Calculations

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Part of Structure Calc. Sheet No.

BOD & Hydraulic Loading Calculations

Drawing Ref. Calculations By Checked By Date

Donal O' Connor 3/14/2008

Ref. Calculations Output

General Details

Current Population = 1400 people Predicted Population in 20 years = 2080 people Allow for 60g BOD / person / day

Allow for 250 l wastewater / person / day

→ Max. BOD Loading = (2080 x 0.06) = 124.8 kg → Min. BOD Loading = (1400 x 0.06) = 84 kg → Max. Hydraulic Loading = (2080 x 0.25) = 520 → Min. Hydraulic Loading = (1400 x 0.25) = 350

Calculation of PE in Riversdale Hotels:

One bed = 1 P.E.

Non-local workers = 0.4 P.E.

Meals = 0.1 P.E.

Sheen Falls Hotel: PEAK Beds = 50 x 1 = 50 Non-local workers = 4 x 0.4 = 1.6 Meals = 100 x 0.1 = 10 Total = 61.6 P.E. OFF - PEAK

The Hotel closes during off-peak times

m3

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The Park Hotel: PEAK Beds = 46 x 1 = 46 Non-local workers = 3 x 0.4 = 1.2 Meals = 80 x 0.1 = 8 Total = 55.2 P.E. OFF - PEAK Beds = 12 x 1 = 12 Non-local workers = 0 = 0 Meals = 12 x 0.1 = 1.2 Total = 13.2 P.E.

Riversdale House Hotel: PEAK Beds = 30 x 1 = 30 Non-local workers = 3 x 0.4 = 1.2 Meals = 60 x 0.1 = 6 Total = 37.2 P.E. OFF - PEAK Beds = 8 x 1 = 8 Non-local workers = 0 = 0 Meals = 8 x 0.1 = 0.8 Total = 8.8 P.E.

→ Total Max. = 154 P.E.

→ Total Min. = 22 P.E.

Guesthouses:

Guest-house occupants = 0.8 P.E.

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Min. occupancy = 10 x 0.8 = 8 P.E.

Restaurants: Meals = 0.1 P.E. Mulcahy's: PEAK Meals = 180 x 0.1 = 18 P.E. OFF - PEAK Meals = 70 x 0.1 = 7 P.E. The Old Dutch:

PEAK

Meals = 120 x 0.1 = 12 P.E. OFF - PEAK

The restaurant closes during off-peak times

→ Total Max. = 30 P.E.

→ Total Min. = 7 P.E.

Hospital & Day Care Centre:

No. of Hospital beds = 35

1 bed = 2 P.E.

Hospital occupants = 0.4 P.E. No. of Hospital staff = 28 No. of Day care children = 30 Day care occupants = 0.3 P.E.

→ Total Max. = ( 35 x 2 ) + ( 28 x 0.4 ) + ( 30 x 0.3 )

(4)

Schools:

School children = 0.3 P.E.

Boys Primary School = 200 pupils Girls Primary School = 220 pupils

Secondary School = 620 pupils ( 366 from outside Riversdale ) → Total Max. = ( 200 x 0 ) + ( 220 x 0 ) + ( 366 x 0.3 )

→ Total Max. = 109.8 P.E.

Industrial Estate:

Total no. of workers = 80 Workers outside town = 72 Workers outside town = 0.4 P.E. Max. no. of workers = 300 Max. no. from outside = 270

→ Total Max. = ( 270 x 0.4 ) = 108 P.E. → Total Min. = ( 72 x 0.4 ) = 28.8 P.E.

Food Industry:

→ Max. BOD Loading = 46 kg / day → Min. BOD Loading = 29.02 kg / day → Max. Hydraulic Loading = 130

→ Min. Hydraulic Loading = 82

Calculation of BOD and Hydraulic Loading in Riversdale

→ Total Max. = 2644 P.E.

→ Total Min. = 1556 P.E.

