final wss and new tank project report

PROPOSAL OF WATER SUPPLY SCHEME AND NEW TANK FOR DHUMMAVAD VILLAGE

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1. INTRODUCTION

The area surveyed for the water supply project is Dhummavad village, which is located at

a distance of 20Km in South-West direction from the centre of the Dharwad city on the

longitude 75’0’54”and latitude15’41’. There exist 250 houses with an average population of

7 persons per house. This village lags behind in supply of potable water. The available water

is inadequate due to rapid urbanization, increasing population etc. The bore wells and open

wells are the main source of drinking water in that area.

The available raw water must be treated and purified before they can be supplied to the

general public for their domestic use. In general the public water supplies are mainly

designed from the view point to the quality requirement of drinking water. The available

water must therefore be made safe and good for domestic use.

The proposed design includes construction of water tank and modeling the pipelines

running through the area. The scheme will be designed for a working period of 30 years. The

aim is to provide these houses safe and potable water and meet their demand efficiently.

1.1 NECESSITY OF THE PROJECT

To provide safe and potable water to the village.

Insufficient water supply. The existing system is supplying the water at a rate of 40 –

50 lpcd, which is insufficient for the people.

Supply of treated water.

1.2 OBJECTIVES OF THE PROJECT

Providing water supply standards to about 100 lpcd.

Treating the water before supply to public use.

To make design and maintenance costs more economical.

To maintain key hygienic issues regarding the water supply system

Design Period is Twenty Years.

S.D.M. COLLEGE OF ENGINEERING AND TECHNOLOGY, DHARWAD


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1.3 KEY MAP OF DHUMMAVAD VILLAGE



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1.4 TOPO SHEET

TOPO SHEET : 48M/3

The highlighted area shows the location of Dhummavad village.



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1.5 PLAN OF DHUMMAVAD VILLAGE



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1.6 CONTOUR MAP OF DHUMMAVAD VILLAGE



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2. POPULATION FORECASTING

A] Arithmetic Increase Method

Year Population Increase

1981 2398 ---------------------

1991 3077 679

2001 3411 334

Cumulative Increase 1013

Average Increase in population = cumulative increase/2 = 1013/2 = 506.5Forecasted Population:

Year Calculation Expected Population

2011 3411+(1*506.5) 3918

2021 3411+(2*506.5) 4424

2031 3411+(3*506.5) 4931

B] Geometric Mean Method

Year Population Increase %Increase1981 2398

1991 3077 679 (679/2398)*100=28.32%2001 3411 334 (334/3077)*100=10.85%

Total = 39.17%

Average percentage increase in population = 39.17/3 = 19.59%

Forecasted Population:

Year Calculation Forecasted Population

2011 3411+(19.59/100*3411) 4079

2021 4079+(19.59/100*4079) 4878

2031 4878+(19.59/100*4878) 5834

select the higher value of population for the design of treatment plant.



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Population by geometric mean method: 5834

Population by arithmetic mean method: 4931

Consider population = 5834 say 6000

2.1 WATER DEMAND



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Rate of water demand: 100 lpcd

The Department Of Rural Development And Panchayati Raj has laid down the

minimum per capita domestic consumption for rural communities with population up to

20,000 should not be less than 70 lpcd.

Peak factor = 1.5

Total quantity of water required

Q = lpcd x population x 1.5

= 100 x 6000 x 1.5

= 900000 lit/day

= 900 m3/day

2.2 GENERAL DETAILS OF THE VILLAGE

Name of the area surveyed : Dhumavad

Number of house existed : 641

Number of vacant plots : 132

Number of members per house : 7 (avg)

No of bore well : 2

Population as per 2001 census : 3411

3. DESIGN OF OVER HEAD TANK

Capacity of OHT = (100 X 6000x1.5) / 1000 = 900m3



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OHT is to be designed for 40% of the capacity of the tank.

