design cal_2300pe.xls

57
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 = 300 mg/l BOD5, BOD = 250 mg/l Influent Ammonia NH3-N = 30 mg/l Effluent BOD5, EBOD = 10 mg/l Effluent SS ESS = 20 mg/l Effluent Ammonia ENH3-N = 10 mg/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 m 3 /c/d m 3 /d m 3 /d m 3 /min

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Page 1: Design Cal_2300PE.xls

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 MAKMURUNTUK 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 = 300 mg/lBOD5, BOD = 250 mg/lInfluent Ammonia NH3-N = 30 mg/l

Effluent BOD5, EBOD = 10 mg/lEffluent SS ESS = 20 mg/lEffluent Ammonia ENH3-N = 10 mg/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

Page 2: Design Cal_2300PE.xls

PRIMARY BAR SCREEN DESIGN

Design Population, PE 2300

Design Flow, Qavg 517.50 0.0060 l/sPeaking 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 = 1Number 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 troughL = 0.50 mW = 0.30 mD = 0.30 m

Volume = 0.045 > 0.109 error

Screen Design

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

Bar Size Bs = 10 mmClear Opening Co = 25 mm

Efficiency Coefficient Eff = Clear OpeningClear Opening + Bar size

= 0.71

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

= 0.026

Total cross sectional area of channel AC = AQpeakEff

= 0.036

Assume, Depth of Flow at Qpeak D = 0.050 m.

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

= 0.72 m

Number of Openings No = Wclr x 1000Co

= 29

m3/day

m3/day

m3/day

m3/day

m3

m3

m3 m3

m2

m2

Page 3: Design Cal_2300PE.xls

Number of Bars Nbars = No - 1= 28

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

= 1.00 m

Set Wch = 0.50 m

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 = 76.75 m

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

m3/day

Page 4: Design Cal_2300PE.xls

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 = 1.80 mW = 1.50 m

Depth between start and stop D = 0.65 mChamfer volume v = 1/2 x ( 2.05 x 0.2 x 0.4 )

= 0.072Therefore effective volume to fill /drain

V = L x W x D - v

= 1.68

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

Pump cycle time at QavgGuideline: 6 min, 15 max @ Qavg

Volume required for pump sump, V = T x q4

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 DISCHARGEVALUE 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 m3.45

TOTAL DYNAMIC HEAD 3.75 m

m3/day or

m3/day or

m3

m3

m3 < m3

Page 5: Design Cal_2300PE.xls
Page 6: Design Cal_2300PE.xls

Size pump for 26.00 L/sec @ 3.75 m TDH

The submersible pump selected is as follows:-Make = TSURUMIModel = 100B42.2Capacity = 18.20 L/secTotal Head = 6.20 mPower = 2.20 kWDischarge Size = 100 mmNo. of Units = 2 (1 duty; 1 standby)

CHANNEL INVERT EL 76.750.0 m

HIGH LEVEL ALARM EL 76.750.15 m

STANDBY PUMP CUT IN EL 76.600.15 m

DUTY PUMP CUT IN EL 76.450.65 m

ALL PUMPS CUT OFF EL 75.800.4 m

INVERT ELEVATION EL 75.40

Pump cycle time at QpeakGuideline: 6 min, 15 max @ Qavg

Time to fill sump = VQpeak

= 1.09 min

Time to empty sump = VQpump - Qpeak

= -3.75 min

Cycle time base on actual operating point = 60 min per hourTime 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 pipe2.32 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

D101
TC Wong: 7.5 kW $7500 Ebara
Page 7: Design Cal_2300PE.xls

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 - 76.00 m= 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 16.03.33 0.11 3.45 0.001 0.004 3.46 14.06.67 0.21 3.45 0.004 0.017 3.47 12.6

10.00 0.32 3.45 0.009 0.038 3.50 10.813.33 0.42 3.45 0.015 0.067 3.53 9.016.67 0.53 3.45 0.022 0.105 3.58 7.020.00 0.64 3.45 0.031 0.151 3.63 5.223.00 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, hst

hf V

D1.167

1000 x 3.142 x 0.10 2 / 4

Thus, losses through the pipe, hf

hm

hm

hst + hf + hm

Pump head (m)

