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Design of Hydraulic Structures 3rd year/ water Resources Engineering
1
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 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.
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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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
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 of Hydraulic Structures 3rd year/ water Resources Engineering
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Design Criteria
1 . W a t e r w a y : 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.
The effective head (He) should be worked out form the formula:Q = C.Be.H
1.5
where
C = coefficient of discharge.
Be = effective length = Bt – 2(N*Kp + Ka)*He
Bt = Net length of crest.
N = number of piers.
Kp = piers contraction coefficient
Ka = abutment coefficient
Table: Coefficients of contraction for piers and abutments.
Type of pier Kp
Square nosed pier 0.02Round nosed pier 0.01Pointed nosed pier 0.01Type of abutment Ka
Square abutment 0.2Round abutment 0.1
Square nosed pier
Pointed nosed pier
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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C = 1.840 (H.R. Crest)
C = 1.705 (C.R. Crest)
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 2/3 He.
6. Level & Length of D/S Floor
They are fixed for the conditions:
a – Full supply, when all the gates of C.R. and H.R. fully opened (generally is
governed).
b – The discharge of the parent channel is insufficient but the off take channel is
running full by partial opening of the C.R. gates or vice versa.
Downstream Floor Level:
For these conditions of flow, the discharge intensity q and HL is worked
out. The corresponding value of Ef2 is read from fig. 13.9. The d/s floor level is
then d/s T.E.L. – Ef2 .
3. Coefficient of discharge (C): The coefficient of discharge C is 1.84 for
crests of width less than or equal to eH3
2. In case of submerged falls, C should
be reduced depending on the drowning ratio, see Fig. 6.5.
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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7. Length of the downstream floor.
Ef1 = Ef2 + HL
The values of D1& D2 are found from energy curve for the corresponding
values of Ef1 & Ef2. The length of the downstream floor = 5 to 6 (D2 – D1). This
length should not be less than 2/3 of the total floor length.
8. 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.
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.
Min. depth of d.s. cut-off
below bed level or G.L.
Up to 3 cumec 1.0 1.03.1 - 30 1.2 1.2
30.1 - 150 1.5 1.5Above 150 1.8 1.8
9. Total floor Length & Exit gradient
If the d is the depth of d/s cut off, and H is the maximum static head when
channel is closed and F.S.L. (Full Supply Level) is maintained in the u/s for
feeding the off take channel (i.e.) H = u/s F.S.L. – d/s floor level.
1
d
HGE ; or
H
dGE
1
Find the value of λ and find α from
2
11 2
2/12112
Therefore the total floor (b) is found from:
b = α*d
If b is large, the depth of d/s cut off should be increased and floor length
correspondingly reduced keeping safe exit gradient.
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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10.Pressure Calculations & Floor Thickness
Use Khosla's method to calculation pressure at key points.
a – Static head pressure
b – Dynamic head pressure
Dynamic head at toe of glacis
H = 50 %*( D2 – D1) + at toe * HL
The minimum thickness floor concrete excluding the thickness of the top
coat shall be as:
For Q > 1.5 cumecs 600 mm
Q < 1.5 cumecs 300 mm
The floor concrete from u/s cut off to d/s cut off should be laid by top coat of:
Q incumecs
Thickness oftop coat (mm)
Up to 150 300150 – 30 20029 – 1.5 150Less 1.5 100
11. Free Board
The following free board shall be adopted:
Capacity of channel (cumecs) F.B. (m)
Less than 1.0 0.31.0 – 10 0.410 – 30 0.630 – 150 0.8Over 150 1.0
12. Protection Works
Properly designed filter loaded by concrete blocks should be provided. The
length of inverted filter is kept 2D (D is the depth of d/s cut off below d/s bed).
Details of minimum thickness of filter are given in table 6.3 (page 214). The
width of gabs between blocks shall not be more than 50 mm which should be
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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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 length
equal to D. The cubic content of material in lunching apron should be equal to
2.25D cu. m/m length.
