CHEMICAL ENGINEERING

MASS TRANSFER 273

or: D^0 D 24. 2 ð 10 ^6 m^2 /s

The mass transfer coefficient is:

D^0

L

(

1

log mean 1 and 0. 8

)

D

24. 2 ð 10 ^6

L

(

0. 2

ln 01. 8

) 1

D

1

L

ð 27 ð 10 ^6 D 0 .030 m/s

and hence: LD 0. 90 ð 10 ^3 m

Case2:

The effective diffusivityD^0 is given by:

1

D^0

D

(

60 / 95

23 ð 10 ^6

)

C

(

35 / 95

52 ð 10 ^6

)

or: D^0 D 28. 9 ð 10 ^6 m^2 /s

The mean transfer coefficient is:

D^0

L

Ð

(

1

logmean1and0. 95

)

D

(

28. 9 ð 10 ^6

0. 9 ð 10 ^3

)(

0. 05

ln 0.^195

) 1

D 0 .033 m/s.

PROBLEM 10.42

State the assumptions made in the penetration theory for the absorption of a pure gas into
a liquid. The surface of an initially solute-free liquid is suddenly exposed to a soluble gas
and the liquid is sufficiently deep for no solute to have time to reach the far boundary of
the liquid. Starting with Fick’s second law of diffusion, obtain an expression for (i) the
concentration, and (ii) the mass transfer rate at a timetand a depthybelow the surface.
After 50 s, at what depthywill the concentration have reached one tenth the value at
the surface? What is the mass transfer rate (i) at the surface, and (ii) at the depthy,if
the surface concentration has a constant value of 0.1 kmol/m^3?

Solution

CA

CAS

Derfc

y

2

p

Dt

(equation 10.108)

Differentiating with respect toy:

1

CAS

∂CA

∂y

D

∂

dy

{

2

p



∫ 1

y/ 2

p

Dt

ey

(^2) / 4 Dt
d

(

y

2

p

Dt

)

D

1

p

Dt

ey

(^2) / 4 Dt
Thus: NA (^) y,tDD

{



1

p

Dt

ey

(^2) / 4 Dt

}

CASD

(√

D

t

ey

(^2) / 4 Dt

)

CAS