Question 31.2: An aeration pond of 20,000 ft³ (566 m³) liquid volume is aer......
An aeration pond of 20,000\,\mathrm{ft}^{3}\ \mathrm{(566m^{3})} liquid volume is aerated with 15 spargers, each using compressed air at a rate of 15 standard {ft}^{3}\operatorname{air}/\operatorname*{min}{(7.08\times10^{-3}\ m^{3}/s)}. The spargers will be located 15.0 ft (4.57 m) below the surface of the pond. Estimate the time required to raise the dissolved oxygen concentration from 2.0 to 5.0\,\mathrm{mg}\ O_{2}/\mathrm{L} if the water temperature is 293 K. At 298 K, Henry’s law constant for dissolution of O_{2} water is 0.826 m³ atm/gmole.
It is first necessary to estimate the mass-transfer coefficient for O_{2} gas–liquid transfer the aeration pond. Using Figure 31.7, the transfer factor, K_{L}(A/V)V, for a single sparger is 1200\,\mathrm{ft}^{3}/\mathrm{h}\mathrm{~at~}Q_{g}=15\,\mathrm{ft}^{3}/\mathrm{min~and~}\mathrm{h}=15\,\mathrm{ft}. For the entire aeration system,
K_{L}\left(\frac{A}{V}\right)=\frac{K_{L}(A/V)V}{V}n=\frac{(1200\;\mathrm{ft}^{3}/\mathrm{h})(1\;\mathrm{h}/3600\,s)(15\;\mathrm{spargers})}{(20,000\,\mathrm{ft}^{3})}=2.50\times10^{-4}\;\mathrm{s}^{-1}
To determine the time required for the aeration process, the dissolved O_{2} concentration at the operating and equilibrium (saturation) conditions must now be specified in consistent units. First, the average hydrostatic pressure of the rising air bubble in the pond is equal to the arithmetic mean of the pressure at the top (liquid surface) and the bottom (submerged discharge point for the sparger):
P_{\mathrm{ave}}={\frac{P_{\mathrm{top}}+P_{\mathrm{bottom}}}{2}}=P_{\mathrm{top}}+{\frac{1}{2}}\rho_{L}g h
P_{\mathrm{ave}}=1.0123\times10^{5}\,{\mathrm{Pa}}+{\frac{1}{2}}\left({\frac{998.2\,{\mathrm{kg}}}{\mathrm{m^{3}}}}\right)\left({\frac{9.81\,{\mathrm{m}}}{{\mathrm{s}}^{2}}}\right)\left({4.55\,{\mathrm{m}}}\right)=1.236\times10^{5}\,{\mathrm{Pa}}
The O_{2} composition in air bubble is 21 mole%. The concentration of dissolved oxygen in equilibrium with O_{2} in the air bubble is determined by Henry’s law
c_{A}^{*}={\frac{p_{A}}{H}}={\frac{y_{A}P_{\mathrm{ave}}}{H}}=\frac{(0.21)(1.236\times10^{5}\,\mathrm{Pa})(1.0\,\mathrm{atm}/1.013\times10^{5}\,\mathrm{Pa})}{(0.826\,\mathrm{m}^{3}\cdot\mathrm{atm}/\mathrm{gmole})}=0.310{\frac{\mathrm{gmole}\,\mathrm{O}_{2}}{\mathrm{m}^{3}}}
The molar concentration of O_{2} dissolved in water at the final condition is
c_{A}=\rho_{A,o}M_{A}=\left(\frac{5.0\,\mathrm{mg}\,\mathrm{O}_{2}}{\mathrm{L}}\frac{1000\,\mathrm{L}}{\mathrm{m^{3}}}\frac{1\,\mathrm{g}}{1000\,\mathrm{mg}}\right)\left(\frac{1\,\mathrm{gmole}}{32\,\mathrm{g\,O}_{2}}\right)=0.156\,\frac{\mathrm{gmole}}{\mathrm{m^{3}}}
Similarly, the molar concentration of O_{2} at the initial condition is c_{A,o}=0.0625\,\mathrm{gmole}\,\,O_{2}/\mathrm{m^{3}}.
Finally, using equation (31-1), the required time is
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}}\right)\left(\frac{1}{K_{L}\left(\frac{A}{V}\right)}\right)=\ln\left(\frac{(0.310-0.0625)\mathrm{gmole}/m^{3}}{(0.310-0.156)\mathrm{gmole}/\mathrm{m^{3}}}\right)\left(\frac{1}{2.50\times10^{-4}\,\mathrm{s}^{-1}}\right)=1900\ s
