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Hydraulic Structures –Head and Cross Regulators December 12, 2010
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Head and Cross Regulators
The supplies passing down the parent canal and off take channel are controlled by cross regulator
and head regulator respectively.
Functions of Cross Regulators
1. Regulation of the canal system.
2. Raising the water level in the main canal in order to feed the off take channels.
3. To facilitate communication by building a road over the cross regulator with little extra cost.
4. To absorb the fluctuations in the canal system.
Functions of Head Regulators
1. To regulate and control supplies entering the off take channel (distributary) from the main
(parent) canal.
2. To control silt entering into the distributary.
3. To serve for measurement of discharge.
Alignment
The best alignment of the off take channel is when it makes angle zero with the parent canal initially
and then separates out in a transition. See Fig. 13.1. In this case there is a transition curve for both
off take and parent channel to avoid silt accumulation.
Another alternative by making both channels an angle with respect to parent channel upstream. Fig.
13.2
Cross Regulator
OFF TAKE CHANNELDistributary channel
Head Regulator
Abutment
Pier
Gate
Hydraulic Structures –Head and Cross Regulators December 12, 2010
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In case of obligatory straight alignment of the parent channel, the usual angle of the off take channel
is 60º to 80º (in most important works needs a model study). For excessive silt entry into the off take
channel. Fig. 13.3.
Design Criteria
1. Waterway
The effective waterway of head regulator should not be less than 60% of bed width of off
taking channel and mean velocity should not exceed 2.5 m/sec.
2. Crest level
Crest level of the distributary head regulator is generally kept 0.3 m to 0.6 m higher than
crest level of cross regulator (C.R.). The crest level of C.R. is provided at bed level of parent
canal.
eH should be worked out from the formula
3/2
e eQ C B H (4.1)
where
C Coefficient of discharge
eB Effective length of crest 2t p a eB N K K H (4.2)
tB Net length of the crest
N Number of piers
pK Pier contraction coefficient
aK Abutment contraction coefficient
Hydraulic Structures –Head and Cross Regulators December 12, 2010
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Table: Coefficients of contraction for piers and abutments.
Type of pier pK
Square nosed pier 0.02
Round nosed pier 0.01
Pointed nosed pier 0.01
Type of abutment aK
Square abutment 0.2
Round abutment 0.1
3. Coefficient of discharge (C)
The coefficient of discharge C is 1.84 for crests of width less than or
equal to 23 eH . In case of submerged falls, C should be reduced
depending on the drowning ratio, see Fig. 6.5.
1.840 (H.R. crest)
1.705 (C.R. crest)
C
C
4. Shape of crest
The u.s face of the crest should be given a slope of 1:1. The d.s. sloping glacis should not be
steeper than 2:1.
5. Crest width should be kept equal to 23 eH .
6. Vertical cut-offs
The cut-off should be provided at the end of u.s. and d.s. floors for safety against scour,
undermining and exit gradient. Due to Lacey’s scour depth
T.E.L.
He
>2/3 He
Broad Crested Weir
Round nosed pier
Pointed nose pier
Square nosed pier
Hydraulic Structures –Head and Cross Regulators December 12, 2010
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Table 6.1 Minimum depth of u.s. and d.s. cut-offs
Canal capacity
cumec
Min. depth of u.s. cut-off
below bed level or G.L.
whichever is lower (m)
Min. depth of d.s. cut-off
below bed level or G.L.
whichever is lower (m)
Up to 3 cumec 1.0 1.0
3.1 - 30 1.2 1.2
30.1 - 150 1.5 1.5
Above 150 1.8 1.8
7. Thickness of top coat
Table 6.2 Thickness of top coat.
Canal capacity (cumec) Thickness of top coat (mm)
1.5Q 100
1.5 30Q 150
30 150Q 200
150Q 300
8. Freeboard Table: Minimum Freeboard
Canal capacity (cumec) Freeboard (m)
1.0Q 0.3
1 10Q 0.4
10 30Q 0.6
30 150Q 0.8
150Q 1.0
9. Protection works
Downstream of floor, properly designed filter loaded by concrete blocks should be provided.
The length of inverted filter is kept equal to 2D (D is the depth of d.s. cut-off below d.s.
bed). Details of minimum thickness of the filter are given in Table 6.3. The width of gabs
between the blocks shall not be more than 50 mm which should be packed with biggest size
of pebbles available. Beyond the filter, an apron of 1.5D length shall be provided. Similar
protection is also provided in the u.s. in a length equal to D. the cubic content of material in
launching apron should be equal to 32.25D m m m length.
Hydraulic Structures –Head and Cross Regulators December 12, 2010
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Design Example
Design a cross regulator and a suitable head for a distributary which takes off at an angle of
60º from a canal which discharges 120 cumec.
