mass transfer (stoffaustausch) fall 2013 - eth z · 1 copy of the book “diffusion” (2nd or 3rd...

1

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

2


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

3



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)

4



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

5



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.

6


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

7


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

8


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

9


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

10


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 )

11


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

12


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.

13


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)

14


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

15


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


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:

17


( ) ( ) ( ) 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

mass transfer (stoffaustausch) fall 2013 - eth z · 1 copy of the book “diffusion” (2nd or 3rd...

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