Conventional ASP design, 0.104 m³/s

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 $BOD_5=S_0=220$ mg/l, effluent BOD $S_e=30$ mg/l, $F/M=0.2$ per day, MLSS $X=3000$ mg/l. The return sludge ratio is also checked (SVI assumed).

Step 1: Design flow

$$Q = 0.104\times86400 = 8985.6\ \text{m}^3/\text{day}$$

Step 2: Volume of aeration tank

$$V = \frac{8985.6\times 220}{0.2\times 3000} = 3294.7\ \text{m}^3$$

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

Step 3: Dimensions

$$\begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 823.7\ \text{m}^2 \end{aligned}$$

Provide 2 tanks (so one can be taken out of service), each with area $411.8$ m² and L : B about 4 : 1:

$$B=\sqrt{\frac{411.8}{4}} = 10.15\ \text{m (adopt 10.5 m)},\qquad L=\frac{411.8}{10.5} = 39.2\ \text{m (adopt 40 m)}$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $40\ \text{m}\times 10.5\ \text{m}\times 4.5\ \text{m}$.

Step 4: Checks

$$\begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{3294.7}{8985.6}\times 24 = 8.8\ \text{h}\ (\text{slightly above the usual 4-8 h; acceptable})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned}$$

BOD removal efficiency $=\dfrac{220-30}{220}\times100=86.4\%$, within the usual 85-95% of a conventional ASP.

$$\begin{aligned} \text{Return sludge ratio: } X_R &= \frac{10^6}{SVI} = \frac{10^6}{100} = 10000\ \text{mg/l}\ (SVI\ \text{assumed } 100\ \text{ml/g})\\ \frac{Q_R}{Q} &= \frac{X}{X_R-X} = \frac{3000}{10000-3000} = 0.429\ \ (\approx 43\%)\\ \text{Horizontal velocity} &= \frac{Q/N}{B\times H} = \frac{8986/(86400\times 2)}{10.5\times 4.0} = 0.0012\ \text{m/s}\ (= 0.07\ \text{m/min}) \end{aligned}$$

Answer: aeration tank volume $\approx 3295$ m³ (detention 8.8 h, depth 4.0 m); provide 2 tanks, each $40\ \text{m}\times 10.5\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).