Separation Process Engineering Includes Mass Transfer Analysis 4th Edition Wankat SOLUTIONS MANUAL Full download: https://testbanklive.com/download/separation-process-engineeringincludes-mass-transfer-analysis-4th-edition-wankat-solutions-manual/ SPE 4th Edition Solution Manual Chapter 2. New Problems and new solutions are listed as new immediately after the solution number. These new problems are:2A8, 2A10 parts c-e, 2A11,2A12, 2A13, 2A14, 2C4, 2D1-part g, 2D3, 2D6, 2D7, 2D11, 2D13, 2D14, 2D20, 2D22, 2D23, 2D31, 2D32, 2E3, 2F4, 2G2, 2G3, 2H1, 2H3, 2H4, 2H5 and 2H6. 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 cpLIQ 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 TF and z on the graph because equilibrium data is at a lower pressure on the graph used for this calculation.
2.A2.
Yes.
2.A3.
The liquid is superheated when the pressure drops, and the energy comes from the amount of superheat.
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. 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.
New Problem in 4th ed.
18
. a. At 100oC and a pressure of 200 kPa what is the K value of n-hexane? 0.29 b. As the pressure increases, the K value a. increases, b. decreases, c. stays constant b c. Within a homologous series such as light hydrocarbons as the molecular weight increases, the K value (at constant pressure and temperature) a. increases, b. decreases, c. stays constant b d. At what pressure does pure propane boil at a temperature of -30oC? 160 kPa
19
2.A9.
a. The answer is 3.5 to 3.6 b. The answer is 36ºC c. This part is new in 4th ed. 102oC
2.A10. Parts c, d, and e are new in 4th ed. a. 0.22; b. No; c. From y-x plot for Methanol x = 0.65, yM = 0.85; thus, yW = 0.15. d. KM = 0.579/0.2 = 2.895, KW = (1 – 0.579)/(1 – 0.2) = 0.52625. e. αM-W = KM/KW = 2.895/0.52625 = 5.501. 2.A11. New problem in 4th edition. Because of the presence of air this is not a binary system. Also, it is not at equilibrium. 2.A12. New problem in 4th edition. The entire system design includes extensive variables and intensive variables necessary to solve mass and energy balances. Gibbs phase rule refers only to the intensive variables needed to set equilibrium conditions. 2A13. New problem in 4th edition. Although V is an extensive variable, V/F is an intensive variable and thus satisfies Gibbs phase rule. 2A14. New problem in 4th edition. 1.0 kg/cm2 = 0.980665 bar = 0.96784 atm. Source: http://www.unit-conversion.info/pressure.html 2.B1.
Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium. Examples:
F, z, Tdrum , Pdrum F, z, y, Pdrum
F, TF , z, p F, TF , z, y
F, z, x, p drum F, z, y, pdrum F, z, x, Tdrum
F, TF , z, x etc. F, TF , z, Tdrum , p drum F, TF , y, p F, TF , y, Tdrum F, TF , x, p F, TF , x, Tdrum F, TF , y, x
Drum dimensions, z, Fdrum , p drum Drum dimensions, z, y, p drum etc.
2.B2.
F, h F , z, p F, h F , z, y
This is essentially the same problem (disguised) as problem 2-D1c and e but with an existing (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 which corresponds to 2-D1c. If F 1000
lb mole , D .98 and L hr
2.95 ft from Problem 2-D1e .
Since D α V and for constant V/F, V α F, we have D α With F = 25,000:
F.
Fnew Fold = 5, D new = 5 Dold = 4.90, and Lnew = 3 D new = 14.7 . Existing drum is too small.
20
Feed
rate
drum can
F α
handle:
2
D.
Fexisting
Dexist
1000
.98
2
4
2
gives
.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
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 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. c1)
V
zA
F
KB 1
2.C2.
