 
LLL F V 1000 250 750 lbmol/h, and W L MW 750 20.82 15, 615 lb/h, 
 
4
VL
lv lv
V L
W 15615 8.89 10
F 0.0744, and n F 2.598
W 6537.5 .925
 
 Then 
4
drum perm 4
.925 8.98 10
K .442, and u .442 14.19 ft/s
8.98 10
 
 
v
2
cs 4 3
perm v
V MW 250 26.15 454 g/lb
A 2.28 ft .
u 3600 14.19 3600 8.98 10 g/ml 28316.85 ml/ft
 
 Thus, csD 4A 1.705 ft. Use 2 ft diameter. 
 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, x0.34, 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 x2, find y2 = 0.62 from equilibrium. From equilibrium point 
plot op. line of slope 
2 2
2
V
L V 1 V F 3 7.
F
 This gives 
2 1z 0.51 x (see Figure). Then from equilibrium, 1y 0.78 . 
 For stage 1, 1 1
1 1
z xV 0.55 0.51
0.148
F y x 0.78 0.51
. 
 
2.D3. a. z 0.4 V F 0.6 V 6.0 k mol h, L 4.0 
 Op. eq. 
L
y x z V F
V
2
y x 2 3
3
 
 See graph: My 0.55 x 0.18 T ~ 82.8 C linear interpretation on Table 2-7 . 
 
25 
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. 
c. 
V
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
Then 
7
z 0.3 0.8 0.545 0.6215.
3
 
Can also draw line of slope 
7
3
 through equil point. 
26 
2.D4. New problem in 3rd edition. Highest temperature is dew point V F 0
Set i i i i iz y . K y x
 Want i i ix y K 1.0
 ref New ref Old i iK T K T y K
If pick C4 as reference: First guess bu tan eK 1.0, T 41 C : C3 C6K 3.1, K 0.125
i
i
y .2 .35 .45
4.0145
K 3.1 1.0 .125
 T too low 
Guess for reference 
C4K 4.014 T 118 C : C3 C6K 8.8, K .9 
i
i
y .2 .35 .45
0.6099
K 8.8 4.0145 .9
C4,NEWK 4.0145 .6099 2.45, T 85 : C2 C6K 6.0, K 0.44
i
i
y .2 .35 .45
1.20
K 6 2.45 .44
C4,NEWK 2.45 1.2 2.94, T 96 C : C3 C6K 6.9, K 0.56
i
i
y .2 .35 .45
0.804 Gives 84 C
K 6.9 2.94 .56
27 
 
 Use 90.5º → Avg last two T C4 C3 C6K 2.7, K 6.5, K 0.49 
 i i
.2 .35 .45
y K 1.079
6.5 2.7 .49
 
 T ~ 87 88º C 
 Note: hexane probably better choice as reference. 
2.D5. a) 
 
 
 b) 1 1
1 1
L F
y x z
V V
 Plot 1st Op line. 
 y = x = z = 0.55 
 to x1 = 0.3 on eq. curve (see graph) 
 Slope 
L 0.55 0.80 .25
0.454545
V .55 0 .55
 
 1 1 1 1
1
L
0.45454 & L V F 1000
V
 
 
 
 
 
 c) Stage 2 2
V L 0.75F
0.25 , 3 , y x z 0.66.
F V 0.25F
 Plot op line 
 At 2
0.66 F z 0.66
x 0, y z V F 2.64. At y 0, x z 0.88
0.25 L L F 0.75
 
 From graph 2 2y 0.82, x 0.63 . 
 2 2
2
V
V F 0.25 687.5
F
 171.875 kmol/h 
 
F1 = 1000 
 
z1 = 0.55 
p1,2 = 1 atm 
x2 
 
x1 = 0.30 
 
v2 
y2 
v1 = F2 
y1 = z2 
2 1 
2
V
0.25
F
 
y1 = 0.66 = z2 
V1 = 687.5 kmol/h = F2 
1
V 687.5
0.6875
F 1000
 
28 
2.D6.
RR eq., i i
i
K 1 z
0
1 K 1 V F
, First guess V/F = 0.6 
1
1.4 .45 0.2 .35 0.7 .2
f 0.0215
1 1.4 .6 1 0.2 0.6 1 0.7 0.6
Use Newtonian Convergence 
2c
i ik
2
i 1 i
K 1 zdf
d V F 1 K 1 V F
k
k 1 k
fV V
dfF F
d V F
T = 50ºC 
P = 200 kPa 
Kc4 = 2.4 
zc4 = 0.45 
F = 1.0 kmol/min 
zc5 = 0.35 
Zc6 = 0.20 L 
V 
Kc5 = 0.80 
Kc6 = 0.30 
29 
2 22
1
2 2 2
1.4 0.45 0.2 0.35 0.7 0.20df
0.570
V 1 1.4 .6 1 0.2 0.6 1 0.7 0.6d
F
2
V 0215
0.6 0.6377
F 0.570
2
1.4 .45 0.2 .35 0.7 0.2
f 0.00028
1 1.4 0.6377 1 0.2 0.6377 1 0.7 0.6377
Which is close enough. 
i i ii
c4
c4
i
y K xz 0.45
x 0.2377, 
V y 2.4 0.2377 0.57051 1.4 .63771 K 1
F
c5 c5
0.35
x 0.4012, y 0.8 0.4012 0.3210
1 0.2 0.6377
 
c6 8
i i
0.20 0.3613 0.1084
x , y 0.30 0.4012
1 0.7 0.6377 x 1.0002 y 0.9998
2.D7. A B
B A
z zV
F K 1 K 1
M PK 5.6 and K 0.21 
V 0.3 0.7
0.2276
F 0.21 1 5.6 1
Eq. (2-38) MM
M
z 0.3
x 0.1466
V 1 4.6 0.22761 K 1
F
P Mx 1 x 0.8534 , M M My K x 5.6 0.1466 0.8208 
P My 1 y 0.1792
2.D8. Use Rachford-Rice eqn: i i
i
K 1 zV
f 0
F 1 K 1 V / F
 . Note that 2 atm = 203 kPa. 
Find iK from DePriester Chart: 1 2 3K 73, K 4.1 K .115
Converge on V F .076, V F V F 152 kmol/h, L F V 1848 kmol/h . 
From ii
i
z
x
V
1 K 1
F
 we obtain 1 2 3x .0077, x .0809, x .9113 
From i i i 1 2 3y K x , we obtain y .5621, y .3649, y .1048 
2.D9. Need hF to plot on diagram. Since pressure is high, feed remains a liquid 
LF P F ref ref
h C T T , T 0 from chart 
L EtOH wP P EtOH P w
C C x C x
30 
Where EtOH wx and x are mole fractions. Convert weight to mole fractions. 
Basis: 100 kg mixture 
30
30 kg EtOH 0.651 kmol
46.07
 
70 kg water 70 18.016 3.885 
Total = 4.536 kmol 
Avg. 
100
MW 22.046
4.536
 Mole fracs: E w
0.6512
x 0.1435, x 0.8565
4.536
. 
Use 
LEtOH
P PC at 100 C as an average C value. 
LP
kcal
C 37.96 .1435 18.0 .8565 20.86
kmol C
Per kg this is LP
avg
C 20.86 kcal
0.946
MW 22.046 kg C
Fh 0.946 2000 189.2 kcal/kg
which can now be plotted on the enthalpy composition diagram. 
Obtain drum E ET 88.2 C, x 0.146, and y 0.617 . 
For F 1000 find L and V from F = L + V and Fz Lx Vy
which gives V = 326.9, and L = 673.1 
Note: If use wt. fracs. 
L LP P avg
C 23.99 & C MW 1.088 Fand h 217.6 . All wrong. 
31 
2.D.10 New Problem. Solution 400 kPa, 70ºC C4z 35 Mole % n-butane C6x 0.7
From DePriester chart C3 C4 C6K 5, K 1.9, K 0.3 
Know ii i i i i i i
i
z
y K x , x , x y 1 z
V
1 K 1
F
R.R. i i C3 C6 C4 C6
i
K 1 z
0 z 1 z z .65 z
V
1 K 1
F
For C6 C6 C6 C6
C6
z z V
0.7 z 0.7 1 0.7
V V F1 K 1 1 0.7
F F
C6
V
z 0.7 0.49
F
RR Eq: C6 C6
4 .65 z 0.9 .35 0.7z
0
V V V
1 4 1 0.9 1 0.7
F F F
2 equations & 2 unknowns. Substitute in for C6z . Do in Spreadsheet. 
Use Goal – Seek to find V F. 
V
0.594
F
 when R.R. equation 0.000881 . 
C6
V
z 0.7 0.49 0.7 (0.49)(0.594) 0.40894
F
2.D11. L F 0.6 V F 0.4 & L V 1.5
Operating line: Slope 1.5, through y x z 0.4 
32 
2.D12. For problem 2.D1c, plot x = 0.2 on equilibrium diagram with feed composition of 0.3. The 
resulting operating line has a y intercept z V / F 1.2 . Thus V F 0.25 (see figure in 
Solution to 2.D1) Vapor mole fraction is y = 0.58. 
Find Liquid Density. 
 L m m w wMW x MW x MW .2 32.04 .8 18.01 20.82
Then, wmL m w
m w
MWMW 32.04 18.01
V x x .2 .8 22.51 ml/mol
.7914 1.00
LL LMW V 20.82 22.51 0.925 g/ml 
Find Vapor Density. vv
p MW
RT
 (Need temperature of the drum) 
v m wm w
MW y MW y MW .58 32.04 .42 18.01 26.15 g/mol 
Find Temperature of the Drum T: 
From Table 2-7 find T corresponding to y .58, x 20, T=81.7 C 354.7K
33 
4
v
ml atm
1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10 g/ml
mol K
Find Permissible velocity: 
perm drum L v v
2 3 4
drum,horizontal drum,vertical lv lv lv lv
u K
K 1.25 K exp A B nF C nF D nF E nF 1.25
Since V V F 0.25 1000 250 lbmol/h,
vvW V MW 250 26.15 lb lbmol 6537.5 lb / h
LLL F V 1000 250 750 lbmol/h, and W L MW 750 20.82 15, 615 lb/h,
4
VL
lv lv
V L
W 15615 8.98 10
F 0.0744, and n F 2.598
W 6537.5 .925
drum,vertical drum,horiz
4
perm 4
K 0.442, and K 0.5525
0.925 8.98 10
u 0.5525 17.74 ft/s
8.98 10
v
cs 4 3
perm v
V MW 250 26.15 454 g/lbm
A
u 3600 17.74 3600 8.98 10 g/ml 28316.85 ml/ft
2
CsA 1.824 ft , 
2
T CsA A 0.2 9.12 ft
With L/D = 4, TD 4A 3.41 ft and L 13.6 ft
2.D13. New Problem. The answer is ycresol = 0.19582
Since pc p p
p
x 1.76 .7
x 0.3, x 0.7, y 0.80418
1 1 x 1 .76 .7
 
c py 1 y 0.19582
Or cp cc
cp c
x
y
1 1 x
, cp
pc
1
0.5682
2.D14. Raoult’s Law: C4C4
Tot
VP
K
P
410 C C 4
10 C6 C6
log VP 4.04615 , VP 11121 mm Hg
log VP 3.2658 , VP 1844.36 mm Hg
i
i
i
z
x 1.0 1.0
1 K 1 V F
 
34 
0.3 0.7
1
11121 1844.36
1 1 0.4 1 1 0.4
P P
 
Solve for Pdrum = 3260 mmHg 
i
i
i
z
x
V
1 K 1
F
C4 C4 C4 C4
.3 11121
x 0.1527, y K x 0.1527 0.52091
11121 32601 1 .4
3260
C6 C4 C6
1844.36
x 1 x 0.84715, y 0.84715 0.47928
3260
Check 1.00019 
2.D15. This is an unusual way of stating problem. However, if we count specified variables we see 
that problem is not over or under specified. Usually V/F would be the variable, but here it 
isn’t. We can still write R-R eqn. Will have three variables: zC2, ziC4, znC4. Need two other 
eqns: iC4 nC4 C2 iC4 nC4z z constant, and z z z 1.0
Thus, solve three equations and three unknownssimultaneously. 
Do It. Rachford-Rice equation is, 
C2 C2 iC4 iC4 nC4 nC2
C2 iC4 nC4
K 1 z K 1 z K 1 z
0
V V V
1 K 1 1 K 1 1 K 1
F F F
Can solve for zC2 = 1 – ziC4 and ziC4 = (.8) znC4. Thus zC2 = 1 – 1.8 znC4 
Substitute for ziC4 and zC2 into R-R eqn. 
C2 iC4 nC4
nC4 nC4 nC4
C2 iC4 nC4
K 1 .8 K 1 K 1
1 1.8 z z z 0
V V V
1 K 1 1 K 1 1 K 1
F F F
Thus, 
C2
C2
nC4
C2 iC4 nC4
C2 iC4 nC4
K 1
V
1 K 1
Fz
K 1 .8 K 1 K 1
1.8
V V V
1 K 1 1 K 1 1 K 1
F F F
Can now find K values and plug away. KC2 = 2.92, KiC4 = .375, KnC4 = .26. 
Solution is znC4 = 0.2957, ziC4 = .8 (.2957) = 0.2366, and zC2 = 0.4677 
2.D16. C1 C4 C5 C6 C1 C4 C5 C6z 0.5, z 0.1, z 0.15, z 0.25, K 50, K .6, K .17, K 0.05
1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom
1
V / F .5 .1 / 3 .53 This first guess is not critical. 
35 
 
 R.R. eq. i i
i
K 1 zV
f 0
F 1 K 1 V F
 
 
49 .5 .4 .1 .83 .15 .95 .25
0.157
1 49 .53 1 .4 .53 1 .83 .53 1 .95 .53
 
 Eq. 3.33 1 2
2 1 i i
2
i
f V FV V
F F z K 1
V
1 K 1
F
 
 where 
1 1
V / F 0.53 and f V / F 0.157 . 
 calculate 
2
V / F .53 0.157 2.92 0.584 
 V .584 150 87.6 kmol/h and L 150 87.6 62.4 
 C1C1
C1
z .5
x 0.016883
1 K 1 (V / F) 1 49 .584
 
 C1 C1 C1y K x 50 0.016883 0.844 
 Similar for other components. 
 
2-D17. a. V 0.4F 400, L 600 Slope L F 1.5 
 Intercepts y = x = z = 0.70. Plot line and find xA = 0.65, yA = 0.77 (see graph) 
b. V = 2000, L = 3000. Rest identical to part a. 
 c. Lowest xA is horizontal op line (L = 0). xA = 0.12 
 Highest yA is vertical op line (V = 0). yA = 0.52. See graph 
 
 
36 
 
 
d. V = 600, L = 400, -L/V = -0.667. 
 Find xA = 0.40 on equilibrium curve. Plot op line & find intersection point with 
 y = x line. zA = 0.52 
2.D18. From ii
i
z
x
V
1 K 1
F
 , we obtain 
h
h
h
z
1
xV
F K 1
 
 Guess drumT , calculate h b pK , K and K , and then determine V F . 
 Check: 1 i
1
K 1 z
0
1 K 1 V F
 ? 
 Initial guess: If hx .85 then drumT must be less than temperature to boil pure hexane 
 hK 1.0, T 94 C . On this basis 85° to 90° would be reasonable. Try 85°C. 
h b pK =0.8, K 4.8, K =11.7 . 
 
0.6
1V 0.85 1.471
F 0.8 1
. Not possible. Must have h
0.6
K 0.706
0.85
 
 Try T 73 C where hK 0.6 . Then b pK 3.8, K 9.9 . 
 
0.6
1V .85 0.735
F .6 1
 
 Check: 
 i i
i
K 1 z 8.9 .1 2.8 .3 .4 .6
0.05276
1 K 1 V F 1 8.9 .735 1 2.8 .735 1 .4 735
 
 Converge on T ~ 65.6 C and V F ~ 0.57 . 
 
2.D19. 90% recovery n-hexane means C6 C60.9 Fz L x 
 Substitute in L F V to obtain C6 C6z .9 1 V F x 
 8C balance: C6 C6 C6 C6 C6 C6z F Lx Vy F V x K Vx 
 or C6 C6 C6 C6z 1 V F x x K V F 
 Two equations and two unknowns. Remove C6x and solve 
 C6C6 6
.9 z KV F
z .93C
1 V F
 
 Solve for 
C6
V .1
V F. 
F .9K .1
. Trial and error scheme. 
 Pick T, Calc C6K , Calc V F, and Check f V F 0 ? 
 If not 
new
ref old
ref
K T
K
1 d f T
 
37 
Try C4 C5 C6 refT 70 C. K 3.1, K .93, K .37 K
V .1
0.231
F .9 .37 .1
. 
Rachford Rice equation 
2.1 .4 .08 .25 .63 .35
f .28719
1 2.1 .231 1 .08 .231 1 .63 .231
ref new
.37
K T 0.28745 use .28
1 0.28719
Converge on New C4 C8T ~ 57 C. Then K 2.50, K .67, and V F 0.293 . 
2.D20. New Problem. The K values are: EK 8.7 , BK 0.54 , PK 0.14 
Can use Eq. (2-40), (2-41) or (2-42). If we use (2-42) the R – R eqn 
i i
F B E B
i
K 1 zV
f 0 Use z 1 z z .8 z
VF 1 K 1
F
Then B B
7.7 .2 .46 z .86 .8 z
RR eq = 
1+7.7 25 1 .46 .25 1 0.86 .25
B B0 0.5265 0.51977 z 0.8764 1.0955z
B B0 .5757z 0.3499 z 0.6078 
2.D21. a.) C2 C5K 4.8 K 0.153
Soln to Binary R.R. eq. A B
B A
z zV
F K 1 K 1
V 0.55 0.45
0.5309
F .153 1 4.8 1
 
C2
C2 C2
C2
z 0.55
x 0.1823, y 0.8749
V 1 3.8 .53091 K 1
F
C5 C5x 0.8177 , y 0.1251 
Need to convert F to kmol. 
Avg MW 0.55 30.07 0.45 72.15 49.17 
kg kmol
F 100,000 2033.7 kmol/h
hr 49.17 kg
V V F F 1079.7, L F V 954.0 kmol/h
b.) L vPerm drum
v
u K
To find LMW 0.1823 30.07 0.8177 72.15 64.48 
38 
 
 VMW 0.8749 30.07 0.1251 72.15 35.33 
 For liquid assume ideal mixture: 
 C2 C51 C2 C2,liq C5 C5,liq C2 C5
C2,liq C5,liq
MW MW
V x V x V x x 
 L
30.07 72.15
V 0.1823 0.8177 103.797 ml/mol
0.54 0.63
 
 LL
L
MW 64.48
0.621 g/ml
V 103.797
 
 For vapor: ideal gas: vv
MW
RT
 
 v
atm g
700 kPa 35.33
101.3 kPa mol 0.009814 g / ml
ml atm
82.0575 303.16K
mol K
 
 drumK : Use Eq. (2-60) with 
vL
lV
V L
W
F
W
 
 V
kmol 64.48 kg
W 997.7 6, 4331.7 kg/h
h kmol
 
 VW 881.5 35.33 31,143.4 kg/h 
 lV
64331.7 0.009814
F 0.2597
31,143.3 0.621
 
 
 
2
drum
3 4
K exp 1.877478 0.81458 n .2597 0.18707 n 0.2597
 0.0145229 n 0.2597 0.0010149 n 0.2597 0.3372
 
 
Perm
0.621 0.009814
u 0.3372 2.6612 ft/s
0.009814
ft 1.0 m
 2.6612 0.8111 m/s
s 3.2808 ft
 
 
2V
C 6 3
Perm v
3 3
kmol kg
1079.7 35.33
V MW h kmolA 1.392 m
u 3600 m s g kg 10 cm
0.8111 3600 0.009814
s h cm 1000g m
 
 CD 4A 1.33 m 
 Arbitrarily 
 L D 4, L 5.32 m 
39 
2.D22. i i iP iP NP NP
i
iP NP
K 1 z K 1 z K 1 zV
f 0
V VF 1 K 1 V F 1 K 1 1 K 1
F F
Solve for V/F. 
NP iP iP iP NP NP
V V
1 K 1 K 1 z 1 K 1 K 1 z 0
F F
iP iP NP NP
NP iP
K 1 z K 1 zV
F K 1 K 1
 where iP NPz z 1.0 
drum totp p 760 mm Hg, T 90 C 
10 NP
1499.2
log VP 7.84767 2.75943
90 204.64
 
NPVP 574.68 mm Hg , NP totK 574.68 p 0.75616
10 iP
1580.9
log VP 8.11778 3.011679
20 219.61
iPVP 1027.256 mm Hg , iPK 1027.256 760 1.35165 
Note: iP NP iPMW MW . z 0.5 in both wt & mol frac., as does NPz . 
0.35165 0.5 0.24384 0.5V
0.629
F 0.24384 0.35165
 
iP
iP
iP
z
x 0.4095
V
1 K 1
F
NP iP ip iP iP NPx 1 x 0.5905; y K x 0.55347 y 0.44653
2.D23. 5. 0°C, 2500 kPa
Fig 2.12: M ethylene Ethane C6K 5.7, K 1.43, K 0.98, K 0.007 
First, try 
1
V
0.6
F
 (equal split ethylene and ethane) 
1
.47 .4 0.43 .05 .02 0.35 0.993 0.2V
f 0.0108
F 1 4.7 .6 1 .43 .6 1 .02 .6 1 .993 6
Eq. (2-46) 1 2
2 1 i i
2
i
V
f
V V F
0.6059
F F z K 1
V
1 K 1
F
Then Eq. (2-38), ii M ethylene
i
z
x . x 0.104, x 0.040
1 K 1 V F
 
ethane C6x 0.354, x 0.502, 1.0001 OK 
Find i i iy K x
40 
2.D24. New Problem. p = 300 kPa
At any T. C3 C3 C3K y x
K’s are known. C6 C6 C6 C3 C3K y x 1 y 1 x
Substitute 1st equation into 2nd C6 C3 C3 C3K 1 K x 1 x
Solve for xC3, C3 C6 C3 C31 x K 1 K x
C3 C3 C6 C6x K K 1 K
C3 C6C6
C3 C3
C3 C6 C3 C6
K 1 K1 K
x & y
K K K K
At 300 kPa pure propane C3K 1.0 boils at -14°C (Fig. 2-11) 
At 300 kPa pure n-hexane C6K 1.0 boils at 110°C 
Check: at -14°C C6C3
C6
1 K
x 1,
1 K
 C6C3
C6
1 1 K
y 1.0
1 K
at 110°C C3
C3
0
x 0,
K
C3
C3
C3
K 0
y 0
K
Pick intermediate temperatures, find C3K & C6K , calculate C3x & C3y . 
T
0ºC 1.45 0.027 0.9915
10ºC 2.1 0.044 0.465 0.976 See
20ºC 2.6 0.069 0.368 0.956 Graph
30ºC 3.3 0.105 0.280 0.924
40ºC 3.9 0.15 0.227 0.884
50ºC 4.7 0.21 0.176 0.827
60ºC 5.5 0.29 0.136 0.75
70ºC 6.4 0.38 0.103 0.659
C6KC3K C3x
1- 0.027 = 0.684
1.45 - 0.027
C3 C3 C3y K x
41 
42 
b. C3x 0.3 , V F 0.4, L V 0.6 0.4 1.5 
Operating line intersects y x 0.3, Slope 1.5
L F
y x z
V V
at 
F 0.3
x 0, y z 0.75
V 0.4
Find yc3 = 0.63 and xC3 = 0.062
Check with operating line: 0.63 1.5 .062 0.75 0.657 OK within accuracy of the graph. 
c. Drum T: C3 C3 C3K y x 0.63 0.062 10.2 , DePriester Chart T = 109ºC 
d. y .8, x ~ .16 
L y .8 .6
Slope 0.45
V x .16 .6
1 f
.45
f
 
V 1
f 0.69
F 1.45
2.D25. New Problem. 20% Methane and 80% n-butane
drumT .50 ºC , 
V
0.40
F
, Find drump
A A B B
BA
K 1 z K 1 zV
0 f
VVF 1 K 11 K 1
FF
Pick drum C4 nC4p 1500 kPa: K 13 K 0.4 
(Any pressure with C1 C4K 1 and K 1.0 is OK) 
Trial 1 1
12 .2 .6 .8
f 0.2178
1 12 .4 1 .6 .4
 Need lower drump
C 4 old
C 4 new
old
K P 0.4
K P 0.511
1 d f P 1 .2138
1.0
new C1P 1160 K 16.5
2
15.5 .2 .489 .8
f 0.4305 .4863 0.055769
1 15.5 .4 1 .489 .4
C4 new
0.511
K P 0.541
1 0.055769
 
new C1P1100 K 17.4 
3
16.4 .2 .459 .8
f 0.0159
1 16.4 .4 1 .459 .4
, OK. Drum pressure = 1100 kPa 
43 
 
 b.) ii C1
i
z 0.2
x , x 0.02645
V 1 16.4 .41 K 1
F
 
 C1 C1 C1y K x 17.4 0.02645 0.4603 
 
2.D26. New Problem. a) Can solve for L and V from M.B. 100 = F = V + L 
 45 Fz 0.8V 0.2162L 
 Find: L = 59.95 and V = 40.05 
 b) Stage is equil. C3C3
C3
y 0.8
K 3.700
x 0.2162
 
 C5
0.2
K .2552
0.7838
 
 These K values are at same T, P. Find these 2 K values on DePriester chart. 
 Draw straight line between them. Extend to drum drumT , p . Find 10ºC, 160 kPa. 
2.D27. New Problem. a.) C5 10
1064.8
VP : log VP 6.853 2.2832
0 233.01
, 
VP 191.97 mmHg 
 b.) VP 3 760 2280 mmHg , 
 10log VP 6.853 1064.8 / T 233.01 
 Solve for T = 71.65ºC 
 c.) totP 191.97 mm Hg [at boiling for pure component totP VP ] 
 d.) C5: 10
1064.8
log VP 6.853 2.8045
30 233.01
 
 VP 637.51 mm Hg 
 C5 C5 totK VP P 637.51 500 1.2750 
 
 C6: 10 C6
1171.17
log VP 6.876 2.2725
30 224.41
 
 C6VP 187.29 mm Hg 
 C6K 187.29 500 0.3746 
 e.) A A A B B B A AK y x K y x (1 y ) / (1 x ) 
 If A BK & K are known, two eqns. with 2 unknowns A AK & y Solve. 
 C6C5
C5 C6
1 K 1 0.3746
x 0.6946
K K 1.2750 0.3746
 
 C5 C5 C5y K x 1.2750 0.6946 0.8856 
 f.) Overall, M.B., F = L + V or 1 = L + V 
 FC5: Fx Lx Vy .75 0.6946 L + 0.8856 V 
 Solve for L & V: L = 0.7099 & V = 0.2901 mol 
 g.) Same as part f, except units are mol/min. 
 
 
44 
 
2.D28. New Problem. 
 
 
 
 
 
 
 
 
 
 
 
 From example 2-4, H drum H Hx 0.19, T 378K, V F 0.51, y 0.6, z 0.40 
 With v
perm v
V MW
h D C, D
u 3600 C
 
 
 C=4, MWv = 97.39 lbm/lbmole (Example 2-4) 
 3v 3 3
1 28316.85ml lbm
3.14 10 g mol 0.198
454g lbm ft ft
 
 Example 2.4 
 
 L vperm drum horiz vertical
V
u K , K 1.25 K 
 From Example 2-4, verticalK 0.4433 , horizK 1.25 0.4433 0.5541 
 
1 2
perm
0.6960 0.00314
u 0.5541 8.231 ft s
0.00314
 [densities from Example 2-4] 
 
V lbmol
V F 0.51 3000 1530 lbmol hr
F hr
 
 
3
lbmol lbm
1530 97.39
h lbmolD 5.067 ft
ft s lbm
8.231 3600 0.1958
s h ft
 
 h 4D 20.27 ft 
 Use 
1
5 20 or 5 22
2
 ft drum. 
 
2.D29. New Problem. The stream tables in Aspen Plus include a line stating the fraction vapor in a given 
stream. Change the feed pressure until the feed stream is all liquid (fraction vapor = 0). For the Peng-
Robinson correlation the appropriate pressure is 74 atm. 
The feed mole fractions are: methane = 0.4569, propane = 0.3087, n-butane = 0.1441, i-butane = 0.0661, 
and n-pentane = 0.0242. 
b. At 74 atm, the Aspen Plus results are; L = 10169.84 kg/h = 201.636 kmol/h, V = 4830.16 kg/h = 
228.098 kmol/h, and Tdrum = -40.22 oC. 
h 
F 
D 
L 
V 
45 
 
The vapor mole fractions are: methane = 0.8296, propane = 0.1458, n-butane = 0.0143, i-butane = 0.0097, 
and n-pentane = 0.0006. 
The liquid mole fractions are: methane = 0.0353, propane = 0.4930, n-butane = 0.2910, i-butane = 
0.1298, and n-pentane = 0.0509. 
c. Aspen Plus gives the liquid density = 0.60786 g/cc, liquid avg MW = 50.4367, vapor density = 
0.004578 g/cc = 4.578 kg/m3, and vapor avg MW = 21.17579 g/mol = kg/kmol. 
 
Since the flow area for vapor = LD and L = 4D, the area for flow = 4D2. Then the equation for the drum 
diameter is 
 
D = {[(MWV) V]/[ρV uperm (L/D)]}0.5 = {[(21.17579 kg/kmol)(228.098 kmol/h)]/[(4.578 kg/m3)(uperm 
ft/s)(1 m/3.281 ft)(3600 s/h)(4)]0.5 
 
where the unit conversions are used to give D in meters. The value of uperm (in ft/s) can be determined by 
combining Eqs. (2-59) and (2-60) for vertical drums with Eq. (2-64a). 
 
Flv = (WL/WV)[ρV/ ρL]0.5 = (10169.84/4830.16)[0.004578/0.60786]0.5 = 0.18272 
 
Resulting Kvertical = 0.378887 , Khorizontal = 0.473608, and uperm = 5.436779 ft/s, and D = 0.4896 m and L = 
1.9585 m. Appropriate standard size would be used. 
 
2.D30. New Problem. a. From the equilibrium data if yA = .40 mole fraction water, then xA = 0.09 mole 
fraction water. 
Can find LA and VA by solving the two mass balances for stage A simultaneously. 
LA + VA = FA = 100 and LA (.09) + VA (.40) = (100) (.20). The results are VA = 35.48 and LA = 64.52. 
 
b. In chamber B, since 40 % of the vapor is condensed, (V/F)B = 0.6. The operating line for this flash 
chamber is, 
y = -(L/V)x + FB/V) zB where zB = yA = 0.4 and L/V + .4FB/.6FB = 2/3. This operating line goes through 
the point y = x = zB = 0.4 with a slope of -2/3. This is shown on the graph. Obtain xB = 0.18 & yB = 0.54. 
LB = (fraction condensed)(feed to B) = 0.4(35.48) = 14.19 kmol/h and VB = FB – LB = 21.29. 
 
c. From the equilibrium if xB = 0.20, yB = 0.57. Then solving the mass balances in the same way as for 
part a with FB = 35.48 and zB = 0.4, LB = 16.30 and VB = 19.18. Because xB = zA, recycling LB does not 
change yB = 0.57 or xA = 0.09, but it changes the flow rates VB,new and LA,new. With recycle these can be 
found from the overall mass balances: F = VB,new + LA,new and FzA = VB,newyB + LA,new xA. Then VB,new = 
22.92 and LA,new = 77.08. 
 
46 
 
Graph for problem 2.D30. 
 
 
47 
 
 
2.E1. New Problem. From Aspen Plus run with 1000 kmol/h at 1 bar, L = V = 500 kmol/h, WL = 9212.78 
kg/h, WV = 13010.57 kg/h, liquid density = 916.14 kg/m3 , liquid avg MW = 18.43, vapor density = 0.85 
kg/m3 , and vapor avg MW = 26.02, Tdrum = 94.1 oC, and Q = 6240.85 kW. 
 
The diameter of the vertical drum in meters (with uperm in ft/s) is 
 
D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = 
{[4(26.02)(500)]/[3600(3.14159)(0.85)(1/3.281)uperm]}0.5 
 
Flv = (WL/WV)[ρV/ ρL]0.5 = (9212.78/13010.57)[0.85/916.14]0.5 = 0.02157 
 
Resulting Kvertical = 0.404299, and uperm = 13.2699 ft/s, and D = 1.16 m. Appropriate standard size would 
be used. Mole fractions isopropanol: liquid = 0.00975, vapor = 0.1903 
 
b. Ran with feed at 9 bar and pdrum at 8.9 bar with V/F = 0.5. Obtain WL = 9155.07 kg/h, WV = 13068.27, 
density liquid = 836.89, density vapor = 6.37 kg/m3 
 
D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = 
{[4(26.14)(500)]/[3600(3.14159)(6.37)(1/3.281)uperm]}0.5 
 
Flv = (WL/WV)[ρV/ ρL]0.5 = (9155.07/13068.27)[6.37/836.89]0.5 = 0.06112 
 
Resulting Kvertical = .446199, uperm = 5.094885 ft/s, and D = 0.684 m. Thus, the method is feasible. 
 
c. Finding a pressure to match the diameter of the existing drum is trial and error. 
 
If we do a linear interpolation between the two simulations to find a pressure that will give us D = 1.0 m 
(if linear), we find p = 3.66. Running this simulation we obtain, WL = 9173.91 kg/h, WV = 13049.43, 
density liquid = 874.58, density vapor = 2.83 kg/m3, MWv = 26.10 
 
D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = 
{[4(26.10)(500)]/[3600(3.14159)(2.83)(1/3.281)uperm]}0.5 
 
Flv = (WL/WV)[ρV/ ρL]0.5 = (9173.91/13049.43)[2.83/874.58]0.5 = 0.0400 
 
Resulting Kvertical = .441162, uperm = 7.742851 ft/s, and D = 0.831 m. 
 
Plotting the curve of D versus pdrum and setting D = 1.0, we interpolate pdrum = 2.1 bar 
 
At pdrum = 2.1 bar simulation gives, WL = 9188.82 kg/h, WV = 13034.53, density liquid = 893.99 , density 
vapor = 1.69 kg/m3, MWv = 26.07. 
 
D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = 
{[4(26.07)(500)]/[3600(3.14159)(1.69)(1/3.281)uperm]}0.5 
 
Flv = (WL/WV)[ρV/ ρL]0.5 = (9188.82/13034.53)[1.69/893.99]0.5 = 0.0307 
 
Resulting Kvertical = .42933, uperm = 9.865175ft/s, and D = 0.953 m. 
 
This is reasonably close and will work OK. Tdrum = 115.42 oC, Q = 6630.39 kW, 
48 
 
Mole fractions isopropanol: liquid = 0.00861, vapor = 0.1914 
 
In this case there is an advantage operating at a somewhat elevated pressure. 
 
 
 
2.E2. This problem was 2.D13 in the 2nd editionof SPE. 
 a. Will show graphical solution as a binary flash distillation. Can also use R-R equation. To 
generate equil. data can use 
 C6 C8 C6 C8 C6 C6 C8 C8x x 1.0, and y y K x K x 1.0 
 Substitute for xC6 C8C6
C6 C8
1 K
x
K K
 
 Pick T, find KC6 and KC8 (e.g. from DePriester charts), solve for xC6. Then yC6 = KC6xC6 
 
T°C KC6 KC8 xC6 yC6 = KC6 xC6 
 125 4 1.0 0 0 
 120 3.7 .90 .0357 .321 
 110 3.0 .68 .1379 .141 
 100 2.37 .52 .2595 .615 
 90 1.8 .37 .4406 .793 
 80 1.4 .26 .650 .909 
 66.5 1.0 .17 1.0 1.0 
 
 Op Line Slope L 1 V F .6 1.5
V V F .4
, Intersection y = x = z = 0.65. 
 See Figure. 
 yC6 = 0.85 and xC6 = 0.52. Thus KC6 = .85/.52 = 1.63. 
 This corresponds to T = 86°C = 359K 
 
 
 b. Follows Example 2-4. 
49 
 
 L C8 C8C6 C8MW x MW x MW .52 86.17 .48 114.22 99.63 
 C6 C8L C6 C8
C6 C8
MW MW 86.17 114.22
V x x .52 .48 145.98 ml/mol
.659 .703
 
 
3
L
L 3
L
MW 99.63 28316 ml/ft lbm
.682 g/ml 42.57
V 145.98 454 g/lbm ft
 
 v C6 C6 C8 C8MW y MW y MW .85 86.17 .15 114.22 90.38 
 v 3v
1.0 90.38 g/molpMW
0.00307 g/ml 0.19135 lbm/ft
ml atmRT 82.0575 359K
mol K
 
 Now we can determine flow rates 
 
V
V F .4 10,000 4000 lbmol/h
F
 
 vvW V MW 4000 90.38 361,520 lb/h 
 LLL F V 6000 lbmol/h, W L MW 6000 99.63 597, 780 lb/h 
 vLlv lv
v L
W 597, 780 0.19135
F 0.111, nF 2.1995
W 361,520 42.57
 
 
2
drum
3 4
K exp 1.87748 .81458 2.1995 .18707 2.1995
 0.01452 2.1995 0.00101 2.1995 0.423
 
 Perm drum L v vu K 0.423 42.57 19135 .19135 6.30 ft/s 
 
v
2
Cs
Perm v
V MW 4000 90.38
A 83.33 ft
u 3600 6.3 3600 0.19135
 
 CsD 4A 4 83.33 10.3 ft. Use 10.5 ft. 
 L ranges from 3 × 10.5 = 31.5 ft to 5 × 10.5 = 52.5 ft. 
 
 Note: This uPerm is at 85% of flood. If we want to operate at lower % flood (say 75%) 
have 
 
75% 85%Perm Perm
u 0.75 0.85 u 0.75 0.85 .63 5.56 
 Then at 75% of flood, ACs = 94.44 which is D = 10.96 or 11.0 ft. 
 
2.F1 
 
xB 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1
yB 0 .22 .38 .52 .62 .71 .79 .85 .91 .96 1 
 Benzene-toluene equilibrium is plotted in Figure 13-8 of Perry’s Chemical Engineers 
Handbook, 6th ed. 
 
2.F2. See Graph. Data is from Perry’s Chemical Engineers Handbook, 6th ed., p. 13-12. 
 
50 
 
 
 
 
 Stage 1) 
1F
2 3
z .4 f 1 3 Slope 2,
1 3
 
 Intercept 1 1 2
.4
1.2 y .872 x .164 z
1 3
 
 Stage 2) 
2F
1 3
z .164 f 2 3 Slope 1 2
2 3
 
 Intercept 2 2 3
.164
.246 x .01 y .240 z
2 3
 
 Stage 3) 
3F
z .240 f 1 2 Slope 1 
 Intercept 3 3
.240
.480 x .022 y .461
1 2
 
 
2.F3. Bubble Pt. At P = 250 kPa. Want 1 1K z 1 
 Guess 1 2 3T 18 C, K 1, K .043, K .00095, .52 
 Converge to T 0 C 
 Dew Pt. Calc. Want 1
1
z
1.0
K
 
 Try 1 2 3T 0 C, K 1.93, K 0.11, K 0.0033, 120.26 
51 
 
 Converge to T 124 C . This is a wide boiling feed. 
 Tdrum must be lower than 95°C since that is feed temperature. 
 First Trial: Guess d,1 1 2 3T 70 C : K 7.8, K 1.07, K .083 
 Guess V F 0.5 . Rachford Rice Eq. 
 
7.8 1 .517 .07 .091 .083 1 .392
f V F .14
1 6.8 .5 1 .07 .5 1 .083 1 .5
 
 V F .6 gives f .6 .101 
 By linear interpolation V F .56. f 0.56 .0016 which is close enough for first 
trial. 
 V V F F 56, L 44 
 ii i i i
i
z
x and y K x
1 K 1 V F
 
 1 2 3x .1075 x .088 x .806 x 1.001 
 1 2 3y .839 y .094 y .067 y .9999 
 Data: Pick refT 25 C . (Perry’s 6
th ed; p. 3-127), and (Perry’s 6th ed; p. 3-138) 
 1 81.76 cal/g 44 3597.44 kcal/kmol 
 2 87.54 cal/g 72 6302.88 kcal/kmol 
 3 86.80 cal/g 114 9895.2 kcal/kmol 
 at pL1T 0 C, C 0.576 cal / (g C) 44 25.34 kcal/(kmol C) . 
 For pL3T 20 to 123 C, C 65.89 kcal/(kmol C) 
 at pL2T 75 C, C 39.66 kcal/(kmol C) . (Himmelblau/Appendix E-7) 
 2pv C a bT cT 
 propane a = 16.26 b = 5.398 × 10-2 c = -3.134 × 10-5 
 n-pentane a = 27.45 b = 8.148 × 10-2 c = -4.538 × 10-5 
 **n-octane a = 8.163 b = 140.217 × 10-3 c = -44.127 × 10-6 
 ** Smith & Van Ness p. 106 
 Energy Balance: E(Td) = VHv + LhL – FhF = 0 
FFh 100 .577 25.34 .091 39.66 .392 65.89 95.25 297,773 kcal/h 
 LLh 44 .1075 25.34 .088 39.66 .806 65.89 70.25 117,450 
 
2
v
2
3
VH 56 .839 3597.4 16.26 5.398 10 45
 0.94 6302.88 27.45 8.148 10 45
 0.67 9895.3 8.163 140.217 10 45 240, 423
 
 drumE T 60,101 Thus, Tdrum is too high. 
 Converge on drum 1 2 3T 57.2 C : K 6.4, K .8, K .054 
 For V F 0.513, f 0.513 0.0027. V 51.3, L 48.7 
52 
 
 1 2 3 1x .137, x .101, x .762, x 1.0000 
 1 2 3 1y .878, y .081, y .041, y 1.0000 
 F L v drumFh 297,773; Lh 90,459; VH 209,999; E T 2685 
 Thus Tdrum must be very close to 57.3°C. 
 1 2 3x .136, x .101, x .762 
 1 2 3y .328, y .081, y .041 
 V 51.3 kmol/h, L 48.7 kmol/h 
 Note: With different data Tdrum may vary significantly. 
 
2.F4. New Problem. yV Lx Fz or 
L F
y x z
V V
 
 
L
V F 0.4, V 4kmol h , L 6, 1.5 slope
V
 
 
F
x 0, y z 2.5 .25 0.625
V
 
 Find: V = 4 kmol/h, L = 6 kmol/h. 
 
 From the graph, x = 0.19 y = 0.34 
 
 Equilibrium is from NRTL on Aspen Plus. 
 
53 
 
 
 
FIGURE 2.F.4. 
 
 
 
2.G1. Used Peng-Robinson for hydrocarbons. 
 
 Find drumT 33.13 C, L 34.82 and V 65.18 kmol/h 
 In order ethylene, ethane, propane, propylene, n-butane, xi (yi) are: 
0.0122 0.0748 , 0.0866 0.3005 , 0.3318 0.3781 , 0.0306 0.0404 , 0.5388 0.2062. 
 
 
2.G2. Used Peng-Robinson. Find drumT 30.11 C, L 31.348, V 68.66 kmol/h. 
 In same order as 2.G1, 
i ix y are: 0.0189 0.1123 , 0.0906 0.3023 , 0.3255 0.3495 , 0.0402 0.0501 , 0.5248 0.1858 .
 
54 
 
2.G3. Used NRTL-2. drumT 79.97 C , M Mx 0.2475, y 0.6287 . Compares to graph with 
M Mx 0.18 and y 0.55 . Different equilibrium data. 
 
2.G4. New Problem. 
 
COMP x(I) y(I) 
METHANE 0.12053E-01 0.84824 
BUTANE 0.12978 0.78744E-01 
PENTANE 0.29304 0.47918E-01 
HEXANE 0.56513 0.25101E-01 
V/F = 0.58354 
 
2.G5. New Problem. Used NRTL. T = 368.07, Q = 14889 kW, 1st liquid/total liquid = 0.4221, 
Comp Liquid 1, x1 Liquid 2, x2 Vapor, y 
Furfural 0.630 0.0226 0.0815 
Water 0.346 0.965 0.820 
Ethanol 0.0241 0.0125 0.0989 
 
2.G6. New Problem. Used Peng Robinson. Feed pressure = 10.6216 atm, Feed temperature = 81.14oC, 
V/F = 0.40001, Qdrum =0. Note there are very small differences in feed temperature with 
different versions of Aspen Plus. 
 
 
COMP x(I) y(I) 
METHANE 0.000273 0.04959 
BUTANE 0.18015 0.47976 
PENTANE 0.51681 0.39979 
HEXANE 0.30276 0.07086 
V/F = 0.40001 
 
 
2.H1. New Problem. The spreadsheet with equations for problem 2.D16 is shown in Appendix B of 
Chapter 2. The spreadsheet with numbers for i-butane replacing n-butane is below. 
 
 
MC flash, HW 2.G.b., MC flash with i-
butane 
 K const. aT1 aT2 aT6 ap1 ap2 ap3 
M -292860 0 8.2445 -0.8951 59.8465 0 
iB -1166846 0 7.72668 
-
0.92213 0 0 
nPentane -1524891 0 7.33129 
-
0.89143 0 0 
nHex -1778901 0 6.96783 
-
0.84634 0 0 
T deg R 509.688 
p 
psia 36.258 F 150 
 zM 0.5 z iB 0.1 z np 0.15 znhex 
V/F 0.602698586 
 
0.25 
55 
 
guess 
KM 51.86751896 
 KiB 0.926804057 
 KnPen 0.175621816 
 KnHex 0.05400053 
 
 
Use goal seek for cell B24 to = 1.0 change B9 
 xM 0.015793905 
 xib 0.104615105 
 xnPen 0.29812276 
 xnHex 0.581601672 
 Sum 1.000133443 
 RR M 0.803396766 
 RR nB -0.007657401 
 RRnP -0.2457659 
 RRnHex -0.550194874 
 sum RR -0.000221409 
 
 
2.H3. New Problem. Use the same spreadsheet as for problem 2H1, but with methane feed mole fraction 
= 0. 
Answer: V/F = 0.8625, 
xib0.08596648 
xnPen 0.203540261 
xnHex 0.710481125 
KiB 3.886544834 
KnPen 1.264637936 
KnHex 0.574940847 
yib = xib Kib = .33411 and so forth 
 
56 
Chapter 3 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 3A7, 3A10, 3A11, 3C3, 3C4, 3D4, 3D8, 3G2. 
3.A7. Simultaneous solution is likely when one of the key variables can be found only from the 
energy balances. For example, if only 1 of D Bx , x , D, B, A distFR are given energy balances 
will be required. This is case for most of the simulation problems and for a few design 
problems. In some simulation problems the internal equations have to be solved also. 
3.B1. a. D Bx , x , opt feed, RebQ
D Bx , x , opt feed, CQ
D Bx , x , opt feed, S (open steam), sat’d vapor steam 
All of above with fractional recoveries set instead of D Bx , x
D, Bx , opt feed, L/D 
b. N, FN , col diameter, frac. recoveries both comp. 
N, FN , col diameter, A oFR dist, L D 
N, FN , col diameter, A RFR dist, Q
N, FN , col diameter, A CFR dist, Q
N, FN , col diameter,  D C Bx , Q or x
N, FN , col diameter, S (sat’d steam), sat’d vapor steam,  D Bx or x
Many other situations are possible [e.g., 2 feeds, side streams, intermediate condensers or 
reboilers etc.] 
3.C1. See solution to problem 3-D2. 
3.C2. See solution to problem 3-D3. 
3.C3. New Problem in 3rd Edition. mix 1 2F F F D B   
mix mix 1 1 2 2 D BF z Fz F z Dx Bx    (Mole frac. MVC) 
1 1 2 2
mix
mix
Fz F zz
F


Now solve like 1 Feed Column  mix mixF & z . From Eq. (3-3), 
mix B
mix
D B
z xD F
x x



 kmol/h. 
mixB F D  kmol/h. 
3.C4. New Problem in 3rd Edition. See solution to 3D4, Part b. 
57 
 
3.D1. 
 
 Mass balance calculation is valid for parts a & b for problem 3G1. 
 a)      o C 0 D 1
L 3, Eq 3-14 Q 1 L / D D h H
D
    
 Dh is a saturated liquid at Dx 0.85 wt. frac. From Fig. 2-4, Dh ~ 45 kcal/kg 
 1H is saturated vapor at D 1 1x y 0.85, H ~ 310   kcal/kg 
    CQ 1 3 764.62 45 310 810,497     kcal/hour 
 EB around column. 
1 21 F 2 F col C R D B
Fh F h Q Q Q Dh Bh      
    
1 2F F
h 81 C, 60 wt% ethanol ~ 190kcal / kg; h 20 C, 10 wt% ethanol ~ 10kcal / kg 
 Bh (sat’d liquid – leaves equil contact, ~ 0 wt% ethanol) ~ 100 kcal/kg, Qcol = 0 (adiabatic) 
            RQ 764.62 45 735.38 100 1000 190 500 10 810,497 657,259        kcal/kg 
 (b) V B 2.5 mass. 
1F
2F 
DD, x
L V
 RQ 
BB, x
oL
1V
1 
CQ
 1 2 BF F B D x 0.0001 0.01%    
1 1 2 2 B D DFz F z Bx D x x 0.85    
avg B
total
D B
z x
Solve D F
x x
 
  
 
 
total 1 2F F F 1500 kg/h   
 
0.43333 0.0001D 1500 764.62 kg/h
0.85 0.0001

 

 
total
kgB F D 1500 764.62 735.38 
h
     
    1 1 2 2
avg
total
 1000 .60 500 0.10Fz F zz 0.43333
F 1500

  
58 
 
Approximately B V Lx ~ y ~ x . Thus BLh h 100  . VH 640kcal kg 
         RQ 1838.45 640 735.38 100 2573.83 100 992,763 kcal / h    
 
 EB. 
1 2C D B col 1 F 2 F R
Q Dh Bh Q Fh F h Q      
 CQ 34407.9 7353.8 190,000 5000 992,763 1,146,001 kcal/h       
 
3.D2. Column: mass bal: F + S = D + B (1) 
 MVC: Fz + S D BSy Dx Bx  (2) 
 Note: Sy 0 
 energy bal: f s C D BFh SH Q Dh Bh    (3) 
 Condenser: mass bal. : 1 oV L D  (4) 
 energy bal.:  1 1 C o DV H Q L D h   (5) 
 Solve Eqs. (1) and (2) to get: 
 
     B B
D B
100 .3 100 .05 100 .05Fz Fx SxD 36.4
x x .6 .05
  
  
 
 
 Note: Not Eq. (3-3). Solve Eqs. (4) and (5) to get: 
   C D 1Q D 1 L D h H   
 Substitute CQ into Eq. (3): 
 
 
 
D B F
D 1
Dh F S D h Fh SHL 1
D D h H
    
 

S 
 From Figure 2-4: F D B 1h 8, h 65, h 92, H 638, H 608 kcal/kg.    S 
 
       
 
36.4 65 163.6 92 100 8 100 638
L D 1 2.77
36.4 65 408
  
  

 
 
3.D3. External balances: F + C = B + D (1) 
 C B DFz Cx Bx Dy   (2) 
 R F C B DQ Fh Ch Bh DH    (3) 
R BL VLh Q VH Bh   
L
 
Reboiler 
Sat’d liqd 
B, Bx =0.0001 
RQ 
(Satd vapor) V 
Sat’d liqd 
L V B L 1838.45 735.38 2573.83     
 
V B 2.5 or V 2.5B 1838.45   
59 
 
 F = 2000, C = 1000, z = .4, C B Dx 1.0, x .05, y .80,     Fh 20 C 30.7,   
  Ch sat 'd liquid 50,    B Dh sat 'd liquid 92, H sat 'd vapor 327 kcal/kg  
 Around reboiler: L V B  
 N Reb BLx Vy Bx  
 N R reb BLh Q VH Bh   
 For a total reboiler: N B N N B N Bx x , y x x , h h 92     
 M.B.:   N R reb BV B h Q VH Bh    
 or R B N
reb N
QV since h h
H h
 

 
 RebH 617 (saturated vapor at N 1y 0.05  ) 
 Solve Eqs. (1) and (2) for B: C D D
B D
Fz Cx Fy CyB
x y
  


 
 Thus 
 
800 1000 1600 800B 800 and D 2200
.05 .8
  
  

 
 From Eq. (3), R B D F CQ Bh DH Fh Ch    
 
            RQ 800 92 2200 327 2000 30.75 1000 50 804,500 cal/h      
 R
reb N
Q 804500V 1532.4 kg/h
H h 617 92
  
 
 
 
3.D4. New Problem in 3rd Edition. F B D  
 MVC B DFz Bx Dy  
 But given recoveries. Thus, use: 
 F,M D,MFz (Frac Rec Methanol in distillate) Dy 
 and F,W B,WFz (Frac Rec water in bottom) Bx 
F,M F,W F,Mz 0.3, z 1 z 0.7    
 D,M B,Wy unknown, x unknown. 
    D,MMethanol 29.7 100 0.3 .99 Dy  
 If 99% methanol recovered in distillate, 1% is in bottoms    B,M0.3 100 0.3 0.01 Bx  
 Water    B,W68.6 100 0.7 0.98 Bx 2%   water in distillate 
   D,W1.4 100 0.7 0.02 Dy  
 Since ix 1 and    i D,i B,iy 1, Dy D, and Bx B     
 Thus, D,M D,WD Dy Dy 29.7 1.4 31.1 kmol h     
 B,M B,WB Bx Bx 0.3 68.6 68.9 kmol h     
 Check: B D 100 F   OK 
60 
 
 D,Md,M
Dy
y 29.7 31.1 0.955
D
   
 B,M B,Mx Bx B 0.3 68.9 0.00435   
a) 0D 31.1 & L D 2.  Thus 0L 62.2 
Reflux liquid is in equilibrium with vapor D,My 0.955 
 From equilibrium data (Table 2-7) M,0x ~ 0.893 (linear interpolation) 
b) E.B. Partial condenser: 1 1 c 0 0 0V H Q DH L h   
1 0V D L 93.3   
 1 1 0 0 0 1 0 0 0 1V y Dy L x y Dy L x V     
    1,My 29.7 62.2 0.893 93.3 0.914;   1,W 1,My 1 y 0.086   
 M 35,270J mol@64.5 C 35,270 kJ / kmol    choose MeOH reference 64.5ºC. 
 40,656J mol@100 C 40,656 kJ/kmol    choose water reference 100ºC. 
 The condenser is at 66.1ºC (linear interpolation Table 2-7). 
    1 M 1,M W 1,W 1,M P,V,M 1 1,W P,V,W 1H y y y C T 64.5 y C T 100       
 1V is at 1T in equilibrium with 1,My 0.914. From Table 2-7 1T ~ 67.6 C 
 Assuming only constant & linear T term are important in P,VC eqs.,  P,V P avgC C T . For 
methanol avg
67.6 64.5T 66.65 C
2

  . For water, avg
67.6 100T 83.8 C
2

  . 
  
 
PV,M
J 1000 mol kJ kJC 42.93 0.08301 66.05 48.41 48.41
mol kmol 1000J kmol C
    
  P,V,W o
J kJC 33.46 0.00688 83.8 34.04 34.04
mol C kmol C
    
 MT 67.6 64.5 3.1    ; WT 67.6 100 32.4     
Then 
            1H 35270 0.914 40656 0.086 48.41 3.1 0.914 34.04 32.4 0.086     
 1H 35775.5 kJ kmol Note λ terms dominate. 
 
 DH is at Dy 0.955@66.1 C 
    D M D,M W D,W D,M P,V,M D D,W P,V,W DH y y y C T 64.5 y C T 100       
  P,V,M P,V,M M,avgC C T . avg,M
66.1 64.5T 65.3
2

  
  P,V,MC 42.93 0.08301 65.3 48.35   
  P,V,W P,V,W W,avgC C T . W,avg
100 66.1T 83.05
2

  
  P,V,WC 33.46 0.00688 83.05 34.03   
 WT 66.1 100 33.9     
61 
 
           DH 35270 .955 40656 0.045 48.35 0.955 1.6 34.03 0.045 33.9     
DH 33682.85 1829.5 73.88 51.91 35534.3     
Reflux liquid at 66.1ºC and M,0 W,0x 0.893, x 0.107  
 ReferenceMeOH 64.5ºC, water reference 100 ºC 
 0 PL,M M,0 M PL,W W,0 Wh C x T C x T    
    PL,M PL,M avg
kJC C T 75.86 0.1683 65.3 86.85
kmol C
    
        0h 86.85 0.893 1.6 75.4 0.107 33.9 149 kJ kmol     
C D 0 0 1 1Q DH L h V H          31.1 35534.3 62.2 149 93.3 35775.5 2,242,030 kJ h      
Overall EB F C R D BFh Q Q DH Bh    
or R D B C FQ DH Bh Q Fh    
Bh is saturated liquid with B,Mx 0.00435 and B,Wx 0.99565 
 Interpolating in Table 2.7 BotT 99.2 C 
     B PL,M PL,Wh C 0.00435 99.2 64.5 C 0.99565 99.2 100    
 avg,M
99.2 64.5T 81.86
2

 
 
 and  PL,MC 75.86 0.1683 81.86 89.64   
 Bh 13.53 60.06 46.5 kJ kmol     
Feed is saturated liquid at M Wz 0.3, z 0.7.  
 From Table 2-7, FT 78 C 
      F P,LM PL,Wh C 0.03 78 64.5 C 0.7 78 100    
 avg,MT 78 64.5 2 71.25   and  P,L,MC 75.86 0.1683 71.25 87.85   
Fh 805.4 kJ kmol  Then 
          RQ 31.1 35534.3 68.9 46.5 2,242,030 100 805.4
 1,105,116 3204 2, 242,030 80,536 3,424,479 kJ h
       
    
 
3.D5. Mass Balances: F = D + S + B, D S BFz Dx Sx Bx   
 Solving simultaneously, B = 76.4 kg/min, D = 13.6 kg/min. 
 Condenser:  C 1 0 1Q V h H  
    1 0V L D L D 1 D 4 13.6 54.4      kg/min 
 From Figure 2-4, 0h 7.7 kcal/kg (x = .9, T = 20°C), 
 1H 290 kcal/kg (y = .9, sat’d vapor). 
 Thus, CQ = 54.4 (7.7 – 290) = -15,357 kcal/min 
 Overall Energy Balance: F R C D S BFh Q Q Dh Sh Bh     
 R D S B F CQ Dh Sh Bh Fh Q     
 From Figure 2-4, 
    S S Fh 61 x .7, sat'd Liq'd ; h 200 z .2, 93 C ,     
62 
 
  B B D oh 99 x .01, sat'd Liq'd , h h 7.7    
 Thus, 
             RQ 13.6 7.7 10 61 76.4 99 100 200 15357 3635.3       kcal/min 
 
3.D6. From Eq. (3-3), D = F B
D B
z x .4 .0022500 998
x x .999 .002
   
    
   
 lbmol/h. 
 Then B = F = 1502. 
 Condenser:    o oV L D L D D D    
    C D V oQ h H D L D 1   
 With 99.9% 5nC have essentially pure 5nC . Thus, it is at its boiling point. 
  D V C5h H 11,369     Btu/lbmol. 
    CQ 11,369 998 4 45,385,048    Btu/h 
 Overall: R D B F CQ Dh Bh Fh Q    
 Distillate is at boiling point of pure 5 C5nC K 1.0 on DePriester Chart) = 35°C. Bottoms 
is at boiling point of  6 C6nC K 1.0   67°C. 
 Converting to °F: 35°C = 95°F, 67°C = 152.6°F, 30°C = 86°F. 
 Note feed is obviously a subcooled liquid. Arbitrarily, pick a liquid at 0°F as reference. 
(This will not affect the result and other values can be used.) 
 
F LC5P C5 P C6 PLC6
C x C z C  
      
FP
C .4 39.7 .6 51.7 46.9   Btu/lbmol °F 
     
FF P F
h C T 0 46.9 86 4033.4    Btu/lbmol 
 Distillate is almost pure 5nC . Liquid at 95°F 
     
LC5D P Dist
h C T 0 39.7 95 3771.5    Btu/lbmol 
 Bottoms is almost pure liquid 6nC at 152.6°F. 
     
pLC6 bot
hC T 0 51.7 152.6 7889.4   Btu/lbmol 
          RQ 998 3771.5 1502 7889.4 2500 4033.4 45,385,048 50,861,491     Btu/h 
3.D7. Eq. (3-3), B
D B
z x 0.7 0.001D F 1000 700.4
x x 0.999 0.001
   
     
   
 kmol/h 
 B F D 299.6   kmol/h 
 Condenser:     o oL L D D 2.8 700.4 1961.1   kmol/h 
 Only this reflux is condensed since product is a vapor. 
  C oQ L  where λ is for essentially pure n-pentane. 
 C
kmol Btu 2.20462 lbmolQ 1966.1 11,369
h lbmol 1 kmol
   
    
   
 
 10C -4
Btu 1 J JQ 49,154,204.85 5.18176 10 
h 9.486 10 Btu h
   

 
63 
 
 From overall balance R D B F CQ DH Bh Fh Q    
 Distillate is vapor at b.p. of pure n-pentane (35°C from DePriester chart, C5K 1.0 ) 
 Bottoms is boiling n-hexane (67°C) 
 Conversions: 35°C = 95°F - distillate & Feed and 67°C = 152.6°F - bottoms 
 
 As reference, arbitrarily choose liquid at 0°F. 
 Feed is subcooled liquid. 
       oPF C5 PLC5 C6 PLC6C z C z C 0.7 39.7 0.3 51.7 43.3Btu lbmol F     
     F PF Fh C T 0 43.3 95 0 4113.5Btu lbmol     
 Distillate  D C5 PLC5 distH C T 0    
   DH 11,369 39.7 95 0 15,140.5 Btu lbmol    
 Bottoms is pure 6C @152.6 F 
    B PLC6 both C T 0 51.7 152.6 0 7889.4 Btu lbmol     
 
R
kmol Btu kmol BtuQ 700.4 15,140.5 299.6 7889.4
h lbmol h lbmol
kmol Btu 2.20462 lbmol Btu1000 4113.5 49,154,204.85
h lbmol kmol h
     
      
     
      
        
      
 
 10R -4
Btu 1 JQ 68,675,167.9 7.240 10 J h
h 9.486 10 Btu
  

 
3.D8. New Problem in 3rd Edition. 
 E wF 300, z .3, z .7   
 98% rec. E in distillate, 81% rec water in bot. 
      D Dist. .98 90 1 .81 300 .7 128.1 kmol/h     
 
  
DE
.98 90
y 0.6885
128.1
  
      B Bottoms .02 90 .81 210 171.9kmol h    
 
 
 
 b. Partial Condenser. 
 
