Complete Water Treatment Plant Project

Planning & designing of Water treatment plant

INTRODUCTION

Water,undubiously is abasic human need.Providing safe and adequate quantities of the same for all rural and urban communities,is perhaps one of the most important undertaking ,for the public works DeptIndeed,the well planned water supply scheme ,is a prime and vital element of a country’s social infrastructure as on this peg hangs the health and wellbeing of it’s people.

The population in India is likely to hundred cores by the turn of this century,with an estimated 40%of urban population.This goes on to say that a very large demand of water supply;fordomestic,industrial,firefighting,public use ,etc;will have to be in accordance with the rising population.Hence,identification of source of water supply,there conservation and optimum utilization is of paramount importance.The water supplied should be ‘Portable’and ‘Wholesome’.Absolute pure water is never found in nature,but invariable contains certain suspended,colloidal,and dissolved impurities (organic and inorganic in nature,generally called solids),in varying degree of concentration depending upon the source.Hence treatment of water to mitigate and/or absolute removal of these impurities (which could be;solids,pathogenic micro-organisms,odor and taste generators,toxicsubstances,etc.)

1N.B.S. Institute of polytechnic, Ausa.


become indispensable. Untreated or improperly treated water becomes unfit for intended use proves to be detrimental for life.

The designed water treatment plant has Reservoir as the basic source of water the type of treatment to be given depends upon the given quality of water available and the quality of water to be served. However such an extensive survey being not possible in the designed water treatment plant. It is assumed that all kinds of treatment processors are necessary and an elaborate design.

The design of water plant for The latitude and longitude of the town corresponding18.2500°N,76.5000°E respectively. The population of the given year 2031 will be 64296. There are many industries like sugar industries and chemical industries in the town so, treated water supply for domestic and industrial uses are very essential.



Basic Data For The Of Water Supply (System)

The given problem includes the design of water treatment plant and distribution system and also the preparation of it’s Technical Report and Engg. Drawing showing the required details of collection and treatment units. The following Table gives the basic necessary data required for the design of water treatment plant.

No Description

1. Name of the place -Ausa

2. District -Latur

3. Location - About 20km from Latur city

4. Latitude (Lat) - 18.2500°N

5. Longitude (Lon) - 76.5000°E



Sr.No Design Consideration Values1. Design Period (years) 202. Average rate of water supply(Ipcd) 1353. Industrial demand (MLD) 0.64. Quality of raw water

1) Ph2) Turbidity (mg/L)

7.550

3) Total Hardness (mg/L)[as CaCo3] 5504) Chlorides(mg/L) 2005) Iron (mg/L) 2.56)Manganese (mg/L) 3.57)Carbonates (mg/L) 1108)M.P.N (No.100ml) 3.5

5. Population of past four decades(in Person)Year 1981 16721Year 1991 23246Year 2001 30876Year 2011 38733

6. F.S.L. of Reservoir (R.L.in mts.) 612.757. Ground level at ; (R.L.in mts.)

a) Jack well site 607.75 b)Location of aeration unit 608.00

8. Invert level of raw material gravity intake pipe(R.L.in mts) 600

9. Invert level of raw water rising main(mts)10. Dead Storage of Reservoir(Million cu.m) 7.38211. Gross Storage of Reservoir(Million cu.m) 27.727

12. Live Storage of Reservoir(Million cu.m) 20.34513. Silt Level of Reservoir in (m) 607.5014. T.B.L. of Reservoir in (m) 615.25



SALIENT FEATURES OF WATER TREATMENT PLANT

A) General

1. Population of the town

Year 2011:38733 Nos.

Year2031 : 64296 Nos

2. Average daily draft (M.L.D) :9.48

Maximum daily draft (M.L.D) :14.22

3. Design period (Years) :20

B) Collection works

Intake works

Intake well

1. No of units :1

2. Dia. Of well (m) : 6.9

3. Ht of intake well :8.0

4. R.L. of bottom well (m) :607.75

5. R.L. of top of well (m) :612.75

6. Detention time (min) :15



Penstock

1. No. of penstock well : 02

2. Dia. Of penstock (mm) :350

Bell mouth strainer

1. No. of bell mouth strainer : 2

2. Dia. (m) :0.95

Gravity main

1. No. of units :1

2. Dia. (mm) :550

3.Invert main (m) :596.89

4. Slope :1.900

Jack well

1.No. of units :1

2.Dia. (m) :6.55

3.Depth of water :4.4

4.Detension time(min) : 15



Rising main and pumping units

Rising:

1.Dia.(m) :0.45

2. Velocity of flow (m/s) :1.0

Pumping unit :

1.Capacity of each pump (HP) :30Hp

2.No.of pumps :2

c) Treatment works

Aeration unit

1. R.L. of aeration unit (m) (top) : 608.90

(Bottom) m : 606.50

2.Dia. of top tray (m) :1.0

3. Dia. Of bottom (m) :5.0

4. Dia. Of each tray decreasing by (m) :1.0

5. Rise of each tray (m) :0.4

6. Tread of each tray (m) :0.5



7. Dia.Of central rising main pipe (m) :1.0

8. No. of Trays :5

Chemical storage house

1.Length (m) : 12

2.Breadth (m) :16

3.Height (m) :3.0

Chemical Dissolving Tank

1.No. of tank :1

2.Length (m) :0.9

3.Breadth (m) :0.9

4.Depth (m) :1.5

Flash Mixer

1. No. of units : 1

2.Dia.(m) : 2.0

3.Detention time (min) : 1

4.Height (m) : 2.8



5.Depth of water (m) : 2.37

Clariflocculator

Flocculator

1. No. of units :1

2.Dia. (m) : 10.5

3.Dia. of inlet pipe (m) :0.45

4.Depth of water flow (m) :3.5

5.Velocity of flow (m/s) :1.0

Clarifire:

1. No. of units : 1

2. Dia. (m) : 24

3. Depth of water (m) : 3.5

4. Overall depth of tank (m) : 4.7

5. Slope of bottom : 8%

Rapid sand filter



1. No. of units : 1

2. Surface aera (sq.m) : 62.48

3. Dimension of unit (m x m) : 8.8 x 7.1

4. Thickness of sand bed (m) : 0.6

5. Thickness of gravel bed (m) : 0.5

6. Die. of manifold (m) : 1

7. Laterals

a) No’s : 120

b) Dia. (m) : 90

c) Length (cm) : 3.05

d) Spacing (cm) : 20

8. No. of orifice : 51

9. Dia. of orifice (mm) : 13

10. Wash water tank : 1

Disinfection House

1. Chlorine required / day (kg) : 19.908

Storage Units



Underground Reservoir

1. No. of units : 1

2. Dimensions :14.1m x 14.1m

4. Depth (m) : 4.5m

5. Compartments : 6

Elevated Service Reservoir (Ausa city)

