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Periodic settling basin with and without turbulence

Short answer

Using Camp velocity 0.1968 m/s and depth 2.4 m, total width is 10.59 m, split into two chambers of about 5.5 m. Without turbulence length is 7.9 m; with turbulence effective fall velocity drops to 0.0432 m/s, giving 10.9 m, so 15 m is adopted.

Compute the dimension of a periodic type settling basin considering and without considering the turbulence effect for a hydropower plant through settling theory. Take settling velocity = 6 cm/sec; discharge = 5 m³/sec; particle size to be removed = 0.2 mm; depth of basin = 2.4 m.

Given: design discharge Q=5.0Q=5.0 m³/s; particles of size d≥0.20d\ge 0.20 mm to be trapped, fall velocity w=6.00w=6.00 cm/s =0.0600=0.0600 m/s (given). Periodic (intermittent) flushing is used, so no extra flushing discharge is added; the basin is designed both without and with turbulence.

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

Take the flow depth H=2.4H=2.4 m in the settling zone (given; sediment storage is extra).

B=QvH=5.00.1968×2.4=10.59 mB=\frac{Q}{vH}=\frac{5.0}{0.1968\times2.4}=10.59\ \text{m}

Provide n=2n=2 chambers so that one can be flushed while the other works: b=B/n=5.29b=B/n=5.29 m each, adopt 5.5 m.

Step 3: Length without turbulence

L0=vHw=0.1968×2.40.0600=7.9 mL_0=\frac{vH}{w}=\frac{0.1968\times2.4}{0.0600}=7.9\ \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=0.132×0.19682.4=0.0168 m/sw′=0.0600−0.0168=0.0432 m/sL=vHw′=0.1968×2.40.0432=10.9 m\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{2.4}}=0.0168\ \text{m/s}\\ w'&=0.0600-0.0168=0.0432\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times2.4}{0.0432}=10.9\ \text{m} \end{aligned}

Adopted dimensions

ItemValue
Flow velocity vv0.197 m/s
Settling depth HH2.4 m (plus about 1.0 m sediment storage in the hopper)
Number of chambers2
Width of each chamber5.5 m
Length of settling zone15 m
Length/width ratio2.7
Inlet transitiongradual, with guide vanes (side slope not steeper than about 1:5)

Total volume of water in the basin ≈396\approx 396 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.

Comparison

CaseEffective fall velocityLength of settling zone
Without turbulencew=0.060w=0.060 m/sL0=7.87L_0=7.87 m
With turbulencew′=0.0432w'=0.0432 m/sL=10.92L=10.92 m

Turbulence increases the length by 39%; the width (10.59 m in total) and the depth are unchanged. For periodic flushing, a sediment storage depth is added below the settling zone, and each chamber is flushed in turn.

Answer: without turbulence, 2 chambers of 5.5 m x about 8 m x 2.4 m; with turbulence, 2 chambers of 5.5 m x about 11 m x 2.4 m (15 m adopted for safety), with v=0.197v=0.197 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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