CHEMICAL ENGINEERING

MASS TRANSFER 229

Solution

The first part of this question is discussed in Section 10.5.2 and the required equation is
presented as equation 10.108.
In Section 10.5.2 the analysis leads to equation 10.113 which expresses the instanta-
neous rate of mass transfer when the surface element under consideration has an age
t,or:

NA (^) tDCAiCAo

√

D/t

The simple penetration theory assumes that each element is exposed for the same time
intervaltebefore returning to the bulk solution. The average rate of mass transfer is then:

NAD

1

te

∫te

0

NA (^) tdtD
CAiCAo
te
∫te
0
D/t
^0.^5 dt
D 2 CAiCAo

√

D/te

and the rate of absorption is proportional to

p

D.

PROBLEM 10.13

Show that in steady-state diffusion through a film of liquid, accompanied by a first-
order irreversible reaction, the concentration of solute in the film at depthybelow the
interface is:

CA

CAi

Dsinh

√

k

D

Ly

sinh

√

k

D

L

Ci

ifCAD0atyDLandCADCAiatyD0, corresponding to the interface.
Hence show that according to the “film theory” of gas-absorption, the rate of absorption
per unit area of interface,NAis given by:

NADkLCAi

ˇ

tanhˇ

whereˇD

p
Dk/kL,Dis the diffusivity of the solute,kthe rate constant of the reaction,
KLthe liquid film mass transfer coefficient for physical absorption,CAithe concentration
of solute at the interface,ythe distance normal to the interface andyLthe liquid film
thickness.

Solution

The basic equation for diffusion through a film of liquid accompanied by a first-order
irreversible reaction is:

Dd^2 CA/dy^2 DkCA or d^2 CA/dy^2 Da^2 CA (equation 10.171) (i)

wherea^2 D

p

k/D.