Horizontal flow grit chamber, Dang

Assumptions (Nepal practice): water supply 135 lpcd, 80% of it reaches the sewer, peak factor 2.5 (assumed, on the safe side; CPHEEO suggests about 2.25 for this population) (so $Q_{avg}=273000\times135\times0.8/1000 = 29484$ m³/day); grit of $d=0.2$ mm, $G=2.65$; $k=0.06$, $f=0.03$; the temperature is 22°C (Dang). Two horizontal-flow units are provided, each able to take the full peak flow.

Step 1: Design flow

$$Q = \frac{273000\times 135\times 0.8\times 2.5}{1000\times 86400} = 0.8531\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 22+70}{100} = 37.52\ \text{mm/s} = 0.0375\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.8531}{0.20} = 4.266\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 4.27\ \text{m (adopt 4.3 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0375} = 5.33\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.66,\ 6.00) = 6.66\ \text{m (adopt 7.0 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{7.0}{0.20} = 35\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.8531\times 86400}{4.3\times 7.0} = 2449\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

 Plan:   inlet ->[ <-- L = 7.0 m --> ]-> outlet (weir)
                 |  B = 4.3 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 7.0\ \text{m}\times 4.3\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.