Contents preface (VII) introduction 1—37

Example 8.9 Solve Example 8.8 considering the entire sediment load as bed load.
Solution: Q_B = 50 × 10^–6 × 9810 × 30 = 14.71 N/s

Silt factor, f₁ = 1.76 d = 1.76 0.3 = 0.964

Since the value of Manning’s n is 0.0225 for f₁ = 1.0, one can take n = 0.022 for f₁ = 0.964
Using Eq. (8.29),

P = 4.75

30.0 = 26.02 m

Let

B = 24.0 m

∴

q

= 14.71

= 0.613 N/m/s

B

24.0

Hence,

φB

=

^qB ^/(^ρs ^g^) =

0.613/(2650 × 9.81)

= 1.128

∆ρs

gd

3

165. × 9.81 × (0.3 × 10^−^3 )^3

ρ

and from Eq. (7.10)

n =

d1/6

s

25.6

=

(0.3 × 10^−^3 )^1/6

25.6

= 0.01

F ns I ^3/2 ρRS

F

0.01 I ^3/2

RS

∴

τ′*

=

G

J

= G

J

∆ρ sd

1.65 × ( 0.3

× 10

−3

)

H

n K

H

0.022K

• 
  619.10 RS
  

∴ Using Meyer-Peter’s equation [Eq. (7.33)],

	φB = 8.0(τ′* – 0.047)^3/2
or	1.128 = 8.0(619.10RS – 0.047)^3/2
∴	RS = 5.135 × 10^–4
Now using the Manning’s equation,
	Q =	1	AR2/3 S1/2
		n
	30 =	1	(26.02 × R^5/3	S1/2)
	0.022

DESIGN OF STABLE CHANNELS				303
∴	R^5/3 S^1/2 = 0.0254
or	R^7/6(5.135 × 10^–4)^1/2 = 0.0254
∴	R = 1.103 m
and	S = 4.66 × 10^–4
	A = Bh +	h^2	= 24h +	h^2	= PR = 26.02 × 1.103
	2	2
or			h^2 + 48h – 57.4 = 0

_{h =} − 48 ± (48)² + 4 × 57.4
2
h = 1.167 m




P = B + 5 h

• 
  24 + 5 × 1.67
  

• 
  26.61 m
  

which is close to Lacey’s perimeter (= 26.02 m) Hence,

B = 24.0 m h = 1.167 m and S = 4.66 × 10^–4