Stp Calculation Ns_2200(1) 35194g

1.0 BASIC DESIGN CRITERIA 1.1 Population Equivalent Total Population Equivalent

=

2300

PE

=

0.225

m3/day.PE

=

517.50

m3/day

=

5.99

x

1.2 Waste Water Flow Waste Water Flow Rate Average Flow (QA)

10-3

m3/sec -0.11

Peak Flow Factor (pf)

= =

Peak Flow (QP)

=

1.3 Waste Water Characteristic Influent BOD5 concentration Influent SS Concentration BOD5 Load ( 0.055 kg/p.day ) SS Load ( 0.068 kg/p.day ) 1.4 Treated Effluent Quality Design Effluent BOD5 Design Effluent SS 1.5 Standard A Effluent Quality Effluent BOD5 Effluent SS

4.70 4.29

x

2300

÷

10-3

m3/sec

1000

2219.31 m3/day

=

25.69

x

= = = =

250 300 126.5 156.400

mg/l mg/l kg/day kg/day

= =

10 20

mg/l mg/l

= =

20 50

mg/l mg/l

Note:The followings are not to be discharged into the system: Oil and grease, chemicals, metals, radio active waste products and excessive detergents.

2.0 DESIGN OF INLET CHAMBER 2.1 Velocity Through Channel Flow velocity through channel Peak flow (QP)

=

1.0

=

25.69

x

10-3

m3/sec

Wetted cross-sectional area

=

25.69

x

10-3

m/sec

Width of the inlet channel

= =

0.0257 0.50

m2 m

= =

0.0257 0.051

m2 m

<

÷ 0.30

0.50 m m (provided)

/

1000

Minimum depth required, d

2.2 Bar Screen and Screen Quantities Adopt clear spacing =

25

m/sec ÷

1.0

….o.k

mm

Volume of screenings

=

0.03

m3

Screenings removed

=

0.0155

m3/day

Volume for 7 days storage Provided collectin bin

= =

0.1087 0.50

m3 m

=

0.045

m

x

0.30 >

3

m3 of wastewater

m

x

0.30 m

0.109 m3

…ok

2.3 Approach & flow through velocity a) Approach Velocity Peak flow Depth of channel Width of channel Cross sectional area of

= = =

25.69 0.30 0.50

m m

channel provided

=

0.15

m2

Approach velocity

= =

25.69 0.171

x m/sec

= = = = = = = =

25 10 35 0.50 14.29 150 0.35 0.30

mm mm mm m mm mm m m

screen available for sewage fl

=

0.35

Hence flow through velocity

= =

25.69 0.245

b) Flow through velocity Clear spacing of bars Width of bars Space between c/c of bars Width of bar screen Hence, no of bars Space occupied by bars Space for sewage flows Depth of channel Cross sectional area of bar

x

m

10-3

m3/sec

10-3 <

m3/sec ÷ 0.15 1.0 m/sec…….o.k

=

x

x m/sec

15

0.30 m 10-3 <

Nos

= 0.105 m2

m3/sec ÷ 0.105 1.0 m/sec…….ok

3.0 DESIGN OF RAW SEWAGE PUMP SUMP 3.1 Raw Sewage Pump Capacity Effective Volume of Sump (wet well) Where

4

V

=

TQ

Q T

= =

Peak flow Cycle time

= 6

### m3/day

V

=

( 6

###

÷

Provide wet well size of 2.50 Required effective depth

m

x = =

x

2.50 2.312

( 4 x 24 x 60 )

2.312

m m3 ÷



0.370

2.50 m 0.45

x

2.50 m

m (provided)

3.2 Pipe Sizing Pump capacity at peak flow

= 25.69 x 10-3 m3/sec = 25.69 l/sec Desirable velocity in the discharge pipe at maximum pump discharge ≤ 2.5 m/sec (MS 1228) Adopt velocity in discharge pipe as 2.0 m/sec Diameter

