Diversion Headworks

Silt excluder and silt ejector

A silt excluder prevents silt from entering the canal; a silt ejector removes silt that has already entered the canal.

Working and sketches

Silt excluder (plan and section through tunnels):

PLAN
 River flow -->
 ____________________________________
 |  under-    | tunnels ->   |  river
 |  sluice    |==============|  (d/s)
 |  pocket    |==============|
 |____________|  head        |
              | regulator    |-> canal
 -------------------------------------
SECTION (through the pocket)
 water level ~~~~~~~~~~~~~~~~~~~~~~~~~
   regulator crest (high)  +-----+
 ------------------------  |     |-> canal
   silty bottom layer -->  [==tunnel==]-> river
 undersluice bed level ____|_________

The tunnels have their roofs at the level of the canal regulator crest. Clear water from the upper layers flows over the tunnel roof into the canal. The bottom layer, which carries the bed load, passes through the tunnels and is released through the undersluices into the river.

Silt ejector (on the canal):

 canal -->  [  excluder at  ] -->  canal continues
            |  ===tunnels===|--> escape to river
 section: bed silt enters the tunnel because the
 tunnel is placed on the canal bed, with a lower
 level outfall into the river.

The canal water passes over tunnel roofs; the bottom silt-laden layer enters the tunnels (formed by a diaphragm wall on the canal bed) and goes by an escape channel back to the river. The tunnels are designed for 20-25% of the canal discharge, with a velocity of 2.5-3 m/s so that the silt keeps moving.

Control of bed load

• The undersluices are kept at a low level so that the flow in the pocket scours the deposits and sends the bed silt downstream.
• The canal head regulator crest is placed about 1.2-2.0 m above the sluice crest, so mainly the upper layer, with less silt, is drawn.
• The silt excluder diverts the bed load in the pocket, and the ejector removes what still enters, so that silting of the canal is reduced and the sediment is sent back to the river.

A diversion headworks is a structure built across a river to raise the water level and divert the flow into a canal. It consists of a weir or barrage and the associated structures.

Layout (plan)