Hence volume of tank: 0.4x900= 360m3

Assume Depth of tank = 3m

Free board = 0.5m

Overall depth = 3.5m

Area of the tank = 360/3.5=102.857m2

Diameter of the tank = d = 11.44m say 12m

Therefore the dimensions of the over head tank are –

Height from ground level = 12m

Diameter = 12m

Overall depth = 3.5m

3.1 DESIGN OF RISING MAIN

From Lake to Water Treatment Plant(WTP)



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Difference in elevation Bottom of lake = 595m

WTP = 606m

Therefore total difference in elevation = 11m

Distance between lake and WTP = 100m

Quantity required/day = 900cum/day

Taking f= 0.075 and pump working for 8hrs.

Quantity of water to be carried by pipe = 900 / (8 X 60 X 60)

=0.03125cum/sec.

Assuming the velocity through the pipe = 1.25m/sec

Cross – sectional area of pipe required A = Q / V

=0.03125 / 1.25

But, A = ∏ d2 / 4 = 0.025, therefore diameter of pipe 20cm.

Provide 25cm diameter CI pipe

Actual area provided = ∏ X (0.25)2 / 4 = 0.0490m2

Actual velocity V = Q / A = 0.03125 / 0.0490 = 0.64 m/sec.

Loss of Head by Hazen William’s Formula

V = 0.85 CH R0.63 S0.54

V = 0.64 , CH = 130 (for new CI pipe) , R = d/4 = 0.25/4 = 0.0625

0.64 = 0.85 X 130 X (0.0625)0.63 X S0.54

S0.54 = 0.64 / (0.85 X 130 X 0.174 )

S = 1.83 X 10-3

Head loss, HL / L = 1.83 X 10-3 = 0.185m

Therefore total head required = 11 + 0.185 = 11.185m.

Assuming efficiency of pump set as 70%

The required HP for pump = w Q H / 75μ

= (1000 X 0.03125 X 11.185) / (75 X 0.70)



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= 6.65HP ≈ 7 HP.

Rising Main from Clear Water Reservoir to Over Head Tank

Difference in elevation = 608 – 606 = 2m.

Height of OH tank = 14m.

Distance = 400 + 500 + 100 =1000m.

Quantity of water required / day = 900cum/day.

Pump works for 8hrs.

Quantity of water to be pumped by pipe, = 900 / (8 X 60 X 60)

= 0.03125cum/sec

Assuming velocity through pipe as 1.25m/sec

C/S area of pipe required, A = Q/V

= 0.03125 / 1.25 = 0.025

But, A = ∏ d2 / 4 = 0.025 , and d = 0.178m = 0.2m = 20cm.

Therefore adopt a PVC pipe of 25cm diameter.

Actual or provided area = ∏ (0.25)2 / 4 = 0.0490m2

Actual velocity, V = Q /A = 0.64m/sec

Loss of Head by Hazen William’s Formula

V = 0.85 CH R0.63 S0.54

V = 0.64, CH = 130 (for new PVC pipe), R = d/4 = 0.25/4 = 0.0625

0.64 = 0.85 X 130 X (0.0625)0.63 X S0.54



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S0.54 = 0.64 / (0.85 X 130 X 0.174)

S = 1.83 X 10-3

Head loss, HL / L = 1.83 X 10-3 X 1000= 1.83m

Therefore total head required = 14 + 1.83 = 15.83m.

Assuming efficiency of pump set as 70%

The required HP for pump,= w Q H / 75μ

= (1000 X 0.03125 X 14.83) / (75 X 0.70)

= 8.827HP ≈ 10HP.

3.2 DESIGN OF GRAVITY MAIN

From OHT to Village

RL of OHT = 608m, RL of end of gravity main = 599.4m

Length of gravity main = 200m

Quantity of water = 0.0104cum/sec and f=0.075

Q = A X V

0.0104 = (∏ d2 / 4) X velocity

0.0104 = 0.785 d2 X velocity, therefore V = 0.0132 / d2

By Darcy – Weisbach Formula,

HL = f l V2 / 2gd

= (0.075 X 300 {0.0132 / d2 }2 ) / (2 X 9.81 X d)

= 1.15 X {0.0132 / d2 }2

Since all the available water head is lost in overcoming the friction

(608 – 599.4) = (0.764 / d) X {0.0132 / d2 }2



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Solving for d, d = 0.118m = 11.8cm

Provide 15cm diameter pipe line.

