Wastewater Treatment

By
Sumarno


Introduction

Biological Wastewater Treatment
Removal the organic contaminats in
domestic and or industrial wastewater by
appropriate treatment process to render the
water suitable for discharge to surface water
Anaerobic, Anoxic and Aerobic Processes

In biological treatment the organic contaminants
serve as the energy source (electron donor) in the
reaction and oxygen, nitrite/nitrate, sulfate or
carbon dioxide serve as the electron acceptor.
Anaerobic processes use sulfate or carbon dioxide
as electron acceptor.
Anoxic processes use nitrite/nitrate as their
electron acceptor.
Aerobic processes use oxygen as electron acceptor


Anaerobic Anoxic Aerobic







- 200 mV 0 mV + 200 mV
Fig 1. ORP Range (mV) for Wastewater Treatment






Methanogenesis
Denitrification
Nitrification
Activated
Sludge
All biological treatment processes are conversion
processes in that they convert the readily bio-
degradable organic contaminants (soluble/colloidal)
into two fraction: (1) a gas which escape from liquid,
(2) the excess biomass.

The Dillema of Biomass Disposal
The cell wall of biomass are very complex and
therefore quite refractory to further biotransformation.
Under aerobic and anaerobic conditions, only 25 – 40%
of resulting biomass which is synthesized may be
further biodegraded, the remaining is so refractory that
it cannot practically be destroyed (Gosset and Belser
1982)
The Excess Biomass
Aerobic treatment normally produced about 10 times
more refractory biomass than anaerobic.
The excess biomass presents an greater dillema due to
(1) need sludge processing area which is may
increasingly be restricted in the future;
(2) its possible listing as hazardous waste classification
which is necessitates special processing.
Treatment Process Selection
Anaerobic and Aerobic Processes Selection
 Awareness of Aerobic Process Disadvantages
Biomass disposal problem caused by large volume of refractory biomass
and possibility listing as hazardous waste
 Alternative Technology Advantages
Capable to biotransformed refractory and or toxic substances
anaerobically
Nutrient requirement of anaerobic processes only 5 – 20% of those for
aerobic processes
 Minimal Waste Biomass
Anaerobic treatment processes give only 5 – 20% as much waste
biomass compared to aerobic treatment processes
 Retention Time Requirement
Volumetric loading rate anaerobic treatment are often > 10 times than
aerobic processes (5 -10 kg/m
3
d for anaerobic and 0.5 – 1.0 kg/m
3
d
for aerobic)
Smaller Reactor Volume
 Energy Requirement and Production
No aeration energy requirement for anaerobic processes
Energy methane production 12,000 BTU/kg COD biodegraded
 Construction
Anaerobic processes construction more simple than aerobic processes
 Achievement of Secondary Effluent Standard
Anaerobic processes incapable of producing an effluent which meets
standard



Anaerobic Wastewater Treatment
Principles of Anaerobic Treatment

Complex Organic Compounds
75% Hydrolysis
Simple Organic Compounds
5% 10% 30% Acidogenesis 35% 20%
Long Chain Fatty Acids
13% 17%

H2 and CO2 Acetate
28% Methanogenesis 72%
CH4 and CO2
Fig 2. Series Metabolisme of Anaerobic Processes


Optimal Anaerobic Treatment Processes
 Proper pH
The pH reactor must range between 6.5 – 8.2
 Adequate macronutrients
Nitrogen concentration in the reactor must ranging 40 – 70 mg/L
Sulfide precursor should be added, commonly in the sulfate form
 Trace metal bioavailability
 Obligate micronutrients requiremen
The obligate requirement for sulfide and trace metal by methanogens
Iron to be required in the highest concentration
 Temperature
Commonly anaerobic reactor are operated at mesophilic temperature
of 30 – 37
0
C

 Toxicity acomodation
The anaerobic process can accomodate toxicity of various forms in
industrial wastewater and even biodegrade certain toxicant e.g. CCL4,
tetrachloroethylene, formaldehyde , acrylate, trichloroethylene,
chloroform , cyanide
 Adequate metabolisme time
Two measured of time are involved Hydraulid Retention Time
(HRT) and Solid Retention Time (SRT)
 Carbon source for synthesis
 Electron donor
Biodegradable COD
 Electron acceptor
CO2 and/or sulfate
Operational Consideration

 BOD/COD assay
BOD assay has value when applied to anaerobically treated effluent
in that it indicate pollutant load for a subsequent treatment unit.
A plot log COD vs 1/HRT for treatment processes indicate the non-
degradable fraction at the intercept
 Biochemical Methane Potential Assay
Assaying the concentration of organic pollutant in a wastewater
which can be aerobically converted to CH4.
Evaluating potential anaerobic process efficiency
Testing for non-biodegradables remaining after treatment
 Anaerobic Toxicity Assay
A simple assay procedure to evaluate the potential toxicity of a
wastewater sample to the anaerobic biomass
Aerobic Wastewater Treatment
Principles of Aerobic Treatment
Fermentation
Aerobic
Organic C + O2 Energy + CO2 + H2O + Residue
microbes

New Aerobic
+ O2 Energy + CO2 + H2O + Residue
microbes

New Aerobic
+ O2 Energy + CO2 + H2O +
microbes Residue

New Cell Biosynthesis
Two kinds of ingredients are required for the
biosynthesis of cell components: (1) precursors that
provide the carbon, hydrogen, nitrogen, and other
elements found in cellular structures, and (2)
adenosine triphosphate (ATP) and other forms of
chemical energy needed to assemble the precursors
into covalently-bonded cellular structure.

