Sewerage Treatment Plant Design Template

ARVINTHRAN RAJA KUMARAN115647

SCHOOL OF CIVIL ENGINEERINGEAP 313/2 - WASTEWATER ENGINEERING

2012/2013 ACADEMIC SESSIONPROF. HAMIDI ABDUL AZIZ

Project Title

PROPOSED HOUSING SCHEME

Data Units Person P.E (recommended)Bungalow 50 - 5Semi-detached house 300 - 5Double storey terrace 500 - 5Single storey terrace 1500 - 5Mosque(2000 people) 1 1000 0.2Day-school (1000 students) 1 1000 0.2

1. Calculation of Population Equivalent (PE)

Population Equivalent (PE) = 50(5) + 300(5) + 500(5) + 1500(5) + 1(1000)(0.2) + 1(1000)(0.2)= 12150

2. Design of Dry Weather Flow & Flow Rate

= 0.23

= 0.230 × 12150

= 2794.50

= 0.032

=

= 4.7 × 12.15 ^ (-0.11)

SEWERAGE TREATMENT PLANT DESIGN PROJECT

The design was executed based on the Guidelines for Developers: Sewage Trearment Plants (1998) Volume IV and MS1228: Code of Practice for Design and Installation of Sewerage Systems (1991)

Your consulting firm has been appointed to design a waste water treatment plant for a new housing scheme in Bayan Baru, Penang. Based on the following data, design a suitable activated sludge treatment plant for this project.

Water consumption rate, q m3 / capita.day

Dry weather flow, DWF

m3 / day

m3 / s

Peak factor, Fp 4.7p-0.11

= 3.57

=

= 9979.27

= 0.116

=

= 2794.50 / 3.57

= 782.55

= 0.009

3. Design of Sewer Pipe

Design criteria

Pipe type = Steel Pipe= 0.01

Slope = 1/600Flow in sewer = 70%

= 0.70

From the Manning formula chart, when d/D = 0.70,

= 0.85

= 1.14

Determination of sewer diameter

= 0.116

= 0.85

= 0.136

Using Manning's equation,

=

0.136 =

0.136 =

0.136 =

D = 0.431 m

= 0.6 m

Check for maximum velocity in sewer system;

=

Maximum flow rate, Qp Fp × DWF

m3 / day

m3 / s

Minimum flow rate, Qmin DWF / Fp

m3 / day

m3 /s

Manning coeficient, n

d/D

q/Qfull

v/Vfull

For maximum flow rate, Qp m3 / s

0.116/Qfull

Qfull m3 / s

Qfull Vfull A

1/n × R2/3 × S1/2 × A

1/n × (D/4)2/3 × S1/2 × πD2/4

D8/3 × (1/n) × (1/4)2/3 × (1/600)1/2 × π/4

Choose design diameter, Ddesign

Vfull 1/n × R2/3 × S1/2

=

= 1.156 m/s

= 1.14

= 1.318 m/s

= 0.009

= 0.090 / 0.136

= 0.067

d/D = 0.18

= 0.53

= 0.53 × 1.156

= 0.613 m / s

4 Design of Primary Screen

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.2.2, Table 5.2

a) Primary screen shall be automatically raked.b) Automatic conveyor is required to transfer screenings to skips.c) Screen motor shall be located above the high water level and access provided for maintenance.

Design criteria:

Raking method = Mechanically RakedClear spacing, S = 25 mmSlope to the vertical =Freeboard = 200 mmStorage periods of screenings = 7 daysWidth of blade, B = 15 mmDepth of chamber, D = 0.8 mRatio of width : length = 1:3Velocity at the feed channel = 1 m /sVelocity at screen face = 1 m /s

(i) Quantity of screening collected in 7 days:

