PROJECT: SALINA SECTION: PLUMBING SECTION DATE: REV.: Assumed Data Calculated data
* You are allowed to change * You are NOT ALLOWED to change
DIMENSION OF SEPTIC TANK *- Dimension of septic tank 1./ Referring to table below: Item 6
Description *-Flow per day in TYPE OF ESTABLISHMENT *-Number of Persons in TYPE OF ESTABLISHMENT
Total flow per day in gallons:
Qty 50 80 4000
Units gallons persons gallons
4125 16
gallons Cu.m
3.472 5.21
m m
2./ The liquid volume of tank in gallons:
Where:
V = 1125+0.75*Q = V: the liquid volume of the tank in gallons Q: the daily sewage flow in gallons 1126 and 0.75 is constant value
3./ 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: Length of Septic Tank : THE TABLE QUANTITIES OF SEWAGE FLOW ITEM
TYPE OF ESTABLISHMENT
1
Small dwellings with seasonal occupancy
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
single family dwellings Multiple family dwellings (Apartment) Rooming houses Boarding houses Hotels without private bath Hotels with private baths (2 persons per room) Restaurants (toilet and kitchen waste per patron) Restaurants (kitchen waste per meals serve) Tourist camps or trailer parks with central bathhouse Tourist courts or mobile home parks with ind. Bath Resort camps night and day with limited plumbing Luxury camps Work or construction camp Day camps no meals serve day schools without cafeterials, gym. Or showers
GALLONS PER PERSON PER DAY 50 75 60 40 50 50 60 7 3 35 50 50 100 50 15 15
17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
day schools with cafeterials, gym. Or showers Day school with but w/o gym or showers Boarding schools Hospitals Institutions other than a hospitals Factories (exclusive of industrial waste) Picnic parks with toilet, bath houses Swimming pools and bath houses Luxury residences Country clubs (per resident member) Motels(per bed space) Motes with bath, toilet, and kitchen, waste Drive in threaters (per car space) Movie theaters (per auditorium seat) Airport (per passenger) Stores(per toilet room) Service stations(per vehicle served) Seft service laundries(gal. per wash per person)
25 20 75100 150 75 15 10 10 100 100 40 50 5 5 3 400 10 50
NOTE: - Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)
LOWED to change
GALLONS PER PERSON PER DAY
1056.688 50 60 10 3
150
4032
25
250 125 35
150
5 12
PROJECT: SALINA SECTION: PLUMBING SECTION DATE: REV.: Assumed Data Calculated data
* You are allowed to change * You are NOT ALLOWED to change
DIMENSION OF SEPTIC TANK *- Dimension of septic tank 1./ Referring to table below: Item 6
Description *-Flow per day in TYPE OF ESTABLISHMENT *-Number of Persons in TYPE OF ESTABLISHMENT
Total flow per day in gallons:
Qty 50 80 4000
Units gallons persons gallons
4125 16
gallons Cu.m
3.472 5.21
m m
2./ The liquid volume of tank in gallons:
Where:
V = 1125+0.75*Q = V: the liquid volume of the tank in gallons Q: the daily sewage flow in gallons 1126 and 0.75 is constant value
3./ 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: Length of Septic Tank : THE TABLE QUANTITIES OF SEWAGE FLOW ITEM
TYPE OF ESTABLISHMENT
1
Small dwellings with seasonal occupancy
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
single family dwellings Multiple family dwellings (Apartment) Rooming houses Boarding houses Hotels without private bath Hotels with private baths (2 persons per room) Restaurants (toilet and kitchen waste per patron) Restaurants (kitchen waste per meals serve) Tourist camps or trailer parks with central bathhouse Tourist courts or mobile home parks with ind. Bath Resort camps night and day with limited plumbing Luxury camps Work or construction camp Day camps no meals serve day schools without cafeterials, gym. Or showers
GALLONS PER PERSON PER DAY 50 75 60 40 50 50 60 7 3 35 50 50 100 50 15 15
17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
