Water tools

Nitrogen and phosphorus requirement of biological treatment

The nutrient a plant needs is the nutrient that leaves in its excess sludge. The familiar 100:5:1 applies only to conventional sludge ages; extended aeration needs about half.

Influent flowm³/d

Influent BOD₅mg/L

Effluent soluble BOD₅mg/L

Sludge aged

MLVSSmg/L

Reactor volumem³

Decay coefficient Kd1/d

Biodegradable fraction of new biomass

Influent TKNmg N/L

Influent total phosphorusmg P/L

Nitrogen

Nitrogen required

26.3kg N/d

Nitrogen in influent

67.5kg N/d

BOD₅ : N

100 : 6.0

Conventional 100:4.0 to 6.0; extended aeration 100:2.5 to 3.5.

Phosphorus

Phosphorus required

5.4kg P/d

Phosphorus in influent

12.0kg P/d

BOD₅ : P

100 : 1.24

Conventional 100:0.9 to 1.2; extended aeration 100:0.5 to 0.6.

Basis

Sludge produced

223kg VSS/d

How it works

The nutrient a plant needs is the nutrient that leaves in its excess sludge. Active biomass (C₅H₇O₂N) is 12.3% nitrogen and 2.6% phosphorus; the endogenous residue about 7% and 1%. The fraction of produced VSS that is still active is f_b/f_b′, so

N = 0.123 \frac{f_{b}}{f_{b}'} P_{x v} + 0.07 (1 - f_{b} / f_{b}') P_{x v}

(1)

P = 0.026 \frac{f_{b}}{f_{b}'} P_{x v} + 0.01 (1 - f_{b} / f_{b}') P_{x v}

(2)

P_{x v} = \frac{X_{v} V}{θ_{c}}

(3)

where

$N, P$: nitrogen and phosphorus required, kg/d
$P_{x v}$: volatile solids produced, kg VSS/d
$f_{b}$: biodegradable fraction of the mixed liquor
$f_{b}'$: biodegradable fraction of new biomass
$X_{v}$: mixed liquor volatile suspended solids, mg/L
$V$: reactor volume, m³
$θ_{c}$: sludge age, d

Nitrogen must be in an assimilable form (ammonia or nitrate); organic nitrogen needs hydrolysis first. A longer sludge age produces less sludge and so needs less nutrient, which is why the familiar 100:5:1 holds only for conventional sludge ages. Industrial wastewaters short of N or P are supplemented or blended with domestic sewage.

Nitrogen and phosphorus requirement of biological treatment

How it works

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