ARVINTHRAN RAJA KUMA 115647 SCHOOL OF CIVIL ENGINEE EAP 313/2 - WASTEWATER ENGI SEWERAGE TREATMENT PLAN PROJECT 2012/2013 ACADEMIC SESS
PROF. HAMIDI ABDUL AZ
The design was executed based on the Guidelines for Developers: Sewage Trearment Plants (1998) V Practice for Design and Installation of Sewerage Systems (1991) Project Title
Your consulting firm has been appointed to design a waste water treatment plant for a new housin Based on the following data, design a suitable activated sludge treatment plant PROPOSED HOUSING SCHEME Data
Units 50 300 500 1500 1 1
Bungalow Semi-detached house Double storey terrace Single storey terrace Mosque(2000 people) Day-school (1000 students)
Person 1000 1000
P.E (recommended) 5 5 5 5 0.2 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 Water consumption rate, q Dry weather flow, DWF
= = = =
Peak factor, F p
Maximum flow rate, Q p
= = = =
0.23 0.230 × 12150 2794.50 0.032 4.7p -0.11 4.7 × 12.15 ^ (-0.11) 3.57 F p × DWF
m3 / capita.day m3 / day m3 / s
= = Minimum flow rate, Q min
= = =
m3 / day
9979.27
3
0.116 DWF / F p 2794.50 / 3.57 782.55
=
0.009
= = = = =
Steel Pipe 0.01 1/600 70% 0.70
m /s
m3 / day m3 /s
3. Design of Sewer Pipe Design criteria Pipe type Manning coeficient, n Slope Flow in sewer d /D
From the Manning formula chart, when d/D = 0.70, q /Q full = 0.85 v /V full = 1.14 Determination of sewer diameter For maximum flow rate, Q p
=
0.116
0.116/Q full
=
0.85
Q full
=
0.136
Q full
=
V full A
0.136
=
1/n × R 2/3 × S 1/2 × A
0.136
=
1/n × (D /4)2/3 × S 1/2 × πD 2/4
0.136
=
D
=
D 8/3 × (1/n ) × (1/4)2/3 × (1/600)1/2 × π /4 0.431 m
3
m /s m3 / s
Using Manning's equation,
Choose design diameter, D design
=
0.6 m
Check for maximum velocity in sewer system; V full = 1/n × R 2/3 × S 1/2 = 1/0.010 × (0.6/4)2/3 × (1/600)1/2 m/s = 1.156 v max /V full
=
1.14
v max
=
1.318
m/s
For minimum flow rate, Q min
=
0.009
m3 / s
0.009/Q full
= =
0.090 / 0.136 0.067
0.8 m/s < vmax < 4.0 m/s
From the Manning formula chart, when Qmin/Qfull = 0.067, d/D = 0.18 vmin/Vfull = 0.53 vmin = 0.53 × 1.156 =
0.613
m/s
0.6m/s < Vmin < 1.0m/s
4 Design of Primary Screen Based on the design, the Population Equivalent, PE = 12150 Design 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 Clear spacing, S Slope to the vertical Freeboard Storage periods of screenings Width of blade, B Depth of chamber, D Ratio of width : length Velocity at the feed channel Velocity at screen face
= = = = = = = = = =
Mechanically Raked 25 mm 45° 200 mm 7 days 15 mm 0.8 m 1:3 1 m /s 1 m /s
(i) Quantity of screening collected in 7 days: From figure 5.3 (Guidelines for developers ), for clear spacing of 25mm, 38/1x106
m3/m3 of sewage
Maximum quantity of screenings collected
=
Quantity of screening in 7days
=
38/106 × Q p × Storage period
=
38/106 × 9979.27 × 7
=
2.65
m3
(ii) Chamber design Width of chamber, W
= = =
(B+S) / S × Q p / VD (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
= =
Surface area of tank, As
= = =
Ratio of width : length Let width, W = x, length,L = 3x , Surface area of tank, As
= = 3x2 x
Width of tank, W Length of tank, L Volume of tank, V
= = = = = =
5 Design of Pump Station Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.3.2, Table 5.3 Design parameters for unit 10000 < PE < 20000 (Option 2) Design criteria:
0.8 + 0.20 1.00
m
Quantities of screenings in 7 days / Depth o 2.654 / 1.0 2.650 m2 1:3 3x2 2.650 0.94 1.00 m 3×1 3.00 m 3 x 1 x (0.8 + 0.2) 3.00 m
