Sewage /Wastewater Treatment
Purpose of Sewage Treatment
The raw sewage must be treated prior its disposal in water body
Degree of treatment depends upon following consideration
i) Characteristics and quality of the sewage
ii) Quality of receiving water body
iii) Assimilative capacity of receiving water body to cope pollution
iv) Intended use of receiving water body
Purposes of Treatment
a) To stabilize the sewage without causing odour and nuisance
b) To protect public health
c) To avoid potential damages to quality of receiving water body
(Rivers, canals & Streams)
d) To protect the aquatic life from the potential threats
Steps in Sewage Treatment
Main steps in for sewage treatment
1. Separation of Solids from liquids
2. Treatment and disposal of liquid
3. Treatment and disposal of solids
The layout of conventional wastewater
treatment is as follows:
1. Preliminary Treatment
Done through physical methods
(Remove coarse, suspended matter, floating materials and O&G)
Treatment units
a) Balancing /Equalization Tank
b) Screens
c) Cutting Screens or Comminutors
d) Grit Chambers
e) Skimming Tanks / Floatation
f) Evaporation
2. Primary Treatment
Done through mostly by physical methods
(Settle-able and Suspended Solids)
The units involved are
a) Sedimentation tank
b) Septic tank
c) Imhoff tank
3. Secondary Treatment
Done through biological processes
(remove organic solids)
a) Trickling Filters
b) Activated Sludge Process
c) Oxidation Ponds
4. Sludge Treatment and Disposal
a) Sludge Digestion
b) Sludge Drying
1. Preliminary Treatment
The process of removal of rags, pieces of wood, plastics, metals,
rubber or fragments of masonry is called preliminary treatment.
Necessary to remove otherwise - interfere the treatment processes
and may cause damage to the installed machinery e.g. pumps and
motors etc.
b) Screening
Types of screens
There are two types of screens
i) Bar Screen or Racks
These can be further of two types
a) Manually / Hand Cleaned (For small STP)
b) Mechanically Cleaned (For large STP
ii) Mesh Screen /Fine Screens
(Not for sewage but for industrial WW i.e for uniform particle sizes)
Contd..
a) Equalization Tank
Figure: A Typical view of Bar Rack Screen
Screening - Design Consideration
Parameter Types of Screen
Manually Cleaned Mech. Cleaned
Bar width(mm) 6 – 15 mm
(1/4” – 5/8”)
6 – 15 mm
(1/4” – 5/8”)
Bar Spacing (mm) 25 -50 15 – 75
Approach Velocity (m/s) 0.3 – 0.6 0.6 – 1.0
Velocity should not be more than 1 m/Sec to avoid Excessive Head
Loss and should not less than 0.6 m/Sec to avoid Sedimentation in
Screen Chambers
Slope <300 >750
Length Approachable
Location Before pumps Before pumps
Number of Screens 02 02
Surface Area perpendicular to
flow
2 times the area of
sewer
Screened material (m3/m3 flow) 1.6x 10-6 – 1.5x10-6
Nature of Screened material Contaminated
Disposal of Screened Material Landfill or Incineration
Instead of removing the large size particles, communitors reduce
them in size so that will not harm the equipment. The chopped or
ground solids are removed in the sedimentation process.
Provided with fixed screen and moving cutter.
Floating comminutors draw the flexible material through screen rather
than chopping them.
(nuisance in un-skimmed clarifiers, trickling filters and aeration basin).
For small plant single unit rated for peak flow may be used in parallel
with manually cleaned bar rack.
For larger plants multiple number of units are used with capacity so
that remaining units may handle the peak flow, with one or two out of
service.
Head loss depends upon the screen details and flow and ranges from
50 -100 mm
c) Comminutors
Figure: A Typical View of Comminutor
d) Grit Chambers
Grit Chambers are used to remove the inert inorganic particles (sand,
metal fragments and eggshells etc) from sewage having size 0.2mm or
larger and specific gravity equal or more than 2.65
Purposes
i. To protect moving mechanical equipment from abrasion and
abnormal wear.
ii. To avoid heavy deposition in pipelines, channels and conduits.
iii. To avoid frequent cleaning of digesters to remove excessive
deposition
Design Considerations
Parameters Value
Velocity (m/s) 0.25 – 0.3
Detention Time (Sec) 40 – 60
Length (m) 10 – 20
Cleaning Intervals (Weeks) 2
Velocity Control Through Proportional Flow Weir OR
Sutro Weir
Figure- A Schematic View of
Used to remove O&G from wastewater
A basin with DT of about 10 minutes is satisfactory.
Includes aerating device which blows air (About 0.1 cubic
feet of air/gallon at a pressure ranging from 40 – 50 Psi.
Rising air coagulates the O&G and cause them to rise to
surface
Removed easily either manually or mechanically.
e) Skimming Tank
Skimming /Floatation Tank
Activation & Sedimentation Tank
Removal of settle-able suspended organic solids and floating
materials.