Discharge of distributary = 10 cumec
Bed width of distributary = 10 m
Water depth of distributary = 1.2 m
Full supply level of distributary = El. 310.2 m
Full supply level of parent channel (F.S.L.) u.s. 311.00
d.s. 310.85
m
m
Bed width of parent channel u.s. 60
d.s. 56
m
m
Depth of water in parent channel u.s. 2
d.s. 2
m
m
Safe exit gradient (G.E.) 1
5
Solution
A. Design of Cross Regulator
1. Fixation of crest level and waterway of cross regulator
Crest level of C.R. F.S.L. of parent channel - water depth
311- 2 309m
u.s. d.s.T.E.L. T.E.L.
311 310.85 0.15
dh
m
u.s.F.S.L. crest level
=311 309 2
eH
m
Degree of submergence
0.150.075
2
d
e
h
H
From Fig. 6.5
0.56, where 1.705
1.705 0.56 0.95
s
s
CC
C
C
3/2
3/2120 0.95 2
e
e
Q CB H
B
44.66 say 45eB m m
Assume 6 bays of 8 m each = 48 m
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2( )
45 2 5 0.01 0.2 2
46 48 O.K.
t e p a eB B N K K H
m m
Provide 5 piers with rounded nose of width 1.6 m each.
Total waterway 6 8 5 1.6 56 O.K.m
2. Level and length of downstream floor
3
120
1202.5 / /
48e
Q cumec
Qq m sec m
B
2
u.s. d.s. u.s. d.s.T.E.L. T.E.L. F.S.L. F.S.L. ( 2 is negligible)
311 310.85 0.15
LH v g
m
From Blench Curves, Fig. 3.5
2 1.435Ef m
d.s. floor level d.s. 2F.S.L.
310.85 1.435 309.415
Ef
m m
Actual cistern level 310.85 2 308.85m lower than 309.415 m O.K.
Cistern length 2 1
2 1
5 firm soil
6( ) weak soil
cL D D
D D
1 2 1.435 0.15 1.585LEf Ef H m
From energy of flow curves (Fig. 2.7)
1 1
2 2
1.575 gives 0.534
1.435 gives 1.3
5 1.3 0.534 3.83c
Ef m D m
Ef m D m
L m
3. Vertical cutoffs
120Q cumec
Referring to table 6.1, the minimum depth of u.s. and d.s. cutoff = 1.5 m.
u.s. cutoff is at El. 309 1.5 307.5m 4. Total floor length and exit gradient
1E
HG
d
u.s. d.s.F.S.L. G.L.
311 308.85 2.15
H
m
Trial and error
d (m) λ α L d (m)
1.5 5.20 9.15 14
1.6 4.57 8.08 12.93
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1 2.15 1, =4.57
5 1.6
1/22
1/22
2 1 1
2 457 1 1 8.08
Total floor length
8.08 1.6 12.93 say 13
L d
m m
d.s. cutoff is at El. 308.85 1.6 307.25m
d.s. floor length = 6.5 m
d.s. glacis length with (2H:1V) slope = 2(309-308.85) = 0.3 m
u.s. floor length = 6.2 m
Total floor length = 13 m
5. Pressure calculation
a. Upstream cutoff
1
1
1.5 , 13
1 1.50.115
13
100 20 80%
100 28 72%
D
C
d m b m
d
b
Correction of C for floor thickness
0.6
80 72 3.2%1.5
D C
t
d
corrected 72 3.2 75.2% of C H
b. Downstream cutoff
1.6 , 13
1 1.60.123
13
31%
22%
E
D
d m b m
d
b
Correction of E for floor thickness 0.6
31 22 3.38%1.6
corrected 31 3.38 27.62% of E H
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6. Floor thickness
a. d.s. floor
1. at 2 m from d.s. end
75.2 27.62%pressure 27.62 1.5
12
33.6%of H
Head % pressure
33.6=2.15 0.722 of water
100
H
m
Minimum concrete thickness min
Head 0.7220.58
2.25 1c w
t m
Provide 0.6 m thick concrete floor for 1.5 m.
2. at 4.5 m from d.s. end
75.2 27.62%pressure 27.62 4.0
12
43.48%of H
Head % pressure
43.48=2.15 0.934 of water
100
H
m
min
Head 0.9340.75
2.25 1c w
t m
Provide 0.8 m thick concrete floor for 2.5 m.
3. at 6.5 m from d.s. end
75.2 27.62%pressure 27.62 6.0
12
54.41%of H
min
2.15 0.51410.9
2.25 1t m
Provide 1.0 m thick concrete floor for the rest of the glacis.
7. Upstream protection
i. Block protection (inverted filter)
Length of protection = depth of cutoff (D) =1.5 m
Provide 2 rows of 0.8 0.8 0.6m m m C.C. blocks over 0.6 m thick inverted
filter.
ii. Launching apron
Volume per meter length 3 22.25D m m m
If thickness is 1.0 m, then the required length2.25 2.25 1.5
3.371.0 1
Dm
Provide 3.5 m long launching apron.
