Define BOD, COD; derive BOD equation

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: $COD-BOD_u$.

Derivation of BOD equation

The oxidation of organic matter by bacteria follows a first-order reaction: the rate of reaction at any time is proportional to the oxygen equivalent of organic matter remaining.

Let $L$ = ultimate first-stage BOD (total organic matter at $t=0$), $L_t$ = BOD remaining at time $t$, $y_t$ = BOD exerted (satisfied) in time $t$, and $K$ = rate constant (day⁻¹).

$$\frac{dL_t}{dt}=-K L_t$$ $$\int_{L}^{L_t}\frac{dL_t}{L_t}=-K\int_{0}^{t}dt\ \Rightarrow\ \ln\frac{L_t}{L}=-Kt\ \Rightarrow\ L_t=L e^{-Kt}$$

In base 10 form, $L_t=L\,10^{-kt}$, where $k=K/2.303$.

BOD remaining after time $t$ is therefore related to the ultimate BOD by

$$\boxed{L_t = L\,e^{-Kt}=L\,10^{-kt}}$$

and the BOD exerted is

$$y_t=L-L_t=L\left(1-e^{-Kt}\right)=L\left(1-10^{-kt}\right)$$

For $k=0.1$ per day at 20°C, $y_5=0.68L$ (about two-thirds of the ultimate BOD is satisfied in 5 days). The curve of $y_t$ against $t$ rises quickly at first, and approaches $L$ slowly.