day schools with cafeterials, gym. Or showers Day school with but w/o gym or showers Boarding schools Hospitals Institutions other than a hospitals Factories (exclusive of industrial waste) Picnic parks with toilet, bath houses Swimming pools and bath houses Luxury residences Country clubs (per resident member) Motels(per bed space) Motes with bath, toilet, and kitchen, waste Drive in threaters (per car space) Movie theaters (per auditorium seat) Airport (per passenger) Stores(per toilet room) Service stations(per vehicle served) Seft service laundries(gal. per wash per person)
25 20 75100 150 75 15 10 10 100 100 40 50 5 5 3 400 10 50
NOTE: - Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)
LOWED to change
GALLONS PER PERSON PER DAY
1056.688 50 60 10 3
150
4032
25
250 125 35
150
5 12
PROJECT: SALINA SECTION: PLUMBING SECTION DATE: REV.: Assumed Data Calculated data
* You are allowed to change * You are NOT ALLOWED to change
DIMENSION OF SEPTIC TANK *- Dimension of septic tank 1./ Referring to table below: Item 6
Description *-Flow per day in TYPE OF ESTABLISHMENT *-Number of Persons in TYPE OF ESTABLISHMENT
Total flow per day in gallons:
Qty 50 164 8200
Units gallons persons gallons
7275 28
gallons Cu.m
6.124 9.19
m m
2./ The liquid volume of tank in gallons:
Where:
V = 1125+0.75*Q = V: the liquid volume of the tank in gallons Q: the daily sewage flow in gallons 1126 and 0.75 is constant value
3./ 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: Length of Septic Tank : THE TABLE QUANTITIES OF SEWAGE FLOW ITEM
TYPE OF ESTABLISHMENT
1
Small dwellings with seasonal occupancy
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
single family dwellings Multiple family dwellings (Apartment) Rooming houses Boarding houses Hotels without private bath Hotels with private baths (2 persons per room) Restaurants (toilet and kitchen waste per patron) Restaurants (kitchen waste per meals serve) Tourist camps or trailer parks with central bathhouse Tourist courts or mobile home parks with ind. Bath Resort camps night and day with limited plumbing Luxury camps Work or construction camp Day camps no meals serve day schools without cafeterials, gym. Or showers
GALLONS PER PERSON PER DAY 50 75 60 40 50 50 60 7 3 35 50 50 100 50 15 15
17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
day schools with cafeterials, gym. Or showers Day school with but w/o gym or showers Boarding schools Hospitals Institutions other than a hospitals Factories (exclusive of industrial waste) Picnic parks with toilet, bath houses Swimming pools and bath houses Luxury residences Country clubs (per resident member) Motels(per bed space) Motes with bath, toilet, and kitchen, waste Drive in threaters (per car space) Movie theaters (per auditorium seat) Airport (per passenger) Stores(per toilet room) Service stations(per vehicle served) Seft service laundries(gal. per wash per person)
25 20 75100 150 75 15 10 10 100 100 40 50 5 5 3 400 10 50
NOTE: - Refence Philippines Plumbing Design and Estimate (MAX B, FAJARDO JR. (Chapter 4 page 91)
LOWED to change
GALLONS PER PERSON PER DAY
1056.688 50 60 10 3
150
4032
25
250 125 35
150
5 12
OPTIMA CONSULTANT PROJECT: KOHSANTEPHEAP SECTION: PLUMBING SECTION DATE: REV.: Assumed Data * You are allowed to change Calculated data * You are NOT ALLOWED to change WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION Waste Water Influent to WWTP Influent quantity Total influent quantity Water consumption
1-
= =
So : Let say Total inluent
=
90% 20 18 18
of water consumption m3/d m3/d m3/d
Influent quality = =
250 mg/l 300 mg/l
Efluent quality = =
20 mg/l 30 mg/l
BOD5 SS 2-
Waste Water Efluent from WWTP BOD5 SS
3-
Waste Water Treatment Plants (WWTP) Diagram Influent Q= 18m3/d BOD=250mg/l
Solid separation tank Q= 18m3/d BOD removal efficiency 30% BOD=175mg/l
