---
title: "1-day BOD 300 mg/l to 5-day at 30°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-1day-300-to-5day-30c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-1day-300-to-5day-30c
site: BCT Engineering
---

# 1-day BOD 300 mg/l to 5-day at 30°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2077 Chaitra, 4 marks
- Video: [Biochemical Oxygen Demand (BOD) | Waste Water Engineering | Estimation of BOD | BOD solved problems](https://www.youtube.com/watch?v=V3PNsNmkdKY) — Civil Engineering Exam
- 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-300-to-5day-30c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

If 1 day BOD of sewage sample at 20°C is 300 mg/l, what will be its 5-day BOD at 30°C? Consider rate constant of 0.1/day (base 10) at 20°C.

## Short answer

Ultimate BOD L =300/(1-10^(-0.1)) =1458.6 mg/L. With temperature coefficient θ=1.047, k30 =0.1×1.047^10 =0.1583/day. Then5-day30°C BOD = L(1-10^(-k30×5)) =1222.9 mg/L.

## Answer

Given: $y_1$ at 20°C = 300 mg/L, $k_{20}=0.1$ day⁻¹ (base 10). Assume $\theta=1.047$ for the temperature correction. The ultimate BOD $L$ is independent of temperature.

#### Step 1: Ultimate BOD
$$
L=\frac{y_1}{1-10^{-k_{20}\,t}}=\frac{300}{1-10^{-0.1\times1}}=\frac{300}{1-0.7943}=1458.6\ \text{mg/L}
$$

#### Step 2: Rate constant at 30°C
$$
k_{30}=k_{20}(1.047)^{30-20}=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}
$$

#### Step 3: 5-day BOD at 30°C
$$
y_5=L\left(1-10^{-k_{30}\times5}\right)=1458.6\left(1-10^{-0.7915}\right)=1222.9\ \text{mg/L}
$$

**Answer: 5-day BOD at 30°C = 1222.9 mg/L.**

## Check yourself

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

1. How does the ultimate BOD L change with temperature?
   A. It stays the same; only the rate constant changes
   B. It increases with temperature
   C. It decreases with temperature
   D. It doubles every 10°C
   Answer: A. L is the total oxidisable organic matter; temperature only changes how fast it is exerted. (Key point: Ultimate BOD unchanged)

2. Which formula corrects the rate constant to 30°C?
   A. k30 = k20 / (1.047)^10
   B. k30 = k20 (1.047)^(30-20)
   C. k30 = k20 x 1.047 x 10
   D. k30 = k20 + 1.047 x 10
   Answer: B. K_T = K20 x theta^(T-20), with theta = 1.047. (Key point: Temperature correction)

3. What is the ultimate BOD from y1 = 300 mg/L and k20 = 0.1 per day?
   A. About 300 mg/L
   B. About 617 mg/L
   C. About 1459 mg/L
   D. About 2430 mg/L
   Answer: C. L = 300 / (1 - 10^(-0.1)) = 300 / 0.2057 = 1458.6 mg/L. (Key point: Step 1 result)

4. Which is the 5-day BOD at 30°C?
   A. About 1459 mg/L
   B. About 997 mg/L
   C. About 300 mg/L
   D. About 1223 mg/L
   Answer: D. k30 = 0.1583, y5 = 1458.6 (1 - 10^(-0.7915)) = 1222.9 mg/L, less than L. (Key point: Final answer)

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