Max. flow rate, Q p Type of station Number of pumps (all identical and work sequentially) Pump design flow Retention time at Qave, t Pass through openings Suction & discharge openings Pumping cycle Lifting device Depth of tank, D Ratio of width : length
0.116 m3/s Wet well & dry well 4 (2 sets), 1 duty, 1 assist, per set (100% standby) each at 0.5Q p
= = = = = = = = = = =
30 75 100 6 to 15 2 1:2
(i) Determining Pump Station tank dimension 0.5Qp each 0.5 × 0.116 x 4 0.231 m3/s Qp x t 0.231 × 30 × 60 415.80 m3 V/D 415.68 / 2 207.90 m2
Pump design flow
= = =
Volume of tank, V
= = =
Surface area of tank, As
= = =
Ratio of width : length Let width, W = x, Length, L = 2x, Surface area of tank, As
=
1 to 2
=
2x²
= = = = = =
207.90 10.20 10 10 2 × 10 20
= = =
L×W×D 20 × 10 ×2 400
=
V/Q
2
2x x x Width of tank, W Length of tank, L
m m m
(ii) Checking the tank designed; Volume of tank, V
Detention time, t
m3
mins mm mm / hour Mechanical m
= =
400 / (0.231 x 60) 28.86 mins
( < 30mins )
6 Design of Secondary Screen Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.4.2, Table 5.4 Design parameters for PE > 10000 Design criteria: Max. flow rate, Q p Clear spacing, S Raking method Screening storage period Slope to the vertical Freeboard Width of Blade, B Depth of chamber,D Ratio of width : length Velocity at screen face, V
= = = = = = = = = =
m3/s mm
0.116 12 Automatic 7 45° 200 15 0.8 1:3 1
days mm mm m m/s
(i) Design of chamber Width of chamber, W
Choose W is divided into 2 because the design requires 2 tanks
= = = = =
(B+S) / S × Q p / VD (15+12) / 12 × 0.115 / (1 × 0.80) 0.32 m 0.40 m 0.20
m
Number of blade, N
= = =
W / (B+S) 0.20 / (0.015+0.012) 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
Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.5.4, Table 5.5 Design parameters for PE > 10000 Design criteria: Grit removal Chamber type Max. flow rate, Q p Detention time at Qp, t Horizontal velocity, Vh Ratio of depth : width Ratio of length : width Estimated grit quantity
= = = = = = = =
Mechanical (Conveyor) Aerated or vortex type 0.116 m3/s 5 mins 0.20 m/s 1:2 2:1 m3 / 103 m3 of sewage 0.03
(i) Determining the dimension of tank; Volume of tank, V
= = =
Quantity of tanks used is 2, therfore V is divided into 2
Qp × t 0.116 × 5 × 60 34.65 m3
=
17.33
Chamber dimension Let depth = D, width = W, length = L D:W L:W L:W:D Volume of tank, V
Depth of tank, D Choose depth of tank, D Width of tank, W
Length of tank, L
= = = =
= = = = = = = = = = =
1:2 1/2 : 1 2:1 4:2:1 4D × 2D × D 8D3 17.33 1.29 2 2D 2×2 4 4D 4×2 8
m m
m
m
m3
(ii) Checking tank design Volume of tank, V
Surface area, As
= = =
D×W×L 2×4×8 64
m3
= = =
L×W 4x8 32
m2
QP / As 9979.27 / 32 311.85 m3/m2/day
Surface loading rate, SLR
= = =
Horizontal velocity, Vh
= = =
QP / Ah 0.116 / (2 × 4) 0.01 m/s
Detention time, t
= = =
V / Qp 64/(0.115 × 60) 9.24 mins
8 Design of Grease Chamber Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.5.4, Table 5.6 Design parameters for PE > 10000 Design criteria: Grease removal Max. flow rate, Q p
= =
Mechanical (conveyor) 0.116 m3/s
Detention time at Qp, t Storage before disposal Ratio of depth : width Ratio of length : width
= = = =
5 7 1:2 2:1
= =
Qp × t 0.116 × 5 × 60
(i) Determining tank dimension Volume of tank, V
mins days
Quantity of tanks used is 2, therfore V is divided into 2
