CHEMICAL ENGINEERING

260 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

The mass transfer rate at timetis:CAkt^1 /^2

Thus the age of surface at which rateDaverage is given by or:

kt^1 /^2 Dk

p

s

tD

1

s

D

100



D 31 .8s

(d)Surface of age less than 10 seconds

The mass transfer taking place into surface of age up to 10 s is given by the same

expression as for the whole surface but with upper limit of 10 s instead of infinity

or: CA 2 k

p

s

∫k 10 t

0

ex

2

dx

whentD10 s:xD

p

stD

p

0. 01 ð 10 D 0. 316

The mass transfer into surface up to 10 s age is then:

CA 2 k

p

s

∫ 0. 316

0

ex

2

dxDCA 2 k

p

s

p



2

erf 0. 316 DCAk

p

sð 0. 345

Thus a fraction: 0.345 is contributed by surface of age 010 sand:

a fraction: 0.655 by surface of age 10 s to infinity.

PROBLEM 10.34

At a particular location in a distillation column, where the temperature is 350 K and the
pressure 500 m Hg, the mol fraction of the more volatile component in the vapour is 0.7
at the interface with the liquid and 0.5 in the bulk of the vapour. The molar latent heat
of the more volatile component is 1.5 times that of the less volatile. Calculate the mass
transfer rateskmol m^2 s^1 of the two components. The resistance to mass transfer in
the vapour may be considered to lie in a stagnant film of thickness 0.5 mm at the interface.
The diffusivity in the vapour mixture is 2ð 10 ^5 m^2 s^1.
Calculate the mol fractions and concentration gradients of the two components at the
mid-point of the film. Assume that the ideal gas law is applicable and that the Universal
Gas ConstantRD8314 J/kmol K.

Solution

In this case:

TD350 K,

PD500 mm HgD

(

500

760

ð 101 , 300

)

D 0. 666 ð 105 N/m^2

DD 2 ð 10 ^5 m^2 /s

and: CTD

P

RT

D

(

0. 666 ð 105

8314 ð 350

)

D 0 .0229 kmol/m^3