Surface Water Resources and Hydrology

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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C. Routing

Channel and Reservoir Routing

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\text{–}5\,\text{mg/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

L_o

exponentially. The rate constant

k

(or

k_d

in the sag model) is base-

e

here — watch for problems that quote a base-10

K

, which differs by a factor of

\ln 10 = 2.303

. The ultimate BOD

L_o

, not the 5-day value, is what drives the deficit model.

\text{BOD}_t = L_o\left(1 - e^{-k t}\right)

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

L_a

and the mixed DO are flow-weighted averages of the two streams. The initial deficit

D_a

is the saturation DO minus the mixed DO — it is the head start the river already has on the sag. Getting

L_a

and

D_a

right is half the problem; a flow-weighting slip propagates through everything downstream.

L_a = \dfrac{Q_r L_r + Q_w L_w}{Q_r + Q_w}, \qquad D_a = \text{DO}_{sat} - \dfrac{Q_r\,\text{DO}_r + Q_w\,\text{DO}_w}{Q_r + Q_w}

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

k_d

drives the deficit up as BOD is exerted; reaeration at rate

k_r

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

\text{DO} = \text{DO}_{sat} - D

.

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}

The critical point: where the sag bottoms out

The minimum DO — the point most likely to violate a standard — occurs at the critical time

t_c

, where the deficit is largest and its rate of change is zero (deoxygenation just balances reaeration). Solving

dD/dt = 0

gives

t_c

in terms of the two rate constants and the initial conditions. At that instant the critical deficit takes a compact form,

D_c = (k_d/k_r)\,L_a\,e^{-k_d t_c}

, and the minimum DO is

\text{DO}_{min} = \text{DO}_{sat} - D_c

. If

\text{DO}_{min}

falls below the standard, the discharge must be reduced (lower

L_a

) or treated.

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

Temperature correction of the rate constants

Both rate constants are reported at

20^\circ\text{C}

and corrected to the stream temperature with the Arrhenius-type relation

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

. Deoxygenation uses

\theta = 1.135

(

4\text{–}20^\circ\text{C}

) or

1.056

(

21\text{–}30^\circ\text{C}

); reaeration uses

\theta = 1.024

. Warmer water makes the sag worse two ways at once:

k_d

rises faster than

k_r

, deepening the deficit, while

\text{DO}_{sat}

itself falls (warm water holds less oxygen). Summer low-flow, high-temperature conditions are therefore the critical design case.

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

L_a

and the mixed DO; (2) form

D_a = \text{DO}_{sat} - \text{DO}_{mix}

; (3) temperature-correct

k_d

and

k_r

if

T \ne 20^\circ\text{C}

(and adjust

\text{DO}_{sat}

); (4) compute

t_c

, then

D_c

, then

\text{DO}_{min}

. Confirm your rate constants are base-

e

— a base-10 value must be multiplied by

2.303

. Keep

k_r > k_d

(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.

Key equations

BOD exertion (first-order)

\text{BOD}_t = L_o\left(1 - e^{-k t}\right)

Oxygen demand exerted by time $t$ (days); $L_o$ = ultimate BOD (mg/L), $k$ = base- $e$ decay constant (d $^{-1}$ ). A base-10 $K$ relates as $k = 2.303\,K$ .

Mixed (initial) ultimate BOD

L_a = \dfrac{Q_r L_r + Q_w L_w}{Q_r + Q_w}

Flow-weighted ultimate BOD entering the reach; $Q$ = flow, $L$ = ultimate BOD, subscripts $r$ = river, $w$ = waste.

Initial DO deficit

D_a = \text{DO}_{sat} - \dfrac{Q_r\,\text{DO}_r + Q_w\,\text{DO}_w}{Q_r + Q_w}

Saturation DO minus the flow-weighted mixed DO (mg/L) at the outfall; the deficit the reach starts with.

Streeter–Phelps deficit

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}

DO deficit (mg/L) at travel time $t$ (days). $k_d$ = deoxygenation, $k_r$ = reaeration (both base- $e$ , d $^{-1}$ ). Then $\text{DO} = \text{DO}_{sat} - D$ .

Critical time

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, where $dD/dt = 0$ . Requires $k_r \ne k_d$ .

Critical deficit

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

Maximum DO deficit (mg/L), at $t_c$ . Minimum dissolved oxygen is $\text{DO}_{min} = \text{DO}_{sat} - D_c$ .

Temperature correction

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

Rate constant at stream temperature $T$ ( $^\circ$ C). $\theta_{k_d} = 1.135$ (4–20°C) or $1.056$ (21–30°C); $\theta_{k_r} = 1.024$ .

Mass loading

M\,(\text{lb/day}) = C\,(\text{mg/L}) \times Q\,(\text{MGD}) \times 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.

Worked examples

Mixing at the outfall

Problem. A river flows at $Q_r = 8.0\,\text{m}^3/\text{s}$ with ultimate BOD $L_r = 2.0\,\text{mg/L}$ and $\text{DO}_r = 8.5\,\text{mg/L}$ . A treatment plant discharges $Q_w = 2.0\,\text{m}^3/\text{s}$ with $L_w = 30\,\text{mg/L}$ and $\text{DO}_w = 1.0\,\text{mg/L}$ . At $20^\circ\text{C}$ , $\text{DO}_{sat} = 9.08\,\text{mg/L}$ . Find the mixed ultimate BOD $L_a$ and the initial deficit $D_a$ .

