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BOD5 given Oxygen Required in Aeration Tank Formula

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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. Check FAQs

BOD 5a = BOD u \cdot \frac{Q s \cdot (Q i - Q)}{O 2 + (1.42 \cdot Q w \cdot X R)}

BOD_5a - BOD5 given Oxygen Required in Aeration Tank?BOD^u - Ultimate BOD?Q_s - Sewage Discharge?Q_i - Influent BOD?Q - Effluent BOD?O₂ - Theoretical Oxygen Requirement?Q_w - Volume of Wasted Sludge per day?X^R - MLSS in Returned or Wasted Sludge?

BOD5 given Oxygen Required in Aeration Tank Example

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Here is how the BOD5 given Oxygen Required in Aeration Tank equation looks like with Values.

Here is how the BOD5 given Oxygen Required in Aeration Tank equation looks like with Units.

Here is how the BOD5 given Oxygen Required in Aeration Tank equation looks like.

13.555 = 2 \cdot \frac{10 \cdot (13.2 - 0.4)}{2.5 + (1.42 \cdot 9.5 \cdot 1.4)}

13.555mg/L = 2mg/L \cdot \frac{10m³/s \cdot (13.2mg/L - 0.4mg/L)}{2.5mg/d + (1.42 \cdot 9.5m³/s \cdot 1.4mg/L)}

13.555 = 2 \cdot \frac{10 \cdot (13.2 - 0.4)}{2.5 + (1.42 \cdot 9.5 \cdot 1.4)}

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BOD5 given Oxygen Required in Aeration Tank Solution

Follow our step by step solution on how to calculate BOD5 given Oxygen Required in Aeration Tank?

FIRST Step Consider the formula

BOD 5a = BOD u \cdot \frac{Q s \cdot (Q i - Q)}{O 2 + (1.42 \cdot Q w \cdot X R)}

Next Step Substitute values of Variables

∴

BOD 5a = 2mg/L \cdot \frac{10m³/s \cdot (13.2mg/L - 0.4mg/L)}{2.5mg/d + (1.42 \cdot 9.5m³/s \cdot 1.4mg/L)}

Next Step Convert Units

∴

BOD 5a = 0.002kg/m³ \cdot \frac{10m³/s \cdot (0.0132kg/m³ - 0.0004kg/m³)}{2.9E-11kg/s + (1.42 \cdot 9.5m³/s \cdot 0.0014kg/m³)}

Next Step Prepare to Evaluate

∴

BOD 5a = 0.002 \cdot \frac{10 \cdot (0.0132 - 0.0004)}{2.9E-11 + (1.42 \cdot 9.5 \cdot 0.0014)}

Next Step Evaluate

∴

BOD 5a = 0.0135550142755365kg/m³

Next Step Convert to Output's Unit

∴

BOD 5a = 13.5550142755365mg/L

LAST Step Rounding Answer

∴

BOD 5a = 13.555mg/L

BOD5 given Oxygen Required in Aeration Tank Formula Elements

Variables

BOD5 given Oxygen Required in Aeration Tank

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.

Symbol: BOD_5a

Measurement: DensityUnit: mg/L

Note: Value should be greater than 0.

Formulas to find BOD5 given Oxygen Required in Aeration Tank

Ultimate BOD

Ultimate BOD is the amount of oxygen required to decompose all of the organic material after infinite time.

Symbol: BOD^u

Measurement: DensityUnit: mg/L

Note: Value should be greater than 0.

Formulas to find Ultimate BOD

Sewage Discharge

Sewage Discharge is the flow rate of sewage when it is being discharged in the river.

Symbol: Q_s

Measurement: Volumetric Flow RateUnit: m³/s

Note: Value should be greater than 0.

Formulas to find Sewage Discharge

Influent BOD

Influent BOD is the total amount of BOD which is present in the incoming sewage.

Symbol: Q_i

Measurement: DensityUnit: mg/L

Note: Value should be greater than 0.

Formulas to find Influent BOD

Effluent BOD

Effluent BOD is the amount of BOD present in the outing sewage.

Symbol: Q

Measurement: DensityUnit: mg/L

Note: Value should be greater than 0.

Formulas to find Effluent BOD

Theoretical Oxygen Requirement

Theoretical Oxygen Requirement is the calculated amount of oxygen required to oxidize a compound to its final oxidation products.

Symbol: O₂

Measurement: Mass Flow RateUnit: mg/d

Note: Value should be greater than 0.

Formulas to find Theoretical Oxygen Requirement

Volume of Wasted Sludge per day

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.

Symbol: Q_w

Measurement: Volumetric Flow RateUnit: m³/s

Note: Value should be greater than 0.

Formulas to find Volume of Wasted Sludge per day

MLSS in Returned or Wasted Sludge

MLSS in Returned or Wasted Sludge is the mixed liquor suspended solids in returned sludge in terms of mg per liter.

Symbol: X^R

Measurement: DensityUnit: mg/L

Note: Value should be greater than 0.

Formulas to find MLSS in Returned or Wasted Sludge

Other formulas in BOD5 category

Go Ratio of BOD to Ultimate BOD

f = \frac{BOD 5}{BOD u}

Go BOD5 given Ratio of BOD to Ultimate BOD

BOD 5r = f \cdot BOD u

How to Evaluate BOD5 given Oxygen Required in Aeration Tank?

BOD5 given Oxygen Required in Aeration Tank evaluator 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 evaluate 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 evaluate BOD5 given Oxygen Required in Aeration Tank using this online evaluator? To use this online evaluator 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.

FAQs on BOD5 given Oxygen Required in Aeration Tank

What is the formula to find BOD5 given Oxygen Required in Aeration Tank?

The formula of BOD5 given Oxygen Required in Aeration Tank is expressed as 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)). Here is an example- 13555.01 = 0.002*(10*(0.0132-0.0004))/(2.89351851851852E-11+(1.42*9.5*0.0014)).

How to calculate BOD5 given Oxygen Required in Aeration Tank?

With 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) we can find BOD5 given Oxygen Required in Aeration Tank using the formula - 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)).

Can the BOD5 given Oxygen Required in Aeration Tank be negative?

No, the BOD5 given Oxygen Required in Aeration Tank, measured in Density cannot be negative.

Which unit is used to measure BOD5 given Oxygen Required in Aeration Tank?

BOD5 given Oxygen Required in Aeration Tank is usually measured using the Milligram per Liter[mg/L] for Density. Kilogram per Cubic Meter[mg/L], Kilogram per Cubic Centimeter[mg/L], Gram per Cubic Meter[mg/L] are the few other units in which BOD5 given Oxygen Required in Aeration Tank can be measured.

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BOD5 given Oxygen Required in Aeration Tank Formula

BOD5 given Oxygen Required in Aeration Tank Example

BOD5 given Oxygen Required in Aeration Tank Solution

BOD5 given Oxygen Required in Aeration Tank Formula Elements

Other formulas in BOD5 category

How to Evaluate BOD5 given Oxygen Required in Aeration Tank?

FAQs on BOD5 given Oxygen Required in Aeration Tank

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