CHEMICAL ENGINEERING

LIQUID MIXING 105

containing 28% by mass of a water soluble solid is agitated with 100 m^3 of water, how
long will it take for all the solute to dissolve assuming conditions are the same as in the
pilot unit? Water is saturated with the solute at a concentration of 2.5 kg/m^3.

Solution

The mass of soluteM, dissolving intsis:

dM/dtDkcsckg/s (i)

For a batch of solution,Vm^3 in volume: dMDVdc

and substituting for dMin (i): Vdc/dtDkcsc/V

Integrating: lncsc 0 /cscDkt/V (ii)

wherec 0 is the concentration of the solute whentD0.
For pure water,c 0 D0 kg/m^3 whentD0 and hence equation (ii) becomes:

cDcs 1 ekt/Vkg/m^3 (iii)

For the pilot test, the batch volume,VD1m^3 ,andcsD 2 .5 kg/m^3 at saturation. When
tD10 s, 75% saturation is achieved or:

cD 2. 5 ð 75 / 100 D 1 .875 kg/m^3

Therefore, in equation (iii):

1. 875 D 2. 5  1 e^10 k/^1 andkD 0. 138

For the full-scale unit,the batch volume,VD100 m^3. Mass of solute presentD 300 ð
28 / 100 D84 kg andcD 84 / 100 D 0 .84 kg/m^3.
Therefore, in equation (iii): 0. 84 D 2. 5  1 e^0.^138 t/^100 andtD297 s

PROBLEM 7.5

For producing an oil-water emulsion, two portable three-bladed propeller mixers are
available; a 0.5 m diameter impeller rotating at 1 Hz and a 0.35 m impeller rotating
at 2 Hz. Assuming turbulent conditions prevail, which unit will have the lower power
consumption?

Solution

Under turbulent conditions, the power requirements for mixing are given by:

PDkN^3 D^5 (equation 7.13)

In this case:P 1 Dk 13 ð 0. 55 D 0. 03125 kandP 2 Dk 23 ð 0. 355 D 0. 0420 k

∴ P 1 /P 2 D 0. 03125 k/ 0. 0420 kD 0. 743

Thus the 0.5 m diameter impeller will have the lower power consumption; some 75% of
that of the 0.35 m diameter impeller.