Rectangular Weir

INTRODUCTION

Surplus weirs are used to dispose of the surplus water from tank to downstream channels, downstream or surplus courses, or drainage channels. It disposes the water from upstream catchments to down stream catchments or downstream tanks. These interlinked tanks in a certain area are called a “Cascade”. Tanks in a cascade all have a common supply channel.

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)

TERMINOLOGY USED

•        Discharge Q

•        Coefficient Of Discharge C_d

•        Maximum Water Level M.W.L (The maximum water level is higher than full tank level in the water spread area of the tank for storing the water. When the inflow is higher than outflow of water in surplus weirs that time the water could reach the maximum water level. It restricted to store the water up to this maximum water level during rainy days.)

•        Full Tank Level F.T.L (The full tank level is the highest level which water could be stored in the tank. Since or any excess water will go out through the surplus weir.₎

•        Catchment Area M

•        Height Of Weir H

•        Head Over The Weir h 

  

METHODOLOGY

Ø Collecting catchment area details

Ø Selection of type of  work

Ø Estimation of flood discharge.

Ø Design of weir.

DESIGN ELEMENTS

1)   Estimation of flood discharge.

2)   Length of surplus weir.

3)   Crest width.

4)   Base width.

5)   Aprons of the weir

6)   Upstream aprons

7)   Downstream aprons

SELECTION  OF TYPE OF WEIR WORK

•        Type A: weir water is dropped immediately below crest wall.

•        Type B: delivered from crest on to a sloping masonry apron.

•        Type C: similar to B but with rough sloping apron.

•        Type D: weir with stepped aprons

RYVE’S FORMULA

Ryve’e Coefficient C	Q=CA2/3
15	15*A2/3
9	9*A2/3
8.1	8.1*A2/3
6.8	6.8*A2/3

§  Q=C*A^2/3

§  A= Area of catchment

§  C=Ryve’s coefficient

NO.	Type of weir	Value of  Cd	Reduced formula for discharge per meter length of weir
1	Weirs with crest width up to one meter	0.625	1.84*H3/2
2	Weirs with over one meter width	0.562	1.66*H3/2
3	Rough stone sloping escapes	0.500	1.48*H3/2
4	Flush escapes	0.437	1.29*H3/2

Q=2/3*Cd*L*H*(2gH)1/2  Where,             L= length of weir             H= height that is (MWL-FTL)              g=9.81m/s2              Q= discharge already calculated above

SAMPLE PROBLEM DESIGN

EXAMPLE:  Design a surplus work of a tank forming part of a chain of tanks. The combined catchment area of the group of tanks is 25,89 sq. kilometers and the area of the catchment intercepted by the upper tanks is 20.71 sq. kilometers.

Design: - Here,

                    MWL (maximum water level) =12.75m

                   FTL (full tank level)                 =12m

Estimation of flood discharge entering the tank:-

Combined catchment area of the group of tanks= 25.89 sq. kms

Intercepted catchment area of the upper tanks   = 20.71 sq. kms

Now,

               Q= CM^2/3-cm^2/3

Where,

               C= 9 (assumed)

               C= 1.50 (assumed)

               Q= 9*25.89^2/3-1.50*20.71^2/3

               Q=67.45 cubic meters

Length of surplus weir:-

FTL                                                     =+12m

Crest level                                             =+12m

Submersion of foreshore i.e. M.W.L. =+12.75 m

Therefore, head of discharge over the weir is 12.75-12 (MWL-FTL) = .75m

Since, temporary regulating arrangements are to be made on top of the weir, to store water at times of necessity, grooved dam stones 15cms x 15 cms, will be fixed in the centre of the crest at one meter interval with top at MWL.

The weir may be assumed as broad crested weir.

So,

      Q=2/3*C_d*h*(2gh)^1/2

Where,

       C_d =0.562

       h   =0.75m

Q=1.66*h2/3

   =1.66*0.752/3

   =1.66*0.65

   =1.08 cubic meters/sec

Clear length of surplus weir required= 67.45/1.08

                                                     =62.5 or say

                                                             =63 meters

Since, the dam stones are to be fixed on top at 1 meter clear intervals, the no. of openings will be 63.

So, the no. of dam stones are required will be 62 Nos.

Size of the dam stones             =15cms x 15cms

Projecting length above crest   =75cms

Therefore, overall length of the surplus weir between the abutments = 63+62*0.15= 72.30meters

Providing overall length of 75meters.

•         Weir:-

Crest level                                            =+12.00 FTL

Top of dam stone (Top of shutter)              =+12.75 MWL

Ground level                                                 =+11.00

Level where hard soil at foundation is met =+9.50

Foundation level is fixed at 9m.

Foundations are about 0.50m deeper into soil.

Thickness of foundation concrete is 0.60m.

So,

Top of foundation concrete                         =+9.60

Height of weir above foundation                =12.00-9.60

                                                                        =2.40meters

Crest width:-

Generally the crest width is assumed as 0.55[(H)^1/2+(h)^1/2]

Where,

               H =height of weir

               H =head over the weir

a=0.55[(2.40)^1/2+(0.75)^1/2]

  =1.3

So crest width is taken as 1.30m.

Base width:-

The maximum overturning moment due to water thrust is equal to

M₀= (H+S)³/6

Where,

               H= height of weir above the foundation

                     S= height of shutters

The slope of weir on either side being the same, the restoring moment M of the weir due to wt. of masonry is

Where,

               H= height of weir

                p= sp. Gravity of masonry,(=2.25)

                a= crest width of weir (in this problem a=1.30)

                b= base width of weir,(=2.40meters)

                 S= height of shutters above weir crest,(=0.75)

The weir will have a trapezoidal profile.

Height of wall above foundation in upstream side= 13.05-9.60= 3.45 meters

Upstream side transition

In order to give an easy approach, the upstream side wing wall are given a splay of 1 in 3.

Downstream side Wings and Returns

As the water after passing over the weir goes down rapidly to normal M.F.L., the wings and returns need not be as high as those on the upstream sides.

The top of wing wall may be fixed at +11.00.

So the height of wall above foundation concrete= 11.00-9.60= 1.40 meters.

Base width required= 1.40 x 0.40=0.56 meters.

Minimum base width= 0.60

So base width= 0.60 meters is adopted.

Downstream transition

The downstream side wing walls are given a splay of 1 in 5.

Aprons of weir

The ground level at site of weir= +11.00

Upstream aprons

Generally no aprons are required on the upstream side of the weir but sometimes puddle apron may be provided.

It is also sometimes provided with a nominal rough stone apron 30 cm thick packed well on puddle clay apron.

Downstream apron

Stepped apron in two stages is provided. The aprons may be designed for hydraulic gradient 1 in 5 which is within the safety limit and will not start undermining the structure.


program to calculate weir discharge 
http://irrigation.wsu.edu/Content/Calculators/Water-Measurements/Rectangular-Contracted-Weir.php