Question 29.4: Awastewater stream is introduced to the top of a mass-transf...
A wastewater stream is introduced to the top of a mass-transfer tower where it flows countercurrent to an air stream. At one point in the tower, the wastewater stream contains 1 \times 10^{-3} \mathrm{~g} \mathrm{~mol} \mathrm{~A} / \mathrm{m}^{3} and the air is essentially free of any A. At the operating conditions within the tower, the film mass-transfer coefficients are k_{L}=5 \times 10^{-4} \mathrm{~kg} / \mathrm{mol} / \mathrm{m}^{2} \cdot \mathrm{s} \cdot\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}\right) and k_{G}=0.01 \mathrm{~kg} \mathrm{~mol} / \mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}. The concentrations are in the Henry’s law region where p_{A, \mathrm{i}}=H c_{A, \mathrm{i}} with H=10 \mathrm{~atm} /\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}\right). Determine
(a) the overall mass flux of A.
(b) the overall mass-transfer coefficients, K_{L} and K_{G}.
At the specified plane,
c_{A, L}=1.0 \times 10^{-6} \frac{\mathrm{kg} \mathrm{mol} \mathrm{A}}{\mathrm{m}^{3}}
and
p_{A, G}=0
A sketch of the partial pressure of A vs. concentration of A reveals this is a stripping operation. By equation (29-17)
\begin{aligned} \frac{1}{K_{L}} & =\frac{1}{H k_{G}}+\frac{1}{k_{L}} \\ & =\frac{1}{\left(10 \frac{\mathrm{atm}}{\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}\right)\left(0.01 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}}\right)}+\frac{1}{5 \times 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}} \end{aligned}
K_{L}=4.97 \times 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}
The equilibrium concentrations are
\begin{aligned} c_{A}^{*}=\frac{p_{A, G}}{H} & =\frac{0 \mathrm{~atm}}{10 \frac{\mathrm{atm}}{\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}}=0 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{3}} \\ p_{A}^{*}=H c_{A, L} & =\left(10 \frac{\mathrm{atm}}{\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}\right)\left(1 \times 10^{-6} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{3}}\right) \\ & =1 \times 10^{-5} \mathrm{~atm} \end{aligned}
The flux of A for this stripping operation is
\begin{aligned} N_{A}=K_{L}\left(c_{A, L}-c_{A}^{*}\right) & =\left(4.97 \times 10^{-4} \frac{\mathrm{m}}{\mathrm{s}}\right)\left(1.0 \times 10^{-6} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{3}}\right) \\ & =4.97 \times 10^{-10} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}} \end{aligned}
The overall mass-transfer coefficient, K_{G} can be determined in two ways.
\begin{aligned} K_{G}=\frac{N_{A}}{p_{A}^{*}-p_{A, G}} & =\frac{4.97 \times 10^{-10} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}}{1 \times 10^{-5} \mathrm{~atm}-0} \\ & =4.97 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}} \end{aligned}
If we multiply both sides of equation (29-17) by Henry’s law constant, H, and relate the results to equation (29-16), we obtain
\frac{H}{K_{L}}=\frac{1}{k_{G}}+\frac{H}{k_{L}}=\frac{1}{K_{G}}
\frac{1}{K_{G}}=\frac{1}{k_{G}}+\frac{m}{k_{L}} (29-16)
then
\begin{aligned} K_{G} & =\frac{K_{L}}{H}=\frac{4.97 \times 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}\right)}}{10 \mathrm{~atm} /\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}\right)} \\ & =4.97 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}} . \end{aligned}