CE 428 Water and Wastewater Treatment Design
Dr. S.K. Ong
Biological Treatment
 Use of microorganisms to degrade the organic materials (both dissolved and suspended organics)
 Microbial metabolism is defined as the diverse reactions by which a cell processes food materials to
obtain energy and compounds from which new cell components are made
 Two types of metabolic processes
catabolism (respiration) – provides energy for the synthesis of new cells as well as maintenance of other cell
functions
anabolism – provides the material necessary for cell growth
endogeneous catabolism – when external food is interrupted microorganisms used stored food for maintenance
energy or when cells die off.
New Cells
Organics +
Microorganism
s
Endogeneous decay
energy
Energy + waste products
O2
CO2, H20 +
organic residue
Availability of Nutrients
 Nutrients must be provided at a minimum level in order to sustain microbial growth. Nitrogen and
phosphorus are the two nutrients required in the largest quantities and are usually the nutrients most likely to
be limiting.
 Microorganisms also require micronutrients such as sulfur, potassium, calcium, magnesium, iron, cobalt
and molybdenum. Essential vitamins, and amino acids are also required.
 The approximate formula for a bacteria cell is C5H7O2NP0.074. As seen from the bacteria cell formula, the
ratio of C:N for cell synthesis is approximately 5:1. However, the C:N ratio for cell synthesis and
as an energy source is about 10:1.
 In a typical municipal wastewater treatment plant, carbon is usually the limiting nutrient. N and P are not
limited.
Availability of Terminal Electron Acceptor
 To facilitate the oxidation reaction, a suitable electron acceptor is needed.
 Oxygen is used as an electron acceptor for aerobic metabolism while nitrate, Mn(IV) and Fe(III), sulfate,
and carbon dioxide may be used for anaerobic metabolism. Table provides the relative order of energy
levels which may be derived by using each electron acceptor.
 The electron acceptor which derives the maximum free energy by the microorganisms will be used first.
 The first dominant microbial community is the aerobic heterotrophs. As oxygen gets used up, denitrifiers using
nitrate as electron acceptors will dominate followed by sulfate reducers, fermenters and finally methanogens.
.
Example:
Glucose C6H12O6
C6H12O6
CO2
O2
H2O
NO3-
N2
Environment
Electron
Acceptors
End Products
of electron acceptors
Process
Aerobic
Anoxic
Anaerobic
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Design of Wastewater Treatment Plants
Generalized Activated Sludge Flow Diagram
Where
Q
So
V
X
S
Xo
Xr
Qw
Qr
Xe
= influent flow
= influent BOD or COD (soluble + suspended = total)
= volume of biological aeration tank
= volatile suspended solids (VSS) of biomass
= soluble effluent BOD or COD
= influent suspended solids
= VSS of recycled sludge
= waste flow rate
= recycled flow rate
= biomass in effluent
Two general approaches
 irrational approach
 rational approach
Irrational Approach (see notes on ten states standards)
Design based on loading factors
 organic loading rate
Completely mixed systems - 40 lbs BOD5/1000 ft3/d
Loading rate
= Q So / V
 Food/Microorganisms (F/M) ratio = 0.2 - 0.5 lbs BOD5/day/lb of MLVSS
(valid for MLVSS between 1000 - 3000 mg/L)
= Q(So- S)/VX
Rational Approach
 based on first principles and mass balances
1. Rate of cell production is given by:
r=X
where

m
S
Ks
Kd
 mS
(Monod's equation)
Ks  S
= maximum growth rate
= substrate concentration (BOD or COD)
= half saturation or half velocity constant (mg/L)
= endogeneous decay rate (time -1)
Net rate of cell growth
rg =  X - kdX
2. Yield coefficient - defined as
Y
therefore

mass of cells produced
mass of substrate used
r = -Y rsu
where rsu = rate of substrate utilization (mg/L/time)
or
rsu = dS/dt = - r /Y
1  m SX
rsu
= 
Y Ks  S
kSX
Ks  S
where k = maximum specific substrate utilization rate (g substrate/g microorganism ● d) = m/Y
= 
3. Hydraulic retention time or residence time (HRT)
V
(book uses )

Q
4. Solids residence time (SRT), sludge age, or mean cells residence time (MCRT) - defined as how long the cells or
biomass is in the aeration tank
biomass in aeration tan k
VX v
(book uses SRT)
c 

biomass wasted
Q w X vr
Minimum MCRT (mc) is a situation whereby the wasting of the biomass is equal or larger than the maximum
growth of the biomass, i.e., wasting the solids too fast such that the microorganisms do not have time to multiplywashout condition. Some have used this mc in their design, i.e., c = SF  mc where SF is a safety factor with
values between 20 - 60.
Typical Values for Municipal wastewater at 20o C
Values
Average
Y
mg VSS/mg BOD5
0.4 - 0.84
0.6
mg VSS/mg COD
0.3 - 0.6
0.4
Ks
mg/L of BOD5
25 - 100
60
mg/L of COD
15 - 70
40
kd
day-1
0.06 - 0.15
0.10
k = m/Y
day-1
2 - 10
5
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Assumptions used in deriving equations for activated sludge plant
- completely mixed system
- negligible biomass in influent and effluent
- no biological activity in sedimentation tank
- steady state conditions
- solids residence time (SRT) based on biomass in aeration tank
- sludge wasted from clarifier
Need to do a mass balance for biomass and substrate. Use
Change of biomass
Or substrate with time
Within system
=
Mass of biomass
or substrate
Inflow
__
For biomass
V
dX
 QX o  Q w X r  (Q  Q w )Xe   Vr g
dt
  mSX
dX
 QX o  Q w X r  (Q  Q w )X e   V 
 kdX
dt
 K s  S
  mS

Q w X r  VX 
 kd 
 K s  S


Qw Xr
1   mS


 kd 
VX
c  K s  S


V



at steady state dX/dt = 0, assume Xe = 0, Xo = 0
dS
 QSo  (Q  Q w )Se  Q wS  Vrsu
dt
Mass balance for substrate
V
V
dS
 mSX
 QS o  (Q  Q w )S  Q w S  V
dt
Y(K s  S )
At steady state, dS/dt = 0
Q(So  S)  V
Q(So  S ) 
 mSX
Y(K s  S )
VX  1


 k d 
Y  c

but =
 mS
Ks  S
with V/Q = 

1
 kd
c
Mass of biomass
or substrate
Outflow
Generation or
Removal
within system
Do a mass balance around secondary clarifier to obtain equation with recycle flow rates
Q+Qr, X
Mass balance
Q - Q w,
Xe = 0
(Q + Qr)X = (Q - Qw)Xe + QrXr + QwXr
remember c =VX/QwXr
Therefore
(Q + Qr)X = QrXr + VX/ c
Q r, X r
Q w, X r
Define r = Qr/Q
1
Then r 

c
Xr
1
X
Xr is usually between 10,000 - 12,000 mg/L