Design Cal_2300PE.xls

(1)

DESIGN OF AN EXTENDED AERATION TREATMENT PLANT

2300

P.E.

CADANGAN PEMBANGUNAN BERCAMPUR DI ATAS TANAH KERAJAAN

SELUAS 30.0EKAR, MUKIM PEDAH, DAERAH JERANTUT, PAHANG DARUL MAKMUR

UNTUK TETUAN DOYENVEST (M) SDN BHD

BASIC DATA:

This is an outline design to produce 10:20 BOD5:SS effluent.

Population, PE = 2300 P.E

Dry Weather Flow, DWF = 0.225

Suspended Solid, SS = 300mg/l

BOD5, BOD = 250mg/l

Influent Ammonia NH3-N = 30mg/l

Effluent BOD5, EBOD = 10mg/l

Effluent SS ESS = 20mg/l

Effluent Ammonia ENH3-N = 10mg/l

LOADING

a. Hydraulic flow, Qavg = PE * DWF

= 517.50

b. Peak Flow, Qpeak = Peak Factor * Qavg

-0.11

Peak Flow Factor Pff = 4.7(P)

= 4.29

Therefore, Qpeak = Pff x Qavg

= 2219.31

= 1.54

c. Suspended Solids Loading Rate = Qavg * SS

= 155.25 kg/d

d. BOD5 Loading Rate = Qavg * BOD

= 129.38 kg/d

e. NH3-N Loading Rate = Qavg * NH3-N

15.53 kg/d m3_/c/d

m3_/d

m3_/d m3_/min

(2)

PRIMARY BAR SCREEN DESIGN

Design Population, PE 2300

Design Flow, Qavg 517.50 0.0060 l/s

Peaking Factor, PF 4.29

Peak Flow, Qpeak 2219.31 25.81 l/s

Guidelines : Quantity of screenings = 30 m3 screening / 10^6 m3 wastewater

Number of Channels = 1

Number of Back-up Channel = 1

Qavg = 517.50

Quantity of Screenings = Qavg x 30 / 1000000

= 0.0155

Provide storage for 7 days

Quantity of Screenings = 7 x Quantity of Screenings per day

= 0.109

Number of Storage Units = 1

Quantity per Unit = 0.109

Dimension of screenings trough

L = 0.50m

W = 0.30m

D = 0.30m

Volume = 0.045 > 0.109 error

Screen Design

Guidelines: Max flow through velocity at Qpeak Vmax = 1.0 m/ses Guidelines: Min approach velocity at Qpeak Vmin = 0.3 m/sec

Bar Size Bs = 10mm

Clear Opening Co = 25mm

Efficiency Coefficient Eff = Clear Opening

Clear Opening + Bar size

= 0.71

Clear area through each screen at Qpe AQpeak = Qpeak Vmax x 24 x 60 x 60

= 0.026

Total cross sectional area of channel AC = AQpeak Eff

= 0.036

Assume, Depth of Flow at Qpeak D = 0.050m

.

Required Width of Clear Opening @ Q Wclr = AC / D

= 0.72 m Number of Openings No = Wclr x 1000 Co = 29 m3_/day m3_/day m3_/day m3_/day m3 m3 m3 _m3 m2 m2

(3)

Number of Bars Nbars = No - 1

= 28

Gross Width of Screen Wch = Wclr + (Nbars x Bs / 1000)

= 1.00 m

Set Wch = 0.50m

Check velocities

Qpeak = 2219.31 0.02569 m3/sec

Velocity of approach Vapp = Qpeak / (Wch x D)

= 1.03 m/sec > = 0.30 m/sec ok

Velocity through screen Vscr = Qpeak / (Wclr x D)

= 0.71 m/sec < = 1.00 m/sec ok

Hydraulic Profile Through Screen

Headloss Through Clean Screen

Formula: HLc = .0729 (V^2 - v^2)

where HLc = ? Clean screen max headloss (m)

V = 0.71 Velocity through screen (m/s) v = 1.03 Approach velocity (m/s)

Therefore HLc = -0.04 m

Elevation of Channel Invert = 76.75m Upstream Water Elevation = 76.76 m Downstream Water Elevation = 76.80 m Headloss Through Half Clogged Screen

Assume velocity through screen is doubled

Formula: HLc = .0729 (V^2 - v^2)

where HLc = ? Clogged screen max headloss (m)

V = 1.429 Velocity through screen (m/s) v = 1.03 Approach velocity (m/s)

Therefore HLc = 0.07 m

Elevation of Channel Invert = 76.75 m Upstream Water Elevation = 76.87 m Downstream Water Elevation = 76.80 m

(4)

PUMP SUMP DESIGN

Influent at Qpeak Influent = 2219.31 25.69 l/s

Influent at Qavg Influent = 517.50 5.99 l/s

Dimension of pump station L = 1.80m

W = 1.50m

Depth between start and stop D = 0.65m

Chamfer volume v = 1/2 x ( 2.05 x 0.2 x 0.4 )

= 0.072

Therefore effective volume to fill /drain

V = L x W x D - v

= 1.68

Pumping rate @ Qpeak P = 26.00l/s >= 25.69 l/s ok

Pump cycle time at Qavg

Guideline: 6 min, 15 max @ Qavg

Volume required for pump sump, V = T x q

4

Assume number of start / stop = 15 times per hr (required 6 - 15 start/hour)

where V = ? Required Volume (m3)

T = 4.0 Cycle Time (minute)

q = 26.00 Pumping Rate (L/sec)

= 1.56 Pumping Rate (m3/min)

Therefore, V = 1.56 1.68 ok

Pump Headloss Calculations

VALVES & FITTINGS K QUANTITY DISCHARGE

VALUE TOTAL

EXIT 1.00 1 1.00

CHECK VALVE 2.50 1 2.50

GATE VALVE 0.20 1 0.20

TEE THRU SIDE 1.80 1 1.80

ELBOW - 90 DEG 0.30 6 1.80

SUM OF K's FOR FITTINGS 7.30

PIPE DIAMETER 0.20 m

PUMPING RATE 26.00 L/sec

V = VELOCITY 0.83 m/sec

F = TOTAL FITTING HEAD LOSS 0.25 m

= (K x V2 / 2g)

L = LINEAR METER OF STRAIGHT PIPE 8.00 m

C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 100.00

P = HEAD LOSS USING HAZEN-WILLIAMS EQATION 0.05 m

TOTAL LOSS = F + P 0.30 m

S = STATIC HEAD LOSS (Worst Case Scenerio) 79.60 - 76.00 m 3.45

TOTAL DYNAMIC HEAD 3.75 m

m3_{/day or} m3_{/day or}

m3

(5)

(6)

Size pump for 26.00 L/sec @ 3.75 m TDH

The submersible pump selected is as

follows:-Make = TSURUMI Model = 100B42.2 Capacity = 18.20 L/sec Total Head = 6.20 m Power = 2.20kW Discharge Size = 100 mm

