SPE 3rd Ed. Solution Manual Chapter 4 New Problems Problems and new new solutions are listed as as new immediately after the solution solution number. These new new problems are: 4A6, 4A13, 4C10, 4C16, 4C16, 4D6, 4D9, 4D13, 4D15, 4D15, 4D18, 4E4, 4E5, 4E5, 4H1 to 4H3. 4H3. 4A1.
Point A: streams leaving stage 2 (L 2, V2) Point B: vapor stream leaving stage 5 (V 5) liquid stream leaving stage 4 (L 4) Temp. of stage 2: know K y 2 / x 2 , can get T from temperature-composition graph or DePriester chart of K = f(T,p). Temp. in reboiler: same as above (reboiler is an equilibr equilibrium ium stage.)
4A2.
4A6. 4A7. 4A13.
4A14.
a. Feed tray = .6, z = 0.51 (draw y = x line), line), y F =0.52, xF = 0.29. b. Two-phase feed. c. Higher her New Problem Problem in 3rd Edition. Answer is a. See Table 11-3 and 11-4 for a partial list. New Problem 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 If feed stage is non-optimum, non-opti mum, the feed conditions can be changed to have an optimum feed location.
4B2.
a. Use columns in parallel. parallel. Lower F to to each column allows for higher higher L/D and may be sufficient sufficient for product specifications. b. Add a reboiler reboiler instead of of steam injection. injection. Slightly less stages stages required and and adds adds 1 stage. c. Make the condenser a partial instead of of a total condenser. Adds a stage. d. Stop removing side stream. Fewer stages 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 energy at highest T (reboilers) or lowest 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 z, yD satisfies operating equation. Solution: Op. Eq.
y
Substitute in
y
yD V But
q
1.0, V y DD
Lz
L V x yD , x
L
D, L Fz
L V 1 xB z
V xB
F, L
V
B
Bx B
Which is external mass balance.
QED.
70
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 Q R Q R / D
4.C16.
V .
With CMO and a saturated saturat ed liquid feed V
V
(1 L / D) D , and then
(1 L / D) .
New Problem Problem in 3rd Edition. Define a fictitious total feed FT , z T , h T
FT
F1
F1z1
F2 , z T
F2 z 2
, hT
F1h F1
F2 h F2
FT FT Intersection of top & bottom operating lines must occur at feed line for fictitious feed F T. (Draw a column with a single mixed feed to prove this.) This feed line goes through y x z T
with slope
where
FT
qT
qT
qT
H m ix
1 hT
H mix
h m ix
and H mix, h mix are saturated
Given p, L/D, saturated liquid reflux, x D , x B
A
z0
B
opt feed locations, z1 , z 2 , F1 , F2 , h F1 , hF2
z2 y
zT
z1 xB x
b.)
Does
qT
q1F1
q 2 F2 FT
71
H mix
check q T
hT
H mix
F1h F1
H mix
h mix
F2 h F2 F1
FT H mix
F2 H mix
h mix
F1 H F1
H mix
F2 h F2
h mix FT
where H mix & h mix are vapor and liquid enthalpies on feed stage of mixed column F1 H mix H m ix
h F1
F2
h mix
qT
H m ix
h F2
H mix
h mix
FT
Usual CMO assumption is λ >> latent heat effects in either vapor or liquid. H mix h F1 H mix h F2 Then q1 a nd q2 H mix h mix H m i x h m ix Thus 4D1.
F1q1
qT
x
When x
if CMO is valid.
FT
a. Top op line: y Intersects y
F2 q 2
L V
xD
x
1
V
x D and
L
L D
1.25
V
1 L D
2 .25
0.5555
0 .9
0, y
1
L
b.Bottom b. Bottom op line: y
x
V Intersects y = x = x B = 0.05 @y
L
1
x
L
xD
V
L V
Plot – See diagram
0.4
1 x B , and
1 0.5 / 2 3 2
L V
V
B V
V B 1
3
V B
2
0.683 this is convenient point to plot
c. See diagram for for stages. Optimum feed stage is #2 above partial reboiler. 5 equilibrium stages + PR is more than sufficient.
72
d. Feed line goes goes from y = x = z = 0.55 to intersection 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 4D2. New Problem Problem in 3rd Edition. Part a. y
x
zE
.6 .6
S lo p e
L
F V
1 V/F
.63
V
V
V F
.37
b.
From Table Table 2-1, at 84.1° 84.1° C
y
c.
liquid at 20°C
H
hF
1.703
.5089
and 40 mole % ethanol. H h The pressure in Figure 2-4 is very close to 1.0 atm, thus it can be used, but must convert to wt frac. q
73
Basis 1kmol feed.
.4 kmole E .6 kmol Water ter From Figure 2-4
H q
.6 MW
39 8 10 3 98
75 q
hF h
0.63 wt frac.
