BA-16m3PROJECT: SALINASECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeDIMENSION OF SEPTIC TANK*- Dimension of septic tank1./ Referring to table below:ItemDescriptionQtyUnits6*-Flow per day in TYPE OF ESTABLISHMENT50gallons*-Number of Persons in TYPE OF ESTABLISHMENT80personsTotal flow per day in gallons:4000gallons2./ The liquid volume of tank in gallons:V = 1125+0.75*Q =4125gallonsWhere:V: the liquid volume of the tank in gallons16Cu.mQ: the daily sewage flow in gallons1126 and 0.75 is constant value3./ To find the dimension of septic tank if the maximum depth is 1.50m.And the width is assumed to be 3m then:Length of block degestion:3.472mLength of Septic Tank :5.21mTHE TABLE QUANTITIES OF SEWAGE FLOWITEMTYPE OF ESTABLISHMENTGALLONS PER PERSON PER DAY1Small dwellings with seasonal occupancy502single family dwellings753Multiple family dwellings (Apartment)604Rooming houses401056.6885Boarding houses506Hotels without private bath505040327Hotels with private baths (2 persons per room)60608Restaurants (toilet and kitchen waste per patron)7109Restaurants (kitchen waste per meals serve)3310Tourist camps or trailer parks with central bathhouse3511Tourist courts or mobile home parks with ind. Bath5012Resort camps night and day with limited plumbing5013Luxury camps10015014Work or construction camp5015Day camps no meals serve1516day schools without cafeterials, gym. Or showers1517day schools with cafeterials, gym. Or showers252518Day school with but w/o gym or showers2019Boarding schools7510020Hospitals15025021Institutions other than a hospitals7512522Factories (exclusive of industrial waste)153523Picnic parks with toilet, bath houses1024Swimming pools and bath houses1025Luxury residences10015026Country clubs (per resident member)10027Motels(per bed space)4028Motes with bath, toilet, and kitchen, waste5029Drive in threaters (per car space)530Movie theaters (per auditorium seat)531Airport (per passenger)3532Stores(per toilet room)40033Service stations(per vehicle served)101234Seft service laundries(gal. per wash per person)50NOTE:- Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)

BB-16m3 PROJECT: SALINASECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeDIMENSION OF SEPTIC TANK*- Dimension of septic tank1./ Referring to table below:ItemDescriptionQtyUnits6*-Flow per day in TYPE OF ESTABLISHMENT50gallons*-Number of Persons in TYPE OF ESTABLISHMENT80personsTotal flow per day in gallons:4000gallons2./ The liquid volume of tank in gallons:V = 1125+0.75*Q =4125gallonsWhere:V: the liquid volume of the tank in gallons16Cu.mQ: the daily sewage flow in gallons1126 and 0.75 is constant value3./ To find the dimension of septic tank if the maximum depth is 1.50m.And the width is assumed to be 3m then:Length of block degestion:3.472mLength of Septic Tank :5.21mTHE TABLE QUANTITIES OF SEWAGE FLOWITEMTYPE OF ESTABLISHMENTGALLONS PER PERSON PER DAY1Small dwellings with seasonal occupancy502single family dwellings753Multiple family dwellings (Apartment)604Rooming houses401056.6885Boarding houses506Hotels without private bath505040327Hotels with private baths (2 persons per room)60608Restaurants (toilet and kitchen waste per patron)7109Restaurants (kitchen waste per meals serve)3310Tourist camps or trailer parks with central bathhouse3511Tourist courts or mobile home parks