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Design of side intake with trashrack

Short answer

Set sill at RL 3316.00 m and opening 1.6 m wide by 1.3 m high, top RL 3317.30 m. Actual rack velocity 0.767 m/s, rack loss 0.0024 m; at the 100-year flood throttle gate to about 0.23 m opening; deck at RL 3320.15 m.

Design and draw a section of a side intake for a project in which the river bed level is 3315.0 masl. Weir crest level is fixed at 3317.5 masl. The highest flood level in 100 years return period is 3319.55 masl. The canal water level is fixed at 3317.3 masl. The turbine discharge of a period is 1.45 m³/s. Assume other suitable data. Take cylindrical trashrack bar with 10 mm thickness and 100 mm spacing.

Given: river bed RL 3315.00 m; weir crest RL 3317.50 m; 100-year flood level RL 3319.55 m; canal water level RL 3317.30 m; Q=1.45Q=1.45 m³/s; circular bars 10 mm thick at 100 mm clear spacing. Assumed: rack velocity 0.8 m/s, rack inclined at 75∘75^\circ, Cd=0.6C_d=0.6 for the gate orifice.

Step 1: Level of the sill and opening

  • Sill level: 1.0 m above the river bed, i.e. RL 3316.00 m, to keep bed load out.
  • Opening height h=1.3h=1.3 m, so the top of the opening is at RL 3317.30 m. This is the canal water level and 0.2 m below the weir crest, so the opening stays submerged at low flow.

Step 2: Rack area and width

Anet=QV=1.450.8=1.812 m2A_{net}=\frac{Q}{V}=\frac{1.45}{0.8}=1.812\ \text{m}^2 Agross=Anet×s+bb=1.812×110100=1.994 m2A_{gross}=A_{net}\times\frac{s+b}{b}=1.812\times\frac{110}{100}=1.994\ \text{m}^2 width =Agrossh=1.9941.3=1.53 m ⇒ adopt 1.6 m\text{width }=\frac{A_{gross}}{h}=\frac{1.994}{1.3}=1.53\ \text{m}\ \Rightarrow\ \text{adopt }1.6\ \text{m}

Adopted gross opening 1.6×1.3=2.081.6\times1.3=2.08 m²; net area =2.08×100110=1.891=2.08\times\frac{100}{110}=1.891 m²; actual velocity V=1.451.891=0.767V=\dfrac{1.45}{1.891}=0.767 m/s (< 1 m/s, so acceptable).

Step 3: Orifice check at normal pond level

Head difference between the pond (crest RL 3317.50 m) and the canal (RL 3317.30 m) =0.2=0.2 m.

Qcap=CdA2gΔh=0.6×2.08×19.62×0.2=2.47 m3/s>1.45Q_{cap}=C_dA\sqrt{2g\Delta h}=0.6\times2.08\times\sqrt{19.62\times0.2}=2.47\ \text{m}^3/\text{s}>1.45

Required head Δh=12g(1.450.6×2.08)2=0.069\Delta h=\dfrac{1}{2g}\left(\dfrac{1.45}{0.6\times2.08}\right)^2=0.069 m. The gate is throttled to keep the canal level at RL 3317.30 m.

Step 4: Trash rack loss (Kirschmer)

ht=β(sb)4/3V22gsin⁡θ=1.79 (0.1)4/30.767219.62sin⁡75∘=0.0024 mh_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta=1.79\,(0.1)^{4/3}\frac{0.767^2}{19.62}\sin75^\circ=0.0024\ \text{m}

which is negligible.

Step 5: Flood condition and levels

At the 100-year flood (RL 3319.55 m) the opening is submerged on both sides (canal WL 3317.30 m), so the head across it is Δh=3319.55−3317.30=2.25\Delta h=3319.55-3317.30=2.25 m. Without control the intake would pass 0.6×2.082g×2.25=8.30.6\times2.08\sqrt{2g\times2.25}=8.3 m³/s, so the gate must be throttled or closed. To pass only 1.45 m³/s, the opening area is 1.450.62g×2.25=0.36\dfrac{1.45}{0.6\sqrt{2g\times2.25}}=0.36 m², i.e. a gate opening of about 0.36/1.6=0.230.36/1.6=0.23 m. Top of the intake wall and hoist deck: flood level + 0.6 m freeboard == RL 3320.15 m.

SECTION (levels in m)
 3320.15 ____ deck
 3319.55 ~~~~~ HFL
 3317.50 ~~~~~~~~~~ weir crest (pond level)
          | rack  |gate|
 3317.30  |###\   |    |~~ canal WL 3317.30
          |###     |    |
 3316.00  |_____sill____|   (1.0 m above bed)
 3315.00 /////////////// river bed

Answer: sill RL 3316.00 m; opening 1.6 m wide x 1.3 m high (top RL 3317.30 m); rack velocity 0.767 m/s; rack loss 0.0024 m; gate opening about 0.23 m at the design flood; deck RL 3320.15 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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