Headworks of Run-of- River (RoR) Plants

Given: design discharge $Q=60.0$ m³/s; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered. (A fall velocity of 1.5 cm/s is lower than the usual value for 0.2 mm sand, so the basin comes out long; the given value is used.)

Step 1: Horizontal velocity (Camp)

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

$$v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}$$

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{60.0}{0.1968\times6.0}=50.82\ \text{m}$$

Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=16.94$ m each, adopt 17.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1968\times6.0}{0.0150}=78.7\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{6.0}}=0.0106\ \text{m/s}\\ w'&=0.0150-0.0106=0.0044\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times6.0}{0.0044}=268.6\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 82620$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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 17.0 m wide x 270 m long x 6.0 m deep (settling zone) with $v=0.197$ m/s.

Given: design discharge $Q=50.0$ m³/s; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered.

Step 1: Horizontal velocity (Camp)

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

$$v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}$$

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{50.0}{0.1968\times6.0}=42.35\ \text{m}$$

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

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1968\times6.0}{0.0150}=78.7\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{6.0}}=0.0106\ \text{m/s}\\ w'&=0.0150-0.0106=0.0044\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times6.0}{0.0044}=268.6\ \text{m} \end{aligned}$$

Adopted dimensions

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

PLAN (flushing gates at the downstream end)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +=================+ /         \
  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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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 $v=0.197$ m/s.

Given: design discharge $Q=40.0$ m³/s; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered; both the plan and the sections are sketched.

Step 1: Horizontal velocity (Camp)

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

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

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 $H=5.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{40.0}{0.1704\times5.0}=46.95\ \text{m}$$

Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=15.65$ m each, adopt 16.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times5.0}{0.0150}=56.8\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{5.0}}=0.0101\ \text{m/s}\\ w'&=0.0150-0.0101=0.0049\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times5.0}{0.0049}=172.5\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 42000$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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 16.0 m wide x 175 m long x 5.0 m deep (settling zone) with $v=0.170$ m/s.

Given: design discharge $Q=25.0$ m³/s; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered.

Step 1: Horizontal velocity (Camp)

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

$$v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}$$

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{25.0}{0.1968\times6.0}=21.17\ \text{m}$$

Provide $n=2$ chambers so that one can be flushed while the others work: $b=B/n=10.59$ m each, adopt 11.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1968\times6.0}{0.0150}=78.7\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{6.0}}=0.0106\ \text{m/s}\\ w'&=0.0150-0.0106=0.0044\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times6.0}{0.0044}=268.6\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 35640$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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: 2 chambers, each about 11.0 m wide x 270 m long x 6.0 m deep (settling zone) with $v=0.197$ m/s.

General arrangement of a typical RoR headworks

PLAN
    river flow ==============================>
        left bank wall
   +-------------------------------------------+
   |  intake  | undersluice | weir (spillway)    |
   | (side)   |  (gates)    | ||||||||||||||||| |
   +----+-----+-------------+--------------------+
        |  gravel trap
        v                         fish ladder
   approach canal --> settling basin --> headrace

SECTION through weir, undersluice and intake
  pond level ~~~~~~~~~~~~~~~~~~~~~~~~~~
        _____      weir crest  ______
  intake|rack |gate          /      \  stilling basin
  sill  |_____|-> canal     /        \____ apron
 ///////////  undersluice /////////  river bed
   (sill 1.0-1.5 m above bed)   flushing gate

Main components:

• Diversion weir / barrage across the river to raise the water level for the intake and pass floods over its crest.
• Undersluice (scouring sluice) at the intake side, with gates, to keep a flushing channel in front of the intake and to pass bed load.
• Intake (side intake with trash rack and gate) on the outer bank, with its sill above the undersluice level.
• Gravel trap and approach canal leading to the settling basin.
• Fish ladder, river training works (guide bunds, flood walls), stilling basin and apron downstream of the weir.

Functional requirements of the headworks

1. Safe passage of floods: the structure must be stable against the design flood (usually 1 in 100 year) and pass it with a limited afflux.
2. Reliable diversion: divert the design discharge at all river stages, including low flow, with the intake always submerged.
3. Minimum sediment entry: bed load must be kept out by a high sill, undersluice and gravel trap, and suspended load limited by the settling basin.
4. Trash and floating debris control: trash rack, trash boom and log deflector.
5. Stability and safety: weir safe against sliding, overturning and uplift; scour protected by cut-offs, aprons and a stilling basin.
6. Environmental flow and fish passage as per law; minimum release downstream.
7. Easy operation and maintenance with small head loss in the intake.

In rivers carrying a high sediment load

• Place the intake at the outer bend, with the sill at least 1.0-1.5 m above the undersluice crest.
• Make the undersluice large enough to give a flushing velocity that carries the boulders and gravel.
• Provide a gravel trap and a settling basin with efficient flushing; allow frequent flushing of the pond in floods.
• Use a flood-operation rule: close the intake during the high sediment-laden flood.