with ind. Bath5012Resort camps night and day with limited plumbing5013Luxury camps10015014Work or construction camp5015Day camps no meals serve1516day schools without cafeterials, gym. Or showers1517day schools with cafeterials, gym. Or showers252518Day school with but w/o gym or showers2019Boarding schools7510020Hospitals15025021Institutions other than a hospitals7512522Factories (exclusive of industrial waste)153523Picnic parks with toilet, bath houses1024Swimming pools and bath houses1025Luxury residences10015026Country clubs (per resident member)10027Motels(per bed space)4028Motes with bath, toilet, and kitchen, waste5029Drive in threaters (per car space)530Movie theaters (per auditorium seat)531Airport (per passenger)3532Stores(per toilet room)40033Service stations(per vehicle served)101234Seft service laundries(gal. per wash per person)50NOTE:- Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)

BC-28m3 PROJECT: SALINASECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeDIMENSION OF SEPTIC TANK*- Dimension of septic tank1./ Referring to table below:ItemDescriptionQtyUnits6*-Flow per day in TYPE OF ESTABLISHMENT50gallons*-Number of Persons in TYPE OF ESTABLISHMENT164personsTotal flow per day in gallons:8200gallons2./ The liquid volume of tank in gallons:V = 1125+0.75*Q =7275gallonsWhere:V: the liquid volume of the tank in gallons28Cu.mQ: the daily sewage flow in gallons1126 and 0.75 is constant value3./ To find the dimension of septic tank if the maximum depth is 1.50m.And the width is assumed to be 3m then:Length of block degestion:6.124mLength of Septic Tank :9.19mTHE TABLE QUANTITIES OF SEWAGE FLOWITEMTYPE OF ESTABLISHMENTGALLONS PER PERSON PER DAY1Small dwellings with seasonal occupancy502single family dwellings753Multiple family dwellings (Apartment)604Rooming houses401056.6885Boarding houses506Hotels without private bath505040327Hotels with private baths (2 persons per room)60608Restaurants (toilet and kitchen waste per patron)7109Restaurants (kitchen waste per meals serve)3310Tourist camps or trailer parks with central bathhouse3511Tourist courts or mobile home parks with ind. Bath5012Resort camps night and day with limited plumbing5013Luxury camps10015014Work or construction camp5015Day camps no meals serve1516day schools without cafeterials, gym. Or showers1517day schools with cafeterials, gym. Or showers252518Day school with but w/o gym or showers2019Boarding schools7510020Hospitals15025021Institutions other than a hospitals7512522Factories (exclusive of industrial waste)153523Picnic parks with toilet, bath houses1024Swimming pools and bath houses1025Luxury residences10015026Country clubs (per resident member)10027Motels(per bed space)4028Motes with bath, toilet, and kitchen, waste5029Drive in threaters (per car space)530Movie theaters (per auditorium seat)531Airport (per passenger)3532Stores(per toilet room)40033Service stations(per vehicle served)101234Seft service laundries(gal. per wash per person)50NOTE:- Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)