Design Example:
Design a cross regulator and suitable head for distributary which take off at
angle 60o from a canal with discharge of 120 m3/sec.
Discharge of distributary = 10 m3/sec
Bed width = 10 m
Water depth = 1.2 m
F.S.L. of distributary = 310.2 m
F.S.L. of parent channel u/s = 311 m
d/s = 310.85 m
Bed width of parent channel u/s = 60 m
d/s = 56 m
Water depth in parent channel u/s = 2.0 m
d/s = 2.0 m
GE = 1/5
Design:
a. Design of Cross Regulator
1. Fixation of crest level and waterway
Crest level of C.R. = F.S.L. of parent channel – water depth
= 311 – 2 = 309 m
The u/s floor shall be provided at the same level.
Degree of submergence = hd / He
hd = 311 – 310.85 = 0.15 m
He = u/s F.S.L. – crest level
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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= 311 – 309 = 2 m
075.02
85.310311
He
hd
From fig. 6.5 page 252
95.0705.1*56.0
56.0
CModified
C
Cs
2/3.. ee HBCQ
Be = effective width
2/32**95.0120 eB
Be = 44.6 m ≈ 45 m
Assume 6 bays of 8 m each.
Use rounded piers & square abutment; kp = 0.01, ka = 0.2, net length
Bt = Be + 2(N*Kp + Ka)*He
= 45 + 2*(5*0.01 + 0.2)*(2)
= 46 m < 48 m O.K.
Provide 5 piers with rounded nose of width 1.6 m each
The total water way = 6.0 * 8 + 1.6 * 5
= 48 + 8 = 56 m
2. Level and length of downstream floor
Q = 120 cumecs
q = discharge intensity between piers.
sec/5.248
120 2mB
Head loss HL = u/s F.S.L – d/s F.S.L.
= 311 – 310.85 = 0.15 m
From fig 2.5 (Blanch curve) for q = 2.5 m2/sec and HL = 0.15 m we get:
Ef2 = 1.435 m
D/s floor Level = d/s F.S.L. – Ef2
= 310.85 – 1.435 = 309.415 m
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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Adopt cistern level = 310.85 – 2 = 308.85 m is lower than 309.415 m.
design d/s floor level on El. = 308.85 m
Ef1 = Ef2 + HL
Ef1 = 0.15 + 1.435 = 1.585 m
From energy curve we get: D2 = 1.3 m
D1 = 0.534 m
Cistern length = 5*(D2 – D1)
= 5*(1.3 – 0.534)
= 3.83 m
3. Vertical Cut offs
Referring ton table 6.1, the min. depth of u/s & d/s cut off for this range of
discharge = 1.5 m.
(u/s cut off is at El. = 309 – 1.5 = 307.5 m, and d/s cut off is at El. 308.85 –
1.6 = 307.25 m)
4. Total floor length and exit gradient
u/s F.S.L = 311 m, d/s floor = 308.85 m
max. static head (H) = F.S.L. u/s – floor level d/s
= 311 – 308.85
= 2.15 m
d = depth of d/s sheet pile (cut off) = 1.6 m
1
d
HGE
1
6.1
15.2
5
1 ; λ = 4.57
2
11 2
22112
2/12112
α = 8.08
Total floor length required b = α*d = 8.08 * 1.6 = 12.98 m.
