CHEMICAL ENGINEERING

MASS TRANSFER 269

From Stokes’ Law, the terminal falling velocity of the droplet is given by:

3 du 0 D



6

d^3 +s+

g

or: u 0 D

d^2 g

18 

+s+

DKd^2

Thus, the time taken for the droplet to travel the depthHof the rising liquid is:

H

1

2 Kd

2

Since the liquid is rising at a velocity of^12 Kd^2 and the relative velocity is

Kd^2 ^12 Kd^2 D^12 Kd^2 , the mass transfer rate (kmol/m^2 s) to droplet at end of travel is:

b

√

K

2 H

d

The mass transfer rate to the drop is:

b

√

K

2 H

dd

^2 Db

√

K

2 H

d^3 Db

√

K

H



1

p

2

d^3

Forcoalesceddrops, the new diameter is: 2^1 /^3 d

The terminal falling velocity is:K 22 /^3 d^2

Its velocity relative to the liquid is:K 22 /^3 d^2 ^12 Kd^2 DKd^2  22 /^3 ^12

Thus: TimeoffallofdropD

H

Kd^2  22 /^3 ^12

Mass transfer rate at end of travelDbd

√

K

H

d

√

 22 /^3 ^12 kmol/m^2 s

Mass transfer rate to dropDbd

√

K

H

d

√

 22 /^3 ^12 

 21 /^3 d

^2 kmol/s

Db

√

K

H

√

 22 /^3 ^12

22 /^3 d^3 kmol/s

The ratio of the mass transfer rate for the coalesced drop to the mass transfer rate for
the single droplet is then:

D

√

 22 /^3 ^12

22 /^3

1 /

p

2

D 2. 34

PROBLEM 10.39

In a drop extractor, a dense organic solvent is introduced in the form of spherical droplets
of diameterdand extracts a solute from an aqueous stream which flows upwards at