SOLUTION MANUAL
for
SEPARATION PROCESS ENGINEERING
5th Edition
by
Phillip C. Wankat
,SPE 5th Edition Solution Manual Chapter 1
New Problems and new solutions are listed as new immediately after the solution number. The new
problems in chapter 1 of the 5th edition are: 1A1, 1D1, 1D2, 1D3, 1D4, and 1D5.
A2. Answers are in the text.
A3. Answer is d
A7. On the internet separation processes often refer to divorce.
A8. New question in 5th edition. Languages with different alphabets will automatically use different
symbols. Even with the same alphabet, words for terms will be different so different notation
automatically appears. Do not assume that the notation in a different source is the same.
B1. Everything except some raw food products has undergone some separation operations. Even the
water in bottles has been purified (either by reverse osmosis or by distillation).
B2. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually
adsorption), or a reverse osmosis system.
B3. For example: the lungs are a gas permeation system, the intestines and kidney are liquid
permeation or dialysis systems.
B4. You probably used some of the following: chromatography, crystallization, distillation,
extraction, filtration and ultrafiltration.
D1. New problem in 5th edition Basis 1h, 1200 kmol feed., 30 mol% ethanol.
(1200 kmol)(0.3)=360 kmol ethanol. Kmol Water = 1200 – 360 E = 840 kmol
Weight ethanol = 360 kmol (46 kg/kmol) = 16,560 kg E
Weight water = 840 kmol (18.016 kg/kmol) = 15,133.4 kg W
Total = 31693.44 kg, & flow rate = 31693.44 kg/h
Wt frac E = 16500/31693.44 = 0.5225 and Wt frac W = 1 – 0.5225 = 0.4775
D2. New problem in 5th edition.
Basis 1200 kg of feed, 30 wt. % EtOH. (0.3) (1200) = 360 kg ethanol. 1200 – 360 = 840 kg water
(360 kg of EtOH)/(46 kg/kmol) = 7.826 kmol EtOH
(840 kg of water)/(18.016 kg/kmol) = 46.625 kmol water
Sum = 54.452 kmol = total molar flow rate in kmol/h
Mole fraction EtOH = (kmol ethanol)/(total kmol) =7.826/54.452 = 0.1437
D3. New problem in 5th edition.
Flow rate of a 30.0 wt. % ethanol and 70.0 wt. % water feed is 1200 kmol/h. What are the ethanol
mole fraction and the total flow rate in kg/h?
This is a problem with mixed units. Fortunately, we can choose any convenient basis to convert
from weight fraction to mole fraction. Thus, the basis of 1200 kg of feed from problem 1.D2
works and the mole fraction of ethanol is the same as in that problem: mole fraction ethanol =
0.1437. Mole fraction water = 1 – 0.1437 = 0.8563.
We found that 1200 kg/h of feed was 54.452 kmol/h. For the same weight fraction in the feed, the
ratio of kg/h to kmol/h will be unchanged. Thus,
kg/h feed = (1200 kmol/h)[(1200 kg/h)/(54.452 kmol/h)] = 26,445 kg/h
D4. New problem in 5th edition. 1900 mm Hg (1.0 atm/760 mm Hg) (101.3 kPa/atm) = 253.25 kPa
D5. New problem in 5th edition. Since 1 kJ/s = 1 kW, have 18.0 kW.
17
, SPE 5th Edition Solution Manual Chapter 2.
New Problems and new solutions are listed as new immediately after the solution number. These new
problems in the 5th edition are: 2D4, 2.D6, 2D13, 2D14, 2D17, 2D23, 2D25, 2D26, 2G1, 2G2, 2G3,
2H5, 2H6, and 2H7. The correlation for sizing flash drums, Eq. (2-62) has been changed to a more recent
correlation. As a result problems 2D1e, 2.D12, 2D21, 2D29, 2E1, 2E2 have different solutions.
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”). Since flashing is an adiabatic process,
the 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
yw zw = 0.965
(pure water)
Flash
.5 operating
line
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. . 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. . __102oC
2.A10. 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. Because of the presence of air this is not a binary system. Also, it is not at equilibrium.
2.A12. 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. Although V is an extensive variable, V/F is an intensive variable and thus satisfies Gibbs phase
rule.
2A14. 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, 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.