D

Provided pipe diameter Actual mean velocity,

V

=

25.69





x

10-3

m3/sec

x

2.0

m/sec

=

0.128

=

150

=

25.69

x

10-3

m3/sec



x m/sec

0.15

m

=

1.45

x

m mm ( DI material or equivalent)

x

x

4

0.15 m

Other losses = pipe + fitting Losses through the pipe hf(Hazen-William Formula) 0.85 x C x R0.63 x (hf/L)0.54 V = Losses through the fitting He, (Hazen-William Formula) K x V2  2g He = V = Mean velocity, 1.45 m/sec C = Hazen-William Coefficient for pipe = 90 for sewage R = Hydraulic Radius = 0.038 m L = Length of pipe = 0.00 m K = Resistance coefficient for fitting & valve 0.60 g = Acceleration due to gravity = 9.81 m/sec (hf/L)0.54 = 1.45 m/sec hf/L hf

= =

0.85

x

0.1503 0.0300

)

90

x

0.00

m

0.038 0.63

1.85

x

= 0.000

He

Total loss Total Head

Provided Total Head

= =

= = = = =

0.60 0.065

x 1.45 )2  m/per unit of fitting

2

x

0.065 m static head + other losses 0.00 m + 0.065 m 0.00 m 0.00 m

Provided 2 numbers of pumps, 1 working and 1 standby

ach pump capacity

=

3.4 Raw Sewage Pump Specification Type = Model = Total Head = 0.00 m Capacity = 0.00 l/sec Power = 0.00 kW

0.00

l/s at

0.00

m total head

0.00

- Refer Appendix A

>

Provided effective volume from 1st pump start level to pump stop level = 2.50

25.69 l/sec

m (L) x >

2.50

=

2.813

m

Retention Time/Time to fill

= =

2.813 7.83

m3 min

Time to empty

= =

2.813 0.00

m3  ( 0.000 min

Actual Pump cycle at ADWF

= =

7.83 7.83

No. of start/stop

= =

60 8

3



…ok

m(W) x 0.45 m(D)

2.312 m3 0.359 m3/min < 30 min…..o.k -

min + min

0.00

min

÷ nos.

7.83

min

0.359 ) m3/min

=

7.7

4.0 DESIGN OF SECONDARY CHAMBER 4.1 Velocity Through Channel Flow velocity through channel Peak flow (QP)

=

1.0

=

25.69

x

10-3

m3/sec

Wetted cross-sectional area

=

25.69

x

10-3

m/sec

Width of the channel

= =

0.0257 0.60

m m

= =

0.0257 0.043

m2 m

<

÷ 0.30

0.60 m m (provided)………….o.k

/

1000

Minimum depth required, d

4.2 Bar Screen and Screen Quantities Adopt clear spacing =

12

m/sec ÷

1.0

2

mm

Volume of screenings

=

0.06

m3

Screenings removed

=

0.0311

Volume for 7 days storage Provided collectin bin

= =

0.217 0.50

m3 m

=

0.080

m

m3 of wastewater

m3/day x

0.40 >

3

m

x

0.40 m

0.217 m3

…ok

4.3 Approach & flow through velocity a) Approach Velocity Peak flow Depth of channel Width of channel Cross sectional area of

= = =

25.69 0.30 0.60

m m

channel provided

=

0.18

m2

Approach velocity

= =

25.69 0.143

x m/sec

= = = = = = = =

12 10 22 0.60 27.27 280 0.32 0.30

mm mm mm m mm mm m m

screen available for sewage fl

=

0.32

Hence flow through velocity

= =

25.69 0.268

b) Flow through velocity Clear spacing of bars Width of bars Space between c/c of bars Width of bar screen Hence, no of bars Space occupied by bars Space for sewage flows Depth of channel Cross sectional area of bar

x

m

10-3

m3/sec

10-3 <

m3/sec ÷ 0.18 1.0 m/sec…….o.k

=

x

x m/sec

28

0.30 m 10-3 <

Nos

= 0.096 m2

m3/sec ÷ 0.096 1.0 m/sec…….ok

5.0 DESIGN OF MANUAL GRIT AND GREASE CHAMBER 5.1 Grit Chamber To provide two units of channel, one working and one standby Peak Flow Flow velocity through chamber

