30 =

=

=

=

= 100 x 61.4 10 -3

=

=

=

=

Calculation of Water Demand

It can be assumed that city is a residential town ( Low Rise Buildings )

Expected Population after

Average Rate of Water Supply / Capita 135

61400

Water required for above purposes for whole town = ( 61400 x 135 )

8.289 MLD

Industrial Demand 0.6 MLD

Fire Requirement :

Maximum Daily Draft = ( 1.5 x 8.889 ) = 13.334 MLD

Coincident Draft = Maximum Daily Draft + Fire Demand

Water for Fire MLD

= 0.78 MLD

Average Daily Draft = ( 8.289 + 0.6 ) = 8.889 MLD

Pipe Main = Maximum Daily Draft = 13.334 MLD

2 x Average Daily 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

Filters and Other Units at Treatment Plant :

( 2 x 8.889 ) = 17.778 MLD

Lift Pump : 2 x Average Daily Demand

( 2 x 8.889 ) = 17.778 MLD

310100 P

S.No.

1

2

3

4 7.0 to 8.5 6.5 to 9.2

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

Turbidity ( Units on J.T.U. Scale )

Colour ( Units on Platinum Cobalt Scale )

Taste & Odour

PH

Zinc ( mg / L as Z n )

Total Dissolved Solids ( mg / L )

Total Hardness ( mg / L as Ca CO 3 )

Chlorides ( mg / L as C 1 )

Sulphates ( mg / L as S O 4 )

Fluorides ( mg / L as F )

Nitrates ( mg / L as N O 3 )

Calcium ( mg / L as Capacity )

Magnesium ( mg / L as Mg )

Iron ( mg / L as Fe )

Manganese ( mg / L as M n )

Copper ( mg / L as C u )

Anionic Detergents ( mg / as MBAS )

Mineral Oil ( mg / L )

Arsenic ( mg / L as A s )

Cadmium ( mg / L as Cd )

Chromium ( mg / L as Hexavalent Cr )

Gross Beta Activity ( ρ Ci / L )

Cause for Rejection

10

25

Unobjectionable

1500

600

1000

400

Cyanides ( mg / L as C ≡ N )

Lead ( mg / L as P b )

Selenium ( mg / L as Se )

Mercury ( mg / L as Hg )

Polynuclear Aromatic Hydrocarbons ( mg/L )

Gross Alpha Activity in pico Curie ( ρ Ci / L )

Phenolic Compounds ( mg / L as Phenol )

0.05

1.5

45

200

150

1

0.5

1.5

15

0.002

1

0.3

0.2

3

30

Acceptable

2.5

5

Unobjectionable

500

200

200

0.01

0.05

0.05

0.1

0.01

0.001

0.001

0.2

0.01

200

1

45

75

30

0.1

0.001

0.2

3

30

Physical & Chemical Standards Of Water

TOXIC MATERIALS

RADIO ACTIVITY

0.05

0.01

0.05

0.05

0.1

0.01

0.05

0.05

5

S.No. Actual Treatment Proposed

1 7.5 7.0 to 8.5 0.25 Hence OK. Not Necessary

2 50Clarifier & Rapid

Sand Filler

3 550 Softening

4 200 -

5 2.5 Aeration

6 3.5 Aeration

7 110 Softening

8 3.5 Chlorination

Particulars

pH

Comparison of Given Data & Standard Data and Treatment Proposed

Standard

Iron

Carbonate

MPN

Turbidity

Total Hardness

Chlorides

Manganese

0

Difference

47.5

350

0

2.4

3.45

110

3.5

2.5

200

200

0.1

0.05

0

S.No.

1

2 5.0 to 10 m ( Maximum = 15 m )

3 10 m

4

5 1 to 3 ( Maximum = 4 )

6

=

=

=

=

= 0.1543 m

0.1543 x 10 x 60

= 92.58 m ³

( 92.58 / 4 )

= 23.15 m ²

23.15 x 4

Π

d = 5.43 m

5.43 m ≡ 5.5 m

S.No. Values

1 1 Unit

2 5.5 m

3 4.0 m

4 24 m

Design Criteria :

Particulars

Detention Time / Period

Diameter of Well

Depth of Well

Velocity of Flow

Number of Units

Free Board

Values

10 minutes

4.0 to

5.0 to

Design of Intake Well

Flow of Water Required = 13.334 MLD / 3600 x 24

Given F.S.L.

Minimum R.L.

Given invert of gravity main

Detention Time

27 m

28 m

24 m

10 minutes

0.6 to 0.9 m / sec.

5 m

Design Assumptions :

3 / sec.