10.8 kg total
29 . 2 k kg g
75, h F 20 C
10
1.20 1.2
Cp, 40% liq 20 84.1 0,
63 wt%, H
H
40%E
398 kcal kg , h
CPvapor st
18.4 kg
6 q 1 .2 40 mole % ethanol boils at 84.1°C (Table 2-1). Then if pick reference as saturated liquid at 40 mole %
Alternate Solution:
40 mole % E
46
18
398 kcal kg , h
Slope
d.
.4 MW
y E CPEvapor
6 5, h F
y w CPw ,vapor
3 98
kcal kg
C p vapor 120 84.1
nd
Assume only 1 and 2 terms in C P equations are significant. From Problem 2.D9
C Pvapor
.4 14.66
0.03758T
. 6 7 .8 8 . 0 0 3 2 T
kcal/kmol T is C which simplifies to
CPvapor
10.592 0.16952T
120
C p dT is equal to CPvapor @ Tavg
For linear 84.1
Tavg
84.1 120 2 102.05 . Then C Pv,avg 39 8
hF
398 15.149
q
e.
q
q
f.
Flash
kcal
hF
V F
398
kg
L
f,
15.1 47
1 3 1 2 , sl o p e
.7,
kcal kmol
kcal 1 kmol kmol 27.2 kg
0. 045.
333 L
12.32
413.15 kcal kg
65
L
0.16952 102.05
12.32 12 1 20 84 .1
413 .15
39 8
10.592
F
L
F
q
q 1
12
13
L
12 13 1 2
1 12
L
1 V F
.3
3
V
V F
.7
7
F
13
See graph for feed lines.
74
Graph for 4.D2
75
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. H v = 395, HL = 65 (from Figure 2-4). hF is liquid at 200°C. Assume C p,liq is not a function of T. Estimate,
Then h F
h L 200 q
h L 60 C
h
C P,liq .63 wt frac ~
T
60
CPL 200
Hv
hF
Hv
hL
h
60
46.1
20
0.691,
65
23
.864 200
60
q
0.691
q 1
0.309
kcal
0.864
80
hL 60 C
395 167.1 395
20
kg C
46.1 167.1 2.24
b. From Figure 2-4 at 50 wt% ethanol H v = 446 and hL = 70. Since CMO is valid obtaining both enthalpies at 50% wt is OK. The feed is a liquid
hF
CP,liq TF
CP,liq
Tref
CP,liq 250 0
CP,EtOH z EtOH
C Pw z w 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 Thus, zW = 0.719 and z E = 0.281
CP,liq
37.96 .281
18.0 .719
C P,liq in kcal kg C hF q
0.911 Hv
hF
Hv
hL
hF
4.D4*. a.
q Slope
25.91 kg/kgmole
H
kg C 446 446
CP
23.61
MWAVG
25.91
250 C 228 70
hF
H
h
25
0.6. y
x
228 0.58
H
300
z
0.911 . Then,
4.D4a
CPv 350 50
H
q q 1
kcal
23.61
y=x
25 300 z
1.5
feed line
0.6 is intersection. .5
q
L
L F where L
slope
q
q 1
L
F where L
b.
c.
q
L
L
0.6F. Then q
L
0.6F L / F
.6 0.6, and
.7
1.5 L
F 5. q
L F5 L F
1 5 , slope
q
q 1
16
76
4.D5*.
h liq
fL
h reflux
H vap
h liq
17500
3100
L0
L0 D
1.1
V1
L0 D 1
2.1
1 fc
L1 V2
3100 1500
L 0 V1
0.1111 0.524
1.1111 .524
1 f c L 0 V1
1
0.55
.111 .524
Alternate Solution H
h0
17500 1500
L0
H
h1
17500
q
Then,
1.1111 L0
L1
L1 V2
4D6.
L1
For subcooled reflux,
qL0
L1 L1
L1 D D
L1 D 1
L1
1.222
V2
2.222
L1
1.111 L0
D
D
a) 175
75 .6
Feed 2.
F2
x
q2 q2
y
z1
L
1
Lx x
L V
B
F2 z 2
2 3 through y
V B 1 V B
At feed 2, L
.4F
B
0, y
x
xB
Bx B
Vy
Bx B
L
y
x
V
F2 z 2
Bx B V
L V
V/B B
2B 181.25
271.875
L or L
V V 0.6F L V 0.961 x
0.4
3 2
Do External Balances and Find D & B. Then V
V
z2
L V 1 x B . Through y
, Slope V Also intersects bot. op. line and Feed line 2.
L
x
V
F2 z 2
0, y
0.4
.06
L V x
F2
D
0.6
0.4
Slope Middle
B
0.1 B 0.9D
60% vapor = 40% liquid q 2
Bottom Op. Line
1.2222
90.625 kmol hr
1, vertical at y
Slope feed line
When
F1
100 0.4
Solve simultaneously. D 84.375 and B b) Feed 1. q1
1.111 1.1
0.55
New Problem in 3rd Edition.
