Design Cal_2300PE.xls

October 6, 2018 | Author: Emily Chay | Category: Pump, Volume, Hydrology, Hydraulic Engineering, Physical Quantities
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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, Dry Weather Flow,

PE DWF

= =

2300 P.E 0.225 m3/c/d

Suspended Solid, BOD5, Influent Ammonia

SS BOD NH3-N

= = =

300 mg/l 250 mg/l 30 mg/l

Effluent BOD5, Effluent SS Effluent Ammonia

EBOD ESS ENH3-N

= = =

10 mg/l 20 mg/l 10 mg/l

LOADING a.

Hydraulic flow,

Qavg

= =

PE * DWF 517.50 m3/d

b.

Peak Flow,

Qpeak

=

Peak Factor * Qavg

Pff

= =

4.7(P) 4.29

Qpeak

= = =

Pff x Qavg 2219.31 m3/d 1.54 m3/min

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

-0.11

Peak Flow Factor

Therefore,

PRIMARY BAR SCREEN DESIGN Design Population,

PE

Design Flow, Peaking Factor, Peak Flow,

Qavg PF Qpeak

2300 517.50 m3/day 4.29 2219.31 m3/day

0.0060 l/s 25.81 l/s

Guidelines : Quantity of screenings = 30 m3 screening / 10^6 m3 wastewater Number of Channels Number of Back-up Channel

= =

1 1

Qavg

=

517.50 m3/day

Quantity of Screenings

= =

Qavg x 30 / 1000000 0.0155 m3/day

Quantity of Screenings

= =

7 x Quantity of Screenings per day 0.109 m3

Number of Storage Units Quantity per Unit

= =

1 0.109 m3

Dimension of screenings trough L W D

= = =

0.50 m 0.30 m 0.30 m

=

0.045 m3

Provide storage for 7 days

Volume

>

0.109 m3

Screen Design Guidelines: Max flow through velocity at Qpeak Guidelines: Min approach velocity at Qpeak

Vmax Vmin

Bar Size Clear Opening

Bs Co

= =

Efficiency Coefficient

Eff

=

10 mm 25 mm Clear Opening Clear Opening + Bar size

=

Clear area through each screen at Qpe AQpeak

Total cross sectional area of channel

AC

0.71

=

Qpeak Vmax x 24 x 60 x 60

=

0.026 m2

=

AQpeak Eff

=

Assume, Depth of Flow at Qpeak

D

= =

0.036 m2

=

0.050 m .

Required Width of Clear Opening @ Q

Wclr

= =

Number of Openings

No

AC / D 0.72 m

=

Wclr x 1000 Co

=

29

1.0 m/ses 0.3 m/sec

error

Number of Bars

Nbars

Gross Width of Screen

Wch

Set

= =

No - 1

=

Wclr + (Nbars x Bs / 1000)

28

=

1.00 m

Wch

=

0.50 m

Qpeak

=

2219.31 m3/day

Vapp

= =

Qpeak / (Wch x D) 1.03 m/sec

>=

0.30 m/sec

ok

= =

Qpeak / (Wclr x D) 0.71 m/sec

=

25.69 l/s

ok

Pump cycle time at Qavg Guideline: 6 min, 15 max @ Qavg Volume required for pump sump,

Assume number of start / stop where

Therefore,

times per hr (required 6 - 15 start/hour) Required Volume (m3) Cycle Time (minute) Pumping Rate (L/sec) Pumping Rate (m3/min) 1.56 m3

<

Pump Headloss Calculations

VALVES & FITTINGS

K VALUE

EXIT CHECK VALVE GATE VALVE TEE THRU SIDE ELBOW - 90 DEG

1.00 2.50 0.20 1.80 0.30

QUANTITY

DISCHARGE TOTAL

1 1 1 1 6

1.00 2.50 0.20 1.80 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 = (K x V2 / 2g)

0.25

m

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

79.60 - 76.00 3.45

m

3.75

m

S = STATIC HEAD LOSS (Worst Case Scenerio)

TOTAL DYNAMIC HEAD

1.68 m3

ok

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 = 2.20 kW Discharge Size = 100 mm No. of Units = 2 (1 duty; 1 standby)

CHANNEL INVERT

EL

76.75

HIGH LEVEL ALARM

EL

76.75

STANDBY PUMP CUT IN

EL

76.60

DUTY PUMP CUT IN

EL

76.45

ALL PUMPS CUT OFF

EL

75.80

INVERT ELEVATION

EL

75.40

Time to fill sump

=

0.0

m

0.15

m

0.15

m

0.65

m

0.4

m

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

= Time to empty sump

= =

Cycle time base on actual operating point

=

V Qpeak 1.09

min

V Qpump - Qpeak -3.75

min

=

60 Time fill + Time empty -22.6

min per hour min per cycle cycle per hour

=

[ 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

6 - 15 cycle / hr error

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

= = Actual velocity in selected pipe

=

0.115 m 100 mm Qpump / cross section area of pipe 2.32 m/s <

>

115 mm

error

2.5 m/s

ok

Determine Total Dynamic Head ( TDH ) TDH

=

hst + hf + hm

= = =

Static head (m) Losses through the pipe ( Hazen - William Formula ) Losses through fittings

Where ;

hst hf hm

A

Thus, static head of pump, hst

B

Hazen - William Formula hf

Where ;

79.60 - 76.00 m 3.45 m

=

6.82

V C

V

=

Velocity m/sec

V = Q/A

=

11.50 L/sec 1000 x 3.142 x 0.10 2 / 4

=

C

= =

1.85

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

Thus, losses through the pipe, h f

=

0.050 m

Losses through fittings hm

=

Where ;

K

=

Head loss coefficeint

7.30

V

=

Velocity m/sec

0.83

G

=

Gravity, m/s²

9.81

hm

=

0.25 m

KV² 2g

Therefore , Total Dynamic Head ( TDH )

