Regulating Structures

Head regulator

A head regulator is a gated structure at the head (off-take) of a canal, distributary or minor, built across the off-taking channel.

Functions

1. Regulates the discharge entering the off-taking channel.
2. Controls silt entry (a high crest and a silt excluder keep the bed load out).
3. Measures the discharge (gauge and rating at the gates).
4. Shuts off the supply during floods, repairs or when no water is needed.
5. Prevents the reverse flow of water from the parent channel in case the off-take is lower.

Cross regulator

A cross regulator is a gated structure built across the parent canal, just downstream of the off-take of a branch or distributary.

Functions

1. Raises (holds up) the water level on its upstream side so that the off-taking channel gets its full supply, even when the parent canal carries less than full discharge.
2. Regulates the discharge passing to the downstream reaches of the parent canal.
3. Helps to close the canal for repairs, and provides a bridge or cross-communication.
4. Allows the supply to be diverted to the off-take in rotation (rotational running).

Sketch of section

 Longitudinal section through a head/cross regulator
  FSL u/s ~~~~~~~~~~~~~|gate|~~~~ FSL d/s
                       | |
                  _____|_|______
  u/s bed _______|  crest wall  |_______ d/s bed
    |u/s cut-off|__ floor ______|d/s cut-off|
        u/s apron   (impervious)  d/s protection:
                                  stilling basin,
                                  inverted filter,
                                  launching apron

Section of the gates: piers carry a bridge deck and a gate hoist; gates (vertical lift steel) slide in grooves cut in the piers.

Need for drop structures

The ground slope of the command is generally steeper than the permissible bed slope of the canal. If the canal followed the ground slope, its velocity would exceed the safe limit and it would scour and erode its bed. If the canal bed were kept at the permissible slope, it would run in deep cutting or lose command over the fields. A drop (fall) structure drops the water level abruptly and dissipates the surplus energy safely. It therefore:

1. Reduces the bed slope and the velocity, preventing erosion of the canal.
2. Limits cutting and filling to economical depths.
3. Maintains command (FSL above the field level).
4. Can also be used for power generation or discharge measurement.

Types of falls (with suitability)

1. Vertical drop fall (Sarda type)

 ~~~~~~~~~~|
           |   crest wall
 ______ ___|   drop
           |\___ cistern ___

Water falls vertically over a wall into a cistern in which a hydraulic jump forms. Suitable for drops up to about 1.5 m and for small and medium canals, and where the soil is good. Simple, cheap, easy to construct.

2. Glacis (sloping) fall

 ~~~~~~~~~~\
            \ glacis 2:1 to 3:1
             \__ jump on floor __

Water slides down a sloping glacis and forms a jump on the floor. Suitable for drops of 1.5 m to about 4 m and for large discharges.

3. Ogee fall

 ~~~~~~~~~~___
              ) ogee curve
              \__ cistern ___

Curved crest joined smoothly to the floor. It reduces the impact and shock; suitable for drops with high discharges.

4. Rapid fall A long sloping channel with the bed slope steeper than the normal; suitable where the ground slope is gentle and uniform, and the drop is large.

5. Stepped fall A series of small vertical drops in succession; suitable where the ground slope is steep and long and the drop is large.

6. Notch fall (rectangular or trapezoidal notch)

 ~~~~|  |~~~~   notches in the crest wall

The crest is divided into notches that are opened or closed, so the u/s depth is held constant over varying discharge. Suitable for small canals with varying supplies.

7. Montague type fall, Inglis (baffle) fall Montague has a parabolic glacis; Inglis fall has baffle wall and glacis; suitable for large drops and high discharges.

Choice depends on the drop height, the discharge, the soil type, and cost.

A canal outlet (module) is a small structure that delivers water from a distributary or minor to the field water course.

Classification

Outlets are classified by how their discharge responds to water levels in the distributary and in the water course.

1. Non-modular outlets Discharge depends on the water level in both the distributary and the water course.

• Open sluice, drowned pipe outlet.
• Simple, cheap, flexible, but discharge varies with water levels in both channels.

2. Semi-modular (flexible) outlets Discharge depends on the water level of the distributary only; it is independent of the water level of the water course provided the minimum working head is available.