= =

25.69 0.20

x m/sec

Surface Loading Rate Detention Time

= =

1500 60.00

m3/m2 day sec ( 1 min)

Required Volume Size of chamber, Leng Width Depth

= = = =

1.541 2.50 1.25 0.65

m3 m m m

=

2.031

m3

=

25.69 1.25 0.0316

x m x <

=

25.69 2.50

x m x

=

0.0082

Adopt grit quantity

=

0.03

m3/103 m3 wastewater

Therefore, grit removed

=

0.03

m3 x 1000

=

0.067

m3/day

=

0.067

m3/day

=

1.997

m

= =

2.031 1.02

m3  times

Horizontal flow velocity

L W D

= Surface Loading Rate

10-3

>

m3/sec

1.541 m3

…ok

10-3 0.65 0.20

m3/sec m m/sec

…ok

10-3 1.25

m3/sec m

m3/m2 day

<

1500 m3/m2 day

5.2 Grit Storage Capacity

Volume of Grit for 30 days

Degritting frequency

### m3/day

x

30

3

1.997 m3

Therefore, degritting work shall be peformed every month

5.3 Grease Chamber Peak Flow

=

25.69

x

=

1.541

m3/min

10-3

m3/sec

days

Detention Time

=

180

Required Volume Size of chamber, Leng Width Depth

L W D

= = = =

4.624 2.50 2.00 0.95

m m m m

Volume

V

=

4.75

m3

Check for actual detention tim

=

4.75

m3

=

1.541 3.08

Volume of grease collected, VG where, QA

sec

(3 min)

3

m3/min min

>

4.62 m3

…ok

>

3.00 min

…ok

= =

( QA x c ) ÷ P

c

=

Grease concentration in water

p VG

=

Density of grease

=

Average Flow

517.50

m3/day 950

=

0.0817

=

2.451

950 x kg/m3

m3/day m3/30 days

0.150 kg/m3

kg/m3 ) 0.15 kg/m3

e detergents.

d)

1

m/sec

….o.k

m2

m2

2

m3

m/sec (MS 1228) 4

0 for sewage

m

9.81

3

m3/min

) m3/min

4

m/sec

d)………….o.k

m2

m2

5

m3/m2 day

6

)

7

STANDARD CALCULATION

1.0 DESIGN DATA 1.1 Design Process

= Extended Aeration

1.2 Population Equivalent

=

2300

1.3 Waste Water Flow Per Person

=

0.225 m3/PE. day (or 225 Ipcd)

1.4 Average Waste Water Flow, QA = 1.5 Peak Flow Factor

PE (maximum allowed PE)

517.50 m3/day

= 4.7 x P-0.11 -0.11

=

4.70

x

2300

÷

1000

=

4.29 -3

Peak Flow, QP

=

###

m3/day

=

1.6 Type of Waste

= Domestic

1.7 Influent BOD5 Influent SS

= =

250 300

ppm ppm

1.8 Effluent BOD5 Effluent SS

≤ ≤

10 20

ppm ppm

1.9 EQA BOD and SS Value for Standard A

= BOD 20 ppm and ss 50 ppm

25.69

x

10

m3/s

Note : The following are not to be discharged into the system : chemical of toxic/corrosive nature, radio active waste products, metal, oil and grease.

6.0 EQUALIZATION TANK Required Volume, V = Qp x 1.50 ÷ 24 QP = Wastewater at Peak Flow ( m3/d ), T = Retention Time = 1.50 hrs VR = ### x 1.50 ÷ 24 = 6.1 Capacity of Provided Tank Tank Diameter Water Depth

138.71 m3

= =

3.40 2.90

m m

Effective Cross Section Area

=

8.85

m2

Effective Volume of End Cap Cylindrical Tank Length No. of Tank No. of end cap Actual Volume of Tank

