Question 11.12: For the plate design in Example 11.11, estimate the plate ef...

Plate will be in the stripping section (see Figure 11.7).

Plate dimensions:

\text { length between downcomers (Figure 11.32) (liquid path, } Z_{L} \text { ) }=0.79-2 \times 0.134= 0.52 m,

weir height = 50 mm.

Flow rates, check efficiency at minimum rates, at column base:

from the MaCable-Thiele diagram (Figure 11.7) at x = 0.05, assuming 60 per cent plate efficiency, y \approx 0.4. The liquid composition, x = 0.05, will occur on around the ninth plate from the bottom, the seventh from the top of the column. The pressure on this plate will be approximately:

say, 1.14 bar

At this pressure the plate temperature will be 79^{\circ} C, and the liquid and vapour physical properties, from PPDS:

liquid

vapour

Average width over active surface = 0.38/0.52 = 0.73 m

N _{G}=\frac{\left(0.776+4.57 \times 10^{-3} \times 50-0.24 \times 2.48+105 \times 4.69 \times 10^{-3}\right)}{\left(\frac{10.0 \times 10^{-6}}{1.35 \times 18.8 \times 10^{-6}}\right)^{1 / 2}} = 1.44 (11.71)

Z_{c}=0.006+0.73 \times 10^{-3} \times 50-0.24 \times 10^{-3} \times 2.48 \times 50+1.22 \times 4.69 \times 10^{-3}=18.5 \times 10^{-3} m ^{3} / m ^{2} (11.75)

t_{L}=\frac{18.5 \times 10^{-3} \times 0.52}{4.69 \times 10^{-3}}=2.05 s (11.73)

N _{L}=\left(4.13 \times 10^{8} \times 4.64 \times 10^{-9}\right)^{0.5} \times(0.21 \times 2.48+0.15) \times 2.05=1.9 (11.72)

D_{e}=\left(0.0038+0.017 \times 2.13+3.86 \times 4.69 \times 10^{-3}+0.18 \times 10^{-3} \times 50\right)^{2} =0.0045 m ^{2} / s (11.77)

P_{e}=\frac{0.52^{2}}{0.0045 \times 2.05}=29.3 (11.76)

From the McCabe-Thiele diagram, at x = 0.05, the slope of the equilibrium line = 1.0.

V/L D 0.032/0.158 = 0.20

So plate efficiency = 71 per cent.

Note: The slope of the equilibrium line is difficult to determine at x = 0.05, but any error will not greatly affect the value of E_{m V}.