mass transfer (stoffaustausch) fall 2013 - eth z · 1 copy of the book “diffusion” (2nd or 3rd...
TRANSCRIPT
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Mass Transfer (Stoffaustausch) Fall 2013
Examination 21. August 2014
Name:______________________________________________
Legi-Nr.:_____________________________________________
Edition “Diffusion” by E. L. Cussler: none 2nd 3rd
Exam Duration: 120 minutes
The following materials are not permitted at your table and have to be deposited in front or back of the examination room during the examination:
bags and jackets
exercises of the mass transfer lecture (also handwritten on summary sheet or textbook)
notebooks, mobile phones, devices with wireless communication ability The following materials are permitted at your table:
1 calculator
1 copy of the book “Diffusion” (2nd or 3rd edition) by E. L. Cussler
1 printout of the lecture script
1 sheet (2 pages) summary in format DIN A4 or equivalent Please read these points:
write your name and Legi-Nr. on each sheet of your solution
begin each problem on a new sheet
write only on the front side of each sheet
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Problem 1 (25 points)
A cylindrical container filled with liquid aniline (molar mass: 93.13 g/mol) is stored in a large
room. The diameter of the container is 45 cm. The ambient air (1 atm) is initially aniline-free
and the room temperature is constant at 20 oC. At t = 0, the lid of the container is removed
and the aniline evaporates slowly.
a) Calculate the diffusion coefficient of aniline in air. Use the empirical equation from Fuller,
Schettler, and Giddings (1966). (8 points)
b) Estimate whether the aniline diffusion has reached steady state at 15 cm above the
surface after 4 min. (3 points)
c) Calculate the concentration of aniline in the air at the position and time of question b.
(6 points)
d) How much mass of aniline will have evaporated after 4 min? (8 points)
Additional Data:
Chemical formula of aniline: C6H5NH2 or
Composition of air: 21 vol% O2, 79 vol% N2
Molar masses: O2=32 g/mol, N2=28 g/mol
Saturation vapor pressure of aniline at 20 oC: 0.5 kPa
R = 8.314 J/mol K
erf 𝑥 =2
√𝜋∫ 𝑒−𝑡
2𝑑𝑡
𝑥
0
Error Function
x erf(x) x erf(x) x erf(x) x erf(x) x erf(x) x erf(x)
0.00 0.00000 0.52 0.53790 1.04 0.85865 1.56 0.97263 2.08 0.99673 2.60 0.99976
0.02 0.02256 0.54 0.55494 1.06 0.86614 1.58 0.97455 2.10 0.99702 2.62 0.99979
0.04 0.04511 0.56 0.57162 1.08 0.87333 1.60 0.97635 2.12 0.99728 2.64 0.99981
0.06 0.06762 0.58 0.58792 1.10 0.88021 1.62 0.97804 2.14 0.99753 2.66 0.99983
0.08 0.09008 0.60 0.60386 1.12 0.88679 1.64 0.97962 2.16 0.99775 2.68 0.99985
0.10 0.11246 0.62 0.61941 1.14 0.89308 1.66 0.98110 2.18 0.99795 2.70 0.99987
0.12 0.13476 0.64 0.63459 1.16 0.89910 1.68 0.98249 2.20 0.99814 2.72 0.99988
0.14 0.15695 0.66 0.64938 1.18 0.90484 1.70 0.98379 2.22 0.99831 2.74 0.99989
0.16 0.17901 0.68 0.66378 1.20 0.91031 1.72 0.98500 2.24 0.99846 2.76 0.99991
0.18 0.20094 0.70 0.67780 1.22 0.91553 1.74 0.98613 2.26 0.99861 2.78 0.99992
0.20 0.22270 0.72 0.69143 1.24 0.92051 1.76 0.98719 2.28 0.99874 2.80 0.99992
0.22 0.24430 0.74 0.70468 1.26 0.92524 1.78 0.98817 2.30 0.99886 2.82 0.99993
0.24 0.26570 0.76 0.71754 1.28 0.92973 1.80 0.98909 2.32 0.99897 2.84 0.99994
0.26 0.28690 0.78 0.73001 1.30 0.93401 1.82 0.98994 2.34 0.99906 2.86 0.99995
