CHEMICAL ENGINEERING

270 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

a velocity equal to half the terminal falling velocityu 0 of the droplets. On increasing
the flowrate of the aqueous stream by 50 per cent, whilst maintaining the solvent rate
constant, it is found that the average concentration of solute in the outlet stream of organic
phase is decreased by 10 per cent. By how much would the effective droplet size have
had to change to account for this reduction in concentration? Assume that the penetration
theory is applicable with the mass transfer coefficient inversely proportional to the square
root of the contact time between the phases and that the continuous phase resistance is
small compared with that within the droplets. The drag forceFacting on the falling
droplets may be calculated from Stokes’ Law,FD 3 duo,whereis the viscosity of
the aqueous phase. Clearly state any assumptions made in your calculation.

Solution

For droplets in the Stokes’ law region, the terminal falling velocity is given by:
/ 6
d^3 +s+
gD 3 du 0

or: u 0 D

d^2 g

18 

+s+

Dkd^2

The mass transfer rate to the droplet isKte^1 /^2 moles per unit area per unit time

/Kt^1 e/^2 moles per unit area in timete

/Kt^1 e/^2 d^2 moles per drop during time of rise

The concentration of solute in drop:/Kt^1 e/^2 d^2 /d^3 /Kte^1 /^2 d^1.

Initial case: u 0 Dkd^2

Rising velocity of liquidD

kd^2

2

Velocity of liquid relative to containerDkd^2 

kd^2

2

D

kd^2

2

Time of exposure in heightHD

H

kd^2 / 2

Dte.

Thus the concentration in the drop/K

√

2 H

kd^2

d^1 /K

√

2 H

k

d^2 DC 1

Second case: New drop diameterDd^0

Rising velocity of liquidD 1. 5 ð

kd^2

2

D

3

4

kd^2

Rising velocity of drop relative to liquidDkd^02

Velocity relative to containerDkd^02 ^34 kd^2

Time of exposureD

H

kd^02 ^34 d^2