2. Septic Tank Calculation

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PROJECT: SALINA

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

* You are allowed to change

Calculated data

* You are NOT ALLOWED to change

DIMENSION OF SEPTIC TANK

*- Dimension of septic tank

1./ Referring to table below:

Item

Description

Qty Units

6 *-Flow per day in TYPE OF ESTABLISHMENT_{*-Number of Persons in TYPE OF ESTABLISHMENT} 50₈₀ _personsgallons

Total flow per day in gallons: 4000 gallons

2./ The liquid volume of tank in gallons:

V = 1125+0.75*Q = 4125 gallons

Where: V: the liquid volume of the tank in gallons 16 Cu.m

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: 3.472 m

Length of Septic Tank : 5.21 m

THE TABLE QUANTITIES OF SEWAGE FLOW ITEM TYPE OF ESTABLISHMENT

1 Small dwellings with seasonal occupancy 50

2 single family dwellings 75

3 Multiple family dwellings (Apartment) 60

4 Rooming houses 40

5 Boarding houses 50

6 Hotels without private bath 50

7 Hotels with private baths (2 persons per room) 60

8 Restaurants (toilet and kitchen waste per patron) 7

9 Restaurants (kitchen waste per meals serve) 3

10 Tourist camps or trailer parks with central bathhouse 35

11 Tourist courts or mobile home parks with ind. Bath 50

12 Resort camps night and day with limited plumbing 50

13 Luxury camps 100

14 Work or construction camp 50

15 Day camps no meals serve 15

16 day schools without cafeterials, gym. Or showers 15

GALLONS PER PERSON PER

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17 day schools with cafeterials, gym. Or showers 25

18 Day school with but w/o gym or showers 20

19 Boarding schools 75100

20 Hospitals 150

21 Institutions other than a hospitals 75

22 Factories (exclusive of industrial waste) 15

23 Picnic parks with toilet, bath houses 10

24 Swimming pools and bath houses 10

25 Luxury residences 100

26 Country clubs (per resident member) 100

27 Motels(per bed space) 40

28 Motes with bath, toilet, and kitchen, waste 50

29 Drive in threaters (per car space) 5

30 Movie theaters (per auditorium seat) 5

31 Airport (per passenger) 3

32 Stores(per toilet room) 400

33 Service stations(per vehicle served) 10

34 Seft service laundries(gal. per wash per person) 50

NOTE:

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* You are NOT ALLOWED to change 1056.688 50 4032 60 10 3 150 GALLONS PER PERSON PER DAY

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25 250 125 35 150 5 12

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PROJECT: SALINA

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

DIMENSION OF SEPTIC TANK

Item

Description

Qty Units

6 *-Flow per day in TYPE OF ESTABLISHMENT_{*-Number of Persons in TYPE OF ESTABLISHMENT} 50₈₀ _personsgallons

V = 1125+0.75*Q = 4125 gallons

4 Rooming houses 40

13 Luxury camps 100

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20 Hospitals 150

NOTE:

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25 250 125 35 150 5 12

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PROJECT: SALINA

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

DIMENSION OF SEPTIC TANK

Item

Description

Qty Units

6 *-Flow per day in TYPE OF ESTABLISHMENT_{*-Number of Persons in TYPE OF ESTABLISHMENT} ₁₆₄50 _personsgallons

V = 1125+0.75*Q = 7275 gallons

4 Rooming houses 40

13 Luxury camps 100

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20 Hospitals 150

NOTE:

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25 250 125 35 150 5 12

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OPTIMA CONSULTANT

PROJECT: KOHSANTEPHEAP

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION

1- Waste Water Influent to WWTP

Influent quantity

Total influent quantity = 90% of water consumption Water consumption = 20 m3/d

So : 18 m3/d

Let say Total inluent = 18 m3/d Influent quality

BOD5 = 250 mg/l SS = 300 mg/l 2- Waste Water Efluent from WWTP

Efluent quality

BOD5 = 20 mg/l SS = 30 mg/l 3- Waste Water Treatment Plants (WWTP) Diagram

Influent Q= 18m3/d BOD=250mg/l

Q= 18m3/d BOD removal efficiency 30% BOD=175mg/l

Q= 18m3/d BOD removal efficiency 25% BOD=131.25mg/l Return sludge

Q=10m3/d

Q= 18m3/d BOD=20mg/l

Solid separation tank

Anaerobic filter tank

Contact aeration tank

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To Public Drainage 4- Design Concept of Solid Separation Tank

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 = 24 hr Therefor, require solid separation tank = 18 m3

Actual tank capacity = m3 > 18 m3

BOD Concept of removal Efficiency

BOD Removal Efficiency = 30 % Influent BOD = 250 mg/l BOD residual next to anaerobic filter tank = 175 mg/l 5- 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 Design concept of media

Influent BOD = 175 mg/l BOD Residual next to contact aerobic tank = 131.25 mg/l Total BOD loading removal = 0.7875 kg/d Use area BOD loading = 0.002571 kg/m2-d Peak flow = 2