Also reduces the load on subsequent biological units.
Sometimes, addition of chemicals to assist in removal of
finally divided and colloidal solids, or to precipitate
phosphorous is carried out.
2. Primary Treatment
Sedimentation Tanks
Separation or removal of suspended solids from the
wastewater by gravity.
If the specific gravity of the SS is more than 1, it will settle with
certain velocity under gravity.
Water is allowed to remain in a basin or tank for certain time, the
SS settle and reach at the bottom of the tanks and thus the
wastewater will become clear
Refers to the settling. of particles in a suspension of low solid conc.
Particles settle as individual (no significant interaction)
It removes grit and sand particles form water.
Particles settles under “Stokes’s law”
Principle of Sedimentation
Discrete Particle Settling
Vs = g (ps – p)d2
18µ
The above equation can be written as
Vs = g (Gs – 1)d2
18 v
Gs = Specific Gravity of the particles
v = Kinematic viscosity
The above law is applicable when RN < 1
Where
Vs = terminal settling velocity (L/T)
ps = the mass density of particles (M/L3)
p = The mass density of fluid (M/L3)
G = Acceleration due to gravity (L/T2)
µ = Absolute viscosity of the liquid (M/LT)
Ideal Sedimentation Basins
Design of the Sedi. basin is based on concept of Ideal Sedi. Basins.
Assumptions
i) Complete mixing and uniform suspension at the inlet zone.
ii) Uniform horizontal velocity in the settling zone.
iii) No flocculation (Discrete free settling)
iv) Particles that reach at bottom are permanently removed.
Particle entering the basin will have a horizontal velocity equal to
the velocity of the fluid
V = Q/A = Q/wh
w = Width of tank &
h = depth of tank
The other velocity is settling velocity “Vs” of that particle defined by
Stoke’s law.
For removal of particle, Settling velocity (Vs) and horizontal
velocity (v) must be such that their resultant velocity ( V ),
must carry it to the bottom of the tank before outlet zone is
reached.
If the particle entering at the top of the basin (point a) is so
removed, all the particles with same velocity will also be removed.
Consider slope of the velocity vector from point “a” to “f” and
dimension of the basin itself. We can write
Surface Overflow Rate (SOR)
Vs = h
V L
Or
Vs = Vh As V = Q/A
L
= Q/WhL
Vs = Q/WL
SOR is numerically equal to the
flow divided by the plan area of the
basin.
It physically represents the settling
velocity of the slowest settling particles
which is 100% removed.
Particles having Vs > SOR will be
entirely removed.
Detention Time
It is defined as the tank volume divided by the flow
Detention Time (DT) = V/Q
Sedimentation Tanks
Used to remove suspended particles upto 50 – 60%.
BOD removal in these tanks is associated with removal of
SS
BOD Removal ranges 25 – 40 %.
Sedimentation tanks are of following types
Types
i) Rectangular Tanks
ii) Circular Tanks
iii) Square Tanks
Horizontal flow pattern. WW flows along the long axis.
Minimizes the effects of inlet and outlet disturbances.
Rectangular Tanks
Sr.No. Parameter Value
i) Design flow Average Daily Flow
ii) Settling Velocity of Particles 0.3 – 0.7 mm/s
iii) Surface Overflow Rate(SOR) 25 – 40 m/day
iv) Detention Time (DT) 1.5- 2.5 hour
v) Depth 2-5 m
vi) Maximum Length 30 m
vii) L : W (ratio) 4 : 1
viii) Depth 2 – 5 m
viii) Weir Loading Not more than 120 m3/m.day
vix) Sludge Accumulation 2.5 Kg wet solids / m3 flow
Secondary Sedimentation Tank
i) Design Flow Average Daily Discharge
ii) SOR 30-40 m/day
iii) Detention Time (DT) 2 – 3 hours
iv) Depth 2.5 – 5 m
Design basis
In circular Tank the flow may either at periphery or at the centre.
Ave. DT in peripheral feed basin is greater.
Diameter of 30 m is provided (generally).
Diameter more than 100 m is not recommended.
Circular Tank
Square basin may be used in situations where land area is
limited.
The length of one side may be 21 m.
Square Basin
Problem: Calculate the settling velocity of a sand particle of o.4
mm in size in water at 10 oC. Specific gravity of sand particle is 2.65. How
much surface area of an ideal settling tank will be required to remove
these particles? Dynamic viscosity of water is 1.31x10-2 poise. The
incoming flow is 28512 m3/day
D = 0.04 mm 0.004 cm
S.G = 2.65
Viscosity (µ) = 1.31x10-2 poise(gm/cm-sec)
Q = 28512 m3/day
Vs = ?
A = ?
Vs =g (Gs – 1)d2
18v
V(Kinematic visco) =µ/p = 1.31x10-2 / 1
p = 1 gm/cm3
Vs =9.81(2.65-1)(0.004)2
18x1.31x10-2= 1.098 x 10-3 m/sec
Now for 100% removal SOR should be equal to the settling velocity i.e.
