Question 11.4: A column is to be designed to separate a mixture of ethylben...

Ethylbenzene is the more volatile component.

P bar, T Kelvin

Material balance, basis 100 kmol feed:

ethylbenzene in the tops = 50 – 0.21 = 49.79 kmol

styrene in tops = 50 – 42.5 = 7.5 kmol

Column bottom temperature, from Antoine equation for styrene

At 93.3^{\circ} C, vapour pressure of ethylbenzene

The relative volatility will change as the compositions and (particularly for a vacuum column) the pressure changes up the column. The column pressures cannot be estimated until the number of stages is known; so as a first trial the relative volatility will be taken as constant, at the value determined by the bottom pressure.

Rectifying section

s=\frac{8}{8+1}=0.89 (11.35)

b=\frac{0.87}{8+1}=0.097 (11.36)

0.89(1.35-1) k^{2}+[0.89+0.097(1.35-1)-1.35] k+0.097=0 (11.29)

k = 0.290

x_{0}^{*}=0.87-0.29=0.58 (11.33)

x_{n}^{*}=0.50-0.29=0.21 (11.34)

c=1+(1.35-1) 0.29=1.10 (11.32)

\beta=\frac{0.89 \times 1.10(1.35-1)}{1.35-0.89 \times 1.1^{2}}=1.255 (11.31)

N=\log \left[\frac{0.58(1-1.255 \times 0.21)}{0.21(1-1.255 \times 0.58)}\right] / \log \left(\frac{1.35}{0.89 \times 1.1^{2}}\right) =\frac{\log 7.473}{\log 1.254}=8.87, \underline{\underline{\text { say } 9}} (11.30)

Stripping section, feed taken as at its bubble point

s=\frac{8 \times 0.5+0.87-(8+1) 0.005}{(8+1)(0.5-0.005)}=1.084 (11.39)

b=\frac{(0.5-0.87) 0.005}{(8+1)(0.5-0.005)}=-4.15 \times 10^{-4}(\text { essentially zero }) (11.40)

k = 0.702

x_{0}^{*}=0.5-0.702=-0.202 (11.37)

x_{n}^{*}=0.005-0.702=-0.697 (11.38)

c=1+(1.35-1) 0.702=1.246 (11.32)

\beta=\frac{1.084 \times 1.246(1.35-1)}{1.35-1.084 \times 1.246^{2}}=-1.42 (11.31)

N=\log \left[\frac{-0.202(1-0.697 \times 1.42)}{-0.697(1-0.202 \times 1.42)}\right] / \log \left(\frac{1.35}{1.084 \times 1.246^{2}}\right) =\frac{\log \left[4.17 \times 10^{-3}\right]}{\log 0.8}=24.6, \text { say } 25 (11.30)