At the critical point:

dD/dt = 0 and the deficit is maximum (DO minimum)

Which makes the sag smaller?

Fast, shallow, turbulent stream (higher k2)

Practice · 6 marks

Oxygen sag curve in a stream

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.

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.

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.

 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

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Oxygen sag curve in a stream

Processes

Zones of a polluted river

Factors affecting the critical point

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