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Hydrology Practice Problems — Rational Method, Curve Number, Tc & Hydrographs (FE & PE)

Nine original hydrology problems with every step shown and every wrong answer explained. Hydrology is the largest single area on the PE Civil: WRE exam (8–12 of 80 questions) and the biggest water topic on the FE Civil exam — and most of the points lost here are lost to unit slips and method mix-ups, not hard math.

Last reviewed: 2026-10-03

Q = CiA  ·  Q = (P − 0.2S)² / (P + 0.8S),  S = 1000/CN − 10  ·  Tc = 0.0078 L0.77 S−0.385 (Kirpich)

Q = CiARational Method peak discharge; i read from the IDF curve at duration = Tc (ac·in/hr → cfs uses ≈ 1.008, usually taken as 1)
CN methodSCS runoff depth in inches; Ia = 0.2S standard — if P ≤ Ia, runoff is zero; S and P in inches
Unit hydrographdirect-runoff ordinate = Σ (excess rain in hour k) × (UH ordinate lagged k steps)

Rational Method

Problem 1 — Weighted runoff coefficient

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A 25-acre watershed drains to a culvert. Land use is 10 ac of commercial development (C = 0.85) and 15 ac of single-family residential (C = 0.45). The 25-year IDF curve gives an intensity of 4.2 in/hr at the watershed's 30-minute time of concentration. Estimate the peak discharge by the Rational Method.

  1. 64 cfs
  2. 68 cfs
  3. 105 cfs
  4. 73 cfs
Full solution
  1. Weight the runoff coefficient by area: C = (10 × 0.85 + 15 × 0.45) / 25 = 15.25 / 25 = 0.61.
  2. Intensity comes from the IDF curve at duration = Tc = 30 min: i = 4.2 in/hr.
  3. Q = CiA = 0.61 × 4.2 × 25 = 64.05 cfs ≈ 64 cfs. (One ac·in/hr ≈ 1.008 cfs; the exam takes it as 1.)

Answer: A — 64 cfs.

Why the wrong answers are wrong: B (68 cfs) straight-averages the two C values (0.65) instead of area-weighting — the single most common Rational Method error. C (105 cfs) uses C = 1.0, the “everything is pavement” panic answer. D (73 cfs) grabs the 15-minute intensity (4.8 in/hr); intensity must be read at duration = Tc.

Problem 2 — Frequency factor on a composite C

PE WRE Hydrology (8–12 of 80) Target pace: ~6 min

A 12-acre commercial site is 8 ac of parking lot (10-year C = 0.90) and 4 ac of lawn (10-year C = 0.30). Size the site outlet for the 100-year storm. The local drainage manual applies a frequency factor of 1.25 to 10-year C values for the 100-year storm (adjusted C capped at 1.0). The 100-year intensity at the 20-minute Tc is 5.5 in/hr. What is the design peak discharge?

  1. 58 cfs
  2. 46 cfs
  3. 50 cfs
  4. 52 cfs
Full solution
  1. Weight the 10-year C by area: C10 = (8 × 0.90 + 4 × 0.30) / 12 = 8.40 / 12 = 0.70.
  2. Apply the frequency factor to the composite: C100 = 0.70 × 1.25 = 0.875 (below the 1.0 cap, so no capping needed).
  3. Q = CiA = 0.875 × 5.5 × 12 = 57.75 cfs ≈ 58 cfs.

Answer: A — 58 cfs.

Why the wrong answers are wrong: B (46 cfs) forgets the frequency factor entirely — the exam always includes the no-factor answer. C (50 cfs) averages C before weighting (0.60 × 1.25 = 0.75), repeating Problem 1's trap under time pressure. D (52 cfs) caps each surface at 1.0 before weighting (parking 0.90 × 1.25 hits the cap); the manual's factor applies to the composite C.

Curve Number (SCS) Runoff

Problem 3 — Runoff depth to volume

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A 40-acre watershed has a weighted curve number CN = 78. A 24-hour design storm drops 5.0 in of rain. Using the standard initial-abstraction convention, estimate the total runoff volume in acre-feet.

  1. 9.0 ac-ft
  2. 10.3 ac-ft
  3. 11.5 ac-ft
  4. 16.7 ac-ft
Full solution
  1. S = 1000/CN − 10 = 1000/78 − 10 = 2.82 in.
  2. Ia = 0.2S = 0.564 in. Since P = 5.0 in > Ia, runoff occurs.
  3. Runoff depth Q = (P − Ia)² / (P + 0.8S) = (4.436)² / (5.0 + 2.256) = 19.68 / 7.256 = 2.71 in.
  4. Volume = 2.71 in × 40 ac / 12 in/ft = 9.04 ac-ft ≈ 9.0 ac-ft.

Answer: A — 9.0 ac-ft.

Why the wrong answers are wrong: B (10.3 ac-ft) uses the newer Ia = 0.05S convention — valid in some software, but the exam default is 0.2S unless stated. C (11.5 ac-ft) drops initial abstraction completely. D (16.7 ac-ft) sets runoff equal to rainfall (5.0 × 40/12), i.e., CN = 100.

