Chapter 5 Liquid-liquid extraction：5章液液萃取.pptVIP

Stage-to-stage calculations for countercurrent extraction. 1. Δ is a point common to all streams passing each other, such as L0 and V1, Ln and Vn+1, Ln and Vn+1, LN and VN+1, and so on. 2. This coordinates to locate this Δ operating point are given for x cΔ and x AΔ in eqn. 5.21. Since the end points VN+1, LN or V1, and L0 are known, xΔ can be calculated and point Δ located. 3. Alternatively, the Δ point is located graphically in the figure as the intersection of lines L0 V1 and LN VN+1. 4. In order to step off the number of stages using eqn. 5.22 we start at L0 and draw the line L0Δ, which locates V1 on the phase boundary. 5. Next a tie line through V1 locates L1, which is in equilibrium with V1. 6. Then line L1Δ is drawn giving V2. The tie line V2L2 is drawn. This stepwise procedure is repeated until the desired raffinate composition LN is reached. The number of stages N is obtained to perform the extraction. * Ex 5.6 Pure isopropyl ether of 450 kg/h is being used to extract an aqueous solution of 150 kg/h with 30 wt% acetic acid (A) by countercurrent multistage extraction. The exit acid concentration in the aqueous phase is 10 wt%. Calculate the number of stages required. Solution: The known values are VN+1 = 450, yAN+1 = 0, yCN+1 = 1.0, L0 = 150, xA0 = 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.10. 1. The points VN+1, L0, and LN are plotted in Fig. below. For the mixture point M, substituting into eqs. 5.12 and 5.13, xCM = 0.75 and xAM = 0.075. 2. The point M is plotted and V1 is located at the intersection of line LNM with the phase boundary to give yA1 = 0.072 and yC1 = 0.895. This construction is not shown. 3. The lines L0V1 and LNVN+1 are drawn and the intersection is the operating point Δ as shown. * * Alternatively, the coordinates of Δ can be calculated from eq. 5.21 to locate point Δ. Starting at L0 we draw line L0 Δ, which locates V1. Then a tie line through V1 locates L1 in equilibrium with V1. (The tie-line data are obtained from an enlarge