mass transfer (stoffaustausch) - eth zexercises of the mass transfer lecture (also handwritten on...

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Prof. Dr. Sotiris E. Pratsinis

Particle Technology Laboratory

Mass Transfer (Stoffaustausch)

Examination 13. August 2013 Name:______________________________________________ Legi-Nr.:_____________________________________________ Edition “Diffusion” by E. L. Cussler: none 2nd 3rd Test 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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Problem 1 (25 points)

A circular pond1 of 5 m diameter and 50 cm depth is filled with water of 30 °C. The air above the pond is stagnant and initially water water-free. Describe the evaporation of water by assuming diffusion only in vertical direction. a) Draw a detailed sketch of the problem and include the concentration profiles.

(4 points)

b) How does the water concentration in the air change as a function of time and

distance from the water surface. Please start with the differential equations and

indicate the boundary and initial conditions.

(6 points)

c) Calculate the diffusion coefficient of water in air at 30 °C using the Chapman-

Enskog equation.

(9 points)

d) How much has the water level dropped after 10 hours?

(6 points)

Data: Molecular weight of water: 18 g/mol Density of water: 1 g/cm3

Saturation pressure of water: ln(psat [Pa]) = 26.3 – 5433/T [K]

1 Pond = Teich (German)

3/17



Solution 1 a)

b) To describe the evolution of the water concentration the semi-infinite slab approach

should be considered:

2

1 1

2

C CD

x t

The initial and boundary conditions are

For t = 0, all z, CA = 0.

For t > 0, z = 0 (water-salt interface), C1 = C1o

For t > 0, z (far away from the water level), C1 = C1∞ = 0.

The solution is:

1 10

1 101/2 1/2

1 10

12 · 2 ·

C C x xerf C C erf

C C D t D t

c) 12 1 21 1

( ) (2.641 3.711) 3.1762 2

12 ( / )( / )B 1 B 2 B/ 809.1 78.6 252.2 k k k

/B

303/252.2 1.2k T

=1.32

2

11

3 322-3 2 -3 2

1 2 2

2

12

1 1 1 11.86 10 1.86 10 303

18 290.221 /

1 3.176 1.32

TM M

D cm sp

d) 5433

ln 26.3 8.369273.15 30

satp

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4.313satp kPa

6

3 3

4.3131.712 10

·8314 303

satsat

p kPa molc

cm kPaRT cmK

mol K

0 satz

Dj c c

t

2H O

dm dV dhA AjM

dt dt dt

2H OMdh

jdt

2H OsatMdh D

c cdt t

Integration:

20 0

1h tH Osat

M Ddh c c dt

t

2

2

6

3

3

2

18 0.221

1.712 10 0 2 36000

1

0.0031

H O

sat

M Dh c c t

g cm

mol mol s sgcm

cm

cm

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A fisherman tries a new cat repellent aroma to keep the stray cats away from his daily catch. The aroma diffuses radially out of a spherical capsule of 5.0 cm3 volume. The aroma saturation concentration at the diffuser surface is 0.015 g/m3. The concentration profile has reached steady state. Cats can tolerate only an aroma

concentration up to 9.8g/m3 and feel uncomfortable if they are closer to the diffuser. a) Draw a detailed sketch of the problem. (5 points) b) Assume that the initial distance between the center of the diffuser and the cat is

1.0 m and the aroma concentration at this point is 0.00012 g/m3. How far does the cat need to move (in radial direction) until it can tolerate the concentration? Start with the complete generalized mass balance and state your assumptions to simplify it.

(12 points) c) Calculate the flux at the surface of the diffuser. How many times a day the diffuser

needs to be replaced to keep cats away if each diffuser contains 0.0045 g of repellent aroma?

