---
title: "Oxygen sag curve in a stream"
subject: "Environment and Pollution Control"
chapter: 4
key: "semester_7/environment_and_pollution_control:oxygen-sag-curve-explain"
source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/oxygen-sag-curve-explain
site: BCT Engineering
---

# Oxygen sag curve in a stream

Environment and Pollution Control · Chapter 4: Water pollution

- Asked in: year not given, 6 marks
- Video: [oxygen sag curve | Deoxygenation | Reoxygenation | BOD | oxygen deficit | zone of self purification](https://www.youtube.com/watch?v=quBzazdeysw) — Civil Engineering Exam
- 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-curve-explain
- Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-4

## Question

What is the oxygen sag curve? Explain the deoxygenation and reaeration processes, the zones of a polluted river and the factors on which the critical point depends. Write the equations for the dissolved oxygen deficit.

## Short answer

The oxygen sag curve shows dissolved oxygen decline downstream of organic waste discharge due to deoxygenation (k₁) and subsequent recovery via reaeration (k₂). The Streeter‑Phelps equation models deficit: Dₜ = (k₁L₀/(k₂−k₁))(e⁻ᵏ¹ᵗ − e⁻ᵏ²ᵗ) + D₀e⁻ᵏ²ᵗ. The critical point where DO is lowest divides degradation, active decomposition and recovery zones.

## Answer

When organic waste enters a river, bacteria oxidise it and use up dissolved oxygen (DO) from the water (**deoxygenation**), while the stream takes oxygen from the atmosphere (**reaeration**). The plot of DO against time or distance downstream of the discharge point is the **oxygen sag curve**. It is a "sag" because DO first falls to a minimum and then recovers.

```text
 DO
 DOsat |------\                      _______ recovery
       |       \                  _--
       |        \              _-
       |         \          _-
 DOmin |...........\______--  <- critical point (tc)
       |  clean | degrad. | active decomp. | recovery | clean
       +---------------------------------------> distance/time
```

#### Processes
- **Deoxygenation rate** $= k_1 L_t$ where $L_t = L_0 e^{-k_1 t}$ is remaining BOD (first order, $k_1$ = deoxygenation constant, $d^{-1}$).
- **Reaeration rate** $= k_2 D$ where $D = DO_{sat} - DO$ is the oxygen deficit (mg/L) and $k_2$ the reaeration constant.
- Net change of deficit: 
$$
\frac{dD}{dt} = k_1 L_0 e^{-k_1 t} - k_2 D
$$
Integrating gives the **Streeter-Phelps equation**:
$$
D_t = \frac{k_1 L_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right) + D_0 e^{-k_2 t}
$$
The critical time (at $dD/dt = 0$) and deficit are:
$$
t_c = \frac{1}{k_2 - k_1}\ln\left[\frac{k_2}{k_1}\left(1 - \frac{D_0 (k_2 - k_1)}{k_1 L_0}\right)\right], \qquad D_c = \frac{k_1}{k_2}L_0 e^{-k_1 t_c}
$$

#### Zones of a polluted river
1. Clean zone (upstream), 2. Zone of degradation (turbid water, DO falling), 3. Zone of active decomposition (DO lowest, may be zero, foul smell, anaerobic), 4. Zone of recovery (DO rising, fish return), 5. Zone of clean water.

#### Factors affecting the critical point
- Waste load ($L_0$) and initial deficit ($D_0$): more waste, lower DO.
- $k_1$ and $k_2$: reaeration is higher in fast, shallow, turbulent streams, so the sag is smaller.
- Temperature (lower solubility and faster decay at high temperature), stream velocity and depth, and DO of the river water upstream.

## Check yourself

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

1. The oxygen sag curve shows:
   A. DO falling to a minimum downstream of a waste discharge and then recovering
   B. DO rising continuously
   C. BOD rising only
   D. DO constant
   Answer: A. Deoxygenation first exceeds reaeration, then reaeration wins and DO recovers. (Key point: Oxygen sag meaning)

2. Deoxygenation rate and reaeration rate are respectively:
   A. k1 L_t and k2 D
   B. k2 L_t and k1 D
   C. k1 D and k2 D
   D. k1 L0 and k2 L0
   Answer: A. Deoxygenation = k1 x remaining BOD (L_t); reaeration = k2 x deficit D. (Key point: Rates)

3. At the critical point:
   A. dD/dt = 0 and the deficit is maximum (DO minimum)
   B. DO equals saturation
   C. Deoxygenation is zero
   D. Reaeration is zero
   Answer: A. At tc deoxygenation equals reaeration, so the deficit is at its maximum. (Key point: Critical point)

4. Which makes the sag smaller?
   A. Higher waste load L0
   B. Fast, shallow, turbulent stream (higher k2)
   C. Higher initial deficit D0
   D. Lower upstream DO
   Answer: B. Higher reaeration in turbulent shallow streams lessens the deficit. (Key point: Factors)

## Related in Environment and Pollution Control

- [Critical DO deficit and location downstream](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Aoxygen-sag-critical-deficit-numerical) — Environment and Pollution Control, chapter 4

## Related in other subjects

- [Critical oxygen deficit and location (table 2)](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-effluent-stream-table-71) — Sanitary Engineering, chapter 6
- [Critical DO deficit, location, sag curve](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asag-curve-effluent-1p5-stream-7p5) — Sanitary Engineering, chapter 6
- [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