20m3OPTIMA CONSULTANTPROJECT: KOHSANTEPHEAPSECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeWASTE WATER TREATMENT PLANTS(WWTP) CALCULATION1-Waste Water Influent to WWTPInfluent quantityTotal influent quantity =90%of water consumptionWater consumption =20m3/dSo :18m3/dLet say Total inluent =18m3/dInfluent qualityBOD5 =250mg/lSS =300mg/l2-Waste Water Efluent from WWTPEfluent qualityBOD5 =20mg/lSS =30mg/l3-Waste Water Treatment Plants (WWTP) DiagramInfluentQ= 18m3/dBOD=250mg/lQ= 18m3/dBOD removal efficiency 30%BOD=175mg/lQ= 18m3/dBOD removal efficiency 25%BOD=131.25mg/lReturn sludgeQ=10m3/dQ= 18m3/dBOD=20mg/lTo Sludge DisposalTo Public Drainage4-Design Concept of Solid Separation TankThis chamber is Design to equalize and homogenizeinfluent and pretreat organic BOD.It provide (12-24) hour of storage time for designed daily waste water flowrate.Hydraulic retention time =24hrTherefor, require solid separation tank =18m3Actual tank capacity =m3>18m3BOD Concept of removal EfficiencyBOD Removal Efficiency =30%Influent BOD =250mg/lBOD residual next to anaerobic filter tank =175mg/l5-Design Concept of Anaerobic Filter TankIn this step, it reduces remaining BOD from solid separation chamber which is anaerobic condition.It uses anaerobic microorganisms to further remove the BOD.5.1Design concept of mediaInfluent BOD =175mg/lBOD Residual next to contact aerobic tank =131.25mg/lTotal BOD loading removal =0.7875kg/dUse area BOD loading =0.0025714286kg/m2-dPeak flow =2Required area media =612.5m2Surface area of media =110m2/m3Required total volume of media =5.57m3Actual volume of media =m3>5.57m35.2Design Volume of anaerobic Filter tankHydraulic retention time =4hrTherefor, required the anaerobic filter tank =3.00m3Actual tank capacity =m3>3.00m3BOD Concept of removal efficiencyBOD Removal Efficiency =25%BOD residual next to contact aerobic tank =131.25mg/l6-Design Concept of contact aeration Tank6.1Fixed film Aerobic BacteriaCalculation of mediaBOD inluent to aeration tank =131.25mg/lBOD loading to be removed =2.00kg/dArea BOD loading =0.0132396694kg/m2-dSafety factor =1.5Required area media =226.875m2Surface area of media =110m2/m3Total required of media =2.06m3Actual volume of media =m3>2.06m36.2Suspended of Aerobic BacteriaDesign Volume of contact Aerobic TankVr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)Where:Tc Mean cell residence time =7dQ Waste water flowrate =18m3/dY Yield factor co.efficient =0.5mg.vss/mg.BodSo BOD Influent =131.25mg/lSe Solube BOD escaped treatment =1.17mg/lEfluent BOD =Influent Solube BOD Escaped treatment+BOD of suspended solid.Determine the BOD of effluent of suspended solidsBiodegratable of efluent solids =19.5mg/lUltimate BOD =27.69mg/lBOD of Suspended solids =18.83mg/lBOD of Escaped treatment =1.17mg/lX MLVSS = 80% MLSS =2000mg/lKd =0.061/dVr =2.89m3Hydraulic retention timeHRT =Vr/Q0.16dHRT =3.85hrRecheck hydraulic retention timeDesign HRT=6hr > 3.85 hr6.3Design Oxygen requirementO2 =a x Lr + b x SaWhere:a BOD eliminated coefficiency0.5kg.O2/kg.BODLr Total BOD load to be treated13kg/db Sludge endogenouse coefficent0.07kg.O2/kg.MLVSS-dSa MLVSS in aeration tank57.52kg/dSo:O2 =10.53kg.O2/dSolubility air in sewage6.5%Oxygen content in air0.277kg.O2/m3 of airVolume of air required584.68m3/dProvide capacity lost20%701.62m3/dDesign Safety21403.23m3/d0.97m3/min6.4Select Air BlowerPower1.5kwDischarge bore40mmPressure0.3kg.f/cm3air discharge0.81m3/minElectricity380/3/50Total2set6.5Design the quantity of excess sludg that must be waste per dayThe observes yields(YOBES) =Y/(1+Kd.Tc)=0.35Kg.VSS/Kg.BODThe mass of activated sludge(Px) =Yobes.Q.