Design of Slow Sand Filter Final

Adavisomanal and 30 Villages

Jal NIrmal Project

Design of Slow sand filter Adavisomanal & 30 Villages As per CPHEEO manual clause the economical cost design of slow sand filter to be adopted Total quantity of water per day to be treated

3600

Cum / day

Since slow sand filters work for 24 hrs a day the quantity to be filter per hour =

3600 24

=

150

Cum / per hour

The rate of filtration for slow sand varies between 0.1 M to 0.2 sedimentation process followed by Raw water storage tank (3 hrs - storage) has been provided, the rate of filtration in this can be adopted 0.150 M / hr. =

Therefore, The area of slow sand filter bed requirement

150 0.150

=

1000

M

per hour. Since

Sqm

As per CPHEEO manual 6 Nos of filter beds are to be provided. As per economical cost design (page - 241) l2 = n b

=

2A +

1

(n

+

1)

Where l = b = A = n =

l

2n A

=

n

l

b

As surface works out Less than 1200 Sqm. As per CPHEEO manual, number filter beds required is

Therefore, l

l

A

=

8

l

x

2

=

2 8

x +

1

2

=

l = Say = Therefore, A = 1000 b = b = Say b

Detalied Scheme Report

Length of wall breadth of wall Total surface area of filter bed Nos of filter beds

=

8

But provide

8 Nos.

b 1000

2000 = 9 14.90711985 15.00 8 x l =

5

222.22222222

M x

x 15.00 1000 15 x 8 8.3333333333 8.5 M

b x

b

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Adavisomanal and 30 Villages

Jal NIrmal Project

Therefore,Area provided is 15.00 x 8.5 Therefore, Each bed will have Length breadth = 8.5

Surface area of each bed

m =

8

No of Beds M

M

=

Total Surface area of

8 15

127.5

bed

=

Sqm 127.5

x

8

x

7

=

1020

>

1000 Sqm

Hence Ok Cheak for Over loading Let us assume 1 Filter bed is Washed at a time Filteration capacity of each filter

= 15

X

150 8.5

= 0.2 Cum /hr

= .2 Cum /hr

Hence Ok The schematic lay out of slow sand filter bed be as under Design of Filter Bed

Let filter bed be as under with over all depth of filter bed =

2.70 mt

0.20mt Free board

1.0mt supernatant water 2.7

1.0mt Initial loading of filter Sand

0.40 mt Gravel

0.10 mt Under drain

Sand Spcification Size of sand

= 0.2- 0.3 mm

Uniformity co-eff =

5

Graval Spcificat = 5.0 layers (0.4 m) Size of garval

= 2.0 mm 40 mm

Size of garval i

= 2.0

5 10

depth in mm

= 80

80

Detalied Scheme Report

20

40

80 80

80

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Adavisomanal and 30 Villages

Jal NIrmal Project

Design of under Drain Area of each filter

=

15.00 x

8.5

ratio of area of perforation to area of filter Provide area of perforation

=

=

127.5

Sq Mts

0.15% to 0.3 %

= 0.15 / 100 x 127.5 =

0.19125

Sqm

1912.5 Sqcm

Ratio of total perforation to total C/S area of latral for perforation of 12 mm

= 2 to 1 % cross sectional area

Total Cross sectional area of latrals

= =

Area of central manifold

=

1.5

=

1.5

x

1913

2868.8 Sqcmt

x

2868.75

4303.125 Sqcmt

Cross section of center manifold with depth of 100 cmt therefore width of central manifold

=

4303 Cm

Provide central manifold of 60 cmt depth ,75 cmt wide Design of Lateral:Assuming 63 mm dia of PVC cmt latral Now internal Dia of 63 mm dia PVC Pipe = No. of Laterals =