No. of Units = 2(1 duty; 1 standby)

CHANNEL INVERT EL 76.75

0.0 m

HIGH LEVEL ALARM EL 76.75

0.15 m

STANDBY PUMP CUT IN EL 76.60

0.15 m

DUTY PUMP CUT IN EL 76.45

0.65 m

ALL PUMPS CUT OFF EL 75.80

0.4 m

INVERT ELEVATION EL 75.40

Pump cycle time at Qpeak

Guideline: 6 min, 15 max @ Qavg

Time to fill sump = V

Qpeak

= 1.09 min

Time to empty sump = V

Qpump - Qpeak

= -3.75 min

Cycle time base on actual operating point = 60 min per hour Time fill + Time empty min per cycle

= -22.6 cycle per hour 6 - 15 cycle / hr error

Determine Size of Force Main

Guideline: 1 - 2.5m/s velocity in pipe

Required size of raw sewage pipe = =

= 0.115 m

= 100 mm > 115 mm error

Actual velocity in selected pipe = Qpump / cross section area of pipe

2.32 m/s < 2.5 m/s ok

[ 4 x Qraw / (2.5 m/s x 3.14] 1/2 [ 4 x 0.0182 / (2.5 m/s x 3.14] 1/2

(7)

Determine Total Dynamic Head ( TDH )

TDH =

Where ; = Static head (m) …………A

= Losses through the pipe ( Hazen - William Formula ) …………B

= Losses through fittings …………C

A = 79.60 - 76.00m

= 3.45 m

B Hazen - William Formula

= 6.82 1.85 x L

C

Where ; V = Velocity m/sec

V = Q / A = 11.50 L/sec = 0.83 m/sec C = Coefficient of roughnes = 100.00 L = Length of pipe, m = 8.00 m D = Diameter of pipe, m = 0.20 m = 0.050 m

C Losses through fittings

= KV²

2g

Where ; K = Head loss coefficeint 7.30

V = Velocity m/sec 0.83

G = Gravity, m/s² 9.81

= 0.25 m

Therefore ,

Total Dynamic Head ( TDH ) =

= 3.76 m

To plot chart for the System Curve & Pump Curve from the above

equitition:-Q (L/s) V (m/s) hst (m) hf (m) hm (m) TDH (m) 0 0.00 3.45 0.000 0.000 3.45 16.0 3.33 0.11 3.45 0.001 0.004 3.46 14.0 6.67 0.21 3.45 0.004 0.017 3.47 12.6 10.00 0.32 3.45 0.009 0.038 3.50 10.8 13.33 0.42 3.45 0.015 0.067 3.53 9.0 16.67 0.53 3.45 0.022 0.105 3.58 7.0 20.00 0.64 3.45 0.031 0.151 3.63 5.2 23.00 0.73 3.45 0.040 0.199 3.69 3

* refer to Chart attached

hst + hf + hm

hst

hf

hm

Thus, static head of pump, h_st

hf V

D1.167

1000 x 3.142 x 0.10 2_{/ 4}

Thus, losses through the pipe, h_f

hm

h_m

hst + hf + hm

Pump head (m)

(8)

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

0.00

5.00

10.00

15.00

20.00 INFLUENT RAW SEWAGE PUMP

(TSURUMI MODEL 100 B42.2)

Flowrate, Q (L/sec)

H

ea

d

(m

)

System Curve

Pump Curve

Pump Operating Point

18.2L/sec @ 5.5 m TDH

(9)

FINE SCREEN DESIGN

Parameters

Population PE = 2,300

Peak Flow Qpeak = 2219.31

= 0.03

Design Flow Qavg = 517.50

= 0.01

Guidelines: Max flow through velocity at Qpeak Vmax = 1.00m/sec Guidelines: Min approach velocity at Qpeak Vmin = 1.00m/sec

Set Elevation of Datum Hd = 0.00m

Design

Bar Size Bs = 10.00mm

Clear Opening Co = 12.00mm

Efficiency Coefficient Eff = Clear Opening

Clear Opening + Bar size

= 0.55

Clear area through each screen at Qpeak, AQpeak = Qpeak Vmax

= 0.0257

Total cross sectional area of channel required = AQpeak Eff

= 0.05

Assume Depth of Flow D = 0.055m - - - 1

Width of Clear Openings Wclr = AQpeak / D

= 0.47 m

Number of Openings No = Wclr x 1000

Co

= 39

Number of Bars Nbars = No - 1

= 38

Width of Frame Wfr = 0.10m

Gross Width of Screen W = Wclr + Wfr + (Nbars x Bs / 1000)

= 0.95 m

Set width of screen W = 0.50m

Determine Width of Downstream Throat

Total Peak Flow, Qpeak = 1.71 B (H^1.5)

Qpeak = 0.026 m3/s

Set depth of flow H = 0.055 m from - - 1

Determine width of throat B = ? m

Therefore B = Qpeak / 1.71 / (H^1.5)

= 1.16 m

= Throat width of grit chamber m3_/day m3_/sec m3_/day m3_/sec m2 m2

(10)

(11)

Energy equation between upstream and downstream of clean screen Velocity at Qpeak through clear openings of screen, Vs

Vs = Qpeak

Wclr x d1

= 0.65 m/sec

Headloss through clean screen hLs = (Vs^2 - v1^2) / 2g x 1/0.7

= 0.0039 m

X2 E2 + d2 + (v2^2 /2g) =

X3 = E3 + d3 + (v3^2 /2g) + hLs

X2 = X3

Where,

E2 = 0.000m Height above datum

Trial and Error d1 = 0.09m Upstream Depth

v1 = 0.60 m/sec Upstream Velocity

d2 = 0.055 m Downstream Depth @ Qpeak

v2 = 0.93 m/sec Channel Velocity

Therefore, X2 = 0.10

X3 = 0.10 ok

0.000 Elevation of Channel Invert = 79.00m Upstream Water Elevation = 79.09 m Downstream Water Elevation = 79.06 m

Energy equation between upstream and downstream of 50% clogged screen

Headloss through clogged screen hLcs = (Vcs^2 - v2^2) / 2g x 1/0.7

Velocity through clogged screen Vcs = Qpeak

(Assume 50% clogging) Wclr x 0.5 x Upstream Depth d"

= 0.92 m/sec hLcs = 0.05 Therefore, X2 =E2 + d" + (v"^2 /2g) X3 = E3 + d3 + (v3^2 /2g) + hLcs X2 = X3 Where,

Trial and Error d" = 0.120m Upstream Depth

v" = 0.43 m/sec Upstream Velocity

d3 = 0.055 m Downstream Depth @ Qpeak

v3 = 0.93 m/sec Channel Velocity

Check:

X2 = 0.13

X3 = 0.15 ok

-0.018 Elevation of Channel Invert = 79.00 m Upstream Water Elevation = 79.12 m Downstream Water Elevation = 79.06 m

(12)