SOR = Vs =
=Q/wL
= 1.098 x 10-3
Surface Area =wL
= 2852 / (1.098 x 10-3x24x60x60)
= 300.5 m2
Problem: A sedimentation basin is 30 m long, 15 m wide and
4m deep and has an overflow rate of 24 m3/day-m2. What is
detention time?.
Solution
L = 30m W = 15m H = 4 m SOR = 24 m3/day-m2. DT = ?
DT V/Q
V(vol) = 30 x 15 x 4 m3 1800 m3
Also
SOR = Q/WL
Q = SOR x WL = 24 x 15x30 m3/day 10800 m3/day
DT = V/Q = 1800 / 10800 day 0.1667 day
DT = 4 hours
It reduces the entrance velocity of effluent
It distributes the water though out the width and depth
of tank and
It mixes it with water already present in the tank to
prevent density currents.
Inlets and outlets
Careful design is important to assure reasonable
performance of ST.
Inlets
Advantages of proper design
Inlet structures
In STs outlets consist of free-falling weirs.
Weir Loading Rates(WLR), specified in volume /unit
length per day, are limited to prevent high approach
velocities near outlets.
WLR calculate the length of the weir but not length over
which overflow occurs.
Outlets are placed as far as possible from the inlets i.e. at
the end of rectangular tanks and towards centre and
along the radii of peripherally fed tanks.
Weirs frequently consist of v-notches approximately 50
mm in depth and placed at 300 mm on centre.
Outlets
Outlet structures
ii) Septic Tanks
These are Primary Sediments basin
Provided in houses with water supply and too low housing density to
allow soakage pits
A minor degree of solid destruction may occur due to anaerobic
digestion. Units are ordinarily sized to provide detention time of 24 hours
at average daily flow.
Usually made of concrete but steel and fibreglass are used.
The effluent is offensive and potentially dangerous.
Effluent needs further treatment (additional process or soil disposal
Ave. BOD ranged from 120 – 270 mg/l & Ave. SS from 44 – 69 mg/l.
De-sludging is required after 1 to 3 years.
Two Compartment ST (Better performance in SS & Pathogen removal)
Septic tanks are more expensive than other on-site WWT system
Low soil permeability,
high groundwater levels and
proximity of wells that supply drinking water.
Factors affecting Sustainability
These include
Single Compartment Septic Tank
iii) Imhoff Tank
Used for separation and digestion of solids in single unit.
Mixing and heating in the digestion zone (in modern units).
widely used in the past, to provide primary treatment.
Old Imhoff tanks have been converted to other uses in recent
years.
Advantages of these units over septic tanks is questionable.
Secondary Treatment
Advanced WWT treatment process
Intends to remove constituents of WW which remain even
after primary treatment.
i) Soluble Organic matter (BOD)
ii) Colloidal SS by improving sedimentation process
iii) Removal of Phosphorous Nitrogen and Phenolic
compounds etc
They employed for
Physical and Chemical processes
i) Coagulation or Chemical Precipitation
Industrial WW may contain very fine particles and heavy metals
May not settle in the reasonable time or stabilize by the Mic. Org.
Certain chemicals are added to WW to be treated
Objectives
a) To improve the performance of sedimentation tank
b) To improve the performance of filtration.
These chemicals are known as coagulants.
Precipitates are formed in WW, which catch the tiny particles and
ions of heavy metals.
PPts increase in size through process of agglomeration and thus
settle down in the sedimentation tanks.
Most Commonly used coagulant is Alum (Aluminium Sulphate), lime
Ferric chloride, Ferric Sulphate, Sodium Dichromate, Ferrous Sulphate
Determine Effective dose ( higher removal efficiency) through “Jar test”.
Addition of coagulants is followed by rapid mixing, flocculation and
final removal in sedimentation tank.
ii) Adjustment of pH
Adjustment of pH is important.
pH of WW effects the operation of WWTP e.g. Coagulation,
biological treatment.
The optimum pH values for various coagulants is as followings
Coagulant Range of pH
Alum 4.0 – 7.0
Ferrous Sulphate 3.5 and above
Ferric Chloride 3.5 – 6.5 and above 8.5
Ferric Sulphate 3.5 – 7.0 and above 9.0
Suspended Solids (SS) 80 – 90 %
Organic Matter 50 - 55 %
Bacteria 80 – 90 %
Removal efficiency of SS, Org. Material and Bacteria through
chemical process /treatment vary and given as below:
iii) Neutralization
Acid or base is added to neutralize the effluent.
Micro. Orga. are sensitive towards the pH values of the
WW to be treated.
Necessary for successful performance of biological
units
Dose (acid or base) is determined in the laboratory.
Biological Treatment
Sec. Treatment is usually understood to employ biological
treatment process.
