SOLUTION MANUAL
for
SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Analysis 3rd Edition (Formerly published as Equilibrium-Staged Separations) by Phillip C. Wankat
SPE 3rd Edition Solution Manual Chapter 1 New Problems and new solutions are listed as new immediately after the solution number. These new problems in chapter 1 are: 1A3, 1A4, 1B2-1B4, 1D1. A2.
Answers are in the text.
A3.
New problem for 3rd edition. Answer is d.
B1.
Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse osmosis or by distillation).
B2.
New problem for 3rd edition. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually adsorption), or a reverse osmosis system.
B3.
New problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.
B4.
New problem for 3rd edition. You probably used some of the following: chromatography, crystallization, distillation, extraction, filtration and ultrafiltration.
D1.
New problem for 3rd edition. Basis 1kmol feed.
.4 kmole E .4 MW 46 18.4 kg
.6 kmol Water .6 MW 18
10.8 kg 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.
17
Chapter 2 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 2A6, 2A9 to 2A16, 2C4, 2C8, 2C9, 2D1.g, 2.D4, 2D10, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3. 2.A1.
Feed to flash drum is a liquid at high pressure. At this pressure its enthalpy can be calculated as a liquid. eg. h TF,Phigh
c p LIQ TF
Tref . 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 T F 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. 1.0 Equilibrium (pure water)
yw
zw = 0.965 Flash operating line
.5
2.A4 0
0
.5
xw
1.0
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.
18
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. b. The answer is
The answer is 36ºC
3.5 to 3.6
2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from the amount of superheat. 2.B1.
Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium. Examples: F, z, Tdrum , Pdrum F, h F , z, p F, TF , z, p
F, z, y, Pdrum
F, TF , z, y
F, h F , z, y
F, z, x, p drum
F, TF , z, x
etc.
F, z, y, p drum
F, TF , z, Tdrum , p drum
F, z, x, Tdrum
F, TF , y, p
Drum dimensions, z, Fdrum , p drum
F, TF , y, Tdrum
Drum dimensions, z, y, p drum
F, TF , x, p
etc.
F, TF , x, Tdrum F, TF , 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. 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. 2 2 Fexisting D exist 4 2 Feed rate drum can handle: F α D. gives 1000 .98 .98 Fexisting 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
19
y = .58, V = .25 (16660) = 4150 16,660
25,000
LTotal x
8340
Since x is not specified, use bypass. This produces less vapor. c) Look at Eq. (2-62), which becomes
V MWv
D
3K drum 3600
L
v
v
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.
2.C5.
a. Start with
V
zA
zB
F
KB 1
KA 1
xi
Fz i L VK i
Fz i
xi
Then y i
From
Kixi
yi
and let V
L
L F
xi
F L Ki
F L or x i
L F
zi 1
L Ki F
K i zi L 1 Ki F
0 we obtain
K i 1 zi L F
L 1 Ki F
0
20
zi
2.C7.
1 V/F f
0 0
Ki 1
.1 -.09
.2 -.1
V F
1 f
.3 -.09
V F
From data in Example 2-2 obtain:
.4 .5 -.06 -.007
.6 .07
.7 .16
.8 .3
.9 .49
1.0 .77
2.C8. New Problem.
p drum F=L+V
Tdrum
Fz y
x
Lx Vy
Solve for L & V Or use lever arm-rule
z
21
2.C9.
New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,
F
V L1
Fzi
L2
L1x i,L1
L2
x i,L2
Vyi
Substituting in equilibrium Eqs. (2-60b) and 2-60c)
Fz i
L1K i,L1
L2
x i,L2
Solving,
x i,L 2
L 2 x i,L2
VK iV
L2
x i,L2
Fz i L1K i,L 2
Fz i
L2
VK i,V
L1K i,L1
L2
F V L1
L2
VK i,V
L2
Dividing numerator and denominator by F and collecting terms.
x i,liq 2
Since
yi
K i,V
L2
x i,L2 , y i
2.D1.
a.
V
i 1
K i,L1
V
L2
x i,liq1
0.4 100
Slope op. line See graph. y b.
K i,L1 x i,L 2
L2
1
K i,L1
L2
C
1,
i 1
L2
yi
0
L 1 1 F
1
F V
K i,V
V F
L2
1
V F
C i 1
L2
C
yi
i 1
x i,L 2
1
L2
L 1 1 F
0
0
V 1 F
(2-62)
K i,L1 L 2 z i L K i,L1 L 2 1 1 K i,V F K i,L1 L 2 1 z i
K i,L1
K i,V
L2
L2
1
V 1 F
V F (2-63)
60 kmol/h
L V 3 2, y x 0.77 and x 0.48 600 and L
1
1 , thus,
C i 1
L2
1 zi
x i,liq 2 , we have x i,liq1
x i,liq 2
K i,V
K i,V L 2 zi L 1 1 K i,V F
L2
K i,V
40 and L
0.4 1500
c. Plot x
i 1
K i,L1
C
which becomes
In addition,
1 C
Stoichiometric equations,
Since x i,liq1
1
zi L 1 1 F
z
0.6
900 . Rest same as part a.
0.2 on equil. Diagram and y
V F z 1.2 0.25 . From equil y d. Plot x 0.45 on equilibrium curve.
x
z
0.3. yint ercept
zF V 1.2
0.58 .
22
L
Slope Plot operating line, y e. Find Liquid Density.
MW L Then, VL
xm
V z at z
x
x m MWm MWm
F V
xw
MWw
.2
w
L
v
RT
.2 32.04 32.04 .7914
MW L VL
P MW
v
.8
4 V V F .2 0.51 . From mass balance F 37.5 kmol/h.
x w MWw
m
Find Vapor Density.
1 V F
.8
.8 18.01
18.01 1.00
1 atm 26.15 g/mol
22.51 ml/mol
20.82 22.51 0.925 g/ml
(Need temperature of the drum)
MW v y m MW m y w MW w .58 32.04 .42 18.01 Find Temperature of the Drum T: From Table 3-3 find T corresponding to y .58, x 20, T=81.7 C 354.7K v
20.82
82.0575
ml atm mol K
354.7 K
26.15 g/mol
8.98 10
4
g/ml
Find Permissible velocity:
23
u perm
K drum
L
K drum
v
exp A
v
B
nFlv V
Since V
Wv L
F V
1000 250
15615
WV
L
6537.5
V MW v
A cs
u perm 3600 Thus, D
MW L
nFlv
lb
4
nFlv
20.82
15, 615 lb/h,
v
14.19 3600 8.98 10
4A cs
2.598
14.19 ft/s
250 26.15 454 g/lb 4
2-60
4
0.0744, and n Flv
.925
8.98 10
E
6537.5 lb / h
lbmol
4
4
3
250 lbmol/h,
750
8.89 10
.925 8.98 10
.442
D
250 26.15 L
V
2
0.25 1000
V MW v
WL
.442, and u perm
nFlv
F
F
750 lbmol/h, and WL
Flv Then K drum
C
g/ml 28316.85 ml/ft
3
2.28 ft 2 .
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, x 0.34, y 0.69. This problem is now very similar to 3-D1c. Can calculate V/F from mass balance, Fz Lx Vy. This is V z y 0.4 0.34 Fz F V x Vy or 0.17 F y x 0.69 0.34 g. Part g is a new problem. V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756. 2-D2.
Work backwards. Starting with x 2, find y2 = 0.62 from equilibrium. From equilibrium point V plot op. line of slope This gives L V 2 1 V F 2 3 7. F 2
z2
0.51 x1 (see Figure). Then from equilibrium, y1
For stage 1, 2.D3.
0.4
z1
x1
0.55 0.51
F
y1
x1
0.78 0.51
V F 0.6 V L y x z V F V Op. eq. 2 y x 2 3 3 See graph: y 0.55 x M 0.18
a.
z
V
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,
y
Lx Vy
4.0
0.665,
F V x Vy
10 V 0.3 0.665V
V 2.985 and V F Can also calculate V/F from slope.
c.
V
F 10,
y If y
L V
F
x
0.3
V
z
7
V F
z
0.8, x Then z
3&L
x
0.2985
7
z 0.3
0.545 @ equil 0.3 0.8
7
0.545 0.6215. 3 7 Can also draw line of slope through equil point. 3
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
yi K i
1.0
K ref TOld K bu tan e
.2
.35
yi K i
1.0,
41 C : K C3
T
yi
.2
.35
.45
Ki
8.8
4.0145
.9
4.0145 .6099
yi
.2
.35
.45
Ki
6
2.45
.44
K C4,NEW
3.1,
K C6
0.125
.45
4.0145 T too low K i 3.1 1.0 .125 Guess for reference K C4 4.014 T 118 C : K C3 8.8,
K C4,NEW
0
yi x i
xi
Want
V F
2.45 1.2 .2
.35
.45
Ki
6.9
2.94
.56
.9
0.6099
2.45, T
85 : K C2
6.0,
K C6
0.44
1.20
2.94, T
yi
K C6
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) v1 = F2 v2
y1 = z2 z1 = 0.55
y2 1
2
F1 = 1000
V F
L
y1
V1
V1
V1
z Plot 1st Op line.
V
0.45454 & L1
V1
F1
1000
V1 = 687.5 kmol/h = F2
V
c) Stage 2 At x
F
0, y
0.25 ,
z V F
From graph y 2
V2
V F
F2
y1 = 0.66 = z2
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
0.25 2
x2
x1 = 0.30
b)
p1,2 = 1 atm
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.
V F = 1.0 kmol/min
T = 50ºC P = 200 kPa
zc4 = 0.45 zc5 = 0.35 Zc6 = 0.20
K i 1 zi 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
df k
F
i 1
V k 1
Fk
0.0215
2
K i 1 zi
c
d V F
V
Kc6 = 0.30
L
RR eq.,
f1
Kc4 = 2.4 Kc5 = 0.80
1
Ki 1 V F
2
fk df d V F
28
df1 V d F V
F2
f2
1.4 2 0.45
0.2 2
1 1.4 .6
0215
0.6
2
0.35
1 0.2 0.6
0.7 2
2
0.20
1 0.7 0.6
2
0.570
0.6377
0.570
1.4 .45
0.2 .35
0.7 0.2
0.00028 1 1.4 0.6377 1 0.2 0.6377 1 0.7 0.6377 Which is close enough. yi K i x i zi 0.45 x c4 0.2377, V 1 1.4 .6377 yc4 2.4 0.2377 1 Ki 1 F 0.35 x c5 0.4012, y c5 0.8 0.4012 0.3210 1 0.2 0.6377
x c6
2.D7.
0.20 1 0.7 0.6377
xi
V
zA
zB
F
KB 1
KA 1
KM
5.6 and K P
1.0002
0.3
0.7
F
0.21 1
5.6 1 zM
0.30 0.4012
yi
0.9998
0.3
xP
1 xM
0.1466 V 1 4.6 0.2276 1 KM 1 F 5.6 0.1466 0.8208 0.8534 , y M K M x M
yP
1 yM
0.1792
Use Rachford-Rice eqn: f
Converge on V F
0.1084
0.2276
K i 1 zi
V F
1
Find K i from DePriester Chart: K1
.076, V
Ki 1 V / F
73, K 2
F V F
4.1 K 3
152 kmol/h, L
0 . Note that 2 atm = 203 kPa.
.115
F V 1848 kmol/h .
zi
we obtain x1 .0077, x 2 .0809, x 3 .9113 V 1 Ki 1 F From yi K i x i , we obtain y1 .5621, y 2 .3649, y3 .1048 Need hF to plot on diagram. Since pressure is high, feed remains a liquid h F CPL TF Tref , Tref 0 from chart From x i
2.D9.
, y8
0.21
V
Eq. (2-38) x M
2.D8.
0.3613
0.5705
CPL
CPEtOH x EtOH
CPw x w
29
Where x EtOH and x w 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 0.6512 100 Avg. MW 0.1435, x w 0.8565 . 22.046 Mole fracs: x E 4.536 4.536 Use CPL at 100 C as an average CP value. EtOH
C PL Per kg this is
37.96 .1435
18.0 .8565
C PL
20.86
MWavg
22.046
0.946
20.86
kcal kmol C
kcal kg C
h F 0.946 2000 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 Lx Vy which gives V = 326.9, and L = 673.1
Note: If use wt. fracs. CPL
23.99 & CPL MWavg
1.088 and h F
217.6 . All wrong.
30
2.D.10 New Problem. Solution 400 kPa, 70ºC
K C3
From DePriester chart Know y i
Kixi ,
xi
zi 1
Ki
V 1 F
5, ,
z C4
35 Mole % n-butane
K C4
1.9,
xi
yi
K C6 1
x C6
0.7
0.3
zi
K i 1 zi 0 z C3 1 z C6 z C 4 .65 z C6 V 1 Ki 1 F zC6 zC6 V For C6 0.7 z C 6 0.7 1 0.7 V V F 1 K C6 1 1 0.7 F F V z C6 0.7 0.49 F 4 .65 z C 6 0.9 .35 0.7z C 6 RR Eq: 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. R.R.
2.D11.
Use Goal – Seek to find V F. V 0.594 when R.R. equation 0.000881 . F V z C6 0.7 0.49 0.7 (0.49)(0.594) 0.40894 F L F 0.6 V F 0.4 & L V 1.5 Operating line: Slope 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 .7914 L
Find Vapor Density.
p MW v
RT
v
.8
.8 18.01 18.01 1.00
MW L VL
20.82
22.51 ml/mol
20.82 22.51 0.925 g/ml
(Need temperature of the drum)
MW v y m MW m y w MW 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 26.15 g/mol
v
82.0575
ml atm mol K
354.7 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
B
0.25 1000
V MW v
1000 250
exp A
nFlv
250 26.15 lb lbmol
V
15615
WV
L
6537.5
A cs A Cs
MW L
8.98 10
0.5525
8.98 10
u perm 3600
Or y c
AT
D
2.D14.
xc 1 xc
cp
Raoult’s Law: K C 4
4
4
A Cs 0.2
1.25
20.82
15, 615 lb/h,
2.598
17.74 ft/s
4A T
4
g/ml 28316.85 ml/ft 3
9.12 ft 2
3.41 ft and L 13.6 ft
1.76 .7 1 .76 .7
1
,
cp
0.80418
0.5682
pc
VPC 4 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
17.74 3600 8.98 10
v
1.824 ft 2 ,
1 yp
3
250 26.15 454 g/lbm
2.D13. New Problem. The answer is ycresol = 0.19582 xp Since x c 0.3, x p 0.7, y p 1 1 xp
yc
4
0.925 8.98 10
V MW v
With L/D = 4,
nFlv
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 WL
Flv
C
zi 1
Ki 1 V F
1.0
33
0.3 0.7 1 11121 1844.36 1 1 0.4 1 1 0.4 P P Solve for Pdrum = 3260 mmHg zi xi V 1 Ki 1 F .3 11121 x C4 0.1527, y C 4 K C 4 x C 4 0.1527 11121 3260 1 1 .4 3260 1844.36 x C6 1 x C 4 0.84715, y C6 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: 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 K nC 4 1 z nC 2 0 V V V 1 K C2 1 1 K iC 4 1 1 K nC 4 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. K C2 1 .8 K iC 4 1 K nC 4 1 1.8 z nC4 z nC 4 z nC 4 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 F F K C2 1 V 1 K C2 1 F Thus, z nC 4 K C2 1 .8 K iC 4 1 K nC 4 1.8 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 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.
0.52091
1 V 1 F
0
1 1
V F
z C1 0.5, z C4 0.1, z C5 0.15, z C6 0.25, K C1 50, K C4 .6, K C5 .17, K C6 1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom V / F 1 .5 .1 / 3 .53 This first guess is not critical.
0.05
34
R.R. eq.
f
K i 1 zi
V F
1
49 .5 1 49 .53 Eq. 3.33
V F
.4 .1
.83 .15
.95 .25
1 .4 .53
1 .83 .53
1 .95 .53
f V F
V F
2
1
zi K i 1
1
1
Ki
V 1 F
2
V/F
1
0.53 and f V / F
calculate
V/F
2
.53 0.157 2.92
x C1
.584 150 1
1
0.157 .
0.584
87.6 kmol/h and L 150 87.6 z C1
K C1 1 (V / F)
y C1 K C1 x C1 50 0.016883 Similar for other components.
0.157
2
where
V
2-D17.
0
Ki 1 V F
.5 1 49 .584
62.4
0.016883
0.844
L F 1.5 V 0.4F 400, L 600 Slope 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 a.
35
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 zh 1 zi V xh 2.D18. From x i , we obtain V F Kh 1 1 Ki 1 F Guess Tdrum , calculate K h , K b and K p , and then determine V F .
K1 1 z i
Check: Initial guess: If x h
Kh
Try T
0 ?
.85 then Tdrum must be less than temperature to boil pure hexane
4.8, K p =11.7 .
0.6 1 V 0.85 1.471 . Not possible. Must have K h F 0.8 1 73 C where K h 0.6 . Then K b 3.8, K p 9.9 . 0.6 1 .85 .6 1
V F
Check:
K i 1 zi 1
K1 1 V F
94 C . On this basis 85° to 90° would be reasonable. Try 85°C.
1.0, T
K h =0.8, K b
1
1
0.85
0.706
0.735
8.9 .1
Ki 1 V F
0.6
2.8 .3
8.9 .735 1
.4 .6
2.8 .735 1
.4 735
0.05276
Converge on T ~ 65.6 C and V F ~ 0.57 . 2.D19.
90% recovery n-hexane means 0.9 Fz C6 Substitute in L
F V to obtain z C6 .9
C 8 balance: z C6 F
Lx C6
z C6
or
Vy C6
L x C6 1 V F x C6
F V x C6
1 V F x C6
K C6 Vx C6
x C6 K C6 V F
Two equations and two unknowns. Remove x C6 and solve
z C6 Solve for V F.
.9 z C6 KV F
.93C 6
V
.1
F
.9K C6
1 V F
.1
. Trial and error scheme.
Pick T, Calc K C6 , Calc V F, and Check f V F If not K ref new
0?
K ref Told 1 df T
36
Try
T
70 C. K C4
V
3.1, K C5
.1
F .9 .37 .1 Rachford Rice equation 2.1 .4 f 1 2.1 .231 1
.37
K ref
0.231 . .08 .25
.63 .35
.08 .231 1
.63 .231
.28719
.37
K ref Tnew Converge on TNew
.93, K C6
0.28745 use .28 1 0.28719 ~ 57 C. Then K C4 2.50, K C8 .67, and V F
0.293 .
2.D20. New Problem. The K values are: K E 8.7 , K B 0.54 , K P 0.14 Can use Eq. (2-40), (2-41) or (2-42). If we use (2-42) the R – R eqn
f
F
Then RR eq =
0
K i 1 zi 0 V 1 Ki 1 F 7.7 .2 .46 z B
V
1+7.7 25
K C2
a.)
zB
K C5
0.153
Soln to Binary R.R. eq.
1
0.6078
zA
zB
F
KB 1
KA 1
0.55
0.45
F
.153 1
4.8 1
0.5309
zC2
0.55
V 1 3.8 .5309 1 K C2 1 F x C5 0.8177 , y C5 0.1251 Need to convert F to kmol. Avg MW 0.55 30.07 0.45 72.15
F 100, 000
V
kmol
K drum
L
0.1823,
yC2
0.8749
49.17
2033.7 kmol/h
hr 49.17 kg
V F F 1079.7,
u Perm
b.)
kg
.8 z B
0.86 .25
V
V x C2
zE
0.8764 1.0955z B
0.3499
4.8
1 zB
.86 .8 z B
.46 .25
0.5265 0.51977 z B
0 .5757z B 2.D21.
1
Use z F
L
F V
954.0 kmol/h
v v
To find
MW L
0.1823 30.07
0.8177 72.15
64.48
37
MW V 0.8749 30.07 For liquid assume ideal mixture: V1
x C2 VC2,liq
0.1251 72.15
x C5 VC5,liq
x C2
MW C2
35.33
x C5
MW C5
C2,liq
VL
0.1823
30.07
MW L
64.48
VL
103.797
L
72.15
0.8177
0.54
0.621 g/ml
v
RT atm g 700 kPa 35.33 101.3 kPa mol v ml atm 82.0575 303.16K mol K WL K drum : Use Eq. (2-60) with FlV WV
K drum
kmol 64.48 kg
WV
997.7
WV
881.5 35.33
exp
h
31,143.3
0.621
n 0.2597
2.6612
AC
D
0.18707
0.0010149
n 0.2597
0.009814 1.0 m
s 3.2808 ft
v
0.8111
4A C
m s 3600 s h
n 0.2597 4
2
0.3372
2.6612 ft/s
0.8111 m/s
1079.7
V MWV u Perm 3600
3
n .2597
0.621 0.009814 ft
L
0.2597
0.81458
0.3372
v
31,143.4 kg/h
0.009814
1.877478
0.009814 g / ml
6, 4331.7 kg/h
kmol
64331.7
0.0145229 u Perm
103.797 ml/mol
0.63
MW v
For vapor: ideal gas:
FlV
C5,liq
kmol h
35.33
0.009814
g cm 3
kg kmol kg 1000g
6
10 cm m3
3
1.392 m 2
1.33 m
Arbitrarily
L D
4,
L
5.32 m
38
2.D22.
K i 1 zi
V
f
F
1
K iP 1 z iP V 1 K iP 1 F
Ki 1 V F
Solve for V/F.
1
V
K NP 1
K iP 1 z iP
F
K iP 1 z iP
V F
p tot
log10 VP
K NP 1 z NP
760 mm Hg,
where z iP
z NP
0
1.0
90 C
1580.9
3.011679 20 219.61 1027.256 mm Hg , K iP 1027.256 760 1.35165
VPiP Note: MWiP
8.11778
MWNP .
z iP
F
0.24384 0.5
z iP 1
x NP
K iP
1 x iP
5. 0°C, 2500 kPa Fig 2.12: K M
5.7,
V
0.4095 V 1 F 0.5905; yip K iP x iP
K ethylene
1.43,
K Ethane
0.98,
y NP
0.44653
K C6
0.007
1
V
.47 .4
0.43 .05
F1
1 4.7 .6
1 .43 .6
V
V
F
F
2
.02 0.35 1
f
xi
x ethane
zi 1
Ki 1 V F
0.354, x C6
0.993 0.2
.02 .6
V F
1
zi K i 1
1
1
yi
0.55347
0.6 (equal split ethylene and ethane)
F
Then Eq. (2-38),
0.629
0.24384 0.35165
x iP
First, try
0.5 in both wt & mol frac., as does z NP .
0.35165 0.5
V
Find
K NP 1 z NP
F
1499.2
7.84767
NP
log10 VPiP
Eq. (2-46)
T
V
0
2.75943 90 204.64 574.68 mm Hg , K NP 574.68 p tot 0.75616
VPNP
f
K iP 1
K NP 1 K iP 1
p drum
2.D23.
1
K NP 1 z NP V 1 K NP 1 F
1
.993 6
0.6059
2 2
Ki 1
V F
. xM
0.104, x ethylene
0.502,
0.0108
0.040
1.0001 OK
Ki xi
39
2.D24. New Problem. p = 300 kPa At any T. K C3
y C3 x C3
K C6
y C6 x C6
K’s are known.
Substitute 1st equation into 2nd
K C6
1 x C3 K C6
Solve for xC3,
x C3 K C3 K C3
At 300 kPa pure propane K C3
x C3
at 110°C
x C3
1 K C3 x C3
1 x C3
1 K C3 x C3
K C6
1 K C6
K C6
&
y C3
1.0 boils at -14°C
At 300 kPa pure n-hexane K C6 at -14°C
1 x C3
1 K C6
x C3
Check:
1 yC3
K C3 1 K C6 K C3
K C6
(Fig. 2-11)
1.0 boils at 110°C 1 K C6 1 K C6 0 K C3
0,
1, y C3
y C3
1 1 K C6
1.0
1 K C6 K C3 0 K C3
0
Pick intermediate temperatures, find K C3 & K C6 , calculate x C3 & y C3 .
T 0ºC 10ºC 20ºC 30ºC 40ºC 50ºC 60ºC 70ºC
K C3
1.45 2.1 2.6 3.3 3.9 4.7 5.5 6.4
K C6
0.027 0.044 0.069 0.105 0.15 0.21 0.29 0.38
x C3 1- 0.027 = 0.684 1.45 - 0.027 0.465 0.368 0.280 0.227 0.176 0.136 0.103
yC3
K C3 x C3
0.9915 0.976 0.956 0.924 0.884 0.827 0.75 0.659
See Graph
40
41
b.
L V 0.6 0.4 1.5 x C3 0.3 , V F 0.4, Operating line intersects y x 0.3, Slope 1.5 L F y x z V V F 0.3 at 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: K C3
d.
y
.8,
y C3 x C3
x ~ .16
0.63 0.062 10.2 , DePriester Chart T = 109ºC
Slope V
L
y
.8 .6
V
x 1
.16 .6
f
F
1.45
1 f
0.45
f
.45
0.69
2.D25. New Problem. 20% Methane and 80% n-butane V Tdrum .50 ºC , 0.40 , Find p drum F K A 1 zA K B 1 zB V 0 f V V F 1 KB 1 1 KA 1 F F Pick p drum
1500 kPa: K C4
13
(Any pressure with K C1 Trial 1
1 and K C4
12 .2
f1
K nC4
.6 .8
1 12 .4 1
1.0 is OK)
0.4
d f Pold
Need lower p drum
0.2178
1 .6 .4
K C 4 Pold
K C 4 Pnew
0.4
1
.2138
0.511
1.0
Pnew
1160
K C1
15.5 .2
f2
.489 .8
1 15.5 .4
K C 4 Pnew
Pnew f3
16.5 1
0.4305
.4863
0.055769
0.511 1
1100
0.541 0.055769 K C1 17.4
16.4 .2 1
.489 .4
16.4 .4
.459 .8 1
.459 .4
0.0159 , OK. Drum pressure = 1100 kPa
42
b.)
xi
yC1 2.D26. New Problem. a)
b) Stage is equil.
K C5
zi 1
0.2
, x C1
Ki
K C1x C1
V 1 16.4 .4 1 F 17.4 0.02645 0.4603
0.02645
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 y C3 0.8 K C3 3.700 x C3 0.2162
0.2
.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 Tdrum , p drum . Find 10ºC, 160 kPa. 2.D27. New Problem. a.)
VPC5 : log10 VP
1064.8
6.853
0 233.01
VP 191.97 mmHg b.)
VP
3 760
2.2832 ,
2280 mmHg ,
log10 VP 6.853 1064.8 / T 233.01 Solve for T = 71.65ºC c.) Ptot 191.97 mm Hg [at boiling for pure component Ptot d.)
C5:
log10 VP
30 233.01 637.51 mm Hg
VP
K C5 C6:
VPC5 Ptot
log10 VPC6
VPC6 K C6 e.)
KA
yA x A
1064.8
6.853
2.8045
637.51 500 1.2750 1171.17
6.876
30 224.41 187.29 mm Hg
187.29 500 KB
2.2725
0.3746 yB x B
(1 y A ) / (1 x A )
If K A & K B are known, two eqns. with 2 unknowns K A & y A
x C5
1 K C6 K C5
K C6
VP ]
1 0.3746 1.2750 0.3746
Solve.
0.6946
y C5 K C5 x C5 1.2750 0.6946 0.8856 f.) Overall, M.B., F = L + V or 1 = L + V C5 : Fx F 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.
43
2.D28. New Problem.
V h
D
F L From example 2-4, x H With h D
C, D
0.19, Tdrum
378K, V F
0.51, y H
0.6, z H
0.40
V MWv u perm 3600
v
C
C=4, MWv = 97.39 lbm/lbmole (Example 2-4)
1
3.14 10 3 g mol
v
28316.85ml
454g lbm
ft
3
0.198
lbm ft 3
Example 2.4
u perm
K drum
L
v
,
K horiz
1.25 K vertical
V
From Example 2-4, K vertical
u perm V
V F
0.5541 F
0.4433 , K horiz
0.6960 0.00314
0.00314 lbmol 0.51 3000 hr
1.25 0.4433
0.5541
12
8.231 ft s [densities from Example 2-4]
1530 lbmol hr
lbmol lbm 97.39 h lbmol ft s lbm 8.231 3600 0.1958 3 s h ft 1530
D
h
5.067 ft
4D
20.27 ft 1 Use 5 20 or 5 22 ft drum. 2 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 PengRobinson 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.
44
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 = 4D 2. 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 u perm (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 x A = 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.
45
Graph for problem 2.D30.
46
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 u perm 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. T drum = 115.42 oC, Q = 6630.39 kW,
47
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 edition of SPE. a. Will show graphical solution as a binary flash distillation. Can also use R-R equation. To generate equil. data can use x C6 x C8 1.0, and yC6 yC8 K C6 x C6 K C8 x C8 1.0 Substitute for xC6
1 K C8
x C6
K C6 K C8 Pick T, find KC6 and KC8 (e.g. from DePriester charts), solve for xC6. Then yC6 = KC6xC6 T°C
KC6
125 120 110 100 90 80 66.5 Op Line Slope
KC8
4 3.7 3.0 2.37 1.8 1.4 1.0
xC6
1.0 .90 .68 .52 .37 .26 .17
L
1 V F
.6
V
V F
.4
0
.0357 .1379 .2595 .4406 .650 1.0
yC6 = KC6 xC6 0 .321 .141 .615 .793 .909 1.0
1.5 , 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.
48
MW L VL
x C8 MW x C6
MW
x C8 MW
C6
MW
x C8
C6
C6
MW v
C8
86.17
.52
.659
C8
MW L
99.63
VL
145.98
L
.52 86.17
C8
yC6 MWC6
.48
114.22
42.57
454 g/lbm
.85 86.17
.15 114.22
1.0 90.38 g/mol 0.00307 g/ml ml atm RT 82.0575 359K mol K Now we can determine flow rates V V F .4 10, 000 4000 lbmol/h F pMW v
L
V MW v
F V
4000 90.38
6000 lbmol/h, WL
Flv
L MW L
v
597, 780
0.19135
Wv
L
361, 520
42.57
exp
1.87748 0.01452
u Perm
K drum
L
.81458
2.1995 v
V MW v
A Cs
4A Cs
v
597, 780 lb/h
0.111, nFlv
2.1995
2.1995 0.00101
.18707 2.1995
4
2.1995
2
0.423
42.57 19135 .19135
6.3 3600 0.19135
4 83.33
90.38
6000 99.63
4000 90.38
u Perm 3600
D
3
0.423
v
ft 3
361, 520 lb/h
WL
K drum
lbm
0.19135 lbm/ft 3
v
Wv
99.63
145.98 ml/mol
.703
28316 ml/ft 3
.682 g/ml
yC8 MWC8
.48 114.22
6.30 ft/s
83.33 ft 2
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 u Perm75% 0.75 0.85 u Perm85% 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
.8
.9
1
yB
0
.22
.38
.52
.62
.71
.79 .85 .91
.96
1
.7
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.
49
Stage 1)
z F1
.4 .4
Intercept Stage 2)
z F2
13
.164
z F3
23
.240
f
.240
Intercept 2.F3.
f .164
Intercept Stage 3)
f
12
13 1.2
y1
23
T
Converge to
T
12
18 C,
T
0 C,
z1 K1
K1
x2
x3
K1z1
K1
0C
x1
1,
.164
13 23 .01
Slope
.480
Dew Pt. Calc. Want Try
.872
.246
2,
13
Slope
Bubble Pt. At P = 250 kPa. Want Guess
23
Slope
z2
12
y2
.240
y3
.461
z3
1
.022
1 K2
.043,
K3
.00095,
.52
K2
0.11,
K3
0.0033,
120.26
1.0
1.93,
50
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 Td,1 70 C : K1 7.8, K 2 1.07, K 3 Guess V F
.083
0.5 . Rachford Rice Eq. 7.8 1 .517
fV F
1
.07 .091
6.8 .5
1
V F .6 gives f .6
.07 .5
1
.14
.083 1 .5
.101
V F .56. f 0.56
By linear interpolation
.083 1 .392
.0016 which is close enough for first
trial.
V
V F F 56, zi
xi
1
Ki 1 V F
x1
.1075
y1
.839
Data: Pick
L
x2 y2
.088 .094
44
and y i
x3 y3
Kixi
.806
x 1.001
.067
y .9999
th
25 C . (Perry’s 6 ed; p. 3-127), and (Perry’s 6th ed; p. 3-138)
Tref 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
T
0 C, CpL1
0.576 cal / (g C) 44
For
T
20 to 123 C, CpL3
at
T
75 C, CpL2
25.34 kcal/(kmol C) .
65.89 kcal/(kmol C)
39.66 kcal/(kmol C) . (Himmelblau/Appendix E-7)
Cpv a bT cT 2 propane a = 16.26 b = 5.398 × 10-2 c = -3.134 × 10-5 -2 n-pentane a = 27.45 b = 8.148 × 10 c = -4.538 × 10-5 -3 **n-octane a = 8.163 b = 140.217 × 10 c = -44.127 × 10-6 ** Smith & Van Ness p. 106 Energy Balance: E(Td) = VHv + LhL – FhF = 0 Fh F 100 .577 25.34 .091 39.66 .392 65.89 95.25 297, 773 kcal/h Lh L
44 .1075 25.34
VH v
56 .839 0.94
.088 39.66
3597.4 16.26 5.398 10
6302.88 27.45 8.148 10
0.67 9895.3 8.163 140.217 10
E Tdrum Converge on For
V F
.806 65.89
2
2
70.25
117, 450
45
45 3
45
240, 423
60,101 Thus, Tdrum is too high. Tdrum
57.2 C : K1
0.513, f 0.513
6.4, K 2
0.0027. V
.8, K 3 51.3, L
.054 48.7
51
x1
.137, x 2
.101, x 3
.762,
x1 1.0000
y1
.878, y 2
.081, y3
.041,
y1 1.0000
Fh F 297, 773; Lh L 90, 459; VH v Thus Tdrum must be very close to 57.3°C. x1 .136, x 2 .101, x 3 .762
209,999; E Tdrum
2685
y1 .328, y 2 .081, y 3 .041 V 51.3 kmol/h, L 48.7 kmol/h Note: With different data T drum may vary significantly. 2.F4.
New Problem.
yV Lx
V F
x
Find:
0.4, V
0, y
Fz 4kmol h , L
F
z 2.5 .25 V V = 4 kmol/h, L = 6 kmol/h.
or
y 6,
L V
1.5
L V
x
F V
z
slope
0.625
From the graph, x = 0.19 y = 0.34 Equilibrium is from NRTL on Aspen Plus.
52
FIGURE 2.F.4.
2.G1. Used Peng-Robinson for hydrocarbons. Find Tdrum 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.
x i yi
Used Peng-Robinson. Find Tdrum 30.11 C, L 31.348, V 68.66 kmol/h. In same order as 2.G1, are: 0.0189 0.1123 , 0.0906 0.3023 , 0.3255 0.3495 , 0.0402 0.0501 , 0.5248 0.1858 .
53
2.G3.
Used NRTL-2.
xM
Tdrum
0.18 and y M
xM
79.97 C ,
0.2475, y M
0.6287 .
Compares to graph with
0.55 . Different equilibrium data.
2.G4. New Problem. COMP METHANE BUTANE PENTANE HEXANE V/F = 0.58354 2.G5.
x(I) 0.12053E-01 0.12978 0.29304 0.56513
y(I) 0.84824 0.78744E-01 0.47918E-01 0.25101E-01
New Problem. Used NRTL. T = 368.07, Q = 14889 kW, 1st liquid/total liquid = 0.4221, Comp Furfural Water Ethanol
Liquid 1, x1 0.630 0.346 0.0241
Liquid 2, x2 0.0226 0.965 0.0125
Vapor, y 0.0815 0.820 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 METHANE BUTANE PENTANE HEXANE V/F = 0.40001
x(I) 0.000273 0.18015 0.51681 0.30276
y(I) 0.04959 0.47976 0.39979 0.07086
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 ibutane 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 p T deg R 509.688 psia 36.258 F 150 zM 0.5 z iB 0.1 z np 0.15 znhex V/F 0.602698586 0.25
54
guess KM KiB KnPen KnHex xM xib xnPen xnHex Sum RR M RR nB RRnP RRnHex sum RR
51.86751896 0.926804057 0.175621816 0.05400053 Use goal seek for cell B24 to = 1.0 change B9 0.015793905 0.104615105 0.29812276 0.581601672 1.000133443 0.803396766 -0.007657401 -0.2457659 -0.550194874 -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, xib 0.08596648 xnPen 0.203540261 xnHex 0.710481125 KiB 3.886544834 KnPen 1.264637936 KnHex 0.574940847 yib = xib Kib = .33411 and so forth
55
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.
3.B1.
Simultaneous solution is likely when one of the key variables can be found only from the energy balances. For example, if only 1 of x D , x B , D, B, FR A dist 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. a.
x D , x B , opt feed, Q Re b
x D , x B , opt feed, Q C x D , x B , opt feed, S (open steam), sat’d vapor steam All of above with fractional recoveries set instead of x D , x B D, x B , opt feed, L/D b. N, N F , col diameter, frac. recoveries both comp. N, N F , col diameter, FR A dist, Lo D N, N F , col diameter, FR A dist, QR N, N F , col diameter, FR A dist, QC N, N F , col diameter, x D , QC
or x B
N, N F , col diameter, S (sat’d steam), sat’d vapor steam, x D or x B 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. Fmix F1 F2 D B
Fmix z mix F1z1 F2 z 2 Dx D Bx B (Mole frac. MVC)
z mix
F1z1 F2 z 2 Fmix
Now solve like 1 Feed Column Fmix & z mix . From Eq. (3-3),
z mix x B Fmix kmol/h. xD xB B Fmix D kmol/h.
D
3.C4.
New Problem in 3rd Edition. See solution to 3D4, Part b.
56
3.D1.
V1 F1
QC
F1z1 F2 z 2 Bx B D x D
1
Lo
F2
D, x D
V QR
xD 0.85
z avg x B Solve D Ftotal xD xB Ftotal F1 F2 1500 kg/h
z avg
L
x B 0.0001 0.01%
F1 F2 B D
D
F1z1 F2 z 2 Ftotal
1000 .60 500 0.10 0.43333 1500
0.43333 0.0001 1500 764.62 kg/h 0.85 0.0001
B Ftotal D 1500 764.62 735.38
B, x B
kg h
Mass balance calculation is valid for parts a & b for problem 3G1. a)
Lo 3, Eq 3-14 QC 1 L0 / D D h D H1 D h D is a saturated liquid at x D 0.85 wt. frac. From Fig. 2-4, h D ~ 45 kcal/kg H1 is saturated vapor at x D y1 0.85, H1 ~ 310 kcal/kg
QC 1 3 764.62 45 310 810, 497 kcal/hour
EB around column.
F1h F1 F2 h F2 Qcol QC QR Dh D Bh B
h F1 81 C, 60 wt% ethanol ~ 190kcal / kg; h F2 20 C, 10 wt% ethanol ~ 10kcal / kg h B (sat’d liquid – leaves equil contact, ~ 0 wt% ethanol) ~ 100 kcal/kg, Qcol = 0 (adiabatic)
QR 764.62 45 735.38 100 1000 190 500 10 810, 497 657, 259 kcal/kg (b) V B 2.5 mass.
57
V (Satd vapor)
Reboiler
QR
L
Sat’d liqd
Lh L QR VH V Bh B L V B L 1838.45 735.38 2573.83
V B 2.5 or V 2.5B 1838.45 B, x B =0.0001 Sat’d liqd
Approximately x B ~ yV ~ x L . Thus h L h B 100 . HV 640 kcal kg
QR 1838.45 640 735.38 100 2573.83 100 992, 763 kcal / h
EB.
QC Dh D Bh B Qcol F1h F1 F2 h F2 QR
QC 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 + SyS Dx D Bx B (2) Note: yS 0 energy bal: Fh f SHs QC Dh D Bh B (3) Condenser: mass bal. : V1 Lo D (4)
energy bal.: V1H1 QC Lo D h D (5)
Solve Eqs. (1) and (2) to get:
D
Fz Fx B Sx B 100 .3 100 .05 100 .05 36.4 xD xB .6 .05
Note: Not Eq. (3-3). Solve Eqs. (4) and (5) to get:
QC D 1 L D h D H1
Substitute Q C into Eq. (3):
L Dh D F S D h B Fh F SH S 1 D D h D H1
From Figure 2-4: h F 8, h D 65, h B 92, HS 638, H1 608 kcal/kg.
L D 3.D3.
36.4 65 163.6 92 100 8 100 638 1 2.77 36.4 65 408
External balances: F + C = B + D
Fz Cx C Bx B DyD
(1) (2)
QR Fh F Ch C Bh B DH D
(3)
58
F = 2000, C = 1000, z = .4, x C 1.0, x B .05, yD .80,
h F 20 C 30.7,
h C sat 'd liquid 50, h B sat 'd liquid 92, H D sat 'd vapor 327 kcal/kg L VB Lx N Vy Re b Bx B
Around reboiler:
Lh N QR VH reb Bh B For a total reboiler: x N x B , y N x N x B , h N h B 92 M.B.:
V B h
N
Q R VH reb Bh B
QR since h B h N H reb h N HReb 617 (saturated vapor at y N 1 0.05 ) Fz Cx C Fy D Cy D Solve Eqs. (1) and (2) for B: B x B yD 800 1000 1600 800 Thus B 800 and D 2200 .05 .8 or
From Eq. (3),
V
QR Bh B DH D Fh F Ch C
QR 800 92 2200 327 2000 30.75 1000 50 804,500 cal/h V 3.D4.
QR 804500 1532.4 kg/h H reb h N 617 92
New Problem in 3rd Edition. F B D MVC Fz Bx B Dy D But given recoveries. Thus, use: Fz F,M (Frac Rec Methanol in distillate) Dy D,M and
Fz F,W (Frac Rec water in bottom) Bx B,W
z F,M 0.3, z F,W 1 z F,M 0.7 yD,M unknown, x B,W unknown.
Methanol 29.7 100 0.3.99 Dy D,M If 99% methanol recovered in distillate, 1% is in bottoms 0.3 100 0.3 0.01 Bx B,M Water
68.6 100 0.7 0.98 Bx B,W 2% water in distillate
1.4 100 0.7 0.02 DyD,W
Since x i 1 and yi 1, Dy D,i D, and Bx B,i B Thus,
D DyD,M DyD,W 29.7 1.4 31.1 kmol h
B Bx B,M Bx B,W 0.3 68.6 68.9 kmol h Check:
B D 100 F
OK
59
yd,M
DyD,M D
29.7 31.1 0.955
x B,M Bx B,M B 0.3 68.9 0.00435
D 31.1 & L0 D 2. Thus L0 62.2
a)
Reflux liquid is in equilibrium with vapor y D,M 0.955 From equilibrium data (Table 2-7) x M,0 ~ 0.893 (linear interpolation) E.B. Partial condenser: V1H1 Qc DH0 L0 h 0
b)
V1 D L0 93.3
V1y1 Dy0 L0 x 0 y1 Dy0 L0 x 0 V1
y1,M 29.7 62.2 0.893 93.3 0.914; y1,W 1 y1,M 0.086 M 35, 270 J
[email protected] C 35, 270 kJ / kmol choose MeOH reference 64.5ºC.
40, 656 J mol@100 C 40, 656 kJ/kmol choose water reference 100ºC. The condenser is at 66.1ºC (linear interpolation Table 2-7).
H1 M y1,M W y1,W y1,M CP,V,M T1 64.5 y1,W CP,V,W T1 100 V1 is at T1 in equilibrium with y1,M 0.914. From Table 2-7 T1 ~ 67.6 C
Assuming only constant & linear T term are important in CP,V eqs., C P,V CP Tavg . For
67.6 64.5 67.6 100 66.65 C . For water, Tavg 83.8 C . 2 2 J 1000 mol kJ kJ 42.93 0.08301 66.05 48.41 48.41 mol kmol 1000J kmol C
methanol Tavg
CPV,M
J kJ 34.04 o mol C kmol C TM 67.6 64.5 3.1 ; TW 67.6 100 32.4
CP,V,W 33.46 0.00688 83.8 34.04 Then
H1 35270 0.914 40656 0.086 48.41 3.1 0.914 34.04 32.4 0.086 H1 35775.5 kJ kmol
Note λ terms dominate.
H D is at yD
[email protected] C
H D M yD,M W yD,W yD,M CP,V,M TD 64.5 yD,W CP,V,W TD 100 66.1 64.5 65.3 2 42.93 0.08301 65.3 48.35
CP,V,M CP,V,M TM,avg . Tavg,M
CP,V,M
100 66.1 83.05 2 33.46 0.00688 83.05 34.03
CP,V,W CP,V,W TW,avg . TW,avg
CP,V,W
TW 66.1 100 33.9
60
H D 35270 .955 40656 0.045 48.35 0.955 1.6 34.03 0.045 33.9 HD 33682.85 1829.5 73.88 51.91 35534.3 Reflux liquid at 66.1ºC and x M,0 0.893, x W,0 0.107 Reference MeOH 64.5ºC, water reference 100 ºC h 0 CPL,M x M,0 TM CPL,W x W,0 TW
kJ kmol C h 0 86.85 0.8931.6 75.4 0.107 33.9 149 kJ kmol
CPL,M CPL,M Tavg 75.86 0.1683 65.3 86.85
QC DH D L0 h 0 V1H1 31.1 35534.3 62.2 149 93.3 35775.5 2, 242, 030 kJ h Overall EB Fh F QC QR DH D Bh B or QR DH D Bh B QC Fh F
h B is saturated liquid with x B,M 0.00435 and x B,W 0.99565 Interpolating in Table 2.7 TBot 99.2 C
h B CPL,M 0.00435 99.2 64.5 CPL,W 0.99565 99.2 100 99.2 64.5 81.86 and CPL,M 75.86 0.1683 81.86 89.64 2 h B 13.53 60.06 46.5 kJ kmo l Tavg,M
Feed is saturated liquid at z M 0.3, z W 0.7. From Table 2-7, TF 78 C
h F CP,LM 0.03 78 64.5 CPL,W 0.7 78 100
Tavg,M 78 64.5 2 71.25 and CP,L,M 75.86 0.1683 71.25 87.85 h F 805.4 kJ kmol Then
Q R 31.1 35534.3 68.9 46.5 2, 242, 030 100 805.4 1,105,116 3.D5.
3204
2, 242, 030
80,536 3, 424, 479 kJ h
Mass Balances: F = D + S + B, Fz Dx D Sx S Bx B Solving simultaneously, B = 76.4 kg/min, D = 13.6 kg/min. Condenser: QC V1 h 0 H1
V1 L0 D L D 1 D 4 13.6 54.4 kg/min From Figure 2-4, h 0 7.7 kcal/kg (x = .9, T = 20°C),
H1 290 kcal/kg (y = .9, sat’d vapor). Thus, Q C = 54.4 (7.7 – 290) = -15,357 kcal/min Overall Energy Balance: Fh F QR QC Dh D Sh S Bh B From Figure 2-4,
QR Dh D ShS Bh B Fh F QC
h S 61 x S .7, sat'd Liq'd ; h F 200 z .2, 93 C ,
61
h B 99 x B .01, sat'd Liq'd , h D h o 7.7 Thus,
QR 13.6 7.7 10 61 76.4 99 100 200 15357 3635.3 kcal/min
3.D6.
z xB .4 .002 2500 998 lbmol/h. .999 .002 xD xB
From Eq. (3-3), D = F
Then B = F = 1502. Condenser: V Lo D Lo D D D
QC h D H V D Lo D 1 With 99.9% nC5 have essentially pure nC5 . Thus, it is at its boiling point.
h D H V C5 11,369 Btu/lbmol. QC 11,369 998 4 45,385, 048 Btu/h
Overall:
QR Dh D Bh B Fh F QC
Distillate is at boiling point of pure nC5 is at boiling point of nC6
K C6 1.0
K C5 1.0 on DePriester Chart) = 35°C.
Bottoms
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.) CPF x C5CPLC5 z C6 CPLC6
CPF .4 39.7 .6 51.7 46.9 Btu/lbmol °F
h F CPF TF 0 46.9 86 4033.4 Btu/lbmol
Distillate is almost pure nC5 . Liquid at 95°F
h D CPLC5 TDist 0 39.7 95 3771.5 Btu/lbmol
Bottoms is almost pure liquid nC6 at 152.6°F.
hC pLC6 Tbot 0 51.7 152.6 7889.4 Btu/lbmol
QR 998 3771.5 1502 7889.4 2500 4033.4 45,385, 048 50,861, 491 Btu/h 3.D7.
z xB 0.7 0.001 1000 700.4 kmol/h 0.999 0.001 xD xB
Eq. (3-3), D F
B F D 299.6 kmol/h Condenser: Lo Lo D D 2.8 700.4 1961.1 kmol/h Only this reflux is condensed since product is a vapor. QC Lo where λ is for essentially pure n-pentane.
kmol Btu 2.20462 lbmol QC 1966.1 11,369 h lbmol 1 kmol Btu 1J J QC 49,154, 204.85 5.18176 1010 -4 h 9.486 10 Btu h
62
From overall balance QR DHD Bh B Fh F QC Distillate is vapor at b.p. of pure n-pentane (35°C from DePriester chart, K C5 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.
CPF z C5CPLC5 z C6CPLC6 0.7 39.7 0.3 51.7 43.3Btu lbmolo F h F CPF TF 0 43.3 95 0 4113.5Btu lbmol
Distillate H D C5 CPLC5 Tdist 0
H D 11,369 39.7 95 0 15,140.5 Btu lbmol Bottoms is pure C6 @152.6 F
h B CPLC6 Tbot 0 51.7 152.6 0 7889.4 Btu lbmol
kmol Btu kmol Btu QR 700.4 15,140.5 299.6 7889.4 h lbmol h lbmol kmol Btu 2.20462 lbmol Btu 1000 49,154, 204.85 4113.5 h lbmol kmol h Btu 1J QR 68, 675,167.9 7.240 1010 J h -4 h 9.486 10 Btu 3.D8.
New Problem in 3rd Edition.
F 300,
z E .3,
z w .7
98% rec. E in distillate, 81% rec water in bot.
D Dist. .98 90 1 .81 300 .7 128.1 kmol/h y DE
.98 90 0.6885
128.1 B Bottoms .02 90 .81 210 171.9 kmol h yD , D, H D b.
Partial Condenser.
Vapor H1 y1 V1
Qc
L0 2, L0 2D 2 128.1 256.2 kmol h. x 0 , L0 , h 0 D x 0 in equilibrium with y 0 , thus from equation data x 0 0.575. Entering vapor y1 (from graph) 0.61
63
V1 L0 D 256.2 128.1 384.3 kmol h. c.
E.B. on PC. V1H1 Qc DHdist L0 h 0 . Can use Figure 2-4 by converting mole fracs to mass fracs. Basis 1 kmole. .6885 mol E MW 46 31.671 kgE Distillate
5.607 kgW 37.28 kg total Mass frac. E = 31.671 37.28 0.8496
.3115 mole W MW 18 Vapor V1
0.61 mole E 46 28.06 kgE
7.02 kgW 35.08 kg total Mass frac E = 28.06 35.08 0.7999 .39 mole W 18
Liquid reflux L 0
0.575 mole E 46 26.45 kgE
7.65 kgE 34.1 total Mass frac E = 26.45 34.1 0.7757 ~ 310 kcal kg, H1 ~ 330 kcal kg, h 0 ~ 65 kcal kg 0.425 mole W 18
From Figure 2-4, Hdist
Q c DH dist L 0 h 0 V1H1 128.1
kmol 37.28 kg kcal 310 256.2 34.1 65 hr kg kg
35.08 kg 330 2, 400,517 kcal hr kmol Fh F QR Qc DHdist Bh B 364.3
Overall EB.
Know Qc 2, 400,517 kcal h and
DHdist 1, 480, 426 kcal h.
To find Fh F and Bh B , need to convert mole frac to wt frac. Basis 1 kmol 30 mole % E: .3 mole 46 13.8 Feed
12.6 total 26.4 kg kmol Mass frac E 13.8 26.4 0.5227 70% W : .7 18
Bottoms
0.01047 mole 46 .48162
17.811 total 18.293 kg kmol Mass frac E 0.48162 18.28 0.0263 0.98953 mole 18
From Figure 2-4
h F satd liqd 70 kcal kg
h B satd liqd 97
64
QR DHdist Bh B Fh F Qc
Then
kmol 18.29316 kg 97 kcal 26.4 kg kcal 300 70 h kmol kg kmol kg 2, 400,517 1, 480, 426 305,525 554, 408 2, 400,517 3,632,069kcal h
QR 1, 480, 426 171.9
3D9. New Problem 3rd Edition. B = (xD – z)/(xD – xB)F = [(0.9999 - 0.76)/(0.9999 – 0.00002)](500) = 120
QR Lh VH Bh B and L V B
Q R L B h Vh V H h V
Assume h h B .
V V B B 1.5 120 180 kmol h. Bottoms is almost pure water. w 9.72 kcal mol 9720 kcal kmol
QR 180 kmol/h 9720 1.750 106 kcal h
3.D10.
2 atm × 101.3 kPa/atm = 202.6 kPa. Pentane Recovery: 0.995 Fz P Dx D
D
0.9951000 0.55 547.6333 0.9993
kmol/h
B = 1000 – 547.6333 = 452.3667
Since
Pentane Recovery Bot F zp , 1 .9951000 0.55 0.006079
Bx B xB
452.3667
mol frac pentane
Distillate is essentially pure Pentane. Bottoms Pure in Hexane. From DePriester Chart K P 1@ p 202.6 kPa when Tdist 59.5 C
K n H 1@ p 202.6 kPa when Tbot 94 C
L QC 1 o D h D H1 D h D pure pentane CPLC5 Tdist Tref choose Tref 25 C
For Total Condenser, Eq. (3-14)
kcal kcal 59.5 25 1369.65 kmol C kmol H1 h D assuming λ is independent of temperature
h D 39.7
Btu 1 lbmol 0.252 kcal kcal 7680.196 lbmol 0.454 kmol Btu kmol kmol kcal Eq. (3-14) is QC 1 2.8 547.6333 6310.5 13,132, 288 h h kcal kcal h B pure hexane CPLC6 Tbot Tref 51.7 94 25 3567.3 kmol C kmol H1 1369.65 11369
Feed is a liquid at 65°C
h F CPC5 zC5 CPC6 zC6 TF Tref h F 39.7 0.55 51.7 0.45 65.25 1804kcal / kmol
65
QR Dh D Bh B Fh F QC
QR 547.63331369.65 452.3667 3567.3 1000 1804 13,132, 288 QR 13, 692, 081 kcal/h. Note that QC and Q R are relatively close. 3.E1.
Was 3.D8 in 2nd Edition.
QC D, x D 0.990 M.B. F + S = D + B F
Fz M SyS,M Dx D Bx B
p = 1.0 atm
100 kmol h z M 0.6
Since steam is pure, yS,M 0 Know F, z M , x D , x B
S B
x B 0.02
Unknowns S, D, B, Need E.B.
Fh F QC SHS Dh D Bh B Pick as basis liquid at 0°C, h W 0 & h M 0 (essentially steam table choice) Assume ideal mixtures.
h F CPavg TP Tref where CPavg 0.4 CPW ,L 0.6 C PM ,L
Felder & Rouseau p. 637 CPW 0.0754 kJ/mol
CPM 0.07586 16.83 105 T kJ/mol
CPM CPM Tavg CPM 0 40 2 CPM 20 C
CPM 0.07586 16.83 105 20 0.079226 CPavg 0.4 0.0754 0.6 0.079226 0.077696
h F 0.077696 40 0 3.1078 kJ/mol feed = 3107.8 kJ/kmol
Can also use steam table for water
H S is sat’d vapor steam 1 atm,
HS 2676.0
kJ 18.0 kg kg kmol
Steam Table F&R, p. 645 H S =48,168 kJ/kmol
h D is sat’d liquid at x D 0.99 . From Table 2-7, T = 64.6°C
h D CPavg 64.6 0 where CPavg 0.01 CPW 0.99 CPM
CPM CPM Tavg CPM 32.3 C 0.07586 16.83 10 5 32.3 0.081296
CPavg 0.01 0.0754 0.99 0.081296 0.08124
66
H D 0.08124 64.6 5.2479 kJ mol 5247.9 kJ kmol h B : Since leaving an equilibrium stage, sat’d liqd. 2% MeOH Table 2-7, T = 96.4°C
h B CPavg 96.4 0 where CP avg 0.98 CPW 0.02 CPM CPM CPM 96.4 0 2 C PM 48.2 C
CPM 0.07586 16.83 105 48.2 0.08397 CPavg 0.98 0.0754 0.02 0.08397 0.07557 kJ mol
h B 0.07557 96.4 7.28509 kJ mol 7285.09 kJ mol Q C do EB around condenser
L QC dist V1 dist Lo D dist o 1 D D dist 0.99 MeOH 0.01 W M 35.27 kJ mol & W 40.656 kJ mol kJ dist 0.99 35.27 0.01 40.656 35.324 35,323,86 kJ kmol mol QC 35,323.86 2.3 1 D 116,568.7D kJ h
Felder & Rousseau:
Plug Q C & 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:
E2.
D = 56.33 kmol/h, B = 211.71 kmol/h
S 168.04 kmol/h, QC 6,566, 000 kJ/h.
Was 3.D9 in 2nd Edition. F+S=D+B
F 500
kmol mol 500, 000 h h
Fz SyS Dx D Bx B
2 eq. 3 unknowns Condenser: QC 1 L0 / D D h o H1 Note Eq (3-14) not valid. For enthalpy pick reference pure liquid water 0°C and pure liquid methanol 0°C. Felder & Rouseau: CPMeOH 75.86 0.01683T at Tavg
64.5 0 3225, CPMeOH 76.4 J mol C. 2
67
Assuming distillate pure methanol, boils at 64.5°C
h D CPMeoh ,liq T Tref 76.4 J/mol 64.5 0 4928.0 J mol H1 h D 1 at 64.5 F 4928 35270 J/mol 40,198 J/mol
QC 4 4928 40198 D 141, 080D J/h where D is mol/h Overall Energy balance: F h F SHS QC DhD BhB Bottoms is essentially pure H 2 O at 100°C
J J h B CPW ,liq T Tref 75.4 100 0 7540 mol C mol HS h B W at 100 C 7540 40656 J/mol 48196 For feed. 60 mole % Methanol boils at 71.2°C (Table 2-7).
h F CPi zi T Tref 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) Fz Dx D Bx B or (500,000) (.6) = 0.998D + 0.0013B
Lo D h D H1 or QC 141, 080D D (4) Fh F Sh S QC Dh D Bh B or (500,000) (54137) + S (48196) (3) QC 1
+ Q C = D(4928) + B (7540) Solve simultaneously: D = 298.98, B = 1245.5, S = 1044.2 kmol/h Q C = - 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.
z xB 0.79 0.004 D 25, 000 19, 788.5 kmol/h F 0.997 0.004 xD xB Note that N 2 mole fractions were used since N 2 is more volatile. B = F – D = 5211.5 From enthalpy comp. diag. h D 0, H1 1350 kcal/kmol, h B 160, h F 1575 . Eq. (3-3)
Then,
QC 1 Lo D D h D H1 519788.5 0 1350 133,572, 000 kcal/h QR Dh D Bh B Fh F QC
QR 0 5211.5160 25, 000 1575 QC 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 NH 3 = 1.765 kmol 70 kg H 2O 3.888 Total 5.653 kmol Thus 1 kmol is 100/5.653 = 17.69 kg
68
Will work problem in weight fractions since data is presented that way. 95% recovery: (0.95) Fz = Dx D or, D = (.95) Fz / x D = (.95) (17.69) (.3)/(.98) = 5.15 kg. B = F – D = 12.54 kg
x B Fz Dx D B 17.69 .3 5.15 98 12.54 0.021
From diagram: h D 55, H1 415, h B 150, h F 5 kcal/kg
Eq. (3-14), QC 1 Lo D D h D H1 3 5.15 55 415 5562 kcal/kmol feed and QR Dh D Bh B Fh F QC
QR 5.15 55 12.54 150 17.69 5 QC 7815 kcal/kmol of feed
G1. a.) Using NRTL. QC 778,863 kcal/h, QR 709,520 kcal/h b.) QC 1, 064,820 kcal/h, QR 995, 478 kcal/h G2.
New Problem in 3rd Edition. ASPENPlus. D = 988, L/D = 3, Peng Robinson, N 40, NF 20 (arbitrary values in Radfrac)
x D 1.000
x DC6 1.211107
x B 0.0013316
QC 4.4426 107 Btu h, QR 4.9852 107 Btu h
69
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 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 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. 4A7. 4A13.
