Download Separations 2 under Chemical Engineering Separations Processes and more Exercises Engineering in PDF only on Docsity!
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
1. Steady state temperature reached by a small amount of liquid evaporating into a large amount of
unsaturated vapor-gas mixture is
a. dry-bulb temperature
b. adiabatic saturation temperature
c. wet bulb temperature
d. gas temperature
2. The operating line for an absorbed is curved when plotted in terms of
a. mole fractions b. vapor pressure c. partial pressure d. mass fractions
3. What is the diffusivity of methanol in water at 20
o
C?
a. 1.28 b. 5.13 c. 1.80 d. 1.
4. The shape of the profiles is such that at any time the effective depth of liquid which contains an
appreciable concentration of solute can be specified is referred to as the.
a. Kick’s Law b. Penetration Theory c. Kinetic Theory d. Fick’s Law
5. A deep pool of ethanol is suddenly exposed to an atmosphere of pure carbon dioxide and
unsteady state mass transfer, governed by Fick’s Law, takes place for 100s. What portion of the
absorbed CO 2 will have accumulated in the 1mm layer closest to the surface in this period?
a. 17% b. 78% c. 37% d. 83%
GIVEN:
t = 100s
y = 1mm
REQUIRED: portion of the adsorbed CO 2 accumulated at t = 100s and y = 1mm
SOLUTION:
D
CO
in ethanol = 4 × 10
m
2
/s
(N
A
t
= C
Ai
𝐷
𝜋𝑡
−𝑦
2
4 𝐷𝑡 (N A
t
= C
Ai
𝐷
𝜋
𝑡
𝑒
0
−
1
2 𝑒
−
𝑦
2
4 𝐷𝑡 𝑑𝑡
Let:
𝑦
2
4 𝐷𝑡
2
−
1
2
2 √𝐷𝑋
𝑦
and dt =
𝑦
2
− 2
4 𝐷𝑋
3
𝑑𝑋 Thus, Integral = ∫ 𝑑𝑦
𝑋
𝑒
∞
𝑦
2
− 2
4 𝐷𝑋
3
2 √
𝐷𝑋
𝑦
−𝑋
2
𝑦
√
𝐷
𝑋
𝑒
∞
− 2
−𝑋
2
Molar transfer per unit area = C Ai
𝐷
𝜋𝑡
𝑦
√𝐷
) {[𝑒
−𝑋
2
− 1
)]
∞
𝑋 𝑒
𝑋 𝑒
∞
[− 2 𝑋𝑒
−𝑋
2
− 1
)]𝑑𝑋}
= C
Ai
𝑦
√𝜋
𝑒
− 1
−𝑋
𝑒
2
∞
𝑋 𝑒
−𝑋
2
𝑑𝑋} = C
Ai
𝑦
√𝜋
2 √
𝐷𝑡 𝑒
𝑦
−𝑦
2
4 𝐷𝑡
𝑒 − √
𝑦
2 √𝐷𝑡 𝑒
= C
Ai
𝐷𝑡 𝑒
𝜋
−𝑦
2
4 𝐷𝑡 𝑒 − 𝑦𝑒𝑟𝑓𝑐
𝑦
2 √
𝐷𝑡 𝑒
Substituting D = 4 × 10
m
2
/s; t = 100s,
@ y = 0: Moles transferred = 2C Ai
( 4 × 10 − 9
𝑚
2
𝑠
)( 100 𝑠)
𝜋
= 7.1365×
C
Ai
@ y = 10
− 3
m: Moles transferred = C Ai
(7.1365×
− 0. 626
− 3
= C
Ai
(3.8161×
– 2.63×
= 1.1861×
C
Ai
Therefore,
Portion of material retained in layer =
( 7. 1365 − 1. 1861 )
- 1365
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
𝑃𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑟𝑒𝑡𝑎𝑖𝑛𝑒𝑑 𝑖𝑛 𝑙𝑎𝑦𝑒𝑟 = 0. 8338 or 83%
6. Material deposits diminishing overall heat transfer coefficient of evaporators.
a. foams b. magma c. salts d. scales
7. When the pressure of the heating steam in an evaporator is increased, the steam consumption for
a given duty and heat transfer area will
a. increases b. decreases c. remains the same d. no answer
8. A solution is to be concentrated from 10 to 65% solids in a vertical long tube evaporator. The solution
has a negligible elevation of boiling point and its specific heat can be taken to be the same as that
of water. Steam is available at 203.6 kPa, and the condenser operates at 13.33 kPa. The feed enters
the evaporator at 295K. The total evaporation is to be 25,000 kg/h of water. Overall heat transfer
coefficient is 2800 W/m
2
- K. Calculate the heat transfer required in kW.
