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SOLUTIONS MANUAL
, 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 T c T T . When pressure is dropped the mixture
is above F,Phigh pLIQ F ref
its bubble point and is a two-phase mixture (It “flashes”). In the flash mixture
enthalpy is unchanged but temperature changes. Feed location cannot be found from
TF and z on the graph because equilibrium data is at a lower pressure on the graph used
for this calculation.
2.A2. Yes.
2.A4.
1.0
Equilibrium
yw (pure water) zw = 0.965
Flash
.5 operatin
g
2.A4
0
0 .5 xw 1.0
18
, 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.
2.A12. New Problem. The answer is b.
2.A13. New Problem. The answer is c.
2.A14. New Problem. The answer is a.
2.A15. New Problem. a. The answer is 3.5 to 3.6
b. The answer is 36ºC
2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from
the amount of superheat.
2.B1. Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium.
Examples:
F, z, Tdrum , Pdrum F, TF , z, p F, h F , z, p
F, z, y, P F, TF , z, y F, h F , z, y
drum
F, z, x, pdrum F, TF , z, x etc.
F, z, y, pdrum F, TF, z, Tdrum , pdrum
F, z, x, Tdrum F, TF , y, p
Drum dimensions, z, Fdrum , F, TF , y, Tdrum F,
pdrum TF , x, p
Drum dimensions, z, y, pdrum F, TF , x, Tdrum F,
etc.
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.
19
, 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 F .
With F = 25,000:
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
Feed rate drum can handle: F α D2. 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
20
, y = .58,
V = .25 (16660) = 4150
25,000 16,660
LTota
8340 lx
Since x is not specified, use bypass. This produces less vapor.
c) Look at Eq. (2-62), which becomes
V MWv
D
3Kdrum 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. V zA zB
F KB 1 KA 1
Fzi
2.C5. a. Start xi and let V F L
L VKi
with
Fz i zi
i
x L F L K or x i
i L 1 L K
i
F F
yi Ki xi Kizi
Then
L L
1 Ki
F F
K i 1 zi
From yi xi 0 we obtain 0
L L
F 1 F Ki
21
, zi V
2.C7. 1 f From data in Example 2-2 obtain:
V F
1 Ki 1
F
V/F 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
f 0 -.09 -.1 -.09 -.06 -.007 .07 .16 .3 .49 .77
2.C8. New Problem.
pdrum
F=L+V
Fz Lx Vy
Tdrum
x y
Solve for L & V
z
22
,2.C9. New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,
F V L L
1 2
Fzi L x xi,L2
1 i,L1 L2 Vyi
Substituting in equilibrium Eqs. (2-60b) and 2-60c)
Fzi L1Ki,L1 L2 xi,L 2 L2 xi,L 2 VKiV L2 xi,L 2
Solving,
xi,L 2 Fzi Fzi
L1 Ki,L 2 L 2 VK i,V L2 L1 Ki,L1 L2 F V L1 VK i,V L2
Dividing numerator and denominator by F and collecting terms.
x
i,liq 2 zi
L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
yi Ki,V L2 x i,L 2 , yi K i,V L2 iz
Since
L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
C C C C
Stoichiometric x i,L 2 1, yi 1 , thus, yi x i,L 2 0
equations,
i 1 i 1 i 1 i 1
C
Ki,V L2
1 zi
which becomes 0 (2-62)
i 1 L1 V
1 Ki,L1 L2 1 K i,V L2 1
F F
K i,L1 L 2 z i
Since xi,liq1 K i,L1 L 2 x i,liq 2 , we have xi,liq1
L V
1 Ki,L1 L 2 1 1 K i,V L2 1
F F
K i,L1
In C
L2 1 zi (2-63)
addition, x i,liq1 xi,liq 2 0
i 1 L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
2.D1. a. V 0.4 100 40 and L F V 60 kmol/h
Slope op.
L V 3 2, y x z 0.6
line See 0.77 and x 0.48
graph. y
b. V 0.4 1500 600 and L 900 . Rest same as part a.
c. Plot x 0.2 on equil. Diagram and y
x z 0.3. yint ercept zF V 1.2
V F z 1.2 0.25 . From equil y 0.58 .
d. Plot x 0.45 on equilibrium curve.
23
, L F V 1 V F .8
Slope 4
V V V F .2
Plot operating line, y x z at z 0.51 . From mass balance F 37.5 kmol/h.
e. Find Liquid Density.