(So-Se)/1000=0.90Kg.VSS/dThe increase in the total mass of MLSS(Px) =Px/80%=1.125Kg.SS/dDesign sludge concentration(Xr) =7000mg/lSo, Volume of excess sludge =0.82m3/d6.6Recheck mean cell Residence Time for controlTc =a/(b+c)Where:Tc Mean cell residence timea weight of biomass aeration tank =57.52kgb weight of sludgr that must be waste =5.73kg/dc weight of sludge in effluent3kg/dSo, Tc =7.00d7-Design Concept of Sedimentation tank7.1Design overflow rateUse overflow rate =24m3/m2-dSurface area of Sedimentation tank =0.75m2Actual surface =m2>0.75m2.7.2Recheck weir overflow rateNormaly weir overflow rate of sedimentation tank is not more than 125m3/m-dLength of construct weir =6.6mSo, weir overflow rate =Flow rate / Length of construct weir=18.94m3/m-d7.3Check volume of sedimentation tankHydraulic retention time =3hrTherefor, required the sedimentation tank =3m37.4Design of return sludge rateQr =QX/(Xr-X)Qr Volume of return sludgeQ Flow rate of waste water =18m3/dX MLSS =2500mg/lXr Sludge concentration in the bottom ofsedimentation tank =7000mg/lVolume of return sludge(Qr) =10m3/dRecheck of return sludge rateNormal of return sludge rate (Qr/Q) =0.25-1So, return sludge rate (Qr/Q) =0.56Air Lift pumpPower =0.75kwDecharge bore =32mmPressure =0.2kg.f/cm2Air decharge rate =0.65m3/min0.42Electricity =380/3/50Total =1set

Solid separation tankAnaerobic filter tankContact aeration tankSedimentation tankStorage tank

30m3OPTIMA CONSULTANTPROJECT: KOHSANTEPHEAPSECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeWASTE WATER TREATMENT PLANTS(WWTP) CALCULATION1-Waste Water Influent to WWTPInfluent quantityTotal influent quantity =90%of water consumptionWater consumption =32.5m3/dSo :29.25m3/dLet say Total inluent =30m3/dInfluent qualityBOD5 =250mg/lSS =300mg/l2-Waste Water Efluent from WWTPEfluent qualityBOD5 =20mg/lSS =30mg/l3-Waste Water Treatment Plants (WWTP) DiagramInfluentQ= 30m3/dBOD=250mg/lQ= 30m3/dBOD removal efficiency 30%BOD=175mg/lQ= 30m3/dBOD removal efficiency 25%BOD=131.25mg/lReturn sludgeQ=16.67m3/dQ= 30m3/dBOD=20mg/lTo Sludge DisposalTo Public Drainage4-Design Concept of Solid Separation TankThis chamber is Design to equalize and homogenizeinfluent and pretreat organic BOD.It provide (12-24) hour of storage time for designed daily waste water flowrate.Hydraulic retention time =24hrTherefor, require solid separation tank =30m3Actual tank capacity =m3>30m3BOD Concept of removal EfficiencyBOD Removal Efficiency =30%Influent BOD =250mg/lBOD residual next to anaerobic filter tank =175mg/l5-Design Concept of Anaerobic Filter TankIn this step, it reduces remaining BOD from solid separation chamber which is anaerobic condition.It uses anaerobic microorganisms to further remove the BOD.5.1Design concept of mediaInfluent BOD =175mg/lBOD Residual next to contact aerobic tank =131.25mg/lTotal BOD loading removal =1.3125kg/dUse area BOD loading =0.0042857143kg/m2-dPeak flow =2Required area media =612.5m2Surface area of media =110m2/m3Required total volume of media =5.57m3Actual volume of media =m3>5.57m35.2Design Volume of anaerobic Filter tankHydraulic retention time =4hrTherefor, required the anaerobic filter tank =5.00m3Actual tank capacity =m3>5.00m3BOD Concept of removal efficiencyBOD Removal Efficiency =25%BOD residual next to contact aerobic tank =131.25mg/l6-Design Concept of contact aeration Tank6.1Fixed film Aerobic BacteriaCalculation of mediaBOD inluent to aeration tank =131.25mg/lBOD loading to be removed =3.34kg/dArea BOD loading =0.0220661157kg/m2-dSafety factor =1.5Required area media =226.875m2Surface area of media =110m2/m3Total required of media =2.06m3Actual volume of media =m3>2.06m36.2Suspended of Aerobic BacteriaDesign Volume of contact Aerobic TankVr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)Where:Tc Mean cell residence time =7dQ Waste water flowrate =30m3/dY Yield factor co.efficient =0.5mg.vss/mg.BodSo BOD Influent =131.25mg/lSe Solube BOD escaped treatment =1.17mg/lEfluent BOD =Influent Solube BOD Escaped treatment+BOD of suspended solid.Determine the BOD of effluent of suspended solidsBiodegratable of efluent solids =19.5mg/lUltimate BOD =27.69mg/lBOD of Suspended solids =18.83mg/lBOD of Escaped treatment =1.17mg/lX MLVSS = 80% MLSS =2000mg/lKd =0.061/dVr =4.81m3Hydraulic retention timeHRT =Vr/Q0.16dHRT =3.85hrRecheck hydraulic retention timeDesign HRT=6hr > 3.85 hr6.3Design Oxygen requirementO2 =a x Lr + b x SaWhere:a BOD eliminated coefficiency0.5kg.O2/kg.BODLr Total BOD load to be treated13kg/db Sludge endogenouse coefficent0.07kg.O2/kg.MLVSS-dSa MLVSS in aeration tank57.52kg/dSo:O2 =10.53kg.O2/dSolubility air in sewage6.5%Oxygen content in air0.277kg.O2/m3 of airVolume of air required584.68m3/dProvide capacity lost20%701.62m3/dDesign Safety21403.23m3/d0.97m3/min6.4Select Air BlowerPower1.5kwDischarge bore40mmPressure0.3kg.f/cm3air discharge0.81m3/minElectricity380/3/50Total2set6.5Design the quantity of excess sludg that must be waste per dayThe observes yields(YOBES) =Y/(1+Kd.Tc)=0.35Kg.VSS/Kg.BODThe mass of activated sludge(Px) =Yobes.Q.(So-Se)/1000=1.40Kg.VSS/dThe increase in the total mass of MLSS(Px) =Px/80%=1.75Kg.SS/dDesign sludge concentration(Xr) =7000mg/lSo, Volume of excess sludge =0.82m3/d6.6Recheck mean cell Residence Time for controlTc =a/(b+c)Where:Tc Mean cell residence timea weight of biomass aeration tank =57.52kgb weight of sludgr that must be waste =5.73kg/dc weight of sludge in effluent =3kg/dSo, Tc =7.00d7-Design Concept of Sedimentation tank7.1Design overflow rateUse overflow rate =24m3/m2-dSurface area of Sedimentation tank =1.25m2Actual surface =m2>1.25m2.7.2Recheck weir overflow rateNormaly weir overflow rate of sedimentation tank is not more than 125m3/m-dLength of construct weir =6.6mSo, weir overflow rate =Flow rate / Length of construct weir=18.94m3/m-d7.3Check volume of sedimentation tankHydraulic retention time =3hrTherefor, required the sedimentation tank =4m37.4Design of return sludge rateQr =QX/(Xr-X)Qr Volume of return sludgeQ Flow rate of waste water =30m3/dX MLSS =2500mg/lXr Sludge concentration in the bottom ofsedimentation tank =7000mg/lVolume of return sludge(Qr) =16.67m3/dRecheck of return sludge rateNormal of return sludge rate (Qr/Q) =0.25-1So, return sludge rate (Qr/Q) =0.56Air Lift pumpPower =0.75kwDecharge bore =32mmPressure =0.2kg.f/cm2Air decharge rate =0.65m3/minElectricity =380/3/50Total =1set

Solid separation tankAnaerobic filter tankContact aeration tankSedimentation tankStorage tank

