Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Referring to fig 844 let the operating point to the

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Unformatted text preview: using P1 until final raffinate RN is reached. The total number of stages is almost 5. Thus, feed F is fed to stage 1 and feed F' is fed to stage 2 of a cascade with almost 5 stages. Analysis: (continued) Exercise 8.17 (continued) Exercise 8.18 Subject: Extraction of trimethylamine (T) from benzene (B) with water (W) in a threeequilibrium stage liquid-liquid extractor shown in Fig. 8.44. Given: Feed of 10,000 kg/h of 40 wt% B and 60 wt% T. Fresh solvent is sent to each of the stages. The solvent flow rate to stage 3, call it S3, is 5,185 kg/h. In Fig. 8.44, V1 is to contain 76 wt% T on a solvent-free basis, and L3 is to contain 3 wt% T on a solvent-free basis. Equilibrium data are given in Exercise 8.14. Find: Using an equilateral-triangle diagram, determine the solvent flow rates to stages 1 and 2. Analysis: The triangular diagram is shown on a following page. Since the extract, V1, is given on a solvent-free basis, its complete composition is obtained by the intersection with the equilibrium curve of a line drawn from the point (76 wt% T and 24 wt% B) to the 100% W point. Similarly, the composition of the raffinate, L3, is obtained by the intersection with the equilibrium curve of a line drawn from the point (3 wt% T and 97 wt% B) to the 100% W point. The resulting compositions of the final raffinate, L3, and the final extract, V1, are as follows: Component T B W Total Raffinate, L3, wt% 3.0 97.0 0.0 100.0 Extract, V1, wt% 39.5 12.5 48.0 100.0 Using these compositions, determine the flow rates of final raffinate and final extract by solving material balances for T and B: T balance: 0.03 L3 + 0.395 V1 = 6,000 B balance: 0.97 L3 + 0.125 V1 = 4,000 Solving, L3 = 2,188 kg/h and V1 = 15,020 kg/h By an overall total material balance, the total solvent water rate is: S3 + S1 + S2 = L3 + V1 - F = 2,188 + 15,020 - 10,000 = 7,208 kg/h where S3 = Solvent in Fig. 8.44, which enters from the right into stage 3. Because S3 is given as 5,185 kg/h, S1 + S2 = 7,208 - 5,185 = 2,023 kg/h From the above results, the overall mass balance is as follows in flow rates in kg/h: Component T B W Total Feed 6,000 4,000 0 10,000 Solvent, S3 0 0 5,185 5,185 L3 66 2,122 0 2,188 V1 5,934 1,878 7,208 15,020 S1 + S2 0 0 2,023 2,023 Exercise 8.18 (continued) Analysis: (continued) The next step is to determine the separate values for S1 and S2. Because fresh solvent enters all three stages, instead of one operating point, P, there will be three operating points, designated here as P1, P2, and P3. Referring to Fig. 8.44, let the operating point to the right of stage 3 be P3. Therefore, from (8-5), P3 = S3 - L3 = 5,185 - 2,188 = 2,997 kg/h This operating point is located along a line from L3 through solvent, S3 to the right of the triangular diagram, as shown on the next page. This operating point is also common to the passing streams, V3 and L2 to the left of Stage 3 in Fig. 8.44. That is, P3 = V3 – L2 = 2,997 kg/h. Using the following compositions from Fig. 8.44 and material balances, the following values are obtained for V3 and L2: Component Wt % in V3 T B W 4.05 0.35 95.60 100.00 Total Flow in V3, kg/h 219 19 5,186 5,424 Wt% in L2 11.75 88.22 0.03 100.00 Flow in L2, kg/h 285 2,141...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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