Design of a Wastewater Treatment Plant
✅ Paper Type: Free Essay  ✅ Subject: Environmental Sciences 
✅ Wordcount: 12184 words  ✅ Published: 23rd Sep 2019 
Design of a Wastewater Treatment Plant
DESIGN OF SECONDARY BIOLOGICAL TREATMENT PLANT
Abstract
TO PROVIDE A DESIGN APPRAISAL FOR INCLUSION OF A SECONDARY TREATMENT PLANT THAT IS CURRENTLY EQUIPED WIH A PRIMARY TREATMENT PLANT
TABLE OF CONTENTS
1. INTRODUCTION
2.INTRODUCTION TO BIOLOGICAL NUTRIENT REMOVAL (BNR)
3. DESIGN DATA & PARAMETERS ASSUMED
3.1 DESIGN FLOW Q
3.2 PROCESSING TEMPERATURE
3.3 INFLUENT TKN
3.4 READILY BIODEGRADABLE CHEMICAL OXYGEN DEMAND
4. DESIGN PROCEDURE
5. BIOWIN MODELLING
6. CONCLUSION
7. REFFERENCES
APPENDEX
Wastewater Treatment Plant Design
1.INTRODUCTION
The Coastal Wastewater treatment plant currently has a Primary Treatment Plant only. A design appraisal has been requested for a secondary treatment plant. This report is to design a draft based on the influent characteristics of the waste water and designed for the give effluent characteristics.
The accumulation of nutrients in the waste water especially nitrogen and phosphorous leads to decrease in water quality. To control and treat the waste water various treatment methods have been tested. After careful consideration of various influent parameters like BOD, COD, AmmoniaN, TSS, Total Nitrogen and Total Phosphorous etc., the “Anerobicanoxicaerobic (A^{2}O)” process was employed due to its ease of working and diverse advantages.
2.Intoduction to Biological Nutrient Removal Process (BNR)
Biological Nutrient Removal Process was developed in the 1960s, it has been used since then due to is various advantages over other Chemical Process. BNR is a modification of Activated Sludge Treatment Process that incorporates an Anoxic and/or anerobic zones to enables removal of nitrogen and phosphorous from the wastewater.
“Anerobicanoxicaerobic” is a type of BNR which has a 3 stage Phoredox AnerobicAnoxicOxic (A^{2}O) system. The stages in this system are Anerobic system, anoxic system, and aerobic system. Each system treats a certain type of effluent. The treatment of phosphorous takes place in the anerobic stage whereas BOD gets treated in the aerobic stage by the process of nitrification. Denitrification process removes the nitrate in the anoxic stage. (Metcalf & Eddy, 2003). Any process has certain limitation, so does A^{2}O process. Some of the limitations of the process are, i) decrease in the reduction of the removal of the Phosphorous due to recycling of RAS along with nitrate in the anerobic stage. ii) decrease in removal of Nitrogen due to internal recycling ratio (Metcalf & Eddy, 2003). iii) reduction in activity of PAOs.
Fig shows the proposed A^{2}O process secondary treatment plant
3. Design data and Parameters assumed.
The designing of the Secondary wastewater Treatment Plant is based on the provided design data, Kinetic data and few important assumptions made. Some of the Kinetic data is incorrect so corrections had to be made to bring them in the recommended range as given in the Metcalf & Eddy, (2003).
3.1 Design Flow Q
To obtain the daily flow rate ‘Q’ for the give proposed capacity of 210,000 Equivalent Population (EP), an average daily consumption of 340 L/ep (Riverina Water County Council, NSW) was taken for calculation along with an 80% recovery factor. A peaking factor of 1.93 was taken (Harmon Coefficient). So, the flow rate, Q= 340 x .8 x 1.93 x 210,000 = 110,241.6 m^{3}/day.
3.2 Processing Temperature
Average processing temperature of 20^{0} was set for easy calculation therefore there was no need for corrections for Kinetic coefficients.
3.3 Influent TKN
The TKN value is taken in the form of NH4N
3.3 Readily Biodegradable Chemical Oxygen Demand (rbCOD)
A major design parameter rbCOD was not given, therefore the value of rbCOD is assumed from Metcalf & Eddy, 2003. 15% – 25% of bCOD was taken at the end. To consider for the rbCOD consumed in anerobic stage, 25% was assigned for removal of Phosphorous removal and 10% was assigned for denitrification.
Table containing the design data and the assumed parameters
INFLUENT PARAMETERS 
EFFLUENT PARAMETERS 