A RoR headworks takes the design discharge from the river without storage, so it must work reliably in both floods and low flows. The minimum standards of a sound headworks are:

1. Stable and safe structure: designed for the 100-year flood (check for a higher flood); stable against sliding, overturning, uplift and piping; strong foundation with cut-offs and protection against scour.
2. Adequate flood passage: spillway, undersluice and gates must pass the design flood with an acceptable afflux, without overtopping the banks or the flood walls.
3. Dependable water supply: the intake must deliver the design discharge during the dry season, with the intake always submerged and no air entry or vortex; an environmental release is also kept in the river.
4. Control of sediment: bed load is excluded by a high-level intake sill, an undersluice, a gravel trap and river training works; suspended sediment above the design size is removed in the settling basin. The sediment must not accumulate in front of the intake.
5. Control of floating debris and trash: trash racks, trash booms, and a raking arrangement.
6. Hydraulic efficiency: small head loss in the intake, a smooth flow transition and a low-turbulence approach.
7. Operation and maintenance: gates and hoists easy to operate in floods, safe access, provision for flushing and inspection, and the arrangement for fish passage.
8. Environment and economy: minimum impact on the river and fish, and a low cost relative to the energy produced.

Intakes commonly used in RoR projects in Nepal

1. Side intake: an opening in the river bank, with a trash rack and gate, built at the side of the weir and undersluice. It is the most common type in Nepal because the river carries a large bed load and the flow can be diverted with simple structures.
2. Bottom (Tyrolean / drop) intake: a trench with a rack across the river bed in steep, boulder-carrying streams; water drops through the bars into a collection chamber (used in small hydropower and micro-hydro projects in the hills).
3. Frontal intake: an opening at the front of the weir facing the flow (less common because it takes much bed load).

Factors for selecting the intake location

• Outer (concave) bank of a stable river bend, where the secondary current moves bed load away from the intake; or a straight stable reach.
• Stable river reach and good foundation: rock or firm soil; no active landslide or erosion.
• Sufficient head and level to give gravity flow to the settling basin and headrace without long canals.
• Away from sediment deposition zones, confluences of tributaries and high debris areas.
• Good access for construction, operation and maintenance.
• Cheap river training and weir construction: a narrow section where the weir is short, and enough space for the gravel trap and settling basin close by.
• Flood safety: high flood level, minimal flood damage to the headworks and powerhouse.
• Environmental and social: minimum effect on existing water users, fish and forest.

Definition

An intake is the structure at the head of the water conductor system that withdraws the design discharge from the river (or reservoir) and admits it into the canal, tunnel or pipe in a controlled way.

Functions

• Diverts the required discharge into the conveyance system at all river levels.
• Controls the flow with a gate; shuts it off for inspection and during floods.
• Excludes the bed load and the floating trash and debris through the sill, trash rack and trash boom.
• Limits the entry of sediment and passes it to the gravel trap and settling basin.
• Provides a smooth, low-loss transition and prevents air entry and vortices.

Sketch of a generalized RoR intake

PLAN
 river ===>  ||||||||||| weir
                 __________________________
 undersluice =|  trash rack  gate   |==> approach
 (flushing)   |_____________________|    canal

SECTION
  pond ~~~~~~~~~~~~~~~~~~
        |    | rack   ____ gate
        |    |##   \ |    |
  sill  |____|##____\|____|=====> canal
  ///////////////// river bed (undersluice lower)

The sill of the intake is kept 1.0-1.5 m above the bed of the undersluice so that bed load cannot enter; the opening is sized so that the velocity at the trash rack is about 0.8-1.0 m/s.

A non-pressurized (free-surface) intake leads water into a canal or a free-surface tunnel, so the flow has a free surface. A pressurized intake leads water into a pressure tunnel or penstock, so the opening is fully submerged below the minimum water level and the conduit runs full.

Given: design discharge $Q=60.0$ m³/s and gross head 300 m (a high-head Pelton plant, so particles above 0.15 mm must be removed); particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered.

Step 1: Horizontal velocity (Camp)

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

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

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{60.0}{0.1704\times6.0}=58.68\ \text{m}$$

Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=19.56$ m each, adopt 20.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\ w'&=0.0150-0.0092=0.0058\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 64800$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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 20.0 m wide x 180 m long x 6.0 m deep (settling zone) with $v=0.170$ m/s.

Given: design discharge $Q=8.0$ m³/s; particles of size $d\ge 0.30$ mm to be trapped, fall velocity $w=4.00$ cm/s $=0.0400$ m/s (assumed for 0.3 mm quartz at about 20 °C, after Rouse). The required trap efficiency is 90%, and turbulence is considered.

Step 1: Horizontal velocity (Camp)

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

$$v=44\sqrt{0.30}=24.10\ \text{cm/s}=0.2410\ \text{m/s}$$

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 $H=3.5$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{8.0}{0.2410\times3.5}=9.48\ \text{m}$$

Provide $n=2$ chambers so that one can be flushed while the others work: $b=B/n=4.74$ m each, adopt 5.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.2410\times3.5}{0.0400}=21.1\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.2410}{\sqrt{3.5}}=0.0170\ \text{m/s}\\ w'&=0.0400-0.0170=0.0230\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.2410\times3.5}{0.0230}=36.7\ \text{m} \end{aligned}$$

For a trap efficiency of 90% (Vetter's relation $\eta=1-e^{-w'L/(vH)}$) the length is multiplied by $\ln\frac{1}{1-\eta}=2.303$:

$$L=2.303\times36.7=84.5\ \text{m}$$

Adopted dimensions

Total volume of water in the basin $\approx 2975$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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: 2 chambers, each about 5.0 m wide x 85 m long x 3.5 m deep (settling zone) with $v=0.241$ m/s.