2.C5.
a. Start with
xi
zB KA 1
Fzi and let V L VK i x
Fz i
F L
or x i
zi
21
i
L
F L Ki
L F
1
L Ki F
22
Then y
K i zi
Kx i
i
i
L F
From yi
xi
1
L K F i
0 we obtain
L
K i 1 zi 0 L 1 Ki F F
2.C4. New Problem. Prove that the intersection of the operating and y = x lines for binary flash distillation occurs at the mole fraction of the feed. SOLUTION: y
L
y
V
F
z ,rearrange: y 1
V
V
therefore
F
L
z , or y
V
F
V L V
z since V + L = F, the result is y = z and
V
x=y=z
(2-18)
The intersection is at the feed composition.
zi
2.C7.
1 V/F f
0 0
.1 -.09
V Ki 1 .2 -.1
1 f
V
From data in Example 2-2 obtain:
F F .3 -.09
.4 .5 -.06 -.007
.6 .07
.7 .16
.8 .3
.9 .49
1.0 .77
23
2.C8.
F
Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are, V L1 L2 and Fzi L1 x i ,L1 L 2 x i ,L 2 Vyi Substituting in Eqs. (2-60b) and 2-60c)
Fz i
L1K i,L1
L2
x i,L 2
L 2 x i ,L 2
VK iV
L2
x i ,L 2
Solving,
Fz i
x i,L 2 L1K i,L 2
Fz i
L2 VK i,V
L1K i,L1
L2
F V L1 VK i,V
L2
L2
Dividing numerator and denominator by F and collecting terms.
zi
x i,liq 2 1
Since
yi
K i ,V
L2
K i,L1
1
L2
K i,V 1
K i,L1
C
x i,L 2
1
L2
C
1,
yi
L1
L2
1
L2
1
C
K
1
L2
L1
V F
zi K i,V
V F
C
1 , thus,
i 1
K i,V 1
L2
F
i 1
which becomes
K i,V
F
x i ,L 2 , yi
Stoichiometric equations,
L1
C
yi
x i,L 2
i 1
i 1
1 zi
0
0
K
1
(2-62)
V
i 1 i,L1 L 2
Since x i,liq1
K i,L1
L2
i,V L 2
F
K i,L1
x i,liq 2 , we have x i,liq1 1
In addition,
F
x i,liq1
x i,liq 2
0
C
1
K i,L1
L2
1
L1 F
L2
zi K i,V
L2
1
V F
K 24
K i,L1
L2
1 zi
(2-63)
L V
i,L1 L 2
1
i 1
2.D1.
a.
V
0.4 100 40 and L
F V
Slope op. line LV 3 2, y See graph. y 0.77 and x 0.48 b.
V
0.4 1500
c. Plot x
1
F
K i,V
L2
1
F
60 kmol/h x
z
0.6
600 and L 900 . Rest same as part a.
0.2 on equil. Diagram and y
x
z
0.3. yint ercept
V F z 1.2 0.25 . From equil y 0.58 . d. Plot x 0.45 on equilibrium curve. L F V 1 V F Slope V V VF
.8 .2
zF V 1.2
4
25
Plot operating line, y e. Find Liquid Density.
MW L Then, V
x
L
x
0.51 . From mass balance F 37.5 kmol/h.
z at z
xm MWm
xw MWw
MWm
MWw
x
m
.8
.7914
w
L
32.04
.2
w m
.2 32.04
MW L VL
.8 18.01
18.01
20.82
22.51 ml/mol
1.00
20.82 22.51 0.925 g/ml
Vapor Density: ρV = p(MW)V,avg/RT (Need temperature of the drum)
MW v
y m MW
m
yw MW
w
.58 32.04
.42 18.01
26.15
g/mol
Find Temperature of the Drum T: From Table 3-3 find T when
y .58, x
20, T=81.7 C 354.7K 82.0575
1 atm 26.15 g/mol
ml atm
v
354.7 K
8.98 10
4
g/ml
mol K
Find Permissible velocity: u
K
,K
exp A
B
nF
C
nF
2
D
nF
3
E
nF
4
2-60
26
perm
V
drum
V F
F
L
v
v
0.25 1000
drum
lv
250 lbmol/h, W v
L F V 1000 250 750 lbmol/h, and WL
lv
V MW
lv
250 26.15 v
L MW L
lv
lb
6537.5 lb / h
lbmol 750 20.82
15, 615 lb/h,
27
4
WL
F lv
Then K drum
15615
V
WV
.442, and u perm
.442
u perm 3600
D
f.