a. 19800 b. 17523 c. 24582 d. 30900
GIVEN:
Ci = 10% Tfeed = 295K
C
f
= 65% H
2
O evaporation (total) = 25,000 kg/h
Psteam = 203.6 kPa U = 2800 W/m
2
- K
P
condenser
= 13.33 kPa
REQUIRED: Q
SOLUTION:
Assume 1-second operation
Let: F = feed; B = thick liquor
*from Steam Table: @ P condenser
= 13.33 kPa; T condenser
= 325K
@ Psteam = 203.6 kPa; Tsteam = 394K
Hv H 2 O @ 325K = 2375 kJ/kg
H
v
steam @ 203.6 kPa = 2198 kJ/kg
*Solids Balance: F(0.10) = B(0.65) (1)
*H 2 O Balance: F(0.90) = B(0.35) + 25,000 kg/h (
1ℎ𝑟
3600 𝑠
Solving 1 and 2 simultaneously:
F = 8.2070 kg/s B = 1.2626 kg/s
Q = m H2O
(H
v
) + m F
Cp∆T
Q = 25,000 kg/h (
1ℎ𝑟
3600 𝑠
)(2375 kJ/kg) + (8.2070 kg/s)(4.187 kJ/kg-K)(325-295)K
Q = 17,523.9368 kJ/s
9. Based from the preceding problem, calculate the steam consumption in kilogram per hour.
a. 28700 b. 35400 c. 40100 d. 43690
REQUIRED: m steam
SOLUTION:
H
v
steam @ 203.6 kPa = 2198 kJ/kg
m steam
𝑄
𝐻𝑣 𝑠𝑡𝑒𝑎𝑚
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
Leaching is concerned with the extraction of a soluble constituent from a solid by means of a
solvent. Extraction of coffee from seed is done by solid-liquid extraction by passing through hot
water(solvent) in the coffee(solute)
14. At minimum reflux ratio, the number of plates is
a. minimum b. maximum c. infinite d. NOTA
Explanation:
The Minimum Reflux Ratio (R min
) is the lowest value of reflux at which separation can be achieved
even with an infinite number of plates. It is possible to achieve a separation at any reflux ratio
above the minimum reflux ratio. As the reflux ratio increases, the number of theoretical plates
required decreases.
15. The Fick’s Law of diffusion gives the rate of diffusion based on driving force.
a. pressure b. temperature c. concentration d. AOTA
16. Raw cotton has been stored in a warehouse at 29C and 50% RH. For 200 kg of cotton from the
warehouse, how many kilograms should appear in the woven cloth, neglecting lintage and thread
losses? Regain of cotton cloth at 70%RH is 8.10.
a. 202.8 b. 202.7 c. 202.6 d. 202.
GIVEN:
X
m
= 0.081 RH
1
= 50%; T = 29℃
GW = 200 kg RH 2 = 70%
REQUIRED: GW
2
SOLUTION:
Gross Weight = Bone Dry Weight + Moisture (1)
But,
M = X
m
(BDW)
Substituting M in 1,
GW = BDW + X
m
(BDW)
GW = BDW(1+Xm)
@ RH
1
= 50% and T = 29℃; X m
BDW = 200 kg / (1 + 0.066)\
BDW = 187.6173 kg
@ RH 2 = 70%
GW = 187.6173 kg (1 + 0.081)
GW = 202.8143 kg
17. Determine the quantity of heat in KJ/min required to raise 14m
3
/min at 80%RH of air from 20°C to
35°C dry bulb.
a. 87.62 b. 247.95 c. 287.62 d. 97.