Parameter 
Unit 
Value 
Parameter 
Unit 
Value 
Alkalinity 
mg as CaCO_{3}/L 
269 
BOD 
mg/L 
21 
COD 
mg/L 
201 
TSS 
mg/L 
4 
pH 
– 
7.78 
AmmoniaN 
mg/L 
1 
TSS 
mg/L 
98 
Total N 
mg/L 
27 
VSS 
mg/L 
80 
Total P 
mg/L 
6 
AmmoniaN 
mg/L 
34.7 

Total N 
mg/L 
48.2 

Total P 
mg/L 
7.63 
PARAMETER 
UNITS 
VALUE 
REMARKS 
Kinetic Data 

Y 
g MLVSS/g BOD removed 
0.60 
Given data 
K_{1} 
kg BOD utilized/kg VSS.day 
10 
Given data 
K_{d} 
day^{1} 
0.1 
Given data 
K_{s} 
mgBODµ/L 
20 
Given data 
Kinetic Constant for Nitrifying Bacteria 

µNmax 
d1 at 20°C 
0.7 
Given data 
K_{N} 
mg NH4 + as N /L 
0.7 
Given data 
Y_{N} 
Kg VSS/ Kg NH4+ as N nitrified 
0.12 
Given data 
Assumptions 

MLSS 
mg/L 
2400 
Given data 
Water consumption/per capita 
L/ep. Day 
340 
(Riverina Water County Council, NSW) 
Recovery factor 
Percentage 
80 

Peaking factor (Harmon coefficient) 
1.93 

rbCOD/bCOD 
— 
0.25 
(Metcalf & Eddy) 
rbCOD/bCOD 
— 
1.6 
(Metcalf & Eddy) 
FS (TKN peak /TKN average) 
— 
1.5 
(Metcalf & Eddy) 
DO 
mg/L 
2 
(Metcalf & Eddy) 
BOD/COD Ratio 
0.5 


Average process temperature 
^{0}C 
20 

K_{o} 
g/m3 
0.5 
(Metcalf & Eddy) 
K_{dn} 
g VSS/g VSS.d 
0.08 
(Metcalf & Eddy) 
F_{d} 
— 
0.15 
(Metcalf & Eddy) 
Nox/TKN 
— 
0.78 
(Metcalf & Eddy) 
VSS/TSS 
— 
0.8 
(Metcalf & Eddy) 
Px,bio = Px,vss 

