Grit chamber for 180 l/s, SOR 1.5

$Q=180$ l/s $=0.18$ m³/s, SOR $=1.5$ cm/s $=0.015$ m/s, detention $=50$ s, particle size 0.25 mm, $k=0.06$, $f=0.03$.

Step 1: Surface area and depth

$$\begin{aligned} Q &= 0.180\ \text{m}^3/\text{s}\\ \text{Surface area } A &= \frac{Q}{SOR} = \frac{0.180}{0.015} = 12.00\ \text{m}^2\\ \text{Volume } V &= Q\,t = 0.180\times 50 = 9.00\ \text{m}^3\\ \text{Depth } H &= \frac{V}{A} = 0.75\ \text{m} \end{aligned}$$

Step 2: Scour velocity for the two kinds of particles

$$v_c=\sqrt{\frac{8k}{f}(G-1)\,g\,d}$$

The horizontal velocity must be above the organic scour velocity (organic matter stays suspended and is carried through) and below the grit scour velocity (settled grit is not re-suspended). Adopt $v_h = 0.25$ m/s, which satisfies 0.089 < 0.25 < 0.254 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.)

Step 3: Dimensions

$$\begin{aligned} L &= v_h\,t = 0.25\times 50 = 12.50\ \text{m}\\ B &= \frac{A}{L} = \frac{12.00}{12.50} = 0.96\ \text{m}\\ \text{Check: } A_c &= B\,H = 0.96\times 0.75 = 0.720\ \text{m}^2 \approx \frac{Q}{v_h} = 0.720\ \text{m}^2 \end{aligned}$$

Adopt $L = 12.5$ m, $B = 1.0$ m, water depth $H = 0.75$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.30 m). Provide two units in parallel.

Answer: grit chamber $L\times B\times H = 12.5\times 1.0\times 0.75$ m, flow velocity 0.25 m/s, detention 50 s, surface overflow rate 1.5 cm/s.