Complete Water Supply Treatment Plant

Calculation of Water Demand Expected Population after 30 Average Rate of Water Supply / Capita

61400 135

= =

Water required for above purposes for whole town = ( 61400 x 135 ) 8.289 MLD = Industrial Demand 0.6 MLD = Fire Requirement : It can be assumed that city is a residential town ( Low Rise Buildings ) Water for Fire

 100  = 100 x

P  10 3

61.4

10

-3

MLD = 0.78 MLD

Average Daily Draft =

( 8.289 + 0.6 ) =

8.889 MLD

Maximum Daily Draft =

( 1.5 x 8.889 ) =

13.334 MLD

Coincident Draft = Maximum Daily Draft + Fire Demand = ( 13.334 + 0.78 ) = 14.114 MLD ( Considering Draft < Maximum Hourly Draft ) Design Capacity For Various Components Intake Structure Daily Draft = 13.334 MLD Pipe Main = Maximum Daily Draft = 13.334 MLD Filters and Other Units at Treatment Plant : = = Lift Pump : = =

2 x Average Daily Demand ( 2 x 8.889 ) = 17.778 MLD 2 x Average Daily Demand ( 2 x 8.889 ) = 17.778 MLD

Physical & Chemical Standards Of Water S.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

Characteristics Acceptable Turbidity ( Units on J.T.U. Scale ) 2.5 Colour ( Units on Platinum Cobalt Scale ) 5 Taste & Odour Unobjectionable PH 7.0 to 8.5 Total Dissolved Solids ( mg / L ) 500 Total Hardness ( mg / L as Ca CO 3 ) 200 200 Chlorides ( mg / L as C 1 ) Sulphates ( mg / L as S O 4 ) 200 1 Fluorides ( mg / L as F ) Nitrates ( mg / L as N O 3 ) 45 Calcium ( mg / L as Capacity ) 75 30 Magnesium ( mg / L as Mg ) Iron ( mg / L as Fe ) 0.1 Manganese ( mg / L as M n ) 0.05 Copper ( mg / L as C u ) 0.05 Zinc ( mg / L as Z n ) 5 0.001 Phenolic Compounds ( mg / L as Phenol ) 0.2 Anionic Detergents ( mg / as MBAS ) Mineral Oil ( mg / L ) 0.01 TOXIC MATERIALS Arsenic ( mg / L as A s ) 0.05 Cadmium ( mg / L as Cd ) 0.01 Chromium ( mg / L as Hexavalent Cr ) 0.05 0.05 Cyanides ( mg / L as C ≡ N ) Lead ( mg / L as P b ) 0.1 Selenium ( mg / L as Se ) 0.01 Mercury ( mg / L as Hg ) 0.001 Polynuclear Aromatic Hydrocarbons ( mg/L ) 0.2 RADIO ACTIVITY Gross Alpha Activity in pico Curie ( ρ Ci / L ) 3 Gross Beta Activity ( ρ Ci / L ) 30

Cause for Rejection 10 25 Unobjectionable 6.5 to 9.2 1500 600 1000 400 1.5 45 200 150 1 0.5 1.5 15 0.002 1 0.3 0.05 0.01 0.05 0.05 0.1 0.01 0.001 0.2 3 30

Comparison of Given Data & Standard Data and Treatment Proposed S.No.

Particulars

Actual

Standard

Difference

1

pH

7.5

2

Turbidity

50

2.5

47.5

3 4 5 6 7 8

Total Hardness Chlorides Iron Manganese Carbonate MPN

550 200 2.5 3.5 110 3.5

200 200 0.1 0.05 0 0

350 0 2.4 3.45 110 3.5

7.0 to 8.5 0.25 Hence OK.

Treatment Proposed

Not Necessary Clarifier & Rapid Sand Filler Softening Aeration Aeration Softening Chlorination

Design of Intake Well Design Criteria : S.No. 1 2 3 4 5 6

Particulars Detention Time / Period Diameter of Well Depth of Well Velocity of Flow Number of Units Free Board

Values 5.0 to 10 minutes 5.0 to 10 m ( Maximum = 15 m ) 4.0 to 10 m 0.6 to 0.9 m / sec. ( Maximum = 4 ) 1 to 3 5m

Design Assumptions : Given F.S.L. Minimum R.L. Given invert of gravity main Detention Time

= = = =

27 m 28 m 24 m 10 minutes

Design Calculations : Flow of Water Required = 13.334 MLD / 3600 x 24 = 0.1543 m 3 / sec. Volume of Well = 0.1543 x 10 x 60 = 92.58 m ³ Cross-sectional Area of Intake Well = ( 92.58 / 4 ) = 23.15 m ² Diameter of Intake Well = 23.15 x 4 Π d= 5.43 m < 10 m ( O.K. ) Hence Diameter of Intake Well = 5.43 m ≡ 5.5 m Summary : S.No. 1 2 3 4

