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BOD5 given Oxygen Required in Aeration Tank Calculator

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BOD5 Important Formulas of Oxygen Requirement of the Aeration Tank BOD to Ultimate BOD D.O Saturation More >>

✖Ultimate BOD is the amount of oxygen required to decompose all of the organic material after infinite time.ⓘ Ultimate BOD [BOD^u]			+10% -10%
✖Sewage Discharge is the flow rate of sewage when it is being discharged in the river.ⓘ Sewage Discharge [Q_s]			+10% -10%
✖Influent BOD is the total amount of BOD which is present in the incoming sewage.ⓘ Influent BOD [Q_i]			+10% -10%
✖Effluent BOD is the amount of BOD present in the outing sewage.ⓘ Effluent BOD [Q]			+10% -10%
✖Theoretical Oxygen Requirement is the calculated amount of oxygen required to oxidize a compound to its final oxidation products.ⓘ Theoretical Oxygen Requirement [O₂]			+10% -10%
✖Volume of Wasted Sludge per day is the amount of sludge that is removed or wasted from a wastewater treatment process on a daily basis.ⓘ Volume of Wasted Sludge per day [Q_w]			+10% -10%
✖MLSS in Returned or Wasted Sludge is the mixed liquor suspended solids in returned sludge in terms of mg per liter.ⓘ MLSS in Returned or Wasted Sludge [X^R]			+10% -10%

✖BOD5 given Oxygen Required in Aeration Tank is the amount of oxygen required by aerobic microorganisms to decompose organic matter in water over a period of five days at a temperature of 20°C.ⓘ BOD5 given Oxygen Required in Aeration Tank [BOD_5a]

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BOD5 given Oxygen Required in Aeration Tank Solution

STEP 0: Pre-Calculation Summary

Formula Used

BOD5 given Oxygen Required in Aeration Tank = Ultimate BOD*(Sewage Discharge*(Influent BOD-Effluent BOD))/(Theoretical Oxygen Requirement+(1.42*Volume of Wasted Sludge per day*MLSS in Returned or Wasted Sludge))
BOD_5a = BOD^u*(Q_s*(Q_i-Q))/(O₂+(1.42*Q_w*X^R))
This formula uses 8 Variables

Variables Used

BOD5 given Oxygen Required in Aeration Tank - (Measured in Kilogram per Cubic Meter) - BOD5 given Oxygen Required in Aeration Tank is the amount of oxygen required by aerobic microorganisms to decompose organic matter in water over a period of five days at a temperature of 20°C.
Ultimate BOD - (Measured in Kilogram per Cubic Meter) - Ultimate BOD is the amount of oxygen required to decompose all of the organic material after infinite time.
Sewage Discharge - (Measured in Cubic Meter per Second) - Sewage Discharge is the flow rate of sewage when it is being discharged in the river.
Influent BOD - (Measured in Kilogram per Cubic Meter) - Influent BOD is the total amount of BOD which is present in the incoming sewage.
Effluent BOD - (Measured in Kilogram per Cubic Meter) - Effluent BOD is the amount of BOD present in the outing sewage.
Theoretical Oxygen Requirement - (Measured in Kilogram per Second) - Theoretical Oxygen Requirement is the calculated amount of oxygen required to oxidize a compound to its final oxidation products.
Volume of Wasted Sludge per day - (Measured in Cubic Meter per Second) - Volume of Wasted Sludge per day is the amount of sludge that is removed or wasted from a wastewater treatment process on a daily basis.
MLSS in Returned or Wasted Sludge - (Measured in Kilogram per Cubic Meter) - MLSS in Returned or Wasted Sludge is the mixed liquor suspended solids in returned sludge in terms of mg per liter.

STEP 1: Convert Input(s) to Base Unit

Ultimate BOD: 2 Milligram per Liter --> 0.002 Kilogram per Cubic Meter (Check conversion here)
Sewage Discharge: 10 Cubic Meter per Second --> 10 Cubic Meter per Second No Conversion Required
Influent BOD: 13.2 Milligram per Liter --> 0.0132 Kilogram per Cubic Meter (Check conversion here)
Effluent BOD: 0.4 Milligram per Liter --> 0.0004 Kilogram per Cubic Meter (Check conversion here)
Theoretical Oxygen Requirement: 2.5 Milligram per Day --> 2.89351851851852E-11 Kilogram per Second (Check conversion here)
Volume of Wasted Sludge per day: 9.5 Cubic Meter per Second --> 9.5 Cubic Meter per Second No Conversion Required
MLSS in Returned or Wasted Sludge: 1.4 Milligram per Liter --> 0.0014 Kilogram per Cubic Meter (Check conversion here)

STEP 2: Evaluate Formula

Substituting Input Values in Formula

BOD_5a = BOD^u*(Q_s*(Q_i-Q))/(O₂+(1.42*Q_w*X^R)) --> 0.002*(10*(0.0132-0.0004))/(2.89351851851852E-11+(1.42*9.5*0.0014))

Evaluating ... ...