WW in addition to organic matter also contains large number of
Micro.Orgs. Able to stabilize the waste in the natural purification
process.
Micro.Orgs. remove soluble and colloidal organic matter from
the waste.
Process requires, large number of microorganisms in relatively
small tank/container to do job in reasonable time.
Basic principle of process is same but the techniques may vary
These processes are designed as
1. Suspended Growth process
2. Attached Growth process
1. Suspended Growth Process
These processes maintain adequate biological mass in
suspension within the reactor by employing either natural
or mechanical mixing.
Generally the required volume is reduced by returning
bacteria form the secondary clarifiers (or sedimentation)
tank in order to maintain a high solid concentration.
The suspended growth processes include
i) Activated Sludge Process (ASP)
ii) Oxidation Ponds
iii) Aerated Lagoons
iv) Sludge Digestion Process
i) Activated Sludge Process (ASP)
A sludge flock (i.e. body of microorganisms gathered in a crowd)
Produced in raw /settled sewage by the growth of bacteria and other
organisms in the presence of DO and accumulated in sufficient conc.
by returning sludge flock previously formed.
devised by Arden and Lockett in Manchester in 1914.
Activated sludge
A mixture of sewage and activate sludge is agitated and aerated in
an Aeration Tank.
Bacteria in the activated sludge aerobically metabolize the Organic
Matter present in the influent.
Process Description
Sedimentation in final clarifiers to separate sludge from the Mixed
Liquor (Mixture of sewage & activated sludge in aeration tank).
A portion of settled sludge is wasted or returned to the aeration tank
as needed. The treated effluent overflows from the final clarifier.
Contd..
A modified form of Conventional Activated Sludge Process.
Requires less aeration period and less population of microorganisms in
the tank.
Ultimately needs less power consumption .
BOD removal is 60 – 70 % as compare to 90 -95% removal in
conventional ASP.
Degree of treatment in ASP depends upon the settle- ability of sludge in
the final clarifier.
Growth of Filamentous Microorganisms
Offers hindrance in sedimentation
Contribute to BOD5 & SS
Sludge Bulking: Excessive carryover of floc, resulting in the Inefficient
Operation of final clarifier is referred as Sludge Bulking.
Act.sludge Contd..
Modified Activated Sludge Process
Increase in biological solids in the system and continuous
seeding with re-circulated sloughed solids.
Maintenance of more uniform hydraulic and organic load,
Dilution of influent with better quality water and
Thinning of biological slime layer
Recirculation rates ranges from 50 – 100 % of the wastewater
with usual rates being 30 – 50 %
Recirculation may not exceed the efficiency in all circumstances
particularly with relatively dilute wastes.
Advantages of Recirculation
These include:
High F:M ratios __ growth of Filamentous Organism--- Sludge Bulking
Conditions Promoting Growth of Filamentous Organisms
Filamentous Organism ----- Sludge Bulking
Conditions Promoting Growth include:
a) Insufficient Aeration
Causes Sludge floatation
Results in DO level less than 2 mg/l
Bacteria in the absence of O2 converts ammonium to nitrate and
finally into N2 gas.
Rising N2 gas causes floatation of Sludge
b) Lack of Nutrients (Nitrogen and Phosphorous)
Problem may be corrected by providing lacking nutrients.
Guiding values of BOD and Nitrogen and Phosphorous are of
the order of 100:5:1
c) Low values of pH
Low pH promotes fungal growth in the basin.
Adjust pH (Neutralize) to cover this problem
d) Overloading
Food to Microorganism Ratio (F:M)
Expressed in terms of Kg of BOD applied per day per Kg of
MLSS. If Q is the sewage flow in m3/day and it has a BOD
expressed in mg/l then
Food = ( Q x BOD ) Kg BOD/day
1000
If “V” is volume of the aeration tank in cubic meter (m3) and it
has an MLSS concentration expressed in mg/l, then
Microorganism = ( V x MLSS ) Kg MLSS
1000
F:M = ( Q x BOD ) per day
( V x MLSS)
As Q/V = t
F:M = ( BOD ) per day
( t x MLSS)
Where “t” is aeration time in days
Design Criterion ( conventional and modified ASP) are as followings:
Parameter Value
Aeration Period 4 – 8 hours
Air Required 90 – 125 m3/kg of BOD5
SS in MLSS 1500 - 2000 – 2500 - 3000 mg/l
F:M 0.5 – 1.0 however 0.25 to 0.5 day-1 is usually
employed & gives sludge settling characteristics
Return Sludge 25% - 100 % of sewage flow
Dissolved Oxygen (DO) 2 mg/l
No. of Aeration Tanks 2 (as minimum)
Depth 3 – 5 m
L :W 5:1
Aeration Systems
a) Diffusers with pressure of 40 KPa
b) Surface Aerators
Qr/Q = Vs/(1000 – Vs) Where
Vs = Volume of Settled Sludge in “ml”
Qr = Flow of Return Sludge
Q = Flow of Sewage
Major characteristics are as followings
Diffuser System
Diffusers size 150mm dia Ceramics Domes
Bubble Size 2.0 – 2.5 mm
Distance b/w Diffusers 0.6 – 1.0 m
Location Bottom of the tank
Operation
Noiseless
Less formation of aerosols
Mechanical surface aerators in Aeration Tank
Spin “partially in” & partially out of the mixed liquor.