Volume of Well =

Cross-sectional Area of Intake Well =

Diameter of Intake Well =

Design Calculations :

Number of Intake Wells

Diameter of Intake Well

Height of Wall

R.L. of bottom of Well

< 10 m ( O.K. )

Hence Diameter of Intake Well =

Summary :

Particulars

a)

b)

= 0.6 to

=

=

c)

=

=

=

=

= 0.103 m ²

0.103 x 4

Π

d)

S.No.

1

2

3

a)

= 0.2 to

= 6.0 to

=

b)

=

= 10.0 mm

c)

( 10 x 10 ) x Π

Calculation

4Area of Each Hole =

0.25 m / sec.

0.75 m / sec.

Less than 1 m

² = 0.79 cm

Area of Strainer 2

Assumptions

Velocity of Flow

Hole Diameter

Design of Bell Mouth Strainer :

Design Criteria

Velocity of Flow 0.3 m / sec.

Hole Diameter 12.0 mm

Number of Pen Stock / Well 2 Units

At Each Level 1 m

Diameter of Pen Stock 0.40 m

Particulars Values

Number of Pen Stock for each Level

Velocity

C / S area of each Pen Stock

2 nos.

0.1543 / ( 0.75 x 2 )

Diameter = 0.3622 m ≡ 0.4 m (Say)

Summary

Diameter of each Pen Stock

Number of Pen Stock for each Intake Well 2 nos.

Design Calculation

Number of Intake Well 1 nos.

Design of Pen Stock & Bell Mouth Strainer

Pen Stock

Design Criteria

Velocity through Pen Stock 1.0 m / sec.

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.

4Aread

4

2d

0.1543 1

( 0.25 x 2 ) 0.7850

= = 6171.98

Π x d 2

4

Π

0.90 m

F.S.L. =

L.W.L. = 26.0 m

5.5 m

24 m

Area of Collection = Area of Pen Stock

0.1543

( 0.25 x 2 ) = 0.7850 x N

²

= 88.67 cm

= 6171.98 cm

N = = 3931.2

Area of Strainer =

Bottom R.L. =

3.0 m

Provide Diameter of for Bell Mouth Strainer.

27.0 m

2

2 x 3931.2 x 0.785

Bell Mouth Strainer

Diameter of Bell Mouth Strainer =

( 6171.98 x 4 ) Diameter = d =

Plan

Section

Regulating Valves Manhole

Gravity Main

4

2d

a)

b)

= 0.3 to

= 0.6 to

= 1 nos.

=

c)

R.C.C. Circular n =

=

= 0.1543

= 0.2204 m ²

Π x d 2

4

Π

» 0.7 x 0.74/3

= 11.6684 x 10 -4

S = 1 : 857

Area 1 D 0.55

Perimeter Π x D 4 4

100

857

( 27 - 3 ) =

d)

S.No.

1

2

3

4

Diameter of Gravity Intake 0.55 m

Invert Level at Intake Well 24.0 m

Invert Level at Jack Well 23.88 m

23.883 m

Summary

Particulars Values

Number of Gravity Intake 1 Units

Head Loss = = 0.117

R.L. of Gravity Main = 24.0 m

R.L. of Gravity Main at Jack Well = ( 24 - 0.117 ) =

=

( 0.55 / 4 ) = 0.001167

4

≡

/ ( 0.70 )

0.55 m

Pipe is used. For this

Conduit Velocity ( Assumed ) 0.70 m / sec.

Using Manning's Formula -

0.013

x 0.013 x 0.013

Π x D x D=Here R =

Diameter = d =0.2204 x 4

= 0.53 m

Velocity of Water 0.9 m / sec.

²

Number of Gravity Main = Number of Intake Well

0.70 m / sec.Assumption Velocity

Design Calculation

Area of Conduit required (A = Q / V )

Diameter of the conduit »= 0.2204 m

Design of Gravity Main

Gravity Main

Design Criteria

Diameter of Gravity Main 1.0 m

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.

2

1

3

21

SRn

V

2

3

2

R

nVS

3

4

22

R

nV

a)

b)

0.5 x ( 3.0 to 15.0 min. )

= 0.5 x 10 minutes

c)

5 minutes

8.0 m

1.0 m

0.5 m

3.12 m

0.1543 x 10 60.0 ³

( 92.58 / 3.12 ) = 29.70 m ²

Π x d 2

4

29.7 x 4

Π

22.88 m

d)

S.No.

