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QUALITY & PURIFICATION OF
DOMESTIC WATER SUPPLY
Group Member
1. Saurabh Singh
2. Brijesh Kumar
3. Ankit Kumar Chaudhari
4. Govind Singh
5. Kamalesh Prasad
Under Guidence
Mr. Anoop Kumar
(Assistant Professor)
INTRODUCTION
The quality of water is a vital concern for mankind, since it is
directly linked with human welfare. At present, the menace of water
borne diseases and epidemics still booms large on the horizons of
developing countries. Polluted water is the culprit in such cases. So this
project conducting some test on quality of specified water samples. In
this project we analyses which source of water is suitable and
economical for water supply.
PROJECT LAYOUT
• Sample collection
• Physical properties of water sample
• Conclusion of physical properties
• Chemical properties of water sample
• Conclusion of chemical properties
• Biological properties of water sample
• Conclusion of biological properties
• Site investigation
• Drawing of the project
• Design of the project
SAMPLE COLLECTION
SURFACE WATER SAMPLE
• Surface water sample ( Ghaghra river) near NTPC Tanda city.
• Surface water sample ( Pond ) near Bahadurpur village , Tanda city.
GROUND WATER SAMPLE
• Ground water sample , Bahadur pur village ,Tanda
• Ground water sample ,Husain ganj village ,Tanda
PHYSICAL PROPERTIES OF WATER
SAMPLE
S NO TEST SURFACE WATER GROUND WATER UNIT
POND
WATER
GHAGHRA
WATER
BAHADURPU
R , TANDA
HUSAINGANG,
TANDA
1. TURBIDIT
Y
25 4 0 1 NTU
2. COLOUR 10-15 5-10 0-5 0-5 HAZEN UNIT
3. SPECIFIC
CONDUCTI
VITY
68.7 265 537 530 μmhos/cm
4. TEMPRATU
RE
22.5 22.6 22.3 22.4 °C
5. TDS 172.7 194.3 330 310 ppm
CONCLUSION
Turbidity:-
It is frequently caused by runoff from disturbed areas around the pond
or from bottom-dwelling fish.
Muddy water is very common in new ponds and usually disappears
the clean water as vegetation grows around the pond.
Deposition of suspended sediment during high flood.
Pond water is very turbid and its turbidity exceed the standard limit so
this water is not healthy for drinking purpose.
The turbidity of Ghaghra river and ground water sample are between
standard limit so these water are usable for drinking use.
Colour:-
Planktonic algae can cause the pond to appear blue/green or green.
 Loose muck and partially digested debris can cause a very dark appearance to a pond
Dissolved organic matter from decaying vegetation or some inorganic colour materials
such as coloured soils, etc., may impart colour to the water
Ghaghra river and Pond water colour exceed the standard limit so this water is not
healthy for drinking purpose.
The colour of ground water samples are between standard limit so these water are usable
for drinking use.
Conductivity:-
Increases in conductivity may be an indicator that polluting discharges have entered the
water.
Conductivity in water is affected by the presence of inorganic dissolved solids such as
chloride, nitrate, sulphate, and phosphate anions (ions that carry a negative charge) or
sodium, magnesium, calcium, iron, and aluminium cations (ions that carry a positive
charge).
The conductivity of all water sample is be between the standard limit.
Temperature:-
The most obvious reason for temperature change in water is the change in
seasonal air temperature.
Water temperature fluctuations in streams may be further worsened by
cutting down trees, which provide shade.
Total dissolve solid:-
Total dissolved solids (TDS) comprise inorganic salts (principally calcium,
magnesium, potassium, sodium, bicarbonates, chlorides, and sulphates) and
some small amounts of organic matter that are dissolved in water
For drinking water, the maximum concentration level set is 500 mg/L
Primary sources for TDS in receiving waters are agriculturel and residential
runoff.
In all water sample the TDS value is under standard limit . So we can use all water
samples for domestic use.