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8. Downstream protection
i. Block protection
Length of protection 2 2 1.6 3.2D m
Provide 4 rows of 0.8 0.8 0.6m m m blocks over 0.6 m inverted filter.
ii. Launching apron
Volume per meter length 3 22.25D m m m
If thickness is 1.0 m, then
Required length2.25 2.25 1.6
3.61.0 1
Dm
Provide 4.0 m long launching apron.
Provide 0.4 m thick and 1.2 deep toe wall between filter and launching apron.
B. Design of distributary head regulator
1. Fixation of crest and waterway
The crest level should be provided 0.5 m higher than upstream floor level.
Crest level 309 0.5 309.5m
311 309.5 1.5eH m
311 310.2 0.8dh m
0.80.533
1.5
d
e
h
H
0.98sC
C (From Fig. 6.5)
1.84 0.98 1.80sC
The effective width of waterway is found by
1.5
1.510 1.8 1.5
3
e e
e
e
Q C B H
B
B m
Provide 60% of distributary width60
10 6100
m
Provide 2 bays of 3 m each separated by 1 m thick pier.
Therefore, the overall waterway 2 3 1 7m
2. Level and length of d.s. floor
10 , waterway 6
101.67
6
Q cumec m
q m sec
Head loss 311 310.2 0.8LH m
Using Blench curves (Fig. 3.5)
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2 2
1 2 1
1.37 gives 1.32
1.37 gives 0.32L
Ef m D m
Ef Ef H m D m
Cistern length
2 15
5 1.32 0.32 5.0
cL D D
m
Provide cistern length = 6m
Floor level required d.s. 2T.E.L.
310.2 1.37 308.83
Ef
m
Provide d.s. floor level at El. 308.8 (Ground Level)
3. Vertical cutoffs
a. u.s. cutoff
Provide u.s. cutoff depth 1.5 m
The bottom of cutoff 309 1.5 307.5m
b. The minimum d.s. cutoff depth = 1.2 m
4. Total floor length and exit gradient
Maximum static head u.s. d.s.F.S.L. F.L.
311 308.8 2.2m
Exit gradient
1E
HG
d
1/2
22 1 1
1 2.2 1
5 1.2
With trial and error
d (m) λ α L d (m)
1.2 8.50 15.99 19.00
1.4 6.26 11.47 16.05
1.6 4.75 8.50 13.62
1.7 4.24 7.42 12.60
Get 13 , 1.7L m d m
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d.s. floor length (cistern length) = 6.0 m
d.s glacis length 2H:1V slope 2 309.5 308.8 = 1.4 m
Crest length 2 23 3
1.5eH = 1.0 m
u.s. glacis length 1H:1V slope 1 309.5 309.0 = 0.5 m
u.s. floor length = 4.1 m
Total floor length = 13 m
5. Pressure calculations
a. Upstream cutoff
1
1
1.5 , 13
1 1.50.115
13
100 20 80%
100 28 72%
D
C
d m b m
d
b
Assuming the u.s. floor thickness = 0.6 m
Correction of C for floor thickness 0.6
80 72 3.2%1.5
corrected 72 3.2 75.2% of C H
b. Downstream cutoff
1.7 , 13
1 1.70.13
13
32%
22%
E
D
d m b m
d
b
Correction of E for floor thickness 0.6
32 22 3.5%1.7
corrected 32 3.5 28.5% of E H
6. Floor thickness
a. d.s. floor
1. at 2 m from d.s. end
75.2 28.5%pressure 28.5 1.5
12
34.33%of H
Minimum concrete floor thickness min
0.3433 2.20.6
2.25 1t m
Provide 0.6 m thick concrete floor for 1.5 m length.
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2. At 4 m from d.s. end
Provide 0.8 m thick concrete floor for 2.5 m.
3. At the toe of glacis (beginning of the hydraulic jump)
5.5
%pressure 28.5 75.2 28.512
49.9%of H
Unbalanced head 0.499 2.2 1.095m
Unbalanced head due to dynamic condition
2 150%
1 49.91.32 0.32 0.8
2 100
0.898
toe LD D H
m
min
1.0950.88
2.25 1t m
Provide floor thickness 1.0 m.
b. d.s Floor thickness
Same as provided in u.s. floor for the cross regulator (minimum thickness of 0.6 m in
the u.s. which should be thickened under the crest).
7. u.s Protection
Same as provided in the u.s. of cross regulator.
8. d.s. protection
i. Block protection (inverted filter)
Length of filter 2
2 1.7 3.4
D
m
Provide 6 rows of 0.6 0.6 0.4m m m C.C. blocks over 0.4 m thick graded filter.
ii. Launching apron
Volume per meter length 32.25 2.25 1.7 3.825D m m
Assume thickness of launching apron = 0.8 m, then
Required length3.8
4.750.8
m
Provide 5.0 m long launching apron.
Masonry toe wall 0.4 m thick and 1.2 deep shall be provided between filter and
launching apron.
Hydraulic Structures –Head and Cross Regulators December 12, 2010
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