Learning objectives

Understand the differences between aerobic

and anaerobic processes

Be able to describe the microbial ecology
and the growth kinetics.

Prepared by Ms Aida Isma

Have broad understanding of the activated
Biological unit process 1 sludge process

Introduction Types of biological treatment

The aim of wastewater treatment is to • Requires oxygen (energy intensive)
• Aerobic heterotrophs
enable wastewater to be disposed safely, Aerobic • Primary significance in biological
without being a danger to public health and treatment of municipal wastewaters)
without polluting watercourses or causing
other nuisance. • Absence of oxygen
• Conversion of pollutant to mainly

Another important aim of wastewater carbon dioxide and methane
treatment is to recover energy, nutrients, • Treatment of higher organic loading
water and other valuable resources from Anaerobic • Low sludge production
wastewater. • Removal of pathogen
• Biogas production
• Low energy consumption

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 Aerobic and Anaerobic Aerobic treatment
Fixed Growth Suspended Growth
• Microbes hang on onto
• Microbes in suspension
media
• Can be grouped into 3 • Types:
general classes: 1. By aerating the WW
1. Non-submerged by pumping in bubbles
attached growth of air or stirring
processes (trickling vigorously (activated
filter). sludge).
2. Suspended growth
2. By relying on algae
processes with fixed-
film packing (RBC). present to produce
3. Submerged attached
oxygen by
growth aerobic photosynthesis
process. (stabilization pond).

Fixed Growth Mechanism -

Fixed Growth Mechanism
Bulk liquid
flow
Liquid
• Diffusion process Sloughing
Wastewater
flow
layer • Factors:
Slime Slime
• Concentration S
thickness: thickness:
• Temperature >0.2 mm ~ 10 mm
• pH
O2
Organics • Bulk liquid flowrate
Filter
packing End
products Sinside ~ 0
Aerobic
Se biodegradation Bacteria enter Bacteria loss the
endogenous ability to cling to
Anaerobic the packing
respiration
biodegradation surface
state.
Biomass layer/ Stagnant liquid Time
biofilm film

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 How does biological treatment
Microbes appetite and needs
work?

With the help of microbes, contaminants in

Different species of microbes
wastewater can be removed by: have different appetite. JUST LIKE HUMAN!

Coagulate and remove nonsettleable
Different needs for survival.
colloidal solids

Removal of carbonaceous BOD
Bacteria

Stabilization of biodegradable material
(also removal of nutrients such as N, P)
Heterotrophs Autotrophs

Remove specific trace organic constituents
and compounds.
Aerobic Anaerobic

Heterotrophs Autotrophs
Nutrients Nutrients Nutrients

Ingestion New cells Ingestion New cells

NH4+ Synthesis New cells
Organic Organic
compounds Bacteria Bacteria
compounds Bacteria
CO2
Energy CO2 + H2O Energy CH4 + CO2 Energy NO2- or NO3-

O2
O2

Aerobic Anaerobic Aerobic

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 Microbes growth process
Microbes Bacteria
Growth rate
Split into 2
ingest food grows bigger

Growth rate of microorganisms is affected by

Substrate concentration
Exponential Stationary
growth phase phase
Availability of other nutrients

Temperature
Concentration

Death
phase
Toxicity

Oxygen: Respiration
Microbes
Substrate growth
Biomass pattern in
Lag batch reactor
phase

Time

Substrate concentration Availability of other nutrients

Besides carbon, growth rate is also
dependent on the availability of nitrogen and
phosphorus.

Optimum ratio of C:N:P in wastewater is
100:5:1.

Other trace components are S, Na, Ca, Mg,
K and Fe.
• Substrate is being
Lack of critical nutrient will cause inhibition of
used for growth.
respiration.
• No growth.

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Temperature Toxicity

Most carbonaceous bacteria optimum

temperature range is between 0 to 30°C.

Growth rate follows the Arhennius rule.

k1T = k120θ T −20

The chemical reactions double in rate for a
10°C in temperature.

Increase in temperature ,increase the growth
rate, increase in O2 requirement of respiration.

FIVE Principles of Activated

Respiration
Sludge Process1

The end products from respiration are
energy, CO2 and H2O.

Energy is required for:

Biosynthesis for growth

Pumping ions out that diffuse through the
cell wall 3
2

Self repair

Rate of respiration is measured from O2
uptake rate using a respirometer. 5
4

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Principles of Activated Sludge Process 1. The workers in mixed liquor
Principle Justification

1 To provide an aerobic condition (aeration reactor) for microbes to

degrade pollutants and grow. The mixture of wastewater and
microbes are called mixed liquor.
2 To provide oxygen (aeration) for microbes and sufficient mixing to
keep the microbes in suspension.

3 To separate microbes from treated water. Microbes settle at the

bottom of clarifier. The watery mixture of microbes is called
activated sludge.
4 To return part of activated sludge into aeration reactor in order to
replenish and maintain the concentration of microbes. This is called
return activated sludge or RAS.
5 To remove part of activates sludge as not to overload (to maintain
F/M ratio) the aeration reactor. The sludge is called waste activated
sludge or WAS.