 Left       <-- river flow: Upstream to Downstream -->
 guide  ______________________________________  Right
 bund  |                                      | guide
       |  Weir   Weir   |D| Under-  |D| Head  | bund
  ~~~> |  (crest) bay   |W| sluices |W| reg.  |
       |=====================|======|====||===|--> canal
       |  Fish ladder   |<-- pocket -->|     |
       |  Divide wall (DW) separates weir/sluices
       |_______ Upstream & downstream aprons ___|
       Marginal / afflux bunds on both banks

(DW = divide wall; the head regulator is located at one end with the silt excluder in front of it.)

Cross-section through the weir

  Pond level ~~~~~~~~~~~~~
  u/s apron     crest|\  d/s glacis
  ______________ ____| \_________ 
  cutoff (sheet pile)       |      d/s pile

Components and their functions

1. Weir or barrage proper – raises the river level to the pond level and passes the flood over (weir) or through gates (barrage). Has a crest, u/s and d/s glacis, and a floor with cut-off walls.
2. Undersluices (scouring sluices) – gated openings at low level next to the canal head, which scour deposits in front of the head regulator and pass low floods.
3. Divide wall – separates the weir portion from the undersluice portion, protects the sluices from cross currents, and forms the silt pocket.
4. Fish ladder – a sloping channel with baffles beside the divide wall, allowing fish to migrate.
5. Canal head regulator – controls the flow into the canal; its crest is higher than that of the undersluices to keep out bed silt.
6. Silt excluder / silt ejector – reduce silt entry into the canal.
7. Guide bunds (guide banks) – train the river to flow axially through the structure.
8. Marginal (afflux) bunds – confine the river's high flood level and prevent outflanking above the structure.
9. Aprons and protection works – u/s and d/s concrete blocks with filters and launching aprons, to protect against scour.
10. Pile cutoffs / curtain walls – at upstream and downstream ends of the floor, to lengthen the creep path and keep exit gradient safe.
11. Piers, bridge, and shutters/gates – for operation of gates and cross-communication.

• Fish ladder. A fish ladder is a sloping channel (generally 1:10 to 1:15) with baffles or steps, built beside the divide wall at the weir. It lets fish swim upstream past the weir to their breeding places. The baffles slow the flow and form resting pools. It is necessary where fish migration must be protected, as in Nepal's rivers.
• Divide wall. A masonry or concrete wall built at right angles to the weir axis between the weir and the undersluices. Its functions: (i) separates the quiet pocket in front of the head regulator from the weir flow so that the silt in the pocket can be scoured by the sluices; (ii) keeps the cross-currents from the weir away from the sluices; (iii) prevents the high-velocity flow over the weir from damaging the sluices; (iv) acts as the side of the fish ladder, if provided; (v) forms a division of the hydraulic jump portion.
• Undersluices (scouring sluices). Gated openings in the weir, next to the head regulator, with crest at a low level (river bed). Functions: (i) scour out the silt deposited in the pocket by opening the gates; (ii) maintain a clear, deep channel along the canal; (iii) pass the low floods without lowering the pond; (iv) induce a curved flow so that the bed silt enters the sluices and the canal gets the upper clear water; (v) to pass part of the high flood.

Why an undersluice is provided

Silt tends to deposit in front of the canal head regulator because the velocity near the pocket is low. An undersluice (scouring sluice) is a set of gated bays at low level beside the head regulator. It is provided to keep a clear, deep channel in front of the regulator so that the canal gets silt-free water.

Major functions

1. Scours out the silt deposited in the pocket in front of the head regulator, by opening the gates (the sluice acts like a drain for the pocket).
2. Maintains a well-defined channel from the river to the regulator, particularly in the dry season when the river flow is low.
3. Passes low floods (and the surplus) without opening the weir shutters, therefore keeping the pond level and the supply to the canal.
4. Draws the bottom, silt-laden layer away from the canal (induced curvature, together with silt excluder).
5. Helps in diverting the river flow to the weir side during construction/ low flow.

Design criteria

• Capacity: pass at least twice the canal discharge (to give a scouring velocity), or the dry-season flow of the river (about 10-15% of the maximum flood), whichever is more.
• Crest level: at the average river-bed level (about 1.2-2.0 m below the canal head regulator crest).
• Waterway: total width about 10-20% of the weir width; gate width 5-8 m with piers 1.5-2.0 m; at least 2 bays.
• Velocity: through the sluices, 2.0-3.0 m/s, enough to carry the bed material.
• Discharge formula: $Q = C\,L\,H^{3/2}$ (free flow, $C \approx 2.0$) or orifice flow when gates are partly open.
• Floor and protection: designed against uplift (Khosla) like the weir, with u/s and d/s aprons, cutoffs, and a stilling basin for the hydraulic jump.

Given: floor length $b = 50$ m, cutoffs at both ends, depth $d = 8$ m below the river bed (the floor top is flush with the bed), head $H = 5$ m, floor thickness $t_1 = 1$ m (u/s end) and $t_2 = 2$ m (d/s end). Khosla's theory is used. The inside corners are the junctions of the floor with the u/s pile ($C_1$) and with the d/s pile ($E_2$).

Khosla's constants

Pressures from the simple profile (thin floor)

Correction for floor thickness

The uplift acts on the underside of the floor, a distance $t$ below the key point, so the pressure is interpolated along the pile:

Uplift pressure heads

Exit gradient