0.28 0.30788 0.80 0.74210 1.32 0.93807 1.84 0.99074 2.36 0.99915 2.88 0.99995
0.30 0.32863 0.82 0.75381 1.34 0.94191 1.86 0.99147 2.38 0.99924 2.90 0.99996
0.32 0.34913 0.84 0.76514 1.36 0.94556 1.88 0.99216 2.40 0.99931 2.92 0.99996
0.34 0.36936 0.86 0.77610 1.38 0.94902 1.90 0.99279 2.42 0.99938 2.94 0.99997
0.36 0.38933 0.88 0.78669 1.40 0.95229 1.92 0.99338 2.44 0.99944 2.96 0.99997
0.38 0.40901 0.90 0.79691 1.42 0.95538 1.94 0.99392 2.46 0.99950 2.98 0.99997
0.40 0.42839 0.92 0.80677 1.44 0.95830 1.96 0.99443 2.48 0.99955 3.00 0.99998
0.42 0.44747 0.94 0.81627 1.46 0.96105 1.98 0.99489 2.50 0.99959 3.02 0.99998
0.44 0.46623 0.96 0.82542 1.48 0.96365 2.00 0.99532 2.52 0.99963 3.04 0.99998
0.46 0.48466 0.98 0.83423 1.50 0.96611 2.02 0.99572 2.54 0.99967 3.06 0.99998
0.48 0.50275 1.00 0.84270 1.52 0.96841 2.04 0.99609 2.56 0.99971 3.08 0.99999
0.50 0.52050 1.02 0.85084 1.54 0.97059 2.06 0.99642 2.58 0.99974 3.10 0.99999
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.00 0.50 1.00 1.50 2.00 2.50 3.00
x
erf
x
2
0
2x
terf x e dt
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Problem 2 (25 points)
Acetone is stored in a vessel with a cylindrical capillary on top, as shown in Setup I (Figure
1). The capillary is 2 mm in diameter and 10 cm in length. The acetone evaporates from the
liquid surface at (1) and flows through the capillary towards (2). There is continuous flow of
fresh air above the capillary that quickly removes the evaporated acetone from (2). Assume
that there is no uptake of air by the acetone. The total pressure of the system is 1 atm.
a) If the temperature of the system is 20 oC, calculate the total molar flow rate (evaporation
rate) of acetone. (9 points)
b) How much faster compared to the question a) is the evaporation rate if the temperature
of the system is 40 oC. (8 points)
c) To minimize the evaporation of acetone at 40 oC, the cylindrical capillary is replaced by a
conical one (see Setup II of Figure 1). The diameter at the top of the conical capillary (2)
is 1 mm. How much slower is the evaporation rate now compared to that of b)?
(8 points)
Additional data:
Temperature, oC 20 40
Saturation pressure of acetone, kPa 25.3 58.7
Diffusion coefficient of acetone in air, cm2/s 0.107 0.120
R = 8.314 J/mol K
Figure 1. Schematic of the acetone vessel with cylindrical capillary (Setup I, questions a-b) and conical one (Setup II, question c)
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Problem 3 (25 points)
A hailstone1 of 8 mm initial diameter is falling from high altitude. As it moves through the air, its size reduces because of sublimation. The local gas-phase mass transfer coefficient for the sublimation of ice changes as a result. Assume that the terminal free fall velocity depends on the diameter of the hailstone as
0 / 0.9m s d m
Make the following assumptions:
I. The hailstone always has the terminal free fall velocity corresponding to its diameter
(pseudo-steady state).
II. The ambient air is at -2 ºC, 1 atm total pressure and 70% relative humidity.
III. The hailstone temperature is constant at -2 ºC.
IV. The hailstone remains spherical all the times.
a) Calculate the initial (for d = 8 mm) mass transfer coefficient. (9 points)
b) Calculate the time required for the initial size of the hailstone to be reduced to 7.99 mm.
(11 points)
c) What is the initial mass transfer coefficient (for d = 8 mm) if we neglect convection?