Problem 4 — Two-subarea weighted CN, carried to volume ★ shown in full

PE WRE Hydrology (8–12 of 80) Target pace: ~6 min

One problem on each page is shown worked in full, so you can judge the quality before trusting the rest. This is that problem.

A 60-acre watershed has two subareas: 25 ac at CN = 85 (commercial) and 35 ac at CN = 70 (open space). The 24-hour design rainfall is 4.5 in. Compute the total runoff volume in acre-feet.

  1. 10.7 ac-ft
  2. 11.3 ac-ft
  3. 10.9 ac-ft
  4. 14.5 ac-ft
Full solution
  1. Weight the CN by area (never a straight average): CN = (25 × 85 + 35 × 70) / 60 = 4,575 / 60 = 76.25.
  2. Potential retention: S = 1000/76.25 − 10 = 13.115 − 10 = 3.115 in.
  3. Initial abstraction: Ia = 0.2S = 0.623 in. Check: P = 4.5 in > 0.623 in, so runoff occurs (if P ≤ Ia, the answer would be zero — always check).
  4. Runoff depth: Q = (P − Ia)² / (P + 0.8S) = (4.5 − 0.623)² / (4.5 + 0.8 × 3.115) = (3.877)² / (4.5 + 2.492) = 15.031 / 6.992 = 2.150 in.
  5. Depth to volume: 2.150 in × 60 ac ÷ 12 in/ft = 10.75 ac-ft ≈ 10.7 ac-ft.

Answer: A — 10.7 ac-ft.

Why the wrong answers are wrong: B (11.3 ac-ft) plain-averages the two CNs (77.5) — the trap this whole problem is built around. C (10.9 ac-ft) is subtle: it runs each subarea through the CN equation separately (2.91 in and 1.67 in) and then area-weights the depths. SCS procedure weights the CN, not the Q — close, but not the method. D (14.5 ac-ft) skips initial abstraction.

Time of Concentration

Problem 5 — Kirpich equation

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

Estimate the time of concentration by the Kirpich equation for a small watershed with a flow-path length of 2,400 ft and an average slope of 0.012 ft/ft.

  1. 17 min
  2. 2.9 min
  3. 42 min
  4. 103 min
Full solution
  1. Kirpich (L in feet, S as a decimal, Tc in minutes): Tc = 0.0078 × L0.77 × S−0.385.
  2. L0.77 = 24000.77 ≈ 400.5; S−0.385 = 0.012−0.385 ≈ 5.49.
  3. Tc = 0.0078 × 400.5 × 5.49 = 17.2 min ≈ 17 min.

Answer: A — 17 min.

Why the wrong answers are wrong: B (2.9 min) enters slope as 1.2 (percent) instead of 0.012. C (42 min) slips a decimal in the slope (0.0012). D (103 min) drops the 0.77 exponent on L — with big exponents missing, always sanity-check the magnitude.

Problem 6 — TR-55 three-segment Tc

PE WRE Hydrology (8–12 of 80) Target pace: ~6 min

A watershed's longest flow path has three segments: (1) 100 ft of sheet flow over dense grass (n = 0.24) at 2% slope; (2) 600 ft of shallow concentrated flow over unpaved ground at 4% slope; (3) 1,200 ft of channel flow at 3.5 ft/s. The 2-year, 24-hour rainfall is 3.5 in. Compute Tc by the TR-55 segmental method.

  1. 22.5 min
  2. 21.8 min
  3. 11.0 min
  4. 16.1 min
Full solution
  1. Sheet flow (Tc equation, result in hours): Tt = 0.007(nL)0.8 / (P20.5 S0.4) = 0.007(24)0.8 / (√3.5 × 0.020.4) = 0.007 × 12.71 / (1.871 × 0.209) = 0.2276 hr = 13.7 min.
  2. Shallow concentrated, unpaved: V = 16.1345√S = 16.1345 × 0.2 = 3.23 ft/s. Tt = 600 / 3.23 = 186 s = 3.1 min.
  3. Channel: Tt = 1,200 / 3.5 = 343 s = 5.7 min.
  4. Tc = 13.7 + 3.1 + 5.7 = 22.5 min.

Answer: A — 22.5 min.

Why the wrong answers are wrong: B (21.8 min) uses the paved-flow velocity (20.3282√S) on unpaved ground — the exam pairs the two constants as a matched trap. C (11.0 min) enters the sheet-flow slope as 2 instead of 0.02. D (16.1 min) forgets the square root on P2 in the sheet-flow denominator.

Unit Hydrographs

Problem 7 — Convolution of a 2-hour storm

PE WRE Hydrology (8–12 of 80) Target pace: ~6 min

The 1-hour unit hydrograph for a watershed has ordinates (cfs) at hours 0–5: 0, 120, 310, 210, 90, 20. A storm produces 1.2 in of excess rainfall in the first hour and 0.8 in in the second hour. What is the peak of the direct-runoff hydrograph?