CHEMICAL PROPERTIES OF WATER
SAMPLE
S NO TEST SURFACE WATER GROUND WATER UNIT
POND
WATER
GHAGHRA
WATER
BAHADURPU
R , TANDA
HUSAINGANG,
TANDA
1. pH 6.73 6.45 6.35 6.58 Mol/liter
2. Fluoride 0-1 1 0-1 1 mg/L
3. Chloride 40 30 50 60 Mg/L
4. Dissolve
oxygen
6.8 8.6 5.8 6.8 ppm
5. Hardness ---- ------ -------- ------- ppm
CONCLUSION
pH Value:-
 pH create acidity in the body! When our blood stream becomes too
acidic, it will steal calcium from the bones to try to balance out the pH
level
The pH level of our drinking water reflects how acidic it is. And
standard value of dirking water is 6.5 to 8.5
In the observation table pond water, Hussainganj village( ground
water) and river water sample ph is between 6.5 to 8.5 .so we use
these sample for drinking purpose.
Fluoride:-
Fluoride is prevent tooth decay.
 Excess of fluoride included unquestionable poisons in water .
Fluoride is not a nutrient; no biochemical process in the human
body needs fluoride
 According to BIS, the standard limit of fluoride in drinking water is 1
mg /L
In the observation table all water sample , fluoride content is
between standard limit so we can use these water sample for drinking
purpose
Chloride :-
Chloride occurs naturally in ground water, but is found in greater
concentrations where seawater and run-off from road .
chlorides are harmless when exceed 250 mg/L. And at low level it
give unpleasant taste.
When chloride excess in water it also damage plumbing, appliances,
and water heaters,
According to BIS standard chloride is not more than 250 mg/L in
drinking water.
In all water sample chloride content is balanced .so we can use this
water for drinking purpose.
Dissolve oxygen
Dissolved Oxygen is the amount of gaseous oxygen dissolved in the water.
Oxygen enters the water by direct absorption from the atmosphere, by rapid
movement, or as a waste product of plant photosynthesis
Temperature affects the dissolved oxygen level in the water.
Cold water can dissolve more oxygen than warm water. Water also holds
less dissolved oxygen at higher elevations because there is less pressure.
A high DO level in a community water supply is good because it makes
drinking water taste better.
high DO levels speed up corrosion in water pipes.
According to BIS standard value of DO for drinking water is 4 to 8 ppm
.so all sample is useful for drinking purpose.
POPULATION OF TANDA CITY
YEAR POPULATION INCREASED
POPULATION IN
EACH DECADE
%INCREASED
INPOPULATION
GROWTH
1981 57855
12750 22.03
1991 70605
12862 18.21
2001 83467
12049 12.61
2011 95516
INTAKE
• intake is structure to help in safely withdrawing water from the source over
predetermined pool levels and then to discharge this water into the
withdrawal conduit (normally called intake conduit), through which it flows
up to water treatment plant.
Design of intake
Data
• RL of Ghaghra river bed = 66 m
• Lowest water flow = 68 m
• Normal level water flow =71m
• Highest flood level = 74 m
• Elevation of Tanda City =78 m
• Average water demand to be pumped = 32.32 MLD
• Pumping hour consider normally 18 hours a day= 0.49 m3
𝑠𝑒𝑐
• Area of opening required each opening
…….. Design continued
• A =
Q
V
=2.7 m2
• Total length of screening = 3.8 m
• Let us provided 2 part at each level
• Size = L×H = 1.9×1 m2
• provided dia of pipe = 0.75 m
• calculate velocity in pipe = 1.10 𝑚
𝑠
• Design of slope using Maning formula
• S = 1 in 307
Design of conduit
Conduits are existing water pipeline used as in public water supply.
Hazen – willian’s formula 𝑉 = 0.85𝐶 𝐻 𝑅0.63
𝑆0.54
CH = 100 ( Taken because new pipe ),𝑅 =
𝑑
4
V= 1.2 𝑚
𝑠
Area of pipe ,𝐴 =
0.49
1.2
= 0.408 𝑚2
Diameter of pipe
𝜋
4
× 𝑑2
= 0.408
d = 0.72 m
Slope S = 1 in 586.8 ≈ 1 𝑖𝑛 587
Design of screening
• screening is the first unit operation used at wastewater
treatment plants . Screening removes objects such as rags, paper,
plastics, and metals to prevent damage and clogging of downstream
equipment, piping, and appurtenances. There are three types ( coarse ,
medium) and fine of screen is used in water treatment plant.
