Download Lecture 7

CE 548
II
Fundamentals of Biological Treatment
1
Modeling Suspended Growth
Treatment Processes
 Description of treatment process:
 All biological treatment reactor designs are based on using
mass balances across a defined volume for each constituent of
interest (i.e., biomass, substrate, etc.)
 Biomass mass balance:
Accumulation = inflow – outflow + net growth
dX
V  QX o  Q  Qw X e  Qw X r   rgV
dt
eq (7 - 32)
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Modeling Suspended Growth
Treatment Processes
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Modeling Suspended Growth
Treatment Processes
Assuming stead-state and Xo = 0, equation 7-32 can be
simplified:
Q  Qw X e  Qw X r  rgV
rg  Yrsu  k d X
Q  Qw X e  Qw X r
VX
SRT 
Eq (7 - 33)
Eq (7 - 21)
rsu
 Y
 kd
X
Eq (7 - 34)
mass of organisms in the reactor
mass of organisms removed form the system each day
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Modeling Suspended Growth
Treatment Processes
VX
SRT 
Q  Qw X e  Qw X r
Eq (7 - 35)
Equation 7-34 can be written as:
rsu
1
 Y
 kd
SRT
X
Eq (7 - 36)
The term 1/SRT is related to µ, the specific biomass growth rate:
1

SRT
Eq (7 - 37)
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Modeling Suspended Growth
Treatment Processes
In Eq. (7-36) the term (-rsu/X) is known as the specific substrate
utilization rate U and can be calculated as the following:
rsu Q( S o  S ) S o  S
U


X
VX
X
Substituting Eq. (7-12) rsu  
1
YkS

 kd
SRT K s  S
Eq (7 - 38)
kXS
Ks  S
into Eq. (7-36) yields:
Eq (7 - 39)
Solving Eq. (7-39) for S yields:
S
K s 1  k d SRT 
SRT Yk  k d   1
Eq (7 - 40)
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Modeling Suspended Growth
Treatment Processes
 Substrate mass balance:
Accumulation = inflow – outflow + generation
0
dS
V  QS o  QS  Qw S  rsuV
eq (7 - 41)
dt
Substituting for rsu and assuming steady-state, Eq. (7-41) can be
written as:
 V  kXS 

So  S   
 Q  K s  S 
 SRT
X 
 
 Y So  S  




1

k
SRT

d

eq (7 - 42)
eq (7 - 43)
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Modeling Suspended Growth
Treatment Processes
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Modeling Suspended Growth
Treatment Processes
 Mixed liquor solids concentration and solids production:
The solids production from a biological reactor represents the
mass of material that must be removed each day to maintain
the process:
X TV
eq (7 - 45)
SRT
where; PX T ,VSS  total solids wasted daily, g VSS/d
PX T ,VSS 
X T  total MLVSS in areation tank, g VSS/m3
Eq. (7-45) can be used to calculate the amount of solids
wasted for any of the mixed liquor components. For the
amount of biomass wasted (PX), the biomass concentration X
can be used in place of XT in Eq. (7-45).
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Modeling Suspended Growth
Treatment Processes
 Mixed liquor solids concentration:
The total MLVSS equals the biomass concentration X plus the
nbVSS concentration Xi :
XT  X  Xi
eq (7 - 46)
A mass balance is needed to determine the nbVSS conc.:
Accumulation = inflow – outflow + generation
dX i / dt V  QX o,i  X iV / SRT  rX ,iV
eq (7 - 47)
Where X o ,i  nbVSS concentrat ion in influent, g/m
3
X i  nbVSS concentrat ion in areation tank, g/m3
rX ,i  rate of nbVSS production from cell debris, g/m3  d
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Modeling Suspended Growth
Treatment Processes
 Mixed liquor solids concentration:
At steady-state
dX i / dt  0 and
substituting Eq. (7-25)
rXd  f d (kd ) X for rX ,i in Eq. (7-47) yields:
X i  X o ,i ( SRT ) /   ( f d )( k d ) X ( SRT )
eq (7 - 49)
Combining Eq. (7-43) and Eq. (7-49) for X and Xi produces the
following equation that can be used to determine XT :
 SRT
XT  
 