Answer: Uplift at the u/s inside corner $\approx 66.4\%$ of $H$ (3.32 m); at the d/s inside corner $\approx 32.3\%$ (1.61 m); exit gradient $G_E \approx 0.104$, which is less than the safe gradient of about 1/6 (0.167) for fine sand, so the structure is safe against piping.

Data and assumptions

$Q = 10000$ m³/s, $d_{50} = 0.33$ mm, flood level $= 207$ m, pond level $= 204$ m, d/s floor level $= 198$ m, safe exit gradient $= 1/6$. The design head for piping is the static head with no tail water, $H = 204 - 198 = 6$ m. The d/s cutoff is taken to a depth of $1.5R$ below the HFL.

1. Silt factor, waterway and scour depth

2. Depth of d/s cutoff

Scour level $= 207 - 1.5R = 191.62$ m. Depth of the d/s pile below the floor:

3. Floor length from the exit gradient

Check: $G_E$ with $b = 34.2$ m is 0.1667 $= 1/6.0$.

Answer: Minimum total impervious floor length (between the end cutoffs) $\approx 34.2$ m, with the d/s cutoff 6.4 m deep. If a deeper d/s cutoff (2R) is used, the exit gradient is satisfied for any floor length, and the length is then fixed by the uplift thickness and the jump.

Conceptual section

  Flood level 207 ~~~~~~~~~~~~~~~~~~~~~~~
  Pond level  204 ~~~~~~~~~|gate|
 u/s block apron  ___crest__|    |__ glacis __ d/s jump floor
 ====================================================== 198
   u/s pile (1.0-1.5R)  <--- b = 34.2 m ---> d/s pile 6.4 m
                                   inverted filter + block apron

Data and assumptions

$Q = 11000$ m³/s, $d_{50} = 0.34$ mm, flood level $= 208$ m, pond level $= 205$ m, d/s floor level $= 199$ m, safe exit gradient $= 1/7$. The design head for piping is the static head with no tail water, $H = 205 - 199 = 6$ m. The d/s cutoff is taken to a depth of $1.5R$ below the HFL.

1. Silt factor, waterway and scour depth

2. Depth of d/s cutoff

Scour level $= 208 - 1.5R = 192.20$ m. Depth of the d/s pile below the floor:

3. Floor length from the exit gradient

Check: $G_E$ with $b = 45.3$ m is 0.1429 $= 1/7.0$.

Answer: Minimum total impervious floor length (between the end cutoffs) $\approx 45.3$ m, with the d/s cutoff 6.8 m deep. If a deeper d/s cutoff (2R) is used, the exit gradient is satisfied for any floor length, and the length is then fixed by the uplift thickness and the jump.

Conceptual section

  Flood level 208 ~~~~~~~~~~~~~~~~~~~~~~~
  Pond level  205 ~~~~~~~~~|gate|
 u/s block apron  ___crest__|    |__ glacis __ d/s jump floor
 ====================================================== 199
   u/s pile (1.0-1.5R)  <--- b = 45.3 m ---> d/s pile 6.8 m
                                   inverted filter + block apron

(a) General layout (plan)

 Marginal bund  <------ river flow ------>
 ___________________________________________
 |                                         |
 | Weir bays | DW | Under- | DW | Head reg. |-> Canal
 |~~~~~~~~~~~|    | sluice |    | (gated)   |
 | Fish ladder     | pocket |  silt excluder |
 |_________________|________|_______________|
 Guide bund (u/s)           Guide bund (d/s)
 Upstream apron / Downstream apron, pile cutoffs

(b) Section of undersluice

  Pond level ~~~~~~~~~~~~~~~~~~~~~
  gate  |#|  pier/bridge deck
  ______|#|___                   ___ d/s apron
 u/s    crest (river-bed level)  |   |
 apron  ------------------------ |   | stilling basin
 sheet pile                      sheet pile

Crest at river-bed level; gate fixed above; u/s floor and cutoff; d/s hydraulic-jump floor with glacis, blocks and apron.

(c) Section of canal head regulator

 Pond level ~~~~~~~~~~~~~~~
          |#| gate        deck
   _______|#|______________
  regulator crest (1.2-2.0 m above sluice crest)
   u/s apron | floor | piles | d/s apron -> canal

Crest is higher than the undersluice crest; piers carry gates and a bridge; floor with u/s and d/s cutoffs.

(d) Section of weir

  HFL  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  Pond level ~~~~~~+--+ shutters
         __________|  |\___
  u/s apron  crest   \ glacis   d/s floor
  ________________________      ______ apron
       sheet pile (cutoff)       pile (cutoff)

Weir crest is raised by shutters above the permanent crest; the d/s sloping glacis dissipates energy by a hydraulic jump on the horizontal floor; u/s and d/s sheet piles control seepage; the aprons and filters protect against scour.

A silt excluder is a set of tunnels built under the canal head regulator in the river pocket that passes the silty bottom water to the river through the undersluices.

Design steps

1. Fix the crest levels. Undersluice crest level (river-bed level) and the head-regulator crest level, normally 1.2-2.1 m higher. The difference gives the available tunnel height (minus roof slab thickness 0.2-0.3 m).
2. Discharge through the tunnels. Take 20% of the canal discharge (Garg), or the silt-carrying capacity needed.
3. Tunnel velocity. Fix 2.5-3 m/s (non-silting, non-erosive). Required area $A = Q_t/V$.
4. Height of tunnels $h =$ (regulator crest level $-$ undersluice crest level) $-$ roof slab thickness. Total tunnel width $= A/h$.
5. Number and width of tunnels. Adopt 2-4 tunnels with 0.5-0.75 m thick piers so that the total width equals the bay of the undersluice.
6. Length of tunnels equals the width of the head regulator plus the transition, with a gently sloping floor to improve flow.
7. Hydraulic checks. The head causing flow comes from the pond to the tailwater in the river (difference of levels), checked by the orifice formula $Q = C_dA\sqrt{2gH}$.
8. Structural design. Roof slab, piers and floor are designed for water load; the tunnel floor, piles and cut-offs are checked for uplift.
9. Entrance and exit. Wing walls with smooth transitions, so that the silt does not enter the canal; d/s outlet to the sluiceway with an apron.

 SECTION
  pond level ~~~~~~~~~~~~~~~~~~~~~~~~~~~
  regulator crest ____________  (roof of tunnel)
  ---> clear water  |  roof slab  |--> canal
  silty water ---> [  tunnel  ]---> to river