1

2

3

4

5

6

7

Design of Jack Well

Head required ( h d ) 4.88 m = ( 8 - 2.12 - 1 )

R.L. of Top of Jack Well 29.88 m

Suction Depth 2.12 m

Top Clearance 0.50 m

Bottom Clearance 1.00 m

Particulars Values

Diameter of Jack Well 6.15 m

R.L. of Bottom of Jack Well 22.88 m

Diameter = d = = 6.15 m

Summary

29.88 m

R.L. of Bottom of Jack Well = ( 23.883 - 1 ) =

R.L. of Bottom of Jack Well when full = ( 22.883 + 7 ) =

Design criteria

Detention Time = ( Detention time of intake Well )

Bottom Clearance =

Top Clearance =

Jack Well

Capacity of Well = = 92.58 m

C / S Area of Well =

Diameter of the Well »

Minimum Depth of Water = ( 26 - 22.883 ) =

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.

= 29.70 m ²

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

( < ) Less than 10 m

Diameter of Well = ( < ) Less than 20 m

Assuming Suction Head =

= 5 minutes

Suction Head =

Design Calculations

Detention Time =

a) Pumps

›

›

→

10 m,

→

→

i)

ii)

iii)

iv)

♥

♥

♥

♥

♥

b)

= 0.1543 m

= ( 0.97 to 1.22 )

= ( 0.97 to 1.22 )

= ( 0.381 to 0.48 )

= 0.43 m Say ≡ 0.45 m

c)

= 0.7 to

= 0.50 m

= 1.00 m

d)

=

0.02 x 190 x ( 0.9 x 0.9 )

2 x 9.81 x 0.45

f = 0.02

L = 190 m

g =

Design Calculation

Hence Provide d =

Design Criteria

Buoyancy Operated pumps

running costs.

The efficiency of the pumps & consequent influence on power consumption and the

Here :

9.81 m / sec.

Present & projected demand & pattern and change in demand.

The details of head & flow rate required.

Suction head should not be greater than

x ( 0.1543 )

Economical Diameter ( d )

( > ) 10 m.

Velocity of flow length 1.1 m / sec.

Top Clearance

Bottom Clearance

= 0.349 m

Frictional Losses in Rising Main -

Velocity ( Assuming ) 0.9 m / sec.

Head Loss

= ( 0.97 to 1.22 )

x ( 0.3930 )

Design Of Pumping System

3 / sec.

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

aeration units.

2

2.5

If head is more than 10 m ,

The following criteria govern pump selection : -

Diameter of Rising Main

Discharge ( Q )

Type of duty required.

Selecting the operating speed of the pump & suitable drive.

Impulse Operated pumps

Velocity Adoptions pumps

Positive Displacement Pumps

water is converted into vapour & thus inspite of creating water head,

head is created & pump ceases to fuction.

The following points are to be stressed upon :

be greater than ( > ) for centrifugal pump.

The suction pumping should be as short & straight as possible. It should not

The following four types of pumps are generally used -

Q

dg

vLfh f

2

2

Head Loss » h f = 0.349 m Say ≡

= 1 m

=

= 8.35 m

Assuming

1000 x 8.35

17.18

0.75

If ƞ = 75 %

e)

S.No.

1

2

Pumps Capacity 25 H.P

Diameter of Pipe 0.45 m

Say ≡= 22.90 H.P

75 = 17.179 H.P

Hence Total Head of Pumping = ( h s + h d + h f + minor losses )

Particulars Values

25 H.P

( 2.12 + 4.88 + 0.35 + 1 )

Minor Losses should be assuming

2 in Parallel is working

0.1543 x

Summary

0.35 m

75

....

HQWPHW

PHWPHB

....

a) General

b)

= 0.9 to

0.9 m

= 0.1543 m 3 / sec.

c)

= ( 0.97 to 1.22 )

= ( 0.97 to 1.22 )

= ( 0.381 to 0.48 )

= 0.43 m Say ≡ 0.45 m

d)

S.No.

1

Particulars Values

Diameter of Mains Pipe 0.45 m

x ( 0.1543 )

x ( 0.3930 )

Hence Provide d =

Total Discharge in Mains

Summary

1.5 m / sec. Permissible Velocity in Mains

Mains Diameter should be less than ( < ) of

Design Calculations

Economical Diameter ( d ) = ( 0.97 to 1.22 )

Design of Rising Main

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.

Design Criteria

These are the pressure pipes used to convey the water from jack well to the treatment units.

Q

i)

ii)

iii)

i)

ii)

iii)

= 4 to 9

= 0.3 to

= 2.0 m

= 0.015 - 0.05 m

Disharge ( Q max. ) = 0.1543 m

= 17.0 m ²

= 5.0 m

= 0.4 m

= 50.0 cm

ɸ 1 = 1.0 m

ɸ 2 = 2.0 m 31.0 m

ɸ 3 = 3.0 m R.L.

ɸ 4 = 4.0 m 30.2 m

ɸ 5 = 5.0 m R.L. 29.8 m

R.L. 29.4 m

R.L. 29.0 m

Inlet

R.L.

30.6 m

R.L.

3 / sec.

