Question : 160 cm^3/s of a solvent S is used to treat 400 cm^3/s of a 1...

(a) Countercurrent operation
Considering the solvent S, 160$cm^{3}$/s = 1.6 × $10^{-4}$$m^{3}$/s

and: mass flowrate = (1.6 × $10^{-4}$ × 800) = 0.128 kg/s

Considering the solution, 400$cm^{3}$/s = 4 × $10^{-4}$ $m^{3}$/s

containing, say, a $m^{3}$/s A and (5 × $10^{-4}$ – a) $m^{3}$ /s B.

Thus: mass flowrate of A = 1200a kg/s

and: mass flowrate of B = (4 × $10^{-4}$ – a)1000 = (0.4 – 1000a) kg/s

a total of: (0.4 + 200a) kg/s

The concentration of the solution is:

0.10 = 1200a/(0.4 + 200a)

Thus: a = 3.39 × $10^{-5}$ $m^{3}$/s

mass flowrate of A = 0.041 kg/s, mass flowrate of B = 0.366 kg/s

and: ratio of A/B in the feed, $X_{f}$ = (0.041/0.366) = 0.112 kg/kg

The equilibrium data are plotted in Figure 13.15 and the value of $X_{f}$ = 0.112 kg/kg is marked in. The slope of the equilibrium line is:

(mass flowrate of B)/(mass flowrate of S) = (0.366/0.128) = 2.86

Since pure solvent is added, $Y_{n+1}=Y_{4}$ = 0 and a line of slope 2.86 is drawn in such that stepping off from $X_{f}$ = 0.112 kg/kg to $Y_{4}$ = 0 gives exactly three stages. When $Y_{4}$ = 0, $X_{n}=X_{3}$ = 0.057 kg/kg,

Thus: the composition of final raffinate is 0.057 kg A/kg B

(b) Multiple contact
In this case, (0.128/3) = 0.0427 kg/s of pure solvent S is fed to each stage.
Stage 1

$X_{f}$= (0.041/0.366) = 0.112kg/kg

and from the equilibrium curve, the extract contains 0.18 A/kg S and (0.18 × 0.0427) = 0.0077 kg/s A.

Thus: raffinate from stage 1 contains (0.041 – 0.0077) = 0.0333 kg/s A and 0.366 kg/s B

and: $X_{1}$ = (0.0333/0.366) = 0.091kg/kg

Stage 2

$X_{1}$ = 0.091kg/kg

and from Figure 13.15 the extract contains 0.14 kg A/kg S

or: (0.14 × 0.0427) = 0.0060 kg/s A

Thus: the raffinate from stage 2 contains (0.0333 – 0.0060) = 0.0273 kg/s A and 0.366 kg/s B

Thus: $X_{2}$ = (0.0273/0.366) = 0.075 kg/kg

Stage 3

$X_{2}$ = 0.075 kg/kg

and from Figure 13.15, the extract contains 0.114 kg A/kg S

or: (0.114 × 0.0427) = 0.0049 kg/s A.

Thus: the raffinate from stage 3 contains (0.0273 – 0.0049) = 0.0224 kg/s A and 0.366 kg/s B

and: $X_{3}$ = (0.0224/0.366) = 0.061 kg/kg

Thus: the composition of final raffinate = 0.061kg A/kg B