(5 points)
Additional Data:
Diffusivity of water vapor in air: 0.178 cm2/s Density of air: 1.29 kg/m3
Density of ice: 917 kg/m3
Molar mass of air: 28.9 g/mol
Molar mass of ice: 18 g/mol
Viscosity of air: 1.7∙10-5 kg/m∙s Saturation vapor pressure of ice at -2 ºC: 0.00511 atm Universal gas constant, R: 0.082 L∙atm/(mol∙K)
Hint:
2
log
( / )
bx a a bxdx
a x b b
1 in german: Hagelkorn
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Problem 4 (25 points)
The catalytic converter of a car contains a monolith (like in Figure 2) that consists of 200
uniform circular channels, internally coated with catalysts. Each channel has inner diameter
of 0.02 cm and a length of 50 cm. The gas mixture reaches the inlet of the catalytic converter
at a flow rate of 5.25∙10-6 m3/s. The gas mixture has a methane (CH4) concentration of
0.001mol/m3 at the inlet of the converter. While the gas mixture passes through the channels,
the CH4 is catalytically oxidized to CO2 and H2O. The reaction is first order with respect to
CH4 and irreversible with rate constant of 4.2·10-4 m/s. The gas flow through each channel is
laminar. Assume excess of oxygen. Assume constant temperature and pressure throughout
the system.
CH4 + 2O2 → CO2 + 2H2O
a) Draw a detailed sketch of the concentration profile and the processes taking place in a
single channel and indicate all important parameters. (5 points)
b) Calculate the overall mass transfer coefficient in one channel. (8 points)
c) Starting with the mass balance over a section of the channel, derive an equation for the
CH4 concentration as a function of the channel length. (6 points)
d) Calculate the concentration of CH4 at the outlet of the channel. By what factor did the
initial concentration decrease? (6 points)
Additional Data:
Diffusion coefficient of CH4 in the gas mixture: 2.01×10-5 m2/s
Figure 2. Typical monolith in the catalytic converter of a car.
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Solution 1
a)
The required gaseous diffusion coefficient can be calculated by the following empirical
equation (Fuller, Schettler, and Giddings, 1966/ Cussler’s book, 3rd ed., Eq.5.1-9):
1.75 1/2-3 1 2
21/3 1/3
i1 i2
T (1/M 1/M )D 10
p V V
i i
(1)
where Vij are the diffusion volumes of parts i of the molecule j, p is in atm and T is in K.
For air (molecule 1):
1M 0.21 32 0.79 28 28.84 / ( 29 / ) g mol g mol
and
i1V 20.1i (Cussler’s book, 3rd ed., Table 5.1-4)
For aniline (molecule 2):
2M 93.13 / / g mol
i2V 6 16.5 7 1.98 1 5.69 20.2 98.35 i
Also:
p = 1 atm
T = 20+273=293 K
So, from Eq.1 we get:
1
21.75
-3
21 1
3 3
1 1293
28.84 93.13D 10
1 20.1 98.35
-2 2D 8.22 10 / cm s
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
b)
21 lcriterion
Fo D t
(1)
2
22
15criterion 11.4
8.22 10 4min 60min
cm
cm
s
s
criterion 1 non steady state
c)