• Pipe outlet (free flow), Kennedy's gauge outlet, Crump's open flume, adjustable orifice semi-module (AOSM).
• They are very common because they are cheap, and give equitable distribution.

3. Rigid modular outlets The discharge is practically constant, independent of the levels in both channels within working limits.

• Gibb's module, Khanna's module, Punjab (Fane) module.
• They give a fixed discharge, but are costly, complicated and need maintenance.

  Distributary                 Water course
 ~~~~~~ FSL ~~~~~~~       ~~~~~~ FSL ~~~~
        (A) Non-modular: q = f(H_d, H_w)
        (B) Semi-modular: q = f(H_d) only
        (C) Modular: q constant

Other types by construction

• Pipe outlet: pipe through bank, simple.
• Scoop (Adjustable) outlet: variable opening.
• Open flume outlet: Crump's.
• Adjustable orifice: roof block moves to vary opening.

Given data and assumptions

• $Q = 7.5$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 156.10$ m, u/s bed $= 155.15$ m, d/s bed $= 154.45$ m.
• Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 155.400$ m); drop in water level $H_L = 0.700$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.25$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.327\ \text{m}^2 \\ v &= Q/A = 3.222\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.529\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). With $L = B = 1.5$ m the head needed would put the crest below the u/s bed. The crest is therefore fixed at the u/s bed level and its length increased (splayed wing walls):

$$\begin{aligned} H &= (FSL_u + h_a) - \text{crest} = 156.629 - 155.150 = 1.479\ \text{m} \\ L &= \frac{Q}{1.838\,H^{3/2}} = 2.27\ \text{m} \end{aligned}$$

Height of crest above d/s bed $d_c = 155.15 - 154.45 = 0.70$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 3.307$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 2.179$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 1.076$ m. Provide $X = 1.10$ m (cistern floor level $= 153.35$ m).
• With this $X$: $y_1 = 0.443$ m, $F_1 = 3.58$, $y_2 = 2.032$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(2.032 - 0.443) = 7.94$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 155.15 - 153.35 = 1.80$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.80 + 1.48}{\sqrt{2.25}} = 2.19\ \text{m}$$

Adopt bottom width $B_w = 2.19$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{3.307/75}} = 5.08\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 7.94$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 5.19$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -8.93\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 14.13$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 11.13$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 4.39$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 4.39}{L_{act}} = 0.483$ m. At the end of the cistern the creep is 12.33 m and $h_2 = 0.089$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.483}{1.25} = 0.51\ \text{m}$$

Adopt floor thickness 0.51 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 155.15 m, crest length 2.27 m, crest wall 0.55 m top and 2.19 m bottom width, cistern depth 1.10 m and length 7.94 m, floor length 11.13 m, floor thickness 0.51 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Given data and assumptions

• $Q = 5.2$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 206.10$ m, u/s bed $= 205.15$ m, d/s bed $= 204.45$ m.
• Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 205.400$ m); drop in water level $H_L = 0.700$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.25$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.327\ \text{m}^2 \\ v &= Q/A = 2.234\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.254\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). With $L = B = 1.5$ m the head needed would put the crest below the u/s bed. The crest is therefore fixed at the u/s bed level and its length increased (splayed wing walls):

$$\begin{aligned} H &= (FSL_u + h_a) - \text{crest} = 206.354 - 205.150 = 1.204\ \text{m} \\ L &= \frac{Q}{1.838\,H^{3/2}} = 2.14\ \text{m} \end{aligned}$$

Height of crest above d/s bed $d_c = 205.15 - 204.45 = 0.70$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 2.430$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.904$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 0.683$ m. Provide $X = 0.70$ m (cistern floor level $= 203.75$ m).
• With this $X$: $y_1 = 0.367$ m, $F_1 = 3.49$, $y_2 = 1.638$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.638 - 0.367) = 6.35$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 205.15 - 203.75 = 1.40$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.40 + 1.20}{\sqrt{2.25}} = 1.74\ \text{m}$$

Adopt bottom width $B_w = 1.74$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{2.430/75}} = 4.71\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 6.35$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.74$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -6.89\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 12.09$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 9.09$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 3.94$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.94}{L_{act}} = 0.472$ m. At the end of the cistern the creep is 10.29 m and $h_2 = 0.104$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.472}{1.25} = 0.50\ \text{m}$$

Adopt floor thickness 0.50 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 205.15 m, crest length 2.14 m, crest wall 0.55 m top and 1.74 m bottom width, cistern depth 0.70 m and length 6.35 m, floor length 9.09 m, floor thickness 0.50 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Given data and assumptions

• $Q = 2.1$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 276.10$ m, u/s bed $= 275.15$ m, d/s bed $= 274.45$ m.
• Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 275.400$ m); drop in water level $H_L = 0.700$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.25$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.328\ \text{m}^2 \\ v &= Q/A = 0.902\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.041\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{2.1}{1.838\times1.5}\right)^{2/3} = 0.834\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 276.100 + 0.041 - 0.834 = 275.31\ \text{m}$$