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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= 12.8 from book
d/s floor length = 6.5 m
d/s glacis length with 2:1 slope
= 2(309 – 308.85) = 0.3 m
u/s floor length including crest = 6.2 m
Total = 13 m
5. Pressure Calculation
a. Upstream Cut off
d = 1.5 m
b = 13 m
115.013
5.11
b
d
D1 = 100 – 20 = 80% of H
C1 = 100 – 28 = 72 % of H
Correction of C for thickness
Assume t = 0.6 m
%2.3)7280(5.1
6.0 CDt
d
tC
Correction due to interference of d/s cut off may be neglected
C corrected = 72 + 3.2 = 75.2 % of H
b. Downstream Cut offd = 1.6 m
b = 13 m
123.013
6.11
b
d
E = 31 % of H
D = 22 % of H
Correction of E for thickness
Assume t = 0.6 m
%38.3)2231(25.2
6.0tC
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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Correction due to interference of u/s cut off may be neglected
E corrected = 31 – 3.38 = 27.62 % of H
6. Floor thickness
a. Downstream floor
I. At 2 m from d/s end
% pressure = 6.33)62.272.75(12
5.162.27 % of H
Unbalanced head = H * % pressure= 2.15 * 0.336 = 0.722
Floor thickness required = 58.025.1
722.0 m of concrete
Provide 0.6 m thick floor
II. At 4.5 m from d/s end
% pressure = 48.43)62.272.75(12
462.27 % of H
Floor thickness = 75.025.1
4348.0*15.2 m of concrete
Provide floor thickness = 0.8 m of concrete
III. At 6.5 m from d/s end (toe of glacis).
% pressure = 41.51)62.272.75(12
662.27
Floor thickness = 88.025.1
5141.0*15.2 m of concrete
Provide floor thickness = 1.0 m of concrete
b. Upstream floor
0.6 m thick shall be provided for the u/s floor. It should thickened to 1 m
under crest
7. Protection
I. Upstream Protection
a. Block Protection
Block protection in u/s is provided in length (L) = D
L = 1.5 m
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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For Q = 120 cumecs
see table 6.3
Provided 2 rows of 0.8 * 0.8 * 0.6 c.c. block
Over 400 (or 600) mm thick apron in length = 1.6 m
b. Lunching apron
Vol. / m length = 2.25*D m3/m
= 2.25*1.5 = 3.37 m3/m
If thickness = 1 m
length required = 3.37 m
Provide launching apron in 3.5 m length
II. Downstream Protection
I. Inverted Filter (D = 1.6 m)
Length of filter (L) = 2*D
= 2*1.6 = 3.2 m
From table 6.3, provide 4 rows of 0.8 * 0.8 * 0.6 c.c. block over 600 mm
thick graded
II. Lunching apron
Vol. / m = 2.25*D m3/m
= 2.25*1.6 = 3.6 m3/m
If the thickness of apron be 1 m, the length required 3.6 m. provide 4 m
length. Provide 0.4 m thick and 1.2 m deep toe wall between inverted filter and
launching apron.
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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b. Design of Distributary head Regulator
1. Fixation of crest level and waterway
The crest level of distributary head shall be provided 0.5 m higher than u/s
floor level
u/s level = bed level of parent channel = 309 m
Crest level = 309 + 0.5
= 309.5 m
He = 311 – 309.5
= 1.5 m
hd = u/s F.S.L. – d/s F.S.L. (distributary)
= 311 – 310.2 = 0.8 m
533.05.1
8.0
He
hd
5.698.0 figfromC
Cs
C = 1.84 for sharp crested weir,
Coefficient of discharge
8.198.0*84.1 Cs
mBe 3)5.1(*8.1
105.1 , This is very small
Provide 60% of distributary width
= 0.6 * 10 = 6 m
Use 2 bays of 3 m each separated 1 m thick piers.
The overall water way = 3 * 2 + 1 = 7 m
2. Level and length of downstream floor
Q = 10 cumecs
L = 6 m (net water way)
sec/67.16
10 2mB
HL = 311 – 310.2 = 0.8 m
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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From Blanch curve for q = 1.67 m2/sec and HL = 0.8 m we get:
Ef2 = 1.37 m.
Floor level required = d.s. F.S.L. – Ef2
= 310.2 – 1.37 = 308.83 m
Provide d/s floor level at El. 308.8 m
Ef1 = Ef2 + HL
Ef1 = 1.37 + 0.8 = 2.17 m
From energy curve we get: D2 = 1.32 m
D1 = 0.32 m
Length of cistern = 5*(D2 – D1)
= 5*(1.32 – 0.32)
= 5 m
Provide cistern length = 6 m
3. Vertical Cut offs
a. U/S cut off
Provide u/s cut off depth = 1.5 m.
