SOLUTION MANUAL
for
SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Transfer Analysis Analysis
3rd Edition Separations) (Formerly published as Equilibrium-Staged Separations
by Phillip C. Wankat
rd
SPE 3 Edition Solution Manual Chapter 1 New Problems Problems and new new solutions are listed listed as new immediately immediately after the solution number. number. These These new problems in chapter chapter 1 are: are: 1A3, 1A4, 1B2-1B4, 1D1.
A2.
Answers are in the text.
A3.
New problem problem for 3rd edition. Answer is d.
B1.
Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse r everse osmosis or by distillation).
B2.
rd exchange), or a filter, or a New problem problem for 3 edition. Many homes have a water softener (ion exchange), carbon water “filter” (actually adsorption), or a r everse osmosis system. system.
B3.
New problem problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.
B4.
rd New problem problem for 3 edition. You probably used some of the following: chromatography, crystallization, crystallization, distillation, distillation, extraction, filtration and ultrafiltration.
D1.
New problem problem for 3rd edition. Basis 1kmol feed.
.4 kmole E
.4 MW
.6 kmol kmol Water ater
46 18.4 kg
.6 MW 18
10.8 kg
total
29.2 kg kg
Weight fraction ethanol = 18.4/29.2 = 0.630 Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.
17
rd
SPE 3 Edition Solution Manual Chapter 1 New Problems Problems and new new solutions are listed listed as new immediately immediately after the solution number. number. These These new problems in chapter chapter 1 are: are: 1A3, 1A4, 1B2-1B4, 1D1.
A2.
Answers are in the text.
A3.
New problem problem for 3rd edition. Answer is d.
B1.
Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse r everse osmosis or by distillation).
B2.
rd exchange), or a filter, or a New problem problem for 3 edition. Many homes have a water softener (ion exchange), carbon water “filter” (actually adsorption), or a r everse osmosis system. system.
B3.
New problem problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.
B4.
rd New problem problem for 3 edition. You probably used some of the following: chromatography, crystallization, crystallization, distillation, distillation, extraction, filtration and ultrafiltration.
D1.
New problem problem for 3rd edition. Basis 1kmol feed.
.4 kmole E
.4 MW
.6 kmol kmol Water ater
46 18.4 kg
.6 MW 18
10.8 kg
total
29.2 kg kg
Weight fraction ethanol = 18.4/29.2 = 0.630 Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.
17
Chapter 2 New Problems Problems and new new solutions are listed as as new immediately after the solution solution number. number. These These new problems are: 2A6, 2A6, 2A9 to to 2A16, 2C4, 2C4, 2C8, 2C9, 2C9, 2D1.g, 2.D4, 2.D4, 2D10, 2D13, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3.
2.A1. 2.A1.
Feed Feed to flash flash drum is a liq liquid uid at hig high h pre pressu ssure. re. At this this pres pressure sure its its ent enthal halpy py can be calcul calculated ated as a liquid. liquid. eg. h TF , Phigh
c p LIQ TF
Tref . When pressure pressur e is dropped the mixture is above
its bubble point point and is a two-phase mixture mixture (It “flashes”). In the flash mixture enthalpy enthalpy is unchanged but temperature changes. changes. Feed location cannot cannot be found from T F and z on the graph because equilibrium data is at a lower lower pressure on the graph used for this calculation. 2.A2.
Yes.
2.A4. !"# 6782)29028: 5'
@' A #"BC%
;.80/ '*1/0< ()*+, "%
-./0*1234
=">? #
#
"%
&'
!"#
2.A6. New Problem. Problem. In a flash drum separating a multicomponent mixture, raising the pressure will: i. Decrease the drum diameter diameter and decrease the relative volatilities. volatilities. Answer is i. i. 2.A8.
a. K incr eases as T increases b. K decreases as P increases increases c. K stays stays same as mole fraction changes changes (T, p constant) -Assumption is no concentration effect in DePriester charts d. K decreases decreases as molecul molecular ar weight weight increases increases
2.A9.
New Problem. Problem. The answer is 0.22
2.A10.
New Problem. Problem. The answer is b.
2.A11.
New Problem. Problem. The answer is c.
18
2.A12.
New Problem. Problem. The answer is b.
2.A13.