X o,i SRT
  Y So  S  

   f d kd  X SRT 

 1  kd SRT 
(A)
(B)
Hetrotroph ic
biomass
Cell debris
eq (7 - 50)
(C)
Nonbiodegr adable
VSS in influent
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Modeling Suspended Growth
Treatment Processes
 Solids production:
The amount of VSS produced and wasted daily is as follows:
PX ,VSS 
QY S o  S 
 f d k d X V   QX o ,i
1  k d SRT
eq (7 - 51)
Eq. (7-43) is substituted for biomass concentration (X) in Eq. (7-51)
to show VSS production rate in terms of the substrate removal,
influent VSS, and kinetic coefficients as follows:
PX ,VSS
QY S o  S  f d k d QY S o  S SRT


 QX o ,i
1  k d SRT
1  k d SRT
(A)
(B)
Hetrotroph ic
biomass
Cell debris
eq (7 - 52)
(C)
Nonbiodegr adable
VSS in influent
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Modeling Suspended Growth
Treatment Processes
 Solids production:
The effect of SRT on the performance of an activated sludge system
for soluble substrate removal is shown in figure 7-13
The total suspended solids (TSS) production can be calculated by
modifying Eq. (7-52) assuming that a typical biomass VSS/TSS ratio
of 0.85 as follows:
PX ,TSS
A
B


 C  Q(TSSo  VSSo )
0.85 0.85
eq (7 - 53)

influent inorganic solids
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Modeling Suspended Growth
Treatment Processes
 The observed yield:
The observed yield for VSS can be calculated by dividing Eq. (7-52)
by the substrate removal rate Q(So-S):
X o ,i
( f d )(kd )(Y ) SRT
Y
Yobs 


1  (kd )SRT
1  (kd ) SRT
So  S
eq (7 - 56)
 Oxygen requirements:
Oxygen used = bCOD removed – COD of waste sludge
Ro  Q( S o  S )  1.42 PX ,bio
eq (7 - 59)
where
Ro  Oxygen required, kg/d
PX ,bio  biomass as VSS wasted per day, kg/d
Study example 7-6
1.42  COD of cell tissues, g O 2 / g cells
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Modeling Suspended Growth
Treatment Processes
 Design and operating parameters:
Following are the design and operating parameters that are
fundamentals to treatment and performance of the process:
 SRT
Food to microorganisms (F/M) ratio
F /M 
total applied substrate rate
total microbial biomass
QS o S o


VX X
eq (7 - 60)
The SRT can be related to F/M by the following equation:
1
E
 Y (F / M )
 kd
SRT
100
eq (7 - 66)
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Modeling Suspended Growth
Treatment Processes
 Design and operating parameters:
Organic volumetric loading rate.
Defined as the amount of BOD or COD applied to the
aeration tank volume per day:
Lorg
QS o

(V )(103 g / kg)
eq (7 - 67)
Where
Lorg  volumetric organic loading, kg BOD/m 3  d
Q  influent flowrate, m3 /d
S o  influent BOD concentrat ion, g/m 3
V  aeration tank volume, m3
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Modeling Suspended Growth
Treatment Processes
 Modeling plug-flow reactors:
Developing a kinetic model for the plug-flow reactor is
mathematically difficult (X vary along the reactor). Two assumptions
are made to simplify the modeling:
 The concentration of microorganisms is uniform along the reactor  X 
This assumption applies only when SRT/  5.
 The rate of substrate utilization is given by:
rsu 
kSX
Ks  S
eq (7 - 72)
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Modeling Suspended Growth
Treatment Processes
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Modeling Suspended Growth
Treatment Processes
 Modeling plug-flow reactors:
Integrating Eq. (7-72) over the retention time in the tank gives:
Yk S o  S 
1

 kd
SRT S o  S   1   K s ln Si / S 
eq (7 - 73)
X 
where;
S o  influent concentrat ion
S  effluent concentrat ion
Si  influent concentrat ion to reactor after dilution with recycle flow