DESIGN OF A CONSTANT VELOCITY GRIT CHANNEL

Guidelines = 0.2 m/s flow through velocity

Number of Channels Provided = 1

Number of Back-up Channels Provided = 1

Constant Velocity Channel Formula

Q = Q = Flow B = Width of Throat m H = Depth of Flow m Therefore, = ( Q / 1.71 / B )^0.67 Area of Parabola, A = 2/3 W H W = Width of Parabola m H = Depth of Flow m Assume: V = 0.2m/s (Per guidelines) Q = A x V = 2/3 W H x V Therefore W = 1.5 Q H x V Assume: B = 0.33m

Qavg per Channel = 517.50 0.0060

FLOW FACTOR Depth Width

( x QAVG) m3/sec H(m) W(m) 0.10 0.001 0.01 0.43 0.25 0.001 0.02 0.58 0.50 0.003 0.03 0.73 1.00 0.006 0.05 0.92 @ Qavg 2.00 0.012 0.08 1.16 3.00 0.018 0.10 1.33 4.00 0.024 0.12 1.46 4.29 0.026 0.13 1.50 @ Qpeak 5.00 0.030 0.14 1.57 6.00 0.036 0.16 1.67 7.00 0.042 0.18 1.76

Determine Length of Channel

Say settlement vel of particle Vs = 0.02m/s

Say velocity of flow V = 0.2 m/s

Depth of channel at peak flow, H = 0.13 m

Width of channel, W = 1.50 m

Length of Channel L = V x H

Vs

L = 1.29 m

Provided Length L = 3.00m

Surface Area of Channel SA = 4.49

Surface Overflow Rate SR = Qpeak / SA

= 494.50 < 1500 1.71 B (H1.5₎ m3_/s m3_/day _m3_/s Q AT DIFFERENT FACTOR m2 m3_/m2_/day _m3_/m2_/day

(13)

Quantity of grit

Guidelines = 0.03 m3 / 1000 m3 of wastewater

Grit Quantity per Channel = Qavg x 0.03/1000

= 0.016

Provide area for 30 days storage

Grit quantity = 0.47

Dimension of hopper at bottom of grit channel

L : W ration 2:1 as per guidelines

L = 3.00 m W = 0.50m D = 0.15m Volume = 0.225 >= 0.466 ok 1500 mm 700 mm 50 mm 450 mm 1000 mm

Check ratio of grit tank dimension

Ratio W: D = 1 : 2 W D

2.00 1

Ratio W: L = 1 : 2 W L

1 2

Hydraulic Retension Time, Tr

Guidelines = 3 min @ Qpeak < 5,000 pe

Area of retangular = 1.05

Area of trapezium = 0.06

Area of grit storage = 0.45

Total Area = 1.56 L = 3.00 m Volume, V = 4.69 = 2219.31 = 1.54 Tr = = 3.04 min >= 3 min ok m3_/day m3 m2 m2 m2 m2 m3 Q_peak _m3_/day m3_/min V / Q_peak

(14)

X-Axis

Y-Axis

-0.837

0.161 -0.787

0.143 -0.748

0.129 -0.731

0.123 -0.664

0.101 -0.580

0.077 -0.460

0.049 -0.365

0.031 -0.290

0.019 -0.214

0.011

0

0.214

0.011

0.290

0.019

0.365

0.031

0.460

0.049

0.580

0.077

0.664

0.101

0.731

0.123

0.748

0.129

0.787

0.143

0.837

0.161 -0.500

-0.400

-0.300

-0.200

-0.100

0.000

0.100

0.200

0.300

0.400

0.500 -0.100

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

CONSTANT VELOCITY FLOW CHANNEL

X - SECTION

Width (m)

D

ep

th

(

m

)

(15)

-0.500

-0.400

-0.300

-0.200

-0.100

0.000

0.100

0.200

0.300

0.400

0.500 -0.100

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

CONSTANT VELOCITY FLOW CHANNEL

X - SECTION

Width (m)

D

ep

th

(

m

)

GRIT STORAGE

(16)

Width

Depth

X - Axis

Y - Axis

0.427

0.011

0.580

0.019

0.731

0.031

0.921

0.049

1.160

0.077

1.328

0.101

1.462

0.123

1.496

0.129

1.575

0.143

1.673

0.161 -0.500

-0.400

-0.300

-0.200

-0.100

0.000

0.100

0.200

0.300

0.400

0.500 -0.100

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

CONSTANT VELOCITY FLOW CHANNEL

X - SECTION

Width (m)

D

ep

th

(

m

)

(17)

-0.500

-0.400

-0.300

-0.200

-0.100

0.000

0.100

0.200

0.300

0.400

0.500 -0.100

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

CONSTANT VELOCITY FLOW CHANNEL

X - SECTION

Width (m)

D

ep

th

(

m

)

GRIT STORAGE

(18)

-0.650

-0.450

-0.250

-0.050

0.150

0.350

0.550 -0.150

-0.050

0.050

0.150

0.250

0.350

0.450

0.550

0.650

0.750

0.850

0.950

1.050

1.150

1.250 CONSTANT VELOCITY FLOW CHANNEL

X - SECTION

Channel Width (m)

D

e

p

th

(

m

)

GRIT

STORAGE

(19)

DESIGN OF A GRIT / GREASE CHAMBER

Guidelines : Detension time T = 360 sec @ Qpeak

Number of Channels Provided = 1

Number of Back-up Channels Provided = 1

Peak Flow Qpeak = 2219.31 m3/day

= 0.026 m3/sec

Detension Time T = 6min

= 360sec

Volume Required V = Qpeak x T

= 9.25 m3

Provide:

Length L = 2.90m

Width W = 1.60m

Depth D = 0.85m

Volume Provided V = 3.94 m3 > 9.25 m3 error

Ratio L : W = 1.81 : 1.0

W : D = 1.88 : 1.0 ok

Determine Length of Channel

Provided Length L = 2.90 m

Width of channel, W = 1.60 m

Surface Area of Channel SA = 4.64 m2

Surface Overflow Rate SR = Qpeak / SA

= 478.30 m3/m2/day

< 1500 m3/m2/day ok

Quantity of grit

Guidelines = 0.03 m3 / 1000 m3 of wastewater

Grit Quantity = Qavg x 0.03/1000

= 0.016 m3/day

Allow for storage of 30 days = 0.47 m3

Quantity of grease

Average quantity of grease S = 8kg / 1000 m3

Specific gravity of grease sg = 0.95

Therefore quantity of grease Qs = Qavg x S

= 4.14 kg/day

Flowrate of Grease Fs = Qs / 1000 / sg

= 0.0044 m3/day

Allow for storage of 30 days Vs = 30 x Fs

= 0.131 m3

Grit/Grease Drying Bed Sizing

Accumulated grease for 30 days Vs = 0.131 m3

Accumulated grit for 30 days = 0.466 m3

Depth of Feed D = 0.25m

Therefore, Area required A1 = 2.39 m2 Grease

Dimension of Grit/Grease Drying Bed

Length = 1.50m

(20)

