King Saud University
Department of Chemical Engineering
February 09, 2009
Final Exam (Part 1) Closed Book
ChE 316
Time: 30min
Name:___________
No:______________
Total Pages: 2
Question 1:
In the adjoining figure, a feed containing two components A and B enters a membrane separator.
The semi-permeable membrane provides easier passage to solute A (PMA > PMB). Here, XF is feed
concentrations. The permeate stream is P, and the retentate stream is R, then
(1)
(a) XFA > XRA
(b) XFA = XRA
Semi-permeable membrane
(easier passage to A)
(c) XFA < XRA
(d) none of these
(2)
(a) XFA > YPA
(b) XFA = YPA
(c) XFA < YPA
(d) none of these
where, X FA is the mole fraction of component A in
the
feed, XRA is the mole fraction of A in the retentate, and
YPA is the mole fraction of A in the permeate. Given XFA =
0.09, the following values could be expected
(3)
(4)
(5)
(6)
α A,B= 1
YPB = 0.62
XRB = 0.62
(a) True
(a) True
(a) True
(b) False
(b) False
(b) False
A high value of separation factor means
(a) XFA << YPA
(c) both (a) and (b) are correct
(b) XFA >> XRA
(d) both (a) and (b) are incorrect
(7)
It is always important to have a membrane of small thickness.
(a) True
(b) False
(8)
Which of the following membrane module, you will not recommend for the case of the gas
permeation
(a) Plate and frame
(b) Spiral wound
(c) Hollow fiber
(d) All of these
(9)
The purpose of cascading the membrane modules in the membrane separation is
(a) to increase the degree of separation
(b) to recycle the permeate to feed
(c) to recycle the retentate to feed
(d) recycle both permeate and retentate
(10)
Using perfect mixing model, the concentration of component A in permeate is found to be
0.40. If the cross flow model is use, one expect the concentration to be
(a) 0.00
(b) 0.30
(c) 0.40
(d) 0.50
(11)
To obtain water for drinking from the seawater, the most common membrane separation
process used is,
(a) gas permeation
(b) dialysis
(c) osmosis
(d) reverse osmosis
(12)
For the separation of two components of a gas mixture using a micro-porous membrane,
their molecular weight should be _________.
(a) equal
(b) closely similar
(c) greatly different
(d) can not say
(13)
To remove urea and other smaller solutes from the blood, the following membrane-based
separation process is used
(a) dialysis
(b) osmosis
(c) reverse osmosis
(d) gas permeation
(14)
For the separation based on the reverse osmosis, the driving force is
(a) difference in salt concentration on the
(b) difference in solvent/water
feed and the permeate side
concentration on feed and permeate side
(c) difference in the pressure on the feed
(d) none of these
and permeate side
(15)
The humidity of a vapor (A)-dry gas (B) mixture is basically,
(a) mass fraction of A on dry basis
(b) mole fraction of A on dry basis
(c) mass fraction of A on wet basis
(d) all of these
(16)
For which of the following system, the wet bulb temperature and the adiabatic saturation
temperature are almost equal.
(a) acetone-air system
(b) benzene-nitrogen system
(c) toluene-nitrogen system
(d) water vapor- air system
(17)
For a unsaturated air-water vapor mixture
(a) Tw > Td
(b) Tw = Td
(c) Tw < Td
(d) none of these
where Td is dry bulb and Tw is wet bulb temperature.
(18)
The concentration polarization effects are desirable (good) for membrane
based separation processes.
(a) true
(b) false
(19)
The cooling tower is used to
(a) heat water
(b) cool water
(c) decrease air humidity (d) increase air humidity
(20)
The air used in cooling tower must be unsaturated
(a) true
(b) false
(21)
To decrease the height of the cooling tower, one should decrease the gas
flow rate.
(a) true
(b) false
(22)
For drying, one should have
(a) dry air
(c) high air velocity
(b) hot air
(d) all of these
(23) The equilibrium moisture content, X*, depends upon the
(a) air humidity
(b) critical moisture content, Xc
(c) both (a) and (b) are correct
(d) both (a) and (b) are incorrect
(24)
The mode of heat transfer in the tray dryer is
(a) Convection
(b) Conduction
(c) Radiation
(d) none of these
(25)
The presence of fan or blower in the dryer enhance the heat transfer by
(a) convection
(b) conduction
(c) radiation
(d) all of these
Answers
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King Saud University
Department of Chemical Engineering
February 09, 2009
Final Exam (Part 2) Closed Book
ChE 316
Time: 150 min
Name:___________
No:______________
Total Pages: 3
Question 2
Part A
An air-water mixture at 1 atm pressure has a humidity of
0.015 kg water vapor per kg dry air. Its dry bulb
temperature is 50 C. Determine
 wet bulb, dew point, and the adiabatic saturation
temperature using the chart.
 Saturation humidity
 Enthalpy using reference temperature 0 degree C.
Part B
An air-water mixture at 150 kPa pressure has a dry bulb
temperature is 50 C. Its wet bulb temperature is 40 C.
Determine its percentage humidity.
Data: Specific heats in (kJ/kg.K) of dry air = 1.005; of water
vapor =1.88, latent heat of vaporization of water at 0 C = 2501
kJ/kg
Question 3
PartA
The bone dry sample weight (dry solid only) of sample of wet solid is 5 kg. Its equilibrium moisture content
