Canal Design

Design of Canals

A network of canals is required to carry water from source ie headworks (storage reservoir or diversion weir) to destination ie fields.

Classification of canals

Based on size/source

1. Main canal

2. Branch canal

3. Major distributaries

4. Minor distributaries

5. Water course/field channel

Based on alignment

1. Watershed canal

2. Contour canal

3. Side slope canal

Based on canal surface

1. Rigid boundary canals

2. Loose boundary canals

Geometric Properties of Canal Sections

Fig. 1 depicts commonly used canal sections. Triangular sections are generally constructed for carrying small discharges. Rectangular sections are constructed for moderate discharges. For carrying large discharges rectangular sections are not preferred. This is on account of stability of side slopes. Vertical side walls require large thickness to resist the earth pressure. Trapezoidal section is better for such cases since sloping side walls require less thickness.

1 yn

m

(a)

yn

b

(b)

2

1

m yn

ϑyn

D/2

b

(c) (d)

TY 1

m

yn

X

(e)

TNGL

h21 ym

r θ θ θh1

(f)

2

3

(g)

(h)

Fig 1. Canal Sections (a) Triangular (b) Rectangular (c) Trapezoidal (d) Circular (e) Parabolic (f) Combined (g) Rounded corner trapezoidal (h) Rounded bottom triangular

For small discharges, semicircular sections are often adopted for irrigation canals. River beds, unlined canals and irrigation furrows all tend to approximate a stable parabolic shape. Therefore, unlined canals can be made more hydraulically stable by initially constructing them in a parabolic shape. Since the channel side slopes along the cross section are always less than the maximum allowable side slope at the water surface, parabolic channels are physically more stable. A lined parabolic channel has no sharp angles of stress concentration where cracks may occur. A parabolic canal is described by:

Y = aX 2

in which Y = ordinate; X = abscissa; and a = parameter. The flow area is⎡ y TA = 2 n − T / 2 ⎤ 2 8

YdX = y T = my 2

⎢ ∫ ⎥ n n

⎣ 0 ⎦ 3 3

where T = top width of the canal at the water surface (m) given by

m m 2

⎛⎠

(1 + m 2 ) − 2(1 − m cot

m

(2 2

2

⎟⎝

= T T⎜⎜ ⎟

4

T = 4myn

in which m = side slope at Y = yn. ⎡ ⎛ ⎞⎤2 2 2 1 1 ⎜ 1 1 ⎟P = ∫ ds

=(dX ) + (dY ) = 2 yn m ⎢⎢⎣ m

1 +2

+ ln⎜ +⎝ 1 + ⎥⎠⎥⎦ (1)

Approximate Perimeter: When m ≥ 1 or 0 < 4 y

approximated to

T ≤ 1, Eq. (1) can satisfactorily be

P = 4 ynm + 2 yn

3m

8 y 2= T + n

3T(2)

Geometric elements of a combined (circular bed trapezoidal) section can be defined as followsθ = cot −1 m ; sin θ =

11 + m 2

;cosθ = m 1 + m 2

y = h1 + h2

= r (1 − cosθ ) +T = 2r sin θ + 2h2 m

(T − 2r sin θ )

2m

r ⎜ T y ⎟ m= ⎛ −⎝ 2m

⎞⎠ 1 + m 2 − mh = ⎛ T − y

⎞ m1 ⎜⎝ 2m

⎟⎠ 1 + m 2

2 y − Th = 1 + m 2 − m)

2

2 1 + m 2 (1 + m 2 − m)A = A + A = T − r (1 − m cot −1 m) =

T

− m(1 − m cot −1

m)⎜ 2

T 2m − y ⎞1 2

4m m 4m⎜

1 + m 2 − m ⎟

P = 2rθ + 2h2

(1 + m 2 ) = T

m(1 + m 2 ) − 2r (1 − m cot

−1 m)m⎛ 2m − y ⎞⎟

m ⎝ 1 + m 2

− m ⎠where, A = flow area of channel section (m2); A1 = area of circular portion (m2); A2 = area oftrapezoidal portion (m2); y = depth of flow (m); h1 = depth of circular part (m); h2 = depth oftrapezoidal part (m); r = radius of bed section or circular part (m); T = top width of channel(m); m = side slope of channel (H:V) at water surface; and θ = side slope angle.

n

⎜ ⎥ n

1

2

1+

1

2

5

Table 1. Geometrical Properties of Canal Sections

Section Shape Flow Perimeter P Area of Flow A

Triangular 2 yn 1 + m 2

my 2

Rectangular

Trapezoidal

Circular

b + 2

yn

b + 2

yn

rϑ

1 + m2

byn(b + myn )yn

0.5r 2 (ϑ − sin ϑ )

