Conventional ASP design, 96 lpcd

Data and design basis

Conventional ASP (plug flow, diffused air). The aeration tank volume follows from the food-to-micro-organism ratio:

$$\frac{F}{M}=\frac{Q\,S_0}{V\,X}\ \Rightarrow\ V=\frac{Q\,S_0}{(F/M)\,X}$$

Data: settled-sewage $S_0=200$ mg/l, $F/M=0.3$/day, MLSS $=2000$ mg/l. Only the aeration tank is designed.

Step 1: Design flow

$$Q = 100000\times96/1000 = 9600.0\ \text{m}^3/\text{day}$$

Step 2: Volume of aeration tank

$$V = \frac{9600.0\times 200}{0.3\times 2000} = 3200.0\ \text{m}^3$$

(units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.)

Step 3: Dimensions

$$\begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 800.0\ \text{m}^2 \end{aligned}$$

Provide 2 tanks (so one can be taken out of service), each with area $400.0$ m² and L : B about 4 : 1:

$$B=\sqrt{\frac{400.0}{4}} = 10.00\ \text{m (adopt 10.0 m)},\qquad L=\frac{400.0}{10.0} = 40.0\ \text{m (adopt 40 m)}$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $40\ \text{m}\times 10.0\ \text{m}\times 4.5\ \text{m}$.

Step 4: Checks

$$\begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{3200.0}{9600.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned}$$

Answer: aeration tank volume $\approx 3200$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $40\ \text{m}\times 10.0\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).