Contents preface (VII) introduction 1—37

8.2. RIGID BOUNDARY CHANNELS CARRYING CLEAR WATER
Channels of this type have rigid, i.e., non-erodible, boundaries like rock cuts or artificial lining. These channels reduce the seepage loss and thus conserve water and reduce the water logging of the lands adjacent to the channel. Cost of operation and maintenance of lined channels are also less. Lined channel sections are relatively more stable.

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DESIGN OF STABLE CHANNELS

279

The design of such channels is based on the Manning’s equation combined with the continuity equation. Thus,

Q =	1	AR2/3 S1/2	(8.1)
n

in which, A is the area of cross-section of flow and the hydraulic radius R = A/P. Here, P is the wetted perimeter. Equation (8.1) can also be written as

Q =	A 5 /3	S1/ 2	(8.2)
nP2 /3

This means that for specified values of Manning’s n and the slope S, the discharge Q is maximum for a given area of cross-section A when the wetted perimeter P is minimum. A channel section with the minimum wetted perimeter for a given cross-sectional area A is said to be the most efficient hydraulic section or, simply, the best hydraulic section. Since a circle has the least perimeter for a given area, the circular section is the most efficient section. But, construction difficulties in having a circular section rule out the possibilities of having a circular channel in most cases. Thus, the problem reduces to determining the geometric elements of the most efficient hydraulic section for a specified geometric shape.
Considering a rectangular section of width B and depth of flow h, one can determine the dimensions of the most efficient rectangular section as follows:

	A = Bh
	P = B + 2h
		=		A	+ 2h
			h
For P to be minimum,	dP	= −		A	+ 2 = 0
dh		h2
i.e.,	A = 2h^2
or	B = 2h
∴	P = 4h
Thus,	R =	h
2

Likewise, such relationships for the geometric elements of other shapes, summarized in Table 8.1, can be determined.

In practice, the sharp corners in a cross-section are rounded so that these may not become zones of stagnation. Sometimes, the side slopes may also have to be adjusted depending upon the type of bank soil. In India, lined channels carrying discharges less than 55 m³/s are generally of triangular section of the permissible side slope and rounded bottom (Fig. 8.1). A trapezoidal section with rounded corners (Fig. 8.2) is adopted for lined channels carrying discharges larger than 55 m³/s. The side slopes depend on the properties of the material through which the channel is to pass. Table 8.2 gives the recommended values of the side slopes for channels excavated through different types of material.
To avoid damage to the lining, the maximum velocity in lined channels is restricted to 2.0 m/s. Thus, the design is based on the concept of a limiting velocity. The expressions for geometric elements of the lined channel section, Figs. 8.1 and 8.2, can be written as follows:

280 IRRIGATION AND WATER RESOURCES ENGINEERING
Table. 8.1 Geometric elements of the most efficient hydraulic sections (2)

Shape of

Area A

Wetted

Hydraulic

Water surface

Hydraulic

cross-section

perimeter P

radius R

width T

depth D

Rectangle

2h^2

4h

0.500h

2h

h

Triangle

h^2

2.83h

0.354h

2h

0.500h

(side slope 1 : 1)

Trapezoid

1.73h^2

3.46h

0.500h

2.31h

0.750h

(half of hexagon)

Semicircle

0.5 πh^2

πh

0.500h

2h

0.250 πh

Parabola

1.89h^2

3.77h

0.500h

2.83h

0.667h

(T = 2 2h )

Table 8.2 Suitable side slopes for channels excavated through

different types of material (3)

Material

Side slopes (H : V)

Rock

Nearly vertical

Muck and peat soil

0.25 : 1

Stiff clay or earth with concrete lining

0.5 : 1 to 1 : 1

Earth with stone lining

1 : 1

Firm clay

1.5 : 1

Loose, sandy soil

2 : 1

0

θ

θ

2θ

h

h

1

h

1

Z

Z

Fig. 8.1 Lined channel section for Q < 55 m³/s

	θ	θ		θ	θ
	h		h
1				h	h		1
	Z				Z

B

Fig. 8.2 Lined channel section for Q > 55 m³s

DESIGN OF STABLE CHANNELS

281

For triangular section, Fig. 8.1,

Area,

Wetted perimeter,

Hydraulic radius

	F	1		2	I	1 2
A = 2	G	2	h		cot θJ +	2	h (2θ)
	H			K

• 
  h² (θ + cot θ) P = 2h cot θ + h(2θ)
  

• 
  2h(θ + cot θ)
  

R =	A	=	h^2 (θ + cot θ)	=	h
P	2h (θ + cot θ)	2

Similarly, for trapezoidal section, Fig. 8.2,

	F	1		2	I
A = Bh + 2	G		h		cot θ J
2
		H				K

= Bh + h²(θ + cot θ) P = B + 2h (θ + cot θ)
Bh + h² (θ + cot θ) ^{R =} B + 2h (θ + cot θ)

F 1		2	I
+ 2 G		h		θJ
2
	H			K

In all these expressions for A, P, and R, the value of θ is in radians. For designing a lined channel, one needs to solve these equations alongwith the Manning’s equation. For given Q, n, S, and A, and R expressed in terms of h for known Q, the Manning’s equation will yield, for triangular section, an explicit relation for h as shown below :

	1		L hO	2 /3
Q =		[h^2	(θ + cot θ)] M		P	S1/2
n
			N	2 Q

∴	h =	L	nQ(2)^2 /3 O	3 /8
M		S (θ + cot θ) ^P	(8.3)
			M	P
		N	Q

However, in case of trapezoidal section, Fig. 8.2, the design calculations would start with an assumed value of velocity (less than the maximum permissible velocity of 2.0 m/s) and the expression for h will be in the form of a quadratic expression as can be seen from the following :
From the Manning’s equation



R =
F Un I ∴ G J

H S K

B =

F Un I ^3 /2
G	J
H S K
3 /2			Q
[B + 2h (θ + cot θ)] = A =
	U
	Q F	S I	3 /2
G	J	– 2h (θ + cot θ)
U H	Un K

On substituting this value of B in the expression for area of flow A, one gets,

h	Q F	S I 3 /2	– 2h^2 (θ + cot θ) + h^2	(θ + cot θ) = A =	Q
G	J	U
	U H		Un K

282

i.e.,

or

IRRIGATION AND WATER RESOURCES ENGINEERING

∴

h^2 (θ + cot

θ) – h

Q F

S I 3 /2

Q

= 0

G

J

+

U

U H Un K

Q F

S I 3 /2

Q^2 F

S I ^3

Q

G

J

±

U

2

G

J

− 4 (θ + cot θ)

U

∴

h =

U H Un K

H

Un K

2(θ + cot θ)

Therefore, in order to have a feasible solution,

Q^2 F

S I ^3

Q

U

2

G

J

≥ 4 (θ + cot θ)

U

H

Un K

S3 /2

4n^3 (θ + cot θ)

U ^4

≥

Q

L

O1/4

QS3/2

U ≤ M

P

(8.4)

4n

3

N

(θ + cot θ) Q

This means that for designing a trapezoidal section for a lined channel, the velocity will have to be suitably chosen so as not to violate the above criterion in order to have a feasible solution (see Example 8.2 for illustration).

Example 8.1 A lined canal (n = 0.015) laid at a slope of 1 in 1600 is required to carry a discharge of 25 m³/s. The side slopes of the canal are to be kept at 1.25 H : 1 V. Determine the depth of flow.
Solution: Since Q < 55 m³/s, a triangular section with rounded bottom, Fig. 8.1, is considered suitable. Here,
cot θ = 1.25
∴ θ = 38.66° or 0.657 radian Thus, from Fig. 8.1, A = h² (θ + cot θ) = h² (0.675 + 1.25)

= 1.925h^2

and

P = 2h(θ + cot θ) = 2h(0.675 + 1.25)

= 3.85h

∴

R =

1.925h^2

=

h

3.85 h

2

From the Manning’s equation,

Q =

1

AR2/3 S1/2

n

1925.h^2

F hI

2/3 F

1 I 1/2

Hence,

25 =

G

J

G

J

0.015

H

2 K

H

1600K

∴

h8 /3 =

25 × 0.015 × 2^2/3

× 40

1925.

∴

h = 2.57 m.

DESIGN OF STABLE CHANNELS