Contents preface (VII) introduction 1—37

Example 8.6 Design a stable channel for carrying a discharge of 30 m³/s using Lacey’s method assuming silt factor equal to 1.0.
Solution: From Eq. (8.29),

From Eq. (8.32),

P = 4.75 Q = 4.75 30.0 = 26.02 m

R = 0.48(Q/f1)	1/3	= 0.48	F	30.0I	1/3
	G		J	= 1.49 m
				H	10. K

298 IRRIGATION AND WATER RESOURCES ENGINEERING
From Eq. (8.35),

19. 
    = 3 × 10^–4 f₁^5/3/Q^1/6 = 3 × 10^–4 (1.0)^5/3/(30)^1/6
    

• 
  1.702 × 10^–4
  

Form Eq. (8.26),

U = 10.8 R^2/3 S^1/3 = 10.8(1.49)^2/3 (1.702 × 10^–4)^1/3 = 0.781 m/s
Assuming final side slope of the channel as 0.5H : 1V (generally observed field value),

	P = B + ( 5 )h = 26.02 m
∴	B = 26.02 –	2.24h
			h^2
and	A = Bh +			= PR = 26.02 × 1.49 = 38.77 m^2
2
∴	26.02h – 2.24h^2 +	0.5h^2 = 38.77
or	1.74h^2 – 26.02h +	38.77 = 0
∴	h =	26.02 ±		(26.02)^2 − 4 × 174. × 38.77
			2 × 174.

₌ 26.02 ± 20.18 3.48
= 13.28 m and 1.68 m
The value of h equal to 13.28 m gives negative B and is, therefore, not acceptable. Hence, h = 1.68 m, and
B = 26.02 – 2.24 × 1.68 = 22.23 m

Example 8.7 An irrigation channel is to be designed for a discharge of 50 m ³/s adopting the available ground slope of 1.5 × 10^–4. The river bed material has a median size of 2.00 mm. Design the channel and recommend the size of coarser material to be excluded or ejected from the channel for its efficient functioning.
Solution: From Eqs. (8.25) and (8.35), f₁ = 1.76 d

• 
  1.76 2
  

• 
  2.49
  

and S = 0.0003 f₁^5/3/Q^1/6

• 
  0.0003 (2.49)^5/3/(50)^1/6
  

• 
  7.15 × 10^–4
  

The computed slope is much large than the available ground slope of 1.5 × 10^–4 which is to be adopted as the channel bed slope. Therefore, the median size of sediment which the channel would be able to carry can be determined by computing the new value of f₁ for S = 1.5 × 10^–4 and the given discharge and then obtaining the value of d for this value of f₁ using Eq. (8.25). Thus,

1.5 × 10^–4 = 0.0003 f₁^5/3/(50)^1/6

DESIGN OF STABLE CHANNELS		299
∴	f1 = 0.976
and	d = (f /1.76)^2	= 0.30 mm
	1

Therefore, the material coarser than 0.30 mm will have to be removed for the efficient functioning of the channel. The hydraulic radius of this channel R is obtained from Eq. (8.32) :

		F	50 I ^1/ 3
	R = 0.48	G		J
		H	0.976K
∴	R = 1.783 m
Using Eq. (8.29)

P = 4.75 50 = 33.59 m

∴B + 5 h = 33.59

or	B = 33.59 – 2.24h
and	A = Bh +	h^2	= PR = 33.59 × 1.783 = 59.89 m^2
2
	or	33.59h – 2.24h^2 + 0.5h^2 = 59.89
or	1.74h^2 – 33.59h + 59.89 = 0
∴	h =	33.59 ±	(33.59)^2 − 4 × 174. × 59.83
			2 × 174.

₌ 33.59 ± 26.67 3.84
= 17.32 m or 1.99 m
Obviously, h = 1.99 m as the other root of h would result in too narrow a channel section.

∴	B = 33.59 – 2.24(1.99)

• 
  29.13 m
  

2. 
   Method of Design of Alluvial Channels Including Sediment Load as a Variable
   

Experiments have revealed (13) that the stable width of a regime channel is practically independent of sediment load. Hence, one can use Lacey’s equation [Eq. (8.29)] for stable perimeter P even when the sediment load is varying. If the bank soil is cohesive, the perimeter may be kept smaller than that given by Eq. (8.29). However, the sediment load is known to affect the regime slope of a channel and, hence, the sediment load should also be included in the design of stable alluvial channels.