1. No. of units : 1

2. Dia. (m) : 12.3

3. Height (m) : 4.3

4. Capacity (Cu. m) : 475



POPULATIONFORECASTING

Design Period:

Water supply project may be designed normally to meet the requirement over a 20 years period after their completion. The time lag between design and completion should be also taken into account. It should not ordinarily exceed 2 years & 5 years even in exceptional circumstances. The 20 years period may however be modified in regard to specific components of the project particularly the conveying mains and trunk mains of distribution system depending on their useful life or the facility for carrying out extension when required, so that expenditure far ahead of utility is avoided. However in our case the design period has been considered as 20 per given data.

Population forecast General consideration

The population to be served during such period will have to be estimated with due regard to all the factors governing the future growth and development of the city in the industrial, commercial, educational, social and administrative spheres. Special factors causing sudden immigration or influx of population should also be foreseen to extent possible.

Calculation of Population with Different Methods 12N.B.S. Institute of polytechnic, Ausa.


1. Arithmetical Increase Method

Sr. No Year Population IncreaseIn pollution

1 2 3 41. 1971 12761 -2. 1981 16721 39603. 1991 23246 65254. 2001 30876 76305. 2011 38733 7857

Total 25972Average 6493

Using the relation

Pn = P + nd

Where,

Pn = Future population after n decades

P = Present population

n = No. of decades

d = Average increase per decade

Given,

n= 3,

p= 38733,

d = 6493



. ۬� . Pn = 38733 + (3*6493)

P2031= 58212(Persons)

2.Geometrical Increase Method

Sr. No

Year Population Increase in population

Percentage increase in population

1 2 3 4 51. 1971 12761 - -2. 1981 16721 3960 18.053. 1991 23246 6525 39.024. 2001 30876 7630 32.825. 2011 38733 7857 25.45

Total 25972 115.34Average per Decade 6493 28.84

Where,

r = Average percentage increase=28.84

n = No. of decades=2

Pn= Population after n decade: 2031

Population in the year 2031



Using the relation

P2031= P [ 1+ r

100¿n

P2031=38733[1+28.84100

¿¿2

= 64296( In persons)

3.Incremental Increase Method

Sr. No


Incremental increase

1. 1971 12761 - -2. 1981 16721 3960 -3. 1991 23246 6525 25654. 2001 30876 7630 11055. 2011 38733 7857 227

Total 25972 3897Average 6493 1299

Using the relation,

Pn = P+ nd + n(n+1)2 ×t

Where,

d = Average increase per decade.



t = Average incremental increase.

n = number of decade.

P2031 = 38733 + ( 2 × 6493 )+ 2(2+1)2

×1299

= 55616( Persons).

4. Decrease Rate of Growth Method

Sr. No


Percentage in population (%)

Decrease in the percentage increase

1. 1971 12761 - - -2. 1981 16721 3960 18.05 -3. 1991 23246 6525 39.02 -20.974. 2001 30876 7630 32.82 +6.25. 2011 38733 7857 25.45 +7.37

Total 25972 115.34 -7.4Average 6493 28.84 -2.47

d = 6493, t = - 2.47

Expected population at the end of 2021

= 38733 + [ (25.45+2.47)100 ]×38733



= 49548 persons

Expected population at the end of 2031

= 49548 + [ (27.92+2.47)100 ]×49548

= 64606 persons

Description of the Various Methods

1.Arithmetic Increase Method

This method is upon assumption that the population increase at a constant rate and rate of growth slowly decreases .In our case also constant rate with slight decrease in growth rate.

Also this method is more suitable for very big and older cities whereas in our case it is relatively smaller and new town.

So result by this method is although good but not as accurate as desired.

2.Geometrical Increase Method

In this method the per decade growth rate is assumed to be constant and which is average of earlier growth rate. The forecasting is done on the basic that the percentage increases per decade will remain same.



This method would apply to cities with unlimited scope for expansion.

3. Incremental Increase Method

These methods include the advantage of both arithmetical as well as geometrical methods. First the average of increase in population calculated according to arithmetical method. And then increase or decrease in the population change for each decade is found out and from these average incremental increases is worked.

4. Decreasing rate method

This method assumes that has some limiting saturation population. The method involves calculation of percentage increase for every decade and then working out the decrease in the percentage increase. The average of decrease in percentage increase is the deducted from the latest percentage increase for each successive future decade.

5. Logistic curve method

This is suitable in cases where the rate of increase or decrease of population while the population growth is likely to reach saturation limit ultimately because of special law factor.

The city shall grow as per the logistic curve, which will plot as a straight line on the arithmetic paper with the time of intervals plotted against population in percentage of solution.



6. Simple graphical method

Since the result obtained by this method is dependent upon the designer, this method of empirical nature and not much reliable. Also this method gives very approximately result. This method is useful to verify the data obtained by some other method.

7. Graphical comparison method

This involves the extension of the population time curve into the future based on a comparison of a similar curve or comparable cities and modified to the extent dictated by the factor governing such predictions.

Calculation of Water Demand

Calculation of different Drafts

Expected population after 20 years

Average rate of water supply

(Including domestic, commercial, public and wastes)

Water required for above purpose for whole town = 64296×135

= 8.68 MLD

Industrial demand = 0.8 MLD

Fire Requirement :



It can be assumed that city is a residential town (low rise building)

Water for fire demand

By using government of India recommendation formula,

Q = 100√ p

Where,

Q = Fire demand in kiloliters/day,

P = Population in thousands.

Q= 100×√64.296

Q = 0.80 MLD

(1) Average daily draft = 8.68 + 0.8

= 9.48 MLD

(2) Maximum daily draft = 1.5 ×9.48

= 14.22 MLD

(3) Coincident draft =maximum daily draft + fire demand

= 14.22+ 0.8

= 15.02 MLD

(Coincident draft < maximum hourly draft)



Design Capacity for Various Components

(1)Intake structure daily draft = 14.22MLD

(2)Pipe main = maximum daily draft = 14.22MLD

(3)Filter and other units at treatment plant

= 2 × average daily demand

= 2× 9.48

= 18.96 MLD

(4) Lift pump = 2 × average daily demand

= 18.96MLD

Standard units of Treatment Plant

Due to previous analysis following units are required to be

design for water treatment plant.

