Surface water resources and hydrology carries 9–14 questions. It follows water from rainfall to runoff, through storage and channels, and asks what that water does to a stream or lake.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam9–14
problems on this page17
areas in the specification15
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What the area covers
The specification breaks the area into five parts:
Runoff: land use and cover, time of concentration, duration, intensity and frequency of rainfall, runoff control and management.
Storage sizing: reservoirs, detention and retention basins.
Routing: through channels and reservoirs.
Water quality and modeling: erosion, channel stability, stormwater quality, wetlands, dissolved-oxygen sag (Streeter-Phelps), eutrophication.
Water budget: evapotranspiration, precipitation, infiltration, soil moisture, storage.
17 runoff, routing and stream problems
0 / 17 answered · 0 correct
Question 1 of 17 · Surface Water Resources and Hydrology · 9–14 items
What does a 25-year storm return period indicate?
Solution
Return period T and annual exceedance probability are reciprocals (Hydrology: design storms).
p = 1/T = 1/25 = 0.04
A 25-year return period means an annual exceedance probability of 1/25 = 0.04.
Return period is a long-run average interval; such storms can come in consecutive years or not at all in 25 years.
Over 25 years the chance of at least one such storm is 1 − 0.9625 ≈ 64%, not 100%.
25% confuses the return period with a percentage; the annual probability is 1/T = 1/25.
Question 2 of 17 · Surface Water Resources and Hydrology · 9–14 items
What is the primary purpose of a stormwater best management practice (BMP) known as a bioretention area?
Solution
Bioretention is a treatment BMP: filtration and biological uptake through engineered soil and plants (Surface water: stormwater management).
Rapid conveyance is the old approach bioretention replaces; it moves pollutants downstream untreated.
Storing water for reuse is rainwater harvesting (cisterns); a bioretention cell drains within a day or two.
A permanent pool is a wet pond feature; bioretention cells are designed to drain and stay mostly dry.
A bioretention cell (rain garden) holds runoff in a planted depression and treats it by filtration, sorption and plant uptake, often with infiltration.
Question 3 of 17 · Surface Water Resources and Hydrology · 9–14 items
For an urban watershed receiving both point and nonpoint source pollution, which integrated management approach is most effective in reducing overall pollutant loads?
Solution
A watershed with both point and nonpoint sources needs a mix of structural and nonstructural measures (Surface water: watershed management).
Detention mainly controls peak flow; it removes some settleable solids but does little for dissolved pollutants or point sources.
Tighter permits help the point sources but leave the nonpoint urban runoff untouched.
Point and nonpoint loads come from different sources, so the most effective plan pairs structural BMPs with source controls such as ordinances, education and good housekeeping.
Street sweeping addresses one nonpoint source and is not enough alone for a mixed point and nonpoint load.
Question 4 of 17 · Surface Water Resources and Hydrology · 9–14 items
Compare how predicted runoff responds to doubling rainfall amount for the Rational method versus the SCS (NRCS) Curve Number method. Which statement best describes their relative linearity?
Solution
Rational: Q = CiA (linear in i). NRCS curve number: Q = (P − 0.2S)2/(P + 0.8S) (nonlinear in P).
This reverses the two: C, i and A enter the rational formula as simple products.
Q = CiA doubles when i doubles, but the curve-number equation subtracts an initial abstraction and divides by P − Ia + S, so runoff more than doubles.
Initial abstraction belongs to the curve-number method; the rational formula has none.
Only the rational method is linear; the curve-number method's initial abstraction makes small storms produce little runoff.
The rational method does double Q for doubled intensity with C and A fixed.
Question 5 of 17 · Surface Water Resources and Hydrology · 9–14 items
A 4.0 square mile watershed receives 0.75 inches of effective rainfall. The resulting hydrograph is approximated as a triangle with a time base of 6.0 hours. Determine the peak discharge.
Given
Watershed area = 4.0 mi2
Effective rainfall = 0.75 in
Hydrograph time base = 6.0 h
Find
Peak discharge (cfs)
Solution
Triangular hydrograph: runoff volume = ½ × time base × peak (Hydrology: hydrographs).
1,290 cfs uses half the time base (the time to peak) as if it were the whole base.
7,740 cfs leaves the 0.75 in of runoff in inches instead of converting it to 0.0625 ft.
38,700 cfs converts 6.0 h to 360 (minutes) instead of 21,600 s.