58.1 mm

Total Area of laterls C/S Area of one Lateral =

28.6875

=

110 Nos.

0.785x0.05812 Spacing of Laterals=

15

=

55 Length of Laterals on each side =

0.273 m Say

250mm C/C

Width of filter bed - Width of Manifold 2 =

8.5 -

0.75

=

3.875m

2 Which is Less Than 3.78(60x.063=3.78m)

Hence Ok

Design of Orifice:Area of Orifice =

0.2

X

15

X

8.5

100 =

0.19125

Sqm

Adopting Spacing Of 18 cm C/C Total Area of Orifice on each Laterals =

3.875

= No. of Orifice = 0.1913 110 Diameter of Orifice d=

x

0.19125

100

387.5 Cm

100 x x

x

100

=

9 Nos.

2 x

4

=

0.49 Cm

3.142

Detalied Scheme Report

Niketan Consultants

Adavisomanal and 30 Villages

Jal NIrmal Project

Thus provide 9mm Orifice at 180 mm C/C

Detalied Scheme Report

Niketan Consultants

Adavisomanal And 30 Villages

Jal Nirmal Project

Design Of Sedimentation tank Total demand With 20 hrs pumping

= = =

Detention provide =

3600 Cum/Day 180 Cum/Day 3

Hrs

Capacity Sedimentation Tank =

3X

180

= 540 Cum

If two units provided capacity of each unit

540 2

=

= 270

Cum

with average water depth in tank as 3.0 mts, Surface area of tank

=

270 3

= 90

Sqm

Length to Width ratio is 1 : 3 =

3B2 = 90 Sqm B = 5.48 Say 5.5 M L = 3 x 5.50 = 16.5Mts So, Provide 16.5 Mts X 5.5 Mts of each unit Total Area Provided = 90.75 Sqm > 90 Sqm Ok Total area provided = 2 (16.5 x 5.5 x 3.0) = 544.5cum > 540 Cum Hence, OK 16.5 Mts

Unit 1 5.5 Mts

Unit 2

5.5 Mts

16.5 Mts

Check for Surface loading =

3600 90.75 x 2.0

=

19.840

Cum/Sqm/Day

<

36 Cum/Sqm/Day

Hence Ok Overall Depth of each Unit with 3.0 mts standing water depth (SWD)

= =

3 + 0.3 0 (FB) + 0.3 (Sluddge) + 3.6 Mts

Inlet And out let Weir Inlet weir (Influent channel with bottom openings ) keeping velocity of water in channel

=

0.3 m/sec

and keeping width of channel as

=

For discharge of 180 Cum/hr i.e.

=

Depth of channel in each unit

=

1/2 x

With FB of 20 cmt , total depth of weir be =

Detailed Scheme Report

0.3 m 0.05 Cum/sec 0.05 0.3 x 0.3 50 cmt

= 0.28 Mts Say 30 cmt

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Adavisomanal And 30 Villages

Jal Nirmal Project

The cross section of Influent channel = 0.3x 0.3 m Check for weir loading Length of wier in eah unit = 5.5 Mts Flow per unit per day = 3600 /2 = 180 Cum /hr Therefore weir loading = 1800/ 5.5 = 327.27 cum / mt /day Between 100 -300 cum /mt / day As it ismore 300Cum/m/ day Let us incrseth length of weir to 6.5 m Now The Modified dimension for the Sedimentation Tank is Length of Sedimentation Tank = Area of C/s 90 13.846 = = Breadth 6.5 Say 14 M Check for weir loading Length of wier in eah unit = 6.5m Flow per unit per day = 3600 /2 = 180 Cum/hr Therefore weir loading = 1800/ 6.5 = 276.93 cum / mt /day Between 100 -300 cum /mt / day Hence Ok 3600 Check for Surface loading = = 19.780 Cum/Sqm/Day < 36 Cum/Sqm/Day 91 x 2.0 Hence Ok Thus the Final Dimension Are = 2 Sets 14 m x 6.5m 14.0 Mts 6.5 Mts

6.5 Mts 14.0 Mts

Bottom floor slope 1 : 12 For Sludge removal, let is divide the Box Slab in each tank into 2 halves longitudinally and slope be provided in word to make into to two hopper of 30 x 30 m in the center and sludge removed by by sludge pipe. To economies construction cost, let the tanks be constructed back to back longitudinally to ensure equal flow into sedimentation tank a distribution box with rectangular weir and flush type inlet valve to each tank.

For uniform dispersal of flow of water provide perforated baffle wall in front of inlet pipe at 1.2 M away from the wall.