Coarse screening
Q = 0.49 𝑚3
𝑠
Let assume velocity in screen = 0.9 𝑚 𝑠
Area of screening =
𝑄
𝑉
=
0.498
0.9
= 0.553 𝑚2
Height of screening = 0.4 m
Length of screening =
0.553
0.03
= 1.38 𝑚
Now assume clear opening width b/w vertival bars = 3 cm
No. of opening. of opening =
1.38
0.03
= 46.11
No. of bar = 216
Dia of bar ∅= 20 mm
Length of occupied the bar = = 47 × 0.02
Total length of screen = 0.92+1.38 = 2.3 m
• Length of screening = 2.3 m
• Height = 0.4 m
• Fine screening = 5 mm
• No. of opening
1.38
0.005
= 276
• Wire dia = 2 mm
• Length of occupied bar = 275×0.002
• L= 0.55 m
• H= 0.4 m
Design of sedimentation tank
A sedimentation tank allows suspended particles to settle out of water
or wastewater as it flows slowly through the tank, thereby providing
some degree of purification.
Particle size in river less than 0.1 mm
Type of particle = discrete (because ghaghra river non alluvial river)
G = 2.65 d = 5×10-3 cm
𝑉𝑆 = 2.23 𝑚𝑚
𝑠
Rectangular tank: -
Maximum daily demand = 32.32 MLD
Velocity of flow maintain in tank
𝑉 = 0.3 𝑚
𝑠
• Length of the tank
= velocity of flow×detention period
= 0.3×6×60
=108 m
Cross section area of tank =
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘
=
8080
108
= 74.81𝑚2
Depth of the tank = 4 m
Width of the tank =
74.81
4
= 18.70 𝑚
Using freeboard = 4 + 0.4 = 4.4 m
Size of rectangular tank= L×B× 𝐻 = 108 × 18.70 × 4.4
Volume of tank =108 × 18.70 × 4.4 = 8886.24 𝑚3
Design of mixing basin :
• It is a basin by which a coagulant is rapidly and uniformly dispersed
through the mass of water. This process usually occurs in a small basin
immediately preceding or at the head of the coagulation basin.
Generally, the detention period is 30 to 60 seconds and the head loss is
20 to 60 cms of water. Here colloids are destabilised and the nucleus
for the floc is formed
Total volume of water to be treated per day = 32.32 MLD
Total volume of water to be treated per day = 32.32 MLD
Assume a detention period 30 minutes in the basin.
The volume of water required to be treated in 30 minutes or ½ hour =
capacity of tank = 673.33 𝑚3
assuming the velocity of flow through the channel as 0.3 m/sec
length of flow = 540 𝑚
The depth of two water in the basin = 2.77 𝑚
The distance b/w the battles is kept equal to the minimum permissible
0.45 m
Clear opening b/w the end of each battle and wall may be taken as equal
1.5times the distance b/w the baffles = 0.70 m
No. of channel required =
540
7.25
= 37.5 ≈
38 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑖𝑛 𝑒𝑎𝑐ℎ ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑡𝑎𝑛𝑘
Clear length of the tank excluding the battle walls and side wall= 17.1 m
Assuming thickness of each baffle = 7.5 cm
Overall inside length of the tank = =19.875 m
Design of settling tank
• Settling tank is that by which particulates settle to the bottom of a
liquid and form a sediment. Particles that experience a force, either
due to gravity or due to centrifugal motion will tend to move in a
uniform manner in the direction exerted by that force.
Capacity of tank required = 5386.66 𝑚3
Assuming over flow rate 1200 𝑙 ℎ𝑟/𝑚2 of plain area
• 𝑄 =
5386.66×103
4
𝑙 ℎ𝑟 = 1346× 103
𝑙 ℎ𝑟
• Plain area = 1200
• BL= 1346 ×
103
1200
= 1122.22 𝑚2
• Using width 12m 𝐿 =
1122.22
12
= 93.51 m
• L= 94 m , B= 12 m, D= 4 m
Provide extra sludge = 94 × 12 × 4.5 𝑚
Design of floc chamber
Floc chamber is used for flocculation. It refers to the process by which
fine particulates are caused to clump together into a floc. The floc may
then float to the top of the liquid (creaming), settle to the bottom of the
liquid (sedimentation)
Depth of flock chamber half of the depth on the tank =
4.5
2
= 2.25
• Detention period 20 mm (15-40 mm)
• Now flow capacity of the chamber = 448.88 𝑚3
• Plain area required =
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
𝑑𝑒𝑝𝑡ℎ
=
448.88
2.25
= 199.50 𝑚2
≈ 200 𝑚2
• Using same width as 12 m gel flock chamber =
200
12
= 16.66 ≈ 17 𝑚
Design of filtration tank
( use rapid sand filter )
• Filters use sieving, adsorption, ion exchanges, biological metabolite
transfer, and other processes to remove unwanted substances from a
quantity of water. And unlike a sieve or screen, a filter can potentially
remove particles. We can use rapid sand filter .
