CHEMICAL ENGINEERING

MASS TRANSFER 251

If the rates from each model are equal, then:

2 Kte^1 /^2 DK

p

s or steD 4 /

PROBLEM 10.29

Ammonia is absorbed in a falling film of water in an absorption apparatus and the film is
disrupted and mixed at regular intervals as it flows down the column. The mass transfer
rate is calculated from the penetration theory on the assumption that all the relevant
conditions apply. It is found from measurements that the mass transfer rate immediately
before mixing is only 16 per cent of that calculated from the theory and the difference has
been attributed to the existence of a surface film which remains intact and unaffected by
the mixing process. If the liquid mixing process takes place every second, what thickness
of surface film would account for the discrepancy? Diffusivity of ammonia in waterD

1. 76 ð 10 ^9 m^2 /s.

Solution

For the penetration theory:
∂CA
∂t

DD

∂^2 CA

∂y^2

(equation 10.66)

When tD 0 ,CAD 0

When t> 0 ,yD 0 ,CADCAiDconstant

When t> 0 ,yD1 CAD 0

As shown earlier in problems 10.19 and 10.21, this equation may be transformed and
solved to give:

CNADAe^2

p

p/D

yCBe 2

p

p/D

y (equation 10.105)

WhenyD 0 , CNADCAi/p

yD1, CNAD 0

and hence: AD0andBDCAi/p

Hence, CNAD

CAi

p

e

p

p/D

y

and:

∂CNA

∂y

DCAi

√

p

D

e

p

p/D

y

At the surface,NA (^) tDD

(

∂CA

∂y

)

yD 0

D

√

D

t

CAiin timet(as in Problem 10.21).

For the film, the origin is taken at the interface between the film (whose thickness isL)
and the mixed fluid.

Again: CNADAe

p

p/D

yCBe

p

p/D

y (equation 10.105)