Surface Water & Groundwater Quality

7% of exam

Stream oxygen dynamics and the Streeter-Phelps DO sag, TMDL and load allocation, and biological and chemical contaminant fate.

3 concepts

Exam areas

A. Stream degradation and oxygen dynamics

The Streeter-Phelps Dissolved-Oxygen Sag

Track dissolved oxygen downstream of a discharge: BOD deoxygenation versus reaeration, the deficit equation, and the critical point that decides whether a stream stays alive.

When organic waste enters a stream, bacteria consume the biodegradable material and, in doing so, draw oxygen out of the water. If they pull it out faster than the atmosphere can put it back, dissolved oxygen (

DO

) falls, and somewhere downstream it reaches a minimum — the bottom of the oxygen ·sag·. If that minimum drops below roughly

4

–

5\,\text{mg/L}

, fish die and the stream is declared impaired. The Streeter-Phelps model is the classic tool that predicts this sag, and on the PE Civil exam it is the centerpiece of stream oxygen dynamics (NCEES PE Civil Reference Handbook §6.7.4 Oxygen Dynamics). Points are lost not in the exponentials but in the bookkeeping: mixing the waste with the river first, keeping deoxygenation (

k_d

) and reaeration (

k_r

) straight, and remembering that the ·deficit· — not the

DO

itself — is what the model actually solves for.

Deficit, not DO, is the state variable

The model works in terms of the dissolved-oxygen ·deficit·

D

, the amount by which the water sits below saturation:

D = DO_{sat} - DO

. A pristine stream has

D = 0

; a stressed stream has a large positive deficit. Two competing first-order processes drive

D

. Deoxygenation by BOD-degrading bacteria adds to the deficit at rate

k_d L

, proportional to the BOD remaining

L

. Reaeration across the water surface removes deficit at rate

k_r D

, proportional to how far below saturation the water already is. The sag is simply the trajectory of

D

as these two race each other downstream.

\dfrac{dD}{dt} = k_d L - k_r D

Mixing the waste into the river first

Before any sag is computed you must blend the discharge with the upstream flow at the outfall — every exam problem hinges on getting this initial condition right. Conserve mass for both BOD and oxygen: the mixed ultimate BOD

L_a

and mixed

DO

are flow-weighted averages of river and waste streams. The initial deficit is then

D_a = DO_{sat} - DO_{mix}

, evaluated at the temperature of the ·mixed· stream. A common slip is to use the river's pre-discharge

DO

D_a

; you must mix in the (usually oxygen-poor) effluent first.

L_a = \dfrac{Q_w L_w + Q_r L_r}{Q_w + Q_r}, \qquad DO_{mix} = \dfrac{Q_w DO_w + Q_r DO_r}{Q_w + Q_r}

The Streeter-Phelps deficit equation

Solving the first-order balance with initial conditions

L(0)=L_a

and

D(0)=D_a

gives the deficit at any travel time

t

downstream. The first term is the deficit generated by BOD that has not yet been satisfied; the second term is the decay of the initial mixing deficit by reaeration. Travel time is the link to geography:

t = x/v

, where

x

is distance downstream and

v

the stream velocity, so convert distances to days before substituting. Note the rate constants in this form are base-

e

(

k = 2.303\,K

, where

K

is the base-10 lab constant). The handbook (§6.7.4.7) prints the same solution with

10^{-K_d t}

and

10^{-K_r t}

exponentials that take the base-10

K

directly — match the constant to the form in front of you, never convert twice.

D = \dfrac{k_d L_a}{k_r - k_d}\left(e^{-k_d t} - e^{-k_r t}\right) + D_a\,e^{-k_r t}

Critical time and critical deficit

The single most-asked output is the worst case: the largest deficit

D_c

and how far downstream it occurs,

t_c

. Set

dD/dt = 0

(deoxygenation just balances reaeration) and solve. The critical time depends on the rate constants and the initial deficit; the critical deficit then follows from a compact relation,

D_c = \tfrac{k_d}{k_r} L_a\, e^{-k_d t_c}

. The minimum dissolved oxygen is

DO_{min} = DO_{sat} - D_c

— the number you compare against the water-quality standard.

t_c = \dfrac{1}{k_r - k_d}\ln\!\left[\dfrac{k_r}{k_d}\left(1 - \dfrac{D_a\,(k_r - k_d)}{k_d L_a}\right)\right]