= = = = =

1.4098 7.60 2 4 8.85

=

140.16 m

>

### m

= = =

5.0 watt/m3 700.80 watt 0.70 kW

x

### m3

<

11.00 kW

Mixing Requirement

= ### m3/day

m3 m unit unit x

15.2

3

+

###

x

4

3

……ok

6.2 Required air Volume Recommended Air

= 1.0 m3 of air/m3 of sewage hour (Ref. Metcalf & Eddy) =

140.16 m3

=

140.16 m3/hr

10% air losses

=

0.234 m /min

Total Air to be supplied

=

2.336 m3/min

= 6.3 Coarse Bubble Air Diff Specification Type Model Capacity Required Air diffs Provided No.

x

1.0 =

m3/hr 2.336 m3/min

3

2.57

+

0.234 m3/min

m3/min

- Refer Appendix B = Coarse Bubble = UNIFLEX PT 3/4 = 200 Nl/min = 2.57 ÷ 0.20 = 14 nos

=

12.85

24

x

6.4 Transfer Pump Total amount of waste water

=

517.50 m3/day

÷

60

to be transferred

=

0.359 m3/min 8

6.5 Pipe sizing Pump capacity

=

0.359 m3/min

= =

5.99 x 5.99 l/sec

10-3

m3/sec

Desirable velocity in discharge pipe at maximum pump discharge ≤ 2.5 m/s (MS 1228) Adopt velocity in discharge pipe as 2.0 m/sec, 2

Diameter, D Provided pipe diameter

= = =

5.99 x 0.062 m 80 mm

Actual mean velocity, V

=

5.99

=

1.19

π

Other losses

=

x x m/sec

10

-3

10-3 0.08

x

4

x x

4 0.08

x

R0.63

÷

π

x

(hf/L)0.54

=

90

= = = =

0.02 3.50 0.60 9.81

x

2

pipe + fitting

Losses through the pipe hf ( Hazen-William Formula ) V =

0.85

x

C

Losses through the fitting He, ( Hazen-William Formula ) where:He V C R

= = = = = L = K = g =

K x V2  2g Mean velocity, 1.19 m/sec Hazen - William Coefficient for pipe Hydraulic Radius 0.08 m ÷ 4 Length of pipe resistance coefficient for fitting & val acceleration due to gravity

for sewage m m m m/sec 0.63

(hf/L)0.54 =

1.19

÷

0.85

x

90

x

0.02

2

x

9.81

+

0.043

1.85

hf/L = hf = =

0.183 0.043 0.151

x

3.50 m 2

Total Head

He = 0.60 x 1.19 ÷ = 0.043 = Static Head + Other Losses = 3.00 m + 0.151 = 3.19 m

Provided Total Head

=

4.00

m

9 6.6 Transfer Pump Specification Type Model Power Capacity Number

- Refer Appendix C = Ebara or Approved Equivalent = 65 DVS 5.75T = 0.75 kW = 6.60 l/sec at 4.00 m head = 2 units (1 working and 1 standby)

>

5.99 l/sec

6.7 Flow Control Box Flow Control Weir Equation Q Q H Design PE Design average flow (Q)

=

1.42

H5/2

= Free Discharge Over 90o V-notch Weir m3/sec = Vertical Distance (head) from weir to the free water surface, m = 2300 PE =

517.50 m3/day

=

5.99

Q

=

1.42

H

= =

x

= 10-3

0.359 m3/min m/sec

H

5/2

2/5

5.99 x 0.112 m

10 or

-3

÷ 112

1.42 m/sec mm/weir

7.0 AERATION TANK 7.1 Design Parameter based on DGSS Guidelines a. F/M Ratio = 0.05 - 0.10 b. HRT/hr. = 18 - 24 c. Sludge Age/day (Qc) = > 20 d. MLSS/ (mg/l) = 2500 - 5000 e. MLVSS = 0.8 MLSS kg O2/BOD5 removed f. Oxygen requirement = 2.0 7.2 Tank Volume VR = (QA x 18) ÷ 24