Anaerobic filter tank Q= 18m3/d BOD removal efficiency 25% BOD=131.25mg/l Return sludge Q=10m3/d
Contact aeration tank Q= 18m3/d BOD=20mg/l
To Sludge Disposal
Sedimentation tank
Storage tank
To Public Drainage Design Concept of Solid Separation Tank 4chamber 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 Therefor, require solid separation tank Actual tank capacity
= = = >
BOD Concept of removal Efficiency BOD Removal Efficiency = Influent BOD = BOD residual next to anaerobic filter tank = 5-
24 hr 18 m3 m3 18 m3
30 % 250 mg/l 175 mg/l
Design Concept of Anaerobic Filter Tank
reduces remaining BOD from solid separation chamber which is anaerobic condition. It uses anaerobic microorganisms to further remove the BOD. 5.1
5.2
Design concept of media Influent BOD BOD Residual next to contact aerobic tank Total BOD loading removal Use area BOD loading Peak flow Required area media Surface area of media Required total volume of media Actual volume of media
175 mg/l = 131.25 mg/l = 0.7875 kg/d = 0.002571 kg/m2-d = 2 = 612.5 m2 = 110 m2/m3 = 5.57 m3 = m3 = 5.57 m3 >
Design Volume of anaerobic Filter tank Hydraulic retention time = Therefor, required the anaerobic filter tank = Actual tank capacity = >
4 hr 3.00 m3 m3 3.00 m3
BOD Concept of removal efficiency BOD Removal Efficiency = BOD residual next to contact aerobic tank = 66.1
Design Concept of contact aeration Tank Fixed film Aerobic Bacteria Calculation of media BOD inluent to aeration tank = BOD loading to be removed = Area BOD loading = Safety factor = Required area media = Surface area of media = Total required of media = Actual volume of media = >
6.2
25 % 131.25 mg/l
131.25 2.00 0.01324 1.5 226.875 110 2.06
mg/l kg/d kg/m2-d
m2 m2/m3 m3 m3 2.06 m3
Suspended of Aerobic Bacteria Design Volume of contact Aerobic Tank Vr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)
Where:
7 d Tc Mean cell residence time = 18 m3/d Q Waste water flowrate = 0.5 mg.vss/mg.Bod Y Yield factor co.efficient = 131.25 mg/l So BOD Influent = 1.17 mg/l Se Solube BOD escaped treatment = Influent Solube BOD Escaped treatment+BO Efluent BOD = Determine the BOD of effluent of suspended solids 19.5 mg/l Biodegratable of efluent solids = 27.69 mg/l Ultimate BOD = 18.83 mg/l BOD of Suspended solids = 1.17 mg/l BOD of Escaped treatment = 2000 mg/l X MLVSS = 80% MLSS = 0.06 1/d Kd = Vr
=
2.89 m3
Hydraulic retention time HRT
=
HRT
=
Vr/Q 0.16 d 3.85 hr
Recheck hydraulic retention time Design HRT=6hr > 3.85 hr 6.3
Design Oxygen requirement O2
Where:
a BOD eliminated coefficiency
= a x Lr + b x Sa 0.5 kg.O2/kg.BOD
13 kg/d 0.07 kg.O2/kg.MLVSS-d 57.52 kg/d
Lr Total BOD load to be treated b Sludge endogenouse coefficent Sa MLVSS in aeration tank So:
O2
10.53 kg.O2/d
=
Solubility air in sewage Oxygen content in air Volume of air required Provide capacity lost Design Safety
6.4
% kg.O2/m3 of air m3/d % m3/d
1.5 40 0.3 0.81 380/3/50 2
kw mm kg.f/cm3 m3/min
m3/d m3/min
Select Air Blower Power Discharge bore Pressure air discharge Electricity Total
6.5
6.5 0.277 584.68 20 701.62 2 1403.23 0.97
set
Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) The mass of activated sludge(Px)
= Y/(1+Kd.Tc) 0.35 Kg.VSS/Kg.BOD = Yobes.Q.(So-Se)/1000 = 0.90 Kg.VSS/d =
The increase in the total mass of MLSS(Px) = Px/80% 1.125 Kg.SS/d = 7000 mg/l Design sludge concentration(Xr) = 0.82 m3/d So, Volume of excess sludge = 6.6
Where:
Recheck mean cell Residence Time for control Tc
Tc Mean cell residence time a weight of biomass aeration tank = b weight of sludgr that must be waste = c weight of sludge in effluent So, Tc
77.1
= a/(b+c)
Design Concept of Sedimentation tank Design overflow rate Use overflow rate Surface area of Sedimentation tank Actual surface
57.52 kg 5.73 kg/d 3 kg/d
=
7.00 d
= = = >
24 m3/m2-d 0.75 m2 m2 0.75 m2 .