  undersluice crest level  ____________

Given and assumptions

Canal discharge $Q = 200$ m³/s; undersluice crest level (tunnel floor) $= 225.25$ m; head regulator crest level $= 227.35$ m (tunnel roof level); bay width of the undersluices $= 10.2$ m.

Assumed: discharge through the tunnels = 20% of the canal discharge; roof slab thickness 0.3 m; tunnel velocity about 2.5 m/s (non-silting); pier thickness 0.6 m.

Design

1. Discharge in tunnels

2. Height of tunnels

Difference of crest levels $= 227.35 - 225.25 = 2.10$ m. Subtract the roof slab, $0.30$ m:

3. Area of tunnels

4. Total width of tunnels

5. Number of tunnels. Providing 3 tunnels of 3.0 m each and 2 piers of 0.6 m:

which fits exactly in the bay width of 10.2 m. Area provided $= 3\times 3.0\times 1.80 = 16.20$ m².

6. Velocity check

This is within the range 2.5-3 m/s, so silt will be carried through the tunnels.

Section

 regulator crest 227.35 -----------------------
        roof slab (0.30)  ===================
        tunnel height 1.80 |  |  |  |  |  |
 undersluice crest 225.25 -----------------------
 3 tunnels x 3.0 m + 2 piers x 0.6 m = 10.2 m

Answer: 3 tunnels, each 3.0 m wide and 1.80 m high (roof 227.35 m, floor 225.25 m), piers 0.6 m, carrying 40 m³/s at about 2.47 m/s.

• Divide wall. A wall between the weir and the undersluices (and the head regulator side). It separates the quiet pocket in front of the regulator from the weir, prevents cross currents, protects the sluices from high-velocity flood flow, and also separates the fish ladder.
• Undersluices. Low-level gated bays beside the head regulator. They scour the silt that settles in the pocket, keep a clear channel in front of the regulator, and pass the low floods.
• Silt excluder. Tunnels built under the head regulator in the pocket. They let the clear upper water go into the canal and pass the silty bottom layer through the undersluices to the river, so the canal water has less bed load.
• Silt ejector (extractor). A device on the canal, a little below the regulator, which removes the silt that has entered the canal. Tunnels carry the bottom layer to a escape channel and back to the river.

A fish ladder is a sloping channel with a series of baffles or steps and pools, built along the divide wall (on the weir side or the river bank side) as shown in the plan below.

 Left  <------ river flow ------>
 bank  ________________________________
  |    | weir bays |DW| sluices |..| head
  |    |           |  | pocket     | reg
  |    |  fish ladder  <-- along DW |
  |____|__________________________|

Necessity

• A weir or barrage across a river blocks the migration of fish, which swim upstream to spawn (e.g. mahseer and snow trout in Nepali rivers). The ladder allows them to pass through the structure.
• Fish need a velocity they can handle, usually below 2-3 m/s; the baffles and pools reduce the velocity and give them resting places.
• It keeps the aquatic ecology and fish population of the river, which provides livelihood and is required in environmental clearance (EIA) of the headworks.
• The ladder is usually 1:10 to 1:15 in slope, with entrance at the lowest point below the weir and exit at the pond level.

The design discharge of the undersluices is fixed on the basis of the scouring needs:

• It should be at least twice the canal discharge (to give enough scouring velocity in the pocket), or
• 10 to 15% of the maximum flood discharge of the river, or
• the dry season (low flood) discharge that the river carries,

whichever is the largest. Sometimes the design discharge is the flow that passes when the pond level is kept and the weir crest is not overtopped. The head is the difference between the pond level and the d/s level at that discharge. Then the waterway of the undersluices is found from the discharge formula for free-flow $Q = C\,L\,H^{3/2}$.

1. Length of waterway (total)

The regime width of the river is found from Lacey's wetted perimeter

where $Q$ is the maximum flood discharge (m³/s). The total waterway of the headworks is taken as $L = 1.0$-$1.2\,P$ (Lacey's width), but not more than the river width if the afflux is allowed. Subtract the sum of piers and bay widths: effective waterway $L_{eff} = L -$ (number of bays $\times$ pier thickness) and also check that afflux (0.5-1.2 m) is within the permissible limit.

Distribution: undersluices occupy about 10-20% of $L$ (sluice bays of 5-8 m); the rest is the weir. The canal head regulator width follows the canal discharge (about 1.5 times canal bed width).

2. Crest levels

• Weir crest level $=$ canal full supply level (FSL) $+$ losses in head regulator (0.3-0.5 m) $+$ pond depth required for command, or crest $=$ FSL $+$ head regulator loss $+$ (depth needed for the weir crest to pass the canal discharge). Pond level (FSL + regulator losses) usually decides it; afflux is determined with the weir equation $Q = C L H^{3/2}$ ($C \approx 1.7$ for a broad crest).
• Undersluice crest level $=$ average river-bed level at the site (or about 1.2-2.0 m lower than the head regulator crest).
• Head regulator crest level $=$ undersluice crest $+ (1.2$-$2.0)$ m, or $\ge$ pond level $-$ canal depth; it must be above the sluice crest to keep bed silt away.
• Weir crest is at the level of the pond (permanent crest) $-$ height of shutters (0.9-2.0 m) if used.

HFL (after construction) $=$ HFL (before construction) $+$ afflux, and the head over the weir and sluices is checked with the discharge formulae.

Why Khosla's theory is more appropriate

• Bligh's theory assumes that the head loss is proportional to the creep length (a straight-line hydraulic gradient), gives equal weight to horizontal and vertical creep, and does not give the exit gradient or the pressure at key points. Lane's theory weights the vertical creep three times the horizontal, but also is empirical.