(X*) is 0.05 kg water per kg dry solid. Its total moisture content (Xt) is 1 kg water per kg dry solid (Use this
for next three questions).
 free moisture content of the solid sample
 total weight of the wet solid
 total weight of the solid sample after long time drying when equilibrium is reached
Part B
In a drying experiment, what will be the % enhancement in the value of the heat transfer coefficient if the
velocity of the air is enhanced by two times for following cases.
 The flow of air is parallel to the drying surface
 The flow of air is perpendicular to the drying surface
Question 4
The wet feed material to a continuous drier contains X 1= 30 wt% water on a dry basis (kg water per kg dry
solid and dried to X2= 10 wt% by countercurrent air flow. The feed rate is Ls = 1000 kg dry solid/h. Fresh air
to the system is at TG1= 25.6 C and has a humidity of H1=0.007 kg water/kg dry air. The moist air leaves the
dryer at TG4 = 37.8 C and H4 =0.022 and a part of it is recirculated and mixed with the fresh air before
entering a heater. The heated mixed air enters the dryer at T G3 = 90 C and H3 =0.01. The solid enters at TS1 =
25 C and leaves at TS2 = 50 C. The heat capacity of the dry solid is 1.5 kJ/(kg dry solid.K) and the heat
capacity of the liquid moisture is 4.2 kJ/(kg H2O.K). Calculate
 the air flowrate in the dryer
 the flowrate of the fresh air and the recycled air
 the heat loss from the dryer in KW
 the flowrate of wet solid feed entering the dryer in kg total solid per hour
Re-circulated air, (5)
Dryer
(6)
(1)
(2)
(3)
Heater
Dried solid, X2
(4)
Wet solid, X1
Note: Number indicates stream numbers. Thus, G1 is dry air flow, H1 is humidity, TG1 is temperature, HG1
is enthalpy of stream 1. Similarly, X (kg water/kg dry solid) is the moisture content of the solid sample.
Question 5
A membrane separation unit is being used to separate component A and B from a feed gas. The semipermeable membrane is more permeable to component A. As a result, permeate is richer in A, and the
retentate is richer in component B. It is desired to obtain a product with a high purity of the component B
using a two stage cascade system. Sketch the proposed cascade system.
Question 6 (Choose one from the following two questions)
A reverse osmosis membrane module is being used with the bulk condition on the feed side being 3.5 wt%
NaCl, 25 C, and 1200 psia and the permeate side being 0.10 wt% NaCl, 25 C, and 50 psia. The permeance
values are 15X10-6 g/cm2-s-atm for water and 20X10-6 cm/s for the salt. Ignore the mass transfer resistances
on both sides of the membrane. Calculate
 Osmotic pressure in atm
 Flux of water in kg/m2-hr
 Flux of salt through the membrane in kg/m2-hr.
(Data: MW of NaCl = 58.5, 1atm = 14.7 psi,)
Formula 1:water flux = permeance X (p – )
Formula 2: = 1.12 mi, where  is in psi, temp. T in K and mi = 2XCNaCl in mol/L for the salt
The feed to a membrane separator consists of 1000 lbmol/h of a mixture of 80% H2 and 20% CH4 at 500 psia.
Permeance values based on a partial pressure driving force are 3.4 X 10 -4 for hydrogen and 5.6 X 10-5 for
methane. Its units are lbmol/(h-ft2-psi). The flow pattern in the membrane separator is such that the permeate
side is well mixed and the feed side is in plug flow. The pressure on the permeate side is constant at 20 psia
and there is no pressure drop on the feed side. It is desired to compute the membrane area and the permeate
purity if 80% of the hydrogen is transferred to permeate. Assume log-mean partial pressure driving force.
 Write down independent equations to solve the problem.
 Identify the total number of variables in the problem.
 Identify the variables of the problem that should be fixed and mention their values.
 Identify the variables to be solved with the help of independent equations
HG  1.0051.88H T T0   0 H
HG is the air enthalpy, hLa is the liquid heat transfer coefficient and KGa is the gas
phase mass transfer coefficient in kg mol/s.m3.Pa, MB (=29) is the air molecular weight, CL (=4.19 kJ/kg.K) is the water
specific heat. Also, latent heat of water vaporization at 0 C is about 2501 kJ/kg.
The humidity, saturation humidity, percent humidity, percent relative humidity, wet bulb temperature eqn and enthalpy are
given as
h k M
H -H W
  c y B ; HG   CPA  H CPB T T0   0 H
T-Tw
w
H  
pA M A 
;
P

pA M B 

Hs  
pA0 M A 
;
P

pA0 M B 



Hp   H Hs 100; R.H .  p A p0A 100
Humid volume, m3/kg dry air   2.83 103  4.56  103   T  K 
Density,
  1     H
The heat transfer coefficient in W/m2K is given as h = 0.0204G0.8 for parallel flow and h = 1.17G0.37 for
perpendicular flow of air. Here, G   v is air flow-rate in (kg/m2h).
h T  Tw 
LS dX
w
A R
Specific heats in (kJ/kg.K) of dry air = 1.005; of water vapor =1.88; Latent heat of vaporization of water at 0 C = 2501 kJ/kg
The rate of drying and the time of drying can evaluated using
Rc  3600
; t
MW of NaCl = 58.5; 1atm = 14.7 psi; water flux = permeance Xp – ) ; Osmotic pr.  = 1.12 mi,
where  is in psia and m is the salt concentration in mol/litre.
Assume any missing data giving proper justification