Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
Mass Transfer (Stoffaustausch) HS 2011
Test 2
22. November 2011
Name:______________________________________________
Legi-Nr.:_____________________________________________
Test Duration: 45 minutes
Permitted materials:
1 calculator
1 copy of Cussler’s book “Diffusion” (2nd or 3rd edition)
1 printout of the lecture script, without notes on the exercises
1 sheet (2 pages) summary
NOT permitted:
exercises (also handwritten on summary or textbook),
notebooks, mobile phones, any device with wireless
communication ability
Problem 1 (50 points)
Reed diffusers are a new eco-friendly way to fill rooms with a scent of different aromas,
e.g. vanillin. The cylindrical reeds are saturated with the vanillin scented oil by capillary
forces, releasing the aroma radially from the surface of the cylinders (curved surface
area) into the air through a stagnant film with thickness 0.2 mm. The upper 13 cm of
the wooden cylinders with a diameter of 0.3 cm are exposed to the air of a large room.
Assume that the air is well ventilated and that the system is dilute.
a) Draw a detailed sketch of the problem. (8 points)
b) Calculate the diffusion coefficient for vanillin in air, using the following equation
(Fuller, Schettler, and Giddings, 1966): (10 points)
~
~
10 -3 T1.75 (1/M 1  1/M 2 )1/2
D
2
p ( i Vi1 )1/3  ( i Vi2 )1/3


c) How many reeds are necessary to maintain a concentration of 60 μg/m3 in the room,
if 2.1 m3/min of the air scented with vanillin is replaced by ventilation with unscented
air?
Write the complete generalized mass balance and state your assumptions to simplify it.
(32 points)
1
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
Data:
Temperature T = 25°C
Vapor pressure of vanillin = 0.3 Pa
Molar mass of vanillin = 152.12 g/mol
Molar mass of air = 28.97 g/mol
Universal gas constant R = 8.314 J/mol K
Chemical formula of vanillin: C8H8O3
2
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
Solution 1
a) Sketch
L=13 cm
cv,sat
croom
r2
r1
8 points
b)
The atomic diffusion volumes are (Table 5.1-4 in Cussler, 2009):
Vair  20.1 A 3
Vvanillin  816.5  81.98  35.48  20.2  144.08 A 3
6 points
With p = 1 atm and T = 298 K we can calculate the diffusion coefficient using the given
correlation:
1
T
D  10
3
1.75
 1
1 2
   

 Mair Mvanillin 
1

p   Vair  3

1
1 2
 1

298 

 28.97 152.12   6.84  10 2 cm2 / s
 10 3
2
1
1
1 2



3
3
1  20.1   144.08  3 
   Vvanillin  



1.75
4 points
3
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
or the diffusion coefficient using wrong vanillin molecular volume:
3
Vvanillin  164.28 A
1
T
D  10 3
1.75
 1
1 2
   

 Mair Mvanillin 
1

p   Vair  3

1
1 2
 1

298 

 28.97 152.12   6.45  10 2 cm2 / s
 10 3
2
1
1
1 2



3
3
3
1
20.1
164.28





   Vvanillin  





1.75
or 3.5 points
c)
The generalized mass balance for cylindrical coordinates is given by:
 1   c V  1  2 c V  2 c V   c V v c V
c V
c 
 D
 2
 2    vr

 v z V   rV
r
2


t
z  
r
z 
r 
 r r  r  r 
Assumptions:
steady state: dcV/dt = 0
symmetric: d/d =0
no reaction: rV=0
no diffusion in axial direction: d/dz = 0
since we assume that the system is dilute, convection can be neglected:
vr  v  vz  0
The mass balance then reduces to:
 1   cV  
0  D
r

 r r  r  
or:
0
  c V 
r
r  r 
6.5 points
The first integration:
K1  r
cV
r
1 point
4
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
The second integration:
cV =K1  ln r  K 2
1 point
The boundary conditions are:
c = cv,sat
c = croom
r = r1
r = r2
2 points
Using these boundary conditions leads to:
K1 
c v,sat  c room
ln r1  ln r2
and K 2  c room  K1 ln r2
2 points
The concentration profile in the stagnant film is:
 ln r  ln r2 
c v (r)   c v,sat  c room  
  c room
 ln r1  ln r2 
1 point
The flux of vanillin molecules from one reed can be calculated with Fick’s first law:
jv   D
j1  
dc v
dr
D   c v,sat  c room  1

ln r1  ln r2
r
1 point
1 point
The saturation concentration, cV,sat, can be calculated from the vapor pressure and the
ideal gas law:
c v,sat 
pv
0.3Pa
mol
mol

