Contents preface (VII) introduction 1—37

Upstream and Downstream Cutoffs:
Concentration factor is accounted for in the computations of scour depth for determination of depths of cutoffs. Using Eq. (13.3), the depth of scour below HFL,

F	1107. × 1107.I	1/ 3
R = 135. G		J	= 6.95 m
0.9
	H	K

^a after taking into account concentration factor of 1.2.

• 
  At the pond level, total discharge through weir and undersluice bays
  

• 
  1.84 (270 – 0.1 × 2 × 17 × 0.8) (0.8)^3/2 + 1.71 (20 – 0.1 × 2 × 1 × 5) (5)^3/2
  
• 
  715.15 m³/s
  

∴	Corresponding river stage	= 251.75 m
∴ D/S flood level (after retrogression)	= 251.25 m
For Q = 715.15 m^3/s,
Lacey’s regime velocity	= 1.267 m/s
∴	Velocity head	= 0.08 m
Level of D/S TEL	= 251.25 + 0.08 = 251.33 m
** Level of U/S TEL	= 254.50 + 0.08 = 254.58 m
*** Total head over the crest (for q = 11.07 m^3/s/m)	= (11.07/1.84)^3/2 = 3.31 m
Level of U/S TEL	= 253.7 + 3.31 = 257.01 m

CANAL HEADWORKS

451

RL of bottom of the scour hole on the upstream side

• 
  Upstream HFL – 1.5R
  

• 
  256.5 – 1.5 × 6.95 = 246.08 m
  

Therefore, provide upstream cutoff up to the elevation of 246.0 m. Further, RL of bottom of the scour hole on the downstream side

• 
  downstream HFL – 2R
  

• 
  255.5 – 2 × 6.95 = 241.6 m
  

Therefore, provide downstream cutoff up to the elevation of, say, 241.5 m. Then, depth of the downstream cutoff, d = 249.5 – 241.5 = 8.0 m

(The permissible exit gradient of 1/7 indicates absence of boulder material in the bed and one may, therefore, provide sheet piles instead of concrete cutoff.)
Maximum static head = Pond level – D/S floor level = 254.5 – 249.5 = 5.0 m

Using Eq. (13.4), the exit gradient, GE =	5.0	1	=	1
		8.0	π λ		7
∴	λ = 1.94

Further,	λ	F	1	+	1 + (b / d)	2	I	= 1.94
= 0.5 H		K
∴	b = 21.61 m
Total impervious floor length:

Downstream horizontal floor = 17 m

Horizontal length of the downstream glacis with a slope of 1(V) : 3(H) = 3 (253.70 – 249.5) = 12.6 m

Let the crest width be 2 m.

Horizontal length of the upstream slope (1(V):2(H)) of the weir = (253.7 – 249.5) × 2 = 8.4 m

After providing these essential floor lengths, the total imervious floor is already 40 m long. Providing 2.5 m long upstream horizontal floor, the total length of the impervious floor is 42.5 m, which is more than the required value of 21.61 m from the exit gradient consideration.

Loose protection on upstream and downstream:
Concentration factor is not to be accounted for in computation of scour depth for designing loose protection.
Discharge intensity of weir at high flood = 1.84 (2.93)^3/2 = 9.23 m³/s/m

F	9.23 × 9.23I	1/ 3
Depth of scour below HFL, R = 135. G	J	= 6.15 m
H	0.9 K

∴ RL of bottom of scour hole on the upstream side

• 
  upstream HFL – 1.5R
  

• 
  256.5 – 1.5 × 6.15 = 247.28 m
  

Depth of scour below river bed = 249.5 – 247.28 = 2.22 m

452 IRRIGATION AND WATER RESOURCES ENGINEERING
Similarly, the depth of scour below the river bed on the downstream

• 
  249.5 – (downstream HFL – 2R)
  

• 
  249.5 – (255.5 – 2 × 6.15) = 249.5 – 243.2 = 6.3 m
  

Thus,
Length of upstream block protection ≥ 2.22 m (Provide 2.3 m) Length of upstream launching apron = 2 × 2.22 = 4.5 m (say)

Length of the downstream block protection ≥ 1.5 × 6.3 m (Provide 10 m) Length of downstream launching apron = 1.5 × 6.3 = 10 m (say)
The line sketch of the designed weir profile is shown in Fig. 13.7. Following the procedure of Example 10.2, one can determine the floor thickness at various locations of the weir section. Similarly, one can determine the profile and design the floor thickness of the undersluice portion.

		256.5U/S HFL																D/S HFL
						Pond level
								253.7
254.5
Launching	Block
apron	protection
249.5		4.5				2.5					8.4			2.0			12.6				17.0				10.0			10.0
				2.3 246.0																			Inverted	Launching
							U/S cutoff																filter			apron
																							Downstream
																											cutoff
																									241.5