(1) Intake Structure :

(a) Intake well

(b) Gravity main

(c) Jack well

(d) Rising main



(e) Pump

(2) Treatment unit :

(a) Aeration unit

(b) Coagulant dose

(c) Lime soda dose

(d) Chemical dissolving tank

(e) Chemical house

(f) Flash mixer

(g)Clariflocculator

(h) Rapid sand filter

(i) Chlorination unit

(3) Storage unit :

(a) Underground storage tank

(b) Elevated storage

A schematic diagram of each of the unit is shown.

Design Of Units :

(a) Intake well

Intake consist of opening. Strainer or grating through which the water enters, end the conduct conveying the water, usually by



gravity to a well or sump From the well, the water is pumped in to the main or treatment plant. Intakes should also be so located and designed that possibility of interference with the supply is minimized and where uncertainty of continuous serviceability exists, intakes should be duplicated. The following must be considered in designing and locating the intakes.

The supply of, whether impounding reservoir. Lake or river (including the possibility of wide fluctuation in water level) .

The character of the intake surrounding, depth of water character or bottom, navigation requirements, the effect of currents, floods and storms upon the structure and in scouring the bottom .

The location with respect to the source of pollution

The prevalence of floating materials, such as ice, logs and

vegetation .

Type of Intakes :

• Wet Intakes : water is up to source of supply .

• Dry Intakes : No water inside it other than in the intake pipe.

• Submerged Intakes : Entirely under the water.

•Movable and Floating Intake : Used where wide variation in surface elevation with sloping blanks.

a)Location of Intakes :

• The location of the best quality of water available.



• Currents that might threaten the safety of the intake structure.

• Navigation channels should be avoided.

• Ice and other difficulties.

• Formation of shoals and bars.

• Fetch of the wing and other condition affection the weight of waves.

•Ice storm.

• Floods.

• Power availability and reliability.

• Accessibility.

•Distance from pumping station.

•Possibilities of damage by moving objects and hazards.

The intake structure used intake is our design is wet-type.

b) Design Criteria

1. Detention time 10 to 15 min2. Diameter 5 to 10 m (max 15

m)3. Depth 4 to 10 m4. Velocity of flow 1.0 to 1.5 m/sec5. Number of units 1 to 3 (max 4 )6. Free board 5 m



(c) Design Assumptions :

Given F.S.L. = 612.75 m

Minimum R.L. = 607.75 m

Given invert level of gravity main = 25 m

Detention time = 15 min

d) Design Calculation

Flow of water required = 14.22MLD / 3600 × 24

= 0.1645cu.m/sec

Volume of well = 0.1645× 15× 60

= 148.05cu.m

Cross-sectional area of intake well = 148.05 / 4

= 37.01sq.m

Diameter of intake well (d) =√4 X 37.01 /π

= 6.86< 10 m (O.K.)

Provide 1 intake well of diameter 6.86 m ≅ 6.9m

(e) Summary

1. Number of intake wells 1 unit



2. Diameter of intake wells 6.9 m3. Height of well 8.0 m4. R.L. of bottom well 607.75m

Design of Pen Stock and Bell Mouth Strainer

(a) Pen stock

This are the pipes provided in intake well to allow water from water body to intake well.These pen stocks are provided atdifferent levels, so as to take account of seasonal variation in water level (as H.F.L., W.L., L.W.L. ).

Trash racks of screens are provided to protect the entry sizeable things which can create trouble in the penstock. At each level more than one penstock is provided to take account of any obstruction during itsoperation. This penstock is regulated by valves provided at the top of intake wells.

(b) Design criteria

Velocity through penstock = 0.6 to 1.0 m/s

Diameter of each penstock = less than 1 m

Number of penstock for each intake well = 2

(c) Design calculation



Number of intake weIl = 1

Number of penstock at each level = 2

Velocity =1.0m/s (assumed)

C/S of each penstock = 0.1645(1.0 ×2)

= 0.082 m2

Diameter = 0.32m~ 0.35 m

(d) summary

1 Number of penstock 2 units 2 At each level 1 m 3 Diameter of penstock 0.35 m

Design of bell mouth strainer

(a) Design criteria

Velocity of flow = 0.2 to 0.3m/s

Hole diameter = 6 to 12 mm

Area of strainer = 2 x diameter of holes

(b) Assumption

Velocity of flow = 0.25 m/s

Hole diameter = 10mm 27N.B.S. Institute of polytechnic, Ausa.


(c) Calculation

Area of each hole = π4d2 =

π4 (102) =78.54 mm2

Area of collection = area of penstock

0.16450.25 ×2

= 7.854 × 10−5N

N = 4188.95Nos

Area of bell moth strainer = 2x area of holes

= 2 x 4188.95 x 7.854 × 10−5

= 0.66 m2

Diameter of bell mouth strainer = 0.92 m ≅ 0.95m

Provide diameter of 0.95 m for bell mouth strainer.

Design of gravity main

(a) Gravity main

The Gravity main connects the intake well to the jack well and water flows through it by gravity. To secure the greatest economy, the diameter of a single pipe through which water flows by gravity should be such that all the head available to



cause flow is consumed by friction. The available fall from the intake well to the jack well and the ground profile in between should generally help to decide if a free flow conduit is feasible. Once this is decided the material of conduit is to be selected keeping in view the local cost and nature of terrain to be traversed. Even when a fall is available, a pumping or force main independently or in combination with gravity main could also be considered. The gravity line should be laid below hydraulic gradient.

(b) Design criteria

Diameter of gravity main = 0.3 to 1m

Velocity of water = 0.6 to 0.9m/s

No of gravity main = 1

No. of intake well = 1

Assumed velocity = 0.7m/s

(c) Design calculation

RCC circular pipe is used.

