Surface Water Resources and Hydrology · Study · FE Environmental · FE → PE Prep
Surface Water Resources and Hydrology
9% of exam
Runoff and the rational method, time of concentration and IDF curves, detention and retention storage sizing, channel and reservoir routing, water-quality modeling (Streeter-Phelps, eutrophication), and the water budget.
5 concepts
A. Runoff calculations
The Rational Method and Peak Runoff
Convert a design storm into a peak discharge with Q = CiA, build a composite runoff coefficient, and read intensity from an IDF curve at the time of concentration.
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B. Water storage sizing
Detention and Retention Storage Sizing
Size a detention basin from the area between the inflow and outflow hydrographs, and size a reservoir from the maximum cumulative deficit on a Rippl mass curve.
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Track a flood wave through a reservoir with the storage-indication (modified Puls) method and through a channel with Muskingum routing, and see how routing attenuates and lags the hydrograph.
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D. Water quality and modeling
The Streeter–Phelps Dissolved-Oxygen Sag
Model the dissolved-oxygen sag below a BOD outfall: deoxygenation vs reaeration, the deficit equation, the critical point, temperature correction, and nutrient limitation.
When organic waste enters a river, bacteria consume oxygen to break it down faster than the atmosphere can resupply it, and the dissolved-oxygen (DO) concentration drops downstream — sags — before recovering. Whether that sag dips below the level fish need (often 4–5mg/L) decides whether a discharge permit is approved. The Streeter–Phelps model, printed in the FE Reference Handbook under Environmental Engineering — Stream Modeling, is the classic tool for predicting that sag, and it is the flagship water-quality calculation on the exam. The whole model is a competition between two first-order processes: deoxygenation by BOD and reaeration from the atmosphere.
BOD as the oxygen demand driving the sag
Biochemical oxygen demand (BOD) is the mass of oxygen microbes will consume to oxidize the organic matter in the water. It is exerted as a first-order decay: the rate is proportional to the BOD remaining, so the oxygen consumed by time t approaches the ultimate demand Lo exponentially. The rate constant k (or kd
BODt=Lo(1−e−kt)
Mixing at the outfall: the initial conditions
Before any modeling, mass-balance the river and the waste stream at the point of discharge to get the conditions entering the reach. The mixed ultimate BOD La and the mixed DO are flow-weighted averages of the two streams. The initial deficit Da is the saturation DO minus the mixed DO — it is the head start the river already has on the sag. Getting La
The deficit equation: deoxygenation vs reaeration
The DO deficit D (saturation minus actual DO) at travel time t downstream is governed by two competing first-order terms. Deoxygenation at rate kd drives the deficit up as BOD is exerted; reaeration at rate kr
The critical point: where the sag bottoms out
The minimum DO — the point most likely to violate a standard — occurs at the critical time tc, where the deficit is largest and its rate of change is zero (deoxygenation just balances reaeration). Solving dD/dt=0 gives tc
Temperature correction of the rate constants
Both rate constants are reported at 20∘C and corrected to the stream temperature with the Arrhenius-type relation kT=k20θT−20
Eutrophication and nutrient limitation
BOD oxygen demand is the acute problem; nutrients are the chronic one. Excess nitrogen and phosphorus fertilize algal blooms whose death and decay impose a delayed BOD load, driving lakes from oligotrophic (clear, low productivity) through mesotrophic to eutrophic (turbid, oxygen-depleted at depth). Phosphorus is usually the limiting nutrient in fresh water — the one in shortest supply relative to need — so controlling phosphorus controls the bloom (the handbook's lake-classification table keys trophic state to total phosphorus, chlorophyll-a, and Secchi depth). This is why the Streeter–Phelps reach analysis is paired with nutrient limits in modern discharge permits.
Exam strategy
Work the problem in order: (1) flow-weight La and the mixed DO; (2) form Da=DOsat−DOmix
Key equations
BOD exertion (first-order)BODt=Lo(1−e−kt)
Worked examples
Mixing at the outfall
Problem. A river flows at Qr=8.0m3/s with ultimate BOD Lr=2.0mg/L
Common pitfalls
•Mixing base-e and base-10 rate constants. The Streeter–Phelps kd, kr are base-e
References
NCEES FE Reference Handbook — Environmental Engineering: Stream Modeling (Streeter–Phelps)
NCEES FE Reference Handbook — Environmental Engineering: Kinetic Temperature Corrections
Davis & Cornwell, Introduction to Environmental Engineering — DO sag derivation and lake trophic classification
Clean Water Act (CWA) — NPDES discharge permitting context
E. Water budget
The Water Budget, Time of Concentration, and Hydrographs
Close the hydrologic budget P = R + ET + ΔS + infiltration, partition rainfall with the NRCS curve number, build a time of concentration from flow segments, and scale a unit hydrograph.