Page 8: Design Cal_2300PE.xls

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 300.00

5.00

10.00

15.00

20.00

INFLUENT RAW SEWAGE PUMP(TSURUMI MODEL 100 B42.2)

Flowrate, Q (L/sec)

Hea

d (m

)

System Curve

Pump Curve

Pump Operating Point18.2L/sec @ 5.5 m TDH

Page 9: Design Cal_2300PE.xls

FINE SCREEN DESIGN

ParametersPopulation 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 = 1.00 m/sec

Guidelines: Min approach velocity at Qpeak Vmin = 1.00 m/sec

Set Elevation of Datum Hd = 0.00 m

Design

Bar Size Bs = 10.00 mmClear Opening Co = 12.00 mm

Efficiency Coefficient Eff = Clear OpeningClear Opening + Bar size

= 0.55

Clear area through each screen at Qpeak, AQpeak = QpeakVmax

= 0.0257

Total cross sectional area of channel required = AQpeakEff

= 0.05

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

Width of Clear Openings Wclr = AQpeak / D

= 0.47 m

Number of Openings No = Wclr x 1000Co

= 39

Number of Bars Nbars = No - 1= 38

Width of Frame Wfr = 0.10 m

Gross Width of Screen W = Wclr + Wfr + (Nbars x Bs / 1000)= 0.95 m

Set width of screen W = 0.50 m

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

Page 10: Design Cal_2300PE.xls

OK

Page 11: Design Cal_2300PE.xls

Energy equation between upstream and downstream of clean screen

Velocity at Qpeak through clear openings of screen, Vs

Vs = QpeakWclr 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 = 0.000 m Height above datum

Trial and Error d1 = 0.09 m Upstream Depthv1 = 0.60 m/sec Upstream VelocityE3 = 0.000 m Height above datumd2 = 0.055 m Downstream Depth @ Qpeakv2 = 0.93 m/sec Channel Velocity

Therefore, X2 = 0.10X3 = 0.10 ok

0.000

Elevation of Channel Invert = 79.00 m

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 = 0.05Therefore,

X2 =E2 + d" + (v"^2 /2g) X3 = E3 + d3 + (v3^2 /2g) + hLcs

X2 = X3

Where,E2 = 0.00 m Height above datum

Trial and Error d" = 0.120 m Upstream Depthv" = 0.43 m/sec Upstream VelocityE3 = 0.00 m Height above datumd3 = 0.055 m Downstream Depth @ Qpeakv3 = 0.93 m/sec Channel Velocity

Check:X2 = 0.13X3 = 0.15 ok

-0.018

Elevation of Channel Invert = 79.00 m

Upstream Water Elevation = 79.12 m

Downstream Water Elevation = 79.06 m

Page 12: Design Cal_2300PE.xls

DESIGN OF A CONSTANT VELOCITY GRIT CHANNEL

Guidelines = 0.2 m/s flow through velocity

Number of Channels Provided = 1Number of Back-up Channels Provided = 1

Constant Velocity Channel Formula

Q =

Q = Flow B = Width of Throat mH = Depth of Flow m

Therefore, = ( Q / 1.71 / B )^0.67

Area of Parabola, A = 2/3 W HW = Width of Parabola mH = Depth of Flow m

Assume:V = 0.2 m/s (Per guidelines)

Q = A x V= 2/3 W H x V

Therefore W = 1.5 QH x V

Assume: B = 0.33 m

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 0.430.25 0.001 0.02 0.580.50 0.003 0.03 0.731.00 0.006 0.05 0.92 @ Qavg2.00 0.012 0.08 1.163.00 0.018 0.10 1.334.00 0.024 0.12 1.464.29 0.026 0.13 1.50 @ Qpeak5.00 0.030 0.14 1.576.00 0.036 0.16 1.677.00 0.042 0.18 1.76