Height of crest above d/s bed $d_c = 275.31 - 274.45 = 0.86$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 1.400$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.691$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 0.208$ m. Provide $X = 0.25$ m (cistern floor level $= 274.20$ m).
• With this $X$: $y_1 = 0.242$ m, $F_1 = 3.74$, $y_2 = 1.168$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.168 - 0.242) = 4.63$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 275.31 - 274.20 = 1.11$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.11 + 0.83}{\sqrt{2.25}} = 1.29\ \text{m}$$

Adopt bottom width $B_w = 1.29$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{1.400/75}} = 4.10\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 4.63$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.29$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -4.72\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 9.92$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 6.92$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 3.49$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.49}{L_{act}} = 0.453$ m. At the end of the cistern the creep is 8.12 m and $h_2 = 0.127$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.453}{1.25} = 0.48\ \text{m}$$

Adopt floor thickness 0.48 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 275.31 m, crest length 1.50 m, crest wall 0.55 m top and 1.29 m bottom width, cistern depth 0.25 m and length 4.63 m, floor length 6.92 m, floor thickness 0.48 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Given data and assumptions

• $Q = 1.8$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 205.95$ m, u/s bed $= 205.05$ m, d/s bed $= 204.35$ m.
• Full supply depth u/s $D_u = 0.900$ m, d/s $D_d = 0.900$ m (d/s FSL $= 205.250$ m); drop in water level $H_L = 0.700$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.2$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.900)(0.900) = 2.160\ \text{m}^2 \\ v &= Q/A = 0.833\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.035\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.8}{1.838\times1.5}\right)^{2/3} = 0.753\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 205.950 + 0.035 - 0.753 = 205.23\ \text{m}$$

Height of crest above d/s bed $d_c = 205.23 - 204.35 = 0.88$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 1.200$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.635$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 0.166$ m. Provide $X = 0.25$ m (cistern floor level $= 204.10$ m).
• With this $X$: $y_1 = 0.209$ m, $F_1 = 4.00$, $y_2 = 1.084$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.084 - 0.209) = 4.38$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.50$ m. Height of wall from cistern floor to crest $H_w = 205.23 - 204.10 = 1.13$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.13 + 0.75}{\sqrt{2.2}} = 1.27\ \text{m}$$

Adopt bottom width $B_w = 1.27$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{1.200/75}} = 3.95\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 4.38$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.27$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -4.45\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 9.65$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 6.65$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 3.47$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.47}{L_{act}} = 0.448$ m. At the end of the cistern the creep is 7.85 m and $h_2 = 0.131$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.448}{1.20} = 0.50\ \text{m}$$

Adopt floor thickness 0.50 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 205.23 m, crest length 1.50 m, crest wall 0.50 m top and 1.27 m bottom width, cistern depth 0.25 m and length 4.38 m, floor length 6.65 m, floor thickness 0.50 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Types of canal outlets (4 marks)

An outlet is a structure that feeds the water course from a distributary. By dependency of its discharge on water levels, it is:

Proportionality (2 marks)

An outlet is proportional if its discharge changes in the same proportion as the discharge in the distributary, so every outlet takes its fair share of any supply. This needs the flexibility to be one.

Flexibility (2 marks)

Flexibility $F$ is the ratio of the rate of change of outlet discharge $q$ to the rate of change of distributary discharge $Q$:

$$F = \frac{dq/q}{dQ/Q}$$

Since $Q$ and $q$ both depend on the water depth $D$ in the distributary, it is also evaluated with respect to depth as $F = \dfrac{D}{q}\dfrac{dq}{dD}$.

• $F = 1$: proportional outlet (shares any change in supply in the same ratio).
• $F > 1$: hyper-proportional (takes more than its share when supply rises).
• $F < 1$: sub-proportional (takes less than its share when supply rises).

Example: a free pipe outlet set low in the bank has $F \approx 0.5$ to 1, a modular outlet has $F = 0$, and an open sluice has $F > 1$.

Non-modular outlet

Its discharge depends on the water levels in both the distributary and the water course: $q = C\,A\sqrt{2g\,(H_d - H_w)}$ when submerged. Any change of either level changes the discharge. Examples: open sluice, drowned pipe outlet.

  distributary FSL ~~~~~|--pipe--|~~~~ water course
                       (submerged)

Semi-modular outlet

The discharge depends only on the water level of the distributary if the minimum working head is available, the water course level not influencing the flow (the jet discharges freely or the flow is made independent by a flume). Examples: pipe outlet with free discharge, Kennedy's gauge outlet, Crump's open flume, adjustable orifice semi-module.

  Crump's adjustable orifice semi-module (AOSM)
 distributary                          water course
 ~~~~~ FSL ~~~~|_roof block_|~~ flume ~~
 ______ bed ___|  orifice   |__expanding__
                free jet, hydraulic jump in flume