rbCOD/NO3N 
g/g 
6.6 
(Metcalf & Eddy) 
rbCOD/P 
g/g 
8 
(Metcalf & Eddy) 
Detention time in the anoxic tank 
H 
1 
(Metcalf & Eddy) 
4. Design Procedure
The various design data was changed to need for calculation purpose. Some of the kinetic constants were assumed. The assumed constants and other design data are mentioned in the above Table. Some of the missing data were also assumed for calculations. All these data and constants were obtained for “Metcalf & Eddy,2003”.
The design process was carried out in 3 stages. The first stage was the design for BOD and Nitrification. The second stage was for removal of Phosphorous. The third stage was designed for Denitrification process. An Activated Sludge process, single sludge bioN removal process and an Anoxic/Aerobic process design was considered. (Metcalf & Eddy,2003).
BOD & Nitrification
The BOD and Nitrification removal process by Activated Sludge process was carried out by the following steps.
1. The DO concentration was assumed to 2 mg/L to calculate growth rate aimed at Nitrification µ_{n }by means of Eq.1. The nitrification factor of 1.5 was taken. Design SRT was computed using Eq.3 (Metcalf & Eddy,2003).
2. Using the Eq.4 the max Specific growth rate µ_{m} was calculated. Based on the Kinetic constants Y & K1, the effluent substrate (S) was calculated by means of Eq.5
3. P_{x},_{bio }(Biomass Production) was calculated by means of Eq.9 grounded on the ideals of A, B and C using Eqs. 6, 7, 8. The NO_{x} was assumed to be approximately equal to 0.8 TKN (Metcalf & Eddy,2003). But the real value was calculated using the Eq. 10. The obtained value was found to be similar to that of the assumed value.
4. The mass of VSS and TSS was calculated using Eqs. 12 and 13. The concentration of VSS was taken to be same as P_{x},_{bio }as there was no info about the _{nb}VSS. The concentration of TSS was calculated using Eq.11 presuming that VSS = 80% TSS (Metcalf & Eddy,2003).
5. The volume of the aeration tank was figured by Eq.14 for the given MLSS. Four number of basins were assumed with a depth of 6.5m and a width ratio of 1.5 : 1(Metcalf & Eddy,2003).
Phosphorous Removal
The removal of Phosphorous takes place in the anerobic stage in the A^{2}O Process. The following are the procedures of the removal:
1. The determination of the _{rb}COD available for the removal of Phosphorous, Eq.25 was used. The nitrate mass balance was performed at the influent using Eq.22 bearing in mind no NO3 concentration in the influent. Taking _{rb}COD/nitrate = 6.6, _{b}COD/BOD= 1.6 and _{rb}COD/_{b}COD = 0.25 equivalent _{rb}COD was computed (Metcalf & Eddy,2003).
2. The P removed by Biological Phosphorous Removal was computed by Eq.26 taking 8g of _{rb}COD/g Phosphorous is removed BPR (Metcalf & Eddy,2003).
3. The concentration of Phosphorous removed was calculated by Eq.29 taking that the concentration of Bio P removed and the concentration of P used for biomass growth.
Denitrification
The anoxic stage is responsible for the denitrification. A recycle stream from aerobic zone with oxygen form nitrification being removed is considered. The design process carried out is as follows;
1. The biomass concentration was calculated by Eq.32, the volume of nitrite nursed into the anoxic tank is computed by Eq.33, Eq.34 and Eq.35. The volume of anoxic zone is computed by Eq.36 for a suitable detention time.
2. The amount of nitrate reduced was calculated by Eq.38. when the nitrate fed is not equal to the nitrate reduced, detention times must be varied. Specific Denitrification Rate as a function of MLSS with observed were computed, the SDNR was got using Eq.39.
3. The oxygen credit and the net oxygen needed was calculated using Eq.40 and Eq.41. The air flow rate was computed using Eq.42 and Eq.43. The equation 43 was used to convert mass flow rate to volumetric flow rate.
4. The required alkalinity was as conc of CaCO_{3} was computed by Eq.46 and Eq.47. It is an important factor as some of the reactions have the propensity to change the pH which impacts the system performance.
Design of Secondary Clarifier
Secondary clarifier was needed to settle and remove the suspended solids that include the nutrients. The design is a 3step process, defining the return sludge recycle ratio, determine the clarifier size and loading volume of solid. The area of clarifier was determined using Eq.54 as a purpose of design flowrate and supposed hydraulic application rate. The diameter of the basin was designed based on the number clarifiers. The loading volume of solids was computed based on the Eq.58. The number of clarifiers varies based on the loading rate of solids.
5.BioWin Modeling:
BioWin is simulator used in designing of a wastewater treatment process, that bonds together the biological, chemical and the process models. BioWin is used all over the world to design, advancement and enhance wastewater treatment plants of all types. The core of BioWin is an exclusive biological model which is accompanied with other process models (e.g. water chemistry models for calculation of pH, mass transfer models for oxygen modelling and other gasliquid interactions).
If you need assistance with writing your essay, our professional essay writing service is here to help!
Essay Writing ServiceThe Biowin modeling was used to validate the theoretical values obtained. It was found that the theoretical values satisfied curtained parameters but did not satisfy other parameter. The model was edited to obtain the required effluent parameters. Parameter like Ammonia and COD obtained by the theoretical calculations were not in the required effluent range. Hence the model was reconfigured, like the theoretical volume of the aeration tank was very less and thus Ammonia couldn’t be reduced to the required effluent range. Therefore, the volume of the Aeration tank was increased by trial and error method to achieve the effluent standards.
The kinetic constants were entered. Mixers were used to mix the influent before it was inlet into the aeration tank. Splitters with a constant of 0.6 were used to recycle the RAS back to the Aeration tank from the Clarifiers. The treated biowaste was collected as sludge and was disposed. All the given Influent parameters were reduced successfully in the Biowin model.
Changes made in the design for attaining the given range was very sensitive. Change in dimensions in the Aeration tank or the clarifiers had subtle changes in the output.
6.Conclusion
To achieve the given effluent values for the influent values an AnerobicAnoxicAerobic process was adopted for the design purpose. The 3stage process consisted of BOD removal, Nitrification, Phosphorous removal & Denitrification in addition to design of clarifiers for settling and removal of biological suspended solids. Due to insufficient data given for designing several kinetic data and certain values were assumed as per requirement.(Metcalf & Eddy 2003). A few changes were made to the obtained values from the theoretical calculations to achieve the required effluent parameters.
The anticipated effluent values were achieved based on the proposed design. The Treatment plant can be implemented based on the proposed design and will have a high success rate in removal of the influent parameters.
APPENDEXA
ASSUMPTIONS 