Particulars Number of Intake Wells Diameter of Intake Well Height of Wall R.L. of bottom of Well

Values 1 Unit 5.5 m 4.0 m 24 m

Design of Pen Stock & Bell Mouth Strainer a) Pen Stock These are the Pipes provided in Intake Well to allow water from water body to intake well. These pen stocks are provided at different 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 pen stock. At each level more than one pen stock is provided to take account of any obstruction during its operations. these pen stocks are regulated by valves provided at the top of intake wells.

b) Design Criteria Velocity through Pen Stock Diameter of each Pen Stock Number of Pen Stock for each Intake Well

= 0.6 to 1.0 m / sec. = Less than 1 m 2 nos. =

c) Design Calculation Number of Intake Well Number of Pen Stock for each Level Velocity C / S area of each Pen Stock Diameter

Area  4  

d

1 nos. = 2 nos. = = 0.75 m / sec. 0.1543 / ( 0.75 x 2 ) = = 0.103 m ² 0.103 x 4 Π

=

0.3622 m ≡ 0.4 m (Say)

d) Summary S.No. 1 2 3

Particulars Number of Pen Stock / Well At Each Level Diameter of Pen Stock

Values 2 Units 1m 0.40 m

Design of Bell Mouth Strainer : a) Design Criteria Velocity of Flow Hole Diameter Area of Strainer

= = =

b) Assumptions Velocity of Flow Hole Diameter

= 0.25 m / sec. = 10.0 mm

c) Calculation Area of Each Hole =

d2  4

( 10 x 10 ) x Π 4

0.2 to 0.3 m / sec. 6.0 to 12.0 mm 2

= 0.79 cm ²

Area of Collection = Area of Pen Stock 0.1543 = 0.7850 x N ( 0.25 x 2 ) 0.1543 1 N= ( 0.25 x 2 ) 0.7850 Area of Strainer = 2 2 x 3931.2 x 0.785 =

=

3931.2

= 6171.98 d Π xd2  = 4 4 ( 6171.98 x 4 ) = 88.67 cm Π for Bell Mouth Strainer. 2

Diameter of Bell Mouth Strainer = Diameter = d = Provide Diameter of

0.90 m

Regulating Valves

6171.98 cm ²

Manhole

F.S.L. = 27.0 m Bell Mouth Strainer

3.0 m L.W.L. = 26.0 m

Gravity Main 5.5 m Bottom R.L. = 24 m

Section

Plan

Design of Gravity Main a)

Gravity Main The Gravity Main connects the Intake Well to the Jack Well & 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 & the ground profile in between should generally help to decide if a free flow conduit is feasible. once this is decided the material of the conduit is to be selected keeping in view the local cost & the nature of the terrain to be traversed. Even when a fall is available, a pumping or force main, independently or in combination with a gravity main could also be considered. Gravity pipelines should be laid below the hydraulic gradient.

b) Design Criteria Diameter of Gravity Main Velocity of Water Number of Gravity Main = Number of Intake Well Assumption Velocity c) Design Calculation R.C.C. Circular Pipe is used. For this n= Conduit Velocity ( Assumed ) Area of Conduit required (A = Q / V ) Diameter of the conduit »

Diameter = d =

V 

2

  V  n V 2  n2 » S   4  23  3  R  R

Area Perimeter

Here R =

0.3 to 1.0 m 0.6 to 0.9 m / sec. 1 nos. 0.70 m / sec.

0.013 = 0.70 m / sec. = 0.1543 / ( 0.70 ) = 0.2204 m ²

Π xd2 = 0.2204 m ² 4 0.2204 x 4 = 0.53 m ≡ Π 2

Using Manning's Formula -

= = = =

0.55 m

1

1  R3 S 2 n

0.7 x 0.7 x 0.013 x 0.013 ( 0.55 / 4 ) 4/3

ΠxDxD 4

100 = 0.117 857 R.L. of Gravity Main = ( 27 - 3 ) = R.L. of Gravity Main at Jack Well =

1 ΠxD

=

=

0.001167

= 11.6684 x 10 S = 1 : 857 D 0.55 = 4 4

Head Loss =

d) Summary S.No. 1 2 3 4

Particulars Number of Gravity Intake Diameter of Gravity Intake Invert Level at Intake Well Invert Level at Jack Well

24.0 m ( 24 - 0.117 ) = Values 1 Units 0.55 m 24.0 m 23.88 m

23.883 m

-4

a)

Design of Jack Well Jack Well This structure serves as a collection of the sump well for the incoming water from the intake well from where the water is pumped through the rising main to the various treatment units. The unit is more useful when number of intake wells are more than one, so that water is collected in one unit and then effected. The Jack well is generally located away from the shore line, so that the installation of pumps, inspection maintenance is made easy.

b)

