Contents preface (VII) introduction 1—37

W	cos α tan θ	1	−	tan^2	α
sl	s	tan^2 θ
				a

(8.7)


On combining Eqs. (8.6) and (8.7)

	k =	^τ sl	= cos α	1	−	tan^2	α
	^τbl	tan^2 θ
i.e.,			τsl = k τbl = 0.9k τc

For non-scouring condition, the design rules become
^τbm ^{≤ τ}bl ^τsm ^{≤ τ}sl

(8.8)

Lane also observed that the channels, which are curved in alignment, scour more readily. He, therefore, suggested some correction factors which should be multiplied with the critical value of tractive stress (1). The values of critical shear stress (and also τ_bl) for bed and sides of curved channels are given in Table 8.3.

DESIGN OF STABLE CHANNELS

287

	1.0
			Trapezoid channel, Z = 2
	0.8
				Trapezoid channel, Z = 1.5
/ghS	0.6
			Rectangular channel
sm	0.4
				h	1
					Z
	0.2			B
	0	2	4		6	8	10
	0

B/h
(a) Channel sides

	1.0
				Trapezoid, Z = 1.5 & 2
	0.8
	0.6		Rectangular
ghSr
	/
bm	0.4
t						1
					h
						Z
	0.2				B
	0	2	4	6		8	10
	0

B/h
(b) Channel bed

Fig. 8.4 Maximum shear stress on (a) sides and (b) bed of smooth channels in uniform flow
Table 8.3 Critical shear stress and the values of τ_bl for curved channels

Type of channel	Critical shear stress (τc)	^τbl
Straight channels		τc	0.900	τc
Slightly curved channels	0.90	τc	0.810	τc
Moderately curved channels	0.75	τc	0.675	τc
Very curved channels	0.60	τc	0.540	τc

288 IRRIGATION AND WATER RESOURCES ENGINEERING

	µ
a	sin
	s
.	W
tsl
	Resultant

h

µ

Fig. 8.5 Forces causing movement of a particle resting on a channel bank
Example 8.4 Design a trapezoidal channel (side slopes 2H : 1V) to carry 25 m³/s of clear water with a slope equal to 10^–4. The channel bed and banks comprise gravel (angle of repose = 31°) of size 3.0 mm. The kinematic viscosity of water may be taken as 10^–6 m²/s.
Solution: From Eq. (7.2),

R0* =

∆ρ s gd^3

(165.) (9.81) (3.0 × 10^−^3 )^3

ρν^2

=

(10^−^6 )^2

∴

R0* = 661.1

From Fig. 7.3,

τc* = 0.045

gd = 0.045 × 1650 × 9.81 × (3.0 × 10^–3)

∴

τ

= (0.045) ∆ρ

s

c

= 2.185 N/m^2

Taking

τbl = 0.9 τc

= 1.97 N/m^2

∴

τ

= 1.97 N/m^2

bm

k = cos α 1 ^−

tan^2 α

2

(0.5)^2

Now,

tan

2

=

5

1 −

(0.6)

2 = 0.494

θ

∴

τ

= 0.494 × τ

= 0.494 × 1.97 = 0.973 N/m^2 = τ

sl

bl

sm

Rest of the computations are by trial and can be carried out as follows : Assume B/h = 10

Therefore, from Fig. 8.4,

^{τ sm} = 0.78

ρghS

0.973

^{h =} _{9810 × 10}−4 _{× 0.78} ^{= 1.27 m}

Also from Fig. 8.4,

^{τ bm} = 0.99

ρghS

DESIGN OF STABLE CHANNELS			289
∴	h =	197.	= 2.03 m
9810 × 10^−^4 × 0.99

Choosing the lesser of the two values of h h = 1.27 m
B = 10h = 12.7 m
A = Bh + 2h² = 12.7 × 1.27 + 2 × (1.27)² = 19.355 m²

	P = B + 2	5 h = 12.7 + (2	5 × 1.27) = 18.38 m
	R = 1.053 m
and	n =	d1/ 6	=	( 3 × 10	−3 )1/6	= 0.0148
25.6	25.6
∴		Q =	1	AR2/3 S1/2 =	1			× 19.355 × (1.053)^2/3 (10^–4)^1/2
	n	0.0148
	= 13.5 m^3/s

Since this value of Q is less than the given value, another value of B/h, say, 20.0 is assumed. Using Fig. 8.4, it will be seen that h = 1.27 m.
B = 25.4 m
A = 35.484 m², P = 31.08 m, R = 1.142 m and Q = 26.195 m³/s

This value of Q is only slightly greater than the desired value 25.00 m³/s.

Hence, B = 25.4 m and h = 1.27 m. The trial calculations can be done in a tabular form as shown below:

B/h		^τ sm			^τbm		h(m)	B(m)	A(m^2)	P(m)	R(m)	Q(m^3/s)
	ρghS			ρghS
10.0		0.78			0.99		1.27	12.7	19.355	18.38	1.053	13.500
20.0		0.78			0.99		1.27	25.4	35.484	32.08	1.142	26.195