1/0.010 × (0.6/4)2/3 × (1/600)1/2

vmax/Vfull

vmax 0.8 m/s < vmax < 4.0 m/s

For minimum flow rate, Qmin m3 / s

0.009/Qfull

From the Manning formula chart, when Qmin/Qfull = 0.067,

vmin/Vfull

vmin

0.6m/s < Vmin < 1.0m/s

45°

From figure 5.3 (Guidelines for developers ), for clear spacing of 25mm,

Maximum quantity of screenings collected =

Quantity of screening in 7days =

=

= 2.65

(ii) Chamber design

Width of chamber, W =

= (15 + 25) / 25 × 0.116 / (1.0 × 0.8)= 0.23 m

Number of blades, N = W / (B+S)= 230 / (15 + 25)= 6 blades

New chamber width, W = N (B +S)= 6 × (10 + 25)= 231 mm= 0.23 m

Height of chamber, T = 0.8 + 0.20= 1.00 m

= Quantities of screenings in 7 days / Depth of tank

= 2.654 / 1.0

= 2.650

Ratio of width : length = 1 : 3Let width, W = x, length,L = 3x ,

=

= 2.650

x = 0.94Width of tank, W = 1.00 mLength of tank, L = 3 × 1

3.00 mVolume of tank, V = 3 x 1 x (0.8 + 0.2)

= 3.00 m

38/1x106 m3/m3 of sewage

38/106 × Qp × Storage period

38/106 × 9979.27 × 7

m3

(B+S) / S × Qp / VD

Surface area of tank, As

m2

Surface area of tank, As 3x2

3x2

5 Design of Pump Station

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.3.2, Table 5.3Design parameters for unit 10000 < PE < 20000 (Option 2)

Design criteria:

= 0.116

Type of station = Wet well & dry well

=

Pump design flow =

= 30 mins

Pass through openings = 75 mmSuction & discharge openings = 100 mmPumping cycle = 6 to 15 / hourLifting device = MechanicalDepth of tank, D = 2 mRatio of width : length = 1 : 2

(i) Determining Pump Station tank dimension

Pump design flow =

= 0.5 × 0.116 x 4

= 0.231

Volume of tank, V =

= 0.231 × 30 × 60

= 415.80

= V / D

= 415.68 / 2

= 207.90

Ratio of width : length = 1 to 2Let width, W = x, Length, L = 2x,

= 2x²

= 207.90

x = 10.20x = 10 m

Width of tank, W = 10 mLength of tank, L = 2 × 10

Max. flow rate, Qp m3/s

Number of pumps (all identical and work sequentially)

4 (2 sets), 1 duty, 1 assist, per set (100% standby)

each at 0.5Qp

Retention time at Qave, t

0.5Qp each

m3/s

Qp x t

m3


m2


2x2

= 20 m

(ii) Checking the tank designed;

Volume of tank, V = L × W × D= 20 × 10 ×2

= 400

Detention time, t = V / Q= 400 / (0.231 x 60)

= 28.86 mins ( < 30mins )

6 Design of Secondary Screen

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.4.2, Table 5.4Design parameters for PE > 10000

Design criteria:

= 0.116

= 12 mmRaking method = AutomaticScreening storage period = 7 daysSlope to the vertical =Freeboard = 200 mm

= 15 mm= 0.8 m

Ratio of width : length = 1 : 3= 1 m/s

(i) Design of chamber

Width of chamber, W =

= (15+12) / 12 × 0.115 / (1 × 0.80)= 0.32 m

Choose = 0.40 m

= 0.20 m

Number of blade, N = W / (B+S)= 0.20 / (0.015+0.012)

m3


Clear spacing, S

45°

Width of Blade, BDepth of chamber,D

Velocity at screen face, V

(B+S) / S × Qp / VD

W is divided into 2 because the design requires 2 tanks

= 7 blades

New chamber width, W = N (B + S)= 7 × (0.015 + 0.012)= 0.20 m

Height of chamber, T = D + Freeboard= 0.80 + 0.20= 1.00 m

7 Design of Grit Chamber


Design criteria:

Grit removal = Mechanical (Conveyor)Chamber type = Aerated or vortex type

= 0.116

Detention time at Qp, t = 5 mins

= 0.20 m/s

Ratio of depth : width = 1 : 2Ratio of length : width = 2 : 1

Estimated grit quantity = 0.03

(i) Determining the dimension of tank;