Design criteria Detention Time = 0.5 x ( Detention time of intake Well ) = 0.5 x 10 minutes = 5 minutes Suction Head = ( < ) Less than 10 m Diameter of Well = ( < ) Less than 20 m

( 3.0 to

15.0 min. )

c) Design Calculations Detention Time = 5 minutes Assuming Suction Head = 8.0 m Bottom Clearance = 1.0 m Top Clearance = 0.5 m Maximum depth of water that can be stored in condition when water is minimum in river . Minimum Depth of Water = ( 26 - 22.883 ) = 3.12 m Capacity of Well = = 92.58 m ³ 0.1543 x 10 60.0 C / S Area of Well = ( 92.58 / 3.12 ) = 29.70 m ² Diameter of the Well » Π xd2 = 29.70 m ² 4 29.7 x 4 Diameter = d = = 6.15 m Π R.L. of Bottom of Jack Well = ( 23.883 - 1 ) = 22.88 m R.L. of Bottom of Jack Well when full = ( 22.883 + 7 ) = 29.88 m d) Summary S.No. 1 2 3 4 5 6 7

Particulars Diameter of Jack Well R.L. of Bottom of Jack Well R.L. of Top of Jack Well Suction Depth Top Clearance Bottom Clearance Head required ( h d )

Values 6.15 m 22.88 m 29.88 m 2.12 m 0.50 m 1.00 m 4.88 m

= ( 8 - 2.12 - 1 )

→ →

Design Of Pumping System Pumps In the water treatment plant, pumps are used to 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 & straight as possible. It should not be greater than ( > ) 10 m, for centrifugal pump. If head is more than 10 m , water is converted into vapour & thus inspite of creating water head, head is created & pump ceases to fuction. 2 2.5

i) ii) iii) iv)

The following four types of pumps are generally used Buoyancy Operated pumps Impulse Operated pumps Positive Displacement Pumps Velocity Adoptions pumps

a) › › →

♥ ♥ ♥ ♥ ♥

The following criteria govern pump selection : Type of duty required. Present & projected demand & pattern and change in demand. The details of head & flow rate required. Selecting the operating speed of the pump & suitable drive. The efficiency of the pumps & consequent influence on power consumption and the running costs.

b) Diameter of Rising Main Discharge ( Q )

=

Economical Diameter ( d )

=

Hence Provide d =

= = = =

0.1543 m 3 / sec. ( 0.97 to 1.22 )  Q ( 0.97 to 1.22 ) x ( 0.1543 ) ( 0.97 to 1.22 ) x ( 0.3930 ) ( 0.381 to 0.48 ) 0.43 m Say ≡ 0.45 m

c) Design Criteria Suction head should not be greater than ( > ) 10 m. Velocity of flow length = 0.7 to 1.1 m / sec. Top Clearance = 0.50 m Bottom Clearance = 1.00 m d) Design Calculation Frictional Losses in Rising Main Velocity ( Assuming ) Head Loss

 hf

Here :

= 0.9 m / sec. 0.02 x 2x

f  L  v2   2 g  d

f L g

= 0.02 = 190 m = 9.81 m / sec.

190 x ( 9.81 x

0.9 x 0.9 ) 0.45

= 0.349 m

» h f = 0.349 m Say ≡ 0.35 m Head Loss Minor Losses should be assuming = 1m Hence Total Head of Pumping = ( h s + h d + h f + minor losses ) = ( 2.12 + 4.88 + 0.35 + 1 ) = 8.35 m 2 in Parallel is working Assuming

W .Q.H  75 W .H .P B.H .P  

W .H .P 



e) Summary S.No. 1 2

1000 x

0.1543 x 75

17.18 = 22.90 H.P 0.75 If ƞ = 75 %

Particulars Pumps Capacity Diameter of Pipe

Values 25 H.P 0.45 m

8.35 Say ≡

= 17.179 H.P 25 H.P

Design of Rising Main a) General These are the pressure pipes used to convey the water from jack well to the treatment units. The design of rising main is dependent on resistance to flow, available head, allowable velocities of flow, sediment transport, quality of water & relative cost. Various types of pipes used are cast iron, steel, R.C.C, P.C.C, asbestos cement, polyethylene, rigid PVC, iron fibre glass pipe, glass reinforced plastic etc. The determination of the suitability in all respects of the pipe of joints for any work is a matter of decision by the engineer concerned on the basis of requirements of the scheme. b)

Design Criteria Permissible Velocity in Mains = 0.9 to 1.5 m / sec. Mains Diameter should be less than ( < ) of 0.9 m Total Discharge in Mains = 0.1543 m 3 / sec.

c) Design Calculations Economical Diameter ( d )

=

( 0.97

Hence Provide d =

= = = =

( 0.97 ( 0.97 ( 0.381 0.43 m

d) Summary S.No. 1

Particulars Diameter of Mains Pipe

Values 0.45 m

to 1.22 )  Q to 1.22 ) x ( 0.1543 ) to 1.22 ) x ( 0.3930 ) to 0.48 ) Say ≡ 0.45 m