 
 
 
 
  0 0
L 2, L 2D 2 128.1 256.2 kmol h.
D
    
 0x in equilibrium with 0y , thus from equation data 0x 0.575. 
 Entering vapor 1y (from graph) 0.61 
D Dy ,D,H 
cQ 
0 0 0x ,L ,h
 
1 1 1H y V 
Vapor 
64 
 
 1 0V L D 256.2 128.1 384.3 kmol h.      
 c. E.B. on PC. 1 1 c dist 0 0V H Q DH L h .    Can use Figure 2-4 by converting mole fracs to mass 
fracs. Basis 1 kmole. 
 Distillate  .6885 mol E MW 46 31.671 kgE  
  
5.607 kgW.3115 mole W MW 18
37.28 kg total
  
 Mass frac. E = 31.671 37.28 0.8496 
 Vapor 1V  0.61 mole E 46 28.06 kgE 
  
7.02 kgW.39 mole W 18
35.08 kg total
 
 Mass frac E = 28.06 35.08 0.7999 
 Liquid reflux 0L  0.575 mole E 46 26.45 kgE 
  
7.65 kgE0.425 mole W 18
34.1 total

 
 Mass frac E = 26.45 34.1 0.7757 
 From Figure 2-4, dist 1 0H ~ 310 kcal kg, H ~ 330kcal kg, h ~ 65 kcal kg 
 
c dist 0 0 1 1
kmol 37.28 kg kcalQ DH L h V H 128.1 310 256.2 34.1 65
hr kg kg
35.08 kg 364.3 330 2, 400,517 kcal hr
kmol
    
  
 
 Overall EB. F R c dist BFh Q Q DH Bh    
 Know cQ 2,400,517 kcal h  
 and distDH 1,480,426 kcal h. 
 To find F BFh and Bh , need to convert mole frac to wt frac. 
 Basis 1 kmol 
 Feed  30 mole % E: .3 mole 46 13.8 
  
12.6 70% W : .7 18
total 26.4 kg kmol
 
 Mass frac E 13.8 26.4 0.5227  
 Bottoms  0.01047 mole 46 .48162 
  
 17.811 0.98953 mole 18
total 18.293 kg kmol
 
 Mass frac E 0.48162 18.28 0.0263  
 From Figure 2-4  Fh satd liqd 70kcal kg 
  Bh satd liqd 97 
 
65 
 
 Then R dist B F cQ DH Bh Fh Q    
R
kmol 18.29316 kg 97 kcal 26.4 kg kcalQ 1,480,426 171.9 300 70
h kmol kg kmol kg
  
   2,400,517 1,480,426 305,525 554,408 2,400,517       3,632,069kcal h 
 
3D9. New Problem 3rd Edition. B = (xD – z)/(xD – xB)F = [(0.9999 - 0.76)/(0.9999 – 0.00002)](500) = 120 
 
 R BQ Lh VH Bh   and L V B  
 Assume Bh h .    RQ L B h Vh V H h V        
     V V B B 1.5 120 180 kmol h.   
 Bottoms is almost pure water. w 9.72 kcal mol 9720 kcal kmol  
    6RQ 180 kmol/h 9720 1.750 10 kcal h   
 
3.D10. 2 atm × 101.3 kPa/atm = 202.6 kPa. Pentane Recovery: P D0.995 Fz Dx  
 
   
 
0.995 1000 0.55
D 547.6333
0.9993
  kmol/h 
 B = 1000 – 547.6333 = 452.3667 
 Since  B pBx Pentane Recovery Bot F z ,  
 
   
B
1 .995 1000 0.55
x 0.006079
452.3667

  mol frac pentane 
 Distillate is essentially pure Pentane. Bottoms Pure in Hexane. From DePriester Chart 
 P distK 1@p 202.6 kPa when T 59.5 C   
 n H botK 1@p 202.6 kPa when T 94 C    
 For Total Condenser, Eq. (3-14)  oC D 1
LQ 1 D h H
D
 
   
 
 
    D PLC5 dist ref refh pure pentane C T T choose T 25 C   
  D
kcal kcalh 39.7 59.5 25 1369.65
kmol C kmol
   
 1 DH h  assuming λ is independent of temperature 
 1
Btu 1 lbmol 0.252 kcal kcalH 1369.65 11369 7680.196
lbmol 0.454 kmol Btu kmol
   
     
   
 
 Eq. (3-14) is     C
kmol kcalQ 1 2.8 547.6333 6310.5 13,132,288
h h
     
      B PLC6 bot ref
kcal kcalh pure hexane C T T 51.7 94 25 3567.3
kmol C kmol
     
 Feed is a liquid at 65°C 
   F PC5 C5 PC6 C6 F refh C z C z T T   
      Fh 39.7 0.55 51.7 0.45 65.25 1804kcal / kmol     
66 
 
 R D B F CQ Dh Bh Fh Q    
           RQ 547.6333 1369.65 452.3667 3567.3 1000 1804 13,132,288     
 RQ 13,692,081 kcal/h. 
 Note that C RQ and Q are relatively close. 
 
3.E1. Was 3.D8 in 2nd Edition. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Pick as basis liquid at 0°C, W Mh 0 & h 0  (essentially steam table choice) 
 Assume ideal mixtures. 
  
avg avg W,L M,LF P P ref P P P
h C T T where C 0.4 C 0.6 C    
 Felder & Rouseau p. 637 PWC 0.0754 kJ/mol 
 5PMC 0.07586 16.83 10 T
   kJ/mol 
       PM PM avg PM PMC C T C 0 40 2 C 20 C    
  5PMC 0.07586 16.83 10 20 0.079226
    
    
avgP
C 0.4 0.0754 0.6 0.079226 0.077696   
  Fh 0.077696 40 0 3.1078   kJ/mol feed = 3107.8 kJ/kmol 
 Can also use steam table for water 
 SH is sat’d vapor steam 1 atm, S
kJ 18.0 kgH 2676.0
kg kmol
 
 Steam Table F&R, p. 645 SH =48,168 kJ/kmol 
 Dh is sat’d liquid at Dx 0.99 . From Table 2-7, T = 64.6°C 
  
avg avgD P P PW PM
h C 64.6 0 where C 0.01 C 0.99 C    
      5PM PM avg PMC C T C 32.3 C 0.07586 16.83 10 32.3 0.081296      
    
avgP
C 0.01 0.0754 0.99 0.081296 0.08124   
Bx 0.02 
B 
CQ
p = 1.0 atm 
Mz 0.6 
100 kmol h
 
F 
DD, x 0.990
 M.B. F + S = D + B 
M S,M D BFz Sy Dx Bx   
Since steam is pure, S,My 0 
Know M D BF, z , x , x 
Unknowns S, D, B, Need E.B. 
F C S D BFh Q SH Dh Bh   
S 
67 
 
  DH 0.08124 64.6 5.2479 kJ mol 5247.9 kJ kmol   
 Bh : Since leaving an equilibrium stage, sat’d liqd. 2% MeOH 
 Table 2-7, T = 96.4°C 
  
avg avgB P P PW PM
h C 96.4 0 where C 0.98 C 0.02 C    
     PM PM PM C C 96.4 0 2 C 48.2 C   
  5PM C 0.07586 16.83 10 48.2 0.08397
    
    
avgP
 C 0.98 0.0754 0.02 0.08397 0.07557 kJ mol   
  Bh 0.07557 96.4 7.28509 kJ mol 7285.09 kJ mol   
 
 CQ  do EB around condenser 
   oC dist 1 dist o dist
LQ V L D 1 D
D
 
        
 
 
 dist MeOH W0.99 0.01     
 Felder & Rousseau: M W35.27 kJ mol & 40.656 kJ mol    
      dist
kJ0.99 35.27 0.01 40.656 35.324 35,323,86 kJ kmol
mol
     
   CQ 35,323.86 2.3 1 D 116,568.7D kJ h     
 
 Plug CQ & numbers into E.B. 
  100 3107.8 116,568.7D 48,168S 5247.9D 7285.09B    
 or 310,780 + 48,168S = 121,816.6D + 7285.09B 
 
 Solve simultaneously with 2 MB. 100 + S = D + B 
 60 + 0 = 0.99D + 0.02B 
 One can use algebra or various computer packages. 
 
 Obtain: D = 56.33 kmol/h, B = 211.71 kmol/h 
 CS 168.04 kmol/h, Q 6,566,000 kJ/h.   
 
E2. Was 3.D9 in 2nd Edition. kmol molF 500 500,000
h h
  
 F + S = D + B 
 S D BFz Sy Dx Bx   
 2 eq. 3 unknowns 
 Condenser:    C 0 o 1Q 1 L / D D h H   
 Note Eq (3-14) not valid. 
 
 For enthalpy pick reference pure liquid water 0°C and pure liquid methanol 0°C. Felder & 
Rouseau: 
MeOHP
C 75.86 0.01683T  at 
MeOHavg P
64.5 0T 3225, C 76.4 J mol C.
2

   
68 
 
 Assuming distillate pure methanol, boils at 64.5°C 
     
Meoh ,liqD P ref
h C T T 76.4 J/mol 64.5 0 4928.0 J mol     
  1 D 1H h at 64.5 F 4928 35270 J/mol 40,198 J/mol     
   CQ 4 4928 40198 D 141,080D    J/h where D is mol/h 
 Overall Energy balance: F S C D BF h SH Q Dh Bh    
 Bottoms is essentially pure 2H O at 100°C 
    
W,liqB P ref
J Jh C T T 75.4 100 0 7540 
mol C mol
 
     
 
 
  S B WH h at 100 C 7540 40656 J/mol 48196     
 For feed. 60 mole % Methanol boils at 71.2°C (Table 2-7). 
          
iF P i ref
h C z T T 0.6 76.459 71.2 0.4 75.4 71.2 5413.7     J/mol 
 Now, Eqs are 
 (1) F + S = D + B or 500,000 + S = D + B 
 (2) D BFz Dx Bx  or (500,000) (.6) = 0.998D + 0.0013B 
 (3)  oC D 1 C
LQ 1 D h H or Q 141,080D
D
 
    
 
 
 (4) F S C D BFh Sh Q Dh Bh    or (500,000) (54137) + S (48196) 
 + CQ = D(4928) + B (7540) 
 Solve simultaneously: D = 298.98, B = 1245.5, S = 1044.2 kmol/h 
 CQ = - 4.218 × 10
+7 kJ/h 
 
3.F1. An enthalpy composition diagram is available on p. 272 of Perry’s Chemical Engineer’s 
Handbook, 3rd ed., 1950. 
 Eq. (3-3)  B
D B
z x 0.79 0.004D F 25,000 19,788.5
x x 0.997 0.004
  
   
  
 kmol/h 
 Note that 2N mole fractions were used since 2N is more volatile. B = F – D = 5211.5 
 From enthalpy comp. diag. D 1h 0, H 1350  kcal/kmol, B Fh 160, h 1575  . Then, 
       C o D 1Q 1 L D D h H 5 19788.5 0 1350 133,572,000       kcal/h 
 R D B F CQ Dh Bh Fh Q    
      R CQ 0 5211.5 160 25,000 1575 Q 95,030,000     kcal/h 
 
3.F2. We will use the enthalpy composition diagram on p. 3-171 of Perry’s 6th edition or p. 3-158 of 
Perry’s 5th ed.Do for 1 kmol of feed: 
 Conversion of feed from kg to moles. Basis 100 kg 
 30 kg 3NH = 1.765 kmol 
 70 kg 2H O  3.888 
 Total 5.653 kmol 
 
 Thus 1 kmol is 100/5.653 = 17.69 kg 
69 
 
 Will work problem in weight fractions since data is presented that way. 
 95% recovery: (0.95) Fz = DDx or, D = (.95) DFz / x = (.95) (17.69) (.3)/(.98) = 5.15 kg. 
 B = F – D = 12.54 kg 
          B Dx Fz Dx B 17.69 .3 5.15 98 12.54 0.021       
 From diagram: D 1 B Fh 55, H 415, h 150, h 5 kcal/kg     
 Eq. (3-14),        C o D 1Q 1 L D D h H 3 5.15 55 415 5562       kcal/kmol feed 
 and R D B F CQ Dh Bh Fh Q    
         R CQ 5.15 55 12.54 150 17.69 5 Q 7815      kcal/kmol of feed 
 
G1. a.) Using NRTL. C RQ 778,863 kcal/h, Q 709,520 kcal/h   
 b.) C RQ 1,064,820 kcal/h, Q 995,478 kcal/h   
 
G2. New Problem in 3rd Edition. 
 ASPENPlus. D = 988, L/D = 3, Peng Robinson, FN 40, N 20  (arbitrary values in Radfrac) 
 7D DC6x 1.000 x 1.211 10
   
 Bx 0.0013316 
 
7
C
7
R
Q 4.4426 10 Btu h,
Q 4.9852 10 Btu h
  
 
 
 
 
 
 
70 
SPE 3rd Ed. Solution Manual Chapter 4 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 4A6, 4A13, 4C10, 4C16, 4D6, 4D9, 4D13, 4D15, 4D18, 4E4, 4E5, 4H1 to 4H3. 
4A1. Point A: streams leaving stage 2 (L2, V2) 
Point B: vapor stream leaving stage 5 (V5) 
 liquid stream leaving stage 4 (L4) 
Temp. of stage 2: know 2 2K y / x , can get T from temperature-compositiongraph or 
DePriester chart of K = f(T,p). 
Temp. in reboiler: same as above (reboiler is an equilibrium stage.) 
4A2. a. Feed tray = .6, z = 0.51 (draw y = x line), yF =0.52, xF = 0.29.
b. Two-phase feed.
c. Higher
4A6. New Problem in 3rd Edition. Answer is a. 
4A7. See Table 11-3 and 11-4 for a partial list. 
4A13. New Problem in 3rd Edition. 
A. Answer is b 
B. Answer is a 
C. Answer is a 
D. Answer is a 
E. Answer is b 
F. Answer is a 
G. Answer is b 
4A14. If feed stage is non-optimum, the feed conditions can be changed to have an optimum 
feed location. 
4B2. a. Use columns in parallel. Lower F to each column allows for higher L/D and may be sufficient
for product specifications. 
b. Add a reboiler instead of steam injection. Slightly less stages required and adds 1 stage.
c. Make the condenser a partial instead of a total condenser. Adds a stage.
d. Stop removing side stream. Fewer stages are now required for the same separation.
e. Remove the intermediate reboiler or condenser and use it at bottom (or top) of column. Fewer
stages, but all energy at highest T (reboilers) or lowest T (condenser) for same separation. 
Many other ideas will be useful in certain cases. 
4C7. Easiest proof is for a saturated liquid feed. Show point Dz, y satisfies operating equation. 
Solution: Op. Eq. By L V x L V 1 x
Substitute in Dy y , x z
D By V Lz L V x
But q 1.0, V D, L F, L V B 
D By D Fz Bx
Which is external mass balance. QED. 
71 
 
 Can do similar for enriching column for a saturated vapor feed. 
 
4.C10. New Problem in 3rd Edition. If we consider λ, the latent heat per mole to be a positive quantity, 
then RQ V . With CMO and a saturated liquid feed (1 / )V V L D D , and then 
/ (1 / )RQ D L D . 
 
4.C16. New Problem in 3rd Edition. Define a fictitious total feed T T TF , z , h 
 T 1 2F F F , 
1 1 2 2
T
T
F z F z
z
F
, 1 21 F 2 FT
T
F h F h
h
F
 
 Intersection of top & bottom operating lines must occur at feed line for fictitious feed FT. 
(Draw a column with a single mixed feed to prove this.) 
 This feed line goes through Ty x z 
 
b.) Does 1 1 2 2T
T
q F q F
q
F
 
x 
Bx 
B 
Tz
 
y 
0z 
A 
2z 
1z 
Given p, L/D, saturated liquid reflux, D Bx , x 
opt feed locations, 1 2 1 2 F1 F2z , z , F , F , h , h 
 
 
 
Plot top op line. Plot all 3 feed lines. Draw 
line from point A to y = x = Bx to obtain 
bot. op. line. Connect pts B & C to get 
middle op. line. 
 
TF
 
with slope T Tq q 1 
where mix TT
mix mix
H h
q
H h
 and mix, mixH h are saturated 
vapor and liquid enthalpies at feed stage of column with 
mixed feed. 
 
 
72 
 
 check 
1 2
1 2
1 F 2 F
mix
1 2 mix 1 F 2 FTmix T
T
mix mix mix mix mix mix T
F h F h
H F F H F H F hFH h
q
H h H h H h F
 
 where mix mixH & h are vapor and liquid enthalpies on feed stage of mixed column 
 
 
1 21 mix F mix F
2
mix mix mix mix
T
T
F H h H h
F
H h H h
q
F
 
 Usual CMO assumption is λ >> latent heat effects in either vapor or liquid. 
 Then 1 2mix F mix F1 2
mix mix mix mix
H h H h
q and q
H h H h
 
 Thus 1 1 2 2T
T
F q F q
q
F
 if CMO is valid. 
4D1. a. Top op line: D
L L
y x 1 x
V V
and L L D 1.25 0.5555
V 1 L D 2.25
 
 Intersects Dy x x 0.9 
 When D
L
x 0, y 1 x 0.4
V
 Plot – See diagram 
b. Bottom op line: B
L L
y x 1 x
V V
, and 
L V B V B 1 3
V V V B 2
 
 Intersects y = x = xB = 0.05 
 
1 0.5 / 2
@ y 1 x 0.683 this is convenient point to plot
3 2
 
c. See diagram for stages. Optimum feed stage is #2 above partial reboiler. 
 5 equilibrium stages + PR is more than sufficient. 
73 
 
 
 
 d. Feed line goes from y = x = z = 0.55 to intersection of two operating lines. 
 
q
Slope 1.0 or q 0.5
q 1
. 
 This is a 2 phase feed which is ½ liquid & ½ vapor. 
 
4D2. New Problem in 3rd Edition. Part a. 
 
E
L F V 1 V / F .63
y x z .6 Slope 1.703
V V V F .37
 
 b. From Table 2-1, at 84.1° C y .5089 
 c. liquid at 20°C F
H h
q
H h
 and 40 mole % ethanol. 
 The pressure in Figure 2-4 is very close to 1.0 atm, thus it can be used, but must convert to wt 
frac. 
74 
 
 Basis 1kmol feed. 
 .4 kmole E .4 MW 46 18.4 kg 0.63 wt frac. 
 
 10.8 kg
.6 kmol Water .6 MW 18
total 29.2 kg
 
 From Figure 2-4 FH 398 kcal kg ,h 75,h 20 C 10 
 
398 10
q 1.20
398 75
 
 
q 1.2
Slope 6
q 1 .2
 
 Alternate Solution: 40 mole % ethanol boils at 84.1°C (Table 2-1). 
 Then if pick reference as saturated liquid at 40 mole % 
 F p,40%liqh C 20 84.1 
 
40%E
h 0, H 
 d. 
vaporF p
kcal
40 mole %E 63 wt%, H 398 kcal kg , h 65, h 398 C 120 84.1
kg
 
 
vapor Evapor w ,vaporP E P w P
C y C y C 
 Assume only 1st and 2nd terms in PC equations are significant. 
 From Problem 2.D9 
 
vaporP
C .4 14.66 0.03758T .6 7.88 .0032T 
 kcal/kmol T is C 
 
 which simplifies to 
vaporP
C 10.592 0.16952T 
 For linear 
120
p
84.1
C dT is equal to 
vaporP avg
C @T 
 avgT 84.1 120 2 102.05 . Then v ,avgP
kcal
C 10.592 0.16952 102.05 12.32
kmol
 
 F
kcal kcal 1 kmol
h 398 12.32 120 84.1
kg kmol 27.2 kg
 
 Fh 398 15.149 413.15kcal kg 
 
398 413.15 15.147
q 0.045.
398 65 333
 
 e. 
F 13
q L L f , L L F L F
12 12
 
 
13 12
q 13 12, slope q q 1 13
1 12
 
 
 f. Flash 
V L 1 V F .3 3
.7, 
F V V F .7 7
 
 See graph for feed lines. 
75 
 
 
Graph for 4.D2 
76 
 
4.D3*. a. Basis 1 mole feed. 
 0.4 moles EtOH × 46 = 18.4 kg EtOH 
 0.6 moles H2O × 18 = 10.8 kg H2O 
 Total = 29.2 
 wt frac 18.4 / 29.2 0.63 wt frac EtOH 
 Calculate all enthalpies at 0.63 wt frac. Hv = 395, HL = 65 (from Figure 2-4). hF is liquid at 
200°C. Assume Cp,liq is not a function of T. Estimate, 
 LP,liq
h 60 C h 20 46.1 23h kcal
C .63 wt frac ~ 0.864
T 60 20 80 kg C
 
 Then 
LF L P L
h h 200 C 200 60 h 60 C .864 200 60 46.1 167.1 
 v F
v L
H h 395 167.1 q 0.691
q 0.691, 2.24
H h 395 65 q 1 0.309
 
b. From Figure 2-4 at 50 wt% ethanol Hv = 446 and hL = 70. Since CMO is valid obtaining both 
enthalpies at 50% wt is OK. The feed is a liquid 
 F P,liq F ref P,liqh C T T C 250 0 
 
wP,liq P,EtOH EtOH P w
C C z C z in Mole fractions 
 Basis 100 kg solution 
 50 kg EtOH 46.07=1.085 kg/kgmole 
 50 kg W 18.016 2.775 kg moles 
 Total 3.860 kg moles 
 Avg M.W. 100 3.86 25.91 kg/kgmole 
 Thus, zW = 0.719 and zE = 0.281 
 P,liqC 37.96 .281 18.0 .719 23.61 
 PP,liq
AVG
C 23.61
C in kcal kg C 0.911
MW 25.91
. Then, 
 F
kcal
h 0.911 250 C 228
kg C
 
 v F
v L
H h 446 228
q 0.58
H h 446 70
 
 4.D4*. a. F Pvh H C 350 50 H 25 300 
 F
25 300H h
q 1.5
H h
 
 Slope q q 1 0.6. y x z 0.6 is intersection. 
 
 
 b. q L L F where L L 0.6F. Then q L 0.6F L / F 0.6, and 
 slope q q 1 1.5 
 c. q L L F where L L F 5. q L F 5 L F 1 5 , slope q q 1 1 6 
z 
y = x 
feed 
line 
.5 .6 .7 
4.D4a 
77 
 
4.D5*. liq refluxL
vap liq
h h 3100 1500
f 0.1111
H h 17500 3100
 
 0 0
1 0
L L D 1.1
0.524
V L D 1 2.1
 
 c 0 11
2 c 0 1
1 f L V 1.1111 .524L
0.55
V 1 f L V 1 .111 .524
 
 Alternate Solution 
 For subcooled reflux, 01
0 1
H hL
q
L H h
17500 1500
1.111
17500 3100
 
 Then, 1 0 0L qL 1.1111 L 
 1 1 1
2 1 1
L L L D
V L D L D 1
, 01
1.111 LL
1.111 1.1 1.2222
D D
 
 1
2
L 1.222
0.55
V 2.222
 
4D6. New Problem in 3rd Edition. a) 1 2175 F F B D 
 85 75 .6 100 0.4 0.1 B 0.9D 
 Solve simultaneously. 
 D 84.375 and B 90.625 kmol hr 
 b) Feed 1. 1q 1, vertical at 1y x z 0.6 
 Feed 2. 60% vapor = 40% liquid 2q 0.4 
 Slope feed line 2
2
q 0.4
2 3
q 1 .06
 through 2y x z 0.4 
 Bottom Op. Line By L V x L V 1 x . Through By x x 
 
V B 1
Slope L V 3 2
V B
 
 Middle 2L F B V 
 2 2 B2 2 B
F z BxL
L x F z Bx V y y x
V V
 
 When 2 2 B
F z Bx
x 0,y , Slope L V
V
 
 Also intersects bot. op. line and Feed line 2. 
 Do External Balances and Find D & B. Then V V / B B 2B 181.25 
 L V B 271.875 
 At feed 2, L .4F L or L L 0.4F 271.875 40 231.875 
 V V 0.6F 181.25 60 241.25 
 L V 0.961 
 
40 9.625
x 0, y 0.126
241.25
 Plot Middle Op Line. 
78 
 
 D
L L
y x 1 x
V V
 
 Know that Dy x x and gives through interaction Middle and Feed line 1. 
 Also, 1L L F 231.875 75 156.875 and V V 241.25 ; thus, 
 L V 156.875 241.25 0.65 
 c) See graph. 
 
Graph for 4D6.
 
79 
 
 
4.D7*. a. Plot top op. line: slope D
L
.8 , x y x .9.
V
 Step off stages as shown on Figure. 
 b. Plot bottom op. line: slope B
L V
1 1 2 , x y x 0.13.
BV
 Step off stages 
(reboiler is an equil stage). Find y2 = 0.515. 
 c. Total # stages = 8 + reboiler 
 Optimum feed plate = 7 or 8 from top. Plot feed line. Goes through x = y = z = .3, and 
intersection of two operating lines. 
 slope 
9 q
4 q 1
 gives q = 0.692. 
 
 
4.D8*. The equilibrium data is plotted and shown in the figure. From the Solution to 4.D7c, 
q 0.692 and q q 1 9 4 
 a. total reflux. Need 5 2/3 stages (from large graph) – 5.9 from small diagram shown. 
 b. 
min
.9 .462
L V 0.660
.9 .236
 (see figure) 
 min
min
min
L V
L D 1.941
1 L V
 
c. In 4.D7, act
L V .8
L D 4
1 L V .2
 
 act minL D Multiplier L D 
 Multiplier = 4/1.941 = 2.06 
80 
 
d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an 
equilibrium contact. Then use MVE AB AC 0.75 (illustrated for the first real stage) 
Stage 1 is the optimum feed stage. 11 real stages plus a partial reboiler are sufficient. 
 
 
 
4D9. New Problem in 3rd Edition. a) 
 1 2 1 1F F D B 100 F 80 B F B 20 
 1 1 2 2 B B 1Fz F z Dx Bx F .42 18 .66 80 0.04 B 
 Solve simultaneously, 1B 113.68, F 93.68 
 b) 
1 L L D 1 2 1
L D , 
2 V 1 L D 3 2 3
 
 
L
L D 40, V L D 120
D
 
 Saturated Liquid Feed V V 120 
 1L L F 40 93.68 133.68, L V 1.114 
 c) Top Op. Line – Normal: Dy L V x 1 L V x 
 Through D
2
y x x , Slope 1 3, y intercept .66 .44
3
 
 Bottom – Normal: B By L V x L V 1 x , through y x x 
 Also through intersection, 2F feed line and middle op. line. 
2
2
F
2
F
L .7
Feed line F slope 
V .3
 
81 
 
 Middle 1 1D
F zD
y L V x x
V V
 (or do around bottom) 
 Slope L V . Through intersection feed line 1F and top op. line. 
 Also, D 1 1
80 .66 93.68 .42Dx F z
x 0, y 0.11212
V 120
 
 d)Opt. Feed 2F stage 1 from bottom, Opt feed 1F , Stage 2. 4 stages + PR more than sufficient.
 
Graph for 4D9. 
82 
 
4.D10*. Operating Line D
L L L D 4
y L V x 1 x , where .8
V V 1 L D 5
 
 Thus, operating line is y = .8x + .192 
a. Equilibrium is 11
1
yy
x or x
1 y 1.79 .76y
 
Start with y1 = .96 = Dx 
Equilibrium: 11
1
y .96
x 0.9317
1.76 .76y 1.76 .76 .96
 
 Operating: 2y .8x .192 .8 .9317 .192 0.93736 
 Equilibrium: 22
2
y .93736
x 0.89476
1.76 .76y 1.76 .76 .93736
 
 Operating: 3 2y .8x .192 .8 .89476 .192 0.9078 
b. Generate equilibrium data from: 1.76xy
1 .76x
 
 
x 1.0 .9 .8 .7 .6 .5 .4
y 1.0 .9406 .8756 .8042 .7253 .6377 .5399 
 
 Plot equilibrium curve and operating line. (See Figure). Slope = L/V = .8, y intercept (x = 0) 
= 0.192, y = x = Dx = 0.96. Find 6x = 0.660. 
 
 
4.D11. a) Same as 4.D2 part g. q = 1.0668, slope feed line = 15.97. 
 b) Top Dy L V x 1 L V x goes through y = x = Dx = 0.99 
 
L D
L V 0.6969
1 L D
 @ x = 0 y = (1-L/V) Dx = (1-0.6969) 0.99 = 0.300 
 Feed line: Slope q q 1 , y x z 0.6 
83 
 
 Bottom Op line: 
 
 y= 0, x = Bx . Also goes through intersection of feed line and top op.line. 
 Stages: Accuracy at top is not real high. (Expand diagram for more occupancy). 
 As drawn opt. Feed = #6. Total = 9 is sufficient, 
 
c. 
min
0.99 0.57
L V Slope 0.4242
0.99 0
 
 
min min
L L V 0.4242
0.73684
D 1 L V 1 0.4242
 
 Actual L/D is 3.12 × this value. 
 
 
 
V L
 
B S 
M,S ByV Lx Sy Bx 
But M,Sy 0 (Pure steam) 
With CMO B L 
B
L L
y x x
V V
 
84 
 
4.D12. L DL V 3 4
1 L D
slope. Top op line goes throug Dy x x 0.998 
 D
L L
y x 1 x @ x 0, y .25 .998 0.2495
V V
 
 Bottom slope 
From Soln to 3.D9 orB 1245167
L V 1.19 
from graph. 1.169S 1044168
 
 Feed line is vertical at z = 0.6. Can also plot top and feed lines, and then find bottom from 2 
points By 0, x x & intersect top & feed . 
 For accuracy – Use expanded portions near distillate & near bottoms. 
 From Table 2-7 from (x = .95, y = .979) 
 Draw straight line to (x = 1.0, y = 1.0) 
 From (x = 0, y = 0) draw straight line to (x = 0.02, y = 0.134) 
 or (x = 0.01, y = 0.067) 
 
Opt feed = # 9 from top. Need 13 equilibrium stages. 
 
 
 
 
85 
 
 
 
 
 
 
 
86 
 
4.D13. New Problem in 3rd Edition. a.) See Figure 
 
 b.) See figure. 
MIN
L 0.665 0.95
0.4385
V 0.30 0.95
 
 
MIN
L L L V 0.4385
0.7808
D V L 1 L V 0.5615
 
 c.) 
MIN
L
2.0 L D 1.5616
D
, L L D 1.5616 0.6096
V 1 L D 2.5616
 
 y intersect D
L
1 y 0.3709
V
. Top operating line D
L L
y x 1 y
V V
 
 Goes through Dy x y 0.95 
 Bottom B
L L
y x 1 x
V V
 
 Goes through By x x & intersection top operating line & feed line. 
 Feed Line: Vertical (saturated liquid, q = 1). Through y x z 0.3 
 Plot & Step off stages. Optimal feed = 5 below PC. 6 + PC + PR more than sufficient. 
 d.) Slope bottom: See figure for parts c & d. 
0.85 0.025
L V 1.941
0.45 0.025
 
 1 1V B V L V 1.0625
L V 1 0.941
. 
87 
 
 
 
Graph for problem 4.D13. 
88 
 
 
 
4.D14a. 
89 
 
 
 
4.D14b. Two approaches to answer. Common sense is all methanol leaks out and MAx 0 . 
McCabe-Thiele diagram: This is enriching column with sz y 0 . Intersection top op. line and 
horizontal feed line is at M,bx x 0 , which is also a pinch point. Thus M,dx 0 also. 
4.D15. New Problem in 3rd Edition. Saturated liquid. 
q
q 1, ,
q 1
 feed line vertical @ z .3 . 
Top operating line D
L L
y x 1 y
V V
, 
L L D 2
Slope
V 1 L D 3
 
 Dy intercept 1 L V y 1 3 0.6885 0.2295 and Dy x y 
 Bottom operating line By L V x L V 1 x goes through By x x 
 And goes through interaction feed line and top operating line. See graph. 
 Optimum feed is stage 2 below partial condenser. Partial condenser + Partial reboiler + 3 
equilibrium stages are more than enough to obtain separation. 
 
 
 
S B 
External M.B. S = B 
 Sys = BxB . Since yS = 0 (pure water) 
 xB = 0 
 
New S.S. 
90 
 
 
 
Graph for problem 4.D15. 
91 
 
 
4.D16*. v F
v L
H hL-L
q
F H h
 
 Using 32°F = 0°C as reference T, Fh 4033.4 Btu/lbmole. 
 v LH h . Approx. at feed conditions. 
 .4 11369 .6 13572 12691 Btu/lbmole 
 For approx. temperature of feed stage, do bubble pt. calc. 
 1 1 1 1 1y 1 K x K z 
 Pick T = 48°C (~ 40% of way between boiling pts.) 
 C5 C6 1 1K 1.5, K .54, K x 1.5 .4 .54 .6 .92 
 C5 new new
.54
K T = =.594, T 50 C 
.92
 
 C5 1 1K 50 C 1.6, K x 1.6 .4 .584 .6 .99 Close enough. 
 v L feedH h 50 C , 50 C 122 F 
 
feed ,Liqv P
H C 122 46.9 122 5721.8 
 P feed,liqNote : C 46.9 is from Prob. 3-D6. 
 vH 5721.8 12691 18412.8 Btu/lbmole 
 V F
V L
H h 18412.8 4033.4
q 1.133
H h 12691
 
 Note: Fh is from Prob. 3.D6. 
4.D17. Top Op. Line: D D
L L
y x 1 x , goes through y x x 0.9
V V
 
 L L D 7 2 7 9
V 1 L D 9 2
 
 D
L 2
x 0, y 1 x .9 =0.2 
V 9
 
 Plot Top. Step off 2 stages. Find Sx ~ 0.81 
 The vertical line at Sx x 0.81 is the withdrawal line. 
 Bot. Op. Line intersects Top at Sx x . 
 Also know it intersects feed line at Bx x (unknown) 
 External Balances F D B S Don’t know BD,B, or x . 
 D B SFz Dx Bx Sx 
 Feed enters as saturated vapor. Thus q0 & V F 
 Bottoms leaves an equilibrium contact, it is saturated liquid L B 
 Do flow balances 
 V F 100 
 V V 100 since S is removed as saturated liquid. 
92 
 
 
L
L V 7 9 100 77.777. D V L 100 77.777 22.222
V
 
 L L S 77.7777 15 62.7777. L V 62.7777 100 0.6278 
 B L 62.777 
 D SB
60 22.222 0.9 15 0.81Fz Dx Sx
x 0.444
B 62.7777
 
 Plot. Op. line 
 Step off stages. 9 is more than sufficient. 
 
 
4.D18. New Problem in 3rd Edition. Feed 1 1F : z 0.6, saturated liquid, q 1,q / (q 1) 
93 
 
 Feed 2 2F : z 0.4, 80% vapor hence 20% liquid Fq L / F 0.2F / F .2 
 
q .2
1 4
q 1 .8
 
 Part a.) Bottom operating line goes through point, By x x 0.04 
 Max L V to point intersection feed F2 line and equilibrium curve. 
 Slope 
max
L 1.0 .04
2.2326
V .47 .04
 
 
min
V V 1 1
0.8113
B L V 1.2326L V 1
 
 
 Part b. 1L F 100 
 
2F
L L L 100 .2 80 116 
 L V B and V B 1.5 
 
116
116 L 1.5B B B 46.4
2.5
 
 
116
V L B 116 46.4 69.6 L V 1.66667
69.6
 
 
2F
D V V V 69.6 .8 80 133.6 
 
L 100
0.7485
V 133.6
 
 Check overall balance 1 2F F D B 180 133.6 46.4 180.0 OK 
 To find Dy use MVC mass balance 
 1 1 2 2 D BFz F z Dy Bx 
 or 1 1 2 2 BD
100 .6 80 .4 46.4 0.4F z F z Bx
y 0.675
D 133.6
 
 Actual bottom op. line: B
L L
y x 1 x
V V
 
 
L V B V B 1 2.5 5
V V V B 1.5 3
 
 Goes through By x x 0.04 , Slope 5 3 
 2nd point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) 
 Plot bottom op. line 
 Top Op. line: 1 1 DyV Fz Lx Dy . 
D 1 1Dy F zLy x
V V
 
 Goes through intersection feed line for F2 and bot. op. line. Does NOT go through 
Dy x y . 
 Since D & F, passing streams, Point 1 Dz , y is on op. line. 
 
94 
 
 
Figure for 4D18 
 
4.D19*. B = 0. Then from external balance F = D + B must have D = F = 1000. Acetone balance 
becomes D DFz Dx or x z 0.75 . 
 To predict Bx need operating lines. Top: D
L L
y x 1 x
V V
 
 D
L L V 2
 and y x x .75
V 1 L D 3
 
95 
 
 Bottom: L V 1.0 . Thus y = x is operating line. From Figure Bx 0.01 to 0.02 
 Feed line can be calculated but is not needed. 
 
 
 
4.D20*. To use enthalpy composition diagram change to wt. fractions. Basis = 1 kg mole 
 
 Distillate: 0.8 ETOH = (.8)(46.07) = 36,856 
 0.2 Water = (.2)(18.016) = 3.6032 
 Total = 40.459 
 Weight Fractions: EtOH = .911, Water = .089 
 
 Feed: 0.32 (EtOH) = (.32)(46.07) = 14.7424 
 0.68 (W) = (.68)(18.016) = 12.25088 
 Total = 26.993 
 Weight Fractions: EtOH = .546, W = .454 
 
 Bottoms: 0.04 EtOH = (.04)(46.07) = 1.8428 
 0.96W = (.96)(18.016) = 17.295 
 Total = 19.1378 
 Weight Fractions: EtOH = .0963, W = .9037 
 
 Condenser Energy Balance is 1 1 c o o DV H Q L h Dh which can be solved for oL D . 
 o c
o 1
L Q
1
D D h H
 
 From chart: D 1h 54 Kcal/kg and H 285 Kcal/kg 
 Need D in weight units. Convert feed to weight units. 
 Ethanol: 
100 kgmoles
.32 46.07 1474.24 kg/hr
hr
 
 Water: (100)(.68)(18.016) = 1225 kg/hr 
 Total: F = 2699.328 kg/hr 
96 
 
 Then, B
D B
F z x 2699.328 .0963
D 1489.93
x x .911 .0963
 kg/hr 
 Then, o o 1
2,065,113
L D Q D h H 1 1 5.0
1489.98 54 285
 
 Now do usual McCabe-Thiele analysis using molar units. Note oL D is the same in mass 
and molar units. 
 Top Operating Line: D
L L L L D
y x 1 x and 
V V V 1 L D
 
 D D
L 5 L 5
; x y x .8, y int ercept x 0 1 x 1 .8 .133
V 6 V 6
 
 Feed Line: Goes through y = x = z = .32 
 Weight fraction of feed = .546. Then, 
1f v 1k
h 15 kcal/kg, H 430.7, and h 69 . 
 V F
V L
430.7 15H h
q 1.149
H h 430.7 69
 Slope 
q 1.149
7.711
q 1 1.149 1
 
 Bottom Operating Line: B
L L
y x 1 x
V V
. Goes through Bx y x and 
intersection top operating line and feed line. 
 
 From Figure need about 8 equilibrium contacts including a reboiler. Stage 1 above reboiler is the 
optimum feed stage location. 
 
 
97 
 
4.D21. Feed 1: 1q 0 , slope feed line = 0 
 Feed 2: 2q 0.9 , slope 2 2q q 1 0.9 0.1 9 
 Top: L L D 1.375 0.579
V 1 L D 2.375
, D
L L
y x 1 x
V V
. 
 Goes through Dy x x , When x = 0, D
L
y 1 x 0.40
V
 
 Bottom: Since 1F is saturated vapor, 1V F 100 kmoles/hr 
 
 L L V V 0.579 108 62.532 
 D V L 108 45.46 
 1 2B F F D 100 80 45.46 134.532 
 
 But B is saturated liquid. L B 134.532 
 Check 2L L .9F 62.532 0.9 80 134.532, OK 
 Draw top op line. Intersects with 2F feed line. Then draw bottom op line with slope 
 L V 1.3453 . 
 Intersection bottom op & q. line gives Bx 0.09 . 
 Check 1 1 2 2 D B B
F z F z Dx 20 36 43.187
x or x 0.095
B 134.532
, OK 
 Check External MB 
 1 2180 F F D B 45.46 134.53 179.99 , OK 
 1 1 2 2 D BFz F z Dx Bx 
 56 20.0 36.0 45.46 0.95 134.53 0.095 55.97 OK 
 See McCabe-Thiele diagram: Optimum feed = 5, 7 equilibrium stages (6.65) more than 
sufficient. fraction ab/ac 0.65 
 
2.1 F
 
L 
V 
2.9 F 
At feed 2F 
2V V 0.1F 100 8 108
 
2L L 0.9 F
 
98 
 
 
4.D22*. Around top of column mass balances are: L D V C and D WLx Dy Vy Cx 
 Solving, D w
L D C
y x y x
V V V
 
 For pure entering water, Wx 1.0 . With saturated liquid entering, L = C. Then from overall 
balance, V = D. Thus L/V = C/D = ¾ and D/V = 1.0. Operating equation becomes 
 y 0.75x .92 .75 0.75x .17 
 Slope D D0.75, y x 0 0.17, y x 1 0.92 y , y x 0.68 yNot 
 
99 
 
4.D23*. Note L/V ≠ C/D since C is subcooled. Let c = amount condensed. The energy required to 
heat stream W to the boiling point must come from this condensation. That is, 
 WH h c h h C 
 pW
C T 18 212 100h h
c C C 0.1154C
H h 17465.4
 
 L C c 1.1154C 
 V D c D 0.1154C 
 
 In addition, C/D = ¾ or D/C = 4/3. 
 L 1.1154C 1.1154 1.1154 0.77
V D .1154C 4 3 .1154 4 3 .1154
 
 This compares to L/V = 0.75 if entering water is a saturated liquid. Very little effect since λ 
is very large. 
 L L D 3.25 0.7647
V 1 L D 4.25
 
 Goes through Dy x y 0.85 
 When Dx 0, y 1 L V y 0.20 
 Bottom Op Line: B
L L
y x 1 x
V V
 
 Through Dy x x 0.05 and intersects top op line @ feed line 
 Opt. Feed is #4 below partial condenser – see diagram. Need 6 equil stages + P.C. (an equil. 
contact) 
 
 Note – Commercial columns usually operate much closer to minimum reflux ratio and have 
many more stages. 
 
 b.) 
min
0.85 0.376
L V 0.558
0.85 0
, Min
min Min
L VL
1.26
D 1 L V
 
 
 c.) Total reflux 5 stages + PC sufficient or 
 4 3 4 equil contacts + PC = 5 3 4 eq. contacts 
 where fraction 
ab 7.6mm
0.74
ac 10.3mm
 or .75 
 
100 
 
 
101 
 
 
 
 
 
4.D25. a. 99.9% methanol is essentially pure. Pure MeOH boils 64.5°C. 
 Eq. (4-66) c o1
2 c o
1 f L DL
V 1 1 f L D
 where c PL BP refluxf C T T 
 For pure MeOH, 5PLC 0.07586 16.83 10 T , average (40 + 64.5)/2 = 52.25°C 
 PLC 0.084654kJ gmole , MeOH BP35.27 kJ mole , T 64.5 
 cf 0.084654 24.5 35.27 0.058804 , 1
2
1.058804 1.2L
0.5596
V 1 1.058804 1.2
 
102 
 
 b. L L D 1.2 0.5454
V 1 L D 2.2
or 2.59% more reflux with 24.5°C cooling! Usually 
subcooling not important. 
4.D26. a) 50% feed: q L L F, L L amt vaporized L F 20 
 
1F
q L L F
20 20
, Slope = q 1 20 0.05 0.0476
q 1 1 20 1 1.05
 
 35% feed: Saturated liquid – vertical feed line. Plot both feed lines. The one with lowest 
intersection point with equilibrium curve will normally control 
min
V B 
 Find 
max min
max
V 1
L V . Then V B
L V L V 1
 
 
max
1 0.1
L V slope 2.497
0.46 0.1
, 
min
V 1
0.6681
B 2.497 1.0
 
 b) 
min
L V 3 V B 2.0043 
 
V B V B 1 3.0043
L V 1.49892
V V B 2.0043
 slope 
 Bot. B
L L
y x 1 x
V V
. Goes through By x x with total reboiler 
 B B
L
y 1, x 1 L V x or x 1 L V 1 x / L V
V
 
 
1
x y 1 1 0.49892 0.1 0.6338
1.49892
 
 Intersects feed line with 50% feed first. 
 Middle operating line: Do mass balance around bottom of column.Mass balance intersects 
streams L & V (in column), F50% and B. 
 50% 50% ByV L x F z Bx 
 50 50 B
F z BxL
y x
V V
 
Intersects bottom operating line & 50% feed line. 
 @ 50 50 Bx 0, y (F z Bx ) / V , Slope L V 
 For 50% feed, 50 50q 0.05 (L L ) / F . 
 External balances: 1 2250 F F D B and 1 50 2 35 D BFz F z Dy Bx 
 Find D = 103.333 and B = 146.666 moles/min 
 Since V B 2.0034, V 293.96 and L V B 440.63 
 Then from 50% 50 50q : L q F L 0.05 100 440.63 445.63 
 50V L F B 445.63 100 146.666 398.964 , Slope L V 1.11698 
 y intercept (x = 0), y [ 100 .5 146.666 0.1 ] / 398.964 0.0885 
 Top Intersects feed line for 35% feed and middle op. line and goes through 
Dy x y 0.85 
103 
 
 
 
Figure for 4D26 
 
actual 
EQ 
op 
eq. change 
Stages 
ML
actual change liquid
0.75 = E =
change at equilibrium
 
104 
 
4.D27*. Top Op. Eqn: D
L L
y x 1 x
V V
 
 D D
L L D 1 L
, y intercept 1 x .46, y x x .92
V 1 L D 2 V
 
 Feed: 
1
L F F L
L L q 5 44q 5 4 , slope 5
F F q 1 1 4
, y x z .48 
 Middle: V B L S , BB s s
BxL
V y Bx L x Sy y 0 or y x
V V
 
 Does not intersect at By x x or at By 0, x x . Does intersect top op. line at feed line. 
Need another point. 
 Bottom: B
L L
y x 1 x
V V
, B
L V B V B 1 1.5
3, y x x 0.08
V B V B .5
 
 The steam is another feed to the column: Sat’d vapor q = 0, q/(q-1) = 0, y = x = z = ys = 0. 
 Middle Op. Line intersects this steam at bottom op. line (see figure). 
 
 This problem is a two-feed column with the lower feed (steam) input at a non-optimum feed 
stage. Otimum feed is 3rd above partial reboiler. Need 5.6 equilibrium stages plus PR. 
 
 
 
 
4.D28. This problem was 4.D35 in 2nd edition. 
Stripping Section: B B
L L
y x 1 x , y x x 0.02
V V
. Feed line is vertical 
105 
 
 
max
min max
1.0 0.02 V V 1 1
L V 1.24, 4.167
0.81 0.02 B L V 0.24L V 1
 
 
min
V V V B V B 1 6.25 1
1.5 6.25, L V 1.16
B B V V B 6.25
 
 Op line
B
L
1 1 x
1 .16 .02V
y 1, x= 0.865
L V 1.16
. Intersection Op & feed lines is Dy 
 Overall Balance: 10,000 = F = D + B 
 D B6000 Fz Dy Bx D .695 B .02 6000 .695D .02D 200 
D 5800 / 0.675 8592.6 kgmoles/day, B 1407.4 . Need a use for impure distillate. 
 
Figure for 4.D28 
106 
 
 
4.D29. D
L L 1
L D 3, 3 4 , 1 x 0.9 0.225
V V 4
 
 Step off 3 stages on top op line. Find Sx 0.76 . Point on top op line at Sx 0.76 is also 
on middle op. line 
 
 
 
 In top: L = 3D = 150.375 and V = L + D = 200.500 
 
 Since saturated liquid withdrawn, V ́= V = 200.500 
 and L ́= L – S = 150.375 – 15 = 135.375 
 
 Middle op. line slope L V 135.375 200.5 0.6752 
 
 Middle Op line: S DyV L x Sx Dx 
 S D
Sx DxL 11.4 45.1125
y x , when x 0, y 0.2819
V V 200.5
 
 
 Feed z = 0.6, 20% vapor = 80% liquid q = 0.8. 
 Feed line slope q q 1 0.8 0.4 4 
 
xD 
10 
xs, S = 15 
V´ 
 z = .6 
3 
6 F = 100 
L´ 
 x6 = .1 
F = D + B + S 
D B SFz Dx Bx Sx 
Solve simultaneously D = 50.125, B = 34.875 
107 
 
 
4.D30*. a. Subcooled Reflux: o o 1 o 2 1
1 o
L L D 3
, L L c and V V c
V 1 L D 4
 
 o
1
c L 500
3
, 
o 1
o1
o2
1 o
1
4
L V
4 L 3L 3
1 L1V V L 1
3 3 V
 
 Substituting in values, we have 1
2
L 1.0 1 4
L 1 1 4 5 4 5
, Top op. line Dy x x 0.92 
 Step off two stages. 2 S Sx 0.62 x y 
 
108 
 
 Mixed feed to column: MF S F 1500 , S M MFz Sy F z 
 Solving for mixed feed, Mz 0.52667 
 Energy balance for mixed feed, F S M FMFh SH F H 
 Since s FH h (sat’d vapor), FM S Fh H h (sat’d vapor), and FMq 0 (horizontal). 
 External Balances: F = D + B, D BFz Dx Bx , Solving simultaneously D = 500 & B = 500 
 o o
4
L 3D 1500, and L L 2000
3
 (subcooled reflux), V L D 2500 
 Middle: L L S 2000 500 1500 , V V 2500 , Slope 
L 1500 3
.6
V 2500 5
 
 Intersects Top Op. line at Sx 
 Plot Bot. from By x x to intersection feed line and middle 
 Step off stages (see figure). Need 4 8/9 equilibrium stages. 
 
b. Mass balances for mixed feed injection: MV F V 
 MV V F 2500 1500 1000 , V B 1000 500 2 
 
 
4.D31*. Was problem 4.D36 in 2nd edition. Solution is trial & error. Need to pick L/V. Final answer 
shown in figure. 
 
L .63 .385 0.245
.389
V .63 0.63
 
 
L V L 0.389
.636
1 L V D 0.611
 
 Note feed stage is not optimum. 
109 
 
 
 
 
Figure for problem 4D31 
 
4.D32*. External Balance: F = B = 50, and BFz Bx . Thus Bx z 0.4 . 
4.D33*. a. 
min
.75 .452
L V 0.397
.75
 (tangent pinch) 
 
min min
L L V
0.659
D 1 L V
 
 b. 
act
L D
L D 1.318. L V 0.569
1 L D
 
 Top operating Line through Dy x y 0.75 
 Bottom through By 0, x y 0.1 and intersection feed and top operating lines. 
 Feed: L L F .25F, q 5 4, slope q q 1 5 
 Optimum feed is 3rd from bottom. Need 9 real stages plus partial condenser (see figure). 
c. From figure slope of bottom operating line L V 2.025 
 Since saturated steam and CMO valid, B S L V 
 Also have mass balances, S + F = B + D 
 S B D sSy Fz Bx Dy y 0 
 Solve 3 eqs. simultaneously. S = 760 lbmoles/hr = 13,680 lb steam/hr. 
 
110 
 
 
4.D34*. Trial and Error, Feed: L L F F 2, q 3 2, Slope 3. 
 
Figure for 4.D34 
4-E1. Find (L/V)min (see diagram) 
 
min
0.95 0.613
L V 0.3547
0.95 0
, min
min min
L VL
0.5497
D 1 L V
 
111 
 
 
act min
L L
2 1.0994
D D
, 
act
L L D
0.5237
V 1 L D
 
 External M.B. F = D + B, D BFz Dx Bx 
 or Eq. (4.3) B
D B
z x 0.6 0.025 kgmol
D F 100 62.162
x x 0.95 0.025 h
, B = F – D = 37.838 
 
 Top op line D
L
y x 1 L F x normal
V
. At x = 0, y = 0.4525 
 1st middle B
L L
y x 1 x
V V
 looks like usual bottom! 
 Goes through By x x , and intersection top & feed line. 
 Slope 
L 168.030
1.2906
V 130.192
 
 At pump-around return, V V 130.192 
 L L P 208.030, L V 208.030 130.192 1.5979 
 At pump-around removal, V V 130.192 , L L P L 168.030 
 Check at bottom L V B or 130.192 168.030 37.838 , OK 
 Bottom Op line B
L L
y x 1 x
V V
, Same as first middle!!! 
 Step off P.R. stage 1 & 2 above. Px is liquid from stage 2, Px 0.335 . Vertical line at Px is 
withdrawal line for pump-around and it is feed-line for return of pump-around. 2nd middle op line 
slope L V intersects Px withdrawal & feed line where bottom & 1
st middle intersect. 
L L D D 68.030 
V = L + D = 130.192 
L L F 168.030 
V L
L F 
V
 
L´ 
L″ 
L
 
V″ 
V´ 
P = 
Px 
B 
D 
At feed V V (sat’d liquid) 
112 
 
 Using MB: P ByV Px Bx L x , 
B
P
BxL P
y x x
V V V
 
 When BP
32838 0.025BxP 40
y 0, x x .335 0.0690
L L 208.030 208.030
 
 Draw, 2nd middle – Step off stage 2 & start 3. 3 is op loc. for feed and where pump-around is 
returned. Need PR + 6 equil stages. (Actually 5 and a large fraction) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
4-E2*. Feed: 
L L q
q , L L 1.5F, q 3 2 , slope 3
F q 1
 
 Bottom op. line: Since steam is saturated vapor S V and B L 
 Thus, 1 S B L V 1.2 . Operating line goes through By 0, x x 0.015 
 Middle op. line: V B Side L S or V L S B Side 
 B side SV y Bx Side x L x Sy 
 Since B sideS
Bx Side xL
y 0 this is, y x
V V
 
 Side stream is removed as a saturated liquid so q = 1. 
 Step off two stages (see figure) and find sidex 0.0975 
113 
 
 Find slope: V S, B L Side 
 L Side B Side B 1 0.4 1 1.68 slope
V S S B 0.833
 
 Draw in the middle operating line. Step off 4 stages. Trial and error to find Dx .85 (see figure 
for final result). 
 
 
 
4-E3*. This column has 4 sections. The exact shape is not known ahead of time. Plot top operating line 
 D
L L D 1.86 L L
0.650, y x 1 x
V 1 L D 2.86 V V
 
 D D
L
1 x .35 .8 .28, y x x 0.8
V
 
 Step off 8 stages and find S Sx 0.495 y . Feed line for this vapor is horizontal. Feed line for 
feed to column is vertical at z = 0.32. From figure the feed is injected below the liquid 
withdrawaland above vapor stream from intermediate reboiler. Can now calculate flows in each 
section of columns. 
 Overall Balance: D BFz Dx F D x 
114 
 
 B
D B
F z x .3 1000
D 385
x x .78
 
 Flows: 
L
L D 716.1, V L D 1101.1 V V , V V S 643.8
D
 
 L L S 258.8, L V 0.235 
 L L L F 1258.8, L V 1.143 
 L V 1.955 (this is a check) 
 To plot: From stage 8 draw line of slope L´/V .́ From intersection of first intermediate operating 
line and feed line draw line of slope L″/V″. Draw line from intersection of second intermediate 
operating line with line sy y to By x x 0.02 . Check if slope L V 1.955 . Optimum 
feed is 10th below condenser while vapor from intermediate reboiler is returned on 11th stage. 
Need 12 ½ stages. 
 Note: Small differences in stepping off stages may change column geometry. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
4.E.4. New Problem in 3rd Edition. External Balance. 
 a) D BF D B, Fz Dx Bx 
 B
D B
z x .25 .025
D F 100 25.7 kmol h
x x .9 .025
, B 74.3kmol h 
 b) V B 1.0, V B 74.3 
 L V B 148.6, L V 2.0 
 At feed, amount condensed = C = F/10 = 10 
 
1
L L F C L F F 148.6 110 38.6
10
 
 V V C 74.3 10 64.3 
115 
 
 At stage 2 RL L L and V V 64.3 
 0L L 21.4 
 
L 21.4
0.333 1 3
V 64.3
 
 RL L L 38.6 21.43 17.17 kgmoles hour 
 c) Top Op. line: R D D
L DL L L
y x x x 1 x
V V V V
 
 
 
 
 Plot 
 Middle: D D
L 38.6
y x 1 L V x . Slope 0.600 y x x 0.9
V 64.3
 
 Bottom: By L V x L V 1 x 
 By x x 
 
V B 1
Slope L V 2
V B
 
 Now have somewhat redundant information. 
 Can plot bottom. 
 Intersection bottom and feed line should also be on middle. – Or use this pt to find middle or 
bottom op. line. 
 
 From graph: Opt. Feed = #4. 
 Need 6 stages + P.R. 
 
 
 
 
 
 
 
 
 
DSlope L V 1 3, y x x 0.9 RL
 
V L 
D 
0L 
Envelope for top 
R
D D R
L D
y x, y x x sin ce V L L D
V L
 
D
L
x 0 , y 1 x 0.6
V
 
116 
 
Graph for 4.E.4. 
 
 
4.E.5. New Problem in 3rd Edition. 
L V B V 1 2.25 9 4 9
V V V B 1.25 5 4 5
 
 Bot. Op. Line: B B
L L
y x 1 x , y x x , Slope L V
V V
 
 External M.B.: D BF D B and Fz Dx Bx , B
D B
z xD .3 0.10 .2 1
F x x 0.90 0.10 .8 4
 
 D 250 kmol day , B 750 kmol day , V V B B 1.25 750 937.5 kmol day 
 L V B 937.50 750 1687.5 kmol day 
 At feed stage: V V 937.5, L L F or L L F 1687.5 1000 687.5 
 
L 687.5
0.733333
V 937.5
 
117 
 
 MB: DL x Dy V y, D V L , D
L L
y x 1 y
V V
 
 Goes through Dy x y , and intersects feed and bot. op. lines. 
 At side withdrawal: 687.5 L L , V S V or V 937.5 200 737.5 
 Op line by intermediate condenser: L D V S 
 D SD S
Dy SxL
L x Dy V y Sx or y x
V V
 
 Find from intersection L V op line @ Sy y plus 
687.5
slope 0.932
737.5
or intersection 
top op line and Sx x 
 At side stream feed point: L L S 487.5, V V 737.5. Thus, 
L 487.5
0.661
V 737.5
 
 Top op. line: D
L L
y x 1 y
V V
Can draw, int ercepty 1 .661 .9 0.305 
118 
 
 
Graph for 4.E5. 
4-F1. 
 
S 
V 
B 
L
 
970.33 Since bottoms are very pure B waterh h @ 212°F 
 SH 1381.4 
EquilH 1192 (in column) 
 Extra heat 189.4 Btu/lb 
 Since y ≈ x, MW are same 
v 
119 
 
 Must vaporize material in column. 
 
extra heat MW 189.4
v S S 0.1952S
MW 970.33
 
 B L v, V S v 
 
L B v B S v S 2 0.1952
1.837
V S v 1 v S 1 0.1952
 
 If super heat not included L V B S 2 , which is incorrect. 
 
4-F3*. An approximate check is to compare molar latent heats of vaporization. Data is available in 
Perry’s and in Himmelblau. 
 a. See Example 4-4. 
b. isopropanol λ = 159.35 cal/g. MW = 60.09, λ = 9.575 kcal/mole. Water λ = 9.72 kcal/mole. 
CMO is OK. 
c. CMO is not valid. AA W5.83 kcal / gmole, 9.72 
d. nC4. λ = 5.331 kcal/mole, MW = 58.12. λ = 0.0917 kcal/g 
 nC5. λ = 6.16 kcal/mole, MW = 72.15. λ = 0.0854 kcal/g 
 Constant mass overflow is closer than constant molar overflow. 
e. benzene. λ = 7.353 kcal/mole, MW = 78.11, λ = 0.0941 kcal/g 
 toluene. λ = 8.00 kcal/mole, MW = 92.13, λ = 0.0868 kcal/g 
 CMO is within about 6%. 
 
4G1. a*. Answer should be close, but not identical, to result obtained in Example 4-4. 
 
4G2. Was 4.G4 in 2nd edition. Used Peng-Robinson. CQ 44,437,300 Btu/hr, 
 RQ 49,859,400 Btu/hr. 
 Optimum FN 17 & N 27 (Total condenser is #1) 
 D Bx 0.9992 and x 0.00187 . 
4G3*. See answers to selected problems in back of book. 
 
4.H.1. New Problem in 3rd Edition. Use VBA program in Appendix B of Chapter 4. 
xd 0.995 xb 0.011 F 250 z 0.4 
L/D 3 q 0 feed stg 4 
partial reboiler total condenser Ethanol-water Prob. 4H1. 
VLE 6th 5th 4th 3rd 2nd 1st constant 
-24.75 85.897 -118.03 82.079 -30.803 6.6048 0 
yeqatxint 0.584177 yint 0.4 xint 0.2016667 
stage x y 
1 0.011 0.069033 
2 0.039445 0.217358 
3 0.112146 0.451872 
4 0.227091 0.607193 
5 0.477924 0.769849 
6 0.694798 0.867005 
7 0.824341 0.919132 
8 0.893842 0.954863 
9 0.941484 0.979257 
120 
 
10 0.974009 0.992216 
11 0.991288 0.996557 
 
4.H.2. New Problem in 3rd Edition. The VBA program is in Appendix B of Chapter 4. With L/D = 6.94 
the result is: 
xd 0.7 xb 0.0001 F 1000 z 0.1 
L/D 6.94 q 0 feed stg 28 
partial reboiler total condenser Ethanol-water Prob. 4H3. 
VLE 6th 5th 4th 3rd 2nd 1st constant 
-47.949 161.42 -212.43 138.68 -46.65 7.9322 0 
yeqatxint 0.099219 yint 0.1 xint 0.0135447 
stage x y 
 Reflux rate too low Reflux rate too low 
 
With L/D = 6.95 the result is given below. With feed stages below 85 the feed stage was too low. 
xd 0.7 xb 0.0001 F 1000 z 0.1 
L/D 6.95 q 0 feed stg 85 
partial reboiler total condenser Ethanol-water Prob. 4H3. 
VLE 6th 5th 4th 3rd 2nd 1st constant 
-47.949 161.42 -212.43 138.68 -46.65 7.9322 0 
yeqatxint 0.100057 yint 0.1 xint 0.0136691 
stage x y 
1 0.0001 0.000793 
2 0.000194 0.001538 
3 0.000295 0.002338 
4 0.000404 0.003197 
5 0.000521 0.004117 
6 0.000646 0.005102 
7 0.000779 0.006154 
8 0.000922 0.007277 
9 0.001075 0.008472 
10 0.001237 0.009742 
11 0.00141 0.011089 
12 0.001593 0.012515 
13 0.001786 0.014021 
14 0.001991 0.015608 
15 0.002206 0.017276 
16 0.002433 0.019025 
17 0.00267 0.020853 
18 0.002919 0.022758 
19 0.003178 0.024739 
20 0.003447 0.026791 
21 0.003725 0.02891 
22 0.004013 0.03109 
23 0.004309 0.033327 
24 0.004613 0.035613 
25 0.004924 0.037941 
26 0.00524 0.040302 
27 0.005561 0.042689 
28 0.005885 0.045091 
121 
 
29 0.006211 0.0475 
30 0.006538 0.049907 
31 0.006865 0.052301 
32 0.00719 0.054674 
33 0.007513 0.057017 
34 0.007831 0.059321 
35 0.008144 0.061579 
36 0.00845 0.063782 
37 0.00875 0.065925 
38 0.009041 0.068002 
39 0.009323 0.070008 
40 0.009595 0.071938 
41 0.009858 0.07379 
42 0.010109 0.075561 
43 0.01035 0.077251 
44 0.010579 0.078857 
45 0.010797 0.08038 
46 0.011004 0.08182 
47 0.0112 0.083179 
48 0.011384 0.084459 
49 0.011558 0.085661 
50 0.011721 0.086787 
51 0.011874 0.087841 
52 0.012018 0.088826 
53 0.012151 0.089743 
54 0.012276 0.090598 
55 0.012392 0.091392 
56 0.0125 0.092129 
57 0.0126 0.092812 
58 0.012693 0.093445 
59 0.012779 0.09403 
60 0.012858 0.094571 
61 0.012932 0.09507 
62 0.012999 0.095531 
63 0.013062 0.095955 
64 0.01312 0.096346 
65 0.013173 0.096705 
66 0.013222 0.097036 
67 0.013267 0.09734 
68 0.013308 0.09762 
69 0.013346 0.097876 
70 0.013381 0.098112 
71 0.013413 0.098328 
72 0.013442 0.098526 
73 0.013469 0.098708 
74 0.013494 0.098874 
75 0.013516 0.099027 
76 0.013537 0.099167 
77 0.013556 0.099295 
78 0.0135730.099412 
122 
 
79 0.013589 0.099519 
80 0.013604 0.099618 
81 0.013617 0.099707 
82 0.013629 0.09979 
83 0.013641 0.099865 
84 0.013651 0.099934 
85 0.01366 0.099997 
86 0.013665 0.100032 
87 0.013706 0.100305 
88 0.014018 0.102401 
89 0.016416 0.11824 
90 0.034534 0.223718 
91 0.155188 0.516574 
92 0.490181 0.652848 
93 0.646064 0.724878 
 
4.H.3. New Problem in 3rd Edition. The Spreadsheet is: 
xd 0.7 xb 0.0001 F 1000 z 0.17 
Multiplier 1.05 q 0.5 feed stg 17 
partial reboiler total condenser Ethanol-water Problem 4.H4. 
VLE 6th 5th 4th 3rd 2nd 1st constant 
-47.949 161.42 -212.43 138.68 -46.65 7.9322 0 
L/Dmin 1.687377 L/Vmin 0.62789 L/D 1.771746 L/V 0.639217 
stage x y 
1 0.0001 0.000793 
2 0.000229 0.001812 
3 0.000418 0.003309 
4 0.000696 0.0055 
5 0.001104 0.008697 
6 0.001698 0.013331 
7 0.002559 0.019994 
8 0.003797 0.029452 
9 0.005555 0.042644 
10 0.008006 0.060585 
11 0.01134 0.08415 
12 0.015719 0.113686 
13 0.021208 0.148522 
14 0.027681 0.186647 
15 0.034766 0.224911 
16 0.041877 0.259918 
17 0.048382 0.28916 
18 0.057276 0.325158 
19 0.113592 0.469947 
20 0.340102 0.575494 
21 0.505222 0.659654 
22 0.636883 0.719124 
Because the multiplier is close to 1.0, this answer is very sensitive to the data fit used. 
One solution to the coding is the following VBA program: 
 
Option Explicit 
123 
 
Sub McCabeThiele() 
' Find minimum reflux ratio assuming it occurs at feed plate. Then 
' L/D actual = L/D min times Multiplier. Steps off stages from the bottom up. 
' Assumes that the feed stage is specified. 
 Sheets("Sheet2").Select 
 Range("A8", "G108").Clear 
Dim i, feedstage As Integer 
Dim D, B, xd, xb, F, z, q, LoverD, LoverV, x, y, xint, yint, yeq As Single 
Dim a6, a5, a4, a3, a2, a1, a0, L, V, LbaroverVbar, LoverDmin As Single 
Dim LoverVmin, LoverVdelta, Multiplier As Single 
' Input values from spread sheet 
 xd = Cells(1, 2).Value 
 xb = Cells(1, 4).Value 
 F = Cells(1, 6).Value 
 z = Cells(1, 8).Value 
 Multiplier = Cells(2, 2).Value 
 q = Cells(2, 4).Value 
 feedstage = Cells(2, 8).Value 
' Fit of equilibrium data to 6th order polynomial to find y. a6 is multiplied 
' by x to the 6th power. 
 a6 = Cells(5, 1).Value 
 a5 = Cells(5, 2).Value 
 a4 = Cells(5, 3).Value 
 a3 = Cells(5, 4).Value 
 a2 = Cells(5, 5).Value 
 a1 = Cells(5, 6).Value 
 a0 = Cells(5, 7).Value 
' Calculate intersection point of two operating lines and use this to find 
' minimum L/D and L/V. Initialize 
 LoverV = 1 
 LoverVdelta = 0.00001 
Do 
 LoverV = LoverV - LoverVdelta 
 xint = ((-(q - 1) * (1 - LoverV) * xd) - z) / (((q - 1) * LoverV) - q) 
 x = xint 
 yint = LoverV * xint + (1 - LoverV) * xd 
' Equilibrium y at value of x intersection. When yint=yeq, have minimum L/V and L/D. 
 yeq = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 
Loop While yint < yeq 
'Print intersection and equilibrium values. 
 LoverVmin = LoverV + LoverVdelta 
 LoverDmin = LoverVmin / (1 - LoverVmin) 
 LoverD = Multiplier * LoverDmin 
 LoverV = LoverD / (1 + LoverD) 
 Cells(6, 2).Value = LoverDmin 
 Cells(6, 4).Value = LoverVmin 
 Cells(6, 6).Value = LoverD 
 Cells(6, 8).Value = LoverV 
' Calculate flow rates and ratios. 
 D = ((z - xb) / (xd - xb)) * F 
 L = LoverD * D 
124 
 
 V = L + D 
 LbaroverVbar = (LoverV + (q * F / V)) / (1 - ((1 - q) * F / V)) 
' Step off stages from bottom up. First stage is partial reboiler. Initialize 
 x = xb 
 i = 1 
' Loop in stipping section stepping off stages with equilibrium and operating eqs. 
Do While i < feedstage 
 y = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 
 Cells(i + 7, 1).Value = i 
 Cells(i + 7, 2).Value = x 
 Cells(i + 7, 3).Value = y 
 i = i + 1 
 x = (y / LbaroverVbar) + (LbaroverVbar - 1) * xb / LbaroverVbar 
Loop 
' Calculations in enriching section continues to Loop While y < xd. 
Do While y < xd 
 y = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 
 Cells(i + 7, 1).Value = i 
 Cells(i + 7, 2).Value = x 
 Cells(i + 7, 3).Value = y 
 i = i + 1 
 x = (y / LoverV) - (1 - LoverV) * xd / LoverV 
 If x < 0 Then 
 Cells(i + 7, 4).Value = "Feed stage too low" 
 Exit Do 
 End If 
 If i > 100 Then 
 Cells(i + 7, 6).Value = "Too many stages" 
Exit Do 
End If 
Loop 
End Sub 
 
125 
SPE 3rd Ed. Solution Manual Chapter 5 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 5A15, 5C1, 5D1, 5D2, 5D9, 5D10, 5H1 to 5H5. Problems and solutions from the first 
edition that were not in the second edition are: 5D6, 5D8, 5D11-5D13, 5E1. 
5.A6.
Ethane is less volatile than methane so it decreases toward distillate. At bottoms it is more 
volatile than propane and butane, so must decrease towards bottoms. Thus ethane 
concentrates within column. 
5.A7. 1. c; 2. c; 3. a (saturated liquid feed); 4. c; 5. B
5.A9. Pure LK cannot be withdrawn because LNK is present. Pure LNK can be removed at 
distillate if all LK is removed in side stream. However, recovery of LNK will be < 100%. 
5.A13. If HNKz F (frac. rec. HK in bot) HKz F, then there is more HNK in bottoms and curves 
cross. 
5A15. New Problem in 3rd Edition. a. a, b. d, c. b, d. c, e. b. 
5.C1. New Problem in 3rd Edition. Start with equilibrium equation, yi,j = Ki,jxi,j and multiply right hand
side by Kref,j /Kref,j. One obtains yi,j = αi-ref,j Kref,jxi,j. Then 1 = Σ y = Σ (αi-ref,j Kref,jxi,j) . Since Kref,j is 
constant, bring it outside the summation and solve for Kref,j = 1/ Σ (αi-ref,j xi,j). This is Eq. (5-29). 
126 
 
5.D1. New Problem in 3rd Edition. 
 
 wt frac. in distillate: M = 0.38399, E = 0.61273, P = 0.00327, B = 0.0. 
 