Parabolic 2 yn2 ⎡

1m ⎢

1 ⎛ 1 ⎞⎤+ + ln⎜ + ⎟ 8 my 2

3

m m ⎝ m m ⎠Rounded Bottom

2 y(ϑ + cot

ϑ ) y 2 (ϑ + cot ϑ )Triangular

Rounded CornerTrapezoidal

b + 2 y(ϑ + cot

ϑ )by + y

2

(ϑ + cot ϑ )

Rigid boundary canals

Rigid boundary canals are designed for uniform flow considering hydraulic efficiency, practicability, and economy. A maximum hydraulic radius results in a section of minimum excavation area and best hydraulic design. When an open channel is constructed, the excavation and the lining constitute major cost. Obviously it is desirable to keep these costs at a minimum by adopting the most economical canal cross section.

Flow Requirements

A rigid boundary is designed to convey the required discharge under uniform flow conditions. The most commonly used uniform flow formula is the Manning equation. Manning’s equation, as an equality constraint function, will be used to provide uniform flow condition in the canal. The uniform flow rate or discharge Q (m3/s) in a canal by Manning’s

1 1= =

equation is

Q = AV = 1 AR 2 / 3

S

1 / 2 ⎛ A ⎞A

2 / 3

S 1 / 2 ⎛ A ⎞A

2 / 3

S 1 / 2

nf

n⎜

P ⎟ f

n⎜

P ⎟ 0

⎝ ⎠ ⎝ ⎠where V = mean velocity of uniform flow (m/s); R = hydraulic radius (m), defined as theratio of flow area to the flow perimeter; n = Manning’s roughness coefficient; Sf = energy slope (dimensionless); and So = bed slope of the canal (dimensionless). For uniform flow Sf = So. In the Manning’s formula all the terms except n can be directly measured. The roughness coefficient is a parameter representing the integrated effects of the channel cross-sectional

6

resistance. The selection of a value of n is subjective, based on experience and engineering judgement. Standard text books list values of n for different conditions of a canal.

Optimal Canal Section

The classical optimal section is the best hydraulic section which has the maximum flow velocity or the minimum flow area and wetted perimeter for a specified discharge and canal bed slope. Mathematically, it could be stated as

Minimize A = A( y, b, m)Subject to φ = Q − 1 A5 / 3

S 1 / 2 = φ (A, P ) = φ ( y, b, m) = 0

n P 2 / 3 0

This is a nonlinear optimisation problem with nonlinear equality constraint. Using Lagrange’s method of undetermined multipliers it can be converted into an unconstrained optimization problem in terms of an auxiliary function as

Minimize F = A + λφwhere λ is the undetermined multiplier. Since F is function of 4 variables (m, b, y, λ) theoptimal conditions are∂F = 0 ⇒ ∂A + λ ∂φ =

0

(a)

∂y ∂y ∂y∂F = 0 ⇒ ∂A + λ ∂φ = 0

(b)

∂m ∂m ∂m∂F = 0 ⇒ ∂A + λ ∂φ = 0

(c)

∂b ∂b ∂b∂F = 0∂λ ⇒ φ ( y, b, m) = 0

(d)

Eliminating λ from (a) & (b) and (a) & (c) and (b) & (c)∂A ∂φ∂y ∂m∂A ∂φ= ∂A ∂φ∂m ∂y

= ∂A ∂φ(e)

(f)∂y ∂b ∂b ∂y∂A ∂φ∂b ∂m

From (d)

= ∂A ∂φ∂m ∂b(g)

∂φ S 1 / 2 A2 / 3 ⎛ ∂P= 0 ⎜ 2 A − 5P∂A ⎞⎟∂b 3 n P 5 / 3 ⎝ ∂b ∂b ⎠

7

∂φ S 1 / 2 A2 / 3 ⎛ ∂P= 0 ⎜ 2 A − 5P∂A ⎞⎟∂y 3 n P 5 / 3 ⎝ ∂y ∂y ⎠∂φ S 1 / 2 A2 / 3 ⎛ ∂P= 0 ⎜ 2 A − 5P∂A ⎞⎟∂m 3 n P 5 / 3 ⎝ ∂m ∂m ⎠

Using these in (e), (f) and (g)∂A ∂P = ∂A ∂P(h)∂y ∂m ∂m ∂y∂A ∂P = ∂A

∂P

(i)

∂y ∂b ∂b ∂y∂A ∂P = ∂A

∂P

(j)

