Hydrology · Study · PE Civil: Water Resources and Environmental · FE → PE Prep
Hydrology
11% of exam
Storm frequency and IDF, time of concentration, Rational and SCS/NRCS runoff, unit and synthetic hydrographs, routing, depletions, and stormwater BMPs.
7 concepts
A. Storm characteristics
Storm Characteristics and Probability of Exceedance
Design-storm depth-duration-frequency, the return period, annual exceedance probability, the risk of one or more exceedances over a design life, and reading an IDF curve.
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B. Runoff analysis
NRCS (SCS) Curve-Number Runoff
Curve number from land use and hydrologic soil group, initial abstraction, the NRCS runoff-depth equation, antecedent moisture adjustment, and area-weighted CN.
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The unit-hydrograph concept, convolution of a storm hydrograph, the S-curve for changing rainfall duration, and synthetic methods (SCS dimensionless and Snyder).
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E. Time of concentration
The Rational Method, Time of Concentration and IDF
Q = CiA for peak discharge on small catchments: composite runoff coefficient, time of concentration by flow segment, and selecting intensity from an IDF curve.
The rational method is the most-used peak-flow tool in civil practice — storm sewers, inlets, small culverts, and parking-lot drainage all start with Q=CiA. Its deceptive simplicity hides three sub-problems that decide whether the answer is right: choosing the runoff coefficient C, computing the time of concentration tc, and reading the correct intensity i from an IDF curve at the design return period. Points evaporate in the unit bookkeeping (A in acres, i in in/hr, answer in cfs) and in the circular dependence between tc and i. The method is set out in PE Civil Reference Handbook §6.5.2.1 (Rational Formula Method) with tc in §6.5.4.
The rational formula
The peak discharge from a small catchment is the product of a dimensionless runoff coefficient C, the design rainfall intensity i, and the drainage area A. The conceptual heart is that the design intensity must have a duration equal to the time of concentration: only when the storm lasts at least tc does the whole watershed contribute to the outlet simultaneously, giving the true peak. In USCS units the conversion is a near-identity — 1acre-in/hr=1.008cfs≈1
The runoff coefficient and composite C
The coefficient C is the fraction of rainfall that becomes peak runoff, lumping infiltration, depression storage, and surface type into one number — near 0.95 for pavement and roofs, 0.10–0.25 for parks and lawns. For a mixed catchment, area-weight the sub-area coefficients into a composite C. Steeper slopes and rarer return periods push C
Time of concentration: the flow path in segments
The time of concentration is the travel time from the hydraulically most distant point to the outlet, and it is built up segment by segment along the flow path. A typical path is sheet (overland) flow, then shallow concentrated flow, then open-channel or pipe flow — each segment has its own travel-time formula, and tc is their sum. Underestimating tc selects too high an intensity and oversizes everything; overestimating does the reverse.
tc=∑ti=Tti+tsc+tch
Sheet, shallow, and channel travel times
Sheet flow over the first ≤300ft uses the kinematic-wave form Tti=i0.4S0.4Ku(nL)0.8
Selecting intensity from the IDF curve
With tc in hand, enter the IDF curve for the design return period at duration =tc and read the intensity. Because the sheet-flow term contains i, the practical procedure is a short iteration: assume an i
Assumptions and limits
The rational method assumes a uniform storm over a small, mostly impervious area with a constant C, and it yields only the peak — no volume, no hydrograph shape. It is generally restricted to drainage areas under about 200 acres (some agencies say 1 mi²); beyond that, spatial rainfall variation and storage make the linearity break down, and you switch to a unit-hydrograph or NRCS approach. Knowing this boundary is itself an exam point: a 500-acre basin is a signal to abandon Q=CiA.
Exam strategy
Work the chain: composite C → segment travel times → tc → IDF read at the return period → Q=CiA. If sheet flow is present, iterate i and tc
Key equations
Rational formulaQ=CiA
Q = peak discharge (cfs); C = runoff coefficient; i = intensity at duration tc
Worked examples
Composite C, time of concentration, and peak flow
Problem. An 8-acre catchment is 5 ac pavement (C=0.90) and 3 ac lawn (C=0.25). The flow path is: 120 ft of sheet flow over short grass (n=0.15, S=0.015
Common pitfalls
•Using a storm duration other than tc to pick the intensity. The rational method requires duration =tc so the whole area contributes at peak.
•Simple-averaging the runoff coefficient instead of area-weighting it; a small paved fraction can dominate a large pervious area.
•
References
NCEES PE Civil Reference Handbook — §6.5.2.1 Rational Formula Method
NCEES PE Civil Reference Handbook — §6.5.4 Time of Concentration
Areal rainfall by arithmetic, Thiessen, and isohyetal methods; stream discharge by the velocity-area mid-section method; rating curves; and double-mass consistency checks.
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G. Depletions
Depletions: Evaporation, Infiltration and Diversions
The hydrologic budget, evaporation/evapotranspiration estimates, infiltration by the Horton and phi-index methods, deep percolation losses, and diversions in a water balance.
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H. Stormwater management and treatment
Detention/Retention Sizing, Routing and BMPs
Detention versus retention, storage-indication (modified Puls) reservoir routing, peak-shaving to a release limit, and water-quality BMPs.