microbes
Simple precursors C
60
H
87
N
12
O
23
P
energy new cell

Endogenous Respiration
Under substrate-limited conditions, microbes will feed
on each other at a higher rate than new cells can be
produced. The aerobic degradation of cellular material
is endogenous respiration

aerobic
C
60
H
87
N
12
O
23
P CO2 + H2O + NH4
+
+ Residue
microbes
Optimal Aerobic Treatment Processes

 Temperature
The rate of bio-oxidation is a function of temperature.
Various microbial species have optimal temperatures for
survival and cell synthesis.
• Psychrophilic microorganisms thrive in a temperature range of
-2° to 30°C . Optimum temperature is 12° to 18°C
• Mesophilic microorganisms thrive in a temperature range of
20° to 45°C . Optimum temperature is 25° to 40°C
• Thermophilic microorganisms thrive in a temperature range of
45° to 75°C. Optimum temperature is 55° to 65°C.

 Food to Microorganisms Ratio (F/M)


 Acid Concentration
The influent pH has significant impact on wastewater
treatment. Benefield and Randall (1985) report that it
is possible to treat organic wastewaters over a wide pH
range, however the optimum pH for microbial growth
is between 6.5 and 7.5.
Anoxic Treatment Processes
Principle of Anoxic Processes
In the wastewater stabilization pond revealed that
the wastewater treatment mechanisms are mainly
based on biochemical reaction in the anoxic system

The algal activities in the presence of light proved to be
important in the surface layer of the pond and
facultative bacteria were working mutually with the
algae. Motile flagellate algae (Euglena and
Chlamydomonas) were the only species found to exist
under anoxic conditions.
It was also determined that sulphate-reducing bacteria
predominated in the lower volume of the anoxic ponds,
rather than acidogenic bacteria, and this caused
sulphide and hydrogen sulphide build-up in the pond’s
contents. The bottom volume of the ponds and the
benthic sludge in anoxic ponds contained acid
producers and methanogenic bacteria causing the
release of biogas.

Which one or combined biological treatment processes
do you choose is depend on strength of organic
pollutants of wastewater, effluent standard and added
value from treatment processes.
DESIGN OF ACTIVATED SLUDGE PROCESS
Process Description
The activated sludge process is an aerobic growth
biological treatment method in which oxygen required
substances are removes by biochemical reactions using
microorganisms.
Bacteria
COHNS + O
2
CO
2
+ NH
3
+ C
5
H
7
NO
2
+ Other
end products

Bacteria

C
5
H
7
NO
2
+ 5 O
2
5 CO
2
+ 7 H
2
O + NH
3
+ energy


V, S, X
Aeration Tank
Sedimentation Tank
Recycle
Q
R
, S
2
, X
R

Q
0
, S
0
Q
0
- Q
W

Fig 3. Flow Scheme of Conventional Activated Sludge
Q
W
, X
W
= X
R

Wastage
Terminology

 ACTIVATED SLUDGE
 floc of microorganisms that form when wastewater is aerated
 MIXED LIQUOR
 mixture of activated sludge and wastewater in the aeration tank
 MIXED LIQUOR SUSPENDED MATTER (MLSS)
 measure of the amount of suspended solids in the mixed liquor
expressed in mg/l
 MIXED LIQUOR VOLATILE SUSPENDED MATTER
(MLVSS)
 proportional to the microorganisms concentration in the aeration
tank
 MEAN CELL RESIDENCE TIME (MCRT)
 the average time a microorganism spends in the treatment process
 FOOD TO MICROORGANISM RATIO (F/M)
 ratio of the amount of food expressed as pounds of COD (or BOD)
applied per day, to the amount of microorganisms, expressed as the
solids inventory in pounds of volatile suspended matter.
 RETURN ACTIVATED SLUDGE (RAS)
 settled mixed liquor collected in the clarifier underflow and
returned to the aeration basin
 WASTE ACTIVATED SLUDGE (WAS)
 excess growth of microorganisms which must be removed to keep
the biological system in balance. Various control techniques
have been developed to estimate the amount of WAS that must
be removed from the process
 COMPLETE MIX ACTIVATED SLUDGE
 an ideal mixing situation where the contents of the aeration tank
are at a uniform concentration
 PLUG FLOW ACTIVATED SLUDGE
 an ideal situation where the contents of the aeration tank flows
along the length of the tank
 WASTE ACTIVATED SLUDGE (WAS)
 excess growth of microorganisms which must be removed to
keep the biological system in balance. Various control
techniques have been developed to estimate the amount of
WAS that must be removed from the process
 COMPLETE MIX ACTIVATED SLUDGE
 an ideal mixing situation where the contents of the aeration
tank are at a uniform concentration
 PLUG FLOW ACTIVATED SLUDGE
 an ideal situation where the contents of the aeration tank
flows along the length of the tank
 BACK MIXING
 mixing the contents of a tank in the longitudinal or flow oriented
direction
 TRANSVERSE MIXING (or CROSS ROLL)
 mixing in a direction across the direction of flow
 SLUDGE REAERATION
 practice of aerating the RAS before it is added to the mixed liquor
 PROCESS LOADING
 organic loading range as measured by the F/M
 CONVENTIONAL LOADING
 process loading of 0.2 to 0.5 lbs BOD applied/lb MLVSS/day
 HIGH RATE LOADING
 process loading of two to three times the conventional loading rate
 EXTENDED AERATION LOADING
 low rate loading that is one half to one tenth of the
conventional loading rate
 SETTLEABILITY
 measure of the volume occupied by the mixed liquor after
settling in a graduated cylinder for 30 minutes
 generally expressed as a percentage based on the ratio of the
sludge volume to the supernatant volume
 SOLIDS INVENTORY (VOLATILE SOLIDS)
• amount of volatile suspended solids in the treatment system
 SOLUBILITY INDEX ( SI )
• soluble BOD per total BOD