Rate constants, reaeration, and self-purification

The deoxygenation constant

k_d

reflects how fast bacteria oxidize the BOD — about

0.23

–

0.70\,\text{day}^{-1}

for municipal wastewater and lower for treated effluent. The reaeration constant

k_r

reflects how fast the channel re-aerates and is governed by velocity and depth: shallow, fast, turbulent streams reaerate quickly. The O'Connor-Dobbins, Churchill, and Langbein-Durum correlations all express

k_r

as increasing with velocity and decreasing with depth (NCEES PE Civil Reference Handbook §6.7.4.6). The stream always recovers eventually — the BOD is finite, so the deficit turns around and decays — but

k_r > k_d

is the hallmark of vigorous self-purification, while a sluggish stream with

k_r < k_d

sags deeper and longer (its

t_c

is still finite). Only

k_r = k_d

breaks the formula's arithmetic, and a non-positive logarithm argument means the minimum DO is at the outfall itself.

Temperature: why summer is the design case

Two temperature effects stack to make warm, low-flow summer conditions the critical design scenario. First, oxygen saturation

DO_{sat}

falls with temperature — from

9.17\,\text{mg/L}

20^\circ\text{C}

8.38\,\text{mg/L}

25^\circ\text{C}

(NCEES PE Civil Reference Handbook §6.7.4.2 saturation table), so the ceiling itself drops. Second, biological deoxygenation speeds up with temperature far more than reaeration does. Correct every rate constant to stream temperature with the Arrhenius form below, using the handbook's banded BOD values

\theta = 1.135

(

T = 4

–

20^\circ\text{C}

) and

\theta = 1.056

(

T = 21

–

30^\circ\text{C}

) for

k_d

, and

\theta = 1.024

for reaeration

k_r

. The result: a stream that passes in winter can fail the same load in August.

k_T = k_{20}\,\theta^{\,T - 20}

Exam strategy

Work the problem in a fixed order: (1) mix waste and river to get

L_a

DO_{mix}

, and

D_a = DO_{sat} - DO_{mix}

; (2) correct

k_d

k_r

, and

DO_{sat}

to the stream temperature if a temperature is given; (3) if asked for a point, convert distance to travel time

t = x/v

and use the deficit equation; (4) if asked for the worst case, compute

t_c

then

D_c

, and report

DO_{min} = DO_{sat} - D_c

. Match the constants to the equation you use: the handbook's printed

10^{-Kt}

form takes the base-10 lab

K

as is, the

e^{-kt}

form needs

k = 2.303\,K

— and the handbook's

t_c

is printed with

\ln

, which balances only with base-

e

constants (or use

\log_{10}

with base-10

K

). Sanity-check sign and magnitude:

D_a

should be small and positive,

t_c

a few days, and

DO_{min}

below

DO_{sat}

but above zero. If your 'minimum DO' exceeds saturation, you mixed up

D

and

DO

Key equations

Dissolved-oxygen deficit

D = DO_{sat} - DO

$D$ (mg/L) is the state variable of the model: how far below saturation the water sits. $DO_{sat}$ is read from the saturation table at the stream temperature.

Streeter-Phelps deficit equation

D = \dfrac{k_d L_a}{k_r - k_d}\left(e^{-k_d t} - e^{-k_r t}\right) + D_a\,e^{-k_r t}

Deficit at travel time $t$ (days). $k_d$ , $k_r$ base- $e$ (day $^{-1}$ ); $L_a$ = mixed ultimate BOD (mg/L); $D_a$ = initial deficit after mixing (mg/L).

Mixed ultimate BOD and DO at outfall

L_a = \dfrac{Q_w L_w + Q_r L_r}{Q_w + Q_r}, \quad DO_{mix} = \dfrac{Q_w DO_w + Q_r DO_r}{Q_w + Q_r}

Flow-weighted mixing of waste ( $w$ ) and river ( $r$ ). Initial deficit $D_a = DO_{sat} - DO_{mix}$ .

Critical time (worst sag location)

t_c = \dfrac{1}{k_r - k_d}\ln\!\left[\dfrac{k_r}{k_d}\left(1 - \dfrac{D_a(k_r - k_d)}{k_d L_a}\right)\right]

Travel time (days) to the minimum DO. Valid for $k_r \neq k_d$ ; the bracket must be positive.

Critical deficit

D_c = \dfrac{k_d}{k_r}\,L_a\,e^{-k_d t_c}

Maximum deficit (mg/L). Minimum dissolved oxygen is $DO_{min} = DO_{sat} - D_c$ .

BOD remaining and exerted

L_t = L_0\,e^{-k_d t}, \qquad BOD_t = L_0\left(1 - e^{-k_d t}\right)

First-order BOD decay. $L_0$ = ultimate BOD (mg/L); exerted BOD is the oxygen consumed up to time $t$ .