4A14.
New Problem in 3rd Edition. Answer is a. See Table 11-3 and 11-4 for a partial list. 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 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 z, y D satisfies operating equation. Solution: Op. Eq.
y
Substitute in
y
yD V But
q 1.0, V
Lz D, L
L V x yD , x
L V 1 xB z
L V xB F, L V
y D D Fz Bx B Which is external mass balance.
B 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 QR V . With CMO and a saturated liquid feed V V (1 L / D) D , and then
QR / D 4.C16.
(1 L / D) .
New 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 1
H mix
hT
H mix
h mix
and H mix , h mix are saturated
vapor and liquid enthalpies at feed stage of column with mixed feed.
A
Given p, L/D, saturated liquid reflux, x D , x B
z0
B
opt feed locations, z1 , z 2 , F1 , F2 , h F1 , h F2
z2 y
zT z1 Plot top op line. Plot all 3 feed lines. Draw
xB
line from point A to y = x = x B to obtain
x b.)
Does
qT
q1F1
op. line. Connect pts B & C to get q 2 Fbot. 2
FT
middle op. line.
71
check q T
H mix
hT
H mix
h mix
F1h F1
H mix
F2 h F2 FT
H mix
F1
F2 H mix
h mix
F1H 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 mix
qT
h F1
F2
h mix
H mix
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 and q2 H mix h mix H mix h mix F1q1 F2 q 2 Thus q T if CMO is valid. FT 4D1.
L
a. Top op line: y Intersects y
x
When x
V
xD
x
L V
x D and
L
L D
1.25
V
1 L D
2.25
0.5555
0.9
0, y
b. Bottom op line: y
1
1
L
x
L
xD
V
L
0.4
1 x B , and
Plot – See diagram
L
V B
V B 1
3
V V V B 2 V V Intersects y = x = xB = 0.05 1 0.5 / 2 @y 1 x 0.683 this is convenient point to plot 32 c. See diagram for stages. Optimum feed stage is #2 above partial reboiler. 5 equilibrium stages + PR is more than sufficient.
72
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.
y
b. c.
x
zE
.6
Slope
L
F V
1 V/F
.63
1.703 V V V F .37 From Table 2-1, at 84.1° C y .5089 H hF liquid at 20°C and 40 mole % ethanol. q 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.
73
Basis 1kmol feed.
.4 kmole E .6 kmol Water From Figure 2-4
H q
Alternate Solution:
40 mole %E
.6 MW
46
0.63 wt frac.
10.8 kg
18
398 kcal kg , h
18.4 kg
total
29.2 kg
75, h F 20 C
10
398 10
1.20 398 75 q 1.2 Slope 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 % h F Cp,40%liq 20 84.1
h d.
.4 MW
0,
63 wt%, H
H
40%E
398 kcal kg , h
CPvapor
y E CPEvapor
65, h F
398
kcal kg
C p vapor 120 84.1
y w CPw ,vapor
Assume only 1st and 2nd terms in C P equations are significant. From Problem 2.D9 CPvapor .4 14.66 0.03758T .6 7.88 .0032T kcal/kmol T is C which simplifies to 120
For linear
CPvapor
10.592 0.16952T
C p dT is equal to CPvapor @ Tavg
84.1
Tavg
84.1 120 2 102.05 . Then C Pv ,avg 398
hF
398 15.149
q e.
q
q
f.
Flash
kcal
hF
V F
kg
10.592
12.32 120 84.1
15.147
398 65
333
f, L
13 12 , slope
.7,
12.32
kcal kmol
kcal 1 kmol kmol 27.2 kg
413.15 kcal kg
398 413.15 L L
0.16952 102.05
L
F
0.045. F
12
q q 1
13
L
12 13 12
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. Hv = 395, HL = 65 (from Figure 2-4). hF is liquid at 200°C. Assume Cp,liq is not a function of T. Estimate, 46.1 23 h h L 60 C h 20 kcal C P ,liq .63 wt frac ~ 0.864 T 60 20 80 kg C Then h F
h L 200 q
Hv
CPL 200 60 hF
.864 200 60
h L 60 C
395 167.1
0.691,
q
46.1 167.1
0.691
2.24 Hv hL 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 h F C P,liq TF Tref CP,liq 250 0
CP,liq CP,EtOH z EtOH CPw 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 25.91 kg/kgmole Thus, zW = 0.719 and zE = 0.281 CP,liq 37.96 .281 18.0 .719 23.61 C P,liq in kcal kg C hF q
hF
4.D4*. a.
q
Hv
hL
446 70
H CPv 350 50
q q 1
x
z
0.911 . Then,
228 0.58
H
25 300
H hF
L L F where L
L L
25.91
Hv
q q
MWAVG
250 C kg C h F 446 228
Slope
slope
23.61
kcal
H h q q 1 0.6. y
b. c.
0.911
CP
4.D4a
y=x
25 300 z
1.5
0.6 is intersection.
L 0.6F. Then q
feed line
L 0.6F L / F
.5 .6 0.6, and
.7
1.5
F where L
L F 5. q
L F5 L F
1 5 , slope
q q 1
16
76
4.D5*.
h liq
fL
h reflux
3100 1500
h liq
17500 3100
L0
L0 D
1.1
V1
L0 D 1
2.1
H vap
1 fc
L1 V2 Alternate Solution
L1
L1 V2
4D6.
q
qL0
L1
0.524
1.1111 .524
1 f c L 0 V1
For subcooled reflux, Then,
L 0 V1
0.1111
1
0.55
.111 .524
L1
H h0
17500 1500
L0
H h1
17500 3100
1.111
1.1111 L0 L1 D
,
L1
1.111 L0
L1 D L1 D 1 D L1 1.222 0.55 V2 2.222
New Problem in 3rd Edition.
1.111 1.1
D
a) 175
F1
F2
1.2222
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. q1 1, vertical at y x z1 0.6 Feed 2.
60% vapor = 40% liquid q 2
Slope feed line Bottom Op. Line
q2
0.4
q2 1
.06
y
L V x
Slope L
Middle
Lx When
x
0, y
F2
L V
0.4 2 3 through y
x
z2
L V 1 x B . Through y
V B 1 V B
0.4
x
xB
32
B V
F2 z 2 F2 z 2
Bx B Bx B
Vy
y
L V
x
F2 z 2
, 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
Bx B V
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 Plot Middle Op Line. 241.25
77
y
L
V Know that y x Also, L
L
x
F1
1
L
xD V x D and gives through interaction Middle and Feed line 1.
231.875 75 156.875 and V
L V 156.875 241.25
V
241.25 ; thus,
0.65
c) See graph. Graph for 4D6.
78
4.D7*. a. Plot top op. line: slope
L V
.8 , x
L
y
.9. Step off stages as shown on Figure.
V B
2 , x y x B 0.13. Step off stages 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 gives q = 0.692. slope 4 q 1 b. Plot bottom op. line:
slope
1 1
xD
4.D8*. The equilibrium data is plotted and shown in the figure. q 0.692 and q q 1 9 4
From the Solution to 4.D7c,
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 V 1 L V
.8 .2
L Dact Multiplier Multiplier = 4/1.941 = 2.06
4
L D
min
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. a) F1 F2 D B
F1z1
F2 z 2
100 F1
Dx B
Bx B
F1 .42
Solve simultaneously, B 113.68, F1 b)
L D L
1 2
,
80 B
F1
B 20
18 .66 80
0.04 B
93.68
L
L D
12
1
V
1 L D
32
3
L
D 40, V L D 120 D V V 120 Saturated Liquid Feed L L F1 40 93.68 133.68, L V c) Top Op. Line – Normal: y Through y Bottom – Normal: y
x
L V x
1.114
1 L V xD
x D , Slope 1 3, y intercept
L V x
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
0, y
Dx D
xD
F1z1
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. Also,
x
D
(or do around bottom) V V L V . Through intersection feed line F1 and top op. line.
L V x
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 x1 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 .8 .9317 y2
x y
1.0 1.0
.192
1.76
.76 .93736
.8x 2 .192 .8 .89476
.9 .9406
0.93736
.93736
1.76 .76y 2
b. Generate equilibrium data from: y
0.9317
.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 x 1 L V x D goes through y = x = x D = 0.99
L V
L D 1 L D
Feed line: Slope
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 Lx Sy M,S Bx B But y M,S 0 (Pure steam) With CMO B L
V S
B
y
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 0 L V
min
1 L V
Actual L/D is 3.12 × this value.
0.4242 0.4242
min
1 0.4242
0.73684
83
4.D12. L V
y
L V
x
L D 1 L D 1
L V
Bottom slope
34
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 x B & 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.)
L
b.) See figure.
V
L
See Figure
0.665 0.95 0.30 0.95
MIN
L
L V
0.4385
0.4385
0.7808 D MIN V L 1 L V 0.5615 L L L D 1.5616 c.) 2.0 L D MIN 1.5616 , D V 1 L D 2.5616 L
y intersect 1
V
yD
0.3709 . Top operating line y Goes through y
Bottom y
L V
x
L V
0.6096 L
x
V x yD
1
L
V 0.95
yD
1 xB
Goes through y
x 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) xB = 0
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
q q 1
L
L D
2
V
1 L D
3
1 3 0.6885 L V x
1,
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.
at feed conditions.
. Approx.
.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 .54, K1x1 1.5 .4 .54 .6 K C5 Tnew =
.54 .92
K C5 50 C Hv
=.594, Tnew
1.6,
Note : CP feed,liq Hv q
1.6 .4 .584 .6
.99 Close enough.
, 50 C 122 F
feed
Hv
50 C
K1 x 1
h L 50 C
.92
CPfeed ,Liq 122
46.9 122
5721.8
46.9 is from Prob. 3-D6. 5721.8 12691 18412.8 Btu/lbmole
HV
hF
18412.8 4033.4
HV
hL
12691
1.133
Note: h F is from Prob. 3.D6. 4.D17.
L
Top Op. Line: y
x
L
V L D
V
1 L D
x
L
1
V 7 2
0, y
1
The vertical line at x
xS
9 2 L
x D , goes through y
x
xD
0.9
7 9 2
xD
.9 =0.2 V 9 Plot Top. Step off 2 stages. Find x S ~ 0.81
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
x B (unknown)
D B S Don’t know D, B, or x B .
Bx B
Sx S
Feed enters as saturated vapor. Thus q
0&V
F
Bottoms leaves an equilibrium contact, it is saturated liquid L Do flow balances V F 100 V V 100 since S is removed as saturated liquid.
B
91
L
L V
V
7 9 100
L
L S
B
L 62.777 Fz Dx D Sx S
xB
77.777. D
77.7777 15 60
V L 100 77.777
62.7777. L V 22.222 0.9
B
Plot. Op. line Step off stages. 9 is more than sufficient.
4.D18. New Problem in 3rd Edition. Feed F1 : z1
62.7777
62.7777 100 15 0.81
22.222
0.6278 0.444
0.6, saturated liquid, q 1, q / (q 1)
92
Feed F2 : z 2
0.4, 80% vapor hence 20% liquid q q
LF / F
0.2F / F .2
.2
14 q 1 .8 Part a.) Bottom operating line goes through point, y
x
xB
0.04
Max L V to point intersection feed F2 line and equilibrium curve.
L
Slope
1.0 .04
V
V B Part b. L
V L
min
F1
1
1
V
L V 1
1.2326
100
.2 80
116
L L F2
L
V B and V B 1.5 L
1.5B B
L B 116 46.4
D
V
V VF2
100
V 133.6 Check overall balance
116
B
V
L
0.8113
100
L
116
2.2326
.47 .04
max
46.4
2.5 69.6
116
L V
69.6 .8 80
69.6
1.66667
133.6
0.7485
F1
F2
D B
180 133.6 46.4 180.0 OK
To find y D use MVC mass balance
F1z1 or
yD
F2 z 2 F1z1
Dy D F2 z 2
Bx B Bx B
100 .6
D
80 .4 133.6
Actual bottom op. line: y
L V
x
L V
46.4 0.4
0.675
1 xB
L
V B
V B 1
2.5
5
V
V
V B
1.5
3
Goes through y x x B 0.04 , Slope 5 3 2nd point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) Plot bottom op. line Dy D F1 z1 L Top Op. line: yV F1z1 Lx Dy D . 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 , y D 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 Feed line can be calculated but is not needed.
4.D20*.
0.01 to 0.02
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 D
Qc
Qc D ho
H1
Lo h o
Dh D which can be solved for L o D .
1
From chart: h D 54 Kcal/kg and H1 285 Kcal/kg Need D in weight units. Convert feed to weight units. 100 kgmoles Ethanol: .32 46.07 1474.24 kg/hr hr Water: (100)(.68)(18.016) = 1225 kg/hr Total: F = 2699.328 kg/hr
95
F z xB
Then,
D
Then,
Lo D
xD
2699.328 .0963
xB
1489.93 kg/hr
.911 .0963
Q D ho
H1
1
2, 065,113 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. L L L L D Top Operating Line: y x 1 x D and V V V 1 L D
L
5
; x
y
xD
.8, y int ercept x
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 69
Bottom Operating Line:
y
x
L
V V intersection top operating line and feed line.
L
1
V
15 kcal/kg, H v
1.149 Slope
L
0
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 Feed 2: q 2 Top:
0 , slope feed line = 0 0.9 , slope
q2
q2 1
L
L D
1.375
V
1 L D
2.375
Goes through y
x
0.9 0.1 L
0.579 , y
V
x D , When x = 0, y
1
Since F1 is saturated vapor, V
Bottom:
x
9 L
1
L V F1
V
xD
xD .
0.40
100 kmoles/hr
L
.1 F2
At feed F2
V
.9 F2
L D B
L V V
V
V
L
L 0.9 F2
0.579 108
0.1F2
100 8 108
62.532
V L 108 45.46 F1 F2 D 100 80 45.46 134.532
But B is saturated liquid. Check L L .9F2
L B 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
F2 z 2
F2
x B or x B
D B Dx D
0.09 . 20 36 43.187 134.532
0.095 , OK
45.46 134.53 179.99 , OK
Bx B
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
97
4.D22*. Around top of column mass balances are: L D Solving,
L
0.75, y x
0
x
D
yD
Vy Cx W
C
xw V V V For pure entering water, x W 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
y
V C and Lx Dy D
0.17, y x 1
0.92
yD , y
x
0.68
Not y D
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 hW C
c
h
hW
Cp
C
T
C
18 212 100
H h L C c 1.1154C V 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
Through 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) Note – Commercial columns usually operate much closer to minimum reflux ratio and have many more stages. b.)
L V
0.85 0.376 min
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 ab 7.6mm where fraction 0.74 or .75 ac 10.3mm
99
100
4.D25.
a. 99.9% methanol is essentially pure. Pure MeOH boils 64.5°C. 1 fc Lo D L Eq. (4-66) 1 where f c CPL TBP Treflux V2 1 1 f c L o D For pure MeOH, CPL
CPL
fc
0.07586 16.83 10 5 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.
4.D26.
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. a) 50% feed: q
L L
F, L
L
amt vaporized
Usually
L F 20
1 F q 1 20 0.05 , Slope = L F 0.0476 20 20 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 V B q
L
min
L V
Find
L V
max
. Then V B
1 0.1
slope
max
b)
0.46 0.1
L V
L V L
Bot. y
V y
L
x 1,
x y
V L V
x
3
V L
min
1 V
2.497 ,
V B
B
1 min
2.497 1.0
0.6681
2.0043
min
V B
V B 1
3.0043
V
V B
2.0043
L V x B or x
1
max
V
1 x B . Goes through y 1
L V
x 1
1.49892
slope
x B with total reboiler 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), F50% and B. yV L x F50% z50% Bx B y
L
x
F50 z 50
Bx B
V V Intersects bottom operating line & 50% feed line. @ x 0, y (F50 z 50 Bx B ) / V , Slope
q 50
For 50% feed,
L V
0.05 (L L ) / F50 .
External balances: 250 F1 F2 D B and F1z50 F2 z35 Dy D Bx B 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 q 50% : L
V
L
F50
B
q 50 F50
L
0.05 100
445.63 100 146.666
440.63
398.964 , Slope
445.63 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 y x y D 0.85
102
EQ
Stages
0.75 = E ML =
actual change liquid
actual
op
change at equilibrium eq. change
Figure for 4D26
103
4.D27*.
Top Op. Eqn:
Feed: q
y
L V
x
1
L
L D
1
V
1 L D
2
L
L
L F
L V
xD
, y intercept
1 F 4 F
1
L V
L
xD q
.46, y
x
xD
.92
54
5 , y x z .48 14 Bx B L Middle: V B L S , V y Bx B L x Sy s y s 0 or y x V V Does not intersect at y x x B or at y 0, x x B . Does intersect top op. line at feed line. Need another point. L V B V B 1 1.5 L L Bottom: 3, y x x B 0.08 y x 1 xB , 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 = ys = 0. Middle Op. Line intersects this steam at bottom op. line (see figure). F
5 4 , slope
q 1
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.
This problem was 4.D35 in 2nd edition. L L Stripping Section: y x 1 xB, y x xB V V
4.D28.
0.02 . Feed line is vertical
104
L V
1.0 0.02 max
0.81 0.02
V B 1 Op line y
1.24,
1, x=
1.5 L V
V B
1 xB
V B
5800 / 0.675
min
L
6.25, L V
1
.16 .02
1 V
L V
max
1 1
0.24
V B
V B 1
6.25 1
V
V B
6.25
min
L V 1.16 Overall Balance: 10,000 = F = D + B 6000 Fz Dy D Bx B D .695
D
V
4.167
1.16
0.865 . Intersection Op & feed lines is y D
B .02
6000 .695D .02D 200
8592.6 kgmoles/day, B 1407.4 . Need a use for impure distillate.
Figure for 4.D28
105
4.D29.
L D
3,
L
3 4,
L
1
1
xD
V V 4 Step off 3 stages on top op line. Find x S on middle op. line
0.9
0.225
0.76 . Point on top op line at x S
0.76 is also
xD
3
xs, S = 15
V´ F = 100
L´
F=D+B+S
6
z = .6
Fz
Dx D
Bx B
Sx S
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
135.375 200.5
L x Sx S y
L V
x
0.6752
Dx D Sx S
Dx D V
Feed z = 0.6, 20% vapor = 80% liquid Feed line slope q q 1 0.8 0.4
, when x
0, y
11.4 45.1125 200.5
0.2819
q = 0.8. 4
106
4.D30*.
a. Subcooled Reflux:
c
1 3
Lo
500 ,
Lo
Lo D
3
V1
1 Lo D
4
L1 V2
Substituting in values, we have Step off two stages.
x2
, L1
4 Lo 3 1 V1 Lo 3 L1 1.0 L2
0.62
1 14
xS
4 3 1
Lo
c and V2
V1
c
L o V1 1 Lo 3 V1 1 4 54
5
, Top op. line y
x
xD
0.92
yS
107
Mixed feed to column: F S Solving for mixed feed, z M
FM
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
0.52667
Energy balance for mixed feed, Fh F Since H s
1500 , Fz SyS
4 3
Lo
SHS HS Dx D
FM H FM
h F (sat’d vapor), and q FM
0 (horizontal).
Bx B , Solving simultaneously D = 500 & B = 500
2000 (subcooled reflux), V
2000 500 1500 , V
V
L D
2500
L
1500
3
V
2500
5
2500 , Slope
Intersects Top Op. line at x S
.6
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 FM
V
V
FM
V
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.
108
Figure for problem 4D31 External Balance: F = B = 50, and Fz
4.D32*. 4.D33*. a.
L V
.75 .452 min
L
b.
L D
act
L V min
0.4 .
0.659
1 L V
1.318. L V
0, x
min
L D 1 L D
Top operating Line through y Bottom through y
z
0.397 (tangent pinch)
.75 D
Bx B . Thus x B
yB
x
0.569
yD
0.75
0.1 and intersection feed and top operating lines.
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 Feed:
Since saturated steam and CMO valid, B S L V Also have mass balances, S + F = B + D SyS Fz Bx B Dy D ys 0 Solve 3 eqs. simultaneously. S = 760 lbmoles/hr = 13,680 lb steam/hr.
109
4.D34*.
4-E1.
Trial and Error, Feed: L
L F F 2, q
Find (L/V)min (see diagram) 0.95 0.613 L V min 0.95 0
3 2, Slope
3.
Figure for 4.D34
0.3547 ,
L V min
L D
min
1 L V min
0.5497
110
L D
L
2
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 V
L D act
Bx B
0.6 0.025 0.95 0.025
100
D L L
V V´
62.162
L
h
, B = F – D = 37.838
L D D
68.030
V = L + D = 130.192
F
At feed V L″
V
kgmol
L
L´
V″
0.5237
1 L D
P=
xP
L
V (sat’d liquid)
L F 168.030
B
Top op line y 1st middle
L V
x
y
1 L F x D normal . At x = 0, y = 0.4525
L
x
L
1 x B looks like usual bottom! V V Goes through y x x B , and intersection top & feed line. Slope
L
168.030
1.2906 130.192 At pump-around return, V V 130.192 L L P 208.030, L V 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 L L Bottom Op line y x 1 x B , Same as first middle!!! V V Step off P.R. stage 1 & 2 above. 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 & 1st middle intersect.
111
Using MB:
yV
When
y
4-E2*. Feed:
q
Px P 0, x
Bx B P
Bx B
L V 40
x
P V
xP
Bx B V 32838 0.025
.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)
L L F
xP
Lx, y
, L
L 1.5F, q
3 2 , slope
q 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 y 0, x
3
xB
0.015
Middle op. line: V B Side L S or V L S B Side V y Bx B Side x side L x SyS Since
yS
0 this is, y
L
x
Bx B
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 B
V
S, B L Side Side B 1 0.4 1
S
SB
0.833
1.68
slope
Draw in the middle operating line. Step off 4 stages. Trial and error to find x D for final result).
.85 (see figure
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 L L 0.650, y x 1 xD V 1 L D 2.86 V V
1
L
xD .35 .8 .28, y x x D 0.8 V Step off 8 stages and find x S 0.495 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 x B
113
D Flows: L
L D
D
F z xB xD
.3 1000
xB
.78
385
716.1, V
L D 1101.1 V
L
L S
258.8, L V
L
L
L
V ,V
V
S
643.8
0.235
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 y y s to y x x B 0.02 . Check if slope L V 1.955 . Optimum feed is 10th 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.
4.E.4. New Problem in 3rd Edition. External Balance. a) F D B, Fz Dx D Bx B
D b)
z xB xD
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 R and V
L
L0
64.3
21.4
L
21.4
V LR
64.3 L L
y
c) Top Op. line:
V
0.333 1 3
38.6 21.43 17.17 kgmoles hour
L V
LR
x
D V
D
L0
L
xD
y
V
x, y
x
L
1
V
LR
D
V
L
xD
xD
x D sin ce V
L
LR
D
Envelope for top
V
L
LR
L
Plot
Middle:
y
Bottom:
y
V L V x
y
x
Slope
Slope
x
1 L V x D . Slope
L V 1 3, y x
38.6 64.3
x x D 0.9 L 0,y 1 xD V
0.600 y
x
xD
0.6
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.
4.E.5. New Problem in 3rd Edition. Bot. Op. Line: External M.B.: F
D
y
V B
At feed stage: L 687.5
V
937.5
V B
V
V
x
V
L V
D B and Fz
250 kmol day , B L
L
L
V
1
2.25
9 4
9
V B
1.25
54
5
1 xB, y Dx D
x
Bx B ,
750 kmol day , V
x B , Slope D F
z xB xD
xB
V B B
L V .3 0.10
.2
1
0.90 0.10
.8
4
1.25 750
937.5 kmol day
937.50 750 1687.5 kmol day V
V
937.5, L
L
F or L
L F 1687.5 1000
687.5
0.733333
116
MB:
L x Dy D
V y, D
V
L , y
L
Top op. line:
y
L V
x
L
1
S
L V
487.5, V
yD
V
1
L
yD V V Goes through y x y D , 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 Dy D Sx S L L x Dy D V y Sx S or y x V V 687.5 Find from intersection L V op line @ y yS plus slope 0.932 or intersection 737.5 top op line and x x S At side stream feed point: L
x
737.5. Thus,
Can draw, yint ercept
L
487.5
V
737.5
1 .661 .9
0.661
0.305
117
Graph for 4.E5.
4-F1.
970.33
L
V
Since bottoms are very pure h B
HS H Equil
v
h water @ 212°F
1381.4 1192
(in column)
Extra heat 189.4 Btu/lb S
B
Since y ≈ x, MW are same
118
Must vaporize material in column. extra heat MW 189.4 v S S MW 970.33
B
L v, V
0.1952S
S v
L
B v
BS vS
2 0.1952
V
S v
1 vS
1 0.1952
If super heat not included L V
BS
1.837
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 5.83 kcal / gmole, W 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. QC
QR
44, 437,300 Btu/hr,
49,859, 400 Btu/hr.
Optimum N F
17 & N
27 (Total condenser is #1)
x D 0.9992 and x B 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
119
10 11
0.974009 0.992216 0.991288 0.996557
4.H.2. New Problem in 3rd Edition. The VBA program is in the result is: xd 0.7 xb 0.0001 F 1000 L/D 6.94 q 0 partial reboiler total condenser Ethanol-water VLE 6th 5th 4th 3rd 2nd 1st -47.949 161.42 -212.43 138.68 -46.65 7.9322 yeqatxint 0.099219 yint 0.1 xint 0.0135447 stage x y Reflux rate too low Reflux rate too low
Appendix B of Chapter 4. With L/D = 6.94 z 0.1 feed stg 28 Prob. 4H3. constant 0
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
120
29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78
0.006211 0.006538 0.006865 0.00719 0.007513 0.007831 0.008144 0.00845 0.00875 0.009041 0.009323 0.009595 0.009858 0.010109 0.01035 0.010579 0.010797 0.011004 0.0112 0.011384 0.011558 0.011721 0.011874 0.012018 0.012151 0.012276 0.012392 0.0125 0.0126 0.012693 0.012779 0.012858 0.012932 0.012999 0.013062 0.01312 0.013173 0.013222 0.013267 0.013308 0.013346 0.013381 0.013413 0.013442 0.013469 0.013494 0.013516 0.013537 0.013556 0.013573
0.0475 0.049907 0.052301 0.054674 0.057017 0.059321 0.061579 0.063782 0.065925 0.068002 0.070008 0.071938 0.07379 0.075561 0.077251 0.078857 0.08038 0.08182 0.083179 0.084459 0.085661 0.086787 0.087841 0.088826 0.089743 0.090598 0.091392 0.092129 0.092812 0.093445 0.09403 0.094571 0.09507 0.095531 0.095955 0.096346 0.096705 0.097036 0.09734 0.09762 0.097876 0.098112 0.098328 0.098526 0.098708 0.098874 0.099027 0.099167 0.099295 0.099412
121
79 80 81 82 83 84 85 86 87 88 89 90 91 92 93
0.013589 0.013604 0.013617 0.013629 0.013641 0.013651 0.01366 0.013665 0.013706 0.014018 0.016416 0.034534 0.155188 0.490181 0.646064
0.099519 0.099618 0.099707 0.09979 0.099865 0.099934 0.099997 0.100032 0.100305 0.102401 0.11824 0.223718 0.516574 0.652848 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
122
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
123
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
124
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 z HNK F cross.
(frac. rec. HK in bot) z HK F, then there is more HNK in bottoms and curves
5A15. New Problem in 3rd Edition. a. a, b. d, c. b, d. c, e. b. 5.C1. New Problem in 3 rd 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 (α i-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).
125
5.D1. New Problem in 3rd Edition.
Dx M ,d
a)
Dx B,d
1.00 2, 000 0.19
2280
0
0
Dx E,d
0.978 2, 000 0.31
Dx P ,d
1 0.994 12, 000 0.27 D
3638.16
x i,d D
19.44 5937.6 kg/h
b) assume NK (Methanol and n-butanol) do not distribute (all methanol in top and all butanol in bottom).
wt frac. in distillate: M = 0.38399, E = 0.61273, P = 0.00327, B = 0.0.
wt. frac Bx M ,B
0
Bx B,B
1.0 12, 000 0.23
Bx E ,B
1 0.978 12, 000 0.31
Bx P ,B
0.994 12, 000 0.27
x i,b
0
B
0
2760.0
0.4553
81.84
0.0135
3220.56
0.5312
x i,b B 6062.4 kg hr
Check B+D=F, OK. 5.D2.
New Problem in 3rd Edition.
y4
yi P
Raoult’s Law Antoine’s Eqn.
log10 VP
Dew Point condition is
xi
yi P VPi
0.30, y 5
xi
A
0.5, y 6
0.20, P
760 mmHg
VPi x i B T C
1
(from Raoult’s Law)
126
B
(from Antoine’s Eqn) T C Combine with Dew Point condition y5P y6P y4P 1 B B B
VPi
10
A4
10 A
4
T C4
10
0.30 760 10
6.809
5
A5
T C5
10
A6
6
T C6
0.50 760
935.86 T 238.73
10
6.853
0.20 760
1064.8 T 233.01
10
6.876
1171.17
1
T 224.41
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. x M,dist .99, x E,dist .01 2195.6 = (.998) (0.22) (10,000) = Dx MD
D
2195.6
2195.6
2217.78 and B x MD .99 Bottoms: MeOH: .0021 (.22) (10,000) = 4.4 4.4 x Mbot 0.0006 7782.22 1.0 .18 10, 000 x n propanol,bot 0.2313 7782.22 1.0 .13 10, 000 x n butanol,bot .1670 7782.22 x EtOH,bot 1 x MeOH,bot x n p,bot x nbut,bot 0.6011 5.D4.
a. .99 F z C5
F D
7782.22
D x C5,dist (1), and .01 Fz C5 =B x C5,bot (2)
.98 F z C6,bot
Bx C6,bot (3), and .02 F z C6 D z C6,dist (4) Assume all heptane in bottoms Fz C7 Bx bot,C6 (5), x dist,C7 0 (6) Take Eqs. 1, 4 & 6: .99 (1000) (.4) = Dx C5d .02 (1000) (.3) = D x C6d 0 = D x C7,dist
Dx l,dist b.
D
402 kg moles/hr
B = F – D = 1000 – 402 = 598 x C7,dist 0
x C5,d x C6,d
.99 1000 .4
0.9851 402 1 0.9851 0.0149
127
x C7,bot
1000 .3 1.0 598 .98 1000 .3
x C6,bot
c.
5.D5.
0.5017
0.4916 598 x C5,bot 1 .5017 .4916 0.0067 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
Assume all methanol and ethanol in distillate. Dx MeOH,dist 0.55 100 0.01 150
Dx EtOH,dist
0.21 100
Dx prop,dist
0.993
Dx bu tan ol,dis
0.03 150 0.23 100
1 0.995
56.5
25.5
0.26 150
0.01 100
61.57
0.70 150
0.53
D 144.1 B Check:
F1
F2
D 105.9
Bx Pr op,bot
1 0.993
Bx but,dist
.23 100 .26 150
0.995 .01 100
0.7 150
0.434
105.47
Check = 105.90
Mole fractions:
x M ,bot
0 , x E ,bot
x but,bot
1 x prop,bot
0.9959
Dx MeOH ,dist
56.5
x M ,dist x E,dist x p,dist
x But,dist
0, x prop,bot
D 144.1 25.5 0.1767 144.1 61.566 0.427 144.1 0.53 144.1 0.0037
Check
= 1.000
0.434 105.90
0.0041
0.392
OK
5.D6. This is 8.D1. in 1st ed.
128
129
5.D7.
Assume 100% recovery C 2 , & propylene in distillate. Assume 100% recovery pentane & hexane in bottoms. Comp.
Distillate
C2 Propylene: n-C3 n-C4 C5 & C6
0.3 (1000) + 0.02 (1500) 0.006 (1000) + 0.001 (1500) (0.991) [1000 (0.45) + 1500 (0.249)] (0.02) [1000 (0.154) + 1500 (0.40)] 0
Bottoms flow rate = F1
x B,C2
0, x B,propylene
F2
D
x B,C5
0, x B,C3
1000 0.09
= 330 = 7.5 = 816.0885 = 15.08 = 15.08 D= = 1168.7
.009 1000 0.45
1500 0.18
1331.33
xD 0.2824 0.0064 0.06983 0.0129 0.0
2500 1168.7 1331.3 kg/h
x B,C4
5.D8.
kg/h
1500 0.249
1331.3 .98 1000 .154 1500 .40 1331.3 0.2704,
x B,C6
0.0056 0.5550
1500 .15 1331.31
0.1690
8.D.6. in 1st edition. Assume all benzene is in the distillate.
130
131
5.D.9. New Problem in 3rd Edition.
132
5.D10. New Problem in 3rd Edition. At the bubble point
x C5
.40 , x C6
.60 , p
yi
760 mmHg , y C5
1.0
y C6
K 5 x C5
K 6 x C6
1.0
133
or
VPC5 Ptot
x C5
0.40 10 TºC 20 51
VPC6 Ptot 6.853
1.0 , VPC5 x C5
x C6
1064.8 T 233.01
0.60 10
(C5 term) x (.4) 441.03 × .4 = 176.4
Final result is:
6.876
VPC6 x C6
Ptot
1171.17 T 224.41
+
1270.095 × .4 = 508.04 +
760
C6 term x (.6) 121.387 × .6 =
SUM 249.23
420.28 × .6 = 252.17
760.21
5.D11. Was problem 6.D2 in 2nd edition SPE. Since x i known, want
yi
1
K i p BP x i
K ref p old
Find new pressure from, K ref p new
Kixi
Use ethane, or n-pentane as reference. First guess: Try K C5 1.0, p 115 kPa
K NC7
0.13
Ki xi
K C2
29 0.1
K rep p new
Ki xi K C5 p new
29
1.0 .35
0.13 .55
1.0
~ 0.3 p new 3.32 8.0 0.1 0.3 .35
3.32 p too low.
440 kPa , K NC7
0.042 0.55
0.042
K C2
8.0
0.927
0.3
.032 p new 400 kPa, K nC7 0.045 K C2 0.927 K i x i 8.7 0.1 0.32 0.35 0.045 0.55 1.004
8.7
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. K C8 b. Lowest B.P. Temp. is pure n-hexane. K C6
1.0, T 174 C
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. K1 1 at T 1 , K 2 1 at T 36 , K 3 1 at T 68 .
Tfirst K1
z1T1 3.6, K 2
.2
1
1.08, K 3
.5 36 0.36.
Choose 2 as ref. Eq. (6-14) is: K 2 Tnew
Tnew
29 C. K1
2.7, K 5
Eq. (6-14) is: K 2 Tnew
Tnew
28.8 K1
K1x1
38
.2 3.6
1.8 1.368
0.26, and
0.789 1.013
2.7, K8
.3 68
Kixi
.5 1.08
.3 .36
1.368
0.789 1.013
0.779
0.255, and
Kx i i
1.006. OK. TBP
28.8 C.
5. E1. This is 8.E4. in 1st edition.
134
135
136
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 Alpha CA-B 3.58 alpha B-B 1.86 B 1 feedstage 8 zA 0.35 z B 0.4 z C 0.25 epsilon (values for 0.00000001 N loop( F 200 q 1 L/D 6 convergence 100 df(HNK frac frac rec B in dist 0.996 guess frac rec C bot 1 recovery) 0.9 frac rec A in dist 0.961 D 67.59011 B 132.4099 L/V 0.857143 Lbar/Vbar 1.279858988 Mass balance xAdist 0.995264 xBdist 0.004734 xCdist 1.56E-06 values Mass balance xAbot 0.020618 xBbot 0.601768 xCbot 0.377614 values
137
Stage by stage calculations i xA yA 1 0.020618 2 0.041225 3 0.072406 4 0.11662 5 0.175066 6 0.245778 7 0.322624 8 0.396885 9 0.515565 10 0.633439 11 0.738112 12 0.822076 13 0.883953 14 0.926687 15 0.954865 16 0.97287 17 0.984143 18 0.991109
0.046992 0.0869 0.143487 0.218289 0.308791 0.407144 0.502187 0.584094 0.685129 0.774848 0.846817 0.899855 0.936484 0.960636 0.97607 0.985732 0.991702 0.995362
xB yB xC yC 0.601768 0.7125981 0.377614 0.688364 0.7538807 0.270411 0.720619 0.7419438 0.206975 0.711292 0.6917342 0.172088 0.672062 0.6158891 0.152873 0.612801 0.5274175 0.141421 0.543675 0.4396808 0.1337 0.475123 0.3632905 0.127992 0.42305 0.2920855 0.061385 0.339977 0.2160685 0.026583 0.251291 0.1497869 0.010597 0.173962 0.098934 0.003962 0.114634 0.0630978 0.001413 0.072825 0.0392226 0.000488 0.044971 0.0238835 0.000164 0.027075 0.0142529 5.45E-05 0.015839 0.0082925 1.77E-05 0.008886 0.0046363 5.57E-06
0.240409437 0.159219617 0.114569647 0.089976459 0.075319919 0.0654388 0.058132319 0.05261578 0.02278591 0.009083172 0.003395923 0.001211305 0.000418127 0.000141178 4.69552E-05 1.54307E-05 4.9941E-06 1.56158E-06
Calc frac recovery C in bottoms 0.9999979 j 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 zA 0.25 zB 0.35 zC 0.4 epsilon (values for F 100 q 1 L/D 0.3 N loop( convergence frac rec B in dist 0.9 frac rec C in bot 0.97 df(LNK frac recovery) Guess: frac rec A in dist 1 D 57.66689 B 42.35298 L/V 0.230769 Lbar/Vbar 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 1 0.229688 0.43295 0.65203 0.546241 0.118282 0.020809168 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
Note x1 = xbot
3
6 0.0001 10 0.9 1.565105
Note y1 = xdist
2
138
5.H.3. New Problem in 3rd Edition. (L/D)min = 0.26761 by trial and error using spreadsheet in Table 5.A1.. 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 zA 0.25 zB 0.35 zC 0.4 epsilon (values for F 100 q 1 L/D 5 N loop( convergence frac rec B in dist 0.583 frac rec C in bot 0.995 df(LNK frac recovery) Guess: frac rec A in dist 0.95 D 44.10818 B 55.89182 L/V 0.833333 Lbar/Vbar 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 1 0.447934 0.532853 0.544443 0.462613 0.007623 0.004534307 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
19 1E-10 100 0.8 1.21119
Note y1 = xdist
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
139
zA 0.38 zB 0.02 zC 0.6 epsilon (values for F 100 q 1 L/D 4 N loop( convergence frac rec B in dist 0.725 frac rec C in bot 0.995 df(LNK frac recovery) Guess: frac rec A in dist 0.99 D 39.37044 B 60.62956 L/V 0.8 Lbar/Vbar 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 1 0.934659 0.95555 0.050434 0.03683 0.014907 0.00761993 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
1E-10 100 0.8 1.308
Note y1 = xdist
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 zA 0.3 zB 0.3 zC 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
140
frac rec A in dist K const. aT1 nB=A -1280557 nPen=B -1524891 nHex=C -1778901
aT2 0 0 0
0.995 aT6 7.94986 7.33129 6.96783
Trebguess ap1 -0.96455 -0.89143 -0.84634
500 ap2 0 0 0
p, psia ap3 0 0 0
14.7
D xAdist xAbot
29.940006 B 0.9969938 xBdist 0.002141 xBbot
70.05999 L/V 0.003006 xCdist 0.42692 xCbot
0.88889 Lbar/Vbar 1.26 2.2E-07 0.57094
stage 1 2 3 4 5 6 7 8 9 10
xA 0.002141 0.0086951 0.0291201 0.0853111 0.214736 0.4743433 0.7600825 0.920961 0.9779073 0.994401
xB 0.42692 0.613018 0.714099 0.717438 0.617813 0.483106 0.233736 0.078451 0.022047 0.005596
xC 0.57094 0.37829 0.25678 0.19725 0.16745 0.04255 0.00618 0.00059 4.6E-05 3.4E-06
yA 0.0103991 0.0361347 0.1069354 0.270011 0.5324156 0.786406 0.9294091 0.980028 0.9946891 0.9986729
Distillate mole fracs = y values Calc frac recovery C in bottoms
yB 0.6614042 0.7887656 0.7929732 0.6674458 0.4297616 0.2080991 0.0700679 0.0199309 0.0053079 0.0013269
yC 0.3282 0.1751 0.10009 0.06254 0.03782 0.00549 0.00052 4.1E-05 3E-06 2.2E-07
T 582.283 570.622 561.854 551.835 536.481 513.748 498.351 491.769 489.686 489.103
KB 1.549247 1.286692 1.110453 0.930318 0.695617 0.430752 0.299774 0.254057 0.240759 0.237137
0.9986729 0.00133 2.2E-07 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
141
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 convergence of 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
142
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
143
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
144
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, Vj y j L j x j Sx S Vj 1 , y j 1 L j 1x j
Fjz j
1
where L j is the flow into the stage below, L j (6-4) to (6-6) are unchanged. Note that the L j b.
Now L jh j
Lj
Vj
1
Fj h Fj
1
L j x j , Eqs.
input into the matrix will be changed.
L jh j S jh j in the E.B. and j
D
j
Fk
k 1 k Substituting into EB we find
DEj
x j . Since L j x j Sx S
L j S, x S
Qj
D hj
j 1 k
1
1
Sk , L j
j
hj
1
SK h j
k j
1
k
1
Vj
1
1
j 1
D
FK h j
k
1
j 1 k
1
j 1
Fk
k
FK h j
1
Sk
1
SK h j
6.C2. Partial condenser mass balances is: Dy1 L1x1 V2 y 2 F1z1 This becomes, DK1 V2 K 2 2 F1z1 1 1 L1 L2 Thus,
B1
1
DK1 L1
, C1
(6-7)
K 2 V2 L2
, and D1
(6-8)
F1z1
Note that only B1 differs. 6.D1.
For n-pentane from Example 6-1, T = 60°C, K C5 Matrix:
L3 1825, L 4 450 B, V2 j = 1 (total condenser),
V3
1.05, L1 V4
L2
825 kmole/hr,
1375.
145
1.05 1375
K 2 V2
C1
L2
j
2, C 2
B2
1
825
j
1
K 3 V3
1.05 1375
L3
1825
V2 K 2
2.75, A 2
L2
j 3, C 3 B3
L4
450
1.791, A 3
L3
1.75
1
2.75
0
6.D2.
1,C5
p = 5 atm: z C2
0 0.791
1
0
1.791
0
to
0.08, z C3
0.33, z C4
yi
1, K C4
5.4 0.08
yi
1.7 0.33
12 , K C2
0
2,C5
0
3,C5
350
4,C5
0
0.49, z C5
0.10
yi
1.0.
506.5 kPa 0.47, K C5
0.47 0.49 Kixi
0.14
0.14 0.10
K C3 20
4.6, K C4
1.7
1.37
1.237
0.35, K C5
1.237
0.10
0.368 0.4521 0.1715 0.0 1.0016 OK
Propane Matrix Analysis: K C3
D3
Fz c3
V2
V3
L1
0
1.7, K C4
K C3 Tnew
Tnew
350
1. DePriester Chart.
5.4, K C3
Need lower T.
0
0
1,C5
x i . Want
Pick C 3 as ref. 5 atm 101.3 kPa atm Try T = 20°C. K C2
1, D 4
0
4.208
As sat’d liquid & for bp calculate z i 1 Guess: Want K C3
1000 0.35
3.208
1
1.67, D1
825
are in Example 6-1.
4,C5
st
Fz C51
4.208, A 4
L4
1.67
L1
550
3.208
1, D 3
1 V4 K 4
1
0
1.05 1375
V3 K 3
D
1
0.791
1, D 2
K 4 V4
4 Reboiler , B 4
Values for
1.75 , B1
1.37, B
330, D1 V4
L1
D L3 L 2 F Total Condenser (1):
D
V5
D2 V6
1025
2025
L4
F D 1000 410 D4
L1 L2
D5
D6
590
L6
0
D 1435 L3
L5
146
B1 1 D L1 1.40, C1 Stage 2.
A2
Stage 3. A 3
1, B2 1, B3
V2 K 2
1
L3
Stage 4:
A4
1, B 4
1
Stage 5:
A5
1, B5
1
Reboiler (Stage 6). A 6
V4 K 4
L5
1, B6
2.918
0
1
1435 1.37
L3
2025
V4 K 4
1435 1.37
L4
2025 V5 K 5
V6 K 6
1435 1.37
V6 K 6 L6
590
0
0
0
0
-0.9708
0
0
0
0
0
0.9708
0
0
1
0
0
0
-1
0
0
0
0
0.9708 0.9708
3.32
=4.32
L6
1.9708
1.918
0.9708
L5
=1.9708, C 5
1
1025
V3 K 3
=1.9708, C 4
L4
Mass balance matrix. 1.40 -1.918
-1
=1.9708, C 3
V5 K 5
1.37 1435
=2.918, C 2
L2
V3 K 3
1
K 2 V2 L 2
1.9708
0.9708
0
1.9708 -3.32 -1
4.32
6.D3. New Problem in 3rd Edition. p = 5 atm = 506.5 kPa z C2 0.08, z C3 0.33, z C4 0.49, z C5 0.10 As sat’d liquid & for bp calculation at z i Result is: Tbp
12 , K C2
4.6, K C3 =1.37, K C4
n-butane Matrix Analysis: K C4
D3
Fz c3
V2
V3
490, D1 V4
V5
x i . Calculation is same as in 5.D11 to obtain T.
0.35, B
D2
V6
D4
L1
0.35, K C5
F D 1000 410
D5
D6
A2
Stage3. A 3 Stage 4.
1, B 2 1, B3
A4
1
1, B 4
1
V2 K 2 L2
V3 K 3 L3 1
=1.49, C 2
=1.2480, C 3
V4 K 4 L4
L6
D 1435
L1
Stage 2.
590
0
D 1025 L 2 , L3 L 2 F 2025 D Total Condenser (1): B1 1 D L1 1.40, C1 K 2 V2 L 2 L1
0.10
=1.2480, C 4
L4
L5
0.35 1435
V3 K 3 L3
1025
1435 0.35 2025 1435 0.35
V4 K 4 L4
2025 V5 K 5 L5
0.49 0.2480 0.2480
0.2480
147
Stage5: A 5
1, B5
V5 K 5
1
L5
Reboiler (Stage 6). A 6
1, B 6
Mass balance matrix. 1.40 -0.49
V6 K 6
1
L6
1435 0.35
V6 K 6
=1.2480, C 5
L6
=1.8513
0
0
0
0
0
0
0
0
0
-1
1.49
-0.248
0
1
1.248
0.248
1
0
0
0
0
0
-1
1.248 -0.851
0
0
0
0
-1
5 60
300
6.D4. New Problem in 3rd Edition. L
1.248
L D D
V L D 360 Saturated liquid feed: V V
360; L
0.8513
590
L F
0.248
0 1.851
400
1
L1 V1
V1
L, L 2 0, V2
L, L3
L, L 4
V, V3
B
V, V4
F D
40
V
2
L2
V3 3 yi
V4
Bubble Pt. Set z i F
1.0 or
Ki xi
L3
xi
yi
Ki xi
1.0 M
3.58,
E
2.17,
NP
1,
NB
0.412
4
b.
yi
xi
i
xi
Eq. (5-30), zi i .3 3.58 Then and
y nP K nP
z NP
i
.25 2.17 NP
zi y x
i
nP
.35 1.0
0.1 .412
2.0077
0.35 1.0
0.1743 2.0077 0.1743 0.4981 0.35 148
KM c.
M NP
K nB
Matrix for n-butanol Stage 1. A1 , B1
1, B 2
L2
300
V4 K 4
C3
F3z nBut
Stage 4. A 4
1.2 1 0 0 d.
y 1 j
2
j=3
1.8468
40
10
1
V4 K 4
2.8468, C 4 , D 4
B 0
0.2463 1.2463
0.1847
1
1.1847
0
1 1.2, V21
V12
B2
A2
D2
C 2 V11
V13
B3
V23 V33
D3
A3
0
2
0
3
10
4
0
1
1
.2463
1.2 1.2463 1
V12
0
1
V13
10
1.8468 1.04105
.20525
.20525 0
0.1847 0.1539 1.2 V3 2 1.1847 1
A3V22
C3 V12
0
0, V31
V3
A 2 V2
V32
1
2.8468
B1
0
0 1.8468
V11
V22
0
1.1847
360 0.2052
1 B4
0.2052
0
0.1847, D 2
400
L3
100 .1
0.2463, D1
360 .2052
V3 K 3
1
K NP
1.2463
L2
L3
L4
D3
V2 K 2
V3 K 3
1, B3
nB NP
1 0.2 1.2,
360 0.2052
1
1.0809, K n-B
0.1
V2 K 2
C2
Stage 3. A 3
1.7832; K E
0.2052, z nB
1 D L1
C1
Stage 2. A 2
K NP
1.04105
0
0.1539
1.0308
1 0 1.0308 9.7014 1.7740
149
V14
j=4
B4
V24
D4
V34
e.
VP
V3
3
A 4 V23
C4 V13
2.8468
V14
0
1.7740
1 9.7014
1.0728
1.0728 9.0428
not needed. 9.0428 (bottoms flow rate)
V24
N
3NB
V23
V33
4
9.7014
2 NB
V22
V33
3
0
0.1539 25.743
3.9619
1NB
V22
V31
2
0
.20525 3.9619
0.8132
VP
VP
nP
Ptot
K nP 760
1.7740 9.0428
n 200
760
0.4981 760
378.5 mmHg
5.2983, n 378.5
1 66.8 273.16 Linear Interpolation: 5.9362
TbP
5.2983
273.16
25.743
in mmHg
Need to interpolate VP data. We know n VP
6.F1.
1
4,NB
K nP
Raoult’s Law
A4
1T 5.9362, n 400
2.9415E 3
5.99146
1 82.0+273.16
2.9415E 3 2.8163E 3 5.2983 5.99146
351.74 or TbP
2.81563E 3
2.9415E 3
0.00284
78.6 C.
Plots of vapor pressure are available in Maxwell (see Table 2-2 for reference) while tabulated values are in Perry’s K VPi p tot . 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 245.7 70 = B 80 2.307 1.042 4 0.534 245.7 150 180 2.307 1.042 3 0.534 245.7 150 180 2.307 1.042 2 0.534 1 245.7 150 30 = D 2.307 1.042 0.534 Stage 4 3 2 1 Stage 4 3 2 1
C -2.705 -2.8404 -2.8404
Benzene B 3.705 3.8404 3.8404 1.2
C -1.191 -1.2504 -1.2504
Toluene B 2.191 2.2504 2.2504 1.2
A -1 -1 -1 A -1 -1 -1 -
D 0 35 0 0 D 0 40 0 0
ℓ 7.9886 29.5978 57.058 135.6569 ℓ 27.1651 59.5188 61.5875 64.1742 150
Stage 4 3 2 1 6.G1.
6.G2.
C -.6103 -.6408 -.6408
Xylenes B 1.6103 1.6408 1.6408 1.2
A -1 -1 -
D 0 25 0 0
ℓ 22.7361 36.6120 21.1971 11.3193
Using Peng-Robinson. Aspen-Plus solution:
Stage
T1°C
L kmol/h
V
C4
C5
C8
1
38.31
825
0
x1 y1
0.360 0.6568
0.6013 0.3424
0.0386 0.00083
2
69.16
557.3
1375
x2 y2
0.0993 0.3601
0.4499 0.6013
0.4508 0.0386
3
107.02
1533.9
1107.3
x3 y3
0.03355 0.2288
0.1731 0.5251
0.7934 0.2461
4
140.92
450
1083.9
x4 y4
0.00436 0.04568
0.0429 0.2271
0.9528 0.7272
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 C2 0.3636, C3 Bottoms product mole fractions C2
1.2 E 13, C3
0.6360, C4
0.0004
0.000492, C4
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 C2 0.9955, C3 0.00448, C4 1.32 E 07 Bottoms product mole fractions C2
0.00112, C3
0.4364, C4
0.5625
151
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 what is 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 152
[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 nbutane. 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_______________________
153
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__
154
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 yi, j
1
K i x i, j 1 . Then substituting into Eq. (7-20), we have Vmin K i x i, j
which is
L min x i, j
Vmin K i x i, j 1
L min x ij
1
Vmin K i K HK
1
Dx i,dist
Dx i,dist where K i
1
Kx i,dist , or L min x i, j
L min
c
Total flow rate L min is L min
i 1
L min x ij
L feed
7.C5.
VF
For saturated vapor feed have
L min
i
i
(A)
1 i
i
(B)
1 i
(C)
c i 1
L min
1/
i 1
Dx i,dist 1/
Fz i
c
i
(7-33 analogue)
i L
i
i
(7-29 analogue)
L
F . For binary system Eq. (7-33) is, z
1
1
Vmin K
L
From Eqs. (A) and (C) we have L min
i
Dx i,dist V K i 1 min HK L min
L min L min Add Eqs. (A) and (B), and use external mass balance, qF
K HK . Rearranging,
c
1
Vmin K HK L min
Vmin K1
L
i
Dx i,dist Vmin K HK L min
1
Bx i ,bot
L min
By a similar analysis obtain,
Let
L min x ij
1
1 1
2
1
z2
2
Clearing fractions we obtain 1
2
z
2
z
1 1
2 2
1
After some algebra this
z
2
Solutions are, For sat’d liq’d
1 1
0 or VF
1
z
2 2
2
1
z
1 1
2
2
0
z2
0 . Clear fractions and equation is linear.
155
x 1 x
n 7.D1.
dist
x 1 x
a. Eq. (7-16), N min
n
bot
.992
.008
.014
.986
10.36
n AB n 2.4 This includes the partial reboiler. Eq. (7-40a) gives, x 1 x
n
z 1 z
N f ,min b. Saturated liquid:
Vf
dist
n
B
B
zB
T
B
,
zT
0
T
1.0 2.4
zB
zB
T
5.97
n 2.4
AB
After clearing fractions and solving for
.008
.4 .6
feed
0 . Eq. (7-33) becomes
T
.992
n
zT
2.4 .4
1.0 .6
1.53846
Which does lie between the α’s of the keys. To use Eq. (7-29) we need D. From mass balances (Eq. (3-3)).
z
D Eq (7-29) is:
x bot
x dist
Vmin
B
x bot
.4 .014
F
.992 .014
Dx Bdist
T
Dx Tdist
B
Vmin
L min c.
L D Using Eq. 7-42b,
1.1 L D N
3.9468 kg moles/hr
.
T
2.4 3.9468 .992 2.4 1.53846 Vmin D 6.9013 .
min
10
1.9234 , x
1.0 3.9468 .008 1 1.53846 L D min 1.75
L D
L D
min
L D 1
10.848
0.0598
N min
0.5563 N 1 .5563 N min Solving for N, N 24.6 (includes reboiler) 1 .5563 N F,min 5.97 From Eq. (7-40b), N F N 24.6 14.2 N min 10.36 Try stage 14 from top for feed stage. 7.D.2.
New problem in 3rd edition.
p
5 atm, z C2
Saturated liquid and for bp. Calc., z i
0.08, z C3
x i . Want
Pick C 3 as reference (this is arbitrary). 5 atm 1st Guess: Want K C3
1 (light key), K C4
yi
101.3
0.33, z C4
0.49, z C5
0.10
1.0 .
kPa
506.5 kPa atm 1 (heavy key). Use DePriester Chart.
156
20 C , K C2
Try T
yi
1.7 , K C4
5.4 , K C3
5.4 0.8
1.7 0.33
0.47 , K C5
0.47 0.49
0.14 0.10
0.432 0.561 0.230 0.014 1.237 K C3 20 1.7 Need lower T. K C3 Tnew 1.237 Kixi Tnew
12 , K C2
yi
n N MIN
b)
C2 C4
0.35, K C5
C4
HK be reference.
0.10
1.37 0.35
C3C4
3.914
0.997 0.998 0.003 0.002
12.01874
n 3.914
0
i
C4 C4
1.0,
3.914 0.33
13.14
i
3.914
Solve for φ. Find φ = 1.74 (Note i
VMIN
0.1 0.35
C5 C4
0.286
VMIN . In Eq (7-33) divide through by F. 13.14 0.08
zi
8.808 (includes PR).
1.36456
4.6 0.35 13.143,
Sat’d liquid feed VMIN
Eq (7-29)
1.37
0.368 0.4521 0.175 0.0 1.0016
For remainder, let
a)
4.6, K C4
0.14
1.0
3.914
LK HK
Dx i,dist
1.0 0.49
0.10 0.286 0.286
HK HK
)
. Assume all C 2 in distillate & all C 5 in bottoms
i
Dx i,C2
Dx i Dx i
80, Dx i,C5 C3
C4
0
0.997 1000 0.33 1 0.998 1000 0.49
D VMIN
c)
13.14 80
329.01 3.914
13.14 1.74 L MIN VMIN
L MIN
Eq (7-42b)
N N
Dx i,d
0.98
409.99
0.98 1.0
0 683.23 kmole h 3.914 1.74 1.0 1.74 D 273.24; L D MIN 0.6664
1.15 L D
Ordinate Gilliland
329.01
MIN
0.7664
0.7664 0.6664 1.7664
N MIN
0.05662
0.545827 0.591422 0.05662
N 1 0.5608 8.808 1 0.5608
0.002743 0.05662
0.56079
21.33 (include PR)
157
Dx LK Dx HK
n N F,MIN Eq. (7-40b)
n
N F,min
NF
N min
N
1 N min AB
x 1 x
n 7.D4.
x 1 x
N min
a.)
dist bot
log
xA xB
n
n3.914
.552
NF
xA xB
d
xA xB
d
xA xB
bot
bot
.01773 0.98227
y
L D
L D
13
.36 .64
1.287
y*
10.02
0.9915 y* 0.9915 0.6 x
1
1 x
(L/V) min = 0.534,
c.) Abscissa =
.545 .455
.9915 .0085
(L/V)min = * z = .6
bot
AB
1 N min
n AB n 2.4 b.) Feed is saturated liquid, feed line is vertical.
y*
xA xB
d
log
xA xB
4.552
21.33 11 (approximate)
8.808
7.D3. At total reflux use Fenske Eq. (7-11). N min
AB
329.01 0.98 0.33 0.49
n
LK HK
log
Rearrange, log
dist
z LK z HK
0.7826 x z 0.6
L D
L V min
1
L V
1.144 min
2.2286 1.144
0.336 L D 1 3.2286 From Eq. (7-42b) N N min 0.002743 0.545827 0.591422 0.336 0.3553 ordinate N 1 0.336 N min ordinate 10.022 0.3553 N 16.1 1 ordinate 1 0.3553 10.022 0.3474 From fitted curve ordinate = 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. min
158
log 7.D5.
Fenske Eq. is: N min
xD 1 xD
xB 1 xB
log
30,
7.D6.
N min
1.30 and x D xD 1 xD
log Fenske Eq.: N min
xD 1 xD
0.984, this is x B
xB 1 xB
log
2.4 .4
x 1
1 x
xD
L V
xD
min
Then,
x
.01 .99
.993 .616
z
.993 .4
1.15 1.746
0.636 ,
0.4 .
L D
L V 1 L V
min
1.746 . min
2.01
act
L D
Gilliland Correlation: Abscissa
L D
2.01 1.746
min
L D 1 3.01 N N min N 10.82 Ordinate 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: Tbp
K C2 Let C 4
4.6, K C3
1.37, K C4
HK
reference.
z
Eq. 7-33, 0
i i i
Want
LK HK
0.35, K C5
0.10
0.02 0.008
8.7121 1.36456
4.6 0.35 13.143,
C2-C4
0.10 0.35
C4 C4
6.38
1.0
0.286
Vmin
13.14 0.08 13.14 3.914
.0878
12 C
n 3.914
C5 C4
Sat’d/liquid feed, Vmin
3.01
3.914 0.35 0.98 0.992
C3 C 4
a) Eq. (7-15), Including PR N min
C4
.264
1.37
HK be reference.
n
b)
10.82
2.4
.616
1.4 .4
y*
L D
1
0.229
.993 .007
log
log
Determine y in equilibrium with feed z
y*
xB 1 xB
xD 1 xD
Solving for x B , we obtain: x B Since N min
xD 1 xD
N min
or
3.914 0.33 3.914
1.0
HK HK
1.0 .49 1.0
. Converge to
0.10 0.286 0.286 = 1.74.