Required area media = 612.5 m2 Surface area of media = 110 m2/m3 Required total volume of media = 5.57 m3

Actual volume of media = m3 > 5.57 m3 5.2 Design Volume of anaerobic Filter tank

Hydraulic retention time = 4 hr Therefor, required the anaerobic filter tank = 3.00 m3

Actual tank capacity = m3 > 3.00 m3 To Sludge

Disposal Sedimentation tank

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BOD Concept of removal efficiency

BOD Removal Efficiency = 25 % BOD residual next to contact aerobic tank = 131.25 mg/l 6- Design Concept of contact aeration Tank

6.1 Fixed film Aerobic Bacteria

Calculation of media

BOD inluent to aeration tank = 131.25 mg/l BOD loading to be removed = 2.00 kg/d Area BOD loading = 0.01324 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 = 18 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+BOD of suspended solid. 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 = 2.89 m3 Hydraulic retention time

HRT = Vr/Q 0.16 d HRT = 3.85 hr

Recheck hydraulic retention time Design HRT=6hr > 3.85 hr 6.3 Design Oxygen requirement

O2 = a x Lr + b x Sa

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Lr Total BOD load to be treated 13 kg/d

b Sludge endogenouse coefficent 0.07 kg.O2/kg.MLVSS-d Sa MLVSS in aeration tank 57.52 kg/d

So: O2 = 10.53 kg.O2/d Solubility air in sewage 6.5 %

Oxygen content in air 0.277 kg.O2/m3 of air Volume of air required 584.68 m3/d

Provide capacity lost 20 %

701.62 m3/d

Design Safety 2 1403.23 m3/d

0.97 m3/min

6.4 Select Air Blower

Power 1.5 kw Discharge bore 40 mm

Pressure 0.3 kg.f/cm3 air discharge 0.81 m3/min

Electricity 380/3/50 Total 2 set

6.5 Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) = Y/(1+Kd.Tc)

= 0.35 Kg.VSS/Kg.BOD The mass of activated sludge(Px) = 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 Design sludge concentration(Xr) = 7000 mg/l

So, Volume of excess sludge = 0.82 m3/d 6.6 Recheck mean cell Residence Time for control

Tc = a/(b+c)

Where: Tc Mean cell residence time

a weight of biomass aeration tank = 57.52 kg b weight of sludgr that must be waste = 5.73 kg/d

c weight of sludge in effluent 3 kg/d So, Tc = 7.00 d 7- Design Concept of Sedimentation tank

7.1 Design overflow rate

Use overflow rate = 24 m3/m2-d Surface area of Sedimentation tank = 0.75 m2

Actual surface = m2

> 0.75 m2 .

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7.2 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

7.3 Check volume of sedimentation tank

Hydraulic retention time = 3 hr Therefor, required the sedimentation tank = 3 m3 7.4 Design of return sludge rate

Qr = QX/(Xr-X) Qr Volume of return sludge

Q Flow rate of waste water = 18 m3/d X MLSS = 2500 mg/l

Xr Sludge concentration in the bottom of

sedimentation tank = 7000 mg/l Volume of return sludge(Qr) = 10 m3/d

Recheck of return sludge rate

Normal of return sludge rate (Qr/Q) = 0.25-1 So, return sludge rate (Qr/Q) = 0.56

Air Lift pump

Power = 0.75 kw Decharge bore = 32 mm

Pressure = 0.2 kg.f/cm2 Air decharge rate = 0.65 m3/min

Electricity = 380/3/50 Total = 1 set

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mg.vss/mg.Bod

Influent Solube BOD Escaped treatment+BOD of suspended solid.

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kg.O2/kg.MLVSS-d

kg.O2/m3 of air

Kg.VSS/Kg.BOD Yobes.Q.(So-Se)/1000

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Flow rate / Length of construct weir

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OPTIMA CONSULTANT

PROJECT: KOHSANTEPHEAP

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION

Total influent quantity = 90% of water consumption Water consumption = 32.5 m3/d

So : 29.25 m3/d

Let say Total inluent = 30 m3/d Influent quality

Efluent quality

Q=16.67m3/d

Q= 30m3/d BOD=20mg/l Solid separation tank

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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 = 24 hr Therefor, require solid separation tank = 30 m3 Actual tank capacity = m3 > 30 m3 BOD Concept of removal Efficiency

In this step, it reduces remaining BOD from solid separation chamber which is anaerobic condition. It uses anaerobic microorganisms to further remove the BOD.