=
34.65
m3
=
17.33
m3
= = = =
1:2 1/2 : 1 2:1 4:2:1
= = = = = = = = = = =
4D × 2D × D 8D3 17.33 1.29 2 2D 2×2 4 4D 4×2 8
= = =
L×W×D 8x4x2 64
Based on Appendix D, Fi 6.2.3 of MS 1
Chamber dimension: Let depth = D, width = W, length = L D:W L:W L:W:D Volume of tank, V
Depth of tank, D Choose depth of tank, D Width of tank, W
Length of tank, L
m m
m
m
(ii) Checking the tank designed; Volume of tank, V
Detention time, t
= = =
m3
V / Qp 64 / (0.116 x 60) 9.24 mins
9 Design of Balancing tank Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.6, Table 5.7 Design criteria: Max. flow rate, Q p
=
0.116
m3/s
Volume of tank, V Detention time at Qp, t
= =
1.5 1.5
Mixing power requirements Aeration Size of tube diffusers medium pore Dead water depth Depth of tank, D Ratio of width : length
= = = = = =
5 1 2.5 1 3 1: 3
hours at Qp hours
W/m3 of sewage m3 of air supply for every m3 of sew mm m m
(i) Determining tank dimension Volume of tank, V
= = =
Qp × t 0.115 × (1.5 × 60 × 60) 623.70 m3
Let depth = D, width = W, length = L Surface area of tank, As
= = = =
L×W V/D 623.70 / 3 207.90
m2
2 balancing tanks are designed because of the large dimension. Let width = W, length = L W:L Let W = x, L = 3x,
=
2 × 3x2 x Choose width of tank, W Length of tank, L
1:3
= = = = =
207.90 5.886 7 3×7 21
= =
21 × 7 × 3 441
m m m
(ii) Checking tank design Volume of 1 tank,V
Volume of 2 tanks,V
= =
Detention time at Qp, t
= =
m3
2 (21 × 7 × 3) 882 m3 V / Qp 882 / (0.115 × 60 × 60)
=
2.12
hours
10 Design of Primary Sedimentation Tank Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.7, Table 5.8 Design criteria: Tank type Maximum flow rate, Q p
= =
Rectangular Tank 0.116 m3 / s
Dry weather flow, Qave
=
0.032
Detention time at Qp, t
=
2
Surface overflow rate at Qp
=
30
Weir loading rate at Qp Upward flow rate at Qp Ratio of length : width Depth of tank, D BOD5 influent
= = = = = =
200 2 3:1 3 250 300
m3 / m / day m / hour
= = =
75 1.03 1
%
Suspended solids (SS) influent
m3 / s hours m3 / m2 / day
m mg / L mg / L
Sludge Production Assume: Percentage removal of sludge Specific gravity of sludge Storage period of sludge,t Dry mass
= = =
Density of sludge, ρ
= =
Volume of sludge produced
= = =
day
DWF × SS × % removal of sludge 2794.50 × 0.300 × 0.75 628.76 kg/day 1.03 × 1000 1030
kg/m3
Dry mass / Density of sludge 628.76 / 1030 0.610 m3/day
Since the storage period of sludge, t = 1 day, therefore, volume for storage of sludge Vsludge = 0.610 × 1 =
0.610
m3
(ii) Determining tank dimension Volume of tank, V
= = =
Surface area of tank, As
= = =
(Qp × t ) + Vsludge (0.116 × 2 × 60 × 60 ) + 0.610 832.22 m3 V/D 832.22 / 3 277.41
m2
The dimension is large, therefore, 2 tanks are designed. Ratio of length : width Let width = W, length = L W = x, L = 3x (x)(3x)(2) 6x2 x Width of tank, W Length of tank, L
=
3:1
= = = = = =
277.41 277.41 6.80 8 3x8 24
m m m m
= = =
L×W×D 24 × 8 × 3 576
m3
(iii) Checking tank design Volume of 1 tank,V
Volume of 2 tanks,Vtotal
= = =
2 (L × W × D) 2 (24 × 8 × 3) 1152 m3
Detention time for 1 tank, t
= = =
V / Qp 576 / (0.116 / 2 ) / 3600 2.76 hours
( > 2 hours)
= = =
V / Qp 1152 / (0.116 × 3600) 2.76 hours
( > 2 hours)
= =
Qp / As (0.116) / (8 × 24 ×2)
Detention time for 2 tanks, t
Surface overflow rate for 2 tanks
=
0.0003
m3 / m2 / s
=
25.99
m / m / day
Horizontal velocity, Vh
= = =
Weir loading rate, WLR
= = =
3
2
