---
title: "BOD1 at 23°C to BOD5 at 30°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-1day-23c-105-to-5day-30c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-1day-23c-105-to-5day-30c
site: BCT Engineering
---

# BOD1 at 23°C to BOD5 at 30°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2070 Magh, 8 marks
- Video: [BOD Calculations](https://www.youtube.com/watch?v=mGaNsGvutSE) — Belinda Sturm
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-1day-23c-105-to-5day-30c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

If one day BOD of a sewage sample at 23°C is 105 mg/l. What will be its five day BOD at 30°C? Assume K20 = 0.1 per day.

## Short answer

1-day BOD at 23°C=105 mg/L, K20=0.1/day. Ultimate BOD=452.1 mg/L. At 30°C, K30=0.1583/day, so 5-day BOD=379.0 mg/L.

## Answer

Given: $y_1$ at 23°C = 105 mg/L, $K_{20}=0.1$ day⁻¹ (base 10), $\theta=1.047$. The ultimate BOD does not change with temperature.

#### Step 1: Ultimate BOD
$$
\begin{aligned}
K_{23}&=0.1\times1.047^{3}=0.1148\ \text{day}^{-1}\\
L&=\frac{105}{1-10^{-K_{23}\times1}}=\frac{105}{1-10^{-0.1148}}=452.1\ \text{mg/L}
\end{aligned}
$$

#### Step 2: 5-day BOD at 30°C
$$
\begin{aligned}
K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\
y_5&=L\left(1-10^{-K_{30}\times5}\right)=452.1\left(1-10^{-0.7915}\right)=379.0\ \text{mg/L}
\end{aligned}
$$

**Answer: Five-day BOD at 30°C = 379.0 mg/L.**

## Check yourself

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

1. What is K23?
   A. 0.1 x 1.047^3 = 0.1148 per day
   B. 0.1 x 1.047^23
   C. 0.1 x 1.047^7
   D. 0.1
   Answer: A. Exponent is 23 - 20 = 3. (Key point: Rate at 23°C)

2. What is L for y1 = 105 mg/L at 23°C?
   A. About 105 mg/L
   B. About 452 mg/L
   C. About 379 mg/L
   D. About 510 mg/L
   Answer: B. L = 105/(1 - 10^(-0.1148)) = 452.1 mg/L. (Key point: Ultimate BOD)

3. Why can the same L be used at 30°C?
   A. Because K is the same at all temperatures
   B. Because 30°C is close to 23°C
   C. Ultimate BOD does not change with temperature
   D. Because y1 equals y5
   Answer: C. Only the rate of exertion changes. (Key point: L independent of T)

4. What is BOD5 at 30°C?
   A. About 452 mg/L
   B. About 105 mg/L
   C. About 250 mg/L
   D. About 379 mg/L
   Answer: D. K30 = 0.1583; y5 = 452.1 (1 - 10^(-0.7915)) = 379 mg/L. (Key point: Final answer)

## Related in Sanitary Engineering

- [BOD1 at 22°C to BOD5 at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-22c-310-to-5day-30c) — 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
- [Temperature effect on BOD; 5-day BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atemperature-effect-decomposition) — 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
- [BOD5 at 30°C converted to 20°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-30c-170-to-20c) — Sanitary Engineering, chapter 5
- [BOD of Bagmati sample at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-thapathali-4day-30c) — Sanitary Engineering, chapter 5

## Related in other subjects

- [BOD versus COD and a BOD and COD calculation](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Abod-cod-difference-numerical) — Environment and Pollution Control, chapter 4