Aeration System
 Aeration is provided by either diffused or mechanical
aeration systems.
 Diffused air systems consist of a blower and a pipe
distribution system that is used to bubble air into the mixed
liquor.
 Mechanical aeration systems consist of pumps or mixers
Diffused Air System
 Diffused air systems are the most common types of aeration
systems used in activated sludge plants.
 Produce fine or coarse bubbles.
Fine Bubble Diffusers
 Fine bubble diffusers are easily clogged by biological growth
and by dirty air, resulting in high maintenance costs.
 The air supply for all fine bubble diffusers should be filtered.


Fig 4. Fine Bubble Diffuser
Surface Aerator
 Mechanical Aerator
• Floating or fixed
• Brush
• Injector
• Others

Fig 5. Mechanical
Aerator
Brush aerator
Injector
aerator
Floating
aerator
DESIGN OF ACTIVATED SLUDGE PROCESS
In the past, designs of biological wastewater
treatment processes were based on the empirical
parameters developed by experience, which
included hydraulic loading, organic loading and
retention time.

Nowadays, the design utilizes empirical as well as
rational parameters based on biological kinetic
equations. Theses equations describe growth of
biological solids, substrate utilization rates, food-
to-microorganisms ratio, and the mean cell
residence time.
DESIGN CRITERIA FOR ACTIVATED SLUDGE PROCESS
 Reactor type selection
 Loading Criteria
 Sludge Production
 Oxygen Demand and Transfer
 Nutrient Requirement
 Control of Filamentous Organism
 Effluents Characteristics
Selection of Reactor Type
Selection of the reactor type requires several
considerations as follows:
 Reactions kinetic
 Oxygen transfer requirements
 Nature of wastewater
 Local environments conditions
 Construction and maintenance costs



Loading Criteria
There are 2 main parameter for the design and control
of the activated sludge process:
 The Food to Microorganism Ratio (F/M ratio)
F/M = total applied substrate rate/total microbial biomass
= Q . S
0
/V . X = S
0
/ θ . X
F/M = food to microorganisms ratio, h
-1
S
0
= Influent BOD or COD concentration, mg/L
Q = Influent wastewater flowrate, m
3
/h
V = Aeration tank volume, m
3

θ = hydraulic detention time, V/Q, h
X = concentration of volatile suspended solid (MLVSS) in
aeration tank, mg/L

 The Mean Cell Residence Time (θ
c
)

θ
c
= V
r
. X /(Q
w
. X
w
+ Q
e
. X
e
)


θ
c
= the mean cell residence time based on aeration tank, d
V
r
= aeration tank volume, m
3
X = concentration MLVSS in aeration tank, mg/L
Q
w
= excess sludge flowrate, m
3
/d
X
w
= concentration of MLVSS in excess sludge, mg/L
Q
e
= effluent flowrate, m
3
/d
X
e
= concentration of MLVSS in effluent, mg/L

The organic loading parameter and others design
parameter for activated sludge processes listed in table 1.


Table1. Design Parameter for Activated Sludge Process
(Metcalf et all, 1991)
Proses θ

hrs
F/M

Kg BOD/
kg MLSS . d
Volumetric
Loading
Kg BOD/m
3
.d

MLSS

mg/L
V/Q

hrs
Q
R
/Q
Conventional

5-15 0.2-0.6 0.32 -0.64 1500-3000 4-8 0.25-0.75
Complete
Mix
5-15 0.2–0.4 0.80-1.92 2500-4000 3-5 0.25-1.0
Step Feed 5-15 0.2-0.4 0.64-0.96 2000-3500 3-5 0.25-0.75
Modified
Aeration
0.2-0.5 1-5-5.0 1.2-2.4 200-1000 1.5-3.0 0.05-0.25
Contact
Stabilization
5-15 0.2-0.6 0.96-1.20 1000-3000
4000-10000
0.5-1.0
3-6
0.5-1.5
Extended
Aeration
20-30 0.05-0.15 0.16-0.40 3000-6000 18-36 0.5-1.5
Proses θ

hrs
F/M

Kg BOD/
kg MLSS . d
Volumetric
Loading
Kg BOD/m
3.
.d
MLSS

mg/L
V/Q

hrs
Q
R
/Q
Krauss Process

5-15 0.3-0.8 0.64-1.6 2000-3000 4-8 0.5-1.5
High-rate
Aeration
5-10 0.4-1.5 1.6-16.0 4000-10000 2-4 1.0-5.0
High Purity
Oxygen
3-10 0.25-1.0 1.6-3.2 2000-5000 1-3 0.25-0.5
Oxidation
Ditch
10-30 0.05-0.3 0.08-0.48 3000-6000 8-36 0.75-1.5
Microorganism and Substrate Balance
The term V . MLSS is function of Solid Retention Time
(SRT) or θ
c
and not Hydraulic Retention Time (HRT) or
return sludge ratio, the FM ratio is also function of SRT.
Therefore, operation of an activated sludge plant at
constant SRT will result in operation in at aconstant FM
ratio.
The mass balance for microorganism in the entire
activated sludge system is expressed as the rate of
accumulation of the microorganisms in the inflow plus
net growth, minus that in outflow.
Mathematically , it is expressed as (Metcalf and Eddy
Inc, 1991)