F/M = Food (Soluble BOD or COD)/Microorganism

Different workers has different • Big workers – eat other little bugs
• Amoeba
job scope… • Flagella
• Small workers – eat dissolved food • Free swimming ciliates
• Carbonaceous bacteria • Crawling ciliates
Crawling ciliates
• Nitrogenous bacteria • Stalked ciliates
• Rotifers Free swimming ciliates

• Nematode
Pseudomonas sp.
• Suctoria
Stalked ciliates

Amoeba

Nitrobacter winogradskyi Nb-

255
Bacteria
Nematode Rotifers
Pseudomonas aeruoginosa
Suctoria

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2. Workers need oxygen! Air diffusers
Fine bubbles

Oxygen (Air)

Required for synthesis and respiration, also for mixing

Target for about 1-2 mg/L dissolved oxygen (DO) in Submersible diffusers

aeration tank
Jet aerators

Target for 0.5 mg/l in clarifier

Can be controlled by Operator
Surface aerators

Workers other requirements Foam in aeration reactor

• Loading – prevent “shock” loading White, stiff, billowing or
Foaming in aeration sudsy foam from very young
• Equalization tank important. reactor is an indication of
• Shock loading will cause disturbance in workers sludge during start up of a
its condition. new plant.
population.
• Need acclimatization period for the population to give the
required efficiency.
• pH
• Optimum range 6.8 – 7.4.
• Toxicity
• Sudden drop in nematode population is often the first
indication of a toxic condition.
• Temperature
• Optimum range 20 - 40°C.
• Nutrients
• Require macronutrient such nitrogen and phosphorus. White foam during start up
• Approximate ratio: C:N:P = 100:5:1. of wastewater treatment
• Require other micronutrients in small amount such as Zn, plant
Mn etc.

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Foam in aeration reactor Foam in aeration reactor Thick, greasy dark-tan to black foam.
Tan foam on healthy biomass
PROBLEM!!!

Modest amount of fresh tan Can be due to many reasons.

foam. A sign of a well • Nocardia species is dominant
operated process. • Low dissolved oxygen
• Low F/M
• Inappropriate sludge age
• Oil and grease carryover

3. Separation of mix liquor Good Biofloc

• To get good separation
• Many good workers are required
• Bad workers are also required but in a small amount

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Poor biofloc Sludge Volume Index Bioflocs are removed using secondary
clarifier.

Settling
problems

Sludge bulking

(settled volume of sludge, mL/L )103 mg/g 

SVI =   = mL
SVI (suspended solids, mg/L ) g
Result indication:
50 watch out for
100 GOOD
F/M 500 BAD
900 VERY BAD
1000 NOT a Clarifier

4. RAS and 5. WAS

• Reason to have both is to maintain optimum condition of the activated sludge
Activated sludge system
Aeration tank
process.
2° Clarifier
• Can control
• F/M ratio
• Sludge age Aeration tank
Influent Effluent
• If both parameters falls out of the optimum range, problems will surface
Q, S0, X0 (Q-Qw),
• Settling problem
• Foaming
Xe, Se
• Low treatment efficiency S, X, V

Return activated sludge (RAS)

Waste Activated
Qr, XR, S
Sludge (WAS)
Qw, XR, S

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F/M and Sludge Age Example: F/M
• The F/M ratio is the measure of food available for biomass.
0 An aeration tank with a volume of 1590 m3 receives a
F Q (S 0 − S e ) (S 0 − S e )
• Degree of starvation.
= =
M VX X (HRT ) primary effluent flowrate of 8900 m3/d. The mixed liquor
S0
= suspended solids concentration is 2480 mg/L. If the primary
X (HRT )
effluent BOD concentration is 184 mg/L, what is the current

• Sludge age is the average amount of time the sludge spends in F/M ratio?
the aeration basin.
• State of starvation. mass of biomass in aeration basin
Sludge age =
sludge removal rate from the system
VX
=
(Q − Qw )X e + Qw X R

Example: Sludge Age RAS

• Return activated sludge (RAS)
is to maintain sufficient
An aeration tank has a MLSS concentration of 2100 mg/L. concentration of activated
sludge in aeration basin.
The volume of the tank is 1850 m3. The plant flowrate is
• The common techniques to
11,200 m3/d. The waste sludge suspended solids is 4900 determine the RAS are
• Settleability
mg/L. The sludge is being removed at a rate of 230 m3/d. • Sludge blanket level control
The secondary effluent suspended solids concentration is • Secondary clarifier mass
balance
12 mg/L. What is the SRT? • Aeration tank mass balance

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 Secondary clarifier mass
Example: RAS (SVI)
balance
Q+Q , X Q ,X e e
R
A mixed liquor is poured into a 1000 ml settleometer. After
30 min., there are 300 ml of sludge on the bottom. If the
plant flowrate is 21.5 ML/d, what should be the return
sludge pumping rate be in L/s?
QR, XR
Qw, XR
0

0 = X (Q + QR ) − QR X R − Qw X R − Qe X e

Aeration tank mass balance WAS

Aeration tank
• Waste activated sludge is to maintain the sludge age.
VX
sludge age = 0
Q Q+QR, X (Q − Qw )X e + Qw X R
VX
=
QR, XR Qw X R
S, X, V
VX
Qw =
X R (sludge age )

0 = X R QR − X (Q + QR )

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Example: WAS
The mixed liquor in a 1900 m3 aeration tank has a
suspended solids concentration (MLSS) of 1800 mg/L. The
waste sludge is being removed at a rate of 470 m3/d and
has a concentration of 3000 mg/L. If the target MLSS is 1750
mg/L, what should be the new waste sludge pumping rate End of chapter
be?

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