Treatment Units - Design Of Aeration Unit Aeration unit Aeration is necessary to promote the exchange of gases between the water & the atmosphere. In water treatment, aeration is practiced for three puposes : i) To add oxygen to water for imparting freshness, e.g. water from under ground sources devoid of or deficient in oxygen. ii) Expulsion of CO 2, H 2 S & other volatile substances causing taste and odour, e.g. water from deeper layers of an impounding reservoir. iii) To precipitate impurities like iron and manganese, in certain forms, e.g. water from some under ground sources. The Concentration of gases in a liquid generally obeys Henry's Law which states that the concentration of each gas in water is directly proportional to the partial pressure or concentration of gas in the atmoshere in contact with water. The saturation concentration of a gas decreases with temperature & dissolved salts in water. Aeration tends to accelerate the gas exchange. The three types of aerators are : i) Water Fall or Multiple Tray Aerators ii) Cascade Aerators iii) Diiffused Air Aerators Design Criteria For Cascade Aerators Number of Trays Spacing of trays Height of the Structure Space Requirement

= = = =

Design Calculations Disharge ( Q max. ) Provide Area at Tray Diameter of bottom most tray Rise of each Tray Tread of each tray

= = = = =

4 to 9 0.3 to 0.75 C / C 2.0 m 0.015 - 0.05 m 2 / m 3 / hr.

0.1543 m 3 / sec. 17.0 m ² 5.0 m 0.4 m 50.0 cm = 0.5 m

ɸ 1 = 1.0 m ɸ 2 = 2.0 m ɸ 3 = 3.0 m ɸ 4 = 4.0 m ɸ 5 = 5.0 m

Inlet

R.L. 31.0 m R.L. 30.6 m R.L. 30.2 m R.L. 29.8 m R.L. 29.4 m R.L. 29.0 m

Design Of Chemical House & Calculation Of Chemical Dose Alum Dose for Coagulation

The terms coagulation & flocculation are used indiscriminately to describe the process of removal of turbidity caused by the suspension colloids & organic colors. The coagulant dose in the field should be judiciously controlled in the light of the jar test values. Alum is used as coagulant. Design Criteria for Alum Dose

Alum 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, & 5 mg / L in the Monsoon Winter Summer seasons respectively. Flow of Water Required Hourly = 0.1543 x ( 60 x

60 )

=

555.48 m 3 / hour

Per day alum required for worst season for intermediate stage 50 x 555.48 x 1000 x 24 = 1000000 = 666.58 Kg / Day ( 666.58 x 180 ) = 119984.40 Kg For 6.0 months ( 180 Days ) = Number of Bags when 1 bag is containing = 50.0 Kg 119984.40 = = 2399.7 Bags 50.0 = 2400 Bags If 15 bags in ( 2400 / 15 ) = each heep = 160.0 no. of heeps 0.2 ² , then total area required = 32.0 m ²

Lime - Soda Process Softening A water is said to be hard, when it does not form leather readily with soap. The hardness of water is due to the presence of Calcium and Magnesium ions in most of the cases. The method generally used are 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, colour and iron salts. Lime -Soda softening con not reduce the hardness to value less ( < 40 mg / L ). Design Criteria For Lime-Soda Process to 200 mg / L total hardness by this process. 30.0 Lime & Soda Required :→ Lime required for alkalinity. Molecular Weight of Ca C O3 Molecular Weight of Ca O

= = =

( 40 + 12 + 16 x 3 ) ( 40 + 12 + 48 ) = ( 40 + 16 ) = 56

100 mg / L of Ca C O3 alkalinity requires

=

110 mg / L of Ca C O3 alkalinity requires

= ( 56 / 100 ) x 110 61.6 mg / L of =

→ Lime required for Magnesium Molecular Weight of Magnesium ( M n ) 24.0 mg / L of Magnesium ( M n ) requires 1.0 mg / L of Magnesium ( M n ) requires 3.5 mg / L of Magnesium ( M n ) requires

= = = =

56 mg / L of

24 56 mg / L of ( 56 / 24 ) ( 56 / 24 ) x 3.5

100

Ca O Ca O

Ca O mg / L of mg / L of

Ca O Ca O

mg / L of Ca O = 8.2 Hence, the total pure lime required ( 61.6 + 8.2 ) = 69.8 mg / L = Also 56 Kg of Pure Lime ( Ca O ) is equivalent to 74 Kg of hydrated lime. Hence hydrated Lime is required 92.24 = ( 69.8 x 74 ) / 56 = ♥ Soda ( Na 2 C O 3 ) Soda is required for non - carbonate hardness, as follows Molecular Weight of Soda ( Na 2 C O 3 ) = ( 2 x 11 + 12 + 16 x 3 ) 48 ) = = ( 22 + 12 + 82 Na 2 C O 3 100 mg / L of Non Carbonate Hardness ( NCH ) requires 106 mg / L of 61.6 mg / L of NCH requires 65.30 mg / L of Na 2 C O 3 = ( 106 / 100 ) x 61.6 = Total Quantity of Lime = =

