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Single chamber settling basin design (15 m³/s)

Short answer

Use design flow 16.5 m3/s. Basin area = 1.5 x 16.5 / 0.02 = 1237.5 m2; with L/B=8 adopt 13 m x 104 m. Depth 4.5 m gives velocity 0.282 m/s. Sediment per 8-h cycle is 380.2 m3, so provide 0.5 m storage.

Design a single chamber settling basin using the following data: design discharge = 15 m³/sec; particle size to settle ≥ 0.2 mm; particle fall velocity = 0.02 m/sec; factor of safety for basin area = 1.5; length to breadth ratio = 8; sediment concentration = 1 kg/m³; density of sediment particles = 2,500 kg/m³; settling time of sediment particles = 8 hours; sediment packing factor = 0.5; additional discharge for flushing = 10%.

Given: Q=15Q=15 m³/s, particles ≥0.2\ge0.2 mm, w=0.02w=0.02 m/s, factor of safety for area =1.5=1.5, L/B=8L/B=8, sediment concentration C=1C=1 kg/m³, particle density 2500 kg/m³, flushing interval 8 h, packing factor 0.5, additional flushing discharge 10%.

Step 1: Design flow

Q′=1.10×15=16.5 m3/sQ'=1.10\times15=16.5\ \text{m}^3/\text{s}

Step 2: Surface area of the basin

For a particle to settle, the surface area must satisfy A≥Q/wA\ge Q/w. With the factor of safety:

A=FS×Q′w=1.5×16.50.02=1237.5 m2A=\frac{FS\times Q'}{w}=\frac{1.5\times16.5}{0.02}=1237.5\ \text{m}^2

Step 3: Width and length (L/B=8L/B=8)

B=A8=1237.58=12.44 m,L=8B=99.5 mB=\sqrt{\frac{A}{8}}=\sqrt{\frac{1237.5}{8}}=12.44\ \text{m},\qquad L=8B=99.5\ \text{m}

Adopt B=13B=13 m and L=104L=104 m (A=1352A=1352 m²).

Step 4: Depth of flow

Limit the horizontal velocity to about 0.3 m/s. Take the flow depth H=4.5H=4.5 m:

v=Q′BH=16.513×4.5=0.282 m/s<0.3 m/s(OK)v=\frac{Q'}{BH}=\frac{16.5}{13\times4.5}=0.282\ \text{m/s}<0.3\ \text{m/s}\quad\text{(OK)}

Step 5: Sediment storage (flushing every 8 h)

Mass of sediment=C Q′ t=1×16.5×(8×3600)=475200 kg\text{Mass of sediment}=C\,Q'\,t=1\times16.5\times(8\times3600)=475200\ \text{kg} Volume of deposit=massρs×p=4752002500×0.5=380.2 m3\text{Volume of deposit}=\frac{\text{mass}}{\rho_s\times p}=\frac{475200}{2500\times0.5}=380.2\ \text{m}^3 Depth of deposit=380.21352=0.28 m ⇒ provide 0.5 m storage\text{Depth of deposit}=\frac{380.2}{1352}=0.28\ \text{m}\ \Rightarrow\ \text{provide }0.5\ \text{m storage}

(All the sediment entering is assumed to be trapped, which is conservative.)

Step 6: Overall dimensions

ItemValue
Length of settling zone104 m
Width13 m
Flow depth4.5 m
Sediment storage0.5 m (hopper with 1:30 slope to the flushing channel)
Free board0.5 m
Total depth at the upstream endabout 5.5 m
PLAN (single chamber, flushing gate at the downstream end)
          inlet           settling chamber          outlet
 canal   transition  +========================+   transition
 =====>  \  guide  / |                        | \  weir   /===> to
         /  vanes  \ |  13 m wide x 104 m long| /         \    headrace
                     +========================+
  trash-                          |  flush gate
  rack                            v  --> flushing channel --> river

Answer: a single chamber of about 13 m x 104 m, with a flow depth of 4.5 m (v=0.282v=0.282 m/s) and 0.5 m of sediment storage, flushed every 8 hours (deposit of 380.2 m³ per cycle).

✓ Answer checked by Claude Opus 5.5 (high effort)

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