QP / Ah 0.116 / (3 × 8 × 2) 0.002 m/s Qp / L 9976.37 / 24 415.803
(iv) Quality Control In primary sedimentation tank, Assume: Total removal of suspended solid influent Total removal of BOD5 influent
= =
Suspended solid remains in the influent
= = = =
BOD5 remains in the influent
75 40 (1 - 75 / 100) × 300 75 (1 - 40 / 100) × 250 150
11 Design of Biological Treatment System Based on the design, the Population Equivalent, PE = 12150 Design Requirements Clause 5.8, Table 5.9 Type of biological treatment system - Extended Aeration Activated Sludge (EAAS) Design criteria : Number of tanks Dry weather flow, Qave
= =
2 0.032
BOD5 influent Suspended solid influent, SS Mixed liquor suspended solid, MLSS
= = =
250 75 xa
Oxygen requirements F : M ratio Ratio of return activated sludge, QR/Q Sludge age
= = = = =
3000 2 0.1 0.25 > 20
m3 / s mg / L mg / L
mg / L kgO2 / kgBOD5 kgBOD5 / kg.MLSS.d days
mg / L
Dissolved oxygen level in tank BOD5/BODL (Domestic) Coefficient of bacteria growth, y Coefficient of bacteria death, kd Rate of microorganism growth, u Hydraulic retention time (HRT) Aerator loading
= = = = = = =
2 0.68 0.5 0.1 0.35 20 0.4
Min. mixing requirement Recirculation ratio, QRAS/QINFLOW Depth of tank, D Ratio of length : width
=
20
W / m3
= = =
1 3 2:1
m
Ratio of return activated sludge, QR/Q
=
0.25
Return activated sludge, QR
=
0.032 × 0.25
=
mg / mg / day day hours kg / m3 / day
(i) Determining tank dimension
Total inflow rate, Qin
= = =
Sludge flow, Qw
= = =
0.0081 Qave + QR 0.032 + 0.0081 0.040 Qave - QR 0.032 - 0.0081 0.024
F:M
=
(Qin × So)/(V × xa)
V
= = =
(Qin × So)/(F:M × xa) (0.040 × 24 × 60 × 60 × 25 2910.94
Volume of tank,
3 tanks designed - 3m in depth each Ratio of length : width Let width = W, length = L W = x, L = 2x 2(x)(2x)(3) 12x2 x
=
2: 1
= = =
2910.94 2910.94 15.57
Width of tank, W Length of tank, L
= = =
16.00 2x 2 x 16
m m
=
32.00
m
= = =
L×W×D 32 × 16 × 3 1536.00
m3
(ii) Checking tank design Volume of 1 tank,V
Volume of 2 tanks,V
= = =
2 (L × W × D) 2 (32 × 16 × 3) 3072.00 m3
= = =
(Qin × So)/(V × xa) 0.040 × 24 × 60 × 60 × 250/1000)/(1536 × 3000/1000) 0.09
Detention time, t
= = =
V / Qin 1536 / (0.040 × 60 × 60) 21.10 hours
Aerator Loading
= = =
(Qin × So) / V (0.040 × 24 × 60 × 60 × 250 / 1000) / 1536 5 3 0.28 kgBOD /m .day
Sludge age, θc
= = =
1 / yu - kd 1 / (0.40 x 0.35) - 0.1 25 days
= = =
(V × MLSS) / (θc × SS) (1536 × 3) / (25 x 0.75 ) 245.76 m3 / day
F:M
Discharge of sludge for 1 tank, Qw
Discharge for 2 tanks, Qw
= =
245.76 x 2 491.52
m3 / day
Increase rate of MLSS, Px
= = =
y / (1+kdθc) × (DWF x So) 0.40 / (1 + (0.1 × 25) × (2794.50 x 0.25) 79.84 kg / day
Oxygen requirement, O2
=
(DWF x S0) / (BOD5 / BODL) - 1.42Px (2794.50 x 0.25) / 0.68 - 1.42 (79.84) 914.02 kg/day
= =
Concentration of sludge return, XR
= = =
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 = 12150 Design Requirements Clause 5.9, Table 5.14 The 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 Tank configuration Maximum flow rate, Qp Dry weather flow, Qave
= =
2 Rectangular 0.116
=
0.032
Minimum hydraulic retention time (HRT) at Qp Minimum side water depth Surface overflow rate at Qp
= = <
2 3 30
Solids loading rate at Qp
<
150
Solids loading rate at Qave
<
50
Weir loading rate at Qave Mixed liquor suspended solid, MLSS
= = = = = = =
150-180 Xa 1600 150 75 2.5 3 to 1