Typical arrangement of the headworks

PLAN
    river flow ==============================>
        left bank wall
   +-------------------------------------------+
   |  intake  | undersluice | weir (spillway)    |
   | (side)   |  (gates)    | ||||||||||||||||| |
   +----+-----+-------------+--------------------+
        |  gravel trap
        v                         fish ladder
   approach canal --> settling basin --> headrace

SECTION through weir, undersluice and intake
  pond level ~~~~~~~~~~~~~~~~~~~~~~~~~~
        _____      weir crest  ______
  intake|rack |gate          /      \  stilling basin
  sill  |_____|-> canal     /        \____ apron
 ///////////  undersluice /////////  river bed
   (sill 1.0-1.5 m above bed)   flushing gate

Importance of the intake

• It is the link between the river and the waterway: the whole plant depends on it for the design discharge.
• It controls the quality of water entering the plant: sediment, gravel and trash that pass the intake reach the settling basin, tunnel and turbines and cause abrasion and blockage.
• It decides the head loss and so the energy generated; a poor intake gives vortices, air entry and heavy losses.
• It provides flow control and safety: gates close during floods and for maintenance.
• A wrongly placed intake can silt up, so its location and design decide the life and reliability of the headworks.

Deciding the location of an intake

1. On the outer bank of a bend (about one-third to one-half way along the bend, downstream of the bend apex), where the helical flow carries bed load away from the opening; on a straight reach put it on the stable bank.
2. In a stable reach with a good foundation, free of landslides, bank erosion and large boulder deposits.
3. At a point that gives sufficient head and an easy alignment of the canal to the settling basin.
4. Upstream of the weir and next to the undersluice so that the flushing current keeps the intake clear.
5. Away from the confluence of tributaries, with good access and space for the gravel trap and settling basin.
6. Where the river is narrow, to reduce the length and cost of the weir, and the flood level is limited.

1. Side intake

The opening is made in the river bank, at an angle (normally 90-110°) to the flow, beside the weir. It has a sill, a trash rack and a gate.

 river ==========================>
        weir crest  ||||||||||||||
 bank   +----------------------+
   rack |  intake    gate  -----+--> canal

• Used where the river carries bed load and boulders: the sill is raised and the undersluice keeps the intake clear. It is the most common in Nepal's RoR plants (steep gravel-boulder rivers).
• The flow must turn 90°, so some sediment enters; a gravel trap is therefore needed.

2. Drop (bottom / Tyrolean) intake

A trench with inclined bars (rack) is built in the river bed across the flow; the water drops through the bars into a chamber and flows into a pipe or canal.

 flow ====>   inclined rack
        ______\\\\\\\\\\______
              |  chamber  |---> conduit

• Used in steep mountain streams with small flow and coarse bed load, where the boulders pass over the rack.
• Needs a large head of fall and takes much fine sediment; it is liable to clogging by leaves and gravel, so it needs regular cleaning.
• Not suited to a large discharge.

3. Frontal intake

The opening is placed in line with the flow (in the weir or the dam face).

 river ===> |intake| ----> conduit

• Entry losses are small and the structure is simple, but it takes in all the bed load; it is used only where the river is clean (e.g. a reservoir or a clean channel) or with a good sediment excluder in front.

Requirements of a good intake

• Pass the design discharge in all river conditions, with the opening always submerged and no vortex or air entry.
• Keep out bed load, sediment and floating trash (high sill, trash rack and boom).
• Low head loss through a smooth, bell-mouthed entry and a short transition.
• Structurally stable in floods; strong foundation; safe against scour and uplift.
• Gates for regulation and shut-off, easy to operate; flushing arrangement in front of the intake.
• Easy access for maintenance and for removal of trash and sediment.
• Economical, with minimal impact on the environment and fish.

Types of intakes used in Nepal

1. Side intake (most common): the opening is in the river bank next to the weir; the undersluice flushes the front.
2. Bottom (Tyrolean) intake: a rack across the stream bed, used in steep streams, e.g. micro and small hydropower projects.
3. Frontal intake (in the dam or weir face): used rarely.
4. Reservoir/tunnel intake (submerged): a pressurised tower or inclined rack on a storage project.

 Side intake (plan)        Bottom intake (section)
 ===> weir  |||||||        flow ==> \\\\\\\\ rack
   rack |  gate -->               |____chamber____|-->

Main parts of a non-pressurized RoR intake

1. Approach channel / river training with guide wall.
2. Intake sill (about 1.0-1.5 m above the undersluice or river bed) to stop bed load.
3. Trash rack (inclined at 70-80°) with a trash boom in front.
4. Gate (regulating gate) with a stop-log slot and hoist deck.
5. Transition / bell-mouth to the gravel trap and the approach canal / settling basin.
6. Flushing (undersluice) gate and flood wall.

