A 20,000-ft³(566 m³) aeration pond is aerated with 15 spargers, each using compressed air at a rate of 15 standard cubic feet per minute (7.08 × 10^-3 m³/s). The spargers will be located 15 ft (4.57 m) belowthe surface of the pond. Find the time required to raise the dissolved oxygen from 2 to

Question

A [latex]20,000-\mathrm{ft}^3\left(566 \mathrm{~m}^3 ight)[/latex] aeration pond is aerated with 15 spargers, each using compressed air at a rate of 15 standard cubic feet per minute [latex]\left(7.08 imes 10^{-3} \mathrm{~m}^3 / \mathrm{s} ight)[/latex]. The spargers will be located 15 ft (4.57 m) belowthe surface of the pond. Find the time required to raise the dissolved oxygen from 2 to 5 mg/L if the water temperature is 293 K.

Accepted Answer

From Figure 31.6, the transfer factor, [latex]K_L(A / V) V[/latex], for a single sparger is [latex]1200 \mathrm{ft}^3 / \mathrm{h}(9.44 imes\left.10^{-3} \mathrm{~m}^3 / \mathrm{s} ight)[/latex] and for the system

[latex]K_L\left(\frac{A}{V} ight)=\frac{\left(9.44 imes 10^{-3} \mathrm{~m}^3 / \mathrm{s} ight)(15 ext { spargers })}{566 \mathrm{~m}^3}=2.50 imes 10^{-4} \mathrm{~s}^{-1}[/latex].

The average hydrostatic pressure of the rising air bubble is equal to the arithmetic mean of the pressure at the top and the bottom of the pond.

[latex]P_{ ext {bottom }}=1 \mathrm{~atm}+\left(15 \mathrm{ft} \mathrm{H}_2 \mathrm{O} ight)\left(0.0295 \mathrm{~atm} / \mathrm{ft} \mathrm{H}_2 \mathrm{O} ight)[/latex]

[latex]=1.44 \mathrm{~atm}\left(1.459 imes 10^5 \mathrm{~Pa} ight)[/latex]

[latex]P_{ ext {mean }}=\frac{1 \mathrm{~atm}+1.44 \mathrm{~atm}}{2}=1.22 \mathrm{~atm}\left(1.236 imes 10^5 \mathrm{~Pa} ight)[/latex]

As the mole fraction of oxygen in air is 0.21, the partial pressure of oxygen within the bubble will equal [latex]y_{\mathrm{O}_2} P=(0.21)\left(1.236 imes 10^5 \mathrm{~Pa} ight)=2.60 imes 10^4 \mathrm{~Pa}[/latex]. The equilibrium concentration of a slightly soluble gas is related to its partial pressure by Henry’s law. At 293 K, this law stipulates for oxygen

[latex]p_{\mathrm{O}_2}=\left(4.06 imes 10^9 \mathrm{~Pa} / ext { mole fraction } ight) \cdot x_{\mathrm{O}_2}[/latex]

Accordingly

[latex]x_{\mathrm{O}_2}=\frac{2.60 imes 10^4 \mathrm{~Pa}}{4.06 imes 10^9 \mathrm{~Pa} / \mathrm{mol} ext { fraction }}=6.40 imes 10^{-6}[/latex]

For 1 L of solution, which is essentially pure water, the equilibrium concentration in milligrams per liter can be calculated

[latex] ext { moles of water }=\frac{\left(1000 \mathrm{~cm}^3 ext { of water } ight)\left(1 \mathrm{~g} / \mathrm{cm}^3 ext { water } ight)}{18 \mathrm{~g} ext { water } / \mathrm{mol}}[/latex]

[latex]=55.6 \mathrm{~mol} ext {. }[/latex]

In the liter of water, the moles of oxygen equal

[latex]x_{\mathrm{O}_2} \cdot( ext { moles of solution })=\left(6.4 imes 10^{-6} ight)(55.6 \mathrm{~mol})=3.56 imes 10^{-4} \mathrm{~mol}[/latex]

The grams of oxygen per liter equal

[latex]\left(3.56 imes 10^{-4} \mathrm{~mol} ight)(32 \mathrm{~g} / \mathrm{mol})=1.139 imes 10^{-2} \mathrm{~g} / \mathrm{L}=11.39 \mathrm{mg} / \mathrm{L}[/latex].

Using equation (31-1), we can solve for the required time

[latex]\ln \left(\frac{c_A^*-c_{A a}}{c_A^*-c_A^i} ight)=K_L a \cdot t[/latex]

or

[latex]c_A=c_A^*-\left(c_A^*-c_{A o} ight) e^{-k_L a \cdot t}[/latex] (31-1)

[latex]t=\ln \left(\frac{c_A^*-c_{A o}}{c_A^*-c_{A t}} ight)\left(\frac{1}{K_L\left(\frac{A}{V} ight)} ight)=\ln \left(\frac{11.39-2}{11.39-5} ight)\left(\frac{1}{2.50 imes 10^{-4 / s}} ight)[/latex]

[latex]t=1540 \mathrm{~s}[/latex].

Question 31.2: A 20,000-ft³(566 m³) aeration pond is aerated with 15 sparge...

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