INTERNSHIP PRESENTATION ON
DESIGN OF THE SEWAGE TREATMENT
PLANT, SHIMOGA
By:
UTTAM UDAY UPARI
8th SEM, B.E (4UB16CV056)
CIVIL ENGINEERING
Under the Guidance of
PROF. MADHUKARAN
ASSISTANT PROFESSOR
CIVIL ENGINEERING DEPARTMENT
OUTLINE:
Introduction
About the Organisation
Background
Objectives of the Internship
Treatment Process
Design of STP Units (Convensional)
Conclusion
2
References
INTRODUCTION:
Sewage is a combination of the liquid carrying wastes removed
from municipalities establishments.
It comprises of 99.9% water and 0.1% solids and is organic
because it consists of carbon compounds.
Fresh sewage is odour less which is not offensive. However,
the decomposition of the organic matter present in the sewage
start within a short period and offensive odour.
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The Karnataka Urban water supply and Drainage Board (KUWS & DB), is
ABOUT THE ORGANISATION:
the nodal Government agency responsible for investigation, planning
preparation and implementation of water supply and drainage schemes for
urban areas in the state of karnataka.
After implementation of the project the scheme is handed over to the local
bodies for maintenance.
The board draws loan from financial institutions like LIC, HUDCO and
other on behalf of the local body. The state government stands guarantee
for all such loan.
The board is now desirous to taking up the work of improving and
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expanding the ground drainage system in Shimoga city.
OBJECTIVES OF THE INTERNSHIP
To reduce the negative impact on Environment.
To promote treated wastewater to the irrigation purpose.
To design a conventional treatment plant units for the treatment of
wastewater.
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TREATMENT PROCESS :
Description of Process
The wastewater treatment is a process which includes receiving of sewage, removal of
inorganic floating material and segregation of waste, biological treatment, disinfection
and then its final disposal. Below outlined steps are carried out during treatment process.
A) Raw Sewage
B) Inlet chamber
C) Screen unit
D) Grit chamber
E) Parshall flume
F) Sedimentation tank
G)Aerated lagoon
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H)Polishing pond
Process Diagram :
Raw Sewage
Screens
Girt Chamber
Parshall Flume
PST
Aerated Lagoon
Polishing Pond
Final Disposal
FIG : FLOW DIAGRAM OF A CONVENTIONAL TREATMENT PLANT
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DESIGN OF CONVENTIONAL STP :
CALCULATION OF DESIGN POPULATION
Year Population Increase in population
(1) (2) (3)
1971 102079
1981 151783 49704
1991 192647 40864
2001 274352 81705
2011 322650 48298
Total 220571
55142
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CALCULATION OF DESIGN POPULATION
The expected population at the end of year 2019
Pn = P 0 +
Pn = 2019
P0 = 322650
55142
P2019 = 322650 +0.8(55142) .
= 366764
Therefore the population at the end of 2019 = 366764.
The expected population at the end of year 2049
Pn = P 0 +
Pn = 2049 P0 = 366764
55142
P2049 = 322650 +3(55142) .
= 532192 9
Therefore the population at the end of 2019 = 532192.
CALCULATION OF SEWAGE GENERATION
Ultimate design period = 30 years
Forecast population = 532192
Per capita water supply =135 LPCD
Average water supply per day = 532192
× 135L
= 71845920 L /day
= 71.84 MLD
Average water supply per day = 72MLD.
Average sewage generation/day = 80% of
water supply
= 0.8 ×72 MLD 10
=57.6MLD
DESIGN OF INLET CHAMBER
Data:-
Design flow = 1.48 m3/sec
Detention time(D.T)= 30 sec
Design:-
Volume required (Vreq) = Flow × Detention
time
= 1.48 × 30
Vreq = 44.4 m3.
Assume the depth of the tank =3m
Area = Volume / Depth
= (44.4)/3
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Area of the inlet chamber = 14.80 m2.