159
Dx i
C2
Dx i
C4
Fzi
0.98 1000 0.33
C3
.008 1000 0.49
13.14 80
Eq. (7-29), Vmin
80, Dx i
323.4 3.914
3.92, Dx i
C5
323.4 0
3.92 1.0
0 671.05 13.14 1.74 3.914 1.74 1.0 1.74 D Dx i,d 407.32, Lmin Vmin D 263.73, L D min
c)
L D 1.2 L D N
Eq. (7-42b),
N min
0.002743
0.545827 0.591422 0.073
0.073
0.073
0.5402
15.05, incl. PR. Nfeed ~ 9 1 0.5402 a. Can do this graphically, or can calculate slope of a line from y x x D .992 to intersection of feed line and equilibrium, or use Underwood. Easiest to calculate slope. Feed line y z F .4 . Equilibrium: x
y y
V
c.
min
N min
.992 .3755
1.2
22.83 N
L
.958 ,
xB 1 xB
D
L V min
L D
27.4 , Abscissa
L D
22.83
1 L V
.992 .005 .008 .995 n 1.11
n This is 95.9 stages plus partial reboiler.
L D
.3755
.4 1.11 .6
.992 .4 xD 1 xD
n b.
.4
1 y
L
min
L D 1
96.9
27.4 22.83 27.4 1
.161
N min
or N 181.9 which includes partial reboiler. N 1 This separation would probably not be done by distillation. LF Fz Feed 80% liquid, L F .8F, , Slope VF .2F. Feed line: y x VF VF Ordinate
7.D9. a.
0.777 0.647 1.777
N 1 0.5402 6.38
N
7.D8.
0.777 . Ordinate Gilliland
min
0.647
See Graph.
.47
L V
L D b. c.
N min
6
L D
L
Min top op line is tangent.
3 4
min
1
min
L V
min
V
slope min
0.5175 1 .5175
.8 0.386 0.8 0
8 2
4
0.5175
1.0725
eq. contacts. See graph.
1.05 1.0725
1.1262 . Abscissa, Gilliland Correlation is
actual
160
L D
L D
min
1.1262 1.0725
0.053666
2.1262
2.1262
L D 1 Ordinate ~ 0.63 from graph. From eq. (7-42b), Ordinate
0.02524
0.545827 0.591422 0.02524
(agrees with graph). N min ordinate 6.75 0.6396 Then N 20.5 1 ordinate 1 0.6396 Need 20 eqs. contacts + P.R. N F,min 6 20.5 18 N F,min from graph = 6. N F N N min 6.75
0.002743 0.02524
0.6396
(7-40b)
7.D10. a) New problem in 3rd edition. Eq. (7-15)
n N MIN
FR E ,dist
FR B,bot
1 FR E ,dist 1 FR B,bot n
EB
161
0.989 0.998
n EB
b)
N MIN
0.011 0.002
N MIN
13.14
n 13.14
4.159 is known..
PB
2.5756
FR B,bot
Dx D
4.159
PB N MIN
1 FR B,bot c)
4.159
3.91. N MIN
Eq. (7-17) FR P ,dist
10.7114
PB
3.91 0.998 3.91 0.002
4.159
0.3677
FR i,dist Fzi
i
Ethane
Dx DE
0.989 100 0.3
29.67
Propane
Dx DP
0.3677 100 0.33
12.134
n-butane
Dx DB
0.002 100 0.37 3
D Also accept D = 0 since total reflux.
i 1
0.074
Dx i,d
41.878
kmol h
7.D.11. New problem in 3rd edition.
D 200
zn
0.35
z iP
0.4
z NP
2
V1
0.25
1 B Use Underwood Eqns. – Case A Assume LNK (propane) is all in distillate. b)
Vfeed
F 1 q
Eq. (7-33). F 1
F since q i
F zi
Dx p,dist
Fz p
20
0
where φ is between α’s of two keys (B and H)
i
1.0 > φ > 0.2. Equation is, 2.04 0.2 1.0 .35 1.0 2.04 1.0 Solving for φ obtain φ = 0.62185.
0.20 0.45 0.20
162
Then
Vmin
Dx B,dist
Dx i,dist
. Find D from fractional recoveries.
0.99 100 .35
34.65
Dx p,dist
20
Dx H ,dist
1 0.98 100 .45
0.9
D
VMIN
L min 7.D12.
2.04 20
55.55
1.0 34.65
0.2 0.9
2.04 0.62188 1.0 0.62188 0.2 0.62188 VMIN D 64.4314 and L D min 1.1599
A = benzene (LK),
AB
2.25, FR A,dist
B = toluene (HK),
BB
1.0, FR B,bot
C = cumene (HNK),
CB
0.210
0.98 0.99
n a. Use Fenske eqn. at total reflux. N min
FR A ,dist
1 FR B,bot
1 FR A ,dist
FR B,bot
n 0.98 0.02
n N min
0.01 0.99
FR A ,dist
AC N min
where
AC
AC
1 FR A ,dist
AB
10.47
n 2.25
N min
FR C,bot
119.98
AB
2.25
CB
0.21
10.71
10.47
10.71 FR C,bot 0.98 10.47 10.71 0.02 (We can also substitute into Eq. (7-17)). N min AC
FR C,dist
FR B,bot
Feed is sat. vapor. q
0,
100 Find
Vfeed
0.21 N min
CB
1 FR B,bot b. Underwood equations – Case B analysis
1 . All cumene goes to bottoms.
F 1 q
.99 .01
F 100 ,
10.47
0.21
Vfeed
2.25 40
1.0 30
0.21 30
2.25
1.0
0.21
8.1 10
10.47
C i 1
i
12
0
Fz1
1
1.6516.
163
C
Vmin
Dx A,dist
100 0.4 0.98
i 1
From mass balance, L min
100 0.3 0.01 1.0 39.2
2.25 1.6516
1 1.6516
1.25 2.71
min
L D
ordinate = 0.46. With N min
xA xB
Underwood:
Vf
V V
Dx B,d
Dx T,d Vmin
L min
dist
D
39.6
min
2.71
feed
AB
n 2.25 N feed N
gives N feed
1 .9899
0.0101
5.30.
10.25. Use stage 10 or 11.
.99 .02 99 , FR C 1 FR C 0.0204 .01 .98 log 99 0.0204 log 4851 5.438 log TC log 1 0.21
2.5 .25
0, 0
1.526 or 0.3374. Use
zA zB
0.9899, xB 39.6 0.9899 0.4 n 0.0101 0.3
min
N min
107.2
0.155 . From Figure 7-3 the
min
39.2
min
1 FR T
L D
n
min
N min Use Fenske eq. FR T
L
min
x A ,dist
7.D13.
39.6
3.39
L D 1
n
N feed
0, D
10.47, we find N = 20.24.
To find N feed , we need N feed
N feed
i,dist
0 146.78
D 146.78 39.6 107.2 ,
abscissa for Gilliland correlation
N feed
Fz1 FR
0.4, Dx C,dist
2.25 39.2
Vmin
L D 1.25 L D
where Dx i,dist
1
39.2, Dx B,dist Vmin
c.
Dx i,dist
i
2.5
1.0 .30 1.0
0.21 45 21
0.3374 as it is between keys. Vmin
Fz B
3 i 1
1
Dx i,d 1
25 (assume all benzene in dist.)
.99 Fz T 2.5 25
29.7, Dx c,d 1.0 29.7
0.02 Fz c
0.9, D
.21 9
2.5 0.3374 1.0 0.3374 0.21 0.3374 V D 16.64 and L D min 0.2993
55.6
72.24
164
N
Gilliland: Ordinate
N min
9 5.438
0.3562
N 1 10 Abscissa ~ .29 (original Gilliland) or .36 (Liddle) L L L If use 0.29 have, 0.29 0.29 D D min D
If use Abscissa = 0.36,
L
.29 0.2993
D L
1 .29 .36 .2993
D
.83 1.03 which are quite different. Safer to use
1 .36
higher value. If 2.25, N min is same. Underwood Eq. gives BT
Vmin
72.68 , L min
29.40 44.78 1.492
Which is 2.7% different than for
BT
1.4666 or 0.3367. Use 0.3367.
V D 17.084 and L D min
0.3073
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
FR B,dist FR C,bot
n 7-D15.
1 FR B,dist
Fenske: Eq. (7-15), N min
N min TC
1 FR C ,bot
FR tol,dist
Dx d ,tol
N F,min
BC
where
FR C ,bot
n
1 FR C,bot
n N min TC
Eq. (7-17), FR T ,dist
1 .21
0.8238 167
Dx d ,benz
0.9992 397
Dx d ,cum
0.0001 436
x LK x HK
dist
n
LK-HK
.9992 .9999 .0008 .0001 2.25 n .21
6.89
1 TC
.21
6.89
1 .21 .9999 .0001
n
z LK z HK
0.8238, and FR tol,bot
6.89
x dist
137.57 396.68
0.2568 0.7418
.0436 D n
0.1762
534.294
0.0008
.7418 .397 .000815 .0436 n 2.25 .21
1.94 .
Underwood: Use a Case C analysis since toluene is a sandwich key. 3 2.25 397 1.0 167 .21 436 1Fz1 Eq. (7-33): 0 VF is, 0 i 1 2.25 1.00 .21 1 1.216 and 0.3373 which lie between α’s.
165
Eq. (7-29):
Vmin
3
Dx i,dist
i 1
Write for For
i
2.25 396.88
becomes Vmin
1.0 Dx tol
2.25
i
.21 0.436
1.0
.21
1.216 and for 0.3373. Obtain 2 eqns and 2 unknowns: Vmin and Dx tol,dist .
1.216 , Vmin
.3373, Vmin
863.525 4.629 Dx tol . For
466.15 1.509 Dx tol .
Solving simultaneously, Vmin
563.84, Dx tol
64.740
D
Dx i,d
396.88 64.74 .436
L min
Vmin
D 101.79 and L D
L D L D min
462.056 min
0.2203
1.2 .2203
0.445 L D 1 2.2 N N min Ord. .245 (Original Gilliland) N 1 Obtain N = 9.45 (includes reboiler) N F,min 2.91 N 9.45 2.66 (Use stage 3) Estimate N F N min 6.89 Gilliland Abscissa
n 7.D16. a. Fenske: N min b. Underwood:
x
x
1 x
dist
Vfeed
EP
2.1 .6
1.0 4
2.1
1.0
To find D: D
1 x
bot
n Fz E
z xD
1.0 F z p 1.0
EP
1
.99 .992 .01 .008 n 2.1
n
,
0 or
xB
F
xB
, zE
.6, z P
1.44 . Use
.6 .008 .99 .008
1000
1.0 .008
D
2.1 1.44
1.00 1.44
c. Use Gilliland: Ordinate
N
N min
Sounds harder than it is: 0
f
Vmin
L
D 1285.3,
D
2.13 min
.30 12.69
VF F
F
3.132
0.558 N 1 31 L D L D min Abscissa ~ 0.8 (Original Gilliland), L D L D 1
7.D17.
V V
602.85 . Then
2.1 .99
1888.12, L min
V feed
1.44 which is between 1.0 and 2.1
Vmin
Vmin
.4,
12.69
i i
zi
,
tol-xy
3.03,
-
xy xy
2.4
1
166
Expand & Solve for
, 0
tol
z tol
xy
tol
xy
Result is linear,
xy
tol
tol z tol
i
1 z tol tol z tol
Dx i,dist
z
D
,
3.03 3.03 .1
0.3
.9
3.03 3.03 .3
0.5
.7
3.03 3.03 .5
.7
.5
3.03 3.03 .7 .3
.9
3.03 .996 3.03
3.03 3.03 .9
.1
3.03 2.51870
1 2.51870
3.03 .996
.004
3.03 1.88316
1 1.88316
D
.004
3.03 1.503722
1 1.50372
3.03 .996
.004
3.03 1.25155
1 1.25155
90.2834
1.071806
54.93685
3.03 .996
70.0405
1.25155
Vmin
L min
.004
49.7976
1.503722
, L min
1
3.03 .996
29.5547
1.88316
0.988
Vmin
9.3117
2.51870
xB
1
xy
z .008
F
1.0 .004
D
z tol 0.1
D
xB
xD
i
Vmin
sin ce
tol
xy z xy
Vmin
Then
z xy
77.6383
D D
3.03 .996
.004
3.03 1.071808
1 1.071808
98.0683 117.739
D
129.08
100 D
L
45.625
D min 4.8997
48.0836
1.6269
48.2707
0.96934
47.6985
0.68101
38.7990
0.429
Check for z = 0.5.
Slope
xD
L V xD
L
y*
V y
y*
.5 = z
min
y*
x D .5
min
where
xF 1
1 xF
0.75186
0.996 0.75186 x
As z
0.996 0.5
L
L
L V
.4922
D
V L
1 L V
1 4922
min
, although L D min , Vmin , thus Qc
Q R,min
min
Vmin
0.4922
0.96934 Perfect
as expected.
also.
167
FR A ,dist FR B,bot
n
1 FR A ,dist
Fenske: Eq. (7-15): N min
7.D18.a.
n
Where A = propane, B = butane,
1 FR B,bot AB
1/ .49
AB
2.04 . (Note value α.)
.9854 .8791
n
.0146 .1209
N min
8.7
n 2.04
For N F,min assume no LNK is bot and no HNK in distillate D = .229 + (.9854) (.368) + (.1209) (.322) = .631 .9854 .368 .1209 .322 x prop 0.575, x C4 .631 .631
x C3 x C4
n N F,min
z C3 zC4
dist
n
.575 .0617
n F
.368 .322
2.94
0.713
C3-C4
Underwood Eqns. (Case A.) 0
0.0617
Vfeed
1
Fz1
for 1.0
.49
1
0
9.92 .229
f
9.92
L min
1.0
9.92 .229
0.6213 , Vmin
Find
1.0 .368
N
.081 0.10
.49
1.0 .363
.10
.49 .0389
9.92 .6213 1.0 .6213 .49 .6213 D 1.057 .631 .426, L D min 0.676
Vmin
Gilliland Correlation (Fig.7-3): abscissa Ordinate
.49 322
N min
L / D (L / D) min L/D 1
1.057
0.33
.32 (Original Gilliland, ~.36 Liddle). Find N = 13.24 (14.13 Liddle) N 1 b. With N = 20, ordinate to Gilliland correlation is, N N min 20 8.7 0.538 N 1 21 Abscissa = 0.1. Since L D min 0.676, solve for L/D = 0.862. c.
FR C 6
FR C 3
dist
1
where
FR C 6
C6 C3
0.0156 1 0.0156
N min C6C3
bot
FR C 3
0.10 dist
N min C6 C3
bot
0.10, FR C3bot
1 FR C3dist
0.0156, and N min
8.7.
8.7
0.10
8.7
0.00000013 , FR C6
bot
0.99999987
168
For all practical purposes all C6 in bottoms at total reflux. d.
FR C3dist
0.999, L / D 1.5, FR B,bot
n N min
1) 2)
0.8791
FR C3dist FR C 4 bot 1 FR C3dist n
.999 .8791
n
1 FR C 4bot
.001 .1209
12.47
0.713
C3C 4
For D assume all LNK in dist, No HNK in dist D = 0.299 + (0.999) (0.368) + (0.1209) (0.322) = 0.6356
f
Now
0
9.92 .229
1.0 .368
9.92
.49 .332
1
.081 .1
.49
.1
Which is same [φ depends only on feed & α’s]. Thus, same φ = 0.6213
Vmin
L min 7.D19.
9.92 0.229
1.0 0.368
0.49 4689 ??
9.92 .6213
1 .6213
.49 .6213
1.0709 0.6356
Use Figure 7-3. Ordinate
L D
N
0.4345
N min
25 11
N 1
L D
L D min
1.0709
0.6848. Very little change.
0.5385
26
0.08 with L D 2.2286 . L D L D 1 Abscissa approximated between original & fitted curves.
Then Abscissa
7.D20. a) Distillate
Dx Bdist
x dist Find
Fz B
min
5, Dx Tdist
1.0 becomes
Fz T
5
15
D
D
min
1.97
15 ,
0.57895 0.07018 1.0
D = 57.001 kmoles/hr, B = 100 – D = 42.999
n b) Can use Fenske eq. (7-11) or alternatives. N min
AB
xylene cumene
Xylene balance, Fz
x x,bot
35
xA xB
dist
xA xB
bot
n
AB
A
xylene
B
cumene
K xy
K xy K tol
xy
0.330
K cum
K cum K tol
cum
0.210
57.00 0.57895
0.0465, x cum,bot
1 .0465
1.57143
42.999 x x,bot 0.9535
169
n N min
0.57895 0.07018 0.04650 0.9534
11.35 n 1.57143 This is # equil contacts at total reflux. Dx x ,dist 57.001 0.57895 c) Alternative: FR xy,dist Fz x 35
1 FR C,bot
Dx C,dist
57.001 0.07018
Fz C
45 n
Use Eq. 7-15.
N min
0.088896 , FR cum,bot
0.91110
FR B,bot
1 FR A dist n
n N min
FR A dist
0.94288
1 FR B,bot AB
,
A
xylene
B
cumene
0.94288 0.91110 0.057122 0.088896 n 1.57143
11.35
7.D21. New problem in 3rd edition. Assume all ethanol in distillate and all n-butanol in bottoms.
Dx E ,dist Dx i
Fz E
100 .3
30
Fz ip Frac Rec iP dist
P,dist
Dx n
P,dist
Fz nP
Dx n
B,dist
0
100 .25 .986
1 Frac Rec nP dist
24.65
100 .35 .008
0.28 0
D
Dx i,dist
54.93
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
Bx E ,bot Bx i
P,bot
0
0 Fz iP 1-Frac Rec iP dist
100 .25 .014
0.35
100 .35 0.992
34.72
Bx n
P,bot
Fz nP Frac Rec nP bot
Bx n
B,bot
Fz n
B
100 0.10
10.0 B = 45.07
x i,bot
Bx i,bot
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
170
FR ip,dist FR nP ,bot
n b.
Fenske eq. (7-15)
1 FR iP ,dist
N Min
n N min
n
iP nP
.986 .992 .014 0.008 n 1.86
This includes PR
x iP x nP
n
N F,MIN Eq. 7 40a , N F,MIN c.
1 FR nP ,bot
n 8733
9.0748
0.62058
0.62058
z iP z nP
dist
n
14.62
0.4488 0.25 0.0051 0.35 0.62058
n
iP-NP
7.76
Underwood Equation: Assume NKs do NOT distribute: Case A. i Fz i Eq. (7-33) Vfeed i
Vfeed
For saturated vapor
z
F divide (7-33) by F. 1
i i
, which becomes
i
1
E nP
zE
iP nP
ENP
1
z iP
nP NP
iP NP
3.58 0.3
z NP
NB NP
NP NP
1.86 0.25
z NB
NB NP
1 0.35
0.412 0.10
3.58 1.86 1 0.412 Solve for φ between α values of keys. LK = i-propanol, HK=n-propanol. 1.0 1.86 . From Goal Seek on spreadsheet 1.48648 i
Then from Eq. (7-29) Vmin
Dx i,dist
where Dx i,dist
Thus, want
values from part a.
i
VMIN
L MIN d.
L D 1.1
VMIN
3.58 30
1.86 24.65
1.0 0.28
0 173.47 3.58 1.48648 1.86 1.48648 1.0 1.48648 D 173.47 54.93 118.54 , L D MIN 118.54 54.93 2.16
L D
Min
Gilliland abscissa, x
or
L D
L D
1 MIN
L D 1
x
L D L D MIN
L D L D MIN
L D MIN L D 1 1 L D
1
1 L D MIN
1 1.1 1 1 1.1 L D MIN
1.1 1 1 1.1 2.16
1
0.0683
171
N
From Eq. 7-42b,
Assuming 7.F2
N MIN
0.5456 N 1 N 1 0.5456 0.5456 N MIN
NF
N F,MIN
N
N MIN
NF
7.76
33.4
33.4
N includes PR
17.7 or Stage 18 below total condenser.
14.62
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. yN2 1 x N2 0.1397 0.9615 x 0.0385, y 0.1397, 4.055 1 y N2 x N2 0.8603 0.0385
x N2
0.4783, y
x N2
.9190, y N2
n Fenske N min
xD
.9770,
4.01 3.744
avg top&bot
b)
0.7893 (needed for part b)
x 1 x x 1 x n
1/ 2
dist
=
.9770 .0810 .0230 .9191
3.744
3.875
n
bot
.998 .002 .001 .999 n 3.875
AB
9.685
z
where x* is in equilibrium with feed y z 0.79 xD x * From equilibrium data x* ~ 0.48. L V min L 0.40 .998 .79 0.66667 L V min 0.40 , D min 1 L V min 1 0.40 .998 .48
L V
min
L / D 1.1
c)
L/D
Gilliland Correlation: abscissa
N
0.7333
min
L D
L D
1 L D
min
0.06666 1.7333
0.0385
N min
0.6 N = 25.7 including PR. Need 25 equil. Stages N 1 7.G.1. New problem in 3rd edition. a. At total reflux N MIN 9 Original correlation, ordinate
b. L D
MIN
0.92
172
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.
y org x org org w in water
p org x org in w
yw x w
At solubility pt. x w in w
pw x w
in w
.975 and x org in w
H org x org x org
H org
VP
VP
w
xw xw
.025, x org in org
Vapor pressures (Perry & Green, 1984). N-butanol: T = 70.1°C 84.3°C l00.8°C VP = 100 mm Hg 200 400
w
VP
org
VP
w
x org x org
in org in w
.573
117.5°C 760 mm Hg
Water: T = 100.8°C, VP = 782 mm Hg, T = 84.3, VP ~ 421.8 mm Hg VPorg VPorg 400 200 at 84.3°C: 0.474 , at 100.8: 0.5115 VPw 421.8 VPw 782 For
org w
use average between 92 and 100°C. Can linearly interpolate at T = 96°C,
VPorg / VPw
0.501 , w org in w
From y w
w o
1
org w in w
1
xw
w o
org w in w
0.501 .573 .025 11.483 1 11.483
0.0871x w
1 xw
1 0.9129x w
generate equilibrium curve,
xw
1.0
.995
yw
1.0
.9495 0.8961
.990
At constant x w , the calculated y w
0.0871
.985
.980
0.8512 0.8102
.975 0.7726
y w ,exp eri min tal . Difference at x B
approximately .7726 .752 .752 100
0.975 is
2.74% .
These equations work better for mixtures which are more completely immiscible.
173
8.D1.
New Problem in 3rd edition.
Top Op. Eq.,
y
L V x
1 L V xD, xD
L
L D
4
V
1 L D
5
xB
0.11 (from diagram). Need 2 equil. Stages.
.8 , y intercept x
0
1
L V
xD
.975
.2 .975
.195
Graph for problem 8.D1. 8.D2. The columns are sketched in the Solution to Problem 8-C2. B1 is butanol phase and B 2 is water product. Two equilibrium diagrams are shown. a. F B1 B2 , Fz B1x B 1 B2 x B2
174
B1 b.
z
x B2
x B1
.28 .995
x B2
L
Col. 1. Bottom Op: y
V
Feed: 70% Liquid, q = .7, Top y
L V
x
V
L
x
V
1
q
7
.3
0.995 .436 0.995 0
min
0
1256.55
1 x B1 , Intersects y
.7
q 1
F B1
3
x
x B1
0.04 .
. Intersects y = x = z = .28.
x B2
V
L
y intercept x
Note that reflux is x 0
L
1
From Figure 8-D2a:
3743.45 , B2
.04 .995
1
L
x B2
V
0.562 ,
L V
1.23 .562
0.69
0.307
0.573 . Optimum feed stage = 3. Need about 3 stages + partial reboiler. L
Stripper (Column 2): y
L V
V
V 2
B V
L
x2
V
2
V B
1
2
V B
2
1 x B2 1.132 .132
8.57
Construction is shown in Figure 8-D2b. Need 1 2/3 eq. contacts or P.R. + 2/3 equil. stage.
175
8.D3.
a)
y = 0.4, x = 0.09 from graph.
B) V/F = 0.3
L
20 = Fz = Vy + Lx
20 = 0.4 V + 0.09 (100 – V) V 11 .31 35.48 kg moles / h L = 64.52 kg moles/h F zw x w 100 0.99 0.999
yw
xw
1 V/F
.7
7
V
V V/F .3 3 L F 7 .99 y x z x V V 3 0.3 See Enlarged figure [Be careful with scales] y w 0.969, x w 0.999
100 = F = V + L
V
F V
0.969 0.999
Use Table 8-2 to find
= 30
V, Tdrum @ y w
kg moles/hr hr
0.40, xw
0.09 , L = 70 and Tdrum ~ 108 C
176
8.D4.
Compositions x
100
0.975 [Shown on Figure 8-D3a]
L and 88 = Fz = L x
L
F z x
L x
100 .88 0.573
= 76.37 kmoles/hr, L 23.63 kmoles/hr x x .975 0.573 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 V B V B 1 5 L L b) L V , Bot. Op. Eq. y x 1 xB . V V B 4 V V
L
8.D5.
F
0.573 , x
Goes through y
x
xB
177
Plot operating line: If y = 1, x
1
0.25 .04
1.25 PR + 2 stages mores than sufficient. (see graph) 0.75 0.04 c) L V Slope 1.332 min 0.573 0.04 V 1 1 L V 3.012 min L V L V 1 0.332
0.808
178
8.D6.
New Problem in 3rd Edition.
F
D, x dist Reflux
L
V
B, x bot 8.D.6 Part a)
F D F .65
D
z
F
V
b)
B .975D .02B
xB
xD
xB
100
.65 .02
V B B 136.124, Reflux
L
L V V
170.155.
Or V F D L 136.124 100 65.969 170.155 L V B V B 1 4 1 1.25 V V V B 4
c)
L
y Goes through
V
y
L
x
x
V
1 xB
xB
Calculate arbitrary point at x .6 y 1.25 .6 .25 .02 See Figure: Need 2 stages + PR d)
65.969
.975 .02
What is V B
MIN
?
V B
.745
V MIN
L V
1 L V
MAX
1
3.0894 On graph.
179
180
8.D7.
y W x W in organic
yA
y A x A in organic
x A in organic
W A in organic
0.9636 0.372 W A in organic
8.D8.
0.0364, y W
1 0.0364
0.628, x W in organic =1-0.628=0.372
44.69
0.0364 0.628
Convert wt frac to mole frac. MW C8 H14 O 72 14 16 102 and MW water Basis 1000 kg 0.994 wt frac. ether:
988 kg W = 54.889 kg moles 12 kg E = 0.118 kg moles x W in organic .998
y = 0.959 ether:
41 kg W = 2.278 kg moles 959 kg E = 9.402 kg moles y W 0.155
y W x W in org W - E in org
.195 .033
yE x E
z(wt) = 0.004 water:
V
min
1
1
y equil w feed xD min
L V
7.026
7.026 0.022 1 x
L V
L D
1
min
.805 .967
4 kg W = 0.222 kg moles 996 kg E = 9.765 kg moles z W 0.022
x
xD
L
18
6 kg W = .333 kg moles 994 kg E = 9.745 Total = 10.078 x W in organic 0.033
0.012 wt frac. ether:
y in equil w feed
0.9636
z
0.138
6.026 0.022
.998 .138 .998 .022
7.467;
min
Generate following equilibrium data using
L D
0.882
11.20; act
w E in org
L V
0.918 act
7.026 :
xW
0
0.01
.022
.033
yW
0
0.066
0.137
0.195 181
Top Op. line:
y
L
x
V Where L/V = 0.918, x D Bottom: From y
x
1
L
xD V .998 , and y = intercept = (1 - .918) (.998) = 0.082
x B to intersection of feed line and top operating line.
xB Obtain x B
0.0004 wt frac.: 0.4 kg W = 0.022 kg moles 999.6 kg E = 9.800 kg mole 0.0023
See plot in Figure: Optimum feed is top stage. Need 4
8.D9.
Convert to Mole fractions: MW C6 H14 O 72 14 16 102; MWwater
3 5
equil. Contacts.
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 0.333 Mole frac. x w in org 0.033 10.078
182
xD
0.998
0.033 Bottom Layer separator is
988 kg W = 54.889 kmol
= 0.012 wt frac. ether
12 kg E = 0.118 kmol Mole frac. x W in org
z = 0.02
0.998
xD
41 kg W = 2.278 kmol Vapor into Condenser is
959 kg E = 9.402 kmol
yazeotrope = 0.959 wt frac. ether Mole frac. y W
y W x W in org W E in organ phase
Feed is y W
in org
L D
act
1 .195
7.026
1 0.033
0.02 mole frac water = z. In equil. With feed:
xD min
0.195 0.033
y E x E in org
y
x *f L V
0.195
xD
0.02 1 y
z x
0.998 0.02
* f
0.988 0.002896
2 L D min
114.36 , L V
Plot on graph, and plot top op. line: x 0, y 0.00868. x 0.04, y
0 0
0.9913 .04
L V
L
0.9828 ,
D
min
L D act
1
0.01 0.066 0.00868 x
x
1
L V
57.18 ,
0.9913
L D
Top Op. y L V x 1 L / V x D,W through y When x = 0, y = 0.00868 7.026x Eq. Data. y . Generate curve, 1 6.026x
x y
0.002896
7.026 6.026 0.02
x D,W
0.022 0.137
0.998 .
0.033 0.195
0.04833 0.03, y .9913 0.03
0.00868
x W ,bot is at intersection y = z = 0.0208 top op. line, x W,bot 0.0123 Step off stages from top down. 1 equil stage is sufficient. But with this very high reflux rate consider alternatives.
183
8.D10.
a.
VP
C10
x C10
VP
W
xW
760
Assume the water layer is pure, x W 1.0. Try 95.5°C, VPC10 60, VPW 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: VPC10 63, VPW 682.07 , (.99) (63) + 682.07 = 744.4 97.5°C: VPC10 T = 97.6 gives VPW
65, VPW
694.57 , (.99) (65) + 694.57 = 758.9
697.1 which will be too high. Thus T = 97.5°C is close enough.
184
b.
nw n org
p tot
VPorg x org VPorg x org
760
65 .99
10.81
65 .99
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.
x F,org 0.9, 95% recovery → 5% left. Octanol left = 0.05 (.9) (1.0) = 0.045 kmol/h Nonvolatiles in bottoms = 0.10 kmol/h
8-D11.
octanol water
x oc tan ol in org
xF
0.045 1.10 0.045
0.3103
W steam
185
a) Water VP can be fit to log10 VP T = 95.5
log 0 645.67
T = 100 log10 760
B
A
A
A
273.16 T B
over short ranges T. T in C, VP is mmHg.
273.16 95.5 B 2.8808 273.16 100
2.8100
A B 368.66
(1)
A B 373.16
(2)
To solve for A and B, subtract 1 from 2 B B B = 2164.42 0.07080 0.00003271B 368.66 373.16 B A 2.8808 8.68105 373.16 Now find T for which p tot VPW x W VPO x O 760 mm Hg where x W 1.0, x O 0.3103 On Spread Sheet find T = 99.782°C VP O x O 19.075 b) y O 0.025098 0.3103 0.007788 p tot 760 754.072 yW x 1.0 0.99220 760 a) Moles octanol = F z O .95 1.0 0.90 0.95 0.855 kmol/h b) Moles water
nW
Check Eq. (8-18): n W
n org
yW y org 0.855
19.075 0.3103
8.D12. New Problem in 3rd edition. All cases
a)
D2
40
1.0 .65 .65 .55
D1
280
.35 .01 D1
B1
0.9922 0.007788 760
60, B2
108.93
19.075 0.3103
40 , D 2
B2
108.93
x P,B2
x P,dist1
x P,dist1
x P,dist 2
140
D1 140 40 180 . b)
0.855
Total feed Col 1 = 246
1400
1440.
Total feed Col 1 = 1500
186
8.D13. New Problem in 3rd edition.
F1
D, x dist
V
L
V
L
F2
B, x bot Part a. F1
F2 D B Water: F1z1 F2 z 2
100 80
D B
Dx dist Bx bot 100 .84 80 .20 Solve simultaneously, D = 99.25 and B = 80.75 V b) V B 121.125, L V B 201.875 B V
Since feed 2 is saturated liquid L L F2 121.875
c)
Doing Mass balance around top V y F1z1 y
L
Lx
Dx dist
Dx dist
F1z1
V
D 0.975
B 0.04
V 121.125
Doing Mass balance around bottom V y Bx bot y
V
L V
x
Lx F2 z 2
F2 z 2 Bx bot
V
These two equations are equivalent.
Slope
L
y
d) Bot. op. line:
L V
V
V B
V
L
x
V
V B
1 x B . Goes through y 1
V B
L
x
xB .
5 3 . Plot Bot Op. line.
121.875
1.0062 V 121.125 At intersection F2 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). Op. line above feed 2:
Slope
187
Graph for 8.D13. 8.D14. Figure is on next page.
F
Part b.
B1
B1 c)
D2
F
B2 ,
Fz
z x B2 x B1 z x b2
x B2
x b1 x b1
D1
D2
B1x EB1
0.85 0.006
F
0.992 0.006
x b2
x d1
x d1
xd2
B2
100
B2 x E 100
B2
85.60 , B2
F B1
14.40 kmol/h
0.85 0.992
0.006 0.449
0.006 0.992
0.449 0.75
21.196
35.596 kmol/h
188
D2
D1
xE
101.3 kPa
1333
F
zE
0.75
0.449
0.85
kPa
ethanol
Water Ethanol
99.4 mole % water
99.2 mole % B1
B2
8.D.15. Part a) New Problem in 3rd Edition. p org VPoc tan ol x oc tan ol =Ptotal where x oc tan ol is mole fraction octanol in organic phase. At 0.05 atm and boiling T, porg From Antoine equation,
log10 VP
oc tan ol
6.8379
0.05 atm. 38 mmHg
1310.62
T 136.05 T 129.8C, VPoctanol 80.905 mmHg
At Since p org
p org
38
0.470 VPbenzene 80.905 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg mol octanol 15 130.23 0.470 15 85 mol octanol mol non-volatiles 130.23 MW
0.470
b)
38 mmHg, x octanol,mole
15
85
15
MW 654.04 130.23 MW 130.23 95% recovery is true on both mass and mole basis. Distillate octanol flow rate 0.95 100 0.15 14.25 kg h. Since MWoc tan ol
130.23, this is 14.25 130.23
In waste there are 0.05 15
0.109 kmol h.
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. 0.75 130.23 Mole frac. octanol in waste 0.0424 0.75 130.23 85 654.04
189
c)
For equilibrium in still pot VP
oct
x oct in org
VPw 1.0
The still pot is perfectly mixed; thus x oct in org Since water boils at 100°C when P
p tot
x oct in waste
760 mmHg .
0.0424 mole frac.
760 mmHg, T < 100°C.
Eq. (8-15) becomes VPoct 0.0424
VPw 1.0 760 Substituting in the Antoine equations for octanol and water and solving with a spread sheet, T = 99.97°C. VPoct 19.27 mmHg and VPw 759.18 mmHg. d)
n oct
From Eq. (8-18),
VPoct x oct
nw
p tot
VPoct x oct
From spread sheet n oct n w Since
n oct
0.109 kmol h, n w
water
101.27
This is a lot of steam! 8.D16.
0.001076 n oct 0.001076 101.27 kmol h
kmol 18.016 kg kmol
1824.5 kg h water in distillate.
kmol
Distillate 1: 0.997 EtOH, 0.0002 solvent. Calculate x d1,W Distillate 2: 0.999 water, 0.00035 solvent. Calculate x D2 E F = 100, x F,E
0.81, sat'd liq'd, x F,solv
1 .9972
0.0028
1 0.99935
0.00065
0
Find D1 , D 2 , M where Makeup is pure solvent. 0 Water: x W M Fz W D1x D1W D 2 x D2W 0 Ethanol: Mx E,M Fz E D1x d1,E D 2 x d 2,E Ethylene Glycol:
Mx Esolv
1.0
Fz solv
Solving water & ethanol balances obtain: D 2 From Ethylene Glycol balance, M 81.2316 0.0002
D1x d1,solv
Dx d 2,solv
18.7913 and D1
81.2316 kmol/h.
18.7913 0.00035 0.02282 kmol/h Can also use overall balance instead of EG bal. Then M D1 D 2 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 x A = FzA/(S+F), xB = FzB/(S+F), xsolvent = S/(S+F). 8.D18.
Ethanol Product: Water Product:
F 100, z E
0.997E, 0.0002 solvent, 0.0028 water 0.9990W, 0.00035 solvent, 0.00065 ethanol
0.20 z W
0.80
190
Water bal: x WM M
Fz W
PE x EP,W
E bal:
x EM M Fz E
EG
0 x M,solvent M Fz solv
PE x EP,E
M 8.D19.
(B)
PW x WP,E
PE x EP,solv
Solve A & B for PE & PW :
(A)
PW x WP,W
PW x WP,solv where x M,solvent
80.0240 kmol/h, PE
PW
80.0240 0.00035
F H
Ethanol:
Fz E 0 B1x E,bot 2 B2 x E,bot 2 0 Fz H H 1.0 B1x H,bot1 B2 0
Hexane:
B2 where H = makeup hexane.
Solving simultaneously, B1 8.D20. New Problem in 3rd edition. a.
1000 0.8094
0.03201 kmoles/h
do M.B. in wts.
M.B. around System. Since everything in wt. units
B1
20.0074 kmol/h
20.0074 0.0002
Overall:
1.0
8000.04, B2
1000
F
Ex E,Ethanol prod
2000.04 and H
0.08 kg/h.
E W
Wx E,wprod
809.4 0.998E 0.0001W 808.3 E 811.0 kmol h 0.9979 W F E 1000 811 189.0 b)
V
V
Fx WF
boilup ratio
L L
Ex W ,Ethanol prod
1000 0.1906
y w ,1 V E Pr od
811.0 0.002
0.300 629.93 811.0
629.93
0.777
V E Pr od. 629.93 811.0 1440.93 L F sat 'd liquid feed 1440.93 1000
440.93
If CMO strictly valid then, L
reflux
440.93
Can also estimate L from _ settler
Pentane flow rate in V1
Ethanol flow rate in V1 Ethanol lost in Water Product.
Pentane flow rate
y P,1V1
629.93 0.6455
406.62
191
Ethanol flow rate in V1
y E,1V1
629.93 0.0555
E in V1 E lost
x E,reflux
L from settler to Col1
Ethanol lost in water product
Lfrom _ settler _ calculation
W prod x E in Wprod
189.0 .0001
1. CMO not totally valid 2. There is some water in reflux 3. K dE value may be incorrect.
Ethanol returned to distillation column V1 y E1 WPr od x E,W Pr od 34.96 0.0189 Using average estimate for L 0 Then 34.91
0.0189
441.6
Match not perfect because:
c.
34.96
440.93 441.6 2
34.19 kmol h.
441.3
441.3 x E in pentane x E,pentane
Then since assume K d
0.0792
x E,Re flux,pantane _ layer
1, x E,Water layer
0.0792
192
d.
x E,water
0.0792
V1 1
V W
0.5
V
L L
283.5
V
94.5
y E1 y E1 W
0.5W
W V
283.5 kmol h.
3
L 0.0792
W .0001 V
283.5 .0792
189.0 .0001
94.5
189.0
x E,W Pr od
94.5 kmol h.
0.237
0.0001
8.D21.
193
L
Bottom:
V
V B 1
1.5
L
L
1 xB .Goes through y V B 0.5 V V Feed line = Horizontal (q = 0). Through y = x = z = 0.4 Top. MB: yV Lx Dx D and V L D y
3, y
L
x
x
x
x B with slope = 3
L
1
x D goes through y x x D 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.
D, x dist Reflux
L
V
B
F Part a. F
D Part b. yV
y
D B & Fz z x bot x dist
x bot
F
Dx dist
Bx bot
0.20 0.08 0.975 0.08
100
13.41 kmol/h and B
Lx Dx dist L V
x
D V
x dist
Substitute in D
Points on operating line: y
x
x dist
F, L
B, thus slope
L V
L
V L to obtain y
0.975 and x
Alternative point is at feed line (y = z = 0.2) & x
V
86.59
x bot
1
0, y intercept
L V 1
x dist L V
x dist
0.08
B F 86.59 100
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
V
x
0.8659
y intercept
0.02 .
194
Figure for problem 8D22.
195
8.D23. New Problem in 3rd edition.
Water phase
XNM = 0.086
N.M. Phase
NM
xw = 0.312
W
F
Water Product
Nitro Methane Product Part a. External balances F
NM Pr od
WPr od
NM :
F .25
NM Pr od
.25
.01
.98 .01
NM Pr od .98
100
WPr od .01
24.74, WProd
75.26
a. W Column:
z = .25, horizontal feed line L L Top y NM x NM 1 x NM ,bot ,col NM Mass balance through top of W column V V and around col. NM. Can easily show that
196
y NM x NM x NM,bot,col NM But do not know L/V so cannot plot yet. Bottom operating line looks familiar:
y NM
x NM
x NM,bot,col w
V
L
x
V
1 x NM ,bot col w
0.01
L
V B 1
54
V
V B
14
col w
L
y NM
5
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 y NM x NM x NM,bot,col w 0.01 See graph. Need PR + ~ 1 2 stage. Build PR + 1 stage. yw
b. NM Column is a stripping column:
L V
To plot,
V 1 B V B
col NM
yw
xw
xw
.3, y w
Need PR + ~ 1 c. W col. Want V. V
1 3
3
B
0.02 1
.3
3
.02
0.3933
xw
0.3 is arbitrary point
stages.
100 V, V= V
1 x w ,bot ,col NM
3
V
B
B V to cond. from Wcol
NM col want V. V=
L V
4
x w,bot,NM _ col 4
L V xw
B
3 NM Pr od
1
WProd
1
75.26 18.81 4 4 118.81 kmol hr 3 24.74
74.22 kmol hr . To condenser.
197
Graph for 8.D23.
198
8.D24. New Problem in 3rd edition. From Equilibrium, y bu tan ol Overall Mass Balance: 100 F V B Butanol MB: 100 0.025 V .092
2.5
.092 V
0.092 at x bu tan ol
0.004
B .004
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)
VPbenzene x benzene where x benzene is
porg
mole fraction benzene in organic. At boiling T, p org From Antoine equation, log10 VP At T
93 C, VPbenzene p org
1.0 atm. 1211.033
6.90565
benzene
T
220.790
1112.44 mmHg . Since
760, x ben,mole
p org
760
0.683 VPbenzene 1112.44 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg 0.683
0.683 b)
Moles benzene Moles benzene + Moles non-volatiles
20
80
20
80 .683
20
78.11
MW
78.11
MW
78.11
1 .683
20 78.11 20 80 78.11 MW
MW
673.2
90% recovery is true on both mass and mole basis. Distillate benzene flow rate 0.9 100 0.2 18.0 kg h . Since MWbenzene 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 78.11 Mole frac. benzene 0.1773 2 78.11 80 673.2
c)
For equilibrium in still pot VP
b
x b in org
VPw 1.0
The still pot is perfectly mixed; thus, x b in org Since water boils at 100ºC when P boils at 80.1ºC, but mole fracs low.
p tot
x b in waste
760 mmHg
0.1773 mole frac.
760 mmHg, T 100 C . Benzene is more volatile and
Antoine equation for water: log10 VPw
8.68105
2164.42 273.16 T 760
VPb 0.1773 VPw 1.0 Substituting in the Antoine equations for benzene and water and solving with a spread sheet, T 92.0411 C . VPben 1082.5 mmHg and VPw 568.1 mmHg. Eq (8-15) becomes
199
d)
n ben
From Eq. (8-18),
VPben x ben
nw
p tot
VPben x ben
From spread sheet n ben n w
n ben
Since
0.337876
0.230 kmol hr, n w
water
0.6807
n ben 0.337876
kmol 18.016 kg kmol
e. To vaporize benzene condense moles water
kmol
n ben
ben
0.6807 kmol hr
12.264 kg h water in distillate.
.
w
This occurs at 92.0411 C
365.1911K; From Perry’s table 2-237, H
hg
h
hf
H
h kJ kg
T = 360
886.7
498.7
388
T = 370
898.6
518.1
380.5
5.1911
0.0911 w
5
x 380.5 388
2265.67 2278.3
Moles water condensed
2278.3
388
384.1
kJ
30, 002 kJ kmol. 10 kg Note: 8th edition, Table 2-193 is very slightly different after unit conversion. T = 360 2663 384.7 2278.3 Water Table 2-352. T = 370 2671 405.88 2265.67 Linear interpolate
2277.8 kJ kg
41,037 kJ kmol
0.230 30, 002
kg h water in waste
0.1682 kmol h water (in waste) 41037 0.1682 18.016 3.029 kg h
200
8.E.1. New Problem in 3rd edition.
Water phase
XNM = 0.086
N.M. Phase
F2
xw = 0.312
NM
W
Water Product
Nitro Methane Product
250
Part a.
F1
F2
NM balance
PNM
PW
F1z1NM
F2 z 2 NM
PMN x
NM Prod
Pw x
NM mol frac
135.5 8 127.5 135.5 PNM Pw b.
F1
PNM
PNM .98
250 PNM
w Pr od NM mol frac
Pw 0.01
0.01
.97PNM
2.5
133 .97 137.11 kmol hr 250 137.11 112.89
Column W – Use y NM vs. x NM (water phase) plot. Top operating line
y NM
L V
Bottom Operating Line y NM
L V
L V
V B
V B 1
43
V
V B
13
water col
water col
x NM
1 L V
x NM
L V
V
4. L
V
wcol
water col
F2 z NM 2
x NM
NM Pr od
Vwcol
1 x NM
Water Pr od
V
B col w 37.63 V . B B col w 150.52 L F1
201
y
Top IS NOT from
x
x NM
0.98 to intersection feed and bottom operating line.
NM Pr od
Instead from intersection of feed and bottom operating line with slope
L V
L F1
V
50.52 37.63 1.32 25 .
Optimum feed is top stage. Need PR + 1 stage. c.
Column NM. Top y Bottom
yw
L V L V
yw
xw
NM col
xw
NMcol
F1z1,NM
1 L V NM x w col
xw
L V
xw
1 xw
NMcol
in WM Prod.
L
0.02.
V
W,W Prod.
V B 1
2.00 V V B Draw bottom operating line. Top is through intersection bottom operating line and feed line F 2 . Slope L V 0.906 (see item d). Need 2 stages + PR. Optimum feed is stage above PR. d.
Column W: PW
V
Bcol w
NM Pr od
112.89
V B Bcol w
1 3 112.89 L
V Bcol w
Saturated liquid feed: V Column NM:
37.63
V
37.63
V B
hr. PNM 1
50.52.
Bcol NM
137.11
2.0
V B
V
L F
kmol
V B 1.0, L V
L
150.52.
V B B 137.11 kmol hr,
L
2.0 137.11
274.22
V 137.11, L L 150 124.22 , L V 0.906. Minimum boilup rate NM column gives combination bottom & top operating lines to go through
Saturated liquid feed V e.
reflux point: y W
0.5, x w
0.312 .
From bottom operating line intersection with feed line
y INTER
L
x
z2
0.15 is
L
1 x W ,NM Prod V V Slope of top operating line to reflux point is L F2 L 0.5 y INTER V
V B Guess V B
z2
0.312 0.15
V
1
L V
L V 1
V
.
Calc V & L & L V
Check is two calculated values L V are same.
V
V B B 137.11 V B .
Calc y int er
Calc L V
L
V
B
L F2 V
202
Spreadsheet.
V B
MIN
0.6105,
L V
0.846
Graph for Problem 8.E1.
203
8.E2.
Balances at mixing point for F & R. To Butanol Column: Overall: FT F R Water:
FT z T,W
100 R
Fz W
Rx W,reflux
R .573
z T,W
FT z T,W
30 0.573R
30
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:
FT
V B 1.90, V 1.90B 135.432 L y
R zT
V B x
FT
206.712,
x W,butonal
F
L V 1.5263
0.02
206.712 100 106.712
106.712 0.573
30
0.4409 206.712 Vertical feed line at z T intersects bot. operating Line at y = 0.67 (see graph) Water Col.
V B
0.1143, y
L V L V x
V B
V B 1
V
V B
9.748
L V 1 x B,water,watercol, y
x
x
B,W watercol
0.995
See graph, y leaving column = 0.8
204
205
8.E3.
Basis: 1000 kg sea water (1 h): 965 kg water kmol 18.016 kg
35 kg NaCl kmol 58.45 kg
53.5635 kmol, x F,W
0.98894
0.5988
kmol, x F,salt 0.011056 Total 54.1623 Water Condensate = (0.60) (53.5635) = 32.1381 kmol/h = n W Water Remaining 53.5635 – 32.1381 = 21.4254 21.4254 kg moles W Waste Water is 0.9728 mole frac. water 21.4254+0.5988 salt a. In still, organic phase is pure decane. VPC10 VPW x W p tot 760 mmHg where x W
0.9728 .
Try T = 99°C. VPC10 ~ 68, VPW 733.2 mm Hg 68 + (0.9728) (433.2) = 781.27 mm Hg, which is too high. Converge to T ~ 98.2°C. 707.27 0.9728 nW yW pW VPW x W b. Distillate: 10.4225 n org y org p org VPorg x org 66 1.0 (This calculation is at 98°C, not 98.2, but will be close.)
206
n C10 8.F1.
32.1381 kmol water/h 10.425 mol water/mol organic
3.0819
kmol h
C10 in distillate
V.P. Data n-nonane (p. 3-59 Perry & Green, 1984)
VP = 20 T = 51.2
40 66
60 75.5
100 88.1
200 107.5
400 128.2
See Solution problem 8.D10 for plot. MWnC9 1984), nonane enthalpies are,
128.25 .
hliquid 671.3 KJ/kg 722.5
Hgas 998.2 1036.5
T 360 K 380 K
760 mm Hg 150.8°C
From p. 3-268 (Perry & Green,
Water VP is given in Problem 8-D10 and on p. 3-45 of Perry and Green (1984). a. Try 95.0°C. VPC9 127 mm Hg, VPW 633.9 . Assume water is pure. Pressure: (.99) (127) + (1.0) 633.9 = 759.6. Close enough and lucky! b.
p tot
nW n org
VPorg x org
760 127 .99 127 .99
VPorg x org
5.045 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: h1 ~ 691.78, Hgas ~ 1013.52. nonane 321.74 KJ/kg Table 3-302 of Perry and Green (1984): h liq,W
397.36, H vap W
2667.8,
W
2270.44 KJ/kg
mol water condensed
C9
321.74 KJ/kg 128.25 kg/kmol
mol C9 vaporized
W
2270.44 KJ/kg 18.016 kg/kmol
b. Now, VPC9
.020
VPW
1.009
760. Temperature will be higher.
Try 99°C: VPC9 149 mm Hg, VPW 733.24 (149) (.020) + 733.24 = 736.22 which is too low. Try 99.9ºC: VPC9 154, VPW 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. p tot VPorg x org 760 154 .02 nW 245.75 n org 154 .02 VPorg x org Need lot more steam!
207
8.F2. n-nonane
water F = 1000 95% n-nonane organic waste steam water
a)
Basis: 1 hour 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) .95 x C9,bot ,org 0.65517 1.45 b) Still T. p W p org p tot 102.633 kPa 770 mm Hg pW
porg
VPW (T)x W where x W
K C9 T x c9,org,bot
K C9 T
1
0.65517 770 mm Hg
Procedure: Guess T, determine VPW & K C8
VP
W
504.481K C9
check if pressure eq. is valid
770 mm Hg?
0.16 (DePriester Chart) 652.62 80.717 738.34 Need higher T T = 97°C, VPW 682.07 and K C9 0.17 682.07 + 85.762 = 767.83 slightly low, but close enough
Try T = 96°C
VPW
504.481 K C9 T
657.62, K C9
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 n org p org 85.762 n org 8.55 d) Eq. (8-18) 0.12567, n W 68.031 kmol water nW p W 682.07 0.12567 0.12567 e) EB simplies to
W
n W,condensed
n W ,condensed
nonane C9 W
n org,dist. where λ’s are at 97°C = 370 K.
n org,dist.
8.55 kmol
C9 W
208
From Perry’s 6th ed. Table 3-268 or 7th ed. Table 2-292, nonane λ’s are: h g h f , @ 360 K, 998.2 671.3 326.9 kJ/kg
380 K, 1036.5 722.5 326.9 314 @ 370 K, λ ≈ 320.45 kJ/kg 2 MW C9 128.258 . Then at 370 K, kJ 128.258 kg
41,100.3 kJ/kmol kg kmol Water 370 K: (Perry’s 6th Ed., Table 3-302). kJ 18.016 kg kJ 2671 405.8 2265.2 40,809.84 W kg kmol kg 41,100.3 n W ,cond 8.55 8.611 kmol water 40,809.84 C9
320.45
314.0 kJ/kg
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.
209
210
211
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 2.4, BB 1.0, BC 0.21. AB A = benzene, B = toluene, C = cumene. Results from Spreadsheet: Stage: Reboiler: x A 0.0003 x B 0.0097 1 0.003298 0.04443 2 0.03137 0.176085 3 0.18019 0.42145 4 0.46126 0.44952 5 0.70274 0.28536 6 0.85421 0.14453 7 0.93403 0.06585 8 0.97145 0.028535 9 0.98791 0.012091 10 0.99493 0.005074 11 0.99788 0.00212 12 0.99912 0.000885
x C 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
212
400 450 475 500 600
0.6740 0.5044 0.3946 0.2696 0.00042
0.0077 0.0089 0.0092 0.0089 0.00095
0.3183 0.4867 0.5962 0.7214 0.9986
Results for other starting conditions are shown in the figure.
213
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
214
Simple distillation calc (residue curves) with BP calcs. aT1 aT6 ap1 -1166846 7.72668 -0.92213 -1280557 7.94986 -0.96455 -1481583 7.58071 -0.93159 -1524891 7.33129 -0.89143 -1778901 6.96783 -0.84634
iB nB iP nP nhex Residue curve calc. x1iB sumx time step 1 2 3 4 5 6 7 8 9 10
0.98 1 xiB 0.98 0.979883 0.979764 0.979645 0.979525 0.979405 0.979283 0.979161 0.979037 0.978913
600 601 602 603 604 605 606 607 608 609 610
0.009754 0.009586 0.009418 0.009253 0.009089 0.008927 0.008766 0.008608 0.008451 0.008296 0.008142
0.071054 0.069204 0.067391 0.065615 0.063874 0.06217 0.060502 0.058869 0.057271 0.055708 0.054179
h x1nB T1guess,R xnB 0.01 0.010032 0.010065 0.010097 0.01013 0.010162 0.010195 0.010228 0.010261 0.010294 0 0 0 0 0 0 0 0 0 0 0
0.01 0.01 500 xiP 0 0 0 0 0 0 0 0 0 0
0.919192 0.92121 0.923191 0.925133 0.927037 0.928903 0.930732 0.932524 0.934279 0.935997 0.937678
N x1iP p,psia xnP 0.01 0.010085 0.010171 0.010258 0.010345 0.010433 0.010522 0.010612 0.010702 0.010793
1000 0 14.7 x nHex 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
542.3988 542.71 543.0165 543.3183 543.6155 543.9078 544.1955 544.4783 544.7563 545.0295 545.2978
epsilon 1E-09 x1nP 0.01 x1nHex 0 TR 472.0604 472.0642 472.068 472.0718 472.0757 472.0797 472.0836 472.0876 472.0916 472.0956
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
215
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
216
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
217
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 F, x F , x Davg , N, L / D, Treflux , P
Single Stage F, x F , x Davg , P
F, x F , x Wf , N, L / D, Tr , P
F, x F , x D tot , P
F, x F , Wfinal , N, L / D, Tr , P
F, x F , x wf , P
F, x F , D tot , N, L / D, Tr , P
F, x Wf , Wfinal , P
F, x F , x Davg , x Wf , N, Tr , P
x F , x Davg , D tot , P
F, x F , x Davg , x Wf , L / D, Tr , P
x F , x Wf , Wfinal , P
x F , x Davg , D tot , N, L / D, Tr , P
x F , x Wf , D tot , P
x F , x Davg , D tot , x Wf , N, Tr , P
F, x F , x Davg , x W
x F , x Davg , D tot , x Wf , L / D, Tr , P
etc.
x F , x Davg , Wfinal , N, L / D, Tr , P x F , x Wfinal , Wfinal , N, L D, Tr , 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 xF
Wfinal
F exp x W,fin
d xW xD
xW
218
Because x D
constant, can integrate analytically. xF
xF
d xW xD
x w ,final
Wfinal
n xD
xW
F exp
Wfinal
xD
xF
xD
F
x W ,final
xD xD
xD xD
xF x W ,final
xF x W ,final
D
Wfinal x W,final
xD
F
xD
Wfinal
Fx F Solve for
n x w ,final
n
F
Mass balances are
xW
D xD
xF x W ,final
Thus results are identical. 9.C2.
External balances over entire cycle Fz Dx Dfinal Wx wAvg and F = D + W a. Ignoring holdup on stages and in reboiler - Out = Accum in accumulator is - x w dD
x w dD
which becomes,
dD
Rearrange,
Ddx D
dx D
D
D final
dD
x Dfinal
D F
D
xD xF
xW
dx D xw
xD
xD
d Dx D
x D dD and integrate,
which is,
n
D final F
x D final
xF
dx D xD
xw
b. Assume CMO and draw mass balance envelop around bottom of the column.
L
y
y
L V L
V
V
x
x
B
Lx
Vy Bx w
B
xw V L
V
1 xw
219
xF
rd
Wfinal
9.D.1. New Problem in 3 Edition. Eq. (9-9)
F exp x W,final
From Simpson’s rule
0.1 0.00346
Area
0.00346 0.05173 0.1
y
4 x
0.00346
y-x
0.03096 .34 0.4416
y-x
1
6
From equilibrium curve (Table 2-1). x y
dx
y
1 x
y
x 0.005173
x
0.1
1 y-x 36.366 3.469 2.9273
0.027498 .28827 .3416
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 0.170 y 0.00346 0.03096 For x = 0.00346, 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 6
Wfinal
0.5 exp
D total
0.5 .2125 F xF
x DAvg
9.D2.
Rayleigh equation is Wfinal
0.8555
0.2125 kmol .5 .1
D avg .75
0.8555
0.2875 kmol
Wfinal x W ,final
F exp -
2.9273
.2125 .00346 0.2875
0.1714
dx
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 dx .2 3.92 4 5.13 6.89 1.044 x 6 .55 y Wfinal
x Davg
100e
Fx F
.55
1.044
35.20, D total
75
Wfinal x final D total
F Wfinal
Operating equation is y
64.8
35.2 0.55 64.8
A graphical integration counting squares gives x Davg 9.D3.
4 3.469
L V x
1
220
L V
0.859
0.861.
x D where L V
L0 D 1 L0 D
0.65
This is y = 0.65 x + 0.35 x D We have two equil. stages (stillpot and one in the column). From McCabe-Thiele diagram we can get the values of x D which are related to x W . Pick x D and get x W from figure. From this we can generate the following table (only two values are shown in Figure).
xD 0.90 0.895 0.85 0.837 0.720 0.70
x w ,final
xW
Raleigh Equation:
D total x Davg
3.205 3.077 2.193 2.096 1.754 1.748
Wfinal
xF
F
x wfinal
100 39.7
Wfinal x wf
0.36 ]
3.077 4 2.096
6
F Wfinal Fx F
0.588 0.570 0.394 0.360 0.150 0.128
.57 .15
dx w xD
1/(xD-xW)
[Midpoint x w
Simpson’s rule (Eq. (9-12)). xf
xW
dx xD
xw
, Wfinal
1.754
Fe
0.925
0.925
39.7 kmol
60.3 kmol
100 0.57
D total
39.7 0.15 60.3
221
0.847
9.D4.
