---
title: "BOD5 and ultimate BOD, 5% dilution"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-dilution-5pct-blank"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-dilution-5pct-blank
site: BCT Engineering
---

# BOD5 and ultimate BOD, 5% dilution

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2072 Asoj, 4 marks
- Video: [17 5 Ultimate BOD](https://www.youtube.com/watch?v=bePvgUvEwUE) — Muzaffar Zainal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-dilution-5pct-blank
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

The following observations were made on 5% dilution of a sewage sample. The DO of blank is 5 mg/L. The DO of diluted sample after 5 days and incubation at 20°C is 1 mg/L. Calculate BOD5 and ultimate BOD of sample. Assume the DO of original sample 0.5 mg/L.

## Short answer

5% dilution, blank DO=5 mg/L, sample DO=0.5 mg/L. Initial mixture DO=4.775 mg/L, after 5 days DO=1 mg/L. BOD5=75.5 mg/L. Assuming K20=0.1/day, ultimate BOD=110.4 mg/L.

## Answer

#### Initial DO of the diluted sample
The dilution water (blank) has DO = 5 mg/L, the raw sample has DO = 0.5 mg/L, and the dilution is 5% ($P=0.05$ sample, 0.95 dilution water). Initial DO of the mixture:

$$
D_1=5(0.95)+0.5(0.05)=4.775\ \text{mg/L}
$$

The DO after 5 days at 20°C is $D_2=1$ mg/L.

#### 5-day BOD
$$
BOD_5=\frac{D_1-D_2}{P}=\frac{4.775-1}{0.05}=75.5\ \text{mg/L}
$$

#### Ultimate BOD
The rate constant is not given, so the usual $K_{20}=0.1$ day⁻¹ (base 10) is assumed:

$$
L=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{75.5}{0.6838}=110.4\ \text{mg/L}
$$

**Answer: BOD₅ = 75.5 mg/L; ultimate BOD ≈ 110.4 mg/L.**

## Check yourself

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

1. How is D1 of the 5% mixture found?
   A. D1 = 5(0.95) + 0.5(0.05) = 4.775 mg/L
   B. D1 = 5 mg/L
   C. D1 = 0.5 mg/L
   D. D1 = 5 x 0.05 = 0.25
   Answer: A. The mixture DO is the weighted average of dilution water and sample DO. (Key point: Initial DO)

2. What is BOD5?
   A. 4.775 - 1 = 3.775
   B. (4.775 - 1)/0.05 = 75.5 mg/L
   C. (5 - 1)/0.95 = 4.2
   D. (4.775 - 1) x 0.05 = 0.19
   Answer: B. Divide the DO drop by P = 0.05. (Key point: BOD5)

3. With K = 0.1 per day, what is L?
   A. About 75.5 mg/L
   B. About 50 mg/L
   C. About 110 mg/L
   D. About 755 mg/L
   Answer: C. L = 75.5/0.6838 = 110.4 mg/L. (Key point: Ultimate BOD)

4. Why is K = 0.1 per day assumed?
   A. BOD5 always uses 0.1
   B. It is derived from the blank
   C. It is the dilution fraction
   D. The rate constant is not given, so the standard K20 is used
   Answer: D. No rate was provided, so the usual value is assumed. (Key point: Rate assumption)

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