The roof block can be raised or lowered to set the opening; the diverging flume gives a free flow.

Modular outlet

Discharge is practically constant and independent of the water levels in both distributary and water course, within a working range. It uses a device (float, plunger or air pocket) to adjust opening automatically. Example: Gibb's module, Khanna's module.

  Gibb's rigid module
  distributary ~~~~ |  float / plunger |~~~~ water course
                    |  varies orifice   |

Modular outlets are accurate but costly and need careful maintenance, so semi-modular outlets are preferred in practice.

Head regulator (4 marks)

A head regulator is a gated structure constructed at the head of an off-taking canal, distributary or minor from a parent canal or river. It admits the required discharge into the off-take and shuts it off when needed.

Functions:

1. Regulates the discharge entering the off-take channel.
2. Controls silt entering the off-take (high crest, silt excluder).
3. Measures the discharge.
4. Closes the supply for repair, flood or when the water is not required.

Importance of outlets (2 marks)

An outlet delivers the water from the distributary to the water course (or field channel).

1. It gives the farmers a fixed and equitable share of water.
2. It allows measurement and control of the discharge to each chak (command area).
3. It makes the best use of the available supply and gives equitable distribution.

Importance of escape structures (2 marks)

An escape is a structure that disposes of the surplus water of a canal into a natural drain or river.

1. Protects the canal against overtopping and breach from sudden excess flow (heavy rainfall or sudden closure of gates).
2. Provides a means to empty the canal quickly for repairs or in an emergency.
3. Also removes silt (scouring escape) and floods entering through a cross-drainage.
4. Placed at the head of the canal, before major structures, and in the upstream of the cross regulator.

An escape is a structure built in the side of a canal to release surplus water into a natural drain or stream. It protects the canal from damage, helps to empty it for repair, and removes silt.

Functions

1. Remove excess water caused by heavy rainfall, sudden closure of the outlets or faulty operation of the head regulator.
2. Empty the canal quickly in an emergency (breach) or for maintenance.
3. Scour out silt deposited at the head of the canal (scouring escape).

Types

1. Weir type escape (surplus escape): a fixed weir wall with its crest at FSL, which automatically spills the excess over its crest (a surplus escape). Used where excess flow comes from rain.
2. Escape with sluice gates: gates at bed level that can be opened to empty the canal quickly or to flush silt. Used at the end of the canal or at silted reaches.
3. Syphon escape (a syphon spillway type): works automatically when water exceeds FSL.

Neat sketch

 Plan                          Section through escape
  canal ==>===+===>==            FSL ~~~~~|
              | escape                 ___|__ crest = FSL
              v                  bed __|    |
        ====== escape                floor, gates
        ====== channel to drain      d/s protection

• The head wall has its crest at the full supply level of the canal.
• A scour sluice at the canal bed level discharges silt and drains the canal.
• A stilling basin and protection work are provided at the toe.
• The escape is located upstream of major structures, near natural drains, and before the cross regulator.

