ENV-25

DO sag deficit

D = [kd La/(ka−kd)](e^{−kd t}−e^{−ka t})+D0 e^{−ka t}. Classic sag.

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Water qualityStreeter–Phelps

Governing equation

D=\dfrac{k_d L_a}{k_a-k_d}(e^{-k_d t}-e^{-k_a t})+D_0 e^{-k_a t}

$t \in \left[0.1, 15\right]$

where

k_d: Deoxygenation (1/d)
k_a: Reaeration (1/d)
L_a: Initial BOD (mg/L)
D_0: Initial deficit (mg/L)
t: Travel time (d)
D: DO deficit (mg/L)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

D=\dfrac{k_d L_a}{k_a-k_d}(e^{-k_d t}-e^{-k_a t})+D_0 e^{-k_a t}

$t \in \left[0.1, 15\right]$

$t = 0.1$ $D = 1.515$

Along the curve

②

Domain of the variable — blue

t_{\min} = 0.1,\quad t_{\max} = 15

$t \in \left[0.1, 15\right]$

$t = 0.1$ $t = 0.1$

Along the curve

③

Response curve — green

D=\dfrac{k_d L_a}{k_a-k_d}(e^{-k_d t}-e^{-k_a t})+D_0 e^{-k_a t}

$t \in \left[0.1, 15\right]$

$t = 0.1$ $D = 1.515$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}D}{\mathrm{d}t}

$t \in \left[0.1, 15\right]$

$t = 0.194$ $D^{\prime} = 4.492$

Along the curve

Lecture brief

Historical brief

Streeter–Phelps (1925) oxygen sag, settling theory and Guldberg–Waage kinetics made water and air quality a rate problem. The lab computes sag, overflow and a snapshot of reactor mass balance. This sheet (ENV-25 — DO sag deficit) is the form associated with Streeter–Phelps. Working symbols: $k_d$ , $k_a$ , $L_a$ , $D_0$ , $t$ $\rightarrow$ $D$ . Coupled dL/dt = −kd L and dD/dt = kd L − ka D. The critical time is when dD/dt = 0.

Purpose

Purpose: compute

D

from

k_d

k_a

L_a

D_0

t

in Environmental via

D=\dfrac{k_d L_a}{k_a-k_d}(e^{-k_d t}-e^{-k_a t})+D_0 e^{-k_a t}

D = [kd La/(ka−kd)](e^{−kd t}−e^{−ka t})+D0 e^{−ka t}. Classic sag. Use it when a real environmental question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

k_d = 0.300\,\mathrm{1/d}

k_a = 0.600\,\mathrm{1/d}

L_a = 20.000\,\mathrm{mg/L}

D_0 = 1.000\,\mathrm{mg/L}

t = 2.000\,\mathrm{d}

, the governing relation

D=\dfrac{k_d L_a}{k_a-k_d}(e^{-k_d t}-e^{-k_a t})+D_0 e^{-k_a t}

yields

D = 5.254\,\mathrm{mg/L}

. A river, a BOD slug, a sag curve. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Deoxygenation k_d0.3 1/dReaeration k_a0.6 1/dInitial BOD L_a20 mg/LInitial deficit D_01 mg/LTravel time t2 d

Outputs

DO deficit D5.254 mg/L

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Narration of this film

A river, a BOD slug, a sag curve.

Coupled dL/dt = −kd L and dD/dt = kd L − ka D. The critical time is when dD/dt = 0.

Reading speed

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