Question

A mixture of n-heptane (A) and ethyl benzene (B) is to be separated in a distillation column operated at 1 atm. The feed flowrate is 300 mol/h and contains 40 mol% n-heptane. 60 % of the feed is liquid. It is required to obtain an overhead product containing 96 mol% n-heptane and a bottom product containing 5 mol% n-heptane. Find the following:

a) The distillate and bottom flowrates

b) The minimum reflux ratio

c) The minimum number of stages

d) The number of stages at a reflux ratio of 1.5R_min

The following is the equilibrium data for n-heptane/ethyl benzene at 1 atm.

xA	yA
0.000	0.000
0.080	0.230
0.250	0.514
0.485	0.730
0.790	0.904
1.000	1.000

Answer 1

a) This part can be solved using material balance equation.

F is the feed flowrate, here 300mol/h,

D is the distillate flowrate,

W is the bottom flowrate,

z_f, x_d,x_w, 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

  

(This equation is obtained by solving the equations of stripping and rectification sections together)

We know q,z_f in this equation.

Feed line originates from the feed point (z_f,z_f) 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 x_d=0.96.

Thus R_min=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.

It can be obtained from the graph by plotting steps starting from D to W. I have got this as 6.

d) At R=1.5R_min, R=2.289

q-line remains same.

Thus the intercept of Rectifying section changes as

Now join D(0.96,0.96) to new intercept at y-axis. Draw a line from W(0.05,0.05) to the meeting point of Rectification line and q line.

Now start drawing no of stages, the steps from D to W as shown.No of stages then=18

for graph, please check.

https://1drv.ms/f/s!Al_FIcMM9Tw1kq1qukaEga4tVPE6Zw