Conduit velocity = 0.7m/s (assumed)



Area of conduit required = 0.1645

0.7

= 0.235m2

Diameter of conduit = 0.55 m

Using manning’s formula

V =R23 X S

12 X

1N

S =N 2 x V 2

R43

S = 1:900

Head loss = 100900

= 0. 111

R.L. of gravity main = 600- 3

= 597 m

R.L. of gravity main at jack well = 597 -0.111

= 596.89m

(d) Summary

1 Number of gravity intake

1 unit

2 Diameter of gravity intake

0.55m

3 Invert level at intake well

597 m



4 Invert level at jack well 596.89mm

Design of jack well

(a) Jack well

This structure serves as a collection of the sump well for incoming water from the intake well from where the water is pumped through the rising main to the various treatment units.

This unit is more useful when numbers of intake wells are than one, so that water is collected in one unit and then affected.

The jack well is generally located away from the shore line, that the installation of pumps, inspection maintenance is made easy.

(b) Design criteria

Detention time = 0.5 x detention time of intake well (3to15 mm)

= 0.5x15

= 7.30 min.

Suction head = <10m

Dia. of well < 20m.



(c) Design calculations

Detention time= 7.30min

Assuming suction head = 8m.

Bottom clearance = 1m

Top clearance = 0.5m

Maximum depth of water that can be stored in condition when water is minimum in reservoir

= 612.75-608.35 = 4.4m

Capacity of well = 0.1645x 15x 60

= 148.05Cu.m

C/S area of well = 148.05

4.4

= 33.65 Sq. m.

Diameter of well = 6.55m

R.L. of bottom of jack well = 608.35m



R.L. of top of jack well when full = 608.35+8

= 616.35 m.

(d) Summery

1. Dia. Of jack well 6.55m

2. R.L. of bottom of jack well 608.35m

3. R.L. on top of jack well 616.75m

4. Suction depth 4.4m

5. Top clearance 0.5m

6. Bottom Clearance 1.0m

Design of Pumping System:

a) Pumps

In the water treatment plant, pumps are used boost the water from the jack well to the aeration units.

The following points are to be stressed upon. The suction pumping should be as short and straight as

possible. It should not be greater than 10m, for



centrifugal pump. If head is more than 10m, water is converted into vapor and thus inspire of creating water head, vapor head is created and pump ceases to function.

The function pipe should be of such size that the velocity should be about 2m/sec.

The delivery pipe should be of such size that the velocity should be about 2.5m/sec.

The following for types of pumps are gradually used.

Buoyancy operated pumps Impulse operated pumps Positive displacement pumps Velocity adoption pumps

The following criteria govern pump selection.

Type of duty required. Present and projected demand and pattern and change in

demand. The details of head and flow rate required. Selecting the operating speed of the pump and suitable

drive. The efficiency of the pumps and consequent influence on

power consumption and the running costs.



b) Diameter of rising main:

Q = 0.1645cu.m/sec

Economical diameter D = 0.97√Q to 1.22√Q

= 0.97√0.1645 to 1.22√0.1645

= 0.393 to 0.494 m

Provide D = 0.45m

c) Design Criteria

Suction head should not be greater than 10m

Velocity of flow length = 0.7 to 1.5 m/s

Top clearance = 0.5 m

Bottom clearance = 1 m

d) Design Calculation

Friction loss in rising main



Assuming velocity = 0.9 m/sec

f = 0.02

hf = fp v2

2 gd =0.02 X 140 X 0.92

2 X 9.81 X 0.45

= 0.25

Total head of pumping =hs+ hd + hf+ minor losses

= 4.12+ 4.88 + 0.25 + 1

= 10.25 m

Assuming two pumps in paralles.

W.H.P = w . Q . H

75 = 1000 X 0.1645 X 10.25

75 = 22.48 HP

B.H.P =w . H . P

n =22.480.75 = 29.97 HP

e) Summery

Provide – 30 HP pumping in parallel

Diameter of pipe 0.45 m



Design of Rising Main

a) General These are the pressure pipes used to convey the water from the jack well to the treatment units. The design of rising main is dependent upon on resistance to flow available head, allowable velocities of flow, sediment transport, quality of water and relative cost. Various types of pipes used are cast iron, steel, reinforced cement concrete; pre stressed concreted, asbestos cement, polyethylene rigid PVC, ductile iron fiber glass pipe, reinforced plastic, fiber reinforced plastic. The determination of suitability in all respect of the pipe of joints for any work is a meter of decision by the engineer concentrated on the basic for the scheme.

b)Design Criteria

Velocity = 0.9 to 1.5 m/sec

Diameter < 0.9 m

c)Design Calculation



Economical diameter, D = 0.97√Q to 1.22√Q

= 0.97√0.1645 to 1.22√0.1645

= 0.393 to 0.494 m

Provide diameter D = 0.45 m

V = Q/A = 1.03 m/sec

d)Summary:

The aim of water treatment is to produce and maintain water that is hygienically safe, aesthetically attractive and palatable, in an economically manner. Albeit the treatment of water would achieve the desired quality, elevation of its quality should not be confined to the end of the treatment facilities but should be extended to the point of consumers use the method of treatment to be employed depends on the characteristics of the raw water and the desired standards of water quality. The unit operation and unit process in water treatment constitute aeration flocculation (rapid and slow mixing) and clarification and may take many different combinations to suit the above requirement.

In the case of ground water the surface water storage which are well protected, where the water has turbidity below 10 JTU


1. Diameter of pipe 0.45


(Jackson candle turbidity units) and is free from odors and color, only only disinfection by chlorination may be necessary.

Where ground water contains excessive dissolved carbon dioxide and odors gases, aeration followed by flocculation and sedimentation rapid gravity or pressure filtration and chlorination may be necessary.

Conventional treatment including pre chlorination, aeration, flocculation and sedimentation rapid gravity filtration and post chlorination are adopted for highly polluted surface waters laden with algae or microscopic organisms.

Based on the data given in second chapter, the following treatment units are designed to meet the quality and quantity requirements of the projects.

Aeration unit

Coagulant dose

Lime soda dose

Chemical dissolved tank

Chemical house

Flash mixer

Clarriflocculator

Rapid sand filter



Chlorination unit

The detail design of the above units are discusses in subsequent section

Design of Aeration unit

Aeration unit

Aeration is necessary to promote the exchange of gases between the water and the atmosphere. In water treatment, aeration is practiced for here purpose.

To add oxygen to water for imparting freshness’s, e.g. water from under ground sources of CO2, H2S and other volatile substance causing taste and odour e.g. water from deeper layers of an impounding reservoir. To precipitated impurities like iron and manganese, in certain forms, e.g. water from some underground sources.