3.3 DESIGN OF INTAKE STRUCTURE

Detention time in the intake well = 20 min

Volume of intake well = Q X D.T.

= 0.0133 X (20 X 60)

= 15.96 m^3

Min height of water required = 1 m

Height of F.R.L. = 3 m

Diameter of the intake well (D) = sqrt (( 15.96 x 4)/ ((22 x 4 )/ 7)))

= 2 m

Total height of the intake well = 3+1

= 4 m

Provide 2 m diameter and height of 4 m to the intake structure

3.4 TEST REPORT OF THE WATER SAMPLE

S.I Parameters Result Permissible Limits



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NO

(as per 10500-1991)

01 pH 7.7 6.5-8.5

02 TURBIDITY 15.9 NTU 10 NTU

03 Conductivity 60 ms/m -------

04 Total dissolved solids 390 mg/lit 2000 mg/lit

05 Hardness 156.4 mg/lit 600 mg/lit

06 Calcium (Ca) 39.7 mg/lit 200 mg/lit

07 Magnesium (Mg) 13.9 mg/lit 75.00 mg/lit

08 Chlorides 38 mg/lit 1000.00 mg/lit

09 Alkalinity 154.0 mg/lit 600.0 mg/lit

10 Sulphates 14.0 mg/lit 400.0 mg/lit

11 Fluorides 0.3 mg/lit 1.5 mg/lit

12 Iron 0.5 mg/lit 1.0 mg/lit

13COLIFORM DENSITY

(MPN)900/100 ml 10/100 ml

Remark: Except Coliform density, all tested parameters are with in permissible limits as per

IS 10500 – 1991

By observing the above results we can conclude that the treatment plant is designed

mainly for ‘Turbidity’ & ‘Coliform density (MPN)’

3.5 LOCATION OF OHT AND WTP

The map shows the satellite image of the Dhummavad village.



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Source: Google Earth

RL 606 Indicates the location of water treatment plant (WTP) and the RL 608

indicates the location of over head tank (OHT).

4. DESIGN OF WATER TREATEMENT PLANT

Components of treatment plant.



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01) Chemical feeding tank

02) Design of approach channel

03) Design of mixing tank

04) Design of flocculating tank

05) Design of settling tank

06) Design of rapid sand filter including wash water tank

07) Disinfection

08) Design of clear water reservoir.

01) Chemical Feeding:

Using alum as a coagulant the optimum quantity of dosage will be determined by

actual jar test in laboratory

Let the optimum dose be 5 to 8 mg/ltr

The quantity of alum req will range from

= 6 x 10 2 x 5 x 10 3

106

= 3 kg/day to 4.8 kg/day

The max requirement in summer will be 1.5 times more i.e. 4.5 to 7.2 kg/day

This quantity of alum shall be first mixed with the water to form a soln of 5%

strength and then added through soln feed devices

The max capacity of feeding devices = (7.2x100)/5 = 1 lit/day = 0.1lit/min

Minimum dose which will feed during avg demand

( assuming max demand = 1.5 x min demand) = 0.1/1.5 = 0.0667 lit/min

Quantity of water to be fed in one shift of 8hr = 0.1x 60 x 8 = 48 lit

Say = 50 lit



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Providing 1 sedimentation tank of capacity equal to 50 lts i.e 0.05 m3

Providing 30 cm depth of sedimentation tank & 10 cm as free board

Area of the tank req = 0.05/0.3 = 0.166 m2

Dimension of Square tank are (40 x 40 x 40)cm

02) Design of Approach channel:

One clarifier has been proposed with one separate mixing coagulation tank

The maximum flow in each channel = 1.04x10-2 m3/sec

Providing a velocity of 25 cm/s the c/s area of approach channel

= (1.04x10-2)/0.25

Area = 0.0416 m2

Say = 0.042 m2

= 420 cm2

= 450 cm2

*Provide (30 x 15)cm

03) Design of Mixing tank:

