Surface Water & Groundwater Quality · Study · PE Civil: Water Resources and Environmental · FE → PE Prep
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
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 4–5mg/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 (
kd
) and reaeration (
kr
) 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=DOsat−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 kdL, proportional to the BOD remaining L. Reaeration across the water surface removes deficit at rate krD, 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.
dtdD=kdL−krD
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 La and mixed DO are flow-weighted averages of river and waste streams. The initial deficit is then Da=DOsat−DOmix, evaluated at the temperature of the ·mixed· stream. A common slip is to use the river's pre-discharge DO as Da; you must mix in the (usually oxygen-poor) effluent first.
Solving the first-order balance with initial conditions L(0)=La and D(0)=Da 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 here are base-e (k=2.303K, where K is the base-10 lab constant).
D=kr−kdkdLa(e−kdt−e−krt)+Dae−krt
Critical time and critical deficit
The single most-asked output is the worst case: the largest deficit Dc and how far downstream it occurs, tc. 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, Dc=krkdLae−kdtc. The minimum dissolved oxygen is DOmin=DOsat−Dc — the number you compare against the water-quality standard.
tc=kr−kd1ln[kdkr(1−kdLaDa(kr−kd))]
Rate constants, reaeration, and self-purification
The deoxygenation constant kd reflects how fast bacteria oxidize the BOD — about 0.23–0.70day−1 for municipal wastewater and lower for treated effluent. The reaeration constant kr 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 kr as increasing with velocity and decreasing with depth (NCEES PE Civil Reference Handbook §6.7.4.6). When kr>kd the stream can recover — the hallmark of self-purification; when kr<kd the deficit keeps growing and the sag formula's tc loses meaning.
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 DOsat falls with temperature — from 9.17mg/L at 20∘C to 8.38mg/L at 25∘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 θ=1.135 (T=4–20∘C) and θ=1.056 (T=21–30∘C) for kd, and θ=1.024 for reaeration kr. The result: a stream that passes in winter can fail the same load in August.
kT=k20θT−20
Exam strategy
Work the problem in a fixed order: (1) mix waste and river to get La, DOmix, and Da=DOsat−DOmix; (2) correct kd, kr, and DOsat 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 tc then Dc, and report DOmin=DOsat−Dc. Confirm the constants are base-e (multiply a base-10 K by 2.303). Sanity-check sign and magnitude: Da should be small and positive, tc a few days, and DOmin below DOsat but above zero. If your 'minimum DO' exceeds saturation, you mixed up D and DO.
Key equations
Dissolved-oxygen deficitD=DOsat−DO
D (mg/L) is the state variable of the model: how far below saturation the water sits. DOsat is read from the saturation table at the stream temperature.
Mixed ultimate BOD and DO at outfallLa=Qw+QrQwLw+QrLr,DOmix=Qw+QrQwDOw+QrDOr
Critical time (worst sag location)tc=kr−kd1ln[kdkr(1−kdLaDa(kr−kd))]
Critical deficitDc=krkdLae−kdtc
BOD remaining and exertedLt=L0e−kdt,BODt=L0(1−e−kdt)
Base-e and base-10 rate constantskd=2.303Kd
Lab BOD data give base-10 K (day−1
Arrhenius temperature correctionkT=k20θT−20
Correct rate constants to stream temperature T
Oxygen saturation (Henry's law)DOsat=HPO2
Travel time from distancet=vx
Convert downstream distance x to travel time t (days) using stream velocity v
Worked examples
Critical point downstream of an outfall
Problem. A treatment plant discharges Qw=0.10m3/s of effluent with ultimate BOD Lw=40mg/L and DOw=2.0mg/L into a river carrying Qr=0.50m3/s at Lr=2.0mg/L and DOr=8.0mg/L. The mixed stream is at 20∘C (DOsat=9.17mg/L), with kd=0.30day−1 and kr=0.55day−1. Find the critical travel time, the critical deficit, and the minimum dissolved oxygen.
