Question

The wastewater inflow to an activated sludge plant is $0.5\, \text{m}^3/\text{s}$, and the plant is to be operated with a food to microorganism ratio of $0.2\, \text{mg/mg-d}$. The concentration of influent biodegradable organic matter (after primary settling) is $150\, \text{mg/L}$, and the mixed liquor volatile suspended solids (MLVSS) to be maintained is $2000\, \text{mg/L}$. Assuming complete removal of biodegradable organics in the tank, the volume of aeration tank (in m$^3$, integer) required is.

Hint:

For activated sludge sizing with complete substrate removal: \(\displaystyle V=\frac{Q\,S_0}{(F/M)\,X}\), using \(\text{kg/m}^3\) units and \(Q\) in \(\text{m}^3/\text{d}\).

Answer 1

Correct Answer: 16200

Step 1: Use F/M definition (per day).
\[ \frac{F}{M}=\frac{Q\,S_0}{V\,X} \] where \(Q\) (m$^3$/d), \(S_0\) (kg/m$^3$), \(X\) (kg/m$^3$), \(V\) (m$^3$).

Step 2: Convert units.
\(Q = 0.5\ \text{m}^3\!/\text{s} = 0.5\times 86400 = 43200\ \text{m}^3/\text{d}\).
\(S_0=150\ \text{mg/L}=0.150\ \text{kg/m}^3\).
\(X=2000\ \text{mg/L}=2.0\ \text{kg/m}^3\).
\(\displaystyle \frac{F}{M}=0.2\ \text{d}^{-1}\).

Step 3: Solve for \(V\).
\[ 0.2=\frac{(43200)(0.150)}{V(2.0)} \;\Rightarrow\; V=\frac{43200\times 0.150}{0.2\times 2.0} =\frac{6480}{0.4} =16200\ \text{m}^3 . \] \[ \boxed{V=16200\ \text{m}^3} \]