SOLUTION:
From psychrometric chart
(80% RH, 20
o
C) Enthalpy=49.92 KJ/kg
Specific volume=0.846 m
3
/kg
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
(RH=32%, 35
o
C) Enthalpy=65.
Specific volume=0.91 m
3
/kg
18. Separation of two or more components of a liquid solution cannot be achieved by
a. fractional crystallization c. absorption
b. liquid extraction d. evaporation
19. A mixture of ammonia and air at a pressure of 745 mm Hg and a temperature of 40 ℃ contains
4.9% NH
3
by volume. The gas phase at a rate of 100 cfm through an absorption tower in which only
ammonia is removed. The gases leave the tower at a pressure of 740 mm Hg, a temperature of
20 ℃ and contain 0.13% ammonia by volume. Calculate the rate of flow of gas leaving the tower
in cfm.
a. 89.5 b. 98.2 c. 107.7 d. 122.
SOLUTION:
𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑔𝑎𝑠
3
3
3
Solving simultaneously:
𝑒𝑥𝑖𝑡 𝑔𝑎𝑠
3
3
20. Based from the preceding data, calculate the weight of ammonia in the tower in lb/min
a. 9.2 b. 6.4 c. 3.2 d. 0.
SOLUTION:
21. What is the relative volatility of benzene to pentane at 100
0
F and 465 psia?
a. 15.2 b. 0.75 c. 3.20 d. 0.
SOLUTION:
𝐴𝐵
81258
8745
𝐴𝐵
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
Ans. T = 102
0
C
25. What is the dew point of an equimolar mixture of benzene and toluene at 101****. 3 kN/m
2
a. 355 K b. 365 K c. 372 K d. 378 K
SOLUTION:
Using the T-x-y Diagram for Benzene-Toluene System at 1 atm:
Ans. Dew Point = 372.15 K
26. If the temperature of the liquid is lower than the saturation temperature for the existing pressure, it
is
a. saturated liquid c. subcooled liquid
b. saturated vapor d. superheated vapor
27. An equimolal mixture of benzene and toluene is subjected to a simple batch distillation at
atmospheric pressure. If the distillation is discontinued when the mols of distillate amount to 60%
of the mols charged, determine the mole fraction of benzene in the residue.
Assume the relative volatility is 2.
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
a. 0.43 b. 0.56 c. 0.62 d. 0.
SOLUTION:
[𝑙𝑛 𝑙𝑛 (
)]
x=0.
x of benzene in residue=1-0.2926= 0.7073 ≈ 0.
Ans. x of benzene in residue= 0.71; D
28. Hydrogen will diffuse compared to oxygen.
a. half as fast b. two times slower c. four times faster d. four times slowe
Explanation:
According to Graham's law of diffusion, the rate of diffusion or movement of gas is inversely
proportional to the square root of its molecular weight. Hydrogen effuses four times as rapidly
as oxygen.
29. The apex of an equilateral triangular coordinate (in ternary liquid system) represents
a. a pure component c. a ternary mixture
c. a binary mixture d. an insoluble binary system
30. It is desired to increase the humidity of the air using a heater and adiabatic humidifier. The ambient
air enters the heater at 80
0
F dry bulb and 76.
0
F dew point temperature. The exhaust from the
adiabatic humidifier has a dry bulb temperature of 110
0
F and 0.0265 lb water/lb dry air humidity.
What is the wet bulb temperature (
0
F) of air entering the heater?
a. 85 b. 80 c. 78 d. 74
SOLUTION:
Using Psychrometric Chart: Wet bulb temperature= 78
0
F; C
31. Based from the preceding problem, what is the percent relative humidity of air entering the
adiabatic humidifier?
a. 15 b. 17 c. 46 d. 90
SOLUTION:
Using Psychrometric Chart: %RH= 16.5%; B
32. Based from the preceding problem, what is the dew point of the air leaving the adiabatic
humidifier?
a. 85 b. 80 c. 78 d. 74
SOLUTION:
Using Psychrometric Chart: Dew Point= 85
0
F; A
33. A single effect evaporator is used to concentrate 7 kg/s of solution from 10% to 50% of solids. The
feed enters the evaporator at 294K and its specific heat is 3.76 kJ/kg-K. Steam is available at 205
kN/m
2
and evaporation takes place at 13.5 kN/m
2
. The specific heat of the thick liquor is 3.14 kJ/kg-
K. The overall heat transfer coefficient is 3 kW/m
2
- K. The condensate leaves the heating space at
352.7K. Calculate the amount of steam used in kg/s.