7.2 Recheck weir overflow rate Normaly weir overflow rate of sedimentation tank is not more than 125m3/m-d 6.6 m Length of construct weir = So, weir overflow rate = Flow rate / Length of construct weir 18.94 m3/m-d = 7.3 Check volume of sedimentation tank 3 hr Hydraulic retention time = 3 m3 Therefor, required the sedimentation tank = 7.4
Design of return sludge rate Qr = QX/(Xr-X) Qr Volume of return sludge 18 m3/d Q Flow rate of waste water = 2500 mg/l X MLSS = Xr Sludge concentration in the bottom of 7000 mg/l sedimentation tank = 10 m3/d Volume of return sludge(Qr) = Recheck of return sludge rate Normal of return sludge rate (Qr/Q) = So, return sludge rate (Qr/Q) =
0.25-1 0.56
Air Lift pump = = = = = =
0.75 32 0.2 0.65 380/3/50 1
Power Decharge bore Pressure Air decharge rate Electricity Total
kw mm kg.f/cm2 m3/min set
allowed to change
NOT ALLOWED to change
of water consumption
mg.vss/mg.Bod
lube BOD Escaped treatment+BOD of suspended solid.
kg.O2/kg.BOD
kg.O2/kg.MLVSS-d
kg.O2/m3 of air
Kg.VSS/Kg.BOD So-Se)/1000
Length of construct weir
0.42
OPTIMA CONSULTANT PROJECT: KOHSANTEPHEAP SECTION: PLUMBING SECTION DATE: REV.: Assumed Data Calculated data
* You are allowed to change * You are NOT ALLOWED to change
WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION Waste Water Influent to WWTP Influent quantity Total influent quantity Water consumption
1-
= =
So :
2-
3-
90% 32.5 29.25 30
of water consumption m3/d m3/d m3/d
Let say Total inluent
=
Influent quality BOD5 SS
= =
250 mg/l 300 mg/l
= =
20 mg/l 30 mg/l
Waste Water Efluent from WWTP Efluent quality BOD5 SS Waste Water Treatment Plants (WWTP) Diagram
Influent Q= 30m3/d BOD=250mg/l
Solid separation tank Q= 30m3/d BOD removal efficiency 30% BOD=175mg/l
Anaerobic filter tank Q= 30m3/d BOD removal efficiency 25% BOD=131.25mg/l Return sludge Q=16.67m3/d
Contact aeration tank Q= 30m3/d BOD=20mg/l
To Sludge Disposal
Sedimentation tank
Storage tank
To Public Drainage 4-
Design Concept of Solid Separation Tank This 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 Therefor, require solid separation tank Actual tank capacity
BOD Concept of removal Efficiency BOD Removal Efficiency Influent BOD BOD residual next to anaerobic filter tank 5-
= = = >
24 hr 30 m3 m3 30 m3
= = =
30 % 250 mg/l 175 mg/l
Design Concept of Anaerobic Filter Tank
In this step, it reduces remaining BOD from solid separation chamber which is anaerobic con It uses anaerobic microorganisms to further remove the BOD. 5.1
5.2
Design concept of media Influent BOD BOD Residual next to contact aerobic tank Total BOD loading removal Use area BOD loading Peak flow Required area media Surface area of media Required total volume of media Actual volume of media Design Volume of anaerobic Filter tank Hydraulic retention time Therefor, required the anaerobic filter tank Actual tank capacity
175 mg/l = 131.25 mg/l = 1.3125 kg/d = 0.004286 kg/m2-d = 2 = 612.5 m2 = 110 m2/m3 = 5.57 m3 = m3 = 5.57 m3 > = = = >
4 hr 5.00 m3 m3 5.00 m3
BOD Concept of removal efficiency BOD Removal Efficiency BOD residual next to contact aerobic tank 66.1
25 % 131.25 mg/l
Design Concept of contact aeration Tank Fixed film Aerobic Bacteria Calculation of media BOD inluent to aeration tank BOD loading to be removed Area BOD loading Safety factor Required area media Surface area of media Total required of media Actual volume of media