Wfinal
2.0 kg moles, x F
0.8, x wf
0.4 . Find F, x DAvg , D total x
1
y(equil)
y 1
16.66
y x
4.76
7.143 .4
.6
.8
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
Can use Simpson’s Rule (Eq. 9-12) or evaluate numerically. xF dx 0.4 4.76 4 7.143 16.666 x 6 x Wfin y
Wfinal
F
Rayleigh eqn
xF
exp
D total
x Wfin
2.0
F Wfin
Fx F
x D AVG
exp
dx y x
3.3333
Wf x Wfin
56.063 kmol
56.068 0.8
2.0 .4
54.063 Wfinal
a)
3.3333
54.063 kmol
D total
9.D.5. New Problem for 3rd Edition
F exp
xF x w ,fin
dx y
x
Can use Simpson’s rule, eq. (9-12) with equilibrium values from plot. 1 1 f x w ,final 2.7397 y x 0.028 0.645 0.28
f
x w ,fin
f xF
xF
1
2
y x
1 0.705 0.52
0.40
1
1
y x
0.735 0.52
0.52
x
3.2787
4.65116
222
0.815
xF
xF
dx
x w ,fin
y x
0.52 0.28
3exp
y D,avg b. Settler:
6
f x w ,fin
2.7397 4 3.2787
6
Wfin
x w fin
0.8202 D V,tot D1
DV,tot y D,avt
Use eq. (9-13)
n
WF F
6
20.5056
1.321 , DV,tot
F Wfin
0.8202
1.6790
0.7088
D2
1.6790
D1x 2
D1
D2 x B
Solving simultaneously, D 2 9.D6.
f xF 0.24
4.65116
3 0.44033
Fz Wfin x w fin D V,tot
4f x w avg
D2
1.6790 0.7088
0.567 and D1 1 1
n
1.112
x W ,F 1 x F x F 1 x W ,f
223
0.573 D1 0.975 D 2
n
1 xf 1 x W ,f
a) F = 1.3, x F n
.6, x Wf
WF
1
1 3
1.4
WF 1.3
Fx F
.3 4
WF
.4
n
.6 7
1.3 0.6
0.3036 0.3
0.6914
1.3 0.3036
0.2335
Wf
0.81725
0.81725 0.3
3.5 0.8125 0.6, x DAVG 0.75
2.0, x F
0.8948 0.5596 1.4544
.7
0.3036 kmol
Wfin x Wf
3.5 0.6
x D AVG
Since
2.4
F Wfin
b) Now Wf 3.5
F
n
0.2335
x DAVG
c)
.3,
F Wfin x D AVG
Fx F
0.6914
x D AVG
Wfin
Wfin x Wf ,
F
x D AVG
xF x W ,f
.
Then Eq. (9-13) becomes
n
0.75 .6 0.75 x Wf
L D 1/ 2, L V
L D
1
1 L D
3
x D AVG
xf
1
x D AVG
x Wf
1.4
F x D AVG
0.5025 . Wfinal
Solution is x w fin 9.D7.
n
x D AVG
xf
xW
0.56
0.11 0.10 0.09
Simpson’s Rule
dx
xF x Wfin
xD
xW
.6 1 x Wf
1 x Wf
1.212 kmol.
.75 0.5025
x Wf 6
1 xD
xW 1
xD
xW
, determine
dx
xF x Wf
xD
xW
f
1/.45 = 2.2222 Interpolate 2.361 1/.40 = 2.50
0.06 0.05 0.02
xF
4
n
slope . Pick series x D values. Plot enriching section op
xD
0.44 0.38 0.26
x Wf .4
2.0 .75 0.60
x Wf
line. Step off two stages. Find x W . Calculate
0.49
n
1 .38
2.6316
1 .33 3.0303 1/.24 = 4.1666
f x WF
0.10 0.02 6
224
0.02
4f x W
4.1666 4 2.6316
0.06 2.361
f xF
0.10
0.22739
WFinal x DAvg
4.0 exp Fx F
0.22739
Wf x Wf
4 0.7966
4 .1
F Wf
3.1864 0.02 4 3.1864
9.D8. New Problem 3rd Edition. a) Op. Eqn.
L
y
v
x
4 L L 1 xd , x d .8@ y D D 5 From McCabe-Thiele plot x w final ~ 0.075 L V
10
b)
F
Wfinal
10 .4 Substitute in for
D tot
D tot
F xF
Wfinal , 4.0
3.1864
1
L V
xd
x , y Intercept
Wfinal
D tot x D
0.075 10 D tot
.8 D tot
.8 .075
225
4.483
Wfinal
1
L V
xD
.2 .8
10 D tot
Wfinal x w final
4.0 10 .075
0.4133
5.517
0.075 Wfinal .8 D tot
.16
9.D8. Figure
9.D8. Part C. Trial & Error to 3 stages ending at x F
y INtercept
0.645
1
L V
xD
1
0.4 (See figure) L V
.8
226
1
L V
9.D9.
a.
L
.645
V
.8
.19375,
.80625
L
L
L V
.15375
D
V L
1 L V
1 .19375
Initial
Mass balances: F
F xF
D total x D
D total
W or 20
D total
W x Wf or 8 .975 D total
0.24
W
.28 W
Solving simultaneously: D total 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 x Wf must be within the two phase region. x Minimum is when y
x . This is x Wf ,min
227
0.21 . See graph.
y
x
9.D10.
L
L
x D , Slope, L V L D V V Plot on McCabe-Thiele graph for series of x D values. Op. Eq., y
Simpson’s rule,
1
1 xD
xW
dx W
xF x W ,fin
x
xd
at x W
or
Wfinal
n
10 e
xW
x D,avg
Wfinal F
1.2491
.52, x W
0.52 0.20 6 0.32
Rayleigh eq.
xF
6 xF x W ,fin
xd
xW
xW
xF
Wfinal
xD
2.67
10 0.52
D total
228
xW
x F x W ,fin 2
xD
1.2491
F Wfinal
2.8677 0.2 7.132
.36
1
Fe
2.8677 and D total
Wfinal x Wfinal
xf ) / 2
4
5.95 4 3.70
dx W
15
.20, and (x Wfin
1 xD
10 0.28677
Fx F
x Wfin
1 LD
7.132
0.648
xW
x W ,fin
xD
xW
0.70 0.65 0.60 0.50 0.40 0.30
0.55 0.405 0.25 0.094 0.055 0.035
x D-x W 0.15 0.245 0.35 0.406 0.345 0.265
229
1/(x D-x W) 6.666 4.082 2.857 2.463 2.899 3.774
1
xW
xD xW 5.95
x F = 0.52 xF
x Wfin
3.70
= 0.36
2 x W ,fin = 0.20
2.67
S
9D11. New Problem in 3rd edition. Eq. (9-17)
x butanol Inital
W
dx pot y
x butanol final
Note W water Equation is in terms of butanol. 1.0 dx but pot S Simpson’s rule – need y at x pot W y but .6
x but
x water
y water
y but
1.0 .8
0 .2
0 0.565
1.0 0.435
.6
.4
0.70
1.0, .8, .6 butanol.
0.30
.4
3.333 4 2.299 1.0 6 0.9019 W 1.804 kmol.
1 y but 1.0 2.2999 3.333
0.90191
S More accurate if done in 2 steps. Thus add points below: x but
xw
yw
y but
.9 .7
.1 .3
.42 0.66
.58 .34
1
0.8 .8
.6
.2 6 .2 6
2.299 4 1.7241
1
3.3333 4 2.9412
2.299
Total Area = .91975. S = 1.8395
230
0.33985 0.5799
1 y but 1.7241 2.9412
xF
9.D12.
Wfinal
dx
F exp -
y x
x Wfinal
xF
xF
dx
x Wfinal
y x
x Wfin
dx y
D tot
0.32 x
6
4
y x
0.48, x Wfinal
x W final
1
1
y x
2
xF
0.16, x avg
x Wfinal
1
0.16
0.36
0.20
0.32
0.545
0.225
4.444
0.48
0.66
0.18
5.555
5.555
Fx F
1.51
(y-x)
y x 5.0
, Wfinal
Wfin x Wfin
3.0 exp
3.0 0.48
D tot
y (from graph)
y-x
0.7 0.61 0.37
0.3 0.37 0.29
0.4 0.24 0.08 Simpson’s Rule xF 0.4 0.08 x w ,fin
F exp
Two Liquids.
0.93826
F xF
x D,AVG b)
3.333 4 2.7027
6
D total x L
L x
0.573 and x L
D total
L x
0.662 k/moles
0.662 0.16
0.571
Wfinal
n
F
xf
x w final
dx y
x
1 y x 3.333 2.7027 3.448
3.448
0.93826
D total
F Wfinal
3.1305
Wfinal x w ,final
1.51
2.338
9.D.13. New Problem for 3rd Edition. a) Rayleigh equation: x
xF
0.32
2
y (from eq. data)
2.338 x DAVG ,
Wfinal
y x
x Wfinal x F
x
5.0 4 4.444
F Wfinal
1
6 xF
xF
x Wfinal
4.8695
0.6057 0.975
L
D total
D total x DAVG
231
L
D total
L .573 L .975
D total .6057
D total .6057 .573
L 9.D14. a.
p org
pw
0.3963
0.975 0.573
L
D total .3963
VPC10 x C10 , and assuming water is pure, p W
760 , p org
4.4732
VPW .
VPC10 x C10 VPW 760 Vapor pressure data for C10 was shown in solution to Problem 8.D10. Guess 99.5ºC. VPW 746.52, VPC10 ~ 70.5 mm Hg 7 + 746.5 = 753.5 < 760 At 100ºC VPC10 ~ 70.5 and VPW 760 . 7.05 + 760 = 767.05 > 760 6.5 By linear interpolation: T 99.5 .5 99.74 ºC 13.55 b. Use Mass balances. Initially 9 moles n-decane, 1 mole non-volatile a .1 Final: a mol n-decane where 1 ; thus, a .111 mol 1 a .9 Wfinal 1.111 mol (Water free) D total
F Wfinal
10 1.111 8.889 kmol
Alternate Solution: Raleigh Eq. with xF
xD
1: Wfinal
F exp x W,final
c.
nW
Wfinal
F exp
D total
F Wfinal
D org
p tot
n .1
dx 1 xW
n .9
F exp
n 1-x W
xF x W ,final
.111 F 1.111
8.889. Same result as mass balance.
VPorg x org in org VPorg x org in org
Should really calculate numerically from integral for most accuracy D
nW
p tot
VPorg x org in org VPorg x org in org
0
dn org
However, estimate at final conditions with VPndecane ~ 70.25 (from part a)
nW
760
70.25 .1
n org 70.25 .1 This should be a good estimate. 9.D15.
107.185
Column is similar to figure in Solution to 9.D13, but with 1 stage in column. a) For finding Wfin & D tot don’t actually need to step off stages. Just want to make sure x Wfinal is obtainable – I checked this at total reflux – It works. Use Mass balances: F x F D tot x d Wfin x Wfin Substitute in F
D tot
Wfinal
232
Then,
Wfinal
Solution is Wfinal
F xD xD
xF
100 0.975 0.48 0.975 0.08
x Wfin
55.307 kmol, D tot
F Wfinal
b) Need to draw operating lines until: initial
44.673
2 stages gives x Feed .
final 2 stages gives x W,final . Then L/V = slope. Initial – There will be a pinch at point reflux is returned. y xd y int ercept 0.975 0.41 L V Initial Slope 0.579 xD 0 0.975 0 Final: A few trials resulted in final result. y xd y int ercept 0.975 0.17 L V final Slope 0.826 xd 0 0.975 0
233
9.D16. a. Need L/V so that 3 stages go from x F error) was used to find this).
L V
.84 .57 a
.84 0
.4 to x D
.3214 and
b. Need L/V so that 3 stages go from x wfinal trial-and-error was used to find line.
L V c.
F
D total
F x feed
.84 .13 b
Wfinal D total x D
.84 0
10
D total
Wfinal x wfinal
.8452 and
0.84 . This is line a in Figure (trial-andL V
L D
a
1
0.08 to x D
D
L V
b
1
a
0.84 (see line b in Figure). Again, L V
L
.4737
a
5.4615
b
L V
b
Wfinal
4 .84 D total
.08 Wfinal
Solving simultaneously, Wfinal = 5.789 kmol and D total 4.211 kmol. The Rayleigh equation could be used, is not needed, but gives the same result.
x init
9.D17. Eq. (9-17),
S W
1.0 . Start 1 kmol and keep 1 kmol. Add water as boil. x W ,initial
x W ,final
dx tan k y
Note balance is on original solvent, methanol.
Use equilibrium data from Table 2-7. Generate table of methanol mole fractions:
x
1.0 0.611 0.222 0.11611 0.01
y
1.0 0.830 .588 .450 0.07
1/y
1.0 1.20489 1.6722 2.222 14.9254
Use Simpson’s rule in two steps. Step 1 (x from 1.0 → 0.222)
234
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. rd
n
9.D.18. New Problem 3 Edition Eq. (9-14)
y D ,final
F D final
y D ,final
F D final
yF
dy y x
dy
exp
y
yF
x
F
D final
y D ,final
dy y x
exp yF
Read x values from equilibrium diagram or interpolate from Table 2-1. y x y-x 1 0.1 0.3 0.5
Area
Ethanol MB:
x C,avg
F yF
0.008 0.045 0.155
0.092 .255 .345
0.5 0.1 6
10.87 4 3.92
0.5
D final
exp
C total
F Dfinal
2.90
1.963
0.0702 kmol
1.963
0.4298 kmol
Dfinal y D,final
C total x C,avg
D final y D,final
0.5 0.1
C total
F yF 0.0702 0.5
0.4298
9.D19. New problem in 3rd edition. VPw x w VPoct x oct Ptot , or in mm Hg, 526.123 1.0 9.D20. Was 9.D18 in 2nd edition. x F,C5 0.35 & x W,final,C5
y x 10.87 3.92 2.90
0.05 : x C5,AVG
10.964 .6
0.20, x C8,AVG
235
0.00346 mole frac ethanol
532.7
0.80, p 101.3 kPa.
yi
B.P.
Ki xi
1.0 . For average mole fractions, the BP calculation converges to T =
84º with K C5 3.7 and K C8 0.30 from the DePriester charts. The close enough to estimate α. K C5 3.7 12.33 C5 C8 K C8 0.30 Eq. (9-13),
n
Wf
1
F
11.33
0.35 0.95
0.5573, Wfin
Wfin F D total
0.05 0.65
n
F Wfin
Fx F
x D,Avg
Wfin x W ,fin
L
x
1
V x D & slope
0.98 which is
0.5847
0.95
0.5573 1.5
1.5 0.8359
0.8359 kmol
0.6641 kmol
1.5 .35
D total
9.D.21. New Problem for 3rd Edition.
y
0.65
n
yi
0.8359 0.05
0.7276
0.6641
L D 1.0
L D
L V
1 L D
12
L
x D , but x D varies. Thus, plot series operating lines of arbitrary V 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 x W for this x D value.
xD
.7 .6 .5 .4 .3 .2 Want to integrate from x W,final Plot
1 xD
xW
x W ,final xF
Area
xW
xD
.067 .05 .038 .027 .018 .01
.633 .55 .462 .373 .282 .19
xW
0.02 to x F
1 xD
xw
1.5798 1.818 2.1645 2.6801 3.5461 5.26316
0.06 and want middle point at x W
vs x W and find values.
0.02,
0.06,
1 xD
xW
1 xD
xW
0.06 0.02 6
3.23; x W,Avg
0.04,
1 xD
1.65
3.23 4 2.1
1.65
236
0.08853
xW
2.1
0.04 .
Wfinal
F exp
Area D total
2.5 exp F
Wfinal
0.08853 0.2118
2.288 x D,Avg
237
F xF
Wfinal x W ,final D total
0.492
9.D.22. New Problem for 3rd Edition. t OP 9.D23.
Prelim. Calc. Feed; Avg MW
MWavg
D tot
0.1 MW
0.1 46
1000 kg 20.8 kg mol L
L D
toperating = __1.49 to 1.50
QR
E
0.9 MW
0.9 18
h
water
20.80 kg/mol
48.0769 kmol
23
2 5 0.4 V 1 L D 53 All op. lines have slope 0.4. Can draw op. line to x W . Ten stages will go from x D to x W because have large number of stages. Thus, do not need to step off stages.
238
From Graph can create table of 1/(xd – 1/xw) versus xW.
xD
xW
xd
xw
1
0.665 0.10 0.630 0.08 0.499 0.052 x Avg
0.565 0.550 0.447
xd xw 1.770 1.8182 2.237
0.440 0.278 0.140 0.057
0.400 0.258 0.130 0.053
2.500 3.876 7.692 18.868
0.040 0.020 0.010 0.004
239
Simpson’s Rule: x F
0.1, x W,final
xF
dx W
x Wfin
xd
0.004, x W,Avg
xF
x Wfin
xW
xd
6
WFinal F
D total
x DAVG
xF
exp x Wfin
F WFinal
Fx F
1
6 0.096
0.104 2
xW
4 x Wfin
18.868 4 2.237
dx W xd
xW
0.052
1 xd
1 xW
1.770
0.62289; Wfinal
x AVG
xd
xW
0.4734
48.0769 0.62289
48.0769 29.9466 18.1303 kmole
WFinal x Wfin
4.80769
D total
29.9466 0.004 18.1303
240
xF
0.2586
29.9466
9.D24. Was 9D22 in 2nd edition. a) F D total Wfin and Fx F =D total x D
Fx F
D total
D total x D
x Wfin +Fx Wfin
0.62 0.45 0.85 0.45
3.0
Wx Wfin D total
xF
x Wfin
xD
x Wfin
1.275 kmol, Wfin
F
F Dtotal
1.725 kmol
b) Want operating line where 2 equil. contacts gives x w fin 0.45 . y
L
L
xD . 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 y x x D . L V
Slope
0.85 0.44 0.85 0
0.482,
Figure for 9.D24. 9.D25. New Problem in 3rd edition. Mix together F F1 F2
2.5
241
L
L
L V
D
V L
1 L V
x
0.932
1
xF
F1x F1
F2 x F2 xF
w FINAL
n
F
xF
xF
x FIN 6
0.8 2.5
dx
xw final
x w Final
2.5 y
1 y
xF
x AVG
x
0.32
x
y
x w ,FIN
y
1 x
x w ,AVG
Wfinal
3.333 4 2.666
D
area
Fe
F Wfinal
Then from 0.2 to 0.1 with F
1.151,
x
WFIN1
Now F x
y
WFIN1
1
y x 0.1 0.4 3.3333 0.15 0.51 2.7777 0.2 0.575 2.6666
F2 0.1 6
Wfinal2
0.6166
Fx F
x DAVG
WFIN x w FIN
.2
2.666 4 2.703
6
Ftot x tot
3.030
0.55026
WFINAL1 = F1e-area = 1.5 .5769 = .865 D1
F1
WFinal,1
0.635, x D,AVG ,1
0.865 1.0 1.865,
xF
3.3333 4 2.7777
2.666
Fe D2
area
F
1.865 0.7518 Wfinal2
Wfin 2 x Wfin
2.5 .32
242
.865 .2
0.635
0.285185
1.4022
1.865 1.4022 1.4022 .1
1.09775
D total Higher distillate mole fraction.
1.5 .4
0.2
Total D total Total x D,AVG
0.578
D
D1
F2
y x 0.2 0.575 2.666 0.3 0.67 2.703 0.4 0.73 3.03
x w ,F
1.349 kmol
1
y
x
Values are from methanol-water equilibrium data.
2.817
2.5 0.5398
First go from 0.4 to 0.2 with F1
Part b)
y
1
y x 0.1 0.4 3.333 0.21 .585 2.666 0.32 .675 2.817
6
0.21
2
4
x
.22
x w FIN
0.46275
.635 .46275 1.09775 0.6010
Part c) Go from 0.4 to 0.1 for F1 1
x y 0.1 0.4 0.25 0.62 0.4 0.73
0.3
y x
6
3.3333 2.7027 3.030
Wfinal
3.3333 4 2.7027 1.5 exp
.8587
3.030
0.8587
0.6356
F1
For F2 go from 0.2 to 0.1. Same as 2nd part of Part b.
Wfinal
area
F2 e
0.28518
1.0 .75187
0.75187
F2
Wfinal total
1.3874
Wfinal
Wfin ,
F1
D total
F1
F2
Wfinal tot
1.11255
F2
Differs from b – Numerical error!
Ftot .32
x DAVG
1.3874 .1
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, Wfinal1 F1 e area 1.5 .433681 0.65052 Then
1.40239 , D total
Wfin total
2.5 1.40239 1.09761
2.5 .32
x D AVG
1.09761
Same as for Part b. 9.D.26. New Problem for 3rd Edition.
L D
4, L V
final
1.40239 .1
a&b) final
L D
45
0.60109
2 3, L V
1 L D
53
0.4
0.7 . Step off 2 stages.
0.7 draw op. line slope = 0.8. Two stages gives
0.11 (See graph (labeled 9.D.c))
x w final
Also, draw a few lines with x D 2 stages. b) Generate Table x D , x w , 1 x D Plot
23
0.8
For Part b, draw op. lines with slope 0.4 for arbitrary x D For Part a, From x D
L D
1 xD
Find 2 areas
xw
0.7 and L V between 0.4 and 0.8 . Find x w values with
xw
vs x w . Note there is a break in curve at x D 1. from x F
2. from x w
0.6 to x w
0.185
0.185to x w final
Area 1. Simpson’s rule. x w avg
0.6 0.185 2
243
0.110
0.3925 .
0.7 x w
0.185 to 0.110
1
From graph
Area 1 Area 2.
0.6 0.185 1 xD
6 xw
xD
x w avg
3.56 4 2.37
1.942
1.0363
curve is straight line. Thus, Area = width x Avg height
1.6949 1.942
0.185 0.110 Total area = 1.1726. Rayleigh eqn.,
c)
2.37 .
xw
n
2
Wfinal F
x w final
1.1726
Wfinal
100 e
D total
F Wfinal
x D,AVG
Fx F
dx w
xF
xD
xw
0.13638
1.1726
3.0955
6.9045
Wfinal x Wfinal D total
6.0
3.0955 0.11 6.9045
244
0.8197
L/D 2/3 2/3 2/3 2/3 2/3 -4.0
xD
xw
0.9 .85 0.8 0.75 0.7 0.7 0.7
0.65 0.48 0.36 0.253 0.185 0.145 0.110
x w ,final
245
1 xD xw 4.0 2.703 2.273 2.012 1.942 1.8018 1.6949
9.D.25. New Problem for 3rd Edition. From the methanol-water equilibrium data, the following table can be obtained.
246
xM 1.0 .8 .6
yM 1.0 .92 .825
1 yM 1.0 1.08696 1.21212 x pot ,Initial
S W
x pot ,final
dx MeOH y MeOH
.4
Simpson’s rule:
1.0 4 1.08696 1.2121 6 0.4373W 0.8747 kmol
S
0.4373
9.E1. octanol water
xF pot
0.90, F=1.0 kmole
1 0.95 .9
Final octanol in pot
1.0 0.1 0.1
Nonvolatiles in pot
x oct,W,final
steam
log10 VPW
0.045 kmol
0.045 0.145
0.3103
2164.42
8.68105
, T C, VPW mmHg 273.16 T 0.3103, x W, in W 1.0, T 99.782 C from
a) Final conditions x oct, in org
problem 8.D11. Initial conditions: x oct,in org
0.90, x W in W
solution
1.0
x i VPi 1.0 atm 760 mm Hg From spread sheet find T = 99.377ºC b)
Wfinal,org
c)
Dorg
F
1 z 1 x Wfin
F Wfinal
0.8550
D org
nW
d) Eq. (9-24)
p tot
Estimate VPoct at average T
p tot VPoct
VP VP
0
nW
1.0
oct
1.0 0.9 1 0.3103
n org oct
x oct
x oct
dn oct
(99.782 99.377) / 2 D org
0
dn oct x oct
0.1450
D org
dn org 0
247
p tot VPoct
99.5795 , VPoc tan ol D org
0
dn oct x oct
D org
18.87 mm Hg
to
F 1.0, n org
F W, x org
dn org
W
W
1
Step-by-Step integration,
n org
F
1
x org
x org
1 x feed
1
0.1 F W
dn org
dn org
x org
dn oct
Avg x org
x org,avg 0
1.0
.9
0.1
.1
.9
.8888
0.2
.1
.8
0.875
0.3
.1
.7
0.8571
0.4
.1
.6
0.8333
0.5
.1
.5
0.8000
0.6
.1
.4
0.750
0.7
.1
.3
0.6666
0.8 0.855
.1 0.055
.2 0.145
>
0.8944
>
0.8819
>
0.8661
>
0.8452
>
0.81666
>
0.775
>
0.7083
>
0.58333
0.50 0.3103 >
0.11181 0.11339 0.11546 0.1183 0.12245 0.12903
0.40517
0.14118 0.17143 0.1357 1.1588
nW
p tot VPorg
Porg
O
dn org x org
D org
760 18.8666
1.15882
0.8550
45.826 kmol
e. Continuous had 108.93 kmoles water/kmole organic fed. The continuous always operates at lowest octanol mole fraction in liquid & thus y oct 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. mol B 78.11 x benz 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
248
dmorg-sum dnw/dnorg 9-23 d)
18 massben-still 2.960203 dnw
2 dmassorg 0.037898
1
Use Spread Sheet for each time dn W dn org Eq. 9-23 step. Did the addition of steps off-line in table below. nW dn W for n org still init 20 78.11 n benz 2 78.11 Step-by-step integration.
still final
Set dm org
1 kg. Values from spreadsheet. Mass benz still, 777 (kg)
dm org kg Initial T
0→1 1→2 2→3 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 → 18 18
final T d)
20 → 19 19 → 18 18 → 17 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3→2 2
dn W divide all values by 78.11 kmol .664431 .677266 .691548 .707537 .725558 .746073 .769462 .79657 .828276 .865852 .91108 .966543 1.036129 1.12596 1.246275 1.415548 1.670846 2.098857 2.960203
TºC from spreadsheet 75.99 76.27 76.577 76.91 77.269 77.66 78.0996 78.58 79.12 79.715 80.39 81.156 82.03398 83.0487 84.2358 85.6435 87.339 89.422 92.04
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 0.5 Steps _ 0
17.2794168
18
1.8122565
n W 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, 18.016 dn W 18 kg benzene overhead 12.3 same as continuous. 78.011 e) To vaporize (n ben ben ) / w varies since T is different. For accuracy, do for each dm org and find λ values at these T. x d ,fin
9.E3.
Simpson’s Rule: xF
dx d xd
xB
0.13 6
1 xd
249
4 xB
x d .63
xd
1 xB
x d 0.565
xd
xB
x D 0.50
Generate table from graph 1 xd xB xd xB 0.3 0.36 3.0303 0.2 0.597 2.5189 0.1 0.565 2.15 0.04 0.500 2.004
x init
9.E4.
Wfinal
F exp x W ,fin
0.13 6
3.0303 4 2.15
D final
Fe
1 e
B
F Dfinal x B,avg
0.2954
2.004
0.2954
7.442 gmole
2.558 mole. F xF
Dx D,fin
5.0 4.6886
B
2.558
dx y x
T & E since dilution effects F & x init . Dilute with 5 kg water → Start with 6, x init Dilute with 4 kg water → Start 5, x init Dilute with 3.5 → Start 4.5, x init
16
15
0.1667
0.2
1 4.5
0.2222
Dilute with 2.75 → Start with 3.75, x init
1 3.75
0.26667
Dilute with 1.75 → Start with 2.75, x init
1 2.75
0.364
250
0.1218
For each dilution want to integrate using Simpson’s rule until find Wfinal need values of 1 y x at x init , x avg , and x final are determined in the following table.
1.0. Thus,
0.01 for each dilution. These values
Table of Values for Integrations Dilute 5 kg Water 4 kg Water 3.5 kg Water 1.75 kg Water 2.75 kg Water Integrations: Dilute 5 kg:
Wfinal Dilute 4 kg:
Wfin Dilute 3.5:
Wfin Dilute 1.75:
Wfin Dilute 2.75:
Wfin
x 0.16667 0.08833 0.01 0.200 0.105 0.01 0.2222 0.11611 0.01 0.364 0.1870 0.01
y 0.537671 0.38708 0.067 0.579 0.428 0.067 0.598 0.450 0.067 0.706 0.563 0.067
y-x 0.3710 0.29875 0.057 0.379 0.3230 0.057 0.376 0.334 0.057 0.342 0.376 0.057
1/(y – x) 2.695412 3.34724 17.544 2.63 3.095975 17.544 2.6596 2.994 17.544 2.923977 2.6596 17.544
0.2666 0.138 0.01
0.636 0.493 0.067
0.369 0.355 0.057
2.7100 2.8169 17.544
Use Simpson’s rule for each addition. 0.15667 2.695412 4 3.34724 17.544 0.87809 6 6 exp -0.87809 2.4934 . Value is too high. Want 1.0 kg.
0.19
32.5579 1.031 6 5 exp -1.031 1.783 too high 0.2122
32.1796 1.138 6 4.5 exp -1.138 1.442 too high .354
31.106 1.833 6 2.75 exp -1.835 0.4389 too low .2566
31.52 1.348 6 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.
251
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 problem in 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),
0
1 AB
1
n
x A ,W ,final 1 x A ,F
1 x A ,F
n
x A ,F 1 x A ,W ,final
1 x A ,W ,final
n
z A 1 Frac. Rec. A.dist x A ,W ,final
Part a. In a spreadsheet Eq. (9-36) is easily solved for xA,Wfinal using Goal Seek. Then W final 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 alpha AC 10.71 frac rec A in distillate xA,Wfin 0.05 9-36 term 1 -2.41222
zA
0.37
The 0.37 is in cell D2
-0.41074
term 3
0.930081 term 2
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).
252
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, 10G110G4. 10.A9.
10.A13.
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. 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.
10.B3.
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. Some possible candidates: Bottle caps Bent bottle caps Natural sponge Synthetic sponge
SOS pads Scotch brite pads Steel wool Cooper cleaning pads
Miscellaneous junk Broken crockery or glassware
Plastic coated wire Tin foil - crushed
Cut up tubing (Tygon) Glass tubing (broken)
String - balls lines tied together
Crushed beer cans Cut up - crushed beer cans
lines twisted lines stretched taught
Coal
Rope
Egg shells Styrofoam packing material Old seat cushions
frayed rope Rope tied into bow ties Porous rock pieces
253
10.C1.
Nuts/bolts/screws/nails Metal filings Wood shavings Kindling Left-over redwood
Pumice Ash from Mt. St. Helens Ashes from coal stove Pieces of cement block Pieces of brick
Staples
Pop-tops
Window screens, rolled up Chicken or barbed wire
Old watch bands - twisted Bent wire coat hangers
Christmas Tree Ornaments Corn cobs
Cookie cutters Combination of the above
L/V constant since L/D const. Thus L/G const. In F v only
n
changes. From Perfect Gas
G
P
MWv . Thus, F v increases as P . However, curve is almost v RT flat in this range and Cs,bflood ~ constant.
Law
G
MWv
u flood
Then
V Thus,
L D
10.C2.
Dia
u flood
P
since
P
1 D and D
F
.5
1
G G
L D
V
Then, F
L
0.2
20 /
Dia 2
.5
C sb flood
F
xD
z
xD
F 1
xB
L
so V
G
.
F
F P
P
P , and exponent = ½
See derivation in solution to Problem 10.C1.
L
V
D
1 D
L D
1 F
xD xD
z xB
Then from Eq. (10-16)
L 4F 1 D
Dia
Since Dia 10.C3.
F1/ 2 , Dia
1/ 2
3600 p fraction u flood
L D 1
Plot points based on ∆p and
xD z xD xB
1/ 2 1/ 2
L
G
G
L
on Figure 10-25. From ordinate values calculate F
for each point. Use an average value.
254
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.
K C6
y/x
and
K C7
1 y
Solve simultaneously. x
1 x
1 K C7 K C6
K C7
KC6
T 149 C 169 171
KC7
1.0 1.3 1.34
193
K C6 x
Bounds are K C6
Pick T and generate equilibrium curve. o
, y
.72 .75 -
-
1.0
x
y
1.0 .483 .424
1.0 .63 0.568
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. 1 1 log10 C6 362.79 .935, T 207.09
log10
n
C7
MIX
436.73
1
1
T
232.53
.5 n .116
1.0 and K C7
.5 n .127
.895,
C6
0.16 .
C7
0.127 .
2.107,
MIX
1.0 .
0.122.
From Eq. (10-6) for 0.217, E o 0.730 . The higher pressure results in higher temperatures and lower viscosities. This increases the predicted column efficiency by 24%. 10.D2.
T v
98.4 273.1 371.5 K (almost pure n-heptane at bottom). 1 100.2 pMW 0.684 62.4 0.205 lb/ft 3 , L RT 1.314 371.5
42.68 lb/ft 3 , 20
Need L V . First, find y at intersection of operating lines. L .999 y Top Operating Line Slope .8 V .999 .5 y .999 .8 .499 0.5998 Then,
L
0.5998 0.001
V
0.5 0.001
1.20
255
L
L MWL
L
G
V MWv
V
Fv
1.2 . This is at bottom where MWL ~ MWv .
0.5
L
v
G
L
0.5
.205
1.2
0.08317
42.68 0.2
From Figure 10-16, Csb
u flood
0.32, C sb
42.68 0.205
0.29
0.32
20
0.205 Saturated liquid feed V V
0.2
12.5
0.29
20
0.5
4.19 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.
L
0.5
0.205
0.0832 , L 1.2 V 3000 V 42.68 For Fig. 10-17 need % flood. In problem 12.D2 designed for 75% flood.
D Use 12 feet:
1.2, F g
1/ 2
const
11.78 , const1/ 2
75% flood
% flood
e
1.2
1/ 2
10.202 /12.0. % flood
0.022 3000
L
A total
12.0
0.9,
A active
v0
t tray
weir
Dia
0.726 ,
0.726 12 ft
weir
90.48 ft 2 , A hole
0.078 in, 3 16 holes, d 0 t tray
v
9.05
h
p ,d ry
Lg
3067.48
4 113.1 ft 2
113.1 1 .2
VMW v 3600
10.20
1 0.9 113.1 11.3 ft 2
Ad Table 10.2,
2
1/ 2
72.3% . Then Fig. 10-17, ψ = 0.022.
67.48 , L e
1 1 .022 This is reasonable amount.
14 gauge,
11.78 .75
37.51 ft s , C0 0.003 37.51
3067.48
2
0.1 90.48 9.05 ft2
2.4
0.759 (unchanged from Ex. 10-3)
0.205
100.2 7.48 42.68
8.71 ft
60
59.87
1 0.01
42.68
0.759
2
2.086 in
897.8 gpm
256
Abscissa
Lg
897.8
2.5 weir
8.71
Parameter
w
Dia
4.01
2.5
0.726, Fweir Fig.10 20
Eq. (10-26),
h crest
0.092 1.03
With 1″ gap,
A du
8.71 12
h du
Eq. (12-27),
h dc
897.8
2.083 in
8.71
0.726 ft 2 2
897.8
0.56
1.03
2/3
4.248 in
449 0.726
2.086 2 2.083 0 4.248 10.417
h dc,aereated
10.417 0.5
11.3 10.417 3600 42.68
t res
3067.48 100.2 12 0.040 12.5
Weeping, Eq. (10-32), h Eq. (10-31), LHS
h
h
2.086 0.0625
2 2.083 0
=4.904 This is OK.
0.0625 in
42.68 3 16
p,dry
x
OK, but close to distance between trays.
20.83
2.148
4.083
RHS 0.10392 0.25119 4.083 0.021675 4.083 LHS > RHS Operation is slightly marginal with high downcomer backup. 1.5 Increase apron gap to 1.5 inches: A du 8.71 ft 1.089 ft 2 12 h du1.5 gap
h dc
0.56
v o,bal
Given Wvalve
2
449 1.089
p,valve
C v Wvalve 2g K vA v
0.08 lb,
Av
Cv 1 12
1.888 in
16.11 inch, OK 1/ 2
(10-36)
v
g
32.3 ft s 2 , K v,closed
Pressure drop in terms of inches of liquid of density
h
0.7682, OK
2.086 2 2.083 1.888 8.05 in
h dc,areated 10.D4.
897.8
2
Wvalve
Kv
v
Av
2g
L
L
0.02182 ft 2 ,
33C v L
1.25,
K v,open
5.5
:
v 02 v
0.1917 lb ft 3 ,
L
41.12 lb ft 3
257
1/ 2
1.25 .08 2 32.2
v o,bal,closed
6.83 ft sec
33 0.2182 .1917 1/ 2
1.25 .08 2 32.2
v o,bal,open
16.73 ft sec
5.5 0.218 .1917
At balance point, h
1.25 0.8 lb
C v Wvalve
p,valve
AL
0.2182 ft 3 41.14 lb ft
L
0.1115 ft closed:
h
33 .1917 p,valve
2.39 10 open:
h
3
v 02 ft
5.5 .1917 p,valve
4
10.D6.
6.83
2.87 10
v 02 for v 0
2 32.2 41.12
3.98 10 10.D5.
1.338 in liquid
v 02 for v 0
2 32.2 41.12
3
v 02 ft
2
v 02 inches
16.73
4.78 10
3
v 02 inches
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 From Fenske eq. and definition of HETP
HETP n
z
x
x
1 x
n n
x 1 x
1 x
dist AB
x dist
1 x
bot
n bot
.987 .013
.008 .992
9.150
z = 3.5 meters. Obtain: a. α = 2.315, HETP = 0.321 b. α = 2.61, HETP = 0.367 c.
AVG
2.315 2.61
1/ 2
2.4581 and HETPAVG
0.344
Can also use McCabe-Thiele diagrams although the solution shown is easier. 10.D7. Current: F v 0.090, 12 spacing, Ordinate 0.2 U nf const., const. 0.2 U nf 0.2 6.0 New: L Vnew 1.11 L V old , then L G new 1.11 L G old , F v,new 1.11F v,old 0.0999 Trays Spacing 24″, Ordinate ~ 0.32, Ordinate = U nf × Const.
258
U Nf 10.D8.
At
Ordinate
0.32
Const
0.2 6
Fv
0.5, Csb,f 1
0.12
0.2
Uf1
C sb
L
Since
nRT,
20
0.5
V
10.D9.
G
0.12
0.5
L
20
RT
G ,new
0.5
1
0.5
0.25; C sb ~ 0.18
4
U f 1,new
C sb,new
G ,old
C sb,new
old
0.18
U f 1,old
C sb,old
G ,new
C sb,old
new
0.12
4
3 2
2
3
3 6 18 ft/s
Mass Balance:
D F D
z xB
0.6 0.01
0.59656 x D x B 0.999 0.01 596.56 kmol/h, B 403.44
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:
At top of col.
V
WL WV 1491.41
kmol
L V since pure MeOH, same mol. wt.
0.6
32.04 kg MeOH
lbm
2.046 lbm
h kmol kg Assume ideal gas. Top of column is essentially pure MeOH. lb 1 atm 32.04 n MWM p lbmol MWM v 3 V RT ft atm 0.7302 607.79 R lbmol R where pure MeOH boils at 64.5 C L
At top
1 0.2
MWv p
G ,old
U fl,new
6.0
G G
n
MWv
G
Fv
New Condition:
L
0.12
G
pV
0.2
6.0
G
20
9.6 ft s
MeOH
h
0.07219
lbm ft 3
1.8 R
607.79 R K 1 kg 2.2046 lbm 28317 cm 3 0.7914 g cm 3 1000g kg ft 3
24 0.0773 64.5
337.66K
105, 346
49.405
lbm ft 3
19.0
259
1/ 2
WL
Fv
WG
0.07219
.6
G L
0.28
0.2
U flood
C sb
20
0.2
19
V
frac u flood 3600
ft 3600 s h s Use either 10 ft (slightly higher frac flood) or 11 ft – (lower frac. flood). 0.90 0.07219 lbm ft 3
D = 10.27ft
3.0, y=
Ref. Bonilla (1993).
L Dact
.9 0.66667 .9 .4
2 L D min
x 1+
-1 x
1.75 ,
L V
0.875D , L avg
Generalize Llow
min
L D
7.25
0.4
x, y feed
L
L V
0.46667
V L 1.75
1 L V
1 .46667
1 L D
act
L V
min
1.75D , Llow
Lavg M where L D
n MWv V
0.875
0.636
2.75
actual
or Llow D
L D
min
0.875
0.5 Lavg
M
10.D11. a. Since Liquid & Vapor have the same mole fractions L G G
23
0.3273
At Minimum (Pinch Point), Llow V
L low
0.75
, at z
0.46667 , L D
1 L V xD
pV = nRT,
7.25 ft s
4 105, 346 lbm h
D
min
0.07219
4V lbm h V
L V
49.405 0.07219
0.28
20
V
Use Eq. 10-14 (Modified), D
10.D10.
0.02294
49.405
Fig 10-16 with 18″ tray spacing: Csb L
1/ 2
p MWv RT , R
L D
min
L V 45.6 cm3 atm gmoles-1
o
R1
260
MWv
.8 46
1 ATM
G
V
L D
L V
.2 18
40.4
3D
6750 lb day , L
4500 6750 .6667 1/ 2
L G
G
45.6 cm3 atm mol 1 R 1
40.4 g mol
H2O
2 F
G
L
1 .82
97
3
1.393 10
3
.82
1/2
.001393 g cm3
4500 lb day
=0.0275
1 g cm3
L
(Table 10-3) .2 G F2 F
G
2 2250
460 R
L G
.6667 1.393 10
L
F = 97 Ordinate This is,
GF
2D
176
.82 g cm3
1 .82
.197 , from Figure 10-25 (flooding line).
gc
.52
0.2
1 g cm 3 62.4 lb ft 3
g cm 3 .82 g cm 3
0.5216 lb (s ft 2 ) 8.64 10 4 s day
2
.197
32.2
45067 lb (day ft 2 ) , G
.75 GF
FIND AREA AND DIAMETER FOR 75% OF FLOODING
AREA
V
.75 G
45067 lb Day ft 2 .75
6750 lb Day
.19970 ft 2
D 2 4 .19970 ft 2 , D .5042 ft 6.05 in COLUMN DIAMETER 6.05 5 8 9.68 which is probably OK. PACKING DIAMETER b. From Fig. 10-25, G 2 F
G
2
97
0.2
1 .82
1.393 10
G
c.
(
G
.52 3
L
0.2
1 62.4
D 2 4 .3503,
L G will be the same; thus
D
2
67500
4 1.998
D
.6679 ft
0.3503 ft 2
8.01 inches
1/ 2
L
G
G
L
V .75 6 ,
.0275
1.927 10 4 lb (day ft )
19267 lb (day ft 2 )
AREA
L
0.036
0.2230 lb (s ft 2 ) 8.64 104 s day 6750 lb day
area
G
2
.82 32.2
AREA
Area
1/2
g c ) .036 at L G
V
will be the same, and G will be the same.
3D as before
.75 .521 8.64 10 4
67,500 lb day
1.998 ft 2 10
earlier value
1/ 2
1.59 ft
19.14 inches
261
10.D12. a.
y
L V x 1 L V x D . When x 0, y 1 L V x D 0.1828 . See figure. Need 2 equilibrium stages. Stop where feed line and operating line intersect. HETP 5 / 2 2.5 ft, x B ~ 0.065
b. M is at x in
y1
.43 L
L M
6.13
L
L M
1.58
L
L
Within accuracy of graph,
3.88
L
L
V L L L L If try a shorter column with same feed won’t work. L L .8 and must adjust column. L V
1
0.8
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. z xB D 0.6 0.01 Mass Balance: 0.59656 F x D x B 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
262
V
L
D
596.56
1255.9 kmol h 0.475 0.475 V D 659.4 kmol h
WL WV
At top of col.
V
1255.9
L V since pure MeOH, same mol. wt.
kmol
32.04 kg MeOH
2.046 lbm
h kmol kg The density and surface tension calculations are the same as in 10.D9. Assume ideal gas. Top of column is essentially pure MeOH. lb 1 atm 32.04 n MWM p lbmol MWM v 3 V RT ft atm 0.7302 607.79 R lbmol R where pure MeOH boils at 64.5 C L
MeOH
337.66K
Fv
G
WG
L
0.2
U flood
C sb
0.07219
.525
Fig 10-16 with 18″ tray spacing: Csb L
20
V
49.405
0.07219
lbm ft 3
49.405
lbm ft 3
1/ 2
0.02007
0.28 19
0.2
0.28
20
V
h
19.0
1/ 2
WL
lbm
1.8 R
607.79 R K 1 kg 2.2046 lbm 28317 cm 3 0.7914 g cm 3 1000g kg ft 3
24 0.0773 64.5
At top
82, 330
0.525
49.405 0.07219 0.07219
7.25 ft s
4V lbm h
Use Eq. 10-14 (Modified), D V
frac u flood 3600
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, V V F 1491.47 1000 491.47 . Since QR V, 3
QR ,liquid _ feed
(1255.9 / 491.7)QR ,vapor _ feed
2.55QR ,vapor _ feed .
Thus, in this case there is a significant energy price for reducing the column diameter by this method. 10.D14. D
z xB xD
xB
F
0.4 0.0001 0.998 0.0001
1000
400.7415 , B 1000 D
599.258 kmol/day
263
or
V
V
V
1202.225
At bottom, L
L D 1 D
3 400.7415
kmol 1 day
1h
1202.225 kmol day 0.013915 kmol s
day 24 h 3600 s
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 y boilup Thus
L G
V in
L
G is lb (s ft 2 )
L V 1.49846
lb
kmol 18.016 kg 2.20462 lb
0.013915
s
density water at 100 C
s
kmol m
3
nRT
where 100ºC
n G
V
Fv
K
L
G
G
L
59.83
3
lbm ft 3
1.0
L
MWw
RT
1.8 R
373.16K
35.31454 ft
1 atm
p
MWw
m3
kg W
0.55268 lb s
1 kg
kg 2.20462 lb
958.365
Data from Perry’s, 7th ed., p. 2-92.
pV
xB
18.016 3
ft atm lbmol R
0.7302
lbm ft 3
0.0367
671.688 R
671.688 R
1/ 2
1/ 2
0.0367
1.49846
0.03713
59.83
From Fig. 10.25, Ordinate at flooding = 0.18 0.18
G flood
F
w
Area
F = 33,
32.2, F
110 Table 10 3
1/ 2
0.15147
0.2
1/ 2
0.3892
lbm s ft 2
0.26 cp from Perry’s p. 2-323.
0.80 G flood
0.80 0.3892
V in lb s G actual lb s ft
Diameter b.
, gc
110 1.0 0.26
100
G actual
1/ 2
gc
0.18 0.0367 59.83 32.2
G flood Where
G L 0.2
0.55268 2
4 area
0.8 0.3892 1/ 2
Diameter Intalox plastic
4
Dia
0.31136 1.77505 ft 2
1.77505 Fint Berl
FBerl
1/ 2
1.503 ft 1/ 4
1.503
33 110
1/ 4
1.112 ft
264
lbm ft 3
Fv
10.D15.
WL
v
Wv
L
0.03713 from prob. 10.D14
From Fig. 10-16 with 12″ tray spacing , Csb,flood
0.21
0.2
K
Csb
where σ = surface tension water at 100ºC.
20
Perry’s, 7th ed., p. 2-306 @ 373.15 K,
K u flood
K
L
4V
0.0367
Dia
lb s
10 5 dynes
1m
1N
100 cm
0.21 1.2411
59.83 0.0367
0.206
v
v
0.2
0.21 58.9 20 v
Dia
0.0589 N m
58.9
dynes cm
0.2606
10.52 ft s , u act
.8 10.52
8.416 ft s
Eq. (10-14) modified for units. V is from Problem 10.D14.
lb ft u act 3 ft s
4 0.55268 0.85 0.0367 8.416
1.637 ft . This can be compared to 1.5 ft for packed.
Tray columns with this small a diameter are seldom used in industry. 10.D16.
F1/4 . F1
From Eq. (10-44). Diameter Diameter (3 )
98 and F3
Diameter (1 )
22 Table 10-3
1/ 4
F3
14.54
F1
22
1/ 4
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. p MW 1 100.2 0.205 lb ft 3 v RT 1.314 371.4 Need L V . Since L V operating lines. Then L V
.8
1 y
where z 1 z 0.6 0 1.2 . .5 0
L
L MWL
G
V MWv 1/ 2
L
v
G
L
1.2
.5, we have y
1.2 1.0
0.6 at intersection of
1.2
0.205 1.684 62.4
1/ 2
0.084
Figure 10-25, Ordinate = 0.05 at ∆p = 0.5
265
0.05 .205 0.684 62.4 32.2
G The value 0.9595 is
V
V
.2
98 .9595 .684 0.205 at 98.4ºC, and
Water
Water
0.375
. Since feed is a saturated liquid
L
0.6944 lbmol/s.
0.6944 100.2
VMWv
Area
G
0.375
185.5 ft 2
1/ 2
D 4 Area 15.37 ft This is somewhat larger than in Example 10-4. Therefore design at bottom. G
G flood
10.D18.
L
gc
ordinate
F
0.2
Assume changing p changes only
G
& ordinate. Then take ratios
G flood ,new
G ,new
G flood ,old
G ,old
ordinate, new ordinate, old
G ,new
p MW
RT
G ,old
p MW
RT
new
p new
old
p old
4 . Assumes small change in
T (in Kelvin). T set by boiling conditions (Vapor Press) not by ideal gas law. Fv
0.5
L
G
G
L
,
0.5
F v ,new
G ,new
p new
F v ,old
L ,old
p old
0.5
2.0 , F v,new
2F v,old
0.4 .
New Ordinate Value ~ 0.5, old value ~ 0.09
G flood ,new
p new
ordinate, new
p old
ordinate, old
G flood ,old
4
0.05 0.09
0.5
0.75
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 = (0.9990.825)/(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 2enthalpy feed. In the middle section of the column at minimum reflux conditions the slope of the middle operating line is L / V (0.825 0.6) / (0.6 0.22) 0.592 . The external mass balances still gives D
Lmin
( L / D) min D
0.7733(596.56)
the saturated liquid feed V
V and L
596.56 kgmoles/hr, B
461.3 and Vmin L
Lmin
D
403.44 . At minimum reflux
461.3 596.56 1057.86 . At
Fliquid . Thus,
[( L Fliquid ) / V ]min ( L / V ) min 0.592 and Fliquid ,min 0.592V L 165.06 . 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, Fliq 165.06 and Fvap 834.94 .
266
At top use saturated liquid reflux ratio L V
0.525, V L D V 0.525V D D 596.56 V 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 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 V V 1255.9 . In the bottom section, 3
V
V
Fvap
1255.9 834.94
420.96 .
Since QR V , QR ,2 enthalpy _ feed (420.96 / 491.7)QR , vapor _ feed 0.86QR , vapor _ feed . 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 C sb . Gas is N 2 . Liquid is ammonia. Since system very dilute, treat as pure ammonia liquid & pure N 2 gas.
L
WL kg h
L kmol h
G
WG kg h
V kmol h MWV
0.61
L
100 cm
1 kg 3
cm 1000 g
m
27.36
RT
L atm mol K
16.642
236
28.08
16.642
610 kg m 3
3
175 atm 28.02 g mol 0.08206
17.03
3
pMW v G
Fv
g
MWL
1000L m
253.2 K
3
kg 1000 g
236.0
kg m3
1/ 2
10.35 610 Off chart. Extrapolate using Eq. (10-10e).
log10 Csb
0.94506 0.70234 log10 10.35
log10 Csb
1.891
20
Assume
u flood D
0.2
v
2
0.01286
1.0. Then
0.01286
610 236 236
4 V MW V v
0.85,
Csb
0.22618 log10 10.35
u op 3600
236.0 kg m3 , u op
0.0162 ft s , u op ,V
0.75 uflood
100 kmol h , MWv
0.01215
28.02 kg kmol
0.01215 ft s , Need to watch units
267
4 100 28.02
D
=1.155 m
0.85 236.0 0.01215 3600 1 3.2808 ft
3.79 ft
Probably use 4 ft diameter column – (standard size) Using larger diameter helps take into account the uncertainty in extrapolating to find C sb . 10.D.21. New Problem in 3rd edition. The mass balance and flow rate calculations are the same as for problem 10.D14.
D
z xB xD
xB
0.4 0.0001
F
V or
400.7415 , B 1000 D
1000
0.998 0.0001
L D 1 D
V
1202.225
3 400.7415
kmol 1 day
599.258 kmol/day
1202.225 kmol day
1h
0.013915 kmol s day 24 h 3600 s L/V = 2/3. Top of column is close to pure methanol Thus L G L V 0.66667 G is lb (s ft 2 )
V in Pure MeOH boils at 64.5 C L
MeOH
lb
kmol 32.04 kg 2.20462 lb
0.013915
s
s
1.8 R
337.66K
K
0.7914 g cm 3
kmol
1 kg
0.98290 lb s
607.79 R 1 kg 2.2046 lbm 28317 cm 3
1000g
kg
ft
49.405
3
lbm ft 3
59.83 / 49.405 1.211 Assume ideal gas. Top of column is essentially pure MeOH. W
L
1 atm
n MWM
p
V
RT
v
Fv
MWM
G
WG
L
lb lbmol
3
ft atm 0.7302 lbmol R
1/ 2
WL
32.04
.66667
0.07219
0.07219
607.79 R
lbm ft 3
1/ 2
0.02548
49.405
From Fig. 10-25, Ordinate at flooding = 0.20 G flood
G flood Where
methanol
G actual
0.20
G L 0.2
F
gc
1/ 2
, gc
32.2, F
0.20 0.07219 49.405 32.2 110 1.211 0.34
64.5 C
110 Table 10 3 1/ 2
0.2146
0.2
1/ 2
0.4633
lbm s ft 2
0.34 cp from Perry’s (8th ed.) p. 2-449.
0.80 G flood
0.80 0.4633
0.3706
268
V in lb s
Area
0.98290
G actual lb s ft
2
Diameter
0.3706 1/ 2
4 area
2.6521 ft 2
4
2.6521
1/ 2
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.
Diameter Intalox
F = 33,
Dia
plastic
10.E1.
D kmol
op time Then
V
L
1.838
FBerl
Berl
where D 18.1303 kmol ,
D kmol hr D
1/ 4
Fint
V 0.4V
L
2
D
3
1/ 4
33
1.360 ft
110
, L V
0.4
D
0.6V and t op
0.6V Use Fig. 10-25 or Eq. (10-39a) to find flooding at the end of the operation at bottom of column. kg MWliq 2 L lbm s ft L kmol h kmol 2 kg G lbm s ft V kmol h MWvapor kmol At end of operation at bottom of column x 0.004, y 0.036 (pinch)
L G
MWavg,liq L
0.004 46
18.128
0.4
0.381
19.023
0.996 18.016
18.128 & MWvap
0.036 46
0.964 18.016
19.023
62.4 lbm ft 3 (Essentially pure water). Boils at ~ 100ºC = 373 K p MW G
Then
1.0 atm v
RT
FV
19.023 3
atm ft 1.314 K lbmol 1/ 2
L
G
G
L
0.381
0.038806
373 K
0.038806 62.4
lbm lbmol
lbm ft 3
1/ 2
0.009501
From Eq. 10-39a.
log10 ordinate
log10 ordinate G
2
From Table 10-3, F
0.6864 & ordinate
0.2059 F
70,
1.6678 1.085 log10 0.009501
G L 0.2
0.29655 log10 F v
2
0.2059 agrees with Fig. 10 25
gc
1.0, g c
32.2,
water
100 C ~ 0.26 c p (Perry’s 5th ed., 3-213)
269
G flood From Eq. (10-41), V
70 1.0 0.26
s
12
Then 10.F1.
lb l bmol
s ft 2
2
0.19635 ft 2 and G act
0.7 G flood
0.0038646 19.023 lbmol 3600s 0.453593 kmol V 0.0038646 6.310665 kmol h hr s h 1.0 lbmol D 18.1303 kmol t op 4.7883 h 287.3 min. 0.6V 0.6 6.310665 yM 1 x M x M 1 yM
0.134 0.98 0.02 0.866
. From Equil. data
0.979 0.05
7.582 , Top 7.582 2.454
0.95 0.021
(Note 40% MeOH is probably wt%) Estimate: n mix x M n M x W n W Feed is 60% M 40% W n mix 0.60 n 0.28 0.40 mix
4.31 0.306
2.454
4.31
Column temperature varies from 64.5º to ~ 98.2 ºC. 64.5+98.2 Avg T= 81.35 C 2 from Perry’s 7th ed., T = 81.35ºC, p. 2-323. liquids
M
n 0.35
0.28 cp,
1.1837 ,
W
mix
0.35 cp
0.306
1.3195
From O’Connell’s Correlation, Fig. 10-14, Eq. (10-6): E o
s ft 2
s kmol
Geometric avg
Then
lbm
lbm
G
4 0.19635 0.53488 0.7
lbmol
Need average
Bot :
MWvapor
0.53488
0.2
Area ft 2
lbmol
Area V
0.5
0.2059 0.03698 62.4 32.2
0.52782 0.27511 log 10 1.3195
Eo
45% 0.044923 log 10
1.3195
2
49.5%
If conservative use 45% 10.F2.
To use O’Connell’s correlation (Fig. 10-14), need α and viscosity of feed. KM yM x M yM 1 x M . Used Table 2-7 for values. MW K W yW x W x M 1 yM Can estimate a geometric average at bottom, feed & top 0.134 0.98 0.729 0.6 7.582 , MW ,feed x .4 4.035 M W ,bot 0.02 1.866 0.4 0.271
270
0.979 0.05
1/ 3
2.454 ,
4.2184 0.75 0.021 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 0.39cp & MeOH 0.30 Note: (MeOH, 40% probably refers to wt % - p. 2-322 Perry’s) Estimate n mix x1 n 1 x 2 n 2 MW ,top
n
mix
.4
avg
n 0.30
10.F3.
z xB F
F
.6 n 0.39
4.2184 0.351 Overall Plate effic. = 43.7% Then
bot
Feed
T
1.0465 ,
mix
0.351
1.481
0.30 0.01 100
36.71 B F D 100 36.71 63.29 lb mol h xD xB 0.8 0.01 At top of column L = D(L/D) = 73.4 and V = L + D = 110.1 L F F3 L Stripping section 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 = D
0.667 and goes through
y
x
xD
0.8 . Bottom operating line goes through
y x x b 0.01 and 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.
271
Overall Efficiency. For O’Connell Correlation, need
yE 1 x E 1 yE x E x
0.019, y
x
0.3273, y
x
.7472, y
AVG
and
Feed
Tcol
. Using Table 2-1 we find α and following mole fraction.
0.170 .981
0.170 :
.830 .019 .5826 .6727
.5826 :
.4174 .3273
.7815 .2528
.7815 :
.2185 .7472 1/ 3
AVG
1
2
10.575 2.87
1.210
3.324
3
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 E o .42 .
N
8.875 .42
Height 22 18
18
21.1 . Thus, need 22 stages plus partial reboiler
disengagement
48 (bottom sump)
38.5 ft
4V MW V
Diameter Calculation Dia v
flooding fraction u flood 3600
Use average values of parameters in stripping and enriching sections.
MW v
MW
eth
stripping section: 18.25
yeth MW
MW
W
y W for both MW V and MW L .
21.5
WL
L MW L
206.75 20
4135 lb h
WV
V MW V
143.46
2869.2 lb h
20
0.96225 g ml 60.07 lb ft 3 Bottom Bottoms T =100ºC = 672ºR L
P MW V
V
Fv
WL WV
V
L
Enriching section:
RT
1.0
4135 2869.2 26.4
MW
20
0.7302 672 R
0.04076 60.07
0.04076 lb ft 3
0.0375, Csb
0.28 .