(21)

DESIGN OF A GREASE CHAMBER

Guidelines : Detension time T = 180 sec @ Qpeak

Number of Chamber Provided = 1

Number of Back-up Chamber Provided = 1

= 0.026

Detension Time T = 180sec

Volume Required V = Qpeak x T

= 4.62 Provide: Length L = 3.80m Width W = 1.20m Depth D = 1.10m Volume Provided V = 5.02 4.62 ok Grease Quantities

Average quantity of grease S = 8

Specific gravity of grease sg = 0.95

Therefore quantity of grease Qs = Qavg x S = 4.14 kg/day Flowrate of Grease Fs = Qs / 1000 / sg

= 0.0044

Allow for storage of 30 days Vs = 30 x Fs

= 0.13

Depth of grease baffle required to contain scum for 30 days

Length L = 3.80 m

Width W = 1.20 m

Depth of water below baffle D = 0.02m

Volume of grease storage provided V = 0.09 0.13 ok

m3_/day m3_/sec m3 m3_> _m3 kg / 1000 m3 m3_/day m3 m3_> _m3

(22)

WEIR BEFORE AERATION BASINS

not use

Design Flow, Qavg 517.50 m3/day 0.01 m3/sec

Peak Flow, Qpeak 2219.31 m3/day 0.03 m3/sec

Number of Aeration Basins, N = 1

Peak Flow to Each Aeration Basin = Qpeak / N

= 0.026 m3/sec

Formula Q(m3/sec) = Cw x Le x H^1.5

where Cw = 1.84 Weir Coefficient

Le = 0.3704 m Length

H = 0.148 m Height

With end contraction

Le = L - (0.1)nH

where L = 0.40m Length

n = 2 Number of side contractions

H = 0.148m Height of Flow

Therefore Le = 0.3704 m

Therefore Q = 0.039 m3/sec >= 0.026 m3/sec ok

Height of water above weir = 102mm Qpeak if both tanks open

Height of weir = 170mm Qpeak if one tank open

Therefore, size opening for 400mm x 250 mm H = 272 mm

RECTANGULAR WEIR AT OUTLET BOX

Design Flow, Qavg 517.50 m3/day 0.01 m3/sec

Formula Q(m3/sec) = Cw x Le x H^1.5

where Cw = 1.84 Weir Coefficient

Le = 0.578 m Length

H = 0.11 m Height

With end contraction

Le = L - (0.1)nH

where L = 0.60m Length

n = 2 Number of side contractions

H = 0.110m Height of Flow

Therefore Le = 0.578 m

Therefore Q = 0.039 m3/sec >= 0.026 m3/sec ok

(23)

90 DEG V-NOTCH WEIR AT OUTLET BOX

Design Flow, Qavg 450.00 m3/day

90 deg V-Notch Weir Formula

5/2

q (m3/sec) = 1.417 H

H (m) = 192mm or 0.192 m at Qpeak

Therefore, q = 0.023 m3/sec >= 0.023 m3/sec ok

(24)

DESIGN OF ANOXIC CHAMBER

Population PE = 2300 Peak Flow = 2,219.31 = 92.47 = 1.54 = 0.026 = 25.69 l/s Design Flow = 517.50 = 21.56 = 0.36 = 0.01 = 5.99 l/s Peaking Factor PF = 4.29

Using the Lodification Ludzack-Ettinger Process

Assume

Influent Ammonia-N Ni = 30mg/l

Effluent Ammonia-N Ne = 10mg/l

Mixed liquor Suspended Solids MLSS = 3600mg/l

Mixed Liquor Volatile Suspended Solids MLVSS = 2880 mg/l (MVLSS = 0.8 x MLSS

Temperature Temp = 28deg C

Dissolved Oxigen Do = 0.1mg/l

Specific Denitrification Rate U = 0.11per day Overall Denitrification rate Ua = ? per day

Residence Time Temp = ? hrs

Calculation Overall Denitrification Rate

Formula : Ua =

0.20 per day

Calculation Residence Time

Formula : T = Ni - Ne

Ua x MLVSS x 24 hrs/day

= 0.81 hrs required

Size anoxic zone for a residence time of Anox = 2 hrs

Determine size of anoxic tank

Anoxic zone volume required, Vol = Anox x Qavg 24

= 43.13

Number of Anoxic Tank, N = 2 nos Volume required per Tank, Vreq = 21.56

Tank Dimension

Length = 2.30 m

Width = 1.50 m

Depth = 4.50 m

=

Anoxic tank volume = 31.05 > 21.56 ok

Actual hydraulic retention time provided = Anoxic tank volume Qavg = 2.88 hrs Q_peak _m3_/day m3_/hr m3_/min m3_/s Q_avg _m3_/day m3_/hr m3_/min m3_/s U x 1.09(Temp-20)X(1-DO) m3 m3 m3 _m3

(25)

EXTENDED AERATION TANK DESIGN

Sizing

Number of Aeration Tank, No. = 2

Minimum Hydraulic Retention Time, HRT = 19.4 hours

Volume Required, V = 418.31

Volume Required Per Tank, Vol = 209.16

Dimension of each tank: Ratio

Area A = 46.48 L : W Length L = 10.60 m 10.60 : 4.40 Width W = 4.40m 2.41 : 1 Depth D = 4.50m Freeboard FB = 0.65 m Volume Vol = 209.88

Total Volume Provided Vp = 419.76 418.31

Therefore, HRT Provided = 19.47 hrs > 19.4 hrs ok

Sludge Age (MCRT)

Guidelines > 20 days

Assume, MLSS = 3600mg/l

SA = Total solids in aeration tank

Excess sludge wasting / day + Solids in effluent Total solids in aeration tank = MLSS x Vp / 1000

= 1511.14 kg

Solids in effluent = 10mg/l

= 5.18 kg/day

Sludge Yield, = 0.40 @ 24 hours HRT = 0.60 @ 18 hours HRT Determine Sy @ Actual HRT

Via Interpolation Therefore actua = (24 -21.9) x (0.6 - 0.4) + 0.4 (24 - 18)

Sy = 0.55 kg / kg BOD removed based on HRT interpolation

Excess sludge wasting / day Sd = QAVG x Sy (BOD5-EBOD5)

(Sludge accumulation per day) = 68.45 kg/day

SA = 20.53 days ok

Waste Activated Sludge (WAS)

Excess Sludge Wasting / day WAS = 68.45 kg/day

Assume underflow concentration = 1% or 10,000 mg/l or 10 kg/m3

Volume of WAS = 6.84 = 6844.61 l/day m3 m3 m2 m3 m3_> _m3 m3_/day

(26)

Return Activated Sludge Flow (RAS)