  1. 500 cfs
  2. 468 cfs
  3. 620 cfs
  4. 372 cfs
Full solution
  1. Lag the second hour's rain one time step, then multiply and add ordinate by ordinate:
t (hr)0123456
1.2 in × UH0144372252108240
0.8 in × UH (lagged)—0962481687216
Direct runoff01444685002769616
  1. The peak is 500 cfs at t = 3 hr — one step after the UH peak, which is the usual pattern.

Answer: A — 500 cfs.

Why the wrong answers are wrong: B (468 cfs) is the t = 2 ordinate — the “stopped one column early” answer. C (620 cfs) multiplies total rain (2.0 in) by the peak ordinate, skipping convolution entirely. D (372 cfs) never applies the second hour of rain.

Problem 8 — Watershed area from a unit hydrograph

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A 1-hour unit hydrograph has ordinates (cfs) at 1-hour intervals: 0, 90, 240, 180, 90, 30, 0. Estimate the watershed area in square miles.

  1. 0.98 mi²
  2. 0.081 mi²
  3. 1.95 mi²
  4. 625 ac
Full solution
  1. A unit hydrograph represents 1 inch of direct runoff. Volume = ΣQ × Δt = (90 + 240 + 180 + 90 + 30) cfs × 1 hr = 630 cfs-hr.
  2. Convert to ft³: 630 × 3,600 = 2,268,000 ft³.
  3. Area = volume / depth = 2,268,000 / (1/12 ft) = 27,216,000 ft² = 27,216,000 / 27,878,400 = 0.976 mi² ≈ 0.98 mi².

Answer: A — 0.98 mi².

Why the wrong answers are wrong: B (0.081 mi²) forgets to divide by the 1-inch depth — it reports volume as if it were area. C (1.95 mi²) treats the ordinate interval as 2 hours. D (625 ac) is the right number in the wrong units (624.8 ac) — the question asks for square miles.

IDF & Precipitation

Problem 9 — Interpolating an IDF table

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

An IDF table gives 50-year intensities of 5.0 in/hr at 20 minutes and 4.2 in/hr at 30 minutes. A watershed's time of concentration is 25 minutes. What 50-year intensity should be used in the Rational Method?

  1. 4.6 in/hr
  2. 5.0 in/hr
  3. 4.2 in/hr
  4. 4.1 in/hr
Full solution
  1. Read intensity at duration = Tc. Tc = 25 min falls between tabulated 20 and 30 min, so interpolate linearly:
  2. i = 5.0 − (5.0 − 4.2) × (25 − 20)/(30 − 20) = 5.0 − 0.8 × 0.5 = 4.6 in/hr.

Answer: A — 4.6 in/hr.

Why the wrong answers are wrong: B (5.0) rounds Tc down to the nearest tabulated duration — conservative, but not what the question asks. C (4.2) rounds up. D (4.1) interpolates correctly but on the 25-year row (4.4/3.7) — always check you are on the design return period.

Which method does the question want?

Rational MethodGives peak discharge. Stem gives C values, an IDF table or intensity, and a Tc. Small watersheds only.
Curve NumberGives runoff depth or volume. Stem gives CN, soil group, or land use plus a rainfall depth.
Unit hydrographGives a full hydrograph. Stem gives UH ordinates and a multi-hour rainfall distribution.
Tc equationsFeeds the other three. Kirpich for small watersheds; TR-55 segmental when the stem lists flow segments; the intensity for Rational always comes from IDF at t = Tc.

Frequently asked questions

When do I use the Rational Method vs. the Curve Number method?

Use the Rational Method (Q = CiA) when the question asks for peak discharge from a small watershed and gives you runoff coefficients and an intensity or IDF table. Use the Curve Number method when it asks for runoff depth or volume from a rainfall depth and gives you CN values, soil groups, or land use. If it hands you a unit hydrograph and a multi-hour storm, convolve — that is the hydrograph method.

Do I area-weight C and CN, or just average them?

Always area-weight. The composite C is the sum of each C times its area, divided by total area — and the same goes for CN. A straight average of the coefficients is one of the most common wrong answers on hydrology questions.

What is Ia in the Curve Number method, and what value does the exam expect?

Ia is initial abstraction — interception, depression storage, and infiltration before runoff begins. The standard SCS convention is Ia = 0.2S, where S = 1000/CN − 10 (inches). Some references use Ia = 0.05S; unless the question says otherwise, use 0.2S. If rainfall P is less than or equal to Ia, runoff is zero.

Which duration do I read intensity from on an IDF curve?

Read intensity at a duration equal to the time of concentration, Tc, for the design return period. A shorter duration gives a higher intensity and a wrong answer; the exam will offer both.

How do I avoid lag mistakes in unit hydrograph convolution?

Write the ordinates in columns and shift each hour of excess rain one time-step to the right before multiplying. The peak of the direct-runoff hydrograph usually lands one time step after the UH peak — if your peak lines up exactly with the UH peak, double-check the lag.

Keep practicing

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