Design of Weirs

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- Overfa

ll Weir (Clear-Overfall)

Q = 2/3 C_d B (2g) ^0.5 H^1.5

·        DSWL is lower than crest level

·        Q is independent of DSWL

·        Q α H

b)    Submerged Weir

Q = 2/3 C_d B (2g) ^0.5 H^1.5 + C_d 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 C_d B (2g) ^0.5 H ^1.5

2 – Submerged Weir

     Q = 2/3 C_d B (2g) ^0.5 h₂^1.5 + C_d B h₁ (2*g*h₂) ^0.5

3 – Broad–Crested Weir

      Q = 1.71 C_d 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……..H_n varies until H_critical (soil particles begin to move)

V_critical = Q_critical / A                    v_cr

v_cr  = k H_cr / L = K i_cr                                                            L = K H_cr / v_cr

k = v_cr L / H_cr         = Q_cr L / A H_cr                                          

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 = v_v/ v_s

e = (1 – v_s) / v_s = (1 / v_s) –1

Or 1+e = 1 / v_s                          or      v_s = 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 = i_cr = ( - 1) / (1 + e)                      can be determined

Safe percolation length     L = H / i_cr

Or L = H / i_cr (F.S.)

Values of i_cr & 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.

C_B = 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` > L_B   (Design is safe, no possibility of undermining)

If L` < L_B   (Design is unsafe, undermining occurs, leads to failure)

       

L` = L + 2 t + 2 S₁ + 2 S₂

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

L` < L_B   (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.

L_L = C_L 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

L_Scour = C_s (H_s) ^0.5

Or

L_Scour = 0.6 C_B (H_s) ^0.5

H_s = USHWL – DSBL – Y_c

      = Scour head; Y_c = critical depth

              &                q = Q / B

where B is the weir length; q is the discharge per unit length

L` = l₁ + l₂ + l_s + 2 t₂

L_B = C_B . 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

H_D

h₂ = H – t₁/C_B – l₁ / C_B

t₂ = t / (γ_m) * Factor of safetyγ

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

t₂ = 1.3. [ h₂ / (γ_m)]

then t₃ = t₂/2    ≥     1 m.

t₃ = F.S. [ h₃ / (γ_m)] 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 H^1.5

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                                                             

Q_A  =   Q_i  +  Q_ii                     &      Q_i   =    2 Q_ii   

Q_A  =  2 Q_ii +  Q_ii

At flood 

Q_A    =   105    =    3 Q_ii 

Q_ii =   35  m³/s                       &   Q_i =  70 m³/s

At summer 

 Q_A    =   45    =    3 Q_ii

Q_ii  =  15 m³/s                       &   Q_i =  30  m³/s  

For  branch  ( i )

Q_max /Q_min         =   (2 B H₁^1.5) / (2 B H₂^1.5)     =   H₁²/H₂² 

 H₁/H₂  =   (Q_max /Q_min )^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)

                           Q_max = 70 = 2 B (1.62)^1.5            B  =  17 m

Q_min  = 30 = 2 B (.92)^1.5              B  = 17 m

                                      B  =  17  m 

   Length  of weir (ii)

                           Q_max = 35 = 2 B (1.62)^1.5            B  =  8.5 m

                           Q_min  = 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 

H_D  =   12.08  -  11       =  1.08

H_D  =   11.38  -  10.76  =  .62

H_D  =   10.46 -  6.76     =  3.7

                                                             take        H_D  =  3.7  m

L_B =  C_B *  H_D     =   16  *  3.7   =   59.2

Assume L₁  =    6 m          L₂  =  6 m

L_S  =  C_S  (H_S)^.5                        C_S  =  .6 C_B

H_S  =   12.08  -   6.76 -  Y_cr           &     H_S   =   4.37       &    L_S  = 20 m 

Assume   t₂  =   2 m

L^\  = 6 + 6 + 20 +  2 * 2 =   36

L^\  <   L_B           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

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