Thrown violently across the surface of the tank to adsorb O2.
These require less maintenance and
Provide visual evidence of break down.
Surface Aerators
Typical Views of Functioning of Diffusers
Measurement of Sludge Settle-ability
Sludge Volume Index (SVI)
It is the volume measured in milliliters (ml) occupied by
one (1) gram of settled suspended solids.
SVI indicates the sludge settling characteristics.
Determination of SVI
Draw sample of mixed liquor at the end of aeration tank
Measure the volume of the settled sludge (Vs) in a one
liter graduated measuring cylinder in 30 minutes.
Measure the Suspended Solid (SS) content in mg/l of the
mixed liquor samples which is called MLSS.
SVI = (Vs x 1000) (ml/g)
MLSSValue of SVI Sludge Characteristics
< 50 Thick and hard sludge difficult to pump
40 – 100 Good sludge
>200 Sludge has bulking tendency ie. Poor settling
characteristics
Types of Activated Sludge Processes (ASP)
i) Conventional (Plug Flow)
Exceed solids are usually wasted from the clarifier underflow.
Arrangement for Separate sludge wasting (possible &
preferred).
Mixing of return sludge with incoming waste.
Uniform Air supply (along the length of the basin through
porous diffusers.
This process consists of
A rectangular basin,
A clarifier, and
A solid return line from the bottom of the clarifier.
F/M ratio varies as the wastewater travels in the tank.
Air is provided uniformly along the length of the basin
through porous diffusers.
Since there is high concentration of BOD as well as
microorganisms, the oxygen at inlet of the tank is consumed
rapidly due to raid exertion of BOD by microorganisms.
Therefore, oxygen demand may be difficult to meet at inlet.
Air supplied may become excess on the other end of tank.
Disadvantages
ii) Tapered Aeration
High Conc. Of MO as well as BOD at inlet, hence rate of
oxidation in Conventional ASP is highest at inlet.
Utilization of O2 at inlet is more.
Utilization of Oxygen by mixed liquor in 6 hours aeration
time (based on various studies) is as following:
Therefore, maintaining aerobic condition throughout the tank
length by providing uniform air distribution, sometime
becomes difficult.
Time BOD Removal (%)
Ist 2 hours 50
Next 2 hours 30
Next 2 hours 20
Similar to the conventional activated sludge process.
Application of diffused air, at varying rate, along the length of
the tank i.e. air supply is gradually reduced along the length
of the basin.
Same volume of air is used but air is concentrated at the inlet
to cope the high demand of microorganism there.
Basic Idea - organism are unable to handle the full load if
all the sewage enters the aeration tank at a same point.
Increment of sewage are added along line of flow of the
aeration tank. Air supply is kept constant along the length of
the tank.
Air diffusers are equally spaced but incoming load (sewage)
is distributed or added in steps along the length of the basin
to have almost constant F:M ratio.
iii) Step Aeration
Step Aeration
Contents of the aeration tank (i.e. Influent sewage and
return sludge) are completely mixed.
This complete mixing is done to achieve constant F:M
ratio though out the tank.
Diffusers or surface aerators are spaced equally in the
tank to result in uniform supply of oxygen.
This process is most widely used.
iv) Completely Mixed Process
Advantages
This process can effectively handle Shock Loads ( i.e. Instant
variations in influent BOD, temperature, pH etc.)
same as complete mix, with just a longer aeration.
Advantage - long detention time in the aeration tank;
provides equalization to absorb sudden/temporary shock
loads. The purpose - to oxidize the sludge to decrease its
volume &
Consequently reduce the capacity (size) of the sludge digester.
Less sludge production because some of the bacteria are
digested in the aeration tank.
v) Extended Aeration
i. Suitable for small sewage flows ( <1MGD or 0.04 m3/s)
ii. F:M ratio is ~ 0.05 to 0.1 per day
iii. Also used to treat industrial wastewater containing soluble
organics that need longer detention times.
iv. Aeration time varies from 24 hours to 36 hours
v. Primary treatment (sedimentation tank) is not provided.
Extended Aeration time
vi) High Rate Activated Sludge Process
Usually employed where high BOD and SS removals are
not required. Size of tank is reduced which increases F:M
ratio i.e.
F:M = BOD/(MLSS x t)
Due to high F:M ratio, the biomass mostly remains in
suspension in the effluent, hence
Effluent quality is poor.
