17
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    10.8 kg
.6 kmol Water .6 MW 18 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.

18
SPE 3rd Edition Solution 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.high LIQF,P p F refh T c T T . 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.A4.
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.
1.0
.5
0 1.0
.5
0 xw
Flash
operating
line
yw
Equilibrium
(pure water)
2.A4
zw = 0.965

19
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:drum drumF, z, T , PFF, T , z, pFF, h , z, pdrumF, z, y, PFF, T , z, yFF, h , z, ydrumF, z, x, pFF, T , z, x
etc.drumF, z, y, pF drum drumF, T , z, T , pdrumF, z, x, TFF, T , y, p
Drum dimensions,drum drumz, F , pF drumF, T , y, T
Drum dimensions,drumz, y, pFF, T , x, p
etc.F drumF, T , x, TFF, T , 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.
Iflb mole
F 1000 , D .98 and L 2.95 ft from Problem 2-D1e
hr .
Since D αV and for constant V/F, V α F, we have D αF .
With F = 25,000:new old new old new newF F = 5, D = 5 D = 4.90, and L = 3 D = 14.7
.
Existing drum is too small.
Feed rate drum can handle: F α D2.2 2
existing exist
F D 4
1000 .98 .98 givesexistingF 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
Since x is not specified, use bypass. This produces less vapor.
c) Look at Eq. (2-62), which becomesv
drum L v v
V MW
D
3K 3600
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.A B
B A
z zV
F K 1 K 1
2.C5. a. Start withi
i
i
Fz
x and let V F L
L VKi i
i i
i
i
Fz z
x or x L LL F L K 1 K
F F
Theni i
i i i
i
K z
y K x L L
1 K
F F
Fromi i
i i
i
K 1 z
y x 0 we obtain 0
L L
1 K
F F
V = .25 (16660) = 4150
LTotal
x
y = .58,
8340
16,660
25,000

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Solution Manual For Separation Process Engineering: Includes Mass Transfer Analysis, 3rd Edition

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