BOD5 influent Suspended solid influent, SS Depth of tank, D Ratio of length : width (i) Formation of sludge Assume: Percentage removal of sludge Specific gravity of sludge Storage period of sludge, t
= = =
Dry mass
= = =
Density of sludge, ρ
=
50 1.03 1
% day
DWF × SS × % removal of sludge 2794.5 × 0.75 x 0.5 kg/day 1047.94 1.03 × 1000
= Volume of sludge produced
= = =
1030
kg/m3
Amount of sludge produced / Density of slu 1047.94 / 1030 1.017 m3/day
Since the storage period of sludge, t = 1 day, therefore, volume for storage of sludge Vsludge = 1.017 × 1 =
1.017
m3
(ii) Determining tank dimension Volume of tank, V
Surface area of tank, As
= = =
(Qp × t ) + Vsludge (0.116 × 2 × 3600 ) + 1.017 832.62 m3
= =
V/D 832.62 / 2.5 333.05
=
m2
The dimension is large, therefore, 2 tanks are designed. Ratio of length : width = 3 : 1 Let width = W, length = L W = x, L = 3x (x)(3x)(2) 6x2 x Width of tank, W Length of tank, L
=
333.05
m2
= = = = =
333.05 7.45 8.00 3×8 24
m2 m m
= = =
L×W×D 24 × 8 × 2.5 480
m
(iii) Checking tank design Volume of 1 tank,V
Volume of 2 tanks,V
= = =
m3
2 (L × W × D) 2 (24 × 8 × 2.5) 960 m3
Detention time of 1 tank, t
= = =
V / Qp 480 / (0.116 / 2) / 3600 2.32 hours
Detention time of 2 tanks, t
= = =
V/Qp 960 / (0.115 × 60 × 60) 2.32 hours
=
Qp/As (0.115 / 2)/(8 × 24) 0.000299479 m3/m2.s 25.88 m3/m2.day
Solid loading rate at Qp
= = =
Qp × MLSS/As (0.115 × 24 × 60 × 60 × 1600 / 1000)/(8 ×2 41.40 kg/m2.day
Solid loading rate at Qave
= = =
Qave× MLSS/As (0.032 × 24 × 60 × 60 × 1600 / 1000) / (24 11.52 kg/m2.day
Weir Loading Rate, WLR
= = =
Qavg (DWF) / Perimeter 0.032 x 2794.5 / (24+8) 2.7945
Surface overflow rate for 1 tank
= = =
(iv) Quality Control In secondary clarifiers, Design is based on Clause 5.12.3, Table 5.6 (Extended Aeration) Assume: Total removal of suspended solid influent Total removal of BOD5 influent Suspended solid remains in the effluent BOD5 remains in the effluent
= =
50 90
= = = =
(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 = 12150 Design Requirements Clause 5.12.3, Table 5.17 The 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. 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. Design criteria: Type of stabilisation Minimum hydraulic retention time (HRT) Detention time, t Depth of sludge bed, D Sludge formed at primary sedimentation tank Sludge formed at secondary clarifiers Density of sludge, ρ Ratio of width : length
= = = = = = = =
(i) Determining tank dimension Total dry weight, W
= =
628.76 + 1047.94 1676.70 kg/day
Flow of sludge, Qs
= =
1676.70 / 1030 1.63 m3 / day
Volume of sludge, Vs
= =
1.63 × 7 11.40
m3
= = =
V/D 11.40 / 0.3 37.98
m2
Surface Area, As
Ratio of length : width Let width = W, length = L W = x, L = 2x
=
(x)(2x) 2x2 x
= = =
2:1
37.98 37.98 4.36
m
Ambient anaerobic digestio 30 7 300 628.76 1047.94 1030 1 to 2
Width of tank, W Length of tank, L
= = =
5 2×5 10
= = =
L×W×D 10 × 5 × 0.3 15
m3
= = =
L×W 10 x 5 50
m2
m m
(ii) Checking tank design Volume of tank, V
Effective sludge feed surface
FLOW DIAGRAM OF S
Steel Sewer Pipe
Grease Chamber 8m x 4m x 2m
Balancing Tank 21m x 7m x 3m
Sludge Dry
NTHRAN RAJA KUMARAN 115647 OL OF CIVIL ENGINEERING - WASTEWATER ENGINEERING GE TREATMENT PLANT DESIGN PROJECT /2013 ACADEMIC SESSION
ROF. HAMIDI ABDUL AZIZ
Trearment Plants (1998) Volume IV and MS1228: Code of Sewerage Systems (1991)
nt plant for a new housing scheme in Bayan Baru, Penang. d sludge treatment plant for this project.