A Sarda-type fall is a vertical-drop fall with a rectangular or trapezoidal crest, designed by Sarda's formula and a cistern, with the floor design based on Bligh's or Khosla's theory. Its design steps are as follows.

Part 1. Hydraulic design (7 marks)

Data: $Q$, u/s and d/s FSL, bed levels, bed width $B$, side slopes, soil.

1. Crest length: $L = B + 0.6D$, where $D$ is the u/s full supply depth.
2. Head over crest: for a rectangular crest $Q = 1.835\,L\,H^{3/2}$, and for a trapezoidal crest $Q = 1.99\,L\,H^{3/2}$. Solve for $H$.
3. Crest level $= FSL_u - H$.
4. Crest width: top width $= 0.55\sqrt{H + d}$ where $d$ = height of the crest above d/s bed. Base width $= (H + d)/\sqrt{G}$ for stability (wall of masonry of specific gravity $G$).
5. Cistern: find the energy available $E_1$ above d/s bed and discharge per metre $q = Q/L$. Choose the cistern depth $X$ below the d/s bed so that the conjugate depth $y_2 = D_d + X$ (the jump forms inside the cistern). Length of cistern $L_c = 5(y_2 - y_1)$.
6. Cistern level $=$ d/s bed $- X$.

Part 2. Floor and protection (6 marks)

7. Bligh's theory: total creep length $L = C\,H_L$ with $H_L$ = difference of u/s and d/s FSL. Provide cut-offs u/s ($d_1$) and d/s ($d_2$).
8. Length of floor: d/s floor length $L_d = 2.21C\sqrt{H_L\sqrt{q/75}}$; u/s floor from the balance of creep: $b_u = CH_L - 2(d_1 + d_2) - B_w - L_d$. If Khosla's method is chosen, find the floor length from the safe exit gradient: $G_E = \dfrac{H}{d}\dfrac{1}{\pi\sqrt{\lambda}}$.
9. Floor thickness: at any section $t = \dfrac{4}{3}\dfrac{h}{G-1}$ where $h$ is the residual uplift head obtained from the creep line.
10. Protection works: d/s block protection of length about $2D$ to $3D$ with an inverted filter, and u/s block protection; splayed wing walls 45° at the upstream and 1:1 or 2:1 downstream, with the wings ending in the bank.
11. Check: stability of the crest wall (no tension, base pressure within safe bearing capacity) and exit gradient.
12. Draw the longitudinal section and plan.

 FSL u/s ~~~~~~~~~|
 u/s bed __________|crest         FSL d/s ~~~~
   cut-off d1 |   |  \___cistern X___ d/s bed
   u/s floor  |wall| Lc |  d/s floor  | d2

Given data and assumptions

• $Q = 1.8$ m³/s, bed width $B = 1.2$ m, side slope 1:1, u/s FSL $= 107.00$ m, u/s bed $= 106.07$ m, d/s bed $= 105.32$ m.
• Full supply depth u/s $D_u = 0.929$ m, d/s $D_d = 0.929$ m (d/s FSL $= 106.247$ m); drop in water level $H_L = 0.750$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.2 + 1\times0.929)(0.929) = 1.978\ \text{m}^2 \\ v &= Q/A = 0.910\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.042\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.8}{1.838\times1.2}\right)^{2/3} = 0.873\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 106.997 + 0.042 - 0.873 = 106.17\ \text{m}$$

Height of crest above d/s bed $d_c = 106.17 - 105.32 = 0.85$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 1.500$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.721$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 0.289$ m. Provide $X = 0.30$ m (cistern floor level $= 105.02$ m).
• With this $X$: $y_1 = 0.255$ m, $F_1 = 3.72$, $y_2 = 1.220$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.220 - 0.255) = 4.83$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.50$ m. Height of wall from cistern floor to crest $H_w = 106.17 - 105.02 = 1.15$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.15 + 0.87}{\sqrt{2}} = 1.43\ \text{m}$$

Adopt bottom width $B_w = 1.43$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.750 = 4.50$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.750\sqrt{1.500/75}} = 4.32\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 4.83$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.43$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -4.76\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 10.26$ m $\ge 4.50$ m, safe. Total floor length $= b_u + B_w + L_d = 7.26$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 3.63$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.63}{L_{act}} = 0.485$ m. At the end of the cistern the creep is 8.46 m and $h_2 = 0.132$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.485}{1.00} = 0.65\ \text{m}$$

