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A Major Project Presentation on Design of Water Treatment Plant By- Batch No-A2Roll No. 17-32Distribution of Earth`s Water

Why Treatment is Required?Following Are The Reason's For Which Water Treatment is Required-To Remove The Turbidity From The Water

To Remove The Bacteria's From The Water

To Remove Acidity/Alkalinity

To Remove the T.D.S.To Remove Hardness From Water

Water Treatment Process1. Population2. Purpose3. Per Capita Demand According To Living StandardFactors Considered in The Designing of Water treatment PlantPopulation ForecastingPopulation Forecasting Can Be Done By Following Three Method's-

Arithmetic MethodGeometric MethodIncremental Increase Method1. Arithmetic methodPn = P0+ nc Where, P0 = Initial population Pn = Population in dh decade n = No. of decades c = Average increase

YearPopulation Difference200123.23 lac-201136.02 lac12.79C=12.79P2021= 36.02 +(1 12.79) = 48.81 lac P2025 = 33.02 + (1.5 12.79) = 55.20 lac 2. Geometric Method Pn =Po (1+r/100)n Pn = Population in the dh decade Po = Population any decade r = Percentage increase N = No. of decade r = {(36.0223.23)/23.23} 100 =55.05% P2021= 36.02 {1+(55.05/100)} 1 = 55.83 lac P2025= 69.50 lac

3. Incremental Increase Method

Pn = Po + (r + i) nr = Average rate of increase in population per decade I = Average rate of incremental increase per decade Po = Population in any decade Pn = Population in n decadeP2021= p0+avg. increase per decade+ avg. incremental increase = 36.02+12.79 = 48.81 lac P2025= 48.81+ 12.79 = 61.6 lacBut On an avg. we take population 60 lac in accordance to JDA 2025 Master Plan For Urban Jaipur Design of Screening

Design ProcedureAssumption a. Manually cleaned screen b. Inclination of bars = 30c. Size of bars = 50 X 6 X 6 mm3d. Clear spacing between two bars = 10mm e. Flow velocity of normal screen = 800mm/sec = 0.8m/s (at peak flow) Design

Population = 7, 50,000 Avg. rate of water Supply = 135 l/p/d Avg. rate of water supply per day = 7, 50,000 X 135 = 101.25 MLD Avg. rate of water supply in cumecs ={7, 50,000 X 135/ 1000 X 24 X 60 X 60 }= 1.17 Cumec (D.W.F) Max. Flow = 3 X DWF = 3.51 Cumecs Or 3 X 101.25 = 303.75 MLD [303.75 X 106/1000 X 24 X 60 X 60] = 3.51 m3/sec. Now, Net submerged area of the screen =[3.51/0.8 ] = 4.4 m2

Submerged area of the screen = 4.4 X Sin 30 = 2.2 m2 Flow velocity in screen = {3.51/2.2} = 1.6 m/s Provide 40 bars = Cross width of screen chamber = 40 X 0.006 (Rod area) + 41 X 0.003 (Gap area) = .363 m Liquid depth = Submerged Screen Area/Cross Width of Screen Chamber = 2.2 /0.363 =6.06 m Providing free board = 0.25m Total depth = 6.06+0.025 = 6.085 m Now Slope Hydraulic mean depth (m) = Area /Wetted perimeter = {6.085 X .363 /(2 X 6.085) + 0.363} = 0.176 m Using mannings formula

1.6 = 1 0.6 X 0.1762/3 X i(1/3) i = 28.56

Settling Tank

Designing of Settling Tank

Volume of water to be treated = 303.75 MLD Detention time = 240 min = 4hr. Velocity of flow = 20cm/min Length of tank = Velocity X Detention time = 0.2 X 240 = 48m Volume of water in 4 hr. ={303.75 X 106 X 4/ 1000 X 24} = 50,625 m3 Cross sect. canal area (A) = V/ L = 50,625/48 = 1054.69 m2 Assume depth = 3 m Width of tank = 1054.69/3 = 351.56 m Provide 1 m extra depth for waste storage & 0.5 m for free board Total depth = 3+1.5 = 4.5m Coagulation There are two methods of feeding coagulants in water:

Dry FeedingWet Feeding Coagulants are: 1. Aluminum sulphate or alum 2. Sodium aluminates Alum Dose Design criteria for alum dose=Alum required in particular season is given that Mansoon = 50mg/l Winter = 20mg/l Summer = 05 mg/l Per day alum required for Summer season = 5 X 10-6 X 7,50,000X 135 X 103 kg/day = 506.25 X 103 kg/day For six months = 506.25 X 103 X 180 = 91.135 X 106 kg No of bags if 1 bag contains 50 kg. Of alum So no. of bags= {91.135 X 106 }/50 = 1.8225 X 106 bag If 15 bags in each heap = 121,500 heaps If area of one heap be 0.2m2 then total area required = 121,500 X 0.2 = 24300 m2 Lime-Soda Process Design criteria Design criteria for lime soda process Molecular weight of Caco3 = 40+12+48 = 100 Cao = 40+16 = 56 100 kg/l of Caco3 alkalinity required = 56mg/l of Cao So 110 mg/l of Caco3 alkalinity required = {56 X 110}/100

=61.6 mg/l of cao Lime required for magnesium 24 mg/l of magnesium required = 56mg/l of cao 1 mg/l of magnesium required = 56/ 24 mg/l of cao So 3.5 mg/l of magnesium required = 56 X 3.5 /24 = 8.2 mg/l of cao Hence the total pure lime required = 61.6+8.2 = 69.8 mg/l Also 56 kg of pure lime (cao) is equivalent to 74 kg of hydrated lime Hence hydrated lime = 69.8 X 74/56 = 92.23 mg/l If one bag contains 50 kg then no of bags = 4482.38 = 4483 bags 15 bags in each heap so no. of heaps = 298.87 ~ 299 heaps If area of one heap is 0.2 m2 then total area required = 2990.2 = 59.8 m2 Soda (Na2Co3) 100 mg/l of NCH required = 106 mg/l of No2Co3 161.6 mg/l of NCH required = {106 X 161.6}/100 = 65.59 mg/l of No2Co3 So Total quantity of lime required = 92.23 X 7,50,000 X 135 X 10-6 kg per day = 9338.28 kg/day So for 180 days = 1.68 X 106 kg If one bag contains 50 kg then no of bags = 33,617.808 ~ 33,618 bags 15 bags in each heap so no. of heaps = 2241.2 ~ 2242 heaps If area of one heap is 0.2 m2 then total area required = 448.24 = 449 m2 Total quantity of soda required for 180 days = 65.59 X 750000 X135 X 180 X10-6 = 6640.98 kg So no of bags required = 6640.98/ 50 = 132.82 bags No of heaps = 133/15 = 8.86 ~9 heaps So area required = 9 X 0.2 = 1.8 m2 Total area required for alum dose + lime dose + soda dose = 449+1.8+59.8 = 510.6 m2 And add 30 % for chlorine storage So = 1.3 X 510.6 = 663.78m2So provide 4 rooms of dimension = 15 X 12 m2 Design of Chemical Dissolving Tank Total quantity of alum, lime and soda = 6640.98+(1.68 X 106)+ (91.135 X 106) = 92.82 X 106 kg For per day requirement = {92.82 X 106}/ 180 = 515666.66 kg/day So daily no. of bags required = 10313.33 bags ~10314 bagsSo no. of heaps = 10314/15 = 687.6 ~ 688 heaps Area required = 688 X 0.2 = 137.6 m2

Clariflocculator Rapid Sand Filter

Disinfection chamber {Chlorination Unit}

Result

Clean drinking water

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