Mechanical flash mixtures will be used for mixing the coagulant solution

with the water assuming the the detention period of = 1 min = 60 sec



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Capacity of flash mixer = 1.04 x 10-2 x 60 = 0.624

Providing depth of 1 m the side of the square plan mixing tank

= 0.624/1 = 0.624 m2

Size = 0.79 x 0.79 x 1 m of mixing tank

04) Design of flocculating tank:

Providing a floculating tank assuming floculating time of = 30 min

The capacity of floculating tank = 1.04 x 10-2 x 10 x 30 x 60

= 18.72 m3

Provide 2 channels laid in parallel to one another capacity of each channel

= 18.72/9.36 = 9.36 m3

Providing 1 m depth of water and 10 m length of channel, its width

= (9.36)/(1x10)

= 0.936 ~ 1m

... Provide channel of side :- (10 x 1x 1)m with water depth 90 cm and 10 cm freeboard

05) Design of settling tank:

Providing rectangular tank with period of 1hr

Capacity of tank = (6x105x1.5)/(24 x 103 )x(1hr) = 37.5 m3



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Assume velocity = 22 cm/min = 0.22m/min

Detention period = 1hr = 60 min

Length of tank = v x d.p = 0.22 x 60 = 13.2 m say 14m

c/s area = capacity/L = 37.5/14 = 2.7 m2

Assuming depth as 2m width of tank = A/Depth = 2.7/2 = 1.35 say 1.5 m

Size of tank 14 x 1.5 x 2.5m with depth of 2.5m including 50 cm freeboard

06) Design of rapid gravity filter:

Quantity of water to be treated = 1.5 x 6 x 105 = 9 x 105

Assuming rate of rapid gravity filter as = 4500 lit/hr/sq.m

and 30 min shall be utilized daily in back washing of filter

Total filter area = (9x105 )/(23.5x4.5x103)

Providing 2 units one as stand with one as stand by

L/B= 1.5

L = 1.5B

Size of tank = LxB = 1.5BxB = 8.5

B = 2.4 m

L = 3.6 m

Let the size of tank be 4X3m

Wash water tank: overhead tank will be provided for back washing of filter.

Assuming quantity of wash water as 4% of total water filtered.

Quantity of wash water = (4/100)x(3x4)x4500 x 23.5



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= 5.08x 104 ltr

= 50.8 m3 ~ 51 m3

Assuming depth of water asa 2m in tank and providing circular tank

Dia of tank = {(51)/(2 x π/4)}1/2

= 5.7 m say 6m

Provide wash water tank of 6m dia and 2.3 m deep with 30 cm freeboard

07) Disinfection:

The disinfection of water will be done by post chlorination using vaccum

vaccum type of chlorinator through which liquid chlorine will be fed in

the water at the end when all other treatment have been completed the

dose of chlorine to be added will be in 1ppm depending on the quality of

of water and chlorine depends on test.

Quantity of chlorine required = (0.9MLD x 1 ppm x 1000)/(24 x 1000)

= 0.0375 kg/hr

= 38 gm/hr

A liquid chlorinator having capacity to feed chlorine at the rate of

38 gm/hr will be installed.

08) Design of Clear water reservoir:

Underground clear water reservoir having a capacity of about 8hr

Quantity of water to be stored =( 0.9x106x8)/(24 x 103)

= 300 m3

Providing a depth of 3 m the area of tank required = 300/3



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= 100 m2

The plan will mainly depend on the area available at the site therefore, reservoir of plan area 100 m2 and depth of 3m shall be provided with additional free board of 50 cm.

4.1 Layout of water treatment plant

4.2 Layout of pipe line



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CONCLUSION:

The project may provide potable water to the Dhummavad village in

accordance with their demands and requirements.



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The provision of such a scheme shall not help only in supplying safe

wholesome water to the people for domestic use. But also keep the diseases

away and there by promoting better health and ensuing better living standards.