Adopt floor thickness 0.65 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 106.17 m, crest length 1.20 m, crest wall 0.50 m top and 1.43 m bottom width, cistern depth 0.30 m and length 4.83 m, floor length 7.26 m, floor thickness 0.65 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Given data and assumptions

• $Q = 1.55$ m³/s, bed width $B = 1.1$ m, side slope 1:1, u/s FSL $= 106.00$ m, u/s bed $= 105.07$ m, d/s bed $= 104.32$ m.
• Full supply depth u/s $D_u = 0.929$ m, d/s $D_d = 0.929$ m (d/s FSL $= 105.247$ m); drop in water level $H_L = 0.750$ m.
• Bligh's coefficient $C = 7$; specific gravity of masonry/concrete $G = 2$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.1 + 1\times0.929)(0.929) = 1.885\ \text{m}^2 \\ v &= Q/A = 0.822\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.034\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.55}{1.838\times1.1}\right)^{2/3} = 0.838\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 105.997 + 0.034 - 0.838 = 105.19\ \text{m}$$

Height of crest above d/s bed $d_c = 105.19 - 104.32 = 0.88$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 1.409$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.713$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 0.247$ m. Provide $X = 0.25$ m (cistern floor level $= 104.07$ m).
• With this $X$: $y_1 = 0.242$ m, $F_1 = 3.77$, $y_2 = 1.176$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.176 - 0.242) = 4.67$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.50$ m. Height of wall from cistern floor to crest $H_w = 105.19 - 104.07 = 1.13$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.13 + 0.84}{\sqrt{2}} = 1.39\ \text{m}$$

Adopt bottom width $B_w = 1.39$ m (not less than the top width), with the d/s face vertical or slightly battered.

Summary

Answer: crest level 105.19 m, crest length 1.10 m, crest wall 0.50 m top and 1.39 m bottom width, cistern depth 0.25 m and length 4.67 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|

Head regulator: a gated structure at the head of an off-taking canal or distributary. It regulates the discharge entering the off-take, controls the entry of silt, measures the discharge and shuts off the supply during floods or repairs.

Cross regulator: a gated structure built across the parent canal just downstream of an off-take. It raises (heads up) the water level in the parent canal so that the off-taking channel gets its full share even when the supply is low, regulates the discharge passing to the downstream reach, and allows the canal to be closed for repair.

Escape: a structure that discharges surplus water of a canal into a natural drain. It protects the canal from overtopping and breaches caused by excess rain-water or faulty operation, empties the canal in emergency or for repairs, and (scouring escape) flushes silt.

Regulating structures control the flow of water in an irrigation system. The main ones are described below.

 Parent canal ===>===[CR]===>===
                |
             [HR] v off-take (distributary)
                |  outlet --> water course --> field

Together these structures give controlled, efficient and equitable distribution.

A canal head regulator is built at the off-take of a canal from the headworks pond (u/s of the weir). A longitudinal section through the axis of the canal shows the following components.

 Pond level ~~~~~~~~|  gate  |~~~~~~~ FSL canal ~~~~~~
                    |  hoist |
                 ___|  bridge|___
 crest level ___/ piers + gates \_____
 |u/s    |  u/s    |  crest     | d/s floor |  d/s
 |cut-off|  floor  |  (raised)  | (stilling)|  protection
 |  d1   |         |            |           | inv. filter
 |       |         |            |           | + apron
 |_______|_________|____________|___________|___ canal bed
                 <----- impervious floor ----->

Components

1. Upstream cut-off (pile) and upstream floor reduce uplift and exit gradient.
2. Crest wall raised 1.2 m above the under-sluice or river bed to exclude bed load.
3. Gates and piers (vertical lift gates) with a bridge and hoist.
4. Downstream floor (stilling basin) to dissipate energy and withstand uplift.
5. Downstream cut-off.
6. Protection works: inverted filter, loose stone apron, and wing walls splayed to the canal.

Sketch of an irrigation headworks (3 marks)

              River flow --->
 left  ___________________________________  right
 bank |  Left  |  Weir/Barrage | Under- |  Right
 -----|  guide |  bays (gated) | sluice |  guide
      |  bund  |===============|========|  bund
 Head |regul-  |               | Fish   |
 reg. |ator    |<-silt excluder| ladder |
 -----+--------+---------------+--------+------
 canal                       Marginal bund

The components are the weir or barrage, undersluice (with divide wall), canal head regulator, silt excluder, fish ladder and guide bunds with marginal bunds.