85
,
1.111
3100
40
L
181.25 9.625
241.25
0.4F 60
271.875
40
231.875
241.25
0.126 Plot Middle Op Line.
77
y
L V
Know that y Also,
L
L L V
x
1
L V
xD
x D and gives through interaction Middle and Feed line 1.
x F1
231.875 75
156.875 241.25
156.875 and V
V
241.25 ; thus,
0.65
c) See graph. Graph for 4D6.
78
L
4.D7*. a. Plot top op. line: slope b. Plot bottom op. line:
V
.8 , x
L
slope
y
1 1
xD
.9. Step off stages as shown on Figure.
V
2 , x y x B 0.13. Step off stages B V (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. 9 q slope gives q = 0.692. 4 q 1
4.D8*. The equilibrium data is plotted and shown in the figure.
q
0.692 and q
q 1
From the Solution to 4.D7c,
9 4
a. total reflux. Need 5 2/3 stages (from large graph) – 5.9 from small diagram shown. .9 .462 b. L V min 0.660 (see figure) .9 .236 L V min L D min 1.941 1 L V min c. In 4.D7, L D act
L D act
L V
.8
1 L V
.2
Multiplier
4
L D
min
Multiplier = 4/1.941 = 2.06
79
d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an equilibrium contact. Then use E MV
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. F1 F2 D
F1z1
a) B
F2 z 2
Dx B
L D L
1 2 L D
, D
Saturated Liquid Feed L L F1
Bottom – Normal: y
Bx B
80
B
F1 .42
F1
B 20
.66 80
V V 120 40 93.68 133.68, L V
1.114
113.68, F1
L
L D
V
1 40,
c) Top Op. Line – Normal: y Through y
F1
18
Solve simultaneously, B b)
100
x
L D V
L
L V x
93.68 12
1
3 2
3
D
120
1 L V xD
x D , Slope 1 3, y intercept
L V x
0.04 B
L V 1 x B , through y
Also through intersection, F2 feed line and middle op. line.
2 3
x
.66
.44
xB
Feed line F2 slope
L F2
.7
VF2
.3
80
Middle
y
Slope Also,
x
0, y
D
F1z1
(or do around bottom) V V L V . Through intersection feed line F1 and top op. line.
L V x
Dx D
xD
F1z1
80 .66
93.68 .42
0.11212 V 120 d)Opt. Feed F2 stage 1 from bottom, Opt feed F1 , Stage 2. 4 stages + PR more than sufficient.
Graph for 4D9.
81
4.D10*. Operating Line y
L V x
1
L
x D , where
V
Thus, operating line is y = .8x + .192 y a. Equilibrium is x or x 1 1 y
L
L D
4
V
1 L D
5
.8
y1 1.79 .76y1
Start with y1 = .96 = x D Equilibrium:
x1
Operating:
y2
Equilibrium:
x2
Operating:
y3
y1
.96
1.76 .76y1
1.76 .76 .96
.8x .192 y2
.8x 2
1.0 1.0
.9 .9406
.192
0.93736
.93736
1.76 .76y 2
b. Generate equilibrium data from: y x y
.8 .9317
0.9317
.192
1.76
.76 .93736
.8 .89476
.192
0.89476
0.9078
1.76x
1 .76x .8 .7 .8756 .8042
.6 .7253
.5 .6377
.4 .5399
Plot equilibrium curve and operating line. (See Figure). Slope = L/V = .8, y intercept (x = 0) = 0.192, y = x = x D = 0.96. Find x 6 = 0.660.
4.D11. a) Same as 4.D2 part g. q = 1.0668, slope feed line = 15.97. b) Top y L V
L V x L D 1 L D
Feed line: Slope
1 L V x D goes through y = x = x D = 0.99 0.6969 @ x = 0
q q 1 ,
y
y = (1-L/V) x D = (1-0.6969) 0.99 = 0.300
x
z
0.6
82
Bottom Op line:
L
yV
Sy M,S Bx B But y M,S 0 (Pure steam) With CMO B L
V
S
B
y
Lx
L V
x
L V
xB
y= 0, x = x B . 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.
L V
min
Slope
L D
0.99
0.57
0.99 L V
min
1
L V
0
0.4242 0.4242
min
1 0 .4242
0.73684
Actual L/D is 3.12 × this value.
83
4.D12. L V y
L V
x
L D
3 4
1 L D 1
L V
Bottom slope
slope. Top op line goes throug y
xD @ x L V
0, y
.25 .998
B
1245167
S
1044168
1.19
x
xD
0.998
0.2495 From Soln to 3.D9 or from graph.
1.169
Feed line is vertical at z = 0.6. Can also plot top and feed lines, and then find bottom from 2 points
y
0, x
xB
& 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.
84
85
4.D13. New Problem in 3rd Edition. a.) b.) See figure.
