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Grit chamber for 50 MLD

Short answer

Design flow 0.5787 m³/s, settling velocity 0.04 m\/ s, horizontal velocity 0.2 m\/ s. Cross-sectional area 2.9 m², depth 1 m, width 2.9 m. Length from v_h H\/ v_s with 25% extra and min 30 s detention gives 6.5 m. Grit chamber: L=6.5 m, B=2.9 m, water depth 1.0 m (total 1.55 m), flow velocity 0.2 m\/ s; provide two parallel units.

Propose the dimensions of grit chamber for a sewage treatment plant with 50 MLD of sewage flow at 25°C to remove 0.2 mm size of grit having specific gravity of 2.65. The specific gravity of organic matter is 1.02. Assume k = 0.06 and f = 0.03.

Grit: d=0.2d=0.2 mm, G=2.65G=2.65 at 25∘25^\circC; organic matter G=1.02G=1.02; k=0.06k=0.06, f=0.03f=0.03.

Step 1: Design flow

Q=50×1061000×86400=0.5787 m3/sQ = \frac{50\times 10^6}{1000\times 86400} = 0.5787\ \text{m}^3/\text{s}

Step 2: Settling velocity of the grit

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

vs=418 (G−1) d2(3T+70100) mm/s=418×(2.65−1)×0.22×3×25+70100=40.00 mm/s=0.0400 m/sv_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 25+70}{100} = 40.00\ \text{mm/s} = 0.0400\ \text{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}

Adopt vh=0.20v_h = 0.20 m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). For the organic matter (G=1.02G=1.02), vc=8kf(1.02−1) g d=0.025v_c=\sqrt{\dfrac{8k}{f}(1.02-1)\,g\,d} = 0.025 m/s, so vh=0.20v_h=0.20 m/s is well above it: the light organic matter stays in suspension and passes on instead of settling.

Step 4: Cross-section and dimensions

Ac=Qvh=0.57870.20=2.894 m2Adopt depth H=1.0 m⇒B=AcH=2.89 m (adopt 2.9 m)Ltheo=vhHvs=0.20×1.00.0400=5.00 mL=max⁡(1.25 Ltheo, 30 vh)=max⁡(6.25, 6.00)=6.25 m (adopt 6.5 m)\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.5787}{0.20} = 2.894\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 2.89\ \text{m (adopt 2.9 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0400} = 5.00\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.25,\ 6.00) = 6.25\ \text{m (adopt 6.5 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=6.50.20=32 s (within 30-90 s)Surface loading=QB L=0.5787×864002.9×6.5=2653 m3/m2/day\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{6.5}{0.20} = 32\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.5787\times 86400}{2.9\times 6.5} = 2653\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}

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

 Plan:   inlet ->[ <-- L = 6.5 m --> ]-> outlet (weir)
                 |  B = 2.9 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×B×H=6.5 m×2.9 m×1.0 mL\times B\times H = 6.5\ \text{m}\times 2.9\ \text{m}\times 1.0\ \text{m} (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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