This document outlines the design of the sewage treatment plant for Cochi India.

1. CIVIL ENGINEERING PROGRAMME
Sewage Treatment Plant
ROYAL UNIVERSITY OF BHUTAN
Jigme Namgyel Engineering College
Department of Civil Engineering and Surveying
Dewathang
Samdrup Jongkhar : Bhutan

2. Design STP for Kochi based on following sewage characteristics, effluent to be discharged in
Marine Costal Areas.
1. Area of Interest:
Kochi also known as Cochin is a major port city on the south-west coast of India bordering
the Laccadive Sea. It is part of the district of Ernakulam in the state of Kerala and is often
referred to as Ernakulam. Kochi is the most densely populated city in Kerala. As of 2011, it
has a corporation limit population of 677,381 within an area of 94.88 km² and a total urban
population of more than of 2.1 million within an area of 440 km², making it the most
populous urban area in Kerala.
Sewage Characteristics
Total Suspended Solids (TSS) = 600 mg/l
Biological Oxygen Demand (BOD) = 600 mg/l
Chemical Oxygen Demand (COD) = 700 mg/l
Oil and Grease = 20 mg/l
2. Design Period
The future period for which the provision is made in designing the capacities of various
components of the sewerage is known as design period. This sewage treatment plant is to
be designed for 30 years.
3. Population Forecast for Kochi City
( Source : http://www.indiaonlinepages.com/population/kochi-population.html )
Here for population forecast Incremental Increase Method is being used.
Kochi

3. Population after nth decade is given by n
( 1)
P . ( . ).
2
n
P n X n Y
+
= + +
Where, Pn = Population after nth decade, X = Average Increase, Y = Incremental Increase
Year Population Increment (X) Incremental Increase (Y)
2011 6,02,046
2012 6,28,464 26,418
2013 6,53,603 25,139 -1,279
2014 6,85,313 31,710 6,571
2015 7,15,809 30,496 -1,214
2016 7,47,662 31,853 1,357
2017 7,77,569 29,907 -1,946
2018 8,15,683 38,114 8,207
2019 8,51,980 36,297 -1,817
2020 8,89,893 37,913 1,616
Total Increment (X) 2,87,847 11,495
Average 31,983 1,437
Average per decade 3,19,830 14,370
Population in the Year 2030, 2030
(1 1)
P 889893 1 x 319830 1 x ) x 14370
2
+ 
= + +  
 
= 1,224,093
Population in the Year 2040, 2040
(2 1)
P 889893 2 x 319830 2 x ) x 14370
2
+ 
= + + 
 
= 1,572,663
Population in the Year 2050, 2050
(3 1)
P 889893 3 x 319830 3 x ) x 14370
2
+ 
= + + 
 
=1,935,603
At design period of 30 years the forecasted population of Kochi City is expected to be
approximately 2 million
Therefore, the population to be considered to sewage design is 2 million.
4. Calculation of Sewage
Ultimate Design Period : 30 years
Population forecasted at 2050 = 2,000,000 persons = 2 million (2M)
Water Demand per Capital = 135 Liter per Capita per Day
Average Water Demand per day = 135 x 2000000 = 270,000000
= 270 MLD ( Millions of Liters per Day)
Average sewage generation per day = 80% of Average Water Demand per day
= 0.8 x 270 = 216 MLD ( Millions of Liters per Day)

4. 216 x 1000000
2500 liter/sec
24 x 60 x 60
= =
= 2.5 m3/sec
Therefore, Sewage Generated = 2.5 m3/s
Taking the peak factor = 2,
Design Discharge = peak factor x Sewage Generated = 5 m3/sec
Layout
Receiving Tank Coarse Screen Grit Chamber Skimming Tank
Primary
Sedimentation Tank
Aeration TankSecondary
Clarifier
Sludge Drying
Bed
Stabilization Tank

5. RECEIVING CHAMBER
This tank receives the sewage from the city and is rectangular in shape. The detention time
assumed for the sewage is 60 seconds.
Design
Design Discharge = 5 m3/s
Detention time = 60 seconds
Required Volume of Receiving Chamber, VR = Design Discharge x Detention Time
VR= 5 x 60 = 300 m3
Providing the Depth = 4 m, Area of the Receiving Chamber, AR =75 m2
Considering the length of the receiving chamber to be twice the width
AR = L x B = 2B x B
75 = 2B2
Therefore, width B = 6.2 m and Length = 12.5 m
Check
Design Volume, VD= 12.5 x 6.2 x 4 = 310 m3 > VR, OK
Size of the Receiving Chamber = 12.5 m x 6.2 m x 4 m
Freeboard of 0.5 m is provided.

