Problem 2: 9.13 A stage extraction process is depicted in Fig. P9.13. In such systems, a stream containing a weight fraction .. of a chemical enters from the left at a mass flow rate of Fr. Simultaneously, a solvent carrying a weight fraction LE of the same chemical enters from the right at a flow rate of Fz. Thus, for stage i, a mass balance can be represented as FX-1+Fxx+1 = FM + FzX (P9.13a) At each stage, an equilibrium is assumed to be established between y, and, as in 1 K=- (P9.13b) . where K is called a distribution coefficient. Equation (P9.13b) can be solved for I; and substituted into Eq. (P9.13a) to yield = F2 1+ KY + 19-1- - (1+x) x + (***) x+1 =0 = (P9.13c) If Fi = 400 kg/h, in = 0.1, F2 = 800 kg/h, x=0, and K= 5, determine the values of you and if a five-stage reactor is used. Note that Eq. (P9.13c) must be modified to account for the inflow weight fractions when applied to the first and last stages. 3 Flow 1 3 FIGURE P9.13 A stage extraction process