= =

hst + hf + hm 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)

0 3.33 6.67 10.00 13.33 16.67 20.00 23.00

0.00 0.11 0.21 0.32 0.42 0.53 0.64 0.73

3.45 3.45 3.45 3.45 3.45 3.45 3.45 3.45

0.000 0.001 0.004 0.009 0.015 0.022 0.031 0.040

* refer to Chart attached

x

hm (m) TDH (m) 0.000 0.004 0.017 0.038 0.067 0.105 0.151 0.199

3.45 3.46 3.47 3.50 3.53 3.58 3.63 3.69

Pump head (m) 16.0 14.0 12.6 10.8 9.0 7.0 5.2 3

…………A …………B …………C

L D1.167

INFLUENT RAW SEWAGE PUMP (TSURUMI MODEL 100 B42.2) 20.00

System Curve

Head (m)

15.00

Pump Operating Point 18.2L/sec @ 5.5 m TDH

10.00

5.00 Pump Curve

0.00 0

2

4

6

8

10

12

14

16

Flowrate, Q (L/sec)

18

20

22

24

26

28

30

FINE SCREEN DESIGN Parameters Population

PE

=

Peak Flow

Qpeak

=

2219.31 m3/day

=

0.03 m3/sec

Qavg

= =

517.50 m3/day 0.01 m3/sec

Guidelines: Max flow through velocity at Qpeak

Vmax

=

1.00 m/sec

Guidelines: Min approach velocity at Qpeak

Vmin

=

1.00 m/sec

Hd

=

0.00 m

Bar Size Clear Opening

Bs Co

= =

10.00 mm 12.00 mm

Efficiency Coefficient

Eff

=

Design Flow

Set Elevation of Datum

2,300

Design

Clear Opening Clear Opening + Bar size

= Clear area through each screen at Qpeak,

AQpeak

=

= Total cross sectional area of channel required

=

0.55 Qpeak Vmax 0.0257 m2 AQpeak Eff

= Assume Depth of Flow Width of Clear Openings

Number of Openings

D

=

Wclr

=

No

0.05 m2 0.055 m

Width of Frame

Nbars

=

0.47 m

=

Wclr x 1000 Co

= =

1

AQpeak / D

= Number of Bars

-----

39 No - 1 38

Wfr

=

0.10 m

Gross Width of Screen

W

= =

Set width of screen

W

=

0.50 m

Qpeak

=

1.71 B (H^1.5)

Qpeak

=

0.026 m3/s

Set depth of flow

H

=

0.055 m from - -

Determine width of throat

B

=

?

Therefore

B

= =

Qpeak / 1.71 / (H^1.5) 1.16 m

=

Throat width of grit chamber

Wclr + Wfr + (Nbars x Bs / 1000) 0.95 m

Determine Width of Downstream Throat Total Peak Flow,

m

1

OK

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

Headloss through clean screen

=

Qpeak Wclr x d1

=

0.65 m/sec

hLs

= =

(Vs^2 - v1^2) / 2g 0.0039 m

X2 X3

E2 = + d2 + (v2^2 /2g) = E3 + d3 + (v3^2 /2g) + hLs

X2

=

E2 d1 v1 E3 d2 v2

= = = = = =

0.000 0.09 0.60 0.000 0.055 0.93

X2 X3

= =

0.10 0.10

x 1/0.7

X3

Where, Trial and Error

Therefore,

m m m/sec m m m/sec

Height above datum Upstream Depth Upstream Velocity Height above datum Downstream Depth @ Qpeak Channel Velocity

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

=

Velocity through clogged screen (Assume 50% clogging)

Vcs

=

hLcs

(Vcs^2 - v2^2) / 2g x 1/0.7 Qpeak Wclr x 0.5 x Upstream Depth d"

=

0.92 m/sec

=

0.05

Therefore, X2 X3

= 2 + d" + (v"^2 /2g) E = E3 + d3 + (v3^2 /2g) + hLcs

X2

=

E2 d" v" E3 d3 v3

= = = = = =

0.00 0.120 0.43 0.00 0.055 0.93

X2 X3

= =

0.13 0.15

X3

Where, Trial and Error

m m m/sec m m m/sec

Check:

-0.018 Elevation of Channel Invert

=

79.00 m

Upstream Water Elevation

=

79.12 m

Downstream Water Elevation

=

79.06 m

ok

Height above datum Upstream Depth Upstream Velocity Height above datum Downstream Depth @ Qpeak Channel Velocity

DESIGN OF A CONSTANT VELOCITY GRIT CHANNEL Guidelines = 0.2 m/s flow through velocity Number of Channels Provided Number of Back-up Channels Provided

= =

1 1

Constant Velocity Channel Formula Q

=

1.71 B (H1.5)

Q B H

= = =

Flow Width of Throat Depth of Flow

=

( Q / 1.71 / B )^0.67

A W H

= = =

2/3 W H Width of Parabola Depth of Flow

V

=

0.2 m/s

Q

= =

A x V 2/3 W H x V

W

=

Therefore, Area of Parabola,

m3/s m m

m m

Assume:

Therefore

Assume:

B Qavg per Channel

= =

(Per guidelines)

1.5 Q H x V

0.33 m 517.50 m3/day

FLOW FACTOR ( x QAVG)

Q AT DIFFERENT FACTOR m3/sec

Depth H(m)

Width W(m)