555.48 x 180 x 1000000 221345.890 Kg

92.24 x

24 x 1000

( One Bag contains = 50.0 Kg ) Number of Bags required =

221345.890 Kg 50.0 Kg

= 4426.9 Bags Say 4427 Bags

If 15 bags in

each heep ( 4427 / 15 ) = = 0.2 ² , then total area required =

Total Quantity of Soda required for 6 months =

295.1 no. of heeps 59.030 m ²

555.48 x 180 x 1000000 156698.69 Kg = 156698.690 Kg Number of Bags required = = 3134.0 Bags 50.0 Kg If 15 bags in each heep ( 3134 / 15 ) = 208.9 = 0.2 ² , then total area required = 41.790 m 65.30 x

Total Area for all Chemicals ( 32 + 59.03 + 41.79 ) = = Add 30 % for chlorine storage, chlorine cylinders etc. hence total Area = Hence Provide room Dimension : Room Area = Area required is ( < ) Greater than

24 x 1000

Say 3134 Bags

no. of heeps ² 132.82 m ² 172.67 m ²

15.00 m Width = 12.00 m Length = ( 15 x 12 ) = 180.00 m ² Room Area provided, hence Ok.

Chemical Dissolving Tanks : Total quantity of Alum, Lime & Soda

= ( 119984.4 + 221345.89 + 156698.69 ) 498028.980 Kg = 498028.980 Kg Total quantity of Alum, Lime & Soda / Day = = 2766.8 Kg 180 2766.830 Kg Number of Bags required = = 55.3 Bags Say 56 Bags 50.0 Kg If 15 bags in each heep ( 56 / 15 ) = 3.733 no. of heeps = 0.2 ² , then total area required = 0.750 m ² Hence Provide room Dimension : 1.50 m Width = 1.50 m Length = Room Area ( 1.5 x 1.5 ) = 2.25 m ² = Area required is ( < ) Greater than Area provided, hence Ok. Chemical Solution Tanks : Total quantity of Alum, Lime & Soda / Day = 2766.8 Kg Hence Solution required per day Hence Solution required per day

Liter / Day = 2766.83 x 20 = 55336.6 55336.6 = = 38.43 Liter / Min 24 x 60 Quantity of solution for 8.0 Hours = 38.43 x 60 x 8 = 18446.4 Liters 18446.4 = = 18.45 m ³ 1000 Assuming Depth of Tank = 1.20 m & Free Board 0.30 m Dimension of Solution Tank = 4.50 m x 3.50 m x 1.50 m Volume of Solution Tank = 23.625 m ³ Summary Particulars Values S.No. Per Day Alum Required 666.58 Kg / Day 1 Hydrated Lime Required 92.24 2 Soda required 65.3 3 Size of Chemical Dissolving tanks 1.50 m x 1.50 m 4 Size of Chemical Solution tanks 4.5 x 3.5 x 1.5 5

Design Criteria for Mechanical Rapid Mix Unit Detention Time = 30 to 60 Sec. Velocity of Flow = 4 to 9 m / sec. Depth = 1 to 3 m Power Required = 0.041 1000 m Impeller Speed = 100 to 250 rpm Loss of Head = 0.4 to 1.0 m Mixing device be capable of creating a velocity gradiend = 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

3

/ Day

Design Calculations Design Flow

= 0.1543 x ( 24 x 60 x 60 ) 3 = 13331.52 m / Day Detention Time = 30 Sec. Ratio of Tank Height to diameter = 1.5 : 1 Ratio of impeller dia. to tank dia. = 0.3 : 1 Impeller Speed = 120 rpm Assume Temperature = 20 0 C i) Dimension of Tank : Volume Diameter D

= =

Height of Tank

=

Tank free board Total Height of Tank

= = =

ii) Power Requirement : Power Spend

=

4.629 m ³ 1.6 m 4.629 1 (Π/4) 1.6 x 1.6 2.37 m (Say) 0.23 m ( 2.37 + 0.23 ) =

= 2.30 m

2.60 m

5.47 KW

iii) Dimensions of Flat Blade & Impeller : Diameter of Impeller = 0.65 m Velocity of Tip Impeller ( V T ) = 4.08 m / sec. Area of Blade AB = Power Spent  Let C D = »

5.47 x

»

