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WEIRS
Classification of Weirs:
Design of Weirs:
Hydraulic Design
Structural Design
Floor Design
Detailed Drawings
Solved Example
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Objectives of Weirs inIrrigation Canals
Proper distribution of water carried by a main canal among the branch canals
depending upon it
Reducing the hydraulic slope (gradient) in a canal (if canal water slope is greater
than the allowable water slope)
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Weirs for reducing water slope in steep lands
Distance between weirs
ac = L * Slope (before)
ab = L * Slope (after)
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rise (R) = acab
= L {Slope (before)slope (after)}
L = distance between weirs
L = R / (natural sloperequired slope)
Classification of Weirs According to Geometrical Shape
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Classification According to Position in Plan
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Classification According to Dimensions of Cross Section
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Classification According to Position of Down-Stream Water Level
a) Free- Overfall Weir (Clear-Overfall)
Q = 2/3 Cd B (2g)0.5
H1.5
DSWL is lower than crest level
Q is independent of DSWL
Q H
b) Submerged Weir
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Q = 2/3 Cd B (2g)0.5
H1.5
+ Cd B h1 (2gh2)0.5
DSWL is higher than weir crest
Q H, h1, h2
Classification According to Crest Length (B)
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Design of Weirs
Design of Weirs is divided to 3 parts:
I. Hydraulic Design(determination of crest level and weirlength according to head)
II. Structural Design(Empirical Dimensioningcheck ofstability)
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IIIDetailed Drawings
For proper Design of Water Structures:
Velocity of Flow:
Must cause minimum Loss in Head
Or minimum Heading Up
Flow of Water in a Channel is controlled either by:
A Weir or
A Regulator
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Weirs: For lands having steep slopes
Regulators: For lands having mild slopes or flat lands
I- Hydraulic Design of Weirs
1- Clear Over fall Weir
Q = 2/3 Cd B (2g)0.5
H1.5
2Submerged Weir
Q = 2/3 Cd B (2g)0.5
h21.5
+ Cd B h1 (2*g*h2)0.5
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3BroadCrested Weir
Q = 1.71 Cd B H1.5
4Fayum Type Weir
Q = 1.65 B H1.5
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5Standing Wave Weir
Q = 2.05 B H1.5
IIStructural Design
1 The super structure
Theoretical Weir Profile
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Velocity distribution through scour hole
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Precautions against scour
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Floor of Heading Up Structures
A weir on solid rock (impervious foundation) does not need long apron (Floor), but needs
sufficient width b to resist soil stresses.
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A weir on pervious soil needs length L to:
a) Cover percolation length,
b) Resist scour from falling water
Definitions
Percolationis the flow of water under the ground surface due to an applied differentialhead
Percolation length(creep length) is the length to dissipate the total hydraulic pressureon the structure
Undermining(Piping) is to carry away (wash) soil particles with flowing water below
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the ground surface causing collapse or failure of the above structure
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Determination of Percolation Length
To determine the critical head:(after which underminingoccurs)
1- Measure Q for different heads
2- H1 ----- Q1, v1= Q1 / A
H2 ------Q2, v2. (k determined)
3- H..Hn varies until Hcritical (soil particles begin to move)
Vcritical = Qcritical / A vcr
vcr = k Hcr / L = K icr L = K Hcr / vcr
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k = vcr L / Hcr = Qcr L / A Hcr
Soil K (cm / min) Type of flowClean gravel 500050 Turbulent
Clean sand 500.05 Turbulent or laminar
Fine sand + silt 0.050.00005 Laminar
Clay < 0.00005 Always laminar
Permeability : (hydr. Conductivity)
Ability of fluid to move in the soil under certain head (dimensions of velocity)
v = k i
i = H / L
v porosity + arrangement of grains
Seepage or percolation below weirs on previous soils:
- a weir may be subject to failure from under seepage
- water head will force (push) the water to percolate through the soil voids
- if water velocity at D.S. end is not safe (> v critical) then undermining occurs, i.e.