4A cs
.925 8.98 10 8.98 10 4
lv
4
14.19 ft/s
250 26.15 454 g/lb v
4
14.19 3600 8.98 10
g/ml 28316.85 ml/ft
2.28 ft 2 .
3
1.705 ft. Use 2 ft diameter. L ranges from 3 D 6 ft to 5
Note that this design is conservative if a demister is used. Plot T vs x from Table 3-3. When T 77 C, x 0.34, y
F V x Vy
V F
z y y x
D=10 ft
0.69. This problem is now
very similar to 3-D1c. Can calculate V/F from mass balance, Fz
Fz
2.598
0.0744, and n F
.925
6537.5
L
V MW v A cs
8.89 10
Lx
Vy.
This is
0.4 0.34 0.17 0.69 0.34
or g. Part g is a new problem. V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756. 2-D2.
Work backwards. Starting with x2, find y2 = 0.62 from equilibrium. From equilibrium point plot op. line of slope
LV
1 2
V
0.51 x1
(see
2
F2 V
Figure). From equilibrium, y1
3 7. Find z 2
VF
z1 x1
0.55 0.51
0.78 .For stage 1,
0.148 . F
y1 x1
0.78 0.51
28
2.D3.
New Problem in 4th edition.. Part a. x ethane T oC
y ethane
0
63.19
0
.025
56.18
0.1610
.05
49.57
0.2970
.10
37.57
0.5060
.15
27.17
0.6503
.20
18.26
0.7492
.25
10.64
0.8175
.30
4.11
0.8652
1.0
-37.47
1.0
b. See Figure. a. If 1 bubble of vapor product (V/F = 0) vapor product, vapor yE = 0.7492 (highest) liquid xE = zE = 0.20 (highest) and T = 18.26 oC. If 1 drop of liquid product (V/F =1) yE = zE = 0.20 (lowest), xE = 0.035, T (by linear interpolation) ~ 56.18 + [(49.57 – 56.18)/(.297 .161)][.2 – 0.16] = 54.2 oC (highest). c. See figure. Slope = -L/V = - (1 – V/F)/(V/F) = - .6/.4 = - 1.5. xE = 0.12, yE = 0.57, T = 33.4oC. d. From equilibrium data yE = 0.7492. For an F = 1, L = 1 – V, Ethane balance: .2L = 1(.3) – 0.7492 V. Solve 2 equations: V/F = 0.1821. Can also find V/F from slope of operating line. e. If do linear interpolation on equilibrium data, x = 0.05 +(45-49.57)(0.1 -0.05)/(37.57 – 49.57) = 0.069. From equilibrium plot y = 0.375. Mass balance for basis F = 1, L = 1 – V and 0.069 L = 0.18 – 0.375 V. Solve simultaneously, V/F = 0.363. 2.D4. New problem in 3rd edition. Highest temperature is dew point Set
zi
yi .
Ki K ref TNew
yi x i . Want x i K ref TOld
i
y
.2
.35
.45
4.0145
K i 1.0
Ki
yi
K bu tan e
If pick C4 as reference: First guess
yi
V F 0
1.0,
T
41 C : K C3
3.1,K C6
0.125
T too low
Ki
3.1 1.0 .125 Guess for reference: K C4 4.014, T 118 C : K C3 y Ki
i
.2 8.8
8.8, K C6
.9
.35 .45 0.6099 4.0145 .9 29
K C4, NEW
4.0145 .6099
2.45, T 85 : K C2
6.0, K C6
0.44
30
y Ki
.2 6
i
K C4,NEW y Ki
.35 .45 1.20 2.45 .44 2.45 1.2 2.94, T 96 C : K C3
.2 6.9
.35 .45 0.804 2.94 .56 Use 90.5º → Avg last two T K C 4 2.7, K C3 .2 .35 .45 i
yi K i
6.5
2.7
.49
6.9, K C6
0.56
Gives 84 C 6.5, K C6
0.49
1.079 , T ~ 87 88º C
Note: hexane probably better choice as reference. 2.D5.