AB=

1  C D  ro  A B  V R 2

1.8 (Flat Blade): 1000 = 1.99 m ²

1 x 1.8 2

and V R = x 1000 x

3

( 3/4 ) x V T A B x ( 3/4 ) x

4.08

Hence Provide 8 Area of Blade Provided = Provide

4 numbers of

Blades of ( 0.50 x 0.50 ) m ( 0.5 x 0.5 ) x 8 = 2.00 m ² length 1.50 m and

projecting

0.2

Provide Inlet & Outlet Pipes of 250 mm diameter. iv) Summary S.No. 1 2 3 4 5 6 7

Particulars Detention Time Speed of impeller Height of Tank ( 0.23 m free board ) Power Required Number of Blade ( 0.50 x 0.50 ) m Number of Baffles ( length 1.50 m ) Diameter of Inlet & Outlet Pipes

Values 30 Sec. 120 rpm 2.60 m 5.47 KW 8 4 250 mm

Design Of Clariflocculator Clariflocculator

The coagulation & sedimentation processes are effectively incorporated in a single unit in the Clariflocculator. Sometimes clarifier & Clariflocculator are designed as separate units. All these units consists of 2 or 4 flocculating paddles placed equidistantly. These paddles rotate on their vertical axis. The flocculating paddles may be of rotor-stator type. Rotating in opposite direction above the vertical axis. The clarification unit outside the flocculation compartment is served by inwardly raking rotating blades. The water mixed with chemical is fed in the flocculator compartment fitted with paddles rotating at low speeds thus forming flocs. 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 & the clear effluent overflows into the peripheral launder.

Design Criteria : ( Flocculator )

Depth of Tank Detention Time Velocity of Flow Total Area of Paddles Range of peripheral velocities of blades Velocity Gradient ( G ) Dimension Less Factor G t Power Consumption Outlet Velocity

= = = = = = = = =

3 to 30 to 0.2 to 10 to 0.2 to 10 to 10 10 to 0.15 to

4.5 m 60 min. 0.8 m / sec. 25 0.6 m / sec. 75

= = = = = = = = =

40 m 3 to 300 m 25 % 1 in 12 or 8 % 1.2 : 1 1 40

3

4

to 10 5 36 KW / MLD 0.25 m / sec.

Design Criteria : ( Clarifier )

Assuming a Surface Overflow rate Depth of Water Weir Loading Storage of Sludge Floor Slope Slope for Sludge Hopper Scraper Velocity Velocity of water at outlet chamber

/ m 2 / Day 4.5 m 3 / m 2 / Day

for mechanically cleaned tank (V:H) 45 to 80 min.

Assumption

Average Outflow from clariflocculator Water Lost in desludging

= =

555.48 m 3 / hour 2%

Design Average Period

= = = = =

Detention Period Average Value of Velocity Gradient

555.48 + 2 % of 555.48 + 11.11 566.59 m 3 / hour 30 min.

555.48

30.0 sec - 1

Design Of Influent Pipe

Discharge ( Q ) = 0.1543 m 3 / sec. Assuming Velocity ( V ) = 1.0 m / sec. Q  A V  A 

Q   Q   d2   V  4  V

Hence Diameter ( d ) = Provide an influent pipes of

»

d 

 Q4    V   

0.1543 x 4 1/2 1 Π 450 mm diameter.

= 0.4434 m

Design Of Flocculatior : Wall

Volume of flocculator =

566.59 x 30 = 283.3 m ³ 60 3.50 m 283.30 = 80.9 m ² 3.50

Provide a Water Depth = Plan Area of flocculator = Diameter of Flocculator ( D )   A d2   d   4 

 A 4      

Diameter of Inlet Pipes ( D P ) Provide a Tank Diameter of 10.20 m

=

80.94 x 4 Π

=

0.45 m

1/2

= 10.15 m

Say 10.20 m

Dimension Of Paddles : Here :

P  G   V

Power dissipated in watts i.e. P = N.m / s μ = Absolute or Dynamic Viscosity of Raw Water in N.s / m2 G = Temporal Mean Velocity Gradient in ( sec -1 ) V = Volume of raw water to which P is applied in m 3 = ( Π / 4 ) x ( 10.2 x 10.2 ) x 3.5 = 285.85 m ³

2

= 30 x 30 x

0.89 1000



x 285.85



=

1  C d    AP  v   3 2

Power Input =

228.97



Cd =

1.8 ρ = 995 Kg / m 3 ( 25 0 C ) v = Velocity of tip of blade = 0.4 m / sec. ν = Velocity of water tip of blade = ( 0.25 x 0.4 ) =

0.1 m / sec.

