Water & Wastewater Treatment

BOD, COD & Oxygen Demand

First-order BOD kinetics, ultimate BOD and the BOD/COD relationship — quantifying organic pollution.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Biochemical Oxygen Demand (BOD) is the oxygen consumed by microorganisms degrading organic matter; it is exerted as a first-order reaction approaching the **ultimate BOD (

L_{0}

). The standard test is BOD₅ at 20 °C. COD measures total chemically oxidisable matter and is always ≥ BOD**; the BOD/COD ratio indicates biodegradability.

The Core Formula Matrix

BOD exerted:

BOD_{t} = L_{0} (1 - e^{- k_{d} t}) = L_{0} (1 - 1 0^{- K t})

Remaining (ultimate) demand:

L_{t} = L_{0} e^{- k_{d} t}

Rate constants:

k_{d} = 2.303 K

(base-e vs base-10).

Biodegradability: BOD ≤ COD; high BOD/COD ⇒ readily biodegradable.

The ‘IIT Traps’

⚠
BOD₅ is not the ultimate BOD. At $k_{d} = 0.23 d^{- 1}$ , BOD₅ ≈ 0.68 $L_{0}$ .
⚠
COD ≥ BOD always — chemical oxidation captures more matter.
⚠
Watch the rate-constant base: $k_{d}$ (ln) vs $K$ (log₁₀) differ by 2.303.

📚 Standard references

Wastewater Engineering: Treatment and Resource Recovery — Metcalf & Eddy
Environmental Engineering Vol. II — S.K. Garg

PART 2

Progressive 3-Tier Question Suite

Q1MEDIUM2 Marks · NAT

A wastewater has ultimate BOD

L_{0} = 300 mg/L

and

k_{d} = 0.23 d^{- 1}

. The 5-day BOD is _____ mg/L.

Q2BASIC1 Mark · MCQ

For the same wastewater, which relationship is always true?

Q3HARD2 Marks · NAT

With

k_{d} = 0.23 d^{- 1}

the fraction of ultimate BOD exerted in 5 days (BOD₅/L₀) is _____.