---
title: "BOD/COD significance; BOD at 30°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-cod-significance-temp-conversion"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-cod-significance-temp-conversion
site: BCT Engineering
---

# BOD/COD significance; BOD at 30°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2080 Chaitra, 2+2+4 marks
- Video: [Biochemical Oxygen Demand (BOD) | Waste Water Engineering | Estimation of BOD | BOD solved problems](https://www.youtube.com/watch?v=V3PNsNmkdKY) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-cod-significance-temp-conversion
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

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.

## 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.

## Answer

**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 (BOD at 30°C)
Given: $y_3$ at 25°C = 250 mg/L, $K_{20}=0.1$ day⁻¹ (base 10). Temperature coefficient $\theta=1.047$. The ultimate BOD $L$ does not change with temperature; only the rate constant changes.

$$
\begin{aligned}
K_{25}&=K_{20}(1.047)^{25-20}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}\\
L&=\frac{y_3}{1-10^{-K_{25}t}}=\frac{250}{1-10^{-0.1258\times3}}=430.5\ \text{mg/L}\\
K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\
y_5(30^\circ\text{C})&=L\left(1-10^{-K_{30}\times5}\right)=430.5\left(1-10^{-0.7915}\right)=360.9\ \text{mg/L}
\end{aligned}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does COD measure?
   A. Only biodegradable matter
   B. Oxygen to oxidise all organic matter by a strong chemical oxidant
   C. Dissolved oxygen
   D. Nitrogen
   Answer: B. Uses potassium dichromate in boiling acid. (Key point: COD)

2. BOD/COD above about 0.5 indicates?
   A. Toxic waste
   B. No organic matter
   C. Waste is easily treated biologically
   D. Sludge
   Answer: C. High ratio means biodegradable. (Key point: Ratio)

3. In converting BOD between temperatures, what stays constant?
   A. K
   B. BOD5
   C. Temperature
   D. Ultimate BOD L; only K changes
   Answer: D. L is independent of temperature. (Key point: Method)

4. The 5-day BOD at 30°C is about?
   A. 360.9 mg/L
   B. 250 mg/L
   C. 430.5 mg/L
   D. 180 mg/L
   Answer: A. L = 430.5, K30 = 0.1583, y5 = 360.9 mg/L. (Key point: Result)

## Related in Sanitary Engineering

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