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.