Contents preface (VII) introduction 1—37

Uc		L Rex O^1/ 8
					= 16.	M		P
			∆ρs	gd	N ^dm Q
			ρ	m
and the limit deposit velocity UL is given as (10)
					UL = F1 8 g Rex ( Ss − 1)
where, Ss =	ρ s
	ρ

ρ_s = the mass density of sediment, ρ = the mass density of water,
R_ex = the hydraulic radius of the excluder tunnel.
(13.5)

(13.6)

and F₁ depends on the concentration and size distribution of sediment as shown in Fig. 13.12. Garde and Pande (8) suggested that for sediment size greater than 0.5 mm, F₁ varies between 0.8 and 1.0, and can be approximated as unity.

If the excluder velocity U_ex is greater than U_L there will be no deposition of sediment in the excluder. However, the value of U_L is generally large and, hence, U_ex is usually less than

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459

U_L and, therefore, the excluder tunnels are partially blocked. The free flow area in such a case can be calculated from

^Ufex	=	UL	(13.7)
4gRfex	4gRex

2.0

1

F

1.0
0.0

0
15%
10% 5%
=2% Cv
C_v = sediment concentration by volume

1 2 3 d, mm

(a) For uniform sediment

1 F

1.2
1.0

0.8
0.6
0.4

0

15%

10

5 C_v = 2

1	2	3	4
	d, mm

(b) For non-uniform sediment

Fig. 13.12 Limit deposit velocity (10)

where,

^Ufex

=

^Qex

(13.8)

^Bex ^Dfex

and

^Rfex ^=

^Bex ^Dfex

(13.9)

2 (Bex + Dfex )

Thus, combining Eqs. (13.6 – 13.9) one gets

2 ^F

^Dfex I

^Qex G

1 +

J

H

Bex K

= 4F1^2(Ss – 1)

(13.10)

gD ^3

B^2

fex

ex

Here, the suffix ‘f’ refers to the free (i.e., unblocked) flow area available, and the suffix ‘ex’ refers to the excluder. B, D, and Q are, respectively, the width, depth, and the discharge of the excluder. Using the above equations, D_fex can be calculated. Hence, the blockage of the tunnels is obtained as D_ex – D_fex. One can also calculate U_fex from Eq. (13.8).
The bed load and suspended load going into the undersluice pocket are worked out by using suitable bed load relation and the velocity and sediment concentration profiles. Thus, the total sediment concentration (by weight) in the tunnels, C_ex, can be obtained. The sediment transport capacity of the tunnels C_t (i.e., sediment concentration by weight) must, obviously, be greater than C_ex. Further, C_t is obtained from (11).

L	I			O^s1
Re f = M				P Ct^1/ 3	(13.11)
d/(4R		)
M	fex		P
N					Q

Here, R_e is the Reynolds number of flow [= U_fex (4R_fex)/ν], f the Darcy-Weisbach resistance coefficient, and I is a suitable function of d as given in Fig. 13.13. Further, the index s₁ is obtained from

460	IRRIGATION AND WATER RESOURCES ENGINEERING
s1 =	1	(13.12)
0.89d^1/ 3

in which, d (in mm) can be either equal to d_m or slightly less than d_m.

1000

100

10.0
I

1.0

0.1

0.01
0.1	0.2	0.4	0.6 0.8 1.0	2.0	3.0
			d in mm

Fig. 13.13 Variation of I with d (11)
A suitable design for an excluder requires that the blockage be limited to a suitable value, say, about 35 per cent and also C_t is greater than C_ex in the excluder tunnels. Ideally, a design which satisfies all the conditions for minimum values of Q_ex, B_ex, and depth of tunnels will be the most suitable. However, there are some constraints on all these parameters. Therefore, one should choose the best combination satisfying all the requirements and constraints. Some margin of safety should always be provided for in order to account for the uncertainties in the sediment load computations. Following example illustrates the procedure for trial design of a sediment excluder based on the method given by Garde and Pande (8).
Example 13.2 Design a suitable sediment excluder using the following data:

River discharge, Q	= 4250 m^3/s
R
River slope	= 1/4000
Depth at pond level	= 4.33 m
Bed width (Barrage width), BR	= 920 m
Canal discharge, Q	= 240.7 m^3/s
C

CANAL HEADWORKS	461
Average grain size of the bed material	= 0.37 mm
Maximum size of the bed material	= 0.60 mm

Solution:
(i) Find shear stress acting on the bed corresponding to the maximum discharge at which pond level is being maintained
τ_o = 1000 × 9.81 × 4.33 × (1/4000) = 10.62 N/m²

Corresponding to this shear stress, the maximum size of the bed material which can be set into motion can be determined using Eq. (7.3) and is equal to 11.00 mm. As such, the maximum size of the bed material (i.e., 0.6 mm) at the site would move under the given conditions.

(ii) Assume some suitable depth of tunnels, t (usually between 1 and 2 m). Let t = 1.6 m
(iii) Assume excluder discharge Q_ex (between 10 and 20 per cent of canal discharge, Q_c).
Let Q_ex = 48m³/s (approx. 20 per cent of Q_c)

(iv) Find the critical velocity U _c and limiting velocity U_L for the maximum size of the bed material which is movable (step (i)). Using Eqs. (13.5) and (13.6), and assuming R_ex = 0.7 m and F₁ = 1.0.

Uc = 1.6 1.65 × 9.81 × 0.6 × 10	−3	L		0.7
		M	0.6	× 10	−3
		N

U_L = 1.0 8 × 9.81 × 0.7 (2.65 − 1) = 9.52 m/s
(v) Choose excluder velocity U_ex such that it is greater than hood of U_L but not exceeding 3 m/s.

