4-49 Chapter 4 Example 4.4-5 ---------------------------------------------------------------------------------- 1A liquid mixture containing 10 mol % n-heptane and 90 mol % n-octane is fed at its boiling point to the top of a stripping tower at 101.32 kPa. Figure E4.4-5 depicts a stripping tower where the feed stream is the saturated liquid and the distillate stream is the saturated vapor. There is no reboiler or condenser in a stripping tower. The bottoms are to contain 98 mol % n-octane. For every 3 mol of feed, 2 mol of vapor is withdrawn as product. Calculate the composition of the vapor and the number of theoretical plates required. The equilibrium data are given below x 0 0.012 0.039 0.067 0.097 0.284 y 0 0.025 0.078 0.131 0.184 0.459 FBV = D Figure E4.4-5 Stripping tower with feed F and product D = V. Solution ------------------------------------------------------------------------------------------ F = 3 = D + B = 2 + B ⇒ B = 1 mol/unit time 50 = 0.90D + 0.10B ⇒ D = 50 kmol/h, B = 50 kmol/h Operating line yV + xBB = xL ⇒ y = xL/V − xBB/V = 1.5x − 0.02×1/2 = 1.5x − 0.01 Overall balance yDD + xBB = xFF ⇒ yD = xFF/D − xBB/D = 0.1×1.5 − 0.02×1/2 = 0.14 The number of theoretical plates required can be determined from the following Matlab codes: 1 Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th ed., Prentice Hall, 2003, p. 7544-50 % Example 4.4-5 clf xB=0.02; V=2;L=3; B=1; % Stripping operating line: y = a*x + b, where a=L/V;b=-xB*B/V; xF=0.10;yF=a*xF+b; xe=[0 0.012 0.039 0.067 0.097 0.284]; ye=[0 0.025 0.078 0.131 0.184 0.459]; pp=spline(xe,ye); x=0:0.01:0.15; y=ppval(pp,x); plot(x,y,[0 .15],[0 .15],[xB xF],[xB yF],[xF xF],[xF yF],'--') xlabel('x');ylabel('y') hold on; xo=xB;yo=xB; for i=1:20; ye=ppval(pp,xo); line([xo xo],[yo ye]) xe=xo; xo=(ye-b)/a; line([xe xo],[ye ye]) if ye>yF, break, end yo=ye; end net=i-1 + (yF-yo)/(ye-yo); fprintf('# of equilibrium stages = %g\n',net) >> E4d4d5 # of equilibrium stages = 3.845574-51 FB, xBV, yDS (Steam)xFLxVy Example 4.4-6 ---------------------------------------------------------------------------------- 2A liquid feed at the boiling point contains 3.3 mol % ethanol and 96.7 mol % water and enters the top tray of a stripping tower shown. Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 99% of the alcohol in the feed. Assume equimolar overflow for this problem. Equilibrium data for mole fraction of alcohol are as follows at 101.32 kPa abs pressure. x 0 0.0080 0.020 0.0296 0.033 y 0 0.0750 0.175 0.250 0.270 (a) For an infinite number of theoretical steps, calculate the minimum moles of steam needed per mole of feed. (Be sure and plot the q-line) (b) Using twice the minimum moles of steam, calculate the number of theoretical steps needed, the composition of the overhead vapor, and the bottoms composition. Solution ------------------------------------------------------------------------------------------ (a) Basis: F = 100 mol/h. For equimolar overflow S = V and B = F = 100 mol/h 2 Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th ed., Prentice Hall, 2003, p. 7544-52 F + S = V + B Ethanol balance: xFF = xBB + yDV (0.033)(100) = xB(100) + yDV From the problem statement: yDV = (0.99)(0.033)(100) and xB(100) = (0.01)(100)(0.033) xB = 0.00033 For infinite number of trays, yD is in equilibrium with xF = 0.033 ⇒ yD = 0.27 yDV = (99)(0.033) ⇒ Vmin = (99)(0.033)/0.27 = 12.1 mol/h = Smin Minimum 0.121 mol steam/mol feed (b) Using twice the minimum moles of steam, calculate the number of theoretical steps needed, the composition of the overhead vapor, and the bottoms composition. S = 2Smin = 24.2 mol/h Since the overhead vapor contains 99% of the alcohol in the feed, xB = 0.00033 Ethanol balance over the top part of the tower: 100(xF − x) = 24.2(yD − y) At the bottom: y = 0, x = xB = 0.00033 ⇒ yD = (100)(0.033 − 0.00033)/24.2 = 0.135 Operating line: 100(0.033 − x) = 24.2(0.135 − y) The number of theoretical steps can be determined from the following Matlab codes: % Example E4d4d6 clf xB=0.00033;xF=0.033;yD = 0.135; % Operating line: 100(0.033 - x) = 24.2(0.135 - y) xe=[0 0.0080 0.020 0.0296 0.033]; ye=[0 0.0750 0.175 0.250 0.270]; pp=spline(xe,ye); x=0:0.001:0.033; y=ppval(pp,x); yF=ppval(pp,xF); plot(x,y,[0 .035],[0 .035],[xB xF],[0 yD],[xF xF],[xF yF],'--') xlabel('x');ylabel('y')4-53 hold on; xo=xB;yo=0; for i=1:20; ye=ppval(pp,xo); line([xo xo],[yo ye]) xe=xo; xo=xF-.242*(0.135-ye); line([xe xo],[ye ye]) if ye>yD, break, end yo=ye; end net=i-1 + (yD-yo)/(ye-yo); fprintf('# of equilibrium stages = %g\n',net) >> s5p3 # of equilibrium stages = 4.73777 Example 4.4-7 ---------------------------------------------------------------------------------- 3(a) For the cascade shown below, calculate the composition of streams V4 and L1. Assume constant molar overflow, atmospheric pressure, saturated liquid and vapor feeds, and the vapor-liquid equilibrium data given below. x 0 0.10 0.30 0.50 0.70 0.90 1.000 y 0 0.20 0.50 0.68 0.82 0.94 1.000 3 J. D. Seader and E. J. Henley, Separation Process Principles, Wiley, 19984-54 V4L11234100 moles30% alcohol70% water100 moles70% alcohol30% waterV4L11234100 moles30% alcohol70% water(a)TotalcondenserD50 moles(b)LR Figure E4.4-7 Vapor-liquid equilibrium cascade. (b) Given the feed compositions in cascade (a), how many stages would be required to produce a V4 containing 85% alcohol? (c) For the configuration in cascade (b), with D = 50 moles what are the composition of D and L1? (d) For the configuration in cascade (b), how many stages are required to produce a D of 50% alcohol? Solution ------------------------------------------------------------------------------------------ Alcohol balance over the lower section of the cascade gives: Lx1 + Vy = Lx + Vy0 Since L = V = 100 moles, we have y = x + y0 − x1 = x + 0.3 − x1 This operating line will terminate on the line y0 = 0.3 and x5 = 0.7 and provide 4 equilibrium stages. Composition of streams V4 and L1 can be determined from the following procedure:4-55 Assume a value for x1 (between 0.15 and 0.30) and step off between the equilibrium curve and the