7. SCREENING
It is used to remove the floating materials such as papers, rages clothes etc., in order the
protect the pumps and other equipment from possible damages.
Coarse Screening:
It is used primarily as protective device and hence used as first treatment unit. Common
type of these screens are bar racks (or bar screen), coarse woven-wire screens, and
comminutors. Bar screens are used ahead of the pumps and grit removal facility. This screen
can be manually cleaned or mechanically cleaned. Manually cleaned screens are used in
small treatment plants. Clear spacing between the bars in these screens may be in the range
of 15 mm to 40 mm.
Design of Coarse Screen
Design Discharge = 5 m3/s
Assuming the velocity of average flow as 0.8 m/s
The required Net Area Screen opening = 5/0.8 = 6.25 m2
Assuming the clear spacing between the bars in these screen as 30 mm = 0.3 m
Assuming the size of the bar as 75 mm x 10 mm,
Assuming the width of channel to be 1 m
The screen bars are placed at 600 to the horizontal
Velocity through the screen at peak flow = 1.6 m/s
Clear Area = 26.25
= 4.51m
1.6 x sin60
No. of clear openings = 4.51 / 0.03 = 150 nos.
Width of the channel = 150 x 10 + 149 x 30 = 5970 mm, i.e. 5.97 m ≈ 6 m
Thus, the coarse screen channel is designed for the size
6 m x 0.75 m ( Side Water Depth ) + 0.5 m ( Free Board )

9. GRIT CHAMBER
Grit chambers are designed to remove grit consists of sand, gravel, cinders or other inert
solid materials that have specific gravity about 2.65, which is much greater than those of the
organic solids in the wastewater.
Design
1. Design Discharge = 5 m3/s
2. Volume of Grit Chamber
Provide two chambers to facilitate periodic cleaning and maintenance
Assuming the detention time of 3 minutes
Volume of each tank = 0.5 x 5 x 3 x 60 = 450 m3
3. Dimension of Aeration Basin
Keeping the Depth to Width ratio as 1:3
Provide Depth, H = 3.5 m, hence Width, B = 9 m
Length, L = V/(B x H) = 15 m
Provide 20% additional length to accommodate inlet and outlet zones
Total Length, L = 1.2 x 15 = 18 m
Thus, the dimensions of the Grit Chamber as per design is 18 m x 9 m x 3.5 m
4. Determination of Air Supply Requirement
Consider 0.3 m3/min meter of length air supply
Air Requirement = 0.3 x 18 = 5.4 m3/min
Provide air swing arrangement at 0.5 m from floor
5. Quantity of Grit
Consider grit collection 0.015 m3/103 m3
Volume of Grit = 5 x 24 x 60 x 60 x 0.015 x 10-3 = 6.48 m3 per day
6. Check for surface overflow rate (SOR)
The settling velocity of the smallest particle = 2.4 cm/sec
The actual SOR in the grit chamber = 5/(2 x 15 x 9) = 0.019 m/sec = 1.85 cm/sec
Since SOR in the grit chamber is less than the settling velocity of the smallest particle
hence design is safe.

11. SKIMMING TANK
This tank is used to remove the oil and grease from the sewage before it enters the primary
sedimentation tank. In this tank, air is blown along with chorine gas by air diffuser placed at
the bottom of the tank. The rising air tends to coagulate and solidify the grease and cause it
to rise to the top of the tank whereas chlorine destroys the protective colloidal effect of
protein which holds the grease in emulsified form The greasy material are collected from
the top of the tank and the collected are skimmed of specially designed mechanical
equipment.
Design
Surface Area of the Skimming tank, A = 0.00622 x Q/Vr
Where Q = Rate of flow of sewage in m3/day
Vr = Min. rising velocity of greasy material to be removed in m/min
Here, Vr is taken as 0.25 m/min
Thus, A =
0.00622 x 5 x 24 x 60 x 60 2687.04
7.47
0.25 x 24 x 60 360
2
7.5 m= = = 
Provide the height of tank, H = 3 m
Taking the L:B ratio as 2:1
A = 2 B x B, A = 2B2
7.5 = 2B2
Thus, B = 2 m
Assuming the width, B = 3 m
Thus, Length, L = 6 m
Thus, the dimensions of Skimming Tank are 6 m x 3 m x 3 m with a free board of 0.5 m.