L D c.)
L D
L
0.665 0.95
V
0.30
MIN
L MIN
See Figure
V
L
2.0 L D
MIN
L
y intersect 1
L V
0.4385
1 L V
0.5615
1.5616 ,
yD
V
0.4385
0.95
L V
0.7808
L D 1
L D
1.5616
0.3709 . Top operating line y Goes through y
Bottom y
L V
L
x
Goes through y
V
x
0.6096
2.5616
x
L V
x
yD
1
L V
yD
0.95
1 xB
x B & 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. 0.85 0.025 d.) Slope bottom: See figure for parts c & d. L V 1.941 0.45 0.025 1 1 V B V L V 1.0625 . L V 1 0.941
86
Graph for problem 4.D13.
87
4.D14a.
88
New S.S.
External M.B. S = B Sys = BxB . Since yS = 0 (pure water)
B
S
4.D14b.
Two approaches to answer. Common sense is all methanol leaks out and x MA
McCabe-Thiele diagram: This is enriching column with z horizontal feed line is at x
x M, b
y intercept
L V
x
L
1
V
1 L V yD
Bottom operating line
y
0 . Intersection top op. line and
0 , which is also a pinch point. Thus x M,d
4.D15. New Problem in 3rd Edition. Saturated liquid. q
Top operating line y
ys
y D , Slope 13 L V x
1,
q q 1
L
L D
2
V
1 LD
3
0.6885
0.
0.2295 and y
0 also.
, feed line vertical @ z
x
L V 1 x B goes through y
.3 .
yD x
xB
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.
89
Graph for problem 4.D15.
90
4.D16*.
q
L-L
Hv
hF
F
Hv
hL
Using 32°F = 0°C as reference T, h F
Hv
hL
4033.4 Btu/lbmole.
. Approx.
at feed conditions.
.4 11369
.6 13572
12691 Btu/lbmole
For approx. temperature of feed stage, do bubble pt. calc.
y1
1
K1x1
K1z1
Pick T = 48°C (~ 40% of way between boiling pts.)
K C5
1.5, K C6
K C5 Tnew =
.54 .92
K C5 50 C Hv
.54,
=.594, Tnew
1.6,
Hv
.92
.99 Close enough.
122 F
CPfeed,Liq 122
46.9 122
5721.8
46.9 is from Prob. 3-D6. 5721.8 12691 18412.8 Btu/lbmole
Hv q
.54 .6
1.6 .4 .584 .6
, 50 C
feed
1.5 .4 50 C
K1 x1
h L 50 C
Note : C P feed,liq
K1 x1
HV
hF
HV
hL
18412.8
4033.4
12691
1.133
Note: h F is from Prob. 3.D6. 4.D17.
L
Top Op. Line: y
V
x
L
1
V
L
L D
7 2
V
1 L D
9 2
x
0, y
1
L V
x D , goes through y
x
xD
0.9
7 9 2
xD
9
.9 =0.2
Plot Top. Step off 2 stages. Find x S ~ 0.81 The vertical line at x
xS
0.81 is the withdrawal line.
Bot. Op. Line intersects Top at x
xS .
Also know it intersects feed line at x
F
External Balances
Fz
Dx D
D
Bx B
B
x B (unknown) S Don’t know D,B, or x B .
SxS
Feed enters as saturated vapor. Thus q
0&V
F
Bottoms leaves an equilibrium contact, Do flow balances
it is saturated liquid L
V
F
100
V
V
100 since S is removed as saturated liquid.
B
91
L
L V
V
L
L S
B
L
xB
Fz
7 9 100
77.777. D
77.7777 15
V
L
62.7777. L V
100
77.777
62.7777 100
22.222
0.6278
62.777 Dx D
Sx S
60
22.222
B
0.9
15
0.81
0.444
62.7777
Plot. Op. line Step off stages. 9 is more than sufficient.
4.D18. New Problem in 3rd Edition. Feed F1 : z1
0.6, saturated liquid,
q
1, q / (q 1)
92
Feed F2 : z 2
0.4, 80% vapor hence 20% liquid q q
.2
q 1
LF / F
0.2F / F
.2
14
.8
Part a.) Bottom operating line goes through point, y
x
xB
0.04
Max L V to point intersection feed F 2 line and equilibrium curve. Slope
L
1.0 .04
V
.47 .04
max
V
V
B
L
min
1
1
V
L V 1
1.2326
100
.2 80
Part b. L
F1
100
L
L
LF2
L
V
B and V B 1.5
116
L
1.5B
V
L
B
D
V
V
L
B
116
VF2
100
2.2326
116
116
B
46.4
2.5
46.4
0.8113
69.6
116
L V
69.6 .8 80
1.66667
69.6
133.6
0.7485
V 133.6 Check overall balance
F1
F2
D
100
.6
B
180
133.6
80 .4
46.4
46.4
180.0 OK
To find y D use MVC mass balance
F1z1 or
yD
F2 z 2 F1z1
DyD F2 z 2
Bx B
D
133.6
Actual bottom op. line: y
L V Goes through
Bx B
V
B V
y
x
L V
L
x
V
0.675
1 xB
V B 1
2.5
5
V B
1.5
3
xB
0.4
0.04 , Slope
5 3
nd
2 point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) Plot bottom op. line Dy D F1z1 L Top Op. line: yV F1z1 Lx DyD . y x V V Goes through intersection feed line for F 2 and bot. op. line. Does NOT go through y x yD . Since D & F, passing streams, Point
z1 , yD is on op. line.
