Design of weir on river cauvery near thottilpatti village
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Transcript of Design of weir on river cauvery near thottilpatti village
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Design of Weiron River Cauvery near Thottilpatti Village
PROJECT BY-VIJAY KR SINGH (13BCL0001)
CIVIL ENGINEERING VIT UNIVERSITY Email Id- [email protected]
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PREFATORY:Vellore District comprises One Corporation (comprising of erstwhile 2 Municipalities, 6 Town Panchayats and 109 Habitations in 9 Village Panchayats), 11 Municipalities, 16 Town Panchayats and 5,688 Rural Habitations in 20 Panchayat Unions.
PRESENT WATER SUPPLY SCENARIO:Palar and Ponnaiar are the two major Rivers traversing through Vellore District. Even though, they are non-perennial, the sub surface water potential fulfills the drinking water requirements.
NEED FOR THE PROPOSAL:Due to discharge of Tannery effluent, sewage and sullage into Palar River from urban areas, the existing Water Supply schemes to Urban Towns and Rural Habitations are adversely affected.
Accordingly, TWAD Board took up Investigation works, to provide a Combined Water Supply Scheme with River Cauvery as source to Vellore Corporation & other local bodies, in Vellore District, as a permanent remedial solution.
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POPULATION & REQUIREMENTThe net demand now works out to 148 mld, 181 mld & 215 mld for the present (2016), intermediate (2027) & ultimate (2042) stages respectively.
DESCRIPTION OF THE PROPOSAL:It is proposed to tap the project ultimate requirement by providing barrage or weiron River Cauvery near Thottilpatti Village, down stream of Mettur Dam & upstream of Chekkanur Barrage. Collected Raw Water will be pumped (23 hours) to the Treatment Plant of 181mld capacity, located at Thottilpatti Village at Head works site owned by TWAD Board (10.50 acres).
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Design Criteria 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
THE ABOVE WEIR IS DESIGNED USING BLIGH THEORY
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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
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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 + LIf 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 S2L` LB (design is safe, no possibility of undermining)L` < LB (design is unsafe, undermining occurs, leads to failure)
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QA = Qi + Qii & Qi = 2 Qii QA = 2 Qii + QiiAt flood QA = 105 = 3 Qii Qii = 35 m3/s & Qi = 70 m3/s
At summer QA = 45 = 3 QiiQii = 15 m3/s & Qi = 30 m3/s For branch ( i )
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Qmax /Qmin = (2 B H11.5) / (2 B H2
1.5) = 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 mH1 = 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 mQmin = 30 = 2 B (.92)1.5 B = 17 m
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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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Distance between successive sheet piles
Distance between sheet piles a-a and b-b d1 + d2· Water percolation length takes the right path -----safe
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• 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
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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)
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Determination of Floor Dimensions
t1 = 0.5 – 1.0 m assumedt2 is taken 2.0 m or t2 = 0.8 (H)0.5
t3 = t2 / 2 1 m and l1 is assumed (1-2) HL2 = is determined according to weir type (3-8) mLScour = Cs (Hs) 0.5
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OrLScour = 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
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Determination of the uplift diagram
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HDh2 = H – t1/CB – l1 / CBt2 = t / (γm) * Factor of safetyγt2 = 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 t3L3 = CB * h3 = x + t3 then get distance x
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Design of weir floor for canal (i) by applying Bligh method BED LEVEL = 10.76 – 4 = 6.76 HD = 12.08 - 11 = 1.08HD = 11.38 - 10.76 = .62HD = 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 mLS = CS (HS).5 CS = .6 CBHS = 12.08 - 6.76 - Ycr & HS = 4.37 & LS = 20 m
Assume t2 = 2 mL\ = 6 + 6 + 20 + 2 * 2 = 36
L\ < LB unsafe use sheet pile d = (59.2 – 36) / 2 = 11.6Use two sheet pile d =7 m & d = 5 m
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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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AUTO –CAD DESIGN OF WEIR
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