13. PRIMARY SEDIMENTATION TANK
After grit removal in grit chamber, the wastewater containing mainly lightweight organic
matter is settled in the primary sedimentation tank . The primary sedimentation tank
generally removes 30 to 40% of the total BOD and 50 to 70% of suspended solids from the
raw sewage. The flow through velocity of 1 cm/sec at average flow is used for design with
detention period in the range of 90 to 150 minutes. This horizontal velocity will be generally
effective for removal of organic suspended solids of size above 0.1 mm. Effluent weirs are
provided at the effluent end of the rectangular tanks, and around the periphery in the
circular tanks.
Design
Assumptions
• For rectangular tank the length more than 40 m is not preferred
• The depth of mechanically cleaned tank should be as shallow as possible, with minimum
2.15 m.
• The average depth of the tank used in practice is about 3.5 m.
• 0.25 m for sludge zone and 0.3 to 0.5 m free board is provided.
• The floor of the tank is provided with slope 2 to 8 % for rectangular tanks.
• The scrapper velocity of 0.6 to 1.2 m/min (0.9 m/min typical) is used in rectangular tank
• The surface overflow rate of 40 m3/m2.d (in the range 35 to 50 m3/m2.d) is used
for design at average flow.
• The detention time in PST could be as low as 1 h to maximum of 2.5 h. Providing
detention time of 1.5 to 2.5 h at average flow is a common practice.
Average Flow Rate, QA = 216 MLD ( Millions of Liters per Day) i.e. 2.5 m3/s
Assume surface settling rate = 50 m3/m2.d
Surface Area of the Tank, A = QA/Surface Setting rate = 216 x 1000/ 50 = 4320 m2
Diameter of the Tank, D =
4 x A 4 x 4320
74.16 75 m= = 
 
Hence, provide two tanks with diameter , D=38 m
Provide Effective depth = 3 m with a freeboard of 0.5 m and 0.5 m space for sludge
Thus, the total depth , D of the tank is 4 m.
Dimension of each Tank = 38 m (Diameter) x 3 m (D), 2 nos.

15. ACTIVATED SLUDGE PROCESS
The activated sludge process consists of an aeration tank, where organic matter is stabilized
by the action of bacteria under aeration and a secondary sedimentation tank (SST), where
the biological cell mass is separated from the effluent of aeration tank and the settle sludge
is recycled partly to the aeration tank and remaining is wasted. Recycling is necessary for
activated sludge process.
AERATION TANK
Provide 4 aeration tank
Average Flow Rate, QA = 216 MLD ( Millions of Liters per Day) i.e. 2.5 m3/s
Average Flow at each tank = 2.5/4 = 0.625 m3/s
Assuming 20% of BOD being removed by the Grit Chamber
Thus, BOD at the inlet of Aeration Tank = 80 % of 600 = 480 mg/l
Assuming the BOD of the effluent to be 20 mg/l
Thus BOD removed at the activated plant = 480-20 = 460 mg/l
Treatment efficiency based on soluble BOD,
460
x100 = 95.83 %
480
 =
Assuming MLSS ( Mixed Liquor Suspended Solids) = 3000 mg/l, F/M ratio = 0.4
3Q x BOD 54000 x 460
, 20700 m
( / M) x MLSS 0.4 x 3000
Volume V
F
= = =
Volume, V = 20700 m3
Assuming the Depth of the tank be 12 m
Let the width to depth ratio be 2.5
Thus, the Width of the tank, B = 30 m
Thus, the Length of the tank, L= 20700/(30 x 12) = 58 m (Approximate)
The Dimension of the tank is 58 m (L) x 30 m (B) x 12 m (D)
Check for Aeration Period/Hydraulic Retention Time
Hydraulic Retention Time, HRT = Q/V
Volume of the Tank = 58 x 30 x 12 = 20880 m3
Discharge in each tank, Q = 54000 m3/day = 2250 m3/hr
Thus, HRT = 20880/2250 = 9.28 hours, The value lies close to 9 hours, Thus OK
Check for Volumetric Loading
-3 -3
3Q x BODx10 54000 x 460 x 10
Volumetric Loading = 1.19 kg BOD m .d
Volume of Aeration Tank 20880
= =

