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BOD/COD significance; BOD at 30°C

Short answer

BOD measures biodegradable organic matter for pollution strength and design; COD measures total organic matter chemically for rapid testing and toxic wastes. For 3-day25°C BOD250 mg/L, using θ=1.047, K25=0.1258/day, L=430.5 mg/L; K30=0.1583/day gives5-day30°C BOD=360.9 mg/L.

Describe about the significance of BOD and COD. If 3 days 25°C BOD of sewage sample is 250 mg/l, what will be its 5 days BOD at 30°C? Assume K20 = 0.1 per day.

BOD (Biochemical Oxygen Demand) is the amount of oxygen required by bacteria to oxidise (stabilise) the biodegradable organic matter in a water or wastewater sample under aerobic conditions at a specified temperature and time (standard: 5 days at 20°C). It is given in mg/L.

COD (Chemical Oxygen Demand) is the amount of oxygen needed to oxidise all organic matter (biodegradable and non-biodegradable) by a strong chemical oxidant, potassium dichromate (K2Cr2O7K_2Cr_2O_7) in boiling acid medium, expressed in mg/L. The test takes about 2-3 hours.

Significance of BOD

  1. It measures the strength (organic pollution load) of sewage; raw domestic sewage has BOD₅ of 200-400 mg/L.
  2. It is used to design treatment plants (trickling filter, activated sludge, ponds) and to find their efficiency.
  3. It is used to find the oxygen demand on a receiving stream, and so the degree of treatment needed before discharge (self-purification calculations).
  4. It is the main parameter in effluent standards (for example, effluent standards in Nepal and India limit BOD₅ to about 30-50 mg/L for discharge to inland surface water).
  5. It measures only the biodegradable organic matter and gives an idea of how much will be decomposed in nature.

Significance of COD

  1. It is a rapid test (hours against 5 days) and can be used for plant control.
  2. It also includes non-biodegradable matter, so it is used for industrial wastes that are toxic to bacteria, where BOD cannot be measured.
  3. COD is always greater than BOD. The ratio BOD/COD shows biodegradability: above about 0.5 the waste is easily treated biologically; below 0.3 it is difficult and contains toxic or refractory matter.
  4. It is used with BOD to estimate the non-biodegradable part: COD−BODuCOD-BOD_u.

Numerical (BOD at 30°C)

Given: y3y_3 at 25°C = 250 mg/L, K20=0.1K_{20}=0.1 day⁻¹ (base 10). Temperature coefficient θ=1.047\theta=1.047. The ultimate BOD LL does not change with temperature; only the rate constant changes.

K25=K20(1.047)25−20=0.1×1.0475=0.1258 day−1L=y31−10−K25t=2501−10−0.1258×3=430.5 mg/LK30=0.1×1.04710=0.1583 day−1y5(30∘C)=L(1−10−K30×5)=430.5(1−10−0.7915)=360.9 mg/L\begin{aligned} K_{25}&=K_{20}(1.047)^{25-20}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}\\ L&=\frac{y_3}{1-10^{-K_{25}t}}=\frac{250}{1-10^{-0.1258\times3}}=430.5\ \text{mg/L}\\ K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ y_5(30^\circ\text{C})&=L\left(1-10^{-K_{30}\times5}\right)=430.5\left(1-10^{-0.7915}\right)=360.9\ \text{mg/L} \end{aligned}

Answer: 5-day BOD at 30°C = 360.9 mg/L (ultimate BOD = 430.5 mg/L).

✓ Answer checked by Claude Opus 5.5 (high effort)

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