This limitation impurities like iron and manganese, in certain forms, e.g. water from some underground sources. This limitation of aeration is that the water is rendered more corrosive after aeration when the dissolved oxygen contents is increased through in earlier circumstance it may otherwise due to removal of aggressive CO2. Also for taste and odour removal, aeration is



not largely effective but can be used in combination with chlorine or activate carbon to reduce their doses.

The concentration of gases in a liquid generally obeys Henry’s law which states that the concentration of each gases in water is directly proportional to the partial pressure, or each gas in water is directly proportional to the partial pressure, or concentration of gases in the atmosphere in contact with water. The saturation concentration of a gas decreases with temperature and dissolved salt in water. Aeration tends to accelerate the gas exchange.

The three types of aerators are:

Waterfall or multiple tray aerators.

Cascade aerators.

Diffused air aerators.

Design Criteria for Cascade Aerators

Number of trays = 4 to 9

Spacing of trays = 0.3 to 0.75 m c/c

Heights of the structure = 2 m

Space requirement = 0.015 – 0.045 sqm/cub m/ hr



Design Calculation

Qmax = 0.1645 m3/sec

Provide area at tray = 17 m2

Diameter of bottom most tray = 5m

Rise of each tray = 0.4 m

Tray of each tray = 50 cm

Summery

Sr. No. Cascade Diameter of tray (m)

R.L. (m)

1. First 1 608.50

2. Second 2 608.10

3. Third 3 607.70

4. Fourth 4 607.30

5. Fifth 5 606.90

R.L. of ground at site = 606.50 m



Design of Chemical House and Calculation of Chemical Dose

The space for storing the chemicals required for the subsequent treatment of water consist of determining space required for storing the most commonly used coagulant alum, lime, chlorine, etc. for the minimum period of three months and generally for six months.

The size of units also depend upon the location, transport facilities, weather condition, distance of production units and availability of chemicals, chemical house should be designed to be free from moisture, sap, etc. these should be sufficient space for handling and moisture chemicals and other related operations. It should be located near to the treatment pant and chemicals should be stored in such size of bags that can be handled easily.

Alum Dose:

Coagulation

The terms coagulation and flocculation are used their indiscriminately to describe the process of removal of turbidity caused by fine suspension colloids and organic colours.



Coagulation describes the effect product by the addition of a chemical to a colloidal dispersion, resulting in particle destabilization. Operationally, this is achieved by the addition of appropriate chemical and rapid intense mixing for obtaining uniform dispersion of the chemical.

The coagulant dose in the field should be judiciously controlled in the light of the jar test valves. Alum is used as coagulant.

Design Criteria for Alum Dose

Alum is required in particular season is given below:

Monsoon = 50 mg/L

Winter =20 mg/L

Summer = 5 mg/L

Alum Required

Let the average dose of alum required be 50 mg/L, 20 mg/L, and 5 mg/L in monsoon, winter, and summer, respectively.

Per day alum required for worst season for intermediate stage



=50 ×10-6 ×592.2 ×103 ×24

= 710.64 kg/day

For six months (180 days) = 710.64 × 180

= 127915.2kg

Number of bags whence 1 bag is containing 50 kg = 2559

If 15 days in each heap = 170.6 heaps

If areas of one each be 0.2 cub m, then total area requied

= 170.6 ×0.2 = 34.12 m2

Lime Soda Process:

Softening

Water is said to be hard, when it does not from leather readily with soap. The hardness of water is due to the presence of calcium ad magnesium ions in most of the cases. The method generally used is lime soda process. Softening with these chemicals is used particularly for water with high initial hardness (> 500 mg/L)

And suitable for water containing turbidity, color and iron salts. Lime soda softening cannot however, reduce the hardness to values less than 40 mg/L.



Design Criteria for Lime-Soda Process:

It should be possible to remove 30mg/L carbonate hardness and 200 mg/L total hardness by this process.

Lime and Soda Required

Lime required for alkalinity

Molecular weight of

CaCo3 = 40 + 12 + 48

= 100

CaO = 40 + 16

= 56

100 mg/L of CaCO3 alkalinity requires = 56 mg/L of CaO

110 mg/L of CaCO3 requires = (56/100) ×110

= 61.6 mg/L of CaO



Lime require for Magnesium

24 mg/L of magnesium requires = 56 mg/L of CaO

1 mg/L of magnesium requires = 56/24 mg/L of CaO

3.5 mg/L of magnesium requires = (56/24) × 3.5

= 8.2 mg/L of CaO

Hence, the total pure lime require = 61.6 + 8.2

= 69.8 mg/L

Also 56 mg of pure lime (CaO) is equivalent to 74 kg of hydrated lime.

Hence, the total pure lime required = (69.8×74)/56

= 92.23 mg/l

= 92.23 ×10-6 Kg/l

Soda (NaCO3)

Soda is required for non-carbonate hardness, as follows.

100 mg/L of NCH requires = 106 mg/L of Na2CO3

161.6 mg/L of NCH requires = (106/100) ×161.6



= 65.59 mg/l of Na2CO3

Total quantity of lime = 92.23 ×10-6 ×592.2×103 ×180 ×24

= 235952.38 kg-day

One bag contain 50 kg

Number of bags required = 4719 Bags

If 15 bags in each heap, number of heaps = 314.6

If area of one heap is 0.2 cub m = 314.6× 0.2

= 62.92 m2

Total quantity of soda required = 65.59 × 10-6 ×592.2×103×24×180

= 167799.15 kg-day

Number of bags = 3356 Bags

If 15 bags in each heaps = 223.73heaps

Total area of heap = 0.2 × 223.73

= 44.75 m2.

Total area for all chemicals = 34.12+62.92+44.75

= 141.79 m2



Add 30% for chlorine storage, chlorine cylinders etc.

Total area = 184.33 m2

Provide room dimension = (12×16) m

= 192m2

Provide dimension = 12 m x 16 m

Chemical Dissolving Tanks:

Total quantity of alum, lime and soda

= 127915.2+235952.38+167799.16

= 531666.74 /180

= 2953.70 Kg

= 60bags

= 4 heaps

Area required = 0.8 m2

Dimensions = 0.9 m x 0.9m

Chemical Solution tanks:

Total quantity of alum, lime and soda required per day

= 2953.70 kg



Assume mix the chemical 1 kg in 20 lit. Water

Hence solution required per day = 59074 lit

= 59074Lit

Quantity of solution for 8 hours = 2461.41 x 8

= 19692 Lit

= 19.692 m3

Assuming depth of tank (1.4 m) and 0.3 m free board

Dimension of solution tank = 3.5 x 3.32 x 1.7

Summary

1. Per day alum required 710.64 kg/day2. Hydrated lime required 1310.85 kg/day

3. Soda required 932.22 kg/day4. Size of chemical

dissolving tanks0.9 x0.9 m

5. Size of chemical solution tank

3.5 x 3.32 x 1.7

Design of Mechanical Rapid Mix Unit

a)Flash Mixer



Rapid mixing is and operation by which the coagulant is rapidly

and uniformly dispersed throughout the volume of water to

create a more or less homogeneous single or multiphase system.