16. Since the value is between 0.8 to 2, Thus OK.
Check for Return Sludge Ratio
MLVSS/MLSS = 0.8
Return sludge SS concentration = 10000 mg/L
MLSS = 3000 mg/l
3000 (Q + QR) = 8000 QR
Thus, Q/Qr = 0.6 or 60 %, lies between 0.5 to 1. Thus OK
Check for Solid Retention Time
C O E
d C
C O E
d C
Q θ α (Y -Y )
V =
MLSS x (1+k θ )
Q θ α (Y -Y )
(1+k θ ) =
MLSS x V
Where α = 0.5 constant for municipal sewage
Q = 54000 m3/day
YO = 480mg/l
YE= 20 mg/l
MLSS = 3000 mg/l
Volume, V = 20880 m3
kd = 0.06 per day
C
C
C C
C
C
54000xθ x 0.5 x (480-20)
(1 0.06θ )
3000 x 20880
(1 0.06θ ) = 0.198 θ
1 = 0.138 θ
θ 7.25 days
+ =
+
=
The value lies between 5 and 8 days. Thus OK
Provide the Dimension of the tank is 58 m (L) x 30 m (B) x 12 m (D) including the FB
BOD5 applied to each tank = 480 mg/l
Average flow in each tank = Q = 54000 m3/day
BOD5 removed in each tank = 54000 x 0.480 = 25920 kg/day = 1080 kg/hr

18. SECONDARY SEDIMENTAION TANK
Number of Secondary Clarifier = 2
Average Flow Rate, QA = 216 MLD ( Millions of Liters per Day) i.e. 2.5 m3/s
Recirculated flow = 53 %
Thus, in each tank= 0.5 x 0.53 x 216000 = 57240 m3/day
Total inflow in each tank = 0.5 x 216000 + 57240 = 180240 m3/day
Providing Hydraulic Detention Pond = 2 hr = 0.083 day
Volume of tank, V = 180240 x 0.083 = 15020 m3
Assuming Liquid Depth = 5 m
Area, A = 3004 m2
Surface loading rate of average flow = 25m3/m2/day
Surface Area Provided = 0.5 x 216000/25 = 4320 m2
Using the greater Value, A = 4320 m2
Diameter, D = 75 m
Check for Solid Loading
Recirculated flow = 57240 m3/day (each tank)
Average Flow Rate in each tank = 108000 m3/d
MLSS in tank = 3000 mg/l
Total Solids in the flow = (57240+108000) x 3 = 495720 kg/day
Solid Loading = 495720/4320 = 114.75 kg/day/m2
Check :
It lies between 100 – 150 kg/day/m2. Hence OK
Provide 2 number of Secondary Clarifier of Sizes 75 m Diameter and Depth of 5 m with 0.5
m Freeboard.
STABILISATION TANK
Total Return flow of two tanks = 0.53 x 216000 = 114480 m3/day = 79.5 m3/min
Detention Time = 15 minutes
Volume of Wet well = 79.5 x 15 = 1192.5 m3
Providing Depth as 4 m and Width as 10 m
Length, L = 30 m
Provide the Stabilization tank of sizes. 30 m (L) x 10 m (B) x 4 m (H) with 0.5 m freeboard

19. Sedimentary Clarifier

20. SLUDGE DRYING BEDS
Assumptions:
Sludge applied to the drying beds at the rate of 100 kg/MLD
Sludge applied = 300 kg/day
Specific gravity = 1.015
Solid content = 2%
Thus, Volume of the Sludge =
Sludge Applied
Specific Gravity x Soid Content x 1000
=
3300
14.78 m /day
1.015 x 0.02 x 1000
=
Assuming 10 days for drying of beds based on the weather condition of Kochi
Number of Cycle in one year = 365/10 = 37 cycles
Period of each cycle = 10 days
Volume of Sludge per cycle = 14.78 x 10 = 147.8 m3
Spreading a layer of 0.3 m per cycle
Area of bed required = 147.8/0.3 = 492.67 m2 = 500 m2
Providing 5 no. of beds
Area of each bed = 100 m2
Taking width of bed as 8 m, the length of bed is 12.5 m
Thus, the size of the bed will be 12.5 m x 8 m to be provided in 5 nos.