---
title: "BOD1 at 22°C to BOD5 at 30°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-1day-22c-310-to-5day-30c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-1day-22c-310-to-5day-30c
site: BCT Engineering
---

# BOD1 at 22°C to BOD5 at 30°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2074 Bhadra, 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-22c-310-to-5day-30c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

BOD1 at 22°C of a sewage sample is 310 mg/l. What will be its BOD5 at 30°C? Assume reaction rate K20 = 0.12 per day.

## Short answer

1-day BOD at 22°C=310 mg/L, K20=0.12/day. Ultimate BOD=1186.3 mg/L. At 30°C, K30=0.1900/day, so 5-day BOD=1053.1 mg/L.

## Answer

Given: $y_1$ at 22°C = 310 mg/L, $K_{20}=0.12$ day⁻¹ (base 10). Temperature coefficient $\theta=1.047$. The ultimate BOD $L$ is independent of temperature.

#### Step 1: Rate constant at 22°C and ultimate BOD
$$
\begin{aligned}
K_{22}&=0.12\times1.047^{2}=0.1315\ \text{day}^{-1}\\
L&=\frac{y_1}{1-10^{-K_{22}\times1}}=\frac{310}{1-10^{-0.1315}}=1186.3\ \text{mg/L}
\end{aligned}
$$

#### Step 2: Rate constant at 30°C
$$
K_{30}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}
$$

#### Step 3: 5-day BOD at 30°C
$$
y_5=L\left(1-10^{-K_{30}\times5}\right)=1186.3\left(1-10^{-0.9498}\right)=1053.1\ \text{mg/L}
$$

**Answer: BOD₅ at 30°C = 1053.1 mg/L.**

## Check yourself

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

1. Which rate constant is used to find L from y1 at 22°C?
   A. K22 = 0.12 x 1.047^2 = 0.1315 per day
   B. K20 = 0.12
   C. K30 = 0.19
   D. K22 = 0.12 x 1.047^22
   Answer: A. L must be computed with the rate constant valid at the test temperature. (Key point: Rate at test temperature)

2. What is L?
   A. About 310 mg/L
   B. About 1186 mg/L
   C. About 1053 mg/L
   D. About 2000 mg/L
   Answer: B. L = 310 / (1 - 10^(-0.1315)) = 1186.3 mg/L. (Key point: Ultimate BOD)

3. What is K30?
   A. 0.12 x 1.047^30 = 0.47
   B. 0.12 x 1.047^8 = 0.17
   C. 0.12 x 1.047^10 = 0.190 per day
   D. 0.12 / 1.047^10 = 0.076
   Answer: C. Exponent is 30 - 20 = 10 (from K20). (Key point: Rate at 30°C)

4. Which is BOD5 at 30°C?
   A. About 1186 mg/L
   B. About 310 mg/L
   C. About 780 mg/L
   D. About 1053 mg/L
   Answer: D. y5 = 1186.3 (1 - 10^(-0.95)) = 1053.1 mg/L, below L. (Key point: Final answer)

## Related in Sanitary Engineering

- [BOD1 at 23°C to BOD5 at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-23c-105-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
- [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
- [Standard rate filter, 3 MLD, 175 mg/l, NRC](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-3mld-175-nrc) — Sanitary Engineering, chapter 7
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