V dX/dt = Q X
0
- V r
g
- (Q
w
X
w
+ Q
e
X
e
)
where
V volume of aeration tank, m
3

dX/dt = rate of change of microorganisms concentration
(VSS), mg /(l . m
3
. d)
Q = influent flow, m
3
/d
X0 = microorganisms concentration (VSS) in influent,
mg/L
X = microorganisms concentration in tank, mg/L
r
g
= net rate of microorganism growth (VSS), mg/(l . d)

The net rate of bacterial growth is expressed as
r
g
= Y r
su
– k
d
X
where
Y = maximum yield coefficient growth, mg/mg over finite
period of log
r
su
= substrate utilization rate, mg/(m3 d)
k
d
= endogenous decay coefficient, per day

Substituting this equation into above equation, and
assuming in the influent is zero and steady-state conditions,
this yields

Q
w
X
w
+ Q
e
X
e
- Y r
su


= - k
d

V X X


1 - Y r
su


= - k
d

θ
c
X

The term 1/θ is the net specific growth rate
The term r
su
can be computed from the following equation

r = Q/V (S
0
- S) = (S
0
- S)/θ

where S
0
- S mass concentration of substrate utilized, mg/L
S
0
substrate concentration in influent, mg/L
S substrate concentration in effluent, mg/L
θ hydraulic retention time, day
Effluent microorganism and substrate concentration
The mass concentration of microorganism X in the aeration
tank
θ Y (S
0
- S) μ
m
(S
0
- S)
X = =
θ (1 + k
d
θ
c
) k (1 + k
d
θ
c
)

Where
μ
m
= maximum specific growth rate, per day
k = maximum rate of substrate utilization per unit mass of
microorganism, per day



The substrate concentration in effluent S can be
determination from the substrate mass balance by the
following equation

K (1 + k
d
θ
c
)
S =
θ
c
(Y k - k
d
) – 1

Where
S = effluent substrate (soluble BOD5) concentration, mg/L
Ks half –velocity constant, substrate concentration at one
half of maximum growth rate, mg/l
Process design and control relationships.
In practice, the relationship between specific substrate
utilization rate , mean cell residence time (θ
c
), and the food
to microorganism ratio (F/M) is commonly used for
activated sludge process design and process control.



Complete Mix Activated Sludge Process
In this modification of activated sludge process, fresh
feed and recycled sludge are combined and then
introduced in several points in the aeration than from a
central channel.
Aerated liquor leaves the reactor from effluent channels
on both side of the aeration tank (fig. )
Oxygen supply and demand are uniform along the tank.
Sludge Recycle
Effluent
Wastage
Influent
Aeration Tank
Clarifier
Fig 6. Flow Scheme of Complete Mix Process
Contact Stabilization Process
Contact stabilization is another modification of the
activated sludge process.
A flow diagram for the system is shown in the figure 4.
Influent wastewater is mixed with stabilized sludge, and
this mixture is aerated in the initial contact tank for
which detention time only 20 -40 minute.
During initial contact an appreciable fraction of
suspended and dissolved BOD is removed by
biosorption after contact with well aerated activated
sludge.
The mixed effluent from the initial contact tank flow to
clarifier.
Clarified effluent is removed and underflow from the
clarifier is taken to stabilized tank, where it is aerated for
a period of 1.5 – 5 hrs.
During the this stabilization period, biosorbed organic
are broken down by aerobic degradation.
Stabilized sludge leaving the stabilization tank is in
starved condition and ready for adsorbed organic waste
Initial Contact Tank Clarifier
Sludge Recycle
Stabilized
sludge
Influent
Effluent
Fig 7. Flow Scheme of Contact Stabilization
Stabilization Tank
Wastage
Step Aeration Process ( Step Feed Process )
Step aeration is modification of the conventional
activated sludge is shown in figure , in which fresh feed
is introduced in several point along aeration tank.
This arrangement is provide for an equalization F/M
ratio along the tank.
The aeration tank is divided by baffles into several
channel.
Each channel constitute of one step of process and the
step are linked together in series
Aeration Tank
Sedimentation Tank
Recycle sludge
Influent
Effluent
Wastage
Fig 8. Flow Scheme of Step Aeration
Tapered Aeration
The purpose of tapered aeration is to match amount of
air supply with oxygen demand along the aeration tank.
Since at the inlet the oxygen demand is the highest,
aerators are space more closely to provide tha higher
oxygenation rate.
Spacing between aerators increase toward the outlet as
oxygen demand decrease.
Aeration Tank
Sedimentation Tank
Recycle sludge
Influent Effluent
Wastage
Fig 9. Flow Scheme of Tapered Aeration
Aeration
Oxidation Ditch
Figure show the diagram of the oxidation ditch.
An essential part of the system is an aeration ditch
provided with an aeration rotor.
This rotor has two function : (1) aeration and (2)
provision of a flow velocity to mixed liquor in the ditch.
Liquid velocity is of the order of 0.3 m/s.
The mixture of wastewater and activated sludge
repeatedly passed over the aeration rotor at short
intervals.
A typical rotor has diameter of approximately 75 cm
revolved at about 75 rpm.

Influent
Aeration rotor
Clarifier
Recycle
sludge
Wastage
Effluent
Fig 10. Flow Scheme of Oxidation ditch
Effluent
Aeration rotor
Sludge Production
Design of the sludge handling and disposal facilities depends
on sludge produced. Quantity of sludge can be estimated by
using the equation

P
x
= Y
obs
. Q . (S
0
- S )

P
x
= excess sludge (kg/d)
Y
obs
= observed yield (g/g)
S
0
= Influent BOD or COD concentration, mg/l
S = Effluent BOD or COD concentration, mg/l
Q = Influent flowrate (m
3
/d)
The observed yield value can be computed using equation
Y
obs
= Y/( 1 + k
d
. θ
c
)


Y = Cell yield coefficient (mg cell produced per mg organic
matter removed)
k
d
= endogenous decay coefficient, day
-1

θ
c
= mean cell residence time, day

Oxygen requirement and transfer
BOD and wasted sludge per day can be used for estimation of
oxygen requirement
bacteria
C
60
H
87
N
12
O
23
P CO2 + H2O + NH4
+
+ energy

cell





Carbonaceous oxygen demand can be defined as
Kg O
2
/d = (total mass of BOD
L
removed, kg/d) – (1.42 .
mass sludge wasted, kg/d)
The BOD of the cell is equal to 1.42 times excess sludge
Kg O
2
/d = (Q . (S
0
- S )/F 10
-3
mg/kg, kg/d) – (1.42 .
mass sludge wasted, kg/d)
F = conversion factor for converting BOD
5
to BOD
L

If nitrification is desired O
2
required to organic nitrogen
oxidation should be added to carbonaceous oxygen
demand
Kg O
2
/d = (Q . (S
0
- S )/F 10
-3
mg/kg, kg/d) – (1.42 . mass
sludge wasted, kg/d) + 4.33 .Q . (N
0
- N) 10
-3
mg/kg, kg/d
N
0
= Influent TKN, mg/l
N = Effluent TKN, mg/l
Air supply must provide :
 Satisfy the BOD of wastewater,
 Satisfy the endogenous respiration by sludge organism,
 Complete mixing,
 Satisfy dissolved oxygen minimum 2 mg/l.


Example 1.
An activated sludge process has a influent BOD concentration of
500 mg/l, influent flow 18,900 m
3
/d and 56,500 kg of suspended
solids under aeration. Calculate the F/M ratio. Assume VSS is
80% of TSS.

Solution
Step 1. Calculate BOD in kg/d
BOD = Q . BOD = 18,900 m
3
/d . 500 mg/l 10
-6
kg/mg . 10
3
l/m
3

= 9,450 kg/d
Step 2. Calculate the VSS under aeration
MLVSS = 56,500 kg . 0.8 = 45,200 kg
Step 3. Calculate the F/M ratio
F/M = 9,450/45,200 = 0,209 kg BOD/d per kg MLSS




Example 2.
Design a complete-mix activated-sludge system.
Given:
Average design flow 0.32 m3/s
Peak design flow 0.80 m3/s
Raw wastewater BOD5 240 mg/L
Raw wastewater TSS 280 mg/L
Effluent BOD5 20 mg/L
Effluent TSS 24 mg/L
Wastewater temperature 30
0
C
Operational parameters and biological kinetic coefficients:
Design mean cell residence time θ
c
= 10 d
MLVSS 2400 mg/L (can be 3600 mg/L)
VSS/TSS = 0.8
TSS concentration in RAS = 9300 mg/L
Y = 0.5 mg VSS/mg BOD
5

k
d
= 0.06/d
BOD
5
/ultimate BOD
U
= 0.67

Assume:
1. BOD (i.e. BOD5) and TSS removal in the primary
clarifiers are 33% and 67%, respectively.
2. Specific gravity of the primary sludge is 1.05 and the
sludge has 4.4% of solid contents
3. Oxygen consumption is 1.42 mg per mg of cell oxidized.

Solution:
Step 1. Calculate BOD and TSS loading to the plant
Design flow Q = 0.32 m3/s . 86,400 s/d
= 27,648 m3/d
Since 1 mg/L = 1 g/m3 = 0.001 kg/m3
BOD loading = 0.24 kg/m3 . 27,648 m3/d
= 6,636 kg/d
TSS loading = 0.28 kg/m3 . 27,648 m3/d
= 7741 kg/d

Step 2. Calculate characteristics of primary sludge
BOD removed = 6636 kg/d . 0.33 = 2190 kg/d
TSS removed = 7741 kg/d . 0.67 = 5186 kg/d
Specific gravity of sludge = 1.05
Solids concentration = 4.4% = 0.044 kg/kg
Sludge flow rate = 5,186 /[(1.05 . 1000 kg/m
3
)0.044]
= 112 m
3
/d

Step 3. Calculate flow, BOD, and TSS in primary effluent
(secondary influent)
Flow = design flow = 27,648 m
3
/d - 112 m
3
/d
= 27,536 m
3
/d = Q for Step 6
BOD = 6636 kg/d - 2190 kg/d
= 4,446 kg/d
BOD = (4446 kg/d . 1000 g/kg)/27,536 m
3
/d
= 161.5 g/m
3
= 161.5 mg/l = S
0

TSS = 7741 kg/d - 5186 kg/d = 2,555 kg/d
= (2,555 kg/d . 1000 g/kg)/27,536 m
3
/d
= 92.8 g/m
3
= 92.8 mg/l

Step 4. Estimate the soluble BOD5 escaping treatment, S, in
the effluent
Use the following relationship
Effluent BOD = influent soluble BOD escaping treatment,
S + BOD of effluent suspended solids
a. Determine the BOD
5
of the effluent SS (assuming 63%
biodegradable)
Biodegradable effluent solids = 24 mg/L . 0.63 = 15.1
mg/L
Ultimate BODu of the biodegradable effluent solids =
15.1 mg/L . 1.42 mg O2/mg cell = 21.4 mg/L
BOD
5
= 0.67 BODu = 0.67 . 21.4 mg/L = 14.3 mg/L
(b) Solve for influent soluble BOD5 escaping treatment
20 mg/L = S + 14.3 mg/L
S = 5.7 mg/L

Step 5. Calculate the treatment efficiency E using Eq.
E = (S0 - S)/S0 . 100%
(a) The efficiency of biological treatment based on soluble
BOD is = [(161.5 – 5.7)/161.5] x 100 = 96.5%
(b) The overall plant efficiency including primary treatment
is = [(240 – 20/240] x 100 = 91.7%

Step 6. Calculate the reactor volume using Eq.
V = [θ
c
Q Y ( S
0
- S)] / [X (1 + k
d
θ
c
)
θ
c
= 10 d
Q = 27,536 m
3
/d (from Step 3)
Y = 0.5 mg/mg
S
0
= 161.5 mg/L (from Step 3)
S = 5.7 mg/L (from Step 4b)
X = 2400 mg/L
k
d
= 0.06 d
-1

(10 d) (27,536 m
3
/d) (0.5) (161.5 – 5.7) mg/l
V =
(2,400 mg/l) ( 1 + 0.006 day
-1
. 10 days)

= 5,586 m
3

Step 7. Determine the dimensions of the aeration tank
Provide 4 rectangular tanks with common walls. Use width-
to-length ratio of 1:2 and water depth of 4.4 m with 0.6 m
freeboard
w . 2 w . (4.4 m) . 4 = 5586 m
3
→ w = 12.6 m
width = 12.6 m and length = 25.2 m
water depth = 4.4 m (total tank depth 5.0 m)

Step 8. Calculate the sludge wasting flow rate from the
aeration tank
total mass SS in reactor V
r
X
θ
c
= =
SS wasting rate Q
w
X
w
+ Q
e
X
e


(5,586 m
3
/d) (2,400 mg/l)
10 days =
Q
w
(9,300 mg/l) + (27,536 m
3
/d) (24 mg/l . 0.8)

Q
w
= 270 m
3
/d

Step 9. Estimate the quantity of sludge to be wasted daily
(a) Calculated observed yield
Y 0.5
Y
obs
= = = 0.3125
1 + k
d
θ
c
1 + 0.06 (10 days)


(b) Calculate the increased in the mass of MLVSS
p
x
= Y
obs
Q (S
0
- S) . (1 kg/1000 g)
= 0.3125 . 27,536 m
3
/d . (161.5 – 5.7) g/m
3
. 0.001 kg/g
= 1341 kg/d

(c) Calculate the increased in MLSS (or TSS)
p
ss
= (1341 kg/d)/0.8 = 1676 kg/d

(d) Calculate lost of TSS in effluent
p
e
= (27,536 – 270) m
3
/d . 24 g/m . Kg/1000 g
= 654 kg/d
Note: Flow is less sludge wasting rate from Step 8.

(e) Calculate the amount sludge that should be wasted
Wastewater sludge = p
ss
- p
e
= (1676 – 654) kg/d
= 1022 kg/d

Step 10. Estimate return activated sludge rate
Using a mass balance of VSS, Q and Qr are the influent and
RAS flow rates, respectively.
VSS in aerator = 2400 mg/L
VSS in RAS = 9300 mg/L . 0.8 = 7440 mg/L
2400 (Q + Q
r
) = 7440 . Q
r

Q
r
/Q = 0.476
Q
r
= 0.4762 . 27,536 m3/d = 13,110 m3/d = 0.152 m3/s

Step 11. Check hydraulic retention time (HRT = θ)
θ = V/Q = 5586 m
3
/(27,536 m
3
/d) = 0.203 d
= 4.87 h
Note: The preferred range of HRT is 5–15 h.

Step 12. Check F/M ratio using U in Eq.
S
0
- S (161.5 - 5.7) mg/l
U = = = 0.32 day
-1

θ X (0.203 day) (2,400 mg/l)
Step 13. Check organic loading rate and mass of ultimate
BODu utilized
Q S
0
27,536 m
3
/d . 161.5 g/m
3

Loading = =
V 5586 m
3
. 1000 g/kg

= 0,80 kg BOD
5
/m
3
d

BOD
5
= 0.67 BOD
u
(given)
BOD
u
used = Q (S0–S)/0.67
= [27,536 m
3
/d. (161.5 - 5.7) g/m
3
]/0.67
= 6403 kg/d


Step 14. Compute theoretical oxygen requirements
The theoretical oxygen required is calculated from Eq.
Q (S
0
– S)
O
2
=
1000 g/kg F
= 6403 kg/d (from Step 13) - 1.42 . 1341 kg/d (from
Step 9b) = 4499 kg/d

Step 15. Compute the volume of air required
Assume that air density 1.202 kg/m
3
and contains 23.2%
mass oxygen, the oxygen transfer efficiency for the aeration
equipment is 8% and a safety factor 2 is used to determined
the actual volume for sizing the blower.
(a) The theoretical air required is
4499 kg/d
Air = = 16,200 m
3
/d
1.202 kg/m
3
. 0.232 g O
2
/g air

(b) The actual air required at an 8% oxygen transfer
efficiency
Air = (16,200 m
3
/d)/0,08 = 02,000 m
3
/d = 140 m/min

(c) The design air required (with a factor of safety 2) is
Air = 140 m3/min x 2
= 280 m3/m
Design Based on Biokinetic Equations.
Microbial Growth
Exponential Phase:
dX/dt = μ X
X = cell concentration (mg dry mass or VSS/L)
dX/dt = volumetric cell production rate (mg/L d)
μ = specific growth rate (h
-1
or d
-1
)
r
g
= dX/dt = μ X
r
g
= volumetric cell production rate (g VSS m
-3
d
-1
)
X = cell concentration (g VSS L
-1
)
μ = the specific growth rate (d
-1
)
Likewise: r
SU
= dS/dt = - q S
r
SU
= volumetric substrate consumption rate (g S m
-3
d
-1
)
S = substrate concentration (g S m
-3
)
q = maximum specific rate of substrate degradation, d
-1

Y = True Yield Coefficient (gVSS/gS)
= biomass produced/substrate consumed
Y = (X – X
0
)/(S
0
– S)  r
g
= - Y r
SU
or r
SU
= - μ X/Y
Monod kinetic model
 = 
m
S/(K
S
+ S)
r
g
=  X = [
m
/(K
S
+ S)]X
μ
m
= maximum specific growth rate (d
-1
)
Ks = saturation or Monod constant (g l
-1
)
S = limiting substrate (g l
-1
)
r
SU
= -r
g
/Y =  X/Y = 
m
S X/[Y(Ks +S)]

The cell growth rate (and therefore the substrate
removal rate) increases with the substrate
concentration, up to a certain level when it stabilize at

m
.

If the limiting substrate concentration is low: conditions
of slow growth.
If part of the biomass produced is degraded with
endogenous decay: r
g
= -Y r
SU
– k
d
X
With k
d
= endogenous coefficient decay (d
-1
)
r
g
=Volumetric biomass production rate (g VSS/m
3
d)
-Y r
SU
= Rate of biomass production from substrate
consumption
k
d
X = Rate of biomass consumption by endogenous
respiration
And  = 
m
S/(Ks + S) – k
d

r
SU
= - r
g
/Y – k
d
X/Y = -Y
-1
( X + k
d
X)
= - 
m
S X/[Y(Ks + S)]
Continuous Treatment in CSTR
Under a steady state in a well mixed reactor (CSTR)
X = X
r
and S = S
r
(X and S = concentrations) and
dX/dt = dS/dt = 0
V , X
r
, S
r

Q, X
0,
S
0

Q, X, S
This also means:
Q (X – X
0
) = P
X
= biomass production rate
Q (S
0
– S) = P
S
= substrate consumption rate

V dX
r
/dt = cell in - cell out + cell produced
= Q X
0
– Q X + r
g
V
For simplification, X
0
= 0 and by definition r
g
= μ X
 X V = X Q   = Q/V
By definition D = dilution rate = Q/V = μ = 1/HRT
(Hydraulic Retention Time): The growth rate is
dictated by the dilution rate.
Monod model: μ = μ
m
S/(K
S
+S)
Under steady state: μ = D, therefore D = μ
m
S/(K
S
+S)
Solving this equation  S = D K
S
/(μ
m
– D)
S mass balance:
V dS/dt = 0 = Q S
0
– Q S + r
SU
V  r
SU
V = - Q (S
0
- S)
Since r
SU
V= (-r
g
/Y)
V = - μ V X/Y = - D V X/Y = - Q X/Y (since D = Q/V)
Q (S
0
– S) = Q X/Y and X = Y(S
0
– S)
Maximum Dilution Rate: D
max
CSTR:  = D = Q/V
At washout condition: S = S
0
= D
max
K
S
/(μ
m
–D
max
)
D
max
= 
m
S
0
/(K
S
+ S
0
) = 
m
/(1 + K
S
/S
0
)
Cell wash-out occurs at too high dilution rates (D >D
max
)
and is especially sensitive at low initial substrate
concentration.
Influence of endogenous decay
Now  = 
m
S/(K
S
+ S) – k
d

S: The biomass balance is identical, hence: D =  =

m
S/(K
S
+ S) – k
d
.
Solving this equation gives:
S = K
S
(D + kd)/(
m
– D – kd)
S mass balance: V dS/dt = 0 = Q S
0
– Q S + r
SU
V
r
SU
= -r
g
/Y – k
d
X/Y = -  X/Y-k
d
X/Y = - X(D + k
d
)/Y
Solving this equation gives:
X = Y D(S
0
– S)/(D + k
d
) = Y(S
0
– S)/(1 + k
d
/D)
D
max
is obtained for S = S
0
= K
S
(D + k
d
)/(
m
– D – k
d
)
Solving this equation gives D
max
= 
m
/(1 + K
S
/S
0
) - k
d