 1.2  10 4 3  1.2  10 10
J
R  T 8.314
m
cm 3
 298K
mol  K
3.5 points
5
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
The release rate of vanillin is:

 vanillin  j1  A  M
m
vanillin  

1
cm 2
 6.84  102
0.15
sec
ln
0.17
D   c v,sat  c room  1

 2r1L  M
vanillin
r1
r
1
ln
r2
mol
mol 
g

  1.2  1010
 3.9  1013
 2  13cm  152.12
3
3 
cm
cm 
mol

 48.73g / min
6 points
Vanillin loss by ventilation:
g
m3
g
L vanillin  60 3  2.1
 126
m
min
min
2 points
Number of reeds:
g
g
 vanillin  126
n reed  L vanillin / m
/ 48.73
 2.59
min
min
Three reeds are necessary to maintain the required concentration.
3 points
2 points
6
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
Problem 2 (50 points)
A cylindrical beaker-glass with a height of 25 cm is filled up to the top with water. One
droplet of butanol (C4H10O), behaving like a rigid sphere, with a diameter of 0.8 cm is
released from the bottom and rises to the water surface at the top with a constant
velocity of 0.1 m/s. While the droplet rises butanol diffuses into the water. The solubility
of butanol in water is 7.4×10−2 g/cm3. Neglect the diffusion of water into the butanol
droplet and assume that the water remains pure at all times.
a) Make a detailed sketch and label all important values. (10 points)
b) What is the initial mass transfer coefficient at t = 0? (10 points)
c) Derive a function that describes the diameter of the butanol droplet as a
function of time t. (15 points)
d) What percentage of mass of the droplet has dissolved into the water at the
timeit reaches the water surface? (15 points)
Data (at the operating conditions):
density of butanol:
0.81 g/cm3
diffusion coefficient of butanol in water: 8.8·10-5 cm2/s
Solution 2
a) Sketch
10 points
7
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
b) The mass transfer correlation for small liquid drops rising in an unstirred solution is:
kd
 dv 
 1.13  
D
D
0.8
At t = 0 the mass transfer coefficient, k, is:
k0  1.13
D  dv 
 
dD
0.8
cm 2
8.8  10
s
 1.13
0.8cm
-5

cm 
 0.8cm  10

s 

2
 8.8  10-5 cm 


s 

0.8
 1.15
cm
s
10 points
c) The mass flux balance is:
m
  kA  coil  cethanol 
t
0.8
  3 
D  dv 
2
 d
  1.13   d   coil  cethanol 
6
t 
dD
0.8
 d
D  dv 
 3d
 1.13   d 2  coil  cethanol 
6 t
dD
2
dt
d
d0
 d  1.13
0.2
2 D 0.2 v 0.8

t
 coil  cethanol    t
0
5 1.2 5 1.2
2 D 0.2 v 0.8
d t  d 0  1.13
 coil  cethanol  t

6
6
5

6
12 D 0.2 v 0.8
d t   d 01.2  1.13
 coil  cethanol  t 
5


15 points
8
Prof. Dr. Sotiris E. Pratsinis
Particle Technology Laboratory
Sonneggstrasse 3, ML F13
Tel.: +41-44-632 25 10
http://www.ptl.ethz.ch
d) The residence time until reaching the water surface is:
t
h 25 cm

 2.5s
v 10 cm
s
The droplet diameter at the water surface is:
5

6
12 D 0.2 v 0.8
d t   d 01.2  1.13
 coil  cethanol  t 
5


5
0.2
0.8

6

cm 2   cm 


12  8.8  105
10
 

s  
s  
1.2




2 g
   0.8cm   1.13
 0  2.5s 
 7.4  10
3
g
cm


5  0.81 3


cm




 0.219cm
The mass loss in percent is:
 d3 
 0.2193t 
m  100  1  t3   100  1 
  99.19%
0.830 
 d0 

15 points
9