water at exit will carry away soil particles
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v = k I (Darcy,s law)
= k dP / dl = k H / L
In practice: icr = vcr / k is unknown
Therefore we carry the 2nd experiment
e = voids ratio
e = vv/ vs
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e = (1vs) / vs = (1 / vs)1
Or 1+e = 1 / vs or vs = 1 / (1+e)
Upward force = H * A
Downward force =
(net weight)
= sp. Gr. Wt. Of soil under water
= ( -1) A L / (1+e)
for stability: H. A. = ( -1) A L / (1 + e)
H / L = icr = ( - 1) / (1 + e) can be determined
Safe percolation length L = H / icr
Or L = H / icr (F.S.)
Values of icr& F.S.
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Soil icr F.S.
Fine gravel 0.250.20 45
Coarse sand 0.200.17 56
Fine sand 0.170.14 67
Silt & clay 0.140.12 78
If I > icr undermining (piping)
i.e. water has v >> to carry away soil particles
Bligh Creep Theory
The length of the seepage path transversed by the water is known as the length of creep
(percolation length).
Bligh supposed that the dissipation of head per unit length of creep is constant throughout the
seepage path.
CB = Bligh coefficient of percolation C B = V/K
Percolation length is the path length from (a) to (b)
LBligh = CBligh . H
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L` = 2 t + L
If L` > LB (Design is safe, no possibility of undermining)
If L` < LB (Design is unsafe, undermining occurs, leads to failure)
L` = L + 2 t + 2 S1 + 2 S2
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L` LB (design is safe, no possibility of undermining)
L` < LB (design is unsafe, undermining occurs, leads to failure)
Lanes Weighted Creep Theory
Lane suggested that a weight of three should be given to vertical creep and a weight of one to
horizontal creep.
LL = CL H
Lane percolation length L` = 1/3 L (horizontal) + L (vertical)
L` = 1/3 L + 2 t + 2 S1 + 2 S2
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Distance between successive sheet piles
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Distance between sheet piles a-a and b-b d1 + d2
Water percolation length takes the right path -----safe
Distance between sheet piles a-a & b-b < d1 + d2;
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Water percolation length takes a short cut from a to b;
Actual percolation length is smaller than designed
unsafe
Design Head for Percolation
H = USHWLDSHWL (1)
H = USLWLDSLWL (2)
H = Crest levelDSBL (3)
Design head H is the biggest of (1), (2), and (3)
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Determination of Floor Dimensions
t1 = 0.51.0 m assumed
t2 is taken 2.0 m or t2 = 0.8 (H)0.5
t3 = t2 / 2 1 m
and l1 is assumed (1-2) H
L2 = is determined according to weir type (3-8) m
LScour = Cs (Hs)0.5
Or
LScour = 0.6 CB (Hs)0.5
Hs = USHWLDSBLYc
= Scour head; Yc = critical depth
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& q = Q / B
where B is the weir length; q is the discharge per unit length
L` = l1 + l2 + ls + 2 t2
LB = CB . H if L` LB no need for sheet pile
If L` < LB unsafe; use sheet pile
Depth of sheet piles = (LB L`) / 2
Sheet pile depth m
Determination of the uplift diagram
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HD
h2 = Ht1/CBl1 / CB
t2= t / (m) * Factor ofsafety
t2 = F.S. [ h2/ (m)] m.; m = 2.2 t/m3
t2 = 1.3. [ h2/ (m)]
then t3 = t2/2 1 m.
t3 = F.S. [ h3/ (m)] m then the head h3 which corresponds to floorthickness t3
L3 = CB * h3 = x + t3 then get distance x
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Precautions Against Percolation
The aprons are of plain concrete blocks of about 1.5 * 1 * 0.75 m deep
For small structure blocks of about 1 * 0.75 * 0.5 m deep may be used
The blocks are placed in rows with (70100) mm open joints filled with broken stone.