PARAMETERS 
UNIT 
VALUE 
REMARKS 
Depth of basin 
m 
6.5 
(Metcalf & Eddy) 
Width to depth ratio 
1.5 : 1 
(Metcalf & Eddy) 

CS,20 
mg/L 
9.08 
(Metcalf & Eddy) 
Α 
— 
0.65 
(Metcalf & Eddy) 
Β 
— 
0.95 
(Metcalf & Eddy) 
F 
— 
0.9 
(Metcalf & Eddy) 
CS,TH=CS,TH=CS,20 
(Metcalf & Eddy) 

Fine bubble ceramic diffusers with an aeration clean water O2 transfer efficiency 
— 
35% 
(Metcalf & Eddy) 
Required alkalinity to transform ammonium to nitrate 
g CaCO3/g NH4N 
7.14 

Residual alkalinity concentration to maintain pH in range 6.87 
g/m3 
80 
(Metcalf & Eddy) 
Influent alkalinity 
g/m3 as CaCO3 
140 
(Metcalf & Eddy) 
Return sludge mass concentration 
g/m3 
8 

Design MLSS XTSS concentration 
g/m3 
3 

Hydraulic application rate 
m3/m2.d 
22 
(Metcalf & Eddy) 
Number of clarifiers 
3 

Phosphorus content of heterotrophic 
g P/g 
0.015 
(Metcalf & Eddy) 
Alkalinity as CaCO3 produced per NO3N Oxidized 
g/g 
3.57 
(Metcalf & Eddy) 
APPENDEXB
SUMMARAY OF RESULTS
PARAMETERS 
UNIT 
VALUE 
REMARKS 
BOD REMOVAL AND NITRIFICATION 

Average wastewater flow 
m3/d 
110,241.6 
1ep =340 L/day 
Average BOD load 
kg/d 
100.5 

Average TKN load 
kg/d 
18 

Aerobic SRT 
d 
3.198 
>7, (Metcalf & Eddy) 
Aeration tanks 
4 
4 tanks (Metcalf & Eddy) 

Aeration tank volume, each 
m3 
4600 

Aeration Tank Length 
M 
31 
approx. 150m per tank, (Metcalf &eddy) 
Aeration Tank Width 
M 
15 
W:D ratio 1.5, (Metcalf & Eddy) 
Aeration Tank Depth 
m 
10 
Assumed 
Hydraulic retention time 
h 
6.5 

MLSS 
g/m3 
2400 

MLVSS 
g/m3 
1582.642 

F/M 
g/g.d 
0.593 
0.041, (Metcalf & Eddy) 
BOD loading 
kg BOD/m3.d 
0.939 
0.041, (Metcalf & Eddy) 
Observed yield 
kg TSS/kg bCOD 
0.810 

kg VSS/kg BOD 
1.296 

Oxygen required 
kg/h 
1393.62 

Air flowrate at average wastewater flow 
m3/min 
396.50 

RAS ratio 
— 
0.6 

Clarifier hydraulic application rate 
m3/m2.d 
22 
1624, (Metcalf & Eddy Table 8.7) 
Clarifiers 
Nos 
3 

