---
title: "BOD5 at 30°C converted to 20°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-1day-30c-170-to-20c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-1day-30c-170-to-20c
site: BCT Engineering
---

# BOD5 at 30°C converted to 20°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2068 Bhadra (old course), 10 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-30c-170-to-20c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

The BOD5 of a sewage incubated for one day at 30°C has been found to be 170 mg/l. What will be the 5 day BOD at 20°C? Assume K = 0.12 per day (base 10) at 20°C.

## Short answer

1-day BOD at 30°C=170 mg/L, K20=0.12/day. Ultimate BOD=479.9 mg/L. At 20°C, K20=0.12/day, so 5-day BOD=359.3 mg/L.

## Answer

The data are read as: the **1-day BOD at 30°C is 170 mg/L**, $K_{20}=0.12$ day⁻¹ (base 10), and the 5-day BOD at 20°C is required. $\theta=1.047$.

#### Step 1: Rate constant at 30°C
$$
K_{30}=K_{20}(1.047)^{30-20}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}
$$

#### Step 2: Ultimate BOD
$$
L=\frac{y_1}{1-10^{-K_{30}t}}=\frac{170}{1-10^{-0.1900\times1}}=\frac{170}{1-0.6457}=479.9\ \text{mg/L}
$$

#### Step 3: 5-day BOD at 20°C
$$
y_5=L\left(1-10^{-K_{20}\times5}\right)=479.9\left(1-10^{-0.12\times5}\right)=479.9\left(1-0.2512\right)=359.3\ \text{mg/L}
$$

**Answer: The 5-day BOD at 20°C = 359.3 mg/L** (ultimate BOD = 479.9 mg/L).

## Check yourself

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

1. Which constant is used with the 1-day BOD at 30°C to find L?
   A. K30 = 0.12 x 1.047^10 = 0.19 per day
   B. K20 = 0.12
   C. K30 = 0.12 / 1.047^10
   D. K = 0.1 (standard)
   Answer: A. The 1-day BOD was exerted at 30°C. (Key point: Rate at test temp)

2. What is L?
   A. About 170 mg/L
   B. About 480 mg/L
   C. About 359 mg/L
   D. About 1000 mg/L
   Answer: B. L = 170/(1 - 10^(-0.19)) = 479.9 mg/L. (Key point: Ultimate BOD)

3. For the final 5-day BOD at 20°C, which rate constant is used?
   A. K30 = 0.19 per day
   B. K = 0.12 x 1.047^5
   C. K20 = 0.12 per day
   D. K = 0.1 per day
   Answer: C. Required BOD is at 20°C, so K20 is used. (Key point: Switch to 20°C)

4. What is y5 at 20°C?
   A. About 480 mg/L
   B. About 170 mg/L
   C. About 420 mg/L
   D. About 359 mg/L
   Answer: D. y5 = 479.9 (1 - 10^(-0.6)) = 359.3 mg/L. (Key point: Final answer)

## Related in Sanitary Engineering

- [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
- [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
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- [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
- [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
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