23:25 6,00 N... (50) a) This part can be solved using material balance equation F- Fy-D F is the feed flowrate, here 300mol/h. D is the distillate flowrate, W is the bottom flowrate, Zf, Xd.xw, are the mole fraction of lighter component, here n-heptane since it is coming abundantly in distillate, and values are 0.4, 0.96 and 0.05 respectively. On Solving these two equations with given parameters we obtain, D=115.385 W=184.615 b) Following three parts has to be solved graphically with the given equilibrium data. Plot the equilibrium data in a graph. Draw the line y=x. (Scale may be taken as I've done, please find the graph attached) Mark the points D(0.96,0.96) and W(0.05,0.05) in the line y=x. It is given that 60% of the feed is liquid. That is feed quality, q, is 0.6. Thus we can plot the q-line aka feed line whose equation is given as A (This equation is obtained by solving the equations of stripping and rectification sections together) We know qiz+ in this equation. + Feed line originates from the feed point (zf,zf) that is (0.4,0-4). Substitute x=0 in the feed line equation, then we will get y=1. Thus we can draw the q-line joining these two points. (0.4,0.4) and (0,1) Through the meeting point of q-line and equilibruim line, draw a line from D(0.96,0.96) to y-axis. The intercept obtained will be From the graph, intercept=0.38 and we know Xd=0.96. Thus Rmin=1.526 c) Minimum no of Stages at Total Reflux (R=infinity). At this condition the equation for rectifying section and stripping section will coincide with y=x. 23:25 6,00 N... (50) a) This part can be solved using material balance equation F- Fy-D F is the feed flowrate, here 300mol/h. D is the distillate flowrate, W is the bottom flowrate, Zf, Xd.xw, are the mole fraction of lighter component, here n-heptane since it is coming abundantly in distillate, and values are 0.4, 0.96 and 0.05 respectively. On Solving these two equations with given parameters we obtain, D=115.385 W=184.615 b) Following three parts has to be solved graphically with the given equilibrium data. Plot the equilibrium data in a graph. Draw the line y=x. (Scale may be taken as I've done, please find the graph attached) Mark the points D(0.96,0.96) and W(0.05,0.05) in the line y=x. It is given that 60% of the feed is liquid. That is feed quality, q, is 0.6. Thus we can plot the q-line aka feed line whose equation is given as A (This equation is obtained by solving the equations of stripping and rectification sections together) We know qiz+ in this equation. + Feed line originates from the feed point (zf,zf) that is (0.4,0-4). Substitute x=0 in the feed line equation, then we will get y=1. Thus we can draw the q-line joining these two points. (0.4,0.4) and (0,1) Through the meeting point of q-line and equilibruim line, draw a line from D(0.96,0.96) to y-axis. The intercept obtained will be From the graph, intercept=0.38 and we know Xd=0.96. Thus Rmin=1.526 c) Minimum no of Stages at Total Reflux (R=infinity). At this condition the equation for rectifying section and stripping section will coincide with y=x