Use rapid sand filter
Water required per day = 32.32 MLD
• Assuming 4% of filtred water is required for washing of the filter
everyday =
32.32
0.96
• = 33.66 𝑀𝐿𝐷 𝑑𝑎𝑦
• Now assuming that 0.55 hour ar ½ hour lost everyday washing of filter
=
33.66
23.5
= 1.43 𝑀𝐿𝐷 ℎ𝑜𝑢𝑟
Now assuming the rate of filtration 6000 𝑙 ℎ𝑟/𝑚2
Area of filter =
1.43×106
6000
= 238.33 𝑚2
• Now assuming filter bed L is 2 times of width of filter bed 2 beds are
provided
3 × 𝐿 × 𝐵 = 238.33 3 𝑢𝑛𝑖𝑡 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
3 × 𝐵 × 𝐵 = 238.33
𝐿 = 2 × 𝐵
𝐿 = 2 × 6.30
𝐿 = 12.60 𝑚
• 3 filter unit each dimension = 2.60× 3.44 𝑚
Design of under drainage system
The under drainage system is an essential part of living in a city or
urban area, as it reduces flood damage by carrying water away.
Total area of the preformation = 0.2 × 𝑓𝑖𝑙𝑡𝑒𝑟 𝑎𝑟𝑒𝑎 = 0.086 m2
• Assuming area of each lateral = 2 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
= 0.173 𝑚2
• Now assuming area of manifold taise the area of lateral
dia of manifold =
𝜋
4
× 𝑑2
= 0.346, d= 0.66 m
using 0.7 m of dia manifold spacing 15 cm
• no. of lateral 15 then given = 12.60 ×
100
15
= 84
other side of manifold =168 in each unit
Continued …….
length of each lateral
=
width of filter
2
−
dia of manifold
2
=
3.2
2
−
0.7
2
= 1.35 m
now adopting 13 mm dai of perforation in the lateral
total area of perforation = 0.086 m2
no. of perforation in each lateral =
650
168
= 3.36 𝑜𝑟 4
• area of perforation each lateral =
𝜋
4
× 1.32 × 4 = 5.30 𝑐𝑚2
• area of each lateral = 2 × 5.30 = 10.60 𝑐𝑚2
• dia of each lateral = 3.67 𝑐𝑚
• 168 lateral 3.67 cm dia @15 c/c each having ≮ppularation 13 mm size
with 0.7 m dia of manifold.
•
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑙𝑎𝑡𝑒𝑟𝑎𝑙
𝑑𝑖𝑎 𝑜𝑓 𝑙𝑎𝑡𝑒𝑟𝑎𝑙
=
1.35 𝑚
3.67𝑐𝑚
=
135
3.67
= 36.78
• less than 60 OK
Let assuming rate of washing of filter be 60 cm/min or 0.60 m/ minute
• Wash water discharge = 0.433 𝑚2 𝑠𝑒𝑐
• Velocity of flow in the lateral wash water = 2.43 𝑚/𝑠𝑒𝑐
SIGNIFICANCE
The water is required for every life and the quality of water is
depend upon the source of the water. Now a days the ground water table
decreases in highly populated cities so new water resource are required
for water supply like surface water. The nearly available surface water
in which the sufficient water available for the supply after treatment. If
the river is located near city then river water can be used after some
treatment.
The significance of this project is based on qualitative purified
river water supply in Tanda.
