Raw Wastewater Parameters
pH, T°C
COD, BOD, BOD/COD (kg O
2/m³)
TSS (kg/m³)
Nutrients (N, P, micronutrients)
Flow Rate (m³/day)
pH
Degree of Acidity
pHwastewater=> pHbioreactorin range 6,5 –8,5
Extreme pH-values = Toxic
pH will change in bioreactor:
In anaerobic Treatment
-pH Degradation VFA, Ammonification
-pH Acidification, Nitrification, CO
2
CODChemical Oxygen Demand
Measure for TotalAmount Organic Contamination
Unit: mg O
2/L Chemical Oxidation
Oxidant: K
2Cr
2O
7
–OXIDATION
Organic matter + O
2-equivalents (k
2Cr
2O
7) CO
2
+ H
2O
BODBiochemical Oxygen Demand
Unit: mg O
2/L Biological Oxidation
Measure for Amount Biodegradable Contamination
Organic matter + O
2 CO
2+ H
2O
Oxidant: Bacteria –Inoculation
Micro-WWTP
BOD/COD
BOD/COD-ratio
Info Biodegradability
BOD/COD > 65 % Easily Biodegradable
BOD/COD < 50 % More Difficult Biodegradable
BOD vs. bCOD: Sludge Adaptation!
Total Suspended Solids (TSS)
TSS = Amount of Un-dissolved Matter (mg/l =
ppm)
Difficult Degradable
To Avoid:
Accumulation in Sludge
Increased Sludge Production
Nutrients
Macro-nutrients:
N, P
BOD/N/P = 100/5/1
Assimilative Removal = Biomass Incorporation
Micro-nutrients
Minerals: Ca, Mg, Na, Fe, K, …
Trace-elements: Co, Cu, Zn, Mn, …
Nitrogen
Kjeldahl-N: organic bound N + NH
4
+
-N
Ammonium-N (NH
4
+
-N) (mgN/l)
Nitrate-N (NO
3
-
-N) (mgN/l)
Nitrite-N (NO
2
-
-N) (mgN/l)
Total Nitrogen = KjN-N + NO
3
-
-N + NO
2
-
-N
Phosphorus
Total P (mgP/l) = organic bound P + PO
4
3—
P
Ortophosphate (PO
4
3-
-P) (mgP/l)
Inhabitant Equivalent
Originates from Municipal WWT
1 P.E.
Hydraulic load = 180 l/day
BOD-load = 54 g BOD = 300 mg O
2/l
COD-load = 135 g COD = 750 mg O
2/l
BOD/COD = 40%
SS-load = 63 g TSS = 350 mg/l
N-load = 9,9 g N = 55 mg N/l
P-load = 2 g P = 11 mg P/l
Why Biological
Wastewater Treatment?
Produces a clear effluent, harmless for
the environment:
Removes organic compoundsfrom the
wastewater
Removes nutrients (eutrofication risk!)
like N and P (partially)
Economical most feasible
Organic Compounds
C
Negative effect of WW on surface water:
Oxygen demand raises
•Microbial growth
•Water organisms die due to O
2-deficiency!
Possibly toxic compounds
Eutrophication
N & P
Negative effect WW on surface water:
Nutrients (N & P) cause algae growth
(autotrophic: C-source = CO
2)
During night these algae consume O
2
Fishes and other water organisms die due
to O
2-deficiency!
Biological WWT
Removal organic contaminants by
bacterial metabolism:
Substrate (= C-source) is “uptaken”
End-products:
•CO
2(anaerobic: CO
2+ CH
4)
•H
2O
•Minerals
•Biomass
Bacterial Metabolism
Substrate degradation:
Catabolism or Dissimilation = Energy Production
Anabolism or Assimilation = Biomass growth
•C
•N, P & Micronutrients
•Energy
Survival
Reproduction
Bacterial Metabolism:
Schematic
End Products
CO2 + H2O + Minerals
Biomass
Proliferation + Reserves
Substrate
C + nutrients
Energy
Cell
Maintenance
Catabolism
Anabolism
Survival
Reproduction
Aerobic versus Anaerobic
Oxygen = final e
—
-acceptor
Molecular: O
2
Bound: NO
3
-
(= de-N)
Activated sludge (flocks)
Final e
—
-acceptor = no O!