= 0.5 m

Spacing of trays 0.75 C / C

Height of the Structure

Space Requirement 2 / m

3 / hr.

Design Calculations

Provide Area at Tray

Diameter of bottom most tray

Rise of each Tray

Tread of each tray

Number of Trays

Aeration is necessary to promote the exchange of gases between the water & the

atmosphere. In water treatment, aeration is practiced for three puposes :

To add oxygen to water for imparting freshness, e.g. water from under ground sources

devoid of or deficient in oxygen.

Expulsion of CO 2, H 2 S & other volatile substances causing taste and odour, e.g.

water from deeper layers of an impounding reservoir.

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 :

Water Fall or Multiple Tray Aerators

Aeration unit

Cascade Aerators

Diiffused Air Aerators

Design Criteria For Cascade Aerators

Treatment Units - Design Of Aeration Unit

=

=

=

50 mg / L, 20 mg / L, & 5 mg / L in the

Monsoon Winter Summer

0.1543 x ( 60 x 60 ) = 555.48 m 3 / hour

50 x 555.48 x 1000 x 24

1000000

= 666.58 Kg / Day

For 6.0 months

50.0 Kg

=

each heep = 160.0

² , 32.0 m ²

5 mg / L

Let the average dose of alum required be

seasons respectively.

Per day alum required for worst season for intermediate stage

=

Alum required

Summer

Flow of Water Required Hourly =

Bags

2400 Bags

( 180 Days ) = ( 666.58 x 180 ) = 119984.40 Kg

Number of Bags when 1 bag is containing =

119984.40=

50.0 = 2399.7

Design Of Chemical House & Calculation Of Chemical Dose

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.

50 mg / L

20 mg / L

Alum Dose for Coagulation

Design Criteria for Alum Dose

Alum required in particular season is given below :

Monsoon

Winter

If 15 bags in ( 2400 / 15 ) =

0.2 then total area required =

no. of heeps

30.0

Ca C O3 = ( 40 + 12 + 16 x 3 )

= ( 40 + 12 + 100

Ca O = ( 40 + 56

Ca C O3 =

Ca C O3 =

=

= 24

requires =

requires = Ca O

requires = Ca O

= 8.2 Ca O

=

74 Kg

=

♥

= ( 2 x 11 + 12 + 16 x 3 )

= ( 22 + 12 + 82

=

= 92.24 x 180 x 24 x 1000

=

Softening

48 ) =

A water is said to be hard, when it does not form leather readily with soap. The hardness of

generally used are Lime-Soda process. Softening with these chemicals is used particularly for

→ Lime required for alkalinity.

Molecular Weight of

Lime & Soda Required :-

Design Criteria For Lime-Soda Process

110 mg / L of alkalinity requires ( 56 / 100 ) x 110

61.6 mg / L of Ca O

Molecular Weight of 16 ) =

100 mg / L of alkalinity requires 56 mg / L of Ca O

56 mg / L of Ca O

1.0 mg / L of Magnesium ( M n ) ( 56 / 24 ) mg / L of

→ Lime required for Magnesium

24.0 mg / L of Magnesium ( M n )

Soda ( Na 2 C O 3 )

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

48 ) =

Also 56 Kg of Pure Lime ( Ca O ) is equivalent to

Hence hydrated Lime is required ( 69.8 x 74 ) / 56 = 92.24

3.5 mg / L of Magnesium ( M n ) mg / L of ( 56 / 24 ) x 3.5

mg / L of

Hence, the total pure lime required ( 61.6 + 8.2 ) = 69.8 mg / L

65.30 mg / L of Na 2 C O 3

555.48 x

1000000

61.6 mg / L of NCH requires ( 106 / 100 ) x 61.6 =

100 mg / L of Non Carbonate Hardness ( NCH ) requires 106 mg / L of Na 2 C O 3

50.0 Kg == 4426.9 Bags Say 4427 Bags

221345.890 Kg

( One Bag contains = 50.0 Kg )

221345.890 Kg Number of Bags required

Lime - Soda Process

of hydrated lime.

water is due to the presence of Calcium and Magnesium ions in most of the cases. The method

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 ).

Molecular Weight of Magnesium ( M n )

to 200 mg / L total hardness by this process.

Total Quantity of Lime

Molecular Weight of Soda ( Na 2 C O 3 )

=

²

= 65.30 x 180 x 24 x 1000

=

=

²

= ²

²

Length = Width =

= ²

Ok.

=

=

180

=

²

Length = Width =

= ²

= 2766.83 x 20 = 55336.6

=

24 x 60

= 38.43 x 60 x 8

18446.4

1000

1.20 m

= 4.50 m x 1.50 m

= 23.625 m ³

S.No.

