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Settling basin 6 m³/s, 0.25 mm, depth 3 m

Short answer

Camp velocity is 44 sqrt(0.25)=0.220 m/s. For Q=6 m3/s and H=3 m total width is 9.09 m; provide two chambers 5 m wide. Turbulence reduces fall velocity to 0.0082 m/s, requiring length 80.2 m, adopt 85 m. Final: two 5 m x 85 m x 3 m chambers.

Design a settling basin for a design discharge of 6 cumec. The basin is designed to remove particle size of 0.25 mm. If the depth of the basin is 3 m and settling velocity 2.5 cm/sec, find the dimensions of the basin considering turbulence.

Given: design discharge Q=6.0Q=6.0 m³/s; particles of size d≥0.25d\ge 0.25 mm to be trapped, fall velocity w=2.50w=2.50 cm/s =0.0250=0.0250 m/s (given). The depth of the basin is 3 m, and turbulence is considered.

Step 1: Horizontal velocity (Camp)

For 0.1<d<10.1<d<1 mm, v=adv=a\sqrt d with a=44a=44 (v in cm/s, d in mm):

v=440.25=22.00 cm/s=0.2200 m/sv=44\sqrt{0.25}=22.00\ \text{cm/s}=0.2200\ \text{m/s}

This is within the usual 0.1-0.4 m/s range, so deposited sediment is not re-suspended.

Step 2: Cross-section and width

Take the flow depth H=3.0H=3.0 m in the settling zone (given; sediment storage is extra).

B=QvH=6.00.2200×3.0=9.09 mB=\frac{Q}{vH}=\frac{6.0}{0.2200\times3.0}=9.09\ \text{m}

Provide n=2n=2 chambers so that one can be flushed while the other works: b=B/n=4.55b=B/n=4.55 m each, adopt 5.0 m.

Step 3: Length without turbulence

L0=vHw=0.2200×3.00.0250=26.4 mL_0=\frac{vH}{w}=\frac{0.2200\times3.0}{0.0250}=26.4\ \text{m}

Step 4: Effect of turbulence

Turbulence reduces the effective fall velocity: w′=w−αw'=w-\alpha, with α=0.132 v/H\alpha=0.132\,v/\sqrt{H}.

α=0.132×0.22003.0=0.0168 m/sw′=0.0250−0.0168=0.0082 m/sL=vHw′=0.2200×3.00.0082=80.2 m\begin{aligned} \alpha&=\frac{0.132\times0.2200}{\sqrt{3.0}}=0.0168\ \text{m/s}\\ w'&=0.0250-0.0168=0.0082\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.2200\times3.0}{0.0082}=80.2\ \text{m} \end{aligned}

Adopted dimensions

ItemValue
Flow velocity vv0.220 m/s
Settling depth HH3.0 m (plus about 1.0 m sediment storage in the hopper)
Number of chambers2
Width of each chamber5.0 m
Length of settling zone85 m
Length/width ratio17.0
Inlet transitiongradual, with guide vanes (side slope not steeper than about 1:5)

Total volume of water in the basin ≈2550\approx 2550 m³.

PLAN (flushing gates at the downstream end)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +=================+ /         \
  trash-                  |  |  flush gates
  rack                    v  v  --> flushing channel --> river
LONGITUDINAL SECTION
 inlet          settling zone (depth H)      outlet
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  WL
 \    --->   --->   --->   --->   --->   --->  |
  \  ______________________________________   | weir
   \/   sediment storage / hopper slope 1:30 \__|__ to headrace
          flush gate -> flushing channel -> river

Flushing arrangement: the floor of each chamber slopes at about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. A flushing velocity of at least 3 m/s is maintained in the channel. The chamber to be flushed is closed at the inlet, drained and flushed to the river, while the other chamber continues to supply the plant.

Answer: 2 chambers, each about 5.0 m wide x 85 m long x 3.0 m deep (settling zone), with v=0.220v=0.220 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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