QRAS = MLSS x QAVG

Cu - MLSS

Cu = Underflow concentration assume at 0.8 % solid or 8,000 mg/l

QRAS = 423.41 4.92 l/s

Set QRAS = 465.75 5.42 l/s

Qavg = 517.50

Ratio of QRAS : QAVG = 0.90 ok

Oxygen Requirements

Oxygen required per kg of BOD5 re OR = 2.00kg O2 / kg BOD5 Actual Oxygen Required AOR = OR x BOD5 removed

= 248.40 kg/day

Maintain DO of 2 mg/l in aeration tank

QAVG = 517.50

Total = 517.50

Oxygen required in tank = Total x 2 / 1000

= 1.04 kg/day

Total AOR = 249.44 kg/day

Oxygen correction factor AOR/SOR

AOR/SOR = (((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20

Where BETA = 0.98 ACF = 0.989 Csf = 7.83 C = 1.5 Css = 9.08 Alpha = 0.75 T = 30 Therefore AOR/SOR = 0.64

Standard Oxygen Required SOR = 391.23 kg/day

O2 transfer efficiency O2eff = 20% @ 6 (See catalog)

Actual O2 transfer efficiency O2act = SOR kg/day

O2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day Total amount of air required

= 4.88 m3/min

Number of diffusers required

Flowrate per diffuser = 0.10 or 6

= 100.00 L/min

Quantity = Total air required / Flowrate

= 48.75 Use 52 ok

Number of diffusers per tank = 26 Nos in each Aeration tank m3_/day m3_/day m3_/day m3_/day m3_/day m3_/hr m3_/min _m3_/hr

(27)

Check F/M Ratio

Guidelines for F/M ratio is between 0.05 to 0.10 kg BOD / kg MLSS

Influent BOD5, = 250 mg/l = 129.38 kg/day MLSS = 3600mg/l Volume provided, Vp = 419.76 F/M Ratio = INBOD / (MLSS x Vp) = 0.086 ok

Check Aeration Loading

AL = INBOD / Vp

= 0.31 ok

Check Mixing Rate

Volume of air from Blower Va = #NAME? (See Blower calculations)

Volume of tank provided Vp = 419.76

Therefore, Mixing Rate = Va / Vp

= #NAME? #NAME?

m3

Guidelines : 0.1 - 0.4 kg/m3_/day

kg/m3_/day

Per EPA guidelines, 0.010 to 0.025 m3_{/min. m}3

m3_/min m3

(28)

CLARIFIER DESIGN

Guidelines : Hydraulic retension time (HRT) = 2 hours @ Qpeak Guidelines : Surface overflow rate = 30 m3/m2/day

Guidelines : Side water depth (SWD) = 3 m

Surface Area =

= 73.98

Number of clarifiers = 2

Surface area required per clarifier = 36.99 Dimension of each clarifier

L = 6.20m

W = 6.20 m

Therefore, actual area provided = 38.44 36.99 ok

Provided Length, L = 6.20 m

Provided Width, W = 6.20 m

Effective Volume = Volume of cone with slope of 60 deg.

SWD in cone = 3.46 m (See attach volume cals)

Cone volume = 65.73 ok Vertical SWD = 1.04m Vertical SWD volume = 39.98 Total Volume = 105.70 Total SWD = 4.50 m HRT

Per guidelines: < 2 hours

=

= 2.29 >= 2 hours ok

Overflow Weir

Per guidelines: 150 - 180 m3/day/m

Max weir loading rate, Wr = 180 Peak Flow, Qpeak = 2219.31

Min Weir Length Required per Clarifier = Qpeak / Wr / No. of Clarifiers

= 6.16 m

Total length of weir provided per Clarifier, Tw = 3.00m

Actual weir loading rate = 369.89 error

Solids Loading Rate (SLR)

Guideliness : < 150 kg/m2/day at Qpeak < 50 kg/m2/day at Qavg

MLSS = 3600mg/l or 3.6

Total Q = Qpeak + QRAS

(For QRAS look under Extended Aeration Tank Sizing)

= #NAME?

SLR (Qpeak) = (Total Q / Actual Surface Area) x MLSS kg/day

= #NAME? #NAME?

SLR (Qavg) = (Qavg + Qras) / Actual Surface Area x MLSS kg/day

= #NAME? #NAME? Qpeak / 30 m3_/m2_/day m2 m2 m2 _> _m2 m3 m3 m3

Total Volume m3_{/ Qpeak m}3_{/day x 24 hours}

m3_/day/m m3_/day m3_/day/m kg/m3 m3_/day kg/m2_/day kg/m2_/day

(29)

Launder Calculation

Overflow from weir to lauder = Qpeak / number of tank

= 1109.66

Launder size

Length = 3.00 m

Width = 0.25m

Flow velocity = 0.80m /s < 1.0 m/s ok

Depth of flow in launder = 21.41 mm < 0.3m as provided

(30)

Determine Volume of Cone

L W L W h h 60 y BW 1 A BW2 B L = 6.20 m W = 6.20 m BW 1 = 2.20 m BW 2 = 2.20 m A = (L - BW1) / 2 = 2.00 B = (W - BW2) / 2 = 2.00 h = height of cone = A x Tan 60 3.46 m Tan y = h /B y = Tan -1 (h/B) = 60.00 degrees

Therefore, Volume of Cone V = [ (1/3 x 2A x 2B x h) +(1/2 x 2A x BW2 x h) + ( 1/2 x (W + BW2) x h) ] V = 65.73 m3 L (m) W (m) h (m) A (m) BW1 (m)

(31)

Scum Withdrawal Airlift Pipe Design

Percent submergence

=

Hs / (Hs + Hl) x 100

Hs

=

Depth (m) of air pipe below water surface

Hl

=

Height (m) of lift

Scum Airlift

Hs

=

3.8 m

Hl

=

0.7 m

Percent Submergence

=

84.44 %

With reference to the attached chart, Discharge

=

45 gpm

2.84 l/s

Velocity

=

2 ft/sec

0.61 m/s

(32)

AEROBIC SLUDGE HOLDING TANK DESIGN

Guidelines: Sludge yield = 0.6 kg/kg BOD5 (Standard "A")

Design Population, PE = 2300

Design Flow, QAVG = 517.5

Sludge Yield Ys = 0.55kg/kgBOD5/day (Calculated earlier)

Sludge accumulation per day Sd = QAVG x Ys (BOD5-EBOD5)

= 68.31 kg

Temperature of Wastewater T = 28 Deg C

Percent Volatile Suspended Solid (VSS) VSS = 75%

Percent VSS Destruction VSSd = 55%

Influent to Digester WAS = 68.31 kg/day

Influent Solid Content Conc = 1%

Total Volatile Solid TVS = VSS/100 x WAS

= 51.23 kg/day Total Volatile Solid Destruction TVSd = VSSd/100 x TVS

= 28.18 kg/day

Therefore, Total Solids Remaining After Digestion TSd = Nonvolatile Solid + VS Remaining = (WAS - TVS) + { (1 - VSSd/100) x TVS } =