Determine Length of Channel

Say settlement vel of particle Vs = 0.02 m/s

Say velocity of flow V = 0.2 m/s

Depth of channel at peak flow, H = 0.13 mWidth of channel, W = 1.50 m

Length of Channel L = V x HVs

L = 1.29 m

Provided Length L = 3.00 m

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

Page 13: Design Cal_2300PE.xls

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 channelL : W ration 2:1 as per guidelines

L = 3.00 mW = 0.50 mD = 0.15 m

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 D2.00 1

Ratio W: L = 1 : 2 W L1 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 = 1.56L = 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

Qpeak m3/day

m3/min

V / Qpeak

Page 14: Design Cal_2300PE.xls

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 00.214 0.0110.290 0.0190.365 0.0310.460 0.0490.580 0.0770.664 0.1010.731 0.1230.748 0.1290.787 0.1430.837 0.161

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

CONSTANT VELOCITY FLOW CHANNELX - SECTION

Width (m)

Dep

th (

m)

Page 15: Design Cal_2300PE.xls

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

CONSTANT VELOCITY FLOW CHANNELX - SECTION

Width (m)

Dep

th (

m)

GRIT STORAGE

Page 16: Design Cal_2300PE.xls

Width DepthX - Axis Y - Axis

0.427 0.0110.580 0.0190.731 0.0310.921 0.0491.160 0.0771.328 0.1011.462 0.1231.496 0.1291.575 0.1431.673 0.161

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

CONSTANT VELOCITY FLOW CHANNELX - SECTION

Width (m)

Dep

th (

m)

Page 17: Design Cal_2300PE.xls

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

CONSTANT VELOCITY FLOW CHANNELX - SECTION

Width (m)

Dep

th (

m)

GRIT STORAGE

Page 18: Design Cal_2300PE.xls

-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 CHANNELX - SECTION

Channel Width (m)

Dep

th (

m)

GRIT STORAGE

Page 19: Design Cal_2300PE.xls

DESIGN OF A GRIT / GREASE CHAMBER

Guidelines : Detension time T = 360 sec @ Qpeak

Number of Channels Provided = 1Number of Back-up Channels Provided = 1

Peak Flow Qpeak = 2219.31 m3/day= 0.026 m3/sec

Detension Time T = 6 min= 360 sec

Volume Required V = Qpeak x T= 9.25 m3

Provide:Length L = 2.90 mWidth W = 1.60 mDepth D = 0.85 m

Volume Provided V = 3.94 m3 > 9.25 m3 error

Ratio L : W = 1.81 : 1.0W : D = 1.88 : 1.0 ok

Determine Length of Channel

Provided Length L = 2.90 mWidth of channel, W = 1.60 mSurface 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 = 8 kg / 1000 m3Specific 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 m3Accumulated grit for 30 days = 0.466 m3

Depth of Feed D = 0.25 m

Therefore, Area required A1 = 2.39 m2 Grease

Dimension of Grit/Grease Drying Bed

Length = 1.50 mWidth = 0.75 m

Page 20: Design Cal_2300PE.xls

Area = 1.13 m2 > 2.39 m2 error

Page 21: Design Cal_2300PE.xls

DESIGN OF A GREASE CHAMBER

Guidelines : Detension time T = 180 sec @ Qpeak

Number of Chamber Provided = 1Number of Back-up Chamber Provided = 1

Peak Flow Qpeak = 2219.31

= 0.026

Detension Time T = 180 sec

Volume Required V = Qpeak x T

= 4.62

Provide:Length L = 3.80 mWidth W = 1.20 mDepth D = 1.10 m

Volume Provided V = 5.02 4.62 ok

Grease Quantities

Average quantity of grease S = 8Specific 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 daysLength L = 3.80 mWidth W = 1.20 mDepth of water below baffle D = 0.02 m