Longitudinal section through the head regulator (3 marks)

 Pond level ~~~~~~~~~|G|~~~~~~~~~~ canal FSL ~~~~~
 u/s cut-off  u/s apron | crest  | d/s floor | d/s
  ___|________ _________|__gate__|___________|cutoff
 |   upstream floor     raised     stilling basin
 |                      crest      with chute blocks
 |___ impervious floor ________|_ inv. filter _|apron_

• Upstream floor: impervious, with cut-off, to reduce uplift and exit gradient.
• Regulator gates: vertical lift gates between piers on a crest raised above the river bed.
• Energy dissipaters: stilling basin/cistern with a hydraulic jump, with chute blocks and end sill, d/s of the gates.
• Protection works: inverted filter, loose stone apron and d/s cut-off to protect against scour and piping.

The design of a canal head regulator proceeds in these steps.

1. Collect data: canal discharge $Q$, river HFL, pond level, canal FSL and bed level, silt factor $f$, safe exit gradient.
2. Crest level: fix the crest 0.6 to 1.2 m above the river bed or under-sluice crest, so that bed load is excluded.
3. Waterway: head over the crest $H =$ pond level $-$ crest; $L = \dfrac{Q}{C\,H^{3/2}}$ with $C = 1.7$ (broad crest). Provide bays and piers.
4. Height of gates: pond level $+$ free-board; top of piers above HFL.
5. Scour depth and cut-offs: $R = 1.35(q^2/f)^{1/3}$; d/s cut-off to $1.5R$ below the d/s water level, u/s cut-off to $1.0R$ or less.
6. Impervious floor length: from Khosla's exit gradient, $G_E = \dfrac{H_s}{d}\dfrac{1}{\pi\sqrt{\lambda}}$, with $H_s$ the maximum static head; $\alpha = b/d$ from $\lambda$; $b = \alpha d$.
7. Floor thickness: from the uplift pressure at key points by Khosla's method, $t = \dfrac{h}{G-1}$.
8. Protection works: d/s inverted filter and launching apron, u/s block protection, wing walls.
9. Check the stability of piers and the exit gradient; prepare drawings.

A cross regulator is placed across the parent canal d/s of an off-take, and a distributary head regulator is placed at the off-take. Both are gated structures with similar design steps.

Cross regulator

1. Data: parent canal discharge $Q_p$, off-take discharge $Q_d$, u/s and d/s FSL, bed levels and widths.
2. Discharge through the regulator: $Q_{cr} = Q_p - Q_d$.
3. Afflux: $\Delta h =$ u/s FSL $-$ d/s FSL (usually 0.15 to 0.3 m).
4. Crest level: at the d/s bed (or raised 0.3 to 0.5 m if the canal is silty). Provide gates and piers.
5. Waterway: from orifice flow, $B = \dfrac{Q_{cr}}{C_d\,D\sqrt{2g\,\Delta h}}$ with $C_d \approx 0.8$, or from the weir formula $Q = 1.7LH^{3/2}$; provide bays and piers.
6. Scour and cut-offs: Lacey's $R = 1.35(q^2/f)^{1/3}$; d/s cut-off at $1.5R$ below the d/s water level and u/s cut-off about $R$.
7. Floor: length by Khosla's exit gradient for the maximum static head $H_s =$ u/s FSL $-$ d/s bed; thickness from uplift.
8. Protection works: inverted filter, stone apron.

Distributary head regulator

1. Data: distributary discharge, parent FSL, distributary FSL and bed level.
2. Sill level: at the distributary bed level (raised 0.3 m to reduce silt entry if needed).
3. Waterway: orifice flow, $B = \dfrac{Q_d}{C_d\,D\sqrt{2g\,\Delta h}}$, $\Delta h = FSL_{parent} - FSL_{dist}$; provide bays and piers.
4. Cut-offs, floor length, thickness and protection as for the cross regulator, using the head $H_s =$ parent FSL $-$ distributary bed.
5. Alignment: the head regulator is set at an angle of about 60° to 90° to the parent canal, and a cross regulator is placed d/s to hold the water level.