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PROPOSAL OF NEW TANK

1. INTRODUCTION



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The arid region of Dharwad district has majority of black cotton soil and farmers

adopt cropping pattern which suits this soil such as cotton, groundnut, wheat, onion, maize,

etc. All these crops require large quantities of water. The New Tank in the village

Dhummavad has been proposed with an emphasis on agriculture.

Bund is a hydraulic structure constructed across a river or stream to store water on its

upstream side. This water is utilized as and when it is needed. As the upstream side is

increased, a large area may be submerged depending upon the water spread of the reservoir

so formed.

1.1 NECESSITY

Basically India is an agricultural country and all resources depend on the agricultural

output, water is the most vital element in the plant life.

To overcome the growing demand of the water for irrigation.

Preservation cultivating of useful aquatic life.

1.2 OBJECTIVES

To construct a tank for the purpose of irrigation in the village.

Providing adequate and quality water for irrigation.

1.3 LOCATION OF THE VILLAGE



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Source: Google Maps

• The village Dhummavad is located on the State Highway 1 (SH1) in between

Dharwad and Kalghatgi. The village is 15km South-west of Dharwad city. The

village has the Latitude of 15W 19’ 0.16’’N and Longitude of 75W 00’ 36.11”E.

2. SALIENT FEATURES



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1 NAME OF PROJECT New Tank Project At Dhummavad

2 LOCATION Village: Dhummavad; Taluk: Kalghatgi;

District: Dharwad

3 LONGITUDE 75 00’ 54’’

4 LATTITUDE 15 19’ 7’’

5 TOPOSHEET 48 M/3

6 WATER SHED DETAILS

5 Arabian Sea

5B Sharavati Basin

5B1 Sharavati To Savitri Basin

5B1A Savitri To Kali River (Sub-Catchment)

5B1A4 Bedti Halla (Water Shed)

7 Catchment Area 4937477.3 m2 = 1220.1 Acres = 4.9 km2

8 Nature Of Catchment AVERAGE

9 Average Rainfall 677.66mm

10 50% Dependable Rainfall 370.75 mm

3. BASIS FOR FORMATION OF TANK



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3.1 AREA OF CATCHMENT BASIN:

Catchment area is the area of the closed traverse obtained by projecting catchment

boundary on to a horizontal plane. It is expressed in Square kilometers.

The catchment area is 4.84 km2, using Toposheet 48-M/3.

3.2 NATURE OF CATCHMENT

The examination of the nature of the catchment is done for computing the run-off

parameters. Generally, catchments with vegetation give good yield and those with porous soil

give more yield. The nature of the catchment is Average since there is no vegetation found

and the soil is loamy clay.

Further the catchment of the proposed tank is examined for the possibility of existing

tanks with ayacuts, their storage capacity, and the ayacuts irrigated by the tank.

3.3 RAINFALL DATA

Rainfall data from the nearest Meteorological Station, i.e. The University of

Agricultural Sciences, Dharwad has been collected for the past 23 years. From the rainfall

data so obtained, the Dependable Rainfall and the Yield of the Catchment has been computed

by using Dependable Percentage value to be 50% of the annual mean rainfall.

The procedure to compute the dependable rainfall is as below

i. The available rainfall for past 23 years is arranged in Descending order by

magnitude.

ii. The order M is given by the equation, M=23*(p/100). The rainfall value

corresponding to this order in the tabulated data represents the required dependable

value.