0.10 0.25 0.50 1.00 2.00 3.00 4.00 4.29 5.00 6.00 7.00

0.001 0.001 0.003 0.006 0.012 0.018 0.024 0.026 0.030 0.036 0.042

0.01 0.02 0.03 0.05 0.08 0.10 0.12 0.13 0.14 0.16 0.18

0.43 0.58 0.73 0.92 1.16 1.33 1.46 1.50 1.57 1.67 1.76

0.0060 m3/s

@ Qavg

@ Qpeak

Determine Length of Channel Say settlement vel of particle

Vs

=

0.02 m/s 0.2 m/s

Say velocity of flow

V

=

Depth of channel at peak flow, Width of channel,

H W

= =

Length of Channel

L

=

L

=

1.29 m

L

=

3.00 m

Surface Area of Channel

SA

=

4.49 m2

Surface Overflow Rate

SR

=

Provided Length

=

0.13 m 1.50 m V x H Vs

Qpeak / SA 494.50 m3/m2/day

< 1500

m3/m2/day

Quantity of grit Guidelines = 0.03 m3 / 1000 m3 of wastewater Grit Quantity per Channel

Provide area for 30 days storage Grit quantity

=

Qavg x 0.03/1000

=

0.016 m3/day

=

0.47 m3

= = = =

3.00 m 0.50 m 0.15 m 0.225

Dimension of hopper at bottom of grit channel L : W ration 2:1 as per guidelines L W D Volume

1500

>=

0.466

ok

mm

700 mm 50 mm

450 mm

1000

mm

Check ratio of grit tank dimension Ratio W: D = 1 : 2

W 2.00

D 1

Ratio W: L = 1 : 2

W 1

L 2

Hydraulic Retension Time, Tr Guidelines = 3 min @ Qpeak < 5,000 pe Area of retangular Area of trapezium Area of grit storage

= = =

1.05 m2 0.06 m2 0.45 m2

Total Area L

= =

1.56 m2 3.00 m

Volume, V

=

4.69 m3

Qpeak

= =

2219.31 m3/day 1.54 m3/min

Tr

= =

V / Qpeak 3.04 min

>= 3 min

ok

X-Axis -0.837 -0.787 -0.748 -0.731 -0.664 -0.580 -0.460 -0.365 -0.290 -0.214 0 0.214 0.290 0.365 0.460 0.580 0.664 0.731 0.748 0.787 0.837

Y-Axis 0.161 0.143 0.129 0.123 0.101 0.077 0.049 0.031 0.019 0.011 0 0.011 0.019 0.031 0.049 0.077 0.101 0.123 0.129 0.143 0.161

CONSTANT VELOCITY FLOW X - SECTION 0.700

)

0.600

0.700

Depth (m)

0.600 0.500 0.400 0.300 0.200 0.100

GRIT STORAG

0.000 -0.100 -0.500

-0.400

-0.300

-0.200

-0.100

0.000

Width (m)

0.10

ITY FLOW CHANNEL ECTION

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

GRIT STORAGE

0.000

Width (m)

0.100

0.200

0.300

0.400

0.500

CONSTANT VELOCITY FLOW CHANNEL X - SECTION 1.250 1.150 1.050

Depth (m)

0.950 0.850 0.750 0.650 0.550 0.450 0.350 0.250 0.150 0.050

-0.650

-0.450

-0.250

-0.050 GRIT STORAGE -0.150 -0.050 Width (m) Channel

0.150

0.350

0.550

DESIGN OF A GRIT / GREASE CHAMBER Guidelines : Detension time T = 360 sec @ Qpeak Number of Channels Provided Number of Back-up Channels Provided

= =

1 1

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

L W D

= = =

2.90 m 1.60 m 0.85 m

V

=

3.94 m3

>

L:W W:D

= =

1.81 1.88

: :

Provided Length Width of channel, Surface Area of Channel

L W SA

= = =

2.90 m 1.60 m 4.64 m2

Surface Overflow Rate

SR

= = <

Qpeak / SA 478.30 m3/m2/day 1500 m3/m2/day

Grit Quantity

= =

Qavg x 0.03/1000 0.016 m3/day

Allow for storage of 30 days

=

0.47 m3

8 kg / 1000 m3 0.95

Peak Flow

Provide: Length Width Depth Volume Provided Ratio

9.25 m3 1.0 1.0

ok

Determine Length of Channel

ok

Quantity of grit Guidelines = 0.03 m3 / 1000 m3 of wastewater

Quantity of grease Average quantity of grease Specific gravity of grease

S sg

= =

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

Accumulated grease for 30 days Accumulated grit for 30 days

Vs

= =

0.131 m3 0.466 m3

Depth of Feed

D

=

0.25 m

Therefore,

A1

=

2.39 m2

= =

1.50 m 0.75 m

Grit/Grease Drying Bed Sizing

Area required

Dimension of Grit/Grease Drying Bed Length Width

Grease

error

Area

=

1.13 m2

>

2.39 m2

error

DESIGN OF A GREASE CHAMBER Guidelines : Detension time T = 180 sec @ Qpeak Number of Chamber Provided Number of Back-up Chamber Provided

= =

1 1

Peak Flow

= =

2219.31 m3/day 0.026 m3/sec 180 sec

Qpeak

Detension Time

T

=

Volume Required

V

=

Qpeak x T

=

4.62 m3

L W D

= = =

3.80 m 1.20 m 1.10 m

V

=

5.02 m3

Average quantity of grease Specific gravity of grease

S sg

= =

8 kg / 1000 m3 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 0.13 m3

Provide: Length Width Depth Volume Provided

>

4.62 m3

ok

0.13 m3

ok

Grease Quantities

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 = 0.02 m Volume of grease storage provided

V

=

0.09 m3

>

WEIR BEFORE AERATION BASINS Design Population, Design Flow, Peaking Factor, Peak Flow,

Number of Aeration Basins,

not use

PE Qavg PF Qpeak

N

=

Peak Flow to Each Aeration Basin

2300 517.50 m3/day 4.29 2219.31 m3/day

Qpeak / N 0.026 m3/sec

Q(m3/sec)

