Wastewater Treatment
Wastewater Treatment
Wastewater Treatment
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
Aerobic
Activated Sludge Nitrification
Denitrification Methanogenesis
- 200 mV
0 mV
+ 200 mV
All biological treatment processes are conversion processes in that they convert the readily biodegradable 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.
Construction
Anaerobic processes construction more simple than aerobic processes
Hydrolysis
Acidogenesis 35% 20%
H2 and CO2
28% Methanogenesis
Acetate
72%
Adequate macronutrients
Nitrogen concentration in the reactor must ranging 40 70 mg/L Sulfide precursor should be added, commonly in the sulfate form
Temperature
Commonly anaerobic reactor are operated at mesophilic temperature of 30 370 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
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 nondegradable fraction at the intercept
Organic C + O2
microbes
New Aerobic
+ O2
microbes
New Aerobic
+ O2
microbes
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
energy
C 60H87 N12O23P
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 60H87 N12O23P
microbes
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.
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 ponds 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.
Aeration Tank Q0 , S0
Sedimentation Tank Q0 - QW
V, S, X
Recycle QR , S2 , XR
Wastage QW, XW = XR
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
(MLVSS)
the average time a microorganism spends in the treatment process 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. settled mixed liquor collected in the clarifier underflow and returned to the aeration basin
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 an ideal mixing situation where the contents of the aeration tank are at a uniform concentration an ideal situation where the contents of the aeration tank flows along the length of the tank
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 an ideal mixing situation where the contents of the aeration tank are at a uniform concentration 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
low rate loading that is one half to one tenth of the conventional loading rate 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
SETTLEABILITY
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
Surface Aerator
Mechanical Aerator Floating or fixed Brush Injector Others
Injector aerator
Brush 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, foodto-microorganisms ratio, and the mean cell residence time.
Effluents Characteristics
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 . S0/V . X = S0 / . X
F/M = food to microorganisms ratio, h-1 S0 = Influent BOD or COD concentration, mg/L Q = Influent wastewater flowrate, m3/h V = Aeration tank volume, m3 = hydraulic detention time, V/Q, h X = concentration of volatile suspended solid (MLVSS) in aeration tank, mg/L
c = Vr . X /(Qw . X w + Qe . Xe ) c = the mean cell residence time based on aeration tank, d Vr = aeration tank volume, m3 X = concentration MLVSS in aeration tank, mg/L Qw = excess sludge flowrate, m3/d Xw = concentration of MLVSS in excess sludge, mg/L Qe = effluent flowrate, m3/d Xe = concentration of MLVSS in effluent, mg/L The organic loading parameter and others design parameter for activated sludge processes listed in table 1.
F/M
Kg BOD/ kg MLSS . d
Volumetric Loading
Kg BOD/m 3 .d
MLSS
mg/L
V/Q
hrs
Q R/Q
5-15 5-15
0.2-0.6 0.20.4
1500-3000 2500-4000
4-8 3-5
0.25-0.75 0.25-1.0
Step Feed
Modified Aeration Contact Stabilization Extended Aeration
5-15
0.2-0.5 5-15 20-30
0.2-0.4
1-5-5.0 0.2-0.6 0.05-0.15
0.64-0.96
1.2-2.4 0.96-1.20 0.16-0.40
2000-3500
200-1000
3-5
1.5-3.0
0.25-0.75
0.05-0.25 0.5-1.5 0.5-1.5
Proses
hrs
F/M
Kg BOD/ kg MLSS . d
Volumetric Loading
Kg BOD/m3..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
High Purity Oxygen Oxidation Ditch
5-10
3-10 10-30
0.4-1.5
0.25-1.0 0.05-0.3
1.6-16.0
1.6-3.2 0.08-0.48
4000-10000
2000-5000 3000-6000
2-4
1-3 8-36
1.0-5.0
0.25-0.5 0.75-1.5
Microorganism and Substrate Balance The term V . MLSS is function of Solid Retention Time (SRT) or 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 X0 - V rg - (Qw Xw + Qe Xe ) where V volume of aeration tank, m3 dX/dt = rate of change of microorganisms concentration (VSS), mg /(l . m3 . d) Q = influent flow, m3/d X0 = microorganisms concentration (VSS) in influent, mg/L X = microorganisms concentration in tank, mg/L rg = net rate of microorganism growth (VSS), mg/(l . d) The net rate of bacterial growth is expressed as rg = Y rsu kd X
where Y = maximum yield coefficient growth, mg/mg over finite period of log rsu = substrate utilization rate, mg/(m3 d) kd = 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 Xw + Qe X e VX = - Y rsu X - kd
1 c
- Y rsu
=
X
- kd
The term 1/ is the net specific growth rate The term rsu can be computed from the following equation r = Q/V (S0 - S) = (S0 - S)/
where S0 - S mass concentration of substrate utilized, mg/L S0 substrate concentration in influent, mg/L S substrate concentration in effluent, mg/L hydraulic retention time, day
Effluent microorganism and substrate concentration The mass concentration of microorganism X in the aeration tank Y (S0 - S) m (S0 - S) X = = (1 + kd c) k (1 + kd c) Where 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 + kd c) S = c (Y k - kd) 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 (c), and the food to microorganism ratio (F/M) is commonly used for activated sludge process design and process control.