1

2

3

4 1.50 m

5

555.48 x Total Quantity of Soda required for 6 months

1000000

156698.69 Kg

( 4427 / 15 ) = 295.1 no. of heeps

² , then total area required = 59.030 m

If 15 bags in

0.2

each heep

If 15 bags in each heep ( 3134 / 15 ) = 208.9 no. of heeps

0.2 ² , then total area required = 41.790 m

Number of Bags required =156698.690 Kg

= 3134.0 Bags Say 3134 Bags50.0 Kg

Hence Provide room Dimension : 15.00 m 12.00 m

Room Area ( 15 x 12 ) = 180.00 m

Total Area for all Chemicals ( 32 + 59.03 + 41.79 ) = 132.82 m

Add 30 % for chlorine storage, chlorine cylinders etc. hence total Area = 172.67 m

Total quantity of Alum, Lime & Soda / Day

Number of Bags required =2766.830 Kg

= 55.3 Bags

Chemical Dissolving Tanks :

Total quantity of Alum, Lime & Soda ( 119984.4 + 221345.89 + 156698.69 )

498028.980 Kg

498028.980 Kg == 2766.8 Kg

0.2 ² , then total area required = 0.750 m

Hence Provide room Dimension : 1.50 m 1.50 m

Say 56 Bags50.0 Kg

If 15 bags in each heep ( 56 / 15 ) = 3.733 no. of heeps

Hydrated Lime Required

Soda required

Size of Chemical Dissolving tanks

Size of Chemical Solution tanks

1.50 m x

Values

666.58 Kg / Day

92.24

65.3

4.5 x 3.5 x 1.5

Particulars

Per Day Alum Required

Dimension of Solution Tank 3.50 m x

Volume of Solution Tank

Summary

= 18.45 m= ³

Assuming Depth of Tank =

Quantity of solution for 8.0 Hours = 18446.4 Liters

Hence Solution required per day Liter / Day

Hence Solution required per day 55336.6

& Free Board 0.30 m

Area required is ( < ) Greater than Area provided, hence Ok.

Area required is ( < ) Greater than Room Area provided, hence

= 38.43 Liter / Min

Room Area ( 1.5 x 1.5 ) = 2.25 m

Chemical Solution Tanks :

Total quantity of Alum, Lime & Soda / Day = 2766.8 Kg

= 30 to 60 Sec.

= 4 to

= 1 to 3 m

= 1000 m

= 100 to

= 0.4 to 1.0 m

= 300 m

0.2 to 0.4 : 1

= 1 to 3 : 1

= 0.1543 x

= 13331.52

= 30 Sec.

= 1.5 : 1

0.3 : 1

= 120 rpm

= 20 0 C

i)

Volume = 4.629 m ³

= 1.6 m

4.629 1

( Π / 4 ) 1.6 x 1.6

= 2.37 m (Say)

= 0.23 m

= 2.60 m

ii)

= 5.47 KW

iii)

= 0.65 m

=

= A B

Let C D =

» 5.47 x 1000 = x 1000 x A B x ( 3/4 ) x 4.08

2

» A B = 1.99 m ²

Design Criteria for Mechanical Rapid Mix Unit

/ sec / m depth

Ratio of Tank Height to diameter

Design Calculations

Impeller Speed

Loss of Head

0.041

250 rpm

Detention Time

Velocity of Flow 9 m / sec.

Depth

Power Required

( 24 x 60 x 60 )

m 3

/ Day

Detention Time

Ratio of Tank Height to diameter

Ratio of impeller dia. to tank dia. =

Impeller Speed

Assume Temperature

Dimension of Tank :

Diameter D

Design Flow

( 2.37 + 0.23 ) =

4.08 m / sec.

=Height of Tank

Tank free board

Total Height of Tank

Power Spent

1.8 (Flat Blade): and V R = ( 3/4 ) x V T

Power Requirement :

Power Spend

Dimensions of Flat Blade & Impeller :

Diameter of Impeller

3 / Day

Ratio of impeller diameter to tank diameter =

Mixing device be capable of creating a velocity gradiend

1 x 1.8

Velocity of Tip Impeller ( V T )

= 2.30 m

Area of Blade

3

2

1RBoD VArC

8 ( 0.50 x 0.50 ) m

2.00 m ²

Provide projecting 0.2

250 mm diameter.

iv)

S.No.

1

2

3

4

5 ( 0.50 x 0.50 ) m

6

7

Speed of impeller 120 rpm

Provide Inlet & Outlet Pipes of

Area of Blade Provided = ( 0.5 x 0.5 ) x 8 =

4 numbers of length 1.50 m and

Hence Provide Blades of

Summary

Particulars Values

Detention Time 30 Sec.