=

Cw x Le x H^1.5

Cw Le H

= = =

1.84 0.3704 m 0.148 m

Le

=

where

L n H

= = =

0.40 m 2 0.148 m

Therefore

Le

=

0.3704 m

Therefore

Q

=

where

0.03 m3/sec

1

= =

Formula

0.01 m3/sec

Weir Coefficient Length Height

With end contraction L - (0.1)nH

0.039 m3/sec

Length Number of side contractions Height of Flow

>=

0.026 m3/sec

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

ok

RECTANGULAR WEIR AT OUTLET BOX Design Population, Design Flow, Peaking Factor, Peak Flow,

Formula

PE Qavg PF Qpeak

2300 517.50 m3/day 4.29 2219.31 m3/day

Q(m3/sec)

=

Cw x Le x H^1.5

Cw Le H

= = =

1.84 0.578 m 0.11 m

Le

=

where

L n H

= = =

0.60 m 2 0.110 m

Therefore

Le

=

0.578 m

Therefore

Q

=

0.039 m3/sec

where

0.01 m3/sec 0.03 m3/sec

Weir Coefficient Length Height

With end contraction

Height of water above weir

=

L - (0.1)nH

110 mm

Length Number of side contractions Height of Flow

>=

@ Qpeak

0.026 m3/sec

ok

90 DEG V-NOTCH WEIR AT OUTLET BOX Design Population, Design Flow,

PE Qavg

2000 450.00 m3/day

Peaking Factor,

PF

Peak Flow,

Qpeak

4.35 1959.73 m3/day

0.023 m3/sec

90 deg V-Notch Weir Formula 5/2

q (m3/sec)

=

H (m)

=

q

=

Therefore,

Set depth of weir @ 250mm

1.417 H 192 mm 0.023 m3/sec

ok

or

0.192 m

at Qpeak

>=

0.023 m3/sec

ok

DESIGN OF ANOXIC CHAMBER Population

PE

=

Peak Flow

Qpeak

= = =

2,219.31 m3/day 92.47 m3/hr 1.54 m3/min

= =

0.026 m3/s 25.69 l/s

= =

517.50 m3/day 21.56 m3/hr

= = =

0.36 m3/min 0.01 m3/s 5.99 l/s

=

4.29

= = = = = = = = =

30 10 3600 2880 28 0.1 0.11

Qavg

Design Flow

Peaking Factor

PF

2300

Using the Lodification Ludzack-Ettinger Process Assume Influent Ammonia-N Ni Effluent Ammonia-N Ne Mixed liquor Suspended Solids MLSS Mixed Liquor Volatile Suspended Solids MLVSS Temperature Temp Dissolved Oxigen Do Specific Denitrification Rate U Overall Denitrification rate Ua Residence Time Temp

? ?

mg/l mg/l mg/l mg/l (MVLSS deg C mg/l per day per day hrs

=

0.8

x MLSS

Calculation Overall Denitrification Rate Formula :

Ua

=

U x 1.09(Temp-20)X(1-DO) 0.20 per day

T

=

Ni - Ne Ua x MLVSS

Calculation Residence Time Formula :

=

Size anoxic zone for a residence time of Anox

x

24 hrs/day

0.81 hrs required

=

2

hrs

=

Anox 24

=

43.13

m3

N

=

2

nos

Vreq

=

21.56

m3

Length Width Depth

= = = =

2.30 1.50 4.50

m m m

Anoxic tank volume

=

31.05

m3

Actual hydraulic retention time provided

=

Determine size of anoxic tank Anoxic zone volume required,

Number of Anoxic Tank, Volume required per Tank,

Vol

x

Qavg

Tank Dimension

=

Anoxic tank volume Qavg 2.88 hrs

>

21.56 m3

ok

EXTENDED AERATION TANK DESIGN Sizing Number of Aeration Tank,

No.

=

Minimum Hydraulic Retention Time,

HRT

=

V

=

418.31 m3

Vol

=

209.16 m3

Area Length Width Depth Freeboard Volume

A L W D FB Vol

= = = = = =

46.48 10.60 4.40 4.50 0.65 209.88

Total Volume Provided

Vp

=

419.76 m3 >

HRT Provided

=

19.47 hrs >

=

3600 mg/l

Volume Required, Volume Required Per Tank,

2 19.4 hours

Dimension of each tank:

Ratio

Therefore,

m2 m m m m m3

L 10.60 2.41

: : :

W 4.40 1

418.31 m3 19.4 hrs

ok

Sludge Age (MCRT) Guidelines > 20 days Assume,

MLSS 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

= =

10 mg/l 5.18 kg/day

Sludge Yield,

= =

0.40 @ 24 hours HRT 0.60 @ 18 hours HRT

Therefore actua

=

Sy

=

Sd

= =

QAVG x Sy (BOD5-EBOD5) 68.45 kg/day

SA

=

20.53 days

WAS

=

68.45 kg/day

Determine Sy @ Actual HRT Via Interpolation

Excess sludge wasting / day (Sludge accumulation per day)

(24 -21.9) (24 - 18)

x

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

= =

+

0.4

0.55 kg / kg BOD removed based on HRT interpolation

Waste Activated Sludge (WAS) Excess Sludge Wasting / day

(0.6 - 0.4)

6.84 m3/day 6844.61 l/day

ok

Return Activated Sludge Flow (RAS) QRAS

=

MLSS Cu - MLSS

Cu

=

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

QRAS

=

423.41 m3/day

4.92 l/s

Set QRAS

=

465.75 m3/day

5.42 l/s

Qavg

=

517.50 m3/day

Ratio of QRAS : QAVG

x

QAVG

=

0.90

ok

OR

=

2.00 kg O2 / kg BOD5

AOR

= =

QAVG Total

= =

517.50 m3/day 517.50 m3/day

= =

Total x 2 / 1000 1.04 kg/day

=

249.44 kg/day

Oxygen Requirements

Oxygen required per kg of BOD5 re Actual Oxygen Required

OR x BOD5 removed 248.40 kg/day

Maintain DO of 2 mg/l in aeration tank

Oxygen required in tank

Total AOR

Oxygen correction factor

AOR/SOR

AOR/SOR = Where

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

= = = = = = =

0.98 0.989 7.83 1.5 9.08 0.75 30

AOR/SOR

=

0.64

Standard Oxygen Required

SOR

=

391.23 kg/day

O2 transfer efficiency

O2eff

=

20 %

Actual O2 transfer efficiency

O2act

=

Therefore

@

6

m3/hr (See catalog)