M ,B
B,B
E,B
P,B
i,b
Bx 0 0
Bx 1.0 12, 000 0.23 2760.0
Bx 1 0.978 12, 000 0.31 81.84
Bx 0.994 12, 000 0.27 3220.56
 B x B 6062.4 kg
i,b
wt. frac
x
0
0.4553
0.0135
0.5312
hr
 
 Check B+D=F, OK. 
 
5.D2. New Problem in 3rd Edition. 4 5 6y 0.30, y 0.5, y 0.20, P 760 mmHg 
 Raoult’s Law i i iy P VP x 
 Antoine’s Eqn. 10
B
log VP A
T C
 
 Dew Point condition is ix 1 
 ii
i
y P
x
VP
 (from Raoult’s Law) 
M ,d
B,d
E,d
P,d
Dx 1.00 2, 000 0.19 2280
Dx 0 0
Dx 0.978 2, 000 0.31 3638.16
Dx 1 0.994 12, 000 0.27 19.44
 i ,d D x D 5937.6 kg/h
 
a) 
b) assume NK (Methanol and n-butanol) do not distribute (all 
methanol in top and all butanol in bottom). 
127 
 
 i
B
VP 10 A
T C
 (from Antoine’s Eqn) 
 Combine with Dew Point condition 
 
4 5 6
4 5 6
4 5 6
5 64
B B B
A A A
T C T C T C
y P y Py P
1
10 10 10
 
 
 
935.86 1064.8 1171.17
6.809 6.853 6.876
T 238.73 T 233.01 T 224.41
0.30 760 0.50 760 0.20 760
1
10 10 10
 
 
 Using Goal Seek in Excel, T = 41.3ºC 
 
5.D3. Assume that ethanol is HK. Then assume that HNK’s are totally in the bottoms. 
 M,dist E,distx .99, x .01 
 2195.6 = (.998) (0.22) (10,000) = MDDx 
 
MD
2195.6 2195.6
D 2217.78 and B F D 7782.22
x .99
 
 Bottoms: MeOH: .0021 (.22) (10,000) = 4.4 
 Mbot
4.4
x 0.0006
7782.22
 
 n propanol,bot
1.0 .18 10,000
x 0.2313
7782.22
 
 n butanol,bot
1.0 .13 10,000
x .1670
7782.22
 
 EtOH,bot MeOH,bot n p,bot nbut,botx 1 x x x 0.6011 
 
5.D4. a. .99 F C5 C5,dist C5 C5,botz D x (1), and .01 Fz =B x (2) 
 .98 F C6,bot C6,bot C6 C6,dist z Bx (3), and .02 F z D z (4) 
 Assume all heptane in bottoms 
 C7 bot,C6 dist ,C7Fz Bx (5), x 0 (6) 
 Take Eqs. 1,4 & 6: .99 (1000) (.4) = C5dDx 
 .02 (1000) (.3) = D C6dx 
 0 = D C7,distx 
 l,distDx D 402 kg moles/hr 
 B = F – D = 1000 – 402 = 598 
 b. C7,distx 0 
 C5,d
.99 1000 .4
x 0.9851
402
 
 C6,dx 1 0.9851 0.0149 
128 
 
 C7,bot
1000 .3 1.0
x 0.5017
598
 
 C6,bot
.98 1000 .3
x 0.4916
598
 
 C5,botx 1 .5017 .4916 0.0067 
 c. L = (L/D)D = (2.5) (402) = 1005 
 V = L + D = 1005 + 402 = 1407 
 At feed stage: L = L + .6F = 1005 + 600 = 1605 
 V = V - .4F = 1407 – 400 = 1007 
 
5.D5. Assume all methanol and ethanol in distillate. 
 MeOH,distDx 0.55 100 0.01 150 56.5 
 EtOH,distDx 0.21 100 0.03 150 25.5 
 
 prop,distDx 0.993 0.23 100 0.26 150 61.57 
 bu tan ol,disDx 1 0.995 0.01 100 0.70 150 0.53 
 
 D 144.1 
 1 2B F F D 105.9 
 Check: Prop,botBx 1 0.993 .23 100 .26 150 0.434 
 but,distBx 0.995 .01 100 0.7 150 105.47 
 
 Check = 105.90 
 Mole fractions: 
 M,bot E,bot prop,bot
0.434
x 0 , x 0, x 0.0041
105.90
 
 but,bot prop,botx 1 x 0.9959 
 
 MeOH,distM,dist
Dx 56.5
x 0.392
D 144.1
 
 E,dist
25.5
x 0.1767
144.1
 
 p,dist
61.566
x 0.427
144.1
 
 But,distx 0.53 144.1 0.0037 
 
 Check = 1.000 OK 
 
5.D6. This is 8.D1. in 1st ed. 
 
129 
 
130 
 
 
 
5.D7. Assume 100% recovery 2C , & propylene in distillate. 
 Assume 100% recovery pentane & hexane in bottoms. 
 
Comp. Distillate kg/h 
Dx 
C2 0.3 (1000) + 0.02 (1500) = 330 0.2824 
Propylene: 0.006 (1000) + 0.001 (1500) = 7.5 0.0064 
n-C3 (0.991) [1000 (0.45) + 1500 (0.249)] = 816.0885 0.06983 
 n-C4 (0.02) [1000 (0.154) + 1500 (0.40)] = 15.08 0.0129 
C5 & C6 0 = 15.08 0.0 
 D = = 1168.7 
 
Bottoms flow rate = 1 2F F D 2500 1168.7 1331.3 kg/h 
 
2 3B,C B,propylene B,C
.009 1000 0.45 1500 0.249
x 0, x 0, x 0.0056
1331.3
 
 
4B,C
.98 1000 .154 1500 .40
x 0.5550
1331.3
 
 
5 6B,C B,C
1000 0.09 1500 0.18 1500 .15
x 0.2704, x 0.1690
1331.33 1331.31
 
 
5.D8. 8.D.6. in 1st edition. Assume all benzene is in the distillate. 
131 
 
 
132 
 
 
 
 
 
5.D.9. New Problem in 3rd Edition. 
133 
 
 
5.D10. New Problem in 3rd Edition. At the bubble point iy 1.0 
 C5 C6x .40 , x .60 , p 760 mmHg , C5 C6 5 C5 6 C6y y K x K x 1.0 
134 
 
 or C5 C6C5 C6
tot tot
VP VP
x x 1.0
P P
, C5 C5 C6 C6 totVP x VP x P 
 
1064.8 1171.17
6.853 6.876
T 233.01 T 224.410.40 10 0.60 10 760 
 
TºC (C5 term) x (.4) C6 term x (.6) SUM 
20 441.03 × .4 = 176.4 + 121.387 × .6 = 249.23 
 Final result is: 
51 1270.095 × .4 = 508.04 + 420.28 × .6 = 252.17 760.21 
 
5.D11. Was problem 6.D2 in 2nd edition SPE. Since ix known, want i i BP iy 1 K p x 
 Find new pressure from, ref oldref new
i i
K p
K p
K x
 
 Use ethane, or n-pentane as reference. 
 First guess: Try 
5C
K 1.0, p 115 kPa 
 
7 2NC C
K 0.13 K 29 
 i iK x 29 0.1 1.0 .35 0.13 .55 3.32 p too low. 
 rep new new
1.0
K p ~ 0.3 p 440 kPa
3.32
, 
7 2NC C
K 0.042 K 8.0 
 i iK x 8.0 0.1 0.3 .35 0.042 0.55 0.927 
 
5C new new
0.3
K p .032 p 400
0.927
 kPa, 
7 2nC C
K 0.045 K 8.7 
 i iK x 8.7 0.1 0.32 0.35 0.045 0.55 1.004 
 Answer (within accuracy DePriester Chart) = 400 kPa 
 
5.D12. Was problem 6.D3 in 2nd edition SPE. a. Highest B.P. Temp. is pure n-octane. C8K 1.0, T 174 C 
 b. Lowest B.P. Temp. is pure n-hexane. C6K 1.0, T 110 C 
 
5.D13. Was problem 6.D6 in 2nd edition SPE. Let 1 = n-butane, 2 = n-pentane and 3 = n-hexane. p = 101.3 kPa. 
 Bubble Point: First guess. 1 2 3K 1 at T 1 , K 1 at T 36 , K 1 at T 68 . 
 first 1 1T z T .2 1 .5 36 .3 68 38 
 1 2 3 1 1K 3.6, K 1.08,K 0.36. K x .2 3.6 .5 1.08 .3 .36 1.368 
 Choose 2 as ref. Eq. (6-14) is: 2 newK T 1.8 1.368 0.789 
 new 1 5 i iT 29 C. K 2.7, K 0.26, and K x 1.013 
 Eq. (6-14) is: 2 newK T 0.789 1.013 0.779 
 new 1 8 BPT 28.8 K 2.7, K 0.255, and K x 1.006. OK. T 28.8 C. i i
 
 
5. E1. This is 8.E4. in 1st edition. 
135 
 
 
136 
 
 
137 
 
 
5.H1. New Problem in 3rd Edition. Same program as 5.H5 except do not list y values as distillate. 
Different input in spreadsheet – see below. 
Ternary Distillation: Constant relative volatility. Step off stages from bottom up. Use whole stages. 
System has A = LK, B = HK and C = HNK 
 
C5H9 
 alpha 
A-B 3.58 alpha B-B 1.86 
Alpha C-
B 1 feedstage 8 
z A 0.35 z B 0.4 z C 0.25 epsilon (values for 0.00000001 
F 200 q 1 L/D 6 
N loop( 
convergence 100 
frac rec B in dist 0.996 guess frac rec C bot 1 
df(HNK frac 
recovery) 0.9 
frac rec A in dist 
 
0.961 
 D 67.59011 B 132.4099 L/V 0.857143 Lbar/Vbar 1.279858988 
xAdist 0.995264 xBdist 0.004734 xCdist 1.56E-06 
Mass balance 
values 
 
xAbot 0.020618 xBbot 0.601768 xCbot 0.377614 
Mass balance 
values 
 
138 
 
Stage by stage calculations 
 i xA yA xB yB xC yC Note x1 = 
1 0.020618 0.046992 0.601768 0.7125981 0.377614 0.240409437 xbot 
2 0.041225 0.0869 0.688364 0.7538807 0.270411 0.159219617 
 3 0.072406 0.143487 0.720619 0.7419438 0.206975 0.114569647 
 4 0.11662 0.218289 0.711292 0.6917342 0.172088 0.089976459 
 5 0.175066 0.308791 0.672062 0.6158891 0.152873 0.075319919 
 6 0.245778 0.407144 0.612801 0.5274175 0.141421 0.0654388 
 7 0.322624 0.502187 0.543675 0.4396808 0.1337 0.058132319 
 8 0.396885 0.584094 0.475123 0.3632905 0.127992 0.05261578 
 9 0.515565 0.685129 0.42305 0.2920855 0.061385 0.02278591 
 10 0.633439 0.774848 0.339977 0.2160685 0.026583 0.009083172 
 11 0.738112 0.846817 0.251291 0.1497869 0.010597 0.003395923 
 12 0.822076 0.899855 0.173962 0.098934 0.003962 0.001211305 
 13 0.883953 0.936484 0.114634 0.0630978 0.001413 0.000418127 
 14 0.926687 0.960636 0.072825 0.0392226 0.000488 0.000141178 
 15 0.954865 0.97607 0.044971 0.0238835 0.000164 4.69552E-05 
 16 0.97287 0.985732 0.027075 0.0142529 5.45E-05 1.54307E-05 
 17 0.984143 0.991702 0.015839 0.0082925 1.77E-05 4.9941E-06 
 18 0.991109 0.995362 0.008886 0.0046363 5.57E-06 1.56158E-06 
 
 
 Calc frac recovery C in bottoms 
 
0.9999979 j 3 
 5.H2. New Problem in 3rd Edition. The spreadsheet is, 
Ternary Distillation with Constant relative volatility. Step off stages from top down. 
System has A = LNK, B = LK and C = HK 
alpha A-B 2.25 alpha B-B 1 Alpha C-B 0.21 feedstage 6 
z A 0.25 z B 0.35 z C 0.4 epsilon (values for 0.0001 
F 100 q 1 L/D 0.3 N loop( convergence 10 
frac rec B in dist 0.9 frac rec C in bot 0.97 df(LNK frac recovery) 0.9 
Guess: frac rec A in dist 1 
D 57.66689 B 42.35298 L/V 0.230769 Lbar/Vbar 1.565105 
xAdist 0.43295 xBdist 0.546241 xCdist 0.020809 Mass balance values 
xAbot 0.001251 xBbot 0.082639 xCbot 0.91611 Mass balance values 
Stage by stage calculations 
 i xA yA xB yB xC yC Note y1 = 
1 0.229688 0.43295 0.65203 0.546241 0.118282 0.020809168 xdist 
2 0.180904 0.386044 0.601681 0.570654 0.217416 0.043302907 
3 0.16005 0.374786 0.537147 0.559034 0.302804 0.066179941 
4 0.147137 0.369973 0.486906 0.544142 0.365957 0.085884852 
5 0.13893 0.366993 0.453607 0.532548 0.407464 0.100458716 
6 0.133981 0.365099 0.433372 0.524864 0.432647 0.1100371 
7 0.062603 0.208987 0.425676 0.631573 0.511721 0.159440004 
8 0.021494 0.097273 0.308019 0.619528 0.670486 0.283199293 
9 0.004909 0.032934 0.146011 0.435383 0.84908 0.531682943 
10 0.000766 0.006976 0.044919 0.181823 0.954315 0.811200627 
 
 
Calc frac recovery A in distillate 0.998702 j 2 
139 
 
5.H.3. New Problem in 3rd Edition. (L/D)min = 0.26761 by trial and error using spreadsheet in Table 5.A-
1.. 
 
5.H.4. New Problem in 3rd Edition. 
Ternary Distillation with Constant relative volatility. Step off stages from top down. 
System has A = LK, B = sandwich and C = HK 5.G.e. 
alpha A-B 1.4 alpha B-B 1 Alpha C-B 0.7 feedstage 19 
z A 0.25 z B 0.35 z C0.4 epsilon (values for 1E-10 
F 100 q 1 L/D 5 N loop( convergence 100 
frac rec B in dist 0.583 frac rec C in bot 0.995 df(LNK frac recovery) 0.8 
Guess: frac rec A in dist 0.95 
D 44.10818 B 55.89182 L/V 0.833333 Lbar/Vbar 1.21119 
xAdist 0.532853 xBdist 0.462613 xCdist 0.004534 Mass balance values 
xAbot 0.026781 xBbot 0.261129 xCbot 0.71209 Mass balance values 
Stage by stage calculations 
 i xA yA xB yB xC yC Note y1 = 
1 0.447934 0.532853 0.544443 0.462613 0.007623 0.004534307 xdist 
2 0.378937 0.462087 0.609404 0.530804 0.011659 0.007108533 
3 0.325116 0.40459 0.658055 0.584939 0.016829 0.010471366 
4 0.284385 0.359739 0.692247 0.625481 0.023368 0.01477984 
5 0.254167 0.325796 0.71427 0.653975 0.031563 0.020228923 
6 0.231958 0.300615 0.726286 0.672327 0.041757 0.027058086 
7 0.215598 0.282107 0.730061 0.68234 0.054342 0.035552935 
8 0.203353 0.268473 0.726901 0.685486 0.069746 0.046040694 
9 0.193893 0.258269 0.717703 0.682853 0.088403 0.058877558 
10 0.18623 0.250387 0.703059 0.675188 0.110711 0.074425299 
11 0.179649 0.244001 0.683384 0.662985 0.136967 0.093014635 
12 0.173652 0.238516 0.65905 0.646589 0.167298 0.114894592 
13 0.167915 0.233519 0.630501 0.62631 0.201584 0.140170911 
14 0.162254 0.228738 0.598352 0.602519 0.239393 0.168742359 
15 0.156599 0.224021 0.563437 0.575729 0.279964 0.200250121 
16 0.150964 0.219308 0.526798 0.546633 0.322237 0.234059083 
17 0.145428 0.214612 0.489617 0.516101 0.364955 0.269286831 
18 0.140099 0.209999 0.453098 0.485116 0.406803 0.304884695 
19 0.135091 0.205558 0.418339 0.454684 0.446571 0.339758041 
20 0.100543 0.157965 0.402362 0.45154 0.497094 0.39049513 
21 0.071479 0.116121 0.372448 0.43219 0.556074 0.45168881 
22 0.048123 0.080918 0.329669 0.395957 0.622209 0.523124245 
23 0.030232 0.05263 0.276757 0.344144 0.693012 0.603226557 
24 0.017191 0.03096 0.2177 0.280057 0.765109 0.688982619 
Mass balance: fraction stage, A, B, C calculated at bottom, % error B 
0.264616 0.026781 0.261129 0.71209 6.65242E-09 
Calc frac recovery A in distillate 0.940127 j 4 
 
 
Part d. Fractional recovery of B in distillate. is 0.725. Note that B goes through a maximum of close to 
1% on stages 7 and 8. 
Ternary Distillation with Constant relative volatility. Step off stages from top down. 
System has A = LK, B = trace sandwich and C = HK 5.G.e. Part d. 
alpha A-B 1.4 alpha B-B 1 Alpha C-B 0.7 feedstage 19 
140 
 
z A 0.38 z B 0.02 z C 0.6 epsilon (values for 1E-10 
F 100 q 1 L/D 4 N loop( convergence 100 
frac rec B in dist 0.725 frac rec C in bot 0.995 df(LNK frac recovery) 0.8 
Guess: frac rec A in dist 0.99 
D 39.37044 B 60.62956 L/V 0.8 Lbar/Vbar 1.308 
xAdist 0.95555 xBdist 0.03683 xCdist 0.00762 Mass balance values 
xAbot 0.00626 xBbot 0.009071 xCbot 0.984668 Mass balance values 
Stage by stage calculations 
 i xA yA xB yB xC yC Note y1 = 
1 0.934659 0.95555 0.050434 0.03683 0.014907 0.00761993 xdist 
2 0.909255 0.938837 0.064694 0.047713 0.026051 0.013449322 
3 0.878109 0.918514 0.079128 0.059121 0.042762 0.02236485 
4 0.839847 0.893598 0.092985 0.070669 0.067168 0.035733649 
5 0.793216 0.862987 0.105202 0.081754 0.101582 0.055258739 
6 0.737611 0.825683 0.114471 0.091527 0.147918 0.082789801 
7 0.673775 0.781199 0.119471 0.098942 0.206753 0.119858439 
8 0.604361 0.73013 0.119295 0.102943 0.276345 0.166926476 
9 0.53382 0.674599 0.113888 0.102802 0.352293 0.222599678 
10 0.467334 0.618166 0.104227 0.098476 0.428438 0.283358088 
11 0.409234 0.564978 0.092025 0.090748 0.498741 0.344274648 
12 0.361849 0.518497 0.079126 0.080986 0.559026 0.400517103 
13 0.325379 0.480589 0.066982 0.070666 0.607639 0.448744495 
14 0.298551 0.451414 0.056436 0.060951 0.645013 0.487634893 
15 0.279454 0.429951 0.047786 0.052515 0.67276 0.517534464 
16 0.266162 0.414673 0.040972 0.045595 0.692866 0.539732259 
17 0.257043 0.40404 0.035754 0.040143 0.707203 0.555816779 
18 0.250839 0.396745 0.031838 0.035969 0.717324 0.567286133 
19 0.246633 0.391781 0.028939 0.032836 0.724428 0.575383054 
20 0.193371 0.320667 0.029598 0.035058 0.777031 0.644274899 
21 0.145301 0.251 0.029111 0.03592 0.825588 0.713080242 
22 0.105056 0.188125 0.027585 0.035283 0.867359 0.776592117 
23 0.073452 0.135485 0.025264 0.033287 0.901284 0.831228269 
24 0.049874 0.094147 0.022436 0.030252 0.92769 0.875601733 
25 0.032959 0.063306 0.019353 0.026552 0.947688 0.910141552 
26 0.02117 0.041182 0.016207 0.02252 0.962623 0.93629802 
27 0.013123 0.025762 0.013125 0.018405 0.973752 0.955833007 
28 0.00771 0.015236 0.010183 0.014373 0.982107 0.970390508 
29 0.004108 0.008156 0.007421 0.010525 0.988471 0.981318291 
Mass balance: fraction stage, A, B, C calculated at bottom, % error B 
0.402462 0.00626 0.009071 0.984668 5.53216E-07 
Calc frac recovery A in distillate 0.990012 j 16 
 
5.H.5. New Problem in 3rd Edition. 
Ternary Distillation with Bubble point Calc. Step off stages from bottom 
up. 
System has A = LK, B = HK and C = HNK 
 feedstage 5 
z A 0.3 z B 0.3 z C 0.4 epsilon 1E-08 
F 100 q 1 L/D 8 N loop 100 
frac rec B in bot 0.997 guess frac rec C bot 1 df 0.9 
141 
 
frac rec A in dist 0.995 Trebguess 500 p, psia 14.7 
K const. aT1 aT2 aT6 ap1 ap2 ap3 
nB=A -1280557 0 7.94986 -0.96455 0 0 
nPen=B -1524891 0 7.33129 -0.89143 0 0 
nHex=C -1778901 0 6.96783 -0.84634 0 0 
 
D 29.940006 B 70.05999 L/V 0.88889 Lbar/Vbar 1.26 
xAdist 0.9969938 xBdist 0.003006 xCdist 2.2E-07 
xAbot 0.002141 xBbot 0.42692 xCbot 0.57094 
 
stage xA yA xB yB xC yC T KB 
1 0.002141 0.0103991 0.42692 0.6614042 0.57094 0.3282 582.283 1.549247 
2 0.0086951 0.0361347 0.613018 0.7887656 0.37829 0.1751 570.622 1.286692 
3 0.0291201 0.1069354 0.714099 0.7929732 0.25678 0.10009 561.854 1.110453 
4 0.0853111 0.270011 0.717438 0.6674458 0.19725 0.06254 551.835 0.930318 
5 0.214736 0.5324156 0.617813 0.4297616 0.16745 0.03782 536.481 0.695617 
6 0.4743433 0.786406 0.483106 0.2080991 0.04255 0.00549 513.748 0.430752 
7 0.7600825 0.9294091 0.233736 0.0700679 0.00618 0.00052 498.351 0.299774 
8 0.920961 0.980028 0.078451 0.0199309 0.00059 4.1E-05 491.769 0.254057 
9 0.9779073 0.9946891 0.022047 0.0053079 4.6E-05 3E-06 489.686 0.240759 
10 0.994401 0.9986729 0.005596 0.0013269 3.4E-06 2.2E-07 489.103 0.237137 
 
Distillate mole fracs = y values 0.9986729 0.00133 2.2E-07 
Calc frac recovery C in bottoms 0.9999998 j 3 
 
Option Explicit 
Sub Ternary_bottom_up_BP() 
' Ternary distillation with constant alpha. Frac recoveries of LK and HK given. 
' There is a HNK present and its frac rec in bottoms is guessed. 
 Sheets("Sheet1").Select 
 Range("A18", "I150").Clear 
 Dim i, j, k, feedstage, N As Integer 
 Dim AaT1, AaT6, Aap1, BaT1, BaT6, Bap1, CaT1, CaT6, Cap1 As Double 
 Dim F, fracBbot, fracCbot, q, LoverD, LoverV As Double 
 Dim LbaroverVbar, D, B, L, V, Lbar, Vbar, Eqsum, fracAdist As Double 
 Dim xA, xB, xC, yA, yB, yC, zA, zB, zC, xAbot, xBbot, xCbot As Double 
 Dim DxA, DxB, DxC, BxA, BxB, BxC, xAdist, xBdist, xCdist As Double 
 Dim fracCbotcalc, difference, epsilon, df As Double 
 Dim T, p, Tinit, KA, KB, KC, sum As Double 
'Input data 
 AaT1 = Cells(9, 2).Value 
 AaT6 = Cells(9, 4).Value 
 Aap1 = Cells(9, 5).Value 
 BaT1 = Cells(10, 2).Value 
 BaT6 = Cells(10, 4).Value 
 Bap1 = Cells(10, 5).Value 
 CaT1 = Cells(11, 2).Value 
 CaT6 = Cells(11, 4).Value 
 Cap1 = Cells(11, 5).Value 
 feedstage = Cells(3, 8).Value 
 F = Cells(5, 2).Value 
142 
 
 q = Cells(5, 4).Value 
 LoverD = Cells(5, 6).Value 
 zA = Cells(4, 2).Value 
 zB = Cells(4, 4).Value 
 zC = Cells(4, 6).Value 
 fracBbot = Cells(6, 3).Value 
 fracCbot = Cells(6, 6).Value 
 fracAdist = Cells(7, 4).Value 
 epsilon = Cells(4, 8).Value 
 N = Cells(5, 8).Value 
 df = Cells(6, 8).Value 
 p = Cells(7, 8).Value 
 Tinit = Cells(7, 6).Value 
' The For loop (remainder of program) is to obtain convergenceof guess of 
' fractional recovery of A in distillate. 
For j = 1 To N 
' Calculate compositions and flow rates. 
 DxA = F * zA * fracAdist 
 DxB = F * zB * (1 - fracBbot) 
 DxC = F * zC * (1 - fracCbot) 
 BxA = F * zA * (1 - fracAdist) 
 BxB = F * zB * fracBbot 
 BxC = F * zC * fracCbot 
 D = DxA + DxB + DxC 
 B = BxA + BxB + BxC 
 xAdist = DxA / D 
 xBdist = DxB / D 
 xCdist = DxC / D 
 xAbot = BxA / B 
 xBbot = BxB / B 
 xCbot = BxC / B 
 L = LoverD * D 
 V = L + D 
 LoverV = L / V 
 Lbar = L + q * F 
 Vbar = Lbar - B 
 LbaroverVbar = Lbar / Vbar 
' Print values of flowrates and mole fractions 
 Cells(13, 2) = D 
 Cells(13, 4) = B 
 Cells(13, 6) = LoverV 
 Cells(13, 8) = LbaroverVbar 
 Cells(14, 2) = xAdist 
 Cells(14, 4) = xBdist 
 Cells(14, 6) = xCdist 
 Cells(15, 2) = xAbot 
 Cells(15, 4) = xBbot 
 Cells(15, 6) = xCbot 
' initialize (reboiler =1) and start loops 
 i = 1 
 xA = xAbot 
143 
 
 xB = xBbot 
 xC = xCbot 
 T = Tinit 
' Calculations in stripping section: equilibrium then operating. 
Do While i < feedstage 
' Bubble point calculaton 
For k = 1 To 10 
 KB = 1 
 KA = Exp((AaT1 / (T * T)) + AaT6 + (Aap1 * Log(p))) 
 KB = Exp((BaT1 / (T * T)) + BaT6 + (Bap1 * Log(p))) 
 KC = Exp((CaT1 / (T * T)) + CaT6 + (Cap1 * Log(p))) 
 yA = KA * xA 
 yB = KB * xB 
 yC = KC * xC 
 sum = yA + yB + yC 
 KB = KB / sum 
 T = Sqr(BaT1 / (Log(KB) - BaT6 - (Bap1 * Log(p)))) 
Next k 
' Print values 
 Cells(i + 17, 1).Value = i 
 Cells(i + 17, 2).Value = xA 
 Cells(i + 17, 3).Value = yA 
 Cells(i + 17, 4).Value = xB 
 Cells(i + 17, 5).Value = yB 
 Cells(i + 17, 6).Value = xC 
 Cells(i + 17, 7).Value = yC 
 Cells(i + 17, 8).Value = T 
 Cells(i + 17, 9).Value = KB 
' Bottom operating line 
 i = i + 1 
 xA = yA / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xAbot 
 xB = yB / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xBbot 
 xC = yC / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xCbot 
Loop 
' Calculations in enriching section 
Do 
For k = 1 To 10 
' Bubble point calculation 
 KA = Exp((AaT1 / (T * T)) + AaT6 + (Aap1 * Log(p))) 
 KB = Exp((BaT1 / (T * T)) + BaT6 + (Bap1 * Log(p))) 
 KC = Exp((CaT1 / (T * T)) + CaT6 + (Cap1 * Log(p))) 
 yA = KA * xA 
 yB = KB * xB 
 yC = KC * xC 
 sum = yA + yB + yC 
 KB = KB / sum 
 T = Sqr(BaT1 / (Log(KB) - BaT6 - (Bap1 * Log(p)))) 
Next k 
' Print values 
 Cells(i + 17, 1).Value = i 
 Cells(i + 17, 2).Value = xA 
144 
 
 Cells(i + 17, 3).Value = yA 
 Cells(i + 17, 4).Value = xB 
 Cells(i + 17, 5).Value = yB 
 Cells(i + 17, 6).Value = xC 
 Cells(i + 17, 7).Value = yC 
 Cells(i + 17, 8).Value = T 
 Cells(i + 17, 9).Value = KB 
' Test for feed stage too low 
If xA < 0 Or xB < 0 Or xC < 0 Then 
 Cells(i + 18, 3) = "Feed stage too low" 
 Exit For 
End If 
 i = i + 1 
' Test for too many stages, which may mean reflux rate is too low. 
If i > 100 Then 
 Cells(i + 18, 2).Value = "Too many stages" 
 Exit For 
End If 
' Top operating line 
 xA = yA / LoverV - ((1 / LoverV) - 1) * xAdist 
 xB = yB / LoverV - ((1 / LoverV) - 1) * xBdist 
 xC = yC / LoverV - ((1 / LoverV) - 1) * xCdist 
' Test for calculations being done. 
Loop While yA < xAdist 
' Fractional recovery of C based on stage-by-stage calculation. 
 fracCbotcalc = 1 - (yC * D) / (F * zC) 
 difference = fracCbot - fracCbotcalc 
If Abs(difference) < epsilon Then Exit For 
 fracCbot = fracCbot + df * (fracCbotcalc - fracCbot) 
' Test if have convergence of fractional recovery of C. 
Next j 
 Cells(i + 19, 1).Value = "Calc frac recovery C in bottoms" 
 Cells(i + 19, 5).Value = fracCbot 
 Cells(i + 19, 6).Value = "j" 
 Cells(i + 19, 7).Value = j 
 Cells(i + 18, 1).Value = "Distillate mole fracs = y values" 
 Cells(i + 18, 5).Value = yA 
 Cells(i + 18, 6).Value = yB 
 Cells(i + 18, 7).Value = yC 
End Sub 
 
145 
Chapter 6 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 6A1, 6A5, 6D3, 6D4, 6G4-6G7. 
6.A1. New Problem in 3rd Edition. Trial and error. Try a feed stage and determine the distillate and
bottoms mole fractions of the key components. Repeat for additional feed stages. The feed stage 
that produces the best separation is the optimum feed stage for this value of N. 
6.A5. New Problem in 3rd Edition. Trial and error. Pick an N that you think is close (a systematic
method to do this is described in Chapter 7). Find the optimum feed stage. If you need more 
separation to meet specifications, increase N, and if you exceed specifications, try decreasing N. 
For an initial estimate of the optimum feed location for the new N, assume that the ratio 
(Optimum feed stage)/(total number of stages) is constant. Continue process until specifications 
are met or slightly exceeded. 
6.C1. a. With a side stream, mass balance is,
j j j j S j 1 j 1 j 1 j 1 j jV y L x Sx V , y L x Fz
 where jL is the flow into the stage below, j j S jL L S, x x . Since j j S j jL x Sx L x , Eqs. 
(6-4) to (6-6) are unchanged. Note that the j 1L input into the matrix will be changed. 
b. Now j j j j j jL h L h S h in the E.B. and 
j j 1 k k
j j
L V D F S
k 1 k 1
, j 1 j k k
j 1 j 1
L V D F S
k 1 k 1
 Substituting into EB we find 
j jE j F j j 1 j K j K j 1
K j 1 K j
j j 1
D F h Q D h h F h F h
k 1 k 1
j 1 j
S h S h
k 1 k 1
6.C2. Partial condenser mass balances is:
1 1 1 2 2 1 1Dy L x V y Fz (6-7) 
This becomes, 
1 2
1 1 2 2 1 1
1 2
DK V
K F z
L L
Thus, 1 2 21 1 1 1 1
1 2
DK K V
B 1 , C , and D F z
L L
(6-8) 
Note that only 1B differs. 
6.D1. For n-pentane from Example 6-1, T = 60°C, C5 1 2K 1.05, L L 825 kmole/hr, 
3 4 2 3 4L 1825, L 450 B, V V V 1375. 
Matrix: j = 1 (total condenser), 
146 
 
 2 21
2
1.05 1375K V
C 1.75
L 825
, 1 1
1
D 550
B 1 1 1.67, D 0
L 825
 
 3 32
3
1.05 1375K V
j 2, C 0.791
L 1825
 
 2 22 2 2
2
V K
B 1 2.75, A 1, D 0
L
 
 4 43
4
1.05 1375K V
j 3, C 3.208
L 450
 
 
1
3 3
3 3 3 C5
3
V K
B 1 1.791, A 1, D Fz 1000 0.35 350
L
 
 4 44 4 4
4
1 V K
j 4 Reboiler , B 4.208, A 1, D 0
L
 
1,C5
2,C5
3,C5
4,C5
1.67 1.75 0 0 0
1 2.75 0.791 0 0
0 1 1.791 3.208 350
0 0 1 4.208 0
 
 
 Values for 1,C5 to 4,C5 are in Example 6-1. 
 
6.D2. p = 5 atm: C2 C3 C4 C5z 0.08, z 0.33, z 0.49, z 0.10 
 As sat’d liquid & for bp calculate i iz x . Want iy 1.0. 
 Pick 3C as ref. 5 atm 101.3 kPa atm 506.5 kPa 
 1st Guess: Want C3 C4K 1, K 1. DePriester Chart. 
 Try T = 20°C. C2 C3 C4 C5K 5.4, K 1.7, K 0.47, K 0.14 
 iy 5.4 0.08 1.7 0.33 0.47 0.49 0.14 0.10 1.237 
 Need lower T. C3C3 new
i i
K 20 1.7
K T 1.37
K x 1.237
 
 new C2 C4 C5T 12 , K 4.6, K 0.35, K 0.10 
 iy 0.368 0.4521 0.1715 0.0 1.0016 OK 
 Propane Matrix Analysis: C3 6K 1.37, B F D 1000 410 590 L 
 3 c3 1 2 4 5 6D Fz 330, D D D D D 0 
 2 3 4 5 6 1V V V V V L D 1435 
 11 2 3
L
L D 1025 L L
D
 
 3 2 4 5L L F 2025 L L 
 Total Condenser (1): 
147 
 
 1 1 2 2 21B 1 D L 1.40, C K V L 1.37 1435 1025 1.918 
 Stage 2. 3 32 22 2 2
2 3
1435 1.37V KV K
A 1, B 1 =2.918, C 0.9708
L L 2025
 
 Stage 3. 3 3 4 43 3 3
3 4
1435 1.37V K V K
A 1, B 1 =1.9708, C 0.9708
L L 2025
 
 Stage 4: 5 54 44 4 4
4 5
V KV K
A 1, B 1 =1.9708, C 0.9708
L L
 
 Stage 5: 5 5 6 65 5 5
5 6
1435 1.37V K V K
A 1, B 1 =1.9708, C 3.32
L L 590
 
 Reboiler (Stage 6). 6 66 6
6
V K
A 1, B 1 =4.32
L
 
 Mass balance matrix. 
 
1.40 -1.918 0 0 0 0
-1 2.918 -0.9708 0 0 0
 
0 1 1.9708 0.9708 0 0
0 0 1 1.9708 0.9708 0
 0 0 0 -1 1.9708 -3.32
 
 0 0 0 0 -1 4.32
 
 
 
6.D3. New Problem in 3rd Edition. p = 5 atm = 506.5 kPa 
C2 C3 C4 C5z 0.08, z 0.33, z 0.49, z 0.10 
 As sat’d liquid & for bp calculation at iiz x . Calculation is same as in 5.D11 to obtain T. 
Result is: bp C2 C3 C4 C5T 12 , K 4.6, K =1.37, K 0.35, K 0.10 
 n-butane Matrix Analysis: C4 6K 0.35, B F D 1000 410 590 L 
 3 c3 1 2 4 5 6D Fz 490, D D D D D 0 
 2 3 4 5 6 1V V V V V L D 1435 
 11 2
L
L D 1025 L
D
, 3 2 4 5L L F 2025 L L 
 Total Condenser (1): 
 1 1 2 2 21B 1 D L 1.40, C K V L 0.35 1435 1025 0.49 
 Stage 2. 3 32 22 2 2
2 3
1435 0.35V KV K
A 1, B 1 =1.49, C 0.2480
L L 2025
 
 Stage3. 3 3 4 43 3 3
3 4
1435 0.35V K V K
A 1, B 1 =1.2480, C 0.2480
L L 2025
 
 Stage 4. 5 54 44 4 4
4 5
V KV K
A 1, B 1 =1.2480, C 0.2480
L L
 
148 
 
 Stage5: 5 5 6 65 5 5
5 6
1435 0.35V K V K
A 1, B 1 =1.2480, C 0.8513
L L 590
 
 Reboiler (Stage 6). 6 66 6
6
V K
A 1, B 1 =1.8513
L
 
 Mass balance matrix. 
 
1.40 -0.49 0 0 0 0
-1 1.49 -0.248 0 0 0
 
0 1 1.248 0.248 0 0
0 0 1 1.248 0.248 0
 0 0 0 -1 1.248 -0.851
 
 0 0 0 0 -1 1.851
 
 
6.D4. New Problem in 3rd Edition. L L D D 5 60 300 
 V L D 360 
 Saturated liquid feed: V V 360; L L F 400 
 
 
 
 
 1 2 3 4V 0, V V, V V, V V 
 
 
 
 Bubble Pt. Set i i i i iz x y K x 
 
 
 iy 1.0 or i iK x 1.0 
 M E NP NB3.58, 2.17, 1, 0.412 
 
 
 
 
 
 b. Eq. (5-30), 
i i
i
i i
x
y
x 
 i iz .3 3.58 .25 2.17 .35 1.0 0.1 .412 2.0077 
 Then NP NPnP
i i
0.35 1.0z
y 0.1743
z 2.0077
 
 and nP nP
0.1743
K y x 0.4981
0.35
 
1 2 3 4L L, L L, L L, L B F D 40
1 
2L 
3L 4
V 
3V 
1V 
F 
4 
2 
3 
149 
 
 M M NP NP E n-B nB NP NPK K 1.7832; K 1.0809, K K 0.2052 
 
c. Matrix for n-butanol nB nBK 0.2052, z 0.1 
 Stage 1. 1 1 1A , B 1 D L 1 0.2 1.2, 
 
 2 21 1
2
360 0.2052V K
C 0.2463, D 0
L 300
 
 
 Stage 2. 2 22 2
2
V K
A 1, B 1 1.2463
L
 
 
 3 32 2
3
360 .2052V K
C 0.1847, D 0
L 400
 
 
 Stage 3. 3 33 3
3
V K
A 1, B 1 1.1847
L
 
 
 4 43
4
360 0.2052V K
C 1.8468
L 40
 
 
 3 3 nButD F z 100 .1 10 
 
 Stage 4. 4 44 4 4 4
V K
A 1 B 1 2.8468, C , D 0
B
 
 
1
2
3
4
1.2 0.2463 0 0 0
1 1.2463 0.1847 0 0
0 1 1.1847 1.8468 10
0 0 1 2.8468 0
 
d. y 1 1 1 1 1
.2463
V1 B 1.2, V2 0, V3 .20525
1.2
 
 j 2 2 2 2 1V1 B A V3 1.2463 1 .20525 1.04105 
 2 2 2 21V2 D A V2 V1 0 1 0 0 
 2 2 1
0.1847
V3 C V1 0.1539
1.2
 
 j=3 3 3 3 2V1 B A V3 1.1847 1 0.1539 1.0308 
 3 3 3 2 3V2 D A V2 V1 10 1 0 1.0308 9.7014 
 3 3 2V3 C V1 1.8468 1.04105 1.7740 
 
150 
 
 j=4 4 4 4 3V1 B A V3 2.8468 1 1.7740 1.0728 
 4 4 4 3 4V2 D A V2 V1 0 1 9.7014 1.0728 9.0428 
 4 4 3V3 C V1 not needed. 
 4,NB N 4V2 9.0428 (bottoms flow rate) 
 3NB 3 3 4V2 V3 9.7014 1.7740 9.0428 25.743 
 2NB 2 3 3V2 V3 0 0.1539 25.743 3.9619 
 1NB 2 1 2V2 V3 0 .20525 3.9619 0.8132 
e. Raoult’s Law 
 nPVP K 760 0.4981 760 378.5 mmHg 
 Need to interpolate VP data. We know n VP 1 T 
 n 200 5.2983, n 378.5 5.9362, n 400 5.99146 
 1 66.8 273.16 2.9415E 3 1 82.0+273.16 2.81563E 3 
 Linear Interpolation: 
 
 bPT 273.16 351.74 or bPT 78.6 C. 
 
6.F1. Plots of vapor pressure are available in Maxwell (see Table 2-2 for reference) while tabulated 
values are in Perry’s i totK VP p . Dew point calculation on feed gives 245.7°F. 
 Overall Mass Balances: D = 30, L = 5D = 150 
 V = L + D = 180, V V F 180 100 80 , L L 150 , B = F – D = 70 
 First Trial Values 
Stage T L V KB KT Kx 
4 245.7 70 = B 80 2.307 1.042 0.534 
3 245.7 150 180 2.307 1.042 0.534 
2 245.7 150 180 2.307 1.042 0.534 
1 245.7 150 30 = D 2.307 1.042 0.534 
 
 Benzene 
Stage C B A D ℓ 
4 - 3.705 -1 0 7.9886 
3 -2.705 3.8404 -1 35 29.5978 
2 -2.8404 3.8404 -1 0 57.058 
1 -2.8404 1.2 - 0 135.6569 
 
 Toluene 
Stage C B A D ℓ 
4 - 2.191 -1 0 27.1651 
3 -1.191 2.2504 -1 40 59.5188 
2 -1.2504 2.2504 -1 0 61.5875 
1 -1.2504 1.2 - 0 64.1742 
in mmHgnP
nP
tot
VPVP
 K
P 760
2.9415E 3 2.8163E 3
5.9362 5.2983 2.9415E 3 0.00284
5.2983 5.99146
151 
 
 Xylenes 
Stage C B A D ℓ 
4 - 1.6103 -1 0 22.7361 
3 -.6103 1.6408 -1 25 36.6120 
2 -.6408 1.6408 - 0 21.1971 
1 -.6408 1.2 - 0 11.3193 
 
6.G1. Using Peng-Robinson. Aspen-Plus solution: 
 
 
Stage T1°C L kmol/h V C4 C5 C8
1 38.31 825 0 x1 0.360 0.6013 0.0386
y1 0.6568 0.3424 0.00083
2 69.16 557.3 1375 x2 0.0993 0.4499 0.4508
y2 0.3601 0.6013 0.0386
3 107.02 1533.9 1107.3 x3 0.03355 0.1731 0.7934
y3 0.2288 0.5251 0.2461
4 140.92 450 1083.9 x4 0.00436 0.0429 0.9528
y4 0.04568 0.2271 0.7272 
 
6.G2. 1. What VLE package did you use? Peng- Robinson. 
 
2. Report the following values: 
 Temperature of condenser = - 2.77 oC 
 Temperature of reboiler = 79.97 oC 
 Distillate product mole fractions 2 3 4C 0.3636,C 0.6360,C 0.0004 
 Bottoms product mole fractions 2 3 4C 1.2 E 13,C 0.000492,C 0.9995 
3. Was the specified feed stage the optimum feed stage? Yes No 
 If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. 
 
4. Which tray gives the largest column diameter (in meters) with sieve trays when one uses the 
originally specified feed stage? Tray # 28 Diameter = 0.792 m. 
 
5. Which components in the original problem are the key components? LK = Propane, HK = 
butane 
 
6. Change one specification in the operating conditions (keep original number of stages, feed 
location, feed flow, feed composition, feed pressure, feed temperature/fraction vaporized 
constant) to make ethane the light key and propane the heavy key. 
 What operating parameter did you change, and what is its new value? D = 20 
 Temperature of condenser = - 31.54 oC 
 Temperature of reboiler = 50.87 oC 
 Distillate product mole fractions 2 3 4C 0.9955,C 0.00448,C 1.32 E 07 
 Bottoms product mole fractions 2 3 4C 0.00112,C 0.4364,C 0.5625 
 
152 
 
6.G3. For column 1 report the following: 
a. Final value of L/D 1.8 
b. Split fractions of ethanol (distillate) 0.9999 and n-propanol (bottoms) 0.9913 
c. Mole fractions in bottoms 1.70 E-5, 0.00871, 0.9913 
d. Mole fractions in distillate 0.4545, 0.5383, 0.00714 
For column 2: 
 a. Optimum feed location in the column. 18 
b. Mole fractions in bottoms 0.00689, 0.9800, 0.0131 
c. Mole fractions in distillate 0.9917, 0.0083, 0.0 
 
6.G4. New Problem in 3rd Edition. 
1. Temperature of condenser = 389.9_ K. Temperature of reboiler = __547.4 K 
Qcondenser = _-772260____cal/sec, Qreboiler = _____912459__cal/sec 
Distillate product mole fractions: B= 0.23529, T= 0.76471, BiP = 0.12E-08_________ 
Bottoms product mole fractions:_B = 0.5 E-10, T = 0.67 E-08, BiP= 1.0000_________ 
 
2. Was the specified feed stage the optimum feed stage? Yes No x 
If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. a__ 
(Note: Do minimum number of simulations to answer these questions. Do not optimize.) 
 
3. Which tray gives the largest column diameter with sieve trays when one uses the originally 
specified feed stage? Aspen Tray #__16______Column Diameter =______2.28____meters 
 [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer 
area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding 
at 0.65. Use the “Fair” design method for flooding.] 
 
4. Which components in the original problem are the key components (label light and heavy 
keys) _____LK = toluene, HK = biphenyl_____________________________________________ 
 
5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed 
composition, feed pressure, feed temperature or fraction vaporized constant at original 
conditions) to make benzene the light key and toluene the heavy key. Also increase the reflux 
ratio to 4.0. 
What operating parameter did you change (not including the reflux ratio), and whatis its new 
value? D = 40________ 
Temperature of condenser = _368.9____ K, Temperature of reboiler = 407.7____ K 
Distillate product mole fractions: _B = 0.9283, T = 0.07173, BiP = 0.8 E-19________ 
Bottoms product mole fractions: _B = 0.01793, T = 0.79457, BiP = 0.1875_________ 
 
6.G.5. New Problem in 3rd Edition. 
1. Temperature of condenser = _121.07___ K. Temperature of reboiler = _166.23___ K 
Qcondenser = ____-757506.6____cal/sec, Qreboiler = ______1058466.75____cal/sec 
Distillate product mole fractions:__B = 0.9779, T = 0.22070, pxy = 0.6004 E-05__ 
Bottoms product mole fractions:___B = 0.0055189, T = 0.55698, pxy = 0.43750___ 
 
2. Was the specified feed stage the optimum feed stage? Yes No x 
If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. a 
(Note: Do minimum number of simulations to answer these questions. Do not optimize.) 
 
3. Which tray gives the largest column diameter with sieve trays when one uses the originally 
specified feed stage? Aspen Tray #____24______Column Diameter =______2.28___meters 
153 
 
 [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer 
area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding 
at 0.7. Use the “Fair” design method for flooding.] 
 
4. Which components in the original problem are the key components (label light and heavy 
keys) ________benzene = LK, toluene = HK______________________ 
 
5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed 
composition, feed pressure, feed temperature or fraction vaporized constant) to make toluene the 
light key and p-xylene the heavy key. 
What operating parameter did you change, and what is its new value?__D=260______ 
Temperature of condenser = _142.2____ K, Temperature of reboiler = _183.98__ K 
Distillate product mole fractions: _B = 0.30769, T = 0.68850, Pxy = 0.003805________ 
Bottoms product mole fractions: __B= 0.3177 E-06, T = 0.007066, Pxy = 0.99293_____ 
 
6.G.6. New Problem in 3rd Edition. Part a. L D 27 . b. L D 60 . c. D = 147, S = 453 (liquid) 
 distillate mole fracs: E = 0.99007, B = 0.00993, P = 0.5 E-9 
 side stream mole fracs: E = 0.0009845, B = 0.98930, P = 0.000854 
 bottoms mole fracs: E = 0.7 E-14, B = 0.00043, P = 0.99957 
 d. distillate : E = 0.89146, B = 0.10854, P = 0.556 E-8 
 side: E = 0.041845, B = 0.95794, P = 0.000218 
 bottoms: E = 0.1 E-14, B = 0.0001095, P = 0.99989 
 Since vapor mole fraction ethane > liquid mole fraction (ethane is LK), have more ethane in 
vapor side stream. 
 e. The separation of n-pentane and n-butane is much more difficult than between ethane and n-
butane. Thus side stream purity is less. Also feed has lot more pentane than ethane, which makes 
side stream below feed less pure. 
 
6.G.7. New Problem in 3rd edition. 
1. Report the following values: 
Temperature of condenser = _373.28____ K. Temperature of reboiler = ___411.75___ K 
Qcondenser = _-829828_____cal/sec, Qreboiler = ____1012650_____cal/sec 
Distillate product mole fractions: M = 0.59998, E = 0.36184, NP = 0.038177, NB = 0.3087E -05 
Bottoms product mole fractions: M= 0.2042E-04, E = 0.03816, NP = 0.46182, NB = 0.50000_ 
 
2. Was the specified feed stage the optimum feed stage? Yes No X 
If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. Answer a 
(Note: Do minimum number of simulations to answer these questions. Do not optimize.) 
 
3. Which tray gives the largest column diameter with sieve trays when one uses the originally specified 
feed stage? Aspen Tray #_____18_____Column Diameter =___1.77____meters 
 [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer area 
(0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding at 0.7. Use 
the “Fair” design method for flooding.] 
 
4. Which components in the original problem are the key components (label light and heavy keys) 
______________LK = ethanol, HK = n-propanol_______________________ 
 
154 
 
5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed 
composition, feed pressure, feed temperature or fraction vaporized constant) to make methanol the light 
key and ethanol the heavy key. 
What operating parameter did you change, and what is its new value?_____D = 60____ 
Temperature of condenser = __368.66__ K, Temperature of reboiler = _404.23___ K 
Distillate product mole fractions: M = 0.97858, E = 0.021417, NP = 0.155 E-07, NB = 0.1 E-10_ 
Bottoms product mole fractions: M = 0.0091787, E = 0.27654, NP = 0.35714, NB = 0.35714__ 
 
 
 
155 
Chapter 7 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 7.A1, 7.A4, 7.D2, 7.D10, 7.D11, 7.D14, 7.D21, 7.G1. 
7.A1. New problem in 3rd edition. f. none of the above.
7.A.4. New problem in 3rd edition. a. estimate fractional recoveries nonkeys at total reflux.
7.C4. Use i, j 1 i i, j 1y K x . Then substituting into Eq. (7-20), we have 
min i i, j 1 min ij i,distV K x L x Dx
which is min i i, j 1 min ij 1 i,distV K x L x Dx where i i HKK K . Rearranging, 
min i HK
min i, j 1 i,dist
min
V K K 1
L x Kx
L
, or i,dist imin i, j 1
min HK
min i
Dx
L x
V K 1
L
Total flow rate minL is 
i,dist i
min min ij 1
min HK
min i
c c Dx
L L x
V K 1i 1 i 1
L
(A) 
By a similar analysis obtain, i ,bot imin
min HK
min i
Bx
L
V K 1
L
(B) 
Let min 1 minL L
min min
V K V K
L L
(C) 
Add Eqs. (A) and (B), and use external mass balance, 
c
i i
feed min min
i 1 i L
Fz
qF L L L
1/
(7-33 analogue) 
From Eqs. (A) and (C) we have 
c i,dist i
min
i 1 i L
Dx
L
1/
 
(7-29 analogue) 
7.C5. For saturated vapor feed have FV F . For binary system Eq. (7-33) is, 
1 1 2 2
1 2
z z
1
Clearing fractions we obtain 
1 2 2 1 1 2 2 1z z
After some algebra this 
2 1 1 2 2 1 2z z 0 
Solutions are, 1 2 1 1 2 20 or z z
For sat’d liq’d FV 0 . Clear fractions and equation is linear.
156 
 
7.D1. a. Eq. (7-16), 
dist
bot
min
AB
x 1 x .992 .008n n
x 1 x .014 .986
N 10.36
n n 2.4
 
 This includes the partial reboiler. Eq. (7-40a) gives, 
 
dist
feed
f ,min
AB
x 1 x .992 .008n n
z 1 z .4 .6
N 5.97
n n 2.4
 
 b. Saturated liquid: fV 0 . Eq. (7-33) becomes B B T T
B T
z z
0 
 After clearing fractions and solving for , 
 T B
B B T T
1.0 2.4z
1.53846
z z 2.4 .4 1.0 .6
 
 Which does lie between the α’s of the keys. To use Eq. (7-29) we need D. From mass 
balances (Eq. (3-3)). 
 bot
dist bot
z x .4 .014
D F 10 3.9468
x x .992 .014
kg moles/hr 
 Eq (7-29) is: B Bdist T Tdistmin
B T
Dx Dx
V . 
 min
2.4 3.9468 .992 1.0 3.9468 .008
V 10.848
2.4 1.53846 1 1.53846
 
 min minL V D 6.9013 . minL D 1.75 
 
 c. 
min
L D 1.1 L D 1.9234 , min
L D L D
x 0.0598
L D 1
 
 Using Eq. 7-42b, min
N N
0.5563
N 1
 
 Solving for N, min
.5563 N
N 24.6
1 .5563
 (includes reboiler) 
 From Eq. (7-40b), F,minF
min
N 5.97
N N 24.6 14.2
N 10.36
 
 Try stage 14 from top for feed stage. 
 
7.D.2. New problem in 3rd edition. C2 C3 C4 C5p 5 atm, z 0.08, z 0.33, z 0.49, z 0.10 
 Saturated liquid and for bp. Calc., i iz x . Want iy 1.0 . 
 Pick 3C as reference (this is arbitrary). 
kPa
5 atm 101.3 506.5 kPa
atm
 
 1st Guess: Want C3K 1 (light key), C4K 1 (heavy key). Use DePriester Chart. 
157 
 
 Try T 20 C , C2K 5.4 , C3K 1.7 , C4K 0.47 , C5K 0.14 
 i
y 5.4 0.8 1.7 0.33 0.47 0.49 0.14 0.10
 0.432 0.561 0.230 0.014 1.237
 
 Need lower T. C3C3 new
i i
K 20 1.7
K T 1.37
1.237K x
 
 new C2 C4 C5T 12 , K 4.6, K 0.35, K 0.10 
 iy 0.368 0.4521 0.175 0.0 1.0016 
 For remainder, let 4C HK be reference. 3 4C C 1.37 0.35 3.914 
 a) MIN
0.997 0.998
n
0.003 0.002 12.01874
N 8.808
n 3.914 1.36456
(includesPR). 
 b) C2 C4 C4 C4 C5 C44.6 0.35 13.143, 1.0, 0.1 0.35 0.286 
 Sat’d liquid feed MIN MINV V . In Eq (7-33) divide through by F. 
 i i
i
13.14 0.08 3.914 0.33 1.0 0.49 0.10 0.286z
0
13.14 3.914 1.0 0.286
 
 Solve for φ. Find φ = 1.74 (Note LK HK HK HK3.914 ) 
 Eq (7-29) i i ,distMIN
i
Dx
V . Assume all 2C in distillate & all 5C in bottoms 
 i,C2 i,C5Dx 80, Dx 0 
 i C3Dx 0.997 1000 0.33 329.01 
 i C4Dx 1 0.998 1000 0.49 0.98 
 i,dD Dx 409.99 
 MIN
13.14 80 329.01 3.914 0.98 1.0
V 0 683.23kmole h
13.14 1.74 3.914 1.74 1.0 1.74
 
 MIN MIN MINL V D 273.24; L D 0.6664 
 c) MIN MINL 1.15 L D 0.7664 
 Ordinate Gilliland 
0.7664 0.6664
0.05662
1.7664
 
 Eq (7-42b) MIN
N N 0.002743
0.545827 0.591422 0.05662 0.56079
N 1 0.05662
 
 
0.5608 8.808
N 21.33
1 0.5608
 (include PR) 
158 
 
 
LK HK dist
LK HK
F,MIN
LK HK
Dx Dx 329.01 0.98n n
z z 0.33 0.49
N 4.552
n n3.914
 
 Eq. (7-40b) F,min F F
min
N N .552
N 21.33 11
N N 8.808
 (approximate) 
 7.D3. At total reflux use Fenske Eq. (7-11). 
A A
B Bd bot
min
AB
x x
log
x x
N
log 
 
 Rearrange, log A AAB
B Bmin d bot
1 x x
log
x xN
 
 
min1 N 1 3
A A
AB
B Bd bot
x x .545 .36
1.287
x x .455 .64
 
7.D4. a.) 
dist
bot
min
AB
x 1 x .9915 .0085n n
x 1 x .01773 0.98227
N 10.02
n n 2.4
 
b.) Feed is saturated liquid, feed line is vertical. 
 
 
 
 
 c.) Abscissa = min
L D L D 2.2286 1.144
0.336
L D 1 3.2286
 
 From Eq. (7-42b) 
 min
N N 0.002743
0.545827 0.591422 0.336 0.3553 ordinate
N 1 0.336
 
 min
N ordinate 10.022 0.3553
N 16.1
1 ordinate 1 0.3553
 
 From fitted curve ordinate = 0.3474, 
10.022 0.3474
N 16.2
1 .3474
 
 Error = 25-16 25 36% low. Aspen Plus equilibrium data is not α = 2.4. Note that α = 
2.24 may be a better fit. 
(L/V)min = 
*0.9915 y
0.9915 0.6
 
*
x z 0.6
x
y 0.7826
1 1 x
 
(L/V) min = 0.534, 
min min
L L V
1.144
D 1 L V
 
y* 
* z = .6 
y 
159 
 
7.D5. Fenske Eq. is: 
D B
ND B min D B
min
D B
x x
log
1 x 1 x x x
N or 
1 x 1 xlog 
 
 Solving for Bx , we obtain: 
D D
B N min
D D
x 1 x
x
x 1 x
 
 Since min D BN 30, 1.30 and x 0.984, this is x 0.229 
 
7.D6. Fenske Eq.: 
D B
D B
min
x x .993 .01log log 
1 x 1 x .007 .99
N 10.82
log log 2.4
 
 Determine y in equilibrium with feed z x 0.4 . 
 *
2.4 .4x
y .616
1 1 x 1 1.4 .4
 
 D
min D
x y *L .993 .616
0.636
V x z .993 .4
, 
min min
L L V
1.746
D 1 L V
. 
 Then, 
act
L
1.15 1.746 2.01
D
 
 Gilliland Correlation: Abscissa min
L D L D 2.01 1.746 .264
.0878
L D 1 3.01 3.01
 
 Ordinate min
N N N 10.82
0.557
N 1 N 1
 
 Clear fractions, and find N = 25.3 (including partial reboiler). 
 