S o  S
1 
  recycle ratio
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Biological Nitrification
 Nitrification is the conversion (by oxidation) of Ammonia (NH4N) to nitrite (NO2-N) and then to nitrate (NO3-N).
 The need for nitrification arises from water quality concerns:
 Effect of ammonia on receiving water; DO demand, toxicity.
 Need to provide nitrogen removal for eutrophication control.
 Need to provide nitrogen removal for reuse applications.
 The current drinking water MCL for nitrate is 45 mg/l as
nitrate or 10 mg/l as nitrogen.
 The total concentration of organic and ammonia nitrogen in
municipal wastewater is typically in the range of 25-45 mg/l as
nitrogen.
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Biological Nitrification
 Process description:
 Nitrification is commonly achieved with BOD removal in the same
single-sludge process.
 In case of the presence of toxic substances in the wastewater, a
two-sludge system is considered.
 Stoichiometry:
Nitritebacteria
( Nitrosomonas )
2 NH 4  3O2    2 NO2  4 H   2 H 2O
Nitratebacteria
( Nitrobacter )
2 NO2  O2    2 NO3
Overall reaction :
NH 4  2 HCO3  2O2  2 NO3  2CO2  3H 2O
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Biological Nitrification
 Process description:
Overall reaction :
NH 4  2HCO3  2O2  2 NO3  2CO2  3H 2O
 The oxygen required for complete oxidation of ammonia is 4.57 g
O2/g N oxidized.
 The alkalinity (alk) requirement is 7.14 g alk as CaCO3 for each g of
ammonia nitrogen (as N).
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Biological Denitrification
 Process description:
 Denitrification is the biological reduction of nitrate (NO3) to nitric
oxide (NO), nitrous oxide (N2O), and nitrogen (N).
 The purpose is to remove Nitrogen from wastewater.
 Compared to alternatives of ammonia stripping, breakpoint
chlorination, and ion exchange, biological nitrogen removal is more
cost-effective and used more often.
 Concerns over eutrophication and protection of groundwater against
elevated NO3-N concentration.
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Biological Denitrification
 Stoichiometry:
 In denitrification, nitrate is used as the electron acceptor instead of
oxygen and the COD or BOD as the carbon source (electron donor).
 The carbon source can be the influent wastewater COD or external
source (Methanol).
Wastewater
C10 H19O3 N  10 NO3  5 N 2  10CO2  3H 2O  NH 3  10OH 
Methanol
5CH 3OH  6 NO3  2 N 2  5CO2  7 H 2O  6OH 
 One equivalent of alkalinity is produced per equivalent of nitrate
reduced. (3.57 g alk per g nitrate)
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Biological Phosphorus Removal
 Process description:
 Phosphorous removal is done to control eutrophication.
 Chemical treatment using alum or iron salts is the most commonly
used technology for phosphorous removal.
 The principle advantages of biological phosphorous removal are
reduced chemical costs and less sludge production.
 In the biological removal of phosphorous, the phosphorous in the
influent is incorporated into cell biomass which is removed by sludge
wasting.
 Phosphorous accumulating organisms (PAOs) are encouraged to
grow and consume phosphorous. Therefore, the system is designed
so that the reactor configuration provides advantage for PAOs to
grow over other bacteria.
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Biological Phosphorus Removal
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Biological Phosphorus Removal
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Anaerobic Fermentation and Oxidation
 Process description:
 Used primarily for the treatment of waste sludge and high strength
organic waste.
 Advantages include low biomass yield and recovery of energy in the
form of methane.
 Conversion of organic matter occurs in three steps:
– Step1 (Hydrolysis): involves the hydrolysis of higher-molecularmass compounds into compounds suitable for use as a source of
energy and carbon.
– Step2 (Acidogenesis): conversion of compounds from step1 into
lower-molecular-mass intermediate compounds.
(nonmethanogenic bacteria)
– Step3 (Methanogenesis): conversion of intermediates into
simpler end products (CH4 & CO2).
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Anaerobic Fermentation and Oxidation
 Process description:
 For efficient anaerobic treatment, the reactor content should be:
– void of O2
– free of inhibiting conc. of heavy metals and sulfides
– pH ~ 6.6 – 7.6
– sufficient alkalinity to ensure pH is not <6.2 (methane bacteria
will not function below 6.2).
 Methanogenic bacteria has slow growth rate, therefore:
– require long detention time for waste stabilization
– yield is low: less sludge production and most organic matter is
converted to CH4 gas.
– sludge produced is stable: suitable for composting
– require relatively high temp for adequate treat.
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