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— so with
A
in acres and
i
in in/hr the answer drops out in cfs.
Q=CiA
upward because losses become a smaller share of a bigger storm. Always weight by area, never simple-average.
Cw=∑Ai∑CiAi
with
Ku=0.933
(USCS, minutes), where
n
is the overland roughness and
S
the slope. Note that
i
appears here — the very intensity you are trying to find — so sheet-flow
tc
is iterative. Shallow concentrated flow uses a velocity
V=kS
(with
k≈20.3
paved,
16.1
unpaved) so
tsc=L/(60V)
; channel/pipe flow uses Manning's velocity and
tch=L/(60V)
, in minutes.
Tti=i0.4S0.4Ku(nL)0.8,tseg=60VL
, compute
tc
, read a new
i
from the IDF, and repeat once or twice until they agree. Many jurisdictions supply a fitted form
i=cTm/(tc+f)e
(
tc
in minutes) so the read becomes a calculation.
i=(tc+f)ecTm
to consistency (usually two passes). Check units relentlessly —
A
in acres,
i
in in/hr,
tc
feeding the IDF in minutes,
Q
in cfs. If the area exceeds ~200 acres or the problem asks for a volume or hydrograph, the rational method is the wrong tool. A quick bound:
Q
can never exceed
CmaxiA
with
C→1
.
(in/hr);
A
= area (acres).
1ac-in/hr≈1cfs
.
Composite runoff coefficientCw=∑Ai∑CiAi
Area-weighted C for a mixed-cover catchment.
Time of concentrationtc=Tti+tsc+tch
Sum of sheet, shallow-concentrated, and channel travel times along the longest hydraulic path.
Sheet-flow travel timeTti=i0.4S0.4Ku(nL)0.8
Kinematic-wave overland time (min); Ku=0.933 USCS, 6.92 SI; n = overland roughness, L≤300ft
Shallow concentrated velocityV=kS
k≈20.3 (paved), 16.1 (unpaved), V in ft/s, S in ft/ft; then tsc=L/(60V) in minutes.
Channel travel timetch=60VL,V=n1.486R2/3S1/2
Manning velocity for the channel/pipe segment (USCS); tch in minutes with L in ft, V in ft/s.
IDF intensityi=(tc+f)ecTm
Design intensity (in/hr); enter at duration =tc (minutes) and the design return period T (yr).
); 400 ft of unpaved shallow concentrated flow at
S=0.02
; and 1000 ft of channel at
V=3.5ft/s
. The 10-year IDF is
i=94/(tc+10)0.79
(in/hr,
tc
in min). Find the 10-year peak discharge.
Solution. Composite coefficient: Cw=80.90(5)+0.25(3)=84.5+0.75=0.656.
Shallow concentrated: V=16.10.02=2.28ft/s; tsc=400/(60⋅2.28)=2.93min.
Channel: tch=1000/(60⋅3.5)=4.76min.
Sheet flow (iterate; first assume i≈4.5): Tti=(4.5)0.4(0.015)0.40.933(0.15⋅120)0.8=1.825⋅0.1860.933(10.10)=0.3409.42=27.7min.
Then tc=27.7+2.93+4.76=35.4min. IDF at 35.4 min: i=94/(45.4)0.79=94/20.4=4.6in/hr, consistent with the assumed value, so no further iteration.
Peak: Q=CwiA=0.656(4.6)(8)=24.1cfs.
Sanity check: a fully paved 8-ac site at this intensity would give 0.95(4.6)(8)=35cfs, so 24.1cfs for a 5/8-paved site is reasonable. Final: Q≈24.1cfs.
Q=0.656(4.6)(8)≈24.1cfs
Two sub-areas with different times of concentration
Problem. A storm drain serves area 1 (6 ac, C=0.80, tc=10min) and area 2 (10 ac, C=0.40, tc=25min) draining to a common outlet. The IDF gives i=120/(tc+15) (in/hr, min). Determine the design peak at the outlet.
Solution. The design tc is the longest path to the outlet, 25min, so all sub-areas contribute at that duration: i=120/(25+15)=120/40=3.0in/hr
Q=0.55(3.0)(16)=26.4cfs
Ignoring the iteration: sheet-flow Tti depends on i, which depends on tc — assume, compute, re-read until consistent.
•Unit slips: A must be in acres and i in in/hr for the answer to come out in cfs; feeding tc to the IDF in hours when the fit expects minutes is a frequent error.
•Applying the rational method to a watershed larger than ~200 ac (or asking it for a runoff volume or hydrograph) — switch to NRCS or a unit hydrograph.
•Forgetting the partial-area check: a smaller, more-impervious sub-area at its shorter tc can produce a higher peak than the whole catchment at the longer tc.
•Capping sheet flow at the wrong length; the kinematic form applies only to the first ~300 ft, after which flow concentrates.
,
S
= slope,
i
= intensity.
.
Composite
C
:
Cw=160.80(6)+0.40(10)=164.8+4.0=0.55
.
Peak:
Q=CwiA=0.55(3.0)(16)=26.4cfs
.
Check the partial-area case: using only area 1 at its shorter