=

517.50

=

388.13 m

7.3 Capacity of Provided Tank Diameter of Tank Water Depth

= =

Cross-Sectional Area

=

x 3

3.40 2.95

m m

8.897 m2

18

÷

24

Volume of End Cap Cylindrical Tank Length Number of tank Number of end cap

= = = =

1.416 11.00 4 8

m3 m unit unit 10

Volume of the provided tank

Check the provided HRT

7.4 Aeration Rate Provided Air Rate

=

8.897

x

=

402.80 m

= =

402.80 ÷ 18.68 hrs

=

3

2.0

44.0

+

>

1.416

x

8

### m

3

### …ok

x

24

kg O2/BOD5 removed

Amount of BOD to be removed =

0.25

kg BOD/m3

-

=

0.24

kg BOD/m of sewage

0.01 kg BOD/m3

3

Total BOD5 per day

= =

517.50 m3/day x 124.2 kg BOD5/day

Amount of O2 to supply (AOR)

= =

124.2 kg BOD/day 248.4 kg O2/day

Mixing Requirement

= = =

20 watt/m3 8055.9 watt 8.06 kW

0.24 kg BOD/m3

x

2.0

x

402.8 m3

<

11.00 kW

kg O2/BOD

……ok

8.5 Field Conditions N(Field) N NS

T-20 xα = NS (CS - CL)  9.17 x (1.024) Oxygen transfered in field condition kg O2/hr = = Oxygen transfer under standard condition kg O2/hr

CS CL

= Dissolve oxygen saturation value at 25oC :- 8.38 mg/l = 1 - 2 mg/l in mixed liquor (assume 2.00 mg/l)

T α Therefore:-

= Temperature (oC), 25oC = Correlation Factor (0.85) 248.4 = Ns = =

7.6 Air Volume Required Assume O2 Weight of 1 m air 3

Ns

x 6.38 248.4 x 9.17 kg O /day 373.36 2

=

23.2

% of air

=

1.20

kg

x ÷ =

1.125 x 0.85 6.38 x 1.125 kg O /min 0.259 2

÷ x

9.17 0.85

Air required normally

=

0.259

But efficiency of diff

= = =

Actual volume of air required =

x

100

÷

23.2

0.93 25 0.93

m /min % x 100

÷

25

3.73

m3/min

x

1.20

3

11 7.7 Fine Bubble Air Diff Specification Type Model Capacity Required Air Diff No. Provided

- Refer Appendix D = Disc Diff = UNIFLEX U-330 = 100 Nl/min = 3.73 ÷ = 40

0.10

=

37.3

7.8 Recirculation Ratio ( QR/Q) Cu = Underflow concentration ( kg/m3 ) MLSS =

4.20

kg/m3

MLSS (Cu-MLSS) = =

4.20 0.72

kg/m3

=

0.72

x

= 374.74 m3/day The allowed recirculation ratio shall be 0.50 - 1.0

÷

10

-

4.20

kg/m3

### m3/day =

0.260 m3/min

7.9 F/M Ratio MLVSS MLSS MLVSS Recheck the F/M Ratio

= = = = =

0.80 x 2500 2000 517.50 x 0.096 (0.05 -

MLSS 5000 4000 0.25 ÷ 0.1)…..ok

3.36

x

###

### Excess Sludge Wastage (SW) Sludge Production = 0.60 kg/kg BOD removed at 18 hrs retention time SW = QA x BOD Removed x Sludge Production = = ### Sludge Age ( θSludge) SS in treated effluent = Solids in treated effluent = = Sludge Age, θSludge =

517.50 m3/day x 0.24 kg/m3 74.52 kg sludge/day

x

20 mg/l 517.50 x 0.02 10.35 kg/day Total solids in aeration tank * MLSS Excess sludge wasting/day + solid in effluent = 402.80 x 4.20 kg

0.60

= ### Waste Activated Sludge Qw WAS = QW =

74.52 20

+ >

402.80

x

7.45

m3/day

10.35 kg/day 20 days …….ok

4.20

÷ 10

20

-

###

=

5.18

x

10-3

x

0.02

m3/min 12

8.0 CLARIFIER TANK (HOPPER) 8.1 Design Criteria Hydraulic detention time at QA >

2.0

hrs

Weir Loading Rate

≤ 150 - 180m3/m.day

Surface overflow

≤

30

m3/m2.day

Solid Loading Rate

≤

50

kg/m2 day

Required Capacity (VSD)