40.4
WL
L MW L
73.42 35
WV
V MW V
110.13 35
2569.7 3854.55 lb h
0.766 g ml 47.92 lb ft 3 Distillate Distillate T = 82ºC = T = 639.6ºR; L
V
P MW V
RT
1 35
0.7302 639.6 R
0.05472 lb ft 3
272
Fv K
Csb
WL WV 20
0.2
V
2569.7 3854.55
L
0.05472 47.92
0.0225, Csb
0.28
, ft/sec. σ, surface tension in dyne/cm. 57th ed. Hdbk of Physics + Chemistry, F-45. Bottoms, σ ~ 46 dyn/cm, Middle, σ ~ 25 dyn/cm, Top σ ~ 18.6 dyn/cm
u flood
0.20
stripping:
K
0.28 46 20
enriching:
K
0.285 18.6 20
K
L
V
V
0.33075
enriching: u flood
0.2809 Dia.
0.2809 ft s
60.07 0.04076
V
12.693 ft s
0.04076
47.90 0.05472
8.31 ft s
0.05472 4V MW
V
enriching: V MW V
0.20
, ft s
stripping: u flood
stripping: V MW V
0.33075 ft s
0.90
V
0.75 u flood
V
3600
1/ 2
143.46 lbmol h
20
0.04076
lb ft 3
70392.5
110.13 lbmol h
35
0.05472
lb ft 3
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. V lbmol s MWV lb lbmol Cross Sectional Area G lb (s ft 2 ) Bottom:
L G
L V
Top:
L G
L V MWL MWV
1” metal Pall rings: F
48,
MWL MWV
0.15,
1.44 ~ 1
1.44
0.667 ~ 1
0.667
0.15
ψ = Density of water/density of liquid At Top: 61 47.92 1.27 At 81ºC ,
w
0.35 cp,
E
y= x d
At top
0.8, n
MWv
y MWE
1/ 2 G
L 2
G F G
0.078 F
G
L 0.2
gc
x1 n
MIX
.8 n .45
MIX
1 y MWw
0.05472 lb ft 3 ,
G
L G
0.45 cp , n
L
G
L
gc
V 110.13 mol h
2
.2 n .35 , and 0.8 46
MIX
.2 18
0.43 40.4
1/ 2
0.02254
0.078 from Figure 10-25
0.078 0.05472 47.92 32.2 48 1.27 0.43
x2 n
47.92
0.667 0.05472 47.92 0.2
1
0.2
1/ 2
0.358 lbm (s ft 2 )
0.0306 lbmol s
273
D2 4
Area
D
4V MW v
1/ 2
V MW v , or G
4 0.0306 40.4
G
1/ 2
0.358
2.096
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 0(base) 10
NF,L --
Two-enthalpy feed: 500 11 6 500 12 6 600 12 6 750 13 6 750 16 7 1000 (all liquid) 9
N 20
dia A Vol tray QR 2.84 6.33 55.0 2 1065
Qc, total -12,330
decr Vol --
20 22 20 20 26 20
2.08 2.07 1.89 1.56 1.56 1.20
-12,340 -12,330 -12340 -12,360 -12,330 -13,680
46.5 % 0.5 % 40.9 % 0 55.8 % 0.5 % 69.6 % 2.3 % 60.3 % 0 82.1% 127 %
3.38 3.38 2.80 1.92 1.91 1.13
29.4 32.5 24.3 16.7 21.8 9.8
2 2 2 2 2 2
1070 1065 1070 1090 1065 2415
Intermediate condenser: FWithdr NF,V NV,with NL,ret N dia A Vol tray 300 11 10 6 20 2.41 4.56 39.6 2 450 11 10 6 20 2.22 3.88 33.7 10/11
QR 1067 1067
Qc, total -12,340 -12,340
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 100 12 6 5 5 20 1.69 2.23 19.4 2 1073 -12,340 80 12 6 13 6 20 1.72 2.33 20.2 2 1079 -12,345
incr QR --
decr Vol incr QR 28.0 % 0.2 % 38.6 % 0.2 % decr Vol incr QR 64.7 % 0.8 % 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 100 12 6 5 5 20 1.51 1.79 15.5 6 1081 -12,350
decr Vol incr QR 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).
274
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 m 2 , Vol = column volume in m 3 = A(N – 2 +1) (tray spacing) where N – 2 is the number of trays and +1 is for disengagement space for liquid and vapor, Q R and Q c = reboiler and condenser duties in kW, p cond condenser pressure
p pressure drop in psi/tray, N feed = optimum feed stage, and condenser is stage 1. Note this solution has a p, each stage. Thus, solution slightly different than students’ solutions. in atm,
N Feed N Dia A Vol tray Q R Q c , L1 D D p cond 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
p 0.1
Table 2. Diameters calculated for standard distillation base cases listed in Table 1. Vapor flow rate Vj and liquid flow rate L j are in kmol/hr, diameter is in m, area is m 2 .
Tray 2 23 24 35
Ethanol-water, vapor Vj Lj Dia 1125 997 2.61 1082 949 1.99 74.3 950 0.71 79.4 956 0.71
Area 5.35 3.11 0.396 0.396
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. N F,V and N F,L 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 Q R are compared to the ethanol-water base case (Table 1). For both runs y D,E N Dia A Vol Tray Q R N F,V N F,L FL 0(base)
FL 600 Qc,total
23
--
36
2.61
2.24
35.8
2
902
N N F,V N F,L 23 17 36 Qc,col Qc,feed condenser
Dia
A
Vol
Tray
1.69
2.24
35.8
2
QR 902
0.7901 and x B,W
0.9986.
Q C,col -10,700
Q C,tot -10,700
Decr Vol 58.2%
Incr Q R 0
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 E10 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.
275
V feed L feed Qc kW QR kW Max Dia Stage yD,C4 xB,C3 Kmol/s Kmol/s m max dia 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=41 NF,liq=18 NFvap=21 N .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.
276
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
L D
MIN
increases, hence L/V increases and Q c increases.
L V increases and Q R decreases. The number of stages probably decreases. (See Figure 7-3. The abscissa increases as L D
min
and L/D increase. Thus ordinate drops.
Since N MIN is constant, N decreases). 11.A.6. New Problem for 3rd edition. b. A liquid side-stream between the feed and the distillate. c. A vapor side-stream between the feed and the bottoms. 11.A.9.
1 point for b. 2 points for c.
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. 0.33 X,T 11.B4. Use heuristics. Hardest separation is xylene-cumene, x ,c 1.57 0.21 C,T Toulene is most abundant. Use heuristics to: Remove toluene early Do hardest separation last One-by-one in overhead If use equil-molal splits: B
T
X
F C Can also derive coupled systems.
277
B
X
F T C
11.B5.
Add in Heuristic 9 for sloppy separation. T
T
B
B X
F
OR
F
X C
C
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 1 K c7 a&b. For binary, x c6 , y c6 K c6 x c6 . Use Fig. 2-11. K c6 K c7 T = 168ºC: K c6 1.29, K c7 0.71, x c6 0.50, y c6 (Guess was aided by solution to Problem 10-D1) .645 .5 1.817 .355 .5
n N MIN
L V
min
0.645
.999 .001 .001 .999 23.13 stages n 1.817 x D y 99 .645 0.7094 x D z .999 .5
278
L
L V
D c.
L D
Gilliland: x
d.
L D
2.441 min
4 2.441
MIN
L D 1
0.3703 23.13
N 1
0.37027
1 0.3703
Fig. 11-1.
D2
5.91 m 2 . Column Vol
4
cos t
$700 m 3 ,
7 bar, Eq (11-5), Fp
2.25 1.82 2.3 1.0
Qc
$844, 000 .5
10.71 0.00756 7
2.30
$ 209,800
4196 50 1 1
total
$1,054,000
500, B 500, L V
Vh D gives, Q c
$4196 tray
7 2.743
.5
L VH v
$131,110
710 5.91 m 2
C P, tray
10.71 0.00756p
69ºC (bp) = 156.2ºF. F 1000, D Energy Bal:
187.3 m 3
3
pD
C BM.tray
Fig.11-9.
m
187.3 m 3
area H
$700
C P,tower
vol Tray cos t Fig. 11-2. $710 m 2 , area Eq. 11-8. CBM,tower 131,110
11.D2.
37.3 contacts or 36.3 stages
N act N eq E o 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 2.743 m 3.2808 ft Tray area
700 kPa
0.3117
5
N N MIN
Eq. (7-42b):
N
1 L V
min
0.8, L D
2000 lbmol h, V
V ho
4 2500
Hv
h D is pure boiling hexane, H v is vapor. Thus, h D
Hv
Qc
2500
Pick 25ºC as basis.
hD
3.39 107 Btu h
13,572
QR
C PL ,c 6 69 25
hB
13,572 Btu/lbmol
c6
Dh D
Bh B
1.8 F C
C PL,c7 98.4 25 1.8
Fh F 51.7
QC 44 1.8
4094.6
50.8 73.4 1.8
Feed is a saturated liquid. From Example 10-1, T = 80ºC
6711.69
Btu lbmol
279
hF
CPL,c6 z c6
CPL,c7 z c7
80 25 1.8
hF
51.7 .5
50.8 .5
55 1.8
QR
500 4094.64
500 6711.69
A
3.423 10 7 Btu h
Q U
1000 5073.75
T
Btu 110 98.4 C h ft 2 F where U is average from Table 11-2. Condenser:
50
U
TAvg
32, 800 ft 2
1.8 F C
3.39 10 7
Qc
A
3.39 10 7
3.423 107 Btu h
QR Reboiler:
5073.75 Btu lbmol
2850 ft 2
110 70 2 Note these areas are very approximate. For detailed design need a much better estimate of U.
A
Costs: Condenser
180 156.2
2850 ft
2
1m
2
3.2808 ft
264.8 m 2
2
Fixed Tube Sheet S&T Fig. 11-3, Cost = $125/m2 Reboiler A
32,800 ft
2
1m
2
3.2808 ft
2
3047 m 2
large because of low U.
Extrapolate $70/m2 1 atm, Fp
Condenser Reboiler 11.D3.
1.0 . Eq. (11-9) CBM
1.0, Fm
C BM
3.29
$125
264.8 m 2
m2
3.29 $70 m 2
C BM
Cp 1.63 1.66 Fm FQ
3.29 C p
$109, 000
$702, 000
304 m 2
$811, 000
Very sensitive to U.
Note: This solution requires the solution to Problem 11-D2. lb Q R 3.423 10 7 lb Steam rate, 35, 704.6 h 958.7 h where Q R is from Solution to problem 11-D2. Steam Cost
$
35, 704.6
h
Cooling water,
lb
$20.00
h
1000 lb
lb
Qc
h
C p w Tw
3.39 10 7 1.0
40
$714 h. 847, 500
lb h
where Q R is from Solution to problem 11-D2. Water Cost
$ h
847, 500
lb
$3.00
1
h
1000 gal
8.3 lb gal
$306 h
280
11.D4.
From Example 11-1 needed 21.09 eq stages + P.R. h pack 21.09 1.1 ft stage 23.2 ft 7.07 m
D2
Vol
4
2
h pack
4
15 23.2
4099.6 ft
3
4099.6 ft
1m
3
3
3.2808 ft
3
116.1 m 3
Cp ~ $250 m3 packing Tower 23.2 ft + 2 ft between sections + 2 ft top + 2 ft top = 29.2 ft 1m h = 29.2 ft 8.9 m , Vessel Vol. 146.1 m 3 3.2808 ft Fig. 11.1 Cp $700 m3 for tower From Fig. 11-2
Fp
1.0 1 atm , Tower FM
1 carbon steel , C BM
C p 2.25 1.82
$700 146.1 4.07 $416, 312
Packing, Fm
4.1. C BM
C p 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 B1 . Thus line LM is extended to 0.0 mole fraction methanol to find location of B1 (0.73333 MB and 0.26667 toluene). We can use mass balance to find point B1 accurately. If D1 is pure methanol, D1 = 50 (all methanol in feed) and B1 = 150. Then from toluene balance 0.2 × 200 = 150 x tol,B1 , which gives x tol,B1 0.266667 . B1 which is toluene product. F Re cycle
F2 which is then split into I (Some of which is recycled) and B 2 100 100
M
Since D1 contains no toluene B1 0.26667
B1
F .4
D1
Re cycle 0
.4
100 150 kmo h. F2 , D1 M B1 0.26667 For Column 2: B 2 essentially pure toluene, B2 0.266667 F2 B1
OK – Satisfies overall external M.B. D 2 Re cycle F2
D2
B2
50.0 kmol h
0.26667 150
40
150 40 110
10.0, which also satisfies external M.B.
281
11-D7. a.)
Proposed Split: Bottoms – Essentially pure toluene Distillate ~ .83 methanol. (See figure for Solution to 11.D7) F 100, z M 0.5, z MB .1, z T .4 Assume all toluene in bottoms & bottoms is pure. B 100 .4 40, D 60
60 x M,dist b.)
50 , x M,dist
0.8333 & xMB,dist
0.166667
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.
282
11.D.8. New Problem for 3rd edition. Part a. Cost in June 2010 $947, 000 556 397
Part b.
F
$1,326, 000
2 x F of Example 11-2.
Since Dia
Tower
F , Dia 2F11
12 ft
2
1m
Diameter
17
Volume
488.2m3
CP,tower
$550 m3
3.2808 ft
2
16.97 or 17 feet.
5.182m, Tray Area
Fig. 11-1, C0p 488.2 m3
21.08m 2 .
Cost vol ~ $550 m3
$268,500
Fig. 11-2. Tray cos t area ~ $750 m3 extrapolate , Cp,tray CBM,tower
$750 m 2
21.08m2
268,500 2.25 1.82 1.0 1.0
$15,800 tray
$1, 093, 000 283
CBM,trays
$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 Foriginal , cost lbmol $1326 lbmol At
11.G1.
2 x Foriginal , cost lbmol
$1164 lbmol
- 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 B1 : 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.
Col 1. N
90, N f
41 ,
L D
8 D
60
Bottom 0.999186 tol 0.000814 MB
Dist. 0.99938 tol, 0.000605 MB, Col 2. N
20, N f
10 , L D
2, B
10 Bot .996679 MB, 0.003088 MeOH, 0.000233 toluene
If increase L/D in column 1, Col 1. L D
Col 2. L D
2, N
24, N f
9, N
90, N f
41, Bot. 0.99941 toluene.
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
284
P=1
NF = 5 NF = 6 NF = 7 NF = 8 NF = 9 NF = 10
Pcol = 3 atm
Pcol = 5 atm
L D 2 NF = 8 NF = 9 NF = 10 L D 2 NF = 8 NF = 9 NF = 10
QR = 52,948 QR = 53,029 53,148 53,175 52896.9 cal/s
QR = 67995.5 QR = 68022.1 67794
QC = -78614.5
Qc = - 74274
QR = 76435.3
7646 Qc
76264
71692.8
xD,E = .72033 .74564 .76332 .77539 .78037 .69964
.75453 0.76027 .69695
.74276 .74840 .69411
xB,E = .031085 .028263 .026298 .024957 .024363← .033373
0.027275 0.026637← .033672
.028582 .027955 ← 0.033988
Size optimal feed columns. Sieve tray 1 uses 18 inch tray spacing at 85% approach flooding, Fair calculation method for flooding. Pcol 5 Max diameter tray 2 0.34867m
Pcol
3
Max diameter tray 2 0.38070m
Pcol
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 twophase 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. Q R = 569,172 cal/s, 285
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 Q C,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. Q c = 243,900 cal/s, L/D = 1.5618, T cond = 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.
286
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 y out , or N for A and y out 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.
Eo
Eq 12 22
N equil N actual n 1 E mV
Eo
(10-1)
Eq 12 34 mV L
1
n ( L mV)
n L mV
n 1 E mV Eo
mV 1 L
n 1 E mV
mV 1 L
(10-4)
n mV L QED
Ref. Lacks, D.J., Chem. Engr. Educ., 302 (Fall 1998).
y
12.C5.
12.C6.
x 1
1 x
When
x
0,
When
x
1,
Apply Kremser as
x
Graphically,
L V
dx
dy
, b
dx dy
1
dx 0 or x
yN MIN
dy
, m
xN
1
, b
1
1 x
2
0
1
1.0 y1
x0
where y N
1
& x N are in equilibrium
287
yN L
m
For absorbers
mV
min
N
Thus as Eq. (12-23) becomes
yN yN
or
mV 1
y1
1
* 1
m
y
p tot
x
0.395
y1
L
x 0 .0001
H
0 min
L
y
3 b 0
x
yN
V
1
xN
min
b
y1
L
x0
V
min
0.395 x
0.0002 unknown
unknown
1
L x0 y1*
mx 0
min
L x0
1.186
where y1*
mV
y1
1
yN 1 b m m which agrees with graphical analysis.
12.D1. New Problem 3rd Edition.
min
L
,
yN
b for linear.
L
&1
V
mx N
1
Gy N
m x0
1
Gy1
L xN
.395 0.0001 0.0000395
N
xN
L
yN
G
1
0.004 Since x N unknown,
unknown
5
1
y N 1 y1* y1 y1*
mV L n
n 5
1
39.5 L
1
unknown.
Eq. (12-30) or (12-31) easiest to use
n Eq. 12-30
y*N
0.004 0.0000395 0.0002 0.0000395 L n 39.5
mV L
mV L 39.5 L
n
1
39.5 L n
24.67
39.5 L
L 39.5
288
T & E → be sure L mV
L 50 75 65 63 62 xN
G
yN
L
1
mV/L 0.79 0.52668 0.60769 0.62698 0.637097 y1
1
L/mV RHS 1.2658 7.58 1.8987 3.90 1.64557 4.67 1.59494 4.895 1.56962 5.015 100 0.004 0.0002 62
x0
OK 0.0001 0.006229
There are alternative solution paths, but L = 39.5 is not valid, it becomes Alternative: Trial-and-error McCabe-Thiele solution. 12.D2.
a)
n1
0
n1
0
.
L & V constant. 97% rec. H2S 3% left in gas. y out 0.03 0.0012 0.000036 Equil.:
y
p H 2S
423x
p tot
2.5
169.2x
M.B. with Const. L & V: 0.0012 V
x out b) See Figure. y c)
L V
0.000036 V
0.0012 0.000036 x
L V
y in
V
x out L
0.001164
L
10
5.82E 6
2000
x out where y in , x out & y out , x in are on op. line .
Eq
yin
169.2
0.000012
LG
x*
7.09E 6
Note that L G min can be calculated from a crude sketch.
min
y
L G y out
L
0.0000036 0
200
G act 0
0.0012 0.000036 7.09 E 6
min
M
164.124
L G
min
M = 1.2186
x
289
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:
Vy IN
Lx IN
y out V Lx , Equil., y
Substitute equil. into M.B. HVx HVx Lx IN Lx p tot p tot Solve p tot b.
y out
H p tot
x out
x 10.96 0.274
100 2 10 0.4 10
6
p tot
x
x 6
10.96 10 0.4 10
HVx L x IN
L x IN
H
6
0.4 10
16.0 10
6
0.274 atm
6
Can also do graphically, with Kremser equation (trial and error) or by solving mass balance first, L y out x in x out 16 10 6 Then p tot H x out yout 0.274 atm. V 12.D4.
Mass bal.
yin Vin
Lin x in
y out V Lx out
290
y out
y in
L V
x in
Op. Eq.,
point yin , x out line. See graph.
L V
x out
L V
x in
y
x out
L
x
12 0.0002 0.00001 y in
0.00228
L
x out V V 0.0, 0.00001 and point y out , x in
0.00228, 0.0002 are on op.
Can also use Kremser equation for this problem.
291
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 Yout , Xin and Yin , X out 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. * YN 1 Yin 0.02, YN+1 mX out 1.5 0.06 0.09
Y1 N
Yout
0.50, Y1*
mX in
n
.02 .09
.5 .6
n
.02 .5
.09 .6
HETP
1.5 .40
0.60
5.88
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 cu ft/lbmol at 60˚C 273.16 N 2 flow rate 2500 437 5.72 lbmol/h Water flow rate: Ignore CO 2 in water. MWw
100, 000 18
18
5560 lbmol/h
292
Equilibrium at 60˚C: H = 3410, y =
L
H PTOT
x
3410x
5560
972.0 . External mass balance is: 5.72 L L L y out x in y in x out x in x out 972. 9 10 6 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. V
Kremser: y N
1
0, y1
0.00875, y1* n
mx 0
3410 9.2 10
0.03137 , m V L =3.508
0 0.03137 0.00875 0.03137
1 3.508
N
6
n 1
3.508 5.07
.4 3.508 1
Probably use 6 real stages.
G Wt
12.D7.
0.828 wt frac air 1050
G mole
Yin
Inlet
869.4 29
y in
kg gas h
29.98 kmol air/h
0.172
0.172
1 y in
1 0.172 .828 29 Yin ,molar 0.2077 0.3543 17 NH3 0.172 1050 180.6 kg NH 3 2% remains in gas
869.4 kg air/h
0.2077
kg NH 3 kg air
3.612 kg NH 3
293
3.612 17
Yout mole
Equilibrium data Table 12-2 - y mole
X wt
kgNH 3 kgW
0.05 0.075 0.100 0.15 0.20 0.25 0.30 0.40 0.50 0.60 M.B.
Ymole
0.007087
869.4 29
p NH 3
p, mmHg
p, mmHg
PTOT
1.30 760
988
p
y mole frac
11.2 17.7 25.1 42.7 64 89.5 119 190 275 380
0.01134 0.017915 0.0254 0.04322 0.06478 0.090587 0.12044 0.19231 0.27834 0.3846
X in L
G mol Ymol
17
L wt 17 G mole
X wt
y
Ymole
1 y
0.011466 0.01824 0.02607 0.04517 0.06926 0.09961 0.13694 0.2381 0.3857 0.6250
LX
wt
GYout ,
kg NH 3 17 kmol NH 3
Yout
Note: Units, although mixed, work in Mass Balance & in Operating equation. See graph. Minimum Solvent
Slope
L wt,min
0.3543 0.007087 0.477 0
= 0.7279=
.7279 17 29.98
Actual Solvent, L wt Op Line Slope
1.5 L min
L wt ,min 17 G mole
370.99 kg W h
556.48
L wt
556.48
17 G mole
17 29.98
kg W h
1.092
294
295
12.D8. New Problem 3rd Edition. X IN
y out
0
0.002
L
Yout
p tot
0.002004
2 atm
760. 2 mmHg
1520 mmHg
L G
G
L
F2
0.5 .475 air 0.05 mole frac HCl
.05
y
0.05263
.95
mol HCl mol air
X out
F1 1.0, G=.8 y1IN
0.20 mole frac HCl
YIN
.2 .8
mol HCl
=0.25
mol air
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 mol air 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 y X p kg water
p p tot
Y
kg HCl
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
296
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
Y
L G(MWHCl )
(L / MWHCl )X GYout , X IN
GY (L / MWHCl )X IN
L
X Yout
0.002004 Goes through (0, 0.002004)
1.275(MWHCl )
Bottom Operating Line GY (L / MWHCl )X out
L
0
GYin
(L / MWHCl )X
L
L 0.8MWHCl 0.25 L 0.8MWHCl X out X out G(MWHCl ) G(MWHCl ) a) Feed line saturated gas at Y 0.05263. Two operating lines intersect at feed line. Y
X YIN
Y1,IN
0.25
X *out
Sketch for Min L determination
Y
0.002 *
0.69 (see figure)
L (G(MW)) L MIN
(G(MWHCl ))
X *OUT
YFIN
0.8(MW) 0.69
L
Yout
(G(MWHCl ))
G(MWHCl ) YF2
YF2
YF2
L
G G
YF2
.25 0.05263
X *OUT
YOUT
G(MWHCl )
X *OUT
X int er sec t
(G(MWHCl )) L
Bottom operating line: YF2
L
0.05263
X
Top operating line:
YF2
YF2
X int er sec t L /(G(MWHCl ))
Yout
From plot X OUT
Solve for X int er sec t
L (G (MWHCl ))
YOUT
X int er sec t
L.
YF2
YFIN
Then
YOUT
L MIN
0
YFIN
G(MW) X
* OUT
YFIN
YF2
G(MW) X *OUT
YF2
YOUT
1.275(MW)
L MIN / MWHCl
0.05263 0.00200 0.69 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
297
298
b. M.B.
F2 YF2 X out
X out Top Bottom
GYIN
(L / MWHCl )X IN
F2 YF2
GYIN
GYout
(L(MWHCl ))
0.475 0.05263
GYout
(L / MWHCl )X out
X IN
0.8 0.25
1.275 0.002
14.584 / 36.461
(L / MW)
0.4
G (L / MW)
1.275 0.4
G
Top goes through
0.8
0
0.5561 kg HCl kg water
0.3137 0.5
X IN
0, Yout
0.002, Slope
(L/MWHCl )
G Arbitrary point for plotting: X .4, Y .3137 .4 0.1255 Y .1255 .002 0.1275 Bottom from X out 0.5561, YIN 0.25 To intersection Top Operating and Feed Line. Need ~ 1.6 stages. Opt. Feed for F2 is Stage 1 (Feed 1 is at bottom.) Check Slope bottom
.25 0 .5561 .06
0.3137
0.504 0K.
299
Figure for 12.D8.
300
12.D9.
Repeat 12.D2 with Kremser.
y
y1*
mx
m x0 n
Eq. (12-22)
m
b
169.2, b
0, x 0
mV
0,
169.2
L 0 , y1
x in
0.0012 0 0.000036 0
1 0.846
N
2000
0.846 , L V
200 y out 0.000036, y N
1
200 10 0.0012
0.846 10.69
n 1 0.846
Graphical solution was 10.4. Pretty close! 12.D10.
Use Kremser equation such as y A out y A* 1 L mV out
y A in
N
y A*
out
4, m 1.414, L V
L mV
.65, y Ain
L mV Equation becomes:
1
y Aout
0, y A*
12.D11.
V L
y out
V L
m x Ain
out
1.414 .02
.02828
.65 1.414 .459 .02828
Overall mass balance: yin V L x in
x out
N 1
y in
x in
.02828 .552
.01267
y out V L x out 1 .65
.01267
0 .02
4.93 10
4
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
L mV becomes, .27 N 1 y N 1 y1* L 1 mV Set up table and try values of m. y1
y1*
1
1 1
1.2 m 1.2 m
5
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.
301
L
X in
.796 1000 .204
796 kg solvent/h ,
.256, X out
.796 x .05 .10 .15 .20 .25
0.025
L G
796 All stages
25,190
0.02564, Y1,in
0.975
0.0316
.0012 1 .0012
.001201
Equilibrium, y = 0.04 x X 0.0526 .11111 .1765 .25 .3333
y .002 .004 .006 .008 .01
Y .002004 .00402 0.00604 .00806 .0101
Plot weight ratios.
Yj
L
Xj
Y0
L
Xj 1 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 Op. Line:
302
12.D13. New Problem 3rd Edition. Strip Vinyl chloride from water at 25ºC and 850 mmHg. H 1243.84x y x 1147.904x p tot 850 760 Want 0.1 ppm water leaving. Entering air is pure, L 1 kmole hr.
y
x0
* out
x IN
y out
y1
y IN
0
1 y
N 1147.904
1
x out
yN
5.0 xN
x out
y1* G b and c. Want
y*out
1147.904 x IN 5739.52 0
x0
5739.52 ppm (mole)
1171.33
5.0 0.1 G 2 G MIN
mix
L G
G MIN
0.00085373
L F
x *N n
0
585.665 1147.904
1
y out
Lx IN
For
L 1 kmol h , G
x IN
h
0.00170746 kmol h air
5.0 0 0.1 0
585.665 1147.904
n 1147.904 585.665 d.
kmol
585.665 (See figure for part b – labeled HW5 Prob 1b). m = 1147.904.
c. Eq. 12-28
N
x IN
1 ppm
L
y out
yN
x out
Gy IN
y IN
Lx out
n 25.0 0.672944
4.78327
G 0.00170746 kmol h
585.665 5.0 0.1
0
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.
303
304
12.D14. a)
95% removal CH4, 5% remains – Constant V: Yout b.
yin
CH 4 Eq.
0.00129
L V
y CH 4
min
x*
y CH 4 y out
0.05 0.00129
0.0000645
CH 4
p CH 4
3600 x CH 4
p TOT
175
y CH 4 in
0.00129
20.5714
20.5714
20.5714 x CH 4
0.00006271
0.000645
CH 4
x CH 4
x in Slope
0.00129 0.0000645
L V
min
L V actual c) Ext. bal.
x out
CH 4
x CH 4 ,out
V
V L
0.00006271 0 1.4 L V min 27.360; L
y in
x CH 4 ,in
V
L 100
CH 4
19.5429
27.360 V
y CH 4 ,out
y CH 4 ,in y CH 4 ,out 0.00129 0.0000645 L 2736 d) Use methane values in Kremser eqn. (12-22) to find N 20.57 100 mV m 3600 / 175 20.57, 0.75183; y1* L 2736 0.00129 0 0.0000645 0
n 1 0.75183 N
2736.0
0.00004479
mx 0
b
0
0.75183 6.11 stages
n 1 0.75183
e) Now use Argon values with N = 6.11 to find y Ar,out & x Ar,out .
m Ar
7700 175
Eq. (12-23)
x Ar ,out
44.00,
y Ar ,in
mV
44 100
L
2736
0.00024
y Ar ,1
0.00024 0
V L
yN
y Ar,out
y Ar,1
y Ar ,in
y Ar ,out
1,Ar
0.00024, x Ar ,in
1.6082, 1 1.6082 1
y1*
mx 0
1 1.6082 1 1.6082
7.11
x Ar ,0
b
Ar
2736
0
7.11
0.60846
0.00024 0.00024 0.60846 100
0.0
0.00024 0.00009397
0.00009397 0.00000534
305
0.00000534 2736
% Argon recovery in liquid 12D15.
100 0.00024
Need equilibrium data. From DePriester chart: K C3 y C3 x C3 1.23 m C3 , K C4
L
Butane is a design problem:
100
y C4 x C4
N
y N 1 y1* y1 y1*
mV L
mV L
n
Propane is simulation: y N
m C4
0.17 C4
m C4 x 0,C4
.0006 0 .0000072 0
.83
L n mV
0.34
mV
5.882,
mV C 4 L 1.2% of the butane leaves as a gas. Thus, * y1,C4 0.006 0.012 0.0000072, y1,C4 1
60.85%
.17 2.39
n 5.882 * 0.0017, y1,C3
1,C3
0,
L
1.626,
mV
C3
0
mV L
0.615 C3
N 1
yN y1
1
y1* y
yN
* 1
1
y1
L 1 mV , y1 1 L mV
12.D16. Was 12.D19 in 2nd edition. .
a.)
Equil.
y CO 2
H CO2 PTOT
yN
0.000298
1
5.7034
x CO 2 , H CO 2 25 C
1640
50 mmHg 50 760 0.06579 atm PTOT 1.0 x CO 2 y CO 2 .00035 Feed H CO 2 1640 Equilibrium in column: y CO 2
H CO 2 PTOT
x CO 2
1640 0.06579
x CO 2
atm mol frac 2.134 10
7
24982 x CO 2
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 CO 2 h in outlet gas. 5% CO2 remains in liquid
x in
x out
.05 2.1341 10
7
outlet liquid mole frac
0.106705 10 7 kmol CO 2 h 0.106705 10 1 kmol h
7
0.106705 10
7
b.)
306
Slope
y max out
24982
y
L V
Slope
yin
24982 2.1341 10
7
5.3314 10
3
y max
y in
5.3314 10
x in
x out
2.1341 0.10605
max
3
0 10
7
L Vmax
0
x out
x
x in
2.62968 10
V
2.62968 10
4
max
1
Vmin
Since L = 1,
L
7
2.1341 10
3.803 10 5 kmol h
4
c.) V 1.5 Vmin 1.5 3.803 105 5.704 10 5 kmol h Conditions for Kremser eq. are satisfied. CO2 Mass Bal: 2.1341 10 7 in 0.106705 10 5 out w. water 5.704 105 yCO2,out
y1
y CO 2 ,out
Eq. 12-29
2.027395 10 5.704 10 N
n
7
3.5543 10
5
x *N
xN
n
yN
x *N
x0
N
0,
1
y1*
y1
x *0
3.5543 10
m
24982
0.106705 10 2.1341 10
7
7
1.4227 10
24982 5.704 10
.0002 1.414
3
1.4227 10
7
0
1.0
n 12D.17.
x *0
L mV
m n
3
.0002828,
7
5.3569
5
L
L
mV 14.14 Can use variety of forms of Kremser equation, but cannot easily use forms with y*N 1 since
y*N
1
mx N and x N is unknown and hard to calculate. Try Eq. (12-21).
yN y1 Do by trial-and-error
L/mV RHS
1
y1*
.0083 .0002828
* 1
.0005 .0002828
y
2 15
3 40
2.9 36.699
L 1 mV 36.91 L 1 mV
N 1
2.91 37.02
307
Linearly interpolate L/mV = 2.907. Then, L = (2.907) (14.14) = 41.10 kmol/h. 12.D18. a.
L, x 0
y
p
11.5
p total
1520
0.00757 , x
y1
0 1
x1
0.0004,
L
y1
VT
x
L
yi
V
xN
Bot. of Column:
yN
1
y
1
y x
0.00757 0.0127
0.596
.00596 L
slope
VN
1
100
VT 150 100 , VF
Ext. MB: Lx 0 150 0.0004
xN
VN
K
.01, y
x0
VF y F 0.003
N
0.0127,
23
50, L 100
VF y F 50 .003
VN 1 y N
1
Lx N
100 .0058
100
VT y1
0.0067
0.0058
L V
x
yN
L 1
V
xN,
Slope
L VN
100 1
100
1.0
308
c. Minimum L.
xN yN
1
Eq.
y
L min
yF
VN
y1 L min
L min
VT
150
x0
L min 100 slope
1
slope x
Pinch is at y F . x F
y F 0.596
L min VT L min
0.003 0.596
Slope min
0.51653 VT
0.00503
0.003 0.0004 0.51653 150
0.00503 0
0.51653
77.48 kmol h .
309
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
.000024
12.D20. a) 99% removal H 2 O , 1% left,
1000 .01
0.00024 moles out in L
0.00024
x H 2Sout
0.00000024 1000 Moles H 2S out in gas = (.000024) (1000) (.99) = 0.02376 0.02376
y H 2Sout b)
H H 2S
Equil. H 2S. y H 2S
p tot
V 26800
x H 2S
15.5
H CO2
y CO2
p tot
0.02376
x H 2S
1729.03 x H 2S , m H 2S
728
x CO 2
0.00691
3.44
15.5
x CO 2
1729.03
46.9677 x CO 2 , m CO 2
46.9677
Can use Kremser eq. for H 2S design [dilute linear system]. For example, Eq. (12-28)
n N
1000
mV
x*N
1729.03 3.44
H 2S
yN
1
n
m
0, x 0
* N * N CO 2
xN
x
x0
x
x N ,CO 2
mV L
1
mV L
1
1
x0 1
0.000024, x N
H 2S
CO 2 N 1
yN
1,CO2
, with N
mCO2
H 2S
0.00000024
H 2S
0.16813 2.4807
0, x 0
2.4807 and
5.9479
L
n 5.9479
unknown, x*N
For CO 2 . x N
mV
0.16813,
.000024 .00000024
1 0.16813
N c)
H 2S
L mV
mV L
n
L
x *N x *N
x0 xN
L mV
1
0.000038 Kremser (12-31) 46.9677 3.44
mV L
CO 2
1000
0.16157
CO 2
mV L N mV L
1
0.000038
1 0.16157 1
.16157
3.4807
0.000031916 CO 2
Little amount of CO 2
310
23.78
12.D21. New Problem 3rd Edition. Abs. y Stripper
Abs
y out
23.78
y
y IN y
Slope
0.2
5 x
x
4.756x
118.90x
0.00098
4.756
x EQ
L MIN G
L MIN
x IN
735.302 Abs
100
0.00001
x
L act
m
y IN
y1* mx IN Eq. (12-22).
N
0.00098, y1
4.756 .00001 y N 1 y1* y1 y1*
mV L n
N abs
459.56 kmol h
735.302 kmol h
V y abs,IN
y abs,out
100 0.00098 0.000079 735.302
Kremser Eq.
1
100 4.5956
4.5956
0.00001
L abs
x abs,out
n
0.000079
0.00020605
1.6 L abs,MIN
L abs x abs,IN
735.302 0.00001
x abs,out
1
0.00098
0.00020605
7.353 x abs,out
External M.B.
yN
0.00098 4.756
Operating line Slope
0.000079
L V
slope equilibrium
x strip,IN
4.756, b
y out
0.00013253 0,
L mV
0.000079
7.353 4.756
1.5460,
mV L
0.64681
0.00004756 mV L
n
L mV
.35319
0.00098 0.00004766 0.000079 .00004756
0.64681
n 1.5460
n 11.1216
2.40889
n 1.5460
0.4357
5.53
311
Stripper 118.90 = Slope Equilibrium
y EQ
y
L
x out
0.015758
= Slope Operating line
VMIN
y IN 0 118.90x N
118.90 0.00013253
0.00001
x IN
x out,abs
0.00013253
x0
118.90
L
0.015758 0
V
VMIN ,Strip
VStrip y Strip out
128.6037
0.00013253 0.00001
MIN Strip
735.302
L Strip 128.6037
1.5 5.718
VStrip y Strip IN
5.718 kmol h.
128.6037 8.576 kmol h.
L Strip x Strip IN
x Strip out
0
735.302 0.00013253 0.00001
VStrip
yStrip,IN
yStrip,out
8.576
x *N
0.010506,
0
yN
1
118.9
Kremser (12-28)
n
1
N
x *N x *N
x0 xN
L mV
L mV
n mV L n
0.00013253 0 0.00001 0
0.27889
N
,
L
735.302
mV
118.90 8.576
0.721106 4.54
n 1.38676
12.D22. New Problem 3rd Edition. K
0.22. y
L 75, V 150, L V External M.B.: Lx N xN
Vy1 V L
Lx 0 yN
1
Vy N y1
0.721106
Kx
0.22x. Plot e.g., at x
0.006, y
0.00132
0.5
1
2 0.003 0.0004
0.0052
Points x 0 , y1
0, 0.0004 , and x N , y N 1 0.0052, 0.003 are on op. line. Plot Op. Line. See graph (labeled 12.D.b). 2 stages more than sufficient.
312
313
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. Yi n .15 .85 0.1765, Yout 0.005 .995
G
1400 .85
Equilibrium: y
0.00502, X in
1190 mol carrier gas/day, L
800 .995
0.005 .995
0.00502
796 mol solvent/day
.5x . Convert to mole ratios
.05 .1 .15 .2
x 0
X 0 .0526 .1111 .1765 .25
Plot ratios on McCabe-Thiele diagram L Absorber: Op. Line: Y X Yout G External balance Absorber: 796 .00502
y 0 .025 .05 .075 .1 L
X in ,
Y 0 .0256 .0526 .0811 .1111 L
0.669 G G 1190 .1765 1190 00502
796 X out
X out Abs 0.2614 Step off stages as shown in figure. Need 8 equilibrium stages. Stripper:
Yin
.00502 same as Yout abs , Xin
0.2614 same as Xout abs
X out 0.00502 same as X in abs , L 796 mol solvent day, have 4 stages Stripper equilibrium: y = 3 x. Convert to mole ratios.
x 0 .025 .05 .075 .1 .15
X 0 .0256 .0526 .0811 .111 .1765
y 0 .075 .15 .225 .3 .45
Y 0 .0811 .1765 .290 .4286 .818
Problem is trial-and-error. Select Yout . Draw operating line from Yout , X in to
Yin , X out .
See if need 4 stages. When need exactly 4 stages, L/G = slope. From Figure, final result shown: G strip
L slope
796 1.0998
723.7 mol carrier gas/day.
Yout strip ~ 0.287 mol ratio
314
315
12.E1.
Convert ppm(wt) to mole fraction: ppm 10 6 wt frac Basis 1000 g. of steam: Feed Liquid 1000 ppm = 0.001 wt frac mole frac
Liquid Water
lg Nitrobenzene
0.00812 gmole
999 g water =
55.4507
F L F C where C
28.1 10
L mV
S Vout
99 61.8
18.0314
37.2g water , C
2.0648 mol
95.7219 2.0648 97.7867 mol h
Outlet: Basis 1000 g 28.1 ppm 28.1 10
S
0.9998535
Total 55.4588 mol 1000g Avg mol. wt 55.4588 1726 g h 95.7219 mol h
L
V
0.0001465
6
wt. frac.
moles
3
Mol frac. 4
g nitrobenzene 2.2825 10 999.9997 18.016 55.5062 1000 28.1 10 3 g water 55.5064 mol 99 18.016 5.495 mol h .
4.11 10
6
~ 1.0
Equil. y mx b, b 0, m H p tot 28.0 Kremser Eq. – Several forms can be used. → Use Eq. 12-28. 97.7867 mV yN 1 b 0 0.63555, 1.5734 , x *N m 28.0 5.495 L
n
1
N
Effic
x0 xN
L mV
x *N x *N
L mV
mV n L
N Eq
5.763
5.763
0.524 N act 11 Ref. Hwang et al, IEC Research, 31 (7) 1992, 1753 & 1759. 12-F1.
H = 59.3 (Perry’s 4th ed., p 14-4).
y
H
59.3
x 11.86x PTOT 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
316
x in
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. 0, X in 0; yin .02, Yin .02041; y out .002; Yout .002003; x out .001, X out See Figure for plot of operating line, equilibrium and stages. N = 3.3 Height 5 ft HETP 1.515 N 3.3 equil stage Yin Yout L .02041 .002003 18.4 G X out X in .001001 0 If use mole fractions find L/V = 18.0
12.F2.
K E 26.0, K p 0.6, K After one pass of mass balance obtain:
.001001
0.019
317
xi,1
ethane 0.032
pentane 0.005
octane 0.963
xi,2 xi,3
0.035 0.033
0.031 0.162
0.934 0.805
yi,1
0.975
0.003
0.021
yi,2 yi,3
0.962 0.885
0.019 0.099
0.019 0.016
For new temperature used multi-variant Newtonian convergence. T1,New 73.90 F, T2,New 80.93 F, T3,New 2.G1.
a) N
4, L
570, y A,out
0.00317
b) N
8, L
500, y A,out
0.00315 while with L
c) N 16, L 490, y A,out d) Have a pinch point.
0.002978 while with L
490, y A 480, y A
99.15 F.
0.00358 0.00337
12.G2. Used Peng-Robinson. a. Total number of stages required 8 b. Feed stage location for the solvent 1 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.
318
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 E N 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:
T, p, z A , z B , F, x Ao , x B , y A N , y BN plus:
y B1 , R, E, N F x B N , R, E, N F x A N , R, E, N F R, E, y B1 , x A N N, N F , y A1 , E N, N F , x B N , R N, N F , x A N , 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 countercurrent 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)
319
13.C.7. Start by defining ∆ and the coordinates of ∆ as: E o R1 , x A E o y Ao R1x A1 , x D E o y Do nd rd Removing ∆ from 2 and 3 equations we obtain
Assume that E o
Rj
xA
E o y Ao
R 1 x A1
Eo
R1
(13-43a)
xD
E o y Do
R 1 x D1
Eo
R1
(13-43b)
R 1 . Next write the three independent mass balances around stages 1 to j.
Ej ,
1
R1x D1
xA
R j 1x A j 1
E j y Aj ,
xD
R j 1x D j 1
E j y Dj
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 R j 1 E j in each of these equations, and finally set the results equal to each other. The development proceeds as follows: Use
Ej
R j 1 to remove ∆ from the other mass balances.
Ej
Rj
1
xA
Solve for R j
1
R j 1x A j 1 Ej ,
Rj Ej
1
E j yA j , E j yAj
xA
xAj 1
xA
,
Rj Rj Ej
1
1
xD
R j 1x Dj 1
yDj
xD
xDj 1
xD
Finally, set these equations equal to each other. yA j x A yDj x D yA j x A rearrange to: xAj 1 xA xDj 1 xD xDj xD
yAj 1
xA
x Dj 1
xD
E j y Dj
This last equation says that the slope of the line between the points
xD , xA
y D j , y A j and
is equal to the slope of the line between the points x D j 1 , x A j 1 and x D , x A
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
320
R
y
x
E 44, R
is on op. line. E
R
1
2.273 0.0037 . Thus, cannot get to x o
Op line intersect equilibrium at x b. Now E
and point x N , y N
x N yN 1 E 100 , Slope R E
0.001, 0.0 0.012 .
74 and R E 1.35 .
44 30
44 0
0.001, but y N
30 0.004
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.
Still want x N
1
c. This alternative (Say use 25 kg/min of 0.004 butanol) 44 0 25 .004 yN 1 0.00145 which is below equilibrium point. 69 Now R E 100 69 1.449 m equil 1.613 Thus, this will work. Obtain
0.00145 69
0.012 100
0.001 100
0.0174 69 Op line closer to equilibrium – require lot more stages. 20 0.004 If use 20 kg/min of 0.004 butanol: y N 1 0.00125 64 0.00125 64 0.011 100 R E 100 64 1.5625 , y1 0.01844 64 Will also work. Becoming close to pinch at top equil y1* 0.019356 y1
If 15 kg/min, y N
15 .004 1
0.0010, R E
59 1.0010 59
y1
0.011 100
100 59 1.6949
0.1964
59
y1*
m eq
0.19356
Won’t work. Thus, there is a small range where option c will work, but with many stages. 13.D2.
R
20, E
20, x IN
Kremser equation
Eq (13-11)
x F , y IN
0, m
R
20
mE
8.333 20
y1 yN
y1 8.3333 x F
y 1
1
* 1
y1*
1
8.3333 x F
8.3333, b
0.12 , y1*
R mE N 1 R mE 1-0.12 1- 0.12
3
0, N
mx 0
b
2 8.3333 x F , y N+1
7.3460 x F , → y1
0
0.9873 x F
321
Rx F
Mass balance,
Recovery = 1 x N x F
Rx N
Ey1 , x N
0.01269 x F 20 0.9873 , which is higher than 0.963 obtained in cross-flow.
13.D3. New Problem in 3rd edition. From M.B.
R .013 Where R = 100 and the unknowns are E and yout.
y out
and Equilibrium:
S
1.613 x out E
E .001
1.613 .007
R .013 .007 y out
20 x F 19.746 x F
R .007
0.01129
100 0.006
0.001
E y out
0.01129 0.001
58.309 kg h
Alternative Solution: 1 Equilibrium Stage
y 1.613x
1.613 y
y out
0.01129 R
from graph
E
E
.001
x out
.007
x
R
0.001 0.01129
E
0.013 0.007 R 1.715
1.715
58.309
.013
Another alternative solution:
322
E, y1
x
y
x0
y2
y1
1
Or
Op. Line:
Eq.
R, x 0
R E
Slope
y1
Points x1 , y 2 ,
R E
x0
x 0 , y1
But x1 and y1 unknown
x where y 2
E
x1
On Op. Line
R, x1
E, y 2
R
y2
y1,in
y1,IN
x1
T & E in this configuration
R Slope known, N = 1 If Eq. line is straight, can Use Kremser with N=1. E 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
R
550 lb h, E
mE R
700 lb h, x 0
1.0538 and R mE * 1
y
b n
0.828 .0097 mE 1 R
yN
R 1
E
R
x0
E
xN n
N
.0003
550
0
1
0.0003,
.00046
0.0075
0.00803
y N 1 y1* y1 y1* n
y1
0.00046, y N
.94893
mx IN
Kremser Eq. (13-11b), N
0.0097, x N
R mE
0.0097
700 .0077316 .0538 .0004716
mE R
550 700
1.5038 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, mE 0.828 560
x0
0.005, x N
0.0003, y N+1
0.0001,
0.862664 mE
1
R
.862664
1.159
323
R
Since
1.0 , can use equation such as 13-11b mE y1* m x 0 b 0.828 0.005 0.00414
n
y N 1 y1* y1 y1*
mE R
1
N
n
n 1 1.1592 N where
y1
yN
R
x0
1
E 1.52637
N
mE R
R mE
0.0001 0.00414 0.0034571 0.00414 xN
1.1592
n0.862664 400 0.0001 0.005 0.0003 560
0.003457
10.332 0.14773 Alternate solution: Eq. (12-28) becomes L
R, V
yN
x *N
1
E , N
b
0.0001
m
b)
n
1
x *N x *N
x0 xN
R mE
0.000120773 , N
0.828
yN
R E
0.0001
1
xN
y1 Part b.
R
x0
E 100
yN
R 1
0.005
E
140 Kremser Eq.
slope x0
0.0003
13.D6. New problem in 3rd edition. Part a. Ey N 1 Rx 0 Ey1 Rx N
y1
min
E
0.005
10.332
.14773
0.828 x 0 y1*
R
Slope = 0.828
n mE R
1.52637
y1*
Equil. y
R mE
min
yN
x0
E MIN
xN
1
0.828 0.005
0.00414
0.00414 0.0001 0.005 0.0003
R 0.85957
400 0.85957
0.85957
465.3 kg h
Ext. M.B.
xN
0.0002
100 140
0.0005
0.003414
324
Convert
x
L
R
x
L
R
y
y
yN
E, x *N
V
1
m
100
0.0002 1.208
0.00016556
0.5913 mV mE 1.208 140 Lots of different forms can be used. n N
For example
n
1
N
0.4087
N Part c.
x *N x *N
L mV
n mV L n
Becomes
x0 x0
L 1 mV
x0 xN
R mE
x *N x *N
R mE
n mE R
0.005 0.00016556 0.0005 0.00016556 1 n 0.5913
0.5913
1.8717 0.5254 Eq.
y EQ
3.6
y 1.208x
1.208x
0.00604
y
R E MIN
y N 1 0.0002 y 1.208x xN R
0.0005
x
slope
x0
0.005
0.00604 0.0002
1.29777 0.005 0.0005 R 100 E MIN 77.05 1.29777 1.29777 Maximum extract out y EQ x 0 0.00604. Part d. The roles of extract and diluents are switched in the two problems, which changes the definitions of y and x. E MIN
13.D7. Equilibrium:
N
30, R
500, y N
y
0.828x, m
1
0.0002, x 0
0.828, x *N
yN
1
0.0111, x N .828
0.00037
0.00024155
325
Since rather dilute and linear equilibrium use one of the Kremser equations.
n N
Where
E 500
x *N x *N
R ME
(12-28 (modified))
mE n R
x0
x *N
xN
x *N
R/mE 1.21
83.756 . Solution is trial-and-error.
Calculated N Negative-Not possible Need
0.8626 .929 .9435 .945015
700 650 640 639
xo xN
R mE
1
R mE
1
17.01 E too high 26.175 E too high 29.84 E too high 30.30 E too low
By linear interpolation need E ~ 639.6 kg/h. Can use other forms of the Kremser equation. Was 13.D10 in 2nd edition. x is raffinate R L Convert Kremser y y extract, V E 13.D8.
a)
Use 12-31
xN
x *N
Other forms OK x 0
* N
x *N
xN
1
x0 1
b)
x
yn
KE R KE R
m
1 mV L mV L
1
K
1
K, b
1 KE R
N 1
1
R E
x0
3
30.488 25 100 30.488 25
1
R x0 xN
0.00001376
3
100
Can use External balance or Kremser to find y out
y1
KE R
1
1
Ey N
0
0
0.00092
3
Ey1
13.D9.
mx
100 25
y1
xN 0.00092 0.00001379
0.003625
Assume very dilute, R = 1500 kg/h, E = 750 kg/h Equil. Y K d X becomes y K d x From Table 13-3. K d,oleic 99% recovery oleic:
m oleic
4.14, K d,linoleic
.99 .0025 1500
md,linoleic
2.17
y1,oleic 750 → y 1,oleic
.00495
326
Use Kremser, Eq. (13-11b).
y1*
m oleic E
4.14 750
R
1500 4.14 .0025
m oleic x 0,oleic n
N
For linoleic acid:
yN
1
0, N
y N 1 y1* y1 y1*
mE R
1
1500
m lin E
750 2.17
Can use Eq. (13-11a):
y1 .00651 Recovery of linoleic:
m L x 0,L
yN
y1*
1
mE R
5.44
.9216 ,
2.17 003 R mE N R 1 mE .07834
.00651
1
y1*
y1
0.01035
R n mE
R
5.44, y1*
2.07
.00651
.40866 Re c .003 1500
1
.00124796 → y1linoleic
0.00526
.00526 750 → Rec = 0.877
th
13.D10. New problem in 4 edition. Analytical or graphical solution OK. Stage 1 F1x F1 E i Equilibrium
x 2,out
1.02 x1
Fx F1
1.02 E R1 x1
Mix with Feed 2
1.02 E
R1
1.02 50
R 1 x1
E2
y 2,in
E 2 y 2,out
y 2,out
1.02 x 2,out
0.0099338 100 R2
R 3,in
R 1 x1
100 0.015
F xF
R 1 x1 1.02 E 2
Ey1,out
y1,out
x1
Stage 2
y1,in
151
100
0.0099338
R 2 x 2,out , R1
0.006579, R 2
R2
R1
F1
F1
100
100
100 70 170
327
x 3,in x 3,in
Stage 3
x 3,out
x 4,out y 4,out
0.006579 100
0.005 70
170
0.0059286
E3
y3,in
E 3 y3,out
R 3 x 3,out
1.02 x 3,out
R 3 x 3,in 1.02 E 3
R 4 x 4,out y 4,out
x F2 F2
R 3,in
R 3 x 3,in y3,out
Stage 4
x 2,out R 2
E4
170 0.0059286 R3
y 4,in
1.02 50
E 4 y 4,out
170
0.00456
R 4 x 4,out , R 4
R3
170
1.02 x 4,out
R 4 x 4,out 1.02 E 4 1.02 x 4out
170 0.00456
R4
51 170
0.003508
0.003578
328
329
13.D11.
R
R
F
2501, E
Equilibrium: K D
E 1000
1.57 . For dilute this becomes m
xN R
Abietic Acid Recovery:
xN
.0475
.0475
R
2501
Top op. Eq.:
y
Goes through pt x 0 Bottom Op. Eq.: y
0.0000190 , y1
R E
x
R
x
yN
R 1
y1
E
x3
1
0.17594 Rx in
.0475
.05 F x F
.05 1.0 .05
E
1000
R
2500
E
1000
0.0000025
y 4 0, y*1
1
1.2399
0.00742
Ey in
2.5
x N through point x N , y N
1 1.2399
0 0.00742
y1 Overall bal.
0.00742
.95 1.0 0.5
x0
E
m 1.613, R mE 1.2399 , y N Eq. (13-11a)
13.D13. a.
R
y1
0, y1 . Slope
E Need 8 ½ stages (see Figure).
13.D12.
.95 F x F
K D in wt. frac. units.
Ey out
mxin
1
0 , R E
0.00742
0.17594
4
0.00742 10 0.0046
R
0.006114 5 0.006114 10
0.1 1000 0.003
xy . 90% recovery, 10% left
x
,out
2.501
0.00154
0.3 kg out
0.0003
330
O xy
For ortho, y max
95% recovery, 5% left
0.15 0.005
0.25 , x O,out
0.00025
0.00075 R E
For para
0.05 1000 0.005
0.00075 0 max,ortho
y max
0.08 0.003
0.00024
R
0.00024 0
E
0.003 0.0003
max,para
b. The p-xylene recovery controls.
E 1.5 11250 16875 ,
0.1579
0.005 0.00025
1000
E min
0.08888
R
0.08888 11, 250
0.0592592 E Can use Kremser eq. (13-11b) for ρ-xy to find N
n N
mE 1 R
y N 1 y1* y1 y1*
mE R
R mE m 0.080, R E 0.0592592, y N 1 0 , y1* mx o,p 0.080 0.003 0.00024 Mass balance: 90% entering ρ-xy leaves w. solvent. 0.9 1000 0.003 y1 0.00016 wt frac 16,875 R 0.0592592 R mE .080 1.35 0.74074, n 0.300106 , R 0.0592592 mE 0.080 mE n
331
n
0 0.00024 0.00016 0.00024
.35
N
1.35
n 0.30
4.012 0.300106 0.30016 Note: Can use other forms of Kremser eq if desired. c. For o-xy check if recovery > 95% R 1 * y1 unknown, y N 1 0 y1 y1 mE Eq. (13-11a) N 1 y1* mx 0 0.15 0.005 0.00075 y N 1 y1* R 1 mE R 0.0592592 0.39506, N 4.012 mE 0.15 y1
yN
1
1
y1* 1
R mE N R mE
External M.B.
y1*
1
Ey1
R xN
Rx 0
xN
0.39506
5.012
0.00075
0.0029194
R x0
5
Ey1 R
a)
1
Ey N 1
% Recovery 13.D14. (was 14.D4. in 2nd ed.)
1 0.39506
0.00075
Ey1 Rx 0
16875 .00029194 1000 100
S
10.0
2
MF
F
15.0
3
SM
7.3584 E 5
98.53%
Once have M, use trial-and-error to find tie through M. (final result is shown). This gives E and R. y A .115, yw 0.04, xA .23, xw .73. b) Plot raffinate, R x A
.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 kg/h. F 15.0 SM
332
13.D15.
Since dilute, use Kremser equations. Assume units are weight fractions. a) Column 1 at 40ºC. x N 0.0008, N 11,, x 0 0.01, E 1000,, R 100 Equilibrium: m
0.1022, thus y1*
mx 0
0.001022. Kremser (Eq. 13-11a):
1 1.022 0.93664 12 y N 1 0.001022 1 1 1.022 This simplifies to: y1 .093664y N 1 .00092628 y1
1
0.001022
External MB: y N 1E Rx o which simplifies to:
yN
y1E Rx N , y N 1
1
1000 Solve 2 eqs and 2 unknowns: y1,coll b) Column 2 at 25ºC: y N
y1,col2
yN
1,col1
1,col2
y1,col1
.6929 10 5 , x 0
1
1000
1 1000 y1
.08
1000 y1 .92
0.00092693, y N+1,coll
0.6929 10
5
0.00092693 ,
0, N
9, m
0.0328, E 1000, y1*
mx 0
0
Use Kremser to solve for R´. This is trial and error. For example, Using Eq. (13-11a), R R 1 1 * 0.0328 1000 y1 y1 mE 0.007475 N 1 10 y N 1 y1* 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 y N 1E R x 0 y1E .92693 0 .006929 xN R 50.35
0.0183
333
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.
R E 10 8 1.25, R mE
13.D16. a.)
* 1
y y1
m x A0
1.613 0.01
0.01613
0.0002 0.01613 xA
1.25 1.613 0.77495 0.01613. Use Eq. (13-11a),
1 0.77495 1 x A0
0.77495 E
yN
E 1
R R b.) Graphical check works fine (not shown)
yj
13.D17.
x6 Note:
x6
R Ej
xj
y IN
0.27044 → y1
7
R E0
y1
x j 1,
7.02498 E
R
10
Ej
2
0.01182 4
5
0.0018 (See graph)
x N,countercurrent
0.000702 even though use more total solvent.
334
13.D18. (was 14.D2. in 2nd ed.) Lever arm rule:
Plot S, F, R and E. Draw lines SF and RE. Intersection is point M.
S
MF
20.3
F
SM
4.5
Or Mass Bal. S + R = M and S y A Solve simultaneously
4.511 → S
F xA
100 4.511
M x A ( S .15
451.1 kg/h
.5 F .21 M )
S = 483.3
335
Difference is due to accuracy in reading numbers. Lever-Arm Rule more accurate!
13.D19.
Equil.
Kd
Acetone
y0 xN
FD
1000 .9
yA x A 0 1
0, x 1
0.10 wt frac
900 kg/h water, FS FD FS
Equil.
Y0
0.287 0.158 1.816
900 1364.1
0.005 X N+1
X1 0.10
.9 1371 .995
0.005 0.995
0.00503
0.1111
1364.1 kg/h chloroform.
0.6598
336
XA
xA 0
yA = 1.86 xA
0
0
YA 0
0.01 0.03
0.0101 0.0309
0.01816 0.05448
0.01850 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.
FD FS
XN
1
Y6
FD FS
X1
YN or YN
0.6598 0.1111
0.06999, y N Results pretty close to 13.D43. 2
1 2
vs 2
2 3
0.6598 .00503
YN 1 YN
0.0655
w i accuracy of graphs.
Note: The graph below should read acetone, not acetic acid as the solute.
337
13.D20. a) Batch Operation – Mix together & settle. Find fraction recovered: R R Operating Eq.: y x x 0 , R 5, S 4, x 0 x F S S Which is, y 1.25 x 1.25 x F Equilibrium
y
8.333 x, m
Eq. (13-21) written for batch
8.333
x
Rˆ Sˆ x 0 m
Frac. Rec 1 0.1304 0.8696 b) Continuous solvent addition: Sˆ 1 n x t ,final x t ,feed Eq. (13-28) Rˆ t m
x t,final x F Recovery = 99.87%.
exp
0.8 8.33
y iN
1.25 x F
Rˆ Sˆ
0.8
0
9.583
1 8.333
n
0.1304 x F
x t ,final xF
0.00127
338
13.D21 (Was 14.D1 in 2nd ed.) a. Let A = methylcyclohexane and D = n-heptane. Mass Balances: F1 F2 S M or M 350
F1 x AF
1
Then
F2 x AF
S y AS
2
x AM
1
F1 x A F
F2 x A F
F1 x D F
M F2 x D F
1
x DM
M x AM , F1 x DF
1
2
2
F2 x DF
S y DS
2
M x DM
S y AS
100 .6
50 .2
0
S y DS
350 100 4 50 .8
0
M
350
0.2 0.229
Plot M. Find tie line through M. (See figure.) This gives location of points E and R. Find x DR 0.48, x AR 0.42, y AE 0.06, y DE 0.05 .
b.
Mass balances: M
E R and Mx AM
Ey AE
Rx AR
Solving simultaneously: E = 214 and R = 136 kg/h 13.D.22. New problem in 3rd edition. 1 Af D s2 4 0.411 and Pperf 2
With interface at center, heavy phase flow area is 1 D s D 5 2.630 2
r
θ
Chord
.1
.4115
Ds 2
0.5115
Center
Interface
α
(length = C) arc
θ
r
.1 α C/2
r
C
2
2
.1
C 2
1.00326 m 339
2
Draw right triangle for interface below center to calculate new perimeter. 0.1 .1 sin .1955 11.274 r .5115 Then angle of arc, 180 2 157.452 3.14159 0.5115 157.452 r Length of arc 1.4056 180 180 Perf C arc length 2.4089m Mensuration formulas are from CRC Standard Mathematical Table. Re settler
4Q
c
Perf
4 0.006 998
c
2.4089 0.95 10
3
10, 466
Interference somewhat more likely than in Example 13-5. 13.D23 (was 14.D7. in 2nd ed.) Pyrdine F x AF Plot M on line FS . y p 0.223,
a) F S 500 300 M S y AO 500 .3 0 M x AM → x AM
150 800
0.1875
By T & E find tie line through M (Use Conjugate line) x p 0.84 ; y w 0.02, x w 0.84 ;
Mass balances: R1
E1
800 , 0.84R
M
0.02E
0.43M
Solve simultaneously, E1 ~ 400, R1 ~ 400 (Note: More accurate than pyrdine values.)