Volume of tank, V =

= 0.116 × 5 × 60

= 34.65

= 17.33

Chamber dimension

Let depth = D, width = W, length = LD : W = 1 : 2

= 1/2 : 1L : W = 2 : 1L : W : D = 4 : 2 : 1


Horizontal velocity, Vh

m3 / 103 m3 of sewage

Qp × t

m3

Quantity of tanks used is 2, therfore V is divided into 2

m3

Volume of tank, V = 4D × 2D × D

=

= 17.33Depth of tank, D = 1.29 mChoose depth of tank, D = 2 mWidth of tank, W = 2D

= 2 × 2= 4 m

Length of tank, L = 4D= 4 × 2= 8 m

(ii) Checking tank design

Volume of tank, V = D × W × L= 2 × 4 × 8

= 64

= L × W

= 4 x 8

= 32

Surface loading rate, SLR =

= 9979.27 / 32

= 311.85

=

= 0.116 / (2 × 4)

= 0.01 m/s

Detention time, t =

= 64/(0.115 × 60)

= 9.24 mins

8 Design of Grease Chamber


8D3

m3

Surface area, As

m2

QP / As

m3/m2/day

Horizontal velocity, Vh QP / Ah

V / Qp

Design criteria:

Grease removal = Mechanical (conveyor)

= 0.116

= 5 mins

Storage before disposal = 7 daysRatio of depth : width = 1 : 2Ratio of length : width = 2 : 1

(i) Determining tank dimension

Volume of tank, V =

= 0.116 × 5 × 60

= 34.65

= 17.33

Chamber dimension:

Let depth = D, width = W, length = LD : W = 1 : 2

= 1/2 : 1L : W = 2 : 1L : W : D = 4 : 2 : 1

Volume of tank, V = 4D × 2D × D

=

= 17.33Depth of tank, D = 1.29 mChoose depth of tank, D = 2 mWidth of tank, W = 2D

= 2 × 2= 4 m

Length of tank, L = 4D= 4 × 2= 8 m

(ii) Checking the tank designed;

Volume of tank, V = L × W × D= 8 x 4 x 2


Detention time at Qp, t

Qp × t

m3

Quantity of tanks used is 2, therfore V is divided into 2

m3 Based on Appendix D, Figure D2, Clause 6.2.3 of MS 1228

8D3

= 64

Detention time, t =

= 64 / (0.116 x 60)

= 9.24 mins

9 Design of Balancing tank

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.6, Table 5.7

Design criteria:

= 0.116

Volume of tank, V = 1.5 hours at Qp

= 1.5 hours

Mixing power requirements = 5

Aeration = 1Size of tube diffusers medium pore = 2.5 mmDead water depth = 1 mDepth of tank, D = 3 mRatio of width : length = 1: 3


Volume of tank, V =

= 0.115 × (1.5 × 60 × 60)

= 623.70

Let depth = D, width = W, length = L

= L × W

= V / D= 623.70 / 3

= 207.90

2 balancing tanks are designed because of the large dimension.

Let width = W, length = LW : L = 1 : 3Let W = x, L = 3x,

m3

V / Qp



W/m3 of sewage

m3 of air supply for every m3 of sewage stored per hour

Qp × t

m3


m2

= 207.90

x = 5.886 mChoose width of tank, W = 7 mLength of tank, L = 3 × 7

= 21 m


Volume of 1 tank,V = 21 × 7 × 3

= 441

Volume of 2 tanks,V = 2 (21 × 7 × 3)

= 882

=

= 882 / (0.115 × 60 × 60)

= 2.12 hours

10 Design of Primary Sedimentation Tank


Design criteria:

Tank type = Rectangular Tank

= 0.116

= 0.032

= 2 hours

= 30

= 200

Upward flow rate at Qp = 2 m / hourRatio of length : width = 3 : 1Depth of tank, D = 3 m

= 250 mg / L

Suspended solids (SS) influent = 300 mg / L

Sludge Production

Assume:Percentage removal of sludge = 75 %

2 × 3x2

m3

m3

Detention time at Qp, t V / Qp

Maximum flow rate, Qp m3 / s

Dry weather flow, Qave m3 / s


Surface overflow rate at Qp m3 / m2 / day

Weir loading rate at Qp m3 / m / day

BOD5 influent

Specific gravity of sludge = 1.03Storage period of sludge,t = 1 day

Dry mass = DWF × SS × % removal of sludge= 2794.50 × 0.300 × 0.75= 628.76 kg/day

= 1.03 × 1000

= 1030

Volume of sludge produced = Dry mass / Density of sludge= 628.76 / 1030

= 0.610

Since the storage period of sludge, t = 1 day, therefore, volume for storage of sludge

= 0.610 × 1

= 0.610

(ii) Determining tank dimension

Volume of tank, V =

= (0.116 × 2 × 60 × 60 ) + 0.610

= 832.22

= V / D

= 832.22 / 3

= 277.41

The dimension is large, therefore, 2 tanks are designed.