∂b ∂m ∂m ∂b

Trapezoidal Section

For a trapezoidal section∂A = b + 2my; ∂A =

y;

∂A = y 2 ;

and

∂P = 2 1 + m 2 ; ∂P = 1;

∂P

=2my

∂y ∂b ∂m ∂y ∂b ∂m 1 + m 2

Applying optimal condition (i)

b + 2my = y

2

1 + m 2

⇒b = 2(y

1 + m 2 − m)Applying optimal condition (j)

y2my

1 + m 2= y 2 ⇒ 1 + m

2

= 2m ⇒ m = 1 / 3

Limits on Side Slope: Many times the value of m is fixed if the side slope is chosen based on the angle of repose of material (See Table 2) for better stability or for vehicles to cross the channel during no flow periods. If restrictions are put on the canal side slope, then theoptimization problem reduces to a case of two variables (y and b) i.e. Minimize A = A( y, b)

,

Subject tohence

φ ( y, b) = 0 . Lagrange’s method for two variables (y and b) gives condition

(j)

b + 2my = y

2

Parabolic shape

1 + m 2

⇒b = 2(y

1 + m 2 − m)Since a parabolic canal is completely described by two independent variables y and m, Substituting A and P in (h)

11

2

2

2

⎜ 11 ⎜

1

⎥

8

⎛ 16 ⎞ ⎡ ⎡ ⎛ ⎞⎤⎤⎟⎛ 8 2 ⎞ ⎡ 2

⎡ 1 ⎛ ⎞⎤⎤⎟

⎜ 3

ym ⎟ ⎢4 ym⎢ln⎜ m

+1 + ⎥⎥ = ⎜ y

m 3

⎟ ⎢2m ⎢m

1 + m 2

+ ln⎜ m

+1 +⎥⎥m⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Simplifying⎡ 13 ln⎢ + 1 + 1 ⎤ =

11 + 1

⎣ m m 2 ⎦ m m 2

Solving by trial and error

m* = 0.514

where superscript * denotes optimum value.

Approximate Perimeter : Substituting the values in (h)⎛ 16 ⎞ ⎡ 2 y

2 ⎤ ⎛ 8 2 ⎞ ⎡ 2 2 ⎤⎜ ym ⎟ ⎢ (6m − 1)⎥ = ⎜ y ⎟ ⎢ (6m + 1)⎥⎝ 3Simplifying

m* = 1

⎠ ⎣ 3m

2 = 0.707

⎦ ⎝ 3

⎠ ⎣ 3m ⎦

Table 2: Recommended Side slopes by IS 10430:2000

S. No. Type of SoilSide Slopes (Horizontal : Vertical)

in Cutting in Embankment

iVery light loose sand to average sandy soil

2:1 to 3:12:1 to 3:1

ii Sandy loam 1.5: 1 to 2: 1 2 : 1

iii Sandy gravel/murum 1.5 : 1 1.5 : 1 to 2: 1

iv Black cotton 1.5 : 1 to 2.5: 1 2: 1 to 3.5 : 1

v Clayey soils 1.5: 1 to 2: 1 1.5: 1 to 2.5: 1

vi Rock 0.25: 1 to 0.5: 1 0.25: 1 to 0.5: 1

Combined Section

Applying optimal conditions on a combined section

T = 2r

Solving for m

1 + m 2

;

T = 2r 1 + m 2

;⎟⎞ h⎜m

9

m* = 0 ;

θ * = π 2

This condition implies that the resultant optimal section is a semicircular section. SolvingManning Eq. with this condition

y * = r * = T *

Therefore

2 = 1.004(Qn / S )3 / 8

A* = (2π )1 4 (Qn

/

S )3 / 4 ; P * = (2π )5 8 (Qn /

S )3 / 8 ≈ 3.1542(Qn

/

S )3 / 8

Limits on Side Slope: the optimization condition for two variables (y and T) gives

T = 2r 1 + m 2

This implies that

y * = r * ;⎛

h* = y⎜1 − ⎟ * = y m1 2⎝ 1 + m 2 ⎠ 1 + m 2

A* = y 2 (m + cot −1 m) = y 2 (θ + cot θ );

P * = 2 y(m + cot −1 m) = 2 y(θ + cot θ )This is the recommended section by Bureau of Indian Standards as shown in Fig 1(h).

