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Settling basin 50 m³/s, 0.2 mm

Short answer

Three chambers, each 14.5 m wide, 270 m long, and 6.0 m deep settling zone. Flow velocity is 0.197 m/s. The basin length includes turbulence correction (effective fall velocity 0.0044 m/s) to remove sediment >0.2 mm with w=1.5 cm/s.

Find out the dimension of a settling basin with turbulence flow for a high head hydropower plant, which utilizes a discharge of 50 m³/sec. The sediment particles coarser than 0.2 mm (w = 1.5 cm/sec) have to be trapped in the basin. Draw plan and section showing major components and flushing arrangement.

Given: design discharge Q=50.0Q=50.0 m³/s; particles of size d≥0.20d\ge 0.20 mm to be trapped, fall velocity w=1.50w=1.50 cm/s =0.0150=0.0150 m/s (given). 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=440.20=19.68 cm/s=0.1968 m/sv=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \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

Assume a flow depth H=6.0H=6.0 m in the settling zone (sediment storage is extra).

B=QvH=50.00.1968×6.0=42.35 mB=\frac{Q}{vH}=\frac{50.0}{0.1968\times6.0}=42.35\ \text{m}

Provide n=3n=3 chambers so that one can be flushed while the others work: b=B/n=14.12b=B/n=14.12 m each, adopt 14.5 m.

Step 3: Length without turbulence

L0=vHw=0.1968×6.00.0150=78.7 mL_0=\frac{vH}{w}=\frac{0.1968\times6.0}{0.0150}=78.7\ \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=0.132×0.19686.0=0.0106 m/sw′=0.0150−0.0106=0.0044 m/sL=vHw′=0.1968×6.00.0044=268.6 m\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{6.0}}=0.0106\ \text{m/s}\\ w'&=0.0150-0.0106=0.0044\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times6.0}{0.0044}=268.6\ \text{m} \end{aligned}

Adopted dimensions

ItemValue
Flow velocity vv0.197 m/s
Settling depth HH6.0 m (plus about 1.0 m sediment storage in the hopper)
Number of chambers3
Width of each chamber14.5 m
Length of settling zone270 m
Length/width ratio18.6
Inlet transitiongradual, with guide vanes (side slope not steeper than about 1:5)

Total volume of water in the basin ≈70470\approx 70470 m³.

PLAN (flushing gates at the downstream end)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +-----------------+ /         \
                     |  chamber 3      |
                     +=================+
  trash-                  |  |  |  flush gates
  rack                    v  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 chambers continue to supply the plant.

Answer: 3 chambers, each about 14.5 m wide x 270 m long x 6.0 m deep (settling zone), with v=0.197v=0.197 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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