40m3OPTIMA CONSULTANTPROJECT: KOHSANTEPHEAPSECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeWASTE WATER TREATMENT PLANTS(WWTP) CALCULATION1-Waste Water Influent to WWTPInfluent quantityTotal influent quantity =90%of water consumptionWater consumption =40m3/dSo :=36m3/dLet say Total inluent =36m3/dInfluent qualityBOD5 =250mg/lSS =300mg/l2-Waste Water Efluent from WWTPEfluent qualityBOD5 =20mg/lSS =30mg/l3-Waste Water Treatment Plants (WWTP) DiagramInfluentQ= 36m3/dBOD=250mg/lQ= 36m3/dBOD removal efficiency 30%BOD=175mg/lQ= 36m3/dBOD removal efficiency 25%BOD=131.25mg/lReturn sludgeQ=20m3/dQ= 36m3/dBOD=20mg/lTo Sludge DisposalTo Public Drainage4-Design Concept of Solid Separation TankThis chamber is Design to equalize and homogenizeinfluent and pretreat organic BOD.It provide (12-24) hour of storage time for designed daily waste water flowrate.Hydraulic retention time =24hrTherefor, require solid separation tank =36m3Actual tank capacity =m3>36m3BOD Concept of removal EfficiencyBOD Removal Efficiency =30%Influent BOD =250mg/lBOD residual next to anaerobic filter tank =175mg/l5-Design Concept of Anaerobic Filter TankIn this step, it reduces remaining BOD from solid separation chamber which is anaerobic condition.It uses anaerobic microorganisms to further remove the BOD.5.1Design concept of mediaInfluent BOD =175mg/lBOD Residual next to contact aerobic tank =131.25mg/lTotal BOD loading removal =1.575kg/dUse area BOD loading =0.0051428571kg/m2-dPeak flow =2Required area media =612.5m2Surface area of media =110m2/m3Required total volume of media =5.57m3Actual volume of media =m3>5.57m35.2Design Volume of anaerobic Filter tankHydraulic retention time =4hrTherefor, required the anaerobic filter tank =6.00m3Actual tank capacity =m3>6.00m3BOD Concept of removal efficiencyBOD Removal Efficiency =25%BOD residual next to contact aerobic tank =131.25mg/l6-Design Concept of contact aeration Tank6.1Fixed film Aerobic BacteriaCalculation of mediaBOD inluent to aeration tank =131.25mg/lBOD loading to be removed =4.01kg/dArea BOD loading =0.0264793388kg/m2-dSafety factor =1.5Required area media =226.875m2Surface area of media =110m2/m3Total required of media =2.06m3Actual volume of media =m3>2.06m36.2Suspended of Aerobic BacteriaDesign Volume of contact Aerobic TankVr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)Where:Tc Mean cell residence time =7dQ Waste water flowrate =36m3/dY Yield factor co.efficient =0.5mg.vss/mg.BodSo BOD Influent =131.25mg/lSe Solube BOD escaped treatment =1.17mg/lEfluent BOD =Influent Solube BOD Escaped treatment+BOD of suspended solid.Determine the BOD of effluent of suspended solidsBiodegratable of efluent solids =19.5mg/lUltimate BOD =27.69mg/lBOD of Suspended solids =18.83mg/lBOD of Escaped treatment =1.17mg/lX MLVSS = 80% MLSS =2000mg/lKd =0.061/dVr =5.77m3Hydraulic retention timeHRT =Vr/Q0.16dHRT =3.85hrRecheck hydraulic retention timeDesign HRT=6hr > 3.85 hr6.3Design Oxygen requirementO2 =a x Lr + b x SaWhere:a BOD eliminated coefficiency0.5kg.O2/kg.BODLr Total BOD load to be treated13kg/db Sludge endogenouse coefficent0.07kg.O2/kg.MLVSS-dSa MLVSS in aeration tank57.52kg/dSo:O2 =10.53kg.O2/dSolubility air in sewage6.5%Oxygen content in air0.277kg.O2/m3 of airVolume of air required584.68m3/dProvide capacity lost20%701.62m3/dDesign Safety21403.23m3/d0.97m3/min6.4Select Air BlowerPower1.5kwDischarge bore40mmPressure0.3kg.f/cm3air discharge0.81m3/minElectricity380/3/50Total2set6.5Design the quantity of excess sludg that must be waste per dayThe observes yields(YOBES) =Y/(1+Kd.Tc)=0.35Kg.VSS/Kg.BODThe mass of activated sludge(Px) =Yobes.Q.