If the sediment load is moving mainly in the form of bed load, one can determine the unknowns R and S for given Q, q_B , d, and n using the Meyer-Peter’s equation [Eq. (7.33)] and the Manning’s equation [Eq.(8.12)]
If the suspended sediment load is also considerable, then one may use the total load equation instead of Meyer-Peter’s equation. Combining Engelund and Hansen equation [Eq. (7.37)] for sediment load,

fφ	= 0.4τ ^5/2	(8.39)
T	*

300	IRRIGATION AND WATER RESOURCES ENGINEERING
and their resistance equation (14),
	U = 10.97 d^–3/4 h^5/4 S^9/8	(8.40)

a design chart (Fig. 8.7) was prepared. This chart can be used to determine h and S, for known values of

L	=	qT /ρs g
^φT M		∆ρs
M		gd^3
M
N		ρ

O	and	q
P		^∆ρs gd3
P
P
Q		ρ

For the width of stable channel, Engelund and Hansen (14) suggested the use of the empirical relation

B =	0.786Q^0.525	(8.41)
d0.316

Equation (8.41) is valid for channels with sandy bed and banks (12). However, because of cohesive soils of India, the width of stable channels is smaller than that given by Eq. (8.41).

q

			3
			2
			6
	10
			8
			6
			4													4
																4×10
															4
			2										3×10
									2×10	4
			5				1.5×10	4
	10
			8				4
3	gd	6		h/d=10
					3
s						8×10 3
Dr			r	4				6×10
					3
					4×10
			2
							3
					3×10
			4			2×10	3
	10				3
			8			1.5×10
			6				3
				h/d=10
			4		8×10	2
				2
						6×10
			2					2
				4×10
						3×10	2
			3
	10		–2									–1
		2	4	6			2
			10		8 10

5

5

5

–

–

–

S=10

.5×10

3×10

–5

1

2×10

5

4×10^–5

h/d=10

6×10^–5

4

8×10

8×10^–5

6×10 ^4

S=10^–4

1.5×10^–4

2×10^–4

3×10^–4

4×10^–4

6×10^–4

8×10^–4

S = 10^–3

1.5×10^–3

2×10^–3
h/d

3×10^–3

S
4×10^–3

4	6	8 10	0	4	6	8 10	1	4	6	2
2	2	8 10

^fT

Fig. 8.7 Engelund and Hansen’s chart for stable channel design

DESIGN OF STABLE CHANNELS

301

Therefore, one can adopt a value of B which is slightly less than or equal to P obtained from Lacey’s equation [Eq. (8.29)].

Equations (8.39) and (8.40) can be solved for h and S to yield the following explicit relations :

	h = 0.17					q20/21	(8.42)
		( gφ T ) ^2 / 7	d^3/7 ( ∆ρ s /ρ)^5/7
	S = 8.4	q 8/9 d2/3					(8.43)
		h^2
or	S = 4.12			(gφ T )^4	/ 7	d^32/21 (∆ρ s /ρ)^10/7	(8.44)
				q64/63

Equations (8.41), (8.42), and (8.43) or (8.44) will yield direct solutions for B, h and S, and avoid interpolation errors involved in the use of the design chart (Fig. 8.7).
In Eqs. (8.40) to (8.44), U is expressed in m/s, Q in m³/s, q in m ²/s, q_T in N/m/s, and h, d, and B in m. It should be noted that Eq. (8.42) is valid for duned-bed conditions.
Example 8.8 Design a channel to carry a discharge of 30 m³/s with sediment load concentration of 50 ppm by weight. The average grain size of the bed material is 0.3 mm.
1 Assume the cross-section of the channel as trapezoidal with side slopes ₂ H : 1V.




Solution: Using Eq. (8.29)




P = 4.75 Q = 4.75 30 = 26.02 m Choose B to be slightly less than P, say B = 24.0 m.

			Q = 50 × 10^–6 × 9810 × 30 = 14.71 N/s
					T
∴			q	= ^14.71 = 0.613 N/m/s
					T		24.0
								30.0
and					q =	24.0 = 1.25 m^2/s
Now			φT	=	0.613/( 2.65 × 9810)
		1.65 × 9.81 × (0.3 × 10^−^3 )^3
							= 1.128
and	q					=	125.
F	∆ρ s I					165. × 9.81 × (0.3 × 10^−^3 )^3
			3
G		J	gd
H	ρ K
								= 59793.2
From Fig. 8.7,
					S = 1.1 × 10^–4
and				h		= 7.2 × 10^3
			d
∴						h = 7.2 × 10^3 × (0.3 × 10^–3) = 2.16 m

302 IRRIGATION AND WATER RESOURCES ENGINEERING

Alternatively, using Eqs. (8.42) and (8.43),

0.17 × (125.)20/21

8. 
   ⁼ (9.81 × 1128.)^2/7 (0.3 × 10⁻³ )^3/7 (165.)^5/7
   

• 
  2.38 m
  

8.4 × (125.)^8/9 × (0.3 × 10⁻³ )^2/3

S =
		(2.38)	2
= 8.1 × 10^–3