REFERENCE
• Indian standard drinking water specification (second revision)
IS 10500:2012, https://law.Resource.Org/pub/in/bis/s06/is.10500.2012.Pdf
• Method of sampling and test (physical and chemical), is 3025
• Water supply engineering environmental engineering vol. 1 by S.. Garg(author)
• Https://en.Wikipedia.Org
• Introduction to water waste water engineering,
http://nptel.Ac.In/courses/105106119/
• Http://fosc.Org/wqdata/wqparameters.Htm
THANK YOU

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Quality and purification of domestic water supply

  • 1. QUALITY & PURIFICATION OF DOMESTIC WATER SUPPLY Group Member 1. Saurabh Singh 2. Brijesh Kumar 3. Ankit Kumar Chaudhari 4. Govind Singh 5. Kamalesh Prasad Under Guidence Mr. Anoop Kumar (Assistant Professor)
  • 2. INTRODUCTION The quality of water is a vital concern for mankind, since it is directly linked with human welfare. At present, the menace of water borne diseases and epidemics still booms large on the horizons of developing countries. Polluted water is the culprit in such cases. So this project conducting some test on quality of specified water samples. In this project we analyses which source of water is suitable and economical for water supply.
  • 3. PROJECT LAYOUT • Sample collection • Physical properties of water sample • Conclusion of physical properties • Chemical properties of water sample • Conclusion of chemical properties • Biological properties of water sample • Conclusion of biological properties • Site investigation • Drawing of the project • Design of the project
  • 4. SAMPLE COLLECTION SURFACE WATER SAMPLE • Surface water sample ( Ghaghra river) near NTPC Tanda city. • Surface water sample ( Pond ) near Bahadurpur village , Tanda city. GROUND WATER SAMPLE • Ground water sample , Bahadur pur village ,Tanda • Ground water sample ,Husain ganj village ,Tanda
  • 5. PHYSICAL PROPERTIES OF WATER SAMPLE S NO TEST SURFACE WATER GROUND WATER UNIT POND WATER GHAGHRA WATER BAHADURPU R , TANDA HUSAINGANG, TANDA 1. TURBIDIT Y 25 4 0 1 NTU 2. COLOUR 10-15 5-10 0-5 0-5 HAZEN UNIT 3. SPECIFIC CONDUCTI VITY 68.7 265 537 530 μmhos/cm 4. TEMPRATU RE 22.5 22.6 22.3 22.4 °C 5. TDS 172.7 194.3 330 310 ppm
  • 6. CONCLUSION Turbidity:- It is frequently caused by runoff from disturbed areas around the pond or from bottom-dwelling fish. Muddy water is very common in new ponds and usually disappears the clean water as vegetation grows around the pond. Deposition of suspended sediment during high flood. Pond water is very turbid and its turbidity exceed the standard limit so this water is not healthy for drinking purpose. The turbidity of Ghaghra river and ground water sample are between standard limit so these water are usable for drinking use.
  • 7. Colour:- Planktonic algae can cause the pond to appear blue/green or green.  Loose muck and partially digested debris can cause a very dark appearance to a pond Dissolved organic matter from decaying vegetation or some inorganic colour materials such as coloured soils, etc., may impart colour to the water Ghaghra river and Pond water colour exceed the standard limit so this water is not healthy for drinking purpose. The colour of ground water samples are between standard limit so these water are usable for drinking use. Conductivity:- Increases in conductivity may be an indicator that polluting discharges have entered the water. Conductivity in water is affected by the presence of inorganic dissolved solids such as chloride, nitrate, sulphate, and phosphate anions (ions that carry a negative charge) or sodium, magnesium, calcium, iron, and aluminium cations (ions that carry a positive charge). The conductivity of all water sample is be between the standard limit.
  • 8. Temperature:- The most obvious reason for temperature change in water is the change in seasonal air temperature. Water temperature fluctuations in streams may be further worsened by cutting down trees, which provide shade. Total dissolve solid:- Total dissolved solids (TDS) comprise inorganic salts (principally calcium, magnesium, potassium, sodium, bicarbonates, chlorides, and sulphates) and some small amounts of organic matter that are dissolved in water For drinking water, the maximum concentration level set is 500 mg/L Primary sources for TDS in receiving waters are agriculturel and residential runoff. In all water sample the TDS value is under standard limit . So we can use all water samples for domestic use.