Allow for 60g BOD / person / day

m3_{/ day}

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Max. BOD Loading:

= ( 2644 x 0.06 ) + BOD Loading from Food Industry = ( 2644 x 0.06 ) + 46

= 204.64 kg BOD / day Min. BOD Loading:

= ( 1556 x 0.06 ) + BOD Loading from Food Industry = ( 1556 x 0.06 ) + 29.02

= 122.38 kg BOD / day Max. Hydraulic Loading:

= ( 2644 x 0.25 ) + Hydraulic Loading from Food Industry = ( 2644 x 0.25 ) + 130

= 791

Min. Hydraulic Loading:

= ( 1556 x 0.25 ) + Hydraulic Loading from Food Industry = ( 1556 x 0.25 ) + 82

= 471

Table of Results:

BOD Loading Hydraulic Loading kg BOD / day

Max. Min. Max. Min.

Population 124.80 84.00 520.00 350.00 Hotels 9.24 1.32 38.50 5.50 Guesthouses 4.32 0.48 18.00 2.00 Restaurants 1.80 0.42 7.50 1.75 Hospital 5.41 5.41 22.55 22.55 Schools 6.59 0.00 27.45 0.00 Industrial Estate 6.48 1.73 27.00 7.20 Food Industry 46.00 29.02 130.00 82.00 Total 204.64 122.38 791 471 m3_{/ day} m3_{/ day} m3_{/ day}

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Assessment of Capacity of Existing Plant

Primary Settling Tank

Length and Breath of Tank = 4 m

→ Area = 16 Velocity = 1.2 m / hour Velocity = 3 DWF Area → 3 DWF = 19.2 → 1 DWF = 6.4

P.E. =

→ P.E. = 614.4 ppl 614 ppl

BOD =

→ BOD = 36.86 kg / day 36.86 kg/day

H. Loading =

→ H. Loading = 153.60 153.60

Height of Pyramidal Section = 8.5 - 4.6 = 3.9 m Volume of Pyramidal Section = 1 / 3 base x height

= 20.8

Volume of Sludge = 1 / 3 of pyramidal section = 6.93 Retention time of 2 hours in the primary settling tank:

Volume of liquid above sludge = 3 DWF @ 2 hours = 38.4

Total Vol. of Liquid = 2 / 3 Vol. of Pyramid + Vol. of Top Section of Tank Vol. of top Section of Tank = 24

Total Volume of Liquid = 37.87

This equates to 2 hours of DWF = 6 x DWF m2 m3_{/ hour} m3_{/ hour} m3_{/ day} _m3_/day m3 m3 m3 m3 m3

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→ DWF = 6.31

P.E. =

→ P.E. = 606 ppl 606 ppl

BOD =

→ BOD = 36.35 kg / day 36.35 kg/day

H. Loading =

→ H. Loading = 151.47 151.47

Trickling Filter

Diameter of Trickling Filter = 10.25 m Depth of Trickling Filter = 2 m

Vol. of Trickling Filter = = 165

= 19.8 kg BOD

Assuming: 45 % removal of BOD in the Primary Settling Tank → 55 % of BOD enters the Tricking Filer

P.E. =

→ P.E. = 600 ppl 600 ppl

BOD =

→ BOD = 36 kg / day 35.99 kg/day

H. Loading =

→ H. Loading = 150 149.95

Humus Tank

Diameter of Humus Tank = 4.3 m

Depth of Humus Tank = 2 m

Vol. of Humus Tank = = 29

Retention Time = 1.5 hrs

Flow Through Tank = 19.35 ( 3 DWF )

→ 1 DWF = 6.45

m3_{/ day}

m3_{/ day} _m3_/day

Π r2_{x d} _m3

The filers can remove 0.12 kg BOD / m3

m3_{/ day} _m3_/day

Π r2_{x d} _m3

m3 _{/ hour}

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P.E. =

→ P.E. = 619 ppl 619 ppl

BOD =

→ BOD = 37.16 kg / day 37.16 kg/day

H. Loading =

→ H. Loading = 154.82 154.82

Upward Velocity in Tank = 1.3 m / hour

Plan Area of Tank = 14.51

Vol. of Wastewater entering tank = 18.87 (3 DWF)

→ 1 DWF = 6.29

P.E. =

→ P.E. = 604 ppl 604 ppl

BOD =

→ BOD = 36.23 kg / day 36.23 kg/day

H. Loading =

→ H. Loading = 150.95 150.95

Table of Results:

