Education

1. Water supply and sanitation
project
By
Attal Thapa c
Niraj Kumar maluwa
Akil anjan
Ankit Kumar Yadav

2. Introduction
Object : To purpose water supply scheme to a village and
provide safe drinking water.
 In order to ensure that sufficient quality of good quantity of water is
available, to plan and build suitable water supply schemes, which may
provide portable water to the various sections of community
accordance with their demands and requirements.
 These schemes shall not only help in supplying safe wholesome water
to the people for drinking, cooking, bathing, etc. so as to keep the
diseases away and providing better health and helps in maintaining
better sanitation and beautification of surrounding, thereby reducing
environmental pollution.

3. Ghati plan

4. Levelling
It is a process to determine the vertical position of different points below on or above the
earth’s surface. This is done by taking measurements in vertical plane. The types of
levelling are :
 Simple levelling
 Fly levelling
 Check levelling
 Profile levelling
 Block levelling
 Cross levelling. The levelling used in this project is given below:
Block levelling – this method is used for a small survey area. This area to be surveyed is
divided into number of squares or blocks and size of each block varies from 5m to 20 m
depending upon nature of contour.
Profile levelling – this method of surveying that has been carried out along central line of
track of land on which linear engineering work is to be laid. The operations involved in
determining elevation of ground surface at small spatial interval along a line is called
profile levelling.

5. Water distribution systems
For efficient distribution system adequate water pressure required at various points. Depending
upon the level of source, topography of the area and other local conditions, the water may be
forced into distribution system by following ways
 Gravity System
 Pumping System
 Combined gravity and pumping system
Gravity system : this system uses gravity to transport water from the source to the user through
a pipe network, bringing water closer to people to reduce time.
Pumping system : In such system, water is supplied by continuous pumping. Treated water is
directly pumped into the distribution main with constant pressure without intermediate storing.
Supply can be affected during power failure and breakdown of pumps. The method is not
general used.
Combined Gravity and Pumping system : Treated water is pumped and stored in an elevated
distribution reservoir. Then supplies to consumer by action of gravity. The excess water during
low demand periods get stored in reservoir and get supplied during high demand period.
Economical, efficient and reliable system.

6. Procedure
 Step 1 : Carry RL from permanent bench mark to reservoir by fly
levelling and RL is established.
 Step 2 : Block levelling for water treatment plant was done neat to
reservoir. The raw water from reservoir will be pumped from reservoir to
water treatment plant and there water will be treated.
 Step 3 : Profile levelling for rising mane was carried out from water
treatment plant to surface elevated tank i.e. rising mane. Plane table
also done for alignment and bearing for direction. Treated water from
plant will be pumped to surface elevated tank.
 Step 4 : Block levelling for surface elevated tank is carried on elevated
area to collect treated water from rising mane, the purpose of
constructing surface elevated tank is to release water in regulated and
controlled manner and also store water.

7.  Step 5 : Profile levelling for water supply line is carried from surface
elevated tank to village to supply treated water from surface elevated
tank to each and every house, temples and shops etc. Plane table was
done for alignment and bearing for alignment and bearing for direction
through main water supply.
 Step 6 : Profile levelling for sanitary line is carried out to align sanitary
pipe line. The used water from village carried through sanitary line and
reaches treatment plant.
 Step 7 : Block levelling for treatment plant, waste water which will be
carried through sanitary line will be treated here and recycled.

8. DESIGN PERIOD
It is utility or useful life period of component during period component should offer
best service.
Design periods for different water supply schemes :
Sl no Component Design period (years)
1 Storage by dams 50
2 Infiltration works 30
3 Pump Sets 45
4 Water Treatment plants 15
5 Pipe connections 30
6 Raw and clear water
conveying mains
30
7 clear water reservoir 15
8 Distribution Systems 30

9. POPULATION FORECAST OF GHATI
YEARS POPULATION INCREMENTAL INCREASE
2004 3000 -
2014 4000 1000
2024 5100 1100
 Total population = 2100
 Average = 2100/2 = 1050
 ARITHMETIC METHOD :
 Pn = Po+nx
 P2034 = 5100+(1*1050) = 6150
 P2044 = 5100+(2*1050) = 7200
 P2054 = 5100+(3*1050) = 8250

10. Determination of the required amount of water
 Total population of town = 8250
 Assuming the rate of water supply as 135 lpcd(100 lpcd-300 lpcd)
 Q max factor = 1.8%
 Efficiency = 75%
 Quantity of water required = Q max factor * population * water demand
= 1.8 * 8250 * 135 =2004750Ltrs/day
Qmax = 2004.75 m^3/day

11. Design of the pumping unit
 Assuming the pump is working 8 hours a day
 Quantity of water to be pumped = 2004.75/(8*60*60)
Q = 0.06960m^3/sec
Using discharge equation Q = A*V
Assuming the velocity = 0.85m/sec
A=Q/V = 0.06960/0.85
A=0.081m^3
 Diameter d = √(4A/π) = √(4*0.081/π)
= 0.32m
Economical diameter of rising main given by lea’s formula
d=1.22*√Q
1.22* √0.06960
= 0.32m =~ 32 cm
Therefore it is economical