P.E (recommended) per house per house per house per house per person per student Total
)(0.2) + 1(1000)(0.2)
m3 / capita.day
P.E (required) 250 1500 2500 7500 200 200 12150
D > 0.2m
O.K!
0.8 m/s < vmax < 4.0 m/s
O.K!
0.6m/s < Vmin < 1.0m/s
O.K!
for maintenance.
max = 25mm (0°-45°) min = 150mm
(max = 1m/s) (max = 1m/s)
m3/m3 of sewage
torage period
0.116 / (1.0 × 0.8)
eenings in 7 days / Depth of tank
ll & dry well uty, 1 assist, per set % standby) at 0.5Q p
chanical
min = 30mins min = 75mm min = 100mm 6 to 15 / hour
( < 30mins )
max = 12mm
(0°-45°) min = 150mm
max = 1.00 m/s
.115 / (1 × 0.80)
12)
O.K!
min = 3 mins max = 0.20m/s
Based on Appendix D, Figure D2, Clause 6.2.3 of MS 1228
( < 1500 m3 / m2 / day )
O.K!
( < 0.20 m / s )
O.K!
( > 3 mins )
O.K!
min = 3 mins
Based on Appendix D, Figure D2, Clause 6.2.3 of MS 1228
( > 3 mins )
O.K!
upply for every m3 of sewage stored per hour (0.6m - 1.0m)
( > 1.5hours )
O.K!
ngular Tank
m3 / m2 / day m3 / m / day
removal of sludge
ity of sludge
100 < WLR < 200 1.2 - 2.0 m / hour
( > 2 hours)
O.K!
( > 2 hours)
O.K!
3
2
( < 30m / m / day )
O.K!
( < 0.015m / s)
O.K!
150 < WLR < 200
% %
(1 - 75 / 100) × 300 mg / L (1 - 40 / 100) × 250 mg / L
min = 2 tanks
kgO2 / kgBOD5 kgBOD5 / kg.MLSS.day
(2500mg / L < xa < 5000mg / L) 1.5 - 2.0 (0.05-0.1) > 20 days
2 mg / L
3
kg / m / day
18 - 24 hours 0.1 - 0.4 kg / m3 / day 0.5 - 1.0 ( < 5m )
0.032 × 0.25 m3/s
0.032 + 0.0081 m3/s
0.032 - 0.0081 m3/s
(Qin × So)/(V × xa)
(Qin × So)/(F:M × xa) (0.040 × 24 × 60 × 60 × 250/1000)/ (0.10 × 1650/1000) m3
)/(1536 × 3000/1000) ( 0.05 < F:M < 0.1 )
O.K!
( 18 hours < t < 24 hours )
O.K!
5
3
( 0.1 - 0.4 kgBOD /m .day )
O.K!
(> 20 days)
O.K!
rectangular tank.
Rectangular m3 / s m3 / s hours m m3 / day / m2 kg / day / m2 kg / day / m2 m3 / day / m mg / L mg / L mg / L m
removal of sludge
e produced / Density of sludge
m3/m2.day
( > 2 hours)
O.K!
( > 2 hours)
O.K!
( < 30m3/m2.day )
O.K!
( < 150kg/m2.day )
O.K!
( < 50kg/m2.day )
O.K!
× 60 × 1600 / 1000)/(8 ×24 × 2)
kg/m2.day
× 60 × 1600 / 1000) / (24 ×8 × 2)
kg/m2.day
150 < WLR < 180
% %
(1 - 50 / 100) × 75 mg / L (1- 90 / 100) × 150 mg / L
Ambient anaerobic digestion with good mixing facility days days mm kg / day kg / day kg / m3
( < 450mm)
FLOW DIAGRAM OF SEWERAGE TREATMENT PROCESS
Primary Screen With 6 blades
Grit Chamber 8m x 4m x 2m
Secondary Screen With
Primary Sedimentation Tank 24m x 8m x 3m
Sludge Drying Bed 10m x 5m x 0.3m
Aeration Tank 32m x
Secondary Sediment
Pump Station
Secondary Screen With 7 Blades
Aeration Tank 32m x 16m x 3m
Secondary Sedimentation Tank 24m x 8m x 2.5m