6.2
= =
131.25 mg/l = 3.34 kg/d = = 0.022066 kg/m2-d 1.5 = 226.875 m2 = 110 m2/m3 = 2.06 m3 = m3 = 2.06 m3 >
Suspended of Aerobic Bacteria Design Volume of contact Aerobic Tank Vr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)
Where:
7 d = 30 m3/d = 0.5 mg.vss/mg.Bod = 131.25 mg/l = 1.17 mg/l = Influent Solube BOD Escaped treatme =
Tc Mean cell residence time Q Waste water flowrate Y Yield factor co.efficient So BOD Influent Se Solube BOD escaped treatment Efluent BOD Determine the BOD of effluent of suspended solids Biodegratable of efluent solids Ultimate BOD BOD of Suspended solids BOD of Escaped treatment X MLVSS = 80% MLSS Kd
= = = = = =
Vr
=
4.81 m3
HRT
=
HRT
=
Vr/Q 0.16 d 3.85 hr
19.5 27.69 18.83 1.17 2000 0.06
mg/l mg/l mg/l mg/l mg/l 1/d
Hydraulic retention time
Recheck hydraulic retention time
6.3
Design HRT=6hr > 3.85 hr Design Oxygen requirement O2
Where:
a BOD eliminated coefficiency
= a x Lr + b x Sa 0.5 kg.O2/kg.BOD
13 kg/d 0.07 kg.O2/kg.MLVSS-d 57.52 kg/d
Lr Total BOD load to be treated b Sludge endogenouse coefficent Sa MLVSS in aeration tank So:
O2
10.53 kg.O2/d
=
6.5 0.277 584.68 20 701.62 2 1403.23 0.97
% kg.O2/m3 of air m3/d % m3/d
1.5 40 0.3 0.81 380/3/50 2
kw mm kg.f/cm3 m3/min
Solubility air in sewage Oxygen content in air Volume of air required Provide capacity lost Design Safety
6.4
Select Air Blower Power Discharge bore Pressure air discharge Electricity Total
6.5
m3/d m3/min
set
Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) The mass of activated sludge(Px)
= Y/(1+Kd.Tc) 0.35 Kg.VSS/Kg.BOD = Yobes.Q.(So-Se)/1000 = 1.40 Kg.VSS/d =
The increase in the total mass of MLSS(Px) = Px/80% 1.75 Kg.SS/d = 7000 mg/l Design sludge concentration(Xr) = 0.82 m3/d So, Volume of excess sludge = 6.6
Where:
Recheck mean cell Residence Time for control Tc Tc Mean cell residence time a weight of biomass aeration tank b weight of sludgr that must be waste c weight of sludge in effluent So, Tc
77.1
Design Concept of Sedimentation tank Design overflow rate Use overflow rate Surface area of Sedimentation tank Actual surface
= a/(b+c)
57.52 kg 5.73 kg/d 3 kg/d
= = = =
7.00 d
= = = >
24 m3/m2-d 1.25 m2 m2 1.25 m2 .
7.2
7.3
7.4
Recheck weir overflow rate Normaly weir overflow rate of sedimentation tank is not more than 125m3/m-d 6.6 m Length of construct weir = So, weir overflow rate = Flow rate / Length of construct weir 18.94 m3/m-d = Check volume of sedimentation tank 3 hr Hydraulic retention time = 4 m3 Therefor, required the sedimentation tank = Design of return sludge rate Qr Qr Volume of return sludge Q Flow rate of waste water X MLSS Xr Sludge concentration in the bottom of sedimentation tank Volume of return sludge(Qr) Recheck of return sludge rate Normal of return sludge rate (Qr/Q) So, return sludge rate (Qr/Q) Air Lift pump Power Decharge bore Pressure Air decharge rate Electricity Total
= QX/(Xr-X) = =
30 m3/d 2500 mg/l
= =
7000 mg/l 16.67 m3/d
= =
0.25-1 0.56
= = = = = =
0.75 32 0.2 0.65 380/3/50 1
kw mm kg.f/cm2 m3/min set
allowed to change
NOT ALLOWED to change
of water consumption
pretreat organic BOD. ater flowrate.
mber which is anaerobic condition.
mg.vss/mg.Bod
lube BOD Escaped treatment+BOD of suspended solid.