An inverted filter of well graded gravel and sand is placed under the blocks in order toprevent the loss of soil through the joints
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EXAMPLE
A canal (A) is divided into two branches (i & ii).The discharge of branch (i)=2Q of branch (ii)
at all times. Two weirs have to be constructed at the entrance of each canal .
Data :-
- Bed width of canals (i & ii ) = ( 23.0 & 8.0 ) m .
- Flood discharge of canal (A) = 105 cum/sec .
- Summer discharge of canal (A) = 45 cum/sec .
- DSHWL in the two canals = ( 11.00 )
- minimum water depth in the two canal branches = 4.0 m .
- Difference between H.W.L & L.W.L in canal(A) = .7 m .
- Submergence in canal (i) = 1/3
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- Bligh coeff. of percolation = 16
- Bed level is constant in canal (A) and its branches .
- Q = 2 B H1.5
If a Board crested weir is constructed at the entrance of the two branches (i&ii) it is required to:-
1- Crest level of weirs ( i & ii ) .
2- Length of each weir .
3- HWL in canals (A) .
4- LWL in canal (A) & (i) .
5- Design of weir floor for canal (i) by applying Bligh method..
solution
QA = Qi + Qii & Qi = 2 Qii
QA = 2 Qii + Qii
At flood
QA = 105 = 3 Qii
Qii = 35 m3/s & Qi = 70 m
3/s
At summer
QA = 45 = 3 Qii
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Qii = 15 m3/s & Qi = 30 m
3/s
For branch ( i )
Qmax /Qmin = (2 B H11.5
) / (2 B H21.5
) = H12/H2
2
H1/H2 = (Qmax /Qmin )2/3
= (70/30)2/3
H1/H2 = 1.527 & H1 = 1.76 H2 (1)
H1 - H2 = .7 (2)
From (1) & (2)
1.76 H2 - H2 = .7 H2 = .92 m
H1 = 1.62 m
h1/H1 = 1/3 h1 = 1.62/3
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1- Crest level of weirs ( i & ii ) = 11 - .54 = ( 10.46 )
2- length of weir (i)
Qmax = 70 = 2 B (1.62)1.5
B = 17 m
Qmin = 30 = 2 B (.92)1.5
B = 17 m
B = 17 m
Length of weir (ii)
Qmax = 35 = 2 B (1.62)1.5
B = 8.5 m
Qmin = 15 = 2 B (.92)1.5
B = 8.5 m
B = 8.5 m
3- HWL in canals (A) = 10.46 + 1.62 = (12.08)
4- LWL in canal (A) = 10.46 + .92 = (11.38)
h2/H2 = 1/3 & h2 = .92/3 = .3
LWL in canal (i) = 10.46 + .3 = ( 10.76 )
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Design of weir floor for canal (i) by applying Bligh method
BED LEVEL = 10.764 = 6.76
HD = 12.08 - 11 = 1.08
HD = 11.38 - 10.76 = .62
HD = 10.46 - 6.76 = 3.7
take HD = 3.7 m
LB = CB * HD = 16 * 3.7 = 59.2
Assume L1 = 6 m L2 = 6 m
LS = CS (HS)
.5
CS = .6 CB
HS = 12.08 - 6.76 - Ycr & HS = 4.37 & LS = 20 m
Assume t2 = 2 m
L\
= 6 + 6 + 20 + 2 * 2 = 36
L\
< LB unsafe use sheet pile d = (59.236) / 2 = 11.6
Use two sheet pile d =7 m & d = 5 m
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h2 = 3.7 - .5/166/16- (2*7)/16 = 2.9
t2 = 2.9 * (1.3/1.2) = 3.1 m
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t3 = t2/2 = 1.6 m > 1
1.6 = 1.3 * h3/1.2 h3 = 1.47
L3 = 16 * 1.47 = X + 2*5 + 1.6 & X = 11.92 m