Since the diffusion takes place at non-steady state, the semi-infinite slab approach should be
used. So, the concentration profile of aniline (species 1) in the air is described by (Cussler’s
book, 3rd ed., Chapter 2.3-Unsteady Diffusion in a Semi-infinite Slab, Eq.2.3-15):
10
1 10
1c cerf
c
(z, t)
c
where:
1c 0
z
4Dt
Thus:
101
zc c 1 erf
4Dt
where:
1sat10 1sat
pc c
RT
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
3
10 3
0.5 10 Pa molc 0.205
J m8.314 293K
mol K
and:
22
z1.69
4Dt cm s4 8.22 10 4min 6
15cm
0s min
erf erf (1.69) 0.98307
So:
1
3 3
1 0 10c c 1 0.98307 0.016 3.48 10 mol9c / m
d)
Since the diffusion takes place at non-steady state, the flux across the interface (at z = 0) as
a function of time is given by (Cussler’s book, 3rd ed., Eq.2.3-18):
1 10 1z 0
Dj c c
t
1 10z 0
Dj c
t
and the flow rate across the interface at z=0 as a function of time is given by:
1 1z 0 z 0J A j
1 10z 0
DJ A c
t
So, the amount of aniline that evaporates after time tf is calculated by integrating1 z 0
J
from
t=0 to t=tf:
ft
zA dtJn
0
01
ft
1 10
0
D 1n A c dt
t
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
ft
1 100
Dn 2A c t
1 10 f
Dn 2A c t
where:
2dA
4
So:
2 22 4
3
223
1
cm m8.22 10 10
ss cm(0.45m) moln 2 0.205 1.64 10 mol
4 m4min 60
min
And:
3
1
gm 1.64 10 mol 93.13
mol
1m 0.152g
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Solution 2
a)
The total flux of acetone in the capillary, which is independent of l, is given by (Cussler’s
book, 3rd ed., Equation 3.3-11):
11
10
1ln
1
lyDcn
L y
where:
1 1( ) 0 ly z L y
1 11 10 1
25.3KPa( 0) 0.250
101.3KPa Ideal gas lawsat sat
sat
tot
c py z y y
c p
where:
5
3 5
3
1.013 1041.6 / 4.16 10
8.314 293
totp Pa molc mol m
JRT cmK
mol K
The total flow rate is therefore:
1 1 N n A
with:
2
22 2 2
dA
4
(2mm)A mm 10 cm
4
So:
25
32 2
1
0.107 4.16 101
ln 1010 1 0.250
cm mol
s cmN cmcm
94.02 10 mol
s
(Intermediate result:
7
1 21.28 10
moln
cm s )
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
b)
It is: 1, 1,
1 1
new newn A n
n A n
At 40 oC, the acetone-air interface (l = 0), the mole fraction of acetone is:
1 11 10 1
58.7KPa( 0) 0.58
101.3KPa
Ideal gas lawsat satsat
tot
c py z y y
c p
where:
53 5
3
1.013 1038.9 / 3.89 10
8.314 313
totp Pa molc mol m
JRT cmK
mol K
Thus:
25
3
1,
0.120 3.89 101
ln10 1 0.58
new
cm mol
s cmncm
7
24.04 10
mol
cm s (7)
Therefore,
7
21, 1,
71 12
4.04 10
3.16
1.28 10
new new
moln A n cm s
moln A n
cm s
times faster
c)
11 1(1 )
dyn y Dc
dz
1 11(1 )
N dyy Dc
A dz
where:
22( )·z
4 4
o LL
d dd zA d
L
So:
1
10
11
10
1·
1
Lyz L
z y
dyN dz D c
A y
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
1 1
00 10
0
4 11· ln
( )( ) 1
L
L
LLL
N L yD c
d dd d yd z
L
1 1
0 0 10
4 11 1· ln
( ) 1
L
L L
N L yD c
d d d d y
11
0 10
14· ln
1
L
L
yLN D c
d d y
0 1
1, 2
10
1ln
4 1
L Lnew
d d yDcN
L y
Thus:
0
1, 2
2
01, 0
14 0.5( ) 2
4
L
new L
new
d dN d mm
dn A d mm
which means 2 times slower.