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

Flowrate per diffuser

= =

0.10 m3/min 100.00 L/min

Quantity

=

Number of diffusers required

= Number of diffusers per tank

=

or

6

Total air required / Flowrate 48.75

Use

52

26 Nos in each Aeration tank

ok

m3/hr

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 F/M Ratio

= = =

419.76 m3 INBOD / (MLSS x Vp) 0.086

ok

Check Aeration Loading Guidelines :

0.1 - 0.4 kg/m3/day AL

=

INBOD / Vp

=

0.31 kg/m3/day

ok

Check Mixing Rate Per EPA guidelines,

0.010 to 0.025 m3/min. m3

Volume of air from Blower

Va

=

Volume of tank provided

Vp

=

Mixing Rate

=

Therefore,

=

#NAME? m3/min

(See Blower calculations)

419.76 m3

Va / Vp #NAME? m3/min. m3

#NAME?

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

= =

Qpeak / 30 m3/m2/day 73.98 m2

Number of clarifiers

=

2

Surface area required per clarifier

=

36.99 m2

L W

= =

6.20 m 6.20 m

Therefore, actual area provided

=

38.44 m2

Provided Length, L Provided Width, W

= =

6.20 m 6.20 m

Effective Volume SWD in cone

= =

Cone volume Vertical SWD

= =

65.73 m3 1.04 m

Vertical SWD volume

=

39.98 m3

Total Volume Total SWD

= =

105.70 m3 4.50 m

Dimension of each clarifier

>

36.99 m2

ok

Volume of cone with slope of 60 deg. 3.46 m (See attach volume cals) ok

HRT Per guidelines: < 2 hours = =

Total Volume m3 / Qpeak m3/day x 24 hours 2.29 >= 2 hours

ok

Overflow Weir Per guidelines: 150 - 180 m3/day/m Max weir loading rate, Wr Peak Flow, Qpeak

= =

Min Weir Length Required per Clarifier

= =

Total length of weir provided per Clarifier, Tw

=

Actual weir loading rate

=

Solids Loading Rate (SLR) Guideliness :

MLSS

180 m3/day/m 2219.31 m3/day Qpeak / Wr / No. of Clarifiers 6.16 m 3.00 m 369.89 m3/day/m

< 150 kg/m2/day at Qpeak < 50 kg/m2/day at Qavg =

3600 mg/l

Total Q = Qpeak + (For QRAS look under Extended Aeration Tank Sizing) = #NAME? m3/day SLR (Qpeak)

= =

SLR (Qavg)

error

= =

or

3.6 kg/m3

QRAS

(Total Q / Actual Surface Area) x MLSS kg/day #NAME? kg/m2/day

#NAME?

(Qavg + Qras) / Actual Surface Area x MLSS kg/day #NAME? kg/m2/day

#NAME?

Launder Calculation Overflow from weir to lauder

Launder size Length Width Flow velocity Depth of flow in launder

= =

= = = =

Qpeak 1109.66

3.00 0.25 0.80 21.41

/ number of tank m3 / launder

m m m /s mm

< 1.0 m/s ok < 0.3m as provided

Determine Volume of Cone L

W

L

W

h

h 60

Therefore, Volume of Cone

BW 1

A

L W BW 1 BW 2

= = = =

6.20 6.20 2.20 2.20

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 y

= = =

h /B Tan -1 (h/B) 60.00 degrees

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

=

L (m) W (m)

h (m)

A (m)

y

BW1 (m)

BW2

B

m m m m

65.73 m3

Scum Withdrawal Airlift Pipe Design Percent submergence Hs Hl

Scum Airlift Hs Hl Percent Submergence

=

Hs / (Hs + Hl)

= =

Depth (m) of air pipe below water surface Height (m) of lift

= = =

x 100

3.8 m 0.7 m 84.44 %

With reference to the attached chart, Discharge Velocity Air Requirement

= = =

45 gpm 2 ft/sec 7 cfm

2.84 l/s 0.61 m/s 0.20 m3/min

AEROBIC SLUDGE HOLDING TANK DESIGN Guidelines: Sludge yield = 0.6 kg/kg BOD5 (Standard "A")

Design Population,

PE

=

Design Flow,

QAVG

=

Sludge Yield

Ys

=

0.55 kg/kgBOD5/day

Sludge accumulation per day

Sd

= =

QAVG x Ys (BOD5-EBOD5) 68.31 kg

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

TSd

=

Nonvolatile Solid + VS Remaining

=

(WAS - TVS) + { (1 - VSSd/100) x

=

(26.73 - 20.05) + { (1 - 55/100) x 20.05)

Temperature of Wastewater

Therefore, Total Solids Remaining After Digestion

TSd

=

Density of water

ρ

=

Specific gravity of sludge

sg

=

2300 517.5 m3/day

1000.00 kg/m3 1.015 3.95 m3/day

T

=

30 days

Volume of Tank Required

Vtank

=

Number of Tank Provided

No.

