2. Shaik Mohammed Sufiyan 20211CIV0031
Syed Abrar 20211CIV0025
Nikhil 20211CIV0022
Swarnima Rai
by
Batch 2
3. • The systematic identification and evaluation of the potential impacts of the propose
d project, plans, programs or legislative actions relative to the physical, chemical, bi
ological, cultural and socioeconomic components of the total environment.
• It is a preventive tool used to evaluate the negative and positive environmental
impacts of policies, plans, programs, and projects; the EIA proposes measures to
adjust impacts to acceptable levels
Introduction
4. PURPOSE OF EIA
To encourage productive and enjoyable harmony between man and his
environment.
To promote efforts which will prevent or eliminate damage to the environment and
biosphere and stimulate the health and welfare of man.
To enrich the understanding of the ecological systems and natural resources import
ant to the Nation.
To implement a strategy of sustainable development and prevent adverse impact on
the environment.
To promote coordinated development of the economy, society, and environment.
5. TYPES OF EIA
• Rapid EIA
• Comprehensive EIA
Rapid Assessment is the initial feasibility study carried out for a minimum
period of 3 months.
After it gets appraisal from MOEF (Ministry Of Environment and Forests),
the next phase, comprehensive assessment is carried out.
It is carried out for a period of 16 months.
6. GUIDING PRINCIPLES OF EIA
Purposive – meeting its aims and objectives
Focused – concentrating on the effects that matter
Adaptive – responding to issues and realities
Participative – fully involving the public
Transparent – clear and easily understood
Rigorous – employing ‘best practicable’ methodology
Practical – establishing mitigation measures that work
Credible – carried out with objectivity and professionalism
Efficient – imposing least cost burden on proponents
7. ROLE OF EIA
Aid to decision making: Clarify trade off associated with the proposed actions.
Aid to formulation of development actions: Systematic consideration of the
action, its impacts and alternatives. Creative process.
Aid to inter disciplinary co operation: EIA requires cooperation between diffe
rent professions.
Aid to public involvement: EIA provides knowledge and frame work for partici
pation in planning and decision making process.
Instrument for sustainable management: Reveal adverse impacts and incon
sistency towards existing environmental policies.
8. EIA FUNDAMENTALS
It assesses the impact of human activities on the environment.
It defines the environment as the integration and relation of social, physical, an
biological systems.
It defines impact as the significant positive or negative alteration of the
environment by human actions.
OBJECTIVES OF EIA
Explicit addressing and incorporation of the environmental considerations into the
decision making process.
To anticipate and avoid, minimize or offset the adverse significant effects of
development proposals.
To protect the productivity and capacity of natural systems and the ecological
processes which maintain their functions.
To promote development that is sustainable and optimizes resource use.
9. IMMEDIATE OBJECTIVES OF EIA
To improve the environmental design of the proposal.
To ensure that resources are used appropriately and efficiently.
To identify appropriate measures for mitigating the potential impacts of the
proposal.
To facilitate informed decision making, including setting the environmental terms
and conditions for implementing the proposal.
10. LONG TERM OBJECTIVES OF EIA
• Toprotect human health and safety.
• To avoidirreversiblechangesandseriousdamagetothe environment.
• To safeguardvaluedresources,naturalareasandecosystem components.
• Toenhance the socialaspects ofthe proposal.
EIA – THREE CORE VALUES
Integrity : The EIA processwill conformtoagreed standards.
Utility : The EIA process will provide balanced, credible information for de
cision_x0002_making.
Sustainability :The EIA processwillresultinenvironmental safeguards
11. • Widespread participation.
• Determination of keypoints.
• Elimination ofinsignificant points.
• Allocation of requirements for study preparation and qualification.Iden
tification of
• the purpose ofthe action and alternative actions.
COMPONENTS OF THE EIA SYSTEM
1. Air Environment 4. Water Environment
2. Biological Environment 5.Land Environment
3. Noise Environment 6.Risk Assessment
SCOPE OF THE EIA
12. • Promotes community participation.
• Protection of environment.
• Optimum utilization of resources.
• Saves overall time and cost of the project.
• Informs decision makers about damages and impacts.
• Maintenance of biodiversity.
• Lays base for environmentally sound projects.
• Increased project acceptance.
• Improved project design.
BENEFITS OF EIA :
13.
14. 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 dem
and
= 1.8 * 8250 * 135 =2004750Ltrs/day
Qmax = 2004.75 m^3/day
DESIGN PERIOD
15. Water distribution systems
For efficient distribution system adequate water pressure required at various points. Depend
ing upon the level of source, topography of the area and other local conditions, the water ma
y 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 thro
ugh 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 s
toring. Supply can be affected during power failure and breakdown of pumps. The metho
d is not general used.
16. Water distribution systems
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 reliab
le system.
18. 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^2
DESIGN PERIOD
19. Design of the pumping unit
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
DESIGN PERIOD
20. 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 =12
50m
• Assuming the co-efficient of friction = 0.01
DESIGN PERIOD
21. Design of the pumping unit
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
DESIGN PERIOD
22. Design of the pumping unit
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 )
DESIGN PERIOD
23. Design of the pumping unit
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 )
DESIGN PERIOD
24. Design of the pumping unit
Total head loss
Ha = Hd + Hf
=34.19+5.75
=39.93m
DESIGN PERIOD
Efficiency of the pump=80%
BHP=W*Q*H/75*η
=1000*0.06960*45.175
/ 75*0.8
=46.31HP
25. 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
Dimension of tank =11m Dia and (3.5+0.5)m deep [Providing a free board of 0.5]
Design of sedimentation tank
Diameter of sedimentation tank
=√(4A/ꙥ)
= √(4*95.49/ꙥ)
= 11m
Provide the tank size of Dia=11m
26. 1. 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
Design Of Slow Sand Filter
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
2. Providing 2 filter units,
Therefore Area of each units
= 556.875/2
=278.437m^2
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
27. 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
DESIGN OF CHLORINATION TANK
28. Quantity of Disinfectant (Bleaching powder)
Max demand of water =2004.750m^3/d
Assuming chloride dosage= 0.5ppm
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
DESIGN OF CHLORINATION TANK
29. 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
Design Of Elevated Surface Tank
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
Area = Qmax/V
0.023/0.85 =0.0272m^2
diameter, d =√4A/ꙥ
=√4*0.0272/ꙥ
= 0.186m
= 19cm.
30. (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
Design Of Elevated Surface Tank
(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
31. 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
Determination Of Gradient
Substitute the values
= 0.85*110*0.0475^0.63*S^0.54
S = 5.800*10^3
=0.0058 (1/0.0058) =172.41
S=1 in 0.0058
S= 1 in 172
32. 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
Total discharge =dry weather flow + wet weather flow
= 0.0556+0.0758
0.1314m^3/sec
Design of Sewer
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
33. 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
Design of Sewer Pipe