2.60 m

Power Required 5.47 KW

8

Height of Tank ( 0.23 m free board )

Number of Blade

Number of Baffles ( length 1.50 m ) 4

Diameter of Inlet & Outlet Pipes 250 mm

= 3 to 4.5 m

= 30 to 60 min.

= 0.2 to

= 10 to

= 0.2 to

= 10 to 75

= 10 4 to 10

5

= 10 to

= 0.15 to

= 40 m

= 3 to 4.5 m

= 300 m

= 25 %

= 1 in 12

= or 8 %

= 1.2 : 1

= 1 45 to 80 min.

= 40

= 555.48 m 3 / hour

= 2 %

Design Of Clariflocculator

0.25 m / sec.

Design Criteria : ( Clarifier )

Assuming a Surface Overflow rate 3 / m

2 / Day

Depth of Water

( V : H )

Scraper Velocity

Assumption

Average Outflow from clariflocculator

Water Lost in desludging

3 / m

2 / Day

Storage of Sludge

Floor Slope

for mechanically cleaned tank

Clariflocculator

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.

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 coagulation & sedimentation processes are effectively incorporated in a single unit in the

Clariflocculator. Sometimes clarifier & Clariflocculator are designed as separate units.

Dimension Less Factor G t

Power Consumption

Total Area of Paddles 25

Range of peripheral velocities of blades 0.6 m / sec.

Depth of Tank

Detention Time

Velocity of Flow 0.8 m / sec.

Design Criteria : ( Flocculator )

36 KW / MLD

Velocity of water at outlet chamber

Weir Loading

Velocity Gradient ( G )

Slope for Sludge Hopper

Outlet Velocity

= 555.48 + 2 % of 555.48

= 555.48 + 11.11

= 566.59 m 3 / hour

= 30 min.

= 30.0 sec - 1

= 0.1543 m 3 / sec.

=

0.1543 x 4 1/2

1 Π

450 mm

= 566.59 x 30 = 283.3 m ³

60

= 3.50 m

= 283.30 = 80.9 m ²

3.50

80.94 x 4 1/2

Π

= 0.45 m

10.20 m

Here :

P = N.m / s

μ =

G =

V =

= ( Π / 4 ) x 285.85 m ³

= 30 x 30 x 0.89 x 285.85 = 228.97

1000

C d = 1.8

ρ = 995 Kg / m 3 ( 25

0 C )

v =

ν = 0.4 ) = 0.1 m / sec.

Diameter of Inlet Pipes ( D P )

Provide a Tank Diameter of

Dimension Of Paddles :

Diameter of Flocculator ( D )

Velocity of water tip of blade = ( 0.25 x

Power Input =

Volume of raw water to which P is applied in m 3

Absolute or Dynamic Viscosity of Raw Water in N.s /

m 2

Temporal Mean Velocity Gradient in ( sec -1

)

Velocity of tip of blade = 0.4 m / sec.

1.0 m / sec.Assuming Velocity ( V )

Discharge ( Q )

= = 10.15 m

= 0.4434 m

Provide an influent pipes of diameter.

Design Of Flocculatior : Wall

Hence Diameter ( d ) =

»

Volume of flocculator

Provide a Water Depth

Plan Area of flocculator

Average Value of Velocity Gradient

Design Of Influent Pipe

Design Average Period

Power dissipated in watts i.e.

Detention Period

Say 10.20 m

( 10.2 x 10.2 ) x 3.5 =

V

Qd

V

QAVAQ

2

4

V

Qd

4

4

4

2 AddA

VGP 2

3

2

1 vAC Pd

» 228.97 3 x A P

» A P = 9.470 m ²

9.47 / ( A P ) x ( 9.11 ( 10.0 to 25 % )

10.5 m ²

A P = 10.5 / Π 10.12 %

( Which is

Provide 5 nos. of paddles

1 ( One ) 5 Paddles.

4

60

0.955 m Say 1 m

r =

Area = 555.48 / = 0.514 m ²

0.51 / ( Π x 10.20 )

= 0.016 m

( 0.25 x 3.5 ) = 0.875 m

8 %

0.29 say ( 0.3 +

= 4.691 m

= 40 m

= 555.48 x 24 = 333.29 m ²

40

D cf =

10.20 x 10.20 ) = 333.29

4

1333.152 + 104.04 1/2

Π

= 22.992 m

= Π x D cf 72.22 m

= 555.48 x 24

72.22

1500

Diametre of Clariflocculator

= ( 1 / 2 ) x 1.8 x 995 x ( 0.4 - 0.1 )

0.3 x 60 x 60

Depth below partition Wall =

25

Provide Slope for Bottom =

Surface Of Clariflocculator

Total Depth of Tank at Partition Wall =

x ( 10.2 - 0.75 ) x 3.5 x 100 =

Acceptable, hence O.K. )

Ratio of Paddles to C / S of Flocculator

10.2 - 0.75 ) x 3.5 x 100 =

Provide A P =

3.5 + 0.016 + 0.875 )

say 4.7 m

Design Of Clarifier

Assuming a Surface Overflow rate 3 / m

2 / Day

0.3 m / sec.