• Khosla's theory solves Laplace's equation for the flow of seepage water below the structure with the real boundary conditions (potential flow). It gives:
  • the pressure at any key point (not a straight line), so uplift and floor thickness can be designed properly;
  • the exit gradient $G_E$, which is the criterion for piping safety;
  • the effect of the depth and position of the sheet piles (that pile at the downstream end is most effective for exit gradient, and the pile at the upstream end for uplift);
  • allowance for complex profiles via the simple profiles and correction for mutual interference, floor thickness and slope.
• Hence it represents the real seepage flow and is more rational and economical.

Limitations of the mutual interference correction

1. It is empirical (based on electrical analogy tests) and approximate; it is valid only when the piles are not too close to each other.
2. It applies only to the pile of smaller depth $d$ due to the influence of a deeper pile $D$; if the interfering pile is shorter than $d$, no correction applies.
3. It accounts for the effect of two adjacent piles only; for three or more piles, it is applied pairwise, so the result is approximate.
4. Piles are assumed vertical, thin, with a horizontal floor of negligible thickness; foundation assumed homogeneous, isotropic, and of infinite depth.
5. The effect varies with the spacing $b'$, so in the formula $b'$ must be taken as the distance between the pile lines, and $D$ and $d$ are measured from the same level; the correction is +ve for the upstream pile (increase pressure) and -ve for the downstream (decrease).

Khosla's method for the design of a barrage floor

Khosla's theory gives the pressure and exit gradient from the flow of seepage water below the floor. The floor is designed in the following steps.

Step 1: Data. Maximum flood discharge, HFL, pond level (PL), d/s tail water level, bed level, silt factor $f$, safe exit gradient $G_E$ (1/5 to 1/7 for sand), specific gravity of floor material $G = 2.4$.

Step 2: Scour depth. Lacey's $R = 1.35\,(q^2/f)^{1/3}$, with $q$ = discharge per metre width. Use it for the cutoff depths: u/s cutoff $\ge 1.0R$-$1.5R$ below HFL and d/s cutoff $\ge 1.5R$-$2R$ below the d/s HFL.

Step 3: Maximum static head $H$ = PL $-$ d/s bed (or floor) level with no flow (usually nil tail water); and under flow with HFL.

Step 4: Exit gradient and length of floor.

Taking $d$ as the d/s cutoff depth, find $\alpha$ (hence floor length $b$) such that $G_E \le$ safe value. Provide the total floor length with some margin.

Step 5: Pressures at key points. Compute $\phi_E,\phi_D,\phi_C$ for each pile line:

• u/s pile: $\phi_{E_1} = 100$, $\phi_{D_1} = 100-\phi_D$, $\phi_{C_1} = 100-\phi_E$.
• d/s pile: $\phi_{E_2} = \phi_E$, $\phi_{D_2}=\phi_D$, $\phi_{C_2}=0$.

Apply corrections: (a) mutual interference of piles $C = 19\sqrt{D/b'}\,(d+D)/b$; (b) floor thickness; (c) slope of the glacis.

Step 6: Floor thickness. Uplift head at a point $h$ (from corrected $\phi\times H$). Required thickness

The thickness is greatest near the toe of the glacis and is decreased in steps (min 0.5-0.6 m).

Step 7: Floor details. The d/s floor length is fixed by the hydraulic jump, and the thickness of the u/s floor is provided with nominal values. Provide the u/s and d/s aprons (blocks with inverted filter) of length 1.5-2 times the scour depth, with launching apron.

Step 8: Check pressure line (HGL), exit gradient, and thickness for both conditions (no flow and flow).

Typical section of a barrage

  Pond level ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        gates (piers)             HFL ~~~~~
 u/s floor |   crest  |\ glacis | jump floor | d/s apron
 ==========|__________|  \______|============|== blocks
   u/s block apron    |         |      filter, launching
 u/s pile (cutoff)  intermediate pile     d/s pile (cutoff)

Components: gates and piers, crest, u/s floor, glacis, horizontal d/s floor (jump), u/s and d/s cutoffs, intermediate piles, aprons with filters and launching aprons, u/s and d/s wing walls.

Bligh's creep theory

Water seeping below the structure follows the contact between the base and the subsoil (creep path). The length of creep $L$ is the sum of the horizontal creep (equal to the horizontal lengths of the contact) and vertical creep (the lengths of the cutoffs, with each counted twice as it goes down and up):

The head loss is assumed to be proportional to the creep length along the path (straight hydraulic gradient line).

   H ~~~~~~~~~|  ____________   ~~~~~~~ (d/s)
   u/s    |   |  |  floor   |  |
   ===+===|===+==+==========+==+====
     pile 1                 pile 2
   point A (start) ... point B (end of creep)
   HGL: straight line from H at A to 0 at B

Safety against piping

with Bligh's coefficient $C$ (e.g. 12 for fine sand, 8 for coarse sand, 5 for boulders and gravel).

Pressure along the foundation

The head loss up to a point is proportional to the creep length up to it:

where $h_x$ is the residual (uplift) pressure head at the point and $L_x$ the creep length from the upstream end to it. The uplift pressure on the floor is $\gamma_w h_x$ (in kN/m²).

Safety against uplift

Let $t$ be the floor thickness and $G$ the specific gravity of the floor material. Safe when the weight of the floor exceeds the uplift:

Usually a 33% increase for safety, i.e., $t = \dfrac{4}{3}\dfrac{h_x}{G-1}$. If the thickness is not enough, thicken the floor or add u/s length or pile, or lengthen the floor.

Four simple Khosla profiles

Khosla's equations exist for simple forms to which a complex weir profile is broken down:

(a) Pile at the upstream end    (b) Pile at the downstream end
 ~~~|                            ~~~~~~~~~~~~~~|
 ___|_______________              _____________|__
    |                                          |
    d  (E1,D1,C1)                              d (E2,D2,C2)