P.E. BOD Loading Hydraulic Loading kg BOD / day

Primary Settling Tank 614 36.86 153.60

2 Hours @ 3DWF 606 36.35 151.47

Trickling Filter 600 36.00 150.00

Humus Tank 619 37.16 154.82

1.5 hr Retention time 604 36.23 150.95

From the above figures it can be clearly seen that the existing Treatment Plant is grossly overloaded and entirely inadequate to deal with the waste being produced by the town of Riversdale.

m3_{/ day} _m3_/day m2 m3_{/ hour} m3_{/ hour} m3_{/ day} _m3_/day m3 _{/ day}

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Sewer Network Design

Catchment Area 1

Area = 0.3 ha

Length of Sewer Pipe = 600 m

Slope = 1 in 231

Choose Pipe Diameter = 225 mm

Flow Chart:

Velocity = 0.83 m / s

Pipe Capacity = 32 l / s

Time of Travel = 12 min

Time of Concentration = Time of Entry + Time of Travel

= 16 min

Figure 5

Return Period of 2 years → Intensity = 34.8 mm

Runoff = 2.78 x A x I = 29.0 l / s

Foul Flow = 791 = 9.15 l / s

Pipe must cater for 2.5 DWF

→ 2.5 DWF = 22.9 l / s

% of Total Foul in Pipe = 11 %

→ Total flow through Pipe = 31.5 l / s 31.5 l / s < 32 l / s

Pipe diameter =

→ Selected pipe is adequate 225 mm

Area = 0.15 ha

Slope = 1 in 156

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Flow Chart:

Time of Travel = 4.1 min

= 8.1 min

Figure 5

Runoff = 2.78 x A x I = 21.4 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Area = 0.2 ha

Slope = 1 in 244

Flow Chart:

Pipe Capacity = 31 l / s m3_{/ day}

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= 11.1 min

Figure 5

Runoff = 2.78 x A x I = 24.1 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

→ Total flow through Pipe = 26 l / s 26 l / s < 31 l / s

Pipe diameter =

Flow at Manhole 4

There are 3 possible TOT's applying T.O.T. 1 = 12 mins

All Areas contribute totally → Area = 0.65 ha

→ Intensity = 34.8 mm (TOC = 16mins)

→ Runoff = 62.9 l / s

Total Foul Flow = 25 % x 22.9 l / s

= 5.7

Total Flow at MH 4 = 68.6 l / s

T.O.T. 2 = 4.1 mins

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Area 1 = ( 4.1 / 12) x 0.3 = 0.1 ha Area 2 = ( 4.1 / 4.1) x 0.15 = 0.15 ha Area 3 = ( 4.1 / 7.1) x 0.2 = 0.12 ha

Total Area = 0.4 ha

→ Intensity = 51.3 mm (TOC = 8.1mins)

→ Runoff = 52.5 l / s

= 5.7 Total Flow at MH 4 = 58.2 l / s T.O.T. 3 = 7.1 mins Area 1 = ( 7.1 / 12) x 0.3 = 0.18 ha Area 2 = ( 7.1 / 7.1) x 0.15 = 0.15 ha Area 3 = ( 7.1 / 7.1) x 0.2 = 0.20 ha Total Area = 0.53 ha

→ Runoff = 63.6 l / s

= 5.7

→ Total Flow at MH 4 = 69.4 l / s

Area = 0.25 ha

Slope = 1 in 250

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= 20.1 min

Figure 5

Runoff = 2.78 x A x I = 75.6 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

→ Total flow through Pipe = 84.5 l / s 84.5 l / s > 32 l / s

→ Selected pipe is inadequate

Flow Chart:

Time of Concentration = Time of Entry + Time of Travel + 12mins

= 27.9 min

Figure 5

Return Period of 2 years → Intensity = 1200 / ( t + 19 )

→ Intensity = 25.6 mm

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Runoff = 2.78 x A x I = 64.03 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Area = 0.5 ha

Slope = 1 in 250

Flow Chart:

= 17.0 min

Figure 5

Runoff = 2.78 x A x I = 46.6 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

m3_{/ day}

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Pipe diameter =

Area = 0.15 ha

Slope = 1 in 179

Flow Chart:

= 18.8 min

Figure 5

Runoff = 2.78 x A x I = 56.7 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