NON-PRESSURIZED INTAKE (section)
 pond ~~~~~~~~~~~~~~~~~~~~~~~
      boom  | hoist deck
   --o--    |  |gate
 trash  |##\ |  |
 rack   |##  \|  |_______ free flow --> canal
 sill   |_______|
 ///////////////////////////// river bed

Main parts of a pressurized intake

1. Approach and trash rack (inclined or vertical, large area, with rake).
2. Bell-mouthed entrance (to reduce entry loss).
3. Minimum submergence below the lowest operating level, with an anti-vortex device.
4. Emergency gate and service gate in a gate shaft, with an air vent behind the gate.
5. Transition from rectangular to the circular pressure tunnel/penstock.

PRESSURIZED INTAKE (section)
 min WL ~~~~~~~~~~~~~~~~~~~~~~~~
        |  air vent
 rack   |##  |gate|
 S      |##  |    |
 _______|bell-|____> pressure tunnel (full)
        mouth

Proportions: rack velocity 0.8-1.0 m/s; the bell-mouth length is about 1.5-2 times the conduit height; the submergence is chosen by the vortex criterion $S\ge C\,V\sqrt D$.

Sketch of a side intake

PLAN
 river ==========================>
   weir |||||||  undersluice
   bank +---------------------+
        |trash rack  | gate   |--> gravel trap --> canal
        +---------------------+

SECTION
  pond ~~~~~~~~~~~~~~~~~~~~~~~~
   boom o  |  rack (75°)  | gate | hoist deck
        ###|##\           |      |
        ###|###\__________|______|======> canal
   sill (1.0-1.5 m above undersluice)
 /////////////////////////////////////////// bed

Components: trash boom, trash rack with rake, intake sill, gate with stop-log slot, hoist deck, flushing undersluice and the transition to the canal.

Head loss at the trash rack (Kirschmer's formula)

$$h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta$$

where $h_t$ is the head loss (m), $\beta$ the bar shape factor (2.42 for rectangular, 1.83 for rectangular bars with round upstream edges, 1.79 for circular bars, 0.76 for streamlined bars), $s$ the bar thickness, $b$ the clear spacing between bars, $V$ the approach velocity (m/s) and $\theta$ the angle of the rack from the horizontal.

For a clogged rack the loss is increased by taking an additional allowance; a clogging factor of 25-50% is used, and the total head loss for the design of the intake should include it.

Example: bars 10 mm thick at 100 mm clear spacing, circular ($\beta=1.79$), $V=0.8$ m/s, $\theta=75^\circ$:

$$h_t=1.79\,(0.1)^{4/3}\frac{0.8^2}{19.62}\sin75^\circ=1.79\times0.0464\times0.0326\times0.966=0.0026\ \text{m}$$

which is very small, so the rack loss is negligible when clean.

The head loss in an intake is the sum of the following parts. Each is computed as a coefficient times a velocity head, except friction.

1. Trash rack loss (Kirschmer):

$$h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V_a^2}{2g}\sin\theta$$

with $\beta$ = bar shape factor (2.42 rectangular, 1.79 circular), $s$ = bar thickness, $b$ = clear spacing, $V_a$ = approach velocity and $\theta$ = rack inclination. For a partly clogged rack, increase the loss using the blocked area (net velocity).

2. Entrance loss: $h_e=k_e\dfrac{V^2}{2g}$, where $k_e=0.5$ for a sharp-edged entrance, 0.2 for a rounded, and 0.05-0.1 for a well-shaped bell-mouth.

3. Gate slot loss: $h_g=k_g\dfrac{V^2}{2g}$, with $k_g\approx0.1$ (nil for a fully open gate with flush slot).

4. Transition loss (contraction or expansion): $h_{tr}=k\dfrac{(V_2-V_1)^2}{2g}$, with $k\approx0.1$ for a gradual contraction and 0.2-0.3 for an expansion.

5. Bend loss: $h_b=k_b\dfrac{V^2}{2g}$, where $k_b\approx0.1-0.25$ depends on the radius and angle.

6. Friction loss in the intake conduit (Manning/Darcy):

$$h_f=\frac{n^2V^2L}{R^{4/3}}\quad\text{or}\quad h_f=f\frac{L}{D}\frac{V^2}{2g}$$

Total loss

$$h_L=h_t+h_e+h_g+h_{tr}+h_b+h_f$$

The loss is minimised by a low velocity at the rack (about 0.8-1.0 m/s), a bell-mouthed entrance, smooth gradual transitions and few gate slots. For a typical RoR intake the total is of the order of 0.1-0.3 m, and it is subtracted from the net head of the plant.

Sketch of a hydropower intake (non-pressurized side intake of a RoR plant)

PLAN
 river ==========================>
   weir |||||||| undersluice
   bank +----------------------+
        | rack    gate  bell   |--> gravel trap --> canal
        +----------------------+

SECTION
 pond ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  boom o  | trash rack | gate | hoist deck
          |###\        |      |
          |####\_______|______|=====> conduit
   sill (1.0-1.5 m above bed)   transition
 ///////////////////////////////////////// bed

Major components: trash boom, trash rack (inclined), intake sill, regulating gate with stop-log slot, hoist deck, bell-mouthed transition, flushing undersluice and the conduit to the gravel trap or settling basin.