a) v1 = F2 v2 =z y
z1 = 0.55 1
2
F1 = 1000
V
0.25
F2 x2
x1 = 0.30
b)
y1
L x1 V
F z Plot 1st Op line. V
1
y1 = 0.66 = z2
1
y = x = z = 0.55 to x1 = 0.3 on eq. curve (see graph) L 0.55 0.80 .25 Slope 0.454545 L
V1
.55 0
V F At x
0, y
0.25 ,
z V F
L
0.75F
V
0.25F
0.66
V
1
.55
V1 = 687.5 kmol/h = F2
c) Stage 2
p1,2 = 1 atm
V
687.5
F1
1000
3, y
2.64. At y
x
0, x 2
F
1
1000
1
0.6875
z2 F
0.66. Plot op line
z
z
0.66
0.88
31
0.25 From graph
y2
0.82, x 2
0.63 . V2
L
V F2 F2
0.25 687.5
L F
0.75
171.875 kmol/h
32
2.D6.
New problem in 4th ed. a.)
The answer is VP = 19.30 mm Hg
1310.62 1.2856 100 136.05 VP 19.30 b.) The answer is K = 0.01693. K log 10 VP
6.8379
Ptot
1.5 760
2.D7. New problem 4th ed. Part a. Drum 1: V1/F1 = 0.3, Slope op line = -L/V = -.7/.3 = -7/3, y=x=z1 =0.46. L1 = F2 = 70. From graph x1 = z2 = 0.395 Drum 2: V1/F1 = 30/70, Slope op line = -L/V = -7/3, y=x=z2 =0.395. L1 = F2 – V2 = 40. From graph x2 = 0.263 Part b. Single drum: V/F = 0.6, Slope op line = -L/V = -40/60 = -2/3, From graph x = 0.295. More separation with 2 drums.
33
2.D8.
Use Rachford-Rice eqn: f
Ki 1 zi
V F
1
Find K i from DePriester Chart: K1 Converge on V F .076, V
zi
From x i
2.D9.
73, K 2
F V F
we obtain x
V K 1 F i
1
Ki 1 V / F
4.1 K 3
. Note that 2 atm = 203 kPa.
.115
152 kmol/h, L F V 1848 kmol/h . .0077, x
1
0
.0809, x 2
.9113 3
From yi K i x i , we obtain y1 .5621, y2 .3649, y3 .1048 Need hF to plot on diagram. Since pressure is high, feed remains a liquid h C T T , T 0 from chart F
CPL
PL
F
ref
ref
CPEtOH x EtOH CPw x w 34
Where x EtOH and x w are mole fractions. Convert weight to mole fractions. 30 Basis: 100 kg mixture: 30 kg EtOH 0.651 kmol 46.07 70 kg water 70 18.016 3.885 Total = 4.536 kmol
35
100 4.536
Avg. MW Use C PL
EtOH
0.6512 4.536
0.1435, x w
0.8565 .
at 100 C as an average C P value.
CPL
Per kg this is
hF
22.046 Mole fracs: x E
37.96 .1435
C PL MWavg
18.0 .8565
20.86 22.046
0.946 2000
0.946
20.86
kcal kmol C
kcal kg C
189.2 kcal/kg
which can now be plotted on the enthalpy composition diagram. Obtain Tdrum 88.2 C, x E 0.146, and y E 0.617 .
F 1000 find L and V from F = L + V and Fz For which gives V = 326.9, and L = 673.1
23.99 & C
Note: If use wt. fracs. C PL
2.D.10 Solution 400 kPa, 70ºC From DePriester chart
MW PL
Vy
1.088 and h
avg
z C4 35 Mole % n-butane K C3 5, K C 4 1.9, K C 6 V
Lx
217.6 . All wrong. F
x C6
0.7
0.3 36
Know y
Kx, i
i
i
zi
x
,
i
x
y i
1
Ki 1
1 i
z i
F
37
K i 1 zi 0
z
1
R.R.
