Give the correct procedure of designing surplus weir.
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Give the correct procedure of designing surplus weir.
aviratdhodare
Surplus weir (waste weir): It is a concrte or masonry structure constructed to dispose off excess water from an irrigation tank. It is a safety device in the tank.
Full tank level (FTL): It is the highest level up to which water could be stored in the tank. Excess water will go out through the surplus weir. Fixation of this level depends on the availability/demand of water.
Max water level (MWL): It is the max level of water allowed in the tank. MWL is higher than FTL. The difference between MWL & FTL is the spillage or head on crest of surplus weir Fixation of this level depends on the submergence of land due to back water.
Tank bund level (TBL): It is the top level of the liqd of the bund & is equal to MWL + freeboard.
Abutment: The walls that flank the edge of a weir and which support the banks on each side of the weir. The length of the abutment is generally kept same as the base width of weir. The top level of the abutment is kept at tank bund level.
Wing wall: A wall on a weir that ties the structure into the bank in continuation of the abutments. Wing walls are provided both on the u/s and d/s sides on both the banks to ensure smooth entry and exit of water away from the tank.
Return wall (Return): These are provided at right angles to the abutment at the end of wing wall and extend into the banks to hold the back-fill.
Splay: Horizontal deviation of wall. Ex: 1 in 3, 1 in 5, etc.
Batter: Vertical deviation of wall. Ex: 1 in 8, 1 in 12, etc.
Hydraulic gradient, Saturation gradient (or) Seepage gradient: It is the head loss
(energy loss) per unit length in the direction of flow traveled by water particle through soil. Ex: Saturation gradient 4:1, it means to dissipate energy of 1m, water should travel a distance of 4 m in the soil
Catchment area(watershed area, drainage area, drainage basin or basin or
catchment): It is a portion of land which catches the rain and produces runoff through a one outlet.
Free catchment: Entire runoff in the catchment will be passed direct to tank. It means water from catchment area is not go to other tank or channels, and it should directly goes to one tank.
Intercepted catchment: Part of runoff will be intercepted and stored by the u/s side tank(s) within the catchment.
Combined catchment: Entire runoff in the catchment will be shared by group of tanks or a chain of tanks which comes under the same catchment.
D/S Apron of the surplus weir: Depending upon the foundation particulars, and the levels of U/S and D/S ground at the location of the work, any one of the following types can be adopted.
Type A → Horizontal masonry apron – when fall height < 75 cm
Type B → Sloping apron
Type C → Similar to B but with rough stone sloping
Type D → Stepped apron – when fall height ⩾ 75 cm
Location of surplus weir: It is desirable to locate the surplus weir at or near the flank of the tank bund and connected to it, and also at a place where it is possible to drain the surplus waters below the work away from the tank bund falling into its natural watercourse. The cost of works should be minimum.
Design a surplus weir for a minor tank forming a group of tanks with the following data:
Combined catchment area = 25.89 km2 (35 km2)
Intercepted catchment area = 20474 km2 (10 km2)
Top width of the bund =2m (2m)
Side slopes of the bund = 2:1 both sides (2:10n both sides)
Top level of bund = +1450 (+ 12.50)
Maximum Water Level (MWL) =+ 12.75 (+ 10.75)
Full Tank Level (FTL) = + 12.00 (+ 10.00)
General ground level at the site =+ 11.00 (+ 9.00)
Ground level slopes off to a level in about 6 m distance) = + 10.00 (+ 8.00 in about 6 m dist)
The foundations are of hand gravel = + 9.50 (+ 7.50)
Saturation gradient = 4:1 with 1 m clearcover (4:1 with 1m clearcover)
Provision is to be made to store water up to MWL in-times of necessity
Components to be designed
(1) Estimation of flood discharge entering the tank (Q) :
Combined catchment area (M) # 25.89 km2
intercepted catchment area (m) = 20.71 km2
Assuming Ryve’s coefficient(C) =9 and c = 1.5
Flood discharge (Q) = CM2/3 – cm2/3
Q = 9 (25.89)2/3 — 1.5 (20.71)2/3 = 78.77 — 11.32
Q = 67.45 m3/s
(2) Length of surplus weir (L):
Assuming the flow over a surplus weir is identical to that of flow over a rectangular weir then discharge is given by Q = 2/3 CdL √2g h3/2
where, Q = 67.45 m3/s, cd = 0.562 (assuming), g = 9.841 m/s2
h = MWL – FTL = 12.75 — 12.00 = 0.75 m, L — Length of the water way
67.45 = 2/3 x 0.562 x L √2×9.81 (0. 7s)3/2 → L=62.75 m ≈ 63.00 m (say)
Since temporary regulating arrangements are to be made on top of weir to store water at times of necessity.
The dam stones of size 15 x 15 x 125 cm are at 1m clear internals keeping top of the stone at M.W.L.
The no. of openings will be = 63, The no. of dam stones required = 62
∴ The overall length of surplus weir between abutments = 63 + (62 x 0.15)
= 72.30 m
However, provide an overall length of 75 m.
(3) Height of the weir (H):
Crest Level = FTL = +12.00
Top of dam stones (top of shutters) = M.W.L = + 12.75
Ground level = + 11.00
Hard soil at the foundation is + 9.50.
However, taking foundations about 0.50 m deep into hard soil and fix up foundation level at + 9.00
Assuming foundation concrete is 60 cm thick
Top of foundation concrete = + 9.60
Height of weir above foundations (H) = 12.00 – 9.60 = 2.4m
(4) Crest width of weir (a):
a = 0.55 (√H + √h) = 0.55(√2.4 + √0.75) = 1.3m
(5) Base width of weir (b):
The base width is determined based on moment considerations. i.e., based on the magnitude of stabilizing and destabilizing moments.