MWL xm MWm x w MWw .2 32.04 .8 18.01 20.82
MWm MW w 32.04 18.01
Then, VL xm xw .2 .8 22.51 ml/mol
m w
.7914 1.00
L
MW L VL 20.82 22.51 0.925 g/ml
P MW
Find Vapor Density. v
v
(Need temperature of the drum)
RT
MWv ym MW yw MW w .58 32.04 .42 18.01 26.15 g/mol
m
Find Temperature of the Drum T:
From Table 3-3 find T corresponding to
y .58, x 20, T=81.7 C 354.7K
ml
1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10
4
g/ml
atm
v
mol K
Find Permissible velocity:
24
, uperm
K drum L v v
2 3 2-60
4
Kdrum exp A B nFlv C nFlv D nFlv E nFlv
V
Since V F 0.25 1000 250
lbmol/h, F
lb
250 26.15 6537.5 lb / h
Wv V MW v lbmol
L F V 1000 250 750 lbmol/h, and WL L MW L 750 20.82 15, 615 lb/h,
Flv WL V 15615 8.89 10 4
0.0744, and n Fl 2.598
WV L 6537.5 .925 v
4
.925 8.98 10
Then K drum .442, and .442 8.98 10 4 14.19 ft/s
uperm
250 26.15 454 g/lb
V MW v 2
A 2.28 ft .
cs 4 3
u perm 3600 v 14.19 3600 8.98 10 g/ml 28316.85 ml/ft
Thus, D 4Acs 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 x2, find y2 = 0.62 from equilibrium. From equilibrium point
V
plot op. line of slope L V 1 V F 3 7. This gives
2 2
F 2
z2 0.51 x1 (see Figure). Then from equilibrium, y1 0.78 .
V z1 x1 0.55 0.51
For stage 1, 0.148 .
F y1 0.78 0.51
x1
2.D3. a. z 0.4
V F 0.6 V 6.0 k mol h, L 4.0
L
x z V FV
y 2
Op. eq. y x 23
3
See y 0.55 xM 0.18 T ~ 82.8 C linear interpretation on Table 2-7 .
graph:
25
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SOLUTIONS MANUAL
, 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 T c T T . When pressure is dropped the mixture
is above F,Phigh pLIQ F ref
its bubble point and is a two-phase mixture (It “flashes”). In the flash mixture
enthalpy is unchanged but temperature changes. Feed location cannot be found from
TF and z on the graph because equilibrium data is at a lower pressure on the graph used
for this calculation.
2.A2. Yes.
2.A4.
1.0
Equilibrium
yw (pure water) zw = 0.965
Flash
.5 operatin
g
2.A4
0
0 .5 xw 1.0
18
, 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.
2.A12. New Problem. The answer is b.
2.A13. New Problem. The answer is c.
2.A14. New Problem. The answer is a.
2.A15. New Problem. a. The answer is 3.5 to 3.6
b. The answer is 36ºC
2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from
the amount of superheat.
2.B1. Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium.
Examples:
F, z, Tdrum , Pdrum F, TF , z, p F, h F , z, p
F, z, y, P F, TF , z, y F, h F , z, y
drum
F, z, x, pdrum F, TF , z, x etc.
F, z, y, pdrum F, TF, z, Tdrum , pdrum
F, z, x, Tdrum F, TF , y, p
Drum dimensions, z, Fdrum , F, TF , y, Tdrum F,
pdrum TF , x, p
Drum dimensions, z, y, pdrum F, TF , x, Tdrum F,
etc.
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.
19
, 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 F .
With F = 25,000:
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
Feed rate drum can handle: F α D2. 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
20
, y = .58,
V = .25 (16660) = 4150
25,000 16,660
LTota
8340 lx
Since x is not specified, use bypass. This produces less vapor.
c) Look at Eq. (2-62), which becomes
V MWv
D
3Kdrum 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. V zA zB
F KB 1 KA 1
Fzi
2.C5. a. Start xi and let V F L
L VKi
with
Fz i zi
i
x L F L K or x i
i L 1 L K
i
F F
yi Ki xi Kizi
Then
L L
1 Ki
F F
K i 1 zi
From yi xi 0 we obtain 0
L L
F 1 F Ki
21
, zi V
2.C7. 1 f From data in Example 2-2 obtain:
V F
1 Ki 1
F
V/F 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
f 0 -.09 -.1 -.09 -.06 -.007 .07 .16 .3 .49 .77
2.C8. New Problem.