Table 3 Limiting Velocities for Channel Surfaces

Channel surfaces Limiting Velocity (m/s)

Sandy soil 0.3 – 0.6

Black cotton soil 0.6 – 0.9

Muram and hard soil 0.9 – 1.1

Firm clay and loam 0.9 – 1.15

Gravel 1.0 – 1.25

Boulder 1.0 – 1.5

Disintegrated rock 1.3 – 1.5

Brunt Clay Tile 1.5 – 2.0

Concrete Tile 2.0 – 2.5

Concrete 2.5 – 3.0

Hard rock 3.0 – 4.0

⎜

10

Design of Loose Boundary Canals

If velocity is low silting will take place and if velocity is high scouring will happen. Therefore design should ensure nonsilting and nonscouring condition in the canal. Design methods

1. Silt theories

a. Kennedy’s method

b. Lacey’s method

2. Tractive force method

Silt Theories

Also known as regime theory. Water carries silt and there is equilibrium between silt load and bed material and hence the canal is nonsilting and nonscouring. The shape of canal section is Trapezoidal in regime canals. In long run the canal side slope acquire 0.5:1.

Kennedy’s method

Nonsilting and nonscouring velocity V = 0.55 CVR y

0.64

Critical velocity ratio = 0.7 (fine bed material) to 1.3 (coarse bed material)

Kutter’s equation is used as uniform flow equation.

1 + 23 + 0.00155

V = n S RS1 + ⎛ 23 + 0.00155 ⎞ n⎟

⎝ S ⎠ R

The side slope is assumed 0.5:1. Still there is shortage of one equation, so either b/y ratio or bed slope S is assumed. For design Wood’s table or Garrets chart is used. Shortcomings – one equation is short so no unique solution and depend upon Kutter’s equation so drawbacks of Kutters equation also involve in Kennedy’s method.

Lacey’s method

Lacey gave Four equations

1. Characteristics of silt f = 1.76 d 50

2. V 2 = 2 fR

5

3. Af 2 = 140V 5

4. Regime flow equation for bed slope V = 10.8R 2 / 3 S 1 / 3

2

) 2

⎜

2

( ⎟

0 ⎝ ⎠

11

In this method sufficient equations are available so all the parameters can be expressed in terms of given/known Q and f

Velocity: multiply both sides of Eq. (3) by V, use AV = Q and then solve for velocity

V = (Qf 2 140)1 / 6

Wetted perimeter: Eliminate f between Eqs (2) and (3), use R = A/P and then P = 4.75 Q

Hydraulic radius: Hydraulic radius is approximately = depth of flow = normal scour depth during flood. Maximum scour depth = 1.5 R to 2 R below HFL. Substitute V into Eq (2) to

5 V 2 1 / 35 ⎛ Qf ⎞ 1 / 3⎛ Q ⎞

get R = = ⎜ ⎟ = 0.481⎜ ⎟ . Alternatively rivers are wide so P ≈ b and2 f 2 f ⎝ 140 ⎠ ⎝ f ⎠

R ≈ y ; so q = Q / P = Q 4.75 Q ⇒ Q = (4.75q)2and hence

1 / 3⎛ q ⎞ 1 / 3⎛ q ⎞R = 0.481⎜⎝ 4.75

f

⎟ = 1.35⎜ ⎟ .⎠ ⎝ f ⎠Canal bed slope: Eliminate V between Eqs (2) and (3), substitute R and then solve for S ie2

fR = 10.82 R 4 / 3 S 2 / 3 ⇒ S = f 3 / 2 = f 5 / 3 f 5 / 3=5 4980R1 / 2 3340Q1 / 6 5813q1 / 3

Tractive Force Method

It is assumed that canal water is clear so no problem of silting. To avoid scouring the velocity should be less than critical (incipient motion) velocity. For details see Appendix. There are two cases (1) Canal shape is trapezoidal and (2) Canal shape is corresponding to maximum efficiency.

Trapezoidal canal: Incipient motion condition (drag force = resistance force) on bed and side slope along with Manning’s equation can be used to find canal dimensions.

Stable hydraulic section for maximum efficiency: In this case incipient motion condition prevails throughout the perimeter. The relationships for this condition can be derived as per Appendix. Resultant shape of the stable canal is cosine curve as⎛ tan φ ⎞

y = y0 cos⎜ x ⎟⎝ y0 ⎠⎛ T ⎞ T π yAs y = 0 at x = T/2 so 0 = y cos⎜ tan φ ⎟ ⇒ tan φ = π

hence top width T = 0 ; Area

T / 2

⎜ y0⎟

2

y0 2 2T / 2

tan φ2Ty

A = 2 ∫ ydx = 0 = 2 y0 and perimeter P = 2 ∫ 1 + (dy / dx)2 dx = 2 y0 E(sin φ ) where

0π tan φ 0 sin φ

E(sin φ) is complete elliptical integral of second kind.

