Dr. R. Nagarajan

Professor

Dept of Chemical Engineering

IIT Madras

Advanced Transport PhenomenaModule 6 Lecture 29

1

Mass Transport: Illustrative Problems

Mass Transport: Illustrative Problems

2

SOLUTION TO THE PROBLEM

3

4

SOLUTION

Catalytic Converter

a. Mechanism of CO(g) transport to the wall

If Re < 2100 (see below),transport to the wall is by

Fick diffusion of CO(g) through the prevailing

mixture.2

1.73 20.216.

300CO mix CO N

T cmD D

p s

SOLUTION

5

Therefore

Analogous heat transfer diffusivity is for gas

mixture

b. Discuss whether the Mass transfer Analogy Conditions(M A

C) and Heat transfer Analogy Conditions (H A C) are met;

implications ?

Since Mmix and Mco are close hence we will

assume

; CO COy

, 1 0.02. CO

SOLUTION

pk c

1.73 20.216 600

600 ,1 0.684 .1 300CO mix

cmD K atm

s

6

c. Sc for the mixture :

Now:

and:

CO mix CO mixSc v D

1700 500

2mix air

43.31 10 poise

43

1 28.975.88 10

82.06 600

pM g

RT cm

SOLUTION

7

therefore

therefore

2

0.563mix

cmv

s

0.5630.823

0.684CO mixSc v D

SOLUTION

8

d. L=? We will need Re

Now:

therefore (laminar-flow regime)

4 0.15 0.1540.15

4 0.15

eff

cm cmAd cm

P cm

310U cm s20.562 v cm s

310 0.15Re 267

0.562

SOLUTION

9

For a square channel

and (used below).

If then the mass-transfer analogy is:

2.976 mNu Re 14.227 . fC

, 0,CO w

,

,

exp 40

.

CO b

mmCO b

zF entrance Nu

SOLUTION

10

where

We estimate at which

If

then

1.

Re .

meff

z

Sc d

z L 0 0.05. b bL

0.05 exp X

SOLUTION

1ln 2.9957

0.05 = X

11

therefore

Tentatively, assume F (entrance =1).Then:

that is,

(at which F (entrance) is indeed ). Solving for L gives: L

=8.3 cm ( needed to give 95 % CO-Conversion).

4 . . 2.9957. m mNu F entrance

01

4 . . 2.976 2.9957;267 0.823 0.15

L

cm

SOLUTION

0 2.9957

0.251664 2.976

m

1

12

e. Discuss underlying assumptions, e.g.,

fully developed flow?

nearly constant thermo physical properties?

no homogeneous chemical reaction?

“ diffusion-controlled” surface reaction?

f. If the catalyst were “ poisoned,” it would not be able to maintain

. This would cause to exceed 8.3 cm. If catalyst

were completely deactivated, then and, of course, , ,CO w CO b

, , ,CO w CO b

SOLUTION

reqL

reqL

13

g. If the heat of combustion is 67.8 kcal/mole CO, how much

heat is delivered to the catalyst channel per unit time?

Overall CO balance gives the CO-consumption

rate/channel:

where

, , 1 . ,CO channel COm m conversion

24 3

2

5.04 10 10 0.15

1.13 10 .

=

m U A

g s per channel

SOLUTION

14

Moreover,

hence,

and

, 1 0.02; 0.95; CO Conversion

2,

4

1.13 10 0.02 0.95

2.156 10

=

CO channelm g s

g s

3167.8 . 2.42 10

28

kcal g mole CO calQ

g mole CO g g

SOLUTION

15

Therefore

The “sensible” heat transfer required to keep the wall

at 500 K can be calculated from a heat balance on the

8.3 cm-long duct- i.e., once we calculate , we

have :

4 32.156 10 2.42 10 0.522 . CO oxid

cal calQ

g s

bT L

0 , - sensible p b bQ m c T T L

SOLUTION

16

where

Mixing cup avg temp at duct outlet ?

21.13 10 ,

0.251

0 700 .

p

b

m g s

c cal g K

T K

exp 4 . . .

0

w b

h hw b

T T LNu F entrance

T T

SOLUTION

17

Again, we see that

Moreover,

1 since;F entrance

1 1 8.3

. ,Re.Pr 267 0.706 0.15

0.293.

heff

h

L

d

2.976;mNu

SOLUTION

18

therefore

and

exp -4 0.293 2.976 0.0304; 0

b w

b w

T L T

T T

5000 0304

700 500506

. ,

=

b

b

T Lor

T L K

SOLUTION

19

Therefore,

2

0

1 135 10 0 251 700 506

.

= . ..

sensible p b bQ m c T T L

g calK

s g K

0 552

0 522 0 552

1 07

.

., . ,

. .

= . .

sens

total CO oxid sens

ie Q cal s

Q Q Q

calper channel

s

SOLUTION

20

h. “Quasi-Steady” Application of These Results? Note that:

and

hence, if the characteristic period of the unsteadiness >> 8.2 ms, the previous results can be used at each flow condition.

flow 3

2 2

diff 2CO mix

L 8 3t 8 3 ms

U 10 cm s

w 2 0 15 2t 8 2 ms

cmD0 684

s

..

.. ;

.

SOLUTION

21

i. Pressure Drop

We have:

Therefore

?p

2

14 227 for square duct, laminar flow.

1 Re2

.wfC

U

21 14 227

2 Re

.. ,w U perimeter mean shear stress

SOLUTION

22

But

Therefore

242 3 2

3 2

1 5 043 1010 =2.98 10 .

2 2

. g cm dyneU

cm s cm

2 2 3

2

14 222 98 10 2 98 10 5 335 10

267

1 345

.( . ) . .

. .

f

w

C

w dyne cm

SOLUTION

23

From overall momentum balance:

4 4 0 15 ( . )w cm

2 2

5 1

. . .

- = 4

= 1.345 4 8.3 0.15 2.48 10

- = 2.93 10 0.003%.

duct channel w

w

p A Perim L

p L w

dyne cm

p p

SOLUTION

24