Aeration Tank
Clarifier
Influent
Effluent
Sludge Recycle
Wastage
Fig 6.
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
Effluent
Stabilized sludge
Stabilization Tank
Sludge Recycle
Wastage
Fig 7.
Recycle sludge
Wastage
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
Wastage
Recycle sludge
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.
Effluent Clarifier
Aeration rotor
Effluent
Fig 10.
Sludge Production
Design of the sludge handling and disposal facilities depends
on sludge produced. Quantity of sludge can be estimated by using the equation Px = Yobs . Q . (S0 - S )
Px = excess sludge (kg/d) Yobs = observed yield (g/g) S0 = Influent BOD or COD concentration, mg/l S = Effluent BOD or COD concentration, mg/l Q = Influent flowrate (m3/d)
The observed yield value can be computed using equation Yobs = Y/( 1 + kd . c ) Y = Cell yield coefficient (mg cell produced per mg organic matter removed) kd = endogenous decay coefficient, day-1 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
Carbonaceous oxygen demand can be defined as Kg O2/d = (total mass of BODL removed, kg/d) (1.42 . mass sludge wasted, kg/d) The BOD of the cell is equal to 1.42 times excess sludge Kg O2/d = (Q . (S0 - S )/F 10-3 mg/kg, kg/d) (1.42 . mass sludge wasted, kg/d) F = conversion factor for converting BOD5 to BODL If nitrification is desired O2 required to organic nitrogen oxidation should be added to carbonaceous oxygen demand Kg O2/d = (Q . (S0 - S )/F 10-3 mg/kg, kg/d) (1.42 . mass sludge wasted, kg/d) + 4.33 .Q . (N0 - N) 10-3 mg/kg, kg/d N0 = 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 m3/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 m3/d . 500 mg/l 10-6 kg/mg . 103 l/m3 = 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 300 C
Operational parameters and biological kinetic coefficients: Design mean cell residence time 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 BOD5 kd = 0.06/d BOD5 /ultimate BODU = 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/m3)0.044] = 112 m3/d
Step 3. Calculate flow, BOD, and TSS in primary effluent (secondary influent) Flow = design flow = 27,648 m3/d - 112 m3/d = 27,536 m3/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 m3/d = 161.5 g/m3 = 161.5 mg/l = S0 TSS = 7741 kg/d - 5186 kg/d = 2,555 kg/d = (2,555 kg/d . 1000 g/kg)/27,536 m3/d = 92.8 g/m3 = 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 BOD5 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 BOD5 = 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 = [c Q Y ( S0 - S)] / [X (1 + kd c ) c = 10 d Q = 27,536 m3/d (from Step 3) Y = 0.5 mg/mg S0 = 161.5 mg/L (from Step 3) S = 5.7 mg/L (from Step 4b) X = 2400 mg/L kd = 0.06 d-1 (10 d) (27,536 m3/d) (0.5) (161.5 5.7) mg/l V = (2,400 mg/l) ( 1 + 0.006 day-1 . 10 days) = 5,586 m3
Step 7. Determine the dimensions of the aeration tank Provide 4 rectangular tanks with common walls. Use widthto-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 m3 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 Vr X c = = SS wasting rate Q w Xw + Q e Xe
Qw = 270 m3/d
Step 9. Estimate the quantity of sludge to be wasted daily (a) Calculated observed yield Y 0.5 Yobs = = = 0.3125 1 + kd c 1 + 0.06 (10 days)
(b) Calculate the increased in the mass of MLVSS px = Yobs Q (S0 - S) . (1 kg/1000 g) = 0.3125 . 27,536 m3/d . (161.5 5.7) g/m3 . 0.001 kg/g = 1341 kg/d (c) Calculate the increased in MLSS (or TSS) pss = (1341 kg/d)/0.8 = 1676 kg/d (d) Calculate lost of TSS in effluent pe = (27,536 270) m3/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 = pss - pe = (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 + Qr) = 7440 . Qr Qr /Q = 0.476
Qr = 0.4762 . 27,536 m3/d = 13,110 m3/d = 0.152 m3/s Step 11. Check hydraulic retention time (HRT = ) = V/Q = 5586 m3/(27,536 m3/d) = 0.203 d = 4.87 h Note: The preferred range of HRT is 515 h. Step 12. Check F/M ratio using U in Eq. S0 - S (161.5 - 5.7) mg/l U = = = 0.32 day-1 X (0.203 day) (2,400 mg/l)
Step 13. Check organic loading rate and mass of ultimate BODu utilized Q S0 27,536 m3/d . 161.5 g/m3 Loading = = V 5586 m3 . 1000 g/kg = 0,80 kg BOD5/m3 d BOD5 = 0.67 BODu (given) BODu used = Q (S0S)/0.67 = [27,536 m3/d. (161.5 - 5.7) g/m3]/0.67 = 6403 kg/d