  • 9. CHEMICAL PROPERTIES OF WATER SAMPLE S NO TEST SURFACE WATER GROUND WATER UNIT POND WATER GHAGHRA WATER BAHADURPU R , TANDA HUSAINGANG, TANDA 1. pH 6.73 6.45 6.35 6.58 Mol/liter 2. Fluoride 0-1 1 0-1 1 mg/L 3. Chloride 40 30 50 60 Mg/L 4. Dissolve oxygen 6.8 8.6 5.8 6.8 ppm 5. Hardness ---- ------ -------- ------- ppm
  • 10. CONCLUSION pH Value:-  pH create acidity in the body! When our blood stream becomes too acidic, it will steal calcium from the bones to try to balance out the pH level The pH level of our drinking water reflects how acidic it is. And standard value of dirking water is 6.5 to 8.5 In the observation table pond water, Hussainganj village( ground water) and river water sample ph is between 6.5 to 8.5 .so we use these sample for drinking purpose.
  • 11. Fluoride:- Fluoride is prevent tooth decay.  Excess of fluoride included unquestionable poisons in water . Fluoride is not a nutrient; no biochemical process in the human body needs fluoride  According to BIS, the standard limit of fluoride in drinking water is 1 mg /L In the observation table all water sample , fluoride content is between standard limit so we can use these water sample for drinking purpose
  • 12. Chloride :- Chloride occurs naturally in ground water, but is found in greater concentrations where seawater and run-off from road . chlorides are harmless when exceed 250 mg/L. And at low level it give unpleasant taste. When chloride excess in water it also damage plumbing, appliances, and water heaters, According to BIS standard chloride is not more than 250 mg/L in drinking water. In all water sample chloride content is balanced .so we can use this water for drinking purpose.
  • 13. Dissolve oxygen Dissolved Oxygen is the amount of gaseous oxygen dissolved in the water. Oxygen enters the water by direct absorption from the atmosphere, by rapid movement, or as a waste product of plant photosynthesis Temperature affects the dissolved oxygen level in the water. Cold water can dissolve more oxygen than warm water. Water also holds less dissolved oxygen at higher elevations because there is less pressure. A high DO level in a community water supply is good because it makes drinking water taste better. high DO levels speed up corrosion in water pipes. According to BIS standard value of DO for drinking water is 4 to 8 ppm .so all sample is useful for drinking purpose.
  • 14. POPULATION OF TANDA CITY YEAR POPULATION INCREASED POPULATION IN EACH DECADE %INCREASED INPOPULATION GROWTH 1981 57855 12750 22.03 1991 70605 12862 18.21 2001 83467 12049 12.61 2011 95516
  • 15. INTAKE • intake is structure to help in safely withdrawing water from the source over predetermined pool levels and then to discharge this water into the withdrawal conduit (normally called intake conduit), through which it flows up to water treatment plant.
  • 16. Design of intake Data • RL of Ghaghra river bed = 66 m • Lowest water flow = 68 m • Normal level water flow =71m • Highest flood level = 74 m • Elevation of Tanda City =78 m • Average water demand to be pumped = 32.32 MLD • Pumping hour consider normally 18 hours a day= 0.49 m3 𝑠𝑒𝑐 • Area of opening required each opening
  • 17. …….. Design continued • A = Q V =2.7 m2 • Total length of screening = 3.8 m • Let us provided 2 part at each level • Size = L×H = 1.9×1 m2 • provided dia of pipe = 0.75 m • calculate velocity in pipe = 1.10 𝑚 𝑠 • Design of slope using Maning formula • S = 1 in 307
  • 18. Design of conduit Conduits are existing water pipeline used as in public water supply.
  • 19. Hazen – willian’s formula 𝑉 = 0.85𝐶 𝐻 𝑅0.63 𝑆0.54 CH = 100 ( Taken because new pipe ),𝑅 = 𝑑 4 V= 1.2 𝑚 𝑠 Area of pipe ,𝐴 = 0.49 1.2 = 0.408 𝑚2 Diameter of pipe 𝜋 4 × 𝑑2 = 0.408 d = 0.72 m Slope S = 1 in 586.8 ≈ 1 𝑖𝑛 587
  • 20. Design of screening • screening is the first unit operation used at wastewater treatment plants . Screening removes objects such as rags, paper, plastics, and metals to prevent damage and clogging of downstream equipment, piping, and appurtenances. There are three types ( coarse , medium) and fine of screen is used in water treatment plant.