→ Total flow through Pipe = 62.5 l / s m3_{/ day}

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62.5 l / s < 75 l / s

Pipe diameter =

There are 2 possible TOT's applying T.O.T. 1 = 25.9 mins

→ Runoff = 110.3 l / s

= 14.7 l / s Total Flow at MH 7 = 124.9 l / s T.O.T. 2 = 16.8 mins Area 1 = ( 2.9 / 12) x 0.3 = 0.07 ha Area 2 = ( 2.9 / 4.1) x 0.15 = 0.11 ha Area 3 = ( 2.9 / 7.1) x 0.2 = 0.08 ha Area 4 = 0.25 ha Area 5 = 0.50 ha Area 6 = 0.15 ha Total Area = 1.16 ha

→ Runoff = 101.3 l / s

= 14.7 l / s

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Area = 0.2 ha

Slope = 1 in 160

Flow Chart:

2 + 12 + 13.9 + 5.1

= 33.0 min

Figure 5

Runoff = 2.78 x A x I = 112.21 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

→ Selected pipe is adequate

2 + 13 + 3.8 + 5.1

= 23.9 min

Figure 5

Return Period of 2 years → Intensity = 1200 / ( t + 19 ) Time of Concentration 1 =

m3_{/ day}

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Runoff = 2.78 x A x I = 66.05 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Note: The total flow to be used is the larger flow = 130.51 l / s

Area = 0.15 ha

Slope = 1 in 136

Flow Chart:

= 8.5 min

Figure 5

Runoff = 2.78 x A x I = 20.7 l / s

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Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Area = 0.3 ha

Slope = 1 in 229

Flow Chart:

Time of Concentration = Time of Entry + Time of Travel + 4.5mins

= 16.2 min

Figure 5

Runoff = 2.78 x A x I = 43.3 l / s

Foul Flow = 791 = 9.15 l / s

m3_{/ day}

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→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Area = 0.2 ha

Slope = 1 in 182

Flow Chart:

= 7.6 min

Figure 5

Runoff = 2.78 x A x I = 29.2 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

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Pipe diameter =

There are 2 possible TOT's applying T.O.T. 1 = 31 mins

→ Intensity = 23.1 mm (TOC = 33mins)

→ Runoff = 153.9 l / s

= 22.9 l / s Total Flow at MH 11 = 176.8 l / s T.O.T. 2 = 21.9 mins 21.9 - 5.1 - 13.9 = 3 mins Area 1 = ( 3 / 12) x 0.30 = 0.08 ha Area 2 = ( 3 / 4.1) x 0.15 = 0.11 ha Area 3 = ( 3 / 6.8) x 0.20 = 0.09 ha Area 4 = 0.25 ha Area 5 = 0.50 ha Area 6 = 0.15 ha Area 7 = 0.20 ha Area 8 = 0.15 ha Area 9 = 0.30 ha Area 10 = 0.20 ha Total Area = 2.02 ha

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(Taken as same proportion of area)

= 19.6 l / s Total Flow at MH 7 = 176.8 l / s T.O.T. 3 = 14.2 mins 14.2 - 5.1 - 3.8 = 5 mins (Area 5) 14.2 - 5.1 = 9.1 mins (Area 4) Area 4 = ( 9.1/13.9) x 0.25 = 0.16 ha Area 5 = ( 5 / 13) x 0.5 = 0.19 ha Area 6 = 0.15 ha Area 7 = 0.20 ha Area 8 = 0.15 ha Area 9 = 0.30 ha Area 10 = 0.20 ha Total Area = 1.36 ha

→ Runoff = 130.4 l / s

(Taken as same proportion of area)

= 13.8 l / s

→ Total Flow at MH 11 = 176.8 l / s

Area = 0.15 ha

Slope = 1 in 80

Flow Chart:

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= 34.8 min

Figure 5

Runoff = 2.78 x A x I = 158.25 l / s

Foul Flow = 791 = 9.15 l / s

→ 2.5 DWF = 22.9 l / s

Pipe diameter =

Summary Table

Sewer Pipe Length Diameter Gradient

(m) (mm) (1 in) MH 1 - MH 4 600 225 231 MH 2 - MH 4 250 225 156 MH 3 - MH 4 330 225 244 MH 4 - MH 7 900 375 250 MH 5 - MH 6 750 300 250 MH 6 - MH 7 250 300 179 MH 7 - MH 11 400 375 160 MH 8 - MH 9 300 225 136 MH 9 - MH 11 550 300 229 MH 10 - MH 11 200 225 182 MH 11 - MH Foul Sump 200 375 80 m3_{/ day}