R 1 S2
b)
R1x A1
S2 y A0 60
x pyr M 2
700
Find tie line by T & E: y pyr2 MB:
R 2xw2
E2 yw 2 R2
Solve simultaneously: E 2
400 300
700
400 0.15
0.053 ; y w 2
M x m2w → 0.945 R 2 M
60
M 2 x AM 2
0.086
0.120; x pyr2
E2
M2
0.005, x w 2
0.005 E 2
0.945
700 0.48
700
346 and R 2
354
340
341
13.D24 (was 14.D10. in 2nd ed.) a) Feed 40% MCH 55% n-heptane, F = 200. Solvent 95% aniline & 5% n-heptane, Stotal 600 . S F M 800
S
Lever arm rule:
3
F
FM 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: y MCH ~ 0.045, Raffinate:x MCH ~ 0.36 wt fracs Mass balance E + R = 800 = M and lever arm rule Solve simultaneously: b)
gives points R1 and E1.
Find:
F S R1
MR
R
ME
. Measure distances on figure.
R = 124.61 kg/h, E = 800 – R = 675.39
2 stage cross flow. Stage 1: F = 200, ρ = 300,
Mass balance 500
E
M
R1
S
3
FM
F
2
MS
. Find point M. Tie line through M
E1 and lever arm rule
207.04 kg h , E1
R1
E 1M 1
E1
M 1R 1
292.95
Note: Isotherms are very sensitive. Thus, calculation is not extremely accurate. Stage 2: Mass balance R 1 S2
M2
507.04
R2
E 2 and lever arm
S2
M 2R1
E1
M 2S 2
Find M 2 and from tie line through M 2 find R 2 . Then can find R2 and E2 from mass balance (given above) and new application of lever arm rule, Solving simultaneously, R 2
R2
E 2M
E2
R 2M
196.16 kg h. E 2
310.88
342
13.D25 (was 14.D9. in 2nd ed.) a. Draw lines from S to F and from R 1 to E N . Intersection gives point M (see Figure). Then from lever-arm,
b.
S
FM
F
SM
1.25 → S
∆ is at intersection of lines E N R N
1
1.25 2000
2500
and E 0 R1 . Then step off stages as shown. Need 2 stages.
343
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 Raffinate Flow Rate:
R1 E 0 EN
Extract Flow Rate:
SM SF 15 57
F S R1
13.D27 (was 14.D12. in 2nd ed.)
y AE
E0
MF
F
E0M
Lever arm rule:
1.112
y wE0
0
R1
M
R1
R 1 5600, R 1
2000 5600 772
6830
0 (Pure solvent)
. Step off stages
211.2 kg/h & lever arm:
Solve simultaneously, R1
770 kg/h.
. Find M. Line RM intersects sat’d extract at E N , y A N
Lines F E N & R1E 0 intersect at M.B. E N
E0
2000 S 2000 → S = 5600 kg/h
64.25, E N
0.18
3 more than enough. Need ~ 2 ¼
EN
MR 1
R1
ENM
2.287 (from graph).
146.95 kg/h
344
345
13.D28 (was 14.D14. in 2nd ed.)
To find ∆: 1) Plot E N and R N
2)
Ej
Rj
EN
1
E N x AN
xA
RN
F
1
1500
1
R N 1x A N 1
0.06666
3) ∆ is on line through points E N and R N 1 . Plot ∆. Or, use lever-rule.
RN
1
R N 1E N EN
1.5
Step off three stages starting at point E N . This gives points R 1 x A1
Mass Balance: E 0
0.275, x D1
RN
E 0 0.13
and
R1
1
E N → E0
1000 0.4
Solving simultaneously, R 1 13.D29 (was 14.D16. in 2nd ed.)
0.675 and E 0 y A1
R1
EN
R 1 0.275
655 kg/h, E 0
.13, and y D 0
RN
1
0.0 .
R 1 1500
2500 0.2
2155 kg/h
a) Plot Points F, S, E N and R 1
Find ∆ point at intersection of lines FE N and R 1S 2 stages is more than enough. (see graph) b)
Draw lines FS and E N NOT calc. value E 2 R1 . Intersection is mixing point M
F
dist. S to M
S F 0.786 1000 0.786 1272 kg/h.
dist F to M
Mass balance Give S
F + S = M and Lever arm
0.786
346
Alternate: Overall MB, F S M and Diluent mass balance, 650 F x F,D S yS,D M x M,D 0.28 M
M
Solve simultaneously:
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. 50 13.D.31. New problem in 3rd edition. Soln. A. Kremser Soln. R mE 0.30998 1.0 161.3 R 50, E 100, m 1.613, b 0, y 2 0.0, x 0 0.01 For example, Use 13-11.
yN
y1 x1 Soln. b.)
1
y1*
y1
* 1
y
1
1
R mE N R mE
y1*
1
mx 0 becomes
0.01613 0.01613 .7633696
y1 m
b
0.01613
y1 0.01613 0 0.01613
1 0.30998 1
0.30998
2
0.7633696
0.00381684
0.00381684
0.0023663 1.613 Do mass balances and equilibrium for single stage.
347
Sy IN Fx F 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. Q sol Q feed Q feed Q tol At feed conditions tol Q sol Q tol Q feed Q feed Q feed Q feed
0.6 .006
Q sol Q feed Q sol .006 .6 .006 1 Q feed Equation 13-48 operation in ambivalent range. a)
.6
tol
.3 L
1 0.3
From Example 13.5
L
L
H
0.375
H
L
0.625
1.6
Q sol Q feed 1 Q sol Q feed .375
865 0.95 10 998. 0.59 10
3
0.3
3
1.10235
0.375
The
1.10235 0.6614 0.625 Either phase can be dispersed. 1.0 b) 0.5 , also ambivalent range d 2.0 .5 Either phase dispersed 1.10235 1.10235 .5 2.0 c) .6667. According to 13-48 at border. d 3.0 .6667 water probably dispersed 1.10235 2.2 .3333 5.0 d) .8333 Equation (13-48), water (heavy) dispersed. d 6.0 0.8333 1.10235 5.5 water dispersed. 0.16667 13.D.33. New problem in 3rd edition. t re s Vliq Qd Qc 1.5 min 90s
Qd
Qc
0.0072 m3 s , Vliq
90 s 0.0072 m3 s
Note that there is a 1 inch air gap at top Vliq H t 0.0254 d 2tan k 4 0.648 , H t
Vliq
2d tan k
0.0254
Using Goal Seek d tan k
d 2tan k 4
0.648m3
2d tan k
0.648
0.7489 and H tan k
1.4978
348
13.D.34. New problem in 3rd edition. N = 500 rpm = 8.335 rps d i 0.20 d tan k 0.2 0.8279 0.16558 m
Use water values for Re L ,estimate
M
d i2 N
998 kg m3 and
w
2
0.16558
L
8.335 998
Curve b in Figure 13-32 again predicts a constant N p0 Then from Equation (13-52), P
P
4.0
M
2
8.335
N P0
.16558
2
M
5
40
d 5i g c where g c
1.0
0.95 10 3 kg m s
w
240, 064
3
0.95 10
L
M
0.034587
N
1.0 and
8.335. (A)
M
will be fairly close to c 998 since Q W 5QToluene (see Equation (13-53)). W 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, d 0.0576 P 0.3 (B) In addition to Equations A and B, we need to solve Equation (13-53) (C) 1 d c 865 d 998 1 d M d d M
Solving equations A, B and C with Goal Seek we obtain Then solving Equation C,
M
d
1
d
d
0.146 and
d
0.146 865
c
d
d,feed
0.874.
0.854 998 978.6
Equation (B) P 0.3 2.876 P 33.84 W. d 0.05076 nd 13.D35 (was 14.D11. in 2 edition) From Eq. (12-46), E1 K 1 E2K 2 B1 1 , C1 , D1 R 0 x 0 R1 R2 (Eq. 6-6) For 1 < j < N A j
1, B j
(Eq. 12-48) For Stage N A N Example 13-4: R 0
1000, x A0
For Acetic Acid, K A j
1, B N
0.35, x D,0
E jK j
1
Rj
ENKN
1
E j 1K j
, Cj
RN
Rj
, DN
6 , EN
0.65, N
1
FN z N 1
, Dj
Fjz j
0
1
E N 1y N
1475, yA,N
1
1
0, yD,N+1
0
y Aj x A j : Use Fig. 14-4 to estimate K A, j .
K A1 K A4
0.03 0.1 0.12
0.5
0.3, K A 2
0.15 0.14
0.33, K A 3
0.09 0.21 0.16
0.43
0.5, K A 5 0.5, K A 6 0.5, 0.24 0.28 0.32 For first guess assume constant E 1475 and R 1000. Then C1 D1
B1
1
E 1K A1 R1
E 2K 2
1475
R2
1000
R D x A ,0
1
0.33
1000 0.35
1475 0.3 1000
1.4425
0.48675 350
349
and so forth with D6 1
DN
475 0
2
0 . Thus matrix for acetic acid is,
3
-0.48675
0
1475
1475
4
0
0
0
0
0
0
0
1.4425
2
-1
3
0
4
0
0
-1
5
0
0
0
-1
6
0
0
0
0
1000
.33
.43
1000 1
-1
1475 1000
1475
.43
0.5
1000
1
6
0
1
1
5
1475 1000
1475
0.5
1000 1475
1
1000
0
0.5
1475
0.5
1000
1
-1
1475 1000
0.5
0.5
13.D.36. Part a. New problem in 3rd edition. See figure
Forg
C Aq ,0
FAq
* org ,1
Min
Forg,Min
Forg
b.
C
0.736 FAq
1.4 147.2
Operating line goes through CAq,N
Corg,1 See Figure.
C Aq ,N C
* org ,N 1
0.10 0.008 0.133 0.008
0.736 200 L h
206.08 ,
147.2
Forg
206.08
FAq
200
0.008 and Corg,N
1
0.736 L h
1.0304
0.008 with slope 1.0304.
0.097
3 stages more than enough.
~2
3 4
stages needed.
350
Part c.
MW Zr NO3
4
91.22 4 14.0067 3 15.994
MW water
2 1.00797
15.994
18.00994
351
Basis 1 liter 0.10 mol Zr NO3
4
Have
33.917g Zr NO3
and
1000 g
33.917
4
966.083 g water
966.083 18.00994 53.64 mol water .1 Mole frac. Zr NO 3 4 0.00186 53.64 .1 33.9179 Mass frac. Zr NO 3 4 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. which is
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 (n 1 = 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. New problem in 3rd edition.
13.D.38.
MWwater
18.02,
F 1.0 kmol hr ,
MWtoluene
S
92.14 , m
0.06 kmol hr.
C toluene
C raffinate
C water
0.00023 ,
x IN
20.8
y IN
0
→ x out Fx in / F Sm Note m m. m is equilibrium in mole fraction units. Assume extract has properties toluene and raffinate properties of water. F x IN
m
Fx out
Sy out and
C extract
kmole benzoic m 3 extract 20.8 kmole benzoic m 3 raffinate
Units on m are
y out
m x out
1 865 kg tol m 3
92.14 kg toluene kmol toluene
1 998 kgW m 3
18.02 kgW kmol W
122.71
kmol benzoic kmol extract kmol benzoic kmol raffinate
352
1.0 0.00023
x out
1.0
0.0000275 , y out
0.06 122.71
If use m
20.8 find x out
13.D.39. Feed is 0.1 1 equil. stage
1 .00023
122.71 0.0000275 1 .06 20.8
0.00337
0.000102, WRONG!
New problem in 3rd edition. CC 4 , 0.9 AA. F 10 kmol h . Solvent pure. S 10 kmol h.
Lever arm:
S
10
F
10
1
FM SM
x F,CC
, Alternatively
x M ,CC
4
x M ,CC
4
x S,CC
4
4
S F
1
Then x M,CC 4 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: x CC 4 0.041, x AA 0.54 . Extract: yCC 4 0.095, y AA 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.
S1
S 2
CCℓ4 1
2
E1
R1 = R2 single stage = 16.6667 Mix with S2 = 10 (pure)
E2
x M 2 ,acetic
16.6667
R1
SM 2
10
S
R 1M 2
x M2,AA
R2
x R 1 ,acetic
x M 2 ,AA
x Sacetic x M 2 acetic
0.54
0
x M 2 ,AA
.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, yCC 4 0.046 y AA 0.065 Raffinate 2,
R2 CC
4
E2
x CC
4
x AA
R 1 S 16.6667 10
balance
Substitution
R2
0.018
19.40 and E 2
0.018R 2 0.018 R 2
0.57
26.6667
0.0046E 2
0.041 16.6667
0.046 26.6667 R 2
0.0 10
0.68333
7.16 kmol h .
353
354
13.D41. New problem in 3rd edition. R N 1 F 10, x CC 4,N 1 0.1, x AA,N
E0
S 14.5,
1
1.0 , y CC
y TEA,0
Mixing. Use lever arm rule.
1.45
x M ,CC
14.5
S
FM
10
F
SM
xN
1,CC
4
0.9 0.091
4,N
xN
1,CC
x M ,CC
S y S,CC F SF 1
4
4
x M ,CC
4
y S,CC
.1
4
4
4
1.45 0
0.041
2.45
Find M. Draw E N MR 1 line. See figure on next page. Raffinate:
x1,CC
4
0.008
x1,AA
.58
Passing Streams E N R N 1 & E O R 1 intersect at . Very close to parallel. Use parallel lines to step off stages. Estimate # Stages = 3. Flow rates 24.5 F S E 3 R 1
CC
4
balance. F .1 E3
S 0
1.0
E 3 .091
R1 0.008
1.0 24.5 0.008
9.69 kmol h , R 1 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 y TEA 0.841 x CC 4 0.0913 y AA 0.067 Raffinate Mole fraction
x TEA Two results are reasonably close.
.418
x CC
4
0.0056
x AA
.577
355
356
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 S FM 0.81 F 1235kg / 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 F, S
Use lever-arm rule to find point M.
E 0 , and R 1 (on saturated raffinate line)
E0M
F
1000
S 1371 FM Line R 1M intersects the saturated extract curve at E N . x acetone
0.067 .
Lines FE N and R 1E 0 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.
357
358
K Dm
13.E1. Since
E
K Do
y m,N Estimate: E
E
0.05, K Do R 1
20, E
200, F 1
1 ortho goes up column and since K D M
yo,N
1
0, x m,0
E .52F and R
200.52 and R
Recoveries:
0.15, R
20.48,
.92 .52 1 x ortho,N
x o,0
0
R .48F R 20
E 200.52 E y ortho,1 or y ortho,1
0.09974 and
E R
1 meta goes down.
R
20.48
E
200
0.1024
0.002386
0.00203
.94 .48 1 Rx meta,N or x meta,N .02179 Plot equilibrium curves and operating lines (see Figure) Feed cannot be 3rd stage since cannot get x m N 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.
359
13.E.2. New problem in 3rd edition. Part a.
x N,p Part b.
xy
0.04 .004
Paraxylene:
96% recovery. 4% p-xy left in diluent
0.00016 wt. frac. y
Ka
0.080
m, E
20, 000, R
x Eq. (12-28) converted to extraction notation is convenient. L
n
x0 xN
R mE
1
N
n
x0 n N Part c.
x *N x *N
mE R
0.00016,
0.004 0.00016
ortho-xy m
.625
xN
xN
x *N
x0
* N
x
R
yN
x *N
V
R
1000
mE
0.080 20, 000
2.3025
m
E
1
0.
Thus
0.006, x *N
0,
0.625
4.899
0.470
0.150, x 0
1000
R mE
n 1 .625
Eq. (12-31) Converted:
Part d. m-xy
0.004, x N
.375
,
mE
0.150 20000
R
1000
3
1 mE R 1
mE R
N 1
1 3
0.006
1.842 E - 5 1 35.899 Alternative Solutions are presented below for meta-xylene. m 0.050, x 0 0.005, x*N 0, N 4.899 E 20,000,
mE R
R
1000, b
0
0.05 20, 000
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). N x x0 x N Solving for N,
N
x0
xN x
Where Δx is determined in same way Δy was determined for Eq. (12-12), L y1 x0 b y1 V x x j x j 1 const x0 L V L V Alternatively,
x
yN
1
L xN V L V
b
yN
1
L V
xN
360
Translating to this extraction problem,
x0
N And solving for xN, x N
xN
L V x0
x x0 0.005 N 1 5.899
R E , yN
1
0,
x
xN
xN xN 0.0008476
Alternative Solution: Redefine terms to match Eq. 12-12 [Relating y to solvent and x to raffinate is arbitrary. Switch these definitions.] y N 1 meta xylene in hexane 0.005
y1 m
1
1
Kd
0.05
mxy out is unknown
20, b yN
1
y1
0, L N y1
E
x 0 is now inlet solvent
20, 000; L V
x0
b
L
20, 000
mV
20 1000
4.899 y1
20 0
0
1 , V 1000 0
361
Solve for y1,
yN
y1
0.005
1
5.899
5.899
0.0008476
This is actually x N in normal notation. Part e. Shown for normal notation.
pxy equil slope = 0.080
y EQ
0.080 .004
0.00032
y
yN
1
0 x 0,pxy
x
x N ,pxy
0.004
0.00016 0.00032 0
Slope Operating line
R
Slope
E MIN
0.004 0.00016 R 0.08333 , E MIN 0.08333
0.08333 1000 0.08333
12, 000 kg h
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 x AR and extract y AE Mass Balances:
0.326
x DR
0.575
0.046
y DE
0.058
E R S E 40 R 40 E Ey AE Rx A,S Sy A,s Fx AF 20 0
Solve simultaneously, E
18.0 kg , R
20 .4
8
22.0 kg.
Part b. First add solvent until reach saturated raffinate curve at intersection with FS line. Initial Raffinate x AR 0.36, x D 0.54
R INIT x AR R init
Fx AF SINIT x AS
8 x AR
8 0.36
20 .4
SINIT 0
8
22.22 kg
SINIT R INIT F 2.22 kg Second, use Eq. (13-27) for the continuous solvent addition batch extraction.
362
x t ,final ,A
S R
x t ,feed ,A
dx t ,A yA
x t,feed,A is the raffinate after solvent addition to form two phases x t,feed,A
0.36 , x t ,final,A
x A,initial raffinate
0.292
From equilibrium find values y A (extract), Approximate values are:
x A,t
yA 0.048 0.046 0.045
0.36 0.326 0.292 0.292
1y 20.8 21.7 22.22
dx A
0.36 0.292
yA
6
0.36
Sadded
1.47R t
20.8 4 21.7
1.47
Eq. A
In the derivation R t is assumed constant, R t
Sadded
22.22
R t,INIT
22.22 kg
32.66 kg
With this approximation E Sadded . 32.66 kg Solute mass balance R t x A,INIT Sadded y A,added R t x A,final
y A,added
y A ,Avg
0, x A,INIT
Ey A,Avg
0.36, x A,final
22.22 0.36 0.292 32.66
0.292
0.046
If we do not assume R is constant, then Eq. (13-27) is x t ,A
Sadded
d R x t ,A
dS added 0
yA
x t ,INITIAL ,raf
We would need to do a numerical integration with a calculation of R x t,A versus y A . this can be done, but is challenging. 13.G.1. New problem in 3rd edition. Extract 1: flow 3.90769, xTRA Raffinate 1: flow Extract 1: Raffinate 2:
0.84986, xcarbontet
16.03923, x TEA
flow flow
0.085102, xAcetic acid
0.41361, x carbontet
11.63396, x TEA
0.91426, x carbontet
14.40527, x TEA
0.065042
0.041332, x Acetic Acid 0.036586, x Acetic Acid
0.54506 0.049149
0.41633, x carbontet =0.016472, x Acetic Acid =0.56719 .
Entering carbon tet 0.10 10 1.0 kmoles hr Leaving in raffinate
0.016472 14.40527
In Out in Raffinate Extracted % extracted = 76.27%
0.23728
0.7627
363
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 Raffinate 3 15.086 6.277 .2758
.4672 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
364
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 where F, U, S, O, R, E are kg Eq. (13-27b) becomes U U y x x F y IN O O U and (13-21) x x IN y IN 1 U O and y = x O Part b. Eq. (13-29b) becomes O 1 n x t ,final x t ,feed U K Where K y x at equilibrium = 1.0 in washing.
n 14.D1. (was 13D29 in 2nd ed.)
a) Translate eq. (12-28),
U mO
1
N
O, R
U, E
x *N x *N
x0 xN
O
U mO
n mO U
Note: x in wt frac. translates to x in kg m 3 if densities are constant. Densities cancel. For washing equilibrium is equal overflow & underflow concentrations. Thus, m = 1, b = 0 yN 1 b H 2SO 4 x *N y N 1 0, x 0 1.0, x N 0.09 m U 40 mO 1 0.8 and 1.25 mO 1.0 50 U 0.8
n
1.0 0 0.09 0
1 0.8
N
0.8 4.96
n 1 0.8
b) HCℓ Use Eq. (12-31) or (14-8)
xN
x0 N
1 1
xN
x *N
x0
* N
mO U mO U
4.96,
N 1
x
1 1
0.75
mO
1.0 50
U
40
mO U mO U
N 1
1 1.25 1 1.255.96 1.25, x *N HC
0.0674 kg m 3 yN
1HC
m
b
0
Alternative: 363
xN
Note:
x0
xN x0
HC
=0.09 H 2 SO 4
Thus, if one is clever and realizes change will be same for HCℓ & H 2SO 4
since
mO U
& N are identical , 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. U U Operating Eq. y j xj 1 y in x out O O U .4 Slope 0.8. Goes through point (y = 0, x = 0.002) O .5 Overall bal. O yin U x in U x out O y out
O
.4 0.035
.4 .002
.5 yout → y out
.0140 .0008
Need 6 2/3 Stages – See Graph (Can also use Kremser eq.)
b. Mass Balance: Op. Eq.:
U xj yj
O j y jin U Oj
xj
U xj y jin
0.0264
O jy j U
Oj
xj
1
U
2 slope , x out 0.002 (see graph) O Obtain approximately same separation, but use much more wash water. (was 13D23 in 2nd ed.) U
14.D3.
1
.5
0.4, O
0.2,
364
y
y
4
y
3
2
y 1
2
1
U=3
x
4
3
U=3
4
O y
Basis:
O 2 3 y 0 in 3
2
4
in 4
0
O y
2
in 2
2 0
O 2 1 y 0 in 1
1 kg CaCO 3 solids
Feed Mole frac. can be arbitrary. Pick x 0
U x iN
M.B.
O yiN U
y out
O y out
U
x iN y iN O O y out , x out at Equil (y = x) line
x in , yin
Point Slope Op line See graph. Find Recovery
x out
U x out
0.01 as basis
1
x in , 0 is on op line
U
3
O
2
x4
0.00127
x0
0.01
0.127
x4
1 0.127 0.873 x0 Recovery is significantly better with counter-current process.
365
14.D4.
(was 13D24 in 2nd ed.)
0.8 0.8 0.2
0.8,
1
4, O
4000 kg/h
366
U F1
1000
kg h
dry solids
U F2 UT In section 2: Slope
U F1
UT
yj
O
Intermediate feed at x
Slope
xj
L liquid L solid
kg
.8
h
.2
U F2 1
1.0
3200
1600
U F1 O
xF xj
1
kg liquid h
kg liquid
2.5 kg liquid 4800 h UT y0 x1 O
h
4800 4000 1.2 Goes through point y0 , x1
UT O
In Section 1: y j
2000
4
kg liquid L liquid kg solid 2.5 L solid
1.0
0, 0.006
0.02 U F1
yN
U F1 O 1600 4000
O
xN
1
y N , x N+1
0.4 . Goes through point
Also intersects Section 2 op. line at feed line. (Or calculate y N from mass balance). Equilibrium is y = x. Step off stages (see Figure). Need 5.4 equilibrium stages. Opt. Feed is 4 th.
14.D5.
(was 13D25 in 2nd ed.)
F1: 1000
kg dry
0.8, 1
0.2
1 1 kg dry L solid h 2.25 0.2 L L under flow kg dry 1 1 F2 : 2000 4000 h 2.5 0.2
2000 L h
L h
F1 5 wt %
F2 2 wt %
367
2000L
Liquid Volumes:
total 0.8 liq
h
3200 L liq
4000 0.8 liq underflow
U0
ON
1
4000 kg h , y N x0
144 4800 Ext. MB,
U
0,
1
f
4000
1.2,
O
4800
ON 1 y N y1
U0 x 0
1
h
1600
U0x0
U
O
O
U
xN
O
mx 0
n 1 0.8333 N
5 wt %
where
kg liq h
1.0 kg L
f
0.05
kg N a 0H kg liq
3200 0.02
0.006
UNx N
Convert to Kremser O V, U L, m 1, y1* Eq. (12-30)
kg liquid
U 0x 0
0.030 , x N specified
h
2 wt %
h 4800
FT : Total liqd h
L liq
1600
O1 y1 x0
x0
xN
4800 4000
0.030 0.006
0.030, mV L
0 0.030 0.0288 0.030 n 1.2
4000 4800
0.0288
0.8333
0.83333 8.83 or 9 stages
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. 1 kg dry solids Part a. 10 kg total 4 kg dry insoluble solids 2.5 kg total
1.5 kg liquid
6 kg liquid. , Ov 10 kg liquid. kg dry solids Before 1st mixing: 0.05 frac BaS 6 kg liquid 0.3 kg BaS 0.3 kg BaS 0.01875 mass frac in U & Ov. 1st Mix: 16 kg liquid total U
4 kg dry solids
Settle – (6 kg liquid in U) 2nd Mix Pure Water
0.01875
0.1125 kg BaS
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 0.0421875kg 0.00264 mass frac BaS in U and Ov. 3rd Mix Pure Water 16 kg liquid Part b. Result is same. Can also be done graphically. Part c. Countercurrent. Easiest solution approach is to use Kremser equation. x N x *N 1 m Ov U N 1 * x 0 x N 1 m Ov U
368
External M.B.
N
3, x 0
xN
x0
x0U
y N 1Ov
y1 14.D7.
0.05, m 1, Ov
1
30 6
1
30 6
U x0
(was 13D27 in 2nd ed.) Operating Eq.:
xNU
xN
U
Ov
Basis 1000 cc wet sand. .4
O
xj
1
m
0
0.0003205
6 0.05 0.0003205 30
xj
y j,in
U O
vol water
xj
y j wt. fractions 1
.4
O = 0.2 kg. Thus, each operating line has slope
y j,in , x j
x0
0.035, y N,in
yN
xN
0.002, yS,in
y N , y 2,in
1,in
0.009936
1000 cm 3 wet sand
vol sand wet
Each op line goes through pt.
yN
y1Ov
Equilibrium:
yj
U
0.05 0.00641
4
6, x *N
30, U
.2
1.0 g
kg
cc
1000 g
0.4 kg
2.
1
yN
2,in
0
y N 1 , y1,in
yN
2
Start at stage N where x N = 0.002. Find y N 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.
369
14.D8.
(was 13D28 in 2nd ed.) Use Kremser equation Fsolv Fsolid .95, y mx is equilibrium with m = 1.18, and N = 11. Recovery is 1 x N
x0 .
Eq. (12-31) becomes x *N
m Fsolv
yN * N * N
1
xN
x
xN
x0
x
x0
0 .
m
m Fsolv Fsolid
1 1
xN
1.18 .95 1.121. Then
Fsolid
N 1
m Fsolv Fsolid
1 1.121
x0
1
12
1.121
0.041
Thus Recovery = 0.959 14.D.9. New problem in 3rd Edition. Assume FSolid and Fsolvent are constant despite removal of sugar from solid.
FSolid
FSolid xF Fsolv
Eq. (13-21) becomes x a.
Fsolv
1.0, x F
3.0,
0.055, y solv,IN
y solv ,IN
mE
FSolid Fsolv
13
0.055 1 1.18 0.01211 , y 3 3 b. x = 0.004. Solve for Fsolv .
xm E
Fsolv Fsolv
FSolid
mE x
L Slurry stream
0.055 0.004
1.0
ySolv,IN
14.D10. (was 13D30 in 2nd ed.)
1.18 .004
G H 2 stream
120 lb h
x 1 x
Yin
G Yin X out
y
Y
lb CH 4 lb H 2
Yout 120
10.805 kg
CONSTANT
CONSTANT
,
100
0
100 lb h of H 2
Operating Line. Must work in weight ratios. Y
x
0.0143 wt frac.
x
x
mE x
1.18 0.01211
y solv,IN
xF xF
1.18,
FSolid Fsolv
x
FSolid
0, m E
1 y in
in
,
30 100
X in
L G
X x in
1 x in
0.30, Yout
L G
X in
Yout
0 lb CH 4 lb H 2
out out
.05 .95
.0527
L X out
0.30 0.0527
0.206 , x out
X
.206
1 X
1.206
0.171 370
Operating line becomes, L Y X Yout where G Equilibrium Curve:
L
120
G
100
y = 1.2 x becomes
x 0 .05 .10 .15 .20 .25
Y Y 1
1.2
X X 1
→ Y
1.2 X 1 .2 X
Plot Y vs X X Y 0 0 .0526 .0038 .1111 .1364 .1765 .2195 .2500 .3158 .3333 .4286
See Figure for Plot. Need 5 1/8 stages. 14.D.11. New problem in 3rd Edition. 10,000 kg h wet solids, 1 frac vol dry solids. Basis 1 m 3 wet solids : Weight liquid + weight solids
1.20 and goes through X in , Yout .
1.0
frac. vol. liquid ,
1000 kg m3 1.0 1
1500 kg m 3
0.4
400 kg Thus
400 1300
900 kg 1300 kg total m 3 wet solids.
of weight is underflow liquid, U
14.D12. (was 13D32 in 2nd ed.)
Fsolv Fsolid
400 1300
10, 000
kg h
3076.9 .
1.36
371
Op. Eq.:
Fsolid
y
Fsolv
x
Fsolid
y1
Fsolv
Where y and x are kg m 3 . y
xF
x0
m E x is equilibrium.
x 0 , x N 1 .975 x F .025 x 0 , y N 1 0, x *N 0, N 5, Fsolv Fsolid Can use any of Kremser equations such as Eq. (12-31). m Fsolv 1 * xN xN Fsolid 1 1.36 m 0.025 N 1 6 * x0 xN 1 1.36 m m Fsolv 1Fsolid
1.36
Which becomes: 0.1582 m 6 1.36 m 0.975 0 Find m = 1.313 1 1.313 1.36 .025 0.025005 which is OK. Check: 6 1 1.313 1.36 14.D13. (was 13D33 in 2nd ed.) a)
x
Use Eq. (13-21),
Rˆ Sˆ 10 12.5
Rˆ Sˆ x 0
0.8, m E
m
1.18
y in
Rˆ Sˆ
g L in liqd
, x
g L in solid Frac. Rec. 1 0.404040 0.5959596 x t ,final Sˆ 1 n b) Eq. (13-29b) mE x t ,feed Rˆ
x t,final
x F exp
1.25 1.18
, equil. y
m E x, y in
0.8 x F
0
1.18 0.8
0.8 1.98
0
xF
0.4040 x F
0.228779 x F , Frac Rec = 1 – 0.228779 = 0.7712
14.D14. (was 13D34 in 2nd ed.)
BaSO 4 coal BaS 2 CO 2 Equil: Soln conc in underflow = soln conc in overflow. Thus really washing Equil : y x, m 1, b 0
U
350
O
2075
kg h
in sol.
kg h
, y in
Eq. (12-29) 14.D.15.
1.5
kg solution kg insoluble solid
0.0, x *N N
0, y1 n
xN
U0 x0
525 kg soln., x in xN
O x *N
x0
n L mV
525 2075
x *0
0.20, x out
0.2 0.00001
0.00001 0.0506, x *0
n 0.00001 .2 .0506 n 525 1.0 2075
New problem in 3rd Edition. With 1000 kg/h dry solids U 1.5 1000
0.0506
6.99 or 7.0
1500 kg h
a) Can use Kremser eq. with large N to find Ov Min or a sketch
372
y1*
Equilibrium is y
0
y1*
U
y yN
x
U 0
0
Ov
b. Ov 1.2 Ov Min
U
1500
Ov
1782
Kremser: Eq. (12-28)
x0
Min
x0
xN
Min
1 .99 .15
0.15
xN
0.15 0.0015
1500
Ov Min
.15
0
10101
1.0101
1485
0.0015
1782
0.84175
y
y
V
Ov
x
x
L
U
xN
0.0015, x *N n
0, x 0
.15, m 1, U Ov x0 xN
U m Ov
1
N
n
x *N x *N
0.84175
U m Ov
m U Ov
0.15 0 .84175 .0015 0 2.81 N 16.33 1 .17227 n .84175 In theory, can use McCabe-Thiele, but it is difficult to accurately step off this large number of stages. U 1500 c. Ov 2000, .75 m 1 Ov 2000 n 1 .75 100 .75 N 11.29 1 n .75 n 1 .84175
N act
15
E overall
N eq
11.29
N sub actual
15
0.753
For m E use N = 15 and change mE with same equation
n N
1
.75 mE n
100
.75 mE
mE .75 373
Vary mE until N = 15. m E .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. U 2 kg, O 2 kg, x IN 0.06, y IN 0 Solution (translation of Eq. (13-21)) is U x x IN y IN 1 U O 1 .06 0 2 .03 O Part b. Want x 0.005 O is unknown, x IN 0.06, y N 0, U 2 Solve for O
x
O
U
U
x
O
O
x IN
y IN ,
0.06 0.005
2
O
x
y IN
U
x IN
O
x
O
14.D.17. New problem in 3rd Edition. K = 1 Eq (13–28) becomes
O
x t ,final
x t ,feed exp
Part b.
U
2, U
x t,final
O 14.D18.
O U
2, x t,feed
O Part c.
2, x t,feed
0.06 e
x IN x
x y IN
22 kg water
0.005 0
Part a.
U
U
0.06 1
n x t ,final x t ,feed
0.02207
0.06, x t,final
0.005
U n x t ,final x t ,feed
2 n
0.005 0.06
4.97 kg
x in Part a.
O normal batch in Part b.
One equilibrium stage. F 1000, x A N+1
E 0 y A,0
.2, S
662, y AS
y DS
0
F x A,N+1
0.12 (same as Example 14-2) E0 F Plot M. By trial and error find tie line through M (Final result shown in Figure). y A1 .238, y D1 0; x A1 .078, x D1 .656 x A,M
Flow rates: Diluent balance:
R1x D1
R1
F x D,N+1 x D1
E1
M R1
F x D,N+1
1219.5
1662 1219.5
442.5
374
14.D19.
This problem is essentially a repeat of Example 14-2, except using exactly 3 stages. Clearly, x A1 0.04 since now have more stages. F, E 0 and M are unchanged. Problem is trial-and-error. Guess location of R 1 . Find E N and ∆. Step off 3 stages and see if have correct location of E N .
x A1
14.D20.
0.026 and y A3
The third and final trial is shown in the figure.
0.38.
Although this is leaching, this cross-flow problem is very similar to cross-flow extraction. We can derive R j 1 x A j 1 E j,in y A j,in x A Mj R j 1 E j,in
M Stage 1:
Rj
R0
1
E j,in where R j
1000, E1,in
M j x A Mj
421, x A0
xAj
yA j yA j
.2 y a1,in
0 , x AM1
200 1421 .1407
Find M on line SR 0 at x AM1 (see Figure). By trial-and-error find tie line through M. 375
This gives E1 and R 1. Find y A1
R1 Stage 2: x A M 2
y A2
.18, x A2 R2
Stage 3: x A M 3
R3
.35, x A1
1421 .1407 .35
0
0.085 1254.9 421 .058, from tie line , M 2
1675.9 0.085 .18 .058 .18 1305 .058
1421
1254.9
.113 .35
1254.9 .113
.113, M1
0
1726
1675.9
1305.0
0.044 , y A,3
1726 0.044 .09 .03 .09
14.D21. a. Basis 1 kg mix in underflow: x NaC values
.09, xA3
.03, M3
1726
1323.3 kg/h
0.8 1.0 crystals
0.2 yNaC
Since crystals are pure NaCℓ, NaOH is in liquid only. Since 20% of the underflow is liquid, x NaOH 0.2 y NaOH . Generate equilibrium table.
376
x NaOH
Soln (y) Mass frac NaOH
0 0.004 0.008 0.012 0.016 0.020 0.024 0.028 0.032 0.036
0 .02 .04 .06 .08 .10 .12 .14 .16 .18
y NaC
x NaC
.270 .253 .236 .219 .203 .187 .171 .156 .141 .126
.854 .8506 .8472 .8438 .8406 .8374 .8342 .8312 .8282 .8252
Feed is 45 wt% NaCℓ crystals. x values: NaCℓ (soln) = 0.5193, NaOH (soln) = 0.099, water 1-0.51930.099 = 0.3817. Since feed is 55% liquid, x F,NaOH 0.55 y NaOH 0.099 y NaOH 0.099 0.55 0.18 , y NaC 0.126 From the equilibrium data F = 100, S = 20, Plot F & S and find M. FM 20 , SM 100 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
R : Raffinate
0.833
E : Extract y NaC
x NaC , 0.026
0.16, y NaOH
x NaOH
0.135
The underflow is z wt frac crystals (Pure NaCℓ) + (1-z) wt frac solution y NaC 0.16 is soln in equil
z 1.0
Thus,
1 z 0.16 z
0.333 0.16
0.84 was 80% solids in problem statement. c.
Same M. Plot
R1
draw line
R1
0.833
M to
EN
EN
R1
80.1% OK
.
2 stages more than sufficient
120
1.137 R 1
EN
R 1M
103.5
R1
ENM
91.0
R1
120
1.137
377
R1
56.14
R1 : x1,NaC
E N : y NaC
E 63.86 kg/min, N kg/min 0.845, x1,NaOH 0.01 0.152 y NaOH
0.147
378
379
14.E1a.
This is difficult part – converting data Basis 1 lb oil-free solids
y oil
ysolvent 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.35 0.3 0.28
0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.70 0.72 Note:
ysolids
1.0
x solids
z 0.20 .242 .283 .339 0.405 0.489 0.600 0.672 0.765 0.810
1 z
0.830 0.80515 0.7794 0.74683 0.71174 0.67159 0.625 0.598086 0.56657 0.552486
y oil z
x oil
1 z 0 0.01948 0.044115 0.07595 0.1153 0.16420 0.2250 0.26124 0.303399 0.3222
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
F x oil,F
S yoil,S
1000
M1x oil,M , x oil,M1
0.252
1500
0.168
See Figure 14.E1a, b, c, d, e. Check: Lever Arm Extract E1 ,
y oil,1
Mass Balances:
M 1S
F
2
M 1F
S
1
0.34; Raffinate: x oil,1
0.092 and x solids,1
1500
0.34 E1 +0.922 R 1
R1
E1
Extract: MB:
R 1 , 252
1040.3,
Finish step c) Stage 2: R1 S2
x oil,M 2
. Find tie line through M1.
M2
E1
459.7 lb
1540.3
, R1 0.092
M2 xoil,M
0.062 . Plot M 2 and find tie line through M 2 .
yoil
2
1540.3
0.115; Raffinate: x oil,2 E2
R 2 , 95.7
0.025 and x solids,2
0.115 E 2
0.80 .
0.025 R 2
R 2 904.8 lb, E 2 635.5 lb d & e – Same answer as b & c but R & E are flowrates. f.
0.730
See Figure 14.E1f. 3 stages is more than enough. Need ~ 2
1 3
equil stages.
380
Lines E N R N
1
and E 0 R 1 intersect at .
381
382
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
x solids
y oil
ysolvent
z 0.20
1.0
x oil
1 z
0.830
y oil z 1 z
0
1.0
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:
S Rt
0
x t ,final
x c ,feed
dx t y
S = Mass Solvent, R t 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. x oil,M
0.21, x solids,M
0.63
Mass balance F + S = M Solids .748F + (0) (S) = 0.63M
M
0.748
F 1187.3 kg R initial 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. 383
x oil
y oil
0.21 0.1625 0.115 0.0675 0.02
1 y oil
.54 .498 .40 .28 0.1
Insoluble Solids M.B. Initial 0.748, F = 100, Final 0.81,
1.852 2.0080 2.50 3.57 10.0
R t final
.81 R t final 748 R tfinal 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. 0.21 0.115 1.852 4 2.008 2.50 6 1
0.115 0.02 6
2
Sadded
2.50 4 3.57
.1961
0.4241
10
0.6202
0.6202 R t , but what is R t ?
Eq. (13-29a) assumes R t
Const.
Use average value of R t .
R t ,avg or
Sadded Stotal
1
R t init R t ,final 1187.3 923.5 1055.4 2 0.6202 R t,avg 0.6202 1055.4 654.6
Initial addition + Sadded
Extract amt S
Stotal
Sremain in raffinate
Oil in extract
x F,0.1 F x final,oil R t,final
187.3 654.6
841.9 157.0
841.9
684.9 by solvent
0.252 1000
0.02 923.5
233.5
Total wt extract 684.9 233.5 918.4 yoil 233.5 918.4 0.254
ysolvent
0.746
14.E.3. New problem in 3rd Edition. Solid Matrix is insoluble. Solids = (.748) 1000 = 748 kg. R t not Constant, but Solid is. Solids Rt x Solids
ydS
d R xA
Solids d
xA x Solids
384
x final ,A x Solids
S Solids
d x A x Solids y
x A ,raf ,init x Solids ,raf ,init
Changes limits integration. x oil 0.21, x Solids
x oil
0.63, x oil x Solids
0.115, x Solids
0.21 0.63 .3333
0.705, x oil x Solids
0.115 .705
0.163
x oil 0.02, x Solids 0.81, x oil x Solids 0.02 0.81 0.0247 Numbers for use in Simpson’s rule are from Solution 14.E.2.
.3333 0.163 6
1
0.163 0.0247 6
2
Sadded Stotal
1.852 4 2.008
2.50
2.50 4 3.57
Solids total integral
10
0.3515
0.6173
Total 0.9688 748 0.9688 724.6 kg
initial added 187.3 724.6
911.9 kg
Extract Amount Solvent
Stotal Sraf ,final 911.9 157 Oil in extract = 0.252 (1000) – 0.02 (923.5) = 233.5 Total weight extract 754.9 233.5 988.4
754.9 kg
wt frac solvent = 0.764, wt frac oil = 0.236
385
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. (1510a) incorrectly predicts molecular diffusion. Equation (15-10b) predicts no molecular diffusion because dyw/dz = 0. B. Generation of Alternatives. 15.B1. For example, 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, 1 AB 2 (1 2 x1 x12 ) x1 x [ B ( A B ) x1 ]2 Then take the derivative and expand all terms. The denominator becomes [ A ( A B) x1 ]3 [ Bx2 Ax1 ]3 and the numerator simplifies to 2 A2 B 2 x2 . Multiply by x1. Q.E.D. * 15.C7. With CMO and y as mole fraction, vmol y Av A yB vB y Av A (1 y A )vB 0 . 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 y A, mass ) / MWB vA vB (Eq. C) y A, mass / MWA
* Then in mass terms vmass y A,mass vA yB ,mass vB which after substituting in Eq. C and simplifying * vmass (1 y A,mass ) ( MWA / MWB )(1 y A,mass ) vB . (Eq. D) * If MWA = MWB, vmass = 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 y A , mass MW A Cm N B (1 y A , mass ) (1 y A , mass ) MW B MW B * vmass (Eq. F) y A , mass / MW A (1 y A , mass ) / MW B * Since yA,mass varies throughout the distillation, vmass is different for each stage.
386
D. Problems. 15.D1. Dprop,water = 0.87E-9 m2/s. Eq. (15-9), J A , z D AB
dC A
( D AB / L )(C A , L C A ,0 ) . If C A,0 = 1.2 dz kg/m3 is the known value, C A, L can be larger or smaller than C A,0 . For smaller C A, L we have
C A, L =1.2 – (0.2E-5)(0.0001)/0.87E-9 = 0.9701 If it is larger value, then C A, L =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[ Eo / (TR )] = 1.52E-09 for T = 335.18K. Flux D (T ) D (298.16) exp[ Eo / (298.16 R )]
J A , z D AB
dC A dz
( D AB / L )(C A , L C A ,0 ) (1.52 10 9 / 0.0001)(0.9701 1.2) 0.35 10 5
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 D (T ) 1 1 E / R ln AB 1 / D AB (T2 ) T2 T1 The constant Do can be found from the known conditions at T 1 Do DAB (T1 ) / exp[ E / ( RT1 )] Or from the known conditions at T 2. 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 C sucrose = 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 o o o and DBA . Solve for the two unknowns. Results are DAB = 0.998×10-9 m2/s and DBA = DAB -9 2 1.308×10 m /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. 387
2 15.D9. Combining Eqs. (15-35b) (15-35d), vvertical ,max,liq 0.5 9 gq / Assume that the bulk is pure 1/3
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, W ,liq exp[52,843 3703.6 / T 5.866 ln T (5.879 1029 )T 10 ] 2.807 104 Pa s The viscosity of the vapor at 372.76K is (p. 2-426) W ,vapor (1.7096 108 )T 1.1146 1.2561105 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). 1/3
9(994.64)(9.81)(7.5 E 6) 2 vvertical ,max,liq 0.5 9 gq / 0.5 0.130 m / s 0.0002807 A check of the units show they work. The modified Reynolds number (using gas properties) is, 2
Re
d tube ( v gas vliq , y ,max )
1/3
(0.10)(0.5903)(0.81 0.130) 1.256 10 5
3195.6
/ The gas phase Schmidt number is Sc gas The viscosity and density were found earlier. D EW gas 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 Scgas are in the range for Eq. (15-47a), the
modified Sherwood number is, k p d tube ( pB )lm 0.0328(Re) 0.77 Scgas 0.33 0.0328(3195.6).77 (1.283).33 17.79 DAB ptot 15.D10. From the Chapman-Enskog theory DNH3-air = 2.05×10-5m2/s at 318.16K and p = 1.2 atm. 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 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.
388
Problem 15.D11, 3rd ed. MW A 28.9 MW B T 396.1642 p sigma A 3.711 sigma B eos A/kB 78.6 eps B/kB kT/EpsAB 1.891168 Col integ D AB 4.11E-05 D desired 4.11E-05 chkB7-B8 chk x E5 Problem MW A T sigma A eos A/kB kT/EpsAB D AB D desired
17 const 1.86E-07 0.9 T^3/2 7885.202 2.9 sigma AB 3.3055 558.3 eps AB/kB 209.4812 1.1069 Linear interpolation table 15-2 1.90E-10 1.90E-05 Goal seek to zero changing B3
15.D11, 3rd ed. 28.9 396.164199034186 3.711 78.6 =B3/F5 =F2*F3*SQRT(1/B2+1/D2)/(D3*F4*F4*D6) 0.00004114
MW B p sigma B eps B/kB Col integ
17 0.9 2.9 558.3 1.1069
const T^3/2 sigma AB eps AB/kB Linear interpolation in Table 15-2
0.0000001858 =B3*SQRT(B3) =0.5*(B4+D4) =SQRT(B5*D5)
chkB7-B8 =B7-B8 chk x E5 =100000*D8 Goal seek to zero changing B3
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 = 0.090241 m/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).
389
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. Sh avg = 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
DEW, m2/s
m ,kmol/m3
γE
DEW , m2/s XE = 0.25 25.0 36.28 0.633×10-9 1.9028 1.495×10-9 -9 XE = 0.35 27.8 31.62 0.625×10 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), 36.28(1.495 10 9 )[1.7083(0.3) 2.1582(0.2)] zN E 9.3445 10 9 1.9028(0.25) From xE = 0.3 to 0.4 (interval is over length δ- Δz) we obtain, 31.62(1.609 10 9 )[1.4338(0.4) 1.7083(0.3)] (0.00068 z ) N E 5.7027 10 9 1.5553(0.35) 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 XE = 0.225 XE = 0.275
DEW, m2/s
m ,kmol/m3
24.3 25.7
37.625 34.99
0.659×10-9 0.624×10-9 z1 N E 5.5371 10 9
γE 2.01976 1.79959
DEW , m2/s 1.482×10-9 1.532×10-9
Equation (15-61c) is now written 3 times: z 2 N E 3.9846 10 9
( z1 z 2 ) N E 5.7027 10 9 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 Dsucrose 0.5228 105 cm 2 / s . Equation becomes
CA C A,0
1 erf
z 4(0.5228 10 5 )t
Numerical values of C A / C A,0 are easily obtained with a spreadsheet or with the use of Table 17-7.
390
z, cm 0 0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 3.5 3.56 4.00 5.0
t = 1000 s 1 0.9221 0.6249 0.3281 0.0505 0.00335 9.16E-05 1.009E-06 7.61E-12
t = 10000 s 1 0.9753 0.8771 0.7571 0.5362 0.3535 0.2161 0.1220 0.6352 0.0304 0.0134 0.00538 0.00198 0.000206 1.49E-05 7.49E-07 6.2E-10
t = 100000 s 1 0.9922 0.9610 0.9221 0.8449 0.7692 0.6957 0.6249 0.5574 0.4936 0.4346 0.3788 0.3281 0.2406 0.1710 0.1177 0.0784 0.0505 0.00335 6.20E-04 4.99E-04 9.16E-05 1.01E-06
Part b. C 1.0 10 6 when C / C0 5.0 104 , 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. C 1.0 10 6 when C / C0 5.0 104 , 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 y A z
m y B z
y y y B C A D AB D AC D AC
yA y A NA D AB D AC
NB
y y y y y B B NA A C B NB DBA DBC DBA DBC DBC
Determine NB from the 1st equation and NA from the second.
391
m y A NB
z
m y B NA
z
y y y B C A D AB D AC D AC yA y A D AB D AC
NA
y y y A C B D BA D BC D BC yB y B D BA D BC
NB
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 N A,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) = -2.80E7 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.
392
HW 15-D22, A=air, B=H2, T
3rd Ed. SPE C = NH3 273.000000000 p
D AB 1 atm D AB
0.000061100 D AC 1 atm 0.000030550 D AC
NA guess
2.000000000
0.000019800 DBC 1 atm 0.000009900 DBC
0.000074800 0.000037400
-0.000059028
del z
0.010000000 ρ
yA z=0 yA z=δ yA avg ρΔyA/δz
0.520000000 0.520000000 0.520000000 0.000000000
NB NA calc chk NA-Nacalc 100000 chk NC
0.000140693 -0.000059028 0.000000000 0.000000000 -0.000081665
y B z=0 yB z=δ yB avg ρΔ yB/δz
0.089278949 0.480000000 y C z=0 0.000000000 yC z=δ 0.240000000 yC avg -4.285389534
0.000000000 0.480000000 0.240000000
kmol/s kmol/s Goal seek to zero change B10
HW 15-D22, A=air, B=H2, T
3rd Ed. SPE C = NH3 273
p
2
D AB 1 atm D AB
0.0000611 =B7/D3
D AC 1 atm D AC
0.0000198 =D7/D3
NA guess
-0.0000590279043468439
del z
0.01
ρ
=D3/(0.0820575*B3)
yA z=0 yA z=δ yA avg ρΔyA/δz
0.52 0.52 =(B14+B15)/2 =-D12*(B14-B15)/B12
y B z=0 yB z=δ yB avg ρΔ yB/δz
0.48 y C z=0 =B9*B11/D3 yC z=δ =(D14+D15)/2 yC avg =-D12*(D14-D15)/B12
NB =(B17+(D16/B8+F16/D8+B16/D8)*B10)/(B16/B8-B16/D8) NA calc =(D17+(B16/B8+F16/F8+D16/F8)*B19)/(D16/B8-D16/F8) chk NA-Nacalc =B20-B10 100000 chk =100000*B21 NC =-B19-B20
DBC 1 atm 0.0000748 DBC =F7/D3
=1-B14-D14 =1-B15-D15 =(F14+F15)/2
kmol/s kmol/s Goal seek to zero change B10
393
SPE 3rd Solution Manual Chapter 16 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 16.A1, 16.A7, 16.A8, 16.C2 16C4, 16.C5, 16.D3, 16.D9, 16.D16 to 16.D22, 16.G116.G3, 16.H1-16.H2. Chapter 16 was chapter 15 in the 2nd edition. Problems from that edition have the same problem number, but the chapter number is now 16 (e.g., problem 15.D6 is now 16.D6). x , Operating y = x, 16.C3. (was problem 15C3 in the 2nd edition.) Equilibrium y 1 1 x y out
n OG y in
dy y
*
y
. Substitute in for y* and let x = y (total reflux operating line)
dy y 1 y
n OG 1 y out
1
n OG
1
1
y y 1
y in
1
1
y out 1
n
y2 y2 1 y
y y 1
y
dy
1
n OG
dy
y out
y out
1 dy y 1
y in
y in
1
y in
1 n
y out
1
y in
1
which becomes Eq. (16-81).
16.C4.
New problem in 3rd edition. With extract dispersed, E MD
Since
y IN
y IN
y M,OUT
y IN
y*M,OUT y M,OUT
0, E MD
y
* M,OUT
Mixer mass balance with y IN Solving for x M,OUT : Then,
y*M ,OUT
E M ,D
0, Fx IN F x IN
x M ,OUT m
F x IN
m x M,OUT
S yM,OUT
F x M,OUT
S y M ,OUT F
S y M ,OUT F
y M ,OUT y
where y*M,OUT
* M ,OUT
y M ,OUT m F x IN
S y M ,OUT F
y M ,OUT
E M ,D m x IN 1
mS E MD F
395
External MB y IN E S,D
0 F x IN
yS,IN
yS,OUT
yS,IN
* S,OUT
y
S yS,OUT
Equil : y*S,OUT
Substituting: y*S,OUT y S,IN
E S,D
Substitute in
y S,IN
y S,IN
y S,OUT
m
E MD m x IN mS 1 E MD F E MD
y S,OUT
m x IN
m x S,OUT S
m x IN
y S,OUT
S y S,OUT F
x IN
y M ,OUT
F x S,OUT
F
yS,OUT
y S,IN 1 E SD 1
m E SD x IN
mS E SD F
and do some algebra to obtain, mS 1 F mS 1 E SD F
E SD 1 E MD mS E MD F
1
The definition of the total stage efficiency is, E total,D
Equilibrium:
x D IN
x D ,S,OUT
x D IN
* D ,S,OUT
x
y*S,OUT
E total,D
m
x IN
y IN
y S,OUT
y S,OUT
y IN
* S,OUT
y *S,OUT
S F
y S,OUT m x IN
y
with y IN
0
yS,OUT . Substitute into definition. y S,OUT x IN
S y S,OUT F
m 1
S y S,OUT F x IN
Which after substituting in yS,OUT and doing some algebra, becomes. E MD E total 1
m SE MD F
1
E SD 1 E MD
m S E MD F
mS F
mS 1 F E MD
E SD 1 E MD
mS 1 F
16.D1. (was problem 15.D1 in the 2nd edition) The corrected value of H G,E
1.41 ft is given in Example 16-2. From the results of Examples 16-1 and 16-2 and from Eq. (16-38), 396
2.61
0.15
0.15
2.61
0.83 0.85 ft 2.2 2.2 mV Eq. (16-27a) H OG H L H G where V L 8 5 L The value of m (the slope of the equilibrium curve) varies throughout the enriching section. From the McCabe-Thiele diagram used to prepare Figure 16-4A values of m were found from y 0.442 y* .63 to y x D 0.8 y* 0.82 . H L,E,cor
y y* m H OG y* - y 1
y* y Arithmetic Average:
H L,E,Initial
0.442 0.63 0.441 2.01
0.5315 0.598 0.406 1.96
0.621 0.66 0.449 2.02
0.7105 .727 .5144 2.11
0.8 .822 .745 2.42
0.188 5.32
0.0665 15.04
0.039 25.641
0.0165 60.606
0.022 45.45
1
H OG
2.01 1.96 2.02 2.11 2.42
5
Geometric Average: H OG
2.01 1.96
2.02
15
2.11 2.42
2.10 2.10
No difference! To find n OG do Simpson’s rule in 2 parts because of the unusual shape of 1 y*
n OG ,1
0.179 6 0.179
n OG ,2
6
5.32 4 15.04
25.641
25.641 4 60.606
y vs. y.
2.718
45.45
9.353
n OG n OG,1 n OG,2 12.072 , h E H OG n OG 25.35 This is reasonably close to the 26.1 ft estimated in Example 16-1. The physical properties and use of Figure 10-25 16.D2. (was problem 15.D2 in the 2nd edition) a. calculated in Example 16-2 are unchanged. Now F = 48.
G flood
2
141,
F 1
55, Dcol
12
48
14
F 2
1
20
G flood 2
F 1
Dia 1 b. In Eq. (16-37)
12
F 2
48
Dia 2
20
1.16
0.749 lb ft 2
4.68
5.83 ft
14
2 same .
Estimate
h p ~ 10 ft (we know it will be less than before), SC,V is unchanged, terms in denominator are unchanged. H G ,E 1
1
hp 1
10
2
hp 2
10
13 13
H G ,E 2
55
10
141
22
13
1.33
0.40 ft
397
1
In Eq. (16-38),
0.045 ,
2
0.07 , h p 1
10 , h p 2
22 .
Cf ,L , and SC,L unchanged. H L,E 1
1
hp 1
2
hp 2
0.15
H L,E 2
0.045
10
0.07
22
16.D3. New problem in 3rd edition. In the enriching section, HETP H OG ln(mV / L) / [(mV / L ) 1] and H OG
0.15
0.83
0.47 ft
H L mV / L H G
With H L 0.827, H G 1.33, m / ( L / V ) 0.63 / (5 / 8), H OG 2.16, and HETP = 2.15 (from Example 16-2). With the same mass transfer coefficients, but m = 0.577, H OG [0.577 / (5 / 8)]0.83 1.33 2.10 and HETP 2.10 ln(0.9232) / (.9232 1) 2.19 ft A 24.4% increase in both mass transfer coefficients gives H L
HG
1.244(0.83) 1.033 and
1.244(1.33) 1.65 .
For m = 0.63, H OG
[0.63 / (5 / 8)]1.033 1.65 2.69 and HETP 2.69 ln[0.63 / (5 / 8)] / [0.63 / (5 / 8) 1] 2.68 . With the same mass transfer coefficients, but with m = 0.577, H OG
2.61 and HETP = 2.71.
A 24.4% decrease in both mass transfer coefficients gives: For m = 0.63, H L 0.63, H G 1.01, H OG 1.65, HETP 1.64 For m = 0.577, H L
0.63, H G
1.01, H OG
1.59, HETP 1.65 .
Clearly, the variation in mass transfer coefficients results in a large range for HETP (from 1.64 to 2.69 feet for m = 0.63) while the small change in m had little effect. To be safe the larger value of HETP = 2.69 would be used. This is a safety factor of 1.20. Note that Bolles and Fair (1982) recommend a safety factor of up to 1.70.
16.D4. (was problem 15.D4 in the 2nd edition)
q
Feed line: Top: y
L V
x
.6,
q
.6
q-1
.4
1
L V
Since we have total reboiler, ys,in
1.5 . Plot this and operating lines (see figure).
x D where y intercept
Bottom goes through
y
x
0.04
xB
L
L D
0.9
V
1 L D
1.9
1
L D
xD
0.474
0.484
0.04 .
398
dy
nG For L HG
V HL
enriching
0.474
y AI y A from
2.568
V HL From figure generate following table.
yA Stripping: .04 .3225 .605 Enriching: .605 .7625 .92 Simpson’s rule, n G ,S
.315 6
.565 6
.605, y E,out
.92
33.33
1 2.26
n
slope
4.17
.8
yAI yAI-yA 1/(yAi-yA) .13 0.09 11.111 .46 .1375 7.2727 .63 .025 40.00 .62 .015 66.66 .80 .0375 26.66 .95 .03 33.333
16.D5. (was problem 15.D5 in the 2nd edition) a) 1
1.3
11.11 4 7.2727
66.66 4 26.66
Eq. (16-81), n OG
y E,in
section, Eq. (16-16) draw line of 1.3 .770 to get y I and x I . For stripping section slope is 0.8
L HG
n G ,E
, where y S,out
7.55 , h strip
40
10.8 , h E
HG n G
Total Reflux. y out
.956 1 .65 .65 1 .956
n
H G n G,S
1.3 10.8 0.956, yin
9.8 ft.
