8.14 y 8.15

Exercise 8.14

Exercise 8.14

Subject: Extraction of Extraction of trimethylamine trimethylamine (TMA) (TMA) from benzene from benzene (C) with (C) with water (S).water (S).

Given:

Given: Three eThree equilibrium quilibrium stages. stages. Solvent-free Solvent-free extract extract to contain to contain 70 wt% 70 wt% TMA. TMA. Solvent-freeSolvent-free raffinate to contain

raffinate to contain 3 wt% TMA. 3 wt% TMA. Liquid-liquid equilibrium Liquid-liquid equilibrium data.data.

Find:

Find: Using the Hunter-Nash method with a right-triangle diagram, find feed composition andUsing the Hunter-Nash method with a right-triangle diagram, find feed composition and water-to-feed ratio.

water-to-feed ratio.

Analysis:

Analysis: The given phase boundary liquid compositions and the equilibrium liquid-liquidThe given phase boundary liquid compositions and the equilibrium liquid-liquid compositions (as dashed tie lines) are plotted on the ri

compositions (as dashed tie lines) are plotted on the right-triangle diagram below. ght-triangle diagram below. Included onIncluded on the diagram are the final extract and raff

the diagram are the final extract and raffinate compositions. inate compositions. The final extract composition isThe final extract composition is obtained by locating a point, P, for 70

obtained by locating a point, P, for 70 wt% TMA, 30 wt% benzene, and 0% wt% TMA, 30 wt% benzene, and 0% water (given solvent-water (given solvent-free composition), and drawing a straight line from P toward the point S

free composition), and drawing a straight line from P toward the point S (pure water) to where(pure water) to where the line intersects

the line intersects the phase boundary. the phase boundary. This is point E for This is point E for the extract. the extract. In a simiIn a similar manner, thelar 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 80oC

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:

Case of 45oC: 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)

Case of 80oC: 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)

Exercise 8.16

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.