TSd = 40.13 kg/day

Density of water ρ = 1000.00 Specific gravity of sludge sg = 1.015

Therefore, TSd in term of volume = 3.95

Storage days provided T = 30days

Volume of Tank Required Vtank = 118.62

Number of Tank Provided No. = 1

Volume of Tank Required V = 118.62

SIZE OF EACH TANK

Depth = 4.5m Width = 3.3m Length = 8.0m Volume = 118.80 118.62 ok m3_/day (26.73 - 20.05) + { (1 - 55/100) x 20.05) kg/m3 m3_/day m3 m3 m3_> _m3

(33)

OXYGEN REQUIREMENT FOR DIGESTION

Guidelines: 1.5kgO2/kgBOD

Wt of sludge digested Ws = Sd - TSd

= 28.18 kg

Amount of oxygen required, (AOR) Ws x 1.5 = 42.27 kg

Oxygen correction factor AOR/SOR

AOR/SOR = (((BETA * ACF * Csf @ 28 deg) - C ) / Css ) * ALPHA * 1.024^T-20

Where BETA = 0.98 ACF = 0.989 Csf = 7.83 C = 1.5 Css = 9.08 Alpha = 0.75 T = 30 Therefore AOR/SOR = 0.64

Standard Oxygen Required SOR = 66.29 kg/day Assume O2 transfer efficiency O2eff = 10%

Total amount of air required = SOR kg/day

O2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day

= 1.65

Number of diffusers required

Flowrate per diffuser = 200.00L/min 0.200 Quantity =

Total air required / Flowrate =

8.26 Use 10 ok

m3_/min

(34)

DRYING BED DESIGN

Digested Sludge From Aerobic Digester DS = #NAME?kg/day (See Aerobic Digester cals)

Concentration of Sludge C = 1%

= 10

Specific gravity of sludge sg = 1.015

Volume of Digested Sludge Vds = #NAME?

Per guidelines: Provide 4 week cycle for 450mm thick feed depth

Volume of Vds required V = Vds(m3/day) x 28days (Based on 21 days drying; 7 days feeding)

= #NAME?

Depth of Feed D = 0.45m

Therefore, Area required A = #NAME?

Provide fully covered drying beds

Therefore actual area required A = #NAME?

Number of Drying Beds provided = 4

Dimension of Drying Beds Dimension of Bed

Length = 16.50m

Width = 2.50m

Area = 41.25

Total area of drying bed provided = nos of bed x bed area

= 165.0 #NAME? ###

Determine Volume of Dewatered Sludge (DS)

Assume final sludge concentration is 25%

DS = Vs x Si x SGi x n

SGo x So

DS = ? Volume of Dewatered Sludge, (m3/day)

Vs = ### Volume of Influent Sludge (m3/day)

Si = 0.01Fractional Percent Solid Content of Influent Sludge So = 0.25Fractional Percent Solid Content of Dewatered Sludge SGi = 1.015Specific Gravity of Sludge Before Thickening

SGo = 1.03Specific Gravity of Sludge After Thickening

n = 0.95Fractional Percent Capture

Therefore

DS = ### @ 25% solid

Provide covered storage for 30 days

Volume required V = DS x 30 days

= #NAME?

Dimension provided Depth = 0.50m

Width = 1.00m Length = 4.60m Volume = 2.30 #NAME? kg/m3 m3_/day m3 m2 m2_{( 1/3 reduction in area)} m2 m2_> _m2 m3_/day m3 m3_> _m3 _ok

(35)

Drying Bed Feed Pump Design

Pump Sizing

Pumping rate per pump P = 9.10L/sec

= 546.00 L/min

Pump Headloss Calculations

VALVES & FITTINGS K QUANTITY SUCTION QUANTITY DISCHARGE

VALUE TOTAL TOTAL

ENTRY 0.50 1 0.50 0 0.00

EXIT 1.00 0 - 1 1.00

CHECK VALVE 2.50 0 - 0 0.00

SLUICE VALVE 0.20 0 - 1 0.20

REDUCER 0.30 0 - 1 0.30

TEE THRU SIDE 1.80 0 - 0 0.00

TEE THRU RUN 0.60 0 - 1 0.60

ELBOW - 45 DEG 0.23 0 - 0 0.00

ELBOW - 90 DEG 0.30 0 - 5 1.50

SUM OF K's FOR FITTINGS 0.50 3.60

PIPE DIAMETER 0.10m 0.10

PUMPING RATE 9.10 L/sec 9.10

V = VELOCITY 1.16 m/sec 1.16

F = TOTAL FITTING HEAD LOSS 0.03 m 0.25

= (K x V2 / 2g)

L = LINEAR METER OF STRAIGHT PIPE 0.00m 15.00

M = MULTIPLYING FACTOR 1.10 1.10

C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 100.00 100.00

P = HEAD LOSS USING HAZEN-WILLIAMS EQATION 0.00 m 0.39

TOTAL LOSS = F + P 0.03 m 0.64

TOTAL SUCTION + DISCHARGE LOSS 0.73 m

S = STATIC HEAD LOSS (Worst Case Scenerio) 4.00m

TOTAL DYNAMIC HEAD 4.73 m

Size pump for 9.10 L/sec @ 4.73 m TDH

The submersible pump selected is as

follows:-Make = Ebara

Model = 65DVS51.5

Capacity = 9.10 L/sec

Total Head = 4.73 m

Power = 1.50kW

Discharge Size = 65mm connect to 100 mm

No. of Units = 1(1 duty)

Determine Size of Force Main Guideline: 1 - 2.5m/s velocity in pipe

Required size of sludge pump =

=

= 0.068 m

= 80 mm > 68 mm ok

Actual velocity in selected pipe = 1.81 m/s > 2.5 m/s ok

[ 4 x Qsludge / (2.5 m/s x 3.14] 1/2

(36)

Determine Total Dynamic Head ( TDH )

TDH =

Where ; = Static head (m) …………A

= Losses through the pipe ( Hazen - William Formula ) …………B

= Losses through fittings …………C

A = 4.00m

B Hazen - William Formula

= 6.82 V 1.85 X L

C

Where ; V = Velocity m/sec 1.16 m/s

V = Q / A = 8 L/sec = 1.16 m/sec C = Coefficient of roughness 0.00 L = Length of pipe, m 15.00 D = Diameter of pipe, m 0.10 = #DIV/0! m

C Losses through fittings

= KV²

2g

Where ; K = Head loss coefficeint 4.10

V = Velocity m/sec 1.16

G = Gravity, m/s² 9.81

= 0.29 m

Therefore ,

Total Dynamic Head ( TDH ) =

= ### m

To plot chart for the System Curve & Pump Curve from the above

equitition:-Q (L/s) V (m/s) TDH (m) Pump head (m) 18.6 4 0 0.00 4 18.6 16.4 4.04 2 0.25 4.04 16.4 14.2 4.15 4 0.51 4.15 14.2 13 4.23 8 1.02 4.57 8.3 8.33 4.57 9.5 1.21 4.80 4.2 4.2 4.8