Base-e and base-10 rate constants

k_d = 2.303\,K_d

Lab BOD data give base-10 $K$ (day $^{-1}$ ); the $e^{-kt}$ form of Streeter-Phelps needs base- $e$ $k$ , while the handbook's $10^{-Kt}$ form takes $K$ directly. Convert once, for the form you use.

Arrhenius temperature correction

k_T = k_{20}\,\theta^{\,T - 20}

Correct rate constants to stream temperature $T$ (°C). Handbook §6.8.3.3 banded BOD values: $\theta = 1.135$ ( $T = 4$ – $20^\circ$ C) and $\theta = 1.056$ ( $T = 21$ – $30^\circ$ C) for $k_d$ ; $\theta = 1.024$ for reaeration $k_r$ .

Oxygen saturation (Henry's law)

DO_{sat} = H\,P_{O_2}

Saturation DO is proportional to the partial pressure of oxygen; in practice read it from the temperature/salinity table. $\approx 9.17\,\text{mg/L}$ at $20^\circ$ C, freshwater, $1\,\text{atm}$ .

Travel time from distance

t = \dfrac{x}{v}

Convert downstream distance $x$ to travel time $t$ (days) using stream velocity $v$ ; mind unit conversion to days.

Worked examples

Critical point downstream of an outfall

Problem. A treatment plant discharges $Q_w = 0.10\,\text{m}^3/\text{s}$ of effluent with ultimate BOD $L_w = 40\,\text{mg/L}$ and $DO_w = 2.0\,\text{mg/L}$ into a river carrying $Q_r = 0.50\,\text{m}^3/\text{s}$ at $L_r = 2.0\,\text{mg/L}$ and $DO_r = 8.0\,\text{mg/L}$ . The mixed stream is at $20^\circ\text{C}$ ( $DO_{sat} = 9.17\,\text{mg/L}$ ), with $k_d = 0.30\,\text{day}^{-1}$ and $k_r = 0.55\,\text{day}^{-1}$ . Find the critical travel time, the critical deficit, and the minimum dissolved oxygen.

Solution. Mix at the outfall.

L_a = \dfrac{(0.10)(40) + (0.50)(2.0)}{0.10 + 0.50} = \dfrac{4.0 + 1.0}{0.60} = 8.33\,\text{mg/L}

DO_{mix} = \dfrac{(0.10)(2.0) + (0.50)(8.0)}{0.60} = \dfrac{0.20 + 4.0}{0.60} = 7.00\,\text{mg/L}

, so

D_a = 9.17 - 7.00 = 2.17\,\text{mg/L}

. Critical time:

t_c = \dfrac{1}{0.55-0.30}\ln\!\left[\dfrac{0.55}{0.30}\left(1 - \dfrac{2.17(0.25)}{0.30(8.33)}\right)\right] = \dfrac{1}{0.25}\ln[\,1.833(1 - 0.217)\,] = 4\ln(1.436) = 1.45\,\text{days}

. Critical deficit:

D_c = \dfrac{0.30}{0.55}(8.33)\,e^{-0.30(1.45)} = (0.545)(8.33)(0.648) = 2.95\,\text{mg/L}

. Minimum DO:

DO_{min} = 9.17 - 2.95 = 6.22\,\text{mg/L}

. Sanity check:

D_a

is small and positive,

t_c

is a couple of days, and

6.22\,\text{mg/L} > 5\,\text{mg/L}

— the stream stays above a typical aquatic-life standard.

DO_{min} = DO_{sat} - D_c = 9.17 - 2.95 = 6.22\,\text{mg/L}

DO at a town downstream

Problem. Using the mixed conditions from the previous problem ( $L_a = 8.33\,\text{mg/L}$ , $D_a = 2.17\,\text{mg/L}$ , $k_d = 0.30$ , $k_r = 0.55\,\text{day}^{-1}$ , $DO_{sat} = 9.17\,\text{mg/L}$ ), a town draws water $26\,\text{km}$ downstream where the mean velocity is $0.30\,\text{m/s}$ . What is the dissolved oxygen at the intake?