Carbon (= fermentation)
CO
2
Granular sludge (UASB, EGSB)
Main difference: O
2
Aerobic versus Anaerobic
Advantages
No post-treatment:
•Low effluent COD
•Bio-N and -P removal
No heating:
•range = 0 -40 °C
Very robust process
Short start-up
Disadvantages
Post-treatment needed?
•COD-removal max. 85-90%
•No bio-N and -P removal
Heating needed?
•range = 25 -40 °C
More sensitive toxic
shocks
Relative long start-up
Aerobic versus Anaerobic
Disadvantages
Higher exploitation cost:
•Aeration energy
–1,1 kWh/kg bCOD
•Sludge disposal
–0,3 kg sludge/kg bCOD
Large footprint
•Relatively low V
b
•Sedimentation tank
No CH
4production
•No energy production
Advantages
Low exploitation cost:
•No aeration
–0,3 kWh/kg bCOD
•Sludge disposal
–0,05 kg sludge/kg bCOD
4 x smaller footprint!
•High V
b
•No sedimentation tank
CH
4production
•Energy source!
Energy Balance:
Schematic
Aerobic Anaerobic
Biochemical reactions with O
2as e
-
-acceptor:
More energy gain for biomass!!!
More sludge growth
No energy-rich end product
Aerobic
Wastewater Treatment
Design Parameters
Flow rate –Hydraulic Retention Time
Temperature 10
O
C –40
O
C
Organic loading (kg BOD or COD/day)
(kg BOD or COD/m³.day)
Sludge loading
(F/M: Food to micro-organisms)
(kg BOD or COD/kg DS or MLSS.day)
MCRT (days) Mean Cell Retention Time
= amount of sludge in reactor (kg)
daily amount of wasted sludge (kg/d)
Temperature 10
O
C –40
O
C
Nitrification / De-nitrification Rate. « mg
N/g MLVSS/hr »
Available MLVSS.
Reactors Volumes (Anoxic & Aeration).
Design parameters « Nutrient Removal
Aerobic WWT
In general:
Wastewater CO
2+ H
2O
Activated sludge Energy
For glucose:
5 C
6H
12O
6+ 30 O
230 CO
2+ 30 H
2O + energy
Micro-organisms use this energyfor:
Cell Processes/Cell Maintenance (Survival)
Biomass Growth (Reproduction)
Substrate Respiration = Oxidation of Substrate
Endogenous Respiration = Oxidation of Biomass Reserves
O
2
Fundamental Process Steps
Feeding + Aeration
Accumulation-Regeneration
•Reduce Filaments
Active Volume
Settling + Discharge
Passive Volume
Extra Process Steps
For biological nutrient removal:
N-removal
•deN in anoxic step
P-removal
•Alternation aerobic-anaerobic steps
For hybrid WWTP’s:
Settling Phase
•intermediate phase before discharge!
Rinse Phase
•sludge evacuation from effluent weir
Nitrification -Denitrification
Nitrification:
NH
4
+
+ 2 O
2 NO
3
—
+ H
2O + 2 H
+
Nitrosomonas & Nitrobacter: Autotrophs
Aeration
Denitrification (heterotrophs):
NO
3
—
+ 5 e-+ 6 H
+
½ N
2+ 3 H
2O
CH
2O + H
2O CO
2+ 4 e-+ 4 H
+
No O
2=No Aeration!!!
Easy biodegradable COD: Feeding + Mixing
Atmosphere
Phosphate Accumulat. Organism
PAO
Acinetobacter Species
Phosphate Release
+ C-uptake
Phosphate Uptake
+ C-metabolism
Net
P-uptake
Phosphate Accumulat. Organism
PAO
Phosphate Accumulat. Organism
PAO
Phosphate Accumulat. Organism
PAO
Following conditions affect the growth of PAO & Bio-P:
•High retention time and longer non aerated zones. Max retention time is
around 1 hour and sludge age of max. 1.5 –2 days (based on MLSS
concentration).
•High Temperature (> 28
O
C).
•Low TKN.
•Low BOD (VFA source). BOD to TP ratio is 20:1 or 7-10 mg/l of VFA are
required per 1 mg of phosphorus removed by bio-P.
•Low pH in aerobic zone (Keep pH > 7.2).
•Secondary release of phosphorus (i.e inside clarifiers) due to high RT.