1 z
Ki 1
C3
F z C6
V 1
K C6 1
F
1 4
C6
z
1 0.7
0.7 1 0.7
C6
V
, z
F
0.7 0.49 C6
V F
F 0.9 .35 V 1 0.9
F
C4
C6
V
4 .65 z C6 V
RR Eq:
.65 z
V
z C6
C6: 0.7
z
F
0.7z C6 V
0
1 0.7
F z . 2 equations & 2 unknowns. Substitute in for C6 Do in Spreadsheet. Use Goal – Seek to find V F. V/F = 0.594 when R.R. equation 0.000881 . V 0.7 (0.49)(0.594) 0.40894 z C6 0.7 0.49 F 2.D11. New Problem 4th ed. Obtain K ethylene = 2.2, K propylene = 0.56 from De Priester chart. KE = yE/xE and KP = yP/xP Since yp = 1 – yE and xp = 1 – xE , Kp = (1 – yE)/(1 – xE). Thus, 2 eqs and 2 unknowns. Solve for yE and xE. xE = (1 – Kp) / (KE – Kp) and yE = KE xE = KE (1 – Kp) / (KE – Kp) xE = (1 – 0.56) / (2.2 – 0.56) = 0.268 and yE = KE xE = (2.2)(0.268) = 0.590 Check: xp = 1 – xE = 1 – 0.268 = 0.732 and yp = 1 – yE = 1 – 0.590 = 0.410 Kp = yp/xp = 0.410/ 0.732 = 0.56 OK 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, V L
xm MWm MWm
x m
xw MWw MWw
x w
m
w
.2 32.04 .2
32.04 .7914 L
.8 18.01 .8
18.01
20.82
22.51 ml/mol
1.00 MW L VL
20.82 22.51 0.925 g/ml
p MW v
Find Vapor Density.
(Need temperature of the drum)
v
38
MW v
RT yw MW
y m MW m
.58 32.04
26.15
g/mol
w
Find Temperature of the Drum T: From Table
y .58, x
.42 18.01
2-7
find
T
corresponding
to
20, T=81.7 C 354.7K 1 atm 26.15 g/mol
82.0575
v
Find Permissible velocity:
ml atm
354.7 K
8.98 10
4
g/ml
mol K u perm
K drum
L
v
v
39
K
1.25 K drum,horizontal
exp A B nF drum , vertical
Since V
Wv
V F
V MW v
lv
0.25 1000
2
C nF
3
D nF
lv
E
nF
lv
4
1.25
lv
250 lbmol/h,
250 26.15 lb lbmol
L F V 1000 250 750 lbmol/h, and WL
6537.5 lb / h L MW L
750 20.82
15, 615 lb/h,
4
WL
F lv
WV
K drum,vertical u perm
15615
V
0.442, and K drum,horiz
u perm 3600
0.5525 4
1.824 ft ,
D
With L/D = 4,
17.74 ft/s
250 26.15 454 g/lbm 17.74 3600 8.98 10 4 g/ml 28316.85 ml/ft 3
v
2
A Cs
2.598
lv
0.925 8.98 10 8.98 10 4
0.5525
0.0744, and n F
.925
6537.5
L
V MW v A cs
8.98 10
2
AT
4A T
A Cs 0.2
9.12 ft
3.41 ft and L 13.6 ft
2.D13. New Problem 4th ed. xbutane = 1 – xE = 0.912, ybutane = 1 – yE = 0.454. KE = yE/xE = 0.546/0.088 = 6.20, Kbutane = yB/xB = 0.454/0.912 = 0.498. Plot KE and Kbutane on DePriester chart. Draw straight line between them. Intersections with T and P axis give Tdrum = 15 oC, and pdrum = 385 kPa from Figure 2-12. Use mass balances to find V/F: F = L + V and FzE = LxE + VyE. Substitute L = F – V into ethane balance and divide both sides by F. Obtain: z = (1 – V/F)x + y(V/F). Solve for V/F = (z-x)/(y-x) = (0.36 – 0.088)/(0.546 – 0.088) = 0.594. Spreadsheet used as a check (using T=15 and p = 385) gave V/F = 0.593. 2.D14.