3.5 m height 0.7

Shaft will support

The Paddles will rotate at an rpm of

V = 2 x Π x r x N = 2 x 3.14 x r x ( 4 / 60 ) = 0.4

» r =

Distance of Paddle from C 1 of vertical shaft.

Let velocity of water below the partition wall between the flocculator & clarifier be

Weir Loading = 184.60 m 3

/ Day / m

According to manual of govt. of india, if it is well clarifier, then it can be exceed upto

Π x ( D cf 2 -

Say 23 m

Length of Weir = ( 3.14 x 23 ) =

=D cf

S.No.

1

2

3

4

5

6

7

8

9

10

11 Diameter of Clariflocculator 23.0 m

No. of Paddles

Distance of Paddle from C.L. of vertical Shaft 1.0 m

Slope of Bottom ( % ) 8 %

Total Depth of Partition Wall 4.7 m

5 nos.

Distance of Shaft from C.L. of Flocculator

Paddles Rotation (RPM) 4

Diameter Of Influent Pipes 0.45 m

Overall Depth of Flocculator 3.5 m

Diameter of Tank 10.20 m

Summary

Particulars Values

Detention Period 30 min.

a)

= 5 to 7 m

=

0.5 m

= 2.6 m

= 0.45 to 0.7

= 1.30 to 1.7

= 2.55 to 2.65

= 2

= 0.6 to 0.75

= 1 to 2 m

b)

= 555.48 m

= 2 %

= 30 min.

= 5 m

= 1.25 to

=

= 13 mm

c)

= 555.48 m

= 24

23.5

= 578.64 m

578.64

5

Using 2.0 Units

116

2

Length x Width = L x 1.25 L = 58.0 m ²

58 1/2

1.25

= 6.8 m

Width = ( 1.25 x 6.8 ) = 8.5 m

Provide 2 Filters 8.6 m x 6.8 m

Design Of Rapid Gravity Filter

Minimum Overall Depth Of Filter Unit Including a Free Board of

Length L

Design Calculations

3 / hour

=Plan Area For Filter = 115.7 m

( > ) 3.0 %

Minimum Number Of Units

Depth Of Sand

Standing Depth of water over the filter

Free Board is less than ( > ) 0.5 m

Time Lost During Backwash

Say 8.6 m

Units, each with a dimension of =

1.33 : 1

116 m ²

=

555.48 x ( 1 + 0.02 ) x

»

Water Flow Required 3 / hour

Design Flow for Filter

3 / m

2 / Day

= 58 m ²

Effective size Of Sand

Uniformity Co-efficient For Sand

Ignition Loss Should Not Exceed ( > ) 0.7 %

Design Criteria : ( Rapid Sand Filter )

Rate Of Filteration

Maximum surface area of One Bed

percent by weight

Specific Gravity

Wearing Loss is not greater than

Design Rate Of Filteration 3 / m

2 / Day

Length & Width Ratio

Under Drainage System

Size of Perforations

Central Manifold With Laterals

Problem Statement :

Net Filtered Water

Quantity of Backwash water used

3 / hour

² ≡

=Hence Plan Area of One Unit

Q x d 3 x h =

ℓ L

Where m

4

10000

h = 2.5 m

Q = 5 x 2 m 3 / m

2 / hr

d = 0.6 mm

4

ℓ 10000

ℓ = 46.04 cm

60.0 cm

40.0 mm

formula : P = 2.54 x

R = 12 mm ( 10 mm to 14 mm )

2 5 10 40

9.2 21.3 30.5 49

9.2 12.1 9.2 9

50.0 cm

= 8.6 m x 6.8 m

= 58.48 m ²

3

1000

²

= ( 3.0 x

= ( 3.0 x 1754.4 ) ²

= ( 2.0 x

= ( 2.0 x 5263.2 ) ²

= 10526.4 x 4

Π

= 20 cm

8.6 100.0

61.2 x 4

86 Units

( Terminal Head Loss )

B x 293223

Q, d, h & ℓ are in 3 / m

2 / hr, mm, m and m respectively

Using Hudson Formula -

Estimation Of Sand Depth :

It is checked against break through of floc.

