Question 31.2: A 20,000-ft³(566 m³) aeration pond is aerated with 15 sparge...
A 20,000-\mathrm{ft}^3\left(566 \mathrm{~m}^3\right) aeration pond is aerated with 15 spargers, each using compressed air at a rate of 15 standard cubic feet per minute \left(7.08 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}\right). 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.
From Figure 31.6, the transfer factor, K_L(A / V) V, for a single sparger is 1200 \mathrm{ft}^3 / \mathrm{h}(9.44 \times\left.10^{-3} \mathrm{~m}^3 / \mathrm{s}\right) and for the system
K_L\left(\frac{A}{V}\right)=\frac{\left(9.44 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}\right)(15 \text { spargers })}{566 \mathrm{~m}^3}=2.50 \times 10^{-4} \mathrm{~s}^{-1}.
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.
P_{\text {bottom }}=1 \mathrm{~atm}+\left(15 \mathrm{ft} \mathrm{H}_2 \mathrm{O}\right)\left(0.0295 \mathrm{~atm} / \mathrm{ft} \mathrm{H}_2 \mathrm{O}\right)=1.44 \mathrm{~atm}\left(1.459 \times 10^5 \mathrm{~Pa}\right)
P_{\text {mean }}=\frac{1 \mathrm{~atm}+1.44 \mathrm{~atm}}{2}=1.22 \mathrm{~atm}\left(1.236 \times 10^5 \mathrm{~Pa}\right)
As the mole fraction of oxygen in air is 0.21, the partial pressure of oxygen within the bubble will equal y_{\mathrm{O}_2} P=(0.21)\left(1.236 \times 10^5 \mathrm{~Pa}\right)=2.60 \times 10^4 \mathrm{~Pa}. 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
p_{\mathrm{O}_2}=\left(4.06 \times 10^9 \mathrm{~Pa} / \text { mole fraction }\right) \cdot x_{\mathrm{O}_2}Accordingly
x_{\mathrm{O}_2}=\frac{2.60 \times 10^4 \mathrm{~Pa}}{4.06 \times 10^9 \mathrm{~Pa} / \mathrm{mol} \text { fraction }}=6.40 \times 10^{-6}For 1 L of solution, which is essentially pure water, the equilibrium concentration in milligrams per liter can be calculated
\text { moles of water }=\frac{\left(1000 \mathrm{~cm}^3 \text { of water }\right)\left(1 \mathrm{~g} / \mathrm{cm}^3 \text { water }\right)}{18 \mathrm{~g} \text { water } / \mathrm{mol}}=55.6 \mathrm{~mol} \text {. }
In the liter of water, the moles of oxygen equal
x_{\mathrm{O}_2} \cdot(\text { moles of solution })=\left(6.4 \times 10^{-6}\right)(55.6 \mathrm{~mol})=3.56 \times 10^{-4} \mathrm{~mol}The grams of oxygen per liter equal
\left(3.56 \times 10^{-4} \mathrm{~mol}\right)(32 \mathrm{~g} / \mathrm{mol})=1.139 \times 10^{-2} \mathrm{~g} / \mathrm{L}=11.39 \mathrm{mg} / \mathrm{L}.
Using equation (31-1), we can solve for the required time
\ln \left(\frac{c_A^*-c_{A a}}{c_A^*-c_A^i}\right)=K_L a \cdot tor
c_A=c_A^*-\left(c_A^*-c_{A o}\right) e^{-k_L a \cdot t} (31-1)
t=\ln \left(\frac{c_A^*-c_{A o}}{c_A^*-c_{A t}}\right)\left(\frac{1}{K_L\left(\frac{A}{V}\right)}\right)=\ln \left(\frac{11.39-2}{11.39-5}\right)\left(\frac{1}{2.50 \times 10^{-4 / s}}\right)
t=1540 \mathrm{~s}.