kg.O2/kg.BOD
kg.O2/kg.MLVSS-d
kg.O2/m3 of air
Kg.VSS/Kg.BOD So-Se)/1000
n 125m3/m-d Length of construct weir
OPTIMA CONSULTANT PROJECT: KOHSANTEPHEAP SECTION: PLUMBING SECTION DATE: REV.: Assumed Data * You are allowed to change Calculated data * You are NOT ALLOWED to change WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION Waste Water Influent to WWTP Influent quantity Total influent quantity Water consumption
1-
= = = =
So : Let say Total inluent Influent quality BOD5 SS 2-
3-
Waste Water Efluent from WWTP Efluent quality BOD5 SS
of water consumption m3/d m3/d m3/d
= =
250 mg/l 300 mg/l
= =
20 mg/l 30 mg/l
Waste Water Treatment Plants (WWTP) Diagram Influent Q= 36m3/d BOD=250mg/l
Solid separation tank Q= 36m3/d BOD removal efficiency 30% BOD=175mg/l
Anaerobic filter tank Q= 36m3/d BOD removal efficiency 25% BOD=131.25mg/l Return sludge Q=20m3/d
90% 40 36 36
Contact aeration tank Q= 36m3/d BOD=20mg/l
To Sludge Disposal
Sedimentation tank
Storage tank
To Public Drainage 4-
Design Concept of Solid Separation Tank This 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 Therefor, require solid separation tank Actual tank capacity
= = = >
BOD Concept of removal Efficiency BOD Removal Efficiency = Influent BOD = BOD residual next to anaerobic filter tank = 5-
24 hr 36 m3 m3 36 m3
30 % 250 mg/l 175 mg/l
Design Concept of Anaerobic Filter Tank
In this step, it reduces remaining BOD from solid separation chamber which is anaerobic con It uses anaerobic microorganisms to further remove the BOD. 5.1
5.2
Design concept of media Influent BOD BOD Residual next to contact aerobic tank Total BOD loading removal Use area BOD loading Peak flow Required area media Surface area of media Required total volume of media Actual volume of media
= 175 mg/l = 131.25 mg/l = 1.575 kg/d = 0.005143 kg/m2-d = 2 = 612.5 m2 = 110 m2/m3 = 5.57 m3 = m3 > 5.57 m3
Design Volume of anaerobic Filter tank Hydraulic retention time = Therefor, required the anaerobic filter tank = Actual tank capacity = >
4 hr 6.00 m3 m3 6.00 m3
BOD Concept of removal efficiency BOD Removal Efficiency = BOD residual next to contact aerobic tank = 66.1
Design Concept of contact aeration Tank Fixed film Aerobic Bacteria Calculation of media BOD inluent to aeration tank BOD loading to be removed Area BOD loading Safety factor Required area media Surface area of media Total required of media Actual volume of media
6.2
25 % 131.25 mg/l
= 131.25 mg/l = 4.01 kg/d = 0.026479 kg/m2-d = 1.5 = 226.875 m2 = 110 m2/m3 = 2.06 m3 = m3 > 2.06 m3
Suspended of Aerobic Bacteria Design Volume of contact Aerobic Tank Vr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)
Where:
Tc Mean cell residence time = 7 d Q Waste water flowrate = 36 m3/d Y Yield factor co.efficient = 0.5 mg.vss/mg.Bod So BOD Influent = 131.25 mg/l Se Solube BOD escaped treatment = 1.17 mg/l Efluent BOD = Influent Solube BOD Escaped treatment Determine the BOD of effluent of suspended solids Biodegratable of efluent solids = 19.5 mg/l Ultimate BOD = 27.69 mg/l BOD of Suspended solids = 18.83 mg/l BOD of Escaped treatment = 1.17 mg/l X MLVSS = 80% MLSS = 2000 mg/l Kd = 0.06 1/d Vr
=
5.77 m3
Hydraulic retention time HRT
=
HRT
=
Vr/Q 0.16 d 3.85 hr
Recheck hydraulic retention time
6.3
Design HRT=6hr > 3.85 hr Design Oxygen requirement O2
Where:
a BOD eliminated coefficiency
= a x Lr + b x Sa 0.5 kg.O2/kg.BOD
Lr Total BOD load to be treated b Sludge endogenouse coefficent Sa MLVSS in aeration tank So:
O2
13 kg/d 0.07 kg.O2/kg.MLVSS-d 57.52 kg/d =
10.53 kg.O2/d
Solubility air in sewage Oxygen content in air Volume of air required Provide capacity lost Design Safety
6.4
% kg.O2/m3 of air m3/d % m3/d
1.5 40 0.3 0.81 380/3/50 2
kw mm kg.f/cm3 m3/min
m3/d m3/min
Select Air Blower Power Discharge bore Pressure air discharge Electricity Total
6.5
6.5 0.277 584.68 20 701.62 2 1403.23 0.97
set
Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) The mass of activated sludge(Px)
= Y/(1+Kd.Tc) = 0.35 Kg.VSS/Kg.BOD = Yobes.Q.(So-Se)/1000 = 1.70 Kg.VSS/d
The increase in the total mass of MLSS(Px) = Px/80% = 2.125 Kg.SS/d Design sludge concentration(Xr) = 7000 mg/l So, Volume of excess sludge = 0.82 m3/d 6.6
Where:
Recheck mean cell Residence Time for control Tc
Tc Mean cell residence time a weight of biomass aeration tank = b weight of sludgr that must be waste = c weight of sludge in effluent So, Tc
77.1
= a/(b+c)
Design Concept of Sedimentation tank Design overflow rate Use overflow rate Surface area of Sedimentation tank Actual surface
57.52 kg 5.73 kg/d 3 kg/d
=
7.00 d
= = = >
24 m3/m2-d 1.50 m2 m2 1.50 m2 .