•Minimize free O
2& NO
3recycle to anaerobic zone.
Formulas for Activated Sludge
Processes
Removal Efficiency (%):
COD, BOD, N, P, …
Correlation amount X Enters WWTP vs. amount X Discharged
X-removal efficiency (%) = (X
influent-X
effluent)/X
influent
MCRT (days):
Residence Time Sludge in Bioreactor
.
³/_/³__
³/³
mkgsludgeDMwasteddaymFlowWasteSludge
mkgDMreactormVtotal
daysMCRT
Formulas for Activated Sludge
Processes )
*08.0(
**
()1(*/
d
20)(T1-
20)(T
TRC
YK
BOD
TSS
YkgBODkgSSYobs
Sludge Production Observed Yield
Where;
: 1.07 (Temp. Coefficient)
Y : Assumed Yield, Normally 0.6 kg DS/kg BOD removed
K
d , T
-1
: Endogenous decaycoefficient = 0.072
TRC, d : Mean Cell Residence Time (Sludge Age)
T,
O
C : Actual Water Temperature
T.S.S., ppm : Total Suspended Solids at the inlet of Bio-Reactor
BOD, ppm : Biological Oxygen Demand
Sludge Production (kg/day)= Sludge Production Observed Yield X BOD removed
Sludge Production (kg/day)= WAS (kg/day)
Formulas for Activated Sludge
Processes
Rule of Thum.
Return Activated Sludge (RAS) = 0.5 to 1.5 Q
inf(m
3
/hr)
Calculation Formula:
Q
RAS= Q
inf* C / (C
r–C)
Cr : Sludge Concentration at the Bottom of the Clarifier.
C : MLSS Concentration in Aeration Tank
Q
inf: Inlet Flow Rate (it may include other flows beside the
raw water flow)
* Mass Balance is a Must for Correct Calculations.
Sludge Recirculation Rates
Formulas for Activated Sludge
Processes
Oxygen Demand BOD, T, Vol., CRT, MLVSS
A) Sustrate Respiration = 0.6 kg O
2/kg BOD * BOD load Removed
B) Endogenous Respiration = (0.0528 kg O
2/kg MLVSS/day)*Vol.*MLVSS*
(T-20)
where = 1.024
Oxygen Calculations in Extd. Aerat. & Low Loaded systems
Actual Oxygen Requirment “A.O.R”= Sub. Resp. + End. Resp.
S.O.T.R = (1/) * [C
s/ ((P(mmHg)/760)*(51.6/(31.6+T))* * (C
s-C
actual))] *
(20-T)
* A.O.R * k
1
Formulas for Activated Sludge
Processes
: 0.65 (Fine Air Diffusers) -0.95 (Surface Aeration)
: 0.95
Water Level : in m
T : 20 °C
C
actual : 1.0 to 2.0 mgO
2/L
C
20 : 9.07 mg O
2/l "Standard"
H.R.T : in hours
k
1 : 1.0 (Function of HRT > 18 hrthen K
1= 1)
: 1.024
Atmospheric Pressure : 760 mm Hg Standard
Atmospheric Pressure : 1013 mbar Standard
Atmospheric Pressure on site: 760 mm Hg Standard
Formulas for Activated Sludge
Processes
Air Required for Aeration = (S.O.T.R) / (Air Oxygen Content) / (S.O.T.E)
Air Oxygen Content = 0.23 kg O
2/m
3
air @ Standard Conditions
S.O.T.E = Supplier Curve / Data Sheet
Note:
-S.O.T.E is a function of water height
Trouble Shooting –Activated Sludge
Sludge Concentration
Degree of Activity
Settling Capacity
Range 3 –6 g/L (3,000 –6,000 mg/L)
•< 3 Stop wasting + Cause
SV/SVI
Prevention Better than Healing!
SVI > 200 ml/g
More Wasting-Longer Settling Phase-Chlorination???
Value (80 –120 ml/g) up to 150 ml/g
Trouble Shooting –Activated Sludge
D.O. ±0 mg O2/l Reduce Feed Flow
Cause?