Diameter ,m 
46 

Alkalinity addition as Na(HCO3) 
kg/d 
27410.47 

BIOLOGICAL PHOSPHOROUS REMOVAL 

P used for biomass growth 
g/m3 
1.22 

P removed 
mg/L 
5.7387 

P content of waste sludge 
% 
7.146 

DENITRIFICATION 

Effluent NO3N 
g/m3 
6 

Internal recycle ratio 
6.266 

RAS recycle ratio 
0.6 

Anoxic volume 
m3 
6429.98 

Overall SDNR 
g NO3N/g MLSS.d 
0.140 

Detention time 
h 
1.2 

Alkalinity required 
kg/d as CaCO3 
16316.08 

EFFLUENT QUALITY PARAMETERS 

Effluent BOD 
mg/L 
19.02 
Target value achieved 
Effluent NH4N 
mg/L 
0.17 
Target value achieved 
Effluent N 
mg/L 
15.37 
Target value achieved 
Effluent P 
mg/L 
3.92 
Target value achieved 
Effluent TSS 
mg/L 
0 
Target value achieved 
APPENDEX – C
NOMENCLATURE
Q: influent wastewater flowrate (m^{3}/d) 
Ko: oxygen inhibition coefficient, g/m^{3} 
µ_{m}: Maximum specific growth rate (d) 
BOD: Biological oxygen demand (mg/L) 
DO: Dissolved oxygen, mg/L 
Fd: cell debris fraction (unitless) 
TSS: Total suspended solids (mg/L) 
µ_{Nmax}: Maximum specific growth rate of nitrifying bacteria, g new cells/g cells . d 
MLSS: mixedliquor suspended solids, mg/L 
TKN: influent TKN concentration (mg/L) 
µ_{m}: Maximum specific growth rate, (d) 
P_{X,bio}: Biomass production (kg VSS/d) 
Ne: effluent NH4N concentration, mg/L 
µn: Specific growth rate for nitrification (d^{1}) 
P_{X,VSS}: Solid production as VSS (kg/d) 
Total P: Total phosphorus (mg/L) 
µ: Specific growth rate (d1) 
P_{X,TSS}: Solid production as TSS (kg/d) 
NH4N: Ammonia as Nitrogen (mg/L) 
SRT: Solid retention time (d) 
Fraction VSS: fraction of VSS over TSS, unitless 
Y: Heterotrophic yield coefficient (kg VSS produced/Kg BOD 
FS: Safety Factor 
X_{VSS} * V: Mass of VSS (kg) 
K1: Maximum specific substrate utilization rate (Kg BOD/Kg VSS.d) 
So: Influent substrate concentration (mg/L) 
X_{TSS} * V: Mass of TSS (kg) 
Kd: Microbial decay coefficient (d1) 
S: Effluent substrate concentration (mg/L) 
V: Total volume of aeration tanks (m3) 
Ks: Saturation coefficient (mg/L) 
A: heterotrophic biomass, kg/day 
Ƭ: Detention time (h) 
Y_{N}: Nitrifier yield coefficient (Kg VSS produced/Kg NH4+ – N nitrified) 
B: cell debris, kg/d 
Lorg: Volumetric BOD (kg/m3.d) 
K_{dn}: Endogenous decay coefficient for nitrifying organisms (g VSS/g VSS∙d) 
C: nitrifying bacteria biomass, kg/day 
Y_{obs,TSS}: Observed yield base on TSS (g TSS/gBOD) 
K_{N}: Halfvelocity constant, substrate concentration at onehalf the maximum specific substrate utilization rate (g/m^{3}) 
D: Nonbiodegradable VSS in influent, kg/day 
Y_{obs,VSS}: Observed yield base on VSS (g VSS/gBOD) 
rbCOD: Readily biodegradable chemical oxygen demand (mg/L) 
TSSo: influent wastewater TSS concentration (mg/L) 
NOx influent: Concentration of NH4N in the influent flow that is nitrified (mg/L) 
N: nitrogen concentration, g/m^{3} 
VSSo: influent wastewater VSS concentration (mg/L) 
β: Salinitysurface tension correction factor (unitless) 
X_{b} : active biomass concentration (mg/L) 
F/M_{b}: BOD F/M ratio based on activated biomass concentration (gBOD/g biomass .d) 
F: fouling factor (unitless) 
k_{d}: endogenous decay coeff. (1/day) 
NOr: nitrate removed (g/d) 
C_{S,T,H}: Oxygen saturation concentration in clean water at temperature t and altitude h (mg/L) 
NO_{x effluent}: nitrogen oxides in the effluent (mg/L) 
SDNR (MLSS): specific denitrification rate referred based on MLSS (g NO3N/g MLVSS. d) 
C_{S,20}: Dissolved oxygen saturation concentration in clean water at 20C and 1 atm or 760 mmHg (mg/L) 
IR : internal recycle ratio (unitless) 
R_{1}: Oxygen credit (kg/h) 
C_{L} : operating oxygen concentration 
Q_{anoxic} : flow rate to anoxic tank (m^{3} /d) 
R_{o}: Net oxygen required (kg/h) 
E: diffusers oxygen transfer efficiency 
NOx feed: NO_{3}– N fed to the anoxic tank (kg/d) 
AOTR: actual oxygen transfer rate under field conditions(kg O_{2}/h) 
Alk _{produceddenitrification}: alkalinity produced in denitrification (g/m^{3}) 
Vnox: volume anoxic tank (m^{3}) 
SOTR: Standard Oxygen Transfer Rate in Tap Water at20°C and zero dissolved oxygen (kg O_{2}/h) 
NOx RAS: nitrogen oxides in the effluent in the return activated sludge (mg/L) 
SDNR: Specific denitrification rate (g NO3N /g MLVSS.d) 
α: Oxygen transfer correction factor for waste (unitless) 
D: diameter (m) 
Qr: RAS flowrate (m3/d) 
R: return activated sludge (RAS) recycle ratio (unitless) 
A: total area of clarifier (m^{2}) 
Xr: Return sludge mass concentration (g/m3) 
X: Mixedliquor suspended solids (mg/L) 