V nV

12

Most efficient stable channel can carry a particular discharge only, if the design discharge is less or more than that then canal area at centre section should be removed or added by maintaining same velocity.

Canal Section Design steps

(A) Design steps for Lined CanalsGiven data Q, S (based on topography), and n & Vper (See Table 3: depending upon

type of lining material)

1. Consider shape of canal and other condition (if given);

2. Use optimal conditions for the shape and condition of Step 1.

3. Calculate y from Manning’sQn =

SA5 / 3

P 2 / 3

4. Find other parameters from optimal & known conditions and then A and P.

5. Compute V = Q / A If V < Vper then optimal design is OK.

6. If V > Vper then optimal design is not possible, so provide nonoptimal section.

Q ⎛ S3 / 2⎞ ⎜ ⎟7. For nonoptimal section one condition is Mannings Eq P = ⎜

per ⎝ ⎟per ⎠ , second

condition is A = Q V per , and other condition may be assumed if required depending

upon the shape of canal. Using these conditions find all the parameters for the section.

(B) Design steps for unlined Canalsi. Kennedy’s method

Given data Q, n, S, CVR and take canal shape is trapezoidal with m = 0.5

1. Assume y;

2. Compute V = 0.55 CVR y 0.64

3. Determine b from A = Q V = by + 0.5 y

2

thus b = (A − 0.5 y 2 ) y4. Calculate P = b + 5 y

5. Compute R = A P and then V from Kutter’s equation.

6. Check Velocities in step 2 and step 5 should be same, if not repeat steps. Increase or decrease y for the next step if V from Kutter’s equation is more or less than from step2.

2

2 2

b

13

ii. Kennedy’s method

Given data Q, n, CVR, b/y and take canal shape is trapezoidal with m = 0.5

1. A = ((b / y) + 0.5)y 2

;

Q = AV = ((b / y) + 0.5)y 2 0.55 CVR y 0.64

2. Solve y 2.64 = Q{0.55 CVR((b / y) + 0.5)} −1

and then b = b y

y

3. Compute R = A = (b / y) + 0.5 y

P (b / y) + 5

4. Determine V = 0.55 CVR y 0.64

5. Calculate S from Kutter’s equation or Manning’s equation.

iii. Lacey’s method

Given data Q, & silt factor f and take canal shape is trapezoidal with m = 0.5

5. Compute V = (Qf 2 140)1 / 6

6. Form one equation A = Q V = by + 0.5 y 2

7. Form second equation P = 4.75 Q = b + 5 y

8. Compute two unknowns b and y from these two equations.

9. Calculate bed slope S = f 5 / 3 3340Q1 / 6

10. Determine R = A P and R = 5 V. Both should be same, if not check calculations.

2 f

iv. Tractive force method

Given data Q, n, S, φ, τb,per and take canal shape is trapezoidal

1. Assume m; m = cot θ θ < φ2. Compute K = cosθ 1 − tan θ =

tan 2 φ 1 − sin θ = τ s

sin 2 φ τ3. τ s ,

per

= K τ b, per

4. Find y from

τ s , per

= 0.75γ w y S

0

0

⎜ QT⎟

14

5. Calculate b from Manning’sQn = A5 / 3 = (by + my 2 )5 / 3

6. V = Q A

S P 2 / 3 (b + 2

y

1 + m 2 )2 / 3

7. Check (a) flow should be subcritical Fr = V gD < 1 and (b) τ b = γ w yS < τ b, per

v. Most efficient stable channel

Given data Q, n, S, φ, and τb,per

1. Compute y0 from τ b,

per

= γ w y0 S

⎛ tan φ ⎞2. Canal profile y = y0 cos⎜ x ⎟⎝ y0 ⎠π y3. Top width T = 0

tan φ4. Area

T / 22Ty

A = 2 ∫ ydx = 0 = 2 y 2

0π tan φ

5. V = 0.9 − 0.8 tan φ y 2 / 3 S 1 / 2

n0

6. Find Q = AV

7. Compare Q with design discharge

8. If design discharge is more than Q then provide additional area at centre of canal of

width T " = n(Q"−Q)

y 5 / 3 S 1 / 2 and hence new top width = T + T”; new area = A + T”y0

9. If design discharge is less than Q then remove some area at centre of canal of width⎛T ' = ⎜1 −⎝

Q' ⎞⎟⎠ and hence new top width = T – T’; new area =

T / 2' ⎛ tan φ T ' ⎞2∫

T '/ 2 y0cos(x tan φ / y0

)dx and new depth at centre y0 = y0 cos⎜ ⎟⎝ y0 2 ⎠

15

Appendix

16

17

18

19

20

21

22

23

24

25

26