•overload
•oil/fat
•sludge concentration too high
•failure aeration system
Aerobic WWTP’s
Lagoon
Biorotor
Biotower/Biofilter
Activated Sludge System
Suspended Growth
Reactor Configurations:
Activated Sludge Systems
Conventional Systems
Batch Reactors (Non-Continuous)
SBR
Hybrid Systems
LUCAS
UNI-TANK
Membrane BioReactor
Submerged MBR
Cross-Flow MBR
Sedimentation
Membrane
Filtration
Reactor Configurations:
Activated Sludge Systems
Plug-Flow
Oxidation Ditch/Carousel
Completely Mixed
Unitank
Lucas
®
…
Conventional
Systems
Hybrid
Systems
Conventional Systems
Fundamental Process Steps Separated in Time:
Separate Bioreactor plug-flow, oxidation ditch, …
Separate Settling Tank: High Footprint
Continuous Influent & Effluent Flow
Constant Water Level (Overflow Weir)
Accumulation-Regeneration: Selector, Reactor Design
Control-In-Place: Limited Flexibility
Sludge Recycle
Conventional Systems
Plug-Flow
Settlers
Plug-Flow
Buffer
Tank
SBR
1 Tank Reactor.
No Settling Tank.
Cyclic Operation
‘Aeration’ & ‘Sedimentation’ separated in time
Control In Time.
Variable Volume.
DEFINETION
Equence Batch reactor in which influent, effluent or both are fed to
the biological system in batches with time intervals.
SBR Advantages
No Sedimentation Tank
Compact System = Small Footprint
Process Flexibility: “Control-in-Time”
Adjustable Phase Length
Possible Continuous Influent & Effluent Flow
No Need for Sludge Recycle
Substrate Gradient by Time
Accumulation-Regeneration
Flock Formers Favored by Integrated selector effect
Modular Design
Reliable and Robust Treatment System
Cyclic Operation
Single cycle or more can be appllied to SBR system.
1 Cycle 2 Cycles 3 Cycles 4 Cycles
F + A 12 –18 hr 6 –8 hr 2 –5 hr 1 –3 hr
A 4 –10 hr 2 –4 hr 1 –4 hr 1 –3 hr
S ½ –1 hr ½ –1 hr ½ hr ½ hr
R 3 –5 min. 3 –5 min. 3 –5 min. 3 –5 min.
D 1 –1½ hr 1 –1½ hr 1 –1½ hr 1 hr
WAS ½ –1 hr ½ –1 hr ½ hr ½ hr
Id 10 –15 min.10 –15 min.5 –10 min.3 –7 min.
24 hr 12 hr 8 hr 6 hr
Multiple SBREffluent
A A
A
Decantationphase
Aerobic phase
MLSS is aerated
Influent
Aerobic phase
MLSS is fed/aerated
ToB
ToB
FromCFromC
PHASE 1
PHASE 2PHASE 3 Effluent
B
B
B
Decantationphase
Aerobic phase
MLSS is aerated
Influent
Aerobic phase
MLSS is fed/aerated
ToC
ToC
FromA
FromA
PHASE 3
PHASE 1
PHASE 2 Effluent
C
C C
Decantationphase
Aerobic phase
MLSS is aerated
Influent
Aerobic phase
MLSS is fed/aerated
ToAToA
FromB
FromB
PHASE 3
PHASE 1
PHASE 2
Thickening & Stabilisation
Thickening: Gravitary Volume Reduction
Stabilisation: Mass Reduction
Aerobic: Aeration (endogenous respiration)
Anaerobic:Sludge Digesters (fermentation)
Thickening & Aerobic Stabilisation can be
Combined in 1 Single Tank (time control)
CASE STUDY
Jubail Export Refinery STP
Design Parameters:
BOD Load = (350 mg/l) * (10,000 m
3
/day) / 1000
= 3,500 kg/day
TSS = 4,000 kg/day
TKN = 600 kg/day
P = 200kg/day
Alkalinity= 2,500 kg/day
CASE STUDY
Jubail Export Refinery STP
Sludge Design Parameters in Extd Aerat. & Low Loaded Systems:
MLSS = 3 g/l (3,000 mg/l)
MLVSS= 2.25 g/l (2,250 mg/l)
MLVSS/MLSS Ratio = 75%
Volume Load = 0.35 -0.40 kg BOD/m
3
/day
(0.3 –0.5 kg BOD/m
3
/day)
Volume Design Parameters:
CASE STUDY
Jubail Export Refinery STP
Design Formulas:
Calculated Volume = BOD Load / Volume Load
= (3,500 kg/day) / (0.35 kg BOD/m
3
/day)
= 10,000 m
3
F/M Ratio = BOD load / (Volume * MLVSS)
= (3,500 kg/day) / (10,000 m
3
* 2.25 kg/m
3
)
= 0.15 d
-1
(0.08 –0.2 d
-1
)
Retention Time = Volume / Flow
= (10,000 m
3
) / (416 m
3
/hr)
= 24 hr (18 –36 hr)
CASE STUDY
Jubail Export Refinery STP
Simulating Different Design Parameters:
Chosen Volume =7,000 m
3
F/M Ratio = BOD load / (Volume * MLVSS)
= (3,500 kg/day) / (7,000 m
3
* 2.25 kg/m
3
)
= 0.22 d
-1
(0.08 –0.2 d
-1
)
Retention Time = Volume / Flow
= (7,000 m
3
) / (416 m
3
/hr)
= 16.8 hr (18 –36 hr)
CASE STUDY
Jubail Export Refinery STP
Sludge Production in Extd Aeration & Low Loaded Systems:
Sludge Production BOD, TSS, T, Vol., CRT, MLSS
Rule of Thumb.
