8.14 y 8.15

Exercise 8.14 Subject: Extraction of trimethylamine (TMA) from benzene (C) with water (S). Given: Three equilibrium equilibrium stages. Solvent-free extract to contain 70 wt% TMA. Solvent-free raffinate to contain 3 wt% TMA. Liquid-liquid equilibrium data. Find: Using the Hunter-Nash method with a right-triangle diagram, find feed composition and water-to-feed ratio. Analysis: The given phase boundary liquid compositions and the equilibrium liquid-liquid compositions (as dashed tie lines) are plotted on the right-triangle right-triangle diagram below. Included on the diagram are the final extract and raffinate raffinate compositions. The final extract composition is obtained by locating a point, P, for 70 wt% TMA, 30 wt% benzene, and 0% water (given solventfree composition), and drawing a straight line from P toward the point S (pure water) to where the line intersects the phase boundary. This is point E for the extract. In a similar similar manner, the raffinate composition, R, is determined.

Exercise 8.14 (continued) On a second diagram, shown below, a trial and error procedure is used to find the operating  point, P' , that will result in the stepping off of three equilibrium stages, as illustrated in Fig. 8.19, to obtain the specified final extract and final raffinate. The final trial is shown on the diagram, where M is the mixing point for extract + raffinate, and for feed + solvent. Assuming water-free feed, the resulting feed composition is at F, with 57.5 wt% TMA and 42.5 wt% benzene. The ratio of mass of solvent to mass of feed is given by the ratio of line lengths = line FM/line MS = 0.56.

Exercise 8.15 Subject: Extraction of diphenylhexane (DPH) from docosane (C) with furfural (U) at 45 and o 80 C Given: Feed, F, of 500 kg/h of 40 wt% DPH in C. 500 kg/h of solvent, S, containing 98 wt% U and 2 wt% DPH. Raffinate to contain 5 wt% DPH. Liquid-liquid equilibrium data. Find:  Number of theoretical stages. DPH in kg/h in the extract. Analysis: o Case of 45 C: The given liquid-liquid equilibrium data are plotted in the right-triangle diagram below. Included on the diagram are composition points F for the feed, R for the raffinate on the equilibrium curve, and S for the solvent. A straight line extends from point F to  point S. Because the mass flow rates of the feed and solvent are equal, the mixing point, M, is located at the midpoint of this line. Another straight line extends from point R to point M, and then to an intersection with the equilibrium curve at point E, which is the final extract. Using the inverse lever-arm rule on line RME, the mass ratio of R to E is 0.445. Combining this with an overall material balance:  F + S = 500 + 500 = 1,000 = R +E  gives R = 308 kg/h and E  = 692 kg/h. From the diagram, the mass fraction of DPH in the extract is 0.281. Therefore, the DPH in the extract is 0.281(692) = 194.5 kg/h, which is 92.6% of the total DPH entering the extractor. On the following page, the equilibrium stages are stepped on another right triangle diagram, as in Fig. 8.17, by determining the operating point P from extensions of lines drawn through points F and E, and S and R, followed by alternating between operating lines and equilibrium tie lines. The result is 5 equilibrium stages.

Exercise 8.15 (continued) Analysis:

o

Case of 80 C: The given liquid-liquid equilibrium data are plotted in the right-triangle diagram below. Included on the diagram are composition points F for the feed, R for the raffinate on the equilibrium curve, and S for the solvent. A straight line extends from point F to  point S. Because the mass flow rates of the feed and solvent are equal, the mixing point, M, is located at the midpoint of this line. Another straight line extends from point R to point M, and then to an intersection with the equilibrium curve at point E, which is the final extract. Using the inverse lever-arm rule on line RME, the mass ratio of R to E is 0.383. Combining this with an overall material balance:  F + S = 500 + 500 = 1,000 = R +E  gives R = 277 kg/h and E  = 723 kg/h. From the diagram, the mass fraction of DPH in the extract is 0.271. Therefore, the DPH in the extract is 0.271(723) = 195.9 kg/h, which is 93.3% of the total DPH entering the extractor. On the following page, the equilibrium stages are stepped on another right triangle diagram, as in Fig. 8.17, by determining the operating point P from extensions of lines drawn through points F and E, and S and R, followed by alternating between operating lines and equilibrium tie lines. The result is 4+ equilibrium stages.

Exercise 8.15 (continued) o

Analysis: (case of 80 C continued)

Exercise 8.16 Subject: Selection of extraction method. Given: Four ternary systems in Fig. 8.45. Diagrams 1, 2, 4 are Type I. Diagram 3 is Type II. Find: Method for most economical process for each system. Methods are: (a) Countercurrent extraction (CE). (b) CE with extract reflux (ER). (c) CE with raffinate reflux (RR). (d) CE with ER and RR. Analysis:  Note that y1 is the composition of the extract. Raffinate reflux (RR) is of little value, so don't use it.

For Type I diagrams, extract reflux is rarely useful. All three Type I diagrams exhibit solutropy, making it almost impossible to obtain a good separation. The Type II diagram uses a poor solvent, making the use of extract reflux questionable. Summary: Use CE for all four systems. However, better solvents should be sought for all four systems.