Question 6 of 17 · Surface Water Resources and Hydrology · 9–14 items
A watershed has the 1-hour unit hydrograph ordinates shown below. A storm produces 0.4 inches of effective rainfall in the first hour and 1.2 inches in the second hour. Calculate the direct runoff discharge at t = 3 hours.
Time (hr)
UH Ordinate (cfs)
0
0
1
20
2
50
3
25
4
10
5
0
Given
UH ordinates: 0 h 0; 1 h 20; 2 h 50; 3 h 25; 4 h 10; 5 h 0 cfs
Effective rainfall hour 1 = 0.4 in
Effective rainfall hour 2 = 1.2 in
Find
Direct runoff at t = 3 h (cfs)
Solution
Convolution: each rainfall pulse is lagged by its start time and scaled by its depth (Hydrology: unit hydrograph).
30 cfs keeps only the second hour's rain on the 3-hour ordinate (1.2 × 25); each pulse uses the ordinate for its own lag.
50 cfs pairs the ordinates the wrong way round (0.4 × 50 + 1.2 × 25).
60 cfs is the second hour's contribution alone (1.2 × 50); the first hour's rain still contributes 0.4 × 25 at t = 3 h.
At t = 3 h the first pulse is 3 h old (0.4 × 25 = 10 cfs) and the second is 2 h old (1.2 × 50 = 60 cfs): 10 + 60 = 70 cfs.
80 cfs lumps both hours of rain (1.6 in) onto one ordinate, 1.6 × 50.
Question 7 of 17 · Surface Water Resources and Hydrology · 9–14 items
An industrial site has a standard Curve Number (CNII) of 88. Due to heavy prior rainfall, Antecedent Moisture Condition (AMC) III applies, increasing the CN to 95. Calculate the runoff depth for a storm of 3.5 inches.
Given
Curve number CNII = 88
Adjusted curve number CNIII = 95
Storm rainfall P = 3.5 in
Find
Runoff depth (in)
Solution
Curve-number runoff with the AMC III curve number (Hydrology: NRCS curve number).
S = 1000/95 − 10 = 0.53 in; Ia = 0.2S = 0.11 in
Q = (P − Ia)2/(P − Ia + S) = (3.39)2/3.92 = 2.94 in
0.53 in is the retention S for CN 95, not the runoff.
2.27 in uses CNII = 88; the wet antecedent condition raises the curve number to 95.
S = 1000/95 − 10 = 0.53 in, Ia = 0.11 in, Q = (3.5 − 0.11)2/(3.5 − 0.11 + 0.53) = 2.94 in.
3.39 in is P − Ia; it leaves out the retention S in the denominator.
3.50 in treats all rainfall as runoff, which only a curve number of 100 would give.
Question 8 of 17 · Surface Water Resources and Hydrology · 9–14 items
The ordinates of a 30-minute unit hydrograph are given in units of m³/s per cm of effective rainfall. The watershed area is 500 hectares. A storm event produces 25 mm of effective rainfall. If the peak ordinate of the unit hydrograph is 12 m³/s per cm, what is the peak direct runoff discharge?
Given
Unit hydrograph duration = 30 min
Watershed area = 500 ha
Effective rainfall = 25 mm
Peak UH ordinate = 12 m3/s per cm
Find
Peak direct runoff (m3/s)
Solution
A unit hydrograph scales linearly with effective rainfall depth in its own unit (Hydrology: unit hydrograph).
Qp = 12 m3/s per cm × 2.5 cm = 30 m3/s
3.0 m3/s scales by 0.25 (25 mm read as 0.25 cm); 25 mm is 2.5 cm.
4.8 m3/s divides the ordinate by the rainfall depth instead of multiplying.
12 m3/s is the response to 1 cm of rain; the storm produced 2.5 cm.
Unit hydrograph ordinates scale with effective rainfall depth: 12 m3/s per cm × 2.5 cm = 30 m3/s.
300 m3/s multiplies by 25 (millimeters) while the ordinate is given for each centimeter.
Question 9 of 17 · Surface Water Resources and Hydrology · 9–14 items
A stormwater bioretention cell treats runoff from a 1-hectare parking lot. During a storm event, the influent volume is 180 m3 with a suspended solids Event Mean Concentration (EMC) of 100 mg/L. The facility infiltrates 40% of the runoff volume into the ground. The remaining volume is discharged with an effluent EMC of 20 mg/L. What is the total mass of suspended solids removed from the surface discharge by the facility?
