Awastewater 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 × 10^-3 g mol A/m³ and the air is essentially free of any A. At the operating conditions within the tower, the film

Question

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 $K_{G}$ .

(a) the overall mass flux of [latex]A[/latex].

(b) the overall mass-transfer coefficients, [latex]K_{L}[/latex] and [latex]K_{G}[/latex].

Accepted Answer

At the specified plane,

[latex]c_{A, L}=1.0 imes 10^{-6} \frac{\mathrm{kg} \mathrm{mol} \mathrm{A}}{\mathrm{m}^{3}}[/latex]

and

[latex]p_{A, G}=0[/latex]

A sketch of the partial pressure of A vs. concentration of A reveals this is a stripping operation. By equation (29-17)

[latex] \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}} ight)\left(0.01 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}} ight)}+\frac{1}{5 imes 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}}} \end{aligned} [/latex]

[latex] K_{L}=4.97 imes 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{kg} \mathrm{mol} / \mathrm{m}^{3}} [/latex]

The equilibrium concentrations are

[latex] \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}} ight)\left(1 imes 10^{-6} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{3}} ight) \ & =1 imes 10^{-5} \mathrm{~atm} \end{aligned} [/latex]

The flux of [latex]A[/latex] for this stripping operation is

[latex] \begin{aligned} N_{A}=K_{L}\left(c_{A, L}-c_{A}^{*} ight) & =\left(4.97 imes 10^{-4} \frac{\mathrm{m}}{\mathrm{s}} ight)\left(1.0 imes 10^{-6} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{3}} ight) \ & =4.97 imes 10^{-10} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}} \end{aligned} [/latex]

The overall mass-transfer coefficient, [latex]K_{G}[/latex] can be determined in two ways.

[latex] \begin{aligned} K_{G}=\frac{N_{A}}{p_{A}^{*}-p_{A, G}} & =\frac{4.97 imes 10^{-10} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}}{1 imes 10^{-5} \mathrm{~atm}-0} \ & =4.97 imes 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}} \end{aligned} [/latex]

If we multiply both sides of equation (29-17) by Henry's law constant, [latex]H[/latex], and relate the results to equation (29-16), we obtain

[latex]\frac{H}{K_{L}}=\frac{1}{k_{G}}+\frac{H}{k_{L}}=\frac{1}{K_{G}}[/latex]

[latex]\frac{1}{K_{G}}=\frac{1}{k_{G}}+\frac{m}{k_{L}}[/latex] (29-16)

then

[latex] \begin{aligned} K_{G} & =\frac{K_{L}}{H}=\frac{4.97 imes 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3} ight)}}{10 \mathrm{~atm} /\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^{3} ight)} \ & =4.97 imes 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}} . \end{aligned} [/latex]

Question 29.4: Awastewater stream is introduced to the top of a mass-transf...

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