Ratio of length : width = 3 : 1Let width = W, length = L W = x, L = 3x

(x)(3x)(2) = 277.41

= 277.41

x = 6.80 mWidth of tank, W = 8 mLength of tank, L = 3 x 8 m

= 24 m

(iii) Checking tank design

Density of sludge, ρ

kg/m3

m3/day

Vsludge

m3

(Qp × t ) + Vsludge

m3


m2

6x2

Volume of 1 tank,V = L × W × D= 24 × 8 × 3

= 576

= 2 (L × W × D)

= 2 (24 × 8 × 3)

= 1152

Detention time for 1 tank, t =

= 576 / (0.116 / 2 ) / 3600

= 2.76 hours ( > 2 hours)

Detention time for 2 tanks, t =

= 1152 / (0.116 × 3600)

= 2.76 hours ( > 2 hours)

Surface overflow rate for 2 tanks =

= (0.116) / (8 × 24 ×2)

= 0.0003

= 25.99

=

= 0.116 / (3 × 8 × 2)

= 0.002 m/s

Weir loading rate, WLR = Qp / L= 9976.37 / 24= 415.803

(iv) Quality Control

In primary sedimentation tank,

Assume:Total removal of suspended solid influent = 75

= 40

Suspended solid remains in the influent = (1 - 75 / 100) × 300= 75

= (1 - 40 / 100) × 250

m3

Volume of 2 tanks,Vtotal

m3

V / Qp

V / Qp

Qp / As

m3 / m2 / s

m3 / m2 / day

Horizontal velocity, Vh QP / Ah

Total removal of BOD5 influent

BOD5 remains in the influent

= 150

11 Design of Biological Treatment System


Design criteria :

Number of tanks = 2

= 0.032

= 250 mg / L

Suspended solid influent, SS = 75 mg / L

Mixed liquor suspended solid, MLSS =

= 3000 mg / L

Oxygen requirements = 2 kgO2 / kgBOD5F : M ratio = 0.1

= 0.25

Sludge age = > 20 daysDissolved oxygen level in tank = 2 mg / L

= 0.68

Coefficient of bacteria growth, y = 0.5 mg / mg

= 0.1 / day

Rate of microorganism growth, u = 0.35 dayHydraulic retention time (HRT) = 20 hours

Aerator loading = 0.4

Min. mixing requirement = 20

= 1

Depth of tank, D = 3 m Ratio of length : width = 2 : 1


= 0.25

= 0.032 × 0.25

= 0.0081

=

= 0.032 + 0.0081

= 0.040

Type of biological treatment system - Extended Aeration Activated Sludge (EAAS)

Dry weather flow, Qave m3 / s

BOD5 influent

xa

kgBOD5 / kg.MLSS.day

Ratio of return activated sludge, QR/Q

BOD5/BODL (Domestic)

Coefficient of bacteria death, kd

kg / m3 / day

W / m3

Recirculation ratio, QRAS/QINFLOW

Ratio of return activated sludge, QR/Q

Return activated sludge, QR

Total inflow rate, Qin Qave + QR

=

= 0.032 - 0.0081

= 0.024

F : M =

Volume of tank, V =

= (0.040 × 24 × 60 × 60 × 250/1000)/ (0.10 × 1650/1000)

= 2910.94

3 tanks designed - 3m in depth each

Ratio of length : width = 2: 1Let width = W, length = L W = x, L = 2x

2(x)(2x)(3) = 2910.94

= 2910.94

x = 15.57 m

Width of tank, W = 16.00 mLength of tank, L = 2x

= 2 x 16= 32.00 m


Volume of 1 tank,V = L × W × D= 32 × 16 × 3

= 1536.00

Volume of 2 tanks,V = 2 (L × W × D)= 2 (32 × 16 × 3)