12. Capacity of the pump
 BHP=W*Q*H/(75*)
 W= Water density = 1000kg/m^3
 Q=max discharged = 0.06960m^3/sec
 L=Length of the pipe through which water has to be supplied =1250m
 Assuming the co-efficient of friction = 0.01
 Head loss due to friction in the pipe
 Hf =4flv^2/2gd
 Hf = 4*0.01*1250*0.85^2/2*9.81*0.32
 Hf = 5.75m
C = HRL – LRL + 4 ( LRL = 886.480 )
C = 916.660 – 886.480 + 4 ( HRL = 916.660 )
C = 34.19 ( 4 = Height of the tank )

13.  Total head loss
Ha = Hd + Hf
=34.19+5.75
=39.93m
 Efficiency of the pump=80%
 BHP=W*Q*H/75*η
=1000*0.06960*45.175/ 75*0.8
=46.31HP

14. Design of sedimentation tank
 Max daily demand=2004.750m^3/day
 Detention time= 4hours
 Capacity tank = Qmax*detention
=2004.750*4/24
= 334.125m^3
 Assume depth of the tank =3.5m
 Area of the tank required = capacity of tank /depth
= 334.125/3.5
= 95.46m^2
 Diameter of sedimentation tank
=√(4A/ꙥ)
= √(4*95.49/ꙥ)
= 11m
 Provide the tank size of Dia =11m
 Dimension of tank =11m Dia and (3.5+0.5)m deep [Providing a free board of 0.5]

15. Design of slow sand filter
Let rate of filtration=150 L/hr/m^2
Qmax = 2004.750m^3
Total surface area required
Q/ROF
= 20004.750/(150*24*10^-3
= 556.875m^2
Provide rectangular tank of length ‘L’ and breath ‘B’ at the ratio of
L:B=2:1 L=2B
and A=L*B =2B*B =2B^2
B=17m L=2B = 2*17
L=34m
Providing 2 filter units, Therefore Area of each units
= 556.875/2
=278.437m^2

16.  A=278.437m^2
278.437=2B^2
B= 12m,L=24m
 Provide 2+1 filter out of which 1 act as stand by of size
24m*12m

17. DESIGN OF CHORINATION TANK
 Quantity of water to be provided =2004.750m^3/day
 Detention period = 15min
 Quantity of water to be treated = 2004.750*15/60*24
= 20.882m^3
 Depth of the tank=2m
 Cross-section area= 20.882/2 =10.441m^2
 Assume =L:B =2:1
A=L*B
A=2B^2
B=2.5m,L=5m
 Quantity of Disinfectant (Bleaching powder)
 Max demand of water =2004.750m^3/d
 Assuming chloride dosage= 0.5ppm

18.  Quantity of chloride required
= 2004.750*10^3*0.5/10^6
=1kg/day
 Bleaching powder is used as disinfectant
 The amount of bleaching powder required = Quantity of
chloride*100/3
= 1 * 100/30
=3.5kg/day

19. Design of elevated surface tank
Max daily demand =2004.750m^3/day
water tank should have storage capacity of locality
Assume height of tank H= 4
Providing circular tank
Dia of tank = √4*Qmax/ꙥ *H
= √ 4*2004.750/ꙥ*4
=25.5m
From elevated surface tank to village
 (a) Design of main line
V=0.85m/sec
Qmax=2004.750m^3/day
=2004.750/60*24*24
=0.023m^2/sec

20. Area = Qmax/V
0.023/0.85 =0.0272m^2
diameter, d =_/4A/ꙥ
=_/4*0.0272/ꙥ
= 0.186m
= 19cm.
(b) Design of branch line
Discharge of branch line = 40% of main line
Q=(40/100)*0.023
Q= 0.0092
area=Q/V
= 0.0092/0.85
= 0.0108m^3
Dia d= √ (4A/ꙥ)
=√ (4*0.0108/ꙥ)
=0.117m =11.7cm =12cm

21. (c) Design of service line
Discharge of service line=30% of main line
Q=(30/100)*0.0092
= 0.00278
Area = Q/V
= 0.0027/0.85
= 0.0031
Dia d= √(4*0.0031/ꙥ)
d=0.0645 = 6.5cm

22. Determination of gradient
By Hazen William’s Formula
V=0.85*CH*R^0.63*S^0.54
Hazen William’s Coefficient CH=110
V=0.85m/sec
R = dmain/4
= 0.190/4
= 0.0475m
Substitute the values
= 0.85*110*0.0475^0.63*S^0.54
S = 5.800*10^3
=0.0058
S=1 in 0.0058
S= 1 in 172

23. Design of Sewer
Design of sewer
Discharge of sewer system
Dry weather flow = discharge*80% of water supply
= 2004.75*0.8/24*60*60
= 0.01856m^3/sec
Max dry weather flow=3*0.01856
=0.05568m^3/sec
Wet weather flow=(A*I*R)/360
Where A=Approx area = 2.6 Hg
I=0.7(Assume)
R=intensity of rainfall
=(2.6*0.7*15)/360
=0.0758m^2/sec
Total discharge =dry weather flow + wet weather flow
= 0.0556+0.0758
0.1314m^3/sec

24. Design of sewer pipe
Assuming that the pipe is running half full,
Longitudinal flow =1in 120
Hydraulic mean radius = (wetted area/wetted perimeter)
= d/4
using Manning’s formula =
V = 1/N *R^2/3*S^1/2
Q = A*1/N*R^2/3 *S^1/2
0.1314 = (ꙥ*d^2/4)*(1/0.012)*(d/4)^2/3*(1/120)^1/2
d = 0.337m = 0.34m
V = Q/A = 0.1314/ꙥ(0.34/4)^2
= 5.78

25. THANK YOU