* refer to Chart attached

hst + hf + hm

hst

hf

hm

Thus, static head of pump, h_st

hf

D1.167

1000 x 3.142 x 0.10 2_{/ 4}

Thus, losses through the pipe, h_f

hm

h_m

(37)

0

2

4

6

8

10

0

2

4

6

8

10

12

14

16

18

20 DRYING BED FEED PUMP

Submersible Pump

EBARA Model 65DVS51.5

Flowrate, Q (L/sec)

H

ea

d

(m

)

System Curve

Pump Curve

Pump Operating Point

9.2 L/sec @ 4.7 m

(38)

X-Axis

Pump

System

Pump

System

0

16.2

4

0

3.3

14.2

4.25

198

6.67

12.4

4.92

400.2

10

11

5.97

600

13.33

9

7.39

799.8

16.67

7.4

9.17 1000.2

19.98

5.6

11.27 1198.8

65DVS

X-Axis

Pump

System

0

18.6

4

2

16.4

4.04

4

14.2

4.15

8

8.33

4.57

9.5

4.2

4.8 3.580986

(39)

Blower- 1 Sizing

Air Requirement for Aeration = #NAME?

Air Requirement for Aerobic Sludge Digestion = #NAME?

TOTAL AIR REQUIREMENT = #NAME?

= Qd x (1.0332 + Pd) x 1.0332

= ? Air flow under standard condition = #NAME? Air flow under discharge condition = 0.41 Discharge static pressure PS = -0.05 Suction static pressure

= 30 Suction temperature = 38 Discharge temperature Therefore, = #NAME? = - 1 x 1.0332 1.0332 + PS = 0.48

Size blower for 9.09 m3/min @ 0.48 kgf/cm2 Provide 2 blowers : 1 duty, 1 standby

Model: Fu Tsu Model TSC 125, 920rpm, 15.0kW

m3_/min m3_/min

Q_d _m3_/min

Determine air flow under standard condition, Q _s

Q_s 273 + S_t 273 + S_d Q_s _(m3_/min) Q_d _(m3_/min) Pd (kgf/cm2) (kgf/cm2₎ S_t o_C S_d o_C Qs m3/min

Determine discharge pressure under standard condition, P _s

P_s 1.0332 + P_d

(40)

Blower-3 Sizing

Air Requirement for Scum Airlift

=

0.20 Air Requirement for RAS

=

#NAME?

Air Requirement for WAS

=

#NAME?

Air Requirement for MLSS

=

#NAME?

TOTAL AIR REQUIREMENT

=

#NAME?

=

Qd x (1.0332 + Pd)

x

1.0332

=

?

Air flow under standard condition

=

#NAME?

Air flow under discharge condition

=

0.45 Discharge static pressure

PS

=

-0.05

Suction static pressure

=

30 Suction temperature

=

38 Discharge temperature

Therefore,

=

#NAME?

=

- 1

x 1.0332

1.0332 + PS

=

0.53 Size blower for 1.76 m3/min @ 0.53 kgf/cm2

Provide 1 blowers : 1 duty

Model:

Fu-Tsu Model TSC 80, 760 rpm @ 5.5 Kw

m

3

_/min

m

3

_/min

m

3

_/min

m

3

_/min

Q

_d

_m

3

_/min

Determine air flow under standard condition, Q

_s

Q

_s

273 + S

_t

273 + S

_d

Q

_s

_(m

3

_/min)

Q

_d

_(m

3

_/min)

P

_d

_(kgf/cm

2

₎

(kgf/cm

2

₎

S

_t o

_C

S

_d o

_C

Q

_s

_m

3

_/min

Determine discharge pressure under standard condition, P

_s

P

_s

1.0332 + P

_d

(41)

Sludge Withdrawal Airlift Pipe Design

Calculate the submerged distance of the air inlet (S) by the following formula: S = Basin SWD - ( 1 + (3 x Pump Dia) / 12 ) where

SWD (ft) = 14.76 ft 4.50m

Pump Dia (in) for RAS = 3.00 in 0.075m

Pump Dia (in) for MLSS & scum = 2.00 in 0.050m

S = 13.01 ft 3.97 m

= h / { C x LOG [ (H + 10.4) / 10.4 ] } where

h = 0.75total lift required (m)

H = 4.50 submergence (m)

C = 10.20constant for less than 15m lift

= 0.47 m3/min of air per m3 of water = 25.00air-lift efficiency (%)

SCUM = 75.0

= 0.0521

=

= 0.0981

RAS = #NAME? See Aeration Tank cals

= #NAME?

=

= #NAME?

WAS = #NAME? See Aeration Tank cals

= #NAME? = = #NAME? MLSS = = 2070.00 = #NAME? MLSS Return = #NAME? = #NAME? = = #NAME?

Calculate the Air Supply Volume (V _air) required: V_air

V_air A eff

Determine Volume of Air Requirement for SCUM, V _SCUM

m3_/day m3_/min V_air-SCUM V_air x RAS x A_eff

m3_/min

Determine Volume of Air Requirement for RAS, V _RAS

m3_/day m3_/min Vair-RAS Vair x RAS x Aeff

m3_/min

Determine Volume of Air Requirement for WAS, V _WAS

m3_/day m3_/min V_air-WAS V_air x WAS x A_eff

m3_/min

Determine Volume of Air Requirement for MLSS Return, V _MR 4 Q_ave - Q_RAS

- 182.3 m3/day m3_/day

m3_/day m3_/min V_air-MLSS V_air x MLSS x A_eff

(42)

Aeration Piping Headloss To Aeration Tanks

Criteria

Ambient air temperature To = 30deg C 303.2 deg K

Ambient barometric pressure Po = 1.00atm

Air supply pressure P = 1.450atm

Blower capacity Qb = #NAME?m3/min

Blower efficiency e = 75%

Equations

Friction factor f = 0.029 x D^0.027

Q^0.148 Temperature in pipe (deg K) T = To x (P/Po)^0.283

Velocity head Hv = 9.82E-8 x TQ^2

PD^4

Headloss (mm) hL = f x (L/D) x Hv or K x Hv

A. Pipe Fittings Losses

No. Valves & Fittings Size (mm) Quantity K Value Q (m3/min) T (deg K) hL, headloss (mm)

1 Check valve 80 1 2.50 #NAME? 336.77 #NAME?

2 Gate valve 80 1 0.80 #NAME? 336.77 #NAME?

9 Reducer 80 0.20 #NAME? 336.77 #NAME?

3 Tee thru side 80 1 1.80 #NAME? 336.77 #NAME?

4 Tee thru run 80 1 0.60 #NAME? 336.77 #NAME?

5 Elbow 90 deg 80 2 0.30 #NAME? 336.77 #NAME?

6 Elbow 90 deg 80 0.30 #NAME? 336.77 #NAME?

4 Tee thru run 50 2 0.60 #NAME? 336.77 #NAME?

9 Reducer 25 2 0.20 #NAME? 336.77 #NAME?

7 Tee thru run 25 0.60 #NAME? 336.77 #NAME?

4 Tee thru run 25 0.60 #NAME? 336.77 #NAME?

5 Elbow 90 deg 25 1 0.30 #NAME? 336.77 #NAME?

8 Elbow 90 deg 25 0.30 #NAME? 336.77 #NAME?

9 Reducer 25 0.20 #NAME? 336.77 #NAME?

10 Gate valve 25 1 0.80 #NAME? 336.77 #NAME?

Subtotal #NAME?