This helps in the formation of micro flocs and results in proper

utilization of chemical coagulant preventing localization of

connection and premature formation of hydroxides which lead to

less effective utilization of the coagulant. The chemical

coagulant is normally introduced at some point of high

turbulence in the water. The source of power for rapid mixing to

create the desired intense turbulence is gravitational and

pneumatic.

The intensity of mixing is dependent upon the temporal mean

velocity gradient ‘G’. This is defined as the rate of change of

velocity per unit distance normal to a section. The turbulence

and resultant intensity of mixing is based on the rte of power

input to the water.

Flash mixture is one of the most popular methods in which the

chemical are dispersed. They are mixed by the impeller rotating

at high speeds.



b)Design Criteria for Mechanical Rapid Mix Unit

Detention time = 30 to 60 sec.

Velocity of flow = 4 to 9 m/sec.

Depth = 1 to3m

Power require = 0.41 KW/1000 cum/day

Impeller speed = 100 to 250 rpm

Loss of head = 0.4 to 1.0

Mixing device be capable of creating a velocity gradient

= 300 m/sec/m depth

Ratio of impeller diameter to tank diameter = 0.2 to 0.4:1

Ratio of tank height to diameter = 1 to 3:1

C) Design Calculation

Design flow = 14212.8 m3/day

Detention time = 30 Sec.

Ratio of tank height to diameter = 1.5:1

Ratio of impeller diameter to tank diameter = 0.3:1



Rotational speed of impeller = 150 rpm

Assume temperature = 200

1. Dimension of tank:

Volume = 6.62 m3

D = 1.8 m

Height = 2.37 / (0.23 m free board)

Total height to tank = 2.6m

2. Power Requirement:

Power spend = 5.82 KW

3. Dimensions of flat bade and impeller:

Diameter of impeller = 0.65 m

Velocity of tip impeller = 4.08 m/sec

Area of blade = A8

Power spent = ½ x CD x r0 x AB x VR3s

Let CD = 1.8 (Flat blade); VR =3/4x VT

5.82 x 103 = ½ x 1.8 x 1000 x AB x 3/4 x 4.08

AB = 2.11 m2

Provide 8 blades of 0.55 x 0.55 m = 2.4m2



Provide 4 numbers of lengths 1.5 m and projecting 0.2 m from the wall.

4. Provide inlet and outlet pipes of 250 mm diameter.

D) Summary

1. Detention time 30 sec

2. Sped impeller 150 rpm

3. Height of Tank (0.23 m free board)

2.6 m

4. Power required 5.825 KW

5. Number of Blade (0.55 m x 0.55 m)

8

6. Number of baffles (length 1.5 m)

4

7 Diameter of inlet and outlet 250

Design of Clariflocculator

a)Clariflocculator

The coagulation and sedimentation processes are effectively

incorporated in a single unit in the clariflocculator. Sometimes

clarifier and clariflocculator are designed as separate units.



All these units consist of 2 or 4 flocculating paddles placed

equidistantly. These paddles rotate on their vertical axis. The

flocculating paddles maybe of rotor-stator type. Rotating in

opposite direction above the vertical axis. The clarification unit

outside the flocculation compartment fitted with paddles rotating

at low speeds thus forming flocks.

The flocculated water passes out from the bottom of the

flocculation tank to the clarifying zone through a wide opening.

The area of the opening being large enough to maintain a very

low velocity. Under quiescent conditions, in the annular setting

zone the floc embedding the suspended particles settle to the

bottom and the clear effluent overflows into the peripheral

launder.

b) Design Criteria: (Flocculator)

Depth of Tank = 3to4.5m

Detention time = 30 to 60 mm

Velocity of flow = 0.2 to 0.8 m/sec

Total area of paddles = 10 to25% of c/s of tank

Range of peripheral velocities of blades= 0.2 to 0.6 m/s



Velocity gradient (G) = 10 to75

Dimension less factor Get = 104 to 105

Power consumption = 10 to 36 KW/mid

Outlet velocity = 0.15 to 0.25 m/sec

c) Design Criteria: (Clarifier)

Surface overflow rate = 40 m3 /m2/day

Depth of water = 3 to 4.5 m

Weir loading = 300m3 /m2/day

Storage of sludge = 25%

Floor slope = 1 in 12 or 8% for

Mechanically cleaned tank

Slope for sludge hopper = 1.2:1 (v:h)

Scraper velocity =1 revolution in 45 to80 minutes.

Velocity of water at outlet chamber = not more than 40 m/sec

d) Assumptions

Average outflow from clariflocculator = 592.2 m3/hr



Water lost desludging = 2 %

Design average period = 604.04 m3/hr

Detention period = 30 min

Average value of velocity gradient = 30 s-1

e)Design of Influent Pipe

Assuming V = 1 m/sec

Dia = 0.447 m

Provide influent pipes of 450 diameter.

f) Design of flocculator :

Volume of fiocculator = 300 m3

Providing as water depth = 3.5 m

Plan area of flocculator = 300/3.5

= 85.71m2

D=diameter flocculator = 10.456 m ≅ 10.5

Dp = diameter of inlet pipes = 0.45 m



g)Dimension of Paddles:

= G2×μν×vol

= 302×0 .89×10−30×( π /4×10−2×3 . 5)

= 229.08

Power Input = ½(cd×p×Ap×(ν−υ )3

Cd = 1.8

P = 995 kg/m (25°c)

V = Velocity of tip of blade = 0.4 m/sec.

V = Velocity of water tip of blade = 0.25 x 0.4

= 0.1 m/sec.