APPENDEX – D
FORMUAL USED
EQUATION NO 
FORMULA 
EQUATION NO 
FORMULA 
BOD REMOVAL AND NITRIFICATION 

1 
$$ =3.198 d Max Specific Growth rate µ_{m} = K_{1} * Y =10 *0.6 = 6 Effluent Substrate Conc (BOD Effluent) $S=\frac{{K}_{s}\left(1+{k}_{d}.\mathit{SRT}\right)}{\mathit{SRT}\left({\mu}_{m}\u2013{k}_{d}\right)\u20131}$ $S=\frac{20\left(1+0.1*3.198\right)}{3.198\left(6\u20130.1\right)\u20131}$ S=1.47726 mg/L Biomass Production A (heterotopic Biomass) $A=\frac{\mathit{QY}({S}_{o}\u2013S(\frac{1\mathit{kg}}{{10}^{3}g})}{1+\left({k}_{d}\right)\mathit{SRT}}$ $=\frac{11024106*0.6(100.5\u20131.4772}{1+\left(0.1\right)3.198}$ =4962.760 Kg/d B (Cell Debris) $B=\frac{\left({f}_{d}\right)\left({k}_{d}\right)\mathit{QY}\left({S}_{O}\u2013S\right)\mathit{SRT}\left(\frac{1\mathit{kg}}{{10}^{3}g}\right)}{1+\left({k}_{d}\right)\mathit{SRT}}$ $B=\frac{\left(0.15\right)\left(0.1\right)110241.6*06\left(100.15\u20131.4772\right)3.198}{1+\left(0.1\right)198}$ =396.772 Kg/d C (Nitrifying Bacteria Biomass) $C=\frac{\mathit{QY}\left({\mathit{NO}}_{x}\right)\left(\frac{1\mathit{kg}}{{10}^{3}g}\right)}{1+\left({k}_{\mathit{dn}}\right)\mathit{SRT}}$ $C=\frac{110241.6*0.12(47.1994)}{1+\left(0.1\right)3.198}$ =477.10325 Kg/d ${P}_{X,\mathit{bio}}=A+B+C$ ${P}_{X,\mathit{bio}}=4962.760+396.772+477.10325$ =5836.6352 Kg VSS/d Oxidation of Ammonia to Nitrate ${\mathit{NO}}_{x}=\mathit{TKN}\u2013{N}_{e}\u20130.12\bullet {P}_{X,\mathit{bio}}/Q$ ${\mathit{NO}}_{x}=48.2\u20131\u20130.12\bullet 5836.6352/110241.6$ =47.1994 g/m^{3} Concentration of VSS and TSS ${P}_{X,\mathit{TSS}}=\frac{A}{0.85}+\frac{B}{0.85}+\frac{C}{0.85}+D+Q\bullet ({\mathit{TSS}}_{o}\u2013{\mathit{VSS}}_{o})$ ${P}_{X,\mathit{TSS}}=\frac{4962.760}{0.85}+\frac{396.772}{0.85}+\frac{477.1832}{0.85}+0+110241.6\bullet (98\u201380/1000)$ =8850.97 Kg/d Mass of VSS and TSS $\left({X}_{\mathit{VSS}}\right)\bullet \left(V\right)=\left({P}_{X,\mathit{VSS}}\right)\bullet \mathit{SRT}$ $\left({X}_{\mathit{VSS}}\right)\bullet \left(V\right)=(5836.6352)\bullet 3.198$ =18665.55 Kg $\left({X}_{\mathit{TSS}}\right)\bullet \left(V\right)=\left({P}_{X,\mathit{TSS}}\right)\bullet \mathit{SRT}$ $\left({X}_{\mathit{TSS}}\right)\bullet \left(V\right)=(8850.97)\bullet 3.198$ =28305.402 Kg Design of Aeration Tank $V=\frac{\left({X}_{\mathit{VSS}}\right)\bullet \left(V\right)}{\mathit{MLSS}}$ $V=\frac{28305.402}{2400}$ =11793.91 m^{3} Assumed no of Tanks = 4 Volume of basin = V/4 =11793.91 / 4 =2948.44 m^{3} Assumed Depth = 6.5 Width to depth ratio = 1.5 : 1 Width = 6.5*1.5 = 9.5 =10 m Length of the tank = Volume of tank /width * depth =2948.44/ 10*6.5 =46 m Detention time $\tau =\frac{V}{Q}$ $\tau =\frac{29900}{110241.6}$ =6.5 h Fraction of VSS $\mathit{Fraction\; VSS}=\frac{\left({X}_{\mathit{VSS}}\right)\bullet \left(V\right)}{\left({X}_{\mathit{TSS}}\right)\bullet \left(V\right)}$ $\mathit{Fraction\; VSS}=\frac{18665.55}{28305.402}$ =0.659 MLVSS MLVSS=Fraction VSS*MLSS =0.659*2400 =1582.642 g/m^{3} Volumetric BOD ${L}_{\mathit{org}}=\frac{Q*{S}_{O}}{V}$ ${L}_{\mathit{org}}=\frac{110241.6*100.5}{11793.91}$ =0.939 Kg/m^{3}.d Observed Yield based on VSS ${Y}_{\mathit{obs},\mathit{TSS}}=\frac{{P}_{X,\mathit{TSS}}}{Q\bullet ({S}_{O}\u2013S)}$ ${Y}_{\mathit{obs},\mathit{TSS}}=\frac{8850.97}{110241.6\bullet \left(100.5\u20131.4774\right)}$ =0.810 gTSS/g bCOD ${Y}_{\mathit{obs},\mathit{VSS}}={Y}_{\mathit{obs},\mathit{TSS}}*\frac{\mathit{VSS}}{\mathit{TSS}}$ ${Y}_{\mathit{obs},\mathit{VSS}}=0.63*0.816$ =0.5142 g VSS/gBOD Oxygen Demand R_{o}=Q (So – S) – 1.42P_{x,bio} + 4.33 Q (NO_{3}) =110241.6(100.51.499) – 1.42*5836.63 + 4.33*110241.6*47.1994 =1393.62 Kg/h Mass of Alk needed as CaCO3 for nitrification Alk used for nitrification = 7.14g CaCO3 g NH_{4}– N*NO_{x } g/m^{3} =7.14*47.1994 =337 g/m^{3} used as CaCO3 Alk to be added = Alkalinity – pH Influent Alk + Alk used =80 – 269 +337 =148 Alk to be added = 110241.6*148/1000 =16315.756 Kg/d as CaCO_{3} Na(HCO_{3}) needed _{ }=Alkalinity needed as CaCO_{3 * }Equivalent amt of Na(HCO_{3}) / Equivalent weight of CaCO_{3} =(163150456 +80)/ 50 =27410.47 Kg/d Na(HCO_{3}) Phosphorous Removal P removal design requirement NO_{3} = 27 mg/L NO_{3}N in RAS NO_{3}N in RAS = NO_{3}_{}N = Total N NH_{4}N =271 =26 mg/L Nitrate mass balance at influent in the reactor NO_{3}Nreact = NO_{3} RAS * Q / Q + Q_{r} =(26*110241.6*0.6) / (110241.6 + (110241.6 * 0.6)) =9.75 mg/L rbCOD available for P removal rbCOD equivalent = NO_{3 react} * 6.6 =9.75*6.6 =64.35 mg/L rbCOD available for P removal rbCOD available for P removal =rbCOD influent – rbCOD equivalent but rbCOD influent = 0.25(1.6*BOD) =0.25(1.6*100.5) =36.15 mg/L Phosphorous removed by BPR mechanism Bio P removal = rbCOD available for P removal / (ratio of rbCOD/BOD) =36.15 / 8 =4.518 mg/L P used for heterotrophic biomass synthesis in addition to phosphorous storage due to BOD P_{x,bio}= A + B = 4962.760 + 477.10325 =5439.86 Kg/d P used for biomass growth = pH content of heterotrophic biomass * P_{x,bio} =0.015 * 5439.8 =1.22 mg/L P removed = Bio p removed + P_{used} for biomass =4.158 + 1.22 =5.7387 mg/L P content of waste sludge Total P in sludge = p removed * Q / 1000 =(5.7387 * 110241.6) / 1000 =632.54 Kg/d P % = ( Total P in sludge/ P_{x,TSS}) * 100 =(632.54 * 100) / 8850.97 = 7.146 % Denitrification Active biomass calculation ${X}_{b}=\left[\frac{Q\bullet \mathit{SRT}}{V}\right]\left[\frac{Y\bullet ({S}_{o}\u2013S}{1+\left({k}_{d}\right)\mathit{SRT}}\right]$ ${X}_{b}=\left[\frac{110241.6\bullet 3.198}{11793}\right]\left[\frac{0.6\bullet (100.5\u20131.4997}{1+\left(0.1\right)3.198}\right]$ =1345.78 mg/L Internal recycle ratio $\mathit{IR}=\frac{{\mathit{NO}}_{X}}{{\mathit{No}}_{x}\mathit{effluent}}\u20131\u2013R$ $\mathit{IR}=\frac{47.1994}{6}\u20131\u20130.6$ =6.266 Amount of NO_{3}N to the anoxic tank Flowrate to anoxic tank = IR * Q + R * Q =6.266*110241.6 + 110241.6 * 0.6 =756980.93 m^{3}/d NO_{x= }feed = Q_{anoxic } * NO_{x }effluent (756980.93 * 6) / 1000 =4541.88 Kg/d Volume of anoxic tank V_{nox} = τ * Q =0.05 * 110241.6 =6429.98 m^{3} SDNR SDNR = 0.42 Amount of NO_{3}N that can be reduced NO_{r }= V_{nox }* SDNR * MLVSS =6429.98 * 0.42 * 1582.642 =7094.68 kg/d Capacity ratio Capacity ratio = NO_{x }/ NO_{x feed } =7097.68 / 4541.88 =1.56 >T =1.4 therefore acceptable Airflow rate $\mathit{SOTR}=\mathit{AOTR}\left[\frac{{C}_{s,20}}{\alpha \bullet F\bullet (\beta \bullet {C}_{S,T,H}\u2013{C}_{L}}\right]$ $\mathit{SOTR}=766.66\left[\frac{9.08}{0.65\bullet 0.9\bullet (0.95\bullet 908\u20132}\right]$ =1795.8 Kg/L $\mathit{Air\; Flow\; Rate}=\frac{\mathit{SORT}}{E*0.27*60}$ $\mathit{Air\; Flow\; Rate}=\frac{1795.8}{0.3*0.27*60}$ =369.50 m^{3}/min Alk to be added = alkalinity – pH – influent Alk + Alk used – Alk produced – denitrification =80 – 269 337.003 =148.003 g/m^{3} Mass of alkalinity needed Mass of alk needed = Alk to be added * Q /1000 =(148.003 * 110241.6) / 1000 =16316.098 Kg/d CaCO_{3} Anoxic zone mixing energy Anoxic zone mixing energy = V * Mixing energy =6429.98 *10 / 1000 =64.29 KW Secondary clarifier Design Return sludge recycle ratio RAS recycle ratio = Q_{r} / Q =(110241.6 * 0.6) /110241.6 =0.6 Size of clarifier Area = Q / hydraulic application rate =110241.6 / 22 =5010.98 m^{3} Area per clarifier = Area / No of clarifier =5010.98 / 3 =1670 m^{3} $D=\sqrt{\frac{4*\mathit{Area}}{\pi}}$ $D=\sqrt{\frac{4*1670}{\pi}}$ =46.12 =46 m $A=\left(\frac{\pi}{4}\right)*{D}^{2}*3$ $A=\left(\frac{\pi}{4}\right)*{46}^{2}*3$ =4983 m^{3} Solid loading $\mathit{Solid\; Loading}=\frac{\left(1+R\right)Q\left(\mathit{MLSS}\right)}{A}$ $\mathit{Solid\; Loading}=\frac{\left(1+0.6\right)110241.6*2400)}{4983*24}$ =3.53 kg MLSS / m^{2 }h APPENDEX F BIOWIN MODEL
7.References
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