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Pumping Station, Rising Main & Wet Well

Rising Main

Main should be able to cater for 6 DWF

6 DWF = 55 l / s

Slope = 1 in 250

Diameter = 250 mm

Cover = 1.2 m

Diameter of Inlet pipe = 375 mm Elevation height of sump = 6.5 m High level cut-in elecrode = 4.68 m Low level cut-in elecrode = 4.18 m

Horizontal distance = 700 m

Add 10% to cater for bends in the pipe

→ Horiz. distance = 770 m

Head loss due to Friction = Horiz. Distance / slope

= 3.08 m

Rising Main enters site of new Treatment Plant at an elevation of 9 m

Static Lift = 4.83 m

Allow for Station Losses = 1.5 m

Total Manometric Head = 9.41 m ~= 10 m

Power of Pumps: Power = Q H 3.67 r Q = 198 l / hr H = 10 m r = 40 % Power = 13.5kW (85 % Efficiency) Power = 15.9kW (100 % Efficiency)

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Retention Time in Rising Main:

Retention time should not be greater than 12 hours to prevent wastewater going septic.

Length of Rising Main = 770 m Radius of Rising Main = 125 mm Volume of Rising Main = 37.8

Wastewater flowing through Rising Main / day = 906.68 Time taken for a plug of sewage to flow through rising main

= 1.00 hrs

= 60 mins

Design of Wet Well

Pump will start approximately 8 times each hour

→ Cycle Time = 60 / 8

= 7.5 mins

~= 8 mins

Pumping Station must be capable of pumping 6 DWF

6 DWF = 55 l / s

55 l / s = 3.3

T = (4 * V) / Q

Q = Pumping Rate = 3.3

V = Capacity of Wet Well = 6.6

Depth of Wet Well = 0.5 m

Plan area of wet well = 6.60 13.2

0.50 m → Let Dimensions of Wet Well = 3m x 4.4m

m3 m3 _{/ day} m3 _{/ min} m3 _{/ min} m3 _{/ min} m3 ₌ _m2

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Design of New Wastewater Treatment Plant

New Treatment Plant must cater for:

Max. BOD Loading = 205 kg BOD / day

Min. BOD Loading = 120 kg BOD / day

Max. Hydraulic Loading = 791 Min. Hydraulic Loading = 471 Design of Oxidation Ditch

Range of operating conditions for the Oxidation Ditch

Minimum Maximum

F / M Ratio 0.05 0.08

MLSS 2200 mg / l 3500 mg / l

BOD 120 kg BOD / day 205 kg BOD / day Minimum BOD Loading :

F / M Ratio = 0.05

BOD = 120 kg BOD / day

→ 120 = 0.05 M → M = 2400 kg → Vol. of ditch = 1091 F / M Ratio = 0.08 → M = 3.5 x 1091 → M = 3818.5 kg → F = 286.4 kg ( Max. Capacity )

The Oxidation Ditch can cover the range of BOD from 120kg - 286.4kg Oxidation Ditch works best at F / M ratio = 0.06

→ 205 = 0.06 M m3_{/ day} m3_{/ day} m3 Capacity at Max. MLSS: (3.5 kg / m 3₎

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→ M = 3417 kg

→ MLSS = 3.13

This MLSS level is within the desirable range Sizing of the Oxidation Ditch :

Depth of Oxidation Ditch = 2 m

Area of the Ditch = Volume / Depth

= 545.45

Area of the Ditch =

= → 545.45 = → = 41.51 Radius of O.D. → r = 6.45 m = 6.45m Width of O.D.