14.1 ft 0.65
.65 1 .956 1
399
n OG
6
1.4904 4.0257 b) Finite reflux. Plot op. line. Find y & y* (see graph). Use Simpson’s rule in 2 parts. y
y*
0.783 0.82225 0.8615 .90075 0.94
0.842 0.8795 .9125 .943 .973
y* y 0.059 0.05725 0.051 0.04225 0.033
4.0257, H OG
1 y* y 16.949 17.467 19.608 23.6686 30.303
n OG1
n OG 2
0.8615 0.783 6 0.94 0.8615 6
16.949 4 17.467
19.608 4 23.6686
19.608
30.303
1.3924 1.89166 n OG
3.2841
c) Changes in L/V in equation connecting HETP and H OG
400
401
16.D6. (was problem 15.D6 in the 2nd edition) a.
n OG
11.11 3.56 14.67
H OG h n OG 7.47 14.67 0.509 m b. From McCabe-Thiele diagram we find the following y y* y* - y 1/(y* - y)
0.016 .0267 0.017 93.46
0.066 0.1067 0.0407 24.57
0.116 0.1815 0.0655 15.27
.494 .623 .129 7.752
.872 .9201 0.0481 20.79
.922 .9523 .0303 33.00
.972 .9832 .0112 89.29
Do integration with Simpson’s rule in three parts. .1 .756 n OG 93.46 4 24.57 15.27 15.27 7.752 4 20.79 6 6 .1 20.79 4 33.0 89.29 15.97 6 The difference is because of inaccuracies in Simpson’s rule. 16.D7. (was problem 15.D7 in the 2nd edition) Bottoms x A
xA
0.1 Distillate
.9 0.1 , equil. y
From equilibrium data in 4.D7. @ x A
y out
partial reboiler
xD
0.262
yin With total condenser,
0.9
yA x A
Find average Bottom
1 yA 0.262 0.1
3.195 ,
0.738 0.9
Geometric Avg., Eq. (16-81),
1 xA
.929 0.9 dist
3.195 1.454
avg
n OG
. From data in 4.D7., y*
1 1 2.155
H OG
h n OG
16.D8. (was problem 15.D8 in the 2nd edition)
.071 0.1
12
0.9.
1.454
2.155
0.9 1 0.262
n
0.929 when x
0.262 1 0.9
n
0.262 1 0.9 1
4.80
0.42 m m
H Ptot
22500 855
26.3
402
L
26.3 .0011
max
V
.0011 .0001 L y out x in V x *A ,out
0,
L
28.93 , x out
26.04 . Basis V = 1, then L = 26.04.
act
V
26.04 .001
L
26.04
mV
26.316
0.9895
Use Colburn Eq. (16-34b), 1 0.0011 0 n OL n 0.010488 0.010488 0.0001 0
h
0.84 9.51
0.02604
.9895
9.51
7.99 m
Can check with Eq. (16-63) and get same n OL 16.D9.
New problem in 3rd edition. For example, if nO ED 1, we have for the perfectly mixed model,
nO
EMD
ED 1/ 2 1 nO ED For the plug flow model, EMD 1 exp( nO ED ) 1 exp( 1)
For same value of nO
ED
0.632
the plug flow model always predicts a higher stage efficiency.
16.D10. (was problem 15.D10 in the 2 nd edition) y in
n OG y out
dy y
y
*
. For both cases y* y in
dy
n OG
y
y out
n
For both countercurrent and cocurrent
0, and y y*
y in
n
y out
0.01 0.0001
y , then 4.6
Note that n OG is same because of irreversible reaction. c. Flow rates enter into solution only as a check that at least the stoichiometric amount of sulfuric acid is available to neutralize the ammonia. 16.D11.
(was problem 15.D11 in the 2nd edition) From Eq. (16-72) with irreversible reaction,
n OG
n
y A in
n
y A out
50.0 ppm 0.01 ppm
8.517 for both cocurrent and countercurrent.
16.D12. a. (was problem 15.D12 in the 2nd edition)
L
.013 .00004
G min L
.0053
G Use Eq. (16-57),
15 L G
yin
min
0.013, y*in
m
H p tot
2.7 1.1 2.4545, y
mx
2.4453 (see Figure)
36.679, x out m x out
y in
y out
L G
0.0003533
0.0008672, yout
0.00004, y*out
0 403
y
y*
.01296
n OG
b.
0.0121328,
in
y*
out
.0121328
n
0.0120928
y
.00004
6.1246 , h
.00004
4.59 ft
Cocurrent. Operating and equilibrium lines shown in figure. Lowest y out is at intersection point = 0.00081. For y out
x out
y in
y out
m
L V 2.4545
L V
36.679
n OG
0.00085 ,
.013 .00085 36.679
0.00033125, y*Aout
mx out
0.000813
0.0669184 . Use Eq. (16-20),
1
0.013 0.000813
1.0669184
1.0669184 h 1.98 ft
0.0669184
0.00085 0.000813
5.496
16.D13. (was problem 15.D13 in 2nd edition) a. Same conditions as Example 16-3. Assume same H OL . If operation is possible, find n OG & y out . Dilute – Use Eq. 16-70. m = 30.36 (Example 16-3) y*out m x out b 30.36 0.001 0.03036
mV L
30.36 15
1 mV 1 L
2.024 , n OG
Lx in
Mass bal.
y out
n OG
L V
Vyin x in
1
y in
n
Lx out L V
mV L
y in
y *out
y out
* out
y
mV L
Vy out
x out
n 3.024
1
15 0 0.03082 15 0.001
0.03082 0.03036
0.01582
mV
3.024 0.01582 0.03036 L Not possible, term inside brackets is negative. b) Same conditions as Example 16-3 except x out 0.002 . Assume same H OL . If operation is possible, then 404
y*out
m x out
30.36 .002
y out
15 0
0.03082
1
n OG
c) Same conditions, except L V * out
y
15 0
n 1.759
1.759
1 exp
Then:
x
x
x
0.8,
0.03082 0.0091 0.01882 0.0091
0.656
0.759
Use equilibrium data shown in Figure 4-16,
n 1 E MV h A active 30 0.8, K y a 1 2 n 1 .77 2 12
30
0.16, K y a
n 1 .69
.8 2 12
m
1
K ya
k xa
k ya
1
0.415
1
330.62
k xa
k ya
0.01, m
0.01882
30 and b
2 12 .
V
K ya
Solving simultaneously, k x a b. x
still not possible
0.759
K y aA active h V [Eq. (16-77)] where V A c
1
Eq. (16-6a),
30.36 40
40 0.0003
16.D14. (was problem 15.D14 in the 2nd edition) a. x 0.8, m 0.415; x 0.16, m 1.5 Mixed: E MV
2.024
0.0003
0.0091 , mV L
0.03082
1
n OG
0.00082
0.00082 0.06072
40 and x out
30.36 0.0003
y out
15 0.002
0.03082 0.06072
n 3.024
3.024
0.06072
, x
0.16,
330.62 where 1 2
263.5
1
1.5
1
263.5
k xa
k ya
1408.19 and k y a
0.8
366.317 142.18 & E MV
6.06. Then from Eq. (16-6a), K y a
0.468 from Eq. 16-77.
16-D15. (was problem 15.D1 in the 2 nd edition) Assume feed to Example 16-4 is sat’d liquid, z = 0.5, & 2.5 .5 separation complete x D ~ 1, x B ~ 0 . x F z 0.5, y f 0.7143 1 1.5 .5
xD
1,
L
1 0.7143
Vmin
1 0.5
At x
0.5714 , x B
.5, y intersection op lines
L V
L V Calculate at x
x
1
0.6 0 actual
0.5 0
L
0, L V
V xD
0.7143 0 max
.8 .5
0.5 0 .2 1.0
1.4286 0.6
1.2
0.1, 0.3, 0.7 in example , 0.9
405
x 0.1
Sect
L/V
Strip
m
1
1 x
1.2
2.5 1 .15
0.3
Strip
1.2
1.8904
2
1.1891
2.5
.8
2.5 1 1.05
0.9 Enrich
.8
const.
V A active h
x 0.1 0.3 0.7 0.9
1.3447 1.767
0.97. Calculate Ept. Solve Eq. (16-78) E pt
and then K y a from Eq. (16-76a), K y a
n
.6348 0.971
1.0092
n
1.3447
n
1.767
0.971
1.0092
n
0.971 1.3447 0.971 1.767
L mV
n
E MV L mV
1
n 1 E pt
E pt 0.6348
mV
1.0092
0.453
2
L 0.6348
0.5949
2
2.5 1 1.35
E mV
Eq. 15 80
2
1 .45 0.7 Enrich
2
1
0.5891
1
0.6802
1
0.731
1
0.7733
K ya 133.42 171.02 196.9 222.65
16.D16. New problem in 3rd edition. Using Simpson’s rule the new value for A1= 7.18, the new value for A2 = 10.85 and the new total area = 18.03. Then the calculated height of the enriching section (0.4054 m)(18.03) = 7.31 m compared to the previous result of 7.95 m. This is an error of 8.1 %. Thus, a rather small error in mole fractions becomes a larger error. 16.D17. New problem in 3rd edition. Relative errors in k G a
24.4% .
Same relative errors in H G and H L . In the enrichment section the slope of the mass transfer line is Slope
L HG
5
0.4054 24.4%
V HL
8
0.253 24.4%
406
a) If H L correct but H G varies by Value at
HG
24.4%, range of slope is from
1.242 to
0.761.
0.4054 is slope = 1.0015. At top:
If equilibrium line straight from azeotrope to point x = 0.7472, y = 0.7815, then can fit this portion of equilibrium as, y
MT line
y
If
s 1.242,
If
s 1.0015,
If
s
0.761,
0.7668x 0.2085 sx
b. Since y
x
0.8, b
0.8 1 s
b 1.7936 b 1.6012 b 1.4088
All calculations at y A
0.8.
mass transfer line, y
sx
y I at intersection equilibrium, y b
x
y
0.7668x 0.2085 and
b
s
407
Substitute in for x,
yI
yI
0.7668
b
0.2085
s
Solve for y I yI
0.7668b s .7668 1 s
0.2085
yI
s 1.242, b 1.7936 :
yI
0.81356
1
y
yI
0.013565
y
73.72
s = 1.0015, b = 1.6012: 0.812426 0.01246 s = 0.761, b = 1.4038: 0.80842 0.008419
80.48* 118.78
*83.3 in Example 16.1 since numbers rounded first. Amount of error depends on distance between equilibrium and operating lines. Less error if closer, but more impact on 1 y AI y I . Assume same relative errors: H L no error, H G higher, H L no error,
1 y AI
yA
80.48 73.72
error
H G lower,
80.48 1 y AI
yA
8.4% 118.78 80.48
error
80.48
47.6%
Assume error in H is same every point. Thus enriching area can be a lot different than calculated. But if H G is down by 24.4%, area is up by 47.6% so there is some cancellation of error. b. If H L & H G both vary could have s 1.634 and b yI
Area H GE
y
0.493,
1 yI
y
2.1072 . Thus, y I
0.81493,
= 66.97
66.97
19.6 16.31. 80.48 1.24(0.4054) 0.503 m, z H G,E n G,E
Since 8.20 m which is a 3.2 % increase.
16.D18. New problem in 3rd edition. a. Equilibrium is y = mx. Value of m is unknown, but Cextract = mconc units Craffinate with
mconc _ units
20.8
kmol _ Benzoic / m 3extract kmol _ Benzoic / m 3raffinate
. We need m in
408
mol Benzoic / mol extract
mmole fraction units
mol Benzoic / mol raffinate
mmole _ fraction _ units
mconc _ units (
MWextract
)(
extract
. The resulting conversion is,
raffinate
MWraffinate
)
Since the system is dilute, extract properties are essentially the same as pure solvent and raffinate properties are essentially the same as diluent. m = (20.8)(92.14/865)(1000/18) = 123.1 Note that y = xD and EMy = EMD. EMy = (yin-yout)/(yin-yout*) where yout* = m xout = (123.1)(1.99E-06) = .000245 EMy = (0-.000230)/(0-.000245) = 0.939 b. From Eq. (16-76b), nO-ED = EMy/(1-EMy) = 0.939/(1 – 0.939) = 15.393 nO KO
KO
ED
QD
ED
a
nO
ED
a Vmixer
D
MWD
ED
QD
D
, Vmixer
h D2 / 4
(0.75)
0.75 2 / 4
0.331m3
/ (Vmixer MWD )
(15.393)(.0012)(865) / [(0.331)(92.14)] 0.524 kmol / s m 2 mol frac disp Note that K O
ED
m 2dispersed / m 3total volume
a is larger than in Example 16-5 because the residence time
Vmixer
37.45s is shorter than the 361 s in Example 16-5. Thus, this problem (Q D Q C ) d requires much more vigorous mixing. t res
c. Differential Model
exp
E MD n OED
n OED KO
ED
a
nO
ED
QD
1 exp 1 E MD
n 0.061 D
n OED 0.061
2.797
n OED
2.797
/ (Vmixer MWD )
(2.7969)(.0012)(865) / [(0.331)(92.14)] 0.0952 kmol / s m 2 mol frac disp
d. Use of mixed models: If use K O nO
ED
KO QD
ED
a Vmixer
D
MWD
ED
a
m 2dispersed / m 3 total volume
0.524 from mixed staged model, then
15.393 and with differential model 409
E MD 1 exp wrong results.
16.D19.
n OED
1 exp( 15.393) .9999998 . Obviously, mixing models gives
New problem in 3rd edition. C D in C D out a. E MD ,Conc with C*D,out * C D in C D ,out
CD,in
0.000 , CD,out
C*D,out
mCD CC,out
nO
.00023
KO
ED
a
nO
ED
QD
.0002578
nO
ED
1 nO
0.00023 , CC,out
0.0481 CC,out
E MD
b. E MD,Conc
ED
ED
D
mCD CC,out 0.00536
0.0002578 .892
E MD,Conc
0.892
1 E MD,Conc
0.108
8.26
/ (Vmixer MWD )
(8.26)(.0012)(865) / [(0.331)(92.14)] 0.2811 kmol / s m 2 kmol/m 3
m 2 dispersed / m 3 total volume
c. Differential Model: E MD,Conc 1 exp n OED → exp n OED KO
ED
a
nO
ED
n .108 QD
D
n OED
2.2256
1 E MD,Conc n OED
0.108
2.2256
/ (Vmixer MWD )
(2.2256)(.0012)(865) / [(0.331)(92.14)] 0.0757 kmol / s m 2 kmol/m 3
16.D.20.
m 2 dispersed / m 3 total volume
d. E MD,Conc 1 exp n OED New Problem in 3rd Edition.
1 exp
8.26
a. Eq. (16-92). Terms:
0.167,
Dbenzoic-water
d
(from Example 16-6) di
2
0.2070 16.6667
g
Note
9.807
0.99974 2.2 10-9
4 3 2 . 6 8
C CD
A
C
2
5.8632
N in rps
410
di
0.2070
dp
0.0002524
820.13
dp
0.0002524
d tank
0.8279
D
d pg
0.00030487
865 0.0002524 9.807
96.447
0.0222
kC
0.00001237
2.2 10
0.0002524
x 0.00030487
12
Significantly higher. But Note: With same k d
0.001905.
96.447
820.13
1
5 4
0.167
12
2
0.014801
0.06, so correlation may not be valid. Same
d
1
1
1
K LD
kD
m CD k C
1
1
K LD
0.001905
1 0.014801 20.8
K LD
0.0005181 & K LD a
E MD
5 12
432.68
1
K OED a
b.
9
1930.2
2.0586 s
1
865 2.0568 92.14 19.13 19.31 20.31
which agrees with guess for residence time.
0.951,
Resistance (dispersed) = Resistance (continuous) Sum of resistances
1 0.001905 1
524.9
1404.5 0.000712 1929.4 Gives K LD 1 1929.4
0.0005181
% resistance from dispersed 524.9 1929.4 100 27.2% This is a significant contribution because of significantly higher predicted continuous phase mass transfer coefficient. c. If k D is ignored , K LD k C m CD 0.014801 20.8 0.000712 Note that this is significantly too high. d. To be safe, use lower estimate of E MD . Note that d is really too large for use of Eq. (16-92). 16.D.21. New problem in 3rd edition. In settler velocity approaches zero ( u t 0.00172 in Example 13-5, but d p 0.000204 not 0.000150 assumed). Thus u t ~ 0.00495 m s and Eq. (16-88a) is reasonable approximation. Sh C
k C d Dbenzoic-water
2.0 where in settler d
0.0002524
411
kC
Using
2.0 2.2 10
K LD ~ k C mCD
9
0.0002524 5
1.7433 10
1.7433 10
1 20.8
5
8.381 10
7
ms
If the interface in the settler is at the centerline, then the volume of aqueous phase is
1
D S2 4 L S . From Example 13-5, DS
2
1.023m and LS
4DS . Then Volume
aqueous phase 1.682 m3 . The water residence time is 280.3s. Entering total velocity is 0.0072 m3 s and
1.682 m3 0.006 m3 s d
0.167 leaving mixer. A drop that starts at the bottom of the water phase travels a
vertical distance of DS 2 0.5115 m to reach the interface. This requires 0.5115 m 0.00495 m s
103.2s.
Assume equal distribution of drops. Everything (half the drops) above interface are collected very quickly. All drops collected in 103.2 280.3 0.368 fraction of settler. If average over this fraction is 0.5 d,IN , then D,avg 0.5 0.167 0.368 0.031. 6
a
6 0.031
d
dp
K OED a
737 and K LD a
0.0002524 D
settler
n OED,settler
K LD a
Vsettler Q D
settler
MWD
0.0058
settler
~ 0.000618
865 0.000618 92.14
3.364 0.0012 0.0058
865 92.14
16.D.22.
1.734
New problem in 3rd edition.
a)
E SD
y M,OUT
yS,OUT
y M,OUT
* S,OUT
y
1.7314.
865 92.14
Since settler is not well mixed, E settler,D 1 exp This is high because of long residence time.
y*M,OUT
0.82,
E MD
0.0058
0.823
y IN
y M ,OUT
y M ,OUT
y IN
* M ,OUT
y*M ,OUT
y
0.794
mx raf ,OUT .
From Example 16.5 in mole fraction units m 123.1. Mixer mass balance with y IN 0 (Example 16.5) is Fx F
Sy M,OUT
Fx C,OUT
In Example 13.5 feed is 0.006 m s and solvent is 0.0012 m3 s 3
S
0.0012 m3 s 865 kg m3 F
92.14 kg kmol
0.006 m3 s 998 kg m3
x M ,OUT
x feed
S F y OUT
0.01127 kmol s 18.02 kg kmol 0.00026
0.01127 0.3323
0.3323kmol s y OUT
Then substitute into E MD y M ,OUT 123.1 0.00026 0.03390 y M ,OUT
0.794 412
Solving, we obtain x M,OUT
y M,OUT
m E MD
For settler, Settler M.B. is,
y M,OUT
0.00589 y*S,OUT
Sy M,OUT x S,OUT
0.00589.
123.1 0.794
m x S,OUT
Fx M,OUT
SyS,OUT
S
x M ,OUT
x S,OUT
6.027E 5.
y S,OUT
F
6.027E 5
Fx S,OUT y M ,OUT
0.03390 yS,OUT
0.00589
Substitute into equation for E SD 0.82
0.00589 yS,OUT 0.00589
123.1
6.027E 5
Solving, we obtain yS,OUT From Mass Balance, x S,OUT
6.027E 5
* S,OUT
m x S,OUT
E Total,D
From 16.C4
E Total,D
0.006174.
0.03390 0.006174 0.00589 y
b)
0.00019967 0.03390y S,OUT
123.1 5.0653E 5
yS,OUT
0.006174
* S,OUT
0.006235
y
mS F
123.1
5.0653E 5
0.01127 0.3323
0.006235
0.99015
4.17495
0.794 0.82 1 0.794 4.17495 1 1
4.1794 0.794
1
4.17495 0.82
4.17495 0.794 0.82 1 0.794 4.17495 1
.99003 16.G1 and 16.G2. New problems in Chapter 16. Aspen Plus runs showed that N =13 (Aspen Plus notation) with feed on Nfeed = 11 (optimum location) gave ethanol mole fractions of xD = 0.7990 and xB = 0.020298. These values are within the specified tolerances. The stripping section starts with the vapor leaving the reboiler (yin,strip = yreb = 0.18709) and ends at the intersection of the two operating lines. This last value can be determined by calculating the points on the operating lines (xn, yn+1). For example n = 1, x1 and y2 are on the enriching section operating line. When the slope changes from 0.61 in the top to 2.07 in the bottom, the intersection point has been passed. This occurs for yout,strip = y11 = 0.44631. The mole fractions of ethanol in the liquid and vapor leaving each stage (Aspen stage notation is used) are: 413
Stage 1 2 3 4 5 6 7 8 9 10 11 12 13
x 0.79904 0.77157 0.74561 0.71971 0.69245 0.66203 0.62575 0.57882 0.51135 0.40050 0.22553 0.10033 0.020289
y 0.81824 0.79904 0.78189 0.76569 0.74957 0.73263 0.71375 0.69129 0.66233 0.62090 0.55339 0.44631 0.18709
Equilibrium: Calculation m The equilibrium parameter m is the average slope of the equilibrium curve from x (calculated at y) to xI. At the reboiler y = 0.18709 for x = 0.020289, and y* = 0.44631 for xoper =0.10033. Then yavg = 0.3167. The slope can be determined by taking the chord from x = 0.04 (y*= 0.29209) to x = .05 (y*= 0.33018). m = (0.33018 – 0.29209)/0.01 = 3.809. The equilibrium values are from Analysis in Aspen Plus. At yout,strip = 0.44631, xoper = 0.22553, y*= 0.55339, and yavg = 0.49985. m = (0.50405 – 0.49482)/(0.15 – 0.14) or m = 0.923.
To use Simpson’s rule for the first integral in Eq. (16-22a) we also need an average m for the y and y* values calculated at the average between yin,strip and yout,stip, which is y = 0.3167. At this y, xop = 0.1659 (determined from the stripping section operating line) and y*=0.5174. The average between y and y* = 0.41705. m = (0.42921 – 0.41012)/(0.09 – 0.08) = 1.909. The second integral in Eq. (16-22a) is the usual estimation from Simpson’s rule of the nOG integral, y A ,out
HG
dy A / ( y A * y A ) = (0.2835/6)(0.44631 – 0.18709)[1/(0.44631 – 0.18709) + 4/(0.5174 - 0.3167) y A ,in
+ 1/(0.55339 – 0.44631)] = 0.4057 m The first integral in Eq. (16-22a) can also be estimated from Simpson’s rule, y A ,out
( H L / ( L / V ))
mdy A / ( y A * y A ) = (0.1067/2.032){[(0.44631 – 0.18709)/6]× y A ,in
[3.8577(3.809) + 4(4.983)(1.909) + 9.3388(0.923)]} = 0.1393 Then h = .4057 + 0.1393 = 0.545 m, which is somewhat more than the 0.507 m calculated in Example 16-1. Note that the Aspen Plus calculation does not assume CMO whereas the calculation in Example 16-1 did, but the calculation here required an assumption of how to calculate m. Thus, it is difficult to say which is more accurate. 16.G2. New problem in 3rd edition. Enriching section yin,enrich = yout,strip = 0.44631, and yout,enrich = xD =x1 = 0.79904. At yin,enrich = yout,strip = 0.44631, the calculation of m is the same as done previously for yout,strip: xoper = 0.22553, y*= 0.55339, and yavg = 0.49985; m = (0.50405 – 0.49482)/(0.15 – 0.14) = 0.923. At yout,enrich = xD =x1 = 0.79904, y* = 0.81824, and yavg = .80864; m = (0.81180 – 0.80481)/0.01 = 0.699. In Example 16-1 the enriching section integration was done with Simpson’s rule in two parts. From yin,enrich to y = 0.725 and from y = 0.725 to yout,enrich. For the calculation here we will use y = 0.71375, which is in the table of data from Aspen Plus, as a convenient break mole fraction to do the integration in two parts. For y = 0.71375, xoper = 0.66203, y*= 0.73263, and to find m, yavg = 0.72319. m = (0.72623 – 0.72101)/(0.01) = 0.522. 414
The average y from yin,enrich = 0.44631 to y = 0.71375 is 0.58003, xoper = 0.44423, y* = 0.63659, and to find m, yavg = 0.60831. m = (0.61020 – 0.6067)/(0.01) = 0.350. For the 2nd integral in Eq. (16-22a) integrated from yin,enrich = 0.44631 to y = 0.71375 we obtain, y A ,out
HG
dy A / ( y A * y A ) = (0.4054/6)[.71375 - .44631][1/(.55339-.44631) + 4/(.63659 -.58003) + y A ,in
1/(.73263-.71375)] = 2.404 m Same integral integrated from y = 0.71375 to yout,enrich = 0.79904. The average y from y = 0.71375 to yout,enrich = 0.79904 is 0.75640, xoper = 0.73068, y* = 0.77246, and to find m, yavg = 0.76443. m = (0.76587 – 0.75985)/(0.01) = 0.602. =(0.4054/6)[.79904 - .71375][ 1/(.73263-.71375) + 4/(.77246 -.75640) + 1/(.81824 - .79904)] = 2.041 Total for integral 2 = 4.445 m For first integral from yin,enrich = 0.44631 to y = 0.71375 we obtain, y A ,out
( H L / ( L / V ))
mdy A / ( y A * y A ) = (0.253/0.625)[(0.71375 – 0.44631)/6][(0.923)(9.388) + y A ,in
4(0.350)(62.2665) + (0.522)(52.966)] = 1.102 Same integral integrated from y = 0.71375 to yout,enrich = 0.79904, y A ,out
( H L / ( L / V ))
mdy A / ( y A * y A ) = (0.253/0.625)[(0.79904 - 0.71375)/6][(0.522)(52.966) + y A ,in
4(.602)(18.2815) + (.699)(52.0833)] = 0.870 Total for integral 1 = 1.972 m Total height of enriching section = 4.445 + 1.972 = 6.417 m Total height of packing = 0.545 (from 16.G1) + 6.417 = 6.962 m. This is less than the total of 8.457 m calculated in Example 16-1. However, it does agree reasonably well with the number of stages (11) in the column since, 11 × HETP (estimated = 0.655 m in Example 16-2) = 7.208. Both this result and Example 16-1 require calculating a small difference and then taking the inverse of this number. This type of calculation can cause very significant errors. The graphical calculation was based on accurate experimental measurements of the equilibrium data, and this data is probably more accurate than the NRTL correlation used in the computer. On the other hand, calculation errors are probably larger in the graphical than in the computer calculation. Both calculations depend significantly on the accuracy of the mass transfer data (HL and HG), which can easily have errors greater than 20%, which can cause even larger errors in calculation of NTU and HTU or in HETP (see problem 16.D16) for the staged calculation. 415
16.G3. New problem in 3rd edition. Part b. Equilibrium stage optimum. (L/D)min = 5.1, L/D = 6.12, N = 33, NF = 14, 1 section, yC4,dist = 0.008556, xC3,bot = 0.005706, Dia = 1.804 m on plate 32, minimum diameter = 1.565m on tray 14. Part c. VPLUG optimum. N = 33, NF = 14, yC4,dist = 0.007953, xC3,bot = 0.005302, Dia = 1.812 m on plate 32, DC backup/tray spacing = 0.391, weir loading = 0.021m2/s. Note that this separation is better than the equilibrium result. Part d. MIXED optimum. N = 49, NF = 22, yC4,dist = 0.007651, xC3,bot = 0.00510, Dia = 1.808 m on plate 48, DC backup/tray spacing = 0.390, weir loading = 0.021m2/s. Note that this separation require significantly more stages than both the equilibrium-stage and the plug flow results. 16.H1 and 16H.2. New problems in 3rd edition. Fit for the ethanol-water VLE was done in Appendix B of Chapter 2. The spread sheet for both problems is given below with a y value chosen in the enriching section (ignore the stripping section operating line values). Overall mass balances to find D and B were done with Eqs. (3-3) and (3-1). L and V were determined at the total condenser and flows in the stripping section were determined at the feed stage with the calculated value of q. L bar = L + q F. The intersection point of the two operating lines was determined from Eq. (4-38). yreb is the start of the stripping section and is the y value in equilibrium (yeq) with y = x = xB. In the stripping section the value of 1/(yeq – y) was determined at the start of the stripping section (yreb), at the end of the stripping section (yintersection), and at the average of these two values. Then Simpson’s rule was used to calculate nOG,strip = 1.57787 from Eq. (16-24b). To determine HOG from Eq. (16-27a) an average slope m of the equilibrium curve is required. The slope of the chord from the equilibrium y at the intersection point of the two operating lines to the reboiler vapor that is in equilibrium with xB, mstrip ( y eq ( xint er sec tion) y reboiler) / ( xint ersec tion x B) = 1.95, and HOG,strip = 1.265. The resulting height of the stripping section, 1.998 feet, is somewhat more than the 1.66 feet determined in Example 16-1. In the enriching section a similar procedure was used except the integration to find nOG,enrich was done in two parts. The average slope of the equilibrium curve was determined from, menrich ( y1 yequilibrium( xint ersec tion)) / ( x D xint ersec tion) = 0.4558 with y1 xD . The resulting value of HOG,enrich = 1.935. The resulting height of the enriching section, 29.35 feet, is somewhat more than the 26.1 feet determined in Example 16-1 and more than the 25.35 feet determined in problem 16.D1. Note: in both 16.H1 and 16.H2 the average slope of the equilibrium curve m has to be calculated with fairly large chords, not by taking the derivative of the 6th order polynomial fit to the VLE. The reason is that the fit oscillates around the experimental data points and the slopes will fluctuate greatly.
416
417
SPE 3rd Edition Solution Manual Chapter 17. New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 17A2, 17A7-17A9, 17B2, 17C4,17D3, 17D8, 17D10-17D13, 17D15b-h, 17D1617D18, 17E1, 17H2-17H7. Chapter 17 was chapter 16 in the 2nd edition. Problems from that edition have the same problem number, but the chapter number is now 17 (e.g., problem 16.D6 is now 17.D6). Figures in the solutions to these old problems still have the designation of chapter 16. 17.A2. New problem in 3rd edition. Change the value of thetatot in the spreadsheet until the area matches the desired value. This can be done as a manual trial and error or a loop can be added to the spreadsheet. 17.A7. New problem in 3rd edition. Part a. Increase the stirrer speed. If no gel, increasing stirring increases mass transfer coefficient k which reduces M and hence xw is reduced. This reduces xp increasing retention R. Part b. Decrease the stirrer speed. This reduces k, increases M and xw. When xw > xgel, a gel forms and R increases (probably to l.0) . 17.A8. New problem in 3rd edition. Since there is a gel the retention of the low molecular weight compound also increases. 17.A9. New problem in 3rd edition. Do not invest. Osmotic pressure can often be ignored in UF because with large molecules with high molecular weight the mole fraction is always low even if the weight fraction is high. With low mole fraction the osmotic pressure is low. If there is a concentrated salt with a low molecular weight the mole fraction will be high and the osmotic pressure cannot be ignored. 17.B1. Look at Suk, D.E. and Matsuura, T. (2006) ‘Membrane-based hybrid processes: a review’, Sep. Sci. Technol. Vol. 41, pp.595–626 for additional processes. 17.B.2. New problem in 3rd edition. One possible approach is as follows: Increase stirring to increase the mass transfer coefficient and reduce the wall concentration to prevent gel formation. Then use a permeate in series cascade with recycle of the retentate from the second module in series back to the feed of the first module. The low molecular weight product is the permeate from the second module. The intermediate molecular weight polypeptide product is the retentate from the first module.
17.D1.
PCO 2
15 10
cc STP cm
10
pr
PCH4
0.48 10
10
PCO2 PCH4 31.25 a) Generate RT curve from Eq. 17-6a. pp yp 1 1 yP pr yr 1 yP AB AB
1 10 6 m
tm
cm 2 s cm Hg
pH
12 atm
pP ˆ CO
AB
1
pL 2
ˆ CH
12 76.0
0.2 atm 4
912.0 cm Hg
15.2 cm Hg
1.0, p p p r
0.016666
y P 1.5042 0.5042 y P 31.25-30.25 y P
418
RT
Curve
yP
yr
Op. Eq., FP FIN
0
0
0.1 0.20
0.00515 0.01114
0.30
0.01830
0.40 0.5
0.02721 0.03882
0.6 0.7
0.05504 0.08000
0.8 0.9 1.0
0.12492 0.2349 1.0
y out
y out
PCO2 t ms
cc STP cm
FP
FP
b.
1000
FIN
gmole
1 hr
h
3600 s
0.0002859
cc STP
J CO 2
Fin Fin,1
FP1
2.125
0.32
Fin
0, y P
.15 .32
yP
pr yr
0.402, y out
0.46875
0.25625
0.0276 CO 2 conc. (17-2b)
pP yP
0.088888
L
0.46875
2.125 .10
15.2 cm Hg 0.402
0.0002859
cc STP cm 2 s
gmol s
1.0 gmole
2
cm s 1000 cc 22.4 LSTP
FP y P,CO 2
FP
1 0.32
912 cm Hg 0.0276
1 10 cm cm s cm Hg
Area
0.15 CO 2 mole frac
0.10, y P
Answer (from graph)
2
J CO 2
FP FIN
Plot two arbitrary points:
4
FP
y IN
Slope
10
15 10
y out
y IN
y out
FP FIN
yP
J CO2 J CO2
1 FP FIN
yP
0.32
1.2764 E 8 gmole s cm 2
2.80 10 6 cm 2
0.32 kgmole/hr , Fout
Fin
FP
1 0.32
0.68 kgmole/hr.
1 kgmole/hr
1 2
FP,part a
Stage 1. FP1 Fin1
1 2
0.32
0.16
0.16 1.00
FP 2 , Fout1
0.16 , yin ,1
Fin2
FP1
1 0.16
0.84
Fin ,2
0.15
RT curve is unchanged! 1 .16 0.84 Op. Line: Slope 5.25 0.16 0.16 Find arbitrary points to plot line: 0.15 If y out ,1 0, y p 0.9375 (off graph). 0.16
419
If
y out,1
0.04, y p
y out,1
If
5.25 0.04
0.08, y p
5.25 0.08
Answer (from graph): Stage 2
FP 2
0.16
Fin 2
0.84
0.7275
0.9375
y P,1
0.5175
0.625, y out,1
0.1905,
0.0595
y in2
0.0595
FP2 Fin2
0.1905
yin,2
0.3123
1 0.1905
4.2500 . Plot curve 0.1905 Answer: y P2 0.250, yout 2 0.015 (see graph) Stage 15 10 10 cm 3 STP 912 0.0595 15.2 0.625 0.0006715 1 10 4 cm 2s 1L 1 mol JCO 2 J CO 2 2.9976 E 8 1000 cc 22.4 cm 2s mol 1 h FP1 0.16 1000 0.04444 mol s h 3600s FP1 0.625 Area 1 92.67m 2 JCO 2 ,1 Slope
J CO 2
0.9375
Stage 2: J CO 2 ,2
15 1010 1 10 JCO 2 ,2
4
912 0.015
13.2 0.250
1L
J CO 2
1
0.0001482
6.6161 E 8
3
1000 cm 22.4 lh FP 2 0.1905 840 mol h 3600 s FP 2 y P 2 Area 2 1, 680, 000 cm 2 JCO 2 ,2
1:
cm 3 STP cm 2s
mol cm 2s
0.04445 mol s 168 m 2
It is interesting to compare parts a and b.
Part a:
1 stage
Area
280m 2
y out,CO 2
0.0276 or 97.24% CH 4
y P,CO 2 Part b:
2 stage
0.402
Total Area
260.67m 2
y out
0.015 or 98.5% CH 4
y P1
0.625
y P2
0.250
420
17.D2. a.
yP Slope
1 FP Fin
y in
, yP FP Fin FP Fin .7 2.333 , When y out .3
y out
y out 0, y P
y in
0.2 CO2 , FP Fin
0.2 0.3
.3
0.6667
421
When
yP
y in
0.2
0.286 1 FP Fin .7 RT curve is same as in Problem 17.D1. Draw op line. From graph: y P,CO2 0.53, y r y out,CO2 b.
0, y out
J CO2
J CO 2
15 10
10
pP yP
t ms cm 3 STP cm 76 cm Hg
1 10 4 cm cm 2s cm Hg J CO 2
A
PA p r y r
0.06
0.002148
atm
cm 3 STP 2
cm s
60 atm 0.06 FP
, FP
0.6 mol s 0.53
Fin
Fin
0.6
1000 cm 3 STP 3
1 mol cm STP 0.002148 22.4 L STP cm 2s
L STP
3.3 0.53 mol s
atm
.
3, 254, 000 cm 2
Or 325.4 m 2 . Very sensitive to y P & y r values. Can also calculate J Check:
Fin Fin x in
FP
Fout , 2 FP y P
J CO 2
0.6 Fout
Fout y out , 0 .4
FP ˆJ
J CH 4 . A
Fout
1.4
2 0.2
0.6 0.53
1.4 0.016
.402 , OK
422
17.D3. New problem in 3rd edition Since no concentration polarization x w
J solv
K solv t ms
pr
pp
a xr
R and M.B.
xp
xp
xp
x out 1
xF
xF
x out
Then,
pp
K solv t ms 1
1 R
xp Then
pr
x out ,
xp
J solv
pr
Solve for
xr
1 R x out
415.4
a xr
xp ,
1 R x out , R ,
1
0.22, x F
K solv t ms
33.29, J solv
415.4
0.9804 0.0077
0.0077 0.22 0.78 1 0.9804 0.22
0.0098175
0.0001924
1.1 15.446 0.0098175 0.0001924 33.29 17.D4. Partially new problem in 3rd edition.
13.72 atm.
423
a.
xw
M
exp ( J solv /
xr
1 xP
Mxr
xP
1
pr
3.6 E 3.6 E
K solv
solv ) k
pP
4.625 / 997000 g/m 3
exp
6.94 10
Mxr
xP a
pr
pp
3.6 E
1.069 0.054 75 2
4
3.6 E
1.069 0.054
J solv
t ms
1.069
xP
4 4
5
4
59.895
4.625
a Mxr
Since
73
xP
1.0689 .054
K solv K A , K A
K solv
and
3.6 E
KA t ms
4
59.895
2.29
1 atm
0.0665
g 2
m s atm
K solv t ms
g KA m s atm 0.029 g t ms 2.29 atm 1 m 2s c. Write Eq. (17-37a) for old and new situations – Divide new by old. Obtain 0.0665
2
0.75
k new
k old
Everything else divides out. Since
new
rpm ,
old
k new 17.D5. a.
K solv t ms
J solv pr
RT eq., R
k old
.75
2000
0.000117m / s
1000
1.5 10
3
g
2
cm s
pP
1 102 atm
1.47 10 5 g
1 FP Fin
1 x P x out , Op eq., x P
cm 2 s atm
x out
x in
FP Fin FP Fin Solve simultaneously & obtain Eq. (17-26), which with M = 1 is 1 R x in 0.003 0.05 xP xP 0.000272 , x out 0.091 1 R 1 0.997 0.45 1 R 1 xP x P pr
xr
pP
xP
xr
xP a
1 0.000272 0.091 0.000272 0.000272 102
0.091 0.000272 59.895
b. Plot the RT curve and operating line xP xP a 1 pr pP xr M 1 a 1 xP See graph. Intersection occurs at x r
J solv
K solv t ms
pr
pP
1
, xP
0.0585, x P a M xr
1
3.44 atm
xr
1
x in
0.000752
xP
424
425
J solv
1.47 10 2
5
g
78 atm
cm s atm FP 1 x P , FP A J solv
2 3
2
Fin
3
4
1000 g kg
g cm 2 s
3.36 10 6 cm 2
2
9.91 10 g cm s
Eq. (17-45) can be written as J solv /
k
solv
n xg
336 m 2
n xr
n x r . Slope = k and intercept = k n x g cm
J solv /
solv
0.052 0.037 0.026 0.0134 See Figure. Plot J solv /
solv
Intercept
Stirred cell data:
cP
J solv /
J solv
g L, solv
wt frac
xr
min
n xg
Slope
J solv
0.000991
3.33 kg s
4
Plot J solvent vs
17.D7.
5 kg s
3.33 kg s 1 7.52 10
A 17.D6.
59.895 atm 3 0.0585-7.52 10 -4
dextran 0.012 0.03 0.06 0.135 versus ln xr
0.0185 0.01596
K solv t ms
J solv p
23.1 / (997 g / L )
n xr 4.423 3.507 2.813 2.00
1.159, which is x g
69.25 3.0
23.1
0.314 .
g 2
m s bar L 0.0232 2 m s bar
426
J A ~ J solv c P Mc
With
0.00696 g (m 2 s) . Also, J A
0.0232 0.30 1, 1 R c
Spiral Wound:
JA
Solving for M c
JA
0.00696
c out J solv
10 (0.0232)
J so l v c P
0.03 → R c
0.97
M c cout 1 R c J solv
cP
1.0
c out 1 R c
8 0.030
K solv
J solv
M c cout 1 R c J solv
t ms
p
4.1660 80.8g / (m 2s)
23.1 3.5
Since osmotic pressure is ignored, M c does not effect solvent flux in UF.
yp
1
He H 2
rd
17.D8. New problem in 3 edition From 17-6b, y rHe
He H 2
Check y rH
2
2
H 2 He
b.
b
Pr
1 y PH
2
1 y PH
H 2 He
y rHe Then
Pp
1
H 2 He
pp pr
1 y rH
2
1
0.975
2
y PHe
y r ,He
0.025 0.07656
c
y F,He (1 yP
1
2.8314 .975
1
.9234
0.07656 OK.
0.05 0.07656
1
.07656
3.8314 1 .2 .025
3.8314
y rout He
1
1
10
y FHe
y F,He 1
1 yp
1 y P He
0.5152
Use solution in Eqs. (17-9) to (17-10e) pp .75 a 1 .2 .261 1 1 pr .25
1
Pr
0.2610 , p p p r 0.2 90.8 10 10 .739 0.2 .975 1 0.021397 0.025 .261 .739 0.25 .279475
y rHe
y PH
23.7 10
PHe PH 2
Pp
.739 .2 3
)
.261 .05
b
b2
4ac
2a
Must use minus sign to have positive y p . y PHe
.25
2.3648 .05
1
.25
.25
1.4874
0.0522
1.4874
2.212359 .49377 2
.15763 4.7296
2.3648
1.4874 1.645 4.7296
0.00333
427
y r,He
y F,He
y PHe
1
1
c. Solve RT eq. (17-6b) for y p : y r
0
yr
pp
1
1 yp
y 2p
pr
pp
1
1
c
yr
yr
pr
y rH2
.9234,
1 yr
pr
y rHe
y PHe
y rHe
FP y PH
FP
A
0.025, 2
.06
.739
.80786
.65264
y p,H 2
0.004842 0.1813
0.05,
y H2 IN
.95
.5152
0.975
FIN
pr yr pp yp tm s 51.52 m3 STP h 1h
h
m3
3600 s
90.8 10
10
cm 3
cm 3 STP
14311.11
14311.11
2
a.
yp yr
PH 2 A
s
cm 3 STP
s STP cm
cm 2s cm Hg
A 17.D9.
1 yp
pr
c
y He,IN
1000000 cm 3
Pp y P,H 2
1
.2956 0.05 0.06 0.004842 0.06
51.52m 3 STP
PH 2 Pr y r,H 2
1
.739 .2 1
.80786
y rHe
0.07656 y p,H 2
FP y PH tm s
pp
1
0.01566
4ac
2a y FHe
Use 17-5a written as
FP
1
100m3 STP / h.
FIN
For Part A
pp
b2
Use + sign for positive y p , d.
y p2
0.1478
.06 .261 b
yp
.739 .2
pr
b
pr
0.10001
b
pp
1
.25 pp
1
a
a
.05
3 .00333
555,186 cm 2
.975 1.0 10 4 cm
380 .9234
76 .975
cm Hg
55.52 m 2
Plot the data on a semilog plot in the form of J solv /
solv
J solv L / (m 2 h)
428
J solv
xr
k n
xg
.
From graph, slope When x r
k n xr.
Intercept
18.3 and k
18.3
k n xg
0, J solv / n xg
k solv
2
=J solv
82.9 L m h
23.0 g m 2s
4.53
5.08 m / s
xg
k n xg L
5.08
2
m h
m
2
s
23.0g / (m s)
k n xg
92.8%
The value of xg is very sensitive. b.
There is only one point further out on the ℓn axis. Any error in point is greatly amplified in the least squares regression. Hence, another point in this region would be most useful. The higher the concentration, the better.
17.D10. a) New problem in 3rd edition
2700 800
3.375 , a
b
1
c
yin 1
yp
3.8884
1
Solve RT eq. and op. eq. simultaneously pp 0.3 .5 1 2.375 1.6116 pr 0.7 2
pp pr
yN 1
1
1
1
3.375 .25 3.8884
2
.7
2.375
.3
.25
1
2
.7
.7
.7
3.8884
1.20536
4 1.6116 1.20536
2 1.6116
.5
, use minus sign to have yp between
0 and 1..
429
yp
3.8884
2 1.6116
From op. eq.: y r b)
Since Fˆp y p,A
A
15.1196 4 1.6116 1.205 y IN
yp
1 PA A ˆ A
t ms
1
PA ˆ A p r y r
pp yp
.365295
.7
pr yr
Fˆ p y p,A t ms
.3
0.365295 .25
0.2006
.7
pp yp
ˆ . Since F IN
1 mol s, Fˆp
FˆIN
0.3 mol s
mol 0.365295 1.2 10 4 cm s 3 cm STP cm 1.0 L STP 1 2 L STP 10 3 cm 3 STP cm s cm Hg 22.4 mol
0.3 A 2700 10
10
76 cm Hg atm A
A
c)
Fˆ p y p,A t ms PA ˆ A p r y r
pp y2
, Fˆ p
Fˆ IN
mol s
0.325 1.2 10 4 cm
9.0823 104 cm 2
17.D11. New problem in 3rd edition.
xp
Gelling occurs at a solvent flux of J solv
J solv
Then x gel
solv
atm
0.4 mol s
1 3 cm STP cm 22.4L STP 1.0 L STP 76 cm Hg 2700 1010 2 cm s cm Hg mol 10 3 cm 3STP atm
A
J solv
.5 .365295
6.569 104 cm 2
0.4 A
2.0 .2006
5200
L 2
m day
x out exp (J solv /
997
g
0 since R 0
x IN 1
0.001 .6
0.5 0.325
atm
0.0016667
5200 L/(m 2 day ) which is
day
L 86400 s
solv ) k
1.0, x out
2.0 0.175
60.0
g m2s
0.0016667 exp
17.D12. New problem in 3rd edition p p / p r
1.0 4.5 .
60.00g / (m 2 s) 997000g / m 3 2.89 10 5 m / s
He H 2
PHe PH 2
0.01334
0.261. Use Eq. (17-6b),
430
y p ,He
p
1
He H 2
1 yp
pr
y r ,He
1
1 yp
Eq. (17-7c)
.261
y FHe
y rHE
.2 .254
y p He
y rHe
.1 .254
a)
If x p
x in , cut
.55 , perfectly mixed
(17-27) 1 R 0.55, what value R required. Find R (including concentration
0.00050 &
polarization effect). From Eq (17-27), x p
xp
0.25446
1 R xn
xp
R x IN
1
0.353
17.D13. New problem in 3rd edition .035 NaCl Rejection
1 1 .1 4.5 .261 1 .1
0.1 .261 1
x IN
xp → R
x IN
xp
x IN
x IN
x IN
0.035 0.0005
xp
xp
Rx IN which gives
0.035
.55 .0005
0.035 0.001
0.9935
b)
If
c)
If R° (inherent rejection coefficient with M = 1) for part b is R 0.992, what was value of M that gave R 0.9869 MA 1 M CaseB . R Case B 1 1 R CaseA . Let A be highly stirred RA R M CaseA
xp
0.0016,
M CaseB
17.D14. wB
0.55 , R
xpR
RT curve: y w
Feed
xp
0.035
M CaseA 1 R CaseB
1.0 1 0.9869
1 R CaseA
1 0.992
wB
1
xw
wB
43 (mole frac). Since x w ,IN
B
x IN
141.6
cal
0.9869
1.6375
43 x w
1 xw
1 42 x w
0.10, only need RT curve below 0.10. Create table and plot
xw
yw
0.10 0.08 0.05 0.03 0.01 0.0025 0.001
0.8269 0.78299 0.6935 0.571 0.3028 0.0973 0.0413
kcal
g 1000 cal
0.9 10.5
.55 .001
74.12 g mol
0.1 9.72
10.5 kcal mol
10.42 kcal mol
431
C PB
0.625
1 1000
0.046 , C PW
74.12
CP,in MW feed
xW , F MWW
a) Assume y P and
C PL,in
0.9 0.046
xB , F MWB
1 1000
0.1 0.018
Tout where Tin
0.5 10.5 Tout
30
then,
x in
1
0.018
0.0435 kcal mol C
0.5 9.72
P
molar ratio. Slope Op line
18.016
0.1(18.016) 0.9(74.12)
0.5 to calculate λp
Tin
1.0
68.51
10.11 0.0435 10.11
30
0.129. This is a
6.75 , and op line goes through point (mole fractions)
0.10
0.775. Plot operating line. From graph, y P 0.57, x out 0.129 (mole fraction water). This value of yp is reasonably close to our assumption. x out
0, y P
Fp / FF
( Fp MW p ) / ( FF MWF )
0.031
( MW p / MWF )
432
MWp
y p ,W MWW
y p , B MWB = 0.57(18.016) + 0.43 (74.12) = 42.13
.129(42.13 / 68.51) Area
b)
Cut
Permeate Rate
0.08
x in,w
Then
P
Tin
c)
Tout
Flux
( 0.0791 100 lb h)
0.2 lb h ft 2
39.53 ft 2
0.10
1 cut
Slope
0.0791 in (lb/h)/(lb/h).
0.92
11.5. Find y P 0.68 from graph. cut 0.08 0.32 10.5 0.68 9.72 9.97
C PL,in
x out
P
0.05, x P FP Fin
P
30
0.08 0.0435kcal / (mol o C)
9.97 kcal mol
48.3 C
0.6935 (From RT table or graph).
x in
x out
0.10 0.05
yP
x out
0.6935 0.05
0.3065 10.5
0.6935 9.72
0.0777
9.959
433
Tin
Tout
C PL,in
17.D15. Parts b to h are new in 3rd edition
xP
0, x r,out
Fout
Fin
Fin
FP
Fin Fout 1
0.0777 0.0435
RT curve:
xP
Fin x r ,in
FP x P
Mass balance perfectly mixed Since
30 C
47.8 C
1 R M xr
0.
Fout x r ,out
x r,in FP
0.8 . Then
Fin
Fin Fin Fout 1 x r ,out 0.10 0.125 , Fout 0.8Fin 0.8 Alternate graphical solution gives same result. 6.5
xP
9.959
1 0.8
80 kg h
Op. line
xP
-4
xr
x r,in 0.10 0.125
xr
x r,out
xP
1
RT curve
FP Fin FP Fin
x r ,out
Slope
x r ,in FP Fin
0.8
4 0.2 When x r,out 0, x P
x r,in
0.10
FP Fin
0.2
b. Area = Fp / Jsolv = (20kg/h)(1 L)/0.997 kg)(24h/day)/ (2500 L/m2 day) = 0.193 m2. c. Gel formation occurs when x w = 0.5 and xw = M xout = 0.125 M. M = 0.5/0.125 = 4.0 d. Gel formation occurs when xw = 0.5 and xw = M xout = M xF / (1 – θ’) = 1.2 (0.1)/(1 – θ’) Then 1 – θ’ = 0.12/0.5 = 0.24 and θ’ = 0.76. e. Gel formation occurs when x w = 0.5 and xw = M xout = 1.2 xF / (1 – θ’) = 1.2 xF / (1 – 0.2). Obtain xF = 0.333. xr,out = xF / (1 – θ’) = 0.3333/0.8 = 0.416 f. We have xr,out = xF /(1 – θ’) = 0.20/0.75 = 0.26667. M = xgel / xr,out = 0.5/0.26667 = 1.875. First occurs when Jsolv = 2500 = k ln (M). Obtain k = 3977 L/(m2 day) = 4.603×10-5 m/s. g. M = 1.875 and k = 3977. Since we change the pressures, J changes which will change M. However with constant stirring k is constant. First, assume no gel and calculate J and M. pr pp K Jsolv 2500 L (m 2 kg) L J solv K solv , solv 2083.33 2 t ms t ms p r pp 2.2 1.0 bar m day bar Then, without gel,
J solv
2083.33 3.4 1.0
5000 L m 2 d ay
From Eq. (17-34) M = exp (Jsolv / k) = 3.516.
434
0.5
0.2. Then, x w
3.516 0.2
Mx F 1
0.878, gel forms.
.8
0.2
xF
With a gel, previous work is incorrect. Set R = 1.0, x p = 0, x r
0.25, 1 .8 And from Eq. (17-45), Jsolv = k ln (xgel/xr) = 3977 ln (0.5/0.25) = 2756.6 L/(m2 day) Note: The same answer is obtained in parts g and h if convert to J´solv and use k in m/s. h.
J solv
k n
x gel
.5
k n
xr
Case C, R C
xp
J solv
MB
x IN
1 R C x out
K solv
pr
tms
1.387 152 1.1
m d ay
pr
pp
1 RB
pr
pp
a M C xout
Equation:
log P
log P
0.27027
2446.6 L (m 2 d ay)
0.27027
1.0, p p,B
1.0 .061 14.1 .024 10.96
B
0.0007505]
1.1, p r,B
12.06
3.27
b
18.71 g m 2s
42 , P 10000,
1.6232 0.27875
1.9
4a b 4a b
1.9020 2b b 0.95100 , a 0.16805 logP 0.951
log
.74
xp
b
a log 10, 000
8, log
C
0.0001,
a log 0.0001
log 1.9
0.5
n
0.2
0.01230
15.446[ 3.27 .01230
log 42
0.2, x r
15.2
1 RC
pp
0.26, x F
0.976, M B
1.1, p r,C
17.D17. New Problem in 3rd edition. a) log a log P b , P
b) If
L 2
0.0093 0.26 .74 1 .939 .26 0.0007505
1
1 RC
3977,
Case B, R 0B
.939, p p,C
For Experiment C. M C
x out
3977
xr
17.D16. a) New problem in 3rd edition
b)
k
Gel forms since it did previously,
.16805
0.90309
log 0.16805
0.28509
.951 0.16805
PO2
1.93 Barrers
435
17.D.18. New problem in 3rd edition. Ideal Gas: Vol% = Mole %
PN2
250, PCO2 FP
2700,
PHe
300, PHe 3
0.4 m3 s. FIN
FIN
550
1.0 m s
10 2 cm
10 3
10 6 cm 3 s , FP
3
m 76 cm Hg
Part A
pr
2.5 atm
Part B
pr
76 cm Hg
t ms
0.8 mil
Eq. (17-11d)
10
190 cm Hg
atm
pp
0.4 76
.00254 cm
FP A
30.4 cm Hg
Need to guess value of FP A or of y P ,
Pi t ms p p
Since CO 2 has highest permeability, CO 2 will be concentrated; thus, guess y p,CO2
y r ,CO 2
Then
y p,CO 2 ,guess
1
FP A guess
Then,
PCO 2
y IN ,CO 2 1
t ms y p,CO 2
76 cm Hg
0.002032 cm
mil
Pi t ms p r
K m,i
pp
.4 106 cm3 s
yCO2,IN
0.4 and y IN,CO2
where p r y r.CO 2
0.40 0.4.
p p y p,CO 2
Use FP A in Eq. (17-11d) to calculate all K mi Then check
y IN,i
y r,i
1
K mi 1
1.0 ?
Put in Spread Sheet. Can use Goal Seek to force Results: a. y P,N2
y r,N2 b.
.15037, y P,CO2
.3164, y r,CO2
.54446, yP,He
.3037, y r,He
y r,i
1.0 as change y P,CO2 .
.3351, yP,H 2
.06099, y r,H2
.3189
.27154,
.999885
1.0000766
Same answers for mole fractions since p r p p is same.
yF N2 F,cm3/s pr, cm Hg P N2 P He Fp yp CO2 Fp/A K N2 K CO2
306 2008
HW 8
Problem 2
0.25 1000000 190 0.000000025 0.00000003 400000 change yp to 0.544455884 0.003983763
yF CO2 tm, cm pp, cm Hg P CO2 P H2
yF 0.4 He 0.002032 theta 76 0.00000027 0.000000055
get sum=1 yr CO2 A, cm 2 A, m 2 0.475237299 y r N2 1.792765611 yr CO2
0.05 yF H2 0.4
0.3
0.303696077 100407575.7 10040.75757 0.316417678 0.303696077
436
K He K H2
0.549397235 yr He 0.851323363 yr H2 Goal seek Sum 0.150373483 0.544455884 0.033509684 0.271546029 0.99988508
yp N2 yp CO2 yp He yp H2 sum
0.060993544 0.318969314 1.000076613
17.E1. New problem in 3rd edition For dilute systems J solvent Transfer Eq. (17-7c)
R
1
xp x out
1
J total solution , FIN
Fp
xF
x out
0.022 0.056
FIN
xp
x out
0.00032 0.056
0.00032
0.9943 , J s u c rose
0.056
J solventx p
Fout , Basis: FIN
Fp
1.0
0.6106
J solvent
x
solution
p
Permeate
0.997 0.4 x p 0.997 0.4 0.00032 0.99713 kg L Initial assumption is OK. J solv J solv / solv (3.923 g / m 2 s) / (997 g / L) 0.003935 L / ( m 2 s) permate
(3.923g / m 2 s)(0.00032)
J sucrose b)
Eq. (17-27)
0.00126g / (m 2s)
1 xp
K water
water solute
K sucrose
x p pr
Mxr
pp
xp
Mxr
xp a
1 .00032 0.056 0.00032 w s
0.00032 60.0 1.1
From Eq. (17-16c),
K water pr
Eq. (17-18) c.
pp
a Mxr
59.895 1.0 .056
K sucrose
M xr
R
1 M 1 R
Then RT equation is
xp
1 R x out
Operating Equation is (17-23)
xp
0.0706 g
0.00126 0.056 0.00032
xp
Solution 1.
xp
0.00032
J sucrose
tms
g sucrose
xp
3.923 60 1.1
g water
J solv
tms
K water t ms
0.056 0.00032 59.895
3.131
1
0.6393 x out
1
2.1 1 0.9943
m 2 s atm
0.0226
g sucrose m 2s wt frac
0.98803
0.01197 x out x out
xF
.39
0.036066
Solve RT & operating equation simultaneously. x out
.61
x out
0.05537 , x p
0.022 0.61
0.000663
437
Check
R
pr
t ms
60 1.1
pp
a M xr
K sucrose
Mx r
t ms
3.67g/(m 2s)
0.000663]
J water x p / (1 x p )
0.00243 g (m2 s)
3.67 0.000663 / (1 .000663)
Alternatively, J sucrose
xp
59.895[ 2.1 0.05537
J sucrose
J sucrose
0K
0.98803
x out
K water
J water = 0.0706
xp
1
xp
0.0226[ 2.1 (0.05537) 0.000663]
0.00261
g m 2s
6.9% different
xp xp
Solution 2. RT Eq. (17-21), x r
2.1 1 xr
Simplifies to, Linearize x r @ x p Slope =
3.140 59.895 0.997
0.0003
xr
0.02509
3.140 59.895 .997
1 xp
0.02509 . Note xp = 0.0003 is an arbitrary point.
0.01196
0.01196 x r or x p Solve simultaneouly with Operating Equation xF 1 xp x out 0.6393 x out 0.036066
0.05538 , x p
1
2.1 391.66 x p
Then linear form of RT equation is x p
x out
3.140 .997 60 1.1
x p 186.5 x p 185.391
0.0003
xp
1
0.01196 x out
0.0006623 . Very close to value obtained with retention analysis.