Minimising head loss in the intake

Head loss reduces the net head and so the energy produced. It is reduced by:

• Low velocity at the trash rack (0.6-1.0 m/s) by providing a large rack area; the loss varies with $V^2$.
• Streamlined bars (rounded or tapered upstream edges, $\beta$ of 0.76-1.8 instead of 2.42), larger spacing, and regular raking to avoid clogging.
• Bell-mouthed, rounded entrance instead of a sharp-edged entry (loss coefficient 0.05-0.1 instead of 0.5).
• Gradual transitions (contraction or expansion not steeper than about 1:5) and smooth curved wing walls.
• Few gate slots and piers, and a fully open gate in operation.
• Smooth lining, short conduit and large bend radius; avoiding sharp bends at the entry.

Why a vortex forms

A vortex forms when water approaches a submerged intake with angular momentum (circulation) and insufficient submergence. As the flow converges to the opening its velocity rises and the surface is drawn down by the low pressure at the centre, so a rotating core develops and may draw air into the intake. Typical causes:

• small submergence of the intake above its crown;
• asymmetric approach flow, flow past piers, abutments or the bank, giving circulation;
• high intake velocity and high Froude number;
• sharp edges, no guide walls, and sudden changes in direction.

Vortices cause air entrainment (reduced capacity and turbine efficiency), vibration, surging, cavitation, and carry floating trash into the intake.

Conditions for no vortex formation

1. Adequate submergence (Gordon's formula):

$$S=C\,V\sqrt{D}$$

where $S$ is the submergence (water surface to the crown of the intake, m), $V$ the velocity in the intake (m/s), $D$ the height or diameter of the intake (m), and $C=0.7245$ for symmetrical approach flow and $0.5434$ for an asymmetrical approach. An alternative criterion (Knauss) is $S/D\ge1+2.3\,Fr$ with $Fr=V/\sqrt{gD}$. 2. Low intake velocity (less than about 1 m/s at the rack, and the Froude number kept low). 3. Symmetrical, smooth approach flow, with guide walls to remove circulation; avoid sharp bends and eddies. 4. Anti-vortex devices: floating rafts or beams, vertical splitter walls or baffles above the intake, and a trash rack that breaks rotation. 5. Bell-mouth entry of a good shape and a sufficient depth of the intake sill above the bed.

Given: $Q=5$ m³/s; weir crest is 2.5 m above the river bed; canal water level is 0.3 m below the crest; intake velocity $\le0.8$ m/s; $C_d=0.6$; bar-contraction coefficient 0.03; trash-rack bar shape factor $\beta=1.83$.

Step 1: Net opening area from the velocity limit

$$A_{net}=\frac{Q}{V}=\frac{5}{0.8}=6.25\ \text{m}^2$$

Step 2: Gross area allowing for the bars

The bars reduce the opening by the contraction coefficient 0.03:

$$A_{gross}=\frac{A_{net}}{1-0.03}=\frac{6.25}{0.97}=6.44\ \text{m}^2$$

Step 3: Dimensions of the opening

Sill level 1.0 m above the river bed (to keep bed load out). Take the height of the orifice $h=1.2$ m so that its top is at 2.2 m above the bed, i.e. at the canal water level (0.3 m below the crest), so the opening stays submerged.

$$b=\frac{A_{gross}}{h}=\frac{6.44}{1.2}=5.37\ \text{m}\ \Rightarrow\ \text{adopt }5.4\ \text{m (2 bays of 2.7 m)}$$

Adopted opening area $A=1.2\times5.4=6.48$ m².

Step 4: Head required across the orifice (check)

$$Q=C_dA\sqrt{2g\,\Delta h}\ \Rightarrow\ \Delta h=\frac{1}{2g}\left(\frac{Q}{C_dA}\right)^2=\frac{1}{19.62}\left(\frac{5}{0.6\times6.48}\right)^2=0.084\ \text{m}$$

Available head difference $=0.3$ m $>0.084$ m, so the opening can pass the design flow; the extra head is absorbed by throttling the gate.

For comparison, the area that would pass 5 m³/s under the full 0.3 m head is $A=\dfrac{5}{0.6\sqrt{2g(0.3)}}=3.43$ m², so the velocity limit (6.25 m²) controls the size.

Step 5: Trash rack loss (Kirschmer)

Assume bars 10 mm thick at 50 mm clear spacing, inclined at $75^\circ$:

$$h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta=1.83\left(\frac{10}{50}\right)^{4/3}\frac{0.8^2}{19.62}\sin75^\circ=0.0067\ \text{m}$$

 SECTION                         PLAN
 crest 2.5 ~~~~~~~~~~~~          river ==>
        |   |gate|  2.2         |rack|gate|--> canal
  rack  |## |    |~~~~ canal    | 5.4 m opening |
 sill   |___|____|              2 bays of 2.7 m
 1.0 m above bed

Answer: orifice opening $5.4\ \text{m}\times1.2$ m with sill 1.0 m above the bed (2 bays of 2.7 m); required head $0.084$ m < 0.3 m available; rack loss $\approx0.0067$ m.