 Plan:   parent canal  ===>===[CR]===>===
                         \ [HR]
                          \==> distributary
 Section: gates | piers | crest | floor | cut-offs

A cross regulator (in the parent canal, downstream of the off-take) holds the water level up for the distributary. The distributary head regulator controls the supply into the distributary. Both are designed as gated structures with Khosla's floor.

Data

Parent canal: $Q = 125$ m³/s, FSL u/s 120.00 m and d/s 119.70 m, depth 3 m, bed width 50 m (u/s) and 45 m (d/s), so bed levels are 117.00 m (u/s) and 116.70 m (d/s). Distributary: $Q = 20$ m³/s, FSL 119.00 m, bed width 20 m, depth 1.8 m, so bed 117.20 m. $f = 0.8$, safe exit gradient 1/6.

A. Cross regulator (6 marks)

1. Discharge through it: $Q_{cr} = 125 - 20 = 105$ m³/s. The permitted afflux (head loss) is $\Delta h = 120.00 - 119.70 = 0.30$ m.

2. Crest and waterway: the crest (sill) is kept at the d/s bed level, 116.70 m. With the gates fully raised the opening acts as an orifice of depth $D = 3.0$ m, $C_d = 0.8$:

$$\begin{aligned} B &= \frac{Q_{cr}}{C_d\,D\sqrt{2g\,\Delta h}} = \frac{105}{0.8\times3.0\times\sqrt{2\times9.81\times0.30}} = 18.03\ \text{m} \end{aligned}$$

Provide 4 bays of 5 m = 20 m clear with 3 piers of 1.5 m; overall width 24.5 m. With 20 m the actual afflux is 0.244 m $< 0.3$ m, safe.

3. Floor: discharge intensity $q = 105/24.5 = 4.29$ m²/s, Lacey scour $R = 1.35(q^2/f)^{1/3} = 3.84$ m. D/s cut-off bottom $= 119.70 - 1.5R = 113.94$ m, so $d = 116.70 - 113.94$, adopt $d = 2.8$ m (u/s cut-off 2.0 m). Maximum static head $H_s = 120.00 - 116.70 = 3.30$ m (d/s dry):

$$\frac{1}{\pi\sqrt{\lambda}} = \frac{d}{6H_s} = \frac{2.8}{6\times3.30} \Rightarrow \lambda = 5.07,\ \alpha = 9.08,\ b = \alpha d = 25.4\ \text{m}$$

Provide floor length 26 m with 1.2 m thickness tapering to 0.8 m, u/s and d/s cut-offs, d/s inverted filter and stone apron.

B. Distributary head regulator (6 marks)

1. Discharge $Q = 20$ m³/s; head loss from the parent FSL 120.00 m to the distributary FSL 119.00 m is $\Delta h = 1.00$ m.

2. Waterway: sill at distributary bed level 117.20 m, opening depth $D = 1.8$ m, $C_d = 0.8$:

$$B = \frac{20}{0.8\times1.8\times\sqrt{2\times9.81\times1.0}} = 3.14\ \text{m}$$

Provide 2 bays of 2.0 m (4.0 m clear, pier 1.2 m; overall 5.2 m). With the gates fully open the loss is only 0.61 m, so the gates throttle the flow to give the required discharge.

3. Floor: $q = 20/5.2 = 3.85$ m²/s, $R = 3.57$ m. D/s cut-off to $119.00 - 1.5R = 113.65$ m, so $d = 3.6$ m below the floor (117.20 m). $H_s = 120.00 - 117.20 = 2.80$ m. For $G_E = 1/6$: $\lambda = 2.21$, $\alpha = 3.26$, floor length $b = 11.7$ m; provide 12 m. Floor thickness 1.0 m.

Answer: cross regulator: crest 116.70 m, clear waterway 20 m (overall 24.5 m), floor 26 m, cut-off depth 2.8 m. Head regulator: sill 117.20 m, clear waterway 4.0 m, floor 12 m, d/s cut-off 3.6 m.

 Plan: parent canal ====>====[HR]====> distributary
                       |  [CR] across parent canal
                       v       (u/s of CR: FSL 120.0)
 Section: gates | crest 116.70 | floor b | d/s cut-off d