DESIGN OF INLET CHAMBER
Provide length to breadth ratio = 2:1
L X B = 2B × B
B= 7.4m
Hence provide breadth of inlet chamber
B=7.4m
Now, length L = 2 × B
L = 2 × 7.4 = 14.8m
Therefore provide length of chamber L =
14.8 m.
Provide 0.5m free board and 0.5 m provision
for sludge accumulation then the total depth
of the tank would be 3+0.5+0.5 = 4m.
Therefore the dimension of the Inlet chamber
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is 14.8m × 7.4m × 4m.
DESIGN OF INLET CHAMBER
Fig 3.1. Diagramatic representation of Inlet chamber
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DESIGN OF SCREEN CHAMBER:-
Assumption1. The velocity through bar screen = 0.8
m/s
2. The screens are placed at 600 to the
horizontal.
Design:-
The maximum discharge (Q)= 1.48 m3/sec.
No. of screens =2
The maximum discharge in one screen = 1.48/2 = 0.74
m3/sec.
Velocity through the bar screen (V) = 0.8 m/s.
The net area for screen opening required will be,
A = Q/V
net
A
net =
A
net = 0.92 m
2
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Using rectangular bars in the screen having 2cm
width placed at a distance of 5cm clear spacing.
DESIGN OF SCREEN CHAMBER
Total inclined area,
Considering m2, when the screens are placed at an angle of
600 to the horizontal then, = 1.47m2.
Assuming depth of screens channel = 1.0m
Gross width =
B=
B= 1.47m
. Compute the number of bars
Provide bars of 20mm x 50mm with a clear spacing of
50mm. Let n be the number of bars then,
Opening(n+1) + size of bars(n)= B(total width)
0.05(n+1) + 0.02n = 1.47
n = 20
Hence providing 20 bars of 20mm x 50mm with 50mm
clear spacing.
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The screen is designed for gross area m2 and of size
1.47m × 1.0m + 0.5 FB.
DESIGN
. OF SCREEN CHAMBER
Fig 3.2: Diagrammatic representation of Bar screen
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DESIGN OF GRIT CHAMBER
Assumption:- 1. Assuming a horizontal flow
velocity of 0.3 m/s
2. Detention time 60 sec
3. Size of the grit particle d = 0.2 mm
Design:- Maximum daily flow of sewage =
1.48 m3/sec
No. of chambers = 2
Maximum daily flow of sewage of each chamber
=1.48/2
= 0.74
Cross sectional area of the tank
Cross sectional area of the tank
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Cross sectional area of the tank m2.
DESIGN OF GRIT CHAMBER
Length of the tank
Length of the tank
Length of the tank
Capacity/ Volume of the tank
Volume of the tank
Volume of the tank 3
Calculating settling velocity by Hazen’s formula
Hazen’s formula for settling velocity Vs = d[3T +70]
Where d = diameter of grit particle = 0.2mm
T = temperature= 200 C
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Settling velocity Vs = 0.02 [ 3×20 +70]
Vs = 2.6 cm/sec.
DESIGN OF GRIT CHAMBER
Depth of the tank required
Liquid depth dL = Settling velocity × detention time
= 2.6 m/s × 60 sec
= 156 cm
Liquid depth dL = 1.56 m
Provide free board of 0.3 m and sludge accumulation
0.14 m
Therefore the total depth of the tank = Liquid depth +
Free board + Sludge accumulation
= 1.56 + 0.3 + 0.14
Total depth of the tank d = 2m
We have width to depth ratio = 2 : 1
The width of the channel = 2d
B= 2 × 2 = 4
Therefore provide width of the channel = 4m 19
Therefore design the rectangular Grit chamber of
size 18m × 4m × 2m of 2 units.
DESIGN OF GRIT CHAMBER
Fig 3.3: Diagramatic representation of grit chamber
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DESIGN OF SEDIMENTATION TANK
Assumption:- 1. Detention period =2 hours
2. Liquid depth
3. Velocity through the tank =
0.3m/minute.
Data:- 1. Average flow = 1.22 m3/sec
2. Population= 532192
3. Water Supply = 135 LPCD
Design:- The tank is designed for peak flow.