7.D7. p 5 atm. From the solution to problem 6.D9: bpT 12 C 
 C2 C3 C4 C5K 4.6, K 1.37, K 0.35, K 0.10 
 Let 4C HK be reference. C3 C4
1.37
3.914
0.35
 
 a) Eq. (7-15), Including PR min
0.98 0.992
n
0.02 0.008 8.7121
N 6.38
n 3.914 1.36456
 
 b) 4 C2-C4 C4 C4C HK reference. 4.6 0.35 13.143, 1.0 
 C5 C4 0.10 0.35 0.286 
 Sat’d/liquid feed, min minV V 
 Eq. 7-33, i i
i
13.14 0.08 3.914 0.33 1.0 .49 0.10 0.286z
0
13.14 3.914 1.0 0.286
 
 Want LK HK HK HK3.914 1.0 . Converge to = 1.74. 
160 
 
 i i iC2 C3Dx Fz 80, Dx 0.98 1000 0.33 323.4 
 i iC4 C5Dx .008 1000 0.49 3.92, Dx 0 
 Eq. (7-29), min
13.14 80 323.4 3.914 3.92 1.0
V 0 671.05
13.14 1.74 3.914 1.74 1.0 1.74
 
 i,d min min minD Dx 407.32, L V D 263.73, L D 0.647 
 c) 
min
L D 1.2 L D 0.777 . Ordinate Gilliland 0.777 0.647 1.777 0.073 
 Eq. (7-42b), min
N N 0.002743
0.545827 0.591422 0.073 0.5402
N 1 0.073
 
 
0.5402 6.38
N 15.05,
1 0.5402
 incl. PR. Nfeed ~ 9 
7.D8. a. Can do this graphically, or can calculate slope of a line from Dy x x .992 to 
intersection of feed line and equilibrium, or use Underwood. Easiest to calculate slope. Feed 
line Fy z .4 . 
 Equilibrium: 
y .4
x .3755
y 1 y .4 1.11 .6
 
 
min
L .992 .4
.958
V .992 .3755
, 
min
L L V
22.83
D 1 L V
 
 b. 
D B
D B
min
x x .992 .005n
1 x 1 x .008 .995N 96.9
n n 1.11
 
 This is 95.9 stages plus partial reboiler. 
 c. L D 1.2 22.83 27.4 , Abscissa min
L D L D 27.4 22.83
.161
L D 1 27.4 1
 
 Ordinate min
N N
.47 or N 181.9
N 1
 which includes partial reboiler. 
 This separation would probably not be done by distillation. 
7.D9. Feed 80% liquid, F FL .8F, V .2F. Feed line: 
F
F F
L Fz
y x
V V
, Slope 
8
4
2 
a. See Graph. Min top op line is tangent. 
min
L .8 0.386
slope 0.5175
V 0.8 0
 
 min
min min
L VL 0.5175
1.0725
D 1 L V 1 .5175
 
 b. min
3
N 6
4
 eq. contacts. See graph. 
 c. 
actual
L
1.05 1.0725 1.1262
D
. Abscissa, Gilliland Correlation is 
161 
 
 min
L D L D 1.1262 1.0725 0.053666
0.02524
L D 1 2.1262 2.1262
 
 Ordinate ~ 0.63 from graph. 
 From eq. (7-42b), Ordinate 
0.002743
0.545827 0.591422 0.02524 0.6396
0.02524
 
(agrees with graph). 
 Then min
N ordinate 6.75 0.6396
N 20.5
1 ordinate 1 0.6396
 
 Need 20 eqs. contacts + P.R. 
 F,minN from graph = 6. 
F,min
F
min
N 6
N N 20.5 18
N 6.75
 (7-40b) 
 
7.D10. a) New problem in 3rd edition. Eq. (7-15) 
 
E,dist B,bot
E ,dist B,bot
MIN
EB
FR FR
n
1 FR 1 FR
N
n 
 
162 
 
 EB 13.14 MIN
0.989 0.998
n
0.011 0.002 10.7114
N 4.159
n 13.14 2.5756
 
 
 b) MIN PBN 4.159 is known.. 3.91. 
 Eq. (7-17) 
MIN
MIN
N
PB
P,dist
NB,bot
PB
B,bot
FR
FR
1 FR
4.159
4.159
3.91
0.3677
0.998
3.91
0.002
 
 c) D i,dist iiDx FR Fz 
 Ethane DEDx 0.989 100 0.3 29.67 
 Propane DPDx 0.3677 100 0.33 12.134 
 n-butane DBDx 0.002 100 0.37 0.074 
 
3
i,d
i 1
kmol
D Dx 41.878 
h
 
Also accept D = 0 since total reflux. 
 
7.D.11. New problem in 3rd edition. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Use Underwood Eqns. – Case A Assume LNK (propane) is all in distillate. p,dist pDx Fz 20 
 b) feedV F 1 q F since q 0 
 Eq. (7-33). F i1
F i
i
z
 where φ is between α’s of two keys (B and H) 
 1.0 > φ > 0.2. Equation is, 
 
2.04 0.2 1.0 .35 0.20 0.45
1.0
2.04 1.0 0.20
 
 Solving for φ obtain φ = 0.62185. 
200 
n
iP
NP
z 0.35
z 0.4
z 0.25
 
1V
 
B 
 
D 
 
1 
 
2 
 
163 
 
 Then i,distmin
Dx
V . Find D from fractional recoveries. 
 
B,dist
p,dist
H ,dist
Dx 0.99 100 .35 34.65 
Dx 20
Dx 1 0.98 100 .45 0.9
 D 55.55
 
 
 MIN
2.04 20 1.0 34.65 0.2 0.9
V 119.98
2.04 0.62188 1.0 0.62188 0.2 0.62188
 
 min MIN minL V D 64.4314 and L D 1.1599 
 
7.D12. A = benzene (LK), AB A,dist2.25, FR 0.98 
 B = toluene (HK), BB B,bot1.0, FR 0.99 
 C = cumene (HNK), CB 0.210 
 a. Use Fenske eqn. at total reflux. 
A,dist B,bot
A ,dist B,bot
min
AB
FR 1 FR
n
1 FR FR
N
n 
 
 min
0.98 0.01
n
0.02 0.99
N 10.47
n 2.25
 
 
N min
AC AB
C,bot ACN minA,dist CB
AC
A,dist
2.25
FR where 10.71
FR 0.21
1 FR
 
 
10.47
C,bot
10.47
10.71
FR 1
0.98
10.71
0.02
. All cumene goes to bottoms. 
 (We can also substitute into Eq. (7-17)). 
 
N 10.47min
12AC
C,dist
N 10.47B,bot min
CB
B,bot
0.21
FR 8.1 10 0
FR .99
0.21
.011 FR
 
 b. Underwood equations – Case B analysis 
 Feed is sat. vapor. feedq 0, V F 1 q F 100 , 
C
i 1
feed
i 1 1
Fz
V 
 
2.25 40 1.0 30 0.21 30
100
2.25 1.0 0.21
 
 Find 1.6516. 
164 
 
 
C i i,dist
min i,dist 1 i,dist
i 1 1
Dx
V where Dx Fz FR 
 A,dist B,dist C,distDx 100 0.4 0.98 39.2, Dx 100 0.3 0.01 0.4, Dx 0, D 39.6 
 min
2.25 39.2 1.0 39.2
V 0 146.78
2.25 1.6516 1 1.6516
 
 From mass balance, min minL V D 146.78 39.6 107.2 , 
min
L 107.2
2.71
D 39.6
 
 c. 
min
L D 1.25 L D 1.25 2.71 3.39 
 abscissa for Gilliland correlation min
L D L D
0.155
L D 1
. From Figure 7-3 the 
 ordinate= 0.46. With minN 10.47, we find N = 20.24. 
 To find feed feedminN , we need N 
 
A A
B Bdist feed
feedmin
AB
x z
n
x z
 N
n 
 
 A,dist B
39.2
x 0.9899, x 1 .9899 0.0101
39.6
 
 feedmin
0.9899 0.4
n
0.0101 0.3
N 5.30.
n 2.25
 
 
feed feedmin
feed
min
N N
 gives N 10.25.
N N
 Use stage 10 or 11. 
 
7.D13. Use Fenske eq. T T
.99
FR 1 FR 99
.01
, C C
.02
FR 1 FR 0.0204
.98
 
 min
TC
log 99 0.0204 log 4851
N 5.438
log log 1 0.21
 
 Underwood: fV V V 0 , 
2.5 .25 1.0 .30 0.21 45
0
2.5 1.0 21
 
 1.526 or 0.3374. Use 0.3374 as it is between keys. 
3 1 i,d
min
i 1 1
Dx
V 
 B,d BDx Fz 25 (assume all benzene in dist.) 
 T,d T c,d cDx .99 Fz 29.7, Dx 0.02 Fz 0.9, D 55.6 
 min
2.5 25 1.0 29.7 .21 9
V 72.24
2.5 0.3374 1.0 0.3374 0.21 0.3374
 
 min minL V D 16.64 and L D 0.2993 
165 
 
 Gilliland: Ordinate min
N N 9 5.438
0.3562
N 1 10
 
 Abscissa ~ .29 (original Gilliland) or .36 (Liddle) 
 If use 0.29 have, 
min
L L L
0.29 0.29
D D D
 
 
L .29 0.2993
.83
D 1 .29
 
 If use Abscissa = 0.36, 
L .36 .2993
1.03
D 1 .36
 which are quite different. Safer to use 
higher value. 
 If BT min2.25, N is same. Underwood Eq. gives 1.4666 or 0.3367. Use 0.3367. 
 minV 29.40 44.78 1.492 72.68 , min minL V D 17.084 and L D 0.3073 
 Which is 2.7% different than for BT 2.5. 
 
7.D14. New problem in 3rd edition. Use Gilliland correlation to find the value of the minimum reflux 
ratio, (L/D)min = 1.4 
7-D15. Fenske: Eq. (7-15), 
B,dist C,bot
B,dist C,bot
min
BC
FR FR .9992 .9999n n1 FR 1 FR .0008 .0001
N 6.89
2.25n n
.21
 
 Eq. (7-17), 
min
min
N
TC
T ,dist TC
C,bot N
TC
C,bot
1
FR where 
FR .21
1 FR
 
 
6.89
tol,dist tol,bot6.89
1
.21FR 0.8238, and FR 0.1762
.9999 1
.0001 .21
 
 
dist
d ,tol
d ,benz
d ,cum
x
Dx 0.8238 167 137.57 
0.2568
Dx 0.9992 397 396.68 0.7418
.0436
Dx 0.0001 436 0.0008
D 534.294
 
 
LK LK
HK HKdist
F,min
LK-HK
x z .7418 .397n nx z .000815 .0436N 1.94
n n 2.25 .21
. 
 Underwood: Use a Case C analysis since toluene is a sandwich key. 
 Eq. (7-33): 
3
1 1
F
i 1 1
2.25 397 1.0 167 .21 436Fz
0 V is, 0
2.25 1.00 .21
 
 1.216 and 0.3373 which lie between α’s. 
166 
 
 Eq. (7-29): 
 
3 i i,dist tol
min min
i 1 i
Dx 2.25 396.88 1.0 Dx .21 0.436
V becomes V
2.25 1.0 .21
 
 Write for 1.216 and for 0.3373. Obtain 2 eqns and 2 unknowns: min tol,distV and Dx . 
 For 1.216 , min tolV 863.525 4.629 Dx . For 
min tol.3373, V 466.15 1.509 Dx . 
 Solving simultaneously, min tolV 563.84, Dx 64.740 
 i,dD Dx 396.88 64.74 .436 462.056 
 min min minL V D 101.79 and L D 0.2203 
 Gilliland Abscissa min
L D L D 1.2 .2203
0.445
L D 1 2.2
 
 Ord. min
N N
.245
N 1
 (Original Gilliland) 
 Obtain N = 9.45 (includes reboiler) 
 Estimate F,minF
min
N 2.91
N N 9.45 2.66
N 6.89
 (Use stage 3) 
 
7.D16. a. Fenske: dist botmin
x x .99 .992n n
1 x 1 x .01 .008N 12.69
n n 2.1
 
 b. Underwood: pEP Efeed
EP
1.0 F zFz
V
1.0
, E P feedz .6, z .4, V V V F 
 
2.1 .6 1.0 4
1
2.1 1.0
, 0 or 1.44 . Use 1.44 which is between 1.0 and 2.1 
 To find D: B
D B
z x .6 .008
D F 1000 602.85
x x .99 .008
. Then 
 min
2.1 .99 1.0 .008V
3.132
D 2.1 1.44 1.00 1.44
 
 min min min
min
L
V 1888.12, L V D 1285.3, 2.13
D
 
 c. Use Gilliland: Ordinate min
N N .30 12.69
0.558
N 1 31
 
 Abscissa ~ min
L D L D
0.8
L D 1
 (Original Gilliland), L D 2.4 
 
7.D17. Sounds harder than it is: F i i tol-xy xy xy
i
V z
0 f , 3.03, 1-F
 
167 
 
 Expand & Solve for , xy xytol tol
tol xy
zz
0 
 Result is linear, xy tol tol
tol tol xy xy tol tol tolz z z 1 z
 xysin ce 1 
 Then i i,distmin
i
Dx
V , B
D B
z x z .008
D F 100
x x 0.988
 
 min min min
3.03 .996 1.0 .004
V D , L V D
3.03 1
 
 
tolz
 
 D 
minV minL 
min
L
D
 
0.1 3.03
2.51870
3.03 .1 .9
 9.3117 3.03 .996 .004 54.93685
3.03 2.51870 1 2.51870
 
45.625 4.8997 
0.3 3.03
1.88316
3.03 .3 .7
 29.5547 3.03 .996 .004 D 77.6383
3.03 1.88316 1 1.88316
 
48.0836 1.6269 
0.5 3.03
1.503722
3.03 .5 .5
 49.7976 3.03 .996 .004 D 98.0683
3.03 1.503722 1 1.50372
 
48.2707 0.96934 
.7 3.03
1.25155
3.03 .7 .3
 70.0405 3.03 .996 .004 D 117.739
3.03 1.25155 1 1.25155
 
47.6985 0.68101 
 
.9 3.03
1.071806
3.03 .9 .1
 90.2834 3.03 .996 .004 D 129.08
3.03 1.071808 1 1.071808
 
38.7990 0.429 
 
 Check for z = 0.5. 
 
 
 
 
min
L L L V .4922
0.96934
D V L 1 L V 1 4922
 Perfect 
 As z , although min min c minminL D , V , thus Q V as expected. 
 R,min Q also. 
y* 
y 
.5 = z 
x 
D
min D
x y*L
V x .5
where 
F
F
x
y* 0.75186
1 1 x
 
 
0.996 0.75186
0.4922
0.996 0.5
 
 
min
Slope L V 
Dx 
168 
 
7.D18.a. Fenske: Eq. (7-15): 
A,dist B,bot
A ,dist B,bot
min
AB
FR FR
n
1 FR 1 FR
N
n 
 
 Where A = propane, B = butane, AB 1/ .49 2.04 . (Note value α.) 
 min
.9854 .8791
n
.0146 .1209
N 8.7
n 2.04
 
 For F,minN assume no LNK is bot and no HNK in distillate 
 D = .229 + (.9854) (.368) + (.1209) (.322) = .631 
 prop C4
.9854 .368 .1209 .322
x 0.575, x 0.0617
.631 .631
 
 
C3 C3
C 4 C 4dist F
F,min
C3-C4
x z .575 .368n nx z .0617 .322
N 2.94
n 0.713
 
 Underwood Eqns. (Case A.) 1 1feed
1
Fz
0 V for 1.0 .49 
 
9.92 .229 1.0 .368 .49 322 .081 0.10
0 f
9.92 1.0 .49 .10
 
 Find 0.6213 , min
9.92 .229 1.0 .363 .49 .0389
V 1.057
9.92 .6213 1.0 .6213 .49 .6213
 
 min min minL V D 1.057 .631 .426, L D 0.676 
 Gilliland Correlation (Fig.7-3): abscissa min
L / D (L / D)
0.33
L / D 1
 
 Ordinate min
N N
.32
N 1
 (Original Gilliland, ~.36 Liddle). Find N = 13.24 (14.13 Liddle) 
 b. With N = 20, ordinate to Gilliland correlation is, 
 min
N N 20 8.7
0.538
N 1 21
 
 Abscissa = 0.1. Since 
min
L D 0.676, solve for L/D = 0.862. 
 c. 
min
min
N
C6 C3
C 6 dist
C3 Nbot
C6C3
C3 bot
FR
FR
1 FR
 
 where 
bot distC6 C3 C3 C3 min
0.10, FR 1 FR 0.0156, and N 8.7. 
 
8.7
C6 dist 8.7
0.10
FR 0.00000013
0.0156
0.10
1 0.0156
, C6 botFR 0.99999987 
169 
 
 For all practical purposes all C6 in bottoms at total reflux. 
 
 d. 
distC3 B,bot
FR 0.999, L / D 1.5, FR 0.8791 
 1) 
dist bot
dist bot
C3 C 4
C3 C 4
min
C3C 4
FR FR .999 .8791n n
1 FR 1 FR .001 .1209
N 12.47
n 0.713
 
 
 2) For D assume all LNK in dist, No HNK in dist 
 
 D = 0.299 + (0.999) (0.368) + (0.1209) (0.322) = 0.6356 
 
 Now 
9.92 .229 1.0 .368 .49 .332 .081 .1
f 0
9.92 1 .49 .1
 
 
 Which is same [φ depends only on feed & α’s]. Thus, same φ = 0.6213 
 
 min
9.92 0.229 1.0 0.368 0.49 4689??
V 1.0709
9.92 .6213 1 .6213 .49 .6213
 
 
 min minL 1.0709 0.6356 0.4345 L D 0.6848. Very little change. 
 
7.D19. Use Figure 7-3. Ordinate min
N N 25 11
0.5385
N 1 26
 
 Then Abscissa min
L D L D
0.08 with L D 2.2286
L D 1
. 
min
L D 1.97 
 Abscissa approximated between original & fitted curves. 
 
7.D20. a) Distillate 
dist distB B T T
Dx Fz 5, Dx Fz 15 , 
 distx 1.0 becomes 
5 15
0.57895 0.07018 1.0
D D
 
 
 Find D = 57.001 kmoles/hr, B = 100 – D = 42.999 
 b) Can use Fenske eq. (7-11) or alternatives.
A B dist
A B bot
min
AB
x x
n
x x A xylene
N
B cumenen 
 
 xy xy tol xyAB xylene cumene
cum cum tol cum
K K K 0.330
1.57143
K K K 0.210
 
 Xylene balance, x,botFz 35 57.00 0.57895 42.999 x 
 x,bot cum,botx 0.0465, x 1 .0465 0.9535 
170 
 
 min
0.57895 0.07018
n
0.04650 0.9534
N 11.35
n 1.57143
 
 This is # equil contacts at total reflux. 
 c) Alternative: x,distxy,dist
x
Dx 57.001 0.57895
FR 0.94288
Fz 35
 
 C,distC,bot
C
Dx 57.001 0.07018
1 FR 0.088896
Fz 45
, cum,botFR 0.91110 
 Use Eq. 7-15. 
dist
dist
A B,bot
A B,bot
min
AB
FR FR
n
1 FR 1 FR A xylene
N , 
B cumenenmin
0.94288 0.91110
n
0.057122 0.088896
N 11.35
n 1.57143
 
 
7.D21. New problem in 3rd edition. Assume all ethanol in distillate and all n-butanol in bottoms. 
 
 
 
E,dist E
i P,dist ip
n P,dist nP
n B,dist
Dx Fz 100 .3 30
Dx Fz Frac Rec iP dist 100 .25 .986 24.65
Dx Fz 1 Frac Rec nP dist 100 .35 .008 0.28
Dx 0 0
 
 
 
 
 
 
 xE,dist = xE,dist/D = 0.5461, xi-P,dist = xi-P,dist/D = 0.4488, xn-P,dist = xn-P,dist/D = 0.0051 
 
 
E,bot
i P,bot iP
n P,bot nP
Bx 0 0
Bx Fz 1-Frac Rec iP dist 100 .25 .014 0.35
Bx Fz Frac Rec nP bot 100 .35 0.992 
n B,bot n B
34.72
Bx Fz 100 0.10 10.0 
 
 
 
 i,bot i,botx Bx B 
xi-P,bot = Bxi-P,bot /B = 0.0108, xn-P,bot = Bxn-P,bot /B = 0.7704, xn-B,bot = Bxn-B,bot /B = 0.2188
 
 
B = 45.07 
 
i,distD Dx 54.93 
171 
 
 b. Fenske eq. (7-15) 
ip ,dist nP,bot
iP,dist nP,bot
Min
iP nP
FR FR
n
1 FR 1 FR
N
n 
 
 min
.986 .992
n
.014 0.008 n 8733 9.0748
N 14.62
n 1.86 0.62058 0.62058
 
 This includes PR 
 F,MINN Eq. 7 40a , 
iP iP
nP nPdist
F,MIN
iP-NP
x z 0.4488 0.25n nx z 0.0051 0.35N 7.76
n 0.62058
 
 
c. Underwood Equation: Assume NKs do NOT distribute: Case A. 
 Eq. (7-33) i ifeed
i
Fz
V 
 For saturated vapor feedV F divide (7-33) by F. 
i i
i
z
1 , which becomes 
 nP NP NP NB NP NBE nP E iP nP iP
ENP iP NP NP NP NB NP
z zz z
1 
 
3.58 0.3 1.86 0.25 1 0.35 0.412 0.10
1
3.58 1.86 1 0.412
 
 Solve for φ between α values of keys. LK = i-propanol, HK=n-propanol. Thus, want 
1.0 1.86 . From Goal Seek on spreadsheet 1.48648 
 Then from Eq. (7-29) i i ,distmin
i
Dx
V where i,distDx values from part a. 
 MIN
3.58 30 1.86 24.65 1.0 0.28
V 0 173.47
3.58 1.48648 1.86 1.48648 1.0 1.48648 
 
MIN MINL V D 173.47 54.93 118.54 , MINL D 118.54 54.93 2.16 
d. 
Min
L
L D 1.1
D
 
 Gilliland abscissa, 
MIN
MIN
MIN
L D 11 1L D L D L D 1.1x
1 1L D 1 1 1
L D 1.1 L D 
 or MIN
MINMIN
L D
1
L D 1.1 1
x 0.0683
L D 1 1
1.1
L D L D 2.16
 
 
172 
 
 From Eq. 7-42b, MIN
N N
0.5456
N 1
 
 MINN 1 0.5456 0.5456 N 33.4 N includes PR 
 
 Assuming F,MINF F
MIN
NN 7.76
N 33.4 17.7
N N 14.62
 or Stage 18 below total condenser. 
 
7.F2 Equilibrium data is available in a variety of sources such as Perry’s Handbook. Data used 
here is from Perry’s (3rd ed.), p. 574. 
a) Need to obtain avg. α from equilibrium data. 
 N 2 N 2
N 2N 2
y 1 x 0.1397 0.9615
x 0.0385, y 0.1397, 4.055
1 y x 0.8603 0.0385
 
 N2x 0.4783, y 0.7893 (needed for part b) 
 N2 N2
.9770 .0810
x .9190, y .9770, = 3.744
.0230 .9191
 
 
top&bot
1/ 2
avg 4.01 3.744 3.875 
 Fenske 
dist
bot
min
AB
x 1 x .998 .002n n
x 1 x .001 .999
N 9.685
n n 3.875
 
 b) D
min
D
x z
L V
x x *
 where x* is in equilibrium with feed y z 0.79 
 From equilibrium data x* ~ 0.48. 
 min
.998 .79
L V 0.40
.998 .48
, min
min min
L VL 0.40
0.66667
D 1 L V 1 0.40
 
 c) 
min
L / D 1.1 L / D 0.7333 
 Gilliland Correlation: abscissa min
L D L D 0.06666
0.0385
1 L D 1.7333
 
 
 Original correlation, ordinate min
N N
0.6
N 1
N = 25.7 including PR. Need 25 equil. Stages 
7.G.1. New problem in 3rd edition. a. At total reflux MINN 9 
 b. 
MIN
L D 0.92 
173 
Chapter 8 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 8.A1, 8.A2, 8A7, 8A12, 8.D1, 8D6, 8D12, 8.D13, 8D15, 8D17, 8D20, 8.D22, 8D23 to 
8.D25, 8.E1, 8G1-8G5, 8.H3.
8.A1. New Problem in 3rd edition. a. 2-pressure distillation
8.A2. New Problem in 3rd edition. b. extractive distillation
8.A7. New Problem in 3rd edition. If there are volatile and non-volatile organics, a single equilibrium
contact gives an organic layer that contains no non-volatiles. Extra stages do not increase the 
separation. If there is entrainment, a second stage may be useful. 
8A.12. New Problem in 3rd edition. Steam distillation is normally operated with 2 liquid phases in the 
still pot and in the settler after the condenser. There is usually no reflux. Azeotropic 
distillation is normally operated with one liquid phase in the column and in the reboiler, but 
with 2 liquid phases in the condenser and settler. One of the liquid phases is refluxed to the 
azeotropic column. 
8.C2. org org org orgorg w in water in w in w 
w w w w
y x p x
y x p x
org org org org org org in org
w w org in ww w w
VPH x x H x
VP x x VP VP x
At solubility pt. w in w org in w org in orgx .975 and x .025, x .573 
Vapor pressures (Perry & Green, 1984). N-butanol: 
T = 70.1°C 84.3°C l00.8°C 117.5°C 
VP = 100 mm Hg 200 400 760 mm Hg 
Water: T = 100.8°C, VP = 782 mm Hg, T = 84.3, VP ~ 421.8 mm Hg 
at 84.3°C: org
w
VP 200
0.474
VP 421.8
, at 100.8: org
w
VP 400
0.5115
VP 782
For org w use average between 92 and 100°C. Can linearly interpolate at T = 96°C, 
org wVP / VP 0.501 , org w in w 0.501 .573 .025 11.483 
w org in w org w in w1 1 11.483 0.0871 
From w o w ww
w o w w
x 0.0871x
y
1 1 x 1 0.9129x
 generate equilibrium curve, 
w
w
x 1.0 .995 .990 .985 .980 .975
y 1.0 .9495 0.8961 0.8512 0.8102 0.7726
At constant wx , the calculated w w,experimin taly y . Difference at Bx 0.975 is 
approximately .7726 .752 .752 100 2.74% . 
These equations work better for mixtures which are more completely immiscible. 
174 
 
8.D1. New Problem in 3rd edition. Top Op. Eq., D Dy L V x 1 L V x , x .975 
 L L D 4 .8
V 1 L D 5
, D
L
y intercept x 0 1 x .2 .975 .195
V
 
 Bx 0.11 (from diagram). Need 2 equil. Stages. 
 
Graph for problem 8.D1. 
8.D2. The columns are sketched in the Solution to Problem 8-C2. 1B is butanol phase and 2B is water 
product. Two equilibrium diagrams are shown. 
 a. 1 2F B B , 1 21 B 2 BFz B x B x 
175 
 
 2
1 2
B
1
B B
z x .28 .995
B 3743.45
x x .04 .995
, 2 1B F B 1256.55 
 b. Col. 1. Bottom Op: 
1B
1
L L
y x 1 x
V V
, Intersects 
1B
y x x 0.04 . 
 Feed: 70% Liquid, q = .7, 
q .7 7
q 1 .3 3
. Intersects y = x = z = .28. 
 Top 
2B
L L
y x 1 x
V V
 
 From Figure 8-D2a: 
min
L 0.995 .436
0.562
V 0.995 0
, 
L
1.23 .562 0.69
V
 
 y intercept 
2B
L
x 0 1 x 0.307
V
 
 
Note that reflux is 0x 0.573 . Optimum feed stage = 3. Need about 3 stages + partial reboiler. 
 Stripper (Column 2): 
22 B
2
L L
y x 1 x
V V
 
 2
2 2
V B 1L V B 1.132
8.57
V V .132V B
 
 Construction is shown in Figure 8-D2b. Need 1 2/3 eq. contacts or P.R. + 2/3 equil. stage. 
 
 
176 
 
8.D3. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
a) y = 0.4, x = 0.09 from graph. B) V/F = 0.3 
 
 20 = Fz = Vy + Lx 
L F V 1 V / F .7 7
V V V / F .3 3
 
 100 = F = V + L 
L F 7 .99
y x z x
V V 3 0.3
 
 20 = 0.4 V + 0.09 (100 – V) See Enlarged figure [Be careful with scales] 
 V 11 .31 35.48 kg moles / h w wy 0.969, x 0.999 
 L = 64.52 kg moles/h 
w w
w w
F z x 100 0.99 0.999 kg moles/hr
V = 30 
y x 0.969 0.999 hr
 
 Use Table 8-2 to find V, drum w wT @ y 0.40, x 0.09, L = 70 and drumT ~ 108 C 
 
177 
 
 
8.D4. Compositions x 0.573 , x 0.975 [Shown on Figure 8-D3a] 
 100 F L L and 88 = Fz = L x L x 
 
F z x 100 .88 0.573
L
x x .975 0.573
= 76.37 kmoles/hr, L 23.63 kmoles/hr 
8.D5. a) Water conc. W is 0.975. 
 200 = F = W + B W = 200 – B 
 Water balance: (200) (.8) = Fz = W(0.975) + B (0.04) 
 Solve: B = 37.433 kmol/hr, W = 162.567 kmol/hr 
 b) 
V B V B 1 5
L V
V V B 4
, Bot. Op. Eq. B
L L
y x 1 x
V V
. 
 Goes through By x x 
178 
 
 Plot operating line: If y = 1, 
1 0.25 .04
x 0.808
1.25
 
 PR + 2 stages moresthan sufficient. (see graph) 
 c) 
min
0.75 0.04
L V Slope 1.332
0.573 0.04
 
 
min
V 1 1
L V 3.012
L V L V 1 0.332
 
 
179 
 
 
8.D6. New Problem in 3rd Edition.
 
8.D.6 Part a) F D B 
 F .65 .975D .02B 
 B
D B
z x .65 .02
D F 100 65.969
x x .975 .02
 
b) V V B B 136.124, Reflux L L V V 170.155. 
 Or V F D L 136.124 100 65.969 170.155 
 c) L V B V B 1 4 1 1.25
V V V B 4
 
 B
L L
y x 1 x
V V
 
 Goes through By x x 
 
 Calculate arbitrary point at x .6 
 y 1.25 .6 .25 .02 .745 
 See Figure: Need 2 stages + PR 
 
 d) What is 
MIN
V B ?
MIN MAX
V V 1
3.0894
B L V L V 1
 On graph. 
 
F 
 
V 
 
L
 
botB, x
 
distD, x 
 
Reflux 
180 
 
 
 
181 
 
8.D7. A WW W in organicW A
in organic A in organic W in organicA A in organic
y 0.0364, y 1 0.0364 0.9636y x
 
x 0.628, x =1-0.628=0.372 y x
 
 
 W A
in organic
0.9636 0.372
44.69 
0.0364 0.628
 
 
8.D8. Convert wt frac to mole frac. 
 8 14MW C H O 72 14 16 102 and MW water 18 
 
 Basis 1000 kg 
 0.994 wt frac. ether: 6 kg W = .333 kg moles 
 994 kg E = 9.745 
 Total = 10.078 
 W in organicx 0.033 
 
 0.012 wt frac. ether: 988 kg W = 54.889 kg moles 
 12 kg E = 0.118 kg moles 
 W in organicx .998 
 
 y = 0.959 ether: 41 kg W = 2.278 kg moles 
 959 kg E = 9.402 kg moles 
 Wy 0.155 
 
W W in org
W - E in org
E E
y x .195 .805
7.026 
.033 .967y x
 
 
 z(wt) = 0.004 water: 4 kg W = 0.222 kg moles 
 996 kg E = 9.765 kg moles 
 Wz 0.022 
 
 in equil w feed
7.026 0.022x
y 0.138
1 1 x 1 6.026 0.022
 
 
 D equil w feed
min D
x yL .998 .138
0.882
V x z .998 .022
 
 
 min
min act actmin
L VL L L
7.467; 11.20; 0.918
D 1 L V D V
 
 
 Generate following equilibrium data using w E in org 7.026 : 
 
xW 0 0.01 .022 .033
yW 0 0.066 0.137 0.195 
182 
 
 Top Op. line: D
L L
y x 1 x
V V
 
 Where L/V = 0.918, Dx .998 , and y = intercept = (1 - .918) (.998) = 0.082 
 
 Bottom: From By x x to intersection of feed line and top operating line. 
 Bx 0.0004 wt frac.: 0.4 kg W = 0.022 kg moles 
 999.6 kg E = 9.800 kg mole 
 Obtain Bx 0.0023 
 See plot in Figure: Optimum feed is top stage. Need 
3
4
5
 equil. Contacts. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
8.D9. Convert to Mole fractions: 
 6 14 waterMW C H O 72 14 16 102; MW 18 
 
 Basis for all conversions is1000 kg soln. 
 Top Layer Separator = 0.994 wt frac. ether 
 6 kg W = 0.333 kmole 
 994 kg ether = 9.745 kmol 
 Total = 10.078 kmol 
 Mole frac. w in org
0.333
x 0.033
10.078
 
 
183 
 
 
 W W in orgW E
in organ phase E E in org
y x 0.195 0.033
7.026
y x 1 .195 1 0.033
 
 Feed is W
in org
y 0.02 mole frac water = z. In equil. With feed: 
 *f
y 0.02
x 0.002896
1 y 7.026 6.026 0.02
 
 D
*min
D f
x z 0.998 0.02
L V 0.9828
x x 0.988 0.002896
,
min
L VL
57.18
D 1 L V
, 
 minactL D 2 L D 114.36 , act
L D
L V 0.9913
1 L D
 
 Top Op. D,Wy L V x 1 L / V x through D,Wy x x 0.998 . 
 When x = 0, y = 0.00868 
 Eq. Data. 
7.026x
y
1 6.026x
. Generate curve, 
 
 
x 0 0.01 0.022 0.033
y 0 0.066 0.137 0.195 
Plot on graph, and plot top op. line: 
x 0, y 0.00868. x 0.04, y 0.9913 .04 0.00868 0.04833 
 x 0.03, y .9913 0.03 0.00868 
 W,botx is at intersection y = z = 0.0208 top op. line, W,botx 0.0123 
 Step off stages from top down. 1 equil stage is sufficient. 
 But with this very high reflux rate consider alternatives. 
Dx 0.998
 
0.033 
z = 0.02 
Bottom Layer separator is 
= 0.012 wt frac. ether 
Mole frac. W in org Dx 0.998 x 
Vapor into Condenser is 
yazeotrope = 0.959 wt frac. ether 
988 kg W = 54.889 kmol 
12 kg E = 0.118 kmol 
41 kg W = 2.278 kmol 
959 kg E = 9.402 kmol 
Mole frac. Wy 0.195 
184 
 
 
8.D10. a. C10 WC10 WVP x VP x 760 
 Assume the water layer is pure, Wx 1.0.Try 95.5°C, C10 WVP 60, VP 645.7. 
 (.99) (60) = 645.7 = 705.1. Too low. Try higher temperature. The attached plot of VP 
vs. T allows estimation of vapor pressure. (Note: a plot of log (VP) vs 1/T will be easier 
to interpolate and extrapolate.) 
 97.0°C: C10 WVP 63, VP 682.07 , (.99) (63) + 682.07 = 744.4 
 97.5°C: C10 WVP 65, VP 694.57 , (.99) (65) + 694.57 = 758.9 
 T = 97.6 gives WVP 697.1 which will be too high. Thus T = 97.5°C is close enough. 
185 
 
 b. tot org orgw
org org org
p VP x 760 65 .99n
10.81
n 65 .99VP x
 
 This is significantly less than in Example 8-2 where 296.8/4.12 = 72.04 mol decane are 
used. Difference is due to higher n-decane concentration in liquid. 
 
8-D11. F,orgx 0.9, 95% recovery → 5% left. Octanol left = 0.05 (.9) (1.0) = 0.045 kmol/h 
 Nonvolatiles in bottoms = 0.10 kmol/h 
 
water 
steam 
octanol 
Fx 
oc tan ol
in org
0.045
x 0.3103
1.10 0.045
 
W 
186 
 
 a) Water VP can be fit to 10
B
log VP A
273.16 T
 over short ranges T. T in C, VP is mmHg. 
 T = 95.5 0
B
log 645.67 A
273.16 95.5
 2.8100 A B 368.66 (1) 
 T = 100 10
B
log 760 A
273.16 100
 2.8808 A B 373.16 (2) 
 
 To solve for A and B, subtract 1 from 2 
 
B B
0.07080 0.00003271B
368.66 373.16
B = 2164.42 
 
B
A 2.8808 8.68105
373.16
 
 Now find T for which tot W W O O W Op VP x VP x 760 mm Hg where x 1.0, x 0.3103 
 On Spread Sheet find T = 99.782°C 
 b) OOO
tot
VP x 19.075
y 0.025098 0.3103 0.007788
p 760
 
 W
754.072
y x 1.0 0.99220
760
 
a) Moles octanol = OF z .95 1.0 0.90 0.95 0.855 kmol/h 
b) Moles water WW org
org
0.9922y
n n 0.855 108.93
y 0.007788
 
 Check Eq. (8-18): W
0.855
n 760 19.075 0.3103 108.93
19.075 0.3103 
 
8.D12. New Problem in 3rd edition. All cases 1 2B 60, B 40 , 
P,B2 P,dist1
2 2
P,dist1 P,dist 2
x x
D B
x x
 
a) 2
1.0 .65
D 40 140
.65 .55
 
 1D 140 40 180 . Total feed Col 1 = 246 
b) 1
.35
D 280 1400
.01
 
 1D 1440. Total feed Col 1 = 1500 
 
 
187 
 
8.D13. New Problem in 3rd edition. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Part a. 1 2F F D B 100 80 D B 
 Water:
 1 1 2 2 dist botFz F z Dx Bx 100 .84 80 .20 D 0.975 B 0.04 
 Solve simultaneously, D = 99.25 and B = 80.75 
 b) 
V
V B 121.125, L V B 201.875
B
 
 Since feed 2 is saturated liquid V V 121.125 
 2L L F 121.875 
 1 1 dist bot 2 2
dist 1 1 2 2 bot
c) Doing Mass balance around top Doing Mass balance around bottom
V y F z L x Dx V y Bx L x F z
Dx F z F z BxL L
y y x
V V V V
 
 
 These two equations are equivalent. 
 d) Bot. op. line: B
L L
y x 1 x
V V
. Goes through By x x . 
 Slope 
V B 1L V B
5 3
V V V B
. Plot Bot Op. line. 
 Op. line above feed 2: Slope 
L 121.875
1.0062
V 121.125
 
 At intersection 2F feed line and bot op. line (at x .2, y 0.306667 ) with 
 slope 1.0062 
 2 stages + PR is more than sufficient (See graph). 
1F
 
2F 
V 
V L
 
L
 
botB, x
 
distD, x 
 
188 
 
 
Graph for 8.D13. 
 
8.D14. Figure is on next page. 
 
Part b. 1 2F B B , B1 B21 E 2 EFz B x B x 
 B21
B1 B2
z x 0.85 0.006
B F 100 85.60
x x 0.992 0.006
, 2 1B F B 14.40 kmol/h 
c) b1 b 2 d12
b 2 b1 d1 d 2
z x x x
D F 
x x x x
0.85 0.992 0.006 0.449
100 21.196
0.006 0.992 0.449 0.75
 
 1 2 2D D B 35.596 kmol/h 
 
189 
 
 
8.D.15. Part a) New Problem in 3rd Edition. 
org oc tan ol oc tan ol total oc tan olp VP x =P where x is mole fraction octanol in organic phase. 
 At 0.05 atm and boiling T, orgp 0.05 atm. 38 mmHg 
 From Antoine equation, 
 10 oc tan ol
1310.62
log VP 6.8379
T 136.05
 
 At octanolT 129.8C, VP 80.905 mmHg 
 Since orgorg octanol,mole
benzene
p 38
p 38 mmHg, x 0.470
VP 80.905
 
 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg 
 
mol octanol 15 130.23
0.470
15 85mol octanol mol non-volatiles130.23 MW
 
 
15 85 15
0.470 MW 654.04
130.23 MW 130.23
 
 b) 95% recovery is true on both mass and mole basis. 
 Distillate octanol flow rate 0.95 100 0.15 14.25 kg h. 
 Since oc tan olMW 130.23, this is 14.25 130.23 0.109 kmol h. 
In waste there are 0.05 15 0.75 kg hr octanol and 85 kg h (organics + solids), or 
85.75 kg h total. Wt frac octanol 0.75 85.75 0.00875. 
 Mole frac. octanol in waste 
0.75 130.23
0.0424
0.75 130.23 85 654.04
 
Water 
99.4 mole % water 
2B 
2D
 
Ex 0.75 
1B
 
101.3 kPa 
F 
Ez 0.85 
Ethanol 
99.2 mole % 
0.449 
ethanol 
1333 
kPa 
1D 
190 
 
 c) For equilibrium in still pot oct in org w totoctVP x VP 1.0 p 760 mmHg . 
 The still pot is perfectly mixed; thus oct in org oct in wastex x 0.0424 mole frac. 
 Since water boils at 100°C when P 760 mmHg, T < 100°C. 
 Eq. (8-15) becomes oct wVP 0.0424 VP 1.0 760 
 Substituting in the Antoine equations for octanol and water and solving with a spread sheet, T = 
99.97°C. 
 oct wVP 19.27 mmHg and VP 759.18 mmHg. 
d) From Eq. (8-18), oct oct oct
w tot oct oct
n VP x
n p VP x
 
 From spread sheet oct wn n 0.001076 
 Since octn 0.109 kmol h, w octn n 0.001076 101.27 kmol h 
 
kmol 18.016 kg
water 101.27 1824.5 kg h
kmol kmol
 water in distillate. 
 This is a lot of steam! 
 
8.D16. Distillate 1: 0.997 EtOH, 0.0002 solvent. Calculate 
1,Wd
x 1 .9972 0.0028 
 Distillate 2: 0.999 water, 0.00035 solvent. Calculate 
2ED
x 1 0.99935 0.00065 
 F = 100, F,E F,solvx 0.81, sat'd liq'd, x 0 
 Find 1 2D , D , M where Makeup is pure solvent. 
 0 
 Water: W W 1 D1W 2 D2Wx M Fz D x D x 
 0 
 Ethanol: E,M E 1 d1,E 2 d2,EMx Fz D x D x 
 
 Ethylene Glycol: 
solvE solv 1 d1,solv d2,solv
Mx 1.0 Fz D x Dx 
 
 Solving water & ethanol balances obtain: 2 D 18.7913 and 1D 81.2316 kmol/h. 
 From Ethylene Glycol balance, 
 M 81.2316 0.0002 18.7913 0.00035 0.02282 kmol/h 
 Can also use overall balance instead of EG bal. Then 
 1 2M D D F 18.7913 81.2316 100 0.2290 , OK 
 
8.D17. New Problem in 3rd edition. Since everything now exits the bottoms, B = S + F, and xA = 
FzA/(S+F), xB = FzB/(S+F), xsolvent = S/(S+F). 
 
8.D18. Ethanol Product: 0.997E, 0.0002 solvent, 0.0028 water 
 
 Water Product: 0.9990W, 0.00035 solvent, 0.00065 ethanol 
 
 E WF 100, z 0.20 z 0.80 
 
191 
 
 Water bal: WM W E EP,W W WP,Wx M Fz P x P x (A) 
 
 E bal: EM E E EP,E W WP,Ex M Fz P x P x (B) 
 0 
 EG M,solvent solv E EP,solv W WP,solvx M Fz P x P x where M,solventx 1.0 
 
 Solve A & B E Wfor P & P : WP 80.0240 kmol/h, EP 20.0074 kmol/h 
 
 M 80.0240 0.00035 20.0074 0.0002 0.03201 kmoles/h 
 
8.D19. M.B. around System. Since everything in wt. units do M.B. in wts. 
 
 Overall: 1 2F H B B where H = makeup hexane. 
 
 Ethanol: E 1 E,bot2 2 E,bot2Fz 0 B x B x 
 0 
 Hexane: H 1 H,bot1 2Fz H 1.0 B x B 0 
 
 Solving simultaneously, 1 2B 8000.04, B 2000.04 and H 0.08 kg/h. 
 
8.D20. New Problem in 3rd edition. a. 1000 F E W 
 E,Ethanol prod E,wprod1000 0.8094 Ex Wx 
 809.4 0.998E 0.0001W 
 
808.3
E 811.0 kmol h
0.9979
 
 W F E 1000 811 189.0 
 
 b) FW W,Ethanol prod
w,1
Fx Ex 1000 0.1906 811.0 0.002
V V 629.93
y 0.300
 
 boilup ratio 
629.93
V E Pr od 0.777
811.0
 
 
 L V E Pr od. 629.93 811.0 1440.93 
 L L F sat 'd liquid feed 1440.93 1000 440.93 
 If CMO strictly valid then, L reflux 440.93 
 
 Can also estimate from _ settler 1L Pentane flow rate in V Ethanol flow rate in 1V 
 Ethanol lost in Water Product. 
 
 Pentane flow rate P,1 1y V 629.93 0.6455 406.62 
192 
 
 Ethanol flow rate in 1V E,1 1y V 629.93 0.0555 34.96 
 
1 lost
E,reflux
from settler to Col1
E in V E
 x
L
 
 
 
prodprod E in W
Ethanol lost in water product W x 189.0 .0001 0.0189 
 from _settler _ calculationL 441.6 
 
 Match not perfect because: 1. CMO not totally valid 
 2. There is some water in reflux 
 3. dEK value may be incorrect. 
 
 c. Ethanol returned to distillation column 
 1 E1 Prod E,W ProdV y W x 34.96 0.0189 34.19 kmol h. 
 Using average estimate for 0
440.93 441.6
L 441.3
2
 
 Then E in pentane34.91 441.3 x 
 E,pentane E,Reflux,pantane _ layerx 0.0792 x 
 
 Then since assume d E,Water layerK 1, x 0.0792 
 
193 
 
 
 d. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
8.D21. 
 
E,waterx 0.0792 
1
V
V 0.5 V 0.5W 94.5 kmol h.
W
 L W V 283.5 kmol h.
 
W 189.0
 
L 283.5
3
V 94.5
 
1E
L 0.0792 W .0001
y
V
 
1E
283.5 .0792 189.0 .0001
y 0.237
94.5
 
E,W Prodx 0.0001 
1 
194 
 
Bottom: B
L V B 1 1.5 L L
3, y x 1 x
V V B 0.5 V V
.Goes through By x x with slope = 3 
 Feed line = Horizontal (q = 0). Through y = x = z = 0.4 
 Top. MB: DyV Lx Dx and V L D 
 D D
L L
y x 1 x goes through y x x 0.975
V V
 
 Intersects Feed line where bottom op line does. 
 Opt. Feed #1 above reboiler. 3 equilibrium stages + PR is sufficient. 
 
8.D22. New Problem in 3rd edition.
 
 
Part a. dist botF D B & Fz Dx Bx 
 bot
dist bot
z x 0.20 0.08
D F 100 13.41 kmol/h
x x 0.975 0.08
and B 86.59 
 
Part b. distyV Lx Dx 
 dist
L D
y x x
V V
 Substitute in D V L to obtain dist
L L
y x 1 x
V V
 
 Points on operating line: disty x x 0.975 and dist
L
x 0, y intercept 1 x
V
 
 Alternative point is at feed line (y = z = 0.2) & botx x 0.08 
 V F, L B, thus slope L V B F 86.59 100 0.8659 
 y intercept 1 0.8659 0.975 0.1307 
Part c. Need 2 Stages. See graph. 
Part d. Pinch at feed line intersection with equilibrium is at x 0.02 . 
 
 
 
V 
 
L
 
B 
distD, x 
Reflux 
F 
195 
 
 
Figure for problem 8D22. 
 
196 
 
 
8.D23. New Problem in 3rd edition. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Part a. External balances Prod ProdF NM W 
 Prod ProdNM : F .25 NM .98 W .01 
 Pr od Prod
.25 .01
NM 100 24.74, W 75.26
.98 .01
 
 
a. W Column: z = .25, horizontal feed line 
Top NM NM NM,bot ,col NM
L L
y x 1 x
V V
 Mass balance through top of W column 
and around col. NM. Can easily show that 
NM 
W F xw = 0.312 
Nitro Methane 
Product 
Water 
Product 
XNM = 0.086 
Water phase 
N.M. Phase 
197 
 
NM NM NM,bot,col NMy x x 
But do not know L/V so cannot plot yet. 
Bottom operating line looks familiar: NM NM,bot col w
L L
y x 1 x
V V
 
 NM NM NM,bot,col wy x x 0.01 
 
col w
L V B 1 5 4
5
V V B 1 4
 
Can plot bottom operating line. Arbitrary point: x 0.2, y 5 0.2 .01 .96 
Now can plot top operating line from intersection of bottom operating line and feed line to point 
 NM NM NM,bot,col wy x x 0.01 
See graph. Need PR + ~ 1 2 stage. Build PR + 1 stage. 
b. NM Column is a stripping column: w w w ,bot ,col NMy L V x L V 1 x 
 
col NM
V
1 4BL V
V 3
B
 
 w w w,bot,NM _ coly x x 0.02 
To plot, w w w
4 1
x .3, y .3 .02 0.3933 x 0.3 is arbitrary point
3 3
 
 Need PR + ~ 
1
1
3
 stages. 
c. W col. Want V. Prod
V 1 1
V 100 V, V= B W 75.26 18.81
B 4 4
 
 V to cond. from colW 118.81 kmol hr 
 NM col want Pr od
V
V. V= B 3 NM 3 24.74 74.22 kmol hr
B
. To condenser. 
 
198 
 
 
Graph for 8.D23. 
 
 
 
199 
 
8.D24. New Problem in 3rd edition. From Equilibrium, bu tan oly 0.092 at bu tan olx 0.004 
Overall Mass Balance: 100 F V B 
Butanol MB: 100 0.025 V .092 B .004 
 2.5 .092 V F V .004 
 2.1 .088V V 23.864, B 76.136 kmol hr 
This problem can also be solved graphically, but using basic mass balances is easier. 
 
8.D25. New Problem in 3rd edition. Part a) org benzene benzenep VP x where benzenex is 
mole fraction benzene in organic. At boiling T, orgp 1.0 atm. 
 From Antoine equation, 10 benzene
1211.033
log VP 6.90565
T 220.790
 
 At T 93 C, benzeneVP1112.44 mmHg . Since 
 orgorg ben,mole
benzene
p 760
p 760, x 0.683
VP 1112.44
 
 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg 
 
 
Moles benzene 20 78.11
0.683
20 80Moles benzene + Moles non-volatiles
78.11 MW
 
 
80 .68320 80 20 20
0.683 1 .683 MW 673.2
78.11 MW 78.11 MW 78.11
 
 
b) 90% recovery is true on both mass and mole basis. 
 Distillate benzene flow rate 0.9 100 0.2 18.0 kg h. 
 Since benzeneMW 78.11, this is 18 78.11 0.230 kmol h 
 In waste there are 2.0 kg/h benzene and 80 kg/h (organics + solids), or 82 kg/h total. 
 Wt frac benzene 2 82 0.0244 
 2 78.11Mole frac. benzene 0.1773
2 78.11 80 673.2
 
c) For equilibrium in still pot b in org w totbVP x VP 1.0 p 760 mmHg 
 The still pot is perfectly mixed; thus, b in org b in wastex x 0.1773 mole frac. 
 Since water boils at 100ºC when P 760 mmHg, T 100 C . Benzene is more volatile and 
boils at 80.1ºC, but mole fracs low. 
 Antoine equation for water: 10 w
2164.42
log VP 8.68105
273.16 T
 
 Eq (8-15) becomes b wVP 0.1773 VP 1.0 760 
 Substituting in the Antoine equations for benzene and water and solving with a spread sheet, 
T 92.0411 C . ben wVP 1082.5 mmHg and VP 568.1 mmHg. 
 
200 
 
d) From Eq. (8-18), ben ben ben
w tot ben ben
n VP x
n p VP x
 
 From spread sheet ben wn n 0.337876 
 Since benn 0.230 kmol hr, w benn n 0.337876 0.6807 kmol hr 
 
 
kmol 18.016 kg
water 0.6807 12.264 kg h
kmol kmol 
water in distillate. 
e. To vaporize benzene condense moles water ben ben
w
n
. 
 This occurs at 92.0411 C 365.1911K; From Perry’s table 2-237, 
 
 
T = 360 886.7 498.7 388
T = 370 898.6 518.1 380.5
fh hgH h H h kJ kg
 
 
 Linear interpolate 
5.1911 kJ
x 380.5 388 388 384.1 30,002 kJ kmol.
10 kg
 
 Note: 8th edition, Table 2-193 is very slightly different after unit conversion. 
 
Water Table 2-352. 
 
 w
0.0911
2265.67 2278.3 2278.3 2277.8 kJ kg 41,037 kJ kmol
5
 
 Moles water condensed 
0.230 30,002
0.1682 kmol h
41037
water (in waste) 
 kg h water in waste 0.1682 18.016 3.029 kg h 
 
 
 
 
T = 360 2663 384.7 2278.3 
T = 370 2671 405.88 2265.67 
201 
 
 
8.E.1. New Problem in 3rd edition. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Part a. 1 2 NM W250 F F P P 
 NM balance 
NM NM1 1 2 2 MN NM Prod w w Pr od
NM mol frac NM mol frac
F z F z P x P x 
 NM w135.5 8 127.5 P .98 P 0.01 
 NM NM NM135.5 P 250 P 0.01 .97P 2.5 
 NMP 133 .97 137.11 kmol hr 
 wP 250 137.11 112.89 
 b. Column W – Use NM NMy vs. x (water phase) plot. 
 Top operating line 22 NMwaterNM NM NMwcolcol NM Pr od wcol
F z
y L V x 1 L V x
V
 
 Bottom Operating Line water waterNM NM NM
col col Water Pr od
y L V x L V 1 x 
 
 col w
col w 1
V
V B 37.63 VV B V B 1 4 3
L V 4. .B
V V B 1 3 L V B 150.52 L F
 
NM 
W xw = 0.312 
Nitro Methane 
Product 
Water 
Product 
XNM = 0.086 
Water phase 
2F 
1F
 
N.M. Phase 
202 
 
 Top IS NOT from NM
NM Pr od
y x x 0.98 to intersection feed and bottom operating line. 
Instead from intersection of feed and bottom operating line with slope 
 1L V L F V 50.52 37.63 1.32 25 . 
 Optimum feed is top stage. Need PR + 1 stage. 
 
 c. Column NM. Top 1 1,NMNMNM w wcolcol W,W Prod.
F z
y L V x 1 L V x
V
 
 Bottom w w wNMcol NMcol
in WM Prod.
y L V x L V 1 x 
 w w w
NM Pr od
y x x 0.02. 
L V B 1
2.00
V V B
 
 Draw bottom operating line. Top is through intersection bottom operating line and feed line F2 . 
Slope L V 0.906 (see item d). Need 2 stages + PR. Optimum feed is stage above PR. 
 
 d. Column W: W col wP B 112.89 
 col wV V B B 1 3 112.89 37.63 
 col wL V B 150.52. L L F 50.52. 
 Saturated liquid feed: 
kmol
V V 37.63
hr.
 
 Column NM: NM col NMV B 1.0, P B 137.11 
 
V B 1
L V 2.0
V B
 
 V V B B 137.11 kmol hr, L 2.0 137.11 274.22 
 Saturated liquid feed V V 137.11, L L 150 124.22 , L V 0.906. 
e. Minimum boilup rate columnNM gives combination bottom & top operating lines to go through 
reflux point: W wy 0.5, x 0.312 . 
 From bottom operating line intersection with feed line 2x z 0.15 is 
 INTER 2 W,NM Prod
L L
y z 1 x
V V
 
 Slope of top operating line to reflux point is 
 INTER 2
0.5 y L FL
V 0.312 0.15 V
 
 
 
V 1
V B .
L V L V 1
 V V B B 137.11 V B .
 
 L V B 
 2int er
L F
Guess V B Calc V & L & L V Calc y Calc L V 
V
 Check is two calculated values L V are same. 
203 
 
 Spreadsheet. 
MIN
V B 0.6105, L V 0.846 
 
 
Graph for Problem 8.E1. 
204 
 
8.E2. Balances at mixing point for F & R. 
 
 To Butanol Column: 
 Overall: TF F R 100 R 
 Water: T T,W W W,reflux T T,WF z Fz Rx F z 30 0.573R 
 T,W
R .573 30
z
100 R
 
 External balances: 100 = W + B, water: 30 = 0.995W + 0.02 B 
 Solve simultaneously: W = 28.72 & B = 71.28 kmol/h 
 
 Butanol Col: V B 1.90, V 1.90B 135.432 
 TF L V B 206.712, L V 1.5263 
 W,butonaly x x 0.02 
 TR F F 206.712 100 106.712 
 T
106.712 0.573 30
z 0.4409
206.712
 
 Vertical feed line at Tz intersects bot. operating Line at y = 0.67 (see graph) 
 
Water Col. 
V B V B 1
V B 0.1143, L V 9.748
V V B
 
 
 B,water,watercol,y L V x L V 1 x B,W
watercol
y x x 0.995 
 See graph, y leaving column = 0.8 
 
 
205 
 
 
 
 
 
 
 
206 
 
 
 
 
8.E3. Basis: 1000 kg sea water (1 h): 
 F,W965 kg water kmol 18.016 kg 53.5635 kmol, x 0.98894 
 F,salt
0.5988
35 kg NaCl kmol 58.45 kg kmol, x 0.011056
Total 54.1623
 
 Water Condensate = (0.60) (53.5635) = 32.1381 kmol/h = Wn 
 Water Remaining 53.5635 – 32.1381 = 21.4254 
 Waste Water is 
21.4254 kg moles W
0.9728
21.4254+0.5988 salt
 mole frac. water 
 
a. In still, organic phase is pure decane. 
 C10 W W tot WVP VP x p 760 mmHg where x 0.9728 . 
 Try T = 99°C. C10 WVP ~ 68, VP 733.2 mm Hg 
 68 + (0.9728) (433.2) = 781.27 mm Hg, which is too high. 
 Converge to T ~ 98.2°C. 
 b. Distillate: W W W W W
org org org org org
707.27 0.9728n y p VP x
10.4225
n y p VP x 66 1.0
 
 (This calculation is at 98°C, not 98.2, but will be close.) 
207 
 
 
 C10 10
32.1381 kmol water/h kmol
n 3.0819 C
10.425 mol water/mol organic h
 in distillate 
 
8.F1. V.P. Data n-nonane (p. 3-59 Perry & Green, 1984) 
 
VP = 20 40 60 100 200 400 760 mm Hg
T = 51.2 66 75.5 88.1 107.5 128.2 150.8°C 
 
 See Solution problem 8.D10 for plot. nC9MW 128.25 . From p. 3-268 (Perry & Green, 
1984), nonane enthalpies are, 
T hliquid Hgas
360 K 671.3 KJ/kg 998.2
380 K 722.5 1036.5 
Water VP is given in Problem 8-D10 and on p. 3-45 of Perry and Green (1984). 
 
 a. Try 95.0°C. C9VP 127 mm Hg, WVP 633.9 . Assume water is pure. 
 Pressure: (.99) (127) + (1.0) 633.9 = 759.6. Close enough and lucky! 
 
 b. tot org orgW
org org org
p VP x 760 127 .99n
5.045
n 127 .99VP x
 mol water/mol nonane 
 
 c. Need to calculate the energy required to vaporize the nonane. T = 273 + 95 = 368 K. 
 By linear interpolation for pure nonane: 1 gas nonaneh ~ 691.78, H ~ 1013.52. 321.74 KJ/kg 
 
 Table 3-302 of Perry and Green (1984): 
Wliq,W vap W
h 397.36, H 2667.8, 2270.44 KJ/kg 
 C9
9 W
321.74 KJ/kg 128.25 kg/kmolmol water condensed
1.009
mol C vaporized 2270.44 KJ/kg 18.016 kg/kmol
 
b. Now, C9 WVP .020 VP 760. Temperature will be higher. 
 Try 99°C: C9 WVP 149 mm Hg, VP 733.24 
 (149) (.020) + 733.24 = 736.22 which is too low. 
 
 Try 99.9ºC: C9 WVP 154, VP 757.29 
 154 (.020) + 7570.29 = 760.37. Close enough. 
 
 The low nonane conc. reduces nonane partial pressure and operation is much closer to 100°C. 
 tot org orgW
org org org
p VP x 760 154 .02n
245.75
n 154 .02VP x
 
 
 Need lot more steam! 
208 
 
8.F2. 
 
 
 Basis: 1 hour 
 a) 
All junk in feed (0.50 kmol) is in bottoms
Organic Bottoms is 0.95 n-C9 0.5 junk 1.45 (see part C)
 
 C9,bot ,org.95
x 0.65517
1.45
 
 b) Still T. W org totp p p 102.633 kPa 770 mm Hg 
 W W W Wp VP (T)x where x 1 
 org C9 c9,org,bot C9 C9p K T x K T 0.65517 770 mm Hg 504.481 K T 
 Procedure: Guess T, determine W C8VP & K check if pressure eq. is valid 
 C9WVP 504.481K 770 mm Hg? 
 Try T = 96°C W C9VP 657.62, K 0.16 (DePriester Chart) 
 652.62 80.717 738.34 Need higher T 
 T = 97°C, W C9VP 682.07 and K 0.17 
 682.07 + 85.762 = 767.83 slightly low, but close enough 
 
 c) 9.50 kmol n-nonane × .90 = 8.55 kmol n-nonane in distillate. 9.5 – 8.55 = 0.95 kmol n-C9 in 
bottoms 
 d) Eq. (8-18) org org orgW
W W
n p n85.762 8.55
0.12567, n 68.031
n p 682.07 0.12567 0.12567
kmol water 
 
 e) EB simplies to W W,condensed nonane org,dist.n n where λ’s are at 97°C = 370 K. 
 C9 C9W,condensed org,dist.
W W
n n 8.55 kmol 
 
water 
F = 1000 
steam 
n-nonane 
water 
organic 
waste 
95% n-nonane 
209 
 
 From Perry’s 6th ed. Table 3-268 or 7th ed. Table 2-292, nonane λ’s are: 
 g f
h h , @ 360 K, 998.2 671.3 326.9 kJ/kg
 380 K, 1036.5 722.5 314.0 kJ/kg
 
 @ 370 K, λ ≈ 
326.9 314
320.45
2
 kJ/kg 
 MW 9C 128.258 . Then at 370 K, 
 C9
kJ 128.258 kg
320.45 41,100.3 kJ/kmol
kg kmol
 
 Water 370 K: (Perry’s 6th Ed., Table 3-302). 
 W
kJ 18.016 kg kJ
2671 405.8 2265.2 40,809.84
kg kmol kg
 
 W,cond
41,100.3
n 8.55 8.611
40,809.84
 kmol water 
 
8.G.1. New Problem in 3rd Edition. 
1. Final makeup solvent flow rate _________0.02__________ kmol/h. 
2. Final value solvent recycle rate (B2) __1400___kmol/h and L/D in col 1 _0.100_. 
3. Final values of flow rates D1 _140.0_, B1 _1460.02_, and D2 __60.02___ kmol/h. 
4. Mole fractions in stream D1 _Pyr=0.0084259, W=0.99157, Bisphenol=.49E-10_ 
5. Mole fractions in stream D2 _Pyr = 0.98001, W = 0.019654, Bisphen = .000333___ 
6. Mole fractions in stream B1 _Pyr=0.040287, W= 0.0008079, Bisphen = 0.95890_ 
7. Mole fractions in stream B2 (solvent recycle stream) Pyr = .526E-8, W = .2E-12, Bisphenol = 
1.0000__ 
8. Heat load in cooler on solvent recycle line__-0.15216E8___ cal/s. 
 
8.G2. New Problem in 3rd edition. Aspen Plus Residue Plot 4.0 atm using NRTL 
 Pressure can have major effect on VLE for non-ideal systems. Compare T-xy diagrams for 
acetone MEK at 1.0 and 4.0 atm. 
 Also compare residue curves for acetone-MEK-MIBK at 1.0 & 4.0 atm. 
 
210 
 
 
 
 
211 
 
 
 
212 
 
8.G3. New Problem in 3rd edition. 
a. Final reflux ratio column 1___0.01_ and final reflux ratio column 2 _0.01_______. If these values are 
not 0.01 you are not finished with Part B. 
b. Flow rates furfural product ___166.0___ kmol/h and water product ___34.0__ kmol/h. 
c. Boilup rate in column 2 _____8.0________ kmol/h. 
d. Mole fraction furfural in furfural product ____0.99816___& mole fraction water in water product 
___0.99102____. 
e. Flow rate of distillate from column 1 _____42.10_____ kmol/h. 
f. Column 1 condenser temperature __370.3___K, & column 1 reboiler temp. __433.59__ K. 
g. Outlet temperature of decanter ____375.2______ K. 
h. Molar ratio of water phase/total liquid in decanter _____0.8393_____ 
 
8.G.4. New Problem in 3rd edition. 
Column 1: a. Bottoms product mole fraction acetonitrile______0.99915______ 
 b. Distillate flow rate ___240____ kmol/h and bottoms flow rate ___170____kmol/h. 
 c. Distillate mole fraction acetonitrile ____0.67910___________ . 
 
Column 2: a. Distillate flow rate ___210_______ kmol/h, and reflux ratio ___1.2____. 
 b. Bottoms product mole fraction water______0.99517______________ . 
 c. Distillate mole fraction acetonitrile _____.77542__________ . 
 