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; $Q=1.45$ m³/s; circular bars 10 mm thick at 100 mm clear spacing. Assumed: rack velocity 0.8 m/s, rack inclined $75^\circ$, $C_d=0.6$ 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 $h=1.3$ m, so the top of the opening is RL 3317.30 m, which 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

$$A_{net}=\frac{Q}{V}=\frac{1.45}{0.8}=1.812\ \text{m}^2$$ $$A_{gross}=A_{net}\times\frac{s+b}{b}=1.812\times\frac{110}{100}=1.994\ \text{m}^2$$ $$\text{width }=\frac{A_{gross}}{h}=\frac{1.994}{1.3}=1.53\ \text{m}\ \Rightarrow\ \text{adopt }1.6\ \text{m}$$

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

Step 3: Orifice check at normal pond level

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

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

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

Step 4: Trash rack loss (Kirschmer)

$$h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta=1.79\,(0.1)^{4/3}\frac{0.767^2}{19.62}\sin75^\circ=0.0024\ \text{m}$$

which is negligible.

Step 5: Flood condition and levels

At the 100-year flood RL 3319.55 m the head on the opening centre (RL 3316.65 m) is $2.90$ m. Without control the intake would pass $0.6\times2.08\sqrt{2g\times2.90}=9.4$ m³/s, so the gate is closed or throttled; for 1.45 m³/s the opening area is only $\dfrac{1.45}{0.6\sqrt{2g\times2.90}}=0.32$ m² (gate opening of about 0.20 m). Top of the intake wall and hoist deck: flood level + 0.6 m free board $=$ 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; deck RL 3320.15 m.

Given: intercepted flow $Q=8$ m³/s; rack width $B=10$ m; rack inclination $\alpha=30^\circ$; bar thickness $s=10$ mm; clear spacing $b=15$ mm; contraction factor $c=0.82$.

Assumptions: the water arrives at the rack at critical depth (over the weir crest), and the specific energy $E=1.5\,h_c$ stays nearly constant along the short rack (Noseda/Mostkow approximation).

Step 1: Unit discharge and energy

$$q=\frac{Q}{B}=\frac{8}{10}=0.8\ \text{m}^2/\text{s},\qquad h_c=\left(\frac{q^2}{g}\right)^{1/3}=0.403\ \text{m},\qquad E=1.5h_c=0.604\ \text{m}$$

Step 2: Opening ratio

$$\varepsilon=\frac{b}{b+s}=\frac{15}{15+10}=0.60$$

Step 3: Discharge passing through the rack per unit plan area

$$q_b=c\,\varepsilon\cos\alpha\sqrt{2gE}=0.82\times0.60\times\cos30^\circ\times\sqrt{2\times9.81\times0.604}=1.467\ \text{m}^3/\text{s per m}^2\ (=\text{m/s})$$

Step 4: Length of the rack

All the unit flow $q$ must pass through the rack:

$$L=\frac{q}{q_b}=\frac{0.8}{1.467}=0.55\ \text{m}$$

Allow 25% extra for partial clogging by gravel and leaves: $L=1.25\times0.55=0.68$ m. Adopt about 0.7 m.

Answer: required rack length $\approx0.55$ m; adopt $0.7$ m (rack 10 m wide, inclined at $30^\circ$).

In a RoR plant on a steep Himalayan river, bed load (gravel, boulders) and floating debris (logs, leaves, trash) must be kept out of the intake. This is done by the arrangement shown below.

PLAN
 river ====================================>
        scour sluice (undersluice)  ||||||| weir
   +---+====================+
   |  flood   divide wall    |
   |  wall   --> gravel trap |--> canal to
   | rack|gate               |    settling basin
   +---------------------------+
   trash boom o-o-o-o (angled)

SECTION (through intake)
 pond ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
   boom o   | rack |gate|
            |###\  |    |==> canal
   sill 1.0-1.5 m above undersluice floor
 //// undersluice floor (low) ////  --> flushing

Control of bed load

1. High intake sill: the intake floor is placed 1.0-1.5 m (or more) above the undersluice crest or river bed, so the bed load rolls past.
2. Undersluice (scouring sluice) next to the intake, with gates, creates a flushing channel in front of it; it is opened during floods to scour deposits, and the divide wall separates the intake from the sluice.
3. Intake location on the outer bank of the bend, where the secondary flow moves bed load to the inner bank.
4. Curtain (skimming) wall or bed-load excluder: only the upper clear water is drawn.
5. Gravel trap after the intake with flushing arrangement, and a settling basin for sand.
6. Guide bunds and training walls keep a stable flow line near the intake.

Control of floating debris

1. Trash boom (floating or fixed), set at an angle to the flow, deflects logs and trash towards the weir spill or sluice.
2. Trash rack (clear spacing about 50-100 mm) with an automatic or manual rake, inclined at about 75°.
3. Skimming walls and a surface spillway pass the floating material over the weir.
4. Low velocity at the rack (0.6-1.0 m/s) so the debris does not hold on the bars.
5. Periodic inspection and cleaning in the monsoon.

Given: design discharge $Q=60.0$ m³/s and gross head 100 m; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (taken for 0.15 mm quartz sand). Turbulence is considered.

Step 1: Horizontal velocity (Camp)

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

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

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{60.0}{0.1704\times6.0}=58.68\ \text{m}$$

Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=19.56$ m each, adopt 20.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\ w'&=0.0150-0.0092=0.0058\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 64800$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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 20.0 m wide x 180 m long x 6.0 m deep (settling zone) with $v=0.170$ m/s.

Given: design discharge $Q=30.0$ m³/s and gross head 100 m; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=2.00$ cm/s $=0.0200$ m/s (given). The result is first found without turbulence, then with turbulence, and checked with Velikanov's method ($\lambda=1.5$).