Volume of grease storage provide V = 0.09 0.13 ok

m3/day

m3/sec

m3

m3 > m3

kg / 1000 m3

m3/day

m3

m3 > m3

Page 22: Design Cal_2300PE.xls

WEIR BEFORE AERATION BASINS not use

Design Population, PE 2300Design Flow, Qavg 517.50 m3/day 0.01 m3/secPeaking Factor, PF 4.29Peak 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 CoefficientLe = 0.3704 m LengthH = 0.148 m Height

With end contractionLe = L - (0.1)nH

where L = 0.40 m Lengthn = 2 Number of side contractionsH = 0.148 m Height of Flow

Therefore Le = 0.3704 m

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

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

Height of weir = 170 mm Qpeak if one tank open

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

RECTANGULAR WEIR AT OUTLET BOX

Design Population, PE 2300Design Flow, Qavg 517.50 m3/day 0.01 m3/secPeaking Factor, PF 4.29Peak Flow, Qpeak 2219.31 m3/day 0.03 m3/sec

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

where Cw = 1.84 Weir CoefficientLe = 0.578 m LengthH = 0.11 m Height

With end contractionLe = L - (0.1)nH

where L = 0.60 m Lengthn = 2 Number of side contractionsH = 0.110 m Height of Flow

Therefore Le = 0.578 m

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

Height of water above weir = 110 mm @ Qpeak

Page 23: Design Cal_2300PE.xls

90 DEG V-NOTCH WEIR AT OUTLET BOX

Design Population, PE 2000

Design Flow, Qavg 450.00 m3/day

Peaking Factor, PF 4.35

Peak Flow, Qpeak 1959.73 m3/day 0.023 m3/sec

90 deg V-Notch Weir Formula5/2

q (m3/sec) = 1.417 H

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

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

Set depth of weir @ 250mm ok

Page 24: Design Cal_2300PE.xls

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 = 30 mg/lEffluent Ammonia-N Ne = 10 mg/lMixed liquor Suspended Solids MLSS = 3600 mg/lMixed Liquor Volatile Suspended Solids MLVSS = 2880 mg/l (MVLSS = 0.8 x MLSSTemperature Temp = 28 deg CDissolved Oxigen Do = 0.1 mg/lSpecific Denitrification Rate U = 0.11 per dayOverall Denitrification rate Ua = ? per dayResidence Time Temp = ? hrs

Calculation Overall Denitrification Rate

Formula : Ua =0.20 per day

Calculation Residence Time

Formula : T = Ni - NeUa 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 Qavg24

= 43.13

Number of Anoxic Tank, N = 2 nos

Volume required per Tank, Vreq = 21.56

Tank DimensionLength = 2.30 mWidth = 1.50 mDepth = 4.50 m

=

Anoxic tank volume = 31.05 > 21.56 ok

Actual hydraulic retention time provided = Anoxic tank volumeQavg

= 2.88 hrs

Qpeak m3/day

m3/hr

m3/min

m3/s

Qavg m3/day

m3/hr

m3/min

m3/s

U x 1.09(Temp-20)X(1-DO)

m3

m3

m3 m3

Page 25: Design Cal_2300PE.xls

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 : WLength L = 10.60 m 10.60 : 4.40Width W = 4.40 m 2.41 : 1Depth D = 4.50 mFreeboard 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 daysAssume, MLSS = 3600 mg/l

SA = Total solids in aeration tankExcess sludge wasting / day + Solids in effluent

Total solids in aeration tank = MLSS x Vp / 1000= 1511.14 kg

Solids in effluent = 10 mg/l= 5.18 kg/day

Sludge Yield, = 0.40 @ 24 hours HRT= 0.60 @ 18 hours HRT

Determine Sy @ Actual HRTVia 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

Page 26: Design Cal_2300PE.xls

Return Activated Sludge Flow (RAS)

QRAS = MLSS x QAVGCu - 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 = 2.00 kg 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 = 0.98ACF = 0.989Csf = 7.83C = 1.5

Css = 9.08Alpha = 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/dayO2eff/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

E128
TC Wong: before 1.5
Page 27: Design Cal_2300PE.xls

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 = 3600 mg/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 = 4.88 (See Blower calculations)

