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

Planning & designing of Water treatment plant

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.

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

Planning & designing of Water treatment plant

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)

5.

Longitude (Lon)

- 18.2500°N -

76.5000°E

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

Planning & designing of Water treatment plant

Sr.No 1. 2. 3. 4.

5.

6. 7.

8. 9. 10. 11. 12. 13.

Design Consideration Design Period (years) Average rate of water supply(Ipcd) Industrial demand (MLD) Quality of raw water 1) Ph 2) Turbidity (mg/L) 3) Total Hardness (mg/L)[as CaCo3] 4) Chlorides(mg/L) 5) Iron (mg/L) 6)Manganese (mg/L) 7)Carbonates (mg/L) 8)M.P.N (No.100ml) Population of past four decades(in Person) Year 1981 Year 1991 Year 2001 Year 2011 F.S.L. of Reservoir (R.L.in mts.) Ground level at ; (R.L.in mts.) a) Jack well site b)Location of aeration unit Invert level of raw material gravity intake pipe (R.L.in mts) Invert level of raw water rising main(mts) Dead Storage of Reservoir(Million cu.m) Gross Storage of Reservoir(Million cu.m) Live Storage of Reservoir(Million cu.m) Silt Level of Reservoir in (m)

Values 20 135 0.6 7.5 50 550 200 2.5 3.5 110 3.5 16721 23246 30876 38733 612.75 607.75 608.00 600 7.382 27.727 20.345 607.50 4

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Planning & designing of Water treatment plant

14.

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 2. Average daily draft (M.L.D) Maximum daily draft (M.L.D) 3. Design period (Years)

: 64296 Nos :9.48 :14.22 :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 5

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Planning & designing of Water treatment plant

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 6

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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 7

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Planning & designing of Water treatment plant

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 8

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Planning & designing of Water treatment plant

4.Height (m) 5.Depth of water (m)

: 2.8 : 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%

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

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

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.

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

Planning & designing of Water treatment plant

Calculation of Population with Different Methods 1. Arithmetical Increase Method Sr. No Year 1

Population

2 1971 1981 1991 2001 2011

1. 2.

3. 4. 5.

3 12761 16721 23246 30876 38733 Total Average

Increase In pollution 4 3960 6525 7630 7857 25972 6493

Using the relation Pn

= P + nd

Where, Pn

= Future population after n decades

d

P

= Present population

n

= No. of decades = Average increase per decade

Given, n= 3, 13 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

p= 38733, d = 6493 . . Pn = 38733 + (3*6493) P2031= 58212(Persons)

2.Geometrical Increase Method

Sr. No 1 1. 2.

3. 4. 5.

Year

Population Increase in population 2 3 4 1971 12761 1981 16721 3960 1991 23246 6525 2001 30876 7630 2011 38733 7857 Total 25972 Average per Decade 6493

Percentage increase in population 5 18.05 39.02 32.82 25.45 115.34 28.84

Where, r

= Average percentage increase=28.84

n

= No. of decades=2

Pn= Population after n decade: 2031 14 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

Population in the year 2031 Using the relation P2031= P [ 1+ P2031=38733[1+

r n ¿ 100 28.84 100 ¿ ¿2

= 64296( In persons)

3.Incremental Increase Method

Sr. No

Year

1. 2.

1971 1981 1991 2001 2011

3. 4. 5.

Population 12761 16721 23246 30876 38733

Total Average

Increase in population 3960 6525 7630 7857 25972 6493

Incremental increase 2565 1105 227 3897 1299

Using the relation, 15 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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) × 1299 2

= 55616( Persons).

4. Decrease Rate of Growth Method

Sr. No

1. 2.

3. 4. 5.

Year Population Increase in Percentage population in population (%) 1971 12761 1981 16721 3960 18.05 1991 23246 6525 39.02 2001 30876 7630 32.82 2011 38733 7857 25.45 Total 25972 115.34

Decrease in the percentage increase -20.97 +6.2 +7.37 -7.4 16

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Planning & designing of Water treatment plant

Average

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

Planning & designing of Water treatment plant

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

Planning & designing of Water treatment plant

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

Planning & designing of Water treatment plant

(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 20 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

(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. 21 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of 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 22 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

(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 . 23 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

• 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 24 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

1. 2. 3. 4. 5. 6.

Detention time Diameter Depth Velocity of flow Number of units Free board

10 to 15 min 5 to 10 m (max 15 m) 4 to 10 m 1.0 to 1.5 m/sec 1 to 3 (max 4 ) 5m

(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 25 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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. 2. 3. 4.

Number of intake wells Diameter of intake wells Height of well R.L. of bottom well

1 unit 6.9 m 8.0 m 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 26 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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 2 3

Number of penstock At each level Diameter of penstock

2 units 1m 0.35 m

Design of bell mouth strainer (a) Design criteria 27 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

Velocity of flow Hole diameter

= 0.2 to 0.3m/s = 6 to 12 mm

Area of strainer

= 2 x diameter of holes

(b) Assumption Velocity of flow Hole diameter

= 0.25 m/s = 10mm

(c) Calculation π 4

d2 =

π 4

(102) =78.54 mm2

Area of each hole

=

Area of collection

= area of penstock 0.1645 0.25× 2

= 7.854 ×

10−5

N

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

Planning & designing of Water treatment plant

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

No of gravity main

= 0.6 to 0.9m/s

= 1 29

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Planning & designing of Water treatment plant

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 2

V = S =

1

R3 X S 2 X

1 N

N2xV2 R

4 3

S = 1:900 Head loss

=

100 900

= 0. 111 R.L. of gravity main

= 600- 3 30

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Planning & designing of Water treatment plant

= 597 m R.L. of gravity main at jack well = 597 -0.111 = 596.89m

(d) Summary 1 2 3 4

Number of gravity intake Diameter of gravity intake Invert level at intake well Invert level at jack well

1 unit 0.55m 597 m 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. 31 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

(b) Design criteria Detention time

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

0.5x15

=

7.30 min.