C PL,in
FP
17.E2. (was 16.D11 in 2nd ed.) Eq. (17-59b):
Fin
Tin
Tout , Tin
Tout
85 25
60
P
Stage 1. Assume y P ~ .05 water, 0.95 ethanol P
0.95
For Feed CPL,in
w
0.05
E
0.1 CPL,w 55 C 0.1 4.1915
2290.3 kJ kg (See Example 17-9)
0.9 CPL,E 0.9 2.7595
kJ kg K
2.903
kJ kg K
Where average temperature from 25 to 85 is 55ºC and C P values are from Perry’s 7th, pp. 2306 and pp. 2-237.
FP1
2.903
Fin
2290.3
60
0.0760 and FP1
0.0760 100
7.60 kg hr .
438
Op. line intersects y P
x out
x IN 1
Slope Op. Eq. is, y P
0.10 (water wt. frac.) 1 0.760
0.1
12.16 x out
12.16
0.0760
12.16 x out 1.32 0.0760 If y P 1, x out 0.32 12.16 0.0263 Plot Op. Line on Figure 16-17a and find intersection: y P1 0.66, x out1 x 2,in 0.055 (water values)
Fout1
Fin 2
100 7.60
92.40 kg h
Stage 1 Trial 2. Since yp ≠ yp, assumed, do a second trial.
yP
0.66 water, 0.34 E, FP1
2.903 60
F1in
1892
FP1
0.0921 100
w
0.34
9.21 kg h , Fout1 9.86 , y P
0.0921 0.086
If y P
0.66
0.66 2359
E
0.34 985
1892 kJ kg
0.0921
1 0.0921
Slope
P
100 9.21 91.79 kg h 0.1
9.86 x out
0.0921
9.86 x out
1.086
1, x out
0.00872 9.86 Plot operating line and determine (from graph)
y P1 0.64, x out1 x in 2 value of yp is close to the assumed value of 0.66. Can proceed to stage 2. Stage 2: Estimate y P For x in 2 2
0.50 (water),
0.050, CPL,in
FP2
2.967
Fin 2
1672
Fout 2
60
0.05 4.1915
0.50
.1065 90.66
0.050
1 0.1065 0.1065
1672
E
2.967 9.64 kg h
8.40
8.40 x out 0.4421 0.1065 0.4421 . Draw op. line. Intersection gives y P 0.34
For x out 2 use MB. x out 2 ,
8.40 x out
w
0.95 2.903
8.10 kg hr , Slope
90.66 9.64
0, y P
0.50
0.1065, Fin 2
yP
If x out
P
0.05 water. This
Fin 2 x in 2 Fout 2
FP2 y P2
90.66 0.050
9.64 0.34
81.00
x out 2 ,w 0.0155 or x out,2 ,ETOH 0.9845. This is a close as we can get graphically. 9.21 0.64 9.64 0.34 Mixed Permeate: y p,mix 0.487 wt frac water 9.21 9.64
439
Area
FP
kg 1000 g h kg
J g h m2
, J from Fig. 16-17b based on x out
Stage 1
J
0.8333 g h m 2 , A1
Stage 2
J
0.208 g h m 2 , A 2
9.34 1000 g h 0.8333 9.64 1000
11, 208 m2
46, 346 m2
0.208 Other flow patterns will reduce area. Area is large because of low flux caused by low ethanol permeation rate.
17.F1.
RT eqn., y
x x 1
1 x
x Benz
0.2,
18.3, y
x Benz
0.1,
6.66, y
Operating equation Slope
1
, x Benz
0.3,
18.3 .2 1 17.3 .2
6.66 .1 1 5.66 .1 .9 .1
16.6, y
16.6 .3 1 15.6 .3
0.87676
0.8266
0.4253 . Plot RT equation.
9 . Plot on graph. Find y PBenz ~ 0.844, x out,Benz ~ 0.238
440
P
CPL,in
y Pbenz
benz
1 y Pbenz
x Pbenxw CPLbenz
iP
0.844 94.27
1 x benz,in CPL ,iP
1 0.844 164
0.3 0.423
.7 0.73
105.15 cal g
0.6379 cal g C
441
Tin
Tout
P
C PL,in
50 C
0.1 105.15
66.48 C
0.6379
17.H1. (was 16.G1 in 2nd edition) This is set up for Area being the unknown and cut being known. Problem 17.H1 Fr,in 10000.000000 yin,A 0.2500 cut=Fp/Fin 0.2500 tmem,cm 0.002540 pr,cm Hg 300.0000 pp,cm Hg 30.0000 yin,B 0.5500 P,A 0.0000000200 Fptot 2500.000000 yin,C 0.2000 P,B 0.0000000050 Fr,out 7500.000000 P,C 0.0000000025 Guess values of A or equivalently Fp/A until sum y,r and sum u,p are = 1.00 Fp/A 0.0007059 (this is final result) KA 2.507328 KB 0.7720 KC 0.4015 sum x eq y,r,A 0.181576 y,p,A 0.455271198 y,r,B 0.583244 y,p,B 0.450268647 y,r,C 0.235190 y,p,C 0.094429331 Area, cm2 3541578.1 sum y,r 1.000010 sum y,p 9.999692E-01 These results agree very well with Geankoplis’ results. 17.H.2. New problem in 3rd edition Part a) y p 0.5243, y r,out 0.0610, A b)
y p,avg
0.6193, y r,out
3, 200,152 cm 2
0.0203, A
2, 636,196 cm 2
17.H3. New problem in 3rd edition Counter –current. Shows final guess for theta. Fin, cm3/s 100000 yin 0.209 thetatot PA/tms 0.003905 pr, cm Hg 114 pp, cm Hg M 15 N 100 yroutguess df 0.9 j=N-i+1 Fr yp yr Area Fp Fp/Fr,j-1 yp Areatot yincalc Fincalc Massbal yrout
100 28600 0.173174301 0.144051968 9710.750234 714 0.024356963 0.235015561 824015.8215 0.208999973 100000 9.09495E-13 0.144051968
99 98 29314 30028 0.1738615 0.174547 0.1447613 0.14547 9664.4075 9618.879 1428 2142 0.0475556 0.069677
97 30742 0.1752323 0.1461768 9574.1438 2856 0.0907935
0.714 PB/tms 76 0.2 erroracc
0.00175 0.0000001
96 95 94 31456 32170 32884 0.175916 0.1765981 0.17727918 0.146883 0.147588 0.14829195 9530.183 9486.978 9444.50995 3570 4284 4998 0.110973 0.1302761 0.14875885
442
17.H4. New problem in 3rd edition The spread sheet equations are shown below for part b. Part a agreed with problem 17.D14. Part b answers: yp,W = 0.412, θ = 0.2122, xout,W = 0.0160, θ’= 0.158, Area = 79.0 ft2. Note that if the starting guess for yp,W is too high, Goal Seek will converge on an answer with yp,W > 1, which is obviously not physically possible.
17.H5.
This problem is very similar to Example 17.7. It is easiest to solve on a spreadsheet, which is
shown below. The results are shown in the spreadsheet. New problem in 3rd edition
443
444
17.H6. New problem in 3rd edition The spreadsheet is similar to that for problem 17.H5 and is shown below,
17.H.7. The same spread sheet that was used in problems 17.H5 is used.
445
446
SPE 3rd Edition Solution Manual Chapter 18. New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 18.A3, 18.A16, 18.B4, 18.C4, 18.C14, 18.D3, 18.D8, 18D9, 18D14, 18D15, 18D18, 18D21, 18D24, 18D25, 18.D29, 18.D30, 18.F1, 18.H1-18.H2. Chapter 18 was chapter 17 in the 2nd edition. Most problems from that edition have the same problem number, but the chapter number is now 18 (e.g., problem 17.D6 is now 18.D6). 18.A1. 1c; 2 b; 3a 18.A.2. 1c; 2a; 3b 18.A.3. New problem in 3rd edition. One barrier is lack of knowledge. Most chemical engineers are not trained in use of adsorption, chromatography, and ion exchange. Thus, they do not think of these processes as a potential solution. A second barrier is the simulation tools are not as developed and widely available as the simulation tools for equilibrium staged separations such as distillation. 18.A4. Regeneration is too difficult. 18.A5. In the SMB the mass transfer zone between the two solutes stays inside the cascade. In a chromatograph the MTZ exits the column and must either be completely separated, which requires a significant amount of desorbent, or recycled appropriately. 18.A7. d 18.A.8. New problem in 3rd edition. The LUB approach assumes constant pattern behavior. Linear systems do not have constant pattern behavior. 18.A9. 18.A10.
d e
18.B.4. New problem in 3rd edition. There are obviously many possibilities. One is to develop sorption processes that use an energy separation processes (e.g., pressure or temperature) to produce purge or desorbent from the feed so that a separate purge or desorbent does not have to be added. 18.C1.
T
1
e
Vavailable P
B
18.C4.
rd
e
1
e
1 f cry
1
e
1
P1
clay
P
e
e
e
P4
1
P1
f
f cry
P2
1 f cry f cry
1 cry
1
e P2
f cry
P2 P1
K di Vcol. 1 f cry
f
f cry
f
P2
(same as 18-3b)
New problem in 3 edition. Amount in mobile phase = e (Vol. Col. Segment) Amount in pores = 0 (no pores) Amount exchanged Δz A c cRT Δy K DE No 1 Obtain, u ion
e
c
e
Δz a c Δx cT
term because c RT is equivalent/L
ε e Δz A c Δx c T v int er
ε e Δz A c c T Δx Δz A c c RT Δy K DE v int er Simplify to, u ion (18-44) c RT y 1 K DE x ecT
445
18.C7.
CA
1
C AF
2
z uAt
1 erf
4E eff t u A v inter
12
Sketch of break through:
erf (a)
.9 a
1.164
95%
t final
5%
tw
t st
t1
erf (a)
At 5% point,
0.90 a 1.164
1.164
L 4E
t st u A v
12
Or let
u A t st
2.328 E t st x1
12 st
t , uAx
2 1
uA
12
x2
By definition, Use
u A t final uA v
12
12
L u A t st
v 2.328 E
12
u 1A2 v1 2
x1
2.328
L 2
0.
E u At v
4Lu A
2u A 2.328
Let
L
4E t final
E 1 2 u 1A2 2.328 v1 2
x1
1.164
and at 95%,
12 fin
t , then x 2 t MTZ
t final
t st
x 22
E 1 2 u 1A2 v1 2
2.328 2 E u A v 2u A
4L u A
x12
sign for both (has to be to have positive times).
446
4u
2 A
4u
2 A
x
2 2
x
2 1
2
2.328 E 1 2 u 1A2 v
2.328E 1 2 u 1A2 v
2
12
2
v
v
t MTZ
x
2 2
x
2 1
E uA
4L u A
2.328
12
2
4L u A
v 2
E uA
v
E uA
v
E uA
2.328
2
2.328
v
2.328E 1 2 u 1A2 v1 2
4
2
12
2.328E 1 2 u 1A2
2
12
2.328
2.328 E 1 2 u 1A2
2.328
2
E uA
v
4L u A
4L u A
4L u A
4u 2A 2
If
4L
2.328 E v t MTZ
2 2.328 E
12
v1 2 u A
18C9. New problem in 3rd edition. For Figure 18-7B, In
L
In – Out = Accumulation t vinter A c CT,after x i,after
Out
Accumulation
very reasonable since E is usually small ,
t vinter A cCT,before x i,before
LA c yi,after
yi,before CRT
LAc x i,after CT,after
Note that C RT is constant. After dividing both sides by
v int er x i ,afterC T ,after x i ,beforeC L
C RT y i ,after
y i ,before
x i,before CT,before
t A c , mass balance is
T ,before
L
x i ,afterC T ,after x i ,beforeC T ,before t t For Figure 18-7B with a total ion wave, L u total ion v int er t The first and third terms in the mass balance cancel each other. Thus, L C RT y i,after y i,before 0 t Which requires, yi,after yi,before 18.C10.
447
1 A
v1
2
v2
v2
F
v3
3
u A ,i
C A ,i L i
u A1
M 1A u port
vF
u BL
C B,i Vi
u B2
M 2 B u port
v B,prod
u A3
M 3A u port
vD
u B4
M 4 B u port
v3 v4
v4
B 4
v A ,prod
v1
CB v2
M 2B u port
CA v2
C A v3
M 3A u port
CA v 2
vF
M 3A u port CA M 2B
If all
18.D1.
Rearrange:
vD
(2)
M 3A u port
CB
M 2B CB
v4
M 3A CA
v1 where v 4
u B4 CB
v1 Thus
(1)
M 2B u port
vF
u port
vD
CB
CA vF
Subtract eq. (2) from (1),
Then
CA
u A1 CA
M 4B u port CB M1A u port CA
M 4B u port
M 1A u port
M 4B
M 1A
CB
CA
CB
CA
M 1A CA
M 4B CB
M 3A CA
D
vD
F
vF
Mi
1.0,
M 4B CB
D F
pA
1
qA
q MAX
1 CB
pA
1 CA
1 q MAX K A
1 CB
1 CA
M 2B CB
vf
M 3A CA
1.0
. Plot p A q A vs. p A
448
296 K p/q 135.863 278.679 478.666 696.073 939.619 1116.143 1189.735
p 275.788 1137.645 2413.145 3757.6116 5239.9722 6274.1772 6687.8589
At 296 K Intercept Slope
KA
0.163636
1 q max
, q max
1 q max
0.163636
1 q max K A
80
80
480 K p/q 1786.943 1709.129 1974.657 2309.538 2778.150 3011.134 3122.979
At 480 K
1
Intercept
q max K A
6.1125
0.00204545
p 637.7598 1296.2036 2378.6716 3709.3486 5329.6030 6246.5981 6687.8589
Slope
KA
1380
0.260606
1
q max K A 1 , q max q max
1 q max
0.260606
1 q max K A
1380
3.8372
0.00018884
449
18.D2.
L soln
a = 22 liter soln/kg ads = 22
1 kg
kg ads 1000 g
b = 375 liter soln/g mole anthracene = 375
q max K A,c C A
q
18.D.3. New problem in 3rd edition.
uj
Part a
1
1
e
p
a
0.022
K A,C
2.104
1
Kd
uj
b
2.104
u
c.
u AN
u DN
e
HETP
L g anth
v super
40
5.671 cm min
L u S,DN
4.576
5.567
u S,AN
L N
g ads
K ij
1782 0.00301
4Ru
N1 2
From (18-83)
2
L
g ads.
40 1 0.69
2.104
g anth
Time AN = L/uAN = 25/5.671 = 4.408 min 40 u DN 5.474, time 1 0.69 1782 0.00316
b.
104.33
u S,AN
N 10885
0.002297 cm 2
From (18-81),
2
width at half height 5.54 peak max N To find width in time units, peak max is in time units = retention L u S,AN 4.40864 min , width 0.09946 min
d. time
0.425 width 1/2 height
t
18.D4.
p
e
1 0.69 1782K j
1 mol
0.10456
s
, u AN
g ads
, vinter
1
e
L
mol 178.22 g
v inter
e
40
L
, thus, K A,C
1 K Ac C A
q max
0.022
v sup er
10.0
v int er
0.43
e
23.26
0.042271 min
cm min
v int er
a) u s
1
1
e
p e
us 0 1
Kd
1
1
e e
(18-15c) p
s
Kx T
f
23.26 0.6027 cm min 0.57 .48 1.0 0.57 0.52 2100 17.46 0.43 0.43 684 t br 200 cm 0.6027 cm min 331.8 min
450
b) Assume wall heat capacity is small: v int er u th 1 e 1 1 e 1 p e
u th 1.636 t th,br c) K x
t br M.B.in
10.0
e
23.26 .57 .52 2100 2000
C ps
F
C pf
5.911
cm min
23.26
1.23 g g @80 C , u s 80
200 cm 5.4868 cm min
A c cm 2
331.8 min 0.684 g cm 3
10.0 A c 0.684
331.8 33.84 .0011
Alternate: Eq. (18-24)
C 80 C 0
0.0011 g tol g fluid
33.84 0.0011
2.611 Cconc → C conc us 0 1 u s 80
0.0011 113.92
5.4868 cm min
1.636 2.603 36.45 min , see figure.
1
C 80
s
.43 684 1841 200 cm 5.911 cm min 33.84 min
out Simplifying:
p
1 u th 1 u th
36.45 33.84 C conc
297.96 .0011
1 0.6027 1 5.4868
2.611 1 5.911 1 5.911
0.1255 wt frac.
113.92
0.1253 wt frac . A very considerable amount of concentration occurs.
451
80º, C = 0 z usol (80ºC)
uth
0.0011
C=0 0.0011 331.8 min
0.0011
t Cconc
33.84 36.45 min 0.1255 Cout 0.0011 0 33.84 18.D5.
36.45
vsuper 20 cm min vint er vsuper e 20 0.4 50 cm min For step input w. unfavorable isotherm, get a diffuse wave. v int er Langmuir formula: u s 1 e 1 p 1 e a 1 Kd p s 1 bc e e But now
us 1
b .6 1.01 .54 .4
0.46
0 50 .6 .46 1.124 kg 1.2 .4 liter 1 0.46 c
c,g/l us, cm/min 18.2437 0 16.676 0.25 14.794 0.50 12.565 0.75 9.997 1.00 7.1813 1.25 1.50 4.3499
2
1.81
time, min
2
50 0.93067 1 0.46 c
2
tout = L/us, min 2.741 min 2.998 3.3797 3.979 5.002 6.9625 11.4944
452
18.D6. a)
f
u th 1
1
e
p
f
e
u th
.57 1 .5 .43
1
e
p
e
C ps
.684 2240
.57 .5 920 1.80 .43
50 cm 12.61 cm min v int er
1
e
p
e
c)
1
C pf
If wall effects are negligible, 0.684 2240 30
b) t thermal,br
u s 300K
C p f v int er
Kd
1
1
e e
p
K xy
s
W C pw eAc
12.61 cm min
3.965 min 30 .57 .57 1 .5 1.0 .5 12.109 .43 .43
u s 350K
6.5298 K xy
t br 300K
50 3.0964 min . Exits at c F
3.0964 cm min
4.423 in same eqn.
0.010 .
453
At t = 20, start hot, t br,hot Feed is concentrated. C 350 C 300
50 12.61 20
1 u s 300
1 u th
C 350
0.010 3.2989
t
1 u s 350
20 L u s 350 K
1 u th
23.965 min
1 3.0964
1 12.61
1 6.5298
1 12.61
3.2989
0.032989 g L . This continues until breakthrough at
20 50 6.5298
27.6572 minutes
0.032989
18.D6.
g/L
0.010
0 t
18.D7.
vint er
vsuper
16.1478
15 0.434
e
1
1
e
p
Kd
1
tr
L us
(A) p s
K 4
e
34.56 0.566 0.43
0.566 0.57 1.0
0.434 60 cm 0.3715 cm min
161.49 minutes. Then exits at C F
1
e
e
1
27.6572
34.56 cm min v int er
a) At 4ºC: u s
us 4 C
23.965
0.3715 cm min
1820 0.08943 0.434 161.49 min . Concentration out is zero from t = 0 to t =
0.01 .
454
v int er
b) u th
1.
1
e
p
1
1
e
e
e
C ps
s
C pf
f
WC pw e
A c C pf
f
34.56 17.293 cm min , 0.566 0.43 0.25 1820 0.434 1.00 1000 60.0 17.293 3.4696 min +1200
u th 1.743
t br,th
p
L u th
Eq. A but with K(60º)
u s 60 C
34.56 0.720258 cm min 0.566 0.43 1.743 1820 0.045305 0.434 t br,conc 60 L u s 60 60.0 0.720258 83.3035 min +1200 C=0 60º 60
60º
z
0º 60º C=0
uth Elution time: 0
c high 60
chigh
cF
CF
1 us 4
3.4696
1 u th
0.01 2.6918 0.05783
83.3
C high
1 u s 60
1 u th
1.38839 0.05783
18.D.8. New problem in 3rd edition. Example 18-3: vinter,F
0.3799 cm min , u s v inter,purge,0 C
u th vinter,purge,
6.466 cm min , u s vinter,purge,80 C
If t purge
t purge
0.019796 kmol m3
18.60 and yinter,purge
u s vinter,F, 0 C
u s v inter,F, 80 C
C=0
25.58 18.60
25.58 cm min .
0.3799
0.5225 cm min
4.343 cm min
18.60
4.343 3.158 cm min 25.58 hot purge time and t F is cold feed time, with
t hot wave breakthrough
vinter,purge
18.56 min (from Example 18-3) then breakthrough
equation is u s v Inter,F, 0 C t F u s v Inter,purge,0 C t
thermal,breakthro ugh
120 cm
455
120
tF
0.5225 18.56
290.35 min 0.3799 The next feed input at 290.35 + 18.56 = 308.91 min. This starts a cold thermal wave at v Inter,F , u th v Inter,F 4.701 cm min which breaks through in another 25.53 min for total time to cold breakthrough of 308.91 + 25.53 = 334.44 min. The solute is hot, first at v Inter,purge u s 80 , v Inter,purge
18.60
u s 80 ,v inter,F
3.158 cm min after 18.56 minutes. Next solute step is
4.343 25.58 u s vinter,purge, 80
t Exit Time Solute
18.56
u s,F 80 , vinter,F
120 4.343 18.56 tF
4.343 cm min and then
t purge
t
3.158
120
12.47 min
290.35 18.56 12.47
334.44
Since Exit Time Solute entire time.
t
321.38 min.
breakthrough cold wave, the solute is at 80°C the
Solute exits from 290.35+18.56=308.91 min to 321.38 minutes = 12.47 minutes & it exits at superficial velocity of 8.0 cm min . Mass Balance All solute in = Solute out t F vsuper A c c IN t out vsuper A c c out,AVG
tF
c out ,AVG
t out
290.35
c IN
0.0009 wt frac
12.47
0.02096 wt frac.
This is same as peak concentration in Example 18-3, but greater than x out,avg 0.00748. To have same concentrations need to recycle the material exiting at feed concentration in counterflow system. NOTE: Counter flow system has advantages of not contaminating the product end of the column and typically has less spreading of the zone. 18.D.9. New problem in 3rd edition. a.
u s,feed,M
e
vinter
vSuper e
e
1
e
p
Kd
0.05 m s. vinter
v inter
1
e
1
p
s
RTK A,p
0.5 0.43
0.1163 m s
0.01712 m s
from Eq. (18-27) is same as Example 18-4, M 0.2128 Pressurization Step Feed end (for pressurization) 0.75m (Measured from closed product end) M
z after which is 0.75 0.5584
y M after
0.003
Feed Step u sfeed
0.01712 m s
0.75 m
4.0 atm
0.2128
1.0 atm 0.1916 m from feed and
4.0
0.5584 m
0.2128 1
1.0 7 sec
0.001007 0.11984 m
0.1916 m for pressurization step
= 0.3114 m. Does not breakthrough in first cycle. From 0 to 0.11984 m, concentration is y F .
456
Blowdown. Measuring from closed top, z before
z after
0.4386 m
0.2128
1.0
0.5890 m 4.0 The far end of the feed wave does not get removed from the bed. 0.11984 or 0.75 0.11984 0.6302 m from closed end has z after
1.0
0.6302
0.4386
0.75 0.3114
0.2128
0.8463 m, so it all exits. The mole fraction of this portion is
4.0
y after
y feed
The close wave
1.0
0.2128 1.0
0.003 2.9781 0.00893 4.0 Part of the feed that was pressurized also exits during blowdown. 1.0
0.2128
1.3431 z before z before 0.5584 m from closed 4.0 (product) end. This is 0.75 0.5584 0.196 m from feed end. This gas entered at an unknown pressure between p L 1.0 and p H 4.0. Can calculate this pressure from Eq. (18-28c) This z after
0.75
p before
z before
p after
z after
1
z before
y after,press
0.003
y after,BD
0.001007
0.5584
4.0
4.0
1 0.2128
1.00003 atm
0.75
0.2128 1
0.001007 1.00003 This gas is depressurized to 1.0 & exits column
Exit from Col y
1
After Pressurization Step.
0.2128 1
0.00300 or essentially the feed composition.
4
0.008933 0 .0030 time
Part b. Want z after blowdown
0.75, then z before
z after
p after p before
0.2128
1.0 z before 0.75 0.5584 4.0 from closed end, which is 0.75 0.5584 0.1916 m from feed end. Want the feed to end at this point. During constant pressure feed step, feed travels u s,feed t F 0.01712 t F . Then for pressurization step z after (from feed end)
0.1916 0.1712 t F . From closed end this is
457
0.5584 0.1712 t F 0.5584 0.1712 t F
z after
z before
p before
0.2128
4
0.5584 or t F 0. 1 Thus, need a purge step if have feed step at constant pressure for complete cleanout. 18.D10. a)
pt.10 : z after
y after
0.4,
0.002
Travels,
0.75
p after
0.2128, p before
A
3.0
0.4 0.5
0.000876
1.015128 0.4 m
25.126s 1.0s for blow-down
0.01592 m s
3.0
.48
1 .2128
1.05128 atm
0.2128 1
3.0
b) Start with Arbitrary point at t = 1 sec (end repress) z after
p before
1 0.2128
2.4763 atm , y after
0.002
26.126s
0.48 (.02 from feed end) 3.0
0.2128 1
0.00172
.5 2.4763 Dist. Traveled @ t = 30s: 0.02 + 0.01592 × 29s = 0.48168 m For blow-down: distance from closed end = 0.01832 cm
z after
0.01832
Purge: u M,purge
0.5
0.5
.2128 1.0
0.026824 , y after ,BD 0.00172 0.007048 3.0 3.0 0.01751 m s . Exits bottom column during purge (point 11)
(distance traveled)/upurge 18.D11.
.2128
31 s +
0.5-0.026824 0.01751
58.023s
If repressurize with product, bed remains clean. Feed step is same as to point 3 (at 0.462 m from feed end) on Figure 18-13. Blowdown then pt. 4 (0.056 m from top) and purge exits at pt. 8 (56.36s) Product gas is cleaner (y = 0), but there is lower productivity – less feed per cycle. See Figure.
458
BD 3
4
y=0
y=0
y = 0.0082
8 18.D12. a) The clean bed receiving feed has a shockwave for Langmuir isotherm.
320 cm 3 min
v sup er
vsup er
r2
A c , where A c
6.366 cm min , vinter
vsup er
4 cm
2
50.2654 cm 2
6.366 10.434 14.669 cm min
e
v int er
u sh 1
1
e
p
Kd
1
p s
e
e
q
q after
q before
c
c after
c before
c after
1
e
where c before
50 mol m 3 , q after
q c
0, q before
0.190 50 1 0.146 50
0
1.1446 mol kg
14.669
u sh 1
0.566 0.57 1.0
0.566 0.43
0.434
0.434
t br
L u sh
1.1446
1820
50 cm 0.5843 cm min
0.5843 cm min
50
85.579 min
Outlet concentration is zero until t br then becomes 50. Concentrated solution eluted by dilute soln. Gives diffuse wave for Langmuir isotherm. v u s u diffuse 1 e 1 p 1 e a 1 K p d p 2 1 bc e e
us 1
0.566 0.57 1.0
14.669 0.566 0.43 1820
0.19
0.434
0.434
1 0.146 c
2
1.74336
14.669 193.92 1 0.146c
2
Create Table.
459
18.D13.
A
c
50, u s
3.218, t
L us
50 3.218 15.537 min
c
0, u s
0.07497 cm min , t
c
40, u s
2.491, t
20.071 min
c
30, u s
1.737, t
28.779 min
c 15, u s
0.7052, t
70.898
c
0.2205, t
226.80
5, u s
de xtran, B
L us
666.93 min
fructose (1)
CA v1
u A1
M1u port
CB v 2
u B2
M 2 u port (2)
C A v3
u A3
M 3u port
CB v 4
u B4
M 4 u port (4)
(3)
v F,sup er 1000 cm 3 min , vF 2 40 e 4 CB vF Solve eqs. (2) and (3) simultaneously, u port CB M2 M3 CA v2
CA
v3
1 1
1
e e
v2
v3 v4
KA
1 .6 1 0.23 .4
0.7435 , C B
M1 CA M2 CB
u port
u port
0.97 0.7435 0.99 0.4914
3.03175 cm min
60 3.03175
M 4 u port
1.03 3.03175
CB
0.4914
KB
0.4914
19.791 min
6.1079 cm min ; V2,sup er
0.7435
1 .6 1 .69 .4
L t sw
3.03175
CA
V1,sup er
1
e
3.955 cm min : V1,sup er
1.01 3.03175
Recycle flow
1 1
3.03175
M 3 u port
1.9894 cm 3 min
e
0.4914 1.9894 cm min 0.4914 0.99 1.01 0.7435 L t sw u port
u port
v1
Vol Feed D2 4
v F , v F,super
v1
D2 e
4
1988.176 cm 3 min
3070.15 cm 3 min
4.1184; V3,sup er
2070.14 cm 3 min
6.3547; V4,sup er
3194.19 cm 3 min
1988.176 cm3 min
460
D
V4,sup er
Check:
3194.19 1988.176 1206.0 cm3 min ,
V1,sup er
VD
D
VF
F
V4
V1
VF
F V2
Extract Product
V4
M 1 u port
CB
CA
vF
M2
V3
M1 CB CA CB M3 CA
.97 .4914 .7435 1.2060 , OK .4914 .99 1.01 .7435 V1 3070.15 1988.18 1081.97 cm 3 min 3194.19 2070.14 1124.05 cm3 min
18.D.14. New problem in 3rd edition. From Eq. (18-40c) K K
KK
Anderson’s data:
1.2060
F
1.03 .4914
D
Raffinate Product
M4
M4 CB
D
2.9 1.3
H
H
KK
Li
KH
Li
2.2308
DeChow’s data: K K H 2.63 1.26 2.0873 For the shockwave Eq. (18-46) holds for K+ Since resin is initially in H+ form, x K,before CK,before CT 0 and y K,before
a)
x K,after
CK,after CT
y K,after
CR ,K,after CRT
CR ,K CRT
0.
1.0 1.0 v inter
u sh ,K
y K ,after 1 C RT K DE ,K x K ,after e CT
1
25 0.42
u sh ,K
y K ,before x K ,before
L
44.84 min 1 2.2 1 0 u sh 1 1.0 0.42 0.1 1 0 Same for both sets of data since K K H does not enter into equation when initial and feed contain only one ion. b) C t 1.0, u Sh,K 9.542 cm min , t sh 5.24 min c)
Ct
1.0, x K,before
yK Anderson’s Data: y K ,before
y K ,after
0.2, x K,after
0.85. y K values depend on equilibrium parameter.
K KH x K 1
K KH 1 x K 2.2308 0.2
1
2.2308 1 0.2
2.2308 0.85 1
1.115 cm min, t sh
2.2308 1 0.85
0.3580 0.9267
461
u sh
25 0.42 1 2.2 1 0.42 1.0
1.0
0.9267 0.3580 0.85 0.2
10.662 , t K
L u sh
4.69 min
DeChow’s data:
y K ,before y K ,after
2.0873 0.2 1
0.3148
2.0873 1 0.2 2.0873 0.85
1
0.9220
2.0873 1 0.85
25 0.42 L u sh 4.95 min 10.100 , t K 1 2.2 0.9220 0.3148 1 1.0 0.42 1.0 0.85 0.2 4.69 4.95 % difference 100 5.55% 4.69 d) There is a difference if either initial or feed contains both ions. System with higher K K H had higher shock velocity. u sh
v
18.D15. New problem in 3rd edition. Part a. u sh ,i
y i,after 1 c RT K DE x i,after e cT
1 For both
Na & K ,
x i,after
y i,before y i,after
t center
0 1.0
v
u i,sh 1
y i,after 1 c RT K DE x i,after e cT
y i,before x i,before
For both Na+ and K+: xbefore = 0.4 and xafter = 0.9. For Na+ K Na H xNa (2.0 / 1.3)(0.4) y Na ,before 1 ( K Na H 1) xNa 1 [(2.0 / 1.3) 1](0.4)
y Na , after
x i,before
25 0.42 5.186 cm min 1 2.2 1 0.42 0.5 L u sh 50 5.186 9.64 min
Thus same u sh , u sh
Part b.
x i,before
y i,before
K Na
H
1 ( K Na
H
xNa 1) xNa
(2.0 / 1.3)(0.9) 1 [(2.0 / 1.3) 1](0.9)
0.506
0.933
462
v
u sh ,Na
y Na ,after 1 c RT K DE x Na ,after e cT
1 t Na
L / ush , Na
50 / 5.98
(25 / 0.42) 1(2.2)(1.0) 0.933 0.506 1 (0.42)(0.5) 0.9 0.4
y Na ,before x Na ,before
5.98
8.36 min .
For K+ we obtain,
y K ,before
y K , after
KK
H
1 (KK
H
KK
H
1 (KK
H
u sh ,K
Part c.
(2.9 / 1.3)(0.4)
1) xK
L / ush , K
xK
(2.9 / 1.3)(0.9)
1) xK v
50 / 7.054
(25 / 0.42) 1(2.2)(1.0) 0.953 0.598 1 (0.42)(0.5) 0.9 0.4
y K ,before x K ,before
7.054
7.09 min .
1 c RT dy K DE dx e cT dy Na
K Na
dx Na
1
dx Na
t Na
K Na
1
L u shNA
KK 1
1
0
xK
KK
K Na x Na
dy K
.9
Li
KH
Li
KH
Li
Li
2
1 x Na
25 0.42 1 2.2 1 0.955 .42 0.5
u Na
5.409
1 xK
H
xK
2
1
KK
0.855, u K
Li
Li
KH
KH
Li
Li 2
1 xK
25 0.42 1 2.2 1 0.855 .42 0.5
5.979
H
2.0 1.3 1.538,
u Na
3.477,
t Na
14.38 min
0
KK xK
2
0
dy Na
dx K
K Na
KK
H
2.9 1.3 1 .5
dx Na
1
0.955,
2
2.9 1.3
dy dx
2
9.244 min
dx K
x Na
1 x Na
H
2.0 1.3 1 .5
dy K
x Na
K Na
H
2.0 1.3
dy Na
Part e.
0.953
1 [(2.9 / 1.3) 1](0.9)
v
u 1
Part d.
0.598
1 [(2.9 / 1.3) 1](0.4)
y K ,after 1 c RT K DE x K ,after e cT
1 tK
xK
H
2.9 1.3
2.231,
uK
2.442,
tK
20.47 min
0
.9
463
dy Na
1.538
dx Na
1
dy K
0.538 .9
2.231
dx K
1
0.502,
2
1.231 .9
0.698, u
2
7.159, t Na
Na
uK
6.984 min tK
9.5075,
5.259 min
Part f. The velocities and hence the derivatives are equal. Thus, K Na
dy Na dx Na
1
K Na
KH
Li
KH
Li
Li
2
1 x Na
KK
dy K
Li
dx K
1
KK
Li
Li
KH
KH
Li
Li
1 xK
2
With xNa = xK. The result from a spreadsheet is x = 0.35056 18.D16.
vint er
vsup er
15 0.40
e
MW p f
37.5 cm s
28.9 g mol 50 kPa
1.0 kg
0.5832 kg m 3
3
m kPa 1000 g 298 K mol K q kg toluene kg carbon . Then, shockwave velocity is is in c kg toluene kg air v int er RT
u sh
0.008314
1
1
e
p
1
Kd
1
e
e
e
q y
s f
37.5 cm s
u sh 1
0.6 0.65 1.0
0.6 0.35
0.4
0.4
1500 kg m 3 q 2 0.5832 kg m 3 y 2
q1 y1
37.5 cm s
u sh
For
1.975 1350.308
q2 y2
u sh ,1 : y1
0, y 2
0, q1
u sh 2 : y1 y2 u sh 2
q1 y1
0.0005, q 2
37.5
u sh ,1
At
p
2000 0.0015 1 2200 0.0015
L min
0.47619 0.104976 cm h
0.69767
37.5 0.69767 0.47619 1.975 1350.308 0.0015 0.0005
L min : u sh1t
1 2200 0.0005 0.00002916 cm s
0.47619 1.975 1350.308 0.0005 0.0005, q1 0.47619
0.0015, q 2
2000 .0005
0.00012539 cm s
0.451393 cm h
u sh 2 t 10 h where t is in hours.
464
Solve for
u sh 2 10
t
0.451393 10
13.03 h 0.451393 0.104976 cm L min u sh1 t 0.104976 13.03 h 1.368 cm h Thus, for any column of partial length we will see a single shockwave exit the column. v sup er 21.0 18.D17. v int er 52.5 cm s 0.4 e pV n RT u sh 2
v
Cinit
Since
u sh1
MW n
MW p
28.9 50
V
RT
1000 g kg 0.008314 298
C F , Get 2 diffuse waves v int er
us 1
1
e
p
Kd
1
1 y 0.0010 0.00075 0.00050 2nd wave (0.00050) 0.00025 0.00
1
e
e
us
p
s
e
f
52.5 0.6 0.35
.6 0.65 1.0 0.4 0.4 q u s cm s y 195.31 0.0001991 284.799 0.0001365 453.515 0.00008573 - add 20 hours 832.466 0.00004671 2000 0.00001914
us y
0.5383 kg m 3
0.001
q y
where
q
2000
y
1 2200y
2
52.5 cm s q 1.975+1350.23 q y y t L u s 25 u s
1500 0.583
125,581s = 34.8835 h 183,117.6s = 50.866 h 291,596.6s = 80.999 h 100.999 h 535,250.5 = 148.681 + 20 = 168.681 h 1285937.96 = 357.205 + 20 = 377.205 h
0.00075
0.00025
z
us y
us y 0
0
0.005 t
2
us y
0.0005
465
80.999
34.88 0.001
50.866
0 0.00075 0.0005
c
·
100.99
· 168.88
0.00025 t
18.D.18. Part a.
. New problem in 3rd edition. u S,G 11.12 S cm min is calculated in Example 18-9.
20 0.61 1.0 0.88
u S,F 1 0
8.416 , u
From Eq. (18-93), N
4Ru u S,G
From Eq. (18-78a)
N
Part b.
1
u S,F
2
L v E eff
u S,F
2
9.771
2
229.465
L
2 229.465 5.0 cm 2 min
2N E eff v
114.73 cm
20 cm min
tG
L u S,G
114.73 11.25 10.20 min
tF
L u S,F
114.73 8.416 13.63 min
Part c. Eq. (18-80a),
K Ag
Li
t
L
1
uS
N
13.63 min
t ,F
K AgK
uS,G
0.39
L
18.D19.
377.2
KK
a) Ion wave: u total con
Li
vint er
8.5 2.9
vsuper
1/ 2
,
10.20 min
t ,G
229.465
1/ 2
0.673 min
1/ 2
1 229.465
2.93 , y Ag e
1
3.0 0.4
Breakthrough of ion wave, 50 cm 7.5 cm min
0.900 min 2.93 x Ag 1 1.93 x Ag 7.5 cm min
6.667 min
466
b) Shock wave,
v int er
u sh
y Ag after 1 C RT KE x Ag after e CT
1
before: x Ag
7.5 cm min 1 2.0 1.0 1.0
u sh 1
0.4
1.2
u s,Ag
x Ag
x Ag
0.5, u s
x Ag
0, u s
From spreadsheet:
1.0 .
y Ag
1.4516 cm min , t sh v int er 1 C RT dy KE dx e CT
1
7.5 cm min 1 2.0 2.93 1 1.0 0.4 1.2 1 1.93 x
1.0, u s
1
7.5 12.208 2.93
50 cm
u sh
1.4516 cm min
1 C RT KE e CT 1
1
34.44 min
1
2 Ag
3.0965
2
cm min
u_dif,Ag 3.097163211 2.852615804 2.598940969 2.337763116 2.071284133 1.802351071 1.53450084 1.27196632 1.019637556 0.782936124 0.567638906
2.0 eq L , c T
0.02 eq L , x Ca
K CaK C RT
0.6183 2.0
CT
0.02 75 cm, vsuper
61.83
K Ag -K
K Ag
1 x Ag
K
2
7.5 12.208 1 1.93 x Ag
L
, t out
us
7.5 50 1.8021 , t out 12.208 1.8021 1 3.86 7.5 L 0.5678 cm min , t out 13.208 us xAg 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Column: L
L
v int er
2
16.147 min
27.745 min
50 0.5678
88.0555 min
t_dif,Ag 16.14380534 17.5277722 19.23860549 21.38796684 24.13961427 27.74154314 32.58388572 39.3091764 49.0370324 63.86217015 88.08416667
Was 18D23 in 2nd edition. Table 18-5, K CaK a.) c RT
x Ag before
1.0
c) Diffuse wave: u s
18.D20.
0 . after: x Ag
y Ag
y Ag before
K Ca
Li
K K Li
0.8 at t
5.2 2
2.9
2
0.6183
0.
shockwave .
20 cm min ,
vinter
20 .4
50 cm min
467
p
Feed:
u sh
0,
0.4, K E
e
v int er , before: x Ca C RT K E y 1 x e CT
from Eq. (18-43) y Ca
u sh
1.0
50 1 2.0 0.971965 1 1.0 0.4 .02 0.8
x Ca
K Ca C RT CT
0.8
0.971965 75
0.16407 cm min , t br
b.) Regenerate: at 500 min → Ion wave at vint er New
0 , after: x Ca
0, y Ca
u sh
50 cm min takes
75 50
457.1 min
1.5 min.
1.2366 y Ca unchanged. Use Eq. (18-43) with new value K Ca C RT CT .
0.9689677 . Obtain diffuse wave. 3
v imter dy Ca 1.2336 1 y Ca 1 x Ca u diffuse where 3 C RT K E dy Ca dx Ca 1 x Ca 1 y Ca 1 C T e dx Ca (Wankat, 1990, Eq. (9-25b)). dy Ca 50 At x Ca 0, 1.2366, u Ca 6.96088 cm min 1.0 2.0 dx Ca 1 1.2366 0.4 1.0 75 cm t out 10.7745 min (slow wave) 6.96088 At x Ca 0.96897, and y Ca 0.971965
dy Ca
1.2336 0.028035
dx Ca
0.03103
3
3
1.96897
0.908386
1.971965
50 9.022 cm min , t out 2.0 1.0 1 0.908386 1.0 0.4 At x Ca 0.5, y Ca 0.534927
75
u Ca
dy Ca
1.2336 0.465073
dx Ca
u Ca
.5
3
18.D.21. New problem in 3rd edition. vF u port , C Tol M2 M3
C Tol
1.5
8.312898 min (fast wave)
0.97013
1.534927
50 2.0 1.0 1 0.97013 1.0 0.4
Cy
3
9.022
75
8.546 cm min , t out
8.546
8.776 min . (in-between)
1 1
1
e e
p
Kd
1
1
e
0.132234 p
K Tol 300
e
468
0.0061 e 2175.2696 300
K Tol 300 K C xy
1
.95 .10007
vF
0.6479 v F 0.6479 cm min , L u port t SW 64.79 cm 1.05 .132234 0.95 .6479 C xy 6.1447 cm min , v3 v 2 v F 5.14476 0.10017 0.6479 C Tol .95 4.6547 0.132234
v2
M 2 u port
v1
M1 u port
v4
M 4 u port C xy
1.05 0.6479
v Tol prod
v2
v1
6.7914 0.10017 6.1447 4.6547 1.4900
v xy,prod
v4
v3
6.7914 5.14476 1.6466
vD Check: vOut 18.D22.
0.10017
1.63627 0.68930 12.1092
u port
E eff
v sup er us
v4
v1
6.7914 4.6547
v tol prod
ED
v xy,prod
u s2
dp 6 k M,c
20 cm 3 min
1 0
v 1
0.0105 e 2115.1052 300 12.1092
8.5972 , K xy 300 K
cm
2
3.1366 , v total in
K
2
1
K
2.035
D F
2.1367
vD
3.1367
vF
, where
6.366 cm min , 15.915
e
2.1367,
dp
1
6 k m,c
k m,c a p
v sup er
v
OK.
15.915 cm min
7.821 cm min
e
E eff
0.15
Eq. (17-69) X
cm 7.821 min
2
0.6 0.69
5.52 min
C
1
CF
2
Argument of erf, a
1
0.4
2
8.063 cm 2 min
z ust
1 erf
4 E eff u s t v int er
12
for step up
200 7.821 t 15.849 t
12
469
Step down:
1
X L, t 8
L u s t 8u s
1 erf
2
12
4 E eff u s t 8 v int er
323.04 7.821 t
Argument of erf , a
15.849 t
Total Solution X X L If t 25.573 min us
X If t
X
1
1
and X
2 L 8u s
2
2
1.998
4 X
31.2835
a
341.89
12
1 2
0.979235 0.0084 cF See also Problem 18.G1.
cF
1 2
0.499 c
1.44473, erf a
12
1 2
50 0.970835
Cinitial
1 .983186
1
e
.998 24.975 .
0.95847
p
Kd
1
CF1 X z, t 17.5
e
1
e
CF1X z, t 28
1.63627 0.68930 12.1092
0.132234 p
K Tol 300
e
8.5972 , K xy 300 K 1
0.0084
48.54
1
0.0061 e 2175.2696 300
C xy (300K)
0.999
0.983186 , X
Cinitial
Tol
K Tol 300 K
c
2.773, erf
0.979236
0.970835 , c
18.D23. Was 18D24 in 2nd edition. Cout
12
532.13
468.663
1.95847
25.0 (for smaller t, can ignore X )
63.96
31.2765
29.575, a
1.69189, erf a
18.D.24. New problem in 3rd edition. vF u port , C Tol (300K ) M2 M3 1 C C
0.50 or c
1 2, a
0, X
c
xy
0 , c cF
7.358
12
279.6
For higher t, X = 1.0,
0.999
Peak at 25.575
12
123.03
0, a
1 1.000
33.575, a
us 1
a
126.792
0.0105 e 2115.1052 300 12.1092
0.10017
470
u port
.90 .10007
vF
1.4812 v F
1.10 .132234
M 2 u port C xy (300K)
v1
M1 u port C Tol (300K)
v4
v2
v1
vD
v4
Check: vOut
v4
v1
13.3084 cm min , v3 0.10017 1.4812 .90 10.0812 0.132234
13.3084 10.0812
v3
v tol prod
v xy,prod
v2
vF
12.3084
16.2655
0.10017 16.2655 12.3084
16.2655 10.0812
148.12 cm
3.2272
1.10 1.4812
M 4 u port C xy (300) v xy,prod
u port t SW
0.90 1.4812
v2
vTol prod
1.4812 cm min , L
3.9571
6.1843,
7.1843 , v total in
vF
D F vD
6.1843 7.1843
OK.
18.D.25. New problem in 3rd edition. Zones 2 & 3 are same as in 18.D.24 since at 300 K u port 0.6479 v F 0.6479 cm min , L u port t SW 64.79 cm
v2
6.1447 , v3
v1
M1 u port CTol 273 K
M1
5.14476
0.5 and M 4
and v 4
M 4 u port C xy 350 K
2.0 (reciprocal values).
K Tol 273K
0.0061 exp 2175.2695 273
17.612
K xy 350K
0.0105 exp 2115.1052 350
4.423
1
C Tol 273K
1
C xy 350K
v1
0.2135
1.63627 0.68930 4.423
0.5 0.6479 0.07259
vTol prod
0.07259
1.63627 0.68930 17.612
v2
4.4627 , v 4
1.6820 , v xy prod
v1
v4
v3
2.0 0.6479 0.2135 0.9260 , v D
v4
v1
6.0707 1.608
D / F 1.608
18.D26. a)
N
u sD
2
4Ru s u sA 1.0 1 5.8
u s,B
, R
0.147059 , u
v
1.5 , u s A 1
1
1.0 s
KA
6.5
0.15385
0.15045
471
4 1.5 0.15045
Need N b) t R ,A
t ,A
CA C A ,max
17689 , L
0.0067873
L
884.45
uA
0.15385
L
1
uA
N
5748.88
95.813 min
t
tR
2
2 t
2
exp
0.05 N
884.45 cm.
95.813 min
12
exp
t,min CA CA,max
12
1
0.7204 min
17689
t 95.813 2 0.7204
2
p
2
90 92 94 95 95.813 96 97 7.27E-15 8.3E-7 0.0421 0.52898 1.00 0.9669 0.2573
CA
0.33 X A L, t
CF
18.D27.
18.D28. a) u p
2
25.0 cm
L t center
35.4 min
b) Large-Scale system
1.0 0.33 X A L, t
1 .55 X A L, t .8t F
0.55 0 X A L, t
0.706 cm min , L MTZ,lab
u pt MTZ
t MTZ, LS
d 2p ,LS D eff
1.0
t MTZ,lab
d 2p ,lab D eff
0.12
t MTZ,LS
Independent of velocity
0.4t F
69.44 2.8
tF
0.706 2.8
1.9774 cm
2
69.44
2
194.44 min
v super u p ,LS
e
u p ,lab
LS
v super e
lab
12
4
9
3
→ u p,Ls
0.706
4 3
0.941 cm min
lab
L MTZ,larg e scale u p t MTZ 0.941 cm min 194.44 min 183.03 cm For frac. bed use = 0.80 & symmetrical pattern, 0.5 183.03 0.5 L MTZ L 457.6 cm 4.576 m , t br t center 1 Frac bed use 1 .8 t center
457.6
L up
486.27 min , t br
486.27
194.44
0.941 2 This is length of feed time if column is completely regenerated. 18D.29. K CaK a.) c RT
K Ca
Li
K K Li
5.2 2
2.9
2.5 eq L , cT
2
t MTZ 2
389.05 min .
0.6183
0.03 eq L , x Ca
0.7 at t
0.
472
K CaK C RT
0.6183 2.5
CT
0.03
Column: L p
Feed:
u sh
90 cm, vsuper
0,
e
shockwave .
51.525
25 cm min ,
0.39, K E
y Ca (1 y Ca )
25 / .39
64.10 cm min
1.0
v int er , before: x Ca C RT K E y 1 x e CT
from Equilibrium,
vinter
0, y Ca
0 , after: x Ca
K Ca K C RT
x Ca
CT
(1 x Ca ) 2
2
0.7
400.75
Solve this for unknown y value. I used a spreadsheet. yCa 0.95128
u sh
64.1 1 2.5 0.95128 1 1.0 0.39 .03 0.7
b.) Regenerate: Ion wave at vint er New
K CaK C RT
(0.6183)(2.5)
CT
1.1
0.22000 cm min , t br
35.0 / 0.39 1.4057 y Ca
old y and with new value K Ca C RT CT 400.75 , find x Ca
90
409.10 min
u sh
89.74 cm min takes
90
1.003 min.
89.74
0.95128 unchanged. Use equilibrium with y Ca (1 y Ca )
2
K Ca K C RT
x Ca
CT
(1 x Ca ) 2
to
0.94251 . Obtain diffuse wave. 3
v imter dy Ca 1.4057 1 y Ca 1 x Ca u diffuse where 3 C RT K E dy Ca dx Ca 1 x Ca 1 y Ca 1 C T e dx Ca (Wankat, 1990, Eq. (9-25b)). dy Ca 89.74 At x Ca 0, 1.4057, u Ca 9.7631 cm min 1.0 2.5 dx Ca 1 1.4057 0.39 1.1 As an alternative can do numerical calculation of derivative. At x = 0, y = 0. x = 0.001, y = 0.001404 and y / x (0.001404 0) / (0.001 0) 1.404 , which is reasonably close.
90 cm
9.22 min (slow wave) 9.7631 At x Ca 0.94251, and y Ca 0.95128 t out
dy Ca
1.4057 0.04872
dx Ca
0.05749
3
3
1.94251
1.95128
0.85169
473
89.74 15.049 cm min , t out 2.5 1.0 1 0.85169 1.1 0.39 From equilibrium, at the arbitrary value x Ca 0.5, y Ca u Ca
dy Ca
1.4057 1 0.55544
dx Ca
3
.5
3
1.5
15.049
5.981min (fast wave)
0.55544
0.95282
1.55544
89.74 2.5 1.0 1 0.95282 1.1 0.39
u Ca
90
13.695 cm min , t out
90 13.695
6.572 min .
This is in-between the other two waves. c. To not have a diffuse wave must have
K CaK C RT
(0.6183)(2.5)
CT
CT
1.0
This requires CT > 1.546. 18.D30. New Problem in 3rd edition. K K
H
KK
Li
KH
Li
DeChow’s data: K K H 2.63 1.26 2.0873 a.) This will be a shock wave since K+ is more concentrated in the feed to the column than it is initially and KK-H > 1. v inter
u sh ,K
y K ,after 1 C RT K E ,K x K ,after e CT
1
Ct
1.0, x K,before
yK
y K ,before y K ,after u sh
tK
0.2, x K,after
1
0.85. y K values depend on equilibrium parameter.
K KH 1 x K
2.0873 1 0.2 2.0873 0.85
1
x K ,before
K KH x K
2.0873 0.2 1
y K ,before
2.0873 1 0.85
0.3148 0.9220
25 0.42 0.9220 0.3148 1.0 0.85 0.2
1 2.2 1 0.42 1.0 L u sh 49.5 min All three times are the same for the shock wave.
10.100 cm/min,
474
b.) This will be a diffuse wave since K+ is less concentrated in the feed to the column than it is initially and KK-H > 1.
v inter
u diffuse,K
1
dy K 1 C RT K E,K dx K e CT
dy K
K KH
At xK = 0.15, dxK
u diffuse,K
25 / 0.42 2.2(1) dy K 1 (0.42)(1.0) dx K
(1 ( K KH
2.0873 1) xK )
2
25 / 0.42 2.2(1) dy K 1 (0.42)(1.0) dx K
1.543
[1 (1.0873)(0.15)]2 59.524 dy 1 5.238 K dx K
59.524 1 5.238(1.543)
6.554cm / min
Thus, at xK = 0.15, tK = L/udiffuse,K = 500/6.554 = 76.29 min. Then at xK = 0.5 we obtain
dy K
K KH
2.0873
dxK
(1 ( K KH 1) xK ) 59.524 u diffuse,K dy 1 5.238 K dx K
2
0.876 [1 (1.0873)(0.5)]2 59.524 10.65cm / min 1 5.238(0.876)
Thus, at xK = 0.5, tK = L/udiffuse,K = 500/10.65 = 46.94 min. Then at xK = 0.8 we obtain
dy K dxK u diffuse,K
K KH (1 ( K KH
59.524 dy 1 5.238 K dx K
1) xK ) 2
0.5970
59.524 1 5.238(0.597)
14.42cm / min
tK = L/udiffuse,K = 500/14.42 = 34.67 min. 18.D31. New problem in 3rd edition. a.
vSuper
10
vint er
10 .4
25,
e
0.4,
L
30.0
475
c RT
2.4,
1.10,
KK
2.9 Na
1
y K ,after 1 c RT K DE x K ,after e cT
b.
u sh,exp t
c.
L MTZ
L t center
u sh t MTZ
L MTZ l arg e scale
1 2.4 1.0 .4 1.1
1
x K ,before
1 0 1 0
7.75 min . , t center ,measured
7.31 min
7.31 7.75
100 6.00% 7.31 30 7.31 4.10 cm min .
4.10 7.57 7.06
Frac. bed use (symmetric wave) d.
25
y K ,before
3.783 cm min, t center,exp ected % error
1.45
2.0
v int er
u sh ,K
u sh,K
cT
L MTZLab
2.093 cm
1 0.5 L MTZ L
d 2p v Super D eff
16 d 2p
l arg e scale
d 2p v Super D eff
0.965 d 2p
Lab
Lab
Lab
200 D eff
L MTZ,Lab
100 D eff
With same beads assume no change in D eff .
L MTZ,larg e scale
16 2 2.093 cm
frac bed use
1 0.5 L MTZ L 1 0.5 66.98 200 t center
Breakthrough start time
v inter,large scale
u sh
u sh ,lab ,exp tl
v inter,lab scale
Breakthrough start time
0.5 t MTZ
L u sh
2580 ft 3 , h=2580/860=3ft.,
End View
0.833
0.5 L MTZ u sh
8.2
[200 0.5(66.98)] / 8.20
18.F1. New problem in 3rd edition. Constraints: w L
wLh
66.98 cm
860 ft 2
p0
20.31min
T max
6 atm
500 C
932 F
88.14 psia
73.14 psig
Weight vessel
Di
ts
L
0.8 D i t s
Seider etal, s
Eq.16 59
h
w Seider etal. (2004), Eq. (16.61)
476
Pd
exp
0.60608 0.91615
Wall thickness
tp
(Eq. 16.60)
s
Relate w to h.:
n p0
96.66 D i
2SE 1.2 Pd
2 13,100 1
2
96.66 psig
3.7057E 3 D i
1.2 96.66
490 lbm ft 3 0.284 lbm in 3 p. 529 (Seider et al, 2004)
cos
cos
90 D
Pd D i
0.0015655
S 13.100 psi p. 529 with SA-387B steel, E = 1.0
where
Weight
n p0
3.7057E 3 D
0.5h
1.5
3
r 0.5w
r w
D w
r
D
D
cos 90
860 D cos 90
cos
D in ft. (In Spreadsheet A cos
cos
1
1
cos
3 D
cos
0.8D
3D
3
1
D
3.7057 E 3 D 490
1
In Spreadsheet, angle is in radius 90 2 D Weight Width L 3.5 33655 1.80 477.0 3.7 31358 2.17 397.1 3.8 30735.9 2.33 368.7 3.9 30325. 2.49 345.1 4 30070.31 2.65 325 4.1 29933 2.7947 307.7 4.2 29889 2.939 292.6 4.3 29918 3.08 279.2 6 35103 5.196 165 8 44837 7.42 115.96 10 56197 9.54 90.15 12 68940 11.62 74.02 14 83144 13.67 62.88 Goal seek L = 60 D = 14.64 ft Weight = 88052.75 Width = 14.333 L = 60 From Seider et al, p. 527: Cp
Cv
(Eq. 16.53) horizontal (Eq. 16.55)
CpL
From p. 531, Fm
Fm C v
Bare module factor, FBm
CBm
0.20294
2
118,323 → Cp
1.0 in 2000
Cp
144, 711 in mid 2000
118323 2724 121047
3.05 for horizontal
Cp Fm
Absorbent: p. 553 Cp
0.04333 n w
2724
1.2 for low-alloy steel, C v
Installed Cost: Calc C p with Fm
0.054266 ft
CPL in mid 2000 (MS = 1103)
exp 8.717 0.2330 n w
1580 D
ts
1.0, 2000
$60 ft 3 , Cp
3.05
1.0 1.0 1.2
60 2580
1
$393, 400
$154,800
477
18.G1. Was 17G1 in 2nd edition. Figures are labeled 17G1.
478
479
480
18.G2.
Was 17G2 in 2nd edition. a.) With QDS with 50 nodes find t center
t MTZ 18.G3.
6.0 3.13
4.52 min 2.87 min
Was 17G3 in 2nd edition.
Find
D F 1.0. D 141.55 E R CA 0.343 and CB 0.219
Eq. (17.31a)
u port
a)
M 2B CB
M 3A CA 141.55 cm 3 min
F
vF e
u port t sw
v1,int er
v1,sup er
Dc
2
4
0.4
10
2
4
4.5057 cm min
4.5057 1 0.219 L u port
CA vint er
Recycle Rate
vF
F.
u A1
2.7295 cm min 1 0.343 50 2.7295 18.32 min
M1u port
2.7295 cm min
2.7295
7.9577 cm min 0.343 0.4 v1,int er 3.18308 cm min
3.18308
10
2
4
250 cm 3 min
Obtained raffinate = 96.6% and extract = 94.3%. b) One approach is to keep a symmetric cycle. Then D = 283.1 and E F E R 212.325 2 Flow optimizer can be used to give t sw ~ 9.1 and Recycle rate ~ 500. Depending on values obtain raffinate and extract > 97%. 18.G4.
Was 17G4 in 2nd edition. Figure below is labeled 17G4.
18G5.
Was 17G5 in 2nd edition. Figure below is labeled 17G5.
481
482
483
18.G6.
Was 17G6 in 2nd edition. a.
k m,a p
1.5 min1 , L
25.0 cm
484
v sup er
20.2 ml min
Eq. (18-66)
2.0 m 2 4
6.366 cm min 6.366 cm min
19.1 1.5 min -1 Satisfied, but close. Thus some bypasses but most undergoes equilibration. 18.G7.
25.0 cm < 4.5
Was 17G7 in 2nd edition. Figure is labeled 17.G7.
485
18H1. New problem in 3rd edition. Spreadsheets with numbers and formulas shown.
486
487
18.H.2. New problem in 3rd edition. The spreadsheets are shown on the next pages. They are based on the previous, but includes both a step up and a step down. Because of the quirk in Excel not allowing negative arguments, it was set up with multiple solution paths. The correct solution occurs when there are numbers. Time, min
15
20
22.5
25
C
0
.0134
1.798
24.96
25.5726 27.5 25.0
30
33.575
42.32 48.52 24.97
35
37.5
11.13 1.114
40 .040
488
489
490