Volume of tank provided Vp = 419.76

Therefore, Mixing Rate = Va / Vp

= 0.012 ok

m3

Guidelines : 0.1 - 0.4 kg/m3/day

kg/m3/day

Per EPA guidelines, 0.010 to 0.025 m3/min. m3

m3/min

m3

m3/min. m3

Page 28: Design Cal_2300PE.xls

CLARIFIER DESIGN

Guidelines : Hydraulic retension time (HRT) = 2 hours @ QpeakGuidelines : Surface overflow rate = 30 m3/m2/dayGuidelines : 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 = 6.20 mW = 6.20 m

Therefore, actual area provided = 38.44 36.99 ok

Provided Length, L = 6.20 mProvided 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 okVertical SWD = 1.04 m

Vertical SWD volume = 39.98

Total Volume = 105.70Total SWD = 4.50 m

HRTPer guidelines: < 2 hours

== 2.29 >= 2 hours ok

Overflow WeirPer guidelines: 150 - 180 m3/day/m

Max weir loading rate, Wr = 180Peak 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 = 3.00 m

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 = 3600 mg/l or 3.6

Total Q = Qpeak + QRAS(For QRAS look under Extended Aeration Tank Sizing)

= 2685.06

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

= 125.73 ok

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

= 46.04 ok

Qpeak / 30 m3/m2/day

m2

m2

m2 > m2

m3

m3

m3

Total Volume m3 / Qpeak m3/day x 24 hours

m3/day/mm3/day

m3/day/m

kg/m3

m3/day

kg/m2/day

kg/m2/day

Page 29: Design Cal_2300PE.xls

Launder Calculation

Overflow from weir to lauder = Qpeak / number of tank

= 1109.66

Launder sizeLength = 3.00 mWidth = 0.25 mFlow velocity = 0.80 m /s < 1.0 m/s okDepth of flow in launder = 21.41 mm < 0.3m as provided

m3 / launder

Page 30: Design Cal_2300PE.xls

Determine Volume of Cone

L

W

L W

h h 60 y

BW 1 A BW2 B

L = 6.20 mW = 6.20 m

BW 1 = 2.20 mBW 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 /By = 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)

Page 31: Design Cal_2300PE.xls

Scum Withdrawal Airlift Pipe Design

Percent submergence = Hs / (Hs + Hl) x 100

Hs = Depth (m) of air pipe below water surfaceHl = Height (m) of lift

Scum AirliftHs = 3.8 mHl = 0.7 m

Percent Submergence = 84.44 %

With reference to the attached chart, Discharge = 45 gpm 2.84 l/sVelocity = 2 ft/sec 0.61 m/sAir Requirement = 7 cfm 0.20 m3/min

Page 32: Design Cal_2300PE.xls

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 = 0.55 kg/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 = 30 days

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 = 4.5 mWidth = 3.3 m

Length = 8.0 m

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

Page 33: Design Cal_2300PE.xls

OXYGEN REQUIREMENT FOR DIGESTIONGuidelines: 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 = 0.98ACF = 0.989Csf = 7.83C = 1.5

Css = 9.08Alpha = 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/dayO2eff/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 = 200.00 L/min

0.200Quantity =

Total air required / Flowrate=

8.26 Use 10 ok

m3/min

m3/min

Page 34: Design Cal_2300PE.xls

DRYING BED DESIGN

Digested Sludge From Aerobic Digester DS = 40.13 kg/day (See Aerobic Digester cals)

Concentration of Sludge C = 1 %

= 10

Specific gravity of sludge sg = 1.015

Volume of Digested Sludge Vds = 3.95

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)