(So-Se)/1000=1.70Kg.VSS/dThe increase in the total mass of MLSS(Px) =Px/80%=2.125Kg.SS/dDesign sludge concentration(Xr) =7000mg/lSo, Volume of excess sludge =0.82m3/d6.6Recheck mean cell Residence Time for controlTc =a/(b+c)Where:Tc Mean cell residence timea weight of biomass aeration tank =57.52kgb weight of sludgr that must be waste =5.73kg/dc weight of sludge in effluent3kg/dSo, Tc =7.00d7-Design Concept of Sedimentation tank7.1Design overflow rateUse overflow rate =24m3/m2-dSurface area of Sedimentation tank =1.50m2Actual surface =m2>1.50m2.7.2Recheck weir overflow rateNormaly weir overflow rate of sedimentation tank is not more than 125m3/m-dLength of construct weir =6.6mSo, weir overflow rate =Flow rate / Length of construct weir=18.94m3/m-d7.3Check volume of sedimentation tankHydraulic retention time =3hrTherefor, required the sedimentation tank =5m37.4Design of return sludge rateQr =QX/(Xr-X)Qr Volume of return sludgeQ Flow rate of waste water =36m3/dX MLSS =2500mg/lXr Sludge concentration in the bottom ofsedimentation tank =7000mg/lVolume of return sludge(Qr) =20m3/dRecheck of return sludge rateNormal of return sludge rate (Qr/Q) =0.25-1So, return sludge rate (Qr/Q) =0.56Air Lift pumpPower =0.75kwDecharge bore =32mmPressure =0.2kg.f/cm20.65m3/minElectricity =380/3/50Total =1set

Solid separation tankAnaerobic filter tankContact aeration tankSedimentation tankStorage tank

53m3OPTIMA CONSULTANTPROJECT:SECTION: PLUMBING SECTIONDATE:REV.:Assumed Data* You are allowed to changeCalculated data* You are NOT ALLOWED to changeWASTE WATER TREATMENT PLANTS(WWTP) CALCULATION1-Waste Water Influent to WWTPInfluent quantityTotal influent quantity =90%of water consumptionWater consumption =58.5m3/dSo :=52.65m3/dLet say Total inluent =53m3/dInfluent qualityBOD5 =250mg/lSS =300mg/l2-Waste Water Efluent from WWTPEfluent qualityBOD5 =20mg/lSS =30mg/l3-Waste Water Treatment Plants (WWTP) DiagramInfluentQ= 36m3/dBOD=250mg/lQ= 36m3/dBOD removal efficiency 30%BOD=175mg/lQ= 36m3/dBOD removal efficiency 25%BOD=131.25mg/lReturn sludgeQ=20m3/dQ= 36m3/dBOD=20mg/lTo Sludge DisposalTo Public Drainage4-Design Concept of Solid Separation TankThis chamber is Design to equalize and homogenizeinfluent and pretreat organic BOD.It provide (12-24) hour of storage time for designed daily waste water flowrate.Hydraulic retention time =24hrTherefor, require solid separation tank =53m3Actual tank capacity =m3>53m3BOD Concept of removal EfficiencyBOD Removal Efficiency =30%Influent BOD =250mg/lBOD residual next to anaerobic filter tank =175mg/l5-Design Concept of Anaerobic Filter TankIn this step, it reduces remaining BOD from solid separation chamber which is anaerobic condition.It uses anaerobic microorganisms to further remove the BOD.5.1Design concept of mediaInfluent BOD =175mg/lBOD Residual next to contact aerobic tank =131.25mg/lTotal BOD loading removal =2.31875kg/dUse area BOD loading =0.0075714286kg/m2-dPeak flow =2Required area media =612.5m2Surface area of media =110m2/m3Required total volume of media =5.57m3Actual volume of media =m3>5.57m35.2Design Volume of anaerobic Filter tankHydraulic retention time =4hrTherefor, required the