---
title: "Ultimate BOD from 5-day BOD at 37°C"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-5day-37c-200"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-5day-37c-200
site: BCT Engineering
---

# Ultimate BOD from 5-day BOD at 37°C

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2072 Magh, 4 marks
- Video: [BOD and Ultimate BOD | BOD problem | how to calculate ultimate BOD | BOD of waste water | sewage BOD](https://www.youtube.com/watch?v=wgg7UpR70Rc) — 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-5day-37c-200
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

If the 5 day BOD at 37°C is 200 mg/l and if the rate of deoxygenation is 0.17/day, calculate the ultimate BOD and BOD remaining after 5 days.

## Short answer

5-day BOD at 37°C=200 mg/L, k=0.17/day (base 10). Ultimate BOD=232.9 mg/L. BOD remaining after 5 days=32.9 mg/L.

## Answer

Given: $y_5=200$ mg/L at 37°C and rate of deoxygenation $k=0.17$ day⁻¹ at that temperature (taken as base 10, the usual form $y_t=L(1-10^{-kt})$, so no temperature correction is needed).

#### Ultimate BOD
$$
L=\frac{y_5}{1-10^{-kt}}=\frac{200}{1-10^{-0.17\times5}}=\frac{200}{1-0.1413}=232.9\ \text{mg/L}
$$

#### BOD remaining after 5 days
$$
L_5=L-y_5=232.9-200=32.9\ \text{mg/L}\quad\left(=L\,10^{-kt}\right)
$$

**Answer: Ultimate BOD = 232.9 mg/L; BOD remaining after 5 days = 32.9 mg/L.**
(If the constant is taken as base $e$, $L=200/(1-e^{-0.85})=349.3$ mg/L.)

## Check yourself

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

1. Why is no temperature correction applied to k = 0.17 per day at 37°C?
   A. The given rate is already for the test temperature
   B. Temperature does not affect k
   C. 37°C is the standard temperature
   D. k must be taken at 20°C only
   Answer: A. The question gives the rate at that temperature. (Key point: No correction needed)

2. What is L with y5 = 200 and k = 0.17 (base 10)?
   A. About 200 mg/L
   B. About 233 mg/L
   C. About 349 mg/L
   D. About 117 mg/L
   Answer: B. L = 200/(1 - 10^(-0.85)) = 232.9 mg/L. (Key point: Ultimate BOD)

3. What is the BOD remaining after 5 days?
   A. 200 mg/L
   B. 232.9 mg/L
   C. L - y5 = about 32.9 mg/L
   D. About 0.17 mg/L
   Answer: C. Remaining = L 10^(-kt) = 232.9 - 200 = 32.9 mg/L. (Key point: BOD remaining)

4. If the constant were base e, how would L change?
   A. It would be the same
   B. It would be smaller
   C. It would be 200 mg/L
   D. L = 200/(1 - e^(-0.85)) = 349 mg/L, larger
   Answer: D. A base e constant of the same number is a slower rate, giving a larger L. (Key point: Base of constant)

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