20
for
SEPARATION PROCESS ENGINEERING
5th Edition
by
Phillip C. Wankat
,SPE 5th Edition Solution Manual Chapter 1
New Problems and new solutions are listed as new immediately after the solution number. The new
problems in chapter 1 of the 5th edition are: 1A1, 1D1, 1D2, 1D3, 1D4, and 1D5.
A2. Answers are in the text.
A3. Answer is d
A7. On the internet separation processes often refer to divorce.
A8. New question in 5th edition. Languages with different alphabets will automatically use different
symbols. Even with the same alphabet, words for terms will be different so different notation
automatically appears. Do not assume that the notation in a different source is the same.
B1. Everything except some raw food products has undergone some separation operations. Even the
water in bottles has been purified (either by reverse osmosis or by distillation).
B2. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually
adsorption), or a reverse osmosis system.
B3. For example: the lungs are a gas permeation system, the intestines and kidney are liquid
permeation or dialysis systems.
B4. You probably used some of the following: chromatography, crystallization, distillation,
extraction, filtration and ultrafiltration.
D1. New problem in 5th edition Basis 1h, 1200 kmol feed., 30 mol% ethanol.
(1200 kmol)(0.3)=360 kmol ethanol. Kmol Water = 1200 – 360 E = 840 kmol
Weight ethanol = 360 kmol (46 kg/kmol) = 16,560 kg E
Weight water = 840 kmol (18.016 kg/kmol) = 15,133.4 kg W
Total = 31693.44 kg, & flow rate = 31693.44 kg/h
Wt frac E = 16500/31693.44 = 0.5225 and Wt frac W = 1 – 0.5225 = 0.4775
D2. New problem in 5th edition.
Basis 1200 kg of feed, 30 wt. % EtOH. (0.3) (1200) = 360 kg ethanol. 1200 – 360 = 840 kg water
(360 kg of EtOH)/(46 kg/kmol) = 7.826 kmol EtOH
(840 kg of water)/(18.016 kg/kmol) = 46.625 kmol water
Sum = 54.452 kmol = total molar flow rate in kmol/h
Mole fraction EtOH = (kmol ethanol)/(total kmol) =7.826/54.452 = 0.1437
D3. New problem in 5th edition.
Flow rate of a 30.0 wt. % ethanol and 70.0 wt. % water feed is 1200 kmol/h. What are the ethanol
mole fraction and the total flow rate in kg/h?
This is a problem with mixed units. Fortunately, we can choose any convenient basis to convert
from weight fraction to mole fraction. Thus, the basis of 1200 kg of feed from problem 1.D2
works and the mole fraction of ethanol is the same as in that problem: mole fraction ethanol =
0.1437. Mole fraction water = 1 – 0.1437 = 0.8563.
We found that 1200 kg/h of feed was 54.452 kmol/h. For the same weight fraction in the feed, the
ratio of kg/h to kmol/h will be unchanged. Thus,
kg/h feed = (1200 kmol/h)[(1200 kg/h)/(54.452 kmol/h)] = 26,445 kg/h
D4. New problem in 5th edition. 1900 mm Hg (1.0 atm/760 mm Hg) (101.3 kPa/atm) = 253.25 kPa
D5. New problem in 5th edition. Since 1 kJ/s = 1 kW, have 18.0 kW.
17
, SPE 5th Edition Solution Manual Chapter 2.
New Problems and new solutions are listed as new immediately after the solution number. These new
problems in the 5th edition are: 2D4, 2.D6, 2D13, 2D14, 2D17, 2D23, 2D25, 2D26, 2G1, 2G2, 2G3,
2H5, 2H6, and 2H7. The correlation for sizing flash drums, Eq. (2-62) has been changed to a more recent
correlation. As a result problems 2D1e, 2.D12, 2D21, 2D29, 2E1, 2E2 have different solutions.
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”). Since flashing is an adiabatic process,
the 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
yw zw = 0.965
(pure water)
Flash
.5 operating
line
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. . 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. . __102oC
2.A10. 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. Because of the presence of air this is not a binary system. Also, it is not at equilibrium.
2.A12. 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. Although V is an extensive variable, V/F is an intensive variable and thus satisfies Gibbs phase
rule.
2A14. 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, 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.
20