  • 21. Coarse screening Q = 0.49 𝑚3 𝑠 Let assume velocity in screen = 0.9 𝑚 𝑠 Area of screening = 𝑄 𝑉 = 0.498 0.9 = 0.553 𝑚2 Height of screening = 0.4 m Length of screening = 0.553 0.03 = 1.38 𝑚 Now assume clear opening width b/w vertival bars = 3 cm No. of opening. of opening = 1.38 0.03 = 46.11 No. of bar = 216
  • 22. Dia of bar ∅= 20 mm Length of occupied the bar = = 47 × 0.02 Total length of screen = 0.92+1.38 = 2.3 m • Length of screening = 2.3 m • Height = 0.4 m • Fine screening = 5 mm • No. of opening 1.38 0.005 = 276 • Wire dia = 2 mm • Length of occupied bar = 275×0.002 • L= 0.55 m • H= 0.4 m
  • 23. Design of sedimentation tank A sedimentation tank allows suspended particles to settle out of water or wastewater as it flows slowly through the tank, thereby providing some degree of purification.
  • 24. Particle size in river less than 0.1 mm Type of particle = discrete (because ghaghra river non alluvial river) G = 2.65 d = 5×10-3 cm 𝑉𝑆 = 2.23 𝑚𝑚 𝑠 Rectangular tank: - Maximum daily demand = 32.32 MLD Velocity of flow maintain in tank 𝑉 = 0.3 𝑚 𝑠 • Length of the tank = velocity of flow×detention period = 0.3×6×60 =108 m
  • 25. Cross section area of tank = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 = 8080 108 = 74.81𝑚2 Depth of the tank = 4 m Width of the tank = 74.81 4 = 18.70 𝑚 Using freeboard = 4 + 0.4 = 4.4 m Size of rectangular tank= L×B× 𝐻 = 108 × 18.70 × 4.4 Volume of tank =108 × 18.70 × 4.4 = 8886.24 𝑚3
  • 26. Design of mixing basin : • It is a basin by which a coagulant is rapidly and uniformly dispersed through the mass of water. This process usually occurs in a small basin immediately preceding or at the head of the coagulation basin. Generally, the detention period is 30 to 60 seconds and the head loss is 20 to 60 cms of water. Here colloids are destabilised and the nucleus for the floc is formed
  • 27. Total volume of water to be treated per day = 32.32 MLD Total volume of water to be treated per day = 32.32 MLD Assume a detention period 30 minutes in the basin. The volume of water required to be treated in 30 minutes or ½ hour = capacity of tank = 673.33 𝑚3 assuming the velocity of flow through the channel as 0.3 m/sec length of flow = 540 𝑚
  • 28. The depth of two water in the basin = 2.77 𝑚 The distance b/w the battles is kept equal to the minimum permissible 0.45 m Clear opening b/w the end of each battle and wall may be taken as equal 1.5times the distance b/w the baffles = 0.70 m No. of channel required = 540 7.25 = 37.5 ≈ 38 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑖𝑛 𝑒𝑎𝑐ℎ ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑡𝑎𝑛𝑘 Clear length of the tank excluding the battle walls and side wall= 17.1 m Assuming thickness of each baffle = 7.5 cm Overall inside length of the tank = =19.875 m
  • 29. Design of settling tank • Settling tank is that by which particulates settle to the bottom of a liquid and form a sediment. Particles that experience a force, either due to gravity or due to centrifugal motion will tend to move in a uniform manner in the direction exerted by that force.
  • 30. Capacity of tank required = 5386.66 𝑚3 Assuming over flow rate 1200 𝑙 ℎ𝑟/𝑚2 of plain area • 𝑄 = 5386.66×103 4 𝑙 ℎ𝑟 = 1346× 103 𝑙 ℎ𝑟 • Plain area = 1200 • BL= 1346 × 103 1200 = 1122.22 𝑚2 • Using width 12m 𝐿 = 1122.22 12 = 93.51 m • L= 94 m , B= 12 m, D= 4 m Provide extra sludge = 94 × 12 × 4.5 𝑚
  • 31. Design of floc chamber Floc chamber is used for flocculation. It refers to the process by which fine particulates are caused to clump together into a floc. The floc may then float to the top of the liquid (creaming), settle to the bottom of the liquid (sedimentation)
  • 32. Depth of flock chamber half of the depth on the tank = 4.5 2 = 2.25 • Detention period 20 mm (15-40 mm) • Now flow capacity of the chamber = 448.88 𝑚3 • Plain area required = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑑𝑒𝑝𝑡ℎ = 448.88 2.25 = 199.50 𝑚2 ≈ 200 𝑚2 • Using same width as 12 m gel flock chamber = 200 12 = 16.66 ≈ 17 𝑚
  • 33. Design of filtration tank ( use rapid sand filter ) • Filters use sieving, adsorption, ion exchanges, biological metabolite transfer, and other processes to remove unwanted substances from a quantity of water. And unlike a sieve or screen, a filter can potentially remove particles. We can use rapid sand filter .