anaerobic filter tank =8.83m3Actual tank capacity =m3>8.83m3BOD Concept of removal efficiencyBOD Removal Efficiency =25%BOD residual next to contact aerobic tank =131.25mg/l6-Design Concept of contact aeration Tank6.1Fixed film Aerobic BacteriaCalculation of mediaBOD inluent to aeration tank =131.25mg/lBOD loading to be removed =5.90kg/dArea BOD loading =0.0389834711kg/m2-dSafety factor =1.5Required area media =226.875m2Surface area of media =110m2/m3Total required of media =2.06m3Actual volume of media =m3>2.06m36.2Suspended of Aerobic BacteriaDesign Volume of contact Aerobic TankVr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)Where:Tc Mean cell residence time =7dQ Waste water flowrate =53m3/dY Yield factor co.efficient =0.5mg.vss/mg.BodSo BOD Influent =131.25mg/lSe Solube BOD escaped treatment =1.17mg/lEfluent BOD =Influent Solube BOD Escaped treatment+BOD of suspended solid.Determine the BOD of effluent of suspended solidsBiodegratable of efluent solids =19.5mg/lUltimate BOD =27.69mg/lBOD of Suspended solids =18.83mg/lBOD of Escaped treatment =1.17mg/lX MLVSS = 80% MLSS =2000mg/lKd =0.061/dVr =8.50m3Hydraulic retention timeHRT =Vr/Q0.16dHRT =3.85hrRecheck hydraulic retention timeDesign HRT=6hr > 3.85 hr6.3Design Oxygen requirementO2 =a x Lr + b x SaWhere:a BOD eliminated coefficiency0.5kg.O2/kg.BODLr Total BOD load to be treated13kg/db Sludge endogenouse coefficent0.07kg.O2/kg.MLVSS-dSa MLVSS in aeration tank57.52kg/dSo:O2 =10.53kg.O2/dSolubility air in sewage6.5%Oxygen content in air0.277kg.O2/m3 of airVolume of air required584.68m3/dProvide capacity lost20%701.62m3/dDesign Safety21403.23m3/d0.97m3/min6.4Select Air BlowerPower1.5kwDischarge bore40mmPressure0.3kg.f/cm3air discharge0.81m3/minElectricity380/3/50Total2set6.5Design the quantity of excess sludg that must be waste per dayThe observes yields(YOBES) =Y/(1+Kd.Tc)=0.35Kg.VSS/Kg.BODThe mass of activated sludge(Px) =Yobes.Q.(So-Se)/1000=2.50Kg.VSS/dThe increase in the total mass of MLSS(Px) =Px/80%=3.125Kg.SS/dDesign sludge concentration(Xr) =7000mg/lSo, Volume of excess sludge =0.82m3/d6.6Recheck mean cell Residence Time for controlTc =a/(b+c)Where:Tc Mean cell residence timea weight of biomass aeration tank =57.52kgb weight of sludgr that must be waste =5.73kg/dc weight of sludge in effluent3kg/dSo, Tc =7.00d7-Design Concept of Sedimentation tank7.1Design overflow rateUse overflow rate =24m3/m2-dSurface area of Sedimentation tank =2.21m2Actual surface =m2>2.21m2.7.2Recheck weir overflow rateNormaly weir overflow rate of sedimentation tank is not more than 125m3/m-dLength of construct weir =6.6mSo, weir overflow rate =Flow rate / Length of construct weir=18.94m3/m-d7.3Check volume of sedimentation tankHydraulic retention time =3hrTherefor, required the sedimentation tank =7m37.4Design of return sludge rateQr =QX/(Xr-X)Qr Volume of return sludgeQ Flow rate of waste water =53m3/dX MLSS =2500mg/lXr Sludge concentration in the bottom ofsedimentation tank =7000mg/lVolume of return sludge(Qr) =29.45m3/dRecheck of return sludge rateNormal of return sludge rate (Qr/Q) =0.25-1So, return sludge rate (Qr/Q) =0.56Air Lift pumpPower =0.75kwDecharge bore =32mmPressure =0.2kg.f/cm20.65m3/minElectricity =380/3/50Total =1set

Solid separation tankAnaerobic filter tankContact aeration tankSedimentation tankStorage tank