(c) Depressed floor             (d) Intermediate pile
 ~~~ __     ___________          ~~~  ______|_______
 ___/  \___/                     ___/       |
        |                                    d

(a) a horizontal floor with a pile at its upstream end; (b) a horizontal floor with a pile at its downstream end; (c) a depressed floor (floor lowered with an upstream and downstream pile); (d) a floor with an intermediate pile. Their pressures at the key points (E, D, C) are given by Khosla's formulas in terms of $\alpha = b/d$ and $\lambda$.

Corrections suggested by Khosla

1. Correction for mutual interference of piles – when one pile of depth $d$ is influenced by another of depth $D$ at distance $b'$:

positive for the pile upstream of the interfering pile and negative for the pile downstream.

2. Correction for thickness of the floor – the key points are at the floor top while the uplift acts at the underside; pressure is interpolated linearly between the points at the pile tip and at the floor level:

3. Correction for the slope of the floor – a sloping floor (glacis) differs from the horizontal floor of the simple profile; a correction (from a table, e.g. 6.5% for a slope 2:1 and 3.3% for 4:1) times the ratio of the slope length to the total floor length is applied: positive for a down slope and negative for an up slope in the direction of flow.

Design of the weir bay of a sloping-glacis weir (Garg / Punmia style).

Step 1: Data. Maximum flood $Q$, HFL, pond level (PL), river bed levels, silt factor $f$, afflux, safe exit gradient.

Step 2: Waterway and discharge intensity. Total waterway $L$ (from Lacey's perimeter, $P = 4.75\sqrt Q$), less the width of the undersluices and piers, gives the weir bay length. Discharge per metre $q = Q/L$.

Step 3: Crest level. Fixed from the pond level and head required to pass the design flood over the crest: $Q = \frac{2}{3}C_d\sqrt{2g}\,L\,H^{3/2}$ (crest level $=$ HFL after afflux $-$ $H$); also crest $=$ canal FSL $+$ regulator losses.

Step 4: Regime (scour) depth. $R = 1.35\,(q^2/f)^{1/3}$.

Step 5: Downstream floor level and length.

• Total energy $E_L =$ (u/s TEL) $-$ (d/s bed or floor level); $H_L =$ (u/s TEL $-$ d/s TEL) and use Blench's curves (or Crump's) to find $E_2$ (the conjugate depth after the jump) for the given $q$ and $H_L$.
• d/s floor level $=$ d/s HFL $-$ (d/s depth required for the jump) $=$ TWL $- \,E_2$ (with a safety margin of 0.5-1 m).
• Length of the horizontal floor $= 5\,(E_2 - E_1)$ (about $5\times$ height of jump) – the glacis length is included with a slope of 3:1.

Step 6: Total length of floor and cutoffs.

• Total length from Bligh's/Khosla's criterion: Bligh $L = C\,H$ with $H$ = max static head; or by Khosla's exit gradient.
• U/s cutoff depth $= 1.0R$-$1.5R$ below u/s HFL; d/s cutoff depth $= 1.5R$-$2R$ below the d/s bed.

Step 7: Thickness of impervious floor. Check Khosla's pressure at key points and take the thickness

at the toe of the glacis (max thickness), provide about 0.6 m at the ends. Taper thickness along the floor. The d/s floor under the jump should also resist the dynamic forces.

Step 8: Protection works. U/s and d/s aprons (blocks and inverted filter) with launching aprons of length $1.5$-$2\times$ scour depth; and check exit gradient $G_E \le$ safe value.

Data and assumptions

$Q = 3200$ m³/s, HFL before construction $= 185.0$ m, d/s bed $= 178.0$ m, canal FSL $= 184.0$ m, afflux $= 1.0$ m, $C = 12$, $f = 1.0$, safe exit gradient 1/6, regulator loss $= 0.5$ m, weir coefficient $1.7$, retrogression $0.5$ m. The u/s bed is assumed equal to the d/s bed (178.0 m).

1. Length of waterway (Lacey)

2. Levels and crest

• Pond level $=$ FSL $+$ regulator loss $= 184.0 + 0.5 = 184.5$ m.
• HFL after construction $= 185.0 + 1.0 = 186.0$ m.
• Discharge per metre $q = 3200/270 = 11.85$ m³/s/m.
• Head over the crest from $Q = 1.7\,L\,H^{3/2}$:

• Crest level (open flood) $= 186.0 - 3.65 = 182.35$ m. The pond level of 184.5 m is held by shutters or gates of height $184.5 - 182.35 = 2.15$ m above this crest.

3. Scour depth and pile depths

• U/s pile bottom $= 186.0 - 1.5R = 175.47$ m, so depth below bed $d_1 = 178.0 - 175.47 = 2.53$ m, adopt 2.5 m.
• D/s pile bottom $= 185.0 - 2R = 170.96$ m. The bed after retrogression is $178.0 - 0.5 = 177.5$ m, so $d_2 = 177.5 - 170.96 = 6.54$ m, adopt 6.5 m.

4. Length of impervious floor (Bligh)

Maximum static head $H_s = 184.5 - 177.5 = 7.0$ m. Total creep length

Adopt a horizontal floor length $b \approx 66$ m between the piles.

5. Exit gradient check

$G_E = 0.145 < 1/6 = 0.167$, so the floor is safe against piping.

Answer: Waterway $\approx 270$ m; crest level $\approx 182.35$ m (shutters to 184.5 m); u/s pile $\approx 2.5$ m deep; d/s pile $\approx 6.5$ m deep; impervious floor $\approx 66$ m between piles (total creep 84 m); exit gradient $= 0.145$ (safe).

Data and assumptions

$Q = 4000$ m³/s, river width $300$ m, Bligh's $C = 12$, Lacey's $f = 1.2$, weir height above low water $3.0$ m, crest width $2.0$ m. Assumed: river bed (low water) level $= 100.0$ m, broad-crested weir with $C_d = 1.7$ ($Q = 1.7LH^{3/2}$), afflux $= 1.0$ m, specific gravity of concrete $G = 2.4$.