Hydraulic Design of baffle wall Flow in each Tank = 180 Cum/hr = 0.05 Cum/Sec Velocity of Flow of flow of water for dispersal = 0.15 /Sec Area of opening required – in baffle wall =

0.15

= 0.333 Sqm

3 0.15 Opening below baffle wall for movement of sludge and water, provided 45 cm considering sludge deposit of 25 cm before flushed one. Therefore opening available below for flow of water (45-25)_ 20 Cm The baffle wall is supported over 20 x 20 Cm columns 3 Nos. Hence length of opening available Area of opening available

=

= 6.25 - (0.2x3) = 5.65 M 5.65 x 0.2 = 1.13 Sqm.

Hence bottom opening adequate. But to make proper dispersal at higher level

Detailed Scheme Report

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Adavisomanal And 30 Villages

Jal Nirmal Project

Let us provide 15 x 15 Cm Square opening in the wall as under.

Detailed Scheme Report

Niketan Consultants

Adavisomanal and 30 Villages

Jal Nirmal Project

Design of Pipe from Raw Water Sump To Distribution Box No. of Filter Beds= 8 Nos Qty of water to be Treated= 150 Cum/Hr Qty of water to be Treated by pipe= 150 Cum/Hr 0.04167 Cum/Sec = Assuming the Velocity of flow In the pipe as= 0.5 M/sec Dia of pipe= 0.04167 x 4 0.5 x 3.142 0.3257 = Provide 250 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= 0.04167 0.785 X 0.25 X 0.25 0.85 M/sec = Hence Ok Calculation Of Head Losses in pipe: Length of pipe= 30 m Hazen Willam Constant= 100 Area of Cross Section in pipe A= 3.142 X 250 X 250 = 4 Wetted perimeter = 3.142 x 250 785.5 mm = Hydraulic Mean Radius r = 62.5 mm 0.0625 m = Now, Accroding to Hazen Willams Formula V= Q/A=0.854XCX r.63xS0.54 0.85 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.164 m Out let RL at Raw water Sump= 595.2 M inlet RL at Distribution Chamber= 595.036 M

49094 mm2

0.63

0.0625

x

0.54

S

Design Of interconnecting From Distribution tank(Inlet Equidistrbution Retangular Weir) For Slow sand Filters Since Separate pipes Are Tobe Give From Distribution Tank to SSF Dischage In each Filter= 0.00521 Cum/Sec Assuming the Velocity of flow In the pipe as= 0.5 M/sec Dia of pipe= 0.00521 x 4 0.5 x 3.142 0.1152 = Provide 100 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= 0.00521 0.785 X 0.1 X 0.1 0.66 = Hence Ok Calculation Of Head Losses in pipe: For SSF 1,SSF 4 on either side Length of pipe= 30 m Hazen Willam Constant= 100 Area of Cross Section in pipe A= 3.142 X 100 X 100 = 7855 mm2 4 Wetted perimeter = 3.142 x 100 314.2 mm = Hydraulic Mean Radius r = 25 mm 0.025 m = V= Q/A=0.854XCX r.63xS0.54 Now, Accroding to Hazen Willams Formula 0.63 0.54 0.66 0.854 x 100 x 0.025 x S = S 0.00498 m/m = Frictional Losses= 0.16434 m Out let RL at Distribution Box= 595.03 M inlet RL at Distribution Chamber= 594.86566 M

Detailed Scheme Report

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Adavisomanal and 30 Villages

Jal Nirmal Project

For SSF 2,SSF 3 on either side Length of pipe= 15 m Hazen Willam Constant= Area of Cross Section in pipe A=

100 3.142 X

100 X

100

=

4 Wetted perimeter = 3.142 x 100 314.2 mm = Hydraulic Mean Radius r = 25 mm 0.025 m = Now, Accroding to Hazen Willams Formula V= Q/A=0.854XCX r.63xS0.54 M 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.08217 m Out let RL at Distribution Box= 595.03 M inlet RL at Distribution Chamber= 594.948 M

7855 mm2

0.63

0.025

x

0.54

S

Design Of interconnecting From Slow sand Filters Out let to Rure Water Sump One of Slow Sand Filter = 8 Nos. Qty of water to be Treated= 143.75 Cum/Hr Qty of water to be Treated by each pipe= 17.9688 Cum/Hr 0.00499 Cum/Sec = a. From SSF 1 to Junction SSF2 Since pipes Are inter connceted from SSF 1 to SSF2 Discharge in pipe = 0.0049913 Cum/Sec Assuming the Velocity of flow In the pipe as= 0.5 M/sec Dia of pipe= 0.00499 x 4 0.5 x 3.142 0.1127 = Provide 100 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= 0.00499 0.785 X 0.1 X 0.1 0.636 M/sec = Hence Ok Calculation Of Head Losses in pipe: Length of pipe= 15 m Hazen Willam Constant= Area of Cross Section in pipe A=

