Consider the remediation trench shown in Figure 30.7, a very simple process to treat contaminated wastewater before discharge to a lake or river. The remediation trench consists of a narrow outdoor open channel with an air sparger aligned along the bottom of the trench. Wastewater containing a

Question

Consider the remediation trench shown in Figure 30.7, a very simple process to treat contaminated wastewater before discharge to a lake or river. The remediation trench consists of a narrow outdoor open channel with an air sparger aligned along the bottom of the trench. Wastewater containing a volatile contaminant dissolved in the water enters one end of the trench. As the wastewater flows down the trench, the aeration gas strips out the dissolved volatile solute and transfers it to the surrounding atmosphere by an interphase mass-transfer process. Consequently, the concentration of the solute in the wastewater decreases down the length of the trench. Remediation trenches can be long, and may extend from a holding pond to the discharge point.

Accepted Answer

We wish to design an aerated remediation trench to treat wastewater contaminated with trichloroethylene (TCE) at a concentration of [latex]50 \mathrm{mg} / \mathrm{L}\left(50 \mathrm{~g} \mathrm{TCE} / \mathrm{m}^{3} ight.[/latex] wastewater). The trench is open duct of width [latex](W) 1 \mathrm{~m}[/latex] and depth [latex](H) 2 \mathrm{~m}[/latex], and the volumetric flow rate of wastewater added to the trench is [latex]0.1 \mathrm{~m}^{3} / \mathrm{s}[/latex]. Air is sparged into the bottom of the duct at a rate that provides a gas holdup of [latex]0.02 \mathrm{~m}^{3}[/latex] of gas per [latex]1 \mathrm{~m}^{3}[/latex] of water, and the average bubble diameter is [latex]1 \mathrm{~cm}(0.01 \mathrm{~m})[/latex]. Determine the length of the trench necessary to reduce the effluent TCE concentration to [latex]0.05 \mathrm{mg} / \mathrm{L}[/latex]. The process temperature is [latex]293 \mathrm{~K}[/latex] and the total system pressure is [latex]1 \mathrm{~atm}[/latex].

Again, the strategy is to develop a material-balance model for the process, and then incorporate the appropriate mass-transfer correlation(s) into the material-balance calculations. The physical system represents a steady-state, continuous flow process where TCE is transferred from the wastewater to the aeration gas. As the aeration gas locally mixes the liquid, assume that the concentration profile is one-dimensional along axial coordinate [latex]z[/latex]. Referring to Figure 30.7, a steadystate material balance for TCE (species A) in the liquid phase of the trench within the differential volume element [latex]W \cdot H \Delta z[/latex] for [latex]R_{A}=0[/latex] is

[latex] \left(\begin{array}{l} ext { rate of TCE into } \ ext { volume element } \ ext { carried by water } \end{array} ight)-\left(\begin{array}{l} ext { rate of TCE exiting } \ ext { volume element } \ ext { carried by water } \end{array} ight)-\left(\begin{array}{l} ext { rate of TCE transferred } \ ext { from water to air by } \ ext { interphase mass transfer } \end{array} ight)=0 [/latex]

or

[latex]\left.W \cdot H\left(v_{\infty} c_{A} ight) ight|_{z}-\left.W \cdot H\left(v_{\infty} c_{A} ight) ight|_{z+\Delta z}-N_{A} \cdot \frac{A_{i}}{V} W \cdot H \Delta z=0[/latex]

where [latex]v_{\infty}[/latex] is the bulk average velocity of the wastewater and [latex]W \cdot H[/latex] is the cross-sectional area of the trench for fluid flow. Dividing by [latex]W \cdot H \Delta z[/latex] and taking the limit as [latex]\Delta z[/latex] goes to zero yields

[latex]v_{\infty} \frac{d c_{A}}{d z}+N_{A} \frac{A_{i}}{V}=0[/latex]

For interphase mass transfer of TCE from water to the gas bubble, [latex]N_{A}[/latex] must be

[latex]N_{A}=K_{L}\left(c_{A}-c_{A}^{*} ight)[/latex]

where [latex]K_{L}[/latex] is the overall interphase mass-transfer coefficient based on the overall liquid phase driving force and [latex]c_{A}^{*}[/latex] is the concentration of TCE in the liquid that is in equilibrium with the partial pressure of TCE vapor in the air bubble. This equilibrium relationship is described by Henry's law

[latex]c_{A}^{*}=\frac{p_{A}}{H}[/latex]

where [latex]H[/latex] is Henry's law constant for TCE in water, which is equal to [latex]9.97 \mathrm{~atm} \cdot \mathrm{m}^{3} / \mathrm{kg}[/latex] mol at [latex]293 \mathrm{~K}[/latex]. As the rate of TCE transfer will be very small relative to the air flow rate bubbled into the trench, [latex]p_{A}[/latex] for TCE inside the air bubble is essentially zero, and so [latex]c_{A}^{*}[/latex] will be essentially zero. Furthermore, as the value of [latex]H[/latex] for TCE in water is very high, TCE is only sparingly soluble in water, and so the interphase mass-transfer process will be liquid-phase controlling. Consequently, the liquid-film mass-transfer coefficient, [latex]k_{L}[/latex], will suffice for the overall mass-transfer coefficient [latex]k_{L}[/latex]. Therefore, the flux equation reduces to

[latex]N_{A}=k_{L} c_{A}[/latex]

The material balance is now

[latex]v_{\infty} \frac{d c_{A}}{d z}+k_{L} c_{A} \frac{A_{i}}{V}=0[/latex]

