Question

3. A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two components, species-A and species-B. A stream of pure species-C is fed into the mixer to drive the extraction. The mixture is then fed to a separator where it is allowed to settle into two phases which are removed in separate streams. Each outlet stream contains all three species. The goal of the process is to produce an outlet stream with a high concentration of species-A. Given the data below, perform a degree- of-freedom analysis and determine the order in which systems of equations must be solved to characterize every stream in the process. Then solve the system for all unknown flow rates and compositions. One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B. One inlet stream to the mixer is pure species-C, and the flow rate is unknown. . One outlet stream from the separator is 85wt% species-A and 10wt% species-B. One outlet stream from the separator is 25wt% species-B and 70wt% species-C.

Answer 1

Simple mass balances are applied on the overall system and individual units as shown below.

Exteact LE) ; WE,A A B MEB NE, C mixer Feed (f) A arra B = F,B Airem, a B m, B C xm, c A B YER,A KRB rer, a Erla, c t RCR, solvent (3) rs, C=1 mixed stream Seponator m Raffinate (R) het rij denotes the male weight fraction of component j in stream i. Degree of freedom Analysis () on orecall system Ft Sa EtR on component A REA F = EXE, AT RURA component C us,cis = Mo: of unknowns 9 [F;S;E;R;35, 9;. KEA; WEB ; URA ; NR, B 3 No. of independent apa 10 to ② on 3

No of 2-o-f. = no. of unknowns independent equations D.O.FI 9-3= 6 On mixer Its =M n component HEA of = xma.M on components res, cis = ren, com No. of unknowns 6 [F, S, M, WF, A ; ANA; Kms 19.00 ] no. of independent egns = 3 D.O.F. = G-3 = 3 (iu) On separator m =E+R component A rema.M E t HR, A'R component C xmc.M MEG E+ KR, c.R KEA: (9) 2 7

No. of unknowns = 9 m, EGR, rem, a Hm, I ; j HEA ; Re, e ; URA ; RR, Mo.of independent equations = 3 *®;) DOF. 9-3 6 Ginen F = 100 kg/h REA=0.85 RFA = 0.35 Xf, B = 6.65 NE, BE E oilo Ne,C=0.05 0.05 YER, A = URB HR, C = 070 0.25 - Balance on A on overall system 0.85 Eto.52 0.358100 (0 35=0.85€ to 056 Balance on B on orecall system = 0.65xico olet 0.25R D 65 = 0.1E +0:25 R

nem, B = 0.235 km, c = 0.637. Multiply eq ② by 5 and subtract eq ④. - ) ito = 4.25 E + 0.25R 0.1E-0.25R 110- 4.15E E= 26.5 kg/h from ® 65 = 2.65 +0.25 R R=249.4 kg/h from @ 100 + s = 26.5 +249.4 S = 175.9 kg/h 175:9 = nm, cim from @ from @; 35 = UmA.M Bosc Also; REB.F xin, B.M 65= Kom; B .M 12 - a nem, c = xm, at um,B =275.9 ma 175:9 + 100 =) kg/h Nm, B= 65 275.9 Um, C = 175.9 - um, a 275.9 .35 275.9 0.127 nem, a =