= 110.71

Depth of Feed D = 0.45 m

Therefore, Area required A = 246.02

Provide fully covered drying beds

Therefore actual area required A = 164.01

Number of Drying Beds provided = 4

Dimension of Drying Beds

Dimension of Bed Length = 16.50 mWidth = 2.50 m

Area = 41.25

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

= 165.0 164.0 ok

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 = 3.95 Volume of Influent Sludge (m3/day)Si = 0.01 Fractional Percent Solid Content of Influent SludgeSo = 0.25 Fractional Percent Solid Content of Dewatered SludgeSGi = 1.015 Specific Gravity of Sludge Before ThickeningSGo = 1.03 Specific Gravity of Sludge After Thickening

n = 0.95 Fractional Percent Capture

Therefore

DS = 0.15 @ 25% solid

Provide covered storage for 30 days

Volume required V = DS x 30 days

= 4.44

Dimension provided Depth = 0.50 mWidth = 1.00 mLength = 4.60 m

Volume = 2.30 4.44

kg/m3

m3/day

m3

m2

m2 ( 1/3 reduction in area)

m2

m2 > m2

m3/day

m3

m3 > m3 ok

Page 35: Design Cal_2300PE.xls

Drying Bed Feed Pump Design

Pump Sizing

Pumping rate per pump P = 9.10 L/sec= 546.00 L/min

Pump Headloss Calculations

VALVES & FITTINGS K QUANTITY SUCTION QUANTITY DISCHARGEVALUE 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 0.10 m 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 0.00 m 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) 4.00 m

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 = EbaraModel = 65DVS51.5Capacity = 9.10 L/secTotal Head = 4.73 mPower = 1.50 kWDischarge Size = 65 mm connect to 100 mmNo. 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

[ 4 x 0.0091 / (2.5 m/s x 3.14] 1/2

Page 36: Design Cal_2300PE.xls

Determine Total Dynamic Head ( TDH )

TDH =

Where ; = Static head (m) …………A= Losses through the pipe ( Hazen - William Formula ) …………B= Losses through fittings …………C

A = 4.00 m

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 40 0.00 4 18.6 16.4 4.042 0.25 4.04 16.4 14.2 4.154 0.51 4.15 14.2 13 4.238 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, hst

hf

D1.167

1000 x 3.142 x 0.10 2 / 4

Thus, losses through the pipe, hf

hm

hm

hst + hf + hm

Page 37: Design Cal_2300PE.xls

0 2 4 6 8 100

2

4

6

8

10

12

14

16

18

20

DRYING BED FEED PUMPSubmersible Pump

EBARA Model 65DVS51.5

Flowrate, Q (L/sec)

Hea

d (m

)

System Curve

Pump Curve

Pump Operating Point9.2 L/sec @ 4.7 m

Page 38: Design Cal_2300PE.xls

X-Axis Pump System Pump System0 16.2 4 0

3.3 14.2 4.25 1986.67 12.4 4.92 400.2

10 11 5.97 60013.33 9 7.39 799.816.67 7.4 9.17 1000.219.98 5.6 11.27 1198.8

65DVS X-Axis Pump System0 18.6 42 16.4 4.044 14.2 4.158 8.33 4.57

9.5 4.2 4.8

3.580986

Page 39: Design Cal_2300PE.xls

Blower- 1 Sizing

Air Requirement for Aeration = 4.88

Air Requirement for Aerobic Sludge Digestion = 1.65

TOTAL AIR REQUIREMENT = 6.53

= Qd x (1.0332 + Pd) x

1.0332

= ? Air flow under standard condition

= 6.53 Air flow under discharge condition

= 0.41 Discharge static pressure

PS = -0.05 Suction static pressure

= 30 Suction temperature

= 38 Discharge temperature

Therefore,

= 8.88

= - 1 x 1.03321.0332 + PS

= 0.48

Size blower for 9.09 m3/min @ 0.48 kgf/cm2Provide 2 blowers : 1 duty, 1 standby

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

m3/min

m3/min

Qd m3/min

Determine air flow under standard condition, Q s

Qs 273 + St

273 + Sd

Qs (m3/min)

Qd (m3/min)

Pd (kgf/cm2)

(kgf/cm2)

