Question

An aeration tank is to be installed for removal of a VOC from water. Flow through the tank $Q=180{,}000$ m$^3$/d. Permissible limit in water $C_{\text{out}}=12\ \mu$g/L. Saturation concentration in water $C^* = 5\ \mu$g/L. Gas-transfer rate constant $k=0.40$ s$^{-1}$ at $25^\circ$C. Initial concentration $C_{\text{in}}=33\ \mu$g/L. Find the required tank volume at $25^\circ$C (in m$^3$, rounded to two decimals).

Hint:

For stripping with nonzero $C^*$, always use the driving force $(C-C^*)$. Compute residence time from the log ratio of inlet/outlet driving forces, then multiply by flow to get volume.

Answer 1

Correct Answer: 7.1

Step 1: Model for continuous aeration/stripping.
For a completely mixed tank with first-order mass transfer to gas phase (with finite equilibrium), the liquid-phase concentration obeys \[ \frac{dC}{dt}=-k\,(C-C^*). \] Integrating from $C_{\text{in}}$ to $C_{\text{out}}$: \[ t=\frac{1}{k}\ln\!\left(\frac{C_{\text{in}}-C^*}{C_{\text{out}}-C^*}\right). \]
Step 2: Compute the necessary hydraulic residence time $t$.
Insert numbers (all in $\mu$g/L): \[ t=\frac{1}{0.40}\ln\!\left(\frac{33-5}{12-5}\right) =2.5\ \ln(4) =2.5\times 1.386294 = \boxed{3.4657\ \text{s}}. \]
Step 3: Convert flow to m$^3$/s and obtain volume.
\[ Q=\frac{180{,}000\ \text{m}^3}{\text{d}}\times \frac{1\ \text{d}}{86{,}400\ \text{s}} = \boxed{2.0833\ \text{m}^3/\text{s}}. \] Tank volume $V=Q\,t=2.0833\times 3.4657=\boxed{7.215\ \text{m}^3} \Rightarrow \boxed{7.22\ \text{m}^3}. \]
Step 4: Sanity checks.
$C_{\text{out}}(=12)$ is above $C^*(=5)$ so removal is feasible by mass transfer; small $t$ arises due to large $k$ (fast gas–liquid transfer). Final Answer:\fbox{7.22 m$^3$}