93
Figure for 4D18 4.D19*.
B = 0. Then from external balance F = D + B must have D = F = 1000. Acetone balance becomes Fz Dx D or x D z 0.75 . L
To predict x B need operating lines. Top: y
V
L
L V
2
V
1 L D
3
x
and y
L
1
V x
xD xD
.75
94
Bottom: L V
1.0 . Thus y = x is operating line. From Figure x B
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:
Weight Fractions: Feed:
Weight Fractions: Bottoms:
Weight Fractions:
0.8 ETOH = (.8)(46.07) = 36,856 0.2 Water = (.2)(18.016) = 3.6032 Total = 40.459 EtOH = .911, Water = .089 0.32 (EtOH) = (.32)(46.07) = 14.7424 0.68 (W) = (.68)(18.016) = 12.25088 Total = 26.993 EtOH = .546, W = .454 0.04 EtOH = (.04)(46.07) = 1.8428 0.96W = (.96)(18.016) = 17.295 Total = 19.1378 EtOH = .0963, W = .9037
Condenser Energy Balance is V1H1
Lo From chart:
Qc
Qc H1
Lo h o
Dh D which can be solved for Lo D .
1
D
D ho
hD
54 Kcal/kg and H1
285 Kcal/kg
Need D in weight units. Convert feed to weight units. 100 kgmoles Ethanol: .32 46.07 hr Water: (100)(.68)(18.016) = 1225 kg/hr Total: F = 2699.328 kg/hr
1474.24 kg/hr
95
F z
xB
2699.328 .0963
xD
xB
.911 .0963
Then,
D
Then,
Lo D
Q D ho
H1
1489.93 kg/hr
2,065,113
1
1489.98 54
285
1
5.0
Now do usual McCabe-Thiele analysis using molar units. Note L o D is the same in mass and molar units. Top Operating Line: L
5
; x
y
L
y xD
V
x
1
L V
.8, y int ercept
V 6 Feed Line: Goes through y = x = z = .32 Weight fraction of feed = .546. Then, h f
q
HV
hF
430.7 15
HV
hL
430.7
Bottom Operating Line:
x
L
x
L
V V intersection top operating line and feed line.
L
L D
V
1 L D
0
L
1
V
15 kcal/kg, H v
1.149 Slope
69 y
x D and
xD
1
5 6
.8
430.7, and h1k1
q
1.149
q 1
1.149 1
1 x B . Goes through x
.133
69 .
7.711
y
x B and
From Figure need about 8 equilibrium contacts including a reboiler. Stage 1 above reboiler is the optimum feed stage location.
96
4.D21. Feed 1:
q1
0 , slope feed line = 0
Feed 2:
q2
0.9 , slope
Top:
q2
q2 1
L
L D
1.375
V
1 L D
2.375
Goes through y
0.9 L
0.579 , y
V
x D , When x = 0, y
x
0.1 x
L
1 L
1
V
xD
V
Since F1 is saturated vapor, V
Bottom:
9
F1
xD . 0.40
100 kmoles/hr
L
.1 F2 At feed F2
D B
L V V
L
L
L 0.9 F2
0.579 108
V L 108 45.46 F1 F2 D 100 80
But B is saturated liquid. Check
V
0.1F2
100 8 108
V
.9 F2
L
V
L .9F2
L
B
62.532 45.46
134.532
134.532
62.532
0.9 80
134.532, OK
Draw top op line. Intersects with F2 feed line. Then draw bottom op line with slope
L V
1.3453 .
Intersection bottom op & q. line gives x B Check
F1z1
F2 z 2
Dx D
B Check External MB 180 F1
F1z1
56
F2 z 2
F2
x B or x B
D Dx D
20.0 36.0
B
0.09 . 20
36
43.187
134.532
0.095 , OK
45.46 134.53 179.99 , OK
Bx B
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
97
4.D22*. Around top of column mass balances are: L Solving,
L
y
For pure entering water, x W
x
D
yD
D
V
C and Lx
DyD
Vy
Cx W
C
xw V V V 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
0.75, y x
0
0.17, y x
1
0.92
yD , y
x
0.68 Not yD
98
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,
H
h c h
c
L V
hW
h
hW C C p
C
T
C
18 212 100
H h C c 1.1154C D c D 0.1154C
17465.4
In addition, C/D = ¾ or D/C = 4/3. L 1.1154C 1.1154
0.1154C
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
y
x
When
x
0, y
1 L V yD
L
L
Bottom Op Line:
y
V
yD
x
0.85
V
0.20
1 xB
y x x D 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) Through
Note – Commercial columns usually operate much closer to minimum reflux ratio and have many more stages.
b.)