1. Waterway and flood levels

Crest level $= 100.0 + 3.0 = 103.0$ m. Discharge per metre $q = 4000/300 = 13.33$ m³/s/m.

2. Scour depth and cutoffs

• U/s pile bottom $= 106.95 - 1.5R = 96.23$ m, so $d_1 = 100 - 96.23 = 3.77$ m, adopt 3.8 m.
• D/s pile bottom $= 105.95 - 2R = 91.66$ m, so $d_2 = 8.34$ m, adopt 8.3 m.

3. Floor length by Bligh's principle

Static head $H = 103.0 - 100.0 = 3.0$ m. Creep length $L_c = CH = 12\times 3 = 36$ m.

4. Khosla's check and design

Adopt a floor length $b = 12$ m between the piles (the crest width 2 m and the glacis fit in this length), pile depths 3.8 m (u/s) and 8.3 m (d/s), floor thickness 1.0 m (u/s) and 1.5 m (d/s) provisionally.

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 12$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 3.00$ m (no flow, pond at crest).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken straight between the corrected key points along the floor).

Exit gradient

Permissible $= 0.1667$. Safe (less than the permissible value).

Required thickness shown in the table is computed at the key points; adopt the larger of the required and the nominal thickness, and taper the thickness between the two ends. Add u/s and d/s aprons (block apron with filter, length $\approx 1.5$-$2$ times the scour depth, say about 15 m).

Sketch

  HFL u/s ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  Crest 103.0  _____ 2.0 m
 u/s apron    |     |   vertical drop     d/s apron
 ============ |_____|=========================
 100.0 bed    |<------ b = 12 m ------>|
   pile 3.8 m                      pile 8.3 m

Answer: Waterway 300 m; crest 103.0 m; u/s cutoff 3.8 m; d/s cutoff 8.3 m; Bligh floor length $\approx 12$ m, adopted 12 m (checked by Khosla: exit gradient 0.098 $< 1/6$).

Assumptions

Afflux $= 1.0$ m, loss through head regulator $= 0.5$ m, weir coefficient $C = 1.7$ ($Q = CLH^{3/2}$), average bed level $= 200.0$ m, undersluice discharge intensity about 10 m³/s/m.

1. Waterway (Lacey)

Discharge intensity $q = 16000/600 = 26.67$ m³/s/m; scour depth $R = 1.35(q^2/f)^{1/3} = 12.05$ m.

2. Levels

• HFL before construction $= 205.7$ m; HFL after $= 205.7 + 1.0 = 206.7$ m.
• Pond level $=$ FSL $+$ regulator loss $= 203.0 + 0.5 = 203.5$ m.
• Head needed to pass the flood over the full waterway: $H = (16000/(1.7\times 600))^{2/3} = 6.27$ m.
• Weir (open-flood) crest level $= 206.7 - 6.27 = 200.43$ m, say 200.4 m (only about 0.4 m above the bed). Therefore the structure is a gated barrage: gates (3.1 m high) retain the pond at 203.5 m.
• Undersluice crest level $=$ river bed $=$ 200.0 m.
• Head regulator crest level $=$ sluice crest $+$ 1.5 m $=$ 201.5 m (the canal FSL 203.0 m gives 1.5 m depth over the crest).

3. Waterway distribution

Undersluice discharge $= 2\times$ canal discharge $= 400$ m³/s, width $= 400/10 = 40$ m, taken as 5 bays of 8 m with 4 piers of 1.5 m: $40 + 6 = 46$ m. Weir bays: $600 - 46 = 554$ m. Head regulator width about 30 m.

Answer: Total waterway 600 m (undersluices 46 m, weir 554 m); weir crest $\approx 200.4$ m (gates raise pond to 203.5 m); undersluice crest 200.0 m; head regulator crest 201.5 m.

Data and assumptions

• Pond level 1135.0 m. For the no-flow condition the d/s water level is taken at the d/s floor level 1123.75 m, so the head $H = 1135.0 - 1123.75 = 11.25$ m.
• The u/s sheet pile (bottom 1120.5 m) is 6.0 m deep below the u/s floor (1126.5 m). The d/s pile (bottom 1118.5 m) is 5.25 m deep below the d/s floor (1123.75 m). Floor length between the piles $b = 40$ m.
• Floor thickness 1.0 m throughout (given); specific gravity of concrete $G = 2.4$.
• Slope correction for 1:3 slope $= 4.5\%$, multiplied by (slope length / $b$): u/s slope length $1.5$ m (up the slope, subtract at $C_1$); d/s glacis length $9.75$ m (down the slope, add at $E_2$).

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 40$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 11.25$ m (no flow).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken straight between the corrected key points along the floor).

Exit gradient

Permissible $= 0.2000$. Not safe; increase the d/s pile depth or lengthen the floor.

Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):

 E1  100.0% |########################################
 D1   76.6% |###############################
 C1   67.6% |###########################
 E2   29.0% |############
 D2   20.0% |########
 C2    0.0% |

The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C of one pile to E of the next).

Conclusion: The thickness required at each location is the "Required t" column (to be compared with the 1.0 m provided). Where the required thickness is greater than 1.0 m, the floor must be thickened (or the u/s floor lengthened and the d/s pile deepened) to resist uplift. The exit gradient is compared with the safe value 1/5 above. The pressure line is drawn from the table in the HGL block.

Data and assumptions

The figure details are only partly legible, so the following reading is adopted: head $H = 5$ m; total floor length $b = 35$ m between the end piles; u/s pile 3 m deep with floor thickness 1 m; d/s pile 6 m deep with floor thickness 2 m (floor top flush with the bed). The inside corners are $C_1$ (floor and u/s pile) and $E_2$ (floor and d/s pile).