= 3072.00

F : M =

= 0.040 × 24 × 60 × 60 × 250/1000)/(1536 × 3000/1000)

= 0.09

Detention time, t =

= 1536 / (0.040 × 60 × 60)

= 21.10 hours

Sludge flow, Qw Qave - QR

(Qin × So)/(V × xa)

(Qin × So)/(F:M × xa)

12x2

m3

m3


V / Qin

Aerator Loading =

= (0.040 × 24 × 60 × 60 × 250 / 1000) / 1536

= 0.28

Sludge age, θc =

= 1 / (0.40 x 0.35) - 0.1

= 25 days

= (V × MLSS) / (θc × SS)= (1536 × 3) / (25 x 0.75 )

= 245.76

Discharge for 2 tanks, Qw = 245.76 x 2

= 491.52

Increase rate of MLSS, Px =

= 0.40 / (1 + (0.1 × 25) × (2794.50 x 0.25)= 79.84 kg / day

== (2794.50 x 0.25) / 0.68 - 1.42 (79.84)= 914.02 kg/day

= MLSS x (1 + R) / R= 3 x (1 + 0.5) / 0.5= 9.00 kg / day

12 Design of Secondary Sedimentation

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.9, Table 5.14The design parameters is based on PE > 5000.Automatic scrapping and desludging devices will be equipped as the design is a rectangular tank.

Design criteria:

Min. number of tanks = 2Tank configuration = Rectangular

Maximum flow rate, Qp = 0.116

= 0.032

= 2

(Qin × So) / V

kgBOD5/m3.day

1 / yu - kd

Discharge of sludge for 1 tank, Qw

m3 / day

m3 / day

y / (1+kdθc) × (DWF x So)

Oxygen requirement, O2 (DWF x S0) / (BOD5 / BODL) - 1.42Px

Concentration of sludge return, XR

Dry weather flow, Qave

Minimum hydraulic retention time (HRT) at Qp

Minimum side water depth = 3

< 30

< 150

< 50

= 150-180

Mixed liquor suspended solid, MLSS = Xa= 1600

= 150

Suspended solid influent, SS = 75Depth of tank, D = 2.5Ratio of length : width = 3 to 1

(i) Formation of sludge

Assume:Percentage removal of sludge = 50 %Specific gravity of sludge = 1.03Storage period of sludge, t = 1 day

Dry mass = DWF × SS × % removal of sludge= 2794.5 × 0.75 x 0.5= 1047.94 kg/day

= 1.03 × 1000

= 1030

Volume of sludge produced = Amount of sludge produced / Density of sludge= 1047.94 / 1030

= 1.017

Since the storage period of sludge, t = 1 day, therefore, volume for storage of sludge

= 1.017 × 1

= 1.017

(ii) Determining tank dimension

Volume of tank, V =

= (0.116 × 2 × 3600 ) + 1.017

= 832.62

V / D

Surface overflow rate at Qp

Solids loading rate at Qp

Solids loading rate at Qave

Weir loading rate at Qave

BOD5 influent


kg/m3

m3/day

Vsludge

m3

(Qp × t ) + Vsludge

m3

Surface area of tank, As =

= 832.62 / 2.5

= 333.05

The dimension is large, therefore, 2 tanks are designed.


(x)(3x)(2) = 333.05

= 333.05

x = 7.45 mWidth of tank, W = 8.00 mLength of tank, L = 3 × 8

= 24 m

(iii) Checking tank design

Volume of 1 tank,V = L × W × D= 24 × 8 × 2.5

= 480

Volume of 2 tanks,V = 2 (L × W × D)= 2 (24 × 8 × 2.5)

= 960

Detention time of 1 tank, t =

= 480 / (0.116 / 2) / 3600

= 2.32 hours

Detention time of 2 tanks, t =

= 960 / (0.115 × 60 × 60)

= 2.32 hours

Surface overflow rate for 1 tank =

= (0.115 / 2)/(8 × 24)

= 0.00029947917

= 25.88

= Qp × MLSS/As

= (0.115 × 24 × 60 × 60 × 1600 / 1000)/(8 ×24 × 2)