L V
0.85 min
0.376
0.85
0
0.558 ,
L V
L D
min
1
Min
L V
1.26
Min
c.) Total reflux 5 stages + PC sufficient or 4 3 4 equil contacts + PC = 5 3 4 eq. contacts where fraction
ab
7.6mm
ac
10.3mm
0.74 or .75
99
100
4.D25.
a.
99.9% methanol is essentially pure. Eq. (4-66)
V2
For pure MeOH, CPL
C PL fc
1
L1 1
fc L o D 1 fc Lo D
Pure MeOH boils 64.5°C. where f c
CPL TBP
Treflux
5 0.07586 16.83 10 T , average (40 + 64.5)/2 = 52.25°C
0.084654 kJ gmole , 0.084654 24.5 35.27
MeOH
35.27 kJ mole , TBP
0.058804,
1.058804 1.2
L1 V2
64.5
1
1.058804 1.2
0.5596
101
b.
L
L D
1.2
0.5454 or 2.59% more reflux with 24.5°C cooling!
V 1 L D 2.2 subcooling not important.
4.D26.
a) 50% feed: q
F
L
q L
20
L 1
F
20
L
F, L
, Slope =
L
amt vaporized
q
1 20
0.05
q 1
1 20 1
1.05
L
Usually
F 20
0.0476
35% feed: Saturated liquid – vertical feed line. Plot both feed lines. The one with lowest intersection point with equilibrium curve will normally control L V
Find
L V
max
. Then
1 0.1
slope
max
V B
0.46
b)
0.1
L V L V L
Bot. y
V y
L
x 1,
x y
x
V
min
L
V
2.497 ,
3
V B
V
B
1
L V
1
max
V
1
B
2.497 1.0
min
0.6681
2.0043
min
V B 1
3.0043
V B
2.0043
V
min
1
1 x B . Goes through y
V L
V
V B
L V x B or x
x
slope
1.49892
x B with total reboiler
1
L V 1 xB / L V
1
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), F 50% and B.
yV
Lx
L
y
F50% z50% F50 z 50
x
Bx B
Bx B
V V Intersects bottom operating line & 50% feed line. @ x 0, y (F50 z50 Bx B ) / V , Slope
q50
For 50% feed, External balances: 250
F1
0.05 F2
L V
(L L ) / F50 .
D
B and F1z50
F2 z35
DyD
Bx B
V
440.63
Find D = 103.333 and B = 146.666 moles/min
V B
Since
2.0034,
Then from q 50% : L
V
L
F50
B
V
q 50 F50
293.96 and L
L
0.05 100
445.63 100 146.666
y intercept (x = 0), y
[ 100 .5
B
440.63
398.964 , Slope
146.666 0.1 ] / 398.964
445.63 L V
1.11698
0.0885
Top Intersects feed line for 35% feed and middle op. line and goes through y x y D 0.85
102
EQ
Stages
0.75 = E ML =
actual
actual change liquid
op
change at equilibrium eq. change
Figure for 4D26
103
4.D27*.
Top Op. Eqn:
Feed: q
L
y
V
x
L
L D
1
V
1 L D
2
L
L
L
F
F
Middle: V
B
L
1
,
1 4
xD
V
y intercept F
Does not intersect at y
S, V y
x
V
xD
L 5 4,
F
L
L
1
Bx B
Lx
x B or at y
slope
Sy s y s
0, x
.46,
y
q
5 4
q 1
14
0
or y
L
x
xD
5, y x
x
.92
z
.48
Bx B
V V x B . Does intersect top op. line at feed line.
Need another point. Bottom:
L
L
1 xB ,
L
V
B
V B 1
1.5
3, y x x B 0.08 V B V B .5 V V The steam is another feed to the column: Sat’d vapor q = 0, q/(q-1) = 0, y = x = z = y s = 0. Middle Op. Line intersects this steam at bottom op. line (see figure). y
x
This problem is a two-feed column with the lower feed (steam) input at a non-optimum feed rd stage. Otimum feed is 3 above partial reboiler. Need 5.6 equilibrium stages plus PR.
4.D28.
nd
This problem was 4.D35 in 2 edition.