7.2
7.3
7.4
Recheck weir overflow rate Normaly weir overflow rate of sedimentation tank is not more than 125m3/m-d Length of construct weir = 6.6 m So, weir overflow rate = Flow rate / Length of construct weir = 18.94 m3/m-d Check volume of sedimentation tank Hydraulic retention time = 3 hr Therefor, required the sedimentation tank = 5 m3 Design of return sludge rate Qr = QX/(Xr-X) Qr Volume of return sludge Q Flow rate of waste water = 36 m3/d X MLSS = 2500 mg/l Xr Sludge concentration in the bottom of sedimentation tank = 7000 mg/l Volume of return sludge(Qr) = 20 m3/d Recheck of return sludge rate Normal of return sludge rate (Qr/Q) So, return sludge rate (Qr/Q)
= =
0.25-1 0.56
Air Lift pump Power Decharge bore Pressure
= = =
Electricity Total
= =
0.75 32 0.2 0.65 380/3/50 1
kw mm kg.f/cm2 m3/min set
allowed to change
NOT ALLOWED to change
of water consumption
d pretreat organic BOD. water flowrate.
amber which is anaerobic condition.
mg.vss/mg.Bod
lube BOD Escaped treatment+BOD of suspended solid.
kg.O2/kg.BOD
kg.O2/kg.MLVSS-d
kg.O2/m3 of air
Kg.VSS/Kg.BOD So-Se)/1000
han 125m3/m-d Length of construct weir
OPTIMA CONSULTANT PROJECT: SECTION: PLUMBING SECTION DATE: REV.: Assumed Data * You are allowed to change Calculated data * You are NOT ALLOWED to change WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION Waste Water Influent to WWTP
1-
Total influent quantity Water consumption So : Let say Total inluent
Influent quantity = = = =
BOD5 SS 2-
of water consumption m3/d m3/d m3/d
Influent quality = =
250 mg/l 300 mg/l
Efluent quality = =
20 mg/l 30 mg/l
Waste Water Efluent from WWTP BOD5 SS
3-
90% 58.5 52.65 53
Waste Water Treatment Plants (WWTP) Diagram Influent Q= 36m3/d BOD=250mg/l
Solid separation tank Q= 36m3/d BOD removal efficiency 30% BOD=175mg/l
Anaerobic filter tank Q= 36m3/d BOD removal efficiency 25% BOD=131.25mg/l Return sludge Q=20m3/d
Contact aeration tank Q= 36m3/d BOD=20mg/l
To Sludge Disposal
Sedimentation tank
Storage tank
To Public Drainage 4-
Design Concept of Solid Separation Tank This 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 Therefor, require solid separation tank Actual tank capacity
= = = >
BOD Concept of removal Efficiency BOD Removal Efficiency = Influent BOD = BOD residual next to anaerobic filter tank = 5-
24 hr 53 m3 m3 53 m3
30 % 250 mg/l 175 mg/l
Design Concept of Anaerobic Filter Tank
In this step, it reduces remaining BOD from solid separation chamber which is anaerobic con It uses anaerobic microorganisms to further remove the BOD. 5.1
5.2
Design concept of media Influent BOD BOD Residual next to contact aerobic tank Total BOD loading removal Use area BOD loading Peak flow Required area media Surface area of media Required total volume of media Actual volume of media
= 175 mg/l = 131.25 mg/l = 2.31875 kg/d = 0.007571 kg/m2-d = 2 = 612.5 m2 = 110 m2/m3 = 5.57 m3 = m3 > 5.57 m3
Design Volume of anaerobic Filter tank Hydraulic retention time = Therefor, required the anaerobic filter tank = Actual tank capacity = >
4 hr 8.83 m3 m3 8.83 m3
BOD Concept of removal efficiency BOD Removal Efficiency = BOD residual next to contact aerobic tank = 66.1
25 % 131.25 mg/l