Operational Control of Activated Sludge Process
Basic parameters used are given as below:
Dissolved Oxygen (DO)
Maintain 1-3 mg/l DO in aeration basin
Mixed Liquor Suspended Solids (MLSS)
70% of Mixed Liquor Suspended Solids
Sludge Volume Index (SVI)
Microscopic examination of sludge to
check filamentous growth
Food : Microorganism (F:M)
Maintain F:M Ratio 0.5 – 1.0 however
preferably 0.25 to 0.5 day-1
Return Sludge
pH
range 6.0 to 9.0 Standard units
Ideal 6.8 to 7.4 Standard Unit
Temperature
Colder water - longer treatment
times
Industrial discharges may increase
the temperature
Most MOs do best under
moderate temperatures (10-25ºC).
Aeration basin temperatures
should be routinely measured and
recorded.
Nitrogen Content
Phosphorus Content
Process and operation are extremely sensitive
/sophisticated
Successful operation requires skilled manpower or
workforce.
Sludge bulking problem is very common
High operating cost
Electric power is required for operations
Not suitable for developing countries facing shortage of
power, revenue and skilled manpower
Disadvantages of ASP
Problem: The Mix Liquor Suspended Solids Concentration in
an aeration tank is 2400 mg/l. The volume of the sludge after
30 minutes settling in one liter cylinder is 220 ml. Calculate
Sludge Volume Index.
Sol.
MLSS
=
2400 mg/l
Vs = 220 ml/l
SVI = ?
SVI = (Vs x 1000)
MLSS
= (Vs x 1000) /2400
=(220 x 1000) / 2400 = 91.67
Prob: Design an Activated sludge process system for wastewater flow of 20,000
m3/day having BOD 200 mg/l and F/m ration as 0.4 and MLSS 2000 mg/l.
Sol
F/m = 0.4 day-1 Q = 20000 m3/day BOD= 200 mg/l MLSS=2000 mg/l
T = ? V = ? Air Supply ?
Deten. Time” t”
F:M = = ( BOD )
( t x MLSS)
= t = BOD/(MLSSxF/m)
t = 200 / (0.4 x 2000) = 0.25 days
Volume
V = t x Q = 5000 m3
Let No. units= 5
Vol.of each unit = 5000/5 = 1000 m3
Dimension
Depth of AT = 4 - 6 m = 5 m
Area = 1000/5 =200 m2
L :W = 1:5 L = 5W
= 5WxW = 200
L = 32 m W = 6.5 m
Air Supply =
= 10m3 of air / m3 of sewage
Q = 20000 m3/day
Air Supply = 20,000 x 10 = 20,0000 m3/day
Air Supply per
tank
= 200000/5 = 40000 m3/day
Return Sludge
Qr/Q = Vs / (1000-Vs)
Sludge volume
Index (SVI)= (Vs x1000) / MLSS
MLSS = 2000 mg/l & Assume SVI = 100
Vs = 200 m3
Qr = (Vs x Q) /(1000 – Vs) (200 x 4000) /(1000-200) =1000 m3/d
Problem: Calculate MLSS concentration required to result in an
operating F/m ratio of 0.4 day-1 for the treatment of 15140 m3/day of
sewage with a BOD of 200 mg/l. Assume detention time in the aeration
tank as 6 hours. Also calculate the volume of the aeration tank.
F/m= 0.4 day-1 Q = 15140 m3/day BO = 200 mg/l DT=6 hrs
MLSS = ? V = ?
F:M= ( Q x BOD )
( V x MLSS)
DT = V/Q
V = DT x Q = 6 x15140
24
= 3785 m3
F:M= ( Q x BOD )
( V x MLSS)
0.4 = (15140 x 200)
3785 x MLSS
MLSS = 2000 mg/l
Sol:
Problem: An Activated Sludge Process system aeration tank has
volume of 900 m3 treating a sewage flow of 4000 m3/d with BOD of 250
mg/l. It is desired to achieve a SVI of 80 mg/l by adopting a re-circulation
ratio (Qr /Q) of 0.25. Calculate F/M ratio at which it should be operated.
F/m = ? Q = 4000 m3/day BOD= 250 mg/l DT = ?
MLSS = ? V = ?
DT =V/Q = 900/4000 = 0.225 days
Qr/Q = Vs /(100-Vs)
Vs = 0.25(1000-Vs) = 250- 0.25 Vs
= Vs
1.25Vs = 250
Vs = 200 m3
SVI = (Vs x 1000) / MLSS
MLSS = = (200x1000)/80 = 2500 mg/l
= 3785 m3
F:M = ( BOD )
( DT x MLSS)
= (250)/(0.225 x 2500)
= 0.44 / day
Sol:
Problem: An ASP is to be designed to treat a sewage of 6
m3 / min with BOD of 200 mg/l using F/M ration 0.4 per day and
MLSS of 3000 mg/l. Calculate volume of aeration tank if SVI of
100 ml /l is maintained. How much sludge should be re-
circulated?