Separation of dependent variable [latex]c_{A}[/latex] from independent variable [latex]z[/latex] followed by integration from the entrance of the trench, [latex]z=0, c_{A}=c_{A o}[/latex], to the exit of the trench, [latex]c_{A}=c_{A L}, z=L[/latex] yields

[latex]-\int_{c_{A o}}^{c_{A L}} \frac{d c_{A}}{c_{A}}=\frac{k_{L}}{v_{\infty}} \cdot \frac{A_{i}}{V} \int_{o}^{L} d z[/latex]

The final design equation is

[latex]L=\frac{\ln \left(\frac{c_{A o}}{c_{A L}} ight)}{\frac{k_{L}}{v_{\infty}} \cdot \frac{A_{i}}{V}}[/latex]

The bulk average velocity of water through the open channel is

[latex]v_{\infty}=\frac{0.1 \frac{\mathrm{m}^{3}}{\mathrm{~s}}}{W \cdot H}=\frac{0.1 \frac{\mathrm{m}^{3}}{\mathrm{~s}}}{1 \mathrm{~m} \cdot 2 \mathrm{~m}}=0.05 \frac{\mathrm{m}}{\mathrm{s}}[/latex]

As the bulk velocity is relatively slow, we assume that the natural convection of the rapidly rising bubbles will dominate the convective mass-transfer process. The term [latex]k_{L} A_{i} / V[/latex] must now be evaluated. For a nonagitated bubble swarm of average bubble diameter [latex]d_{b}[/latex], the interphase masstransfer area per unit liquid volume is

[latex]\frac{A_{i}}{V}=\frac{V_{g}}{V} \cdot \frac{6}{d_{b}}=\left(\frac{0.02 \mathrm{~m}^{3} ext { air }}{1 \mathrm{~m}^{3} ext { water }} ight) \frac{6}{0.01 \mathrm{~m}}=12 \frac{\mathrm{m}^{2}}{\mathrm{~m}^{3}}[/latex]

where [latex]V_{g} / V[/latex] is the aeration gas volume per unit volume of liquid, commonly called the gas holdup ratio. This value for [latex]A_{i} / V[/latex] is much higher than the open surface area per unit volume, which is [latex]0.5 \mathrm{~m}^{2} / \mathrm{m}^{3}[/latex] of liquid. Equation (30-14a)

[latex]\mathrm{Sh}={\frac{k_{L}d_{b}}{D_{A B}}}=0.31\,\mathrm{Gr}^{1/3}\,\mathrm{Sc}^{1/3}[/latex] (30-14a)

for mass transfer of gas bubble swarms into a nonagitated liquid is appropriate

[latex]\mathrm{Sh}=\frac{k_{L} d_{b}}{D_{A B}}=0.42 \mathrm{Gr}^{1 / 3} \mathrm{Sc}^{1 / 2} ext { for } d_{b}>2.5 \mathrm{~mm}[/latex] (30-15)

with

[latex]\mathrm{Gr}=\frac{d_{b}^{3} ho_{L} g \Delta ho}{\mu_{L}^{2}}=\frac{(0.01 \mathrm{~m})^{3}\left(998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ight)\left(9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}} ight)(998.2-1.19) \frac{\mathrm{kg}}{\mathrm{m}^{3}}}{\left(9.93 imes 10^{-4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}} ight)^{2}}=9.89 imes 10^{6}[/latex]

and

[latex]\mathrm{Sc}=\frac{\mu_{L}}{ ho_{L} D_{A B}}=\frac{9.93 imes 10^{-4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}}{\left(998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ight)\left(8.9 imes 10^{-10} \frac{\mathrm{m}^{2}}{\mathrm{~s}} ight)}=1118[/latex]

where values for [latex] ho_{L}[/latex] and [latex]\mu_{L}[/latex] for water and [latex] ho[/latex] for air at 1 atm and [latex]293 \mathrm{~K}[/latex] were obtained from Appendix I. The molecular diffusion coefficient of TCE in water [latex]\left(D_{A B} ight)[/latex] at [latex]293 \mathrm{~K}[/latex] was determined by the HaydukLaudie correlation. Finally

[latex] \begin{aligned} k_{L} & =\frac{D_{A B}}{d_{b}} 0.42 \mathrm{Gr}^{1 / 3} \mathrm{Sc}^{1 / 2}=\frac{8.9 imes 10^{-10} \frac{\mathrm{m}^{2}}{\mathrm{~s}}}{0.01 \mathrm{~m}} 0.42\left(9.89 imes 10^{6} ight)^{1 / 3}(1118)^{1 / 2} \ & =2.683 imes 10^{-4} \frac{\mathrm{m}}{\mathrm{s}} \end{aligned} [/latex]

The channel length necessary to reduce the dissolved TCE concentration from [latex]c_{A o}=50 \mathrm{mg} / \mathrm{L}[/latex] to [latex]c_{A L}=0.05 \mathrm{mg} / \mathrm{L}[/latex] is

[latex] L=\frac{\ln \left(\frac{c_{A o}}{c_{A L}} ight)}{\frac{k_{L}}{v_{\infty}} \cdot \frac{A_{i}}{V}}=\frac{\ln \left(\frac{50 \mathrm{mg} / \mathrm{L}}{0.05 \mathrm{mg} / \mathrm{L}} ight)}{\frac{\left(2.683 imes 10^{-4} \frac{\mathrm{m}}{\mathrm{s}} ight)\left(12 \frac{\mathrm{m}^{2}}{\mathrm{~m}^{3}} ight)}{0.05 \frac{\mathrm{m}}{\mathrm{s}}}}=107.3 \mathrm{~m} [/latex]

Question 30.6: Consider the remediation trench shown in Figure 30.7, a very...

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