Given
Influent volume = 180 m3
Influent TSS EMC = 100 mg/L
Infiltrated fraction = 40 %
Effluent EMC = 20 mg/L
Find
TSS mass removed from surface discharge (kg)
Solution
Load = volume × EMC; removal from the surface discharge = load in − load out (Surface water: stormwater BMPs).
Min = 180 m3 × 100 g/m3 = 18.0 kg
Mout = 180 × 0.60 m3 × 20 g/m3 = 2.16 kg
Mremoved = 18.0 − 2.16 = 15.8 kg
2.16 kg is the mass still discharged at the surface, not the mass removed.
7.20 kg is only the load infiltrated (72 m3 × 100 mg/L); treatment of the discharged water adds more.
8.64 kg applies the 80 mg/L concentration drop to the 108 m3 discharged only and ignores the infiltrated load.
14.4 kg applies the concentration drop to the full 180 m3, as if nothing infiltrated.
Load in = 180 m3 × 100 g/m3 = 18.0 kg; out at the surface = 108 m3 × 20 g/m3 = 2.16 kg; removed = 15.8 kg.
Question 10 of 17 · Surface Water Resources and Hydrology · 9–14 items
A stream hydrograph and corresponding suspended sediment concentrations (SSC) are measured at 30-minute intervals. Estimate the total sediment mass load transported during the 60-minute period using the trapezoidal rule.
Time (min): 0, 30, 60
Flow Q (m3/s): 10, 20, 10
SSC (mg/L): 50, 200, 50
Given
Time = 0, 30, 60 min
Flow Q = 10, 20, 10 m3/s
SSC = 50, 200, 50 mg/L
Find
Total sediment mass load (kg)
Solution
Sediment load = ∫Q·SSC dt; integrate the product with the trapezoidal rule (Surface water: sediment transport).
M = 2 × [(500 + 4,000)/2 × 1,800 s] = 8.1 × 106 g = 8,100 kg
8.10 kg divides by 1,000 one time too many; g/s × s gives grams, and one ÷1,000 gives kilograms.
135 kg integrates over 30-minute steps in minutes instead of 1,800 s.
4,050 kg integrates only the first 30-minute interval.
6,750 kg multiplies average flow by average concentration; the load is the integral of their product, which peaks together.
Load rates are 500, 4,000 and 500 g/s; each trapezoid holds (500 + 4,000)/2 × 1,800 s = 4,050 kg, so the total is 8,100 kg.
Question 11 of 17 · Surface Water Resources and Hydrology · 9–14 items
A rectangular river channel is 20 m wide and 2.0 m deep. The bed slope is 0.001 and the Manning's roughness coefficient n is 0.03. Using the O'Connor-Dobbins empirical equation k2 = 3.9 v0.5 H−1.5 (where v is velocity in m/s, H is depth in m, and k2 is in day−1), determine the reaeration coefficient k2.
Given
Channel width = 20 m
Depth H = 2.0 m
Bed slope S = 0.001
Manning's n = 0.03
Find
Reaeration coefficient k2 (d−1)
Solution
Manning velocity, then the O'Connor–Dobbins reaeration rate (Surface water: DO sag).
R = A/P = (20 × 2.0)/(20 + 2 × 2.0) = 1.67 m; v = (1/0.03)(1.67)2/3(0.001)1/2 = 1.48 m/s
0.30 day−1 uses S instead of √S in Manning's equation, so the velocity comes out about 30 times too low.
1.4 day−1 leaves out the velocity term, as if v = 1 m/s.
R = 40/24 = 1.67 m, v = (1/0.03)(1.67)2/3(0.001)1/2 = 1.48 m/s, k2 = 3.9 × 1.480.5/2.01.5 = 1.7 day−1.
2.0 day−1 uses v rather than √v in the O'Connor–Dobbins equation.
13 day−1 multiplies by H1.5 instead of dividing.
Question 12 of 17 · Surface Water Resources and Hydrology · 9–14 items
In a Streeter-Phelps analysis of a river, if the reaeration rate constant (k2) increases while the deoxygenation rate constant (k1) and initial loading remain constant, how do the critical dissolved oxygen deficit (Dc) and the time to reach critical deficit (tc) change?