= =

QA

8.2 Provided tank dimensions Diameter Water Depth No. of Tanks

x x

2.0 2.0

hrs ÷

÷ 24

m2 m

x x

4 4

= =

36.32 m2 36.45 m

4

=

71.48 m3

517.50

=

43.13 m3

= = =

3.40 m 3.250 m 4 unit

Surface Area Weir Length

= =

Tank Capacity

=

17.87 m3

x

Provided Capacity

=

71.48 m3

>

Actual Detention Time

= =

71.48 x 3.32 hrs

2

Surface Overflow Rate (SOR)

= =

Weir Overflow Rate (WOR)

QA

=

QA

= =

43.13 m3 hrs

÷

Area

517.50 m /day 14.25

=

÷

÷

3

=

=

Solid Loading Rate (SLR)

9.08 9.11

36.32 m2

m3/m2.day ÷

WL

517.50 m3/day

÷

36.45 m

14.20 m /m.day 3

QA 517.50

x x

2.00 2.00

24

÷ ÷

Area 36.32

43.13

=

28.50 kg/m2day

13 8.3 Scum Skimmer Design To provide operated scum skimmer to remove scum accumulated on the surface of the Clarifier Tank to Sludge holding Tank Thickness of scum to be removed = 75 mm Volume of scum to be removed

=

0.075

x

=

2.724 m3

=

2.724 m3

36.32 m2

Assume skimmer to operate 40 min/day , Flow rate

÷

40.0 min

= 0.068 m /min is calculated as follows:= h ÷ c log [ (H +10.4) ÷ 10.4] = volume of free air = Total lift, m = Constant, 10.2 = Submergence, m = 1.000 m = 2.200 m = 1.000 10.2 log ( 2.20 + 3

Air Required for airlift pumping Va Va h C H For h H Air required, Va

Adopt efficiency of air lift Air required for 2 units scum skimmer/tank Qsc Total air required in Clarifier

= =

10.4 ) ÷

10.4

0.927 m / m of scum 25% 3

3

= Flow rate of scum removed/25% removal efficiency =

0.252 m /min

=

1.010 m3/min

3

x Va

14 9.0 SLUDGE HOLDING TANK Sludge accumulation rate for Extended Aeration Process is 0.60 kg solid/kg BOD with 1% concentration of transfer sludge. (Ref. : Guidelines on Design of Sewerage Treatment Works, DGSS, July 1995) Anaerobic storage tank is designed to provide a minimum of 30 days solid storage capacity. Amount of BOD Loading to be removed

= =

517.50 m3/day x 124.20 kgBOD/day

Sludge accumulation rate

= =

124.20 kgBOD/day 74.52 kg solids/day

VSS digested = Specific gravity of digested slu =

50 % 1.015

Density of water

=

1000

Digested VSS volume

= =

74.52 kg solids/day 37.26 kg/day

0.24 kgBOD/m3

x

0.60 kg solids/kg BOD

x

0.50

kg/m3

SS remained in Sludge Holding = Tank =

74.52 37.26 kg/day 37.26 kg/day

Volume of sludge remained in

=

37.26 kg/day

Sludge Holding Tank

=

3.67

m /day

Required volume for 30 days

=

3.67

m3/day

(1 month) storage

=

9.1 Provided Capacity Diameter of Tank Water Depth

= =

110.13 m

3

3.40 2.95

÷

1.015

x

30

x

3

m m

Cross-Sectional Area

=

8.897 m2

Volume of End Cap

=

1.416 m3

days

0.01

x

###

Cylindrical Tank Length Number of tank Number of end cap Total Tank Capacity

= = = =

Actual Storage Capacity

12.20 m 1 unit 2 unit 8.897 x

12.20 >

+

=

111.38 m

= =

111.38 ÷ ### m3 1.01 months

3

1.416

x

2

1

month

### m

3

x

15 9.2 Sludge Pump Total amount of waste water

=

37.26 m3/day

x

to be transferred

= =

260.82 m / 7 days 260.82 ÷ 24

=

181.13

x

=

301.88

=

301.88

7

days

x

60

3

10

-3

m /min

x

10

-5

m3/sec

x

10-5

m3/sec

3

9.3 Pipe sizing Pump capacity

or

3.019 l/sec

Desirable velocity in discharge pipe at maximum pump discharge ≤ 2.5 m/s (MS 1228) Adopt velocity in discharge pipe as 2.0 m/s, 0.5

