Grit chamber for 190 l/s

The flow is $Q=190$ l/s $=0.19$ m³/s. The particle size is read as 0.21 mm (the "0.21 m" in the question is a typing slip). $k=0.06$, $f=0.03$, SOR (surface overflow rate) $=2$ cm/s $=0.02$ m/s, detention time $=60$ s.

Step 1: Surface area and depth

$$\begin{aligned} Q &= 0.190\ \text{m}^3/\text{s}\\ \text{Surface area } A &= \frac{Q}{SOR} = \frac{0.190}{0.020} = 9.50\ \text{m}^2\\ \text{Volume } V &= Q\,t = 0.190\times 60 = 11.40\ \text{m}^3\\ \text{Depth } H &= \frac{V}{A} = 1.20\ \text{m} \end{aligned}$$

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 $v_h = 0.20$ m/s, which satisfies 0.081 < 0.20 < 0.233 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.)

Step 3: Dimensions

$$\begin{aligned} L &= v_h\,t = 0.20\times 60 = 12.00\ \text{m}\\ B &= \frac{A}{L} = \frac{9.50}{12.00} = 0.79\ \text{m}\\ \text{Check: } A_c &= B\,H = 0.79\times 1.20 = 0.950\ \text{m}^2 \approx \frac{Q}{v_h} = 0.950\ \text{m}^2 \end{aligned}$$

Adopt $L = 12.0$ m, $B = 0.8$ m, water depth $H = 1.20$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.75 m). Provide two units in parallel.

Answer: grit chamber $L\times B\times H = 12.0\times 0.8\times 1.20$ m, flow velocity 0.20 m/s, detention 60 s, surface overflow rate 2 cm/s.