»

228.97

= ( 1 / 2 ) x 1.8 x 995 x ( 0.4 - 0.1 ) 3 x A P » AP = 9.470 m ²

Ratio of Paddles to C / S of Flocculator ( A P ) x ( 10.2 - 0.75 ) x 3.5 x 100 = 9.47 / 9.11 ( 10.0 to 25 % ) Provide A P = 10.5 m ² AP= x ( 10.2 - 0.75 ) x 3.5 x 100 = 10.5 / Π 10.12 % ( Which is Acceptable, hence O.K. ) 3.5 m height 0.7 Provide 5 nos. of paddles Shaft will support 1 ( One ) 5 Paddles. The Paddles will rotate at an rpm of 4 V= 2xΠxrxN = 2 x 3.14 x r x ( 4 / 60 ) = 0.4 60 »r= 0.955 m Say 1 m r = Distance of Paddle from C 1 of vertical shaft. Let velocity of water below the partition wall between the flocculator & clarifier be 0.3 m / sec. 0.3 x 60 x 60 Area = 555.48 / = 0.514 m ² Depth below partition Wall = 0.51 / (Πx 10.20 ) = 0.016 m 25 ( 0.25 x 3.5 ) = 0.875 m Provide Slope for Bottom = 8% Total Depth of Tank at Partition Wall = =

0.29 say ( 0.3 + 3.5 + 0.016 + 0.875 ) 4.691 m say 4.7 m

Design Of Clarifier

Assuming a Surface Overflow rate Surface Of Clariflocculator

= =

Diametre of Clariflocculator D cf = Π x ( D cf 2 - 10.20 x 10.20 ) 4

40 m 3 / m 2 / Day 555.48 x 24 = 333.29 m ² 40

= 333.29

1333.152 + 104.04 1/2 Π Say 23 m = 22.992 m Length of Weir = ( 3.14 x 23 ) = = Π x D cf 72.22 m Weir Loading = 555.48 x 24 = 184.60 m 3 / Day / m 72.22 According to manual of govt. of india, if it is well clarifier, then it can be exceed upto 1500 D cf

=

Summary S.No. 1 2 3 4 5 6 7 8 9 10 11

Particulars Detention Period Diameter Of Influent Pipes Overall Depth of Flocculator Diameter of Tank No. of Paddles Distance of Shaft from C.L. of Flocculator Paddles Rotation (RPM) Distance of Paddle from C.L. of vertical Shaft Slope of Bottom ( % ) Total Depth of Partition Wall Diameter of Clariflocculator

Values 30 min. 0.45 m 3.5 m 10.20 m 5 nos. 4 1.0 m 8% 4.7 m 23.0 m

Design Of Rapid Gravity Filter a) Design Criteria : ( Rapid Sand Filter ) Rate Of Filteration = 5 to 7 m Maximum surface area of One Bed = Minimum Overall Depth Of Filter Unit Including a Free Board of 0.5 m = 2.6 m Effective size Of Sand = 0.45 to 0.7 Uniformity Co-efficient For Sand = 1.30 to 1.7 Ignition Loss Should Not Exceed ( > ) 0.7 % percent by weight Specific Gravity = 2.55 to 2.65 Wearing Loss is not greater than ( > ) 3.0 % Minimum Number Of Units = 2 Depth Of Sand = 0.6 to 0.75 Standing Depth of water over the filter = 1 to 2 m Free Board is less than ( > ) 0.5 m b) Problem Statement : Net Filtered Water Quantity of Backwash water used Time Lost During Backwash Design Rate Of Filteration Length & Width Ratio Under Drainage System Size of Perforations c) Design Calculations Water Flow Required Design Flow for Filter

2.0

Provide

2 Filters

= 555.48 m 3 / hour = 2% = 30 min. = 5 m 3 / m 2 / Day = 1.25 to 1.33 : 1 Central Manifold With Laterals = = 13 mm

= 578.64 m 3 / hour 578.64 = 115.7 m = 5

²

24 23.5 ≡

116 m ²

Units 116 = 58 m ² 2 L x 1.25 L = 58.0 m ² 1/2 58 » Length L = 1.25 = 6.8 m Width = ( 1.25 x 6.8 ) = 8.5 m Units, each with a dimension of = 8.6 m x 6.8 m

Hence Plan Area of One Unit Length x Width =

2

/ m / Day

= 555.48 m 3 / hour = 555.48 x ( 1 + 0.02 ) x

Plan Area For Filter Using

3

=

Say 8.6 m

Estimation Of Sand Depth : It is checked against break through of floc. Using Hudson Formula 3

Where Assume

B= h= Q= d=

»

Q x d x h = B x 293223 L ℓ Q, d, h & ℓ are in m 3 / m 2 / hr, mm, m and m respectively 4 ( Poor Response ) < Average degree of pre-treatment 10000 ( Terminal Head Loss ) 2.5 m 5 x 2 m 3 / m 2 / hr ( Assuming 100 % overload of filter ) ( Mean Diameter ) 0.6 mm

0.6 x 0.6 x 0.6 x 2.5 4 = x 293223 10000 ℓ = 46.04 cm { or ( > ) greater than } ℓ Hence provide depth of sand bed = 60.0 cm