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 35$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.00$ m (head H = 5 m).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Exit gradient

Permissible $= 0.1667$. Safe (less than the permissible value).

Answer: Uplift at the u/s inside corner $C_1 \approx 78.7\%$ of $H$ = 3.93 m; at the d/s inside corner $E_2 \approx 32.4\%$ = 1.62 m; exit gradient $G_E = 0.143$.

Data and assumptions

Corrected pressures: $\phi_{C1} = 82\%$ (u/s pile junction), $\phi_{C2} = 35\%$ (d/s pile junction). Full reservoir level 505 m, cistern floor 499 m, no tail water, so the head is $H = 505 - 499 = 6$ m. Floor length between piles $b = 40$ m, cistern length 15 m at the d/s end, $G = 2.4$.

Pressure along the floor

The HGL is a straight line between the junctions. The cistern starts $40 - 15 = 25$ m from the u/s pile:

Uplift heads above the cistern floor level:

Thickness of the cistern floor

Exit gradient

The depth of the d/s pile is found from $\phi_{C2} = \phi_E$ of the d/s pile (the junction pressure):

Answer: Minimum cistern floor thickness $\approx 2.26$ m at the beginning (u/s end of the cistern) and $\approx 1.50$ m at the end; exit gradient $\approx 0.156$ (about 1 in 6.4), below the usual safe value of 1/6.

The figure is not available, so the weir profile of the standard case (pond 102.50 m, 2:1 slopes, three pile lines) is used to show the method.

Data and assumptions

The figure is read as follows. Pond level 102.50 m; with no flow the d/s water level is taken at the d/s floor level 97.00 m, so $H = 5.50$ m. Piles (positions measured from the u/s pile): u/s pile at 0 m, floor level 100.00 m, bottom 94.00 m ($d = 6$ m); intermediate pile at 15 m, floor level 99.00 m, bottom 94.00 m ($d = 5$ m); d/s pile at 45 m, floor level 97.00 m, bottom 92.00 m ($d = 5$ m). Total floor length $b = 45$ m. Floor thickness is not given, so the thickness correction is neglected. The 2:1 slope correction is 6.5%, applied in proportion to the slope length / $b$: d/s glacis (101.00 to 98.00 m, 6 m long, down slope: $+6.5\times 6/45$ at $E_3$) and u/s slope (2 m, up slope: $-6.5\times 2/45$ at $C_1$).

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 45$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.50$ m (no flow).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken straight between the corrected key points along the floor).

Exit gradient

Permissible $= 0.2000$. Safe (less than the permissible value).

Answer: The uplift pressures are the corrected values in the table (as % of H and in metres of water); the exit gradient is as computed above.

Data and assumptions

The figure is read as follows. Pond level 102.50 m; with no flow the d/s water level is taken at the d/s floor level 97.00 m, so $H = 5.50$ m. Piles (positions measured from the u/s pile): u/s pile at 0 m, floor level 100.00 m, bottom 94.00 m ($d = 6$ m); intermediate pile at 15 m, floor level 99.00 m, bottom 94.00 m ($d = 5$ m); d/s pile at 45 m, floor level 97.00 m, bottom 92.00 m ($d = 5$ m). Total floor length $b = 45$ m. Floor thickness is not given, so the thickness correction is neglected. The 2:1 slope correction is 6.5%, applied in proportion to the slope length / $b$: d/s glacis (101.00 to 98.00 m, 6 m long, down slope: $+6.5\times 6/45$ at $E_3$) and u/s slope (2 m, up slope: $-6.5\times 2/45$ at $C_1$).

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 45$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.50$ m (no flow).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Exit gradient

Permissible $= 0.2000$. Safe (less than the permissible value).

Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):

 E1  100.0% |########################################
 D1   77.8% |###############################
 C1   67.6% |###########################
 E2   55.9% |######################
 D2   49.5% |####################
 C2   44.6% |##################
 E3   28.6% |###########
 D3   18.7% |#######
 C3    0.0% |

The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C of one pile to E of the next).

Data and assumptions

The figure is partly unclear. Reading adopted: pond level 103 m; two pile lines with floor length $b = 15 + 20 = 35$ m. U/s pile: floor level (point C) 99.5 m, bottom 92 m ($d = 7.5$ m). D/s pile: floor level 97 m, bottom 92 m ($d = 5$ m). With no flow the d/s water level is the d/s floor level 97 m, so $H = 6$ m. Point B is on the depressed floor at 15 m (floor top 96 m, bottom 95 m, thickness 1.0 m); point A is at the d/s end (floor top 97 m, bottom 94.5 m, thickness 2.5 m). Floor thickness 0.5 m at the u/s end. Safe exit gradient 0.2.

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 35$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 6.00$ m (no flow).

Corrections: int = mutual interference of piles, thk = floor thickness, slope = slope correction.

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken straight between the corrected key points along the floor).

Exit gradient

Permissible $= 0.2000$. Safe (less than the permissible value).