(75d) On the Use of Co-Solvent in Counter Current Multistage Continuous Liquid Extraction Operations

The use of co-solvent is described to improve the yield in multistage counter-current liquid extraction operations. For the case when the equilibrium relationship is linear a co-solvent factor is defined in addition to the extraction factor and a analytical solution is derived for the number of ideal stages required to complete a desired level of separation of solute from the feed to the extract phase. A co-solvent can be used in addition to the solvent inorder to extract more of the solute from the raffinate phase during continuous co-current multi-stage continuous liquid extraction operations. Fresh solvent and co-solvent is assumed without any solute in them. For the special case when the raffinate flow, RNP = RNp-1 = °K.= R2 = R1 = F = R and when the two extract phases flow rate is constant in all the stages and is equal to B and Bs respectively. The equilibrium relationships between the distribution of solute between the extract and raffinate phases can be assumed to be linear. This is so the case when the solutions are dilute. Thus,

yBk* = m1xk (1) yBsk* = m2xk (2)

A solute component mass balance on stage 1 can be written as;

x0R + ybs2Bs + yb2B = x1R1 + ybs1Bs + yb1B (3)

Dividing throughout the Eq. [3] by B and using the Eqs. [1,2] to reduce the x1 and x0 values in terms of yb1 and yb0 values;

Ryb0/Bm1 + ybs2Bs/B + yb2 = Ryb1 /Bm1 + ybs1Bs/B + yb1 (4)

The extraction factor, Af can be defined as Af = R/m1B. Further it can be seen that, ybs2 = m2x2 = (m2/m1) yb2

Af yb0 + yb2Bsm2/Bm1 + yb2 = Af yb1 + yb1Bsm2/Bm1 + yb1 (5)

A co-solvent factor can be defines as,

Cs = Bsm2/Bm1

Af yb0 + Cs yb2 + yb2 = Af yb1 + Cs yb1 + yb1 (6)

Or (1 + Cs) yb2 = yb1(1 + Af + Cs) - Afyb0 (7)

yb2 = yb1(Af/(1+Cs) +1) - Afyb0/(1+Cs) (8)

In a similar fashion

y3 = = yb1 (1 + Af /(1+Cs) + (Af /1+ Cs)2 ) - y0(Af /(1+Cs) + (Af /1+ Cs)2 ) ) (9) Thus,

yN+1= y1(1 + Af/(1+ Cs)+°K+.AfNp/(1 + Cs)Np - y0(Af /(1+ Cs + +°K.+AfNp)/(1 + Cs)Np (10)

(Af /(1 + Cs) -1)yNp +1 = y1 ((Af /(1 + Cs)) Np+1 -1) - y0Af/(1+Cs) ((Af /(1 + Cs)Np -1) (11)

Eq. [11] is applicable when Af ?j 1. In a similar fashion the raffinate compositions can be derived;

xNp = x1((Af /(1 + Cs)Np -1 )/(Af-1) - x0Af((AfNp-2 -1 )/(Af-1) (12)