229.08 = ½ x 1.8 x 995 x Ap x (0.4-

0.1)3

∴ Ap = 9.47m2

Ratio of paddles to c/s of flocculator

9.47π (10.5−0.75 ) 3.5

x100 = 8.83 % < 10 to 25 %

Provide Ap = 11.00m2



Ap = 11.00

π (10.5−0.75 ) 3.5x100 = 10.26 % ……. ok

Which is acceptable (within 10 to 25 %?)

Provide 5 no of paddles of 3 m height and 0.7 m width

One shaft will support 5 paddles

The paddles will rotate at an rpm of 4

V = 2×π×r×π /60

0.4 = 2×π×r×4 /60

r = 0.96m ¿1m

r = distance of paddle from C1. Of vertical shaft

Let velocity of water below the partition wall between the

flocculatorand clarifier be 0.33/sec.

Area = 592.2 /(0.3 x 60 x 60) = 0.55 m2

Depth below partition wall = 0 .55/( π×10 . 5)

= 0.016 m

Provided 25 % of storage of sludge = 0.25 x 35

= 0.875m



Provide 8% slope for bottom

Total depth of tank at partition wall = 0.3+3.50.016+ 0.875

= 4.69m¿4.7m

h)Design of Clarifier

Assuming a surface overflow rate of 40 m3/m2/day

Surface of clariflocculator = (592.2 x 24)/40

= 355.32 m2

Dcf = Dia. Of Clariflocculator

π /4 [ Dcf2 - (10.5)2] = 355.32

DCf = 23.72¿ 24m

Length of weir = π×D cf = 75.39 m

Weir loading = (592.2 x 24)/75.39

= 188.52 m3/day/m

According to manual of Govt. of India. If it is a well clarifier, It

can exceed upto 1500 m3/day/m

i) Summary (Clariflocculator)

1. Detention Period 30 min



2. Diameter of influent pipe 450 mm

3. Overall depth of flocculator 3.5 m

4. Diameter of tank no. 10.5 m

5. No. of paddles (3m height and 0.7m width)

5

6. Distance of shaft from C.L. of flocculator

1m

7. Paddles rotation (RPM) 4

8. Distance of paddle from C.L. vertical shaft

1 m

9. Slope of bottom (%) 8

10.

Total depth of partition wall 4.7 m

11.

Diameter of clariflocculator 24 m

Design of Rapid Gravity Filter

a. Rapid Sand Filter

The rapid sand filter comprises of a bed of sand serving as a

single medium granular matrix supported on gravel overlying an

under drainage system, the distinctive features of rapid sand

filtration as compared to slow sand filtration include careful pre-

treatment of raw water to effective flocculate the colloidal



particles, use of higher filtration rates and coarser but more

uniform filer media to utilize greater depths of filter media to

trap influent solids without excessive head loss and back

washing of filer bed by reversing the flow direction to clear the

entire depth of river.

The removal of particles within a deep granular medium filter such as rapid sand filter occurs primarily within the filter bed and is referred to as depth filtration. Conceptually the removal of particles takes place in two distinct slips as attachment step. In the first step the impurity particles must be brought to the surface of him medium of the previously deposited solids on the medium. Once the particles come closer to the surface as attachment step is required to retain in on the surface instead of letting it flow down the filter.

The transport step may be accomplished by straining gravity, setting, impaction hydrodynamic and diffusion and it may be aided by flocculation in the interstices of the filter.

b.Design Criteria: (Rapid Sand Filter) Rate of filtration = 5 to 7.5 m

cub Max surface area of one bed = 100 m

squ



Min. over all depth of filter including a free board of 0.5 m=2.6m

Effective size of sand = 0.45 to 0.7

Uniformity co-efficient for sand = 1.3 to 1.7 Lenition loss should not exceed 0.7 % by weight Silica content should not be less than 90% Specific gravity = 0.55 to

2.65 Wearing loss is not greater than 3% Minimum number of units = 2 Depth of sand = 0.6 to

0.75 Standing depth of water over the filter = 1to 2 m Free board is not less than 0.5 m

c. Problem statement Net filtered water = 592.2 m3/hr Quantity of backwash eater used = 2% Time lost during backwash = 30 min Design rate of filtration = 5 m3 /m2/hr Length-width ratio = 1.2 to 1.33 Under drainage system = central manifold with

laterals Size of perforations =13 mm

d.Design Calculation

Solution required flow of water = 592.2 m3/hr



Design flow of filter = 592.2 x (1+0.02) x 2/23.5

= 616.896 m3/hr

Plan area for filter = 616.896 / 5

= 123.379 m3 ≅ 124 m3

Assume depth 2m

Using 1 units,

Plan area = 62 m2

Length x width = L x 1.25L

L = 7.1 m

Provide 1 filter units, each with dimension 8.8 m x 7.1 m

Estimate of Sand Depth:

It is checked against breakthrough of floc.

Using Hudson Formula:

Q x d x h/l=B x 293223/1

Where, Q, d, h and 1 are in mm, m, m/hr respectively.



Assume, B = 4 x10-4 (poor response) < average degree of pretreatment

h = 2.5 m (terminal head loss)

Q = 5 x 2 cub m/ hr (assume 100% overload of filter)

d = 0.6 mm (mean dia)

10 x (0.6)3 x 2.5/1 = 4 x 10-4 x 293223

L > 46 m

Provide depth of sand bed = 60 cm

Estimation of Gravel and Size Gradation:

Assume size gradation of 2 mm at 40 mm at bottom using empirical formula:

P = 2.54 R (load d)

Where, R = 12 (10 to 14)

The unit of L and cm and mm, respectively.

Size 2 5 10 20 40

Depth(cm) 9.2 21.3 30.5 40 49

Increment 9.2 12.1 9.2 9.5 9 65N.B.S. Institute of polytechnic, Ausa.


Provide 50 cm depth gravel.

Design of Under Drainage System:

Plan area of each filter = 8.8 x 7.1

=62.48 m2

Total area of perforation = 13x10-3x 62.48

=0.81224 m2

= 8122.4 cm2

Total cross section area of laterals perforation =3 x area of perforation

= 3 x 8122.4

= 24367.2 cm2

Diameter of central manifold = √ 24367.2 x 4π

=176.14 cm

Providing a diameter of 150 cm



Assuming spacing for laterals = 20 cm

Number of laterals = 8.8 x 150/20

= 66 on 8.8m side

& =7.1 x (150/20)

= 54 On 7.1m Side

Total Number of laterals = 54+66 =120 Nos.