Step 1: Horizontal velocity (Camp)

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

$$v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}$$

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 $H=5.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{30.0}{0.1968\times5.0}=30.49\ \text{m}$$

Provide $n=2$ chambers so that one can be flushed while the others work: $b=B/n=15.25$ m each, adopt 15.5 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1968\times5.0}{0.0200}=49.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{5.0}}=0.0116\ \text{m/s}\\ w'&=0.0200-0.0116=0.0084\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times5.0}{0.0084}=117.4\ \text{m} \end{aligned}$$

Step 5: Check by Velikanov's method

$$L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w^2}=\frac{1.5^2\times0.1968^2\times(\sqrt{5.0}-0.2)^2}{7.51\times0.0200^2}=120.2\ \text{m}$$

The Velikanov length (120.2 m) is 2% greater than the turbulence length (117.4 m). The larger of the two is adopted and rounded up, giving 125 m.

Adopted dimensions

Total volume of water in the basin $\approx 19375$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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: 2 chambers, each about 15.5 m wide x 125 m long x 5.0 m deep (settling zone) with $v=0.197$ m/s.

Given: design discharge $Q=60.0$ m³/s and gross head 300 m; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (taken for 0.15 mm sand, as in similar designs). The basin has continuous flushing, so 10% extra flow (6 m³/s) is added. The trap efficiency required is 95% for particles larger than 0.15 mm, and the turbulence effect is included. Velikanov's relation is checked with $\lambda=1.5$ and silty water of density 1.105 t/m³.

Flushing water of 10% is added for continuous flushing, so the design flow of the basin is $Q'=60.0+6.0=66.0$ m³/s.

Step 1: Horizontal velocity (Camp)

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

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

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 $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{66.0}{0.1704\times6.0}=64.55\ \text{m}$$

Provide $n=3$ chambers so that one can be flushed while the others work: $b=B/n=21.52$ m each, adopt 22.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\ w'&=0.0150-0.0092=0.0058\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m} \end{aligned}$$

Step 5: Check by Velikanov's method

The silty water has density 1.105 t/m³ (sediment 2.65 t/m³), which reduces the effective fall velocity by the buoyancy ratio $\frac{2.65-1.105}{2.65-1}=0.936$, so $w_s=1.405$ cm/s.

$$L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w_s^2}=\frac{1.5^2\times0.1704^2\times(\sqrt{6.0}-0.2)^2}{7.51\times0.01405^2}=223.2\ \text{m}$$

The Velikanov length (223.2 m) is 27% greater than the turbulence length (175.8 m). The larger of the two is adopted and rounded up, giving 225 m.

Adopted dimensions

Total volume of water in the basin $\approx 89100$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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.

The adopted length is greater than the Velikanov length, so more than 95% of the particles coarser than 0.15 mm are expected to be trapped. Continuous flushing is done through a longitudinal slot (or hoppers) below each chamber, carrying 6 m³/s of flushing water back to the river.

Answer: 3 chambers, each about 22.0 m wide x 225 m long x 6.0 m deep (settling zone) with $v=0.170$ m/s.

Given: $Q=15$ m³/s, particles $\ge0.2$ mm, $w=0.02$ m/s, factor of safety for area $=1.5$, $L/B=8$, sediment concentration $C=1$ kg/m³, particle density 2500 kg/m³, flushing interval 8 h, packing factor 0.5, additional flushing discharge 10%.

Step 1: Design flow

$$Q'=1.10\times15=16.5\ \text{m}^3/\text{s}$$

Step 2: Surface area of the basin

For a particle to settle, the surface area must satisfy $A\ge Q/w$. With the factor of safety:

$$A=\frac{FS\times Q'}{w}=\frac{1.5\times16.5}{0.02}=1237.5\ \text{m}^2$$

Step 3: Width and length ($L/B=8$)

$$B=\sqrt{\frac{A}{8}}=\sqrt{\frac{1237.5}{8}}=12.44\ \text{m},\qquad L=8B=99.5\ \text{m}$$

Adopt $B=13$ m and $L=104$ m ($A=1352$ m²).

Step 4: Depth of flow

Limit the horizontal velocity to about 0.3 m/s. Take the flow depth $H=4.5$ m:

$$v=\frac{Q'}{BH}=\frac{16.5}{13\times4.5}=0.282\ \text{m/s}<0.3\ \text{m/s}\quad\text{(OK)}$$

Step 5: Sediment storage (flushing every 8 h)

$$\text{Mass of sediment}=C\,Q'\,t=1\times16.5\times(8\times3600)=475200\ \text{kg}$$ $$\text{Volume of deposit}=\frac{\text{mass}}{\rho_s\times p}=\frac{475200}{2500\times0.5}=380.2\ \text{m}^3$$ $$\text{Depth of deposit}=\frac{380.2}{1352}=0.28\ \text{m}\ \Rightarrow\ \text{provide }0.5\ \text{m storage}$$

(All the sediment entering is assumed to be trapped, which is conservative.)

Step 6: Overall dimensions

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

Answer: single chamber of about 13 m x 104 m, flow depth 4.5 m ($v=0.282$ m/s) with 0.5 m sediment storage, to be flushed every 8 hours (deposit 380.2 m³ per cycle).