pdrum
F=L+V
Fz Lx Vy
Tdrum
x y
Solve for L & V
z
22
,2.C9. New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,
F V L L
1 2
Fzi L x xi,L2
1 i,L1 L2 Vyi
Substituting in equilibrium Eqs. (2-60b) and 2-60c)
Fzi L1Ki,L1 L2 xi,L 2 L2 xi,L 2 VKiV L2 xi,L 2
Solving,
xi,L 2 Fzi Fzi
L1 Ki,L 2 L 2 VK i,V L2 L1 Ki,L1 L2 F V L1 VK i,V L2
Dividing numerator and denominator by F and collecting terms.
x
i,liq 2 zi
L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
yi Ki,V L2 x i,L 2 , yi K i,V L2 iz
Since
L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
C C C C
Stoichiometric x i,L 2 1, yi 1 , thus, yi x i,L 2 0
equations,
i 1 i 1 i 1 i 1
C
Ki,V L2
1 zi
which becomes 0 (2-62)
i 1 L1 V
1 Ki,L1 L2 1 K i,V L2 1
F F
K i,L1 L 2 z i
Since xi,liq1 K i,L1 L 2 x i,liq 2 , we have xi,liq1
L V
1 Ki,L1 L 2 1 1 K i,V L2 1
F F
K i,L1
In C
L2 1 zi (2-63)
addition, x i,liq1 xi,liq 2 0
i 1 L1 V
1 K i,L1 L2 1 K i,V L2 1
F F
2.D1. a. V 0.4 100 40 and L F V 60 kmol/h
Slope op.
L V 3 2, y x z 0.6
line See 0.77 and x 0.48
graph. y
b. V 0.4 1500 600 and L 900 . Rest same as part a.
c. Plot x 0.2 on equil. Diagram and y
x z 0.3. yint ercept zF V 1.2
V F z 1.2 0.25 . From equil y 0.58 .
d. Plot x 0.45 on equilibrium curve.
23
, L F V 1 V F .8
Slope 4
V V V F .2
Plot operating line, y x z at z 0.51 . From mass balance F 37.5 kmol/h.
e. Find Liquid Density.
MWL xm MWm x w MWw .2 32.04 .8 18.01 20.82
MWm MW w 32.04 18.01
Then, VL xm xw .2 .8 22.51 ml/mol
m w
.7914 1.00
L
MW L VL 20.82 22.51 0.925 g/ml
P MW
Find Vapor Density. v
v
(Need temperature of the drum)
RT
MWv ym MW yw MW w .58 32.04 .42 18.01 26.15 g/mol
m
Find Temperature of the Drum T:
From Table 3-3 find T corresponding to
y .58, x 20, T=81.7 C 354.7K
ml
1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10
4
g/ml
atm
v
mol K
Find Permissible velocity:
24
, uperm
K drum L v v
2 3 2-60
4
Kdrum exp A B nFlv C nFlv D nFlv E nFlv
V
Since V F 0.25 1000 250
lbmol/h, F
lb
250 26.15 6537.5 lb / h
Wv V MW v lbmol
L F V 1000 250 750 lbmol/h, and WL L MW L 750 20.82 15, 615 lb/h,
Flv WL V 15615 8.89 10 4
0.0744, and n Fl 2.598
WV L 6537.5 .925 v
4
.925 8.98 10
Then K drum .442, and .442 8.98 10 4 14.19 ft/s
uperm
250 26.15 454 g/lb
V MW v 2
A 2.28 ft .
cs 4 3
u perm 3600 v 14.19 3600 8.98 10 g/ml 28316.85 ml/ft
Thus, D 4Acs 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 x2, find y2 = 0.62 from equilibrium. From equilibrium point
V
plot op. line of slope L V 1 V F 3 7. This gives
2 2
F 2
z2 0.51 x1 (see Figure). Then from equilibrium, y1 0.78 .
V z1 x1 0.55 0.51
For stage 1, 0.148 .
F y1 0.78 0.51
x1
2.D3. a. z 0.4
V F 0.6 V 6.0 k mol h, L 4.0
L
x z V FV
y 2
Op. eq. y x 23
3
See y 0.55 xM 0.18 T ~ 82.8 C linear interpretation on Table 2-7 .
graph:
25
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TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