(8 points) Data: Diffusion coefficient of repellent aroma in air 0.259 cm2/s

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Solution 2 a) Sketch

b) The generalized mass balance for spherical coordinates is given by:

22

r2 2 2 2 2

vvc 1 c 1 c 1 c c c cD r (sin ) v r

t r r r r sin r sin r r r sin

Assumptions:

steady state: dc/dt = 0

symmetric: d/d, d/d =0

no reaction: r =0

since we assume that the system is dilute, convection can be neglected:

0 r zv v v

The mass balance then reduces to:

2

2

1 c0 D r

r r r

The first integration:

2

1

cK r

r

Cd

r Spherical cat repellent diffuser

rt

csat

Ct

ri

rd

C∞

ri

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The second integration:

12

-Kc = K

r

The boundary conditions are:

r = ri c = csat r = rd c = cd

Using these boundary conditions:

sat d 11 2 d

d

d i

(c - c ) KK and K c

r1 1

r r

We need to calculate the initial radius of the sphere from the given volume:

3

i

4V r

3

3

3

i

3 V 3 5 cmr

4 4

ir 1.06 cm

34 2

1

(0.015 - 0.00012)g/mK = -1.59 10 g / m

1 1 100 cm

100 cm 1.06 cm 1 m

4 25 3

2

-1.48 10 g / mK +0.00012 = -3.9 10 g / m

1 m100 cm

100 cm

The concentration profile as a function of radius (r):

1 2

1c(r) K K

r

1t

t 2

Kr

c(r ) K

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4 2

t 6 3 5 3

( 1.594 10 g / m )r 3.24 m = 324 cm

9.8 10 g / m ( 3.94 10 g / m )

Therefore the cat has to move ~224 cm away from the initial position.

c) Flux on the surface of the sphere is calculated using Fick’s first law:

dcj D

dr

12

Kc(r) K

r

12

Kd{ K }

rj Ddr

1 2

1j D K

r

Flux on the tube surface at r = ri

ir r 1 2

i

1j D K

r

Plugging all the values we get,

i

2 4 2 4 2 2

r r 2

1j 0.259 cm / s 1.59 10 g / m 10 m / cm

(1.06 cm)

i

9 2

r rj 3.67 10 g / cm s

The mass flowrate from radial direction:

i i

2

r r r r iJ j A j 4 r

9 2 2J 3.67 10 g / (cm s) 4 (1.06 cm)

8 6J 5.18 10 g / s 3.11 10 g / min

The aroma mass that one aroma sphere can produce is 0.0045 g,

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The diffusor last for following time,

6time 0.0045 g / 3.11 10 g / min 1446 min 24 hrs

Therefore for we need only one diffusor a day.

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The off-gas of a reactor is flowing upwards through a vertical wetted-wall column (“scrubber”). Thereby a water-soluble chemical is removed from the off-gas. The inner wall of the cylindrical column is wetted by initially pure water that counter-flows from the top to the bottom at 2 cm/s forming a film of constant film thickness . The column length is 8.5 cm and the film thickness is much smaller than the column diameter. The diffusion coefficient of the chemical in water is 2.1∙10-5 cm2/s. At the bottom of the column the concentration of chemical in water is 15% of the saturation concentration. Assume that the chemical in the off-gas is always higher than the saturation concentration in water. a) Draw a sketch of the wetted-wall column for gas scrubbing, showing the important

problem parameters. (5 points) b) Calculate the film thickness . (20 points)

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Solution 3

(See Example 8.3-1) a)

b) Mass balance of water-soluble vapor in the falling water-film over a differential column length Δz: (accumulation) = (flow in minus flow out) + (amount absorbed by water per time)

))(()()(0 11interf10

1

0 csatckAcvAcvAzz

crossz

cross

(1)

where Across: the cross section towards the flow, v

0: the average film velocity, c1: the concentration of the water-soluble vapor in water, k: the local mass transfer coefficient for the absorption of the vapor by the water, Ainterf: the interfacial area for the absorption and c1sat: the saturation concentration of the vapor at the gas-liquid interface. The cross section of the annular water film is given by:

dddrrrrrrrrA outoutinoutinoutinout since ,)())(())(()(22

cross

while the interfacial area for the absorption is: zdA interf .

Thus, eq. (1) becomes:

)()()( 1,1110 cczdkccvd satzzz (2)

)()()(

1,10

11cc

v

k

z

cc

sat

zzz

(3)

or:

1 [21]

1 [21]

1 [21]

off-gas

pure water

purified off-gas

water withdissolved vapor

L=7.5 cm

d

ɭ ɭ

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)( 1,101 cc

v

k

dz

dcsat

(4)

The mass transfer correlation for falling films is (Table 8.3-2/Cussler, 3rd ed):

5.00

69.0

D

zv

D

kz (5)

or:

