Contents preface (VII) introduction 1—37

Table 13.3 Thickness of pitching (T) for loose stone protection in millimetres (2)

Type of river bed material			River slope (m/km)
	0.05	0.15		0.20	0.30	0.40
Very coarse	400	500		550	650		700
Coarse	550	650		700	800		850
Medium	700	800		850	950		1100
Fine	850	950		1000	1100	1150
Very fine	1000	1100		1150	1250	1300

The depth of scour holes, calculated from Lacey’s equation, is to be modified according to Table 13.2 for different locations of launching aprons. No allowance in the form of concentration factor need be made for computing normal scour depth for the upstream and downstream protection works (2).

The total quantity of stone for launching apron, worked out as mentioned above, should be laid in a length of about 1.5 D to 2.5 D. Here, D is the depth of scour below the floor level. The higher values are to be used for flatter launched slope. The thickness of loose stone at the inner edge should correspond to the required quantity of stone for thickness of launched apron equal to T. The material required for extra thickness of 0.25 T on launched slope should be distributed over the length of apron in the form of a wedge with increasing thickness towards the outer edge.

448	IRRIGATION AND WATER RESOURCES ENGINEERING
Example 13.1 For the following data related to a canal headworks, design the profile of
the weir section:
Flood discharge	= 3000 m^3/s
Maximum winter flood discharge	= 300 m^3/s
HFL before construction of weir	= 255.5 m
RL of river bed	= 249.5 m
Pond level	= 254.5 m
Lacey’s silt factor (for the river bed material) = 0.9
Permissible afflux	= 1.0 m
Bed retrogression	= 0.5 m
Offtaking canal discharge	= 200 m^3/s
Looseness factor	= 1.1
Concentration factor	= 1.2
Permissible exit gradient	= 1/7

The stage-discharge curve of the river is as shown in Fig. 13.6.

Riverstage(metres)

256

255
254

253
252
251
250
249
248

0 1000 2000 3000

Q (cumecs)

Fig. 13.6 Stage-discharge curve for Example 13.1
Solution:

Lecey’s waterway,

Actual width of waterways

Lacey’s regime velocity

P = 4.75 Q = 4.75 3000 = 260.17 m

• 
  P × Losseness factor
  

• 
  260.17 × 1.1 = 286.19 m
  

F Qf12 I ^1/ 6	L 3000	× 0.9	× 0.9 O^1/ 6
^= G		J	^= M	140	P
H	140 K	N	Q

= 1.61 m/s

CANAL HEADWORKS				449
Velocity head	=	161. × 161.	= 0.13 m
2 × 9.81
Level of d/s TEL	= 255.5 + 0.13
	= 255.63 m (255.13 m after retrogression)
Level of u/s TEL	= d/s TEL + Afflux = 255.63 + 1.0
	= 256.63 m (256.13 m after retrogression)
Level of u/s HFL	= 256.63 – 0.13 = 256.5 m (256.0 m after retrogression)

The undersluice bays should carry maximum of the following three discharges: (i) Twice the discharge of the offtaking canal = 2 × 200 = 400 m³/s

(ii) 20% of the design flood discharge = 0.2 × 3000 = 600 m³/s (iii) Maximum winter flood discharge = 300 m³/s

∴ Undersluice discharge = 600 m³/s

Providing the undersluice crest at the river bed level (i.e., 249.5 m), the total head on the undersluice crest during high flood condition = 256.63 – 249.5 = 7.13 m.
Discharge intensity q through the undersluice portion = 1.71 (7.13)^3/2 = 32.56 m³/s/m. On providing two bays of 10.0 m each (and assuming pier contraction coefficient = 0.1), Discharging capacity of undersluices = 1.71 (20 – 0.1 × 2 × 1 × 7.13) (7.13)^3/2

= 604.77 m³/s = 604 m³/s (say)

It should be noted that the pier contraction coefficient would depend on the shape of the pier and may be as small as 0.01 for rounded nose piers.
The weir portion has to carry the remaining flood discharge
= 3000 – 604 = 2396 m³/s On providing 18 bays of 15 m each for the weir portion,

Discharge intensity for weir	=		2396	= 8.87 m^3/s/m
18 × 15
Height of TEL over the weir crest	F	8.87I ^2 /3
	= G	J	= 2.85 m
	H	184.K

Here, the formula q = 1.84 k^3/2 has been used since the weir crest is usually not wide enough to behave as a broad crest.
Required level of the weir crest = 256.63 – 2.85 = 253.78 m
Provide weir crest at 253.70 m so that the total discharge through the weir and undersluice
= 1.84 (270 – 0.1 × 17 × 2 × 2.93) (2.93)^3/2 + 604
= 2401 + 604
= 3005 m³/s against the required value of 3000 m^3/s
and clear waterway = 270 + 20
= 290 m against the required value of 286.19 m
Thus, the requirements of discharge capacity and clear waterway are met by providing 18 weir bays (with weir crest at 253.70 m) of 15 m each and two undersluice bays (with crest at 249.5 m) of 10 m each.

450	IRRIGATION AND WATER RESOURCES ENGINEERING
Required height of undersluice gates	= 254.5 – 249.5 = 5.0 m
Required height of weir shutters	= 254.5 – 253.7 = 0.8 m
Hydraulic jump calculations for weir sections:
Quantity	High flood condition		Pond level condition
Discharge intensity, q (in m^3/s/m)	1.84 (2.93)^3/2 × 1.2		1.84 (0.8)^3/2 × 1.2
	= 11.07^a		= 1.58^a
Critical depth, hc	2.32		0.63
D/S flood level (after retrogression) (m)	255.00		251.25*
D/S TEL (m)	255.13		251.33*
U/S TEL (m)	257.01***		254.58**
Head loss, HL (m)	1.88		3.25
HL/hc	0.81		5.16
E2/hc (from Table 9.1)	2.01		2.74
E2 (m)	4.66		1.73
E1 = E2 + HL (m)	6.54		4.98
Pre-jump depth, h1(m)	1.07		0.162
Post-jump depth, h2 (m)	4.32		1.69
Height of the jump = h2 – h1 (m)	3.25		1.528
Length of concrete floor required	16.25		7.64
(= 5 × height of the jump) (m)
Level of the jump = D/S TEL – E2 (m)	250.47		249.60

The level of the downstream floor should be at or lower than 249.60 m and the length of the downstream horizontal floor should be equal to or more than 16.25 m. However, the river bed is at 249.0 m. Therefore, provide downstream horizontal floor of length 17 m at 249.5 m.