  • 34. Use rapid sand filter Water required per day = 32.32 MLD • Assuming 4% of filtred water is required for washing of the filter everyday = 32.32 0.96 • = 33.66 𝑀𝐿𝐷 𝑑𝑎𝑦 • Now assuming that 0.55 hour ar ½ hour lost everyday washing of filter = 33.66 23.5 = 1.43 𝑀𝐿𝐷 ℎ𝑜𝑢𝑟 Now assuming the rate of filtration 6000 𝑙 ℎ𝑟/𝑚2 Area of filter = 1.43×106 6000 = 238.33 𝑚2
  • 35. • Now assuming filter bed L is 2 times of width of filter bed 2 beds are provided 3 × 𝐿 × 𝐵 = 238.33 3 𝑢𝑛𝑖𝑡 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 3 × 𝐵 × 𝐵 = 238.33 𝐿 = 2 × 𝐵 𝐿 = 2 × 6.30 𝐿 = 12.60 𝑚 • 3 filter unit each dimension = 2.60× 3.44 𝑚
  • 36. Design of under drainage system The under drainage system is an essential part of living in a city or urban area, as it reduces flood damage by carrying water away.
  • 37. Total area of the preformation = 0.2 × 𝑓𝑖𝑙𝑡𝑒𝑟 𝑎𝑟𝑒𝑎 = 0.086 m2 • Assuming area of each lateral = 2 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 = 0.173 𝑚2 • Now assuming area of manifold taise the area of lateral dia of manifold = 𝜋 4 × 𝑑2 = 0.346, d= 0.66 m using 0.7 m of dia manifold spacing 15 cm • no. of lateral 15 then given = 12.60 × 100 15 = 84 other side of manifold =168 in each unit
  • 38. Continued ……. length of each lateral = width of filter 2 − dia of manifold 2 = 3.2 2 − 0.7 2 = 1.35 m now adopting 13 mm dai of perforation in the lateral total area of perforation = 0.086 m2 no. of perforation in each lateral = 650 168 = 3.36 𝑜𝑟 4 • area of perforation each lateral = 𝜋 4 × 1.32 × 4 = 5.30 𝑐𝑚2 • area of each lateral = 2 × 5.30 = 10.60 𝑐𝑚2 • dia of each lateral = 3.67 𝑐𝑚
  • 39. • 168 lateral 3.67 cm dia @15 c/c each having ≮ppularation 13 mm size with 0.7 m dia of manifold. • 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑎 𝑜𝑓 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 = 1.35 𝑚 3.67𝑐𝑚 = 135 3.67 = 36.78 • less than 60 OK Let assuming rate of washing of filter be 60 cm/min or 0.60 m/ minute • Wash water discharge = 0.433 𝑚2 𝑠𝑒𝑐 • Velocity of flow in the lateral wash water = 2.43 𝑚/𝑠𝑒𝑐
  • 40. SIGNIFICANCE The water is required for every life and the quality of water is depend upon the source of the water. Now a days the ground water table decreases in highly populated cities so new water resource are required for water supply like surface water. The nearly available surface water in which the sufficient water available for the supply after treatment. If the river is located near city then river water can be used after some treatment. The significance of this project is based on qualitative purified river water supply in Tanda.
  • 41. REFERENCE • Indian standard drinking water specification (second revision) IS 10500:2012, https://law.Resource.Org/pub/in/bis/s06/is.10500.2012.Pdf • Method of sampling and test (physical and chemical), is 3025 • Water supply engineering environmental engineering vol. 1 by S.. Garg(author) • Https://en.Wikipedia.Org • Introduction to water waste water engineering, http://nptel.Ac.In/courses/105106119/ • Http://fosc.Org/wqdata/wqparameters.Htm