Stripping Section: y
L V
x
L V
1 xB, y
x
xB
0.02 . Feed line is vertical
104
L V
1.0 max
1.24,
0.81 0.02
V B 1
Op line y
0.02
1, x=
1.5 L V
V B
V B
V L
min
6.25, L V
V
V
1
L V
.16
.02
1.16
L V
B V
min
1 xB
1
1
max
1
0.24
V B 1
6.25 1
V B
6.25
4.167
1.16
0.865 . Intersection Op & feed lines is y D
Overall Balance: 10,000 = F = D + B
6000 D
Fz
5800 / 0.675
Dy D
Bx B
D .695
8592.6 kgmoles/day, B
B .02
6000
.695D .02D
200
1407.4 . Need a use for impure distillate.
Figure for 4.D28
105
4.D29.
L D
3,
L
3 4,
V
L
1
V
1
xD
4
0.9
Step off 3 stages on top op line. Find x S
0.225
0.76 . Point on top op line at x S
0.76 is also
on middle op. line
xD
3
xs, S = 15
V´
L´
F = 100
F=D+B+S
6
z = .6
Fz
Dx D
Bx B
SxS
Solve simultaneously D = 50.125, B = 34.875 10 x6 = .1
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
L V
Middle op. line slope Middle Op line: yV
Lx y
135.375 200.5
SxS L V
x
Dx D Sx S
Dx D V
Feed z = 0.6, 20% vapor = 80% liquid Feed line slope q
q 1
0.8
0.6752
0.4
, when x
0, y
11.4
45.1125 200.5
0.2819
q = 0.8.
4
106
4.D30*.
a.
Subcooled Reflux:
Lo
Lo D
3
V1
1 Lo D
4
, L1 4
c
1 3
Lo
500 ,
4 Lo 3
L1 V2
V1
Substituting in values, we have Step off two stages.
x2
1 3
3
Lo
1
Lo
c and V2
1 Lo 3 V1
1.0
1
4
L2
1 14
5 4
5
xS
c
L o V1
L1
0.62
V1
, Top op. line
y
x
xD
0.92
yS
107
Mixed feed to column:
F S
Solving for mixed feed, z M
1500 , Fz SyS
FM
0.52667
Energy balance for mixed feed, Fh F Since H s
h F (sat’d vapor), h FM
External Balances: F = D + B, Fz Lo
3D
Middle: L
1500, and L
L S
FM z M
4 3
Lo
2000 500
SHS HS
FM H FM
h F (sat’d vapor), and q FM
Dx D
0 (horizontal).
Bx B , Solving simultaneously D = 500 & B = 500
2000 (subcooled reflux), V
1500 , V
V
L
2500 , Slope
D
2500
L
1500
3
V
2500
5
.6
Intersects Top Op. line at x S Plot Bot. from y
x x B to intersection feed line and middle Step off stages (see figure). Need 4 8/9 equilibrium stages. b. Mass balances for mixed feed injection: V
V
V
FM
2500 1500
FM
V
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.
108
Figure for problem 4D31 4.D32*.
External Balance: F = B = 50, and Fz
4.D33*. a.
L V
.75 .452 min
L
b.
L D
act
L V
1.318. L V
Feed:
L
L
min
L D 1 L D
Top operating Line through y Bottom through y
0.4 .
0.659
1 L V
min
z
0.397 (tangent pinch)
.75 D
Bx B . Thus x B
0, x
x
yB
F .25F, q
0.569
yD
0.75
0.1 and intersection feed and top operating lines. 5 4 , slope
q
q 1
5
rd
Optimum feed is 3 from bottom. Need 9 real stages plus partial condenser (see figure). c. From figure slope of bottom operating line
L V
Since saturated steam and CMO valid, B S Also have mass balances, S + F = B + D
SyS
Fz
Bx B
DyD ys
2.025
L V
0
Solve 3 eqs. simultaneously. S = 760 lbmoles/hr = 13,680 lb steam/hr.
109
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) L V
0.95 min
0.95
0.613 0
0.3547 ,
L
L V min
D
1 L V min
min
0.5497
110
L
L
2
D
1.0994 ,
D
act
min
External M.B. F = D + B, Fz or Eq. (4.3) D
z
xB
xD
Dx D
F
xB
L
L D
V
1 L D
act
0.5237
Bx B
0.6
0.025
0.95
0.025
100
62.162
D
kgmol
L
h
, B = F – D = 37.838
L D D
68.030
L
V
V = L + D = 130.192
F V´
L´
″
V
″
L
P=
xP
At feed V
V (sat’d liquid)
L
168.030
L
F
L
V
B
Top op line y 1st middle
L V
x
y
1 L F x D normal . At x = 0, y = 0.4525
L V
L
x
V
1 x B looks like usual bottom!
Goes through y Slope At pump-around return, V
x
x B , and intersection top & feed line.
L
168.030
V
130.192 130.192
V
1.2906
L
L
P
V
V
130.192 , L
Check at bottom
L
V
B or
Bottom Op line
y
At pump-around removal,
Step off P.R. stage 1 & 2 above.