No Pre-Treatment >> Sludge Yield = 0.9 kg DS/kg BOD remov.
Pre-Treatment >> Sludge Yield = 0.6 –0.7 kg DS/kg BOD remov.
Notes:
-Coarse screening & Equalization are not Considered as
pretreatment steps.
-Pre-treatment (grit removal, primary clarification, multiflow,...)
CASE STUDY
Jubail Export Refinery STP
Sludge Production in Extd Aeration & Low Loaded Systems:
Sludge Production (WAS*):
= (0.9 kg DS/kg BOD remov.) *(3,500 kg BOD/day)
= 3,150 kg DS/day.
Sludge Concentration = 5.5 g/l (at the bottom of the clarifier)
Sludge Capcity = (3,150 kg DS/day)/(5.5 kg/m
3
) = 572.7 m
3
/day
WAS = 23.8 m
3
/hr 24 m
3
/hr
* WAS = Waste Activated Sludge
CASE STUDY
Jubail Export Refinery STP
Sludge Recirculation Rates:
Rule of Thum.
Return Activated Sludge (RAS) = 1 to 1.5 Q
inf(m
3
/hr)
Calculation Formula:
Q
RAS= Q
inf* C / (C
r–C)
Cr : Sludge Concentration at the Bottom of the Clarifier.
C : MLSS Concentration in Aeration Tank
Q
inf: Inlet Flow Rate (it may include other flows beside the
raw water flow)
* Mass Balance is a Must for Correct Calculations.
CASE STUDY
Jubail Export Refinery STP
Sludge Recirculation Rates:
Q
RAS= Q
inf* C / (C
r–C)
Q
RAS= (416 m
3
/hr) * (3 g/l) / (5.5 g/l –3 g/l)
= 499 m
3
/hr
Q
RAS= 1.2* Q
inf
% Re-circulation= 120%
Total Sludge in Sludge Draw-off Pipe = RAS + WAS
= 499 m
3
/hr + 24 m
3
/hr = 523 m
3
/hr
3 Duty Pumps each 185 m
3
/hr (RAS + WAS)
CASE STUDY
Jubail Export Refinery STP
Oxygen Calculations in Extd. Aerat. & Low Loaded systems:
Oxygen Demand BOD, T, Vol., CRT, MLVSS
A) Sustrate Respiration = 0.6kg O
2/kg BOD * BOD load Removed
=(0.6kg O
2/kg BOD) * (3,500 kg BOD/day) = 2,100 kg O
2/day
B) Endogenous Respiration = (0.1 kg O
2/kg MLVSS/day)*Vol.*MLVSS
= (0.1 kg O
2/Kg MLVSS/day)*(10,000 m
3
)*(2.25 kg MLVSS/m
3
)
= 2,250 kg O
2/day
C) Nitrification Respiration = (4.57 kg O
2/kg TKN) * (TKNload)
= (4.57 kg O
2/kg TKN)*(32.5 mg/l * 10,000 m
3
/day/1000)
= 1,485 kg O
2/day
TKN = TKN
inf–TKN
eff–Assim.TKN = 60 –10 –0.05 BOD
Actual Oxygen Requirment “A.O.R”=
Sub. Resp. + End. Resp. + Nitrif. Resp. –Denitrif. Resp.