Given: $Q=16$ m³/s, particle $\ge0.2$ mm with $w=0.022$ m/s, horizontal velocity $v=0.2$ m/s, two basins, factor of safety 1.5, $L/B=6$, $C=1$ kg/m³, $\rho_s=2650$ kg/m³, settling (flushing) time 1 day, packing factor 0.5, extra discharge for flushing 10%.

Step 1: Design flow

$$Q'=1.10\times16=17.6\ \text{m}^3/\text{s}$$

Step 2: Surface area

$$A_{total}=\frac{FS\,Q'}{w}=\frac{1.5\times17.6}{0.022}=1200.0\ \text{m}^2\quad\Rightarrow\quad A_{basin}=\frac{1200.0}{2}=600.0\ \text{m}^2$$

Step 3: Plan size of one basin

$$B=\sqrt{\frac{A_{basin}}{6}}=\sqrt{\frac{600.0}{6}}=10.00\ \text{m},\qquad L=6B=60.0\ \text{m}$$

Adopt $B=10$ m, $L=60$ m.

Step 4: Depth of flow

Flow in one basin $=Q'/2=8.8$ m³/s; with $v=0.2$ m/s:

$$H=\frac{Q'/2}{B\,v}=\frac{8.8}{10\times0.2}=4.40\ \text{m}\ \Rightarrow\ \text{adopt }4.5\ \text{m}$$

Step 5: Sediment storage (1 day)

$$\begin{aligned} \text{Mass}&=C\,Q'\,t=1\times17.6\times86400=1520640\ \text{kg}\\ \text{Volume}&=\frac{1520640}{2650\times0.5}=1148\ \text{m}^3\ \text{(both basins)}\\ \text{Depth per basin}&=\frac{1148/2}{600.0}=0.96\ \text{m} \end{aligned}$$

Provide about 1.0 m of sediment storage in the hopper below the flow depth (or flush more often, as in practice).

Answer: 2 basins, each 10 m x 60 m in plan with a flow depth of 4.5 m and 1.0 m storage depth.

Given: $Q=32$ m³/s, particles coarser than 0.15 mm, $w=0.025$ m/s, $\lambda=1.5$ (turbulence/safety coefficient). High head plant, so two chambers are provided, one flushed while the other works.

Step 1: Horizontal velocity (Camp)

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

Step 2: Surface area with turbulence coefficient

$$A=\frac{\lambda Q}{w}=\frac{1.5\times32}{0.025}=1920\ \text{m}^2$$

Step 3: Depth, width and length

Assume a flow depth $H=5$ m:

$$B=\frac{Q}{vH}=\frac{32}{0.1704\times5}=37.56\ \text{m}\ (\text{two chambers of }18.78\ \text{m}),\qquad L=\frac{A}{B}=\frac{1920}{37.56}=51.1\ \text{m}$$

Check: $L=\dfrac{\lambda vH}{w}=\dfrac{1.5\times0.1704\times5}{0.025}=51.1$ m. Adopt $L=55$ m, chamber width $19.0$ m, giving $L/b=2.9$.

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 to a central flushing channel (slope 1:30) and a flushing gate at the downstream end; the inlet gate of the chamber is closed, the chamber is flushed, and the other chamber carries the flow meanwhile. Flushing water runs through the flushing tunnel to the river.

Answer: 2 chambers, each 19.0 m wide x 55 m long x 5 m deep (plus storage).

Given: design discharge $Q=10.0$ m³/s; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=5.00$ cm/s $=0.0500$ m/s (given). The basin is of the continuous flushing type, so a flushing discharge of 10% (1.0 m³/s) is added. Turbulence is included, and other data are assumed. (A fall velocity of 5 cm/s is high for 0.15 mm sand, so the length is short; the given value is used.)

Flushing water of 10% is added for continuous flushing, so the design flow of the basin is $Q'=10.0+1.0=11.0$ m³/s.

Step 1: Horizontal velocity (Camp)

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

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

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 $H=3.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{11.0}{0.1704\times3.0}=21.52\ \text{m}$$

Provide $n=2$ chambers so that one can be flushed while the others work: $b=B/n=10.76$ m each, adopt 11.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times3.0}{0.0500}=10.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{3.0}}=0.0130\ \text{m/s}\\ w'&=0.0500-0.0130=0.0370\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times3.0}{0.0370}=13.8\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 990$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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: 2 chambers, each about 11.0 m wide x 15 m long x 3.0 m deep (settling zone) with $v=0.170$ m/s.

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

Step 1: Horizontal velocity (Camp)

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

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

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

Assume a flow depth $H=3.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{6.0}{0.2200\times3.0}=9.09\ \text{m}$$

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

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.2200\times3.0}{0.0250}=26.4\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.2200}{\sqrt{3.0}}=0.0168\ \text{m/s}\\ w'&=0.0250-0.0168=0.0082\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.2200\times3.0}{0.0082}=80.2\ \text{m} \end{aligned}$$

Adopted dimensions

Total volume of water in the basin $\approx 2550$ 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 about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. Flushing velocity of at least 3 m/s is kept 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: 2 chambers, each about 5.0 m wide x 85 m long x 3.0 m deep (settling zone) with $v=0.220$ m/s.