=

Volume of Tank Required

V

=

Storage days provided

TVS }

40.13 kg/day

=

Therefore, TSd in term of volume

(Calculated earlier)

118.62 m3 1 118.62 m3

SIZE OF EACH TANK Depth Width Length

= = =

Volume

=

4.5 m 3.3 m 8.0 m 118.80 m3

>

118.62 m3

ok

OXYGEN REQUIREMENT FOR DIGESTION Guidelines: 1.5kgO2/kgBOD Wt of sludge digested

Ws

Amount of oxygen required, (AOR) Ws x 1.5

Oxygen correction factor AOR/SOR Where

= =

Sd - TSd 28.18 kg

=

42.27 kg

AOR/SOR =

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

BETA ACF Csf C Css Alpha T

= = = = = = =

0.98 0.989 7.83 1.5 9.08 0.75 30

AOR/SOR

=

0.64

Standard Oxygen Required

SOR

=

Assume O2 transfer efficiency

O2eff

=

Therefore

Total amount of air required

=

66.29 kg/day 10 % SOR kg/day O2eff/100 x 1.201 kg/m3 x 0.232 kg O2/kg air x 1440 min/day

=

1.65 m3/min

Flowrate per diffuser

=

200.00 L/min 0.200 m3/min

Quantity

=

Number of diffusers required

Total air required / Flowrate = 8.26

Use

10

ok

DRYING BED DESIGN Digested Sludge From Aerobic Digester Concentration of Sludge

DS

=

C

=

#NAME? kg/day 1% 10 kg/m3

= Specific gravity of sludge Volume of Digested Sludge

sg

=

Vds

=

(See Aerobic Digester cals)

1.015 #NAME? m3/day

Per guidelines: Provide 4 week cycle for 450mm thick feed depth Volume of Vds required (Based on 21 days drying; 7 days feeding)

V

=

Vds(m3/day) x 28days

=

#NAME? m3

Depth of Feed

D

=

Therefore,

A

=

#NAME? m2

A

=

#NAME? m2 ( 1/3 reduction in area)

Area required

0.45 m

Provide fully covered drying beds Therefore actual area required Number of Drying Beds provided

=

4

= = =

16.50 m 2.50 m 41.25 m2

Dimension of Drying Beds Dimension of Bed Length Width Area Total area of drying bed provided

=

nos of bed x bed area 165.0 m2

=

>

#NAME? m2

###

Determine Volume of Dewatered Sludge (DS) Assume final sludge concentration is 25% DS

=

Vs x Si x SGi x n SGo x So

DS Vs Si So SGi SGo n

= = = = = = =

Therefore DS

=

### m3/day

V

=

DS x 30 days

=

#NAME? m3

= = = =

0.50 1.00 4.60 2.30

? ### 0.01 0.25 1.015 1.03 0.95

Volume of Dewatered Sludge, (m3/day) Volume of Influent Sludge (m3/day) Fractional Percent Solid Content of Influent Sludge Fractional Percent Solid Content of Dewatered Sludge Specific Gravity of Sludge Before Thickening Specific Gravity of Sludge After Thickening Fractional Percent Capture

@ 25% solid

Provide covered storage for 30 days Volume required

Dimension provided

Depth Width Length Volume

m m m m3

>

#NAME? m3

ok

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

ENTRY EXIT CHECK VALVE SLUICE VALVE REDUCER TEE THRU SIDE TEE THRU RUN ELBOW - 45 DEG ELBOW - 90 DEG

K VALUE

QUANTITY

SUCTION TOTAL

QUANTITY

DISCHARGE TOTAL

0.50 1.00 2.50 0.20 0.30 1.80 0.60 0.23 0.30

1 0 0 0 0 0 0 0 0

0.50 -

0 1 0 1 1 0 1 0 5

0.00 1.00 0.00 0.20 0.30 0.00 0.60 0.00 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 = (K x V2 / 2g)

0.03 m

0.25

L = LINEAR METER OF STRAIGHT PIPE

0.00 m

15.00

M = MULTIPLYING FACTOR

1.10

1.10

100.00

100.00

C = HAZEN-WILLIAMS ROUGHNESS COEFFICIENT 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

The submersible pump selected is as follows:Make = Ebara Model = 65DVS51.5 Capacity = 9.10 L/sec Total Head = 4.73 m Power = 1.50 kW Discharge Size = 65 mm No. of Units = 1 (1 duty)

TDH

connect to

100

mm

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

=

[ 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

=

Actual velocity in selected pipe

0.068 m

=

80 mm

=

1.81 m/s

>

>

68 mm

ok

2.5 m/s

ok

Determine Total Dynamic Head ( TDH ) TDH

=

hst + hf + hm

= = =

Static head (m) Losses through the pipe ( Hazen - William Formula ) Losses through fittings

Where ;

hst hf hm

A

Thus, static head of pump, h st

B

Hazen - William Formula hf = 6.82

Where ;

=

V C

1.85

X

=

Velocity m/sec

V = Q/A

=

8 L/sec 1000 x 3.142 x 0.10

1.16 m/s

2

/ 4

1.16 m/sec

C

=

Coefficient of roughness

L

=

Length of pipe, m

D

=

Diameter of pipe, m

Thus, losses through the pipe, hf

C

L D1.167

V

=

4.00 m

0.00 15.00 0.10

=

#DIV/0! m

Losses through fittings hm

=

KV² 2g

Where ;

K

=

Head loss coefficeint

4.10

V

=

Velocity m/sec

1.16

G

=

Gravity, m/s²

9.81

hm

=

0.29 m

Therefore , Total Dynamic Head ( TDH )

= =

hst + hf + hm ### m

To plot chart for the System Curve & Pump Curve from the above equitition:Q (L/s)

V (m/s)

TDH (m)

0 2 4 8 9.5

0.00 0.25 0.51 1.02 1.21

4 4.04 4.15 4.57 4.80

* refer to Chart attached

Pump head (m) 18.6 16.4 14.2 8.3 4.2

18.6 16.4 14.2 13 8.33 4.2

4 4.04 4.15 4.23 4.57 4.8

…………A …………B …………C

DRYING BED FEED PUMP Submersible Pump EBARA Model 65DVS51.5 20 Pump Curve

18

Head (m)