Let	Uex = 2.5 m/s.
∴	Bex =	48	= 12.0 m
2.5 × 16.
	^Rex ^=	12 × 16.		= 0.706 m
	2 (12 + 16.)

_O1/ 8

P = 0.38 m/s

Q
U_c and in the neighbour-

∴ Revised	UL = 1.0	8 × 9.81 × 0.706 (2.65 − 1) = 9.56 m/s
and revised	Uc = 1.6	1.65 × 9.81 × 0.6 ×	10	−3	L	0.706	−3	O	1/ 8
	M	0.6 × 10	P	= 0.493 m/s
						N			Q

(vi) Determine the width B_RE which contributes the discharge to the undersluice pocket using the relation

B =	(Q	+ Q	ex	)	BR
c
RE						QR

920

= (240.7 + 48) ₄₂₅₀ = 62.5 m

(vii) Determine the amount of bed load coming into the tunnels fed by the stream width B_RE as shown in the following steps:

462

IRRIGATION AND WATER RESOURCES ENGINEERING

1

L

920 × 4.33

O

2 /3F

1 I 1/2

4250 =

(920

× 4.3) M

P

G

J

n

∴

N

(920 + 2 × 4.33) Q

H

4000K

n = 0.039

(0.37 × 10^−^3 )^1/ 6

and

n =

= 0.0105

25.6

From Eq. (7.31),

s

3 /2 F

I F

F

I

F

0.0105I

920 ×

4.33

1 I

1

G

J

G

J G

J

G

J

165. × 0.37 × 10

−3

H

0.039 K

H 920 + 2

× 4.33K H

4000K

H

K

q 2 /3

= 0.047 + 0.25

B

(2.65 × 1000)^2 /3 (9.81)(0.37 × 10^−^3 )(165.)^1/ 3

or

0.2454 = 0.047 + 0.3044q ^2/3

∴

B

qB = 0.526 N/m/s

Total bed load coming into the excluder pocket = 0.526 × 62.50 = 32.875 N/s
(viii) Determine the suspended load coming into the tunnels of the excluder. An approxi-mate simple equation of Engelund’s curve (12) relating suspended load with the river discharge is

	Qs		= 0.48 × 10–4G^F ^u^* J^I ^4 .36
	QR
					H w K
		u = τ o / ρ =	10.62	= 0.103 m / s
	*				1000
	w = fall velocity of the bed material
			= 0.05 m/s (for d = 0.37 mm)
∴	Qs = 4250 × 0.48 × 10	–4^F	0.103I ^4.36	3
G		J	= 4.765 m /s
								H	0.05 K
Suspended load going into the pocket, Qsp is obtained from
	Q	=	Qs	× BRE
	sp		BR

= ⁴₉₂₀^.765 × 62.5 = 0.324 m³/s

For determining the fraction of suspended load Q_sp entering the tunnel, i.e., Q_st, one can use curves of Fig. 13.14 (8). For the present problem,

t

=

16.

= 0.37

D

4.33

z =

w

=

0.05

= 1.2

u k

0.103 ×

0.4

*

k

u

= 0.4

4250 / (920 × 4.33)

= 4.14

u*

0.103

∴

^Qst

= 0.73

^Qsp

CANAL HEADWORKS 463

	1.0														1.0
	t/D = 0.2													t/D = 0.3			8
	0.8														0.8								5	4	.0
			8
										3													=	3
						5				=
SP							4								SP						u / u *
								u	/u *											k
Q													Q
	0.6							k							0.6
	/																/
ST																ST
Q														Q
	0.4														0.4
	0.2														0.2
	0.4	0.8			1.2		1.6	2.0	0.4	0.8									1.2	1.6
	1.0														1.0
	t/D = 0.4													t/D = 0.5			8
	0.8														0.9								5			4
																					3	.0
																							=
SP				8									SP
																u	/	u	*
/Q					4									/Q						k
0.6															0.8
ST													ST
						3
Q														Q
					=
				u/
				k	u*
	0.4														0.7
	0.2															0.6
	0.4	0.8			1.2		1.6	2.0	0.4	0.8								1.2	1.6
					Z = w/ku*												Z = w/ku*

2.0

2.0

Fig. 13.14 Suspended load entering excluder tunnels

∴	Qst = 0.73 × 0.324 × 2.65 × 9810 = 6148.7 N/s
	C	=	(6148.75 + 32.875)	= 0.013 (by weight)
	48 × 9810
		ex

Using Eq. (13.10), one gets

^{12 +} ₃^Dfex _{= 48.56}

^Dfex

D_fex = 0.64 m

R_fex = (12 × 0.64)/[2 (12 + 0.64)] = 0.304 m
R_e = [48/(12 × 0.64)] [4 × 0.304] [10⁶] = 7.6 × 10⁶ Taking d = 0.5 mm (less than d_m which is 0.6 mm)

1

^s1 ⁼ _{0.89 (0.5)}1/ 3 ^{= 1.416 and I = 5.2}

Hence, from Eq. (13.11)

	6		L				5.2	O1. 416	1/ 3
7.6 × 10		0.02	= M					P	^Ct
	0.5	× 10	−3
				N		/ ( 4 × 0.304) Q

464	IRRIGATION AND WATER RESOURCES ENGINEERING
∴	Ct = 4.66
∴	^Ct ^> ^Cex

Since this trial design has yielded relatively large blockage one needs to make another trial. The final design should be such that the sum of the clear waterway, i.e., B_ex and thickness of tunnel walls is equal to the width of whole number of undersluice bays.