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Question

# An aeration basin is used to oxygenate wastewater at 283 K. The basin is filled to a depth of 4.55 m with $283 \mathrm{m}^{3}$ wastewater of 0.050 mmole/L initial dissolved oxygen $\left(\mathrm{O}_{2}\right)$ cocentration. The filled basin is then aerated with six six spargers, with each sparger delivering compressed air to bottom of the basin at a volumetric flow rate of $7.08 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s}$. The dissolved solids content will be low enough that Henry's law will be obeyed, with a Henry's law constant of $H=3.27 \times 10^{4} \mathrm{atm}$ $\left(p_{A}=H \cdot x_{A}^{\circ}\right)$ at 283 K. For purposes of this calculation, it may be assumed that the rate of $\mathrm{O}_{2}$ transferred to the liquid is small relative to molar flow rate of $\mathrm{O}_{2}$ in the aeration gas. a. Determine the dissolved oxygen concentration after a 5.0 min aeration time. b. Determine the time required to bring the dissolved oxygen concentration up to a concentration of 0.20 mmole/L (1 mmole/L = 1 gmole/m$^{3}$).

Solution

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Step 1
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Known:

• $T=283 \text{ K}$
• $V=283 \text{m}^3\approx 10000 \text{ft}, h=4.55$ m $\approx 15$ ft
• $c_{A,o}=0.05 \frac{\text{ mmol}}{\text{L}}$
• number of spargers: $n=6$
• $Q_g=7.08\frac{\text{ m}^3}{\text{s}}$
• $H=3.27 \cdot 10^4 \text{ atm}$ ($p_A=H \cdot x^*_A$)
• assumed that the rate of $O_2$ transferred to the liquid is small relative to molar flow rate of $O_2$ in the aeration gas
• a) $t=5$ min
• b) $c_A=0.2 \frac{\text{ mmol}}{\text{L}}$

Start of by converting the gas flow to feet per minute in order to use Figure 31.7 to find the oxygen transfer factor. and with it, the capacity coefficient. Use the Henry's law to find the average pressure which is later used to find the concentration of dissolved oxygen. With it found, use the equation which connects concentrations of oxygen and the capacity coefficient (Equation 31-1) to find the the dissolved oxygen concentration. In the second case, use the same equation but in the opposite way.

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