BOD, COD; ultimate BOD of 480 mg/l

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 ($K_2Cr_2O_7$) 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: $COD-BOD_u$.

Numerical

Given: $y_5=480$ mg/L, $K=0.25$ day⁻¹ (base $e$). First-stage BOD: $y_t=L\left(1-e^{-Kt}\right)$.

$$L=\frac{y_5}{1-e^{-Kt}}=\frac{480}{1-e^{-0.25\times5}}=\frac{480}{1-0.2865}=672.7\ \text{mg/L}$$

BOD remaining (unsatisfied) after 20 days: $L_{20}=Le^{-Kt}=672.7\times e^{-0.25\times20}=672.7\times0.00674=4.53$ mg/L.

$$\text{Proportion un-oxidised}=\frac{L_{20}}{L}=e^{-5}=0.00674\ (\approx0.67\%)$$

Answer: Ultimate BOD = 672.7 mg/L; proportion remaining after 20 days = 0.00674 (0.67%), i.e. about 4.53 mg/L.