---
title: "Critical DO deficit and location downstream"
subject: "Environment and Pollution Control"
chapter: 4
key: "semester_7/environment_and_pollution_control:oxygen-sag-critical-deficit-numerical"
source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/oxygen-sag-critical-deficit-numerical
site: BCT Engineering
---

# Critical DO deficit and location downstream

Environment and Pollution Control · Chapter 4: Water pollution

- Asked in: year not given, 8 marks
- Video: [Critical DO deficit | Sanitary numerical | 2019 PU | TU.PU.KU.POU | MOST IMPORTANT](https://www.youtube.com/watch?v=ZxXckyhmRNQ) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/oxygen-sag-critical-deficit-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-4

## Question

A town discharges treated sewage into a river. After mixing, the BOD ultimate of the river water is L0 = 20 mg/L and the initial oxygen deficit is D0 = 2 mg/L. The saturation DO is 9.1 mg/L, deoxygenation constant k1 = 0.2 per day and reaeration constant k2 = 0.4 per day (base e, at the river temperature). The river flows at 0.3 m/s. Calculate (a) the critical time, (b) the critical deficit and the minimum DO, (c) the distance downstream at which the minimum DO occurs, (d) the DO deficit after 1 day.

## Short answer

Using Streeter‑Phelps: critical time = 2.94 days, critical deficit = 5.56 mg/L, so minimum DO = 9.1 – 5.56 = 3.54 mg/L. At velocity 0.3 m/s the critical point occurs 76 km downstream. After 1 day, deficit = 4.31 mg/L (DO = 4.79 mg/L).

## Answer

Given: $L_0 = 20$ mg/L, $D_0 = 2$ mg/L, $DO_{sat} = 9.1$ mg/L, $k_1 = 0.2\ d^{-1}$, $k_2 = 0.4\ d^{-1}$, $u = 0.3$ m/s.

#### (a) Critical time
$$
t_c = \frac{1}{k_2 - k_1}\ln\left[\frac{k_2}{k_1}\left(1 - \frac{D_0(k_2 - k_1)}{k_1 L_0}\right)\right]
$$
$$
\begin{aligned}
\frac{D_0(k_2 - k_1)}{k_1 L_0} &= \frac{2\times0.2}{0.2\times20} = 0.1 \\
t_c &= \frac{1}{0.2}\ln\left[2\,(1 - 0.1)\right] = 5\ln(1.8) = 5\times0.5878 = 2.94\ \text{days}
\end{aligned}
$$

#### (b) Critical deficit and minimum DO
$$
D_c = \frac{k_1}{k_2}L_0\,e^{-k_1 t_c} = \frac{0.2}{0.4}\times20\times e^{-0.2\times2.94} = 10\times e^{-0.5878} = 10\times0.5556 = 5.56\ \text{mg/L}
$$
$$
DO_{min} = DO_{sat} - D_c = 9.1 - 5.56 = 3.54\ \text{mg/L}
$$

#### (c) Distance
Velocity $= 0.3\times86\,400 = 25\,920$ m/day.
$$
x_c = 25\,920 \times 2.94 = 76\,200\ \text{m} \approx 76\ \text{km}
$$

#### (d) Deficit after 1 day
$$
\begin{aligned}
D_1 &= \frac{0.2\times20}{0.4 - 0.2}\left(e^{-0.2} - e^{-0.4}\right) + 2\,e^{-0.4} \\
&= 20\,(0.8187 - 0.6703) + 2\times0.6703 = 2.968 + 1.341 = 4.31\ \text{mg/L}
\end{aligned}
$$
DO after 1 day $= 9.1 - 4.31 = 4.79$ mg/L.

| Time (days) | 0 | 1 | 2.94 (critical) |
|---|---|---|---|
| Deficit D (mg/L) | 2.00 | 4.31 | 5.56 |
| DO (mg/L) | 7.10 | 4.79 | 3.54 |

**Answer: $t_c$ = 2.94 days; $D_c$ = 5.56 mg/L; minimum DO = 3.54 mg/L; at about 76 km downstream; deficit after 1 day = 4.31 mg/L.** The minimum DO is below the 4-5 mg/L usually needed for fish, so further treatment would be needed.

## Check yourself

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

1. In the oxygen sag calculation, what does the critical time t_c represent?
   A. The time at which BOD is fully exhausted
   B. The time at which the oxygen deficit is greatest, i.e. DO is lowest
   C. The time at which the deficit becomes zero
   D. The time at which reaeration stops
   Answer: B. t_c is the time when the rate of deoxygenation equals the rate of reaeration, so the deficit D reaches its maximum and DO its minimum. (Key point: Critical time formula)

2. With k1 = 0.2/d, k2 = 0.4/d, L0 = 20 and D0 = 2 mg/L, the term D0(k2-k1)/(k1 L0) = 0.1. What is t_c?
   A. 1.0 day
   B. 5.0 days
   C. 2.94 days
   D. 0.59 days
   Answer: C. t_c = (1/0.2) ln[2(1-0.1)] = 5 ln(1.8) = 5 x 0.5878 = 2.94 days. (Key point: Value of t_c)

3. How are the critical deficit and minimum DO obtained in this answer?
   A. D_c = (k1/k2) L0 e^(-k1 t_c); DO_min = DO_sat - D_c = 3.54 mg/L
   B. D_c = (k2/k1) L0 e^(-k1 t_c); DO_min = D_c = 5.56 mg/L
   C. D_c = D0 e^(-k2 t_c); DO_min = DO_sat + D_c
   D. D_c = L0 - D0; DO_min = 9.1 - 2 = 7.1 mg/L
   Answer: A. At the critical point D_c = (k1/k2) L0 e^(-k1 t_c) = 5.56 mg/L, and the minimum DO is the saturation DO minus this deficit. (Key point: Critical deficit and minimum DO)

4. How is the distance of the minimum DO point found for a river flowing at 0.3 m/s?
   A. Divide t_c by the velocity in m/day
   B. Convert velocity to m/day (0.3 x 86 400) and multiply by t_c, giving about 76 km
   C. Multiply t_c in days by 0.3 directly
   D. Divide the critical deficit by the velocity
   Answer: B. Distance = velocity x time = 25 920 m/day x 2.94 days, about 76 200 m. (Key point: Distance downstream)

## Related in Environment and Pollution Control

- [BOD versus COD and a BOD and COD calculation](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Abod-cod-difference-numerical) — Environment and Pollution Control, chapter 4
- [Oxygen sag curve in a stream](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Aoxygen-sag-curve-explain) — Environment and Pollution Control, chapter 4

## Related in other subjects

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- [Critical oxygen deficit and location (table)](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-effluent-stream-table-78) — Sanitary Engineering, chapter 6
- [Critical DO deficit, 300 l/s into river](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-city-300ls-48kmday) — Sanitary Engineering, chapter 6
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