Design Concept of contact aeration Tank Fixed film Aerobic Bacteria Calculation of media BOD inluent to aeration tank = 131.25 mg/l BOD loading to be removed = 5.90 kg/d Area BOD loading = 0.038983 kg/m2-d Safety factor = 1.5 Required area media = 226.875 m2 Surface area of media = 110 m2/m3 Total required of media = 2.06 m3 Actual volume of media = m3 > 2.06 m3
6.2
Suspended of Aerobic Bacteria Design Volume of contact Aerobic Tank Vr = Tc x Q x Y x (So-Se) / X(1+Kd x Tc)
Where:
Tc Mean cell residence time = 7 d Q Waste water flowrate = 53 m3/d Y Yield factor co.efficient = 0.5 mg.vss/mg.Bod So BOD Influent = 131.25 mg/l Se Solube BOD escaped treatment = 1.17 mg/l Efluent BOD = Influent Solube BOD Escaped treatment+ Determine the BOD of effluent of suspended solids Biodegratable of efluent solids = 19.5 mg/l Ultimate BOD = 27.69 mg/l BOD of Suspended solids = 18.83 mg/l BOD of Escaped treatment = 1.17 mg/l X MLVSS = 80% MLSS = 2000 mg/l Kd = 0.06 1/d Vr
=
8.50 m3
Hydraulic retention time HRT
=
HRT
=
Vr/Q 0.16 d 3.85 hr
Recheck hydraulic retention time
6.3
Design HRT=6hr > 3.85 hr Design Oxygen requirement O2
Where:
a BOD eliminated coefficiency
= a x Lr + b x Sa 0.5 kg.O2/kg.BOD
Lr Total BOD load to be treated b Sludge endogenouse coefficent Sa MLVSS in aeration tank So:
13 kg/d 0.07 kg.O2/kg.MLVSS-d 57.52 kg/d
O2
=
10.53 kg.O2/d
Solubility air in sewage Oxygen content in air Volume of air required Provide capacity lost Design Safety
6.4
% kg.O2/m3 of air m3/d % m3/d
1.5 40 0.3 0.81 380/3/50 2
kw mm kg.f/cm3 m3/min
m3/d m3/min
Select Air Blower Power Discharge bore Pressure air discharge Electricity Total
6.5
6.5 0.277 584.68 20 701.62 2 1403.23 0.97
set
Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) The mass of activated sludge(Px)
=
Y/(1+Kd.Tc) = 0.35 Kg.VSS/Kg.BOD = Yobes.Q.(So-Se)/1000 = 2.50 Kg.VSS/d
The increase in the total mass of MLSS(Px) = Px/80% = 3.125 Kg.SS/d Design sludge concentration(Xr) = 7000 mg/l So, Volume of excess sludge = 0.82 m3/d 6.6
Where:
Recheck mean cell Residence Time for control Tc Tc Mean cell residence time a weight of biomass aeration tank b weight of sludgr that must be waste c weight of sludge in effluent So, Tc
77.1
Design Concept of Sedimentation tank Design overflow rate Use overflow rate Surface area of Sedimentation tank Actual surface
= a/(b+c)
= =
57.52 kg 5.73 kg/d 3 kg/d
=
7.00 d
= = = >
24 m3/m2-d 2.21 m2 m2 2.21 m2 .
7.2
7.3
7.4
Recheck weir overflow rate Normaly weir overflow rate of sedimentation tank is not more than 125m3/m-d Length of construct weir = 6.6 m So, weir overflow rate = Flow rate / Length of construct weir = 18.94 m3/m-d Check volume of sedimentation tank Hydraulic retention time = 3 hr Therefor, required the sedimentation tank = 7 m3 Design of return sludge rate Qr Qr Volume of return sludge Q Flow rate of waste water X MLSS Xr Sludge concentration in the bottom of sedimentation tank Volume of return sludge(Qr)
= QX/(Xr-X) = =
53 m3/d 2500 mg/l
= =
7000 mg/l 29.45 m3/d
Recheck of return sludge rate Normal of return sludge rate (Qr/Q) So, return sludge rate (Qr/Q)
= =
0.25-1 0.56
Air Lift pump Power Decharge bore Pressure
= = =
Electricity Total
= =
0.75 32 0.2 0.65 380/3/50 1
kw mm kg.f/cm2 m3/min set
owed to change
OT ALLOWED to change
f water consumption
pretreat organic BOD. water flowrate.
mber which is anaerobic condition.
mg.vss/mg.Bod
be BOD Escaped treatment+BOD of suspended solid.
g.O2/kg.BOD
g.O2/kg.MLVSS-d
g.O2/m3 of air
g.VSS/Kg.BOD
n 125m3/m-d
ength of construct weir
