Contents preface (VII) introduction 1—37

7.3.2. Prediction of Regimes of Flow
There are several methods for the prediction of regimes. The method described here has been proposed by Garde and Ranga Raju (8).
The functional relationship for resistance of flow in alluvial channels was written, following the principles of dimensional analysis, as follows:

U

L R

S

g1/ 2 d3 /2

(∆ ρ

/ρ) gR

= f M

,

,

d

∆ ρ

/ ρ

ν

s

M

s

N

O

P (7.11)

PQ

260 IRRIGATION AND WATER RESOURCES ENGINEERING
Here, S is the slope of the channel bed. Since resistance to flow and the regime of flow are closely related with each other, it was assumed that the parameters on the right-hand side of Eq. (7.11) would predict the regime of flow. The third parameter (i.e., g^1/2 d^3/2 /ν) was dropped from the analysis on the plea that the influence of viscosity in the formation of bed waves is rather small. The data from natural streams, canals, and laboratory flumes in which the regimes had also been observed, were used to develop Fig. 7.5 on which lines demarcating the regimes of flow have been drawn. The data used in developing Fig. 7.5 cover a wide range of depth of flow, slope, sediment size, and the density of sediment.
It should be noted that the lines of 45° slope on Fig. 7.5 – such as the line demarcating

F	ρ RS I
‘no motion’ and ‘ripples and dunes’ regimes – represent a line of constant value of τ* G =		J ^.
H	∆ ρsdK

This means that different regimes of flow can be obtained at the same shear stress by varying suitably the individual values of R and S. Therefore, shear stress by itself cannot adequately define regimes of flow.

	10	–2
						Data from different sources
						Ripples dunes
						Transition
		+			Antidunes	Antidunes
		+		No motion
		+
			+
				+
	10		–3
	)	+
	/ρ
S		S
	∆ρ
	(					Transition
	10	–4		No motion
					Ripples and dunes
2 x 10	–5
		10		100	1000	10,000	100,000
						R/d

Fig. 7.5 Predictor for regimes of flow in alluvial channels (8)
The method of using Fig. 7.5 for prediction of regimes of flow consists of simply calculating the parameters R/d and S/(∆ ρ_s/ρ) and then finding the region in which the corresponding point falls. One obvious advantage of this method is that it does not require knowledge of the mean velocity U and is, therefore, suitable for prediction of regimes for resistance problems.
Example 7.2 An irrigation canal has been designed to have R = 2.5 m and S = 1.6 × 10^–4. The sediment on the bed has a median size of 0.30 mm. Find: (i) the bed condition that may be

HYDRAULICS OF ALLUVIAL CHANNELS

261

expected, (ii) the height and spacing of undulations, and (iii) the advance velocity of the undulations. Assume depth of flow and mean velocity of flow to be 2.8 m and 0.95 m/s, respectively.

Solution:

	R	=			2.5	= 8333.33
	d	0.3	× 10^−^3
S	=	16.	× 10^−^4	= 9.7 × 10^–5
∆ ρ s / ρ			165.

From Fig. 7.5, the expected bed condition would correspond to ‘ripples and dunes’ regime. From Eq. (7.10),

n		=	d1/ 6	=	(0.3	× 10	−3 )1/ 6	= 0.01
s	25.6		25.6
and from the Manning’s equation [Eq. (7.9)],

U =

1

R′2 /3 S1/ 2

n

F

s

L

O3 /2

Uns I

3 /2

0.95 ×

0.01

∴

R′ =

G

J

= M

P

= 0.651 m

−4 )1/ 2

H

S^1/ 2 K

N

( 1.6 × 10

Q

ρ R′ S

0.651 × 16. ×

10−4

τ ′ =

=

= 0.21

∆ ρsd

165. × 0.3 × 10^−^3

Using Eq. (7.6),

*

H I ^F

U I ^3

F

U

= 6500 (τ*′)^8/3

G

J G

J

(∆ ρ s / ρ) gd

H

d K H

gR K

H

F

0.95

I ^3 F

0.95

I

= 6500 (0.21)^8/3

−3

G

J

G

J

0.3

× 10

9.81 ×

G

9.81 × 165. × 0.3

× 10

−3 J

H

2.5 K

H

K

∴

H = 0.316 m

Similarly, from Eq. (7.7)

L

O3

×

^L = 1.8 × 10^8 (τ ′)^10/3

M

U

P

U

M ( ∆ ρ s / ρ) gd P

gR d

*

N

Q

L

0.95

O3

0.95

L

M

P

8

10/3

i.e.,

M

9.81 × 165. × 0.3 × 10

−3

P

9.81 × 2.5

^× 0.3 × 10^−^3

= 1.8 × 10

(0.21)

N

Q

∴

L = 0.611 m

U

L

O^3

^w = 0.021 M

U P

U

M

gh P

N

Q

∴

L

0.95

O

3

Uw = 0.95 × 0.021 M

9.81 × 2.8

P

= 0.43 m/hr

M

P

N

Q