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Multiple Feed and Side Stream. Multiple Feed. F 1 + F 2 = B + D, F 1 x F1 + F 2 x F2 = Bx B + Dx D. There are three operating lines. a) Rectifying section. n. x D. D x D. L. V. n+1. F 1 x F1. q 1. k. b) Intermediate line. k+1. F 2 x F2. q 2. m. m+1. B x B.
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Multiple Feed and Side Stream • Multiple Feed F1 + F2 = B + D, F1xF1 + F2xF2= BxB + DxD • There are three operating lines a) Rectifying section n xD D xD L V n+1 F1 xF1 q1 k b) Intermediate line k+1 F2 xF2 q2 m m+1 B xB ChE 334: Separation Processes Dr Saad Al-Shahrani
xF1 xF2 Distillation of Binary Mixture c) Stripping line ChE 334: Separation Processes Dr Saad Al-Shahrani
Binary Multistage Distillation • Feed with side stream F = D + Ls +B a) xs > xF xD D xD L 1. Rectifying line V Ls xs k k+1 F xF 2. intermediate line B xB ChE 334: Separation Processes Dr Saad Al-Shahrani
xs xF Binary Multistage Distillation 2. Stripping line Note: to obtain the locus of intersections of rectifying and intermediate lines, put yn+1=yk+1, xn=xk in equaton (2) and subtract equation (2) from equation (1). y = yn+1=yk+1, x = xn=xk and intersect the 45 oC line at xs,xs ChE 334: Separation Processes Dr Saad Al-Shahrani
Binary Multistage Distillation b) xF >xs 1. Rectifying section The feed stream and the top of column as shown in figure. L xD D xD V F xF k 2. Intermediate section k+1 Ls xs B xB ChE 334: Separation Processes Dr Saad Al-Shahrani
xF xS Binary Multistage Distillation 3. Bottom section ChE 334: Separation Processes Dr Saad Al-Shahrani
V L F xF Ls xs V L F xF Vs xs Binary Multistage Distillation • Side stream condition a) Side stream is saturated liquid b) Side stream is saturated vapor ChE 334: Separation Processes Dr Saad Al-Shahrani
Binary Multistage Distillation V L c) Side stream is (vapor + liquid) F xF SV+SL xs ChE 334: Separation Processes Dr Saad Al-Shahrani
McCabe Thiele Graphical Equilibrium-Stage Example: mixture of water and ethyl alcohol containing 0.16 mole fraction alcohol is continuously distilled in a plate fractionating column to give a product containing 0.77 mole fraction alcohol and a waste of 0.02 mole fraction alcohol. Its proposed to withdraw 25% of the alcohol in the entering stream as a side stream with a mole fraction of 0.5 alcohol. Determine the no. of theoretical plates required and the plate from which the side stream should be withdrawn if the feed is liquid at the boiling point and RD= 2.0 Solution Basis = 100 k mol of feed 16 k mol of alcohol enter, and 25% (4 k mol) are to be removed in the side stream. As the side stream composition is to be 0.5, there are 8 k mol in the side stream ChE 334: Separation Processes Dr Saad Al-Shahrani
McCabe Thiele Graphical Equilibrium-Stage Total balance: F=D+B+S 100=D+B+8 92=D+B (1) xD L D xD=0.77 V Sat. liq. Component balance on alcohol 100*0.16=0.77D+0.02B+4 12=0.77D+0.02B (2) Ls= 8 Xs=0.5 F xF=0.16 By solving equations (1) and (2) D= 13.55 K mol B= 78.45 K mol B xB=0.02 ChE 334: Separation Processes Dr Saad Al-Shahrani
McCabe Thiele Graphical Equilibrium-Stage a) Rectifying section R=L/D=2.0, L=2*13.55=27.1 K mol xD L V=L+D=27.1+13.55 = 40.65 K mol D xD=0.77 V Sat. liq. L/V = 27.1/40.65 = 0.67 Ls= 8 Xs=0.5 b) Intermediate section F xF=0.16 B xB=0.02 ChE 334: Separation Processes Dr Saad Al-Shahrani
McCabe Thiele Graphical Equilibrium-Stage b) Bottom section xD L D xD=0.77 V Sat. liq. Ls= 8 Xs=0.5 F xF=0.16 B xB=0.02 ChE 334: Separation Processes Dr Saad Al-Shahrani
Mol fraction alcohol in liquid McCabe Thiele Graphical Equilibrium-Stage Number of ideal plates= 8 + reboiler Side stream plate = 4 from the top Feed plate= 7 from the top. ChE 334: Separation Processes Dr Saad Al-Shahrani