StoC

SdoC

Qs m3/min

Determine discharge pressure under standard condition, P s

Ps 1.0332 + Pd

kgf/cm2

Page 40: Design Cal_2300PE.xls

Blower-3 Sizing

Air Requirement for Scum Airlift = 0.20

Air Requirement for RAS = 0.6091

Air Requirement for WAS = 0.0090

Air Requirement for MLSS = 2.10

TOTAL AIR REQUIREMENT = 2.91

= Qd x (1.0332 + Pd) x

1.0332

= ? Air flow under standard condition

= 2.91 Air flow under discharge condition

= 0.45 Discharge static pressure

PS = -0.05 Suction static pressure

= 30 Suction temperature

= 38 Discharge temperature

Therefore,

= 4.08

= - 1 x 1.03321.0332 + PS

= 0.53

Size blower for 1.76 m3/min @ 0.53 kgf/cm2Provide 1 blowers : 1 duty

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

m3/min

m3/min

m3/min

m3/min

Qd m3/min

Determine air flow under standard condition, Q s

Qs 273 + St

273 + Sd

Qs (m3/min)

Qd (m3/min)

Pd (kgf/cm2)

(kgf/cm2)

StoC

SdoC

Qs m3/min

Determine discharge pressure under standard condition, P s

Ps 1.0332 + Pd

kgf/cm2

Page 41: Design Cal_2300PE.xls

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 4.50 mPump Dia (in) for RAS = 3.00 in 0.075 m

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

S = 13.01 ft 3.97 m

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

where

h = 0.75 total lift required (m)H = 4.50 submergence (m)C = 10.20 constant for less than 15m lift

= 0.47 m3/min of air per m3 of water

= 25.00 air-lift efficiency (%)

SCUM = 75.0

= 0.0521

=

= 0.0981

RAS = 465.8 See Aeration Tank cals

= 0.3234

=

= 0.6091

WAS = 6.84 See Aeration Tank cals

= 0.0048

=

= 0.0090

MLSS == 2070.00

= 1604.25

MLSS Return = 1604.3

= 1.1141

=

= 2.0984

Calculate the Air Supply Volume (V air) required:

Vair

Vair

A eff

Determine Volume of Air Requirement for SCUM, V SCUM

m3/day

m3/min

Vair-SCUM Vair x RAS x Aeff

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

Vair-WAS Vair x WAS x Aeff

m3/min

Determine Volume of Air Requirement for MLSS Return, V MR

4 Qave - QRAS

- 182.3 m3/day

m3/day

m3/day

m3/min

Vair-MLSS Vair x MLSS x Aeff

m3/min

Page 42: Design Cal_2300PE.xls

Aeration Piping Headloss To Aeration Tanks

CriteriaAmbient air temperature To = 30 deg C 303.2 deg K

Ambient barometric pressure Po = 1.00 atm

Air supply pressure P = 1.450 atm

Blower capacity Qb = 6.53 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 6.53 336.77 59.322 Gate valve 80 1 0.80 6.53 336.77 18.989 Reducer 80 0.20 6.53 336.77 0.003 Tee thru side 80 1 1.80 6.53 336.77 42.714 Tee thru run 80 1 0.60 6.53 336.77 14.245 Elbow 90 deg 80 2 0.30 6.53 336.77 14.246 Elbow 90 deg 80 0.30 6.53 336.77 0.004 Tee thru run 50 2 0.60 4.35 336.77 82.939 Reducer 25 2 0.20 1.09 336.77 27.647 Tee thru run 25 0.60 1.09 336.77 0.004 Tee thru run 25 0.60 1.09 336.77 0.005 Elbow 90 deg 25 1 0.30 1.09 336.77 20.738 Elbow 90 deg 25 0.30 1.09 336.77 0.009 Reducer 25 0.20 1.09 336.77 0.0010 Gate valve 25 1 0.80 1.09 336.77 55.29

Subtotal 336.08

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 1130.94 8.88 0.02000 336.77 28.802 4.00 50 3015.84 5.92 0.02100 336.77 215.033 7.00 50 1507.92 2.96 0.02300 336.77 103.044 4.00 25 3015.84 1.48 0.02500 336.77 511.98