16 Pump Operating Point 9.2 L/sec @ 4.7 m

14 12 10 System Curve 8 6 4 2 0 0

2

4

6 Flowrate, Q (L/sec)

8

10

X-Axis 0 3.3 6.67 10 13.33 16.67 19.98 65DVS

X-Axis 0 2 4 8 9.5

Pump System 16.2 4 14.2 4.25 12.4 4.92 11 5.97 9 7.39 7.4 9.17 5.6 11.27 Pump System 18.6 4 16.4 4.04 14.2 4.15 8.33 4.57 4.2 4.8

3.580986

Pump 0 198 400.2 600 799.8 1000.2 1198.8

System

Blower- 1 Sizing Air Requirement for Aeration

=

#NAME? m3/min

Air Requirement for Aerobic Sludge Digestion

=

#NAME? m3/min

=

#NAME? m3/min

Qd

TOTAL AIR REQUIREMENT

Determine air flow under standard condition, Qs Qs

=

Qd x (1.0332 + Pd)

273 + St

x

273 + Sd

1.0332

Qs

=

?

Air flow under standard condition

(m3/min)

Qd

=

#NAME?

Air flow under discharge condition

(m3/min)

Pd

= =

0.41 -0.05

Discharge static pressure Suction static pressure

(kgf/cm2) (kgf/cm2)

=

30

Suction temperature

o

=

38

Discharge temperature

o

=

#NAME? m3/min

PS St Sd

C C

Therefore, Qs

Determine discharge pressure under standard condition, Ps Ps

=

=

1.0332 + Pd 1.0332 + PS

-

0.48 kgf/cm2

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

1

x

1.0332

Blower-3 Sizing Air Requirement for Scum Airlift

=

0.20 m3/min

Air Requirement for RAS

=

#NAME? m3/min

Air Requirement for WAS

=

Air Requirement for MLSS

=

#NAME? m3/min #NAME? m3/min

=

#NAME? m3/min

Qd

TOTAL AIR REQUIREMENT

Determine air flow under standard condition, Qs Qs

=

Qd x (1.0332 + Pd)

273 + St

x

273 + Sd

1.0332

Qs

=

?

Air flow under standard condition

(m3/min)

Qd

=

#NAME?

Air flow under discharge condition

(m3/min)

Pd

=

0.45

Discharge static pressure

(kgf/cm2)

PS St

=

-0.05

Suction static pressure

(kgf/cm2)

=

30

Suction temperature

o

Sd

=

38

Discharge temperature

o

=

#NAME? m3/min

C C

Therefore, Qs

Determine discharge pressure under standard condition, P s Ps

=

=

1.0332 + Pd 1.0332 + PS

-

0.53 kgf/cm2

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

1

x

1.0332

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) Pump Dia (in) for RAS Pump Dia (in) for MLSS & scum

= = =

14.76 ft 3.00 in 2.00 in

4.50 m 0.075 m 0.050 m

S

=

13.01 ft

3.97 m

Calculate the Air Supply Volume (Vair) required: Vair

=

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

where h H C

= = =

Vair

=

A eff

=

0.75 total lift required (m) 4.50 submergence (m) 10.20 constant for less than 15m lift 0.47 m3/min of air per m3 of water 25.00 air-lift efficiency (%)

Determine Volume of Air Requirement for SCUM, VSCUM SCUM

= =

Vair-SCUM

= =

75.0 m3/day 0.0521 m3/min Vair x RAS x Aeff 0.0981 m3/min

Determine Volume of Air Requirement for RAS, VRAS RAS

= =

Vair-RAS

= =

#NAME? m3/day #NAME? m3/min

See Aeration Tank cals

Vair x RAS x Aeff #NAME? m3/min

Determine Volume of Air Requirement for WAS, VWAS WAS

= =

Vair-WAS

= =

#NAME? m3/day #NAME? m3/min

See Aeration Tank cals

Vair x WAS x Aeff #NAME? m3/min

Determine Volume of Air Requirement for MLSS Return, VMR MLSS

= = =

MLSS Return

= =

Vair-MLSS

= =

4 Qave - QRAS 2070.00 - 182.3 m3/day #NAME? m3/day #NAME? m3/day #NAME? m3/min Vair x MLSS x Aeff #NAME? m3/min

Aeration Piping Headloss To Aeration Tanks Criteria Ambient air temperature

To

=

Ambient barometric pressure

Po

=

1.00 atm

Air supply pressure

P

=

1.450 atm

Blower capacity

Qb

=

#NAME? m3/min

Blower efficiency

e

=

75 %

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

Headloss (mm)

hL

=

f x (L/D) x Hv

Size (mm) 80 80 80 80 80 80 80 50 25 25 25 25 25 25 25

Quantity 1 1

K Value 2.50 0.80 0.20 1.80 0.60 0.30 0.30 0.60 0.20 0.60 0.60 0.30 0.30 0.20 0.80

30 deg C

303.2 deg K

Equations

x

TQ^2 PD^4 or

K x Hv

A. Pipe Fittings Losses No. 1 2 9 3 4 5 6 4 9 7 4 5 8 9 10

Valves & Fittings Check valve Gate valve Reducer Tee thru side Tee thru run Elbow 90 deg Elbow 90 deg Tee thru run Reducer Tee thru run Tee thru run Elbow 90 deg Elbow 90 deg Reducer Gate valve

1 1 2 2 2

1

1

Q (m3/min) #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME?

T (deg K) 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 336.77 Subtotal

hL, headloss (mm) #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME?

B. Straight Pipe Losses No. 1 2 3 4

Length (m) 8.00 4.00 7.00 4.00

DIA (mm) 100 50 50 25

Velocity (m/min) #NAME? #NAME? #NAME? #NAME?

Q (m3/min) #NAME? #NAME? #NAME? #NAME?

f, fric. factor #NAME? #NAME? #NAME? #NAME?

