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Rectangular grit chamber for 10 MLD

Short answer

Given v_s=0.02 m/s, v_h=0.30 m/s. A_c=0.386 m², H=0.6 m, B=0.7 m. L_theo=9.0 m, L=max(1.25×,30 s)=11.5 m. Grit chamber: 11.5 m × 0.7 m × 0.6 m water depth, total depth 1.15 m, flow velocity 0.30 m/s.

Design a rectangular grit chamber for maximum wastewater flow for 10 MLD to remove particles up to 0.2 mm diameter having sp. gr. 2.65. Settling velocity of grits is found to be 0.02 m/s in average and maintain a flow velocity of 0.3 m/s constant through a flow weir.

Grit particles of d=0.2d=0.2 mm and G=2.65G=2.65 are to be removed from the maximum flow of 10 MLD. The settling velocity (0.020.02 m/s) and the flow velocity (0.30.3 m/s, maintained constant by a proportional weir at the outlet) are given. k=0.06k=0.06 and f=0.03f=0.03 are assumed only for the scour check.

Step 1: Design flow

Q=10×1061000×86400=0.1157 m3/sQ = \frac{10\times 10^6}{1000\times 86400} = 0.1157\ \text{m}^3/\text{s}

Step 2: Settling velocity of the grit

The settling velocity is given: vs=0.02v_s = 0.02 m/s.

Step 3: Horizontal (flow-through) velocity

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

vc=8kf(G−1) g d=8×0.060.03(2.65−1)×9.81×0.2×10−3=0.228 m/sv_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}

The flow velocity is given: vh=0.3v_h = 0.3 m/s. (It is slightly above the Camp-Shields scour value for 0.2 mm grit, but a proportional weir holds it constant and the usual 0.3 m/s limit applies.)

Step 4: Cross-section and dimensions

Ac=Qvh=0.11570.30=0.386 m2Adopt depth H=0.6 m⇒B=AcH=0.64 m (adopt 0.7 m)Ltheo=vhHvs=0.30×0.60.0200=9.00 mL=max⁡(1.25 Ltheo, 30 vh)=max⁡(11.25, 9.00)=11.25 m (adopt 11.5 m)\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.1157}{0.30} = 0.386\ \text{m}^2\\ \text{Adopt depth } H &= 0.6\ \text{m} \Rightarrow B = \frac{A_c}{H} = 0.64\ \text{m (adopt 0.7 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.30\times 0.6}{0.0200} = 9.00\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(11.25,\ 9.00) = 11.25\ \text{m (adopt 11.5 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

Detention time=Lvh=11.50.30=38 s (within 30-90 s)Surface loading=QB L=0.1157×864000.7×11.5=1242 m3/m2/day\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{11.5}{0.30} = 38\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.1157\times 86400}{0.7\times 11.5} = 1242\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}

Total depth =H+grit storage 0.25+free board 0.30=1.15= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.15 m.

 Plan:   inlet ->[ <-- L = 11.5 m --> ]-> outlet (weir)
                 |  B = 0.7 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 0.6 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×B×H=11.5 m×0.7 m×0.6 mL\times B\times H = 11.5\ \text{m}\times 0.7\ \text{m}\times 0.6\ \text{m} (water depth), total depth about 1.15 m, with flow velocity 0.30 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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