Stabilizing moments are caused by self weight of the weir which is given by
M = γw /12 = [{(G+15)H + 2.5S}b2 + a(GH – H – S)b – ½a2 (H +3S)]
Where, γw = Unit weight of water = 1000 kg/m3
G = Specific gravity of masonry = 2.25
H = Height of the weir = 2.40 m
a = Crest width of weir = 1.30 m
b = Base width of the weir = ?
S = h = height of shutter above weir crest = 12.75 – 12.00 = 0.75 m
Destabilizing moments (M,)
Mr = γw (H + S)3 / 6
Equating both the moments: M,=M
Mr = (2.4 + 0.75)3 / 6 = 1 /12 [{2.25 + 1.5)2.4 + 2.5 x 0.75} b2 + 1.3 (2.25 x 2.4 – 2.4 – 0.75)b – ½ (1.3)2 (2.4 + 3 x 0.75)]
Solving, b = 2.4 m
(6) Abutments, Wing walls and Returns:
The top width of abutments, wing walls & returns will all be uniformly 0.50 m with a front batter of 1 in 8. Diag in attachment.
Abutment (AB)
Length of the abutment = width of bund = 2m
The top level of the abutment is kept at TBL = + 14.50
Bottom level of the abutment = top of foundation level = + 9.60
Height of the abutment = 14.50 — 9.60 = 4.90 m
Bottom width= 0.4 x height = 0.4 x 4.90 = 1.96 m = 2.00 m (say)
Top width 2 0.5 m (assuming), Front batter = 1 in 8
Wing walls:
U/S Wing Wall:
BD is called u/s wing wall
Section at B:
Same as the section of abutment
Wing wall from B to C is sloping and
Top level of C = M.W.L + 30 cm = 12.75 + 0.30 = 13.05
Section at C:
Top Level at C = 13.05
Bottom level = 9.60
Height of wing wall = 13.05 – 9.60 = 3.45 m
Bottom width = 0.4 x height = 0.4 x 3.45 = 1.38 = 1.40 m (say)
Top width from B to C is the same as 0.5 m.
But, bottom width gets slowly reduced
from 2.00 m at section at B to 1.40 m at Section C:
From C to D wing wall is horizontal. Therefore, Section at D = Section at C
U/S Return (DE):
Section at E = Section at D
U/S transition:
In order to give an easy approach, the u/s side wing wall may be splayed at 1 in 3.
D/S wing wall:
AF is called d/s wing wall.
Section at A: Same as the section of abutment. The Wing wall from A to F will slope down till the top reaches the ground level at F.
Section at F:
Top of wing wall at F = + 11.00
Bottom of wing wall = + 9.60
Height = 11.00 – 9.60 = 1.40 m
Bottom width = 0.4 x 1.4 = 0.56 m
However, provide a minimum of 0.6 m
D/S return (FG):
The same section at F is continued for FG also
D/S transitions:
Provide a splay of 1 in 5.
(7) Aprons of the weir:
i). U/S Apron: Though apron is not required on the u/s side of the weir, a puddle clay apron is usually provided to minimize the seepage under the weir.
ii).D/S Apron: Since the ground level is falling down to +10.00 in a distance of about 6m. Then, the fall is (12.00 – 10.00) = 2.00 m > 0.75 m therefore provide a stepped apron (Type D) Diagram in attachment. The stepping may be done in two stages.
(a) The length of the Apron: The length of the apron should be adequate to avoid piping problem.
[Maximum uplift will be occurred when water level on U/S is up to top of dam stone (M.W.L.) and no water on D/S (+10.00))
Max. Uplift head = 12.75 – 10.00 = 2.75 m (max. energy to be dissipated)
Assuming a hydraulic gradient of 1 in 5
The length of the creep required = 2.75 x 5 = 13.75 m
The length and thickness of apronts to be designed.
The length of the creep = AB + BC + CD + DE + EF = 1.40 + 0.60 + 3.00 + DE + 1 (Assuming EF = 1 m)
This length should not be less than 13.75 m, if the structure is to be safe.
13.75 = 1.40 + 0.60 + 3.00 + DE + 1 → DE = 7.75 m = 8.0 m (say)
Provide total length of solid apron ts 8 m.
First step in 3 m and second step in 5 m length.
(b) Thickness of solid apron: The maximum uplift on the apron is felt immediately above the point D. (i.e., at point K)
Assuming the thickness of apron at point K = 80 cm = 0.80 m.
Then the level of K = 11.00 – 0.80 = 10.20
The length of the creep from A to K = 1.4 + 0.6 + 3 + 0.6 + (10.20 – 9.60) = 6.20 m
Head loss in percolation along the path up to the point K = 6.20/5 = 1.24 m
Residual head exerting uplift under the apron at point K = 2.75 – 1.24 = 1.51 m
Thickness of apron required = Residual head / Sp. gravity = 1.51/2.25 = 0.67 m
Provide 20% of more thickness as a safety
Then thickness of apron required = 0.80 m
So, provide the first solid apron as 80 cm thick.
The second apron can be similarly checked for a thickness of 50 cm.
8) Talus: At the end of d/s side apron, a nominal 3 to § m length of Talus (i.e., rough stone apron) with a thickness of 50 cm may be provided as a safety mechanism.