T (deg K) 336.77 336.77 336.77 336.77 Subtotal

hL, headloss (mm) #NAME? #NAME? #NAME? #NAME? #NAME?

C. Supply Pressure At The Blower 1 2 3 3 4 5 6

Losses in piping Losses in pipe fittings Losses in air filter Losses in silencer Losses in blower Losses in diffusers Static head

= = = = = = =

Total

#NAME? #NAME? 50.00 50.00 150.00 160.00 4100.00

mm mm mm mm mm mm mm

#NAME? mm

Therefore, the absolute supply pressure

=

@

#NAME? m

#NAME? atm

D. Power Requirement of Blower, P (kw)

where

Therefore,

P

=

R

=

8.314 kJ/k mole deg K

w

=

air mass flow, kg/s

P

=

#NAME? Kw

=

5.5 Kw ok

Select next bigger size motor

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

w RTo 8.41 e

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

or

#NAME? HP 7 HP

CHLORINATION TANK DESIGN Qpeak

=

2219.31 m3/day

=

1.54 m3/min

Detension Time at Qpeak

t

=

Volume of Tank

V

=

15.00 min Qpeak x t

=

23.12 m3

Number of Tanks

N

=

1

Number of Bays per Tank

n

=

4

Dimension of Tank Provided Depth Wetted depth Width Length Number of pass

H h W L n

= = = = =

1.80 m 1.50 0.75 m 4.50 m 4

Vp

=

Volume Provided

24.30 m3 >=

ok

max 3m

23.12 m3

Check: Ratio

Ratio Wetted depth 1.50 2 Length 4.50 6

Detension Time at Qpeak

t

= =

ok : : : : : :

Width 0.75 1 Width 0.75 1

Vp / Qpeak 15.77 min >

15 min ok

PARSHALL FLUME FOR DISCHARGE Design Flow, Peaking Factor Peak Flow

Qavg PF Qpeak

= = =

517.5 m3/day 4.29 2219.31 m3/day

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

Flow thru 1" PF,

Q

=

0.338 x H1.55

Flow tru PF

Q

= =

flow in PF in cfs 0.9071091 cfs = Qpeak

Head over flume

H

=

in ft

H

=

1.8263509 ft 556.7 mm

where

For Q > Qpeak, water level in PF

90 DEGREE V-NOTCH WEIR AT OUTLET BOX Design Population , Design Flow, Peaking Factor Peak Flow

PE Qavg PF Qpeak

= = = =

2300 517.5 4.29 2219.31

m3/day m3/day

90 Deg V-Notch Weir Formula

1.417H 5/2

q(m3/sec)

=

H (m)

=

150

Therefore, q

=

0.012348

Set depth of weir @

200

mm

ok

mm or m3/sec > =

0.15

m at Qpeak

0.02569

m3/sec ok

INFLUENT Flow (m3/d) BOD (kg/day) TSS(kg/d)

INFLUENT

517.50 129.38 155.25

Flow (m3/d) BOD (kg/day) TSS(kg/d)

INFLUENT FLOW

INFLUENT

#NAME? 129.38 #NAME?

Flow (m3/d) BOD (kg/day) TSS(kg/d)

SECONDARY EFFLUENT

#NAME? 109.97 #NAME?

SCREEN CHAMBER *

MLSS Flow (m3/d)

Flow (m3/d)

#NAME?

BOD(kg/d) TSS(kg/d)

PUMP STATION

ANOXIC TANK

AERATION BASINS

#NAME? #NAME? #NAME?

SECONDARY CLARIFIERS

FINAL EFFLUENT 10 mg/L BOD 20 mg/L TSS

LAST MANHOLE RAS RAS = 1% Flow (m3/d) TSS(kg/d)

423.41 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) TSS(kg/d)

#NAME? #NAME? DIGESTED SLUDGE Flow (m3/d) TSS(kg/d)

DEWATERED SLUDGE

3.95 #NAME?

Flow (m3/d) TSS(kg/d)

AEROBIC SLUDGE HOLDING TANK (75% VSS) (55% VSS Destruction)

SLUDGE DRYING BEDS (25-40% SOLIDS) (95% CAPTURE)

SUPERNATANT Flow (m3/d) TSS(kg/d)

SOLIDS BALANCE @ Qavg

#NAME? #NAME?

#NAME? #NAME?

INFLUENT Flow (m3/d) BOD (kg/day) TSS(kg/d)

INFLUENT FLOW

INFLUENT

689.63 172.41 206.89

Flow (m3/d) BOD (kg/day) TSS(kg/d)

INFLUENT

697.82 172.41 236.97

Flow (m3/d) BOD (kg/day) TSS(kg/d)

SECONDARY EFFLUENT

683.87 146.55 201.43

Flow (m3/d) BOD(kg/d) TSS(kg/d)

SCREEN CHAMBER *

AERATION BASINS

683.82 6.84 13.68

SECONDARY CLARIFIERS

FINAL 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) TSS(kg/d)

9.10 91.03 THICKENED SLUDGE Flow (m3/d) TSS(kg/d)

GRAVITY SLUDGE THICKENER TANK (3% Solid)

SUPERNATANT Flow (m3/d) TSS(kg/d)

DIGESTED SLUDGE

2.09 62.79

Flow (m3/d) TSS(kg/d)

AEROBIC SLUDGE HOLDING TANK (80% VSS) (55% VSS Destruction)

SOLIDS BALANCE @ Qavg

Flow (m3/d) TSS(kg/d)

SLUDGE DRYING BEDS (25-40% SOLIDS) (95% CAPTURE)

SUPERNATANT

7.01 28.24

DEWATERED SLUDGE

1.23 36.89

Flow (m3/d) TSS(kg/d)

1.18 1.84

0.05 35.05

SLUDGE HOLDING AREA (30 DAYS)

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