A.O.R = 2,100 + 2,250 + 1,485 = 5,835 kg O
2/day
Correction Factor (C.F.)
Based on water salinity, type of wastewater, temp., water
height in aeration tanks, alklinity, retention time and altitude
this correction factor can be determined.
As rule of thum.
CASE STUDY
Jubail Export Refinery STP
Oxygen Calculations in Extd. Aerat. & Low Loaded systems:
Surface Aeration >>> 1.5
Air Diffusing System >>> 2
CASE STUDY
Jubail Export Refinery STP
Oxygen Calculations in Extd. Aerat. & Low Loaded systems:
A.O.R = 2,100 + 2,250 + 1,485 = 5,835 kg O
2/day
Saturation Oxygen Transfer Rate = S.O.T.R = A.O.R * C.F.
S.O.T.R = 5,835 kg O2/day * 1.5 = 8,752 kg O
2/day= 364 kg O
2/hr
Aerator Efficiency:
-High Speed >> 1.5 –1.8 kw/kg O
2/hr (floating or fixed)
-Low Speed >> 1.8 –2.2 kw/kg O
2/hr (floating or fixed)
Aeration Power= (364 kg O
2/hr) / (1.5 kw/kg O
2/hr)= 243 kw
Total Aerators No. = 8 Aerators
Aerator Power = 243 kw / 8 aerators = 30 kw/aeartor
Mixing Power = 30 kw/aerator * 30% extra = 39 kw/aerator
Based on the Market AvailabilityNearest Aerator was 45 kw.
CASE STUDY
Jubail Export Refinery STP
Pre-Treatment Design Key Factors:
1. Hourly Peak Flow
Hourly Peak Flow m
3
/hr. = Avg. m
3
/hr x Peak Factor
Avg. Flow : 10,000 m
3
/day >>>> 416.6 m
3
/hr
Peak Factor: 2
Peak Flow = (416.6 m
3
/hr) x (2) = 833 m
3
/hr
Why Peak Flow is Important?
>>> To determine the sizing of the following:
-Screening Channels and Screens,
-Equalization Tank HRT, Volume and Air Required.
-Grit Removal Units Volume and Surface Loads (not existing)
-Primary Clarifiers Area and Rising Velocities, (not existing)
-................etc
CASE STUDY
Jubail Export Refinery STP
Pre-Treatment Design Key Factors:
2. Total Suspended Solids Load
TSS Loads = (TSS in mg/l) X (Peak Flow m
3
/hr) / 1000 = kg/hr
Why TSS LOAD is Important?
>>> To determine the sizing of the following:
-Types of Screens (Bar, Coarse, Fine)
-Equalization Tank Aeration Type (Diffused Air, Surface Aeration,....etc)
-Grit Removal Units Volume and Surface Loads to determine Grit/TSS
ratio (N/A)
-Primary Clarifiers Type (wiht coag/flocc., with or without lamella
settlers, dense sludge recirculation,.....etc), (N/A)
-Primary Clarifiers Area and Solid Rate, (N/A)
-................etc
CASE STUDY
Jubail Export Refinery STP
Pre-Treatment Design Key Factors:
3. Fat, Oil & Grease (F.O.G)
F.O.G. Loads = (F.O.G in mg/l) X (Peak Flow m
3
/hr) / 1000 = kg/hr
Why F.O.G TYPE & LOAD are Important?
>>> To determine the following:
-Higher F.O.G Loads Require F.O.G Removal System. (N/A)
-Types of F.O.G Removal System (DAF, Static Floatation,...etc)(N/A)
-Equalization Tank Aeration Type (Diffused Air, Surface Aeration,....etc)
-Primary Clarifiers Type (wiht acidific./coag/flocc., with or without lamella
settlers, dense sludge recirculation,.....etc), (N/A)
-Primary Clarifiers Area and Solid Rate, (N/A)
-................etc
CASE STUDY
Jubail Export Refinery STP
4-Type of Suspended Solids.
5-Temperature (max. 40
O
C) >>>>> Material Selection (Industrial Waste)
6-pH >>>>> Material Selection (Industrial Waste)
N/A: Indicates that the specified unit or application does not exist in this case
study plant.
Pre-Treatment Design Key Factors: