Contents preface (VII) introduction 1—37

Example 11.1 Design a suitable cross-drainage structure for the following data:

Discharge in the canal,	= 357.0 m^3/s
Bed width of the canal, Bc	= 23.0 m
Side slope of the canal, m0	= 2.0
Bed level of the canal,	= 267.0 m
Depth of flow in the canal, hc	= 6.7 m
Bed width of the flume, Bf	= 15.0 m
High flood discharge of the stream	= 500 m^3/s
High flood level of the stream	= 268.0 m
Bed level of the stream	= 265.0 m

Solution: Using Eq. (11.21), the length of expansion transition,
L = 2.35 (B_c – B_f) + 1.65 m₀h_c
= 2.35 (23 – 15) + 1.65 × 2 × 6.7 = 40.91 ≅ 41 m
Let the transition be divided into 9 sub-reaches of 4.0 m each and the remaining 5 m length be considered as the tenth sub-reach at the end.
Bed width, b is calculated using Eqs. (11.19) and (11.20).

∴

or

or

n = 0.8 – 0.26 m₀^1/2
= 0.8 – 0.26 (2)^1/2 = 0.432

B − Bf

x

L

F

x I ^n O

=

M1

− G 1

−

J

P

Bc − Bf

^L M

H

LK

P

N

Q

B − 15

x

L

F

x I

0.432 O

=

M1

− G 1

−

J

P

23 − 15

41 M

H

41K

P

N

Q

	8x	L	F		x I	0.432 O
B = 15 +	41	M1	− G 1	−		J	P
	M	H		41K	P
		N					Q

For example, at section 1-1,

		8	× 4 ^L	F		4 I	0.432 O
B	= 15 +		41	M1	− G 1	−		J	P
1 – 1			M	H		41K	P
	= 15.034 m	N					Q

Values of B for different values of x have been similarly calculated and are as shown in col. 3 of Table 11.5.

400

IRRIGATION AND WATER RESOURCES ENGINEERING

Side slopes at various sections are calculated using Eq. (11.22). Thus

m

F

x I 1/ 2

= 1 − G

1 −

J

m0

H

LK

L

F

x

I

1/ 2 O

or

m = 2 M1

− G 1

−

J

P

M

H

41K

P

For example, at section 1-1,

N

Q

L

1/ 2 O

F

4

I

m

= 2 M1

− G 1

−

J

P = 0.100

1 – 1

M

H

41K

P

N

Q

Values of m for different values of x have been similarly calculated and are tabulated in col. 4 of Table 11.5.
For constant specific energy condition:

v 2

E = E = h +

c

2g

f

c

c

and

vc = 357/[6.7 (23 + 2 × 6.7)] = 1.464 m/s

∴

E = E = 6.7 +

(1464.)

2

= 6.809 m

f

c

2 × 9.81

hf +

Q^2

= 6.809

2gB

^2 h

2

f

f

or

hf +

(357)^2

= 6.809

2g(15)

2

2

hf

Solving for hf , one obtains hf = 6.006 m

Similarly,

^E1 – 1 ^= ^h1 – 1 ^+

Q^2

= Ec = 6.809 m

2g(b

− 1

+ m

h

− 1

)^2 h^2

1

1 −

1 1

1 − 1

∴

^h1 – 1 ^+

(357)^2

= 6.809

2 × 9.81 (15.034 + 0.1 h

1

)

2 h2

1 −

1 − 1

On solving this equation,

h1 – 1 = 6.088 m

Further,

vf

=

357

= 3.963 m/s

15

× 6.006

and

^v1 – 1

=

357

= 3.749 m/s

(15.034

+

0.1 × 6.088) 6.088

Using Eq. (11.10),

hL

= 0.3 ×

1

(3.963^2

− 3.749^2) = 0.025 m

f , 1 − 1

2

×

9.81

Using Eq. (11.23),

∆zf,1 – 1 = 0.025 m

The positive value of ∆z indicates fall in the bed elevation with respect to the flume bed. Values of ∆z for other sections can be similarly obtained. The values of ∆z and h have been tabulated in cols. 5 and 6 of Table 11.5.

CROSS-DRAINAGE STRUCTURES								401