8G5. New Problem in 3rd edition. 
Results are residue curves and profiles of mole fraction vs plate location. For an equal molar feed, N = 10 
does not give the desired purity even if L/D = 10. N = 50 does work with L/D = 2, but not for L/D = 1.0. 
 
8.H1. Part b. Was 8.D12 in 2nd edition of SPE. Use Eq. (8-25b) with 
AB BB BC2.4, 1.0, 0.21. 
 A = benzene, B = toluene, C = cumene. Results from Spreadsheet: 
Stage: Reboiler: 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
Ax 0.0003 
 0.003298 
0.03137 
0.18019 
0.46126 
0.70274 
0.85421 
0.93403 
0.97145 
0.98791 
0.99493 
0.99788 
0.99912 
Bx 0.0097 
0.04443 
 0.176085 
0.42145 
0.44952 
0.28536 
0.14453 
0.06585 
 0.028535 
 0.012091 
 0.005074 
0.00212 
 0.000885 
Cx 0.990 
0.95227 
0.79255 
0.39836 
0.08923 
0.01189 
0.001265 
0.000121 
1.102 E-5 
9.802 E-7 
8.637 E-8 
7.580 E-9 
 6.641 E-10 
 
8.H2. Was 8.D13 in 2nd edition of SPE. Use a spreadsheet with Eq. (8-30) as recursion equation. 
 Result is shown in Figure. The VBA program was given in Example 8-3. The results obtained 
for the starting conditions given are: 
k xA xB xC 
1 0.990 0.001 0.009 
100 0.9763 0.0017 0.0220 
200 0.9431 0.0029 0.0534 
300 0.8630 0.0049 0.1331 
213 
 
400 0.6740 0.0077 0.3183 
450 0.5044 0.0089 0.4867 
475 0.3946 0.0092 0.5962 
500 0.2696 0.0089 0.7214 
600 0.00042 0.00095 0.9986 
 
Results for other starting conditions are shown in the figure. 
 
214 
 
Figure for problem 8H2. 
 
 
8H.3. New Problem in 3rd edition. The spread sheet including the first 10 time steps and time steps 600 
to 610, and the VBA program are listed. 
Part a 
 
 
 
215 
 
 Simple distillation calc (residue curves) with BP calcs. 
 aT1 aT6 ap1 
iB -1166846 7.72668 -0.92213 
nB -1280557 7.94986 -0.96455 
iP -1481583 7.58071 -0.93159 
nP -1524891 7.33129 -0.89143 
nhex -1778901 6.96783 -0.84634 
 
Residue curve calc. h 0.01 N 1000 epsilon 1E-09 
x1iB 0.98 x1nB 0.01 x1iP 0 x1nP 0.01 
sumx 1 T1guess,R 500 p,psia 14.7 x1nHex 0 
time step xiB xnB xiP xnP x nHex T R 
1 0.98 0.01 0 0.01 0 472.0604 
2 0.979883 0.010032 0 0.010085 0 472.0642 
3 0.979764 0.010065 0 0.010171 0 472.068 
4 0.979645 0.010097 0 0.010258 0 472.0718 
5 0.979525 0.01013 0 0.010345 0 472.0757 
6 0.979405 0.010162 0 0.010433 0 472.0797 
7 0.979283 0.010195 0 0.010522 0 472.0836 
8 0.979161 0.010228 0 0.010612 0 472.0876 
9 0.979037 0.010261 0 0.010702 0 472.0916 
10 0.978913 0.010294 0 0.010793 0 472.0956 
 
600 0.071054 0.009754 0 0.919192 0 542.3988 
601 0.069204 0.009586 0 0.92121 0 542.71 
602 0.067391 0.009418 0 0.923191 0 543.0165 
603 0.065615 0.009253 0 0.925133 0 543.3183 
604 0.063874 0.009089 0 0.927037 0 543.6155 
605 0.06217 0.008927 0 0.928903 0 543.9078 
606 0.060502 0.008766 0 0.930732 0 544.1955 
607 0.058869 0.008608 0 0.932524 0 544.4783 
608 0.057271 0.008451 0 0.934279 0 544.7563 
609 0.055708 0.008296 0 0.935997 0 545.0295 
610 0.054179 0.008142 0 0.937678 0 545.2978 
 
Option Explicit 
Sub Residue_Curve_BPcalc() 
' K value data for nbutane,ibutane, ipentane,npentane and nhexane included. 
' Only want 3 for residue curve. Thus, set x values = 0 for 2 components. 
' The reference component is nbutane. 
Dim i, N, j As Integer 
Dim h, epsilon, xiB, xnB, xiP, xnP, xnHex As Double 
Dim T, p, aT1iB, aT6iB, ap1iB, aT1nB, aT6nB, Ap1nB As Double 
Dim aT1iP, aT6iP, ap1iP, aT1nP, aT6nP, ap1nP, aT1nHex, aT6nHex, ap1nHex As Double 
Dim KiB, KnB, KiP, KnP, KnHex, Ksum, chksum, inside As Double 
Dim yiB, ynB, yiP, ynP, ynHex As Double 
 
Sheets("Sheet1").Select 
216 
 
Range("A15", "G1045").Clear 
aT1iB = Cells(5, 2).Value 
aT6iB = Cells(5, 3).Value 
ap1iB = Cells(5, 4).Value 
aT1nB = Cells(6, 2).Value 
aT6nB = Cells(6, 3).Value 
Ap1nB = Cells(6, 4).Value 
aT1iP = Cells(7, 2).Value 
aT6iP = Cells(7, 3).Value 
ap1iP = Cells(7, 4).Value 
aT1nP = Cells(8, 2).Value 
aT6nP = Cells(8, 3).Value 
ap1nP = Cells(8, 4).Value 
aT1nHex = Cells(9, 2).Value 
aT6nHex = Cells(9, 3).Value 
ap1nHex = Cells(9, 4).Value 
h = Cells(11, 4).Value 
N = Cells(11, 6).Value 
epsilon = Cells(11, 8).Value 
xiB = Cells(12, 2).Value 
xnB = Cells(12, 4).Value 
xiP = Cells(12, 6).Value 
xnP= Cells(12, 8).Value 
xnHex = Cells(13, 8).Value 
T = Cells(13, 4).Value 
p = Cells(13, 6).Value 
For i = 1 To N 
j = i + 1 
Do 
KiB = Exp((aT1iB / (T * T)) + aT6iB + (ap1iB * Log(p))) 
KnB = Exp((aT1nB / (T * T)) + aT6nB + (Ap1nB * Log(p))) 
KiP = Exp((aT1iP / (T * T)) + aT6iP + (ap1iP * Log(p))) 
KnP = Exp((aT1nP / (T * T)) + aT6nP + (ap1nP * Log(p))) 
KnHex = Exp((aT1nHex / (T * T)) + aT6nHex + (ap1nHex * Log(p))) 
Ksum = KiB * xiB + KnB * xnB + KiP * xiP + KnP * xnP + KnHex * xnHex 
KnB = KnB / Ksum 
inside = aT1nB / (Log(KnB) - aT6nB - (Ap1nB * Log(p))) 
T = Sqr(inside) 
chksum = Ksum - 1 
Loop While Abs(chksum) > epsilon 
Cells(13 + i + 1, 1).Value = i 
Cells(13 + i + 1, 2).Value = xiB 
Cells(13 + i + 1, 3).Value = xnB 
Cells(13 + i + 1, 4).Value = xiP 
Cells(13 + i + 1, 5).Value = xnP 
Cells(13 + i + 1, 6).Value = xnHex 
Cells(13 + i + 1, 7).Value = T 
yiB = xiB * KiB 
217 
 
ynB = xnB * KnB 
yiP = xiP * KiP 
ynP = xnP * KnP 
ynHex = xnHex * KnHex 
xiB = xiB + (h * (xiB - yiB)) 
xnB = xnB + (h * (xnB - ynB)) 
xiP = xiP + (h * (xiP - yiP)) 
xnP = xnP + (h * (xnP - ynP)) 
xnHex = xnHex + (h * (xnHex - ynHex)) 
Next i 
End Sub 
 
 
218 
Chapter 9 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 9.A4, 9.A5, 9C1, 9D1, 9.D5, 9.D8, 9D11, 9D13, 9D18, 9D19, 9D21, 9D22, 9D25, 9E2, 9.H1. 
9.A4. New Problem in 3rd edition. Answer is g.
9.A5. New Problem in 3rd edition. Answer is c.
9.A6. b. The same.
9.B1.
Multi Stage Single Stage 
F Davg refluxF, x , x , N,L / D,T ,P F DavgF, x , x ,P
F Wf rF, x , x , N,L / D,T ,P F DtotF, x , x ,P
F final rF, x , W , N,L / D,T ,P F wfF, x , x ,P
F tot rF, x ,D , N,L / D,T ,P fW finalF, x , W ,P
fF Davg W r
F, x , x , x , N,T ,P F Davg totx , x ,D ,P
fF Davg W r
F, x , x , x ,L / D,T ,P
fF W final
x , x , W ,P
F Davg tot rx , x ,D , N,L / D,T ,P fF W totx , x ,D ,P
F Davg tot Wf rx , x ,D , x , N,T ,P F Davg WF, x , x , x
F Davg tot Wf rx , x ,D , x ,L / D,T ,P etc. 
F Davg final rx , x , W , N,L / D,T ,P
finalF W final r
x , x , W , N,L D,T ,P
etc. 
9.B2. a. Replace the column with one containing more trays or more packing.
b. Retray or repack existing column.
c. Run a batch in several steps. For example, take the feed and operate so that the desired bottoms
concentration is met. Collect all the distillate and use this as the feed for a second batch. Operate
so that the distillate for this run meets specifications. The bottoms from this run can be used as
feed for a 3rd run, or it can be mixed with the next feed batch.
An alternate is to first collect distillate of desired purity. Then collect distillate which does not
meet purity requirements while bottoms is reduced to the desired purity. The material not
meeting requirements is then mixed with fresh feed for next batch.
Other operating variations are possible.
d. Hook up two batch stills in series – Either to run 1 batch or to run separate batches (second still
takes distillate from first as the charge).
e. See if product specifications can be relaxed.
f. Reducing the pressure increases the relative volatility and may help. However, one must watch
for earlier flooding.
9.C.1. Rayleigh eqn
F
W,fin
x
W
final
D Wx
d x
W F exp -
x x
 219 
Because Dx constant, can integrate analytically. 
 
F
F
w ,final
w ,final
x
x
W D F
D W
D W D W ,finalx x
d x x x
n x x n
x x x x
 
 D F D Ffinal
D W ,final D W ,final
x x x x
W F exp n F
x x x x
 
 Mass balances are finalF W D 
 F final W,final DFx W x D x
 
Solve for D Ffinal
D W ,final
x x
W F
x x
 
 Thus results are identical. 
 
 
9.C2. External balances over entire cycle 
 
finalD wAvg
Fz Dx Wx and F = D + W 
 a. Ignoring holdup on stages and in reboiler 
 - Out = Accum in accumulator is - w Dx dD d Dx 
 which becomes, w D Dx dD Ddx x dD 
 Rearrange, D W D
dD
dx x x
D
 and integrate, 
 
final Dfinal
D F
D x
D
D F x x w D
dxdD
D x x
 which is, 
xD final
F
final D
x D w
D dx
n
F x x
 
 
b. Assume CMO and draw mass balance envelop around bottom of the column. 
 
 L V B wLx Vy Bx 
 w
L B
y x x
V V
 
 w
L L
y x 1 x
V V
 
 220 
9.D.1. New Problem in 3rd Edition. Eq. (9-9) 
F
W,final
x
final
x
dx
W F exp -
y-x
 
From Simpson’s rule 
0.00346 x 0.005173 0.1
0.1 0.00346 1 4 1
Area
6 y x y x y x
 
 From equilibrium curve (Table 2-1). 
x y y-x 1
y-x
 
0.00346 0.03096 0.027498 36.366 
0.05173 .34 .28827 3.469 
0.1 0.4416 .3416 2.9273 
 
 The y values are found by linear interpolation of data in Table 2-1. For example, at x = 0.00346, 
linearly interpolated first 2 data pts Table 2-1. 
 
0.170
y x for x 0.19.
0.019 
 For x = 0.00346, 
0.170
y 0.00346 0.03096
0.019 
 For y at x = 0.1, 
 y = 0.4375 +[ (.4704 - .4375)/(.1238 - .0966)](.10 - .0966) = 0.4416 
 
 [Alternatively, could fit equilibrium data to constant α.] 
 
0.1 0.00346
Area 36.366 4 3.469 2.9273 0.8555
6 
 finalW 0.5 exp 0.8555 0.2125 kmol 
 totalD 0.5 .2125 0.2875 kmol 
 
Avg
F final W,final
D
avg
F x W x .5 .1 .2125 .00346
x 0.1714
D 0.2875
 
9.D2. Rayleigh equation is 
.75
final
.55
dx
W F exp -
y-x
 
 Most of values of 1/(y – x) are listed in Example 9-1. From Table 2-7 can easily generate 
values for x = .55: y = 0.805, y – x = .255. (y – x)-1 = 3.92. The mid-point for Simpson’s 
rule is at x = .65. Then from Eq. (9-12) and values in Example 9-1, 
 
.75
.55
dx .2
3.92 4 5.13 6.89 1.044
y x 6
 
 1.044final total finalW 100e 35.20, D F W 64.8 
 F final finalDavg
total
75 35.2 0.55Fx W x
x 0.859
D 64.8
 
 A graphical integration counting squares gives Davgx 0.861. 
9.D3. Operating equation is 0D
0
L D
y L V x 1 L V x where L V 0.65
1 L D
 
 221 
 This is y = 0.65 x + 0.35 Dx 
 We have two equil. stages (stillpot and one in the column). From McCabe-Thiele diagram 
we can get the values of Dx which are related to Wx . Pick Dx and get Wx from figure. 
From this we can generate the following table (only two values are shown in Figure). 
 
 
 
 
 
 
 
 
 
 
 Simpson’s rule (Eq. (9-12)). [Midpoint wx 0.36 ] 
 
f
w ,final
x
w
x D W
.57 .15dx
3.077 4 2.096 1.754 0.925
x x 6
 
 Raleigh Equation: 
F
wfinal
x
final
x D w
W dx
F x x
, 0.925finalW Fe 39.7 kmol 
 total finalD F W 100 39.7 60.3 kmol 
 F final wfDavg
total
100 0.57 39.7 0.15Fx W x
x 0.847
D 60.3
 
 
xD xW 1/(xD-xW) 
0.90 0.588 3.205 
0.895 0.570 3.077 
0.85 0.394 2.193 
0.837 0.360 2.096 
0.720 0.150 1.754 
0.70 0.128 1.748 
 222 
9.D4. final F wfW 2.0 kg moles, x 0.8, x 0.4 . Find F, DAvg totalx ,D 
 
 
 
 
Can use Simpson’s Rule (Eq. 9-12) or evaluate numerically. 
 
F
Wfin
x
x
dx 0.4
4.76 4 7.143 16.666 3.3333
y x 6
 
 Rayleigh eqn 
F
Wfin
final
x
x
W 2.0
F 56.063 kmol
exp 3.3333dx
exp
y x
 
 total finD F W 54.063 kmol 
 
 fin
AVG
F f W
D
total
Fx W x 56.068 0.8 2.0 .4
x 0.815
D 54.063
 
 
9.D.5. New Problem for 3rd Edition a) 
F
w ,fin
x
final
x
dx
W F exp
y x
 
 Can use Simpson’s rule, eq. (9-12) with equilibrium values from plot. 
 w,final
0.028
1 1
f x 2.7397
y x 0.645 0.28
 
 w,fin F
0.40
x x 1 1
f 3.2787
2 y x 0.705 0.52
 
 F
0.52
1 1
f x 4.65116
y x 0.735 0.52
 
 
x y(equil) 1
y x
 
0.80 0.86 16.666 
0.70 0.80 10 
0.60 0.74 7.143 
0.50 0.67 5.882 
0.40 0.61 4.76 
16.66 
.8 
7.143 
.4 .6 
4.76 
x 
1
y x
 
 
 223 
 
 
 
F
fin
avg
w ,fin
x
F w
w,fin w F
x
x xdx
f x 4f x f x
y x 6
 
 
0.52 0.28
2.7397 4 3.2787 4.65116
6
0.24
20.5056 0.8202
6
 
 finW 3exp 0.8202 3 0.44033 1.321 , V,tot finD F W 1.6790 
 finfin wD,avg
V,tot
Fz W x
y 0.7088
D
 
 b. Settler: V,tot 1 2 1 2D D D 1.6790 D D 
 V,tot D,avt 1 2 2 B 1 2D y D x D x 1.6790 0.7088 0.573 D 0.975 D 
 Solving simultaneously, 2 1D 0.567 and D 1.112 
9.D6. Use eq. (9-13) W ,F FF f
W ,fF W ,f
x 1 xW 1 x1
n n n
F1 1 xx 1 x
 
 224 
a) F = 1.3, F Wfx .6, x .3, 2.4 
 F
.3 4W 1 .4
n n n
1 3 1.4 .6 7 .7
0.8948 0.5596 1.4544 
 F FW 1.3 0.2335 W 0.3036 kmol 
 f
AVG
F fin W
D
fin
Fx W x 1.3 0.6 0.3036 0.3
x 0.6914
F W 1.3 0.3036
 
 b) Now f fW 3.5 0.2335 W 0.81725 
 
AVGD
3.5 0.6 0.81725 0.3
x 0.6914
3.5 0.8125
 
 c) 
AVGF D
F 2.0, x 0.6, x 0.75 
 Since AVG
AVG f
AVG
D Ffin
fin D F fin W
D W ,f
x xW
F W x Fx W x , 
F x x
. 
 Then Eq. (9-13) becomes 
 AVG f
f AVG f ff
D f W
W D W WW
x x x .40.75 .6 1 4
n n n n
0.75 x x x 1.4 1 x.6 1 x
 
 Solution is 
finw
x 0.5025 . AVG
AVG f
D f
final
D W
F x x 2.0 .75 0.60
W 1.212
x x .75 0.5025
 kmol. 
9.D7. L D 1L D 1/ 2, L V slope
1 L D 3
. Pick series Dx values. Plot enriching section op 
line. Step off two stages. Find Wx . Calculate 
F
Wf
x
xD W D W
1 dx
, determine 
x x x x
 
Dx 
 
 
 0.56 
 
 0.49 
Wx 
 
 
0.11 
0.10 
 0.09 
D W
1
f
x x
 
 
1/.45 = 2.2222 
Interpolate 2.361 
 1/.40 = 2.50 
 
0.44 
0.38 
0.26 
 
0.06 
0.05 
 0.02 
 
1 .38 2.6316 
1 .33 3.0303 
1/.24 = 4.1666 
 Simpson’s Rule 
 
F
f
F
Wfin
x
F W
W W F
x D W
x xdx
f x 0.02 4f x 0.06 f x 0.10
x x 6
 
 
0.10 0.02
4.1666 4 2.6316 2.361 0.22739
6
 
 225 
 FinalW 4.0 exp 0.22739 4 0.7966 3.1864 
 
 f
Avg
F f W
D
f
Fx W x 4 .1 3.1864 0.02
x 0.4133
F W 4 3.1864
 
 
 
 
 
9.D8. New Problem 3rd Edition. a) Op. Eqn. d
L L
y x 1 x
v V
 
 d d
4L L
L V 1 x , x .8@ y x
D D 5
, Intercept D
L
y 1 x .2 .8 .16
V
 
 From McCabe-Thiele plot 
finalw
x ~ 0.075 
 
 b) final tot final tot10 F W D W 10 D 
 
finalF final w tot D final tot
10 .4 F x W x D x 0.075 W .8 D 
 Substitute in for finalW , tot tot4.0 0.075 10 D .8 D 
 tot
4.0 10 .075
D 4.483
.8 .075
 
 finalW 5.517 
 
 226 
9.D8. Figure 
 
 
 
 
9.D8. Part C. 
 Trial & Error to 3 stages ending at Fx 0.4 (See figure) 
 INtercept D
L L
y 0.645 1 x 1 .8
V V
 
 227 
 
L .645
1 .80625
V .8
 
 
Initial
L L L L V .15375
.19375, 0.24
V D V L 1 L V 1 .19375
 
 
 
9.D9. a. Mass balances: total totalF D W or 20 D W 
 
fF total D W total
F x D x W x or 8 .975 D .28 W 
 Solving simultaneously: totalD 3.453 and W 16.547 
 Can also use Rayleigh equation to obtain same result. (Use of the Rayleigh equation for this 
type problem is illustrated in the Solution to Problem 9-D14.) 
 
 b. Vapor in equilibrium with 
fW
x must be within the two phase region. x y x 
 Minimum is when y x . This is 
f ,minW
x 0.21 . See graph. 
 228 
 
 
9.D10. Op. Eq., D
L L
y x 1 x
V V
, Slope, L V L D 1 L D 1 5 
 Plot on McCabe-Thiele graph for series of Dx values. 
 Simpson’s rule, 
D W
1
x x
 at 
fin finW F W W W f
x x .52, x x .20, and (x x ) / 2 .36 
 
F
F W ,finW ,fin
F W ,fin
x
W
x xx d W D W D W D Wx x
2
dx 0.52 0.20 1 4 1
x x 6 x x x x x x
 
 
0.32
5.95 4 3.70 2.67 1.2491
6
 
 Rayleigh eq. 
F
W ,fin
x
final W
final
x d W
W dx
n W Fe
F x x
 
 or 1.2491finalW 10 e 10 0.28677 2.8677 and total finalD F W 7.132 
 finalF final WD,avg
total
Fx W x 10 0.52 2.8677 0.2
x 0.648
D 7.132
 
 
 229 
 
 
 
 
 
 
 
 
 
 
xD xW xD-xW 1/(xD-xW)
0.70 0.55 0.15 6.666
0.65 0.405 0.245 4.082
0.60 0.25 0.35 2.857
0.50 0.094 0.406 2.463
0.40 0.055 0.345 2.899
0.30 0.035 0.265 3.774
 230 
Wx 
D W
1
x x
 
Fx = 0.52 5.95 
finF W
x x
2
 = 0.36 
3.70 
W,finx = 0.20 2.67 
 
9D11. New Problem in 3rd edition. Eq. (9-17) 
butanol Inital
butanol final
x 
pot
x
dxS
W y
 
 Note W water 
 Equation is in terms of butanol. 
 
1.0 
but pot
but.6
dxS
W y
 Simpson’s rule – need y at potx 1.0, .8, .6 butanol. 
 
butx waterx watery buty 
but
1
y
 
1.0 0 0 1.0 1.0 
.8 .2 0.565 0.435 2.2999 
.6 .4 0.70 0.30 3.333 
 
 
.4
3.333 4 2.299 1.0 0.90191
6 
 S 0.9019 W 1.804 kmol. 
 More accurate if done in 2 steps. Thus add points below:
 
 
 
butx wx wy buty 
but
1
y
 
.9 .1 .42 .58 1.7241 
.7 .3 0.66 .34 2.9412 
 
 
1
0.8
.2
2.299 4 1.7241 1 0.33985
6 
 
.8
.6
.2
3.3333 4 2.9412 2.299 0.5799
6
 
 
 Total Area = .91975. S = 1.8395 
 
 231 
9.D12. 
F
Wfinal
x
final
x
dx
W F exp -
y x
F
final
W Ffinal
W W Ffinal final
x
F W
x x
x x x
2
x xdx 1 1 1
4
y x 6 y x y x y x
 
 final
final
F W
F W avg
x x
x 0.48, x 0.16, x 0.32
2
 
 
x 
 
y (from eq. data) 
 
(y-x) 
1
y x
 
0.16 0.36 0.20 5.0 
0.32 0.545 0.225 4.444 
0.48 0.66 0.18 5.555 
F
Wfin
x
x
dx 0.32
5.0 4 4.444 5.555 1.51
y x 6
 
final, W 3.0 exp 1.51 0.662 k/moles 
 tot finalD F W 2.338 , 
fin
AVG
F fin W
D
tot
Fx W x 3.0 0.48 0.662 0.16
x 0.571
D 2.338
 
9.D.13. New Problem for 3rd Edition. a) Rayleigh equation: 
f
wfinal
x
final
x
W dx
n
F y x
 
x y 
(from graph) 
y-x 1
y x
 
 0.4 0.7 0.3 3.333 
0.24 0.61 0.37 2.7027 
0.08 0.37 0.29 3.448 
 
 
 Simpson’s Rule 
 
F
w ,fin
x
x
0.4 0.08
3.333 4 2.7027 3.448 0.93826
6
 
 finalW F exp 0.93826 3.1305 total finalD F W 4.8695 
 F final w ,finalD,AVG
total
F x W x
x 0.6057
D
 
 b) Two Liquids. x 0.573 and x 0.975 
 total totalL L D L D L 
 
 total DAVG total totalL x L x D x D L .573 L .975 D .6057 
 
 
 232 
 total
D .6057 .573
L 0.3963
0.975 0.573
 totalL D .3963 4.4732 
 
9.D14. a. org wp p 760 , org C10 C10p VP x , and assuming water is pure, W Wp VP . 
 C10 C10 WVP x VP 760 
 Vapor pressure data for C10 was shown in solution to Problem 8.D10. Guess 99.5ºC. 
 W C10VP 746.52, VP ~ 70.5 mm Hg 
 7 + 746.5 = 753.5 < 760 
 At 100ºC C10 WVP ~ 70.5 and VP 760 . 
 7.05 + 760 = 767.05 > 760 
 By linear interpolation: 
6.5
T 99.5 .5 99.74
13.55
ºC 
b. Use Mass balances. Initially 9 moles n-decane, 1 mole non-volatile 
Final: a mol n-decane where 
a
1
1 a
; thus, 
.1
a .111 mol
.9
 
 finalW 1.111 mol (Water free) 
 total finalD F W 10 1.111 8.889 kmol 
 
 Alternate Solution: Raleigh Eq. with 
 
F
F
W ,final
W,final
x
x
D final W x
Wx
dx
x 1: W F exp - F exp n 1-x
1 x
 
 finalW F exp n .1 n .9 .111 F 1.111 
 total finalD F W 8.889. Same result as mass balance. 
 c. tot org org in orgW org
org org in org
p VP x
n D
VP x
 
 Should really calculate numerically from integral for most accuracy 
 
D
tot org org in org
W org
0 org org in org
p VP x
n dn
VP x
 
 However, estimate at final conditions with ndecaneVP ~ 70.25 (from part a) 
 W
org
760 70.25 .1n
107.185
n 70.25 .1
 
 This should be a good estimate. 
 
9.D15. Column is similar to figure in Solution to 9.D13, but with 1 stage in column. 
 
a) For finding fin totW & D don’t actually need to step off stages. Just want to make sure Wfinalx 
is obtainable – I checked this at total reflux – It works. Use Mass balances: 
 
finF tot d fin W
F x D x W x 
 Substitute in tot finalF D W 
 233 
 Then, 
fin
D F
final
D W
F x x 100 0.975 0.48
W
0.975 0.08x x
 
 Solution is finalW 55.307 kmol, tot finalD F W 44.673 
 b) Need to draw operating lines until: Feedinitial 2 stages gives x . 
W,finalfinal 2 stages gives x . Then L/V = slope. 
 Initial – There will be a pinch at point reflux is returned. 
 d int ercept
Initial
D
y x y 0.975 0.41
L V Slope 0.579
x 0 0.975 0
 
 Final: A few trials resulted in final result. 
 d int ercept
final
d
y x y 0.975 0.17
L V Slope 0.826
x 0 0.975 0
 
 
 
 
 234 
9.D16. a. Need L/V so that 3 stages go from F Dx .4 to x 0.84 . This is line a in Figure (trial-and-
error) was used to find this). 
 a
a a a
L VL .84 .57 L
.3214 and .4737
V .84 0 D 1 L V
 
 
b. Need L/V so that 3 stages go from wfinal Dx 0.08 to x 0.84 (see line b in Figure). Again, 
trial-and-error was used to find line. 
 b
b b b
L VL .84 .13 L
.8452 and 5.4615
V .84 0 D 1 LV
 
 c. total finalF D W total final10 D W 
 feed total D final wfinalF x D x W x total final4 .84 D .08 W 
 Solving simultaneously, finalW = 5.789 kmol and totalD 4.211 kmol. 
 The Rayleigh equation could be used, is not needed, but gives the same result. 
 
9.D17. initx 1.0 . Start 1 kmol and keep 1 kmol. Add water as boil. 
 Eq. (9-17), 
W ,initial
W ,final
x
tan k
x
dxS
W y
 Note balance is on original solvent, methanol. 
 Use equilibrium data from Table 2-7. Generate table of methanol mole fractions: 
x y 1/y
1.0 1.0 1.0
0.611 0.830 1.20489
0.222 .588 1.6722
0.11611 .450 2.222
0.01 0.07 14.9254 
 
 Use Simpson’s rule in two steps. Step 1 (x from 1.0 → 0.222) 
 235 
 
0.778
1.0 4 1.20489 1.6722 0.97143.
6
 
 Step 2 (x from 0.222 → 0.01) 
 
0.216
1.6722 4 2.222 14.9254 0.9005.
6
 
 Total = 1.87194 = S/W with W = 1. 
9.D.18. New Problem 3rd Edition Eq. (9-14) 
D ,final
F
y
final y
F dy
n
D y x
 
 
 
D ,final
F
y
final y
F dy
exp
D y x
 
 
D ,final
F
final y
y
F
D
dy
exp
y x
 
 Read x values from equilibrium diagram or interpolate from Table 2-1. 
y x y-x 1
y x
 
0.1 0.008 0.092 10.87 
0.3 0.045 .255 3.92 
0.5 0.155 .345 2.90 
 
 
0.5 0.1
Area 10.87 4 3.92 2.90 1.963
6
 
 final
0.5
D exp 0.0702 kmol
1.963
 
 total finalC F D 0.4298 kmol 
 Ethanol MB: 
 final D,final total C,avg FD y C x F y 
 
 F final D,finalC,avg
total
F y D y 0.5 0.1 0.0702 0.5
x 0.00346 mole frac ethanol
C 0.4298
 
 
 
9.D19. New problem in 3rd edition. 
w w oct oct totVP x VP x P , or in mm Hg, 526.123 1.0 10.964 .6 532.7 
 
9.D20. Was 9.D18 in 2nd edition.
 F,C5 W,final,C5 C5,AVG C8,AVGx 0.35 & x 0.05 : x 0.20, x 0.80, p 101.3 kPa. 
 236 
 B.P. i i iy K x 1.0 . For average mole fractions, the BP calculation converges to T = 
84º with C5K 3.7 and C8K 0.30 from the DePriester charts. The iy 0.98 which is 
close enough to estimate α. 
 C5C5 C8
C8
K 3.7
12.33
K 0.30
 
 Eq. (9-13), f
0.05 0.65W 1 0.65
n n n 0.5847
F 11.33 0.35 0.95 0.95
 
 finW F 0.5573, finW 0.5573 1.5 0.8359 kmol 
 total finD F W 1.5 0.8359 0.6641 kmol 
 F fin W,finD,Avg
total
Fx W x 1.5 .35 0.8359 0.05
x 0.7276
D 0.6641
 
9.D.21. New Problem for 3rd Edition. 
L D
L D 1.0 L V 1 2
1 L D
D D
L L
y x 1 x , but x varies.
V V
 Thus, plot series operating lines of arbitrary 
Dx & slope 1 2. With a total of 21 equilibrium contacts there will be a pinch where the operating line 
intersects the equilibrium curve. This intersection is Wx for this Dx value. 
Dx Wx D Wx x
 D w
1
x x
 
.7 .067 .633 1.5798 
.6 .05 .55 1.818 
.5 .038 .462 2.1645 
.4 .027 .373 2.6801 
.3 .018 .282 3.5461 
.2 .01 .19 5.26316 
 
 Want to integrate from W,finalx 0.02 to Fx 0.06 and want middle point at Wx 0.04 . 
 Plot 
D W
1
x x
 vs Wx and find values. 
 W,final W,Avg
D W D W
1 1
x 0.02, 3.23; x 0.04, 2.1
x x x x
 
 F
D W
1
x 0.06, 1.65
x x
 
 
0.06 0.02
Area 3.23 4 2.1 1.65 0.08853
6 
 237 
 
finalW F exp Area 2.5 exp 0.08853 2.288 
 total finalD F W 0.2118 
F final W,final
D,Avg
total
F x W x
x 0.492
D
 
 238 
9.D.22. New Problem for 3rd Edition. totOP
R
D
t
Q
 toperating = __1.49 to 1.50 h 
9.D23. Prelim. Calc. Feed; Avg MW 
E water
0.1 MW 0.9 MW 
 avgMW 0.1 46 0.9 18 20.80 kg/mol 
 1000 kg 48.0769
20.8 kg mol
kmol 
 L L D 2 3 2 5 0.4
V 1 L D 5 3
 
 All op. lines have slope 0.4. 
 Can draw op. line to Wx . Ten stages will go from D Wx to x because have large number 
of stages. Thus, do not need to step off stages. 
 
 
 
 239 
 
 
From Graph can create table of 1/(xd – 1/xw) versus xW. 
 
 
 
 
 
 
 
 
 
 
 
Dx Wx d wx x 
d w
1
x x
 
0.665 0.10 0.565 1.770 
0.630 0.08 0.550 1.8182 
0.499 0.052 Avgx 0.447 2.237 
0.440 0.040 0.400 2.500 
0.278 0.020 0.258 3.876 
0.140 0.010 0.130 7.692 
0.057 0.004 0.053 18.868 
 240 
 
Simpson’s Rule: F W,final W,Avgx 0.1, x 0.004, x 0.104 2 0.052 
 
F
fin
W W AVG Ffin fin
x
F WW
d W d W d W d Wx x x x
x xdx 1 1 1
4
x x 6 x x x x x x
 
 
0.096
18.868 4 2.237 1.770 0.4734
6
 
 
F
Wfin
x
Final W
final
d Wx
W dx
exp 0.62289; W 48.0769 0.62289 29.9466
F x x
 
 total FinalD F W 48.0769 29.9466 18.1303 kmole 
 fin
AVG
F Final W
D
total
Fx W x 4.80769 29.9466 0.004
x 0.2586
D 18.1303
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 241 
9.D24. Was 9D22 in 2nd edition. 
 a) 
fintotal fin F total D W
F D W and Fx =D x Wx 
 fin
fin fin
fin
F W
F total D W W total
D W
x x
Fx D x x +Fx D F
x x
 
 total
0.62 0.45
D 3.0 1.275
0.85 0.45
 kmol, fin totalW F D 1.725 kmol 
b) Want operating line where 2 equil. contacts gives 
finw D
L L
x 0.45 . y x 1 x
V V
. 
Surprisingly, with 2 contacts T & E not needed. – Start stepping off stages from top & from 
bottom simultaneously. The intersection point must be on op. line as is Dy x x . 
 L 0.85 0.44 L L L VSlope 0.482, 0.932
V 0.85 0 D V L 1 L V
 
 
 
Figure for 9.D24. 
 
 
9.D25. New Problem in 3rd edition. 
Mix together 1 2F F F 2.5 
 242 
 
1 2F 1 F 2 F
x F x F x 2.5 0.8 2.5 0.32 
 
F
FIN
final
x
F wFINAL
AVG
xw
x xw dx
n x 0.21
F y x 2
 
 
F
FIN
w w ,FIN w ,AVG w ,FFinal
x
F
x x x x
x x 1 4 1
6 y x y x y x
 
 
x y 1
y x
 
 0.1 0.4 3.333 
0.21 .585 2.666 
0.32 .675 2.817 
 
 
.22
3.333 4 2.666 2.817 0.6166
6
 
 areafinalW Fe 2.5 0.5398 1.349 kmol 
 FINF FIN wfinal DAVG
Fx W x
D F W 1.151, x 0.578
D
 
Part b) First go from 0.4 to 0.2 with 1 1F D 
 
 Then from 0.2 to 0.1 with FIN1 2F W F 
 
 
 
 
 
 
 
 
 
 Now FIN1 2 FF W F 0.865 1.0 1.865, x 0.2 
 
0.1
3.3333 4 2.7777 2.666 0.285185
6
 
area
final2W F e 1.865 0.7518 1.4022 
 
 2 final2D F W 1.865 1.4022 0.46275 
 Total totalD .635 .46275 1.09775 
 Total fintot tot fin 2 WD,AVG
total
F x W x
x
D
2.5 .32 1.4022 .1
1.09775
 0.6010 
 Higher distillate mole fraction. 
x y 1
y x
 
 0.2 0.575 2.666 
 0.3 0.67 2.703 
 0.4 0.73 3.03 
 
x y 1
y x
 
 0.1 0.4 3.3333 
0.15 0.51 2.7777 
 0.2 0.575 2.6666 
 
Values are from methanol-water 
equilibrium data. 
.2
2.666 4 2.703 3.030 0.55026
6
 
1
-area
FINAL 1W = Fe = 1.5 .5769 = .865 
 
1 1 Final,1 D,AVG,1
1.5 .4 .865 .2
D F W 0.635, x
0.635
 
 243 
Part c) Go from 0.4 to 0.1 for 1F 
 
0.3
3.3333 4 2.7027 3.030 0.8587
6
 
final
F1
W 1.5 exp .8587 0.6356 
 
 For 2F go from 0.2 to 0.1. Same as 2
nd part of Part b. 0.28518 
 areafinal 2
F2
W F e 1.0 .75187 0.75187 
 final total final fin
F1 F2
W 1.3874 W W , total 1 2 final totD F F W 1.11255 
 Differs from b – Numerical error! 
 totDAVG
F .32 1.3874 .1
x 0.59436
1.11255 
 Should be same as part b. There are numerical errors in use of Simpson’s rule. 
 
 More accurate for .4 to .1 is .4 → .2 (Area = .55026) + .2 →.1 (Area = .285185) 
 Total Area = 0.835445, 
1
area
final 1W F e 1.5 .433681 0.65052 
 Then fin totalW 1.40239 , totalD 2.5 1.40239 1.09761 
 
AVGD
2.5 .32 1.40239 .1
x 0.60109
1.09761
 
 Same as for Part b. 
 
9.D.26. New Problem for 3rd Edition. a&b) L D 2 3L D 2 3, L V 0.4
1 L D 5 3
 
 
final final
L D 4, L V 4 5 0.8 
 For Part b, draw op. lines with slope 0.4 for arbitrary Dx 0.7 . Step off 2 stages. 
 For Part a, From Dx 0.7 draw op. line slope = 0.8. Two stages gives 
 
finalw
x 0.11 (See graph (labeled 9.D.c)) 
 Also, draw a few lines with Dx 0.7 and L V between 0.4 and 0.8 . Find wx values with 
2 stages. 
 b) Generate Table D w D wx , x , 1 x x 
 Plot w
D w
1
vs x
x x
. Note there is a break in curve at D wx 0.7 x 0.185 to 0.110 
 Find 2 areas 
final
F w
w w
1. from x 0.6 to x 0.185
2. from x 0.185to x 0.110
 
 Area 1. Simpson’srule. 
avgw
0.6 0.185
x 0.3925
2
. 
x y 
 0.1 0.4 3.3333 
0.25 0.62 2.7027 
 0.4 0.73 3.030 
 
1
y x
 244 
 From graph 
wavg
D w x
1
2.37
x x
. 
 Area 1 
0.6 0.185
3.56 4 2.37 1.942 1.0363
6
 
 Area 2. 
D w
1
x x
curve is straight line. Thus, Area = width x Avg height 
 
1.6949 1.942
0.185 0.110 0.13638
2
 
 Total area = 1.1726. 
 Rayleigh eqn., 
F
wfinal
x
final w
x D w
W dx
n 1.1726
F x x
 
 1.1726finalW 100 e 3.0955 
 total finalD F W 6.9045 
 c) finalF final WD,AVG
total
Fx W x 6.0 3.0955 0.11
x 0.8197
D 6.9045
 
 245 
 
 
 
L/D 
Dx wx 
D w
1
x x
 
2/3 0.9 0.65 4.0 
2/3 .85 0.48 2.703 
2/3 0.8 0.36 2.273 
2/3 0.75 0.253 2.012 
2/3 0.7 0.185 1.942 
-- 0.7 0.145 1.8018 
4.0 0.7 0.110 1.6949 
 
w,finalx 
 
 246 
 
 
9.D.25. New Problem for 3rd Edition. From the methanol-water equilibrium data, the following table can 
be obtained. 
 247 
 
Mx My M1 y 
1.0 1.0 1.0 
 .8 .92 1.08696 
 .6 .825 1.21212 
 
pot ,Initial
pot ,final
x
MeOH
MeOHx
dxS
W y
 
Simpson’s rule: 
.4
1.0 4 1.08696 1.2121 0.4373
6
 
 S 0.4373W 0.8747 kmol 
 
9.E1. 
 
 10 W W
2164.42
log VP 8.68105 , T C, VP mmHg
273.16 T
 
 a) Final conditions oct, in org W, in Wx 0.3103, x 1.0, T 99.782 C from solution to 
problem 8.D11. Initial conditions: oct,in org W in Wx 0.90, x 1.0 
 i ix VP 1.0 atm 760 mm Hg 
 From spread sheet find T = 99.377ºC 
 b) 
fin
final,org
W
1 z 1.0 0.9
W F 1.0 0.1450
1 x 1 0.3103
 
 c) org final orgD F W 0.8550 n 
 d) Eq. (9-24) 
orgD
tot octoct
W oct
oct0 oct
p VP x
n dn
VP x
 
 Estimate octVP at average T (99.782 99.377) / 2 99.5795 , oc tan olVP 18.87 mm Hg 
 
org org orgD D D
tot oct tot oct
W org org
oct oct oct oct0 0 0
p dn p dn
n dn D
VP x VP x
 
water 
octanol 
pot 
steam 
Fx 0.90, F=1.0 kmole 
Final octanol in pot 1 0.95 .9 0.045 kmol 
Nonvolatiles in pot 1.0 0.1 0.1 
oct,W,finalx 0.045 0.145 0.3103 
 248 
 org org feed
F
F 1.0, n F W, x 1 1 x 1 0.1 F W
W
 
 Step-by-Step integration, orgorg
org org
dn1
dn
x x
 
orgn orgdn W orgx
 orgAvg x oct
org,avg
dn
 
 x
 
 0 1.0 .9 
 > 0.8944 
0.1 .1 .9 .8888 0.11181 
 > 0.8819 
0.2 .1 .8 0.875 0.11339 
 > 0.8661 
0.3 .1 .7 0.8571 0.11546 
 > 0.8452 
0.4 .1 .6 0.8333 0.1183 
 > 0.81666 
0.5 .1 .5 0.8000 0.12245 
 > 0.775 
0.6 .1 .4 0.750 0.12903 
 > 0.7083 
0.7 .1 .3 0.6666 0.14118 
 > 0.58333 
0.8 .1 .2 0.50 0.17143 
0.855 0.055 0.145 0.3103 > 0.40517 
 0.1357 
 1.1588 
 
Porg
orgtot
W org
org orgO
dnp
n D
VP x
760
1.15882 0.8550 45.826
18.8666
 kmol 
 
e. Continuous had 108.93 kmoles water/kmole organic fed. The continuous always operates at 
lowest octanol mole fraction in liquid & thus octy is always at lowest value. Thus, requires 
more water to carry over octanol then the batch operation. 
 
9.E2. New Problem in 3rd edition. Parts a & b. See solution to problem 8.D25. 
 
benz
mol B 78.11
x
mol B 78.11 80 673.2
 
c. Find T from Eq. 8-15 With Spread Sheet: 
 
SPE, Problem 9E5. Solution for temperature 
T deg C 92.04234 Do step by step Antoine VP values 
VPW A,B,C 8.68105 2164.42 273.16 2.754418 568.0906 
VPben A,B,C 6.90565 1211.033 220.79 3.034461 1082.583 
X ben 0.17727 xw 1 ptot 760 
Eq8 -5E-05 Goal seek B6 to zero changing B2 
massbeninit 20 massbenfin 2 
 249 
dmorg-sum 18 massben-still 2 dmassorg 1 
dnw/dnorg 9-23 2.960203 dnw 0.037898 
 
d) Step-by-step integration. W orgdn dn Eq. 9-23 Use Spread Sheet for each time 
step. Did the addition of steps off-line in table below. 
 
still init still final
W W org benzn dn for n 20 78.11 n 2 78.11 
 Set orgdm 1 kg. Values from spreadsheet. 
 orgdm kg 
Mass benz still, 
777 (kg) 
Wdn divide all 
values by 78.11 
kmol 
TºC 
from spreadsheet 
 
Initial T 0 → 1 20 → 19 .664431 75.99 
 1 → 2 19 → 18 .677266 76.27 
 2 → 3 18 → 17 .691548 76.577 
 3 17 .707537 76.91 
 4 16 .725558 77.269 
 5 15 .746073 77.66 
 6 14 .769462 78.0996 
 7 13 .79657 78.58 
 8 12 .828276 79.12 
 9 11 .865852 79.715 
 10 10 .91108 80.39 
 11 9 .966543 81.156 
 12 8 1.036129 82.03398 
 13 7 1.12596 83.0487 
 14 6 1.246275 84.2358 
 15 5 1.415548 85.6435 
 16 4 1.670846 87.339 
 17 → 18 3 → 2 2.098857 89.422 
final T 18 2 2.960203 92.04 
 
d) For accuracy add time steps 1 to 17 + (time steps 0 & 18)/2. This is then identical to the use of 
average values of dm for every step. 
 Steps_1 to 17 17.2794168
 
 
0.5 Steps _ 0 18 1.8122565
 
 
Wn 19.0916733 78.11 18.016 4.4034 kg water 
 Compared to continuous, nw = 12.3, batch requires less because benzene mole fraction is higher 
in batch operation for most of batch. If we take final cut and used that value for entire operation, 
 W
18.016
dn 18
78.011
 kg benzene overhead 12.3 same as continuous. 
 e) To vaporize ben ben w(n ) / varies since T is different. 
 For accuracy, do for each orgdm and find λ values at these T. 
9.E3. Simpson’s Rule: 
d ,fin
F d d D
x
d
d B d B d B d Bx x .63 x 0.565 x 0.50
dx 0.13 1 4 1
x x 6 x x x x x x
 
 250 
 Generate table from graph 
 
0.13
3.0303 4 2.15 2.004 0.2954
6
 
 0.2954finalD Fe 1 e 7.442 gmole 
finalB F D 2.558 mole. 
 F D,finB,avg
F x Dx 5.0 4.6886
x 0.1218
B 2.558
 
 
 
 
9.E4. 
init
W ,fin
x
final
x
dx
W F exp
y x
 
 T & E since dilution effects F & initx . 
 Dilute with 5 kg water → Start with 6, initx 1 6 0.1667 
Dilute with 4 kg water → Start 5, initx 1 5 0.2 
 Dilute with 3.5 → Start 4.5, initx 1 4.5 0.2222 
Dilute with 2.75 → Start with 3.75, initx 1 3.75 0.26667 
Dilute with 1.75 → Start with 2.75, initx 1 2.75 0.364 
Bx dx 
d B
1
x x
 
0.3 0.36 3.0303 
0.2 0.597 2.5189 
0.1 0.565 2.15 
0.04 0.500 2.004 
 251 
 For each dilution want to integrate using Simpson’s rule until find finalW 1.0. Thus, 
need values of 1 y x at init avg finalx , x , and x 0.01 for each dilution. These values 
are determined in the following table. 
 
Table of Values for Integrations 
 
Dilute x y y-x 1/(y – x) 
5 kg 
Water 
0.16667 
0.08833 
0.01 
0.537671 
0.38708 
0.067 
0.3710 
0.29875 
0.057 
 2.695412 
 3.34724 
17.544 
4 kg 
Water 
0.200 
0.105 
0.01 
0.579 
0.428 
0.067 
0.379 
0.3230 
0.057 
 2.63 
 3.095975 
17.544 
3.5 kg 
Water 
0.2222 
0.11611 
0.01 
0.598 
0.450 
0.067 
0.376 
0.334 
0.057 
2.6596 
 2.994 
17.544 
1.75 kg 
Water 
0.364 
0.1870 
0.01 
0.706 
0.563 
0.067 
0.342 
0.376 
0.057 
 2.923977 
 2.6596 
17.544 
2.75 kg 
Water 
0.2666 
0.138 
0.01 
0.636 
0.493 
0.067 
0.369 
0.355 
0.057 
 2.7100 
 2.8169 
17.544 
 
 Integrations: Use Simpson’s rule for each addition. 
 Dilute 5 kg: 
0.15667
2.695412 4 3.34724 17.544 0.87809
6
 
 finalW 6 exp -0.87809 2.4934 . Value is too high. Want 1.0 kg. 
 Dilute 4 kg: 
0.19
32.5579 1.031
6
 
 finW 5 exp -1.031 1.783 too high 
 Dilute 3.5: 
0.2122
32.1796 1.138
6
 
 finW 4.5 exp -1.138 1.442 too high 
 Dilute 1.75: 
.354
31.106 1.833
6
 
 finW 2.75 exp -1.835 0.4389 too low 
 Dilute 2.75: 
.2566
31.52 1.348
6
 
 finW 3.75 exp -1.348 0.9740 
 Close to desired 1.0 kg. Thus, 2.75 kg water. The final still pot is 99% water so have (.99) 
(.974) = 0.964 moles water remaining. Moles of water distilled off is 2.75 – 0.964 = 1.786. 
9.H.1. New Problem in 3rd edition. This problem is challenging for students because they must first 
derive the forms of the equations they need to use. 
 252 
A. Define. The system is the simple still pot shown in Figure 9-1. Find Wfinal, D, xA,Wfinal, and xA,dist,avg. 
B. Explore. At first it may appear that the problemin Part a is under specified since there are now five 
unknowns. However, in specifying the problem based on the fractional recovery of benzene in 
the distillate we have added the equation for the definition of fractional recovery of A in the 
distillate. This equation is most conveniently written as, 
 
 FzA (1 – Frac. Rec. A in distillate) = Wfinal xA,Wfinal (9-35a) 
 
which becomes, Wfinal = FzA (1 – Frac. Rec. A in distillate)/ xA,Wfinal (9-35b) 
 
C. Plan. If we write Eq. (9-13) for A and substitute in Eq. (9-35b) we obtain Eq. (9-36), 
 A,W ,final A,F A,F A
AB A,W ,finalA,F A,W ,final A,W ,final
x 1 x 1 x z 1 Frac. Rec. A.dist1
0 n n n
1 xx 1 x 1 x
Part a. In a spreadsheet Eq. (9-36) is easily solved for xA,Wfinal using Goal Seek. Then Wfinal can 
be determined from Eq. (9-35b). Then DTotal is determined from Eq. (9-11) and xA,dist,avg is 
determined from Eq. (9-10) written for component A or from the fractional recovery. 
 
Part b. Now solve Eq. (9-36) for frac. rec. of A in distillate using Goal Seek. For both parts a and b can 
use fractional recovery values and DTotal to find xA,dist,avg = FzA(Frac Rec. A in distillate)/ DTotal 
 
D. Do It. Because Eq. (9-36) for xA,Wfinal is nonlinear, it is easiest to solve this problem with a spreadsheet 
and use Goal Seek to solve Eq. (9-36). The spreadsheets are shown below. 
Part a 
F 5 zA 0.37 The 0.37 is in cell D2 
alpha AC 10.71 
frac rec A in distillate 0.75 
xA,Wfin 0.143185 
9-36 term 1 -1.25687 term 2 -0.3075 term 3 0.436926 
 
Eq 9-36 -1.7E-05 Use Goal seek 
Wfinal from 9-35b 3.230095 D total 1.769905 xAdist,avg 0.78394 
Part a. Use Goal Seek for cell B8, setting it equal to zero by varying cell B5 (xA,Wfin). 
Part b. Use Goal Seek for cell B8, setting it equal to zero by varying cell C4. 
 
Part b 
F 5 zA 0.37 The 0.37 is in cell D2 
alpha AC 10.71 
frac rec A in distillate 0.930081 
xA,Wfin 0.05 
9-36 term 1 -2.41222 term 2 -0.41074 term 3 0.658942 
 
Eq 9-36 -0.00023 Use Goal seek 
Wfinal from 9-35b 2.586992 D total 2.413008 xAdist,avg 0.713073 
Part a. Use Goal Seek for cell B8, setting it equal to zero by varying cell B5. 
Part b. Use Goal Seek for cell B8, setting it equal to zero by varying cell C4 (frac rec A in distillate). 
 
253 
SPE 3rd Edition Solution Manual Chapter 10 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 10A4, 10A5, 10A7, 10A12, 10A17, 10C4, 10C5, 10D13, 10D19, 10D21, 10G1- 
10G4. 
10.A9. A good packing will have: good contact between liquid and vapor, high surface area, low 
pressure drop, inert, inexpensive, and self-wetting. 
Marbles have low surface area, poor contact and relatively high ∆p. 
10.A13. Figure 10-25 shows that if viscosity increases the ordinate increases and ∆p/foot increases.
10.A14. a.Want low F since this gives low ∆p.
b. F decreases as size increases.
c. Ceramics have much thicker walls than metal or plastic. Ceramics are used in corrosive
environments.
10.A16. 1. a. fewer; 2.b. larger; 3. a. lower
10.A.17. Answer is c.
10.B1. The trick is to have maximum and minimum positions of the valve with a larger area for 
vapor flow at the maximum position. 
a. Use a cage to prevent excess movement.
b. Use feet.
c. Have a flap that moves up and down.
d. Use a spring to provide force and maximum position.
e. Sliding valve controlled by an external feedback mechanism.
f. Two flaps to make a roof.
Many other ideas can be generated. 
10.B3. Some possible candidates: 
Bottle caps SOS pads 
Bent bottle caps Scotch brite pads 
Natural sponge Steel wool 
Synthetic sponge Cooper cleaning pads 
Miscellaneous junk Plastic coated wire 
Broken crockery or glassware Tin foil - crushed 
Cut up tubing (Tygon) String - balls 
Glass tubing (broken) lines tied together 
Crushed beer cans lines twisted 
Cut up - crushed beer cans lines stretched taught 
Coal Rope 
Egg shells frayed rope 
Styrofoam packing material Rope tied into bow ties 
Old seat cushions Porous rock pieces 
254 
 
Nuts/bolts/screws/nails Pumice 
Metal filings Ash from Mt. St. Helens 
Wood shavings Ashes from coal stove 
Kindling Pieces of cement block 
Left-over redwood Pieces of brick 
Staples Pop-tops 
Window screens, rolled up Old watch bands - twisted 
Chicken or barbed wire Bent wire coat hangers 
Christmas Tree Ornaments Cookie cutters 
Corn cobs Combination of the above 
 
10.C1. L/V constant since L/D const. Thus L/G const. In vF only G changes. From Perfect Gas 
Law G v v
n P
MW MW
v RT
. Thus, vF increases as P . However, curve is almost 
flat in this range and 
floods,b
C ~ constant. 
 Then flood
.5 .5
sb L G
flood L G0.2
G
C 1
u since 
P20 /
. 
 D
D B
x zL
V L D 1 D and D F so V F
D x x
 
 Thus, 
V flood
F F
Dia 
1u P
P
 F 
P
 
 Then, F 2Dia P , and exponent = ½ 
 
10.C2. See derivation in solution to Problem 10.C1. 
 D
D B
x zL L
V 1 D 1 F
D D x x
 
 Then from Eq. (10-16) 
 
1/ 2
D
D B
flood
x zL
4F 1
D x x
Dia
 3600 p fraction u
 
 Since 1/ 21/ 2Dia F , Dia L D 1 
10.C3. Plot points based on ∆p and 
1/ 2
G
L
L
G
 on Figure 10-25. From ordinate values calculate F 
for each point. Use an average value. 
 
255 
 
10.C4. New Problem in 3rd edition. Part of the operating lines become closer to the equilibrium curve. 
Thus, for the same separation more stages are needed. Fortunately, this effect is often small. 
 
10.C5. New Problem in 3rd edition. You can show this by proving that the minimum reflux ratio (Figure 
10-18A) or the minimum boilup ratio (Figure 10-19A) must increase compared to the base 
cases. These ratios increase because cooling the entire feed (Figure 10-18A) or heating the 
entire feed (Figure 10-19A) automatically changes the feed line and moves the minimum 
reflux (Figure 10-18A) or minimum boilup (Figure 10-19A) operating lines towards the y = x 
line. This means larger minimum external reflux ratio or larger minimum boilup ratio. 
 
10.D1. C6 C7K y / x and K 1 y 1 x 
 Solve simultaneously. C7 C6
C6 C7
1 K
x , y K x
K K
 
 Pick T and generate equilibrium curve. C6 C7Bounds are K 1.0 and K 1.0 . 
T KC6 KC7 x y
149oC 1.0 - 1.0 1.0
169 1.3 .72 .483 .63
171 1.34 .75 .424 0.568
193 - 1.0 0 0 
 Average temperature is T = 171ºC = 444.1 K. 
 
y x .568 .424
1.786
1 y 1 x .432 .576
 
 Viscosity equation and terms are given in Example 10-1. 
 10 C6 C6
1 1
log 362.79 .935, 0.16
T 207.09
. 
 10 C7 C7
1 1
log 436.73 .895, 0.127
T 232.53
. 
 MIX MIXn .5 n .116 .5 n .127 2.107, 0.122. 
 From Eq. (10-6) for o0.217, E 0.730 . The higher pressure results in higher 
temperatures and lower viscosities. This increases the predicted column efficiency by 24%. 
 
10.D2. T 98.4 273.1 371.5 K (almost pure n-heptane at bottom). 
 3v
1 100.2pMW
0.205 lb/ft
RT 1.314 371.5
, 3L 0.684 62.4 42.68 lb/ft , 20 
 Need L V . First, find y at intersection of operating lines. 
 Top Operating Line Slope 
L .999 y
.8
V .999 .5
 
 y .999 .8 .499 0.5998 
 Then, 
L 0.5998 0.001
1.20
V 0.5 0.001
 
256 
 
 L
v
MWL L L
1.2
G V MW V
. This is at bottom where L vMW ~ MW . 
 
0.5 0.5
v
v
L
L .205
F 1.2 0.08317
G 42.68
 
 From Figure 10-16, 
0.2 0.2
sb sb
12.5
C 0.32, C 0.32 0.29
20 20
 
 
0.5
flood
42.68 0.205
u 0.29 4.19
0.205
 
 Saturated liquid feed V V 2500. Use η = .90 
 
1/ 2
4 2500 1.314 371.5
D 12.35 ft
.90 3600 1 .75 4.19
 
 Somewhat larger. Would design at bottom of column. Use a 12 ft. diameter column. 
 
10.D3. New 12’ dia col. First, redo entrainment calculation. 
 
0.5
g
L 0.205
1.2, F 1.2 0.0832
V 42.68
, L 1.2 V 3000 
 For Fig. 10-17 need % flood. In problem 12.D2 designed for 75% flood. 
 
1/ 2const
D 11.78
75% flood
, 1/ 21/ 2const 11.78 .75 10.20 
 Use 12 feet: 1/ 2% flood 10.202 /12.0. % flood72.3%. Then Fig. 10-17, ψ = 0.022. 
 
0.022 3000L
e 67.48
1 1 .022
, L e 3067.48 
 This is reasonable amount. 
 2 2totalA 12.0 4 113.1 ft 
 2dA 1 0.9 113.1 11.3 ft 
 Table 10.2, weir0.9, Dia 0.726 , weir 0.726 12 ft 8.71 ft 
 2activeA 113.1 1 .2 90.48 ft , 
2
holeA 0.1 90.48 9.05 ft 
 14 gauge, tray 0 trayt 0.078 in, 3 16 holes, d t 2.4 
 v0
v
VMW
v 37.51 ft s
3600 9.05
, 0C 0.759 (unchanged from Ex. 10-3) 
 
2
p,d ry 2
59.87 1 0.01
h 0.003 37.51 0.205 2.086 in
42.68 0.759
 
 g
100.2 7.48
L 3067.48 897.8 gpm
42.68 60
 
257 
 
 Abscissa g 2.52.5
weir
L 897.8
4.01
8.71
 
 Parameter w weirDia 0.726, F Fig.10 20 1.03 
 Eq. (10-26), 
2 / 3
crest
897.8
h 0.092 1.03 2.083 in
8.71
 
 With 1″ gap, 2duA 8.71 12 0.726 ft 
 Eq. (12-27), 
2
du
897.8
h 0.56 4.248 in
449 0.726
 
 dch 2.086 2 2.083 0 4.248 10.417 
 dc,aereatedh 10.417 0.5 20.83 OK,but close to distance between trays. 
 res
11.3 10.417 3600 42.68
t =4.904
3067.48 100.2 12
 This is OK. 
 Weeping, Eq. (10-32), 
0.040 12.5
h 0.0625 in
42.68 3 16
 
 Eq. (10-31), LHS p,dryh h 2.086 0.0625 2.148 
 x 2 2.083 0 4.083 
 2RHS 0.10392 0.25119 4.083 0.021675 4.083 0.7682, OK 
 LHS > RHS 
 Operation is slightly marginal with high downcomer backup. 
 Increase apron gap to 1.5 inches: 2du
1.5
A 8.71 ft 1.089 ft
12
 
 
1.5 gap
2
du
897.8
h 0.56 1.888 in
449 1.089
 
 dch 2.086 2 2.083 1.888 8.05 in 
 dc,areatedh 16.11 inch, OK 
 
10.D4. 
1/ 2
v valve
o,bal
v v v
C W 2g
v
K A
 (10-36) 
 Given 2valve v,closed v v,openW 0.08 lb, g 32.3 ft s , K 33C 1.25, K 5.5 
 Pressure drop in terms of inches of liquid of density L : 
 2valve v vp,valve v 0
v L L
W K
h C v
A 2g
 
 2v
1
A 0.02182 ft
12
, 3 3v L0.1917 lb ft , 41.12 lb ft 
258 
 
 
1/ 2
o,bal,closed
1.25 .08 2 32.2
v 6.83 ft sec
33 0.2182 .1917
 
 
1/ 2
o,bal,open
1.25 .08 2 32.2
v 16.73 ft sec
5.5 0.218 .1917
 
 At balance point, v valvep,valve 3 3
L L
1.25 0.8 lbC W
h
A 0.2182 ft 41.14 lb ft
 
 0.1115 ft 1.338 in liquid 
 closed: 2p,valve 0 0
33 .1917
h v for v 6.83
2 32.2 41.12
 
 3 2 2 20 02.39 10 v ft 2.87 10 v inches 
 open: 2p,valve 0 0
5.5 .1917
h v for v 16.73
2 32.2 41.12
 
 4 2 3 20 03.98 10 v ft 4.78 10 v inches 
 
10.D5. Do calculation at total reflux. From a McCabe-Thiele diagram (not shown). Total # 
Contacts = 4.2 N = Total – 1 (P.R.) = 3.2 
 
Length 1
HETP 0.31 m
N 3.2
 
 
10.D6. From Fenske eq. and definition of HETP 
 
dist bot
AB
z
HETP
x x
n
1 x 1 x
n 
 
 
dist bot
x x .987 .008
n n 9.150
1 x 1 x .013 .992
 
 
 z = 3.5 meters. Obtain: 
 a. α = 2.315, HETP = 0.321 
 b. α = 2.61, HETP = 0.367 
 c. 
1/ 2
AVG AVG2.315 2.61 2.4581 and HETP 0.344 
 Can also use McCabe-Thiele diagrams although the solution shown is easier. 
 
10.D7. Current: 
v nf nfF 0.090, 12 spacing, Ordinate 0.2 U const., const. 0.2 U 0.2 6.0 
 New:
 new v,new v,oldold new oldL V 1.11 L V , then L G 1.11 L G , F 1.11F 0.0999 
 Trays Spacing 24″, Ordinate ~ 0.32, Ordinate = nfU × Const. 
259 
 
 Nf
Ordinate 0.32
U 9.6 ft s
Const 0.2 6
 
 
10.D8. At v sb,f1F 0.5, C 0.12 
 
0.2 0.2
L G L G
f 1 sb 0.2
G G G
L
6.0 6.0 1
U C
20 0.12 20
0.12
20
 
 Since vG v
MW pn
pV nRT, MW
V RT
 
 New Condition: 
0.5 0.5
G ,new
v sb
G ,old
1
F 0.5 0.5 0.25; C ~ 0.18
4
 
 f 1,new sb,new G ,old sb,new old
f 1,old sb,old G ,new sb,old new
U C C 0.18 3
4 2 3
U C 0.12 2C
 
 fl,newU 3 6 18 ft/s 
 
10.D9. Mass Balance: B
D B
z xD 0.6 0.01
0.59656
F x x 0.999 0.01
 
 D 596.56 kmol/h, B 403.44 
 At top: L V 0.6, V L D V D 0.6V 
 
D 596.56
V 1491.47 kmol h
0.4 0.4
 
 L V D 491.41 kmol h 
 At top of col. L VW W L V since pure MeOH, same mol. wt. 0.6 
 
kmol 32.04 kg MeOH 2.046 lbm lbm
V 1491.41 105,346
h kmol kg h
 
 Assume ideal gas. Top of column is essentially pure MeOH. 
 Mv M 33
lb
1 atm 32.04
n MW p lbmlbmolMW 0.07219
V RT ftft atm
0.7302 607.79 R
lbmol R
 
 where pure MeOH boils at 
1.8 R
64.5 C 337.66K 607.79 R
K
 
 
3
3
L 3 3
1 kg 2.2046 lbm 28317 cm lbm
MeOH 0.7914 g cm 49.405
1000g kg ft ft
 
 At top 24 0.0773 64.5 19.0 
260 
 
 
1/ 2 1/ 2
GL
v
G L
W 0.07219
F .6 0.02294
W 49.405
 
 Fig 10-16 with 18″ tray spacing: sbC 0.28 
 
0.2 0.2
L V
flood sb
V
19 49.405 0.07219
U C 0.28 7.25 ft s
20 20 0.07219
 
 Use Eq. 10-14 (Modified), 
V flood
4V lbm h
D
frac u 3600
 
 
3
4 105,346 lbm h
D
ft
0.90 0.07219 lbm ft 0.75 7.25 3600 s h
s
 
 D = 10.27ft Use either 10 ft (slightly higher frac flood) or 11 ft – (lower frac. flood). 
 