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Solution 3
a)
Forced convection around a sphere
3/12/1
6.00.2
D
vd
D
dk rel
(1)
1/2 1/3
1/2 1/3
2.0 0.6
2 0.90.6
reld vDk
d D
Dk D
d v D
(2)
2Dk C
d (3)
1/2 1/30.9
0.6 0.25 /C D cm sv D
kinematic viscosity:
55 2
3
1.7 10 / ( )1.32 10 /
1.29 /
kg m sm s
kg m
1/31/22 2 2
2 2
2 0.178 / 0.178 / 0.8 0.72 / 0.132 /0.6
0.8 0.8 0.132 / 0.178 /
cm s cm s cm cm s cm sk
cm cm cm s cm s
0.697 /k cm s
b)
The mass transfer equation for the sublimation of the naphthalene sphere is:
1 1i 1N A k (c c ) A (4)
where c1i and c1 are the concentrations of water vapor at the interface and in the bulk phase
of air, with:
sat1i
1 1i
P 0.00511atmc 0.23mol
RT 0.082L atm / mol / K 271K
c RH c 0.7 0.23mol 0.161mol
(5)
and A is the interfacial area, given by:
2A d (6)
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
The dissolution rate of the hailstone is also given by:
3
21
dm dd 6M ddMN A ddt dt dt2M
(7)
So:
2 21i
ddd k c d
dt2M
(8)
f
0
dt
1i 10 d
1dt dd
k2M (c c )
(9)
applying eq. 3:
f
0
dt
1i 10 d
1dt dd
2D2M (c c )C
d
(10)
ff 0
01i 1
2D Cdt C d d 2Dlog
2D Cd2M (c c )
(11)
3
3 2
22
2
917kg / mt
2 18 10 kg / mol (0.23 0.161)mol(0.25cm / s)
2 0.178cm / s 0.25cm / s 0.799cm0.25cm / s 0.799 0.8 cm 2 0.178cm / s log
2 0.178cm / s 0.25cm / s 0.8cm
t 18min
c)
Mass transfer coefficient of pure diffusion:
22 2 0.178 /
0.8
0.445 /
D cm sk
d cm
k cm s
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
Solution 4
a)
Note: In the solution, concentration of CH4 = CA
b)
The overall mass transfer correlation is given by (Cussler, 3rd ed., Eq.16.3-7):
1 2 3 2
1
1 1 1
K
k k K
(1)
Since we have an irreversible reaction, it is 2 0 and therefore:
22
2
K
(2)
Now, we get following expression for overall mass transfer correlation:
1 2
1
1 1
K
k
(3)
Where k1 is the local mass transfer coefficient of CH4 (A) and κ2 is the reaction rate constant.
For k1 we can find the appropriate mass transfer correlation for fluid-solid interfaces, which is
laminar flow through a circular tube (Cussler, 3rd ed., Table 8.3-3):
1/32
1 1.62
d u Dk
LD d (4)
16
Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
where D is the given diffusion coefficient of CH4 in exhaust gas and u is the gas velocity,
which can be calculated from the exhaust gas feed rate in single channel, Q, and the tube
cross sectional area, Across, as:
cross
Qu
A (5)
Exhaust gas flowrate in single channel is calculated using cross section area of the channel
and the gas flowrate per unit cross section area of the channel.
38 2
2 cross section 3.14 10 0.833cross
mQ A flowrate per unit area of a channel m
s m
382.62 10
mQ
s
Now the velocity of the gas through the channel using equation 5.
38
2 2
2.62 10
0.833 /
0.00024
m
su m s
m
So, from Eq.4 we get:
1/32
2 2 5
1 25
0.0002 0.83 2.01 10
1.62 0.02430.0002
0.50 2.01 10
m mm
ms skm m s
ms
Consequently, the overall MTC from Eq.1 is:
4
41 2
1 14.13 10
1 1 1 1
0.0243 4.2 10
mK
sm mk
s s
(6)
c)
/i)
The mass balance over a section of the pipe is:
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Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch
( ) ( ) ( ) cross A A circ AA u c z c z z K A c z (7)
2 ( ) ( ) ( )4
A A Ad u c z c z z K d z c z
( ) ( ) 4 ( ) A A Ad u c z c z z K z c z
( ) ( ) 4( )
A A
A
c z z c z Kc z
z d u
4(z)
AA
dc Kc
dz d u (8)
Boundary conditions: 00: A Az c c ; : ( )A Az c c z (9)
After integration equation 8 using boundary conditions from equation 9, we obtain the
concentration profile of CH4 (A) as function of z:
0
4( ) exp
A A
Kc z c z
d u (10)
/ii)
For z L :
4
6
3 3
4 4.13 10
0.001 exp 0.5 7.05 10
0.0002 0.833
AL
m
mol molsc mmm m
ms
/iii)
Thus, the conversion of CH4 (A) is given by:
30
6
3
0.001
142
7.05 10
A
AL
molc m
molc
m
times lower