5.1 Design concept of media

Influent BOD = 175 mg/l BOD Residual next to contact aerobic tank = 131.25 mg/l Total BOD loading removal = 1.3125 kg/d Use area BOD loading = 0.004286 kg/m2-d Peak flow = 2

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6.1 Fixed film Aerobic Bacteria Calculation of media

BOD inluent to aeration tank = 131.25 mg/l BOD loading to be removed = 3.34 kg/d Area BOD loading = 0.022066 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

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 = 4.81 m3

Hydraulic retention time

HRT = Vr/Q 0.16 d HRT = 3.85 hr Recheck hydraulic retention time

Design HRT=6hr > 3.85 hr 6.3 Design Oxygen requirement

O2 = a x Lr + b x Sa

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701.62 m3/d

Design Safety 2

1403.23 m3/d

0.97 m3/min

Power 1.5 kw

Discharge bore 40 mm

Pressure 0.3 kg.f/cm3

air discharge 0.81 m3/min

Electricity 380/3/50

Total 2 set

6.5 Design the quantity of excess sludg that must be waste per day

The observes yields(YOBES) = Y/(1+Kd.Tc)

= 1.75 Kg.SS/d Design sludge concentration(Xr) = 7000 mg/l So, Volume of excess sludge = 0.82 m3/d 6.6 Recheck mean cell Residence Time for control

Tc = a/(b+c) Where: Tc Mean cell residence time

a weight of biomass aeration tank = 57.52 kg b weight of sludgr that must be waste = 5.73 kg/d c weight of sludge in effluent = 3 kg/d

So, Tc = 7.00 d 7- Design Concept of Sedimentation tank

Use overflow rate = 24 m3/m2-d Surface area of Sedimentation tank = 1.25 m2 Actual surface = m2

> 1.25 m2 .

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Normaly weir overflow rate of sedimentation tank is not more than 125m3/m-d Length of construct weir = 6.6 m

Q Flow rate of waste water = 30 m3/d X MLSS = 2500 mg/l Xr Sludge concentration in the bottom of

sedimentation tank = 7000 mg/l Volume of return sludge(Qr) = 16.67 m3/d

Air Lift pump

Power = 0.75 kw Decharge bore = 32 mm Pressure = 0.2 kg.f/cm2 Air decharge rate = 0.65 m3/min Electricity = 380/3/50

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mg.vss/mg.Bod

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kg.O2/kg.MLVSS-d

kg.O2/m3 of air

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OPTIMA CONSULTANT

PROJECT: KOHSANTEPHEAP

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION

Total influent quantity = 90% of water consumption Water consumption = 40 m3/d

So : = 36 m3/d Let say Total inluent = 36 m3/d Influent quality

Efluent quality

Q=20m3/d

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Influent BOD = 175 mg/l BOD Residual next to contact aerobic tank = 131.25 mg/l Total BOD loading removal = 1.575 kg/d

Use area BOD loading = 0.005143 kg/m2-d Peak flow = 2

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6.1 Fixed film Aerobic Bacteria Calculation of media

BOD inluent to aeration tank = 131.25 mg/l BOD loading to be removed = 4.01 kg/d

Area BOD loading = 0.026479 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

O2 = a x Lr + b x Sa

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701.62 m3/d

Design Safety 2

1403.23 m3/d

0.97 m3/min

Power 1.5 kw

Total 2 set

6.5 Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) = Y/(1+Kd.Tc)

Tc = a/(b+c)

Where: Tc Mean cell residence time

> 1.50 m2 .

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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

Air Lift pump

Power = 0.75 kw Decharge bore = 32 mm Pressure = 0.2 kg.f/cm2 0.65 m3/min Electricity = 380/3/50 Total = 1 set

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mg.vss/mg.Bod

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kg.O2/kg.MLVSS-d

kg.O2/m3 of air

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OPTIMA CONSULTANT

PROJECT:

SECTION: PLUMBING SECTION

DATE:

REV.:

Assumed Data

Calculated data

WASTE WATER TREATMENT PLANTS(WWTP) CALCULATION

Total influent quantity = 90% of water consumption Water consumption = 58.5 m3/d

So : = 52.65 m3/d Let say Total inluent = 53 m3/d

Influent quality

Efluent quality

Q=20m3/d

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Influent BOD = 175 mg/l BOD Residual next to contact aerobic tank = 131.25 mg/l Total BOD loading removal = 2.31875 kg/d

Use area BOD loading = 0.007571 kg/m2-d Peak flow = 2

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6.1 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

O2 = a x Lr + b x Sa

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701.62 m3/d

Design Safety 2

1403.23 m3/d

0.97 m3/min

Power 1.5 kw

Total 2 set

6.5 Design the quantity of excess sludg that must be waste per day The observes yields(YOBES) = Y/(1+Kd.Tc)

Tc = a/(b+c) Where: Tc Mean cell residence time

> 2.21 m2 .

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Q Flow rate of waste water = 53 m3/d X MLSS = 2500 mg/l Xr Sludge concentration in the bottom of

sedimentation tank = 7000 mg/l Volume of return sludge(Qr) = 29.45 m3/d

Air Lift pump

Power = 0.75 kw Decharge bore = 32 mm Pressure = 0.2 kg.f/cm2 0.65 m3/min Electricity = 380/3/50 Total = 1 set

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mg.vss/mg.Bod

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kg.O2/kg.MLVSS-d

kg.O2/m3 of air

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