B. Straight Pipe Losses

No. Length (m) DIA (mm) Velocity (m/min) Q (m3/min) f, fric. factor T (deg K) hL, headloss (mm)

1 8.00 100 #NAME? #NAME? #NAME? 336.77 #NAME?

(43)

C. Supply Pressure At The Blower

1 Losses in piping = #NAME? mm

2 Losses in pipe fittings = #NAME? mm

3 Losses in air filter = 50.00mm

3 Losses in silencer = 50.00mm

4 Losses in blower = 150.00mm

5 Losses in diffusers = 160.00mm

6 Static head = 4100.00mm

Total #NAME? mm @ #NAME? m

Therefore, the absolute supply = #NAME? atm pressure

D. Power Requirement of Blower, P (kw)

P = w RTo x [ (P/Po)^0.283 - 1 ]

8.41 e

where R = 8.314 kJ/k mole deg K

w = air mass flow, kg/s

Therefore, P = #NAME? Kw or #NAME? HP

Select next bigger size motor = 5.5Kw 7 HP

ok

Size blower for 9.09 m3/min @ 0.48 kgf/cm2 Fu Tsu Model TSC 125, 920rpm, 15.0kW Provide 2 blowers; 1 running, 1 standby

(44)

CHLORINATION TANK DESIGN

= 2219.31

= 1.54

Detension Time at Qpeak t = 15.00min

Volume of Tank V =

= 23.12

Number of Tanks N = 1

Number of Bays per Tank n = 4

Dimension of Tank Provided

Depth H = 1.80m ok max 3m Wetted depth h = 1.50 Width W = 0.75 m Length L = 4.50 m Number of pass n = 4 Volume Provided Vp = 24.30 23.12 Check: Ratio Ratio ok

Wetted depth : Width

1.50 : 0.75

2 : 1

Length : Width

4.50 : 0.75

6 : 1

Detension Time at Qpeak t =

= 15.77 min > 15 min ok

PARSHALL FLUME FOR DISCHARGE

Design Flow, Qavg = 517.5

Peaking Factor PF = 4.29

Formula for flow calculation with diffrent throat width of Parshall Flume by Harlan Bengtson

Flow thru 1" PF, Q = where

Flow tru PF Q = flow in PF in cfs = 0.9071091 cfs = Qpeak

Head over flume H = in ft

water level in PF H = 1.8263509 ft 556.7 mm Q_peak _m3_/day m3_/min Qpeak x t m3 m3_>= _m3 Vp / Q_peak m3_/day m3_/day 0.338 x H1.55 For Q > Qpeak,

(45)

90 DEGREE V-NOTCH WEIR AT OUTLET BOX

Design Population , PE

=

2300

Design Flow,

Qavg

=

517.5 m3/day

Peaking Factor

PF

=

4.29 Peak Flow

Qpeak

=

2219.31

m3/day

90 Deg V-Notch Weir Formula

q(m3/sec)

=

H (m)

=

150 mm or

0.15 m at Qpeak

Therefore, q

=

0.012348

m3/sec > = 0.02569

m3/sec ok

Set depth of weir @

200 mm

ok

(46)

INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT Flow (m3/d) 517.50 Flow (m3/d) #NAME? Flow (m3/d) #NAME? MLSS Flow (m3/d) #NAME?

BOD (kg/day) 129.38 BOD (kg/day) 129.38 BOD (kg/day) 109.97 Flow (m3/d) #NAME? BOD(kg/d) #NAME?

TSS(kg/d) 155.25 TSS(kg/d) #NAME? TSS(kg/d) #NAME? TSS(kg/d) #NAME?

INFLUENT SCREEN PUMP ANOXIC TANK AERATION SECONDARY FINAL

FLOW CHAMBER * STATION BASINS CLARIFIERS EFFLUENT

LAST 10 mg/L BOD MANHOLE 20 mg/L TSS RAS RAS = 1% Flow (m3/d) 423.41 TSS(kg/d) 4234.09 LIQUID FLOW SOLID FLOW WAS * Assume 15% of BOD/TSS is removed from screenings and grit

* Assume 2% of flow is removed from screenings WAS = 1% Flow (m3/d) #NAME?

TSS(kg/d) #NAME?

DIGESTED SLUDGE DEWATERED SLUDGE Flow (m3/d) 3.95 Flow (m3/d) #NAME?

TSS(kg/d) #NAME? TSS(kg/d) #NAME?

AEROBIC SLUDGE

SLUDGE HOLDING DRYING BEDS TANK (75% VSS) (25-40% SOLIDS) (55% VSS Destruction) (95% CAPTURE) SUPERNATANT Flow (m3/d) #NAME? TSS(kg/d) #NAME?

(47)

INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT

Flow (m3/d) 689.63 Flow (m3/d) 697.82 Flow (m3/d) 683.87 Flow (m3/d) 683.82

BOD (kg/day) 172.41 BOD (kg/day) 172.41 BOD (kg/day) 146.55 BOD(kg/d) 6.84

TSS(kg/d) 206.89 TSS(kg/d) 236.97 TSS(kg/d) 201.43 TSS(kg/d) 13.68

INFLUENT SCREEN AERATION SECONDARY FINAL

FLOW CHAMBER * BASINS CLARIFIERS EFFLUENT

10 mg/L BOD 20 mg/L TSS RAS

LIQUID FLOW SOLID FLOW

* Assume 15% of BOD/TSS is removed from screenings and grit * Assume 2% of flow is removed from screenings

WAS

WAS = 1%

Flow (m3/d) 9.10

TSS(kg/d) 91.03

THICKENED SLUDGE DIGESTED SLUDGE DEWATERED SLUDGE

Flow (m3/d) 2.09 Flow (m3/d) 1.23 Flow (m3/d) 0.05

TSS(kg/d) 62.79 TSS(kg/d) 36.89 TSS(kg/d) 35.05

GRAVITY AEROBIC SLUDGE SLUDGE

SLUDGE THICKENER SLUDGE HOLDING DRYING BEDS HOLDING

TANK TANK AREA

(3% Solid) (80% VSS) (25-40% SOLIDS) (30 DAYS)

(55% VSS Destruction) (95% CAPTURE)

SUPERNATANT SUPERNATANT

Flow (m3/d) 7.01 Flow (m3/d) 1.18

TSS(kg/d) 28.24 TSS(kg/d) 1.84