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RAINFALL DATA

SL NO YEAR RAIN FALL1 1991 1104.82 2005 1011.13 2007 10004 2006 8495 1992 824.26 1993 806.37 2004 7718 1997 760.39 1998 753.810 1988 74911 1994 741.512 1995 731.913 1990 730.514 1996 686.615 1987 62616 1986 594.717 1985 551.318 1989 54519 2000 528.720 1999 435.721 2002 362.322 2001 24723 2003 175.5

3.4 CONVERTING DEPENDABLE RUNOFF TO DEPENDABLE YIELD

Average Rain Fall = 15586.2/23 =677.66mm

M= 23*(50/100) =11.5

M= 11 and Dependable rain fall is 741.5mm

50% Dependable rain fall

=(50/100)x741.5 = 370.75mm

3.5 YIELD OF THE CATCHMENT

From Strange’s table



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Yield of catchment for 370.75mm rain fall

By interpolation, for 370.75mm

= ((15.15x4.735x10-3)/25.4)+0.022129= 0.02495= 24950

Yield of the catchment = 24950 x4.9 = 122255m3

3.6 DESIGN OF NEW TANK

Assumptions:

Number of filling per year = 2

Utilization of the yield per filling = 50%

Capacity of the tank = 50% of yield = 0.5 x 122255 = 61127.5 m3

Utilization of the yield per year = 2 x 61127.5 = 122255 m3



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3.7 CONTOUR MAP



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3.7 CAPACITY TABLE

Sl no

R.L Area (m2) √(A1xA2)Contour interval

Capacity by Prismoidal formula

=[h/3(A1+A2+√(A1xA2))]

Accmulation

(m3)

1 590 57500 1

2 591 34200 44345.24 1 45348.41 45348.41

3 592 58200 44614.35 1 45671.45 91019.86

4 593 71900 64688.33 1 64929.44 155949.30

5 594 67200 69510.29 1 69536.76 225486.06

6 595 75500 71229.21 1 71309.74 295795.81

7 596 12650 30904.29 1 39684.76 335480.57

8 597 11440 12029.80 1 12039.93 347520.5

9 598 101700 34109.35 1 49083.12 396603.62

3.8 AREA ELEVATION CURVE

The areas enclosed by the successive contours were calculated using Auto Cad. A close observation shows that as the value of elevation increases the area also increases a curve is plotted between area and elevation as shown above.



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3.9 STORAGE AREA CURVE

The bund capacity or the volume of storage corresponding to a given area in reservoir may be calculated either by trapezoidal or prismoidal formula. If ‘V’ is the volume of storage and ‘h’ is the contour interval, then by trapezoidal formula.

V = h[(A1+An)/2+A2+A3+A4+……..An-1]

A curve is plotted between area and volume as shown below.

4.0 FEATURES OF THE BUND

An earthen bund must be safe and stable during phases of construction and operation.

The following recommendations are generally used for selecting the suitable values of top

width, free board, upstream and downstream slopes, drainage arrangement etc.

4.1 TOP WIDTH

The top width of the earthen bund depends upon the following criteria

Width of the road on top of the bund

Practicability of the construction

Protection against earthquake forces



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Following are some of the empirical expressions for top width ‘b’ of the earthen bund in

terms of the height ‘H’ of the bund.

b=H/5+3 (this is applicable when the height of the bund is low)

b=1.65(H+1.5) ^1/3 (this is given by USBR)

Top width is assumed to be 2.0m (according to TABLE NO. 2)

4.2 FREE BOARD

Free board is the vertical distance between the horizontal crest of the embankment

and the water level. Sufficient free board must be provided so that there is no possibility what

so ever of the embankment being over topped.

USBR suggests the following free boards:

Nature of spillway Height of dam Free board

Free AnyMin 2m and max of 3m over

the flood level

Controlled Less than 60m 2.5m above top of the gates

Therefore free board of 2.5m is provided

4.3 PITCHING

The upstream slope of the bund is protected from the wave wash by a layer of stones called as stone pitching. A thin layer of spauls is laid on the upstream slope and on this a layer of big size, hand packed and roughly hammered dressed stones is laid. The voids between these big size stones are filled with small stones which are hammered in.

4.4 UPSTREAM AND DOWNSTREAM SLOPES

Using Terzaghi’s table no. 10.2 (c) page no. 409 (Irrigation engineering B.C.Punmia)



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Sl No.