BOD_5a = 0.0135550142755365

STEP 3: Convert Result to Output's Unit

0.0135550142755365 Kilogram per Cubic Meter -->13.5550142755365 Milligram per Liter (Check conversion here)

FINAL ANSWER

13.5550142755365 ≈ 13.55501 Milligram per Liter <-- BOD5 given Oxygen Required in Aeration Tank

(Calculation completed in 00.004 seconds)

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< BOD5 Calculators

BOD5 given Oxygen Required in Aeration Tank

LaTeX Go BOD5 given Oxygen Required in Aeration Tank = Ultimate BOD*(Sewage Discharge*(Influent BOD-Effluent BOD))/(Theoretical Oxygen Requirement+(1.42*Volume of Wasted Sludge per day*MLSS in Returned or Wasted Sludge))

BOD5 given Ratio of BOD to Ultimate BOD

LaTeX Go BOD5 given Ratio of BOD to Ultimate BOD = Ratio of BOD to Ultimate BOD*Ultimate BOD

BOD5 when Ratio of BOD to Ultimate BOD is 0.68

LaTeX Go 5 Days BOD = Ultimate BOD*0.68

< Important Formulas of Oxygen Requirement of the Aeration Tank Calculators

BOD5 given Oxygen Required in Aeration Tank

BOD5 given Ratio of BOD to Ultimate BOD

LaTeX Go BOD5 given Ratio of BOD to Ultimate BOD = Ratio of BOD to Ultimate BOD*Ultimate BOD

Ratio of BOD to Ultimate BOD

LaTeX Go Ratio of BOD to Ultimate BOD = BOD of 5 days at 20° C/Ultimate BOD

BOD5 when Ratio of BOD to Ultimate BOD is 0.68

LaTeX Go 5 Days BOD = Ultimate BOD*0.68

BOD5 given Oxygen Required in Aeration Tank Formula

LaTeX Go

What is BOD?

Biochemical oxygen demand is the amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period.

How to Calculate BOD5 given Oxygen Required in Aeration Tank?

BOD5 given Oxygen Required in Aeration Tank calculator uses BOD5 given Oxygen Required in Aeration Tank = Ultimate BOD*(Sewage Discharge*(Influent BOD-Effluent BOD))/(Theoretical Oxygen Requirement+(1.42*Volume of Wasted Sludge per day*MLSS in Returned or Wasted Sludge)) to calculate the BOD5 given Oxygen Required in Aeration Tank, BOD5 given Oxygen Required in Aeration Tank formula is defined as a measure of the amount of oxygen required to remove organic matter in wastewater during the aeration process, which is essential for the design and operation of aeration tanks in wastewater treatment plants. BOD5 given Oxygen Required in Aeration Tank is denoted by BOD_5a symbol.

How to calculate BOD5 given Oxygen Required in Aeration Tank using this online calculator? To use this online calculator for BOD5 given Oxygen Required in Aeration Tank, enter Ultimate BOD (BOD^u), Sewage Discharge (Q_s), Influent BOD (Q_i), Effluent BOD (Q), Theoretical Oxygen Requirement (O₂), Volume of Wasted Sludge per day (Q_w) & MLSS in Returned or Wasted Sludge (X^R) and hit the calculate button. Here is how the BOD5 given Oxygen Required in Aeration Tank calculation can be explained with given input values -> 13555.01 = 0.002*(10*(0.0132-0.0004))/(2.89351851851852E-11+(1.42*9.5*0.0014)).

FAQ

What is BOD5 given Oxygen Required in Aeration Tank?

BOD5 given Oxygen Required in Aeration Tank formula is defined as a measure of the amount of oxygen required to remove organic matter in wastewater during the aeration process, which is essential for the design and operation of aeration tanks in wastewater treatment plants and is represented as BOD_5a = BOD^u*(Q_s*(Q_i-Q))/(O₂+(1.42*Q_w*X^R)) or BOD5 given Oxygen Required in Aeration Tank = Ultimate BOD*(Sewage Discharge*(Influent BOD-Effluent BOD))/(Theoretical Oxygen Requirement+(1.42*Volume of Wasted Sludge per day*MLSS in Returned or Wasted Sludge)). Ultimate BOD is the amount of oxygen required to decompose all of the organic material after infinite time, Sewage Discharge is the flow rate of sewage when it is being discharged in the river, Influent BOD is the total amount of BOD which is present in the incoming sewage, Effluent BOD is the amount of BOD present in the outing sewage, Theoretical Oxygen Requirement is the calculated amount of oxygen required to oxidize a compound to its final oxidation products, Volume of Wasted Sludge per day is the amount of sludge that is removed or wasted from a wastewater treatment process on a daily basis & MLSS in Returned or Wasted Sludge is the mixed liquor suspended solids in returned sludge in terms of mg per liter.

How to calculate BOD5 given Oxygen Required in Aeration Tank?

BOD5 given Oxygen Required in Aeration Tank formula is defined as a measure of the amount of oxygen required to remove organic matter in wastewater during the aeration process, which is essential for the design and operation of aeration tanks in wastewater treatment plants is calculated using BOD5 given Oxygen Required in Aeration Tank = Ultimate BOD*(Sewage Discharge*(Influent BOD-Effluent BOD))/(Theoretical Oxygen Requirement+(1.42*Volume of Wasted Sludge per day*MLSS in Returned or Wasted Sludge)). To calculate BOD5 given Oxygen Required in Aeration Tank, you need Ultimate BOD (BOD^u), Sewage Discharge (Q_s), Influent BOD (Q_i), Effluent BOD (Q), Theoretical Oxygen Requirement (O₂), Volume of Wasted Sludge per day (Q_w) & MLSS in Returned or Wasted Sludge (X^R). With our tool, you need to enter the respective value for Ultimate BOD, Sewage Discharge, Influent BOD, Effluent BOD, Theoretical Oxygen Requirement, Volume of Wasted Sludge per day & MLSS in Returned or Wasted Sludge and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

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