Resolver os Exercícios 8 12 e 8 15

Exercise 812 Subject Extraction of acetone A from water C by 112trichloroethane S at 25C Given 1000 kgh of 45 wt A in C Liquidliquid equilibrium data 10 wt A in raffinate Find Using a right triangle diagram a Minimum flow rate of S b Number of equilibrium stages for solvent rate of 15 times minimum c Flow rate and composition of each stream leaving each stage Analysis Using the given equilibrium data in weight fractions the righttriangle diagram is shown on the next page where a solid line is used for the equilibrium curve and dashed lines are used for the tie lines only three of which are given Additional tieline locations can be made using either of the two techniques illustrated in Fig 816 a The minimum solvent flow rate corresponds to an infinite number of equilibrium stages To determine this minimum 1 Points are plotted on the triangular diagram for the given compositions of the solvent S pure S feed F 45 wt A 55 wt C and raffinate RN 10 wt A on the equilibrium curve followed by drawing an operating line through the points S and RN and extending it only to the right because the tie lines slope down from left to right 2 Because the pinch region where equilibrium stages crowd together may occur anywhere additional potential operating lines are drawn through the tie lines extended to the right until they cross the operating line at the solvent end drawn in 1 where these intersections for four tie lines the three given and one other that extends through the feed point are denoted P1 P2 P3 and P4 3 The intersection point farthest from the triangular diagram is the one used to determine the minimum solvent rate where here it is P1 Pmin corresponding to a pinch point at the feed end of the cascade 4 An operating line is drawn from P1 through F to the determine the extract E1 followed by drawing lines from E1 to RN and from S to F with the intersection of these two lines determining the mixing point M The wt A at the mixing point is 365 From Eq 810 SminF xAF xAM xAM xAS 045 0365 0365 00 0233 Therefore Smin 0233F 02331000 233 kgh b Solvent rate 15 Smin 15233 350 kgh With this rate the total feed solvent component flow rates becomes A 450 kgh C 550 kgh and S 350 kgh giving a total of 1350 kgh Thus the mixing point is at xAxM 4501350 0333 xCxM 5501350 0408 and xSxM 3501350 0259 This point is plotted as M on the second triangular diagram on a following page A line from RN through M to an intersection with the equilibrium curve determines the extract point E1 as illustrated in Fig 814 The operating point P is now determined as in Fig 817 by finding the intersection of lines drawn through S and RN and through F and E1 The number of equilibrium stages is now stepped off as in Fig 817 giving a value of just slightly greater than Nv 5 Exercise 812 continued Analysis a Righttriangle plot continued a Determination of minimum solvent rate Mass fraction of acetone Mass fraction of water Exercise 812 continued Analysis b Righttriangle plot continued b Determination of stages for 15 times minimum solvent rate Mass fraction of acetone Mass fraction of water Exercise 812 continued Analysis continued c First compute the two product flow rates using the following product compositions read from the plot Mass fractions Component Acetone Water 112 TCE Feed 0450 0550 0000 Solvent 0000 0000 1000 Raffinate 0100 0895 0005 Extract 0509 0040 0451 By overall total material balance F S 1000 350 1350 RN E1 1 By overall acetone balance 045F 450 0100RN 0509E1 2 Solving Eqs 1 and 2 simultaneously E1 770 kgh and RN 580 kgh The composition of each stream leaving each stage can be read from the righttriangle plot prepared in part b Let x be mass fraction in the raffinate and y be mass fraction in the extract The stages are arbitrarily numbered from the feed end The flow rate of each stream leaving each stage is best obtained by total material balances around groups of stages and the inverse lever arm rule using the operating lines A total balance for the first n 1 stages is F EN 1000 E1 RN 1 E1 RN 1 770 or RN 1 EN 1000 770 230 3 By the inverse lever arm rule with the righttriangle diagram RN 1 EN EP RN 1 P 4 where the lines are measured from the righttriangle diagram Eqs 3 and 4 are solved simultaneously for RN 1 and EN for n 2 to 6 The following results are obtained Raffinate Extract kgh Stage xA xC xS yA yc yS R E 1 0390 0581 0029 0509 0040 0451 935 770 2 0325 0651 0024 0440 0033 0537 822 706 3 0260 0724 0016 0360 0024 0616 738 592 4 0190 0800 0010 0270 0017 0713 656 508 5 0100 0895 0005 0165 0011 0824 580 426 Operações Unitárias 2 03092024 2000 às 2100 Alternativa a finalizada Carta psicrométrica pra vc riscar tb Ao final mando a carta com todas as marcações Exercise 812 Exercise 815 continued Analysis case of 80C continued Exercise 815 Subject Extraction of diphenylhexane DPH from docosane C with furfural U at 45 and 80C Given Feed F of 500 kgh of 40 wt DPH in C 500 kgh of solvent S containing 98 wt U and 2 wt DPH Raffinate to contain 5 wt DPH Liquidliquid equilibrium data Find Number of theoretical stages DPH in kgh in the extract Analysis Case of 45C The given liquidliquid equilibrium data are plotted in the righttriangle 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 leverarm rule on line RME the mass ratio of R to E is 0445 Combining this with an overall material balance F S 500 500 1000 R E gives R 308 kgh and E 692 kgh From the diagram the mass fraction of DPH in the extract is 0281 Therefore the DPH in the extract is 0281692 1945 kgh which is 926 of the total DPH entering the extractor On the following page the equilibrium stages are stepped on another right triangle diagram as in Fig 817 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 815 continued Analysis Case of 80C The given liquidliquid equilibrium data are plotted in the righttriangle 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 leverarm rule on line RME the mass ratio of R to E is 0383 Combining this with an overall material balance F S 500 500 1000 R E gives R 277 kgh and E 723 kgh From the diagram the mass fraction of DPH in the extract is 0271 Therefore the DPH in the extract is 0271723 1959 kgh which is 933 of the total DPH entering the extractor On the following page the equilibrium stages are stepped on another right triangle diagram as in Fig 817 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