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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 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
yw zw = 0.965
(pure water)
Flash
.5 operating
line
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. 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, 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 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
2
Fexisting D exist 4
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
8340 x
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.
V zA zB
2.C2.
F KB 1 KA 1
Fz i
2.C5. a. Start with xi and let V F L
L VK i
Fz i zi
xi or x i
L F L Ki L L
1 Ki
F F
K i zi
Then y i Kixi
L L
1 Ki
F F
K i 1 zi
From yi xi 0 we obtain 0
L L
1 Ki
F F
20
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 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
yw zw = 0.965
(pure water)
Flash
.5 operating
line
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. 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, 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 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
2
Fexisting D exist 4
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
8340 x
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.
V zA zB
2.C2.
F KB 1 KA 1
Fz i
2.C5. a. Start with xi and let V F L
L VK i
Fz i zi
xi or x i
L F L Ki L L
1 Ki
F F
K i zi
Then y i Kixi
L L
1 Ki
F F
K i 1 zi
From yi xi 0 we obtain 0
L L
1 Ki
F F
20