Peak flow = 1.4 × Avg flow
= 1.4 × 0.66 m3/sec
= 0.924 m3/sec
= 79833.6 m3/day
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Peak flow = 79833.6 m3/day.
DESIGN OF SEDIMENTATION TANK
No. of tanks = 2
Flow per tank = 79833.6/2
= 39916 m3 / day
The quantity of sewage to be treated in 2 hours i.e.
the capacity of the tank required.
Capacity = flow × Detention time
=39916 × 2/24 per hour
Capacity = 3326.33m3.
The length of the tank required Length = flow
velocity × Detention time = 0.3 × 120
Length L = 36 m
Cross sectional area required Ac/s=
Capacity/length 22
Ac/s= (3326.33)/36
Ac/s= 92.39 m^2
DESIGN OF SEDIMENTATION TANK
Breadth of the tank required
Assuming an overflow rate of 40,000 lt/m2/day
Surface Overflow Rate Vo =
=27.69 m
Provide breadth B = 28m
Depth of the tank required
Depth
= 3.29 m
Provide depth d = 3.3 m.
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DESIGN OF SEDIMENTATION TANK
Assuming free board of 0.5m and since the
tank is mechanically cleaning no extra
bottom space is required for sludge
accumulation, however provide 0.2m for
sludge zone.
Therefore total depth D=3.3m+0.5m+0.2m
Therefore total depth D = 4.0m
The overall size of the sedimentation 24
tank designed is 36m × 28m × 4.0m.
DESIGN OF SEDIMENTATION TANK
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Fig 3.5: Diagrammatic Representation of Primary Sedimentation tank
DESIGN OF AERATED LAGOON
Assumption:- 1. Detention period =5 day
2. Effective depth =3.9m
3. Number of lagoons to be provided = 2.
Data:- 1. Average flow = 130000m3/day
2. Sludge volume = 0.03cum/capita/year
3. Desludging frequency = 3 years
Design:- Flow per each lagoon = 130000/2 26
= 65000 m3/ day
DESIGN OF AERATED LAGOON
Volume of each lagoon with 5 days=5*65000
=325000 m3
Sludge generation=0.03 cum/capita/year
Total sludge generation= 0.03*3*532192
=47897.28cumec
Thus sludge volume for each lagoon= 47897.28/2
=23948.64cumec
The total volume of each lagoon=325000+23948.64
=348948.64cumec
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DESIGN OF AERATED LAGOON
Let the free board be =0.5m
Surface area required for each
lagoon=348948.64/3.9
=89474 m2
Therefore provide a lagoon area =895*100 m2
Therefore provide a lagoon size
=895*100*(3.9+0.5) m3
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DESIGN OF AERATED LAGOON
Fig3.6:Diagrammatic representation of Aerated lagoon
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CONCLUSION
o Design parameters should be carefully considered for each
part of the STP
o In this Design, the population density used was “532192"
which resulted in large and added basins were used.
o Convectional treatment plant Technology is better than the
SBR Technology, it requires less operation and maintainance
cost and high effluent quality,possibility of producing
electrical energy from biogas and high treatment efficiencies
possible for BOD,COD,TSS,N,P.
o The treated effluent is used for irrigation purpose near by the
lands.
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REFERENCE
Data collected through Treatment Plant visit and
Data log sheet available with the Management.
Niraj S.Topare, S. J. Attar and Mosleh M. Manfe
(2011), “Sewage/Wastewater Treatment
Technologies : A Review”, Scientific. Reviews &
Chemecal Communications: 1(1) page no: 18-24.
Chandaluri Subba Rao (2011), “Determination of
Water quality of Some Rural areas of Guntur district,
Andhra Pradesh, India”, Dept. of Environmental
Science: Page no: 185-217.
Vigneswara S, Sundaravadivel M and Chaudhary D S
(2007), “Sequencing Batch Reactors: Principles,
Design/Operation and Case Studies”, University of
Technology Sydney, Australia: Page no. 77-86. 31
THANK YOU
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