Subtotal 858.84

Page 43: Design Cal_2300PE.xls

C. Supply Pressure At The Blower

1 Losses in piping = 858.84 mm2 Losses in pipe fittings = 336.08 mm3 Losses in air filter = 50.00 mm3 Losses in silencer = 50.00 mm4 Losses in blower = 150.00 mm5 Losses in diffusers = 160.00 mm6 Static head = 4100.00 mm

Total 5704.92 mm @ 5.70 m

Therefore, the absolute supply = 1.55 atmpressure

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 = 7 Kw or 9 HP

Select next bigger size motor = 5.5 Kw 7 HPok

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

Page 44: Design Cal_2300PE.xls

CHLORINATION TANK DESIGN

= 2219.31

= 1.54

Detension Time at Qpeak t = 15.00 min

Volume of Tank V =

= 23.12

Number of Tanks N = 1

Number of Bays per Tank n = 4

Dimension of Tank ProvidedDepth H = 1.80 m ok max 3mWetted depth h = 1.50Width W = 0.75 mLength L = 4.50 mNumber of pass n = 4

Volume Provided Vp = 24.30 23.12

Check:

Ratio Ratio okWetted depth : Width

1.50 : 0.752 : 1

Length : Width4.50 : 0.75

6 : 1

Detension Time at Qpeak t == 15.77 min > 15 min ok

PARSHALL FLUME FOR DISCHARGE

Design Flow, Qavg = 517.5Peaking Factor PF = 4.29

Peak Flow Qpeak = 2219.31

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 ft556.7 mm

Qpeak m3/day

m3/min

Qpeak x t

m3

m3 >= m3

Vp / Qpeak

m3/day

m3/day

0.338 x H1.55

For Q > Qpeak,

Page 45: Design Cal_2300PE.xls

90 DEGREE V-NOTCH WEIR AT OUTLET BOX

Design Population , PE = 2300Design Flow, Qavg = 517.5 m3/dayPeaking Factor PF = 4.29Peak 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

1.417H 5/2

Page 46: Design Cal_2300PE.xls

INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT

Flow (m3/d) 517.50 Flow (m3/d) 521.31 Flow (m3/d) 510.88 MLSS Flow (m3/d) 510.73

BOD (kg/day) 129.38 BOD (kg/day) 129.38 BOD (kg/day) 109.97 Flow (m3/d) 1604.25 BOD(kg/d) 5.11

TSS(kg/d) 155.25 TSS(kg/d) 157.26 TSS(kg/d) 133.67 TSS(kg/d) 10.21

INFLUENT SCREEN PUMP ANOXIC TANK AERATION SECONDARY FINALFLOW CHAMBER * STATION BASINS CLARIFIERS EFFLUENT

LAST 10 mg/L BODMANHOLE 20 mg/L TSS

RASRAS = 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) 6.84

TSS(kg/d) 68.45DIGESTED SLUDGE DEWATERED SLUDGE

Flow (m3/d) 3.95 Flow (m3/d) 0.15

TSS(kg/d) 40.13 TSS(kg/d) 38.13

AEROBIC SLUDGESLUDGE HOLDING DRYING BEDS

TANK(75% VSS) (25-40% SOLIDS)

(55% VSS Destruction) (95% CAPTURE)

SUPERNATANT

Flow (m3/d) 3.81

TSS(kg/d) 2.01

SOLIDS BALANCE @ Qavg

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Page 47: Design Cal_2300PE.xls

INFLUENT INFLUENT INFLUENT SECONDARY EFFLUENT

Flow (m3/d) 689.63 Flow (m3/d) 697.82 Flow (m3/d) 683.87 Flow (m3/d) 683.82BOD (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 FINALFLOW CHAMBER * BASINS CLARIFIERS EFFLUENT

10 mg/L BOD20 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) 91.03THICKENED 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 SLUDGESLUDGE 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

SOLIDS BALANCE @ Qavg

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