= 41.40

m2

m2

6x2 m2

m3

m3

V / Qp

V/Qp

Qp/As

m3/m2.s

m3/m2.day

Solid loading rate at Qp

kg/m2.day

= Qave× MLSS/As

= (0.032 × 24 × 60 × 60 × 1600 / 1000) / (24 ×8 × 2)

= 11.52

Weir Loading Rate, WLR = Qavg (DWF) / Perimeter= 0.032 x 2794.5 / (24+8)

= 2.7945

(iv) Quality Control

In secondary clarifiers,

Assume:Total removal of suspended solid influent = 50

= 90

Suspended solid remains in the effluent = (1 - 50 / 100) × 75= 37.5

= (1- 90 / 100) × 150

= 15

Thus, standard A has been achieved for both parameters in the effluent.

13 Design of Sludge Drying Bed

Based on the design, the Population Equivalent, PE = 12150Design Requirements Clause 5.12.3, Table 5.17The design parameters is based on PE > 2000.Drying bed must be designed to support mechanical/machine lift.The design should be based on one full time working shift only.

Design criteria:

Type of stabilisation = Ambient anaerobic digestion with good mixing facilityMinimum hydraulic retention time (HRT) = 30

Solid loading rate at Qave

kg/m2.day

Design is based on Clause 5.12.3, Table 5.6 (Extended Aeration)

Total removal of BOD5 influent

BOD5 remains in the effluent

The drying bed is designed to handle a maximum of 7 days continuous feed with the next feed being after a minimum of 21 days from the last feed

The sludge from primary sedimentation and secondary clarifier are disposed and dried at the sludge drying bed.

Detention time, t = 7Depth of sludge bed, D = 300Sludge formed at primary sedimentation tank = 628.76Sludge formed at secondary clarifiers = 1047.94

= 1030

Ratio of width : length = 1 to 2


Total dry weight, W = 628.76 + 1047.94= 1676.70 kg/day

Flow of sludge, Qs = 1676.70 / 1030

= 1.63

Volume of sludge, Vs = 1.63 × 7

= 11.40

Surface Area, As = V / D= 11.40 / 0.3

= 37.98


(x)(2x) = 37.98

= 37.98

x = 4.36 mWidth of tank, W = 5 mLength of tank, L = 2 × 5

= 10 m


Volume of tank, V = L × W × D= 10 × 5 × 0.3

= 15

Effective sludge feed surface = L × W= 10 x 5

= 50


m3 / day

m3

m2

2x2

m3

m2

FLOW DIAGRAM OF SEWERAGE TREATMENT PROCESS

INFLUENT

Steel Sewer Pipe

Grease Chamber 8m x 4m x 2m

Balancing Tank 21m x 7m x 3m

EFFLUENT

Sludge Drying Bed 10m x 5m x 0.3m

EFFLUENT

ARVINTHRAN RAJA KUMARAN115647

SCHOOL OF CIVIL ENGINEERINGEAP 313/2 - WASTEWATER ENGINEERING

2012/2013 ACADEMIC SESSIONPROF. HAMIDI ABDUL AZIZ

P.E (recommended) P.E (required)per house 250per house 1500per house 2500per house 7500per person 200per student 200

Total 12150

50(5) + 300(5) + 500(5) + 1500(5) + 1(1000)(0.2) + 1(1000)(0.2)

SEWERAGE TREATMENT PLANT DESIGN PROJECT

The design was executed based on the Guidelines for Developers: Sewage Trearment Plants (1998) Volume IV and MS1228: Code of Practice for Design and Installation of Sewerage Systems (1991)

Your consulting firm has been appointed to design a waste water treatment plant for a new housing scheme in Bayan Baru, Penang. Based on the following data, design a suitable activated sludge treatment plant for this project.

m3 / capita.day

D > 0.2m O.K!

O.K!

