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Minimum vapor rate for tower 1:

The minimum vapor corresponds to the minimum reflux ratio, and there are two possibilities in this case: (1) the pinched-in region could occur at the feed plate, or (2) it could occur at the top of the tower.

If the feed plate is the limiting condition, then

(0.0138 is the vapor in equilibrium with a liquid of 0.01).

If the top plate is the limiting condition, then the concentration on thio plate must be equal to the reflux concentration, and the top vapor will be in equilibrium with this composition, thus,

Since the top condition requires more vapor, it is the limiting condition and (VnUin = 755.

Minimum vapor rate for tower 2:

Tower 2 will pinch at top, the liquid on the top plate will have a composition of 0.331, and the equilibrium vapor composition will equal 0.0403.

Solution of Part 2 Tower 1:

0.0403(F)min - 0.3310 - 0.9999W2 » 0.331 F - 0.66891^2

(Fn)aot - «(755) - 1,007; On+i - 1,006 Vm = 1,007; Om+1 - 1,106

Above the feed plate,

1,007yn - 1,006®»+1 + 0.9999 0.9999 - yn - 0.999(0.9999 - sn+1)

Below the feed plate, l,007yw = l,106xm+i - 0.00001(99) ym - 0.00001 - 1.098(sTO+i - 0.00001)

Tower 2:

Operating line for phenol,

Operating line for water,

These operating lines are plotted in Figs. 8-7 and 8-8. Logarithmic plotting is used to facilitate the calculations at the low concentrations. In the case of tower 1, the operating lines for phenol are used; for tower 2, the operating line for water is used.

According to the diagram, tower 1 requires a still and 15 theoretical plates. Since the bottoms of this tower are essentially water, live steam could be used, but if the same phenol recovery (99.99 per cent) were obtained, a total of 20 theoretical plates would be required. The large increase in the number of plates is due to dilution by the large amount of vapor used. If the same bottoms composition had been maintained instead of the same recovery, 16 theoretical plates would be required and the recovery would be 98.9 per cfent.

The steps for the region x = 0.00001 to 0.001 of this tower could be calculated by Eq. (7-77) since the equilibrium curve is y — 2x.

In the case of tower 2 using mol fractions of water instead of phenol, the plates are stepped up the operating line from xw = 0.0001 to xf =» 0.669. In Fig. 8-8 a still and six theoretical plates would give a reflux composition of 0.54, and seven theoretical plates give 0.71. Thus a still and seven theoretical plates would give a slightly better separation than desired.

Solution of Part 3. At total reflux, the plates for tower 1 correspond to the steps between the equilibrium curve and the y — x line from xw 5=3 0.00001 to the reflux composition, xr = 0.0168. From the diagram it is found that a still and between 11 and 12 theoretical plates are needed.

For tower 2, the steps at total reflux go from x 0.0001 to 0.669. In this case, a still and five theoretical plates are required.

It might be thought that the effectiveness of the fractionation would be improved by refluxing to tower 1 not only the 1.68 per cent phenol layer but also a portion of the 33.1 per cent layer. This would result in an increased overhead vapor composition but would require additional plates for the same vapor rate. Thus for the same separation and vapor rate, more plates are required, indicating that only

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