Consider the liquid/liquid stage extraction process with an immiscible solvent s

Question

Consider the liquid/liquid stage extraction process with an immiscible solvent shown in Figure. Note that stream W enters the first stage containing X_in weight fraction of material A and solvents S enters the last stage (N) containing Y_in weight fraction of material A. As the solvent flows through the process, it retains more A, thus extracting A from W. Countercurrent extraction process. For each stage (i), we assume equilibrium between the weight fraction of A in S(y_i) and the weigh fraction of A in W (x_i). Assuming that each stage is operated at the same temperature, the relationship between X_i and Y_i can be given as Y_i = K X_i Now performing a mass balance on A for the ith stage, assuming that X_i and Y_i are small so that S and W remain constant through the process: X_i-1 W + Y_i-1 S = X_i W + Y_i S Using the equilibrium relationship, assuming K remains constant and rearranging results in X_i-1 - (1 + KS/W) X_i + (KS/W) X_i+1 = 0 Applying this equation to the first stage results in - (1 + KS/W) X_1 + (K/W) X_2 = X_in And for the last stage X_N-1 - (1 + KS/W) X_N = (KS/W)Y_in S = 1000 Kg/hr X_in = 0.05 Y_in = 0.0 K = 10 N = 10

Explanation / Answer

If a solute is extracted from an aqueous phase into an organic phase

Saq Sorg   

then the partition coefficient is

KD = [Sorg][Saq]

distribution ratio, D, as the ratio of the solute’s total concentration in each phase.

D=[Sorg]total / [Saq]total] KD = [Sorg]total / [Saq]total

For calculating the efficiency

(qaq)1   = (molesSaq)1 / (molesSaq)0 = Vaq/ (DVorg+Vaq)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---