10.D10. Ref. Bonilla (1993). feed
x
3.0, y= , at z 0.4 x, y 2 3
1+ -1 x
 
 
min
.9 0.66667
L V 0.46667
.9 .4
, 
min
L L V 0.46667
L D 0.875
V L 1 L V 1 .46667
 
 act minL D 2 L D 1.75 , act
L D 1.75
L V 0.636
1 L D 2.75
 
 D1 L V x 0.3273 
 At Minimum (Pinch Point), low minL V L V or low minL D L D 0.875 
 lowL 0.875D , avgL 1.75D , low avgL 0.5 L 
 Generalize low avg actual minL L M where L D M L D 
 
 
 
10.D11. a. Since Liquid & Vapor have the same mole fractions L G L V 
 pV = nRT, G v vn MW V p MW RT , 
3 -1 o 1R 45.6 cm atm gmoles R 
261 
 
 vMW .8 46 .2 18 40.4 
3 1 1 3
G 1 ATM 40.4 g mol 45.6 cm atm mol R 176 460 R .001393 g cm 
V L D 3D 6750 lb day, L 2D 2 2250 4500 lb day 
 L V 4500 6750 .6667 L G 
 
1/21/ 2 3
G LL G .6667 1.393 10 .82 =0.0275 
 
2
3 3
H O L 1 g cm .82 g cm 1 .82 
 F = 97 (Table 10-3) 
 Ordinate 2 .2F G L cG F g .197 , from Figure 10-25 (flooding line). 
 This is, 
 
23
0.22
F 3
3 3 3
1 1 g cm
G 97 .52
.82 62.4 lb ft
.197
1.393 10 g cm .82 g cm 32.2
 
 2 4 2FG 0.5216 lb (s ft ) 8.64 10 s day 45067 lb (day ft ) , FG .75 G 
 
 FIND AREA AND DIAMETER FOR 75% OF FLOODING 
 2 2AREA V .75 G 6750 lb Day 45067 lb Day ft .75 .19970 ft 
 2 2D 4 .19970 ft , D .5042 ft 6.05 in 
 
COLUMN DIAMETER
6.05 5 8 9.68
PACKING DIAMETER
 which is probably OK. 
 
 b. From Fig. 10-25, 1/22 0.2 G L c G LG F ( g ) .036 at L G .0275 
 
0.2 22
3
1
G 97 .52 1 62.4
.82 0.036
1.393 10 .82 32.2
 
 22 4 4G 0.2230 lb (s ft ) 8.64 10 s day 1.927 10 lb (day ft ) 
 2 2AREA 6750 lb day 19267lb (day ft ) 0.3503 ft 
 2AREA D 4 .3503, D .6679 ft 8.01 inches 
 c. L G will be the same; thus 
1/ 2
G
L
L
G
 will be the same, and G will be the same. 
 Area V .75 6 , V 3D as before 67,500 lb day 
 4 2area 67500 .75 .521 8.64 10 1.998 ft 10 earlier value 
 
1/ 2
4 1.998
D 1.59 ft 19.14 inches 
 
262 
 
10.D12. a. Dy L V x 1 L V x . When Dx 0, y 1 L V x 0.1828 . See figure. Need 
2 equilibrium stages. Stop where feed line and operating line intersect. 
 BHETP 5 / 2 2.5 ft, x ~ 0.065 
b. M is at in 1x y .43 
 
L L M 6.13
3.88
L 1.58L M
 
 Within accuracy of graph, 
L L LL
0.8
V L L L L 1
 
 If try a shorter column with same feed won’t work. 
 
L L
.8
L V
 and must adjust column. 
 
10.D.13. New Problem in 3rd edition. Saturated vapor feed in problem 10.D.9 has minimum L/V = 
(0.999-0.6)/(0.999-0.22)=0.5122. This is (L/D)min = 1.05. The actual L/V = 0.6, which is an 
L/D = 1.5. Thus, the multiplier M of the minimum was M = (L/D)/(L/D)min = 1.5/1.05 = 
1.43. 
 For a saturated liquid feed (L/V)min = (0.999-0.825)/(0.999-0.6) = 0.4361, which corresponds to 
(L/D)min = 0.7733. If we use the same multiplier, L/D =1.43(0.7733) =1.106 and L/V = 
0.525. 
 Mass Balance: B
D B
z xD 0.6 0.01
0.59656
F x x 0.999 0.01
 
 D 596.56 kmol/h, B 403.44 . These are same as in 10.D9. 
 At top: L V 0.525, V L D V 0.525V D 
263 
 
 
D 596.56
V 1255.9 kmol h
0.475 0.475
 
 L V D 659.4 kmol h 
 Attop of col. L VW W L V since pure MeOH, same mol. wt. 0.525 
 kmol 32.04 kg MeOH 2.046 lbm lbmV 1255.9 82,330
h kmol kg h
 
 The density and surface tension calculations are the same as in 10.D9. 
 Assume ideal gas. Top of column is essentially pure MeOH. 
 Mv M 33
lb
1 atm 32.04
n MW p lbmlbmolMW 0.07219
V RT ftft atm
0.7302 607.79 R
lbmol R
 
 where pure MeOH boils at 
1.8 R
64.5 C 337.66K 607.79 R
K
 
 
3
3
L 3 3
1 kg 2.2046 lbm 28317 cm lbm
MeOH 0.7914 g cm 49.405
1000g kg ft ft
 
 At top 24 0.0773 64.5 19.0 
 
1/2 1/2
GL
v
G L
W 0.07219
F .525 0.02007
W 49.405
 
 Fig 10-16 with 18″ tray spacing: sbC 0.28 
 
0.2 0.2
L V
flood sb
V
19 49.405 0.07219
U C 0.28 7.25 ft s
20 20 0.07219
 
 Use Eq. 10-14 (Modified), 
V flood
4V lbm h
D
frac u 3600
 
 
3
4 82,330 lbm h
D 9.08ft
ft
0.90 0.07219 lbm ft 0.75 7.25 3600 s h
s
 
Probably use 9 ft, which is a slightly higher fraction of flooding. This compares with 10.27 ft for the 
saturated vapor feed. The smaller diameter column will be less expensive. 
With a saturated liquid feed and CMO, the vapor flow rate in the bottom of the column is the same as in 
the top, V = 1255.9 kmol/hr. For problem 10.D9 with a saturated vapor feed, 
1491.47 1000 491.47V V F . Since RQ V , 
, _ , _ , _(1255.9 / 491.7) 2.55R liquid feed R vapor feed R vapor feedQ Q Q . 
 
Thus, in this case there is a significant energy price for reducing the column diameter by this method. 
 
10.D14. B
D B
z x 0.4 0.0001
D F 1000 400.7415
x x 0.998 0.0001
 , B 1000 D 599.258 kmol/day 
264 
 
 V V L D 1 D 3 400.7415 1202.225 kmol day 
 or 
kmol 1 day 1 h
V 1202.225 0.013915 kmol s
day 24 h 3600 s
 
 At bottom, L V B 1202.225 599.258 1801.483 kmol day 
 
L 1801.483
1.49846
V 1202.225
 
 Bottom of column is essentially pure water. Also boilup By x 
 Thus 2L G L V 1.49846 G is lb (s ft ) 
 
lb kmol 18.016 kg 2.20462 lb
V in 0.013915 0.55268 lb s
s s kmol 1 kg
 
 
3
L 3 3 3
kg 2.20462 lb m lbm
 density water at 100 C 958.365 59.83
m kg 35.31454 ft ft
 
 Data from Perry’s, 7th ed., p. 2-92. W L 1.0 
3
G w w 33
lbm
1 atm 18.016 
n p lbmftpV nRT MW MW 0.0367 
V RT ftft atm
0.7302 671.688 R
lbmol R
where 100ºC 
1.8 R
373.16K 671.688 R
K
 
 
1/ 2 1/ 2
G
v
L
L 0.0367
F 1.49846 0.03713
G 59.83 
 From Fig. 10.25, Ordinate at flooding = 0.18 
 
1/ 2
G L c
flood c0.2
0.18 g
G , g 32.2, F 110 Table 10 3
F
 
 
1/ 2
1/ 2
flood 0.2 2
0.18 0.0367 59.83 32.2 lbm
G 0.15147 0.3892
s ft110 1.0 0.26
 
 Where w 100 0.26 cp from Perry’s p. 2-323. 
 actual floodG 0.80 G 0.80 0.3892 0.31136 
 22
actual
V in lb s 0.55268
Area 1.77505 ft
G lb s ft 0.8 0.3892
 
 
1/ 2 1/ 24 area 4 1.77505
Diameter 1.503 ft 
 b. F = 33, 
plastic
1/ 4 1/ 4
int
Intalox Berl
Berl
F 33
Diameter Dia 1.503 1.112 ft
F 110
 
265 
 
10.D15. vLv
v L
W
F 0.03713 from prob. 10.D14
W
 
 From Fig. 10-16 with 12″ tray spacing , sb,floodC 0.21 
 
0.2
sbK C 20
 where σ = surface tension water at 100ºC. 
 Perry’s, 7th ed., p. 2-306 @ 373.15 K, 
510 dynes 1 m dynes
0.0589 N m 58.9 
1N 100 cm cm
 
 0.2K 0.21 58.9 20 0.21 1.2411 0.2606 
 L vflood
v
59.83 0.0367
u K 0.206 10.52 ft s
0.0367
, actu .8 10.52 8.416 ft s 
 
v act3
lb
4V
sDia
lb ft
u
ft s
 Eq. (10-14) modified for units. V is from Problem 10.D14. 
 
 
4 0.55268
Dia 1.637 ft
0.85 0.0367 8.416
. This can be compared to 1.5 ft for packed. 
Tray columns with this small a diameter are seldom used in industry. 
 
10.D16. From Eq. (10-44). Diameter 1/4 F . 1 3F 98 and F 22 Table 10-3 
 
1/ 4
3
1
F
Diameter (3 ) Diameter (1 )
F
 
1/422
14.54 10.0 ft
98
 
 Can also repeat entire calculation which is a lot more work. 
 
10.D17. At the bottom of the column have essentially pure n-heptane. Then, following Example 10-4, 
we have. 
 3v
p MW 1 100.2
0.205 lb ft
RT 1.314 371.4
 
 Need L V . Since 
1 y
L V .8 where z .5, we have y 0.6 
1 z
at intersection of 
operating lines. Then 
0.6 0
L V 1.2
.5 0
. 
 L
v
MWL L
1.2 1.0 1.2
G V MW
 
 
1/ 21/ 2
v
L
L 0.205
1.2 0.084
G 1.684 62.4
 
 Figure 10-25, Ordinate = 0.05 at ∆p = 0.5 
266 
 
 .2
0.05 .205 0.684 62.4 32.2
G 0.375
98 .9595 .684 0.205
 
 The value 0.9595 is Water at 98.4ºC, and Water L . Since feed is a saturated liquid 
V V 0.6944 lbmol/s. 
 2v
0.6944 100.2VMW
Area 185.5 ft
G 0.375
 
 1/ 2D 4 Area 15.37 ft 
 This is somewhat larger than in Example 10-4. Therefore design at bottom. 
 
10.D18. G L cflood 0.2
g ordinate
G
F
 
 Assume changing p changes only G & ordinate. Then take ratios 
 flood ,new G ,new
flood ,old G ,old
G ordinate, new
G ordinate, old
 
 G ,new new new
G ,old oldold
p MW RT p
4
pp MW RT
. Assumes small change in 
T (in Kelvin). T set by boiling conditions (Vapor Press) not by ideal gas law. 
 
0.5
G
v
L
L
F
G
, 
0.5 0.5
v ,new G ,new new
v,old L ,old old
F p
2.0
F p
, v,new v,oldF 2F 0.4 . 
 New Ordinate Value ~ 0.5, old value ~ 0.09 
 newflood,new flood,old
old
ordinate, newp 0.05
G G 4 0.5 0.75
p ordinate, old 0.09
 
 
10.D.19. New Problem in 3rd edition. Saturated vapor feed in problem 10.D.9 has minimum L/V = 
(0.999-0.6)/(0.999-0.22)=0.5122. 
This is (L/D)min = 1.05. The actual L/V = 0.6, which is an L/D = 1.5. Thus, the multiplier M of the 
minimum was M = (L/D)/(L/D)min = 1.5/1.05 = 1.43. For a saturated liquid feed (L/V)min = (0.999-
0.825)/(0.999-0.6) = 0.4361, which corresponds to (L/D)min = 0.7733. If we use the same multiplier as in 
10.D9, L/D =1.43(0.7733) =1.106 and L/V = 0.525. This is the slope we use in the top section for the 2-
enthalpy feed. In the middle section of the column at minimum reflux conditions the slope of the middle 
operating line is / (0.825 0.6) / (0.6 0.22) 0.592L V . 
 
The external mass balances still gives D 596.56 kgmoles/hr, B 403.44 . At minimum reflux 
min min( / ) 0.7733(596.56) 461.3L L D D and min min 461.3 596.56 1057.86V L D . At 
the saturated liquid feed V V and liquidL L F . Thus, 
min min[( ) / ] ( / ) 0.592liquidL F V L V and ,min 0.592 165.06liquidF V L . 
Since the total feed rate is 1000, the fraction liquefied is 0.16506. The same fraction can be liquefied at 
the finite reflux ratios. Thus, 165.06liqF and 834.94vapF . 
267 
 
At top use saturated liquid reflux ratio L V 0.525, V L D V 0.525V D 
 D 596.56V 1255.9 kmol h
0.475 0.475
 
This is the same as for problem 10.D13 and the remainder of the calculation of the diameter is identical to 
that calculation. The result of the calculation at the top of the column is 
 
3
4 82,330 lbm h
D 9.08ft
ft
0.90 0.07219 lbm ft 0.75 7.25 3600 s h
s
 
We now need to calculate the vapor flow rate in the bottom. Assuming CMO, in the middle section 
1255.9V V . In the bottom section, 
1255.9 834.94 420.96vapV V F . 
Since RQ V , ,2 _ , _ , _(420.96 / 491.7) 0.86R enthalpy feed R vapor feed R vapor feedQ Q Q . 
Thus, in this case the two-enthalpy feed design results in the same reduction in diameter as liquefying the 
entire feed, and it has energy savings compared to the vapor feed. However, the two enthalpy feed 
system will require more stages than the other systems. A complete economic analysis is required to 
determine the most economical system. 
 
10.D20. Use Fig. 10-16 to find sbC . Gas is 2N . Liquid is ammonia. 
 Since system very dilute, treat as pure ammonia liquid & pure 2N gas. 
 L L
G V
W MWL kg h L kmol h 17.03
27.36 16.642
G W kg h V kmol h MW 28.08
 
 
3
3
L 3 3
100 cmg 1 kg
0.61 610 kg m
cm 1000 g m
 
 vG 3 3
175 atm 28.02 g molpMW 1000L kg kg
236.0 
L atmRT m 1000 g m0.08206 253.2 K
mol K
 
 
1/ 2
v
236
F 16.642 10.35
610
 
 Off chart. Extrapolate using Eq. (10-10e). 
 210 sb 10 10log C 0.94506 0.70234 log 10.35 0.22618 log 10.35 
 10 sb sblog C 1.891C 0.01286 
 Assume 0.220 1.0. Then 
 flood
610 236
u 0.01286 0.0162 ft s
236
, op floodu 0.75 u 0.01215 
 V
v op
4 V MW
D
u 3600
 , vV 100 kmol h , MW 28.02 kg kmol
3
v op0.85, 236.0 kg m , u 0.01215 ft s , Need to watch units 
268 
 
 
4 100 28.02
D
0.85 236.0 0.01215 3600 1 3.2808 ft
 =1.155 m 3.79 ft 
 Probably use 4 ft diameter column – (standard size) 
 Using larger diameter helps take into account the uncertainty in extrapolating to find sbC . 
 
10.D.21. New Problem in 3rd edition. The mass balance and flow rate calculations are the same as for 
problem 10.D14. 
 
B
D B
z x 0.4 0.0001
D F 1000 400.7415
x x 0.998 0.0001
 , B 1000 D 599.258 kmol/day 
 V L D 1 D 3 400.7415 1202.225 kmol day 
 or 
kmol 1 day 1 h
V 1202.225 0.013915 kmol s
day 24 h 3600 s
 
 L/V = 2/3. Top of column is close to pure methanol 
 Thus 2L G L V 0.66667 G is lb (s ft ) 
 
lb kmol 32.04 kg 2.20462 lb
V in 0.013915 0.98290 lb s
s s kmol 1 kg
 
Pure MeOH boils at 
1.8 R
64.5 C 337.66K 607.79 R
K
 
 
3
3
L 3 3
1 kg 2.2046 lbm 28317 cm lbm
MeOH 0.7914 g cm 49.405
1000g kg ft ft
 
 W L 59.83 / 49.405 1.211 
Assume ideal gas. Top of column is essentially pure MeOH. 
 Mv M 33
lb
1 atm 32.04
n MW p lbmlbmolMW 0.07219
V RT ftft atm
0.7302 607.79 R
lbmol R
 
 
1/2 1/2
GL
v
G L
W 0.07219
F .66667 0.02548
W 49.405
 
 From Fig. 10-25, Ordinate at flooding = 0.20 
 
1/2
G L c
flood c0.2
0.20 g
G , g 32.2, F 110 Table 10 3
F
 
 
1/2
1/2
flood 0.2 2
0.20 0.07219 49.405 32.2 lbm
G 0.2146 0.4633
s ft110 1.211 0.34
 
 Where methanol 64.5 C 0.34 cp from Perry’s (8
th ed.) p. 2-449. 
 actual floodG 0.80 G 0.80 0.4633 0.3706 
269 
 
 2
2
actual
V in lb s 0.98290
Area 2.6521 ft
G lb s ft 0.3706
 
 
1/2 1/24 area 4 2.6521
Diameter 1.838 ft 
Note that this is larger than the calculation of 10.D14 at the bottom of the column. Thus, do calculations 
at top of column. 
 b. F = 33, 
plastic
1/4 1/4
int
Intalox Berl
Berl
F 33
Diameter Dia 1.838 1.360 ft
F 110
 
 
 
10.E1. op time 
D kmol
 where D 18.1303 kmol
D kmol hr
, 
L 2
, L V 0.4
D 3
 
 Then op
D
D V L V 0.4V 0.6V and t
0.6V
 
 Use Fig. 10-25 or Eq. (10-39a) to find flooding at the end of the operation at bottom of column. 
 
2 liq
2
vapor
kg
MWL lbm s ft L kmol h kmol
kgG lbm s ft V kmol h MW
kmol
 
 At end of operation at bottom of column x 0.004, y 0.036 (pinch) 
 
L 18.128
0.4 0.381
G 19.023
 
avg,liq vapMW 0.004 46 0.996 18.016 18.128 & MW 0.036 46 0.964 18.016 19.023 
3
L 62.4 lbm ft (Essentially pure water). Boils at ~ 100ºC = 373 K 
 vG 33
lbm
1.0 atm 19.023p MW lbmlbmol 0.038806
RT ftatm ft
1.314 373 K
K lbmol
 
 Then 
1/ 2 1/ 2
G
V
L
L 0.038806
F 0.381 0.009501
G 62.4
 
 From Eq. 10-39a. 
 210 10 10 vlog ordinate 1.6678 1.085 log 0.009501 0.29655 log F 
 10log ordinate 0.6864 & ordinate 0.2059 agrees with Fig. 10 25 
 2 G L c0.2
0.2059 g
G
F
 
 From Table 10-3, c water pF 70, 1.0, g 32.2, 100 C ~ 0.26 c (Perry’s 5
th ed., 3-213) 
270 
 
 
0.5
flood 0.2 2
0.2059 0.03698 62.4 32.2 lbm
G 0.53488
s ft70 1.0 0.26
 
 From Eq. (10-41), 
2
2
vapor
Area ftlbmol lbm
V G
lbs s ftMW
l bmol
 
 
2
2
act flood
1 2
Area 0.19635 ft and G 0.7 G
4
 
 
0.19635 0.53488 0.7lbmol
V 0.0038646
s 19.023
 
 
kmol lbmol 3600s 0.453593 kmol
V 0.0038646 6.310665 kmol h
hr s h 1.0 lbmol
 
 Then op
D 18.1303 kmol
t 4.7883 h 287.3 min.
0.6V 0.6 6.310665
 
10.F1. Need average M M
M M
y 1 x
.
x 1 y
 From Equil. data 
 
0.134 0.98
Bot : 7.582
0.02 0.866
, 
0.979 0.05
Top 2.454
0.95 0.021
 
 Geometric avg 7.582 2.454 4.31 
 Column temperature varies from 64.5º to ~ 98.2 ºC. 
 
64.5+98.2
Avg T= 81.35 C
2
 
 liquids from Perry’s 7
th ed., T = 81.35ºC, p. 2-323. M W0.28 cp, 0.35 cp 
 (Note 40% MeOH is probably wt%) 
 Estimate: mix M M W Wn x n x n 
 Feed is 60% M 40% W 
 mixn 0.60 n 0.28 0.40 n 0.35 1.1837 , mix 0.306 
 Then mix 4.31 0.306 1.3195 
 From O’Connell’s Correlation, Fig. 10-14, oE 45% 
 Eq. (10-6): 
2
o 10 10E 0.52782 0.27511 log 1.3195 0.044923 log 1.3195 49.5% 
If conservative use 45% 
 
10.F2. To use O’Connell’s correlation (Fig. 10-14), need α and viscosity of feed. 
 M MM M MMW
W W W M M
y 1 xK y x
K y x x 1 y
. Used Table 2-7 for values. 
 Can estimate a geometric average at bottom, feed & top 
 M W,bot
0.134 0.98
7.582
0.02 1.866
, MW ,feed
0.729 0.6
x .4 4.035
0.4 0.271
 
271 
 
 MW,top
0.979 0.05
2.454
0.75 0.021
, 1/3avg bot F T 4.2184 
 Averages can be calculated many other ways. 
 The feed is saturated liquid. From Table 2-7, T = 75.3ºC 
 Viscosities from Perry’s, p. 2-323, W MeOH0.39cp & 0.30 
 Note: (MeOH, 40% probably refers to wt % - p. 2-322 Perry’s) 
 Estimate mix 1 1 2 2n x n x n 
 mix n .4 n 0.30 .6 n 0.39 1.0465 , mix 0.351 
 Then Feed 4.2184 0.351 1.481 
 Overall Plate effic. = 43.7% 
 
10.F3. B
D B
z x F 0.30 0.01 100
D 36.71
x x 0.8 0.01
B F D 100 36.71 63.29 lb mol h 
 At top of column L = D(L/D) = 73.4 and V = L + D = 110.1 
 Stripping section 
L F F 3 L
L L qF where q 4 3
F
 
 L 206.7, V L B 143.3 
 Feed line has slope 4/3 and goes through y x z .3 . Top operating line has slope L/V = 
0.667 and goes through Dy x x 0.8 . Bottom operating line goes through 
by x x 0.01and the intersection of top operating line and feed line. McCabe-Thiele 
solution is shown in Figure. Optimum feed is 8th from top. Need 8 7/8 equilibrium stages 
plus partial reboiler. 
 
 
272 
 
 Overall Efficiency. For O’Connell Correlation, need AVG Feed col and T 
 E E
E E
y 1 x
1 y x
. Using Table 2-1 we find α and following mole fraction. 
 
0.170 .981
x 0.019, y 0.170 : 10.575
.830 .019
 
 
.5826 .6727
x 0.3273, y .5826 : 2.87
.4174 .3273
 
 
.7815 .2528
x .7472, y .7815 : 1.210
.2185 .7472
 
 1/3AVG 1 2 3 3.324 
 Can estimate μ from p. 99 Ethyl Alcohol Handbook at z .3 .523 wt. 
 frac., 0.55 cp. Thus αμ = 1.83. From O’Connell Correlation oE .42 . 
 N 8.875 .42 21.1 . Thus, need 22 stages plus partial reboiler 
 Height 22 18 18 disengagement 48 (bottom sump) 38.5 ft 
 
 Diameter Calculation 
V
v flood
4V MW
Dia
flooding fraction u 3600
 
 Use average values of parameters in stripping and enriching sections. 
 v V Leth Weth WMW MW y MW y for both MW and MW . 
 stripping section: 18.25 MW 21.5 
 LLW L MW 206.75 20 4135 lb h 
 VVW V MW 143.46 20 2869.2 lb h 
 3L 0.96225 g ml 60.07 lb ft Bottom 
 Bottoms T =100ºC = 672ºR 
 3VV P MW RT 1.0 20 0.7302 672 R 0.04076 lb ft 
 v V L sbL VF W W 4135 2869.2 0.04076 60.07 0.0375, C 0.28 . 
 Enriching section: 26.4 MW 40.4 
 LLW L MW 73.42 35 2569.7 
 VVW V MW 110.13 35 3854.55 lb h 
 3L 0.766 g ml 47.92 lb ft Distillate 
 Distillate T = 82ºC = T = 639.6ºR; 
 3VV P MW RT 1 35 0.7302 639.6 R 0.05472 lb ft 
273 
 
 v L V V LF W W 2569.7 3854.55 0.05472 47.92 0.0225, sbC 0.28 
 
0.2
sbK C 20 , ft/sec. σ, surface tension in dyne/cm. 57
th ed. Hdbk of Physics + Chemistry, F-45. 
 Bottoms, σ ~ 46 dyn/cm, Middle, σ ~ 25 dyn/cm, Top σ ~ 18.6 dyn/cm 
 stripping: 0.20K 0.28 46 20 0.33075 ft s 
 enriching: 0.20K 0.285 18.6 20 0.2809 ft s 
 flood L V Vu K , ft s 
 stripping: flood
60.07 0.04076
u 0.33075 12.693 ft s
0.04076
 
 enriching: flood
47.90 0.05472
u 0.2809 8.31 ft s
0.05472
 
 
1/ 2
V floodV
Dia. 4V MW 0.90 0.75 u 3600 
 stripping: 3V VV MW 143.46 lbmol h 20 0.04076 lb ft 70392.5 
 enriching: 3V VV MW 110.13 lbmol h 35 0.05472 lb ft 70441.37 
 Diameters: stripping section: Dia = 1.7 ft and enriching section: Dia = 2.1 ft 
 Probably use 2.5 ft diameter since there is little if any cost penalty. 
 
10.F4. Numbers from solution of Problem 10.F3 are used.Cross Sectional Area V
2
V lbmol s MW lb lbmol
G lb (s ft )
 
 Bottom: L VL G L V MW MW 1.44 ~ 1 1.44 
 Top: L VL G L V MW MW 0.667 ~ 1 0.667 
 1” metal Pall rings: F 48, 0.15, 0.15 
 ψ = Density of water/density of liquid At Top: 61 47.92 1.27 
 At 81ºC , w E0.35 cp, 0.45 cp , MIX 1 1 2 2n x n x n 
 At top y= d MIX MIXx 0.8, n .8 n .45 .2 n .35 , and 0.43 
 v E wMW y MW 1 y MW 0.8 46 .2 18 40.4 
 3G L0.05472 lb ft , 47.92 
 1/ 2 1/ 2G LL G 0.667 0.05472 47.92 0.02254 
 2 0.2 G L cG F g 0.078 from Figure 10-25 
 
1/2
2G L c
0.20.2
0.078 0.05472 47.92 32.20.078 g
G 0.358 lbm (s ft )
F 48 1.27 0.43
 
 V 110.13 mol h 0.0306 lbmol s 
274 
 
 v2
V MW
Area D 4 , or
G
 
 
1/ 2 1/ 2
v4V MW 4 0.0306 40.4
D 2.096
 G 0.358
 
 Probably use 2 or 2.5 foot diameter columns. 
 The calculation at the bottom of the column gives a smaller diameter. 
 HETP N Height of Packing, or 1.2 ~ 10 12.0 ft 
 
10.G.1. New Problem in 3rd edition. The result from Wankat (2007a) is listed in the following Table: 
Results for distillation of vapor feed 5 mole % methanol, 95 mole % water. Distillate is 0.9543 mole 
fraction methanol and bottoms is 0.9976 mole fraction water. Tray spacing = 0.4572 m. Base case 
conditions are listed in Tables 1 and 2 in Wankat (2007a). When two trays are listed, they have the same 
diameters. The decrease in volume and increase in QR are compared to the base case. 
 
FL NF,V NF,L N dia A Vol tray QR Qc, total decr Vol incr QR 
0(base) 10 -- 20 2.84 6.33 55.0 2 1065 -12,330 -- -- 
 
Two-enthalpy feed: 
500 11 6 20 2.08 3.38 29.4 2 1070 -12,340 46.5 % 0.5 % 
500 12 6 22 2.07 3.38 32.5 2 1065 -12,330 40.9 % 0 
600 12 6 20 1.89 2.80 24.3 2 1070 -12340 55.8 % 0.5 % 
750 13 6 20 1.56 1.92 16.7 2 1090 -12,360 69.6 % 2.3 % 
750 16 7 26 1.56 1.91 21.8 2 1065 -12,330 60.3 % 0 
1000 (all liquid) 9 20 1.20 1.13 9.8 2 2415 -13,680 82.1% 127 % 
 
Intermediate condenser: 
FWithdr NF,V NV,with NL,ret N dia A Vol tray QR Qc, total decr Vol incr QR 
300 11 10 6 20 2.41 4.56 39.6 2 1067 -12,340 28.0 % 0.2 % 
450 11 10 6 20 2.22 3.88 33.7 10/11 1067 -12,340 38.6 % 0.2 % 
 
Two-enthalpy feed (FL = 600 kmol/hr) plus one intermediate condenser: 
Fwithdr NF,V NF,L NV,with NL,ret N dia A Vol tray QR,total Qc decr Vol incr QR 
100 12 6 5 5 20 1.69 2.23 19.4 2 1073 -12,340 64.7 % 0.8 % 
80 12 6 13 6 20 1.72 2.33 20.2 2 1079 -12,345 63.2 % 1.3 % 
 
Two-enthalpy feed (FL=600 kmol/hr, NF,V =12, NF,L=6, N=20) plus two intermediate condensers: 
Fwthd1 NVwth1 NLret1 Fwthd2 NVwth2 NLret2 dia A Vol tray QR,total Qc decr Vol incr QR 
100 5 4 80 13 8 1.50 1.77 15.4 2/6 1081 -12,350 72.1 % 1.6 % 
 
Two-enthalpy feed (FL = 680 kmol/hr) plus one intermediate condenser: 
Fwithdr NF,V NF,L NV,with NL,ret N dia A Vol tray QR,total Qc decr Vol incr QR 
100 12 6 5 5 20 1.51 1.79 15.5 6 1081 -12,350 71.7 % 1.6 % 
 
With constant Qc and QR, the two-enthalpy feed with FV = 750 and N = 26 appears to be best. 
 
 
10.G.2. New Problem in 3rd edition. Results are from Wankat, P. C., "Balancing Diameters of 
Distillation Column with Vapor Feeds," Ind. Engr Chem. Research, 46, 8813-8826 (2007). 
275 
 
 
Table 1. Simulation conditions and results for base cases. F 1000 kmol / h, D = distillate flow rate 
kmol/h, N = number of trays + condenser + reboiler, tray spacing = 18 inches = 0.4572 m, 
operation at 80% of flooding, dia = maximum calculated diameter in m, tray = tray at which max 
diameter occurred. A = max calculated column area in 2m , Vol = column volume in 3m = A(N 
– 2 +1) (tray spacing) where N – 2 is the number of trays and +1 is for disengagement space for 
liquid and vapor, RQ and cQ = reboiler and condenser duties in kW, condp condenser pressure 
in atm, p pressure drop in psi/tray, feedN = optimum feed stage, and condenser is stage 1. 
Note this solution has a p, each stage. Thus, solution slightly different than students’ solutions. 
 
FeedN N Dia A Vol tray RQ cQ , 1L D D condp p 
 
 
Ethanol (10 mole %). Water (90 mole%) Vapor Feed Base Case: 
23 26 2.61 5.35 85.6 2 902 -10,700 8.0 125 1.80 0.1 
 
 
 
 Table 2. Diameters calculated for standard distillation base cases listed in Table 1. Vapor flow 
rate jV and liquid flow rate jL are in kmol/hr, diameter is in m, area is 
2m . 
Tray Vj Lj Dia Area
2 1125 997 2.61 5.35
23 1082 949 1.99 3.11
24 74.3 950 0.71 0.396
35 79.4 956 0.71 0.396
Ethanol-water, vapor
 
 
 Table 3. Simulation conditions and results for a distillation column separating a vapor 10 mole % 
ethanol, 90 % water feed (see Table 1 for base conditions). Partial condenser is stage 1. F,VN and F,LN 
are optimum feed locations for vapor and liquid portions of the feed, respectively. Decrease in column 
volume Vol (equal to change in area when the number of stages is unchanged) and increases in RQ are 
compared to the ethanol-water base case (Table 1). For both runs D,Ey 0.7901 and B,Wx 0.9986. 
LF F,VN F,LN N Dia A Vol Tray RQ C,colQ 
0(base) 23 -- 36 2.61 2.24 35.8 2 902 -10,700 
 
LF F,VN F,LN N Dia A Vol Tray RQ C,totQ Decr Vol Incr RQ 
600 23 17 36 1.69 2.24 35.8 2 902 -10,700 58.2% 0 
c,total c,col c,feed condenserQ Q Q 
 
10.G3. New Problem in 3rd edition. Part a. S Dia = 2.032 m. Distillate mole fractions (vapor) = 0.22222 
Eth, 0.77765 Propane, 0.12383 E-03 B, and 0.28503 E-9 pentane. Bottoms mole fractions = 0.14843 E-
10 Eth, 0.10132 E-03 Propane, 0.81808 Butane and 0.18182 pentane. Other values are in Table for 10.G4. 
Part b. Worst backup is 0.232 m on stages 30 and 31. Maximum weir loading is 0.0204 m2/s on plate 31. 
Part c. Same mole fractions, same Qc and QR. Max backup 0.1614 m in panel A on stages 29 to 32. 
Maximum weir loading is 0.01183 m2/s on plate 31 of panel A which is acceptable. 
 
10.G4. New Problem in 3rd edition. 
 
276 
 
V feed 
Kmol/s 
L feed 
Kmol/s 
Qc kW QR kW Max Dia 
m 
Stage 
max dia 
yD,C4 xB,C3 
0* 1 pass .1(NF=16) -1463 2827 2.032 31 .000124 .000101 
0 part d .1(NF=15) -1463 2827 2.032 30 .000115 .0000942 
0* 2 pass .1(NF = 16) -1463 2827 2.032 31 .000124 .000101 
.01 .09 -1463 2600 1.956 31 .000145 .000119 
.02 .08 -1463 2373 1.876 31 .000199 .000163 
.03 .07 -1463 2147 1.792 31 .000300 .000246 
.04 .06 -1464 1921 1.905 32 .000525 .000429 
.05 .05 -1466 1696 1.650 18 .00112 .000915 
.06 .04 -1472 1474 1.600 18 .003188 .002609 
b. Change 
N 
N=41 NF,liq=18 NFvap=21 
.03 .07 -1463 2146 1.793 34 .0000817 .0000668 
 * Values from problem 10.G3. 
 Part c. Tray rating program with Dia = 1.793 m and defaults for tray spacing (0.6096m) & for DC 
clearance (0.0373m) obtain 0.2207 m backup on tray 34, which is acceptable. Maximum weir loading is 
0.01887 m2/s on tray 34 which is acceptable. 
 Part d. Shown above, plus maximum backup is 0.2320 on plate 31 (acceptable) and maximum 
weir loading is 0.0204 m2/s on plate 31, which is marginal. 
277 
SPE 3rd Edition Solution Manual Chapter 11 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 11.A19-11.A22,11.B6, 11.C3, 11.D5,11.D8, 11.G3-11.G4. 
11.A3. As feed temperature increases
MIN
L D increases, hence L/V increases and cQ increases. 
L V increasesand RQ decreases. The number of stages probably decreases. 
(See Figure 7-3. The abscissa increases as 
min
L D and L/D increase. Thus ordinate drops. 
Since MINN is constant, N decreases). 
11.A.6. New Problem for 3rd edition.
b. A liquid side-stream between the feed and the distillate. 1 point for b. 
c. A vapor side-stream between the feed and the bottoms. 2 points for c. 
11.A.9. If the feed rate is consistently one half the design capacity, the entire economy of scale will be
lost. In addition, distillation columns probably operate at lower than design efficiencies. 
11.A11. Column 2 would have to be at a lower pressure than column 1.
11.A.12. New Problem for 3rd edition. The heuristics to not do would include items such as:
1. Remove dangerous, corrosive, and reactive components last.
2. Use distillation only for very difficult separations (α < 1.10). And so forth.
11.B4. Use heuristics. Hardest separation is xylene-cumene, X,Tx,c
C,T
0.33
1.57
0.21
Toulene is most abundant. Use heuristics to: 
Remove toluene early 
Do hardest separation last 
One-by-one in overhead 
If use equil-molal splits: 
Can also derive coupled systems. 
B XT
F
C
278 
 
 
 
11.B5. Add in Heuristic 9 for sloppy separation. 
 
11.C.1. New Problem for 3rd edition. Take the log of both sides of Eq. (11-2). 
 
 log (cost A/cost B) = (exponent) log (size A/size B) 
 
 log (cost A) – log (cost B) = (exponent) [log (size A) – log (size B)] 
 
 exponent = [log (cost A) – log (cost B)]/ [log (size A) – log (size B)] 
 
11.D1. NOTE: This solution requires the solution to Problem 10-D1. Estimate α at feed composition 
x = z = 0.5 
 a&b. For binary, c7c6 c6 c6 c6
c6 c7
1 K
x , y K x .
K K
 Use Fig. 2-11. 
 T = 168ºC: c6 c7 c6 c6K 1.29, K 0.71, x 0.50, y 0.645 
 (Guess was aided by solution to Problem 10-D1) 
 
.645 .5
1.817
.355 .5
 
 MIN
.999 .001
n
.001 .999N 23.13 stages
n 1.817
 
 D
min D
x yL 99 .645
0.7094
V x z .999 .5
 
 
X 
T 
F 
C 
B 
X 
F 
C 
B 
OR 
T 
X 
T 
F 
C 
B 
279 
 
 
min min
L L V
2.441
D 1 L V
 
 c. Gilliland: MIN
L D L D 4 2.441
x 0.3117
L D 1 5
 
 Eq. (7-42b): MINN N N 1 0.37027 
 N 0.3703 23.13 1 0.3703 37.3 contacts or 36.3 stages 
 d. act eq oN N E 36.3 0.73 49.7 or 50 stages 
 24” tray spacing × 50 trays = 100 ft 
 + 4 disengagement 
 H = 104 ft = 31.7 m 
 
m
D 9 ft 2.743m
3.2808 ft
 
 Tray area 
2
2 3D 5.91 m . Column Vol area H 187.3 m
4
 
 Fig. 11-1. 3 3P,tower 3
cos t $700
$700 m , C 187.3 m $131,110
vol m
 
 Fig. 11-2. 2 2P,tray
Tray cos t
$710 m , C 710 5.91 m $4196 tray
area
 
 Eq. 11-8. BM,towerC 131,110 2.25 1.82 2.3 1.0 $844,000 
 p
7 2.743pD
700 kPa 7 bar, Eq (11-5), F .5 .5 2.30
10.71 0.00756p 10.71 0.00756 7
 
 Fig.11-9. BM.tray
 $ 209,800
C 4196 50 1 1
total $1,054,000
 
 
11.D2. 69ºC (bp) = 156.2ºF. F 1000, D 500, B 500, L V 0.8, L D 4 
 L 2000 lbmol h, V 2500 
 Energy Bal: v c D c o vVH Q Vh gives, Q V h H 
 D vh is pure boiling hexane, H is vapor. 
 Thus, D v c6h H 13,572 Btu/lbmol 
 7cQ 2500 13,572 3.39 10 Btu h 
 R D B F CQ Dh Bh Fh Q 
 Pick 25ºC as basis. 
 D PL,c6
1.8 F
h C 69 25 51.7 44 1.8 4094.6
C
 
 B PL,c7
Btu
h C 98.4 25 1.8 50.8 73.4 1.8 6711.69
lbmol
 
 Feed is a saturated liquid. From Example 10-1, T = 80ºC 
280 
 
 F PL,c6 c6 PL,c7 c7h C z C z 80 25 1.8 
 Fh 51.7 .5 50.8 .5 55 1.8 5073.75 Btu lbmol 
 7RQ 500 4094.64 500 6711.69 1000 5073.75 3.39 10 
 7RQ 3.423 10 Btu h 
 Reboiler: 
7
2
2
Q 3.423 10 Btu h
A 32,800 ft
U T Btu 1.8 F
50 110 98.4 C
h ft F C
 
 where U is average from Table 11-2. 
 Condenser: 
7
c 2
Avg
3.39 10Q
A 2850 ft
U T 110 70
180 156.2
2
 
 Note these areas are very approximate. For detailed design need a much better estimate of U. 
 Costs: Condenser 
2
2 2
2
1 m
A 2850 ft 264.8 m
3.2808 ft
 
 Fixed Tube Sheet S&T Fig. 11-3, Cost = $125/m2 
 Reboiler 
2
2 2
2
1 m
A 32,800 ft 3047 m
3.2808 ft
large because of low U. 
 Extrapolate $70/m2 
 1 atm, p mF 1.0, F 1.0 . Eq. (11-9) BM p m Q pC C 1.63 1.66 F F 3.29 C 
 Condenser 2BM 2
$125
C 3.29 264.8 m $109,000
m
 
 Reboiler 2 2BM
$702,000 
C 3.29 $70 m 304 m
$811,000 Very sensitive to U. 
 
 
11.D3. Note: This solution requires the solution to Problem 11-D2. 
 Steam rate, 
7
RQlb 3.423 10 lb35,704.6
h 958.7 h
 
 where RQ is from Solution to problem 11-D2. 
 Steam Cost 
$ lb $20.00
35,704.6 $714 h.
h h 1000 lb
 
 Cooling water, 
w
7
c
p w
Qlb 3.39 10 lb
847,500 
h C T 1.0 40 h
 
 where RQ is from Solution to problem 11-D2. 
 Water Cost 
$ lb $3.00 1
847,500 $306 h
h h 1000 gal 8.3 lb gal
 
 
281 
 
11.D4. From Example 11-1 needed 21.09 eq stages + P.R. 
 packh 21.09 1.1 ft stage 23.2 ft 7.07 m 
 
2
2 3
pack
D
Vol h 15 23.2 4099.6 ft
4 4
3
3 3
3
1 m
4099.6 ft 116.1 m
3.2808 ft
 
 From Fig. 11-2 3pC ~ $250 m packing 
 Tower 23.2 ft + 2 ft between sections + 2 ft top + 2 ft top = 29.2 ft 
 h = 
1 m
29.2 ft 8.9 m
3.2808 ft
, Vessel Vol. 3146.1 m 
 Fig. 11.1 3pC $700 m for tower 
 
p M BM pF 1.0 1 atm , Tower F 1 carbon steel , C C 2.25 1.82 $700 146.1 4.07
 $416,312
Packing, mF 4.1. 
 
BM pC C 1.63 1.66 4.1 250 116.1 8.436 244,855
 Total $661,000 
 
 Does not include cost distributors, supports, hold down plates, etc. 
 
11.D.5. New Problem for 3rd edition. n = [log (cost A) – log (cost B)]/[ log (size A) – log (size B)] 
Let size A = 10 m2 and size B = 1.0 m2. The cost A = $400/m2 = ($400/m2)(10 m2) = 4000, and 
cost B = $2100. 
 n = [log (4000) – log (2100)]/[log (10) – log (1)] = [3.602 – 3.322]/[1 - 0] = 0.28 
 
11.D6. See residue curves in Figure. The recycle is pure MB. Mixing point is determined in same way 
as in Fig. 11-11. Now mixing point splits into light (L) component methanol on 1B . Thus line 
LM is extended to 0.0 mole fraction methanol to find location of 1B (0.73333 MB and 0.26667 
toluene). We can use mass balance to find point 1B accurately. If 1D is pure methanol, 1D = 50 
(all methanol in feed) and 1B = 150. Then from toluene balance 0.2 × 200 = 150 1tol,Bx , which 
gives 
1tol,B
x 0.266667 . 1 2B F which is then split into I (Some of which is recycled) and 2B
which is toluene product. 
 1 1F Recycle 100 100 M B D 
 Since 1D contains no toluene 1B 0.26667 F .4 Recycle 0 
 1 2
.4
B 100 150 kmo h. F
0.26667
, 1 1D M B 50.0 kmol h 
 For Column 2: 2B essentially pure toluene, 2 2B 0.266667 F 0.26667 150 40 
 OK – Satisfies overall external M.B. 
 2 2 2D Recycle F B 150 40 110 
 2D 10.0, which also satisfies external M.B. 
 
282 
 
 
 
 
11-D7. a.) Proposed Split: Bottoms – Essentially pure toluene 
 Distillate ~ .83 methanol. (See figure for Solution to 11.D7) 
 M MB TF 100, z 0.5, z .1, z .4 
 Assume all toluene in bottoms & bottoms is pure. 
 B 100 .4 40, D 60 
 M,dist60 x 50 , M,dist MB,distx 0.8333 & x 0.166667 
b.) Proposed Split. Distillate pure M 
 Bottoms 0.2 MB, 0.8 T (See figure). 
 D 100 .5 50, B 50 
 Note – Doubtful this will work. 
 
283 
 
 
 
 
11.D.8. New Problem for 3rd edition. Part a. 
 Cost in June 2010 $947,000 556 397 $1,326,000 
Part b. 
 F 2 x F of Example 11-2. 
 Since Dia F , 11 2Dia 2F 12 ft 2 16.97 or 17 feet. 
 21mDiameter 17 5.182m, Tray Area 21.08m .
3.2808 ft
 
 Tower 3Volume 488.2m Fig. 11-1, 0 3pC Cost vol ~ $550 m 
 3 3P,towerC $550 m 488.2 m $268,500 
 Fig. 11-2. 
3 2 2
p,trayTraycos t area ~ $750 m extrapolate, C $750 m 21.08m $15,800 tray 
 BM,towerC 268,500 2.25 1.82 1.0 1.0 $1,093,000 
284 
 
 BM,traysC $15,800 36 1.0 1.0 $569,000 
 Total bare module cost September 2001 = $1,662,000 
 In June 2010, Cost $1,662,000 556 397 $2,328,000 
Part c. Original feed rate 1000 lbmol h . 
 At originalF , cost lbmol $1326 lbmol 
 At original2 x F , cost lbmol $1164 lbmol 
 
11.G1. - Use Fig. 11-10b as flowsheet. Use NRTL. 
 Feed : 1 atm, sat'd Liq, 100 kmol/h, 0.5 MeOH, 0.4 T, 0.1 MB 
 Fed to Stage 30 
 Int. Recycle sat’d Liq, 100 kmol/h, 100% MB, fed to Stage 20 
 
 Block 1B : N = 46, total condenser, partial reboiler, D = 50, L/D = 3 
 Dist : 0.999759 MeOH, 0.000241 tol, 1.537 E 0.8 MB 
 Bot : 8.02 E 0.5 MeOH, 0.266586 tol, 0.73333 MB 
 
 Block B2: N = 85, L/D = 9, B = 40, feed = 41 
 Dist: 0.0001094 MeOH, 0.0008684 Tol, 0.999022 MB 
 Bot : 2.855 E 35 MeOH, 0.99731 tol, 0.0026888 MB 
 Thus, this is feasible. 
 
11.G2. a) For Fig. 11-10A. 
 f
Bottom 0.999186 tolL
Col 1. N 90, N 41 , 8 D 60
 0.000814 MBD
 
 
 f
Dist. 0.99938 tol, 0.000605 MB,
Col 2. N 20, N 10 , L D 2, B 10 Bot .996679 MB, 0.003088 MeOH,
0.000233 toluene
 
 If increase L/D in column 1, fCol 1. L D 9, N 90, N 41, Bot. 0.99941 toluene. 
 fCol 2. L D 2, N 24, N 12, Dist. 0.99954 MeOH, Bot. 0.9976 MB 
 
 Which now meets specifications. Thus Figure 11-10a without recycle is feasible. 
 
 b.) For Fig. 11-10B – converged N = 30, L/D = 6 
 Dist. Col 1. 88.7% M & 11.3% T – No MB 
 (azeotrope) 
 Would not converge higher N. 
 Does not appear to work; thus, not feasible. 
 
11.G.3. New Problem for 3rd edition. F 100, 10% Ethanol, 5.0 atm, Sat’d liquid feed 
 N = 10 includes partial reboiler, total condenser, D = 10, L D 2 
 
285 
 
P = 1 NF = 5 QR = 52,948 xD,E = .72033 xB,E = .031085 
 NF = 6 QR = 53,029 .74564 .028263 
 NF = 7 .76332 .026298 
 NF = 8 53,148 .77539 .024957 
 NF = 9 53,175 QC = -78614.5 .78037 .024363← 
 NF = 10 52896.9 cal/s .69964 .033373 
 
 
 
Pcol = 3 atm L D 2 
 NF = 8 QR = 67995.5 .75453 0.027275 
 NF = 9 QR = 68022.1 Qc = - 74274 0.76027 0.026637← 
 NF = 10 67794 .69695 .033672 
 
 
Pcol = 5 atm L D 2 
 NF = 8 QR = 76435.3 .74276 .028582 
 NF = 9 c7646 Q 71692.8 .74840 .027955 ← 
 NF = 10 76264 .69411 0.033988 
 
Size optimal feed columns. Sieve tray 1 uses 18 inch tray spacing at 85% approach flooding, Fair 
calculation method for flooding. 
 colP 5 Max diameter tray 2 0.34867m 
 colP 3 Max diameter tray 2 0.38070m 
 colP 1 Max diameter tray 2 0.46429m 
 
Part d. D1. 1.0 atm gives the best separation because the relative volatility is highest. 
D2. The lowest Qreboiler is 1.0 atm. The effect of pressure on Qreboiler in this problem occurs 
because the feed is always a saturated liquid at 5.0 atm. For the 5 atm column this feed remains a 
saturated liquid and the feed line is vertical. At lower column pressures the feed flashes and is a two-
phase feed in the column. These feed lines have a negative slope. For the feed lines at lower pressures 
the slopes of the bottom operating lines are steeper, which means lower boilup ratios, Vreboiler/B. This 
means lower Qreboiler at the lower pressures. Another way to think about this is the flashing feed produces 
vapor and thus less vapor is required from the reboiler. 
D3. The lowest absolute value of Qcondenser is 5 atm. All columns have the same D and L/D. 
Thus, V entering the condenser is the same. At higher pressures the latent heat of vaporization λ is lower. 
Since Qcondenser = Vλ, the result is a lower absolute value of Qcondenser at the higher pressures. 
D4. The smallest diameter column is at 5 atm. Vapor density is highest. 
 
Part e. Increasing pressure above 1 atm for the same purity requires more stages, but smaller column 
diameter. Thus capital cost initially goes down. Above 8 atm the column must be designed for high 
pressure operation, which makes it more expensive. Operating cost may go up if a higher L/D is required 
to achieve the desired purity. 
 
11.G.4. New Problem for 3rd edition. 
 a. (L/D)min = 1.3962 → L/D = 1.5358. Obtain N (Aspen Notation) = 19 with Nfeed = 9 (on stage). 
Distillate is 0.75056 ethanol and bottoms is 0.00005987 mole fraction ethanol. QR = 569,172 cal/s, 
286 
 
Qc = - 443187 cal/s, Dia = 1.0755 m, A = 0.90843 m2, H = 11.5824 m, Vol = 10.52 m3 
 
Part b. Note that balancing the flow rates to achieve the desired separation in each column is trial and 
error. The easiest was to do this is to first find a flow rate that works for the high pressure column (note 
that D changes every time the flow rate is changed). This gives a value for QC,high pressure = - QR,low pressure. 
Then design the low pressure column with this QR. Check that both columns work. The correct flow rate 
into the high pressure column is between 570.6 and 577.275. The results are reported below with the 
latter value: 
High pressure column: F = 577.275 kmol/h, QR = 0.7555(569172) = 430,000 cal/s. N = 20, Nfeed = 9, 
distillate is 0.75004 mole fraction ethanol and bottoms is 0.00009733 mole fraction ethanol. Qc = -
243,900 cal/s, L/D = 1.5618, Tcond = 382.16 K, Dia = 0.79443 m, A = 0.49568, H = 12.192 m, Vol = 
6.043 m3. 
 
Low pressure column: F = 422.725 kmol/h, QR = 243,900 cal/s, N = 19, Nfeed = 9, distillate is 0.75087 
mole fraction ethanol and bottoms is 0.00003829 mole fraction ethanol. Qc = -190,627 cal/s, L/D = 
1.5803, Tcond = 351.56 K, Treb =373.16 K, Dia = 0.7014 m, A = 0.3864, H = 11.5824 m, Vol = 4.475 m3. 
 
Part c. Energy requirement of multieffect distillation system was set at 75.55 % of the single column. 
This is not optimized, but was set because it was known to work. Cooling is only in the low pressure 
column of the multieffect system, and is significantly less than with only one column. The total volume 
of columns is the same; however, this is misleading because volume in the single column would have 
been less if it was designed at 3.0 atm. The capital cost will be higher for building two columns than one 
larger one, but energy costs are less. 
 
 
287 
SPE 3rd Edition Solution Manual Chapter 12 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 12.A5, 12.A6, 12.C2, 12.D1, 12.D3, 12D8, 12D13, 12D19, 12D21, 12D22, 12.G3. 
12.A1. By raising T or dropping p can make gas desorb. The direction of transfer of solute 
controls whether a column is a stripper or absorber. If operating line (on Y vs X) is 
below equilibrium have a stripper. 
12.A5. New Problem in 3rd edition. c. AspenPlus 
12.A6. New Problem in 3rd edition. A. a. B. d. C. e. D. h. E. i 
12.B1. Calculate: N or outy , or N for A and outy for B, or m, or N for A and m for B, or L/V, 
or feed composition, or b, or m and b from 2 experiments, or Overall Efficiency. 
12.B2. Two feeds, Sidestream, Reboiled absorber, Coupled absorber and stripper (see Figure 
12-2), Interstage coolers (absorption) or heaters (strippers), Packed columns, Two 
different solvents, Two different stripping gases, Add solid adsorbed to solvent (see 
Chapt. 14). Cross-flow, Co-current flow, Combinations of flow patterns, etc. 
12.B3. See Isom and Rogers (1994). 
12.C4. o equil actual
Eq 12 22
E N N
Eq 12 34
(10-1) 
mV mV
o
mV mV
n 1 E 1 n 1 E 1
L L
E
n ( L mV) n L mV
mV
o
mV
n 1 E 1
L
E
n mV L
(10-4) 
QED 
Ref. Lacks, D.J., Chem. Engr. Educ., 302 (Fall 1998). 
12.C5. 2
x dy
y , m
1 1 x dx 1 1 x
When 
dy
x 0, , b 0
dx
When 
dy 1 1
x 1, , b
dx
Apply Kremser as x 0 or x 1.0
12.C6. Graphically, N 1 1
MIN
N 0
y y
L V
x x
 where N 1 Ny & x are in equilibrium288 
 
 N 1 Ny mx b for linear. 
 For absorbers 
min min
L L
m & 1
V mV
 
 Thus as 
min
L
N , 0
mV
 
 Eq. (12-23) becomes *N 1 1 1 0*
minN 1 1
y y L
 where y mx b
y y mV
 
 or N 1 1 N 1 1
N 1 min minN 0
0
y y y yL L
y b V x x Vx
m m
 
 which agrees with graphical analysis. 
 
12.D1. New Problem 3rd Edition. 
tot
H 1.186
y x x 0.395 x
p 3
 
 m 0.395 b 0 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Eq. 12-30 
*
N 1 1
*
1 1
y ymV mV
n 1
L y y L
5
mV
n
L
 
 
 
39.5 0.004 0.0000395 39.5
n 1
L 0.0002 0.0000395 L
5
L
n
39.5
 
39.5 39.5
n 1 24.67
L L
L
n
39.5
 
*
N 1y unknown.
 
1y 0.0002 
0 N 1 1 NL x Gy Gy L x
 
unknown 
 
Nx
 
G 
unknown 
1 
N 
*
1 0y m x .395 0.0001 0.0000395 
L 
0x .0001 
L 
unknown 
Since Nx unknown, 
N 1y 0.004 
Eq. (12-30) or (12-31) easiest to use 
289 
 
 
 T & E → be sure L mV 1 
 
L mV/L L/mV RHS
50 0.79 1.2658 7.58
75 0.52668 1.8987 3.90
65 0.60769 1.64557 4.67
63 0.62698 1.59494 4.895
62 0.637097 1.56962 5.015 OK 
 N N 1 1 0
G 100
x y y x 0.004 0.0002 0.0001 0.006229
L 62
 
 There are alternative solution paths, but L = 39.5 is not valid, it becomes 
n 1 0
n 1 0
. 
 Alternative: Trial-and-error McCabe-Thiele solution. 
 
12.D2. 
 a) L & V constant. 97% rec. H2S 3% left in gas. 
 outy 0.03 0.0012 0.000036 
 Equil.: 2H S
tot
p 423x
y 169.2x
p 2.5
 
 M.B. with Const. L & V: out0.0012 V 0.000036 V x L 
 out
10V
x 0.0012 0.000036 0.001164 5.82E 6
L 2000
 
 b) See Figure. in out in out out in
L L
y x y x where y , x & y , x are on op. line
V V
. 
 c) 
 
 
 
iny 0.000012 
y 
Eq Note that minL G can 
be calculated from a 
crude sketch. 
x* 7.09E 6 
min
L G 
min
L 0.0012 0.000036
164.124
G 7.09 E 6
 
0 
outy 0.0000036 
minact
L L
200 M
G G
 
0 
x 
M = 1.2186 
169.2 
290 
 
 d) L/G too high – L too high – Liquid too dilute. 
 Need much better solvent (e.g., MEA solution). 
 
12.D3. New Problem 3rd Edition. Mass Balance: 
 IN IN outVy Lx y V Lx , Equil., 
tot
H
y x
p
 
 Substitute equil. into M.B. 
 IN IN
tot tot
HVx HVx
Lx Lx L x x
p p
 
 Solve 
6
tot 6 6
IN
10.96 10 0.4 10HVx
p 0.274 atm
L x x 100 2 10 0.4 10
 
 b. 6 6out out
tot
H 10.96
y x 0.4 10 16.0 10
p 0.274
 
 Can also do graphically, with Kremser equation (trial and error) or by solving mass balance first, 
 6out in out
L
y x x 16 10
V
 Then tot out outp H x y 0.274 atm. 
 
 
12.D4. Mass bal. in in in in out outy V L x y V Lx 
 
291 
 
 outy y in out in outin
L L L
x x x x
V V V
 12 0.0002 0.00001 0.00228 
 Op. Eq., 
L
y x y
V outin
L
x 
V
 
 in outpoint y , x 0.0, 0.00001 and out inpoint y , x 0.00228, 0.0002 are on op. 
line. See graph. 
 
 Can also use Kremser equation for this problem. 
 
 
292 
 
 
 
12.D5. Since equilibrium is linear in weight ratio units can do either McCabe-Thiele or 
Kremser solutions. For McCabe-Thiele solution know that points 
out in in outY ,X and Y ,X are on operating line. McCabe-Thiele diagram is shown in 
Figure. N = 5.9 stages. HETP = 10/.59 = 1.69 ft. 
 
 
 
 Kremser: Several different forms can be used. We will illustrate with Eq. (12-26) 
written in ratio units. 
 *N 1 in N+1 outY Y 0.02, Y mX 1.5 0.06 0.09 
 *1 out 1 inY Y 0.50, Y mX 1.5 .40 0.60 
 
n .02 .09 .5 .6
N 5.88
n .02 .5 .09 .6
 
 HETP h N 10 5.88 1.70 feet 
 
12.D6. First, assume Nitrogen is an ideal gas: 1 lbmol = 359 cu ft at 0˚C and 1 atm 
 
333.16
359 437
273.16
cu ft/lbmol at 60˚C 
 2N flow rate 2500 437 5.72 lbmol/h 
 Water flow rate: Ignore 2CO in water. wMW 18 
 
100,000
18
5560 lbmol/h 
293 
 
 Equilibrium at 60˚C: H = 3410, y = 
TOT
H
x 3410x
P
 
 L 5560 972.0
V 5.72
. External mass balance is: 
 6out in in out in out
L L L
y x y x x x 972. 9 10 0.00875
V V V
 
 Can solve with either McCabe-Thiele diagram or with Kremser Eq. (12-34). The 
McCabe-Thiele solution is shown in the figure. Note different scales on axes. Need 5 
real stages. 
 
 
 
Kremser: * 6N 1 1 1 0y 0, y 0.00875, y mx 3410 9.2 10 0.03137 , m V L =3.508 
 
 
0 0.03137
n 1 3.508 3.508
0.00875 0.03137
N 5.07
n 1 .4 3.508 1
 
 Probably use 6 real stages. 
 
12.D7. WtG 0.828 wt frac air 
kg gas
1050 869.4
h
 kg air/h 
 mole
869.4
G 29.98
29
 kmol air/h 
 3inin
in
kg NHy 0.172 0.172
Y 0.2077
1 y 1 0.172 .828 kg air
 
 in ,molar
29
Y 0.2077 0.3543
17
 
 Inlet 3NH 0.172 1050 180.6 kg 3NH 
 2% remains in gas 33.612 kg NH 
294 
 
 
moleout
3.612 17
Y 0.007087
869.4 29
 
 Equilibrium data Table 12-2 - 3NHmole
TOT
p p, mmHg p, mmHg
y
P 1.30 760 988
 
 
3
wt
kgNH
X
kgW
 
 
p 
 
y mole frac mole
y
Y
1 y
 
0.05 11.2 0.01134 0.011466 
0.075 17.7 0.017915 0.01824 
0.100 25.1 0.0254 0.02607 
0.15 42.7 0.04322 0.04517 
0.20 64 0.06478 0.06926 
0.25 89.5 0.090587 0.09961 
0.30 119 0.12044 0.13694 
0.40 190 0.19231 0.2381 
0.50 275 0.27834 0.3857 
0.60 380 0.3846 0.6250 
 
 M.B. in
X L
wt
mol mol
3
3
LX
G Y
kg NH17 17
kmol NH
outGY , 
 wtmole wt out
mole
L
Y X Y
17 G
 
 
 Note: Units, although mixed, work in Mass Balance & in Operating equation. See graph. 
 
 Minimum Solvent 
 
0.3543 0.007087
Slope 
0.477 0
wt,min
mole
L
= 0.7279=
17 G
 
 wt,minL .7279 17 29.98 370.99kg W h 
 Actual Solvent, wt min
kg W
L 1.5 L 556.48
h
 
 Op Line Slope wt
mole
L 556.48
1.092
17 G 17 29.98
 
 
295 
 
 
 
 
296 
 
 
12.D8. New Problem 3rd Edition. 
 
 
 
 
 
 
 
 
 
 
Assume Water (not total liquid) flow rate is constant in both sections. Assume air flow rate (not total gas) 
constant is each section. In bottom section G 0.8 molair h . In top section 
G 0.8 0.5 .95 1.275 mol air h. Keep X as kg HCl kg water (from equil. data). 
Convert p to y (mole fraction) to Y (mole ratios). 
kg HCl
X
kg water
 p 
tot
p
y
p
 
kg HCl y
Y
kmol air 1 y
 
0 0 0 0 
0.0870 0.000583 3.8355 E-7 3.8355E-7 
0.1905 0.016 0.00001053 0.00001053 
0.316 0.43 0.0002829 0.0002830 
0.47 11.8 0.007763 0.0078240. 
0.563 56.4 0.037105 0.038535 
0.667 233 .15329 .18104 
0.786 840 .55263 1.2353 
 
2F 0.5 .475 air 
1F 1.0, G=.8
 
.05 mol HCl
0.05 mole frac HCl y 0.05263
.95 mol air
 
out outy 0.002 Y 0.002004 
L 
L 
G 
G 
totp 2 atm 760. 2 mmHg 
1IN IN
.2 mol HCl
y 0.20 mole frac HCl Y =0.25 
.8 mol air
 
1520 mmHg 
outX 
INX 0
L
 
297 
 
 Y vs X equilibrium data is curved. 
 Using these units, we need L in kg water hr and G and G in kmol air hr , and we need to 
convert the X terms from kg HCl/h to kmol HCl/h. 
 Top Operating Line HCl IN HCl outGY (L / MW )X (L / MW )X GY , INX 0 
 out
HCl
L
Y X Y
G(MW ) HCl
L
0.002004
1.275(MW )
 Goes through (0, 0.002004) 
 Bottom Operating Line HCl out in HClGY (L / MW )X GY (L / MW )X 
IN out
HCl HCl
L L
Y X Y X
G(MW ) G(MW )
 HCl HCl outL 0.8MW 0.25 L 0.8MW X 
a) Feed line saturated gas at Y 0.05263. Two operating lines intersect at feed line. 
 
 
 
 
 
 
 
 
 
 From plot *OUTX 0.69 (see figure) 
 Top operating line: 
2out int er sec t F
HCl
L
Y X 0 Y
(G(MW ))
 
 Bottom operating line: 
2 IN
*
F OUT int er sec t F
HCl
L
Y X X Y
(G(MW ))
 
 Solve for 
2int ersec t HCl F OUT
X G(MW ) Y Y L. Then 
2 2 IN
* HCl
F OUT F OUT F
HCl
G(MW )L
Y X Y Y Y
(G(MW )) L
 
IN 2 2 IN 2 2
*
OUT F F F OUT MIN F F F OUT* *
OUT OUT
L G G(MW) G(MW)
X Y Y Y Y L Y Y Y Y
(G(MW)) G X X
MIN
0.8(MW) 1.275(MW)
L .25 0.05263 0.05263 0.00200
0.69 0.69
 
 MIN HClL / MW 0.22883 0.09355 .3224 kg water h 
 
 Since MWmin = 36.461, Lmin = 11.755 kg/h, L = 1.2407 Lmin = 14.584 kg/h. L/(MW)HCl = 0.40 
 
*
outX 1,INY 0.25 
out0.002 Y 
HClL (G (MW )) 
Y 
int ersec tX 
X 
 
F2Y 0.05263HClL /(G(MW )) 
Sketch for Min L determination 
298 
 
 
299 
 
 
 b. M.B. 
22 F IN HCl IN out HCl out
F Y GY (L / MW )X GY (L / MW )X 
 22 F IN outout IN
HCl
F Y GY GY
X X
(L(MW ))
 
 out
0.475 0.05263 0.8 0.25 1.275 0.002
X 0 0.5561 kg HCl kg water
14.584 / 36.461
 
 
 Top 
(L / MW) 0.4
0.3137
G 1.275
 
 Bottom 
(L / MW) 0.4
0.5
G 0.8
 
 Top goes through HClIN out
(L/MW )
X 0, Y 0.002, Slope 0.3137
G
 
 Arbitrary point for plotting: X .4, Y .3137 .4 0.1255 
 Y .1255 .002 0.1275 
 Bottom from out INX 0.5561, Y 0.25 
 To intersection Top Operating and Feed Line. 
 Need ~ 1.6 stages. Opt. Feed for 2F is Stage 1 (Feed 1 is at bottom.) 
 Check Slope bottom 
.25 0
0.504
.5561 .06
 0K. 
 
 
300 
 
 
Figure for 12.D8. 
301 
 
12.D9. Repeat 12.D2 with Kremser. 
 , 
 , 
 
Eq. (12-22) 
 
 Graphical solution was 10.4. Pretty close! 
 
 
12.D10. Use Kremser equation such as 
 
*
out out
*
in out
A A
N 1
A A
y y 1 L mV
y y 1 L mV
 
 
 *
in inout
A AA
N 4, m 1.414, L V .65, y 0, y m x 1.414 .02 .02828 
 L mV .65 1.414 .459 
 Equation becomes: 
outA
y .02828 .02828 .552 .01267 
 Overall mass balance: in in out outy V L x y V L x 
 
 4out out in in
V V 1
 x y y x .01267 0 .02 4.93 10
L L .65
 
 
12.D11. Any of the vapor forms of Kremser equation can be used but problem is trial and error. 
For example, use Eq. (12-21) inverted for L m V 1 
 
*
1 1
N 1*
N 1 1
L
1y y mV
y y L
1
mV
 becomes, 5
1.2
1
m.27
1.2
1
m
 
 Set up table and try values of m. 
 
m 1.0 1.2 1.3 1.4 1.41 1.415 
RHS .1344 0/0 .233 .2658 .2691 .2706 
 
 By linear interpolation m = 1.414. Note that m = 1.2 is a trap for the unsuspecting 
student since L/(mV) = 1.0 and special form of Kremser is required. 
 
12.D12. Note this requires information in Section 13.4. 
mV 169.2
y m x m 169.2, b 0, 0.846
L 200
2000
L V 200
10
*
1 0 0 iny m x b 0, x x 0 1 out N 1y y 0.000036, y 0.0012
0.0012 0
n 1 0.846 0.846
0.000036 0
N 10.69
n 1 0.846
302 
 
 L .796 1000 796 kg solvent/h , 
All stages
L 796
0.0316
G 25,190
 
 
 in out 1,in
.204 0.025 .0012
X .256, X 0.02564, Y .001201
.796 0.975 1 .0012
 
 
x X y Y
.05 0.0526 .002 .002004
.10 .11111 .004 .00402
.15 .1765 .006 0.00604
.20 .25 .008 .00806
.25 .3333 .01 .0101
Equilibrium, y = 0.04 x
 
 
 Plot weight ratios. 
 Op. Line: j j 0 j 1
L L
Y X Y X
G G
 
 Slope = - L/G = - 0.0316. Step off stages backwards (start w. stage N) since it is 
different than other stages and we wouldn’t be sure when to switch if stepping off 
forwards. Need 4 equilibrium stages. 
 Note: Can also plot y vs X, since y ~ Y and G ~ V 
 
 
 
 
303 
 
12.D13. New Problem 3rd Edition. Strip Vinyl chloride from water at 25ºC and 850 mmHg. 
 
tot
H
y x
p
 1243.84x 1147.904x
850 760
 
 Want 0.1 ppm water leaving. Entering air is pure, L 1 kmole hr. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 * *1 out INy y 1147.904 x 5739.52 ppm (mole) 
 MIN
mix
L 5739.52 0 kmol
1171.33 G 0.00085373
G 5.0 0.1 h
 
 b and c. Want MING 2 G 0.00170746 kmol h air 
 L G 585.665 (See figure for part b – labeled HW5 Prob 1b). m = 1147.904. 
 c. Eq. 12-28 *Nx 0 
 
585.665 5.0 0 585.665
N n 1
n 25.01147.904 0.1 0 1147.904
4.78327
n 1147.904 585.665 0.672944
 
 
d. out IN IN outy Lx Gy Lx G 
 For L 1 kmol h , G 0.00170746 kmol h 
out IN out IN
L
y x x y 585.665 5.0 0.1 0
F
 
2869.76 ppm 0.00286976 mole frac. 
 Probably send waste gas to incinerator. Will require additional fuel to burn. 
e. All concentrations are dilute enough that L G and equilibrium are straight and operation is very 
close to isothermal. 
 
 
 
 
*
outy 
outx 
y
 
IN 05.0 x x 
Ny 
N outx x 1 ppm 
1147.904 
out 1y y 0 IN
x x 
IN N 1y 0 y 
N 
1 
*
outy 
*
outy 
304 
 
 
305 
 
 
12.D14. a) 95% removal CH4, 5% remains – Constant V: 
CH4
outY 0.05 0.00129 0.0000645 
 b. 
 
 Slope
min
0.00129 0.0000645
L V 19.5429
0.00006271 0
 
 actual minL V 1.4 L V 27.360; L 27.360 V 2736.0 
 c) Ext. bal. 
4,in 4
CHCH 44
out in CH CH ,out
V V
x y x y
L L
 
 
4 out 4 ,in 4 ,outCH , CH CH
V 100
x y y 0.00129 0.0000645 0.00004479
L 2736
 
 d) Use methane values in Kremser eqn. (12-22) to find N 
 *1 0
20.57 100mV
m 3600 /175 20.57, 0.75183; y mx b 0
L 2736
 
 
 
0.00129 0
n 1 0.75183 0.75183
0.0000645 0
N 6.11 stages
n 1 0.75183
 
 e) Now use Argon values with N = 6.11 to find Ar,out Ar,outy & x . 
 Ar Ar,in N 1,Ar Ar,in Ar,0
7700
m 44.00, y y 0.00024, x x 0.0
175
 
 *1 0 Ar
44 100mV
1.6082, y mx b 0
L 2736
 
 Eq. (12-23) 
7.11
Ar ,1
7.11
0.00024 y 1 1.6082 1 1.6082
0.60846
0.00024 0 1
1
1.6082
 
 Ar,out Ar,1y y 0.00024 0.00024 0.60846 0.00009397 
 Ar,out Ar,in Ar,out
V 100
x y y 0.00024 0.00009397 0.00000534
L 2736
 
4CH Eq. 
4 4
4 4
CH CH
CH CH
TOT
p 3600 x
y 20.5714 x
p 175
 
4inCH
y 0.00129
x* 0.00006271
20.5714 20.5714
 
inx
 
min
L V 
CH4
outy 0.000645 
4CH
y 
iny 0.00129
 
4CH
x 
306 
 
 % Argon recovery in liquid 
0.00000534 2736
100 60.85%
100 0.00024
 
 
12D15. Need equilibrium data. From DePriester chart: 
 C3 C3 C3 C3 C4 C4 C4 C4K y x 1.23 m , K y x 0.34 m 
 Butane is a design problem: 
C4 C4
L mV
5.882, 0.17
mV L
 
 1.2% of the butane leaves as a gas. Thus, 
 *1,C4 1,C4 C4 0,C4y 0.006 0.012 0.0000072, y m x 0 
 
 
*
N 1 1
*
1 1
y ymV mV .0006 01 n .83 .17
L y y L .0000072 0
N 2.39
L n 5.882n
mV
 
 Propane is simulation: *N 1,C3 1,C3
C3 C3
L mV
y 0.0017, y 0, 1.626, 0.615
mV L
 
 
N 1
*
N 1 1 N 1
*
1 1 1
L
1
y y y mV
y y y 1 L mV
, N 11
y
y 0.000298
5.7034 
 
12.D16. Was 12.D19 in 2nd edition. 
. a.) Equil. 2
2 2 2
CO
CO CO CO
TOT
H atm
y x , H 25 C 1640
P mol frac
 
 50 mmHg 50 760 0.06579 atm 
 
2 2Feed
2
7TOT
CO CO
CO
P 1.0
x y .00035 2.134 10
H 1640
 
 Equilibrium in column: 2
2 2 2 2
CO
CO CO CO CO
TOT
H 1640
y x x 24982 x
P 0.06579
 
 
 Basis: L = 1 kmol total/h. Assume L & G constant. Input = 2.1341 × 10-7 kmol CO2/h. 95% 
removal = (.95) (2.1341 × 10-7) = 2.027395 × 10-7 kmol 2CO h in outlet gas. 
 5% CO2 remains in liquid 7 7.05 2.1341 10 0.106705 10 kmol 2CO h 
 in outx x outlet liquid mole frac 
7
70.106705 10 0.106705 10
1 kmol h
 
 
 b.) 
307 
 
 
 c.) 5 5minV 1.5 V 1.5 3.803 10 5.704 10 kmol h 
 Conditions for Kremser eq. are satisfied. 
 CO2 Mass Bal: 
2,out
7 5 5
CO2.1341 10 in 0.106705 10 out w. water 5.704 10 y 
 
2,out
7
3
1 CO 5
2.027395 10
y y 3.5543 10
5.704 10
 
 Eq. 12-29 
* *
N N 0 0n x x x x
N
L
n
mV
 
 
3
* * 7N 1 1
N 0
y y 3.5543 10
x 0, x 1.4227 10
m m 24982
 
 
7
7 7
5
0.106705 10 0
n
2.1341 10 1.4227 10
N 5.3569
1.0
n
24982 5.704 10
 
12D.17. *1
L L
y .0002 1.414 .0002828, 
mV 14.14
 
 Can use variety of forms of Kremser equation, but cannot easily use forms with *N 1y since 
*
N 1 N Ny mx and x is unknown and hard to calculate. Try Eq. (12-21). 
 
N 1
*
N 1 1
*
1 1
L
1
y y .0083 .0002828 mV36.91
Ly y .0005 .0002828 1
mV
 
Do by trial-and-error 
L/mV 2 3 2.9 2.91 
RHS 15 40 36.699 37.02 
Slope 24982 
max
Slope 
 L V
 
Since L = 1, 
4
max
L
2.62968 10
V
 
5
min 4
1
V 3.803 10 kmol h
2.62968 10
 
3
max in
7
max in out
y yL 5.3314 10 0
V x x 2.1341 0.10605 10
 
outx
 
iny 0
 
y 
x 
out
7 3
maxy 24982 2.1341 10 5.3314 10 
7
inx 2.1341 10 
308 
 
 
 Linearly interpolate L/mV = 2.907. Then, L = (2.907) (14.14) = 41.10 kmol/h. 
 
12.D18. a. 
total
p 11.5
y 0.00757
p 1520
, 
0.00757
x 0.0127, K y x 0.596
0.0127
 
 
 
 
0L, x 0 
N 
Nx 
1 i 0
T
L L
y x y x
V V
 slope 
T
L 100
2 3
V 150
 
N 1V 100 , FV 50, L 100 
1x .01, y .00596
 
N 1 N
L L
y x y x , 
V V
 
N 1y 0.0058 
N
150 0.0004 50 .003 100 .0058
x0.0067
100
 
0 F F N 1 N 1 N T 1Ext. MB: Lx V y V y Lx V y 
1y 0.0004,
 
N 1
L 100
Slope 1.0
V 100
 
FV 
Fy 0.003 
N 1V 
1 
Bot. of Column: 
309 
 
 
 
 c. Minimum L. 
 
 Pinch is at Fy . F Fx y 0.596 0.003 0.596 0.00503 
 min T min L V Slope 0.003 0.0004 0.00503 0 0.51653 
 min TL 0.51653 V 0.51653 150 77.48kmol h . 
 
N 1y
 
min
min
N 1
L
L 100 slope
V
 
Fy 
x 
1y 
min min
T
L L
 slope
V 150
 
Eq. 
Nx 
y 
0x
 
310 
 
12.D19*. New Problem 3rd Edition. Found m and L/V in Example 12-1. 
 L/(mV) = (L/V)/m = 133/105.6 = 1.259, (mV)/L = 0.794, y1* = 105.6×0 = 0 
 
 If use Eq. (12-22), N = {ln[(1 - .794)((100-0)/(10-0)) + .794]}/ln[1.259] = 4.5 
 
12.D20. a) 99% removal 2H O , 1% left, .000024 1000 .01 0.00024 moles out in L 
 
2 outH S
0.00024
x 0.00000024
1000
 
 Moles 2H S out in gas = (.000024) (1000) (.99) = 0.02376 
 
2 outH S
0.02376 0.02376
y 0.00691
V 3.44
 
b) Equil. 2
2 2 2 2 2
H S
2 H S H S H S H S H S
tot
H 26800
H S. y x x 1729.03 x , m 1729.03
p 15.5
 
 2
2 2 2 2 2
CO
CO CO CO CO CO
tot
H 728
y x x 46.9677 x , m 46.9677
p 15.5
 
 Can use Kremser eq. for 2H S design [dilute linear system]. For example, Eq. (12-28) 
 2
*
0 N
*
N N H S
x xL L
n 1
mV x x mV
N
mV
n
L
 
 
2 2H S H S
L 1000 mV
0.16813, 5.9479
mV 1729.03 3.44 L
 
 
 
2 2
*
N N 1 0 NH S H S
x y m 0, x 0.000024, x 0.00000024 
 
.000024
n 1 0.16813 0.16813
.00000024
N 2.4807
n 5.9479
 
c) For 2CO . 2 2
*
N N N 1,CO CO 0x unknown, x y m 0, x 0.000038 Kremser (12-31) 
2
22
2
*
CON N
N 1*
CO0 N CO
CO
mV
1
L 46.9677 3.44x x mV
, with N 2.4807 and 0.16157
x x L 1000mV
1
L
 
 2
N ,CO 0 2N 1 3.4807
2
mV
1 1 0.16157Lx x 0.000038 0.000031916 CO 
1 .16157mV
1
L
Little amount of CO
 
311 
 
 
12.D21. New Problem 3rd Edition. Abs. 23.78y x 4.756x
5
 
 Stripper 
23.78
y x 118.90x
0.2
 
 
 
 Abs 
 
 
 
 
 
 
Abs
735.302
L V 7.353
100
 
 External M.B. abs abs,IN abs,IN abs,outabs,out
abs
L x V y y
x
L
 
 abs,out
735.302 0.00001 100 0.00098 0.000079
x
735.302
 
 
 abs,out strip,INx x 0.00013253 
 Kremser Eq. 
7.353 mV
m 4.756, b 0, L mV 1.5460, 0.64681
4.756 L
 
 N 1 IN 1 outy y 0.00098, y y 0.000079 
 
 *1 INy mx 4.756 .00001 0.00004756 
 Eq. (12-22).
 
*
N 1 1
*
1 1
y ymV mV 0.00098 0.00004766n 1 n .35319 0.64681
L y y L 0.000079 .00004756
N
L n 1.5460n
mV
 
 
 abs
n 11.1216 2.40889
N 5.53
n 1.5460 0.4357 
 
y 
x 
EQx 0.00098 4.756 0.00020605 
4.756 slope equilibrium INy 0.00098 
outy 0.000079 
act abs,MINL 1.6 L 735.302 kmol h
 
INx 0.00001 
MINL 100 4.5956 459.56 kmol h
 
0.00098 0.000079
Operating line Slope 4.5956
0.00020605 0.00001
 
MINSlope L G 
312 
 
Stripper 
 
 
 
 
 
 
 
MIN
Strip
L 0.015758 0
128.6037
V 0.00013253 0.00001
 
 MIN,Strip Strip
735.302
V L 128.6037 5.718 kmol h.
128.6037
 StripV 1.5 5.718 8.576 kmol h. 
 
Strip Strip IN Strip Strip IN Strip out
Strip out
Strip
V y L x x
y
V
 
0 735.302 0.00013253 0.00001
8.576
 
 *Strip,IN Strip,out N N 1y y 0.010506, x 0 y 118.9 
 
Kremser (12-28) 
 
*
0 N
*
N N
x xL L
n 1
mV x x mV L 735.302
N , 0.721106
n mV L mV 118.90 8.576
 
 
0.00013253 0
n 0.27889 0.721106
0.00001 0
N 4.54
n 1.38676
 
 
12.D22. New Problem 3rd Edition. K 0.22. y Kx 0.22x. Plot e.g., at x 0.006, y 0.00132 
 L 75, V 150, L V 0.5 
 External M.B.: 
 
 N 1 0 N 1Lx Vy Lx Vy 
 N N 1 1
V
x y y 2 0.003 0.0004 0.0052
L
 
 Points 0 1 N N 1x , y 0, 0.0004 , and x , y 0.0052, 0.003 are on op. line. 
 Plot Op. Line. See graph (labeled 12.D.b). 2 stages more than sufficient. 
EQy 118.90 0.00013253 0.015758 
y 
INy 0
118.90 
IN out,abs 0x x 0.00013253 x
118.90 
MIN
L
V
 
N outx x 0.00001
 
118.90 = Slope Equilibrium 
= Slope Operating line 
313 
 
 
 
 
314 
 
 
12D23. Apparatus similar to Figure 12-2, except part of treated gas is heated and used as stripping gas. 
 
 Absorber: Work in terms of mole ratios. 
 in out inY .15 .85 0.1765, Y 0.005 .995 0.00502, X 0.005 .995 0.00502 
 G 1400 .85 1190 mol carrier gas/day, L 800 .995 796mol solvent/day 
 Equilibrium: y .5x . Convert to mole ratios 
x X y Y
0 0 0 0
.05 .0526 .025 .0256
.1 .1111 .05 .0526
.15 .1765 .075 .0811
.2 .25 .1 .1111 
 
 Plot ratios on McCabe-Thiele diagram 
 Absorber: Op. Line: out in
L L L
Y X Y X , 0.669
G G G
 
 External balance Absorber: out796 .00502 1190 .1765 1190 00502 796 X 
 out AbsX 0.2614 
 Step off stages as shown in figure. Need 8 equilibrium stages. 
 
 Stripper: in out absY .00502 same as Y , in out absX 0.2614 same as X 
 out in absX 0.00502 same as X , L 796 mol solvent day, have 4 stages 
 Stripper equilibrium: y = 3 x. Convert to mole ratios. 
 
x X y Y
0 0 0 0
.025 .0256 .075 .0811
.05 .0526 .15 .1765
.075 .0811 .225 .290
.1 .111 .3 .4286
.15 .1765 .45 .818 
 
 Problem is trial-and-error. Select outY . Draw operating line from out inY ,X to in outY ,X . 
See if need 4 stages. When need exactly 4 stages, L/G = slope. 
 
From Figure, final result shown: stripG L slope 796 1.0998 723.7 mol carrier gas/day. 
 out stripY ~ 0.287 mol ratio 
 
315 
 
 
 
 
 
 
 
316 
 
 
 
12.E1. Convert ppm(wt) to mole fraction: 6ppm 10 wt frac 
 Basis 1000 g. of steam: Feed Liquid 1000 ppm = 0.001 wt frac 
 
mole frac
lg Nitrobenzene 0.00812 gmole 0.0001465 
 999 g water = 55.4507 0.9998535 
 Total 55.4588 mol 
 
1000g
Avg mol. wt 18.0314
55.4588
 
 Liquid Water F 1726 g h 95.7219 mol h 
 outL F C where C S V 99 61.8 37.2g water , C 2.0648 mol 
 L 95.7219 2.0648 97.7867 mol h 
 Outlet: Basis 1000 g 
 628.1 ppm 28.1 10 wt. frac. moles Mol frac. 
 3 428.1 10 g nitrobenzene 2.2825 10 64.11 10 
 31000 28.1 10 g water 
999.9997 18.016 55.5062
 55.5064 mol
 ~ 1.0 
 V S 99 18.016 5.495 mol h . 
 Equil. toty mx b, b 0, m H p 28.0 
 Kremser Eq. – Several forms can be used. → Use Eq. 12-28. 
 
L 97.7867 mV
0.63555, 1.5734
mV 28.0 5.495 L
, * N 1N
y b
x 0
m
 
 
*
0 N
*
N N
x xL L
n 1
mV x x mV
N 5.763
mV
n
L
 
 Eq
act
N 5.763
Effic 0.524
N 11
 
 Ref. Hwang et al, IEC Research, 31 (7) 1992, 1753 & 1759. 
 
12-F1. H = 59.3 (Perry’s 4th ed., p 14-4). 
TOT
H 59.3
y x 11.86x
P 5
 
 To be absolutely correct should convert this to mole ratios, although at these low 
concentrations could use mole fractions with small error. 
 
y x
Y , X
1 y 1 x 
 
317 
 
x y X Y 
0 0 0 0 
.001 .01186 .001001 .012 
.0015 .01779 .001501 .01810 
.002 .02372 .002003 .0243 
 Change specified conditions to mole ratios. 
in in in in out out out outx 0, X 0; y .02, Y .02041; y .002; Y .002003; x .001, X .001001 
 See Figure for plot of operating line, equilibrium and stages. N = 3.3 
 
Height 5 ft
HETP 1.515
N 3.3 equil stage
 
 in out
out in
 Y YL .02041 .002003
18.4
G X X .001001 0
 
 If use mole fractions find L/V = 18.0 
 
12.F2. E pK 26.0, K 0.6, K 0.019 
 After one pass of mass balance obtain: 
318 
 
ethane pentane octane
xi,1 0.032 0.005 0.963
xi,2 0.035 0.031 0.934
xi,3 0.033 0.162 0.805
yi,1 0.975 0.003 0.021
yi,2 0.962 0.019 0.019
yi,3 0.885 0.099 0.016 
 
 For new temperature used multi-variant Newtonian convergence. 
 1,New 2,New 3,NewT 73.90 F, T 80.93 F, T 99.15 F. 
 
2.G1. a) A,outN 4, L 570, y 0.00317 
 b) A,out AN 8, L 500, y 0.00315 while with L 490, y 0.00358 
 c) A,out AN 16, L 490, y 0.002978 while with L 480, y 0.00337 
 d) Have a pinch point. 
 
12.G2. Used Peng-Robinson. 
 a. Total number of stages required 8 
 b. Feed stage location for the solvent1 
 c. Feed stage location for stream A 8 
 d. Feed stage location for stream B 6 
 e. Outlet mole fractions of gas stream leaving absorber 0.9991009, 0.00024691, 
 0.00019067, 0.0004617 
 f. Outlet mole fractions of liquid leaving absorber 0.012878, 0.0310992, 0.0198928, 
 0.93612992 
 g. Outlet gas flow rate 161.6478 kmol/h 
 h. Outlet liquid flow rate 213.352 kmol/h 
 i. Highest temperature in column 19.1287 ˚C and stage it occurs on 8 
 
12.G.3. New Problem 3rd Edition. Used NRTL. 
Column pressure = 1.0 atm. Feed gas flow rate = 752 kmol/h. Feed gas temperature = 100oC. 
Liquid feed temperature = 75oC. Recovery of isopropyl alcohol = 0.98000. 
T1 = 302.5K, T2 = 298.6K, T3 = 299.9K, T4 = 301.9K, T5 = 303.3K. 
Leaving gas: G = 802.2 kmol/h, Mole fractions: IPA = 0.02443, W = 0.033836, N2 = 0.93721 
Leaving liquid: L = 149.8 kmol/h, Mole fractions: IPA = 0.002671, W = 0.99638, N2 = 0.000950 
Column diameter = 1.5455 m. 
 
 
 
319 
SPE 3rd Edition Solution Manual Chapter 13 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are: 13.A12, 13.A13, 13.D3, 13.D5, 13D6, 13D10, 13D22, 13D30-13D34, 13D36-13D42, 
13.E2, 13.E3, 13.G1, 13.G2 . Chapters 13 and 14 from the 2nd edition were rearranged to place all the
extraction material into chapter 13 and the material for other separations in Chapter 14. Thus, the 
numbers of many problems have changed. 
13.A3. The amount of solvent should be increased. This will decrease F/S and move the mixing 
point M towards S. As a result the saturated extract product NE will be moved down 
(less solute). The difference point ∆ will be moved towards the triangular diagram. The 
combined effect will be that fewer stages are required. By adjusting F/S a condition 
requiring exactly two stages can be found. 
13.A5. The vertical axis will be the extract phase and the hypotenuse will be the raffinate phase. 
These will be connected by tie lines. Usual procedure can be used. 
13.A7. Situation where E = R and ∆ point is at infinity. All operating lines are parallel. 
However, this does not correspond to minimum number of stages in extraction. 
13.A.11 c.
13.A12. a. C will be spread out and go into both raffinate and extract streams.
b. C will concentrate around the feed edge. If C is very dilute in the feed, can concentrate C.
Then by stopping the feed but continuing to flow solvents, solutes A and B can be
removed. Solute C can now be collected by withdrawing a stream near the feed stage.
13.B1. Specify: 
o N NA B A B A B
T, p, z , z , F, x , x , y , y plus: 
1
N
N
1 N
1
N
N
B F
B F
A F
B A
F A
F B
F A
y , R, E, N
x , R, E, N
x , R, E, N
R, E, y , x
N, N , y , E
N, N , x , R
N, N , x , R, etc.
 Could also not be given one of standard variables (such as solvent concentration). 
13.B2. a). One can build stages which are cross-flow (e.g. see Figure 12-12) within a counter-
current cascade. This effectively increases stage efficiency. Not that upward flowing less dense 
liquid will be mixed. 
b.) Build chambered stages within a counter-current cascade to prevent mixing of the 
dense liquid and give better cross flow on each stage. 
c.) Put in baffles to prevent MIXING of both less and more dense liquids. This will be 
more effective if counter-current is arranged so that flow across stages is always in same 
direction (see sketch) 
320 
 
 
13.C.7. Start by defining ∆ and the coordinates of ∆ as: 
 o 1E R , o 1A o A 1 Ax E y R x , o 1D o D 1 Dx E y R x 
 Removing ∆ from 2nd and 3rd equations we obtain 
 
o 1A o A 1 A o 1
x E y R x E R (13-43a) 
 
o 1D o D 1 D o 1
x E y R x E R (13-43b) 
 Assume that o 1E R . Next write the three independent mass balances around stages 1 to j. 
 j 1 jR E , j 1 jA j 1 A j Ax R x E y , j 1 jD j 1 D j Dx R x E y 
 These equations are now in a form similar to the form of the mixing equations developed 
previously. To develop the three point form of a straight line use the first equation to remove ∆ 
from the other two equations, solve for j 1 jR E in each of these equations, and finally set the 
results equal to each other. The development proceeds as follows: 
 
 Use j j 1E R to remove ∆ from the other mass balances. 
 
j 1 jj j 1 A j 1 A j A
E R x R x E y , 
j 1 jj j 1 D j 1 D j D
E R x R x E y 
 Solve for j 1 jR E , 
j
j 1
A Aj 1
j A A
y xR
E x x
, j
j 1
D Dj 1
j D D
y xR
E x x
 
 Finally, set these equations equal to each other. 
 j j
j 1 j 1
A A D D
A A D D
y x y x
x x x x
 rearrange to: j j 1
j j 1
A A A A
D D D D
y x y x
x x x x
 
 This last equation says that the slope of the line between the points 
j jD A
y , y and 
D Ax , x is equal to the slope of the line between the points j 1 j 1D Ax , x and D Ax , x and 
thus the lines are colinear. Furthermore, the lever-arm rule is valid for this system. 
 
13.D1. a. If we have a single column with only pure solvent then 
321 
 
 N N 1
R R
y x x y
E E
 and point N N 1x , y 0.001,0.0 
 is on op. line. E 44, R 100 , Slope R E 2.273 
 Op line intersect equilibrium at x 0.0037 . Thus, cannot get to ox 0.012 . 
b. Now E 44 30 74 and R E 1.35 . 
Still want Nx 0.001, but N 1
44 0 30 0.004
y 0.00162
74
 
At x 0.001, equilibrium value of y 1.613 x 0.001613 . Alternative works, but 
have pinch point and need very large number of stages. 
 
 c. This alternative (Say use 25 kg/min of 0.004 butanol) 
 N 1
44 0 25 .004
y 0.00145
69
 which is below equilibrium point. 
 Now equilR E 100 69 1.449 m 1.613 Thus, this will work. Obtain 
1
0.00145 69 0.012 100 0.001 100
y
69
0.0174 
 Op line closer to equilibrium – require lot more stages. 
 If use 20 kg/min of 0.004 butanol: N 1
20 0.004
y 0.00125
64
 
 R E 100 64 1.5625 , 1
0.00125 64 0.011 100
y 0.01844
64
 
 Will also work. Becoming close to pinch at top *1equil y 0.019356 
 If 15 kg/min, N 1 eq
15 .004
y 0.0010, R E 100 59 1.6949 m
59
 
 *1 1
1.0010 59 0.011 100
y 0.1964 y 0.19356
59
 
 Won’t work. 
 Thus, there is a small range where option c will work, but with many stages. 
 
13.D2. IN F INR 20, E 20, x x , y 0, m 8.3333, b 0, N 2 
 Kremser equation 
R 20
0.12
mE 8.333 20
, *1 0 F N+1y mx b 8.3333 x , y 0 
 Eq (13-11) 
*
1 1
N 1*
N 1 1
R
1y y mE
y y R
1
mE
 
 1 F F F3
1-0.12
y 8.3333 x 8.3333 x 7.3460 x
1- 0.12
, → 1 Fy 0.9873 x 
322 
 
 Mass balance, F N 1Rx Rx Ey , 
F F
N F
20 x 19.746 x
x 0.01269 x
20
 
 Recovery = N F1 x x 0.9873 , which is higher than 0.963 obtained in cross-flow. 
 
13.D3. New Problem in 3rd edition. 
 From M.B. outR .013 E .001 R .007 E y 
Where R = 100 and the unknowns are E and yout. 
 
and Equilibrium: out outy 1.613 x 1.613 .007 0.01129 
 
 
out
R .013 .007 100 0.006
S E 58.309 kg h
y 0.001 0.01129 0.001 
 
Alternative Solution: 1 Equilibrium Stage 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Another alternative solution: 
y 1.613x 
1.613 
R
E
 
x 
outx .007 
.001 
outy 0.01129
 
y 
from graph
R 0.001 0.01129
1.715
E 0.013 0.007
R
E 58.309
1.715
 
.013 
323 
 
 
 If Eq. line is straight, can Use Kremser with N=1. 
 Both representations are correct. Treating similar to a flash is easier. 
 
 
13.D4. Since concentrations are low, use wt. fractions and total flow rates. 
 Equilibrium: y 0.828 x or m 0.828 
 0 N N 1
R 550 lb h, E 700 lb h, x 0.0097, x 0.00046, y 0.0003,
mE R 1.0538 and R mE .94893
 
 *1 INy mx b 0.828 .0097 0 0.00803 
 Kremser Eq. (13-11b), 
*
N 1 1
*
1 1
y ymE mE
n 1
R y y R
N
R
n
mE
 
 1 N 1 0 N
R R
y y x x
E E
550 550
.0003 0.0097 .00046 0.0075
700 700
 
 
.0077316
n .0538 1.5038
.0004716
N 33.6
n .94893 
 
13.D5. New problem in 3rd edition. 
 
 Part a) Can do with Kremser eq or graphically. 
 y m x b, m 0.828, b 0 
 
R 400
R 400, E 560,0.862664
mE 0.828 560
 
 0 N N+1
mE 1
x 0.005, x 0.0003, y 0.0001, 1.159
R .862664
 
Or 
 1E, y 
1R, x
 
 
2E, y 
1 
0R, x 
1x 
R E Slope 
2 1,INy y 
0x x 
x 
y 
1y 
Op. Line: 
2 1 1 0
R R
y x y x
E E
 
Points 1 2 0 1x , y , x , y 
On Op. Line 
But 1 1x and y unknown 
T & E in this configuration 
Slope 
R
E
known, N = 1 
where 2 1,iny y 
Eq. 
324 
 
 Since 
R
1.0
mE
, can use equation such as 13-11b 
 *1 0y m x b 0.828 0.005 0.00414 
 
*
N 1 1
*
1 1
y ymE mE
n 1
R y y R
N
R
n
mE
 
 
0.0001 0.00414
n 1 1.1592 1.1592
0.0034571 0.00414
N
n0.862664
 
 where 1 N 1 0 N
R 400
y y x x 0.0001 0.005 0.0003 0.003457
E 560
 
 
1.52637
N 10.332
0.14773
 
 Alternate solution: Eq. (12-28) becomes 
 
*
0 N
*
N N
x xR R
L R, V E , N n 1 n mE R
mE x x mE
 
 * N 1N
y b 0.0001
x 0.000120773
m 0.828
, 
1.52637
N 10.332
.14773
 
 b) 
 
 
13.D6. New problem in 3rd edition. 
 Part a. N 1 0 1 NEy Rx Ey Rx Ext. M.B. 
 1 0 N 1 N
R R
y x y x
E E
 
 1
100 100
y 0.005 0.0002 0.0005 0.003414
140 140
 
Part b. Kremser Eq. 
*
1 0y 0.828 x 0.828 0.005 0.00414 
*
1 N 1
min 0 N
y yR 0.00414 0.0001
0.85957
E x x 0.005 0.0003
 
MIN
400
E R 0.85957 465.3kg h
0.85957
 
Equil. 
Slope = 0.828 
min
R E slope
 
0x 0.005 Nx 0.0003
 
N 1y 0.0001
 
y 
325 
 
 Convert * N 1N
y 0.0002
x x L R y y V E, x 0.00016556
m 1.208
 
 
L R 100
0.5913
mV mE 1.208 140
 
 Lots of different forms can be used. 
 For example 
*
0 N
*
0 N
x xL L
n 1
mV x x mV
N
n mV L
 
 Becomes 
*
0 N
*
N N
x xR R
n 1
mE x x mE
N
n mE R
 
 
0.005 0.00016556
n 0.4087 0.5913
0.0005 0.00016556 1.8717
N
1 0.5254n
0.5913
 3.6 
Part c. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MIN
R 0.00604 0.0002
1.29777
E 0.005 0.0005
 
 MIN
R 100
E 77.05
1.29777 1.29777
 
 Maximum extract out EQ 0y x 0.00604. 
Part d. The roles of extract and diluents are switched in the two problems, which changes the definitions 
of y and x. 
 
13.D7. 
NN 1 0N 30, R 500, y 0.0002, x 0.0111, x 0.00037 
 Equilibrium: *N N 1y 0.828x, m 0.828, x y .828 0.00024155 
0x 0.005 
MIN
R
slope
E
 
Eq. y 1.208x 
x 
Nx 0.0005
 
N 1y 0.0002
 y 1.208x 
y 
EQy 1.208x 0.00604
 
326 
 
 Since rather dilute and linear equilibrium use one of the Kremser equations. 
 
*
o N
*
N N
x xR R
n 1
mE x x ME
N
mE
n
R
 (12-28 (modified)) 
 Where 
*
0 N
*
N N
x x
83.756
x x
. Solution is trial-and-error. 
 
 
 
E R/mE Calculated N 
500 1.21 
Negative-Not possible Need 
R
1
mE
 
700 0.8626 17.01 E too high 
650 .929 26.175 E too high 
640 .9435 29.84 E too high 
639 .945015 30.30 E too low 
 
 By linear interpolation need E ~ 639.6 kg/h. Can use other forms of the Kremser equation. 
 
13.D8. Was 13.D10 in 2nd edition. 
Convert Kremser 
x is raffinate R L
y mx m K, b 0
y extract, V E 
 
 a) 
*
N N
N 1 3*
0 N
Use 12-31 x x 1 mV L 1 K E R
Other forms OK x x mV KE
1 1
L R
 
 *N n 1x y K 0 
 N 0 3 3
30.488 25KE 11
100Rx x 0.00092 0.00001376
KE 30.488 251 1R 100
 
 b) Can use External balance or Kremser to find out 1y y 
 1 N 1 0 NEy Ey R x x 
 1 0 N
R 100
y x x 0.00092 0.00001379 0.003625
E 25
 
 
13.D9. Assume very dilute, R = 1500 kg/h, E = 750 kg/h 
 Equil. d dY K X becomes y K x 
 From Table 13-3. d,oleic oleic d,linoleic d,linoleicK m 4.14, K m 2.17 
 99% recovery oleic: 1,oleic.99 .0025 1500 y 750 → 1,oleicy .00495 
327 
 
 Use Kremser, Eq. (13-11b). oleic
4.14 750m E
2.07
R 1500
 
 *1 oleic 0,oleicy m x 4.14 .0025 0.01035 
 
*
N 1 1
*
1 1
y ymE mE
n 1
R y y R
N 5.44
R
n
mE
 
 For linoleic acid: 
lin
R 1500
.9216
m E 750 2.17
, 
 N 1y 0 , 
*
1 L 0,LN 5.44, y m x 2.17 003 .00651 
 Can use Eq. (13-11a): 
*
1 1
N 1*
N 1 1
R
1y y mE
y y R
1
mE
 
 1
.07834
y .00651 .00651 .00124796
.40866
 → 
linoleic1
y 0.00526 
 Recovery of linoleic: Rec .003 1500 .00526 750 → Rec = 0.877 
13.D10. New problem in 4th edition. 
 
Analytical or graphical solution OK. 
Stage 1 
11 F i 1,in 1,out 1 1
F x E y Ey R x 
 Equilibrium 1,out 1y 1.02 x 
 
1F 1 1
Fx 1.02 E R x 
 F1
1
100 0.015F x
x 0.0099338
1.02 E R 1.02 50 100
 
 Stage 2 1 1 2 2,in 2 2,out 2 2,out 1 2 1R x E y E y R x , R R F 100 
 2,out 2,outy 1.02 x 
 
 1 12,out 2 1 1
2 2
0.0099338 100R x
x 0.006579, R R F 100
1.02 E R 151
 
 Mix with Feed 2 3,inR 100 70 170 
328 
 
 22,out 2 F 23,in
3,in
x R x F 0.006579 100 0.005 70
x
R 170
 
 3,inx 0.0059286 
 Stage 3 3 3,in 3 3,in 3 3,out 3 3,outR x E y E y R x 
 3,out 3,outy 1.02 x 
 3 3,in3,out
3 3
R x 170 0.0059286
x 0.00456
1.02 E R 1.02 50 170
 
 Stage 4 4 4,out 4 4,in 4 4,out 4 4,out 4 3R x E y E y R x , R R 170 
 4,out 4,outy 1.02 x 
 4 4,out4,out
4 4
R x 170 0.00456
x 0.003508
1.02 E R 51 170 
 4,out 4outy 1.02 x 0.003578 
 
329 
 
 
 
 
330 
 
 
 
13.D11. R R F 2501, E E 1000 
 Equilibrium: DK 1.57 . For dilute this becomes Dm K in wt. frac. units. 
 Abietic Acid Recovery: N Fx R .95 F x .95 1.0 0.5 .0475 
 N
.0475 .0475
x 0.0000190
2501R
, F1
.05 1.0 .05.05 F x
y 0.0000025
E 1000
 
 Top op. Eq.: 1 0
R R
y x y x
E E
 
 Goes through pt 0 1
R 2500
x 0, y . Slope 2.5
E 1000
 
 Bottom Op. Eq.: N 1 N
R R
y x y x
E E
 through point N N 1x , y 0 , R E 2.501 
 Need 8 ½ stages (see Figure). 
 
 
 
13.D12. m 1.613, R mE 1.2399 , *N 1 4 1 iny y 0, y mx 0.00742 
 Eq. (13-11a) 1 4
y 0.00742 1 1.2399
0.17594
0 0.00742 1 1.2399
 
 1y 0.17594 0.00742 0.00742 0.006114 
 Overall bal. in in out3
10 0.0046 5 0.006114Rx Ey Ey
x 0.00154
R 10
 
 
13.D13. a. xy . 90% recovery, 10% left 0.1 1000 0.003 0.3 kg out 
 ,outx 0.0003 
331 
 
 O xy 95% recovery, 5% left 0.05 1000 0.005 0.25 , O,outx 0.00025 
 
 
 
 
For ortho, maxy 0.15 0.005 0.00075 
 
max,ortho
R 0.00075 0
0.1579
E 0.005 0.00025
 
 For para maxy 0.08 0.003 0.00024 
 
max,para
R 0.00024 0
0.08888
E 0.003 0.0003
 
 b. The p-xylene recovery controls. min
1000
E 11, 250
0.08888
 
 E 1.5 11250 16875 , 
R
0.0592592
E
 
 Can use Kremser eq. (13-11b) for ρ-xy to find N 
 
*
N 1 1
*
1 1
y ymE mE
n 1
R y y R
N
R
n
mE
 
 N 1m 0.080, R E 0.0592592, y 0 , 
*
1 o,py mx 0.080 0.003 0.00024 
 Mass balance: 90% entering ρ-xy leaves w. solvent. 
 1
0.9 1000 0.003
y 0.00016
16,875
 wt frac 
 
R 0.0592592 R
0.74074, n 0.300106
mE 0.080 mE
, 
mE .080
1.35
R 0.0592592
 
332 
 
 
0 0.00024
n .35 1.35
n 0.300.00016 0.00024
N 4.012
0.300106 0.30016
 
 Note: Can use other forms of Kremser eq if desired. 
 c. For o-xy check if recovery > 95% 
 Eq. (13-11a) 
*
1 N 11 1
*N 1*
1 0N 1 1
R
1 y unknown, y 0y y mE 
y mx 0.15 0.005 0.00075y y R
1
mE
 
 
R 0.0592592
0.39506, N 4.012
mE 0.15
 
* *
1 N 1 1 1N 1
R
1
mEy y y y
R
1
mE
5.012
1 0.39506
0.00075 0.00075 0.0029194
1 0.39506
 
 External M.B. 
N 11 N 0
Ey R x Ey R x 
 0 1N
5 16875 .00029194Rx Ey
x 7.3584 E 5
R 1000
 
 1
0
Ey
% Recovery 100 98.53%
Rx
 
13.D14. (was 14.D4. in 2nd ed.) a) 
S 10.0 2 MF
F 15.0 3 SM
 
 
 Once have M, use trial-and-error to find tie through M. (final result is shown). This 
gives E and R. A w A wy .115, y 0.04, x .23, x .73. 
b) Plot raffinate, AR x .1 . Find tie line through this point (not trial-and-error). This gives 
E. Draw Line ER. Intersection with line SF gives M. 
 
S S MF
. Find S 85.7
F 15.0 SM
kg/h. 
333 
 
 
13.D15. Since dilute, use Kremser equations. Assume units are weight fractions. 
 a) Column 1 at 40ºC. N 0x 0.0008, N 11,, x 0.01, E 1000,, R 100 
 Equilibrium: *1 0m 0.1022, thus y mx 0.001022. Kremser (Eq. 13-11a): 
 1 1 2
N 1
1
1y 0.001022 1.022 0.93664
y 0.0010221
1
1.022
 
 This simplifies to: 1 N 1y .093664y .00092628 
 External MB: N 1 o 1 Ny E Rx y E Rx , N 1 1y 1000 1 1000 y .08 
 which simplifies to: N 1 1
1
y 1000 y .92
1000
 
 Solve 2 eqs and 2 unknowns: 51,coll N+1,colly 0.00092693, y 0.6929 10 
 b) Column 2 at 25ºC: N 1,col2 1,col1y y 0.00092693 , 
 5 *1,col2 N 1,col1 0 1 0y y .6929 10 , x 0, N 9, m 0.0328, E 1000, y mx 0 
 
 Use Kremser to solve for R .́ This is trial and error. For example, Using Eq. (13-11a), 
 
*
1 1
N 1 10*
N 1 1
RR 11 0.0328 1000y y mE0.007475
y y R R
1 1
mE 32.8
 
 
R 50 60 50.5 50.35
RHS 0.007855 0.001981 0.007307 0.007467
 
 Within error R ́= 50.35 
 N 1 0 1N
y E R x y E .92693 0 .006929
x 0.0183
R 50.35
 
 
334 
 
c) Could be practical if m’s were larger, and have bigger shift in m. A similar scheme is used 
commercially for citric acid. Not practical here since have to pump around too much 
solvent. In addition, benzene is carcinogenic and would probably not be used as solvent. 
 
13.D16. a.) R E 10 8 1.25, R mE 1.25 1.613 0.77495 
 
0
*
1 Ay m x 1.613 0.01 0.01613. Use Eq. (13-11a), 
 1 7
y 0.01613 1 0.77495
0.27044
0.0002 0.01613 1 0.77495
→ 1y 0.01182 
 
0A A N 1 1
E E
x x y y 7.02498 E 4
R R
 
b.) Graphical check works fine (not shown) 
 
13.D17. j j IN j 1
j 0 j
R R R 10
y x y x , 5
E E E 2
 
 6x 0.0018 (See graph) 
 Note: 6 N,countercurrentx x 0.000702 even though use more total solvent. 
 
335 
 
 
 
 
13.D18. (was 14.D2. in 2nd ed.) Plot S, F, R and E. Draw lines SF and RE. Intersection is point M. 
 
 Lever arm rule: 
S MF 20.3
4.511
F 4.5SM
 → S 100 4.511 451.1kg/h 
 
 Or Mass Bal. S + R = M and A A AS y F x M x (S .15 .5 F .21 M ) 
 
 Solve simultaneously S = 483.3 
 
336 
 
 Difference is due to accuracy in reading numbers. Lever-Arm Rule more accurate! 
 
 
 
 
13.D19. Equil. d A AK y x 0.287 0.158 1.816 
 Acetone 0 0 1 1
0.005
y 0 Y 0, x 0.005 X 0.00503
0.995
 
 N 1 N+1
0.10
x 0.10 wt frac X 0.1111
.9
 
 DF 1000 .9 900 kg/h water, SF 1371 .995 1364.1 kg/h chloroform. 
 D S
900
F F 0.6598
1364.1
 
 Equil. 
337 
 
xA XA yA = 1.86 xA YA
0 0 0 0
0.01 0.0101 0.01816 0.01850
0.03 0.0309 0.05448 0.0576
0.05 0.0526 0.0908 0.09987
0.07 0.0753 0.1271 0.1456
0.09 0.9890 0.1634 0.1954
0.1 0.1111 0.1816 0.2219 
 
 External M.B. 
 D N 1
S
F
X Y
F
D
6 1 N N
S
F
X Y or Y 0.6598 0.1111 0.6598 .00503
F
 
 NN
N
Y
0.06999, y 0.0655
1 Y
 
 Results pretty close to 13.D43. 
1 2
2 vs 2 w i
2 3
accuracy of graphs. 
 
Note: The graph below should read acetone, not acetic acid as the solute. 
 
338 
 
 
 
 
13.D20. a) Batch Operation – Mix together & settle. Find fraction recovered: 
 Operating Eq.: 0 0 F
R R
y x x , R 5, S 4, x x
S S
 
 Which is, Fy 1.25 x 1.25 x 
 Equilibrium y 8.333 x, m 8.333 
 Eq. (13-21) written for batch 
0 iN F
F
ˆR̂ S x y 1.25 x 0
x 0.1304 x
ˆˆ 9.583m R S
 
 Frac. Rec 1 0.1304 0.8696 
b) Continuous solvent addition: 
Eq. (13-28) t ,finalt ,final t ,feed
Ft
ˆ xS 1 1
n x x 0.8 nˆ m 8.333 xR
 
 t ,final Fx x exp 0.8 8.33 0.00127 
Recovery = 99.87%. 
 
339 
 
13.D21 (Was 14.D1 in 2nd ed.) a. Let A = methylcyclohexane and D = n-heptane. 
 Mass Balances: 1 2F F S M or M 350 
 
F F S M1 2
1 A 2 A A AF x F x S y M x , F F S M1 21 D 2 D D DF x F x S y M x 
 
 Then F F S1 2
M
1 A 2 A A
A
F x F x S y 100 .6 50 .2 0
x 0.2
M 350
 
 F F S1 2
M
1 D 2 D D
D
F x F x S y 100 4 50 .8 0
x 0.229
M 350
 
 
 Plot M. Find tie line through M. (See figure.) This gives location of points E and R. 
 Find DR AR AE DEx 0.48, x 0.42, y 0.06, y 0.05 . 
 
 
 b. Mass balances: 
M E RA A A
M E R and Mx Ey Rx 
 Solving simultaneously: E = 214 and R = 136 kg/h 
 
13.D.22. New problem in 3rd edition. With interface at center, heavy phase flow area is 
2
f s
1
A D 4 0.411
2
 and 
fper s 5
1
P D D 2.630
2
 
 
 
 
 
 
 
 
 
 
 
 θ 
 
 
 
θ 
2
2 2 Cr .1
2
 
C 1.00326 m
 
r 
.1 
C/2 
α 
α 
sr D 2 0.5115 
Center 
.1 
Chord 
arc 
.4115 Interface 
 
(length = C) 
 
340 
 
 Draw right triangle for interface below center to calculate new perimeter. 
 
0.1 .1
sin .1955
r .5115 
 11.274 
 Then angle of arc, 180 2 157.452 
 Length of 
3.14159 0.5115 157.452r
arc 1.4056
180 180 
 
fer
P C arc length 2.4089m 
 Mensuration formulas are from CRC Standard Mathematical Table. 
 
f
3
settler c er cRe 4Q P 4 0.006 998 2.4089 0.95 10 10, 466 
 
 Interference somewhat more likely than in Example 13-5. 
 
 
13.D23 (was 14.D7. in 2nd ed.) a) F S 500 300 M 
 Pyrdine 
F O MA A A
F x S y 500 .3 0 M x → 
MA
x 150 800 0.1875 
 Plot M on line FS . By T & E find tie line through M (Use Conjugate line) 
 p py 0.223, x 0.84 ; w wy 0.02, x 0.84 ; 
 Mass balances: 1 1R E M 800 , 0.84R 0.02E 0.43M 
 Solve simultaneously, 1 1E ~ 400, R ~ 400 (Note: More accurate than pyrdine values.) 
 b) 1 2 2R S 400 300 700 M 
 
1 0 21 A 2 A 2 AM
R x S y 400 0.15 60 M x 
 
2pyr M
60
x 0.086
700
 
 
 Find tie line by T & E: pyr2 pyr2y 0.120; x 0.053 ; w 2 w 2y 0.005, x 0.945 
 
 MB: 2 w 2 2 w 2 m2wR x E y M x → 0.945 2 2R 0.005 E 700 0.48 
 
 2 2R E M 700 
 
 Solve simultaneously: 2E 346 and 2R 354 
 
341 
 
 
342 
 
13.D24 (was 14.D10. in 2nd ed.) a) Feed 40% MCH 55% n-heptane, F = 200. Solvent 95% aniline 
& 5% n-heptane, totalS 600 . S F M 800 
 Lever arm rule: 
S FM
3 .
F MS
 Find M (Easy way is divide line FS into 4 parts) 
 
 Use tie line through M to find points E & R (T & E) 
 
 Extract: MCH MCHy ~ 0.045, Raffinate:x ~ 0.36 wt fracs 
 
 Mass balance E + R = 800 = M and lever arm rule 
E MR
.
R ME
 Measure distances on figure. 
 
 Solve simultaneously: R = 124.61 kg/h, E = 800 – R = 675.39 
 
b) 2 stage cross flow. Stage 1: F = 200, ρ = 300, 
S 3 FM
F 2 MS
. Find point M. Tie line through M 
gives points R1 and E1. 
 Mass balance 1 1500 F S M R E and lever arm rule 
1 1 1
1 1 1
R E M
E M R
 
 Find: 1 1R 207.04 kg h , E 292.95 
 
 Note: Isotherms are very sensitive. Thus, calculation is not extremely accurate. 
Stage 2: Mass balance 1 2 2 2 2R S M 507.04 R E and lever arm 
2 2 1
1 2 2
S M R
E M S
 
 Find 2M and from tie line through 2 2M find R . Then can find R2 and E2 from mass balance 
(given above) and new application of lever arm rule, 2 2
2 2
R E M
E R M
 
 Solving simultaneously, 2 2R 196.16 kg h. E 310.88 
 
343 
 
 
 
 
13.D25 (was 14.D9. in 2nd ed.) a. Draw lines from S to F and from 1 NR to E . Intersection gives point 
M (see Figure). Then from lever-arm, 
 
 
S FM
1.25
F SM
 → S 1.25 2000 2500 
 
b. ∆ is at intersection of lines N N 1 0 1E R and E R . Then step off stages as shown. Need 2 stages. 
 
 
344 
 
 
13.D26. (was 14.D6. in 2nd ed.) Guess a value for M and step off stages. Repeat until need 3 stages. 
After three trials found M shown in Figure. This required 3 1/10 stages which is close enough. 
 Extract Composition: Acetic Acid = 10.5%, Water = 3.5%. 
 Raffinate Composition: Acetic Acid = 5%, Water = 93% 
 Solvent Flow Rate: F S F SM SF 15 57 2000 S 2000 → S = 5600 kg/h 
 Raffinate Flow Rate: 1 0 0 1 1 1R E E R R 5600, R 770 kg/h. 
 Extract Flow Rate: N 1E F S R 2000 5600 772 6830 
 
13.D27 (was 14.D12. in 2nd ed.) 
E 00
A wEy y 0 (Pure solvent) 
Lever arm rule: 0
0
E MF
1.112
F E M
. Find M. Line RM intersects sat’d extract at 
NN A
E , y 0.18 
 Lines N 1 0F E & R E intersect at . Step off stages 3 more than enough. Need ~ 2 ¼ 
 M.B. N 1E R M 211.2 kg/h & lever arm: 
N 1
1 N
E MR
2.287
R E M
 (from graph). 
 Solve simultaneously, 1 NR 64.25, E 146.95 kg/h 
345 
 
 
346 
 
13.D28 (was 14.D14. in 2nd ed.) Tofind ∆: 1) Plot N N 1E and R F 
 2) j j 1 N N 1E R E R 1500 
 N N 1N A N 1 AA
E x R x
x 0.06666 
3) ∆ is on line through points N N 1E and R . 
 Plot ∆. Or, use lever-rule. 
 N 1 NN 1
N
R E
R 1.5
E
 
 Step off three stages starting at point NE . This gives points 
 
1 1 1 01 A D 0 A D
R x 0.275, x 0.675 and E y .13, and y 0.0 . 
 
 Mass Balance: 0 N 1 1 NE R R E → 0 1 N N 1 1E R E R R 1500 
 and 0 1E 0.13 1000 0.4 R 0.275 2500 0.2 
 Solving simultaneously, 1R 655 kg/h, 0E 2155 kg/h 
 
13.D29 (was 14.D16. in 2nd ed.) a) Plot Points N 1F, S, E and R 
 Find ∆ point at intersection of lines N 1FE and R S 
 2 stages is more than enough. (see graph) 
 
 b) Draw lines N 2 1FS and E NOT calc. value E R . 
 Intersection is mixing point M 
 Mass balance F + S = M and Lever arm 
F dist. S to M
0.786
S dist F to M
 
 Give S F 0.786 1000 0.786 1272kg/h. 
347 
 
 Alternate: Overall MB, F S M and Diluent mass balance, 
 F,D S,D M,D650 F x S y M x 0.28 M 
 Solve simultaneously: M 2321 and S 1321 kg/h. But this is less accurate. 
 
 
 
13.D.30. New problem in 3rd edition. Equation (13-59) becomes Qc /Ai < ut /(1 + safety factor). 
Using the equals sign and solving for the safety factor Sf we have, 
 Sf = ut Ai / Qc -1 = 0.00172 (1.0)(4.0)/.006 – 1 = 0.1467 
 where Ai = Ds Ls. Thus safety factor is 14.67% instead of 20%. This may still be acceptable. 
13.D.31. New problem in 3rd edition. Soln. A. Kremser Soln. 
50
R mE 0.30998 1.0
161.3
 
 2 0R 50, E 100, m 1.613, b 0, y 0.0, x 0.01 
 For example, Use 13-11. *1 0y mx b 0.01613 
 
*
1 1
N 1*
N 1 1
R
1y y mE
y y R
1
mE
becomes 1 2
y 0.01613 1 0.30998
0.7633696
0 0.01613 1 0.30998
 
 1y 0.01613 0.01613 .7633696 0.00381684 
1 1
0.00381684
x y m 0.0023663
1.613
 
 Soln. b.) Do mass balances and equilibrium for single stage. 
348 
 
 IN FSy Fx Sy Fx 
 0 0.5 100y 50x 
 also y x 1.613 . Solve simultaneously and obtain identical result. 
 Soln. c. Do graphically as single stage system. 
 Soln. d. Do graphically as counter-current system, N=1. Solution is valid, but awkward. 
 
13.D.32. New problem in 3rd edition. Fixed Dispersed Phase. 
 At feed conditions sol feed feedtol sol feedtol
soltol feed sol feed
feed feed
feed
Q Q QQ Q Q
QQ Q 1 Q QQ Q
Q
 
 a) sol feedtol
sol
feed
0.6 .006 Q Q .6
.375
Q.006 .6 .006 1.61
Q
 
 Equation 13-48 operation in ambivalent range. 
 
0.3.3 3
L L H
3
L H L
865 0.95 100.375
1 0.625 998. 0.59 10 
 From Example 13.5 0.3 1.10235 
 The 
0.375
1.10235 0.6614
0.625
 
 Either phase can be dispersed. 
 b) d
1.0
0.5
2.0
, also ambivalent range 
 
.5
1.10235 1.10235 
.5
 Either phase dispersed 
 c) d
2.0
.6667.
3.0
 According to 13-48 at border. 
 
.6667
1.10235 2.2 
.3333 
water probably dispersed 
 d) d
5.0
.8333
6.0
Equation (13-48), water (heavy) dispersed. 
 
0.8333
1.10235 5.5 water dispersed.
0.16667
 
13.D.33. New problem in 3rd edition. re s liq d ct V Q Q 1.5min 90s 
 
3
d cQ Q 0.0072m s , 
3 3
liqV 90 s 0.0072m s 0.648m 
 Note that there is a 1 inch air gap at top 
 2liq t tan kV H 0.0254 d 4 0.648 , t tan kH 2d 
 2liq tan k tan kV 2d 0.0254 d 4 0.648 
 Using Goal Seek tan k tan kd 0.7489 and H 1.4978 
 
349 
 
13.D.34. New problem in 3rd edition. N = 500 rpm = 8.335 rps 
 i tan kd 0.20 d 0.2 0.8279 0.16558 m 
 
3 3
M w M wUse water values for 998kg m and 0.95 10 kg m s 
 
22
i L
L,estimate 3
L
0.16558 8.335 998d N
Re 240, 064
0.95 10
 
 Curve b in Figure 13-32 again predicts a constant 
0p
N 40 
 Then from Equation (13-52), 
0
2 5
P M i cP N d g where cg 1.0 and N 8.335. 
 2 5M MP 4.0 8.335 .16558 1.0 0.034587 (A) 
 M will be fairly close to c W 998 since W TolueneQ 5Q (see Equation (13-53)). 
The series of messy terms for Equation (13-56a) can be calculated. Since the tank dimensions and 
physical properties are the same as in Example 13-5, the only term on the RHS of Equation (13-56a) that 
is different is P. Thus the result in the same as Equation B in Example 13-5, 0.3d 0.0576 P (B) 
 In addition to Equations A and B, we need to solve Equation (13-53) 
 M d d d c d d1 865 998 1 (C) 
 Solving equations A, B and C with Goal Seek we obtain d d d,feed0.146 and 0.874. 
 Then solving Equation C, M d d d c1 0.146 865 0.854 998 978.6 
 Equation (B) 0.3 dP 0.05076 2.876 P 33.84 W. 
13.D35 (was 14.D11. in 2nd edition) From Eq. (12-46), 
1 1 2 2
1 1 1 0 0
1 2
E K E K
B 1 , C , D R x
R R
 
 (Eq. 6-6) For 1 < j < N j j j 1 j 1j j j j j j
j j 1
E K E K
A 1, B 1 , C , D F z 0
R R
 
 (Eq. 12-48) For Stage N N NN N N N N N 1 N 1
N
E K
A 1, B 1 , D F z E y
R
 
Example 13-4:
00 A D,0
R 1000, x 0.35, x 0.65, N 6 , N 1 A,N 1 D,N+1E 1475, y 0, y 0 
 For Acetic Acid, 
j jA Aj A
K y x : Use Fig. 14-4 to estimate A, jK . 
 
1 2 3
4 5 6
A A A
A A A
1 A1
1
1
0.03 0.5 0.09
K 0.3, K 0.33, K 0.43
0.1 0.15 0.21
0.12 0.14 0.16
K 0.5, K 0.5, K 0.5, 
0.24 0.28 0.32
For first guess assume constant E 1475 and R 1000.
1475 0.3E K
Then B 1 1 1.4425
R 1000
C 2 21
2
1 D A ,0
E K 1475
0.33 0.48675
R 1000
D R x 1000 0.35 350
 
350 
 
 and so forth with 6 ND D 475 0 0 . Thus matrix for acetic acid is, 
 
13.D.36. Part a. New problem in 3rd edition. See figure 
 org Aq,0 Aq,N* *
Aq org ,1 org ,N 1Min
F C C 0.10 0.008
0.736
F C C 0.133 0.008
 
 org,Min Aq
L
F 0.736 F 0.736 200 L h 147.2
h
 
 b. orgF 1.4 147.2 206.08 , 
org
Aq
F 206.08
1.0304
F 200
 
 Operating line goes through Aq,N org,N 1C 0.008 and C 0.008 with slope 1.0304. 
 org,1C 0.097 
 See Figure. 3 stages more than enough. 
3
~ 2
4
 stages needed. 
 
1.4425 -0.48675 0 0 0 0 
1475
1 .33
1000
 
1475
.43
1000
 - 1 
0 
0 0 0 
- 1 
1475
1 .43
1000
 
1475
0.5
1000
 
0 0 
0 - 1 0 
3 
1 
2 
4 
5 
6 
0 0 0 - 1 
1475
1 0.5
1000
 
1475
1 0.5
1000
 
1475
0.5
1000
 
0 
1475
0.5
1000
 
0 0 0 0 - 1 
1 2 3 4 5 6 
 14751 0.5
1000
 
351 
 
Part c. 3 4MW Zr NO 91.22 4 14.0067 3 15.994 
 MW water 2 1.00797 15.994 18.00994 
352 
 
 Basis 1 liter 0.10 mol 3 4Zr NO 
 Have 3 433.917g Zr NO 
 and 1000 g 33.917 966.083 g water 
 which is 966.083 18.00994 53.64 mol water 
 Mole frac. 3 4
.1
Zr NO 0.00186
53.64 .1
 
 Mass frac. 3 4
33.9179
Zr NO 0.033917
1000 g
 
 System is dilute if consider mole fraction, less so if use mass fractions. If densities are constant, 
then constant flow rates is valid. Even with variable density, solving problem with mole fractions and 
constant molar flow rates would be accurate. This would require converting equilibrium data to mole 
fractions. Use of fractions with concentrations in mol/L is NOT correct. 
 
13.D.37. New problem in 3rd edition. 
 
 Part a. HETPlarge-scale = HETPpilot (Dlarge/Dpilot)0.38 = (0.24 m) (1.1 m/.05 m).038 = 0.78 m 
 
 flarge-scale = fpilot (Dpilot/Dlarge)0.14 = (1.4 s-1)(.05 m/1.1 m)0.14 = 0.91 s-1 
 
Part b. HETPlarge-scale = HETPpilot (Dlarge/Dpilot)0 = HETPpilot = 0.24 m 
 
 flarge-scale = fpilot (Dpilot/Dlarge)0 = fpilot = 1.4 s-1 
 
c. Use of the more conservative design developed for difficult systems (n1 = 0.38, n2 = 0.14) results in a 
much higher HETP and thus a much taller column and more expensive column than use of the design 
procedure for easy systems (n1 = 0, n2 = 0). Considerably more data is needed for a large variety of 
systems to determine best design practice. If a variable speed motor is used in the large-scale system the 
difference in predicted optimum frequency is not as serious because the system can tuned to find the 
optimum frequency. 
 
13.D.38. New problem in 3rd edition. 
 waterMW 18.02, tolueneMW 92.14 , extract toluene
raffinate water
C C
m 20.8
C C
 
 F 1.0kmol hr , S 0.06 kmol hr. INx 0.00023 , INy 0 
 IN out outF x Fx Sy and out outy m x → out inx Fx / F Sm 
 Note m m. m is equilibrium in mole fraction units. Assume extract has properties 
toluene and raffinate properties of water. 
 
33
3 3
1 92.14 kg toluenekmole benzoic
865 kg tol m kmol toluenem extractm 20.8 122.71
kmole benzoic 1 18.02 kgW
m raffinate 998 kgW m kmol W
 
 Units on m are kmol benzoic kmol extract
kmol benzoic kmol raffinate
 
353 
 
 out
1.0 0.00023
x 0.0000275
1.0 0.06 122.71
, outy 122.71 0.0000275 0.00337 
 If use m 20.8 find outx 1 .00023 1 .06 20.8 0.000102, RONG!W 
13.D.39. New problem in 3rd edition. 
 Feed is 0.1 4CC , 0.9 AA. F 10 kmol h . Solvent pure. S 10 kmol h. 
 1 equil. stage 
 Lever arm: 
S 10 FM
1
F 10 SM
, Alternatively 4 4
4 4
F,CC M ,CC
M ,CC S,CC
x x S
1
x x F
 
 Then 
4M,CC
x 0.05 Find Mixing Point M. 
 [The figure is shown at the end of problem 13D39 as the single stage mixing line.] 
 Phases split along the line –TE to find the line through M 
 Rafinate: 
4CC AA
x 0.041, x 0.54 . Extract: 
4CC AA
y 0.095, y 0.07 
 Overall Balance: E+R+=F+S+=20 
 CCℓ4 Balance: .095E+0.041R = (0.0) S+0.1 for F=1.0 
 Solve simultaneously, R 16.6667, E 20 R 3.3333 
 NOTE: Since CCℓ4 mole fracs can be read more accurately, the CCℓ4 balance is 
probably more accurate than the acetic acid balance equations. 
 
 
13.D.40. 
 
CCℓ4 
 
 
 
 
 
 
S1 S
2 
1 2 
E1 E2 
R2 R1 = R2 single stage = 16.6667 
Mix with S2 = 10 (pure) 
 
 
 2 ,AA2 acetic
1 2 2 ,AA
M 02 M ,acetic S1
1 2 R ,acetic M acetic M
xx xR16.6667 SM
10 S R M x x 0.54 x
 
 
2,AAM
x .54 1.6667 2.6667 .3375 
 Find M2 and by trial and error find a tie line though M2. See figure on next page. 
 Extract 2, 
4CC AA
y 0.046 y 0.065 
 Raffinate 2, 
4CC AA
x 0.018 x 0.57 
 2 2 1R E R S 16.6667 10 26.6667 
 4CC balance 2 20.018R 0.0046E 0.041 16.6667 0.0 10 
 Substitution 2 20.018 R 0.046 26.6667 R 0.68333 
 2 2R 19.40 and E 7.16 kmol h . 
 
354 
 
 
355 
 
13.D41. New problem in 3rd edition. 
N 1R F 10, CC 4,N 1x 0.1, AA,N 1x 0.9 
 0E S 14.5, TEA,0y 1.0 , CC 4,Ny 0.091 
 
Mixing. Use lever arm rule. 
 4 4
4 4
N 1,CC M ,CC
M ,CC S,CC
x x14.5 S FM
1.45
10 F x ySM
 
 
4 4
4
N 1,CC S,CC
M,CC
S
x y
.1 1.45 0Fx 0.041
S F 1 2.45
 
 
 Find M. Draw N 1E MR line. See figure on next page. 
 Raffinate: 
41,CC 1,AA
x 0.008 x .58 
 Passing Streams N N 1 O 1E R & E R intersect at . 
 Very close to parallel. Use parallel lines to step off stages. 
 Estimate # Stages = 3. 
 Flow rates 3 124.5 F S E R 
 4 3 1CC balance. F .1 S 0 1.0 E .091 R 0.008 
 3 1
1.0 24.5 0.008
E 9.69 kmol h , R 24.5 9.69 14.81
.091 .008
 
 Can compare to 13.G.2 Part c. 
 Extract 10.066 and Raffinate 14.433 
 Extract Mole fraction 
4TEA CC AA
y 0.841 x 0.0913 y 0.067 
 Raffinate Mole fraction 
4TEA CC AA
x .418 x 0.0056 x .577 
 Two results are reasonably close. 
356 
 
 
357 
 
 
 
13.D42. a. First, plot points EN and R1 on the saturated extract and saturated raffinate curves, 
respectively. 
Second, Find point Δ at the intersection of lines FENΔ and R1SΔ. 
Third, step off equilibrium stages. Need about 3. See graph. 
Part b. Easiest: use the lever-arm rule. Find mixing point M at the intersection of lines FS and ENR1.Then 
0.81 1235 /
S FM
F kg h
F SM
 
 Can also write 3 mass balances (overall, pyridine, and water) and solve for the unknown flow 
rates F, EN and R1. Unfortunately, this will not be very accurate because it is difficult to read the water 
values accurately. 
 
 
13D.43 (was 14.D5. in 2nd ed.) Plot points for 0 1F, S E , and R (on saturated raffinate line) 
 Use lever-arm rule to find point M. 0
E M F 1000
S 1371FM
 
 Line 1R M intersects the saturated extract curve at N acetoneE . x 0.067 . 
 Lines N 1 0FE and R E intersect at ∆ (a second piece of paper was attached to find ∆ 
accurately). Step off stages. 3 is more than sufficient. Need about 2 & 2/3 stages. This 
is close to the 2 + ½ estimated in problem 13.D19 with a McCabe-Thiele analysis. 
 
358 
 
 
359 
 
 
13.E1. 
m oD D
K 0.05, K 0.15, R 20, E 200, F 1 
 Since 
oD
E
K 1
R
ortho goes up column and since 
MD
E
K 1
R
meta goes down. 
 m,N 1 o,N 1 m,0 o,0y y 0, x x 0 
 Estimate: E E .52F and R R .48F 
 
R 20 R 20.48
E 200.52 and R 20.48, 0.09974 and 0.1024
E 200.52 E 200
 
 Recoveries: ortho,1 ortho,1.92 .52 1 E y or y 0.002386 
 ortho,Nx 0.00203 
 meta,N meta,N.94 .48 1 Rx or x .02179 
 Plot equilibrium curves and operating lines (see Figure) 
 Feed cannot be 3rd stage since cannot get 
Nm
x desired. Cannot be 5 as will be past 
intersection of R E and meta op lines. 
 Thus feed must be 4th stage. Do not get match of total number of stages. 
 Need 8 1/3 for ortho and ~ 5 2/3 for meta. 
 A very slight adjustment of recovery meta will change this. (Meta is approaching a pinch 
point at feed stage). 93% recovery was not enough. Therefore, need ~ 93.5% recovery 
with ~ 8 stages. 
 
 
360 
 
13.E.2. New problem in 3rd edition. Part a. 96% recovery. 4% p-xy left in diluent , 
 
 N,p xyx 0.04 .004 0.00016 wt. frac. 
 Part b. Paraxylene: a
y
K 0.080 m, E 20,000, R 1000
x
 * N 1N
y
x 0.
m
 Thus 
Eq. (12-28) converted to extraction notation is convenient. L R V E 
 
 
*
0 N
*
N N
x xR R
n 1
mE x x mE
N
mE
n
R
 
 0 N
R 1000
x 0.004, x 0.00016, 0.625
mE 0.080 20,000
 
 
0.004
n .375 .625
0.00016 2.3025
N 4.899
n 1 .625 0.470
 
 Part c. ortho-xy *0 Nm 0.150, x 0.006, x 0 , 
0.150 20000mE
3
R 1000
 
 Eq. (12-31) Converted: 
*
N N
N 1*
0 N
x x 1 mE R
x x mE
1
R
 
 N 5.899
1 3
x 0.006 1.842 E - 5
1 3
 
 Part d. Alternative Solutions are presented below for meta-xylene. 
 m-xy *0 Nm 0.050, x 0.005, x 0, N 4.899 E 20,000, R 1000, b 0 
 
0.05 20,000
mE R 1
1000
 
 Must use special form. But the L mV 1 form in terms of x is not available. Thus, need to 
derive, or translate or find in another source. Looking at development of Eq. (12-12). 
 0 NN x x x 
Solving for N, 
0 Nx xN
x 
Where Δx is determined in same way Δy was determined for Eq. (12-12), 
 
1 0
1
j j 1 0
L
y x b yVx x x const x
L V L V 
Alternatively, 
N 1 N
N 1
N
L
y x b yVx x
L V L V
 
361 
 
Translating to this extraction problem, N 1 NL V R E , y 0, x x 
 0 N 0 N
N
x x x x
N
x x 
And solving for xN, 0N
x 0.005
x 0.0008476
N 1 5.899 
 
Alternative Solution: Redefine terms to match Eq. 12-12 [Relating y to solvent and x to raffinate is 
arbitrary. Switch these definitions.] 
 N 1y meta xylene in hexane 0.005 
 1 0y mxy out is unknown x is now inlet solvent 0 
 
d
1 1
m 20, b 0
K 0.05
, 
L 20, 000
L E 20, 000; 1
mV 20 1000
, V 1000 
 N 1 1 1 0 1
L
y y N y x b 4.899 y 20 0 0
V 
 
362 
 
Solve for y1,
 
 N 11
y 0.005
y 0.0008476
5.899 5.899
 
 
 This is actually Nx in normal notation. 
 
Part e. Shown for normal notation. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0.00032 0
Slope Operating line 0.08333
0.004 0.00016
 
 
MIN
R
Slope 0.08333
E
, MIN
R 1000
E 12,000 kg h
0.08333 0.08333 
 
13.E.3. New problem in 3rd edition. Part a. Plot the equilibrium data and points F and S. Straight line 
from power F to point S passes through mixing point M. Since amounts of F and S are equal, 
M is at the half-way point of the line. Find tie line through M by trial-and-error. This is 
difficult since tie line is very sensitive. 
 
 Approximately, raffinate AR DRx 0.326 x 0.575 
 and extract AE DEy 0.046 y 0.058 
 
Mass Balances: E R S E 40 R 40 E 
 AE A,S A,s AFEy Rx Sy Fx 20 0 20 .4 8 
Solve simultaneously, E 18.0 kg , R 22.0 kg. 
 
 Part b.First add solvent until reach saturated raffinate curve at intersection with FS line. 
 Initial Raffinate AR Dx 0.36, x 0.54 
 INIT AR AF INIT AS INITR x Fx S x 20 .4 S 0 8 
 init ARR 8 x 8 0.36 22.22 kg 
 INIT INITS R F 2.22 kg 
 Second, use Eq. (13-27) for the continuous solvent addition batch extraction. 
pxy equil slope = 0.080 
EQy 0.080 .004 0.00032 
0,pxyx
0.004
 x 
N,pxyx
0.00016
 
N 1y 0 
y 
363 
 
 
t ,final ,A
t ,feed ,A
x
t ,A
Ax
dxS
R y
 
 t ,feed,Ax is the raffinate after solvent addition to form two phases 
 t ,feed,A A,initial raffinatex x 0.36 , t ,final,Ax 0.292 
 From equilibrium find values Ay (extract), Approximate values are: 
A,tx Ay 1 y 
0.36 0.048 20.8 
0.326 0.046 21.7 
0.292 0.045 22.22 
 
 
0.292
A
A0.36
0.36 0.292dx
20.8 4 21.7 22.22 1.47
y 6
 
 added tS 1.47R Eq. A 
 In the derivation tR is assumed constant, t t ,INITR R 22.22 kg 
 addedS 32.66 kg 
 With this approximation addedE S . 32.66 kg 
 Solute mass balance 
 t A,INIT added A,added t A,final A,AvgR x S y R x Ey 
 A,added A,INIT A,finaly 0, x 0.36, x 0.292 
 A,Avg
22.22 0.36 0.292
y 0.046
32.66
 
 
 If we do not assume R is constant, then Eq. (13-27) is 
 
 
t ,Aadded
t ,INITIAL ,raf
xS
t ,A
added
A0 x
d R x
dS
y
 
 We would need to do a numerical integration with a calculation of t ,AR x versus Ay . this 
can be done, but is challenging. 
 
13.G.1. New problem in 3rd edition. 
 Extract 1: TRA carbontet Acetic acidflow 3.90769, x 0.84986, x 0.085102, x 0.065042 
 Raffinate 1: TEA carbontet Acetic Acidflow 16.03923, x 0.41361, x 0.041332, x 0.54506 
 Extract 1: TEA carbontet Acetic Acidflow 11.63396, x 0.91426, x 0.036586, x 0.049149 
 Raffinate 2: TEA carbontet Acetic Acidflow 14.40527, x 0.41633, x =0.016472, x =0.56719 . 
 Entering carbon tet 0.10 10 1.0 kmoles hr 
 Leaving in raffinate 0.016472 14.40527 0.23728 
 Extracted In Out in Raffinate 0.7627 
 % extracted = 76.27% 
364 
 
 
13.G.2. New problem in 3rd edition. 
 Part a, 3 stage cross-flow. All flow rates are kmol/h 
 Total Flow rate TEA flow CCl4 flow Acetic flow 
Extract 1 3.961 3.366 .3371 .2576 
Extract 2 11.634 10.637 .4256 .5718 
Extract 3 11.052 10.419 .1554 .4777 
Raffinate 3 13.353 5.580 .0819 7.693 
Carbon tet remaining in raffinate 3 is 0.0819 kmol/h. Since carbon tet feed was 1.0 kmol/h, 0.9181 
kmol/h was extracted. Fraction extracted = 0.9181/1.0 = 0.9181. 
 
Part b. 3 stage counter-current with S = 10 kmol/h. 
Extract 1 4.9142 3.723 .7242 .4672 
Raffinate 3 15.086 6.277 .2758 8.533 
 
Carbon tet remaining in raffinate 3 is 0.2758 kmol/h. Since carbon tet feed was 1.0 kmol/h, 0.7242 
kmol/h was extracted. Fraction extracted = 0.7242/1.0 = 0.7242. 
 
Part c. 3 stage counter-current with S set to give same fraction extracted as in part a (0.9181) and outlet 
raffinate carbon tet flow rate of 0.0819 kmol/h. This is trial-and-error. 
First trial: S = 20 and CCl4 raf 3 flow rate = 0.0289 
Second trial: S = 18 and CCl4 raf 3 flow rate = 0.04045 
Third trial: S = 16 and CCl4 raf 3 flow rate = 0.0590 
Fourth trial: S = 14 and CCl4 raf 3 flow rate = 0.0908 
Fifth trial: S = 14.5 and CCl4 raf 3 flow rate = 0.08104 
This is close enough. 
Final Results: 
Extract 1 10.066 8.469 0.9189 .6786 
Raffinate 3 14.433 6.031 0.0810 8.321 
 
363 
SPE 3rd Edition Solution Manual Chapter 14 
New Problems and new solutions are listed as new immediately after the solution number. These new 
problems are:14.A3, 14.A4, 14.C5, 14.D6, 14.D9, 14.D11, 14.D15-14D17, 14.E2, 14.E3. Chapters 13 
and 14 from the 2nd edition were rearranged to place all the extraction material into chapter 13 and the 
material for other separations in Chapter 14. Thus, the numbers of many problems have changed. 
14.C.5. New problem in 3rd edition. Part a. y y, x x, m 1, F U, S O, R U, E O
where F, U, S, O, R, E are kg 
Eq. (13-27b) becomes 
F IN
U U
y x x y
O O
and (13-21) IN IN
U
x x y 1 U O
O
 and y = x 
Part b. Eq. (13-29b) becomes 
t ,final t ,feed
O 1
n x x
U K
Where K y x at equilibrium = 1.0 in washing. 
14.D1. (was 13D29 in 2nd ed.) a) Translate eq. (12-28),
*
0 N
*
N N
x xU U
n 1
mO x x mO
N
n mO U
Note: x in wt frac. translates to x in 3kg m if densities are constant. Densities cancel. For 
washing equilibrium is equal overflow & underflow concentrations. Thus, m = 1, b = 0 
* N 1
2 4 N N 1 0 N
y b
H SO x y 0, x 1.0, x 0.09
m
U 40 mO 1
0.8 and 1.25
mO 1.0 50 U 0.8
 
1.0 0
n 1 0.8 0.8
0.09 0
N 4.96
n 1 0.8
b) HCℓ Use Eq. (12-31) or (14-8)
*
N N
N 1*
0 N
1 mO Ux x
x x 1 mO U
N 0 N 1 5.96
1 mO U 1 1.25
x x 0.75 0.0674
1 1.251 mO U
 3kg m
HC
HC
N 1*
N
y b1.0 50mO
N 4.96, 1.25, x 0
U 40 m
Alternative: 
364 
 
 Note: 
2 4
N N
0 0HC H SO
x x
=0.09 
x x
 
 Thus, if one is clever and realizes change will be same for HCℓ & 2 4H SO
mO
since & N are identical
U
, don’t need to use Kremser eqn for part b. 
 
14.D2. (was 13.D22 in 2nd ed.) a. 1000 cc sand = 400 cc underflow liquid. This is about 400 g = 0.4 kg 
liquid. Equil: y = x. Use nomenclature of Table 13-4. 
 Operating Eq. j j 1 in out
U U
y x y x
O O
 
 Slope 
U .4
0.8.
O .5
 Goes through point (y = 0, x = 0.002) 
 Overall bal. in in out outO y U x U x O y 
 outO .4 0.035 .4 .002 .5 y → out
.0140 .0008
y 0.0264
.5
 
 Need 6 2/3 Stages – See Graph (Can also use Kremser eq.) 
 
 b. Mass Balance: 
inj 1 j j j j j
U x O y U x O y 
 Op. Eq.: 
inj j j j 1
j j
U U
y x y x
O O
 
 
U
U 0.4, O 0.2, 2 slope
O
, outx 0.002 (see graph) 
 Obtain approximately same separation, but use much more wash water. 
14.D3. (was 13D23 in 2nd ed.) 
365 
 
 
 
 Basis: 1 kg 3CaCO solids 
 Feed Mole frac. can be arbitrary. Pick 0x 0.01as basis 
 M.B. iN iN out outU x O y U x O y 
 out out iN iN
U U
y x x y
O O
 
 out outy , x at Equil (y = x) line 
 Point in in inx , y x ,0 is on op line 
 Slope Op line 
U 3
O 2
 
 See graph. 
 Find 4
0
x 0.00127
0.127
x 0.01
 
 Recovery 4
0
x
1 1 0.127 0.873
x
 
 Recovery is significantly better with counter-current process. 
4 3 2 1 
y4
 
U = 3 
U = 3 
y2
 
x4
 
y1
 y3
 
O 24
 
O 22
 
O 23
 O 21
 
4
y 0in
 
3
y 0in
 
2
y 0in
 
1
y 0in 
366 
 
 
 
14.D4. (was 13D24 in 2nd ed.) 
 0.8 0.8 0.2 0.8, 4, O 4000
1
kg/h 
367 
 
 
1F
kg liquid
1.0
kg L liquid kg liquidL liquidU 1000 dry solids 4 1600
kg solidh L solid h2.5
L solid
 
 
2F
kg .8 1.0 kg liquid
U 2000 3200
h .2 2.5 h
 
 
1 2T F F
kg liquid
U U U 4800
h
 
 In section 2: T Tj j 1 0 1
U U
y x y x
O O
 
 Slope TU O 4800 4000 1.2 Goes through point 0 1y , x 0, 0.006 
 Intermediate feed at Fx x 0.02 
 In Section 1: 1 1F Fj j 1 N N 1
U U
y x y x
O O
 
 Slope 
1F
U O 1600 4000 0.4 . Goes through point N N+1y , x 
 Also intersects Section 2 op. line at feed line. (Or calculate Ny from mass balance). Equilibrium 
is y = x. Step off stages (see Figure). Need 5.4 equilibrium stages. Opt. Feed is 4th. 
 
 
 
14.D5. (was 13D25 in 2nd ed.) 0.8, 1 0.2 
 
 1 1
kg dry 1 1 2000 L
F : 1000 F 5 wt %
kg dry L solidh h2.25 0.2
L L under flow
 
 2 2
kg dry 1 1 L
F : 2000 4000 F 2 wt %
h 2.5 0.2 h
 
368 
 
 Liquid Volumes: 
2000L L liq
total 0.8 liq 1600 5 wt %
h h
 
 
3200 L liq
4000 0.8 liq 2 wt %
h
 
 T
underflow
F : Total liqd h 4800 
 0 f f
kg liquid
U 4000 4800 where 1.0 kg L
h
 
 aN 1 N 1 0 0
kg N 0HU kg liq
O 4000 kg h , y 0, 1.2, U x 1600 0.05 3200 0.02
O h kg liq
 
 0x 144 4800 0.030 , Nx specified 0.006 
 Ext. MB, N 1O y 0 0 N N 1 1N 1 U x U x O y 
 0 01 N 0 N
U x U U 4800
y x x x 0.030 0.006 0.0288
O O O 4000
 
 Convert to Kremser
 *1 0 0O V, U L, m 1, y mx x 0.030, mV L 4000 4800 0.8333 
 Eq. (12-30) 
0 0.030n 1 0.8333 0.83333
0.0288 0.030
N 8.83 or 9 stages
n 1.2
 
 Use 2 feeds! 
 
14.D.6. New problem in 3rd Edition. 2.5 kg wet is 1 kg dry solids-insoluble, and 1.5 kg underflow liquid. 
 Part a. 
1 kg dry solids
10 kg total 4 kg dry insoluble solids
2.5 kg total
 
 
1.5 kg liquid
U 4 kg dry solids 6 kg liquid.
kg dry solids
, Ov 10 kg liquid. 
 Before 1st mixing: 0.05 frac BaS 6 kg liquid 0.3 kg BaS 
 1st Mix: 
0.3 kg BaS
0.01875 mass frac in U & Ov.
16 kg liquid total 
 
 Settle – (6 kg liquid in U) 0.01875 0.1125 kg BaS 
 2nd Mix Pure Water 
0.1125 kg BaS
0.00703 mass frac in U & Ov.
16 kg liquid
 
 Settle – (6 kg liquid in U) 0.00703 0.0421875 kg BaS 
 3rd Mix Pure Water 
0.0421875kg
0.00264 mass frac BaS in U and Ov.
16 kg liquid
 
 Part b. Result is same. Can also be done graphically. 
Part c. Countercurrent. Easiest solution approach is to use Kremser equation. 
 
*
N N
N 1*
0 N
x x 1 m Ov U
x x 1 m Ov U
 
369 
 
 *0 N N 1N 3, x 0.05, m 1, Ov 30, U 6, x y m 0 
 N 0 4
1 30 6
x x 0.05 0.00641 0.0003205
1 30 6
 
 External M.B. 0 N 1 N 1x U y Ov x U y Ov 
 1 0 Ny U x x Ov 6 0.05 0.0003205 30 0.009936 
 
14.D7. (was 13D27 in 2nd ed.) Equilibrium: j jx y wt. fractions 
 Operating Eq.: j j j,in j 1
U U
y x y x
O O
 
 Basis 1000 cc wet sand. 
 3
vol water 1.0 g kg
U .4 1000 cm wet sand 0.4 kg
vol sand wet cc 1000 g
 
 O = 0.2 kg. Thus, each operating line has slope 
.4
2.
.2
 
 Each op line goes through pt. j,in j 1y , x 
 0 N,in N 1,in N 2,inx 0.035, y y y 0 
 N S,in N 2,in N 1 1,in N 2x 0.002, y y , y y , y y 
 Start at stage N where Nx = 0.002. Find Ny then work backwards to stage N-2. This 
gives inlets for first 3 stages so can then work forward (see Figure). Note: that stages 5 and 
N-2 are not connected. 8 stages gives more than enough separation, but 7 is not enough. 
 
 
 
370 
 
14.D8. (was 13D28 in 2nd ed.) Use Kremser equation 
 solv solidF F .95, y mx is equilibrium with m = 1.18, and N = 11. 
 Recovery is N 01 x x . 
 Eq. (12-31) becomes *N N 1x y m 0 . 
 
solv
*
N N N solid
N 1*
0 N 0 solv
solid
m F
1
x x x F
x x x m F
1
F
 
 solv N 12
solid 0
m F x 1 1.121
1.18 .95 1.121. Then 0.041
F x 1 1.121
 
 Thus Recovery = 0.959 
 
14.D.9. New problem in 3rd Edition. Assume Solid solventF and F are constant despite removal of sugar from solid. 
 Solid F solv,IN EF 1.0, x 0.055, y 0, m 1.18, 
Eq. (13-21) becomes Solid F solv ,IN E Solid solv
solv
F
x x y m F F
F
 
 a. solv Solid solvF 3.0, F F 1 3 
 
0.055 1
x 1.18 0.01211
3 3
, Ey m x 1.18 0.01211 0.0143 wt frac. 
 b. x = 0.004. Solve for solvF . 
 E solv,INSolid
solv F
xm yF
F x x
 
 Fsolv Solid
E Solv,IN
x x 0.055 0.004
F F 1.0 10.805 kg
m x y 1.18 .004 0
 
 
 
14.D10. (was 13D30 in 2nd ed.) 2 2G H stream 100 lb h of H CONSTANT 
 L Slurry stream 120 lb h CONSTANT 
 Operating Line. Must work in weight ratios. in out
L L
Y X X Y
G G
 
 inin
in
xx y
x , Y , X 0
1 x 1 y 1 x
 
 44 outinin out
2 2in out
lb CHlb CH 30 .05
Y 0.30, Y .0527
lb H 100 lb H .95
 
 in out outG Y Y L X 
 out
100
X 0.30 0.0527 0.206
120
, out
X .206
x 0.171
1 X 1.206
 
371 
 
 Operating line becomes, 
 out
L L 120
Y X Y where 1.20
G G 100
 and goes through in outX , Y . 
 Equilibrium Curve: y = 1.2 x becomes 
Y X
1.2
Y 1 X 1
→ 
1.2 X
Y 
1 .2 X
 
x X Y
0 0 0
.05 .0526 .0038
.10 .1111 .1364
.15 .1765 .2195
.20 .2500 .3158
.25 .3333 .4286
Plot Y vs X
 
 
 
See Figure for Plot. Need 5 1/8 stages. 
14.D.11. New problem in 3rd Edition. 10,000 kg h wet solids, frac. vol. liquid , 
1 frac vol dry solids. 
 Basis 31 m wet solids : 
 Weight liquid + weight solids 3 31.0 1000 kg m 1.0 1 1500 kg m 
 0.4 
 3400 kg 900 kg 1300 kg total m wet solids. 
 Thus 
400
1300
of weight is underflow liquid, U 
400 kg
10,000 3076.9
1300 h
. 
 
14.D12. (was 13D32 in 2nd ed.) solv solidF F 1.36 
372 
 
 Op. Eq.: solid solid1 0
solv solv
F F
y x y x
F F
 
 Where y and x are 3kg m . Ey m x is equilibrium. 
 *F 0 N F 0 N 1 N solv solidx x , x 1 .975 x .025 x , y 0, x 0, N 5, F F 1.36 
 Can use any of Kremser equations such as Eq. (12-31). 
 
solv
*
N N solid
N 1 6*
0 N solv
solid
m F
1
x x F 1 1.36 m
0.025
x x 1 1.36 mm F
1-
F
 
 Which becomes: 60.1582 m 1.36 m 0.975 0 Find m = 1.313 
 Check: 6
1 1.313 1.36
.025 0.025005
1 1.313 1.36
 which is OK. 
14.D13. (was 13D33 in 2nd ed.) 
a) Use Eq. (13-21), 
0 in
E in
ˆR̂ S x y
x , equil. y m x, y 0ˆˆm R S
 
E
g L in liqdˆR̂ S 10 12.5 0.8, m 1.18
g L in solid
, F F F
0.8 x 0 0.8
x x 0.4040 x
1.18 0.8 1.98
 
 Frac. Rec. 1 0.404040 0.5959596 
 b) Eq. (13-29b) t ,final
E t ,feed
ˆ xS 1
nˆ m xR
 
 t ,final F Fx x exp 1.25 1.18 0.228779 x , Frac Rec = 1 – 0.228779 = 0.7712 
 
14.D14. (was 13D34 in 2nd ed.) 4 2coalBaSO BaS 2 CO 
 Equil: Soln conc in underflow = soln conc in overflow. Thus really washing 
 Equil : y x, m 1, b 0 
in out 
kg kg solution
U 350 in sol. 1.5 525 kg soln., x 0.20, x 0.00001
h kg insoluble solid
 
0 0 N* *
in N 1 0
U x xkg 525
O 2075 , y 0.0, x 0, y 0.2 0.00001 0.0506, x 0.0506
h O 2075
 
 Eq. (12-29) 
* *
N N 0 0n x x x x n 0.00001 .2 .0506
N 6.99 or 7.0
n L m V n 525 1.0 2075
 
14.D.15. New problem in 3rd Edition. With 1000 kg/h dry solids U 1.5 1000 1500 kg h 
a) Can use Kremser eq. with large N to find MinOv or a sketch 
373 
 
 
 
 
 
 
 
 
 
 
 
 
 
Equilibrium is y x 
Min
U
0
 
N 0y 0 
0x .15
 
Nx 1 .99 .15 0.0015 
Min
1500
Ov 1485
10101
 
*
1
Min 0 N
y 0U 0.15
1.0101
Ov x x 0.15 0.0015
 y 
 b. MinOv 1.2 Ov 1782 
 
U 1500
0.84175
Ov 1782
 
 Kremser: 
y y V Ov
x x L U
 
 Eq. (12-28) *N N 0x 0.0015, x 0, x .15, m 1, U Ov 0.84175 
 
*
0 N
*
N N
x xU U
n 1
m Ov x x m Ov
N
m
n
U Ov
 
 
0.15 0
n 1 .84175 .84175
.0015 0 2.81
N 16.33
1 .17227n
.84175
 
 In theory, can use McCabe-Thiele, but it is difficult to accurately step off this large number of 
stages. 
 c. 
U 1500
Ov 2000, .75 m 1
Ov 2000
 
 
n 1 .75 100 .75
N 11.29
1
n
.75
 
 eqact overall
sub actual
N 11.29
N 15 E 0.753
N 15
 
 For Em use N = 15 and change mE with same equation 
 E E
E
.75 .75
n 1 100
m m
N
m
n
.75
 
*
1y 
374 
 
 Vary mE until N = 15. Em .911 
 On a McCabe-Thiele diagram this is trial and error. Kremser is much easier. 
 
14.D.16. New problem in 3rd Edition. Part a. IN INU 2 kg, O 2 kg, x 0.06, y 0 
 Solution (translation of Eq. (13-21)) is 
 IN IN
U
x x y 1 U O 1 .06 0 2 .03
O
 
 Part b. Want x 0.005 O is unknown, IN Nx 0.06, y 0, U 2 Solve for O 
 IN IN
U U
x x x y
O O
, IN IN
IN IN
x y x xO
 O U
U x x x y
 
 
0.06 0.005
O 2 22 kg water
0.005 0
 
14.D.17. New problem in 3rd Edition. K = 1 Eq (13–28) becomes t ,final t ,feed
O
n x x
U
 
 Part a. t ,feedO 2, U 2, x 0.06 
 1t ,final t ,feed
O
x x exp 0.06 e 0.02207
U
 
 Part b. t ,feed t,finalU 2, x 0.06, x 0.005 
 t ,final t ,feed
0.005
O U n x x 2 n 4.97 kg
0.06
 
 Part c. t ,finalx x in Part a. 
 normal batchO O in Part b. 
 
14.D18. One equilibrium stage. 
N+1 S SA A D
F 1000, x .2, S 662, y y 0 
 0 A,0 A,N+1A,M
0
E y F x
x 0.12
E F
 (same as Example 14-2) 
 Plot M. By trial and error find tie line through M (Final result shown in Figure). 
 
1 1 1 1A D A D
y .238, y 0; x .078, x .656 
 Flow rates: Diluent balance: 
11 D D,N+1
R x F x 
 
11 D,N+1 D
R F x x 1219.5 
 1 1E M R 1662 1219.5 442.5 
375 
 
 
 
14.D19. This problem is essentially a repeat of Example 14-2, except using exactly 3 stages. 
Clearly, 
1A
x 0.04 since now have more stages. F, 0E and M are unchanged. Problem 
is trial-and-error. Guess location of 1R . Find NE and ∆. Step off 3 stages andsee if 
have correct location of NE . The third and final trial is shown in the figure. 
A1x 0.026 and A3y 0.38. 
 
 
14.D20. Although this is leaching, this cross-flow problem is very similar to cross-flow extraction. 
We can derive 
 j 1 j,in
Mj
j 1 A j,in A
A
j 1 j,in
R x E y
x
R E
 
 j 1 j,inM R E where 
Mj j
j j
j A A
j
A A
M x y
R
x y
 
 Stage 1: 
0 1,in0 1,in A a
R 1000, E 421, x .2 y 0 , 
M1A
x 200 1421 .1407 
 Find M on line 
M10 A
SR at x (see Figure). By trial-and-error find tie line through M. 
376 
 
 This gives 1 1E and R . Find 1 1A A 1y .35, x .113, M 1421 
 1
1421 .1407 .35
R 1254.9
.113 .35
 
 Stage 2: 
M 2A
1254.9 .113 0
x 0.085
1254.9 421
 
2 2A A 2
y .18, x .058, from tie line , M 1675.9 
 2
1675.9 0.085 .18
R 1305.0
.058 .18
 
 
 Stage 3: 
M 3A
1305 .058 0
x 0.044
1726
, 
3A,3 A 3
y .09, x .03, M 1726 
 
 3
1726 0.044 .09
R 1323.3
.03 .09
 kg/h 
 
 
 
 
14.D21. a. Basis 1 kg mix in underflow: NaC NaC
crystals
x values 0.8 1.0 0.2 y 
 
 Since crystals are pure NaCℓ, NaOH is in liquid only. Since 20% of the underflow is liquid, 
 NaOH NaOHx 0.2 y . Generate equilibrium table. 
 
377 
 
 
 Soln (y) 
NaOHx Mass frac NaOH NaCy NaCx 
0 0 .270 .854 
0.004 .02 .253 .8506 
0.008 .04 .236 .8472 
0.012 .06 .219 .8438 
0.016 .08 .203 .8406 
0.020 .10 .187 .8374 
0.024 .12 .171 .8342 
0.028 .14 .156 .8312 
0.032 .16 .141 .8282 
0.036 .18 .126 .8252 
 
Feed is 45 wt% NaCℓ crystals. x values: NaCℓ (soln) = 0.5193, NaOH (soln) = 0.099, water 1-0.5193-
0.099 = 0.3817. Since feed is 55% liquid, 
 F,NaOH NaOH
x 0.55 y 0.099
, NaOH
y 0.099 0.55 0.18
 
 From the equilibrium data NaC
y 0.126
 
 F = 100, S = 20, Plot F & S and find M. 
 
FM 20
,
100SM Tie line through M gives E & R. 
 
E RM
1.119
R EM (measured on figure) 
 E R 120 1.119 R R 120 
 R 56.63 kg/min, E 63.37 kg/min 
 NaC NaOH
R : Raffinate 0.833 x , 0.026 x
 
 NaC NaOH
E : Extract y 0.16, y 0.135
 
 
 The underflow is z wt frac crystals (Pure NaCℓ) + (1-z) wt frac solution 
 NaC
y 0.16 is soln in equil
 
 Thus, 
z 1.0 1 z 0.16 0.833
 
 
0.333 0.16
z 80.1% OK
0.84 
 was 80% solids in problem statement. 
 
 c. Same M. Plot 1
R
 draw line 1
R
M to N
E
. 
 
 2 stages more than sufficient 
 N 1 1 1
E R 120 1.137 R R 120
 
 
N 1
1 N
E R M 103.5
1.137
R E M 91.0 
 
378 
 
 1
R 56.14
kg/min, N
E 63.86
 kg/min 
 1 1,NaC 1,NaOH
R : x 0.845, x 0.01
 
 N NaC NaOH
E : y 0.152 y 0.147
 
 
 
379 
 
 
 
 
 
 
380 
 
14.E1a. This is difficult part – converting data 
 Basis 1 lb oil-free solids 
 
 
oily 
 
solventy 
 
z solids
1.0
x
1 z
 oiloil
y z
x
1 z
 
0 1.0 0.20 0.830 0 
0.1 0.9 .242 0.80515 0.01948 
0.2 0.8 .283 0.7794 0.044115 
0.3 0.7 .339 0.74683 0.07595 
0.4 0.6 0.405 0.71174 0.1153 
0.5 0.5 0.489 0.67159 0.16420 
0.6 0.4 0.600 0.625 0.2250 
 0.65 0.35 0.672 0.598086 0.26124 
 0.70 0.3 0.765 0.56657 0.303399 
 0.72 0.28 0.810 0.552486 0.3222 
 
 Note: solidsy 0 for all streams, Z = lb solution/lb oil free solids. 
 
 Plot data on triangular diagram. See Figure 14.E1a, b, c, d, e. 
 
 b&c. F + S = M1 = 1500 
 
 oil,F oil,S 1 oil,MF x S y M x , 1oil,M
1000
x 0.252 0.168
1500
 
 
 See Figure 14.E1a, b, c, d, e. 
 
 Check: Lever Arm 1
1
M S F 2
.
S 1M F
 Find tie line through M1. 
 
 Extract 1E , oil,1 oil,1 solids,1y 0.34; Raffinate: x 0.092 and x 0.730 
 
 Mass Balances: 1 11500 E R , 1 1252 0.34 E +0.922 R 
 1 1R 1040.3, E 459.7 lb 
 Finish step c) Stage 2: 1 2 2R S M 1540.3 , 1 2 oil,MR 0.092 M x 
 
2oil,M
x 0.062 . Plot 2M and find tie line through 2M . 
 Extract: oil oil,2 solids,22y 0.115; Raffinate: x 0.025 and x 0.80 . 
 MB: 2 21540.3 E R , 2 295.7 0.115 E 0.025 R 
 2 2R 904.8 lb, E 635.5 lb 
 d & e – Same answer as b & c but R & E are flowrates. 
f. See Figure 14.E1f. 3 stages is more than enough. Need ~ 
1
2
3
 equil stages. 
 
381 
 
 Lines N N 1 0 1E R and E R intersect at . 
 
 
 
 
 
 
 
 
 
 
382 
 
 
 
 
 
 
 
 
 
383 
 
 
14.E.2. New problem in 3rd Edition. Converting data is the difficult part, but is obviously identical to 
Problem 14.E.1. 
 Basis 1 kg oil-free solids 
 
oily 
 
solventy 
 
z 
solids
1.0
x
1 z
 oiloil
y z
x
1 z
 
0 1.0 0.20 0.830 0 
0.1 0.9 .242 0.80515 0.01948 
0.2 0.8 .283 0.7794 0.044115 
0.3 0.7 .339 0.74683 0.07595 
0.4 0.6 0.405 0.71174 0.1153 
0.5 0.5 0.489 0.67159 0.16420 
0.6 0.4 0.600 0.625 0.2250 
 0.65 0.35 0.672 0.598086 0.26124 
 0.70 0.3 0.765 0.56657 0.303399 
 0.72 0.28 0.810 0.552486 0.3222 
 
Approximate solution, use Eq. (13-29a) Oil balance: 
t ,final
c ,feed
x
t
t x
dxS
R y
 
 
 S = Mass Solvent, tR Mass raffinate (solids + solute) 
 x = Mass frac. solute (oil) in raffinate 
 y = Mass frac. solute (oil) in raffinate in extract (solvent) 
 
a) M is now at saturated raffinate curve. oil,M solids,Mx 0.21, x 0.63 
Mass balance F + S = M 
Solids .748F + (0) (S) = 0.63M 
 
 initial
0.748
M F 1187.3 kg R
0.63
 
 S 187.3 kg 
b) Now mixing is from S to a point on raffinate curve. 
From equilibrium curve in solution to 14.E.1. 
384 
 
 
oilx oily oil1 y 
0.21 .54 1.852 
0.1625 .498 2.0080 
0.115 .40 2.50 
0.0675 .28 3.57 
0.02 0.1 10.0 
 
 Insoluble Solids M.B. 
 Initial 0.748, F = 100, Final 0.81, 
finalt
R 
 .81 
final finalt t
R 748 R 923.5 kg 
 Raffinate is 0.81 solids, 0.02 oil and 0.17 solvent 
 Solvent remaining in raffinate is 0.17 923.5 157.0 kg 
 Needs to be recovered by evaporation. 
 Do Simpson’s rule in 2 parts. 
 
1
0.21 0.115
1.852 4 2.008 2.50 .1961
6
 
 
2
0.115 0.02 0.4241
2.50 4 3.57 10
6 0.6202
 
 added tS 0.6202 R , but what is tR ? 
 Eq. (13-29a) assumes tR Const. 
 Use average value of tR . 
 
initt ,avg t t ,final
1
R R R 1187.3 923.5 1055.4
2
 
 or added t,avgS 0.6202 R 0.6202 1055.4 654.6 
 total addedS Initial addition + S 187.3 654.6 841.9 
 Extract S total remain in raffinateamt S S 841.9 157.0 684.9 by solvent 
 Oil in extract F,0.1 final,oil t ,finalx F x R 0.252 1000 0.02 923.5 233.5 
 Total wt extract 684.9 233.5 918.4 
 oily 233.5 918.4 0.254 
 solventy 0.746 
 
14.E.3. New problem in 3rd Edition. Solid Matrix is insoluble. Solids = (.748) 1000 = 748 kg. tR not 
Constant, but Solid is. 
 t
Solids
Solids
R
x
 
 AA
Solids
x
ydS d R x Solids d 
x
 
 
385 
 
 
final ,A Solids
A ,raf ,init
Solids ,raf ,init
x x
A Solids
x
x
d x xS
Solids y
 
 Changes limits integration. 
 oil Solids oil Solidsx 0.21, x 0.63, x x 0.21 0.63 .3333 
 oil Solids oil Solidsx 0.115, x 0.705, x x 0.115 .705 0.163 
 oil Solids oil Solidsx 0.02, x 0.81, x x 0.02 0.81 0.0247 
 Numbers for use in Simpson’s rule are from Solution 14.E.2. 
 
 
1
.3333 0.163
1.852 4 2.008 2.50 0.3515
6
 
 
 
2
0.163 0.0247
2.50 4 3.57 10 0.6173
6
 Total 0.9688
 
 addedS Solids total integral 748 0.9688 724.6 kg 
 totalS initial added 187.3 724.6 911.9 kg 
 Extract Amount Solvent total raf ,final S S 911.9 157 754.9 kg 
 Oil in extract = 0.252 (1000) – 0.02 (923.5) = 233.5 
 Total weight extract 754.9 233.5 988.4 
 
 wt frac solvent = 0.764, wt frac oil = 0.236 
 
 
 
 
 
386 
Chapter 15 Solution Manual 
Since this is a new chapter, all problems are new. 
A. Discussion Problems. 
15.A1. The mole fraction water is constant but since the temperature within the vessel varies the total
molar density Cm varies and the water concentration = Cw = ywCm also varies. Thus, Eq. (15-
10a) incorrectly predicts molecular diffusion. Equation (15-10b) predicts no molecular 
diffusion because dyw/dz = 0. 
B. Generation of Alternatives. 
15.B1. Forexample, one could operate with both inflow and outflow at the bottom of the tube. If flow is
controlled with a constant head tank, the height of liquid in the tube will be very close to 
constant. 
C. Derivations. 
15.C4. Substitute in q = (μ Re)/(4ρ) into Eq. (15-35d) and obtain δ = [(3μ2Re)/(4ρ2g)]1/3.
15.C5. Start with Eq. 15-52a), set vB=0 and solve for yAvA. Then NA = Cm yAvA. Substitute in the
expression for yAvA and Eq. (15-52e) for JA. This gives the desired result. 
15.C6. This problem is included to show that one can derive the expressions in books. There is a lot of
algebra, but the derivation works. First, can expand the derivative, 
2 2
1 1 1
2
1 1
(1 2 )
[ ( ) ]
AB x x
x x B A B x
    
  
    
Then take the derivative and expand all terms. The denominator becomes 
3 3
1 2 1[ ( ) ] [ ]A A B x Bx Ax    and the numerator simplifies to 
2 2
22 .A B x Multiply by x1. Q.E.D. 
15.C7. With CMO and y as mole fraction, * (1 ) 0mol A A B B A A A Bv y v y v y v y v      . Since NA = -NB, 
CAvA = -CBvB and for an ideal gas Ci = yi Cm. The total molar concentration Cm is constant. 
 Then, vA = -(1-yA)vB/yA (Eq. A) 
In terms of mass fractions yA =(yA,mass/MWA)/[yA,mass/MWA + (1 – yA,mass)/MWB]. (Eq. B) 
Substitute Eq. B into Eq. A and simplify. 
,
,
(1 ) /
/
A mass B
A B
A mass A
y MW
v v
y MW

  (Eq. C) 
Then in mass terms * , ,mass A mass A B mass Bv y v y v  which after substituting in Eq. C and simplifying 
*
, ,(1 ) ( / )(1 )mass A mass A B A mass Bv y MW MW y v      . (Eq. D)
If MWA = MWB, *massv = 0. We can write 
vB = NBCB = NByBCm = NBCm(1 – yA) (Eq. E) 
where the y are mole fractions. Substituting Eq. B into Eq. E and then substituting this into Eq. D, 
we obtain 
,
, ,
*
, ,
1
(1 ) (1 )
/ (1 ) /
A mass A
m B A mass A mass
B B
mass
A mass A A mass B
y MW
C N y y
MW MW
v
y MW y MW
   
      
   

 
 (Eq. F) 
Since yA,mass varies throughout the distillation, *massv is different for each stage. 
387 
 
 
D. Problems. 
15.D1. Dprop,water = 0.87E-9 m2/s. Eq. (15-9), , , ,0( / )( )
A
A z AB AB A L A
dC
J D D L C C
dz
     . If ,0AC = 1.2 
kg/m3 is the known value, ,A LC can be larger or smaller than ,0AC . For smaller ,A LC we have 
 ,A LC =1.2 – (0.2E-5)(0.0001)/0.87E-9 = 0.9701 
 If it is larger value, then ,A LC =1.2 +(0.2E-5)(0.0001)/0.87E-9 = 1.430 
 
15.D2. Taking the ratio of Eq. (15-23c) at the unknown T and at T =298.16, 
 
exp[ / ( )]
( ) (298.16)
exp[ / (298.16 )]
o
o
E TR
D T D
E R



= 1.52E-09 for T = 335.18K. Flux 
9 5
, , ,0( / )( ) (1.52 10 / 0.0001)(0.9701 1.2) 0.35 10
A
A z AB AB A L A
dC
J D D L C C
dz
 
          
 
 The temperature can be found with Goal Seek from a spread sheet, but one has to trick Goal Seek 
into working. Multiply the desired and the calculated fluxes by 1,000,000 and have Goal Seek 
match these two values. 
 
15.D3.a. 0.181cm2/s, b. 0.198 cm2/s, c. 0.0725 cm2/s, d. 0.198 cm2/s 
 
15.D4. a. 0.0875 cm2/s, b. 0.096 cm2/s, c. 0.175 cm2/s, d. 0.096 cm2/s. 
 
15.D5. Use Arrhenius form in Eq. (15-23c) but for mole fraction 0.0332 instead of infinite dilution. Write 
the equation for both known temperatures and divide one of these equations by the other. The 
constant Do divides out. Take the natural log of both sides and solve for E/R. The result is 
1
2 2 1
( ) 1 1
/ ln /
( )
AB
AB
D T
E R
D T T T
   
    
   
 
 The constant Do can be found from the known conditions at T1 
 1 1( ) / exp[ / ( )]o ABD D T E RT  
 Or from the known conditions at T2. The results are: E/R = 1348.3, E = 2677.6 cal/mol, DAB 
(x=0.0332, T=300) = 1.313×10-9m2/s. 
 
15.D6. Same equations as in 15.D5. At 298.16 K for the infinite dilution value set Csucrose = 0. Final 
results are Eo = 4953.8 cal/mol, DAB(infinite dilution, T = 320K) = 0.925×10-9m2/s. 
 
15.D7. For an ideal solution the term in brackets in Eq. (15-22) is equal to 1.0. Write this equation for 
two of the xA values with the corresponding diffusivities (e.g., x = 0.0332 with D = 1.007×10-9 
m2/s and x = 0.7617 with D = 1.226×10-9m2/s). Then have two equations with the two unknowns: 
o
ABD and 
o
BAD . Solve for the two unknowns. Results are 
o
ABD = 0.998×10
-9 m2/s and oBAD = 
1.308×10-9 m2/s. Check results with the other two mole fractions and find that the fit is good. 
 
15.D8. From http://www.engineeringtoolbox.com/ the density of methanol at bp is 750.5 kg/m3 (used a 
linear interpolation), which means partial molar volume = 1/(density/MW)= 0.0426 kg/m3. 
Viscosity of water is 1.0 cp = 0.001 Pa s = 0.001 kg/(m∙s). 
 a. With φB = 2.26, DAB = 1.43×10-9 m2/s. 
 b. With φB = 2.26, DAB = 1.56×10-9 m2/s. 
388 
 
 
15.D9. Combining Eqs. (15-35b) (15-35d),  
1/32
,max, 0.5 9 /vertical liqv gq  Assume that the bulk is pure 
water with infinite dilution of ethanol. From Perry’s Chemical Engineer’s Handbook, 8th edition, 
(p. 2-305) at 1.0 bar (0.1 MPa) water has ρW,m,liq = 55.212 kmol/m3 →ρW,liq=994.64 kg/m3 and 
ρW,m,vapor = 0.032769kmol/m3 → ρW,vapor = 0.5903 kg/m3. The water boils at 372.76K. At this 
temperature, from p. 2-432, the viscosity of liquid water in Pa∙s is, 
 29 10 4, exp[ 52,843 3703.6 / 5.866ln (5.879 10 ) ] 2.807 10W liq T T T Pa s
 
         
The viscosity of the vapor at 372.76K is (p. 2-426) 
 8 1.1146 5, (1.7096 10 ) 1.2561 10W vapor T Pa s
 
     or kg/(m∙s). 
 Now we can calculate the vertical velocity of the liquid water for q = 7.5×10-6m2/s (remember to use 
liquid properties). 
  
1/321/32
,max,
9(994.64)(9.81)(7.5 6)
0.5 9 / 0.5 0.130 /
0.0002807vertical liq
E
v gq m s 
 
   
 
 
 A check of the units show they work. The modified Reynolds number (using gas properties) is, 
 
, ,max
5
( ) (0.10)(0.5903)(0.81 0.130)
Re 3195.6
1.256 10
tube gas liq yd v v


 
   

 
 The gas phase Schmidt number is 
/
gas
EW gas
Sc
D
  
  
 
The viscosity and density were found earlier. 
The diffusivity of ethanol and water in the vapor phase at 372.76K and 1.0 bar = 0.98717 atm can be 
estimated from the Chapman-Enskog theory with the parameters in Table 15-2. This value of DEW = 
1.658×10-5 m2/s. Then Scgas = 1.283. Since both Re and gasSc are in the range for Eq. (15-47a), the 
modified Sherwood number is, 
 0.77 0.33 .77 .33
( )
0.0328(Re ) 0.0328(3195.6) (1.283) 17.79p tube B lm gas
AB tot
k d p
Sc
D p
   
 
15.D10. From the Chapman-Enskog theory DNH3-air = 2.05×10-5m2/s at 318.16K and p = 1.2 atm. 
 
 
 The concentration at z = L is CNH3 (L) = CNH3 (z = 0) + JNH3L/DNH3-air. Results are 0.0002483 kmol/m3 
and 0.0002117 kmol/m3. 
 
15D11. D = JL/ΔC = 4.114×10-5m2/s. Set up spreadsheet to obtain this value. Since the collision integral 
was entered manually, had to do several iterations. After 6 iterations T = 396.2K (see 
spreadsheet, and note that collision integral does not exactly match the value of kT/εAB. 
Problem 15.D10a, 3rd ed.
MW A 28.9 MW B 17 const 1.86E-07
T 318.16 p 1.2 T^3/2 5675.033
sigma A 3.711 sigma B 2.9 sigma AB 3.3055
eos A/kB 78.6 eps B/kB 558.3 eps AB/kB 209.4812
kT/EpsAB 1.5188 Col integ 1.197 Linear interpolation table 15-2
D AB 2.05E-05
D, cm^2/s 2.05E-01
389 
 
 
 
 
 
15.D12*. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). 
At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(ms). 
 Calculate δ = 0.000115282 m, vy,avg = 0.04338 m/s, Re = 19.966. This is a long residence time 
with Re< 20 so there are no ripples. Shavg = 3.41 and kavg = 3.295E-05 m/s, and 0.000168 kg/s 
carbon dioxide are absorbed. 
 
15.D13. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 
298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). 
 Calculate δ = 0.0001663 m, vy,avg = 0.090241m/s, Re = 59.898. This is a long residence time, 
laminar flow, with no surfactant so there are ripples. Shavg = 5.8 and kavg = 3.89E-05 m/s, and 
0.000198 kg/s carbon dioxide are absorbed 
 
15.D14. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 
298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). 
 Calculate δ = 0.0007717 m, vy,avg = 1.9441 m/s, Re = 5989.8. This is turbulent flow with 1300 < 
Re < 8300. Scliq = 899.2, Shavg = 255.5 and kavg = 0.0003689 m/s, and 0.00188 kg/s carbon 
dioxide are absorbed 
 
15.D15. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa s = 0.001 kg/(m∙s). At 
298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). 
 Calculate δ = 0.000115282 m, vy,avg = 0.04338 m/s, Re = 19.966. This is a short residence time 
with Re< 20 so there are no ripples. Shavg = 9.942 and kavg = 9.61E-05 m/s, 7.851E-09 kg/s 
carbon dioxide are absorbed. 
 
15.D16. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 
298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). 
Problem 15.D11, 3rd ed.
MW A 28.9 MW B 17 const 1.86E-07
T 396.1642 p 0.9 T^3/2 7885.202
sigma A 3.711 sigma B 2.9 sigma AB 3.3055
eos A/kB 78.6 eps B/kB 558.3 eps AB/kB 209.4812
kT/EpsAB 1.891168 Col integ 1.1069 Linear interpolation table 15-2
D AB 4.11E-05
D desired 4.11E-05 chkB7-B8 1.90E-10
chk x E5 1.90E-05 Goal seek to zero changing B3
Problem 15.D11, 3rd ed.
MW A 28.9 MW B 17 const 0.0000001858
T 396.164199034186 p 0.9 T^3/2 =B3*SQRT(B3)
sigma A 3.711 sigma B 2.9 sigma AB =0.5*(B4+D4)
eos A/kB 78.6 eps B/kB 558.3 eps AB/kB =SQRT(B5*D5)
kT/EpsAB =B3/F5 Col integ 1.1069 Linear interpolation 
D AB =F2*F3*SQRT(1/B2+1/D2)/(D3*F4*F4*D6) in Table 15-2
D desired 0.00004114 chkB7-B8 =B7-B8
chk x E5 =100000*D8 Goal seek to zero 
changing B3
390 
 
 Calculate δ = 0.0001663 m, vy,avg = 0.090241 m/s, Re = 59.898. This is a long residence time, 
laminar flow, with surfactant so there are no ripples. Shavg = 3.41 and kavg = 2.28E-05 m/s, and 
0.0001165 kg/s carbon dioxide are absorbed. 
 
15.D17. Used a spreadsheet set up to solve Example 15-6. For δ = 0.001 meter one obtains xNH3 = 
0.04988, yNH3,surface = .21593, Nwater = 0.5393, NNH3 = 0.0228307. The concentrations are the same 
as in Example 15-6, but the fluxes are 10× larger. 
 
15.D18. Part a. For two part solution need values at xE = 0.25 and 0.35. The average molecular weights 
are calculated as in Example 15-5, and are used to determine the average molar densities. The 
Fickian diffusivities are estimated by interpolating between values given in the Table in Example 
15-5. The activity coefficients are determined in the same way as in Example 15-5. Then the 
Maxwell-Stefan diffusivities are found by the same method. The values are listed below 
 
 MWavg m ,kmol/m
3 DEW, m2/s γE EWD , m
2/s 
XE = 0.25 25.0 36.28 0.633×10-9 1.9028 1.495×10-9 
XE = 0.35 27.8 31.62 0.625×10-9 1.5553 1.609×10-9 
Write Eq. (15-61c) for both intervals. For Δz from xE = 0.2 to 0.3 we obtain (values at xE = 0.2 and 0.3 
are in Example 15-5), 
9
936.28(1.495 10 )[1.7083(0.3) 2.1582(0.2)] 9.3445 10
1.9028(0.25)E
zN

 
     
From xE = 0.3 to 0.4 (interval is over length δ- Δz) we obtain, 
 
9
931.62(1.609 10 )[1.4338(0.4) 1.7083(0.3)](0.00068 ) 5.7027 10
1.5553(0.35)E
z N

 
       
Adding the two equations to remove the unknown Δz and then solving for NE and Δz,, we obtain 
 NE = -2.128×10-5kmol/s and Δz = 0.0004223m 
 
Part b. Since the interval Δz is greater than the interval δ – Δz = 0.0002577m, we subdivide the interval 
from xE = 0.2 to 0.3 into 2 parts. The values needed are given below. 
 
 MWavg m ,kmol/m
3 DEW, m2/s γE EWD , m
2/s 
XE = 0.225 24.3 37.625 0.659×10-9 2.01976 1.482×10-9 
XE = 0.275 25.7 34.99 0.624×10-9 1.79959 1.532×10-9 
Equation (15-61c) is now written 3 times: 
9
1
9
2
9
1 2
5.5371 10
3.9846 10
( ) 5.7027 10
E
E
E
z N
z N
z z N



   
   
      
 
and solved for the 3 unknowns Δz1, Δz2, and NE. Obtain NE = -2.2389×10-5kmol/s, Δz1 = , 0.0002473, and 
Δz2 = 0.0001780m. 
 
15.D19 (Optional, Unsteady diffusion) At the average C = 0.001 mol/L 5 20.5228 10 cm / ssucroseD

  . 
Equation becomes 
5
,0
1
4(0.5228 10 )
A
A
C z
erf
C t
 

 
Numerical values of ,0/A AC C are easily obtained with a spreadsheet or with the use of Table 17-7. 
 
391 
 
z, cm t = 1000 s t = 10000 s t = 100000 s 
0 1 1 1 
0.01 0.9221 0.9753 0.9922 
0.05 0.6249 0.8771 0.9610 
0.1 0.3281 0.7571 0.9221 
0.2 0.0505 0.5362 0.8449 
0.3 0.00335 0.3535 0.7692 
0.4 9.16E-05 0.2161 0.6957 
0.5 1.009E-06 0.1220 0.6249 
0.6 0.6352 0.5574 
0.7 7.61E-12 0.0304 0.4936 
0.8 0.0134 0.4346 
0.9 0.00538 0.3788 
1.0 0.00198 0.3281 
1.2 0.000206 0.2406 
1.4 1.49E-05 0.1710 
1.6 7.49E-07 0.1177 
1.8 0.0784 
2.0 6.2E-10 0.0505 
3.0 0.00335 
3.5 6.20E-04 
3.56 4.99E-04 
4.00 9.16E-05 
5.0 1.01E-06 
 
Part b. 61.0 10C   when 40/ 5.0 10C C

  , which for t = 100000 s occurs for a thickness of <3.56 
cm (Goal Seek gives 3.559 cm). Thus, at this time a layer 3.56 cm or thicker appears to be 
infinitely thick. 
Part c. 61.0 10C   when 40/ 5.0 10C C

  , which for  =0.10 cm occurs at t = 78.938 s (done with 
Goal Seek on spreadsheet). 
 
H. Spreadsheet Problems 
 
15H1. Let A = air, B = hydrogen, and C = ammonia. Then NC = -NA – NB. Substitute this expression 
into Eqs. (15-65a, b) 
 
m A CB A A A
A B
AB AC AC AB AC
m B CB B A B
A B
BA BC BA BC BC
y yy y y y
N N
z D D D D D
y yy y y y
N N
z D D D D D


   
        
    
   
       
    
 
 
Determine NB from the 1st equation and NA from the second. 
 
392 
 
m A CB A
A
AB AC AC
B
A A
AB AC
m B CA B
B
BA BC BC
A
B B
BA BC
y yy y N
z D D D
N
y y
D D
y yy y N
z D D D
N
y y
D D


 
   
  

 
 
 
 
   
  

 
 
 
 
 
Put these equations and the values for mole fractions at the boundaries, diffusivities, ρm, and Δz = δ into a 
spread sheet. Guess a value for NA,guess, calculate NB and NA,calc, and use Goal Seek to make 
NA,guess - NA,calc = 0 by changing the value of NA,guess. 
Results Nair = -6.209E-5, NH2 = 14.026E-5, and NNH3 = -7.817E-5 kmol/s. As expected hydrogen 
diffuses in the positive direction and ammonia in the negative direction. The surprise is the 
substantial negative diffusion rate of air. (Spreadsheet shown in 15.H4, but with different 
numbers.) 
 
15.H2. Used a spreadsheet set up to solve Example 15-6. For a bulk gas that is 40% air, 15 % NH3 and 
45% water obtain xNH3 = 0.05596, yNH3,surface = .24225, Nwater = 0.032964, NNH3 = 0.001955 kmol/s 
 
15.H3a. See solution to 15.H1 for procedure and 15.H4 for example spreadsheet. 
Results: Nair = -1.87E-8, NH2 = 2.98E-7, NNH3 = -2.80E-7 kmol/s. 
b. For ammonia Deff = 1.5656E-5 m2/s. Estimating dC/dz with the difference approximation for a very 
dilute solution, N = J = -Deff ρm Δy/(Δz)= -(1.5656E-5)(0.08928 kmol/m3)(.002)/(.01m) = -2.80E-
7 kmol/s. Thus, this is accurate. For hydrogen and air Dair-H2 = 3.0550E-5 m2/s. Then 
Nair = Jair = (3.0550E-5)(0.08928 kmol/m3)(-0.001)/(.01m) = 2.73E-7 kmol/s. The same value is 
obtained for hydrogen. The hydrogen value is close, but the air value is not close. Conclusion: 
Use the Maxwell-Stefan approach. 
 
15.H4. See solution to 15.H1 for procedure. Results: Nair = -5.903E-5, NH2 = 14.069E-5, NNH3 = -
8.166E-5 kmol/s. Note the substantial negative diffusion of air despite the zero “driving force.” 
The air is dragged along with the ammonia. The spreadsheet is given below (labeled as 15.D22), 
first with the numbers, and then with the formulas. 
393 
 
HW 15-D22, 3rd Ed. SPE
A=air, B=H2, C = NH3