F/m = 0.4 Q = 6 m3/min = _ m3/day BOD = 200 mg/l
MLSS = 3000 mg/l SVI = 100 ml/l
Qr/Q = ? DT = ?
F:M= ( BOD )
( DT x MLSS)
DT =(200) / (DT x 3000) = 0.16 = 230.4 min
V = Q x DT = 6 x 230.4 = 1382.4 m3
SVI = (Vs x 1000) / MLSS
Vs = 300 m3
Qr/Q = Vs /(1000-Vs)
Qr = (Vs x Q) / ((1000-300) = 2.5 m3 /min
Sol:
Considered to be completely mixed biological reactors without
solids return.
Mixing by natural processes (wind, heat, fermentation).
Augmentation of mixing by mechanical or diffused aeration.
Also known as stabilization ponds or wastewater lagoons.
Controlled shape impart treatment by biological process.
Oxidation Ponds
Oxidation Pond is an open, flow through earthen basin
of controlled shape specially design and constructed to treat
sewage and bio-degradable industrial waste by natural
processes involving bacteria and in many instances algae.
Oxidation Ponds
Suitable - Small communities (Cheep land)
Capital cost is low
Min. mechanical equipment are required.
No imported parts / machinery is required
Easy to operate and maintain
Less O & M cost
No skilled manpower is required for O&M
Negligible odour problems
Advantages
Poor Fecal coliform removal
Needs final clarifiers
Sludge handling problems
Not suitable for extreme climate areas (MO are sensitive
towards extreme Temp. conditions i.e. cold & hot)
Disadvantages
Maintenance of Ponds
General repair and maintenance
Increases the life of structural components
Side dressing
Maintains shape of ponds and side slopes etc
Periodic removal of vegetative growth
Which may otherwise reduce capacity of ponds and
treatment efficiency
Under severe conditions, may increase BOD of the effluent
Periodic de-sludging of ponds
Which may otherwise reduce capacity of ponds and treatment
efficiency
Mechanical Aeration (Surface)
Aeration through Diffusers
Bottom view
Surface view
Screened Material
Collection Well
WW Dist. ChannelAccess Road
Views of Components of WWTP FaisalabadViews of Components of WWTP Faisalabad
WW Dist.chamber
Oxidation Ponds
Chamber to
distribute treated
WW among Ponds
Maintenance of WWTP, Faisalabad
Non removal of Vegetative Growth
Non removal of produced sludge
Non removal of dead vegetation & other
floating matter
Disposal Channel of Treated WW
Final receiving water body
Pharang Drain, Faisalabad
Collapsed Deteriorated Appurtenances
Poor Maintenance of Structures
Collapsed & filled manholes
Production of Algal Blooms in Oxidation Ponds
Most probable cause?????
Detergents rich effluent
from Textile Sector in FSD
Classification of Oxidation Ponds
The oxidation ponds are classified on the basis of biological actions
that occur in these ponds.
Anaerobic Ponds
Aerobic Ponds
Anaerobic Ponds
Devoid of O2 throughout their depth except top thin layer
Depth ranges from 2-6 meters.
Useful to treat strong organic wastes or wastes having high conc. of SS.
Remove high vols. of Org. Matter in relatively short detention time.
(1-5 days).
Most of the SS along with worms, eggs and pathogenic bacteria settle to
Gases dispersed into atmosphere.
Floating material forms scum to which also absorbs the escaping gases.
A liquid DT 1-5 days is generally employed.
Effluent of anaerobic ponds require treatment in facultative ponds
before disposal
The sludge accumulation rate is very slow and as such they require
de-sludging every three to five years.
BOD removal efficiency is better at higher temperature.
BOD removal efficiencies at 20 0C are :
Detention Time (days) 01 2.5 5
Removal Efficiency (%) 50 60 70
Parameter Value
Detention Time 1-5 days
Volumetric loading 100-300 gm / m3-day
These ponds are designed on the basis of detention of time and
volumetric loading.
For the design of ponds select a detention time and check for loading.
Then adopting suitable depth, calculate the a
Design Criteria
Cause Odour Problems
Formation of H2S gas
H2S gas, in the presence of sulphate ions (SO4-2) in
concentration more than 100 mg/l causes odour.
Require Suitable Location
Ponds should be located at least 1 Km away from
residential areas to avoid nuisance.
Effluent needs further treatment
Effluent of these ponds needs further treatment in
other types of ponds before final disposal.
Disadvantages of Anaerobic Ponds
Problem: Design an anaerobic pond treat a flow of
2000 m3/day of sewage with BOD of 600 mg/l.
Design of ponds - select a DT & check for loading.
Select suitable depth & calculate the area.
Design of ponds
Facultative Ponds
O2 persists throughout the liquid depth but absent near the
bottom of these ponds.
Deposition of Settle-able particles at bottom
OM in sludge layer undergoes anaerobic decomposition
Soluble and colloidal organics in the incoming wastes
decompose aerobically
Prolific (excessive/ plentiful) growth of algae (plants)
Production of O2 by algae through photosynthesis to
bacteria.
