Question 31.4: In Problem 30.10, the production of ultra pure silicon tetra...
In Problem 30.10, the production of ultra pure silicon tetrachloride for the chemical vapor deposition of silicon was described. Hydrogen chloride, HCl, was produced in the elimination of trichlorosilane, \mathrm{SiHCl}_3, and the HCl was stripped from the liquid silicon tetrachloride into an \mathrm{N}_2 carrier gas stream.
(a) If the HCl concentration was reduced from 1.8 to 0.2 mol % by its transfer into the 298 K and 1.013 \times 10^5 \mathrm{~Pa} nitrogen gas carrier stream, determine the \left(L_s / G_s\right)_{\min } ratio. At the involved low concentration levels, the \mathrm{HCl}-\mathrm{H}_2 system obeys Henry’s law with the Henry’s constant, H’, equal to 47.5 atm/mol fraction HCl in the liquid.
(b) Determine the HCl concentration in the exiting \mathrm{N}_2 stream when 40 times the minimum G_s is used in the stripper.
The compositions of the known streams, L_2, L_1, and G_1, on a solute-free basis are evaluated from the known mole fractions as
X_{\mathrm{HCl}, 2}=\frac{x_{\mathrm{HCl}, 2}}{1-x_{\mathrm{HCl}, 2}}=\frac{0.018}{0.982}=0.0183X_{\mathrm{HCl}, 1}=\frac{x_{\mathrm{HCl}, 1}}{1-x_{\mathrm{HCl}, 1}}=\frac{0.002}{0.998}=0.002
Y_{\mathrm{HCl}, 1}=\frac{y_{\mathrm{HCl}, 1}}{1-y_{\mathrm{HCl}, 1}}=0
Note that the solute-free compositions are essentially equal to the mole-fraction compositions for low concentrations.
The equilibrium concentration is evaluated with Henry’s law
p_{\mathrm{HCl}}(\mathrm{atm})=\left(47.5 \frac{\mathrm{atm}}{\text { mole fraction }}\right)\left(x_{\mathrm{HCl}}\right)At end 2 of the tower
p_{\mathrm{HCl}, 2}=\left(47.5 \frac{\mathrm{atm}}{\text { mole fraction }}\right)(0.018)=0.855 \mathrm{~atm}and
Y_{\mathrm{HCl}, 2}^*=\frac{p_{\mathrm{HCl}, 2}}{P-p_{\mathrm{HCl}, 2}}=\frac{0.855}{0.145}=5.90Figure 31.13 illustrates the operating lines for both G_{S, \text { min }} \text { and } G_{S, \text { actual }} for this example. As one notes, the operating lines for the stripping process are below the equilibrium line.
\left(\frac{L_S}{G_S}\right)_{\min }=\frac{Y_2^*-Y_1}{X_2-X_1}=\frac{5.90-9}{0.0183-0.002}=362and
\left(\frac{L_S}{G_S}\right)_{\text {actual }}=\frac{L_S}{(G s)(40)}=\frac{362}{40}=9.05\left(\frac{L_S}{G_S}\right)_{\text {actual }}=9.05=\frac{Y_2-Y_1}{X_2-X_1}=\frac{Y_2-0}{0.0183-0.002}
Y_2=0.147
The mole fraction of the exiting gas stream is
y_{\mathrm{HCl}, 2}=\frac{Y_2}{1+Y_2}=\frac{0.147}{1.147}=0.128.