Solution. Flow-weight the BOD:

L_a = \dfrac{8.0(2.0) + 2.0(30)}{8.0 + 2.0} = \dfrac{16 + 60}{10} = 7.60\,\text{mg/L}.

Flow-weight the DO:

\text{DO}_{mix} = \dfrac{8.0(8.5) + 2.0(1.0)}{10} = \dfrac{68 + 2}{10} = 7.00\,\text{mg/L}.

Initial deficit:

D_a = 9.08 - 7.00 = 2.08\,\text{mg/L}.

Sanity check: the mixed values lie between the two streams and are pulled toward the larger river flow, as expected. Final:

L_a = 7.60\,\text{mg/L}

,

D_a = 2.08\,\text{mg/L}

.

L_a = \dfrac{16 + 60}{10} = 7.60\ \text{mg/L}, \qquad D_a = 9.08 - 7.00 = 2.08\ \text{mg/L}

Critical time, critical deficit, and minimum DO

Problem. Continue the previous reach with $L_a = 7.60\,\text{mg/L}$ , $D_a = 2.08\,\text{mg/L}$ , $\text{DO}_{sat} = 9.08\,\text{mg/L}$ , $k_d = 0.30\,\text{d}^{-1}$ , and $k_r = 0.55\,\text{d}^{-1}$ (both base- $e$ , $20^\circ\text{C}$ ). Find the critical travel time, the critical deficit, and the minimum DO. Does it meet a $5.0\,\text{mg/L}$ standard?

Solution. Critical time, with

k_r - k_d = 0.25\,\text{d}^{-1}

and

k_r/k_d = 1.833

:

t_c = \dfrac{1}{0.25}\ln\!\left[1.833\left(1 - \dfrac{2.08(0.25)}{0.30(7.60)}\right)\right] = 4.0\,\ln\!\left[1.833(1 - 0.228)\right].

1.833(0.772) = 1.415

;

\ln 1.415 = 0.347

;

t_c = 4.0(0.347) = 1.39\,\text{d}

. Critical deficit:

D_c = \dfrac{0.30}{0.55}(7.60)\,e^{-0.30(1.39)} = (0.545)(7.60)(0.660) = 2.73\,\text{mg/L}.

Minimum DO:

\text{DO}_{min} = 9.08 - 2.73 = 6.35\,\text{mg/L} > 5.0\,\text{mg/L}\ \checkmark.

Sanity check: substituting

t_c = 1.39\,\text{d}

back into the full deficit equation also gives

D = 2.73\,\text{mg/L}

, confirming it is the maximum. Final:

t_c \approx 1.39\,\text{d}

,

D_c \approx 2.73\,\text{mg/L}

,

\text{DO}_{min} \approx 6.35\,\text{mg/L}

— the standard is met.

t_c = 1.39\ \text{d}, \quad D_c = 2.73\ \text{mg/L}, \quad \text{DO}_{min} = 6.35\ \text{mg/L}

Temperature correction to summer conditions

Problem. The rate constants $k_d = 0.30\,\text{d}^{-1}$ and $k_r = 0.55\,\text{d}^{-1}$ are reported at $20^\circ\text{C}$ . Correct them to a summer stream temperature of $25^\circ\text{C}$ . Use $\theta = 1.056$ for deoxygenation (21–30°C) and $\theta = 1.024$ for reaeration.

Solution. Deoxygenation:

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

Reaeration:

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

Sanity check: both rates increase with temperature, but

k_d

rises by

31\%

versus

k_r

by only

13\%

, so the sag deepens — and the saturation DO drops as well — confirming summer as the critical case. Final:

k_d(25) \approx 0.394\,\text{d}^{-1}

,

k_r(25) \approx 0.619\,\text{d}^{-1}

.

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

Common pitfalls

•Mixing base- $e$ and base-10 rate constants. The Streeter–Phelps $k_d$ , $k_r$ are base- $e$ ; a tabulated base-10 $K$ must be multiplied by $\ln 10 = 2.303$ before use.
•Using the 5-day BOD instead of the ultimate BOD $L_o$ for $L_a$ . The model needs the ultimate demand; convert $\text{BOD}_5$ to $L_o$ via $L_o = \text{BOD}_5/(1 - e^{-k\cdot 5})$ if only the 5-day value is given.
•Forgetting to flow-weight at the outfall. $L_a$ 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: $\text{DO}_{min} = \text{DO}_{sat} - D_c$ , not $D_c$ itself.
•Ignoring temperature. Warm water raises $k_d$ faster than $k_r$ AND lowers $\text{DO}_{sat}$ ; both deepen the sag, so $20^\circ\text{C}$ tabulated values understate the summer critical case.
•Assuming $k_r > k_d$ always holds. A sluggish, deep stream can have $k_r < k_d$ ; and if $k_r = k_d$ 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.

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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