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in the sag model) is base-
e
here — watch for problems that quote a base-10
K
, which differs by a factor of
ln10=2.303
. The ultimate BOD
Lo
, not the 5-day value, is what drives the deficit model.
and
Da
right is half the problem; a flow-weighting slip propagates through everything downstream.
pulls the deficit back down toward zero as the atmosphere resupplies oxygen. The Streeter–Phelps solution superposes them — the first term is the BOD-driven rise-and-fall, the second is the decay of the initial deficit. The actual DO is then
DO=DOsat−D
.
D=kr−kdkdLa(e−kdt−e−krt)+Dae−krt
in terms of the two rate constants and the initial conditions. At that instant the critical deficit takes a compact form,
Dc=(kd/kr)Lae−kdtc
, and the minimum DO is
DOmin=DOsat−Dc
. If
DOmin
falls below the standard, the discharge must be reduced (lower
. Warmer water makes the sag worse two ways at once:
kd
rises faster than
kr
, deepening the deficit, while
DOsat
itself falls (warm water holds less oxygen). Summer low-flow, high-temperature conditions are therefore the critical design case.
kT=k20θT−20
; (3) temperature-correct
kd
and
kr
if
T=20∘C
(and adjust
DOsat
); (4) compute
tc
, then
Dc
, then
DOmin
. Confirm your rate constants are base-
e
— a base-10 value must be multiplied by
2.303
. Keep
kr>kd
(reaeration usually outpaces deoxygenation); if they are equal the deficit formula is indeterminate and a separate limiting form applies. Always report the minimum DO, not the deficit, when the question asks whether a standard is met.
Oxygen demand exerted by time t (days); Lo = ultimate BOD (mg/L), k = base-e decay constant (d−1). A base-10 K relates as k=2.303K.
Travel time (days) to the minimum DO, where dD/dt=0. Requires kr=kd
Critical deficitDc=krkdLae−kdtc
Maximum DO deficit (mg/L), at tc. Minimum dissolved oxygen is DOmin=DOsat−Dc
Temperature correctionkT=k20θT−20
Rate constant at stream temperature T (∘C). θkd=1.135 (4–20°C) or 1.056 (21–30°C); θkr=1.024.
Mass loadingM(lb/day)=C(mg/L)×Q(MGD)×8.34
Converts a concentration and flow to a daily mass load; the 8.34 factor carries units lb·L/(mg·MG). Used to set BOD load limits.
and
DOr=8.5mg/L
. A treatment plant discharges
Qw=2.0m3/s
with
Lw=30mg/L
and
DOw=1.0mg/L
. At
20∘C
,
DOsat=9.08mg/L
. Find the mixed ultimate BOD
La
and the initial deficit
Da
.
Solution. Flow-weight the BOD:
La=8.0+2.08.0(2.0)+2.0(30)=1016+60=7.60mg/L.
Flow-weight the DO:
DOmix=108.0(8.5)+2.0(1.0)=1068+2=7.00mg/L.
Initial deficit:
Da=9.08−7.00=2.08mg/L.
Sanity check: the mixed values lie between the two streams and are pulled toward the larger river flow, as expected. Final: La=7.60mg/L, Da=2.08mg/L.
La=1016+60=7.60mg/L,Da=9.08−7.00=2.08mg/L
Critical time, critical deficit, and minimum DO
Problem. Continue the previous reach with La=7.60mg/L, Da=2.08mg/L, DOsat=9.08mg/L, kd=0.30d−1, and kr=0.55d−1 (both base-e, 20∘C). Find the critical travel time, the critical deficit, and the minimum DO. Does it meet a 5.0mg/L standard?
Solution. Critical time, with kr−kd=0.25d−1 and kr/kd=1.833
tc=1.39d,Dc=2.73mg/L,DOmin=6.35mg/L
Temperature correction to summer conditions
Problem. The rate constants kd=0.30d−1 and kr=0.55d−1 are reported at 20∘C. Correct them to a summer stream temperature of 25∘C. Use θ=1.056 for deoxygenation (21–30°C) and θ=1.024 for reaeration.
•Using the 5-day BOD instead of the ultimate BOD Lo for La. The model needs the ultimate demand; convert BOD5 to Lo via Lo=BOD5/(1−e−k⋅5) if only the 5-day value is given.
•Forgetting to flow-weight at the outfall. La and the mixed DO are flow-weighted averages of river and waste; using the raw waste concentration overstates the load.
•Reporting the deficit when the question asks for DO. The standard is on dissolved oxygen: DOmin=DOsat−Dc, not Dc itself.
•Ignoring temperature. Warm water raises kd faster than kr AND lowers DOsat; both deepen the sag, so 20∘C tabulated values understate the summer critical case.
•Assuming kr>kd always holds. A sluggish, deep stream can have kr<kd; and if kr=kd the deficit formula is indeterminate and a limiting form is required.
•Confusing nitrogen and phosphorus limitation. In fresh water phosphorus is usually limiting, so phosphorus control governs eutrophication — nitrogen often limits in estuaries instead.