L V
x
208.030, L V
L V
L
P
208.030 130.192 L
130.192 168.030
1.5979
168.030 37.838 , OK
1 x B , Same as first middle!!!
x P is liquid from stage 2, x P
0.335 . Vertical line at x P is
withdrawal line for pump-around and it is feed-line for return of pump-around. 2nd middle op line
slope
L V
intersects x P withdrawal & feed line where bottom & 1 st middle intersect.
111
Using MB:
yV
When
y
4-E2*. Feed:
q
Px P
Bx B P
0, x
Bx B
V
P
x
40
V
xP
L F
, L
L 1.5F, q
.335
Bx B V
32838
0.025
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)
L
xP
L
L x, y
3 2, slope V and B
Bottom op. line: Since steam is saturated vapor S Thus, 1 S B
L V
Middle op. line: V
1.2 . Operating line goes through y B
Vy Since
yS
Side
Bx B
0 this is, y
L
S or V
Side x side L
x
Bx B
L
Lx
q q 1
3
L
0, x
xB
0.015
S B Side
SyS
Side x side
V V Side stream is removed as a saturated liquid so q = 1. Step off two stages (see figure) and find x side 0.0975
112
Find slope:
V
L
Side
V
S
B
S, B
L
Side
Side B 1
0.4 1
S B
0.833
1.68
slope
Draw in the middle operating line. Step off 4 stages. Trial and error to find x D
.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
L
L D
1.86
V
1 L D
2.86
1
L V
xD
.35 .8
Step off 8 stages and find x S
0.495
0.650, y .28, y
L V x
x
1 xD
L V
xD
0.8
yS . 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 withdrawal and above vapor stream from intermediate reboiler. Can now calculate flows in each section of columns. Overall Balance:
Fz
Dx D
F D xB
113
D L
Flows: L
D
D
F z
xB
.3 1000
xD
xB
.78
716.1, V
L
L
L S
258.8, L V
L
L
L
D
F
1101.1
385 V
V ,V
V
S
643.8
0.235
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 x x B 0.02 . Check if slope L V 1.955 . Optimum feed is 10 below condenser while vapor from intermediate reboiler is returned on 11 th stage. Need 12 ½ stages. Note: Small differences in stepping off stages may change column geometry. operating line with line y
y s to y
th
rd
4.E.4. New Problem in 3 Edition. External Balance. a) F D B, Fz Dx D Bx B D b)
z xD
xB xB
F
.25 .025 .9 .025
V B 1.0, V
B
100
25.7 kmol h , B
74.3 kmol h
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 10 V V C 74.3 10 64.3
38.6
114
At stage 2
L
L
L
L0
L
L R and V
64.3
21.4
21.4
0.333
V 64.3 L R L L
y
c) Top Op. line:
V
38.6 L V
13
21.43 L R
x
D
L
xD
V
D
17.17 kgmoles hour
y
V
x
x, y
L0
L
1
V
L R
D
V
L
xD
xD
x D sin ce V
L
L R
D
Envelope for top
V
L
L R
Slope
L V
1 3, y x
x
xD
0,y
1
0.9 L V
xD
0.6
Plot Middle:
y
Bottom:
y y
L V
x
1 L V x D . Slope
L V x x
Slope
38.6 64.3
0.600 y
x
xD
0.9
L V 1 xB
xB L V
V B 1
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.
115
Graph for 4.E.4.
L
4.E.5. New Problem in 3rd Edition.
Bot. Op. Line: External M.B.: F
D
D
V L
y
V
V
B V
L
x
V
B and Fz
V
1
2.25
9 4
9
V B
1.25
5 4
5
1 xB, y
Dx D
Bx B ,
250 kmol day , B
750 kmol day , V
L
750
V
B
At feed stage: L
687.5
V
937.5
937.50 V
V
x
x B , Slope D
z
F
xD
xB xB
V B B
L V .3
0.10
0.90
1.25 750
0.10
.2
1
.8
4
937.5 kmol da
1687.5 kmol day
937.5, L
L
F or L
L
F 1687.5 1000
687.5
0.733333
116
MB:
Lx
Goes through y
Dy D
V y,
D
V
L , y
L V
x
1
L
yD
V
y D , and intersects feed and bot. op. lines.
x
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 L Dy D Sx S L x Dy D V y Sx S or y x V V Find from intersection L V top op line and x
yS plus slope
687.5 737.5
0.932 or intersection
xS
At side stream feed point: L Top op. line:
op line @ y
y
L V
x
L 1
S L V
487.5, V yD
V
737.5. Thus,
Can draw, yintercept
L
487.5
V
737.5
1 .661 .9
0.661
0.305
117
Graph for 4.E5.
4-F1.
970.33 L
Since bottoms are very pure h B
HS
v
H Equil
h water @ 212°F
1381.4 1192
(in column)
Extra heat 189.4 Btu/lb S
B
Since y ≈ x, MW are same
118
