Top Operating Line: The top operating line or rectification line is a material balance for a

section above the feed plate. Total material balance is given as

Whereas individual (more-volatile component) balance is as follows:

Now these equations can be combined to give the top section operating line.

This line passes through (x_D, x_D) and has a slope of x_D/(R+1).

Feed quality line: The mass balance at the feed plate is represented by feed quality (q) line.

This line passes through (x_F, x_F) and has a slope of q/(q-1). It is more convenient to draw q-line by joining (x_F, x_F) and (x_F/q, 0). There may be five different conditions of the feed.

(1) If feed is cold, q > 1, q-line is tilted to the right,

(2) If feed is boiling, q = 1, q-line is vertical,

(3) If feed is partially vapor, 1>q>0, q-line is tilted to the left,

(4) If feed is saturated vapor, q = 0, q-line is horizontal,

(5) If feed is superheated vapor, q<0, q-line is directing south-west.

Intersection of q-line and top-operating line defines the rectification section below the rectification section is stripping section.

Intersection of q-line with the rectification-line: The intersection of these two lines can occur at any of the following points:

(1) At diagonal line: This is a situation when the tower is operating at total reflux. This gives us a way to determine the minimum number of stages that are required when no product is withdrawn.

(2) Above diagonal line and below equilibrium curve: This corresponds to normal operation of a distillation column.

(3) On the equilibrium curve: This is a situation when the tower is operating at minimum reflux. An infinite number of stages are required to achieve the desired separation.

(4) Above the equilibrium curve: This is a situation when an infinite number of stages will be insufficient to achieve the desired separation.

Minimum Reflux: One of the important information about a column could be the value of minimum reflux. Knowing its value we can get an idea of the operating reflux and an estimate of the number of stages. The easiest method is to draw the equilibrium curve and locate the intersection of the q-line and the equilibrium curve. Draw a line originating from (x_D, x_D) and passing through this intersection point. Extend this line to y-axis. The steps involved in finding the minimum reflux ratio are

(1) Find the slope of the q-line = q/(q - 1).

(2) Draw point (1) having coordinates (x_F/q, 0).

(3) Draw point (2) having coordinates (x_F, x_F).

(4) Join points (1) and (2) and extend the line (1-2) to intersect the equilibrium curve and obtain point (3).

(5) Draw point (4) having coordinates (x_D, x_D).

(6) Now join point (4) with (3) and extend the line (4-3) to intersect at y-axis at point (5). Point (5) has the coordinates (0, x_D/(R_m+1)).

(7) Calculate R_m given by

Example 6.9: Calculate the minimum reflux ratio for a column handling a feed that contains binary mixture. The more-volatile component is present at a composition of 36 % on molar basis. The distillate withdrawn is 91.5 % more-volatile component and 8.5 % less-volatile component on molar basis. The equilibrium relationship can be very well represented by a fairly constant value of relative volatility, α, of 4.12. Feed enters the column at condition, which can be represented by a q-value of 1.04.

Solution:

C Slope of the q-line = q/(q - 1) = 26. The q-line is tilted to the right.

C Draw point (1) having coordinates (x_F/q,0) = (0.346, 0).

C Draw point (2) having coordinates (x_F,x_F) = (0.36, 0.36).

C Join points (1) and (2) and extend the line (1-2) to intersect the equilibrium curve and obtain point (3) = (0.373, 0.71).

C Draw point (4) having coordinates (x_D, x_D) = (0.915, 0.915).

C Now join point (4) with (3) and extend the line (4-3) to intersect at y-axis at point (5) (0, x_D/(R_m+1)) = (0, 0.57).

Minimum reflux ratio can be found using Equation 6.33