---
title: "BOD, COD; ultimate BOD of 480 mg/l"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-cod-ultimate-bod-480"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-cod-ultimate-bod-480
site: BCT Engineering
---

# BOD, COD; ultimate BOD of 480 mg/l

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2079 Chaitra, 4+4 marks
- 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-ultimate-bod-480
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

Describe briefly about BOD and COD with their significance. The five days BOD of sewage was measured as 480 mg/l. If the base 'e' constant 'K' = 0.25 day, what is the ultimate BOD of the sewage? What proportion of BODu would remain un-oxidised or unsatisfied after 20 days?

## Short answer

BOD quantifies biodegradable organic matter for pollution assessment; COD measures total organic matter chemically, used for quick monitoring. Given5-day BOD480 mg/L and K=0.25/day (base e), ultimate BOD L =480/(1-e^(-1.25)) =672.7 mg/L. The proportion unxdised after20 days is e^(-5) =0.00674 (0.67%), about4.53 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
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.**

## Check yourself

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

1. Which statement correctly defines BOD?
   A. Oxygen used by bacteria to stabilise biodegradable organic matter under aerobic conditions, standard 5 days at 20°C
   B. Oxygen needed by potassium dichromate to oxidise all organic matter in boiling acid
   C. Oxygen present in the water sample after 5 days at 20°C
   D. Oxygen used by bacteria in 20 days at 5°C
   Answer: A. BOD is the oxygen consumed by bacteria in oxidising biodegradable organic matter, measured at 20°C for 5 days. (Key point: Definition of BOD)

2. Why is COD always greater than BOD?
   A. COD is measured over a longer time than BOD
   B. COD oxidises both biodegradable and non-biodegradable organic matter, while BOD covers only the biodegradable part
   C. COD uses bacteria that work faster than those in BOD
   D. COD is measured at a higher temperature than 20°C only
   Answer: B. The dichromate oxidant attacks all organic matter, whereas bacteria in the BOD test attack only the biodegradable part. (Key point: COD vs BOD)

3. Which formula gives the ultimate BOD L from the 5-day BOD y5 with base e constant K?
   A. L = y5 x (1 - e^(-5K))
   B. L = y5 x e^(-5K)
   C. L = y5 / (1 - e^(-5K))
   D. L = y5 / e^(-5K)
   Answer: C. Since y_t = L(1 - e^(-Kt)), rearranging gives L = y_t / (1 - e^(-Kt)). (Key point: Ultimate BOD formula)

4. With K = 0.25 per day, what fraction of the ultimate BOD remains unsatisfied after 20 days?
   A. 1 - e^(-5) = 99.3%
   B. e^(-0.25) = 0.78
   C. e^(-20) which is practically zero
   D. e^(-5) = 0.00674 (about 0.67%)
   Answer: D. Remaining fraction = e^(-Kt) = e^(-0.25 x 20) = e^(-5) = 0.00674. (Key point: Fraction remaining)

## Related in Sanitary Engineering

- [Rate constant K at 25°C; ultimate BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-k-ultimate-293-56) — Sanitary Engineering, chapter 5
- [BOD/COD significance; BOD at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-cod-significance-temp-conversion) — Sanitary Engineering, chapter 5
- [Ultimate BOD from 5-day BOD at 37°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-5day-37c-200) — Sanitary Engineering, chapter 5
- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [Ultimate BOD and rate constant](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-ultimate-rate-constant-100-210) — Sanitary Engineering, chapter 5
- [BOD5 and ultimate BOD, 5% dilution](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-dilution-5pct-blank) — Sanitary Engineering, chapter 5

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