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Desilting basin 32 m³/s, 0.15 mm

Short answer

Use Camp velocity v=44 sqrt(0.15)=0.1704 m/s. With two chambers and H=5 m, adopt 19 m width per chamber. Required length is 57.7 m by Velikanov, so adopt 60 m. Final: two chambers 19 m x 60 m, depth 5 m plus sediment storage.

Determine the dimensions of a desilting basin for a high head hydropower plant with a discharge of 32 m³/s. The basin must trap sediment particles coarser than 0.15 mm. Take w = 0.025 m/s and λ = 1.5. Additionally, draw a plan and longitudinal section of the desilting basin, illustrating the major components and the flushing arrangement.

Given: Q=32Q=32 m³/s, particles coarser than 0.15 mm, w=0.025w=0.025 m/s, λ=1.5\lambda=1.5 (Velikanov's coefficient for turbulence and the required efficiency). For a high-head plant, two chambers are provided so that one can be flushed while the other works.

Step 1: Horizontal velocity (Camp)

v=440.15=17.04 cm/s=0.1704 m/sv=44\sqrt{0.15}=17.04\ \text{cm/s}=0.1704\ \text{m/s}

Step 2: Depth and width

Assume a flow depth H=5H=5 m:

B=QvH=320.1704×5=37.56 m (two chambers of 18.78 m)B=\frac{Q}{vH}=\frac{32}{0.1704\times5}=37.56\ \text{m}\ (\text{two chambers of }18.78\ \text{m})

Adopt a chamber width of 19.0 m.

Step 3: Length by simple settling theory

Without turbulence: L0=vHw=0.1704×50.025=34.1L_0=\dfrac{vH}{w}=\dfrac{0.1704\times5}{0.025}=34.1 m.

Using λ\lambda as a factor on the surface area:

A=λQw=1.5×320.025=1920 m2,L=AB=192037.56=51.1 mA=\frac{\lambda Q}{w}=\frac{1.5\times32}{0.025}=1920\ \text{m}^2,\qquad L=\frac{A}{B}=\frac{1920}{37.56}=51.1\ \text{m}

Step 4: Check by Velikanov's method

LV=λ2v2(H−0.2)27.51 w2=1.52×0.17042×(5−0.2)27.51×0.0252=57.7 mL_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w^2}=\frac{1.5^2\times0.1704^2\times(\sqrt{5}-0.2)^2}{7.51\times0.025^2}=57.7\ \text{m}

The larger length governs. Adopt L=60L=60 m, giving L/b=60/19=3.2L/b=60/19=3.2.

ItemValue
Number of chambers2
Width of each chamber19.0 m
Length60 m
Flow depth5 m (+ 1.0 m sediment storage in the hopper)
Velocity0.1704 m/s
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: each chamber has a hopper floor sloping (about 1:30) to a central flushing channel, with a flushing gate at the downstream end. To flush a chamber, its inlet gate is closed and the deposit is flushed out while the other chamber carries the flow. The flushing water runs through the flushing tunnel to the river.

Answer: 2 chambers, each 19.0 m wide x 60 m long x 5 m deep (plus storage).

✓ Answer checked by Claude Opus 5.5 (high effort)

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