D = √61.2x 4 / π

=8.83cm =90cm

Number of perforations/ laterals = 1/2 width of filter-1/2 dia of manifold

= ½ 7.1 – ½ x 1

= 3.05m

Let n be the total no. of perforation of 13 mm dia

There for,

Total area perforation

8122.4 = n xπ4

x1.32

N = 6119.38 say 6120 nos



No. of perforation/ laterals = 6120/120

= 51

Spacing of perforation = 3.05 x 100/51

=59.80 cm c/c say 60 cm c/c

Provide 51 perforations of 13 mm diameter at 60 mm c/c

Computation of wash water troughs:

Wash water rate = 36 m3/m2/hr

Wash water discharge for one filter = 36 x 62.48

= 2249.2 m3/ hr

= 0.6248 m3/sec

Assuming a spacing of 1.8 m for wash water through which will run parallel to the longer dimension of the filter unit.

No. of through = 7.1/ 1.8

= 3.94 say 4

Discharge per unit through = 0.6248/4

= 0.1562 cub m/sec



For a width of 0.4 m the water depth at upper end is given by

Q = 1.376 x bh3/2

0.1562 = 1.376 x 0.4(h) 3/2

H = 0.43 Say 0.45m

Freeboard = 0.1 m

Provide 4 trough of 0.4 m wide x 0.5 deep n each filte.

Total Depth of Filter Box:

Depth of tilter box = depth of under drain + gravel + sand + water depth + free board

= 900 + 500 + 600 + 2200 + 300

= 4300

Designs of filter air wash:

Assume rate at which air is supplied = 1.5 m3/m2/min

Duration of air wash = 3 min

Total quantity of air required per unit bed = 1.5 x 3 x 8.8 x 7.1



= 281.16 m3

d. Summary

1. Number of units 1

2. Size of units 8.8 x 7.1 m

3. Depth of sand bed 60 cm

4. Depth of gravel 50cm

5. Diameter of perforation 13 mm

6. Diameter of central manifold 150 cm

7. Spacing for laterals 20 cm

8. Number of laterals 120

9. Diameter of laterals 90mm

10. Number of perforation 51

11. Number of trough 4

12. Size of trough 0.4 x 0.5 m

13. Total depth of filter box 4300 mm



14. Duration of air wash 3 min

15. Total quantity of air required per unit bed 281.16 m3

Disinfection Unit

a. Chlorination

Treatment method such as aeration, plain sedimentation, coagulation, filtration, would render the water chemically and aesthetically acceptable with some reduction in the pathogenic bacterial content. However, the foregoing treatment methods do not ensure 100% removal of pathogenic bacteria, and hence it becomes necessary to ‘disinfect’ the water to kill the pathogenic bacteria.

Disinfection should not only remove the existing bacteria from water but also ensure their immediate killing even afterwards, in the distribution system. The chemical which is used as disinfectant must therefore be able to give the ‘residual sterilsing effect’ for a long period, thus affording some protection against recontamination. In addition to this, it should be harmless, unobjectionable to taste, economical and measurable by simple taste. ‘Chlorine’ satisfies the above said more than any other disinfectant and hence is widely used.



b. Design Criteria (Chlorination)

Chlorine dose =1.4 mg/L (rainy season) = 1 mg/L (winter season)

= 0.6 mg/L (summer season)

Residual chlorine = 0.1 to 0.2 mg/L (minimum)

c. Design calculation

Rate of chlorine required, to disinfect water be 2 p.p.m.

Chlorine required. Per day = 14.22 x 106 x 1.4 x10-6

= 19.908 kg

For 6 months = 19.908 x 180

= 3583.44 kg



Number of cylinder (one cylinder contain 16 kg)

= 3583.44 x 2/16

= 447.93

Number of cylinder used per day = 2 of 16 kg

e. Summery

1 Chlorine required per day 19.908 kg

2 Number of cylinder required per day 2 of 16 kg

Storage Tank

Distribution reservoir also called service reservoir are the storage reservoir which store the treated water for supplying the same during emergencies and also help in absorbing the hourly fluctuation in water demand. Depending open their elevation with respect to the ground they are classified as



underground reservoir and elevated reservoir both of these reservoir designed for this project.

Storage Capacity

Ideally the total storage capacity of a distribution capacity of distribution reservoir is the summation of (1) Balancing reserve (2) Breakdown reserve and (3) Fire reserve. The balancing storage capacity of a reservoir can be worked out from the data of hourly consumption of water for the town/city by either the mass curve method or analytical method. In absence of availability of the date of hourly demand of water the capacity of reservoir is usually ¼ to 1/3 of the daily average supply.

Underground Storage Reservoir (U.S.R)

a. General

The reservoir is used for storing the filtered water which is now fit for drinking. From this, the water is pumped to E.S.R. normally the capacity of reservoir depend open the capacity of the pumps and hour of pumping during a day. If



the pumps work for 24 minutes the capacity of this reservoir may be between 30 minutes to 1 hour.

b.Design Criteria (U.S.R.)

1. Detention time 1 to 4 hr

2. Freeboard 0.4 to 0.6 m

c. Design Calculation

Assuming that all pumping are working for 4 hours.

Capacity of underground reservoir

= 6 hr capacity of average demand

= 19 MLD x 106 x 6 x 10-3 /24

= 4750 m3



Assuming 6 compartments

Let Depth = 4m

Area = 1187.5 m2

Area of each compartment = 198 m2

Dimension = 14.1 m x 14.1 m

Freeboard = 0.5 m

Provide 6 compartment of 14.1m x 14.1m x 4.5m

d. Summery


1. Capacity of reservoir 4750 m3

2. Total depth 4.5 m

3. Compartments 6

4. Size 14.1m x14.1m x 4.5m

5. Detention time 4 hr


Elevated Service Reservoir (ESR)

a. General

Where the areas to be supplied with treated water are at higher elevation than the treatment plan site, the pressure requirements of the distribution system necessitates the construction of ESR. The treated water from the underground reservoir is pumped to the ESR and than supplied to the consumers.

b.Design Calculation

Assuming capacity of ESR = 1/10 underground

Storage

= 475 m3



Free board = 0.3 m

Overall depth = 4m

Diameter = √ 475 x 4π x 4

There for d =12.296 m

Provide 1 ESR of overall height = 4.3 m and diameter = 12.296m

C. Summery

1. Number of tanks 1

2. Depth of tank 4.3 m

3. Diameter of tank 12.296m


Complete Water Treatment Plant Project

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