Suction head =

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

Planning & designing of Water treatment plant

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 47

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Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

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 49

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Planning & designing of Water treatment plant

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 50

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Planning & designing of Water treatment plant

= 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. 2. 3. 4.

Per day alum required Hydrated lime required Soda required Size of chemical dissolving tanks 5. Size of chemical solution tank

710.64 kg/day 1310.85 kg/day 932.22 kg/day 0.9 x0.9 m 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 51 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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 52

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

Planning & designing of Water treatment plant

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 14212.8 m3/day

Design flow

=

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: =

6.62 m3

D

=

1.8 m

Height

=

2.37 / (0.23 m free board)

Volume

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

Planning & designing of Water treatment plant

Total height to tank 2.

2.6m

=

5.82 KW

Power Requirement: Power spend

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 54

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

Planning & designing of Water treatment plant

2. Sped impeller

150 rpm

3. Height of Tank (0.23 m free 2.6 m board) 4. Power required 5.825 KW 5. Number of Blade (0.55 m x 8 0.55 m) 6. Number of baffles (length 4 1.5 m) 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.

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

Planning & designing of Water treatment plant

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

Velocity of flow

=

0.2 to 0.8 m/sec

Total area of paddles

=

10 to25% of c/s of tank

60 mm

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) 56 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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

Planning & designing of Water treatment plant

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

Dp = diameter of inlet pipes

=

0.45 m

≅

10.5

g)Dimension of Paddles: = = =

G 2    vol 30 2  0.89 10 30  ( / 4 10 2  3.5)

229.08 58

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

Planning & designing of Water treatment plant

(cd  p  Ap  (   ) 3

Power Input

=

½

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

229.08

=

0.1 m/sec.

=

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 9.47 x 100 π ( 10.5−0.75 ) 3.5

=

Provide Ap Ap =

=

11.00 x 100 π ( 10.5−0.75 ) 3.5

8.83 % < 10 to 25 %

11.00m2 =

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

Planning & designing of Water treatment plant

The paddles will rotate at an rpm of 4

2    r   / 60

V

=

0.4

=

r

=

0.96m 1m

r

=

distance of paddle from C1. Of vertical shaft

2    r  4 / 60 

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)

Depth below partition wall

= =

0.55 m2

=

0.55 /(  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 60 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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



/4 [ Dcf2 - (10.5)2]

=

DCf

355.32

= 23.72



  Dcf

24m

Length of weir

=

= 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 61

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

Planning & designing of Water treatment plant

5. 6. 7. 8. 9.

No. of paddles (3m height and 5 0.7m width) Distance of shaft from C.L. of 1m flocculator Paddles rotation (RPM) 4 Distance of paddle from C.L. 1 m vertical shaft Slope of bottom (%) 8

10 Total depth of partition wall . 11. Diameter of clariflocculator

4.7 m 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 pretreatment 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 62 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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

Planning & designing of Water treatment plant

 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 Design flow of filter

= 592.2 m3/hr = 592.2 x (1+0.02) x 2/23.5 64

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

Planning & designing of Water treatment plant

= 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 65 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

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 4 π

=176.14 cm Providing a diameter of 150 cm Assuming spacing for laterals

= 20 cm

Number of laterals

= 8.8 x 150/20 67

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

Planning & designing of Water treatment plant

= 66 on 8.8m side & =7.1 x (150/20) = 54 On 7.1m Side Total Number of laterals = 54+66 =120 Nos. D

= √ 61.2 x 4 /π =8.83cm =90cm

Number of perforations/ laterals of manifold

= 1/2 width of filter-1/2 dia = 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 =

nx

π x 1.32 4

N = 6119.38 say 6120 nos No. of perforation/ laterals

= 6120/120 = 51 68

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

Planning & designing of Water treatment plant

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 69

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

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

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 71

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

Planning & designing of Water treatment plant

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) 72 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

 Chlorine dose

=1.4 mg/L (rainy season) = 1 mg/L (winter season) = 0.6 mg/L (summer season)

 Residual chlorine (minimum)

= 0.1 to 0.2 mg/L

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

Planning & designing of Water treatment plant

= 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 74 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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.) 75 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

Area

= 1187.5 m2

= 198 m2

Area of each compartment

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) 77 N.B.S. Institute of polytechnic, Ausa.

Planning & designing of Water treatment plant

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

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

Planning & designing of Water treatment plant

Diameter

=

There for

d

Provide 1 ESR of overall height 12.296m

√

475 x 4 π x4

=12.296 m

= 4.3 m and diameter =

C. Summery 1.

Number of tanks

1

2.

Depth of tank

4.3 m

3.

Diameter of tank

12.296m

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