Solution. Travel time:

t = \dfrac{x}{v} = \dfrac{26{,}000\,\text{m}}{0.30\,\text{m/s}} = 86{,}667\,\text{s} = 1.00\,\text{day}

. Deficit:

D = \dfrac{(0.30)(8.33)}{0.25}\left(e^{-0.30} - e^{-0.55}\right) + 2.17\,e^{-0.55}

= (10.0)(0.7408 - 0.5769) + 2.17(0.5769) = (10.0)(0.1639) + 1.252 = 1.64 + 1.25 = 2.89\,\text{mg/L}

DO = 9.17 - 2.89 = 6.28\,\text{mg/L}

. Sanity check: the intake at

t = 1.0\,\text{day}

is just upstream of the critical point (

t_c = 1.45\,\text{day}

), so its DO (

6.28

) should be slightly above the minimum (

6.22\,\text{mg/L}

) — it is.

D = \dfrac{k_d L_a}{k_r - k_d}\left(e^{-k_d t} - e^{-k_r t}\right) + D_a\,e^{-k_r t}

Temperature correction to summer conditions

Problem. Laboratory tests at $20^\circ\text{C}$ give $k_d = 0.30\,\text{day}^{-1}$ and a reaeration estimate $k_r = 0.55\,\text{day}^{-1}$ . Correct both to a summer stream temperature of $25^\circ\text{C}$ (use $\theta = 1.056$ for BOD in the $21$ – $30^\circ\text{C}$ band and $\theta = 1.024$ for reaeration) and state the design $DO_{sat}$ .

Solution. Deoxygenation:

k_d(25) = 0.30\,(1.056)^{25-20} = 0.30\,(1.056)^{5} = 0.30(1.313) = 0.394\,\text{day}^{-1}

. Reaeration:

k_r(25) = 0.55\,(1.024)^{5} = 0.55(1.126) = 0.619\,\text{day}^{-1}

. Saturation: from the table,

DO_{sat}(25^\circ\text{C}) = 8.38\,\text{mg/L}

(down from

9.17

20^\circ

C). Sanity check: deoxygenation rose by

31\%

but reaeration by only

13\%

, and the saturation ceiling dropped

0.79\,\text{mg/L}

— every effect pushes the summer sag deeper, confirming why warm low-flow conditions govern the discharge permit.

k_d(25) = 0.30\,(1.056)^{5} = 0.394\,\text{day}^{-1}, \quad k_r(25) = 0.55\,(1.024)^{5} = 0.619\,\text{day}^{-1}

Common pitfalls

•Treating the river's pre-discharge DO as the initial deficit. You must flow-mix waste and river first, then $D_a = DO_{sat} - DO_{mix}$ — the oxygen-poor effluent lowers $DO_{mix}$ and raises $D_a$ .
•Confusing the deficit $D$ with the dissolved oxygen $DO$ . The model solves for $D$ ; the answer the standard is compared against is $DO_{min} = DO_{sat} - D_c$ .
•Mismatching the constant and the exponential. The $e^{-kt}$ form needs base- $e$ $k = 2.303\,K$ ; the handbook's $10^{-Kt}$ form takes the base-10 $K$ directly. Converting for one and using the other is off by 2.303 in the exponent.
•Forgetting to convert distance to travel time. The equation runs on $t$ (days); use $t = x/v$ and watch the seconds-to-days conversion.
•Reading $DO_{sat}$ at the wrong temperature (or ignoring temperature entirely). Saturation falls with warming and salinity, and warm water is usually the critical case — correct $DO_{sat}$ , $k_d$ , and $k_r$ to stream temperature.
•Misreading the $t_c$ formula's limits. $k_r < k_d$ still gives a valid, finite $t_c$ (the sag is just deeper and later); only $k_r = k_d$ breaks the arithmetic, and a zero or negative logarithm argument means the deficit is already past its peak at the outfall — the minimum DO is at $t = 0$ .
•Swapping $k_d$ and $k_r$ in the prefactor $k_d L_a/(k_r - k_d)$ . The numerator is deoxygenation; the denominator is reaeration minus deoxygenation — order matters for both sign and magnitude.

References

NCEES PE Civil Reference Handbook — §6.7.4 Oxygen Dynamics (Streeter-Phelps)
NCEES PE Civil Reference Handbook — §6.7.3 Biochemical Oxygen Demand
NCEES PE Civil Reference Handbook — §6.8.3.3 Kinetic Temperature Corrections
Davis & Cornwell, Introduction to Environmental Engineering — Derivation of the sag and critical-point relations

B. Total maximum daily load (TMDL)

TMDL and Load Allocation

Set the pollutant budget a waterbody can absorb: convert flow and concentration to mass load, split the load among sources with a margin of safety, and back out a permit limit.

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C. Biological and chemical contaminants

Biological and Chemical Contaminants: Fate, Partitioning and Risk

Where contaminants go and what they do: indicator organisms, partition coefficients, sorption and volatilization, first-order decay, and the exposure math behind drinking-water risk.

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