( Assuming 100 % overload of filter )

( Mean Diameter )

0.6 x 0.6 x 0.6 x 2.5= x 293223

( Poor Response ) < Average degree of pre-treatment

Where :

Estimation Of Gravel & Size Gradation :

Assuming size gradation of 2.0 mm to at bottom using empirical

10 x »

{ or ( > ) greater than }

Hence provide depth of sand bed =

R x ( log d )

²

= 1754.40 cm

Total Cross Section Area of Laterals Area of perforation )

x 58.48=Total Area of perforation = 0.17544 m

Estimation Of Under Drainage System :

Plan Area of each filter

Hence provide depth of gravel.

The Units of L & d are cm & mm, respectively.

Depth (cm )

Increment

Size 20

40

9.5

B =Assume

Providing a commercially available diameter of = 100.0 cm

Assuming spacing for laterals

20

= 5263.20 cm

Area of Central Manifold Area of Laterals )

= 10526.40 cm

Diameter of Central Manifold = 115.80 cm

Number of perforations / laterals =

= 43 on either side =Number of Laterals

Π=D = 8.8 cm say 90 mm

2 2

= ( 1 x 6.8 ) - ( 1 x 1.0 ) = 2.9 m

2 2

» n x Π x ( 1.3 ) ² =

4

» n = 1322.43 say 1322

1322.0

86.0

2.90 x 100

16.0

Provide 16

= 36 m

= ( 36 x 58.48 ) = 2105.28 m

= 0.5848 m 3 / sec.

1.80 m for

6.8

1.8

0.5848

4.0

0.40 m

0.1462 = 1.376 x 0.40 x ( h ) 3/2

0.1462 2/3

1.376 x 0.40

Freeboard = 0.1 m

Provide 4.0 troughs of 0.5 m 0.50 m

= ( 900 + 2200 + 300 )

= 4500 mm

= 1.5 m

= 3.0 min.

= 1.5 x 3.0 x 8.6 x 6.8

= 263.16 m ³

Depth of filter box = ( depth of under drain + gravel + sand + water depth + free board )

500 + 600 +

Design Of Filter Air Wash :

Assume Rate at which air is supplied 3 / m

2 / min.

Duration of Air Wash

Total Quantity of air per unit bed

=h = 0.41 m

Wide &

Wash Water discharge for one filter

Let n be the total number of perforation of 13.0 mm diameter

Total Area of perforation

say 0.5

= 18.13 cm C / C

perforations of 13.0 mm diameter at 180 cm C / C

Computation Of Wash Water Troughs :

Wash Water Rate 3 / m

2 / hour

=Spacing of Perforation

Say 180.00 mm C / C

1754.40

= 15.37 say 16=Number of Perforation or Laterals

deep in each filter.

Total Depth Of Filter Box :

=Discharge per unit trough = 0.146 m 3 / sec.

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

3 / hour

Assuming a spacing of wash water trough which will run parallel to the longer

dimension of the filter unit.

=Number of trough = 3.78 say 4

Length of One lateral = ( 1 ) width of filter - ( 1 ) diameter of manifold

2

3

376.1 hbQ

S.No.

1

2 8.60 m x 6.8 m

3

4

5

6

7

8

9

10

11

12 0.40 m x 0.5 m

13

14

15 263.2 m ³

Duration of Air Wash 3 min.

Total Quantity Of Air Required Per Unit Bed

Diameter Of Perforation

Diameter Of Laterals 90 mm

Number Of Perforations 16

Number Of Trough 4

Size Of Trough

Total Depth Of Filter Box 4500 mm

Depth Of Gravel 50.0 cm

13 mm

Diameter Of Central Manifold 100.0 m

Spacing For Laterals 20 cm

Number Of Laterals 86 Units

Summary

Particulars Values

Number Of Units 2 min.

Size Of Unit

Depth Of Sand Bed 60.0 cm

a)

b)

= 1.4 mg / L

= 1 mg / L

= 0.6 mg / L

= 0.1 to 0.2 mg / L

= 20 to 30 min.

c)

2 p.p.m.

= 13.33 x 1000000 x 1.40 x 1

1000000

= 18.662 Kg

= ( 18.66 x 180 )

16.0 Kg ) = ( 3359.16 x 2 ) = 419.895

16

= 2 Cylinders of 16.0 Kg

d)

S.No.

1

2 2 Cylinders of 16.0 mNumber Of Cylinders required per day

Design Of Disinfection Unit

Number Of Cylinders used per day

Summary

Particulars Values

Chlorine required per day 18.662 Kg

Rate of Chlorine required, to disinfect water be =

Chlorine required Per Day

For 6 Months = 3359.16 Kg

Number of Cylinder ( One Cylinder contain

( Winter Season )

( Summer Season )

→ Residual Chlorine ( Minimum )

→ Contact Period

Design Calculations

Chlorination

→ Chlorine Dose

Disinfection should not only remove the existing bacteria from water but also ensure their

immediate killing even afterwards, in the distribution system.

( Rainy Season )

Design Criteria ( Chlorination )