Contents preface (VII) introduction 1—37

7.4.2. Resistance Relationship Based on Division of Resistance
In dealing with open channel flows, hydraulic radius R of the flow cross-section is taken as the characteristic depth parameter. The use of this parameter requires that the roughness over the whole wetted perimeter is the same. Such a condition can be expected in a very wide channel with alluvial bed and banks. However, laboratory flumes with glass walls and sand bed would have different roughnesses on the bed and side walls. In such cases, therefore, the hydraulic radius of the bed R_b is used instead of R in the resistance relations. The hydraulic radius of the bed R_b can be computed using Einstein’s method (14) which assumes that the velocity is uniformly distributed over the whole cross-section. Assuming that the total area of cross-section of flow A can be divided into areas A_b and A_w corresponding to the bed and walls, respectively, one can write
A = A_w + A_b
For rectangular channels, one can, therefore, write

	(B + 2h) R = 2 hRw + B Rb
∴	F	1 +	2hI	R −	2h R
	Rb = G		J		B	w
	H		B K

• 
  (B + 2h)(R/B) – 2hR_w/B
  
• 
  (PR/B) – 2hR_w/B
  
• 
  (A/B) – 2hR_w/B
  

= h – 2hRw/B	(7.20)
Using Manning’s equation for the walls, i.e.,
U =	1		R 2 /3 S1/ 2	(7.21)
		w
n
		w

one can calculate the hydraulic radius of the wall R_w if the Manning’s coefficient for the walls, n_w is known. Using Eq. (7.20), the hydraulic radius of the bed R_b can be computed.
Example 7.4 A 0.40m wide laboratory flume with glass walls (n_w = 0.01) and mobile bed of 2.0 mm particles carries a discharge of 0.1 m³/s at a depth of 0.30m. The bed slope is 3 × 10^–3. Determine whether the particles would move or not. Neglect viscous effects.
Solution:

Hydraulic radius,

R =

0.3 × 0.4

= 0.12 m

0.4 + 2 × 0.3

Mean velocity of flow in the flume =

0.1

= 0.833 m/s

0.3 × 0.4

Using Eq. (7.21),

R

F Unw I ^3 /2

w

= G

S

1/ 2 ^J

H

K

F

0.833 × 0.01 I ^3 /2

= G

J

× 10

−3

)

1/ 2

H (3

K

= 0.0593 m

266			IRRIGATION AND WATER RESOURCES ENGINEERING
Using Eq. (7.20),	R	b	= 0.3 − ^2 ^× ^0.3	× 0.0593
			0.4
			= 0.211 m
∴ Bed shear,	τb = ρgRb S

• 
  9810 × 0.211 × 3 × 10^–3
  
• 
  6.21 N/m²
  

On neglecting viscous effects and using Yalin and Karahan’s curve,

	τ c	= 0.045
	∆ρs gd
∴ Critical shear,	τ	= 0.045 × 1.65 × 9810 × 2 × 10^–3
	c	= 1.457 N/m^2

Since τ_b > τ_c, the particles would move.
Einstein and Barbarossa (15) obtained a rational solution to the problem of resistance

in alluvial channels by dividing the total bed resistance (or shear) τob into resistance (or shear)
due to sand grains τ′	ob	and resistance (or shear) due to the bed forms τ ″, i.e.,
						ob
		τ		= τ ′ + τ ″	(7.22)
		ob	ob	ob
or		ρg R S = ρg R′S + ρg R″S
		b			b	b
i.e.,		R	b	= R ′ + R″	(7.23)
				b	b

where R ′ and R ″ are hydraulic radii of the bed corresponding to grain and form resistances (or

b b

roughnesses).

For a hydrodynamically rough plane boundary, the Manning’s roughness coefficient for the grain roughness n_s is given by the Strickler’s equation i.e.,

	d 1/6
ns =	65	(7.24)
24.0

Here, d₆₅ (in metres) represents the sieve diameter through which 65 per cent of the sediment will pass through, i.e., 65 per cent of the sediment is finer than d₆₅. Therefore, Manning’s equation can be written as

	U =	1 R′ ^2 /3S1/ 2
				b
	n
					s
	U =	24		R ′ ^2 /3S1/ 2	(7.25)
	d	1/ 6
						b
					65
Since	U		′ =	τ	^′ /ρ =	gR^′ S
	*				ob			b
	U				24		R^′	^2 /3S1/ 2
	U*^′ ^= d65^1/ 6		b
				gRb^′ S
	U					F		R^′	I 1/ 6
or			=	7.66 G		b	J	(7.26)
U′ *
						H ^d65 K

Einstein and Barbarossa (15) replaced this equation with the following logarithmic relation having theoretical support.

HYDRAULICS OF ALLUVIAL CHANNELS		267
	U	12.27 R^′
		= 5.75 log	b	(7.27)
	U^′	^d65
*

Equation (7.27) is valid for a hydrodynamically rough boundary. A viscous correction factor x (which is dependent on d₆₅/δ′, Table 7.1, Fig. 7.8) was introduced in this equation to make it applicable to boundaries consisting of finer material (d₆₅/δ′ < 10). The modified equation is (15)

U

12.27 R ^′

x

5.75 log

b

U′ ^=

^d65

(7.28)

*