Influence of non-biodegradable VSS (nbVSS )
An amount of non biodegradable, also called inert,
VSS is introduced into the reactor in the wastewater.
This amount is not degraded biologically and
therefore, at a steady state, the nbVSS concen-
trations in the effluent and reactor are similar to the
nbVSS concentration in the influent (X
0,i
)
The total mass of VSS in the Bioreactor includes the
biomass produced (r
g
), the nbVSS introduced (X
0,i
)
and the debris released from the endogenous decay:
r
VSS
= total VSS production rate
= r
g
+ Q X
0,i
/V + f
d
(k
d
) X
r
g
= -Y r
SU
– kdX = biomass production from bCOD (-Y
r
SU
) minus endogenous decay (kd X)
f
d
(k
d
) X = rate of cell debris production
with f
d
= fraction of biomass remaining as cell debris.
The cell debris production is directly proportional to the
biomass concentration and k
d
Q X
0.i
/V = Amount of nbVSS in the influent
Q = influent flow rate,
X
0,i
= influent nbVSS and
V = reactor volume).
Solid Retention Time (SRT)
The SRT is defined as the average time the solids
stay inside the aeration tank.
When VSS = active cells (X), the SRT is also called
Mean Cell Retention Time (MCRT) with:
SRT = Amount of active biomass = VX (kg)
The production rate can be obtained from the
biomass mass balance under steady state as
X
out
– X
in
= Production, with
X
out
= Q
e
X
e
+ Q
w
X
w
, X
in
= Q
0
X
0

Note Xr = Xw
Often Xw >> Xe and Qw Xw + Qe Xe  Qw Xw
Other name of SRT: Sludge Age
Expression of S
Biomass balance:
V dX/dt = 0 = Q
0
X
0
- (Q
e
X
e
+ Q
w
X
w
) + V r
g

(Q
e
X
e
+ Q
w
X
w
) – Q
0
X
0
= V μ X
From the definition of SRT:
(Q
e
X
e
+ Q
w
X
w
) – Q
0
X
0
= (Q
e
X
e
+ Q
w
X
w
)= V X/SRT
1/SRT = μ
μ = μ
m
S/(Ks + S) – kd
S = K
S
(1 + (k
d
) SRT)/[SRT(μ
m
– k
d
) – 1]
Expression of X
X can be obtained from S mass balance:
V dS/dt = Q
0
S
0
– (Q
e
S
e
+ Q
w
S
w
) + r
SU
V = 0
With the assumptions that S = S
w
= S
e
= S
r
(S = bsCOD)
and since (Q
e
+ Q
w
) = Q
0
, this becomes:
Q
0
(S
0
– S) = - r
SU
V

Since r
g
= -Y r
SU
– k
d
X; r
g
= μ X = X/SRT and HRT = V/Q
X = (SRT/HRT) Y(S0 – S)/(1 + (kd)SRT)
Expression of U
1 S
0
- S
= Y U - k
d
= Y - k
d

SRT  X

 X K
S
1 1 1
= + =
S
0
- S k S K U

where:
SRT: Solids retention time, d
Y: Biomass yield, mg VSS/mg sCOD
U : Substrate utilization rate, mg sCOD/mg VSS.d
k
d
: Endogenous decay coefficient, 1/d
S
0
: Influent substrate concentration, mg sCOD/l
S : Effluent substrate concentration, mg sCOD/l
X : Biomass concentration, mg VSS/l
 : Hydraulic retention time, d
K
S
: Half-velocity constant, mg sCOD/l
k : Maximum rate of substrate utilization,
mg sCOD/mg VSS . d
Plotting 1/SRT versus U, the biokinetic coefficients Y
can be determined from the slope and k
d
from the
intercept of the equation.
Plotting 1/U or  X/(S
0
- S) versus 1/S, the biokinetic
coefficients K
S
can be determined from the slope and k
from the intercept of the equation.

Equations describing the performance of the system are
the mass balance equations of both the biomass and
substrate. The biomass balance can be expressed by:
[rate of change of biomass in the reactor] = [rate of
increase due to growth] − [rate of loss due to endogenous
respiration] − [delibarate wastage]
which can be mathematically expressed as:
V dX/dt = μ X V − k
d
X V − Q
w
X
where
V = reactor volume (l);
X = biomass concentration in the reactor (mg/l);
k
d
= biomass decay coefficient (day
-1
);
Q
w
= wastage flow rate (s
-1
);
t = time (s).

At steady-state conditions, dX/dt = 0, hence, Eq. above
becomes:
μ = k
d
+ Q
W
V (a)
Since the solid retention time (SRT) is defined as:
total mass of organisms in the reactor
SRT =
total mass of organisms leaving the system/day
SRT = V X/Q
W
X = V/Q
W
(b)
Substituting Eq. (b) into Eq. (a) results in:
μ = k
d
+ 1/SRT (c)
Substituting for the value of μ from Eq. (c) into Monod
model: μ = μ
m
S/(K
S
+S) yields the following equation that
describes the steady-state condition for substrate
concentration in the reactor:

SRT K
S
1 1
= + (d)
1 + SRT x k
d

m
S 
m


.
Substituting the value of K
S
and k
d
in Eq. (d) and
plotting SRT/[1 + (SRT x k
d
)] versus 1/S, the biokinetic
coefficients, μ
m
can be determined from the Y-intercept
of the equation
In continuous-flow and completely-mixed reactor,
determination of the biokinetic coefficients K
S
, k, μ
m
, Y,
and k
d
is usually achieved by collecting data from lab-
scale or pilot-scale experimental setups operated at
various hydraulic retention times (HRTs) and/or at
various sludge retention times (SRTs), and by allowing
steady-state condition to prevail for each HRT or SRT
under investigation.
Determination of μ
m