CHEMICAL ENGINEERING

274 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

WhentD50 s andCA/CASD 0 .1, then:

0. 1 Derfc

y

2

p

10 ^9 ð 50

or: 0. 9 Derf

y

2

p

Dt

From Table 13 in the Appendix of Volume 1, the quantity whose error fractionD 0 .9is:

y

2

p

Dt

D 1. 16

At the surface: NA (^) yD 0 ,tD

√

D

t

CAS

D

√

10 ^9

ð 50

ð 0. 1

D 0. 252 ð 10 ^6 kmol/m^2 s.

At a depthy: NAD 0. 252 ð 10 ^6 ðe^1.^16

2

D 0. 0656 ð 10 ^6 kmol/m^2 s.

PROBLEM 10.43

In a drop extractor, liquid droplets of approximately uniform size and spherical shape are
formed at a series of nozzles and rise countercurrently through the continuous phase which
is flowing downwards at a velocity equal to one half of the terminal rising velocity of the
droplets. The flowrates of both phases are then increased by 25 per cent. Because of the
greater shear rate at the nozzles, the mean diameter of the droplets is, however, only 90
per cent of the original value. By what factor will the overall mass transfer rate change?
It may be assumed that the penetration model may be used to represent the mass
transfer process. The depth of penetration is small compared with the radius of the droplets
and the effects of surface curvature may be neglected. From the penetration theory, the
concentrationCAat a depthybelow the surface at timetis given by:

CA

CAS

Derfc

[

y

2

p

Dt

]

where erfcXD

2

p



∫ 1

X

ex

2

dx

whereCASis the surface concentration for the drops (assumed constant) andDis the
diffusivity in the dispersed (droplet) phase. The droplets may be assumed to rise at their
terminal velocities and the drag forceFon the droplet may be calculated from Stokes’
Law,FD 3  du.

Solution

Case1: For a volumetric flowrateQ 1 , the numbers of drops per unit time is:

Q 1

^16 

d^3

D

6 Q 1

d^31