5.0069.0 zDvk (6)

where k: the local mass transfer coefficient, z: the position along the film, v0 the superficial velocity, D the diffusion coefficient of the material being transferred and v0: the average film velocity. Substituting k in eq. (4) by eq.(6), we have:

dzzvD

cc

dc

dzzv

Dv

cc

dc

sat

sat

5.00

1,1

1

5.0

0

0

1,1

1

/69.0

)(

69.0

)(

(7) The boundary conditions are:

z = 0, 01 c

z = L, satL

ccc,111

15.0

If we integrate eq.(7) using the B.C.s, we have:

Lc

c sat

sat dzzvD

cc

ccdsat

sat0

5.00

85.0

1,1

1,1 /69.0

)(

)(,1

,1

(8)

Lc

csatz

vDcc

sat

sat0

5.00

85.0

1,12

/69.0)ln(

,1

,1 (9)

5.00

,1,1

/38.1)ln()85.0ln( L

vDcc

satsat (10)

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cm

s

cms

cmcm

vLD

08.0

)85.0

1ln(

0.2

101.25.8

38.1

)85.0

1ln(

/38.1

25

0

(11)

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The irreversible gas phase reaction A + B C takes place on the surface of spherical catalyst particles (diameter dp = 1.5 cm) that are distributed in a packed bed reactor (diameter dreactor = 0.4 m). The gas at the inlet consists of A with excess B, and has a flow rate of 40’000 L/min. The reaction is first order with respect to A and has a reaction rate constant of 83 cm/s. Assume that the catalyst bed is porous and does not affect the gas velocity.

a) Draw a sketch of the problem and indicate all important processes, dimensions and concentrations. (5 points)

b) Select the appropriate mass transfer correlation and calculate the overall mass transfer coefficient. Is the reaction mass transfer-limited or reaction-limited? (8 points)

c) Derive an equation for the concentration of A as function of the reactor length. Start with a mass balance over a differential reactor volume dV. (8 points)

d) How long should the reactor be to react 99.5% of A? (4 points)

Data: Average gas viscosity: 1.48·10-5 m2/s Diffusion coefficient of A: 0.212 cm2/s Specific surface area of the catalyst: 140 m2 catalyst / m3 reactor

15/17



Solution 4 – Chemical Reactions

a)

A+B

A+B+C

L

dbedz=0

z=L

dp

A+B

Ck1

κ2

v0

b)

The gas velocity can be found from the gas feed rate:

3

6

0

2 2

1 1 min40'000

min 10 60 sec5.3052 /

0.44

cross

L m

mlQv m s

Am

For an irreversible, first order heterogeneous reaction the overall MTC can be found:

1 2

1

1 1K

k

Where k1 is the resistance to mass transfer and κ2 is the reaction rate constant.

For k1 we can find the appropriate mass transfer correlation, which is for a packed bed:

0.42 2/30

01.17pd v D

k v

Entering the given values gives:

0.42 2/35 2

5 2 5 2

2.12 10 /0.015 5.3052 /1.17 5.3052 / 0.214 /

1.48 10 / 1.48 10 /

m sm m sk m s m s

m s m s

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Entering the values in the overall MTC:

1 2

1 10.170 /

1 1 1 1

0.214 / 0.83 /

K m s

k m s m s

This reaction is neither mass transfer limited nor reaction limited since both play a

more or less equal role (same order of magnitude).

c)

The mass balance over reactor volume dV is

0 ( ) ( )cross A A exchange A AiA v c z c z z K A c c

where

exchange bed crossA dV A dz

0 ( ) ( )cross A A cross A AiA v c z c z z K A dz c c

0

( ) ( )A AA Ai

c z c z z Kc c

dz v

0A

A Ai

dc Kc c

dz v

The boundary conditions are:

0

,

0

0

A A

A AL

A i

z c c

z L c c

all z c

After integration we obtain the concentration profile of A at the reactor outlet as

function of L:

0 0exp AAL A

K cc c L

v

d)

For 99.5% conversion:

0

0

100% 99.5%A AL

A

c c

c

0

0.005AL

A

c

c

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0

0

ln AL

A

cv

cL

K

Filling in the numbers:

2 3

ln 0.005 5.3052 /1.18

0.170 / 140 /cat reactor

m sL m

m s m m

mass transfer (stoffaustausch) - eth zexercises of the mass transfer lecture (also handwritten on...

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