O.K!

c) Screen motor shall be located above the high water level and access provided for maintenance.

max = 25mm(0°-45°)

min = 150mm

(max = 1m/s)(max = 1m/s)

0.8 m/s < vmax < 4.0 m/s

0.6m/s < Vmin < 1.0m/s

(15 + 25) / 25 × 0.116 / (1.0 × 0.8)

Quantities of screenings in 7 days / Depth of tank

m3/m3 of sewage

× Storage period

Wet well & dry well

min = 30minsmin = 75mmmin = 100mm6 to 15 / hour

Mechanical

4 (2 sets), 1 duty, 1 assist, per set (100% standby)

each at 0.5Qp

( < 30mins ) O.K!

max = 12mm

(0°-45°)min = 150mm

max = 1.00 m/s

(15+12) / 12 × 0.115 / (1 × 0.80)

7 × (0.015 + 0.012)

min = 3 mins

max = 0.20m/s

Based on Appendix D, Figure D2, Clause 6.2.3 of MS 1228

O.K!

( < 0.20 m / s ) O.K!

( > 3 mins ) O.K!

( < 1500 m3 / m2 / day )

min = 3 mins

Based on Appendix D, Figure D2, Clause 6.2.3 of MS 1228

( > 3 mins ) O.K!

(0.6m - 1.0m)

of air supply for every m3 of sewage stored per hour

( > 1.5hours ) O.K!

Rectangular Tank

100 < WLR < 2001.2 - 2.0 m / hour

m3 / m2 / day

m3 / m / day

DWF × SS × % removal of sludge

Dry mass / Density of sludge

( > 2 hours) O.K!

( > 2 hours) O.K!

O.K!

( < 0.015m / s) O.K!

150 < WLR < 200

%

%

(1 - 75 / 100) × 300mg / L

(1 - 40 / 100) × 250

( < 30m3 / m2 / day )

mg / L

min = 2 tanks

kgO2 / kgBOD5 1.5 - 2.0(0.05-0.1)

> 20 days2 mg / L

18 - 24 hours

0.5 - 1.0( < 5m )

0.032 × 0.25

0.032 + 0.0081

(2500mg / L < xa < 5000mg / L)

kgBOD5 / kg.MLSS.day

kg / m3 / day 0.1 - 0.4 kg / m3 / day

m3/s

m3/s

0.032 - 0.0081

(0.040 × 24 × 60 × 60 × 250/1000)/ (0.10 × 1650/1000)

0.040 × 24 × 60 × 60 × 250/1000)/(1536 × 3000/1000)

( 0.05 < F:M < 0.1 ) O.K!

( 18 hours < t < 24 hours ) O.K!

m3/s


(Qin × So)/(F:M × xa)

m3

O.K!

(> 20 days) O.K!

Automatic scrapping and desludging devices will be equipped as the design is a rectangular tank.

Rectangular

hours

( 0.1 - 0.4 kgBOD5/m3.day )

m3 / s

m3 / s

m

mg / L

mg / L

mg / Lm

DWF × SS × % removal of sludge

Amount of sludge produced / Density of sludge

m3 / day / m2

kg / day / m2

kg / day / m2

m3 / day / m

( > 2 hours) O.K!

( > 2 hours) O.K!

O.K!

(0.115 × 24 × 60 × 60 × 1600 / 1000)/(8 ×24 × 2)

O.K!

m3/m2.day ( < 30m3/m2.day )

kg/m2.day ( < 150kg/m2.day )

(0.032 × 24 × 60 × 60 × 1600 / 1000) / (24 ×8 × 2)

O.K!

150 < WLR < 180

%

%

(1 - 50 / 100) × 75mg / L

(1- 90 / 100) × 150

mg / L

Ambient anaerobic digestion with good mixing facilitydays

kg/m2.day ( < 50kg/m2.day )

daysmm ( < 450mm)

kg / daykg / day

kg / m3

FLOW DIAGRAM OF SEWERAGE TREATMENT PROCESS

Primary Screen With 6 blades

Pump Station

Secondary Screen With 7 Blades

Grit Chamber 8m x 4m x 2m

Primary Sedimentation Tank 24m x 8m x 3m

Aeration Tank 32m x 16m x 3m

Secondary Sedimentation Tank 24m x 8m x 2.5m

Sludge Drying Bed 10m x 5m x 0.3m

Pump Station

Secondary Screen With 7 Blades

Aeration Tank 32m x 16m x 3m

Secondary Sedimentation Tank 24m x 8m x 2.5m

Sewerage Treatment Plant Design Template

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