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**Oxygen Requirement**

Usually we can compute the Oxygen Requirement by following three ways:

*1. Theoretical
Oxygen Demand*

The Theoretical Oxygen Requirements can compute by the
following Eq:

C5H7NO2
+ 5 O2 ---> 5CO2 + 2 H2O + NH3 +Energy

( Source from Metcalf & Eddy)

Basically you can assume all BOD5 can be converted to end products. The total
oxygen demand would be computed by converting BOD5 to BODL. The BODL of one mole
of cells is equal to 1.42 times the concentration of cells.

The Theoretical Oxygen Requirements can be computed as following:

Kg, O2/day = [Q(S'-S) / (
1000g/Kg) X f] - 1.42 (Px)

f: Conversion factor for converting
BOD5 to BODL ( you

can assume the
f=0.68)

Q: influent wastewater flowrate (m3/day)

S': influent BOD or COD concentration,
(mg/L)

S: effluent BOD or COD concentration,
(mg/L)

Px: net waste activated sludge produced
each day

measured in
terms of volatile SS (Kg/day)

If we consider the nitrification, the O2 required for nitrogen conversion( from ammonia to nitrate) need to be added. The Eq as following:

Kg, O2/day = [Q(S'-S) / (
1000g/Kg) X f] - 1.42 (Px) + 4.57 Q ( N'-N)/(1000g/Kg)

N': influent TKN, mg/L

N: effluent TKN, mg/L

4.57: conversion factor for amount of
oxygen required for complete oxidation of TKN

*2. Experimental
Oxygen Demand*

The Experimental Oxygen Requirements can compute by the
following Eq:

U= a'Y + b'Z

U: Oxygen demand( Kg/day)

Y: BOD removal amount (Kg/day) = (S'rQ)/1000 (Kg/day)

Q: as defined previously

S': as defined previously

r: BOD removal rate

Z: MLSS amount in aeration tank (Kg)

a': 0.35 - 0.5 (Kg O2/ Kg BOD)

b': 0.05 - 0.24( Kg O2/ Kg MLSS-day)

If we consider the nitrification, the O2 required for nitrogen conversion( from ammonia to nitrate) need to be added. The Eq as following:

U= a'Y + b'Z + 4.57 Q ( N'-N)/( 1000 g/Kg/)

N': as defined previously

N: as defined previously

**Above Oxygen Demand Eqs base on
activated sludge****process.**

**3. ****For BioFilm Aeration process, you can compute Oxygen Demand by the
following Eq:**

U=[ (Q X Y)/ ( BOD5 / BODL)
] - 1.42 X to be wasted

sludge

U: as defined previously

Q: as defined previously

Y: as defined previously

BOD5 / BODL: Assumeing 0.68

to be wasted sludge: Assumeing 0 ( Usually the
BioFilm Aeration process is less

than activated sludge for wasted sludge )

**-------- 06/22/2000**

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