Page 179 - Separation process engineering

The y = x intercept is

(4-52)

This point can be plotted if S, x , D, and x are known. Derivation of Eqs. (4-51) and (4-52) is left as
S D
Problem 4.C3.
A second point can be found where the side stream is withdrawn. A saturated liquid withdrawal is
equivalent to a negative feed of concentration x . Thus there must be a vertical feed line at x = x . The top
S
S
and middle operating lines must intersect at this feed line.
Side-stream calculations have one difference that sets them apart from feed calculations. The stage must
hit exactly at the point of intersection of the two operating lines. This is illustrated in Figure 4-22B. Since
the liquid side stream is withdrawn from tray 2, we must have x = x . If the stage location is given, x S
S
2
can be found by stepping off the required number of stages.
For a liquid withdrawal, a balance on the liquid gives

(4-53)

while vapor flow rates are unchanged, V = V′. Thus slope, L′/V′, of the middle operating line can be
determined if L and V are known. L and V can be determined from L/D and D, where D can be found from
external balances once x is known.
S
For a vapor side stream, the feed line is horizontal at y = y . A balance on vapor flow rates gives
S

(4-54)

while liquid flow rates are unchanged. Again L′/V′ can be calculated if L and V are known.

If a specified value of x (or y ) is desired, the problem is trial and error. The top operating line is
S
S
adjusted (change L/D) until a stage ends exactly at x or y .
S S
Calculations for side streams below the feed can be developed using similar principles (see Problem
4.C4).