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PROJECT #4
BIOLOGICAL TREATMENT DESIGN
GENERAL
- Why to use the biological treatment processes?
 Primary treatment usually reduce:
BOD5:
25 to 40%
250 mg/L  175 mg/L
SS:
50 to 70%
250 mg/L  100 mg/L
 Effluent requirements:
BOD5:
20 to 25 mg/L
SS:
20 to 25 mg/L
 Normally use biological processes for remaining
removal, in addition to low costs and proven
successes
- Basic Principles
Use the living microbes to break down waste organic
substrates, and then separate the grown microbes from
effluents usually by settling to produce treated effluents
› Aerobic Processes
Microbes oxidize the substrates in the presence of O2
Organics+ O2  CO2 + H2O + New Cells + Energy
› Anaerobic Processes
Microbes degrade the substrates in the absence of O2
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› Facultative Processes
Microbes use O2 if available but can also respire and
multiply in its absence
› Sludge Composition and Nutrient Requirements
Typical composition:
C5H7NO2P0.0074
C5H7NO2 +5O2 → 5CO2 + 2H2O + NH3
 Nutrient requirements:
COD:N:P = 100:5:1
 Oxygen requirements:
1.42 mg COD/mg VSS
TREATMENT PROCESSES

Suspended Growth Systems
› Biological mass suspended free in the liquid
› Biological mass may be recycled to keep a high conc.
of active biomass
› Most common type is Activated Sludge Processes
which use aeration to supply oxygen for degradation
› Many variations for AS processes
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
Fixed Film or Attached Growth Systems
› Microbes grow in a layer or a film on some types of inert
support media (i.e., rocks, peat, plastics etc)
› Most important design parameter is the surface area
› Common types are trickling filter and rotating biological
contactor (RBC) systems
- For trickling filters
 WW sprinkled over surface
 Use effluent recycle to maintain adequate hydraulic
loading and reduce the influent strength
 Develop biological film on the media
 Film extracts organic from WW
 Oxygen provided by ventilation due to temperature
difference between air and WW
- For RBC
 Film is normally aerated during travel through air
 No effluent recycle required
 Normally built in doors and used in small
communities

Lagoon and oxidation ponds
- Aerobic
- Facultative
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ACTIVATED SLUDGE REACTOR DESIGN
 General Growth Pattern in Pure Culture

Phases # of Cells
Cell Mass
1
Lag
Lag
2
Log-growth
Log-growth
3
Stationary
Declining-growth
4
Log-death
Endogenous
Microbial Growth Rates
 dX 

  X ,
dt

g
  m
S
Ks  S
where:  = specific growth rate, s-1
m = maximum specific growth rate, s-1
Ks = saturation constant, mg/L
S = conc. of growth-limiting substrate, mg/L
X = conc. of biomass, mg MLVSS/L
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In AS, the endogenous decay occurs at:
 dX 

  Kd X
 dt  d
where: Kd = microbial decay rate constant, s-1
 Net microbial growth rate rgnet is
 dX 
rgnet  

net
 dt  g

 XS
 dX 
 dX 

 Kd X
 
  m
dt
dt
K

S

g 
d
s
Substrate Removal
 m XS
kXS
 dS 

  
 dt u Y ( K s  S ) ( K s  S )
Monod Equation
where: Y = growth yield coefficient (  dX / dt g /dS / dt u
To account for the effect of endogenous decay on the
net biomass growth rate, the observed growth yield
coefficient Yobs is:
Yobs 
where:
(dX / dt ) net
g
(dS / dt )u

Y rsu / X   K d
rsu / X 
rsu / X  = specific utilization rate, U
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
Lawrence & McCarty Model
S0, Q0, X0
S, X
V
(1+α)Q0
S, (Q0- Qw), Xe
S, X
Qr, Xr, S
Qw, S, Xr
α
Q0
S0
X0
S
=
=
=
=
=
X =
=
Qw =
Qr =
Xr =
Xe =
Qr/Q0
inflow to the process, m3/d
substrate concentration in the inflow, mg/L or g/m3
suspended solids in the inflow, mg/L or g/m3
substrate concentration in the reactor, effluent and recycle
flow, mg/L
suspended solids in the aeration tank, mg/L or g/m3
mixed liquor suspended solids (or volatile)
flow to waste, mg/L or g/m3
recycle flow, mg/L or g/m3
suspended solids in the recycle flow, mg/L or g/m3
effluent suspended solids, mg/L or g/m3
- Some definitions
mass of organisms in the reactor
VX

mass of organism wasted per time (Q0  Qw ) X e  Qw X r
Volume of reactor
V
td 

Influent flowrate
Q0
mass of substrate into the reactor per time So  S
F/M 

mass of organisms in the reactor
X
c 
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- Substrate in the Effluent
Apply the biomass balance at steady state:
V
dx
S
 Q0 X 0  YXK
V  K d XV  Q0  Qw X e  Qw X r 
dt
Ks  S
Note the steady state condition (dx/dt = 0) and  = V/Qo
0
X0

X
0 0

S
 (Q  Qw ) X e  Qw X r 
 Kd X  X  0

Ks  S
VX


S
X
 YXK
 Kd X 
Ks  S
c
 YXK
Because X0  0
1
r
 Y su  Kd
c
X
1
or
c
 YU  K d
Alternatively,
0  S (YXK  Kd X 
S
Ks X (Kd 
1
c
X (YK  K d 
or
S 
X
c
)  K s ( K d X 
)
X
c
)
K s ( K d  c  1)
 c (YK  K d )  1
- Reactor Volume
r
 Y su  K d
c
X
1
Solve for X 
Yrsu
1


K
d

 c

X
c
)
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From the definition, note rsu 
X 
or
td 

Y S 0  S 
1

td   K d 
 c

Y c S 0  S 



Q
S0  S   S0  S and substitute:
V
td
Y c S 0  S 
t d 1  K d  c 
V
X 1  K d  c  Q0
Q Y S  S 
V  0 c 0
X 1  K d  c 

- Sludge Production Rate
Px  Yobs Q( S 0  S ) 
YQ ( S 0  S )
1  K d c
- Oxygen Utilization Rate
 Q( S 0  S ) 
OUR  
  1.42Px
f


where:
f
= conversion factor (BOD5 to BODL) (=0.68)