Diameter, D Provided pipe diameter

= = =

301.88 x 0.044 m 100 mm

Actual mean velocity, V

=

301.88

= Other losses

=

x π x 0.384 m/sec

10

-5

x

4

÷

π

x

2

( D.I pipe or equivalent ) 10-5 0.10

x x

4 0.10

x

R0.63

x

(hf/L)0.54

K x V2 2g Mean velocity, 0.384 m/sec Hazen - William Coefficient for pipe Hydraulic Radius 0.10 m ÷ 4

=

90

pipe + fitting

Losses through the pipe hf ( Hazen-William Formula ) V =

0.85

x

C

Losses through the fitting He, ( Hazen-William Formula ) where:He V C R

= = = = =

=

for sewage

0.025 m

L = Length of pipe K = resistance coefficient for fitting & val g = acceleration due to gravity

= = =

10.00 m 0.60 m 9.81 m/sec 0.63

(hf/L)0.54 =

0.3843

÷

0.85

x

90

x

0.025

÷

2

x

9.81

1.85

hf/L = hf = =

0.0513 0.0041 x 0.0411 m

He = =

0.60 x 0.0045 m

10.00 m 2

0.384

16 Total Head

Provided Total Head 9.4 Sludge Pump Specification Type Model Power Capacity Number

= Static Head + Other Losses = 3.20 m + 0.041 = 3.25 m = 4.00 m

m

- Refer Appendix C = Ebara or Approved Equivalent = 50 DVS 5.4T = 0.40 kW = 3.80 l/sec at 4.00 m head = 1 unit

+

### m

>

3.02 l/sec

### SLUDGE DRYING BED Volume of sludge remain in Sludge Tank

=

Required volume for 30 days

=

Required area based on 450m

=

37.26 kg/day 3.67

m3day

÷

1.015

x

30

x days

0.01

x

###

=

### m3

110.13 m3

÷

0.45 m

=

### m2

Area Required for covered bed = Provided dimension for covered =

244.73 m2 7.00 m (L)

x x

2/3 6.00 m (W)

= x

### m2 4 units

Sludge Drying Bed Provided nos. of sludge drying bed

= =

168.00 m2 4

>

### m2

=

517.50 m3/day

=

0.359 m3/min

### CHLORINATION CHAMBER Average Flow, QA

Hydraulic Retention Time

=

Total Volume of Chamb =

30

minutes

0.359 m3/min

x

30

min

Let liquid depth

= =

10.78 m 1.00 m

Surface Area Required

=

10.78 m3

÷

1.00 m

Provided surface area

= =

10.78 m 3.50 m

x

3.00 m

= = =

10.50 m 3.50 m 4 es

> ÷

10.78 m2 0.875 m

Number of es

3

2

2

17 ### AIR BLOWER SIZING The required air blower's pressure is determined by summing up the following: Static water pressure over the = 2950 diff Pressure loss through distributi = 100 pipe Air filter resistance = 100 Air diff resistance = 100 Surplus pressure = 100 3350 mm H2O Blower's pressure

= = =

3350 mm H2O 3350 mm H2O 32.86 kPa

x

9.81 kPa

### Total air requirement or air blower: 1. Homogenisation Tank

=

2.570 m3/min

2. Aeration Tank

=

3.725 m3/min

3. Clarifier Tank

=

1.010 m3/min

Total air required,

= =

Air Flow @ std condition

= = =

### Air Blower Specification Type Model

7.30

m3/min

438.27 m3/hr Qn

x

###

470.35 m /hr 3

7.84

m3/min

- Refer Appendix E = Futsu (2 units, working alternately) = TSC 100

÷

1000 mm H2O

Pressure

=

Capacity Power Speed

= = =

0.35

kgf / cm2

8.49 m3 / min 11.00 kW 1450 rpm

18