Solution. Mix at the outfall. La=0.10+0.50(0.10)(40)+(0.50)(2.0)=0.604.0+1.0=8.33mg/L
DOmin=DOsat−Dc=9.17−2.95=6.22mg/L
DO at a town downstream
Problem. Using the mixed conditions from the previous problem (La=8.33mg/L, Da=2.17mg/L
Temperature correction to summer conditions
Problem. Laboratory tests at 20∘C give kd=0.30day−1
Common pitfalls
•Treating the river's pre-discharge DO as the initial deficit. You must flow-mix waste and river first, then Da=DOsat−DOmix — the oxygen-poor effluent lowers DOmix and raises Da.
•Confusing the deficit D with the dissolved oxygen DO. The model solves for D; the answer the standard is compared against is DOmin=DOsat−Dc
•Using base-10 lab constants K directly in the exponentials. Streeter-Phelps needs base-e: k=2.303K. Skipping this underestimates the rates by more than half.
•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 DOsat at the wrong temperature (or ignoring temperature entirely). Saturation falls with warming and salinity, and warm water is usually the critical case — correct DOsat, kd
•Applying the tc formula when kr≤kd. With insufficient reaeration the deficit never turns around, and the logarithm's argument goes non-physical; recognize a continuously declining sag instead.
•Swapping kd and kr in the prefactor kdLa/(kr−kd)
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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Deficit at travel time t (days). kd, kr base-e (day−1); La = mixed ultimate BOD (mg/L); Da = initial deficit after mixing (mg/L).
Flow-weighted mixing of waste (w) and river (r). Initial deficit Da=DOsat−DOmix.
Travel time (days) to the minimum DO. Valid for kr=kd; the bracket must be positive.
Maximum deficit (mg/L). Minimum dissolved oxygen is DOmin=DOsat−Dc.
First-order BOD decay. L0 = ultimate BOD (mg/L); exerted BOD is the oxygen consumed up to time t.
); the Streeter-Phelps exponentials need base-
e
k
. Always convert.
(°C). Handbook §6.8.3.3 banded BOD values:
θ=1.135
(
T=4
–
20∘
C) and
θ=1.056
(
T=21
–
30∘
C) for
kd
;
θ=1.024
for reaeration
kr
.
Saturation DO is proportional to the partial pressure of oxygen; in practice read it from the temperature/salinity table. ≈9.17mg/L at 20∘C, freshwater, 1atm.
— the stream stays above a typical aquatic-life standard.
,
kd=0.30
,
kr=0.55day−1
,
DOsat=9.17mg/L
), a town draws water
26km
downstream where the mean velocity is
0.30m/s
. What is the dissolved oxygen at the intake?
Solution. Travel time: t=vx=0.30m/s26,000m=86,667s=1.00day.
Deficit: D=0.25(0.30)(8.33)(e−0.30−e−0.55)+2.17e−0.55.
=(10.0)(0.7408−0.5769)+2.17(0.5769)=(10.0)(0.1639)+1.252=1.64+1.25=2.89mg/L.
DO=9.17−2.89=6.28mg/L.
Sanity check: the intake at t=1.0day is just upstream of the critical point (tc=1.45day), so its DO (6.28) should be slightly above the minimum (6.22mg/L) — it is.
D=kr−kdkdLa(e−kdt−e−krt)+Dae−krt
and a reaeration estimate
kr=0.55day−1
. Correct both to a summer stream temperature of
25∘C
(use
θ=1.056
for BOD in the
21
–
30∘C
band and
θ=1.024
for reaeration) and state the design
DOsat
.
Solution. Deoxygenation: kd(25)=0.30(1.056)25−20=0.30(1.056)5=0.30(1.313)=0.394day−1.
Reaeration: kr(25)=0.55(1.024)5=0.55(1.126)=0.619day−1.
Saturation: from the table, DOsat(25∘C)=8.38mg/L (down from 9.17 at 20∘C).
Sanity check: deoxygenation rose by 31% but reaeration by only 13%, and the saturation ceiling dropped 0.79mg/L — every effect pushes the summer sag deeper, confirming why warm low-flow conditions govern the discharge permit.