WEIRS design1

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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

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