Extraction - Principles of Chemical Separations with Environmental Applications - page 150

Example 5.7

Problem:
100 kg hr of waxed paper is to be dewaxed in a continuous countercurrent leaching

process. The paper contains 25 wt% wax and 75 wt% paper pulp. The paper pulp
should contain less than 0.3 kg of wax per 100 kg paper pulp. Entering solvent contains
0.045 kg of wax per 100 kg solvent. The paper pulp retains 1.5 kg solvent for every kg
of entering diluent.
Extract should contain 5 kg wax for every 100 kg solvent. How many equilibrium-
limited stages are required?
Solution:
Begin with a mass balance. Assume that all of the solvent retained by the paper pulp
is done so in the first contact stage and, further, that no separation occurs in this stage,
only mixing and solvent retention.
Figure 5.29 is a schematic flow diagram for the solvent. Since 1.5 kg solvent are
retained for every 1 kg diluent, we can say that 100 kg diluent

/

150 kg solvent exit
Stage 1 for a total of 250 kg. Wax and paper pulp are then diluted as follows:
25 kg

+

/ hr wax enters Stage 1

/

25 kg hr

=

hr =

. .

Leaving Stage 1: x 0 1

250 kg /

Next, a mass balance gives us S , the solvent flowrate:

10 − 4 Wax in: 100 kg / hr (0 . 25) + 4 . 5 × S

10 − 3 )

Wax out: 100 kg / hr (3 . 0 × + ( S − 150) kg / hr (0 . 05) .

Solving, S hr.

Figure 5.30 is a redrawn schematic diagram for the solvent. In the figure, the ∗ denote
concentrations in equilibrium, i.e., raffinate concentration in equilibrium with solvent
concentrations.
We need to calculate x a , the exiting wax concentration in the solvent prior to Stage 1.
Since we assume no separation occurs in Stage 1, we can estimate it as a function of
solvent flowrates and wax concentration leaving in the extract

= 650 kg /

(0 . 05)(500)

650

10 − 2

x a = = 3 . 9 × .

We must do the same to get x b . Problem statement says 0.3 kg wax per 100 kg paper
pulp. Total stream flow is 250 kg / hr. So,

003 100
250

x b 10 − 3

= 0 . = 1 . 2 × .