Step 14. Compute theoretical oxygen requirements The theoretical oxygen required is calculated from Eq. Q (S0 S) O2 = 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/m3 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 m3/d 1.202 kg/m3 . 0.232 g O2/g air (b) The actual air required at an 8% oxygen transfer efficiency Air = (16,200 m3/d)/0,08 = 02,000 m3/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
Microbial Growth
Exponential Phase: dX/dt = X X = cell concentration (mg dry mass or VSS/L) dX/dt = volumetric cell production rate (mg/L d)
rg = dX/dt = X rg = volumetric cell production rate (g VSS m-3 d-1) X = cell concentration (g VSS L-1) = the specific growth rate (d-1)
If part of the biomass produced is degraded with endogenous decay: rg = -Y rSU kd X With kd = endogenous coefficient decay (d-1) rg =Volumetric biomass production rate (g VSS/m3 d) -Y rSU = Rate of biomass production from substrate consumption kd X = Rate of biomass consumption by endogenous respiration
And = m S/(Ks + S) kd
Under a steady state in a well mixed reactor (CSTR) X = Xr and S = Sr (X and S = concentrations) and dX/dt = dS/dt = 0
Q, X0, S0
Q, X, S
V , Xr , Sr
This also means: Q (X X0) = PX = biomass production rate Q (S0 S) = PS = substrate consumption rate
V dXr/dt = cell in - cell out + cell produced = Q X0 Q X + rg V For simplification, X0 = 0 and by definition rg = X X V = X Q = Q/V By definition D = dilution rate = Q/V = = 1/HRT (Hydraulic Retention Time): The growth rate is dictated by the dilution rate.
Maximum Dilution Rate: Dmax CSTR: = D = Q/V At washout condition: S = S0 = Dmax KS/(mDmax) Dmax = m S0/(KS + S0) = m/(1 + KS/S0)
Cell wash-out occurs at too high dilution rates (D >Dmax) and is especially sensitive at low initial substrate
concentration.
Now = m S/(KS + S) kd S: The biomass balance is identical, hence: D = = m S/(KS + S) kd. Solving this equation gives: S = KS (D + kd)/(m D kd) S mass balance: V dS/dt = 0 = Q S0 Q S + rSU V rSU = -rg/Y kd X/Y = - X/Y-kd X/Y = - X(D + kd)/Y Solving this equation gives: X = Y D(S0 S)/(D + kd) = Y(S0 S)/(1 + kd/D) Dmax is obtained for S = S0 = KS (D + kd)/(m D kd) Solving this equation gives Dmax = m/(1 + KS /S0) - kd
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 concentrations in the effluent and reactor are similar to the nbVSS concentration in the influent (X0,i) The total mass of VSS in the Bioreactor includes the biomass produced (rg), the nbVSS introduced (X0,i) and the debris released from the endogenous decay: rVSS = total VSS production rate = rg + Q X0,i/V + fd (kd) X
rg = -Y rSU kdX = biomass production from bCOD (-Y rSU) minus endogenous decay (kd X) fd (kd) X = rate of cell debris production with fd = fraction of biomass remaining as cell debris. The cell debris production is directly proportional to the biomass concentration and kd
Q X0.i/V = Amount of nbVSS in the influent Q = influent flow rate, X0,i = influent nbVSS and V = reactor volume).
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 Xout Xin = Production, with Xout = Qe Xe + Qw Xw , Xin = Q0 X0
Expression of X
X can be obtained from S mass balance:
V dS/dt = Q0 S0 (Qe Se + Qw Sw) + rSU V = 0 With the assumptions that S = Sw = Se = Sr (S = bsCOD)
- kd 1 U
KS k
1 S
1 K
U : Substrate utilization rate, mg sCOD/mg VSS.d kd : Endogenous decay coefficient, 1/d S0 : Influent substrate concentration, mg sCOD/l S : Effluent substrate concentration, mg sCOD/l X : Biomass concentration, mg VSS/l : Hydraulic retention time, d KS : 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 kd from the intercept of the equation. Plotting 1/U or X/(S0 - S) versus 1/S, the biokinetic coefficients KS 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 = X V kd X V Qw X where V = reactor volume (l);
kd = biomass decay coefficient (day-1 ); Qw = wastage flow rate (s-1 ); t = time (s). At steady-state conditions, dX/dt = 0, hence, Eq. above becomes: = kd + QW V (a) Since the solid retention time (SRT) is defined as:
SRT = V X/QW X = V/QW (b) Substituting Eq. (b) into Eq. (a) results in: = kd + 1/SRT (c) Substituting for the value of from Eq. (c) into Monod
describes the steady-state condition for substrate concentration in the reactor: SRT 1 + SRT x kd . = KS m 1 S + 1 m (d)
plotting SRT/[1 + (SRT x kd)] versus 1/S, the biokinetic coefficients, m can be determined from the Y-intercept of the equation
In
continuous-flow
and
completely-mixed
reactor,
and kd is usually achieved by collecting data from labscale 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 m