New Problem. Problem. The answer is c.
2.A14.
New Problem. Problem. The answer is a.
2.A15.
New Problem. Problem. a. b. The answer is
The answer is 36ºC
3.5 to 3.6
Problem. The liquid is superheated when the pressure drops, and the energy comes fr om the 2.A16. New Problem. amount of superheat.
2.B1. 2.B1.
Must be Must be sure you you don don’t ’t violate violate Gibbs Gibbs phase phase rule rule for inten intensive sive variable variabless in in equi equili libri brium. um. Examples: F, TF , z, z, p F, z, Tdrum , Pdrum F, h F , z, z, p
F,z,y,Pdrum
F, TF , z, z, y
F, h F , z, z, y
F,z,x,p drum
F, TF , z, z, x
etc.
F,z,y,p drum
F, TF , z, Tdrum , p drum
F,z,x,Tdrum
F, TF , y, y, p
Drum dimensions, z, Fdrum , p drum
F, TF , y, y , Tdrum
Drum dimensions, z,y,p drum
F, TF , x, x, p
etc.
F, TF , x, x , Tdrum F, TF , y, y, x
2.B2. 2.B2.
This This is is essen essentia tially lly the same problem problem (disgui (disguised sed)) as proble problem m 2-D1c and e but but with with an exis existin ting g (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 0. 25 which corresponds corresponds to 2-D1c. lb mole If F 1000 , D .98 and L 2.95 ft from Problem 2-D1e . hr Since D α V and for constant V/F, V α F, we have D α With F = 25,000:
F.
Fnew Fold = 5, Dnew = 5 Dold = 4.90, and Lnew = 3 Dnew = 14.7 . Existing drum is too small. Feed
Fexisting
rate
drum
can
handle:
F
α
2
D.
Fexisting
D exist
1000
.98
2
4
2
.98
gives
16,660 lbmol/h
Alternatives Alternatives a) Do drums drums in parallel. Add a second second drum drum which can handle remaining 8340 lbmol/h. b) Bypass with liquid liquid mixing
19
5 A "%GH D A "=% ;!CCC#< A ?!%# !CHCC#
=%H###
EF-1*) &
GI?#
Since x is not specified, specifi ed, use bypass. This produces less less vapor. c) Look Look at Eq. Eq. (2-6 (2-62), 2), which which becom becomes es
V MW v
D
3 K d r um 3 6 0 0
L
v
v
Bypass reduces V c1)
K drum improvements can be made with a better demister demister drum is already 0.35. Perhaps small improvements → Talk to the manufacturers. c2) ρv can be increased increased by increasing increasing pressure. pressure. Thus operate at higher pressure. pressure. Note this this will change change the equilibrium equilibrium data data and raise temperature. Thus a complete complete new calculation needs to be done. d) Try bypass bypass with with vapo vaporr mixin mixing. g. e) Other Other alte alterna rnativ tives es are pos possi sible ble.. V
2.C2.
2.C5.
zA
F
a. Start with
xi
KB
zB 1
Fz i L
KA
and let V
VK i
Fz i
xi
1
L
F
F
L
or x i L Ki
zi L F
Then y i
Kixi
yi
L F
K i
K i zi L F
From
1
xi
1
L F
K i
0 w e o b ta i n
Ki L F
1
1 zi L F
0 K i
20
zi
2.C7.
1 V/F f
0 0
Ki
.1 -.09
1
V
1
f
V F
From data in Example 2-2 obtain:
F
.2 .2 -.1
.3 -.09
.4 .5 .5 -.06 -.007
.6 .07
.7 .16
.8 .8 .3
.9 .49
1.0 .77
2.C8. New Problem. Problem.
p drum (AEJD
Tdrum
Fz
Lx
Vy
5
&
@
21
2.C9. New Problem. of Eqs. (2-62) and (2-63). Overall and component component mass balances are, Problem. Derivation of
F
V L1
Fz i
L2
L1x i ,L1
L2
x i ,L 2
Vyi
Substituting in equilibrium equilibrium Eqs. (2-60b) and 2-60c)
Fz i
L1K i ,L1
L2
x i ,L 2
L2 x i ,L 2
VK iV
L2
x i ,L 2
Solving,
x i,L2
Fz i L1K i ,L 2
Fz i
L2
VK i ,V
L1K i ,L1
L2
F V
L2
L1
VK VK i ,V L 2
Dividing numerator numerator and a nd denominator denominator by F and collecting terms.
zi L1
x i,liq2 1
Since
yi
K i ,V
K i , L1
C
Stoichiometric Stoichiometric equations, equations,
i 1
K i ,L1
x i,L 2
K i,V
C
which becomes i 1
K i , L1
i 1
1
K i , L1
L2
1
yi
K i,V
2.D1.
a.
V
0.4 100
See graph. y b.
V
c.
Plot x
V F
L2
1 C i 1
x i ,liq 2
40 and L
0
L1
K i ,V
F
F V
L V 3 2, y 0.77 and x 0.48
0.4 1500
600 and L
V F
K i , L1
K i , L1
L2
i 1
x i ,L 2
0
0
(2-62)
F
1
L2
1
C
yi
K i , L1
K i , L1 1
F
V
1
L2
C i 1
V
1 zi
L2
x , we have x i ,liq1 L 2 i , liq 2
x i ,liq1
Slope op. line
1
1 , thus,
1 In addition,
L2
zi
F
C
1,
L2
L1
1
L2
K i,V
F
K i ,V
, y i L 2 x i ,L 2
1
Since x i , liq1
1
L2
L2
L1 F
L2
zi
L1
K i,V
F 1 zi K i ,V
L2
1
L2
1
V
V F (2-63)
F
60 kmol/h x
z
0 .6
900 . Rest same as part a.
0.2 on equil. Diagram and y
z 1. 1.2 0.25 . From equil y d. Plot x 0.45 on equilibrium curve.
x
z
0.3. y intercept
zF V
1.2
0. 5 8 .
22
L
Slope Plot operating line, y e. Find Liquid Density.
MW L Then, VL
x
MWm
0.51 . From mass balance F
MWw w
m
RT
y w MW
.2 32.04 32.04
w
v
.8
.7914
MW L VL
P MW
y m MW
.2
.2
4
z at z
L
MW v
.8
V F
x w MWw
xw
v
1 V F
V
m
Find Vapor Vapor Density.
V
V
x m MWm
xm
F
37.5 kmol/h.
.8 18.01
18.01 1.00
20.82
22.51 ml/mol
20.82 22.51 0.925 g/ g/ml
(Need temperature of the drum)
.58 32.04
.42 18 18.01
26.15 g/mol
Find Temperature Temperature of the Drum T: From Table 3-3 find T corresponding to
y v
1 at m
.58, x
20, T= T=81.7 C
26.15 g/mol
354.7 K
82.0575
ml atm mol K
354.7 K
8.98 10
4
g/m l
Find Permissible Permissible velocity: velocity:
23
u perm
K drum
L
K dr u m
v
exp A
v
B
nFlv V
Since V
Wv
L
F
V
1000
250
WL WV
L
6 5 3 7 .5
u perm 3600 Thus, D
MW L
8.98 10
1 4 .1 9
4A cs
3 6 0 0 8 .9 8 1 0
3
E
nFlv
6537.5 lb / h
lbmol
2 0 .8 2
1 5, 615 lb/h,
4
0.0744, and n Flv
2.598
4
14.19 ft/s
4
454 g/ g/lb 4
2-60
4
250 lbmol/h, lb
.9 2 5
2 5 0 2 6 .1 5 v
nFlv
750
8 .8 9 1 0
.925 8.98 10
.442
V MW v A cs
250 26.1 5
L
15615
D
0.25 1000
V MW v
V
.442, and u perm
nFlv
F
F
750 lbm ol/h, and W L Flv
Then K drum
C
2
g/ml
2
28316 .85 ml/ft
2.28 ft .
3
1.705 ft. Use 2 ft diameter.
L ranges from 3 D 6 ft to 5 D=10 ft Note that this this design is conservative if a demister is used. f.
Plot T vs x from Table 3-3. When T 77 C, x 0.34, y 0.69. This problem is now very similar similar to 3-D1c. Can calculate calculate V/F from mass balance, balance, Fz Lx Vy. This is
Fz
V x
V
Vy or
z
y
0.4 0.34
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.
F
Work backwards. Starting with x 2, find y2 = 0.62 from equilibrium. equilibrium. From equilibrium equilibrium point plot op. line of slope
z2
0 .5 1
For stage 1,
2.D3.
a.
z
L V
2
1
V F
equilibrium, m, y1 x 1 (see Figure). Then from equilibriu
V
z1
x1
0 .55 0.51
F
y1
x1
0.7 8 0.51
0. 6
V
0. 4
V F y
Op. eq.
y See graph: y
L
x
z V F
x
2 3
0.55
xM
V 2 3
0.18
V F
2
3 7.
This
gives
2
0.78 .
0.148 .
6.0 k mol h, L
4 .0
T ~ 82.8 C linear interpretation on Table 2-7 .
24
b. Product
78.0 C
x
Mass Bal: Fz or
0.30, Lx
4.0
y
Vy
0.665, F V x
Vy
10 V 0.3 0.665V
V
2.985 and V F
0.2985
Can also calculate V/F from slope.
c.
F
y If y
V
10,
F
L V
x
0.3
V
z
7
V F
z
0.8, x Then z
3&L
x
7
z 0.3
0.545 @ equil 0.3 0.8
Can also draw line of slope
7 3 7 3
0.545
0.6215.
through equil point.
25
2.D4. New problem in 3rd edition. Highest temperature is dew point Set
zi
yi .
Ki
K ref TNew If pick C4 as reference: First guess
yi Ki Guess for reference
K C4
yi K i
K bu tan e .35
.45
3.1
1.0
.125
.2
.35
.45
Ki
8.8
4.0145
.9
4.0145 .6099
yi
.2
.35
.45
Ki
6
2.45
.44
2.45 1.2 .2
.35
.45
Ki
6.9
2.94
.56
3.1,
K C6
0.125
4.0145 T too low
8.8,
K C6
.9
0.6099
2.45, T
85 : K C2
6.0,
K C6
0.44
1.20
2.94, T
yi
41 C : K C3
T
118 C : K C3
T
yi
K C4,NEW
yi K i
1.0,
.2
K C4,NEW
1.0
K ref TOld
4.014
0
yi x i
xi
Want
V F
0.804
96 C : K C3
6.9,
K C6
0.56
Gives 84 C
26
K C4
Use 90.5º → Avg last two T
yi K i
2.D5.
2.7, K C3
.2
.35
.45
6.5
2.7
.49
6.5, K C6
0.49
1.079
T ~ 87 88º C Note: hexane probably better choice as reference. a) L! A (= L= @! A #"%% !
.!H= A ! *1:
=
(! A !###
V F
L
y1
V1
V1
V1
V
0.45454 & L1
V1
F1
1000
D! A CGQ"% R:-)S, A (=
V
c) Stage 2 At x
0, y
From graph
V2
0.25 ,
F
V F
z V F
y2 F2
5! A #"CC A @=
st z Plot 1 Op line.
y = x = z = 0.55 to x1 = 0.3 on eq. curve (see graph) 0.55 0.80 .25 0.454545 .55 0 .55
L
Slope
L1
F
x1
2
&=
&! A #"I#
b)
0.25
L
V
687.5
F
1000
1
0.75F
3, y x V 0.25F 0.66 2.64. At y 0, x 2 0.25
0.82, x 2 0.25 687.5
0.6875
0.66. Plot op line
z2
F L
z
z
0.66
L F
0.75
0.88
0.63 . 171.875 kmol/h
2
27
2.D6.
D WX? A ="?
( A !"# R:-)S:23 F A %#TU
WX% A #"G#
V A =## RV*
WXC A #"I#
@X? A #"?% @X% A #"I% YXC A #"=#
E
K i 1 zi
RR eq.,
1
0 , First guess V/F = 0.6
Ki 1 V F
1.4 .45
0.2 .35
0.7 .2
1 1.4 .6 1 0.2 0.6 Use Newtonian Convergence
1 0.7 0.6
f1
c
df k d V F
V F
i 1
V k 1
Fk
2
Ki
1
Ki
1 V F
1
0.0215
zi 2
f k df d V F
28
1.4
df 1 d
V
2
0.45
F2
1 1.4 .6
0215
0.6
2
1 0.2 0.6
0.2 .35
1 1.4 0.6377 Which is close enough.
2
0.20
0.570
2
1 0.7 0.6
1
Ki
V
1
1
0.4012, y c5
zA KB
KM
xi
1
KA
0.3
0.7
F
0.21 1
5.6 1 zM
xM 1
1.0002
, y8
0.3210 0.1084
0.30 0.4012
yi
0.9998
1
KM
0.2276 0.3 1
xP
1 xM
0.8534 , y M
yP
1 yM
0.1792
Use Rachford-Rice eqn: f
V
1
Converge on V F
KM xM
V
0.1466
0.2276
5.6 0.1466 Ki
1 zi
Ki
1 V/F
73, K 2
4.1 K 3
V F
.076, V
zi
4.6
F
1
Find K i from DePriester Chart: K1
1
0.5705
0.21
V
From x i
2.4 0.2377
zB
5.6 and K P
Eq. (2-38)
0.8 0.4012
0.3613
1 0.7 0.6377
F
y c4
Kixi
F
0.20
V
0.00028
yi
0.2377,
1.4 .6377
1 0.2 0.6377
x c6
0.2
1 0.7 0.6377
0.45
0.35
x c5
0.7
1 0.2 0.6377
zi
x c4
2.D9.
0.7
0.6377
0.570
1.4 .45
f2
2.D8.
2
0.35
F
V
2.D7.
2
0.2
F V F
152 kmol/h, L
we obtain x1
.0077, x 2
0.8208
0 . Note that 2 atm = 203 kPa.
.115
F V
.0809, x 3
1848 kmol/h . .9113
Ki 1 F From y i K i x i , we obtain y1 .5621, y 2 .3649, y 3 .1048 Need hF to plot on diagram. Since pressure is high, feed remains a liquid
hF CPL
CPL TF
Tref , Tref
C PEtOH x EtOH
0 from chart
C Pw x w
29
Where x EtOH and x w are mole fractions. Convert weight to mole fractions. Basis: 100 kg mixture
30 kg EtOH
30
0.651 kmol 46.07 70 kg water 70 18.016 3.885 Total = 4.536 kmol 100 Avg. MW 22.046 Mole fracs: x E 4.536 Use C P at 100 C as an average C P value.
0.6512
0.1435, x w
4.536
0.8565 .
L EtOH
C PL Per kg this is
hF
37.96 .1435
18.0 .8565
C PL
20.86
MWavg
22.046
0.946 2000
0.946
20.86
kcal kmol C
kcal kg C
189.2 kcal/kg
which can now be plotted on the enthalpy composition diagram. Obtain Tdrum
88.2 C, x E
0.146, and y E
0.617 .
For F 1000 find L and V from F = L + V and Fz which gives V = 326.9, and L = 673.1
Note: If use wt. fracs. C P L
23.99 & C PL MWavg
Lx
Vy
1.088 and h F
217.6 .
All wrong .
30
2.D.10 New Problem. Solution 400 kPa, 70ºC
Kixi ,
zi
xi 1
Ki
R.R.
1 For C6
V
1
K C4
1.9,
,
xi
yi
z C3
1 zC6
K C6 1
0.7
0.3
zi
zC4
.65 z C6
F
zC6 K C6
0.7
5,
x C6
F
0
V
1
0.7
z C6
Ki
1 zi
Ki
1
35 Mole % n-butane
K C3
From DePriester chart Know y i
z C4
0.49
z C6 1
V F
1 0.7
V
z C6
0.7 1 0.7
V F
F
V F
4 .65 z C6
0.9 .35
0.7z C6 0 V V V 1 4 1 0.9 1 0.7 F F F 2 equations & 2 unknowns. Substitute in for z C6 . Do in Spreadsheet. RR Eq:
Use Goal – Seek to find V F.
V F z C6 2.D11.
0.594 when R.R. equation 0.7
0.49
V
0.7 F L F 0.6 Operating line: Slope
0.000881 .
(0.49)(0.594)
0.40894
V F 0.4 & L V 1.5 1.5, through y x z 0.4
31
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.
MW L Then, VL
x m MWm xm
MWm
x w MWw xw
MWw
m
w
.2 32.04 .2
32.04
p MW Find Vapor Density.
MW v
y m MW
v
m
v
RT
y w MW
.8
.7914 L
.8 18.01 18.01
22.51 ml/mol
1.00
MW L VL
20.82
20.82 22.51 0.925 g/ml
(Need temperature of the drum) w
.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
32
1 atm
v
26.15 g/mol
82.0575
ml atm
354.7 K
mol K
8.98 10
4
g/ml
Find Permissible velocity:
u perm
K drum
K drum,horizontal
L
v
1.25 K drum ,vertical
Since V Wv L
F
V
v
V F
250
B
0.25 1000
V MW v
1000
exp A
nFlv
250 26.15 lb lbmol
V
15615
WV
L
6537.5
MW L 8.98 10
0.5525
A Cs
8.98 10
17.74
v
2 1.824 ft ,
AT
D
Or y c 2.D14.
1 yp
xc 1 xc
cp
Raoult’s Law: K C4
1.25
A Cs 0.2
4A T
15, 615 lb/h, 2.598
4
4
9.12 ft
17.74 ft/s 454 g/lbm 4
g/ml
28316.85 ml/ft
3
2
3.41 ft and L 13.6 ft
1.76 .7 1 .76 .7
1
,
cp
0.80418
0.5682
pc
VPC4 PTot
4.04615 ,
VPC 4
11121 mm Hg
log 10 VPC 6
3.2658 ,
VPC 6
1844.36 mm Hg
1.0
4
0.5525
log 10 VPC 4
xi
nFlv
0.19582 cp
1
E
0.0744, and n Flv
3600 8.98 10
2.D13. New Problem. The answer is ycresol = 0.19582 x p Since x c 0.3, x p 0.7, y p 1 1 x p
yc
3
20.82
250 26.15
u perm 3600
With L/D = 4,
nFlv
4
0.925 8.98 10
V MW v A cs
750
.925
0.442, and K drum ,horiz
u perm
D
6537.5 lb / h
L
WL
K drum ,vertical
2
nFlv
250 lbmol/h,
750 lbmol/h, and W L Flv
C
zi 1
Ki 1 V F
1.0
33
0.3
0.7
11121
1
1 0.4
P
1
1 x C4
0.84715,
yC4
yC6
11121
K C4 x C4
1844.36 3260
3260
0.84715
0.1527
0.52091
0.47928
1.00019
Check 2.D15.
1 0.4
P
Solve for P drum = 3260 mmHg zi xi V 1 Ki 1 F .3 xC4 0.1527, 11121 1 1 .4 3260
x C6
1
1844.36
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: z C2, ziC4, z nC4. Need two other eqns: z iC4 z nC4 constant, and z C2 z iC4 z nC4 1.0 Thus, solve three equations and three unknowns simultaneously. Do It . Rachford-Rice equation is, K C2 1 zC2 K iC 4 1 z iC 4
1
K C2
V
1
K nC 4
1 z nC 2
V
V
0
1 K iC 4 1 1 K nC 4 1 F F F Can solve for z C2 = 1 – ziC4 and ziC4 = (.8) znC4. Thus zC2 = 1 – 1.8 znC4 Substitute for z iC4 and zC2 into R-R eqn. K C2 1 .8 K iC 4 1 K nC4 1 1 1.8 z nC4 z nC 4 z nC 4 V V V 1 K C2 1 1 K iC 4 1 1 K nC 4 1 F F F K C2 1 1 Thus,
z nC4 1.8 1
K C2
1
K C2
1
K C2 .8 K iC 4
V
1
K iC 4
1
V F 1 1
V
1
K nC 4
1
K nC 4
1
F F Can now find K values and plug away. K C2 = 2.92, K iC4 = .375, K nC4 = .26. Solution is z nC4 = 0.2957, ziC4 = .8 (.2957) = 0.2366, and zC2 = 0.4677 2.D16.
z C1
0.5, z C4
0.1, z C5
0.15, z C6
0.25, K C1
0
50, K C4
.6, K C5
V F
.17, K C6
0.05
st
1 guess. Can assume all C1 in vapor, ~ 1/3 C 4 in vapor, C 5 & C6 in bottom
V/F
1
.5
.1 / 3 .53 This first guess is not critical.
34