Type of material Upstream Downstream

1 Homogenous W.G material 2:1 2:1

2 Homogenous coarse silt 3:1 2.5:1

3 Homogenous silty clay

a) Height <15m 2.5:1 2:1

b) Height >15m 3:1 2.5:1

4 Silt or sand and gravel with clay core

3:1 2.5:1

5 Sand or gravel with RCC core wall 2.5:1 2:1

4.5 DRAINAGE OF EARTHEN BUND

The water percolates through every earthen bund however it stagnates in the downstream portion of the bund and over-saturates it. This may result in slip of material and also results in the failure of the embankment. The draining of the water from the downstream portion so as to avoid slip is called Artificial Drainage of the earthen bund. This is necessary when the material in the downstream portion of the bund does not have proper natural drainage.

In case of drains, a few longitudinal stone drains are provided parallel to the bund length and its bottom. The cross drains takes the water percolating in longitudinal drain to a downstream longitudinal drain at some distance away from the toe of the bund. Water from the downstream drain is collected by clear drains and is taken away from the bund.

The materials provided in the drains for drainage purpose being more permeable than the material of the dam. As long as the drainage system works properly and efficiently, the ground on the vicinity of the toe of the dam should be in dry condition.



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5. DESIGN OF WASTE WEIR

Catchment Area, M = 1.89m2

Ryves constant, C =650

Q=C*M(2/3)

= 650x(1.89) 2/3

=27.648m3

Height of the weir , H=3 feet.

Q=2/3X(Cd*L*√2g*h3/2)

27.648=2/3X(0.62XLX√2X9.81X(0.9144) 3/2

L=17.03

Provide length of weir =20m

Crest / Top width of the Waste Weir

Crest width,a,=√d*√(H/1.8)+1

Or a minimum of 3d/2f

d= Depth of water over weir = 1m

f=Specific Gravity of body Wall = 2.40

a=√1 x √(3/1.8)+1 =1.63m

Or ( 3x1)/(2x2.40) = 0.625m

Crest Width ,a=0.625m

Provide Crest width of 0.625m for Stability

Bottom Width =(H+D)/√f

=(3+1)/√2.40 = 2.58m

Provide Bottom width of 5.0m for Stability

Depth of Water Cushion = ((1/2)xd)/(H-d)

=((1/2)x1)/(3-1) =0.25m

Provide Water Cushion of 0.3m



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Proposed Basin floor level =594.00

Discharge per meter=28/20 =1.4 m3/sec

Vth=√2g((H+h)/2) h=3m,H=4m

= √2x9.81x((4+3)/2) = 8.28 m/sec≈8.3m/sec

Vact=0.98x Vth = 0.98x8.3 = 8.134m/sec

Froudes number = Vact/gd1

D1=1.56/8.134 = 0.1917

= 8.134/(9.81X0.1917) = 4.325

5.1 Stability of Weir



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Sl no. Force(kN) Distance from Toe(m) Moment(kNm)

1 W1=(0.625x2x1)x25= 31.250

4.375+(0.625/2)=4.687

146.468

2 W2=((1/2)x4.374x2x1)25= 109.375

(2/3)x4.375 =2.916

318.937

3 Pw=γH2/2=9.81x32/2 =44.145

(1/3)x3 =1m 44.145

∑V=31.25+109.375 =140.625

∑M = (146.468+318.937)-44.145 = 421.272

Step 1:

140.625*X- =421.272

X= 421.272/140.625 = 2.995

Step 2:- Stability against Overturning

=Stability moment/Overturning moment

= (146.468+318.937)/44.145

=10.542 >2 …safe

Step 3:- Stability against Sliding

= Resisting Force/Sliding Force =μ(w+w2)/44.145 =0.6(31.250+109.375)/44.145

= 1.911>1.5. Therefore, Safe.

Step 4:

Pmax = ∑V/B (1+6e/B) = (140.625/5)x(1+(6x0.495)/5) = 44.831

Pmin = ∑V/B (1-6e/B) =(140.625/5)x(1-(6x0.495)/5) = 11.418


final wss and new tank project report

Documents

available water

water supply project

supply of potable water

insufficient water supply

supply of treated water

water supply standards

average population

public water supplies