The bottom R.L. of cut off = 309 – 1.5 = 307.5 m
b. D/S cut off
The min. d/s cut off depth (see table 6.1) = 1.2 m below d.s. floor.
Provide d/s cut off depth = 1.6 m from safe exit gradient consideration.
4. Total floor length and exit gradient
Max static head (H) = F.S.L. u/s – floor level d/s
= 311 – 308.8
= 2.2 m
Depth of d/s cut off (d) = 1.6
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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G.E. = 1/5
1
d
HGE
1
6.1
2.2
5
1 ; λ = 4.79
2/12 )1)1*2(( = 8.52
Total floor length required (b) = α*d = 1.6 * 8.52 = 13.63 m
13bال نأخذ لالستمرار مع نتائج المث =
d/s floor length = 6 m
D/s glacis length 2:1 slope = 2(309.5 – 308.8) = 1.4 m
Crest width = 1m
U/s glacis length with 1:1 slope = 1(309.5 – 309) = 0.5 m
U/s floor = 4.1 m
Total length = 13 m
5. Pressure Calculation
a – Upstream cut off
d = 1.5 m
b = 13 m
115.013
5.11
b
d
D1 = 80 % of H
C1 = 72 % of H
Correction due to thickness
Assume floor thickness t = 0.6 m
%2.3)7280(5.1
6.0tC
C1 corrected = 72 + 3.2 = 75.2 % of H
b. Downstream Cut off
d = 1.6 m
b = 13 m
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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13.013
6.11
b
d
E = 32 % of H
D = 22 % of H
Correction due to thicknessAssume t = 0.6 m
%75.3)2232(6.1
6.0tC
E corrected = 32 – 3.75 = 28.25 % of H
6. Floor Thickness
a. D/S floor thickness
Max H = 311 – 308.8
= 2.2 m
I. at 2 m from d/s end
% pressure = 12.34)25.282.75(12
5.125.28 % of H
Min floor thickness = 6.024.1
22.0*3412.0 m of concrete
Provide floor thickness = 0.6 m.
II. at 4 m from d/s end
% pressure = 5.41)25.282.75(12
5.325.28 % of H
Min floor thickness = 73.024.1
22.0*415.0 m of concrete
Provide floor thickness = 0.8 m of concrete
III. at toe of glacis or 6 m from d/s end
% pressure = 77.49)25.282.75(12
5.525.28 % of H
Unbalanced head = 0.4977 * 2.2 = 1.095 m
Unbalanced head due to dynamic condition =50 %*( D2 – D1) + at toe * HL
= 8.0*100
77.49)32.032.1(
100
50 = 0.898 m
Design as static head = 1.095 m
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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Min fool thickness = 88.024.1
095.1 m of concrete
Provide 1 m thickness under toe.
b. U/S floor thicknessMin. thickness of 0.6 m shall be provided in the upstream which should be
thickened under the crest.
7. Protection
a . Upstream Protection
Same as provided in u/s of cross regulator
b. Lunching apron
I. Inverted filter (Block protection)
Length of filter = 2D
= 2*1.7 = 3.4 m
Provide 5 rows of 0.6 * 0.6 * 0.4 m c.c. blocks over 0.4 m thick graded
filter.
II. Launching apron
Vol. / unit length = 2.25*D m3/m
= 2.25*1.7 = 3.83 m3/m or (m2)
Assume the thickness of apron = 0.8 m
Length required = m5.48.0
6.3
Provide 5 m length of lunching apron.
Masonry wall = 0.4 * 1 deep
Shall be provide between filter and lunching apron
Design of Hydraulic Structures 3rd year/ water Resources Engineering
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Design of Hydraulic Structures 3rd year/ water Resources Engineering
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