New Problem 4th ed. DePriester chart, Fig. 2-12: KC1 = 50, KC4 = 1.1, and KC5 = 0.37; z1 = 0.12, z4= 0.48, z5 = 0.40
Rachford-Rice equation:
K C2 1 z C1 1
Equation becomes:
5.88
K C1 1
K iC 4 1 z nC4 V
1
F 0.048
K nC 4 1
V
F 0.252
K nC 4 1 z nC5 0 V 1 K nC5 1 F 0 40
1 49(V / F ) 1 0.1(V / F ) 1 0.63(V / F ) Trials: V/F = 0.4, Eq. = -.005345; V/F = 0.39, Eq. = 0.004506; V/F = 0.394, Eq. = 0.000546, which is close enough with DePriester chart. Liquid mole fractions: xC4 = 0.4618, xC5 = 0.5321, and ∑xi = 0.9998 z C1 .12 x 0.00591; C1
1
K C1 1 (V / F)
1 49 .394
Vapor mole fractions: yi = Ki xi : yC1 = 50(0.00591) = 0.2955, yC4 = 0.5080, yC5 = 0.1969, ∑yi = 1.0004.
41
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 iC 4 z nC 4 constant, and z C 2 z iC 4 z nC 4 1.0 Thus, solve three equations and three unknowns simultaneously. Do It. Rachford-Rice equation is,
K C2 1 z C2
K iC 4 1 z iC4 V
KC2 1
1
1
K nC 4 1 z nC2 V
K iC 4 1
F
0 V
K nC 4 1
1
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 1
1 1.8 z V
K C2 1
.8 K iC 4 1 nC4
1
K iC 4 1
F
z V
K nC4 1
z nC 4
nC 4
1
K nC 4 1
F
V
0
F
K C2 1 1 z nC 4
Thus,
K C2 1
KC2 1
1.8 1
.8 K iC 4 V
KC2 1
V F 1
K iC 4 1
1
F
V
K nC 4 1
1
K nC 4 1
F
V 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.
z C1
0.5, z C4
0.1, z C5
0.15, z C6
0.25, K C1
50, K C 4
.6, K C5
.17, K C6
0.05
1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom
V/F
R.R. eq.
f
1
.5
K i 1 zi
V F
1
49 .5 1 49 .53
V
.1 / 3 .53 This first guess is not critical.
V
Ki 1 V F
0
.4 .1
.83 .15
.95 .25
1 .4 .53
1 .83 .53
1 .95 .53
0.157
f VF 42
1
Eq. 3.33
F 2
F
z K 1
i
V/F V/F V .584 150
i
K
V 1 F
2
0.157 .
0.53 and f V / F 1
calculate
2
i
1 where
1
2
1
.53 0.157 2.92 0.584
87.6 kmol/h and L 150 87.6 62.4
43
x C1
yC1 2-D17.
K C1 x C1
1
z C1 K C1 1 (V / F)
50 0.016883
.5 1 49 .584
0.016883
0.844
Similar for other components. a. V 0.4F 400, L 600 Slope
LF
1.5
Intercepts y = x = z = 0.70. Plot line and find xA = 0.65, yA = 0.77 (see graph) b. V = 2000, L = 3000. Rest identical to part a. c. Lowest xA is horizontal op line (L = 0). xA = 0.12 Highest yA is vertical op line (V = 0). yA = 0.52. See graph
d.
2.D18.
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
zi
From x i
1
Ki 1
, we obtain
V F
V
zh 1 xh
F
Kh 1
Guess Tdrum , calculate K h , K b and K p , and then determine V F .
K1 1 zi
33
Check:
1
K1 1 V F
0 ?
Initial guess: Tdrum must be less than temperature to boil pure hexane
33
Kh
1.0, T
94 C . Try 85°C as first guess (this is not very critical and the calculation
will tell us if there is a mistake). K h =0.8, K b
0.6 V F Try T
4.8, K p =11.7 .
1
0.85 0.8 1
73 C where K h
V F
1.471 . Not possible. Must have K h
0.6. Then K 0.6 1 .85 .6 1
b
3.8, K p
0.6 0.85
0.706
9.9 .
0.735
Check:
K i 1 zi 1
8.9 .1
Ki 1 V F
1
2.8 .3
8.9 .735 1
.4 .6
2.8 .735 1
0.05276 .4 735
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
C8 balance: z C6 F Lx C6 z C6
or
VyC6
L x C6 1 V F x C6
F V x C6 K C6 Vx C6
1 V F x 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 1 VF
.93C 6
V
.1
F
.9K C6
. Trial and error scheme.