Bacteria breakdown of organic matter and release CO2
Algae utilize CO2 released by bacteria in breaking down
the organic components of the wastewater
Algae-Bacterial Symbiosis
Design Criteria of facultative Ponds
Parameter Value
Detention Time 20-40 days
Depth of water 1.5 – 2 meter
Treated sewage from facultative pond is expected to have a
BOD between 75 – 85 mg/l
These ponds require desludging every 10-15 years , thus
Pose minimum health risk to handler.
Maturation Ponds
Fully aerobic ponds (usually 1.5m deep)
Used as polishing stage after facultative ponds
Main /principle function - destruction of pathogens.
Further reduction in BOD is also achieved
Designed for removal of pathogens (Indicator -faecal coliforms)
Detention time of 7-14 days (depends upon nature of reuse of
treated effluent form system.
Treated sewage can possibly have an effluent with
BOD around 30 mg/l,
faecal coliform content > 1000 per 100 ml and
helminth count of less than 1 nematode egg per liter.
Aerobic Pond
O2 is present throughout the pond depth.
Aerobic decomposition of Organic matter.
Depth - maximum of two feet deep to permit sunlight
throughout the entire depth of the pond.
Sunlight supports growth of algae grow throughout depth.
O2 is given off by algae for living of aerobic microorganisms.
Aerobic ponds are not used in colder climates because they
will completely freeze in the winter
Dissolved Oxygen level
(wind current, vertical mixing of O2 etc.)
pH value
Diurnal Variations in Aerobic Ponds
Parameters Values
Depth (m) 0.15- 0.5
Retention time (day) 2 – 6
BOD5 loading( lb/acre day) 100-200
BOD5 removal(%) 80-90
Algae concentration(mg/l) 100-200
Effluent SS Conc.(mg/l) 150-350
Design Criteria of Aerobic Ponds
Q-1 A rectangular sedimentation basin 24 feet long, 6 feet wide and 10
feet deep. The flow into the basin is 0.5 MGD. Is the SOR is within
the recommended limits.
Q-2 A sedimentation basin has a recommended detention time of flow
is 0.7 MGD. What should be the volume of the tank in ft3.
Q-3 A rectangular sedimentation basin has a volume of 19600 ft3.
What is depth of basin.
Q-4 The volume of rectangular tank is 19,600 ft3. the depth is 12 feet.
The length of the tank is four times the width of tank. What is the
tank’s width.
Q-5 Design a circular settling tank unit for the primary treatment of
sewage at 13.5 million liters per day. The detention time is be
about 2 hours, and the surface loading rate is 45,000 liters per
square meter.
Q- A circular sewer is to carry 2.5 m3/min of sanitary sewage when
flowing full. Conditions are such that minimum allowable grade must be
adopted. Taking n=0.015 determine the commercial pipe size
Q- A 300 mm sewer is laid at minimum slope. What is maximum
population that can be served by the sewer if the average water
consumption is 300 lpcd.
Q- BOD remaining in a sewage sample after 3 days and 10 days is
150 and 30 mg/l respectively at 20 oC . Calculate 5 days BOD of the
sample at temperature of 24 oC
Q- BOD remaining in a sewage sample after 5 days and 10 days at
20 oC is 100 and 70 mg/l respectively. Calculate 7 days BOD of the sample
at temperature of 30 oC
Q- Find the diameter required for a single stage trickling filter to
yield an effluent BOD5 30 mg/l when treating settled sewage with BOD5 of
180 mg/l. The wastewater flow is 2500 m3/day and re-circulation ratio
rate is 7500 m3/day. Assume filter depth of 1.5 meter. What are the
hydraulic and organic loading rate for filter.
Q An activated sludge process is to treat a domestic sewage flow of
6000 m3/day with BOD of 240 mg/l. The F.M ratio is to be maintained at
0.4 Kg BOD/Kg MLSS day. The sludge re-circulation ratio is 0.25 and it is
desired to achieve and SVI of 100. calculate the mixed liquir suspended
solids in aeration and the size of the aeration tank.
Q- Design a system of anaerobic, facultative and maturation
ponds to be operated in series for the treatment of 1500 m3/day of
sewage slow with BOD 400 mg/l. Use volumetric loading of 300 g/m3-
day for anaerobic pond, organic loading of 200 Kg/hac-day for
facultative pond and detention time of 10 days for maturation pond.
What will be the percentage coliform removal if coliform decay rate is 2.6
per day.
Q- Calculate the volume of an aeration for the deign of an activated
sludge plant treating a sewage flow 3x104 m3/day. The BOD of raw
sewage is 240 mg/l.
It is desired to maintain and MLSS concentration of 2000 mg/l and F/M
ratio of 0.4 Kg BOD/Kg MLSS. day. How much sludge should be
returned to the process if the sludge Volume Index od 120 is desired.