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
34. Based from the preceding problem, estimate the heating surface required in m
2
a. 38 b. 52 c. 60 d. 69
SOLUTION:
2
2
35. A sheet material measuring 3 ft square and 2 inches thick is dried from 50 to 2 percent moisture
content (wet basis) under constant drying conditions. The dry density of the material is 30 lb/ft
3
and
its equilibrium moisture content is negligible. Experiment showed that the rate of drying under the
correct constant conditions was constant at 1.0 lb/ft
2
- h between moisture contents 50 and 25%.
Below 25% the rate decreased. Calculate the time in hours for CRP.
a. 1.67 b. 1.56 c. 3.34 d. 3.
SOLUTION:
𝑠
30 ∗ 3 ∗ 2
12
1
- 5
1 − 0. 5
2
- 02
1 − 0. 02
X
𝑐
- 25
1 − 0. 25
𝑐
15
3 ∗ 1
𝑐
36. Based from the preceding problem, calculate the time for FRP.
a. 1.7 hrs b. 3.4 hrs c. 2.3 hrs d. 4.6 hrs
𝑓
𝑓
37. The liquid mixture of 60% benzene and 40% toluene is charged into a still pot where differential
distillation is carried at 1.2 atm absolute pressure. How much of the charge must be boiled away to
leave a liquid mixture containing 80% toluene? Assume the relative volatility to be 2.
a. 14.41% b. 41.14% c. 59.85 d. 85.59%
SOLUTION:
B=14.4075 V=85.
Ans. % evaporated =85.5925%; D
38. An ideal mixture of A and B is to be distilled continuously. If the relative volatility is constant at 1.
and feed is saturated liquid with 50 mol% A, feed rate is 200 lbmols/h. Distillate composition of 90%
A and bottoms composition of 10% A. Find the minimum number of stages.
a. 4.5 b. 5.8 c. 6.8 d. 7.
SOLUTION:
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
39. What is the BPR of a 20% LiNO 3 solution at 220
0
F.
a. 6
o
F b. 8
o
F c. 10
o
F d. 11
o
F
SOLUTION:
40. A single effect evaporator is operating under the following conditions: feed = 40,000 lb/h; x F
solid, TF=
0
F; steam temperature=
0
F and xP=45% solid. Separation temperature of the vapor
going into the condenser is 125
0
F. specific heat of all solution is 1 cal/g
0
C. Overall heat transfer
coefficient is 500 Btu/h- ft
2
0
F. Assume no boiling point rise, calculate the pounds of evaporation
per hour.
a. 44,000 b. 40,000 c. 35,600 d. 33,
SOLUTION:
Let m v
= mass flowrate of water evaporated
mp = mass flowrate of product
TMB:
𝑉
𝑝
SOLIDS BAL:
𝑝
𝑝
𝑉
𝑙𝑏
ℎ𝑟
𝑉
41. The diffusivity of ammonia (ft
2
/s) in air at 25
0
C and 1 atm is
a. 0.164 b. 0.229 c. 0.240 d. 0.
SOLUTION:
From Perry’s Chemical Engineering Handbook: Ans. Diffusivity= 0.229; B
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
SOLUTION:
- 8822 ∗. 3369
700 ∗. 235
45. If the temperature of ethanol – air mixture is increased to 100
O
F at constant pressure, what is its
mass diffusivity?
a. 0.56 b. 0.24 c. 0.15 d. 0.
SOLUTION:
𝑇 2
− 5
- 9278
300
3 / 2
𝑇 2
− 5
46. Using Antoine’s equation, the vapor pressure of tetrahydrofuran at 80
O
C is estimated to be psi.
a. 49.55 b. 38.68 c. 29.36 d. 22.