100 3.142 X

100 X

100

=

4 Wetted perimeter = 3.142 x 100 314.2 mm = Hydraulic Mean Radius r = 25 mm 0.025 m = .63 0.54 Now, Accroding to Hazen Willams Formula V= Q/A=0.854XCX r xS 0.64 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.08217 m Out let RL at Slow Sand Filter = 593.88 M RL at Outlet of SSF 2 = 593.798 M

b. From SSF 2 to Junction SSF3 Since pipes Are inter connceted from SSF 2to SSF3 Discharge in pipe = 0.0099826 Cum/Sec Assuming the Velocity of flow In the pipe as= Dia of pipe= =

Detailed Scheme Report

0.5 M/sec 0.00998 0.5 0.1594

x x

7855 mm2

0.63

0.025

x

0.54

S

4 3.142

Niketan Consultants

Adavisomanal and 30 Villages

Jal Nirmal Project

Provide 150 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= =

0.00998 0.785 X 0.15 0.565 M/sec Hence Ok

X

0.15

Calculation Of Head Losses in pipe: Length of pipe= 15 m Hazen Willam Constant= Area of Cross Section in pipe A=

100 3.142 X

150 X

150

=

4 Wetted perimeter = 3.142 x 150 471.3 mm = Hydraulic Mean Radius r = 37.5 mm 0.0375 m = .63 0.54 Now, Accroding to Hazen Willams Formula V= Q/A=0.854XCX r xS 0.57 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.08217 m RL at Outlet of SSF 2 = 593.798 M RL at Outlet of SSF 3= 593.716 M c. From SSF 3 to Junction SSF4 Since pipes Are inter connceted from SSF3 to SSF4 Discharge in pipe = 0.014974 Cum/Sec Assuming the Velocity of flow In the pipe as= Dia of pipe= =

0.5 M/sec 0.01497 0.5 0.1953

x x

4 3.142

X

0.15

17674 mm2

0.63

0.0375

x

0.54

S

Provide 150 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= =

0.01497 0.785 X 0.15 0.848 M/sec Hence Ok

Calculation Of Head Losses in pipe: Length of pipe= 30 m Hazen Willam Constant= Area of Cross Section in pipe A=

100 3.142 X

150 X

150

=

4 Wetted perimeter = 3.142 x 150 471.3 mm = Hydraulic Mean Radius r = 37.5 mm 0.0375 m = .63 0.54 Now, Accroding to Hazen Willams Formula V= Q/A=0.854XCX r xS 0.85 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.16434 m Out let RL at Raw water Sump= 593.716 M inlet RL at Distribution Chamber= 593.551 M

Detailed Scheme Report

17674 mm2

0.63

0.0375

x

0.54

S

Niketan Consultants

Adavisomanal and 30 Villages

Jal Nirmal Project

d. From SSF 4 to pure Water Sump Since pipes Are inter connceted from SSF3 to SSF4 Discharge in pipe = 0.0199653 Cum/Sec Assuming the Velocity of flow In the pipe as= Dia of pipe= =

0.5 M/sec 0.01997 0.5 0.2255

x x

4 3.142

X

0.2

Provide 200 mm Dia CI Pipe Now, Velocity in pipe V=Q/A V= =

0.01997 0.785 X 0.2 0.636 M/sec Hence Ok

Calculation Of Head Losses in pipe: Length of pipe= 30 m Hazen Willam Constant= Area of Cross Section in pipe A=

100 3.142 X

200 X

200

=

4 Wetted perimeter = 3.142 x 200 628.4 mm = Hydraulic Mean Radius r = 50 mm 0.05 m = .63 0.54 V= Q/A=0.854XCX r xS Now, Accroding to Hazen Willams Formula 0.64 0.854 x 100 x = S 0.00498 m/m = Frictional Losses= 0.16434 m RL at Outlet of SSF 3 = 593.551 M RL at inlet Level of Sump = 593.387 M

Detailed Scheme Report

31420 mm2

0.63

0.05

x

0.54

S

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