WEIRS

  

Classification of Weirs:

 

Design of Weirs:

                    Hydraulic Design

                           Structural Design

                           Floor Design

                           Detailed Drawings

                    Solved Example

 

 

 

Objectives of Weirs in Irrigation 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)

 

 

 

Reducing head on existing structures

 

  Collecting sediments at US of structures (sand strap)

 Weirs for reducing water slope in steep lands

Distance between weirs

 

 

 

 

 

 

ac = L * Slope (before)

ab = L * Slope (after)

 

rise (R)  = ac – ab

                   = L {Slope (before) – slope (after)}

 

L = distance between weirs

L = R / (natural slope – required slope)

Classification of Weirs According to Geometrical Shape    

 

 

 

 

 

 

 

Classification According to Position in Plan

 

 

Classification According to Dimensions of Cross Section

 

 

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 α H

 

 

b)    Submerged Weir

 

Q = 2/3 Cd B (2g) 0.5 H1.5 + Cd B h1 (2gh2)0.5

 

·        DSWL is higher than weir crest

 

Q α H, h1, h2

 

Classification According to Crest Length (B)

 

 

Design of Weirs

Design of Weirs is divided to 3 parts:                               

                              I.      Hydraulic Design (determination of crest level and weir length according to head)

                           II.      Structural Design (Empirical Dimensioning – check of stability)

 

 

 

III – Detailed 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

 

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 H 1.5

 

 

 

 

2 – Submerged Weir

     Q = 2/3 Cd B (2g) 0.5 h21.5 + Cd B h1 (2*g*h2) 0.5

 

 

 

 

3 – Broad–Crested Weir

      Q = 1.71 Cd B H 1.5

 

     

 

 

 

4 – Fayum Type Weir

      Q = 1.65 B H 1.5

 

 

 

 

 

5 – Standing Wave Weir

Q = 2.05 B H 1.5

 

 

 

II Structural Design

 

1  The super structure                                                                

 

Theoretical Weir Profile

 

 

Scour Length of Weir Floor

 

Scour may be defined as deepening and widening of water channel under the influence of the flowing water with high velocities.

 

     The scour continues until the energy of the flowing water reaches the normal channel energy.

 

 

 

Velocity distribution through scour hole

 

 

Precautions against scour                                            

 

 

 

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.

 

A weir on pervious soil needs length “L” to:

 

a)      Cover percolation length,

b)     Resist scour from falling water

  

Definitions

            Percolation is the flow of water under the ground surface due to an applied differential head

 

            Percolation length (creep length) is the length to dissipate the total hydraulic pressure on the structure

 

            Undermining (Piping) is to carry away (wash) soil particles with flowing water below the ground surface causing collapse or failure of the above structure

 

 

 

Determination of Percolation Length

To determine the critical head: (after which undermining occurs)       

 

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

 

k = vcr L / Hcr         = Qcr L / A Hcr                                          

 

Soil

K (cm / min)

Type of flow

Clean gravel

5000 – 50

Turbulent

Clean sand

50 – 0.05

Turbulent or laminar

Fine sand + silt

0.05 – 0.00005

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 α 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

 

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

 

e = (1 – vs) / 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.

 

 

Soil

icr

F.S.

Fine gravel

0.25 – 0.20

4 –5

Coarse sand

0.20 – 0.17

5 – 6

Fine sand

0.17 – 0.14

6 – 7

Silt & clay

0.14 – 0.12

7 – 8

 

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

 

   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

L`  LB   (design is safe, no possibility of undermining)

L` < LB   (design is unsafe, undermining occurs, leads to failure)

   

 Lane’s 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

 

 

Distance between successive sheet piles

 

·        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;

Water percolation length takes a short cut from a to b;

Actual percolation length is smaller than designed

     unsafe

 

Design Head for Percolation                                        

 

H = USHWL – DSHWL                 (1)

H = USLWL – DSLWL                  (2)

H = Crest level –DSBL                   (3)

 

Design head H is the biggest of (1), (2), and (3)

 

 

 

Determination of Floor Dimensions

 

t1 = 0.5 – 1.0 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 = USHWL – DSBL – Yc

      = Scour head; Yc = critical depth

 

              &                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

 

HD

h2 = H – t1/CB – l1 / CB

t2 = t / (γm) * Factor of safetyγ

t2 = F.S. [ h2 / (γm)] m.;   γm = 2.2 t/m3

t2 = 1.3. [ h2 / (γm)]

then t3 = t2/2    ≥     1 m.

t3 = F.S. [ h3 / (γm)] m               then the head h3 which corresponds to floor thickness t3

L3 = CB * h3 = x + t3           then get distance x

 

 

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 (70 – 100) mm open joints filled with broken stone.

  • An inverted filter of well graded gravel and sand is placed under the blocks in order to prevent the loss of soil through the joints

 

 

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

- 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 m3/s

At summer 

 QA    =   45    =    3 Qii

Qii  =  15 m3/s                       &   Qi =  30  m3/s  

For  branch  ( i )

Qmax /Qmin         =   (2 B H11.5) / (2 B H21.5)     =   H12/H22 

 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

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 )

 

Design of weir floor for canal (i) by applying Bligh  method

 

BED LEVEL  =  10.76 – 4   =   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 = (59.2 – 36) / 2  = 11.6

Use two sheet pile  d =7 m   & d = 5 m

 

h2  =  3.7 - .5/16 – 6/16- (2*7)/16  =   2.9

t2  =   2.9 *  (1.3/1.2)     =   3.1  m      

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