10 x

Estimation Of Gravel & Size Gradation : Assuming size gradation of 2.0 mm to 40.0 mm at bottom using empirical formula : P = 2.54 x R x ( log d ) Where : R = 12 mm ( 10 mm to 14 mm ) The Units of L & d are cm & mm, respectively. Size 20 2 5 10 40 Depth (cm ) 40 9.2 21.3 30.5 49 Increment 9.5 9.2 12.1 9.2 9 Hence provide depth of gravel. 50.0 cm Estimation Of Under Drainage System : Plan Area of each filter Total Area of perforation Total Cross Section Area of Laterals

= = = = = = = =

8.6 m x 6.8 m 58.48 m ² 3 x 58.48 1000

= 0.17544 m ²

= 1754.40 cm ² Area of perforation ) = 5263.20 cm ² 1754.4 ) Area of Laterals ) = 10526.40 cm ² 5263.2 ) x4 = 115.80 cm

( 3.0 x ( 3.0 x Area of Central Manifold ( 2.0 x ( 2.0 x Diameter of Central Manifold 10526.4 Π Providing a commercially available diameter of = 100.0 cm Assuming spacing for laterals = 20 cm 8.6 100.0 Number of Laterals = 20 61.2 x 4 D = Π Number of perforations / laterals = 86 Units

= 43 on either side = 8.8 cm

say 90 mm

Length of One lateral =

( 1 ) width of filter - ( 1 ) diameter of manifold 2 2 = (1 x 6.8 ) - ( 1 x 1.0 ) = 2.9 m 2 2 Let n be the total number of perforation of 13.0 mm diameter Total Area of perforation » nxΠ x ( 1.3 ) ² = 1754.40 4 » n = 1322.43 say 1322 1322.0 Number of Perforation or Laterals = = 15.37 say 16 86.0 2.90 x 100 Spacing of Perforation = = 18.13 cm C / C 16.0 Say 180.00 mm C / C perforations of 13.0 mm diameter at 180 cm C / C Provide 16 Computation Of Wash Water Troughs : Wash Water Rate = 36 m 3 / m 2 / hour Wash Water discharge for one filter = ( 36 x 58.48 ) = 2105.28 m 3 / hour = 0.5848 m 3 / sec. Assuming a spacing of 1.80 m for wash water trough which will run parallel to the longer dimension of the filter unit. 6.8 Number of trough = = 3.78 say 4 1.8 0.5848 Discharge per unit trough = = 0.146 m 3 / sec. 4.0 For a width of the water depth at upper end is given by : 0.40 m 3

Q  1.376  b  h 2 1.376 x 0.40 x ( h ) 3/2 2/3 0.1462 h = = 0.41 m say 0.5 1.376 x 0.40 Freeboard = 0.1 m Wide & deep in each filter. 0.5 m 0.50 m 0.1462 =

Provide

4.0

troughs of

Total Depth Of Filter Box : Depth of filter box = ( depth of under drain + gravel + sand + water depth + free board ) 500 + 600 + = ( 900 + 2200 + 300 ) = 4500 mm Design Of Filter Air Wash : Assume Rate at which air is supplied = 1.5 m 3 / m 2 / min. Duration of Air Wash = 3.0 min. Total Quantity of air per unit bed = 1.5 x 3.0 x 8.6 x 6.8 = 263.16 m ³

Summary S.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Particulars Number Of Units Size Of Unit Depth Of Sand Bed Depth Of Gravel Diameter Of Perforation Diameter Of Central Manifold Spacing For Laterals Number Of Laterals Diameter Of Laterals Number Of Perforations Number Of Trough Size Of Trough Total Depth Of Filter Box Duration of Air Wash Total Quantity Of Air Required Per Unit Bed

Values 2 min. 8.60 m x 6.8 m 60.0 cm 50.0 cm 13 mm 100.0 m 20 cm 86 Units 90 mm 16 4 0.40 m x 0.5 m 4500 mm 3 min. 263.2 m ³

Design Of Disinfection Unit a) Chlorination Disinfection should not only remove the existing bacteria from water but also ensure their immediate killing even afterwards, in the distribution system. b) Design Criteria ( Chlorination ) → Chlorine Dose

→ Residual Chlorine → Contact Period

= = = = =

1.4 mg / L ( Rainy Season ) 1 mg / L ( Winter Season ) 0.6 mg / L ( Summer Season ) 0.1 to 0.2 mg / L ( Minimum ) 20 to 30 min.

c) Design Calculations Rate of Chlorine required, to disinfect water be = Chlorine required Per Day = 13.33 x

2 p.p.m. 1000000 x

1.40 x

= For 6 Months = Number of Cylinder ( One Cylinder contain Number Of Cylinders used per day d) Summary S.No. 1 2

1 1000000

18.662 Kg = 3359.16 Kg ( 18.66 x 180 ) 16.0 Kg ) = ( 3359.16 x 2 ) = 419.895 16 = 2 Cylinders of 16.0 Kg

Particulars Chlorine required per day Number Of Cylinders required per day

Values 18.662 Kg 2 Cylinders of 16.0 m