GIVEN:
A = 6.
B = 1202.
C = 226.
REQUIRED: P
𝐴−
𝐵
𝐶+𝑇 ; 𝑃 = 10
99515 −
29
254 + 80 = 1173. 1577 𝑚𝑚 𝐻𝑔 ; 𝑃 = 1173. 1577 𝑚𝑚 𝐻𝑔 (
7 𝑝𝑠𝑖
760 𝑚𝑚 𝐻𝑔
47. A vessel with a volume of 1m
3
contains liquid water and water vapor in equilibrium at 600 kPa. The
liquid water has a mass of 1 kg. Calculate the mass of the water vapor.
a. 0.99 kg b. 1.57 kg c. 1.89 kg d. 3.16 kg
GIVEN:
V = 1m
3 m
L
= 1 kg P = 600 kPa
REQUIRED: mv
SOLUTION:
𝑓
3
𝑔
3
48. One half cubic meter per second of air at 15°C dry bulb and 13°C wet bulb temperature are mixed
with 0.20m
3
per second of air at 25°C dry bulb and 18°C wet bulb temperatures. Determine the dry
bulb temperature of the mixture
a. 20.
o
C b. 17.
o
C c. 19.
o
C d. 21.
o
C
GIVEN:
V 1 = 0.5 m
3
V 2 = 0.20m
3
REQUIRED: T
db
SOLUTION:
From the Psychrometric Chart
First Stream: Second Stream:
T
db
= 15°C T
db
= 25°C
Twb1 = 13°C Twb2 = 18°C
SLU | SEA | ChE 520L | 8:30 – 11:30 TTh| Coverage: Separations 2
W
1
= 0.0085 kg/kg W 2
= 0.011 kg/kg
v 1 = 0.825 m3 /kg v 2 = 0.858 m3 /kg
1
𝑉 1
𝑣 1
5 𝑚 3
825 𝑚 3 /𝑘𝑔
𝑘𝑔
𝑠
2
𝑉
2
𝑣
2
20 𝑚 3
858 𝑚 3 /𝑘𝑔
𝑘𝑔
𝑠
3
1
2
𝑘𝑔
𝑠
𝑑𝑏 3
𝑚 1
( 𝑇 𝑑𝑏 1
) +𝑚 2
( 𝑇 𝑑𝑏 2
)
𝑚 3
- 606
𝑘𝑔
𝑠
( 15°𝐶
)
𝑘𝑔
𝑠
(25°𝐶)
- 839
𝑘𝑔
𝑠
𝑑𝑏 3
49. Which of the following is NOT usually a factor in water flux computations?
a. Cp b. Tw c. kw d. Cv
50. Oxygen is diffusing through carbon monoxide under steady state conditions with CO non-diffusing.
The total pressure is 1 atm and the temperature 0
0
C. The partial pressure of oxygen at two planes
2 mm apart is 0.128 atm and 0.0642 atm, respectively. Calculate the rate of diffusion of oxygen in
kg-mols/m
2
- s.
a. 1.55 x 10
b. 2.16 x 10
c. 2.91 x 10
d. 1.98 x 10
GIVEN:
P = 1 atm
T = 0°𝐶 = 273.15 K
R = 8314 m
3
. Pa/kg-mole. K
( z 2 - z 1 ) = 2.0 mm = 2 x 10
m
P
A
= 0.128 atm
PA2 = 0.0642 atm
D
AB
= 1.87 x 10- 5 m
2
/s
SOLUTION:
𝐴
𝐷 𝐴𝐵
(𝑃)
𝑅𝑇(𝑧 2
−𝑧 1
)
𝑃−𝑃 𝐴 2
𝑃−𝑃 𝐴 1
𝐴
- 87 𝑥 10 − 5 𝑚 2 /𝑠( 1 ∗ 101 , 325 )
8314 ( 273. 15 )( 2 𝑥 10 − 3 )
1 − 0. 0642
1 − 0. 128
𝐴
− 5
2