.1
Pick T, Calc K C6 , Calc V F , and Check f V F If not K refnew
0?
K ref Told 1 df T
T 70 C. K C4 3.1, K C 5 .93, K C 6 .37 K ref V .1 0.231. F .9 .37 .1 Rachford Rice equation .08 .25 .63 .35 2.1 .4
Try
34
f
1
2.1 .231 1 K ref Tnew
.08 .231 1 .37 1 0.28719
Converge on TNew ~ 57 C. Then K C 4 2.D20.
.63 .231
.28719
0.28745 use .28
2.50, K C8
.67, and V F
0.293 .
New Problem 4th ed.
35
2.D21.
a.)
K C2
Soln to Binary R.R. eq.
4.8
K C5
V
zA
F
KB 1
0.153 zB KA 1
,
V
0.55
F
.153 1
0.45
0.5309
4.8 1
x C2 36
0.8749 , x
C5
z C2
,
0.8177
0.1251
yC5
0.55
0.18 23, y V 13 .8 .530 9C2 1
K C2 1
F
37
Need to convert F to kmol. Avg MW
F 100, 000
kg kmol hr 49.17 kg
2033.7 kmol/h , V
u
b.)
0.55 30.07
L
K Perm
0.45 72.15
V F F 1079.7,
49.17
L F V
954.0 kmol/h
v
drum v
To find
MW L
0.1823 30.07
0.8177 72.15
64.48
MW V
0.8749 30.07
0.1251 72.15
35.33
For liquid assume ideal mixture:
V
x V
x V
C2
C5
1
C 2,liq
MW C2
x
C5,liq
C2
C5 C 2,liq
VL
MWL VL
L
For vapor: ideal gas:
0.1823
0.8177
0.54
72.15
82.0575
mol K
K drum : Use Eq. (2-60) with FlV
997.7
kmol 64.48 kg h kmol
FlV K drum
6, 4331.7 kg/h , WV
64331.7 0.009814 31,143.3 0.621
exp 1.877478
0.0145229
0.81458
n 0.2597
3
C5,liq
103.797 ml/mol
0.009814 g / ml
303.16K
WL W V
WL
0.63
64.48 0.621 g/ml 103.797 atm g 700 kPa 35.33 101.3 kPa mol ml atm
MW v RT
v
30.07
MW C 5
x
v
L
881.5 35.33
31,143.4 kg/h
0.2597 n .2597
0.18707
0.0010149
n 0.2597
n 0.2597
4
2
0.3372
u Perm 38
0.621
0.3372
0.8111 m/s
0.009814 ft
1.0 m 0.0098 14 s 3.2808 ft
AC
V MWV u Perm 3600
1079.7 v
0.8111
m
3600 s
D
4A C
s
kmol h
35.33
0.009814 g h
kg kmol kg
cm3 1000g
1.33 m . Arbitrarily L D 4,
2
106 cm 3
1.392 m
m3
L 5.32 m
2.D22. New problem in 4th edition. a. V = F – L = 50 – 20 = 20 kmol/h. V/F = 3/5, Slope operating line = -L/V = -20/30 = - 2/3, zM = 0.7
39
From graph, y = 0.8, x = 0.54. b. From graph of T vs. xM, Tdrum = 72.3oC. (see graph).
2.D23. New Problem 4th ed. Part a. Fnew = (1500 kmol/h)(1.0 lbmol/(0.45359 kmol)) = 3307 lb mol/h. V, WV, L, and WL are the values in Example 2-4 divided by 0.45359. The conversion factor divides out in Flv term. Thus, Flv, Kdrum, and uperm are the same as in Example 2-4. The Area increases because V increases: Area = AreaExample 2-4/0.45359 = 16.047/0.45359 = 35.38 ft2. Diameter 6.71 feet 4 Area / Probably round this off to 7.0 feet and use a drum height of 28 feet. b.
Fparallel = 3307 – 1500 = 1807 lbmol/h.
Flv, Kdrum, and uperm are the same as in Example 2-4. Vparallel = (V/F) Fparallel = 0.51 (1807) = 921.6 kmol/h.
40
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