Width of Oxidation Ditch = 2 x r = 12.9 m = 12.9m

Length of Wall

Length of Centre Wall = 5 x r = 32 m = 32m

Retention Time:

Max. Hydraulic Loading = 791

Retention Time in Tank = Volume / Loading

= 1.38 days

= 33 hours

This Retention Time is sufficient as it is greater than 24 hours Power of Rotor: 15 Watts: Power = 16.36 kW 20 Watts: Power = 21.82 kW kg / m3 m2 π r2_{+ ( 5r x 2r )} 13.14 r2 13.14 r2 r 2 m3_{/ day}

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Choose Mixing Power = 20 kW

Oxygen Required:

= 259 mg/l

% BOD removal in Oxidation Ditch = 92 %

Oxgen Content ( OC ) / BOD Content = 0.8 + 2.2 * (0.6 - F / M) F / M Ratio = 0.06

→ OC / BOD = 2

For every kg of BOD removed, 2 kg of Oxygen is required

Kg of BOD entering Ditch per hour = 205 = 8.5 kg BOD/ hour 24

→ 17 kg of Oxygen required per hour 1 kW will transfer 2 kg of Oxygen per hour

→ 8.5 kW is required to transfer Oxygen per hour Therefore an extra 11.5 kW is needed in the tank

→ Install an additional 4 x 3 kW submerged mixers Rotors :

Lane width in Oxidation Ditch = 6.2 m Immersion depth of Rotors = 80 mm From Fig. 16 (Page 36 of Notes)

80mm = 1.6 kg / m hour 17 kg of Oxygen required per hour

→ Length of Rotors req. = 17 = 10.6 m

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1.6 There are 4 rotors in the Oxidation Ditch

Length of Rotor

→ Approx. length of rotor = 2.50 m = 2.5m

Design of Final Settling Tank

Upward Velocity in Tank = 0.9 m / hour A flow of 3 DWF goes through tank every day

→ 1 DWF = 791

→ 3 DWF = 98.88

Area of Settling Tank = 3 DWF = 110

Velocity

Area =

→ = 110

Radius of FST

→ r = 5.92 m ~= 6 m = 6m

Allow 2 hours retention time

→ Volume =

= 197.75

Plan Area of Tank = 113.04 115

Depth of Tank = Volume / Plan Area = 1.72 m Depth of FST

~= 2 m = 2m

→ New Retention Time = 2.33 hours → New Volume = 230

Design of Sludge Holding Tank For every 1 kg BOD removed

→ 0.7 kg of dry solids generated m3_{/ day} m3_{/ hour} m2 Π x r2 Π x r2 _m2 2 hr x 98.88 m3_/hr m3 m2_~= _m2 m3

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Max. BOD Loading = 205 kg BOD / day In the Oxidation Ditch there is a 92% reduction in BOD

→ BOD Loading = 188.6 kg BOD / day

→ Dry Solids = 132.02 kg → Total removed = 6.6 7 days of storage = 7 x 6.6 = 46.21 Depth of SHT Depth of Tank = 3 m = 3m → Plan Area = 46.21 = 15.4 3 → r = 2.2 m

Phosphorus levels must be less than 1 mg / l

Wastewater contains approximately 10 mg / l of phosphorous The Sludge Holding Tank removes 60 % = 6 mg / l

→ 4 mg / l of phosphorous is left Need to remove 3 mg / l

→ = 790, 000 l

→ 790,000 x 3 = 2.37 kg of Phosphorous 2.2 kg of Fe is needed to remove 1kg of Phosphorous

→ 5.21 kg of Fe per litre

→ 5.214 x 2.5 = 13.04 kg of extra dry solids Total amount of dry solids = 132.02 + 13.04

= 145.06 kg → Total removed = 7.25 m3 m3 m2 790 m3 m3

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7 days of storage = 7 x 7.25 = 50.77

Depth of Tank = 3 m

→ Plan Area = 50.77 = 16.9

3

→ r = 2.32 m

The larger radius must be used

Radius of SHT

→ Radius of Sludge Holding Tank = 2.32 2.5m = 2.5m

BOD Level at Outflow

BOD of effluent and river water mixture = X + YZ Z + 1 where

X = BOD of effluent mg/l = 20 mg / l

Y = BOD or river water above outfall mg/l = 1.1 mg / l Z = Dilution factor (river water to effluent)

Z = Rate of flow of receiving water = 24

Rate of Discharge

BOD of river after discharge = X + YZ = 1.856 mg / l Z + 1

Increase in level of BOD in receiving waters = 0.756 mg / l 0.756 mg / l < 1 mg / l

→ This level of Effluent Discharge is Acceptable

m3

m2

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