Solution
Streeter–Phelps: tc = [1/(k2 − k1)]·ln{(k2/k1)[1 − Da(k2 − k1)/(k1La)]} and Dc = (k1/k2)Lae−k1tc; both fall as k2 rises.
Faster reaeration replaces oxygen sooner, so the sag bottoms out earlier and less deeply.
Dc = (k1/k2)Lae−k1tc falls as k2 rises, so it cannot stay constant.
The sag is shallower, but faster reaeration also moves the critical point upstream (earlier), not later.
A larger k2 speeds recovery; it cannot deepen the sag or delay its low point.
The low point comes sooner, but it is shallower, not deeper.
Question 13 of 17 · Surface Water Resources and Hydrology · 9–14 items
A reservoir (surface area 40 acres) receives a constant stream inflow of 5.00 cfs and releases 3.20 cfs for 10 days. Total precipitation on the surface during the period is 0.75 in and total evaporation is 0.40 in. Compute the change in reservoir water surface elevation (ft) over the 10‑day period.
Given
Surface area = 40 acres
Inflow = 5.00 cfs
Release = 3.20 cfs
Duration = 10 days
Precipitation = 0.75 in
Evaporation = 0.40 in
Find
Change in water surface elevation (ft)
Solution
Water budget on the reservoir surface (Hydrology: reservoir storage).
0.0292 ft is the precipitation-minus-evaporation term alone; the 1.80 cfs net inflow adds far more.
0.893 ft counts the net inflow but leaves out the 0.35 in of net precipitation on the surface.
Net inflow 1.80 cfs × 864,000 s = 1.555 × 106 ft3 over 1.742 × 106 ft2 = 0.893 ft, plus (0.75 − 0.40)/12 = 0.029 ft, gives 0.922 ft.
0.955 ft adds the 0.75 in of precipitation but not the 0.40 in lost to evaporation.
0.989 ft adds evaporation as a gain instead of a loss.
Question 14 of 17 · Surface Water Resources and Hydrology · 9–14 items
Compute the seasonal crop water requirement (total volume) for a 12-hectare field if the daily reference evapotranspiration ET0 = 5.0 mm/day and the crop coefficient Kc is 0.70 for 20 days, 1.15 for 60 days, and 0.85 for 30 days.
Given
Field area = 12 ha
Reference evapotranspiration ET0 = 5.0 mm/day
Kc = 0.70 for 20 days
Kc = 1.15 for 60 days
Kc = 0.85 for 30 days
Find
Seasonal crop water requirement (m3)
Solution
Crop ET by growth stage, summed and converted to volume (Water resources: irrigation demand).
Question 15 of 17 · Surface Water Resources and Hydrology · 9–14 items
Route the inflow hydrograph below through a level-pool reservoir using the modified Puls method with 10-min time steps (Δt = 600 s) and Ī − Ō = ΔS/Δt. Inflow is 5.0, 8.0, 12.0, 9.0 and 6.0 m3/s at t = 0, 10, 20, 30 and 40 min. Use the storage–outflow relation in the table and an initial outflow of 5.0 m3/s. What is the outflow at t = 30 min?
Outflow (m3/s)
Storage (m3)
4
2560
6
3640
8
4720
10
5800
12
6880
Given
Δt = 600 s
Initial outflow O0 = 5.0 m3/s
Inflow: 5.0, 8.0, 12.0, 9.0, 6.0 m3/s at 0, 10, 20, 30, 40 min
Storage–outflow relation as tabulated
Find
Outflow at t = 30 min (m3/s)
Solution
Level-pool routing with a linear storage–outflow relation (Hydrology: reservoir routing).
S = 400 + 540O, so each step solves Ī − (O1 + O2)/2 = 0.9(O2 − O1); O = 6.07, 8.88 and then 10.04 m3/s at 30 min.
9.46 m3/s uses the end-of-step inflow instead of the average over each step.
9.00 m3/s sets outflow equal to the inflow at 30 min, ignoring the storage that delays and lowers the peak.
8.88 m3/s is the outflow at t = 20 min; one more 10-minute step is needed.
5.22 m3/s puts Δt in minutes (10) instead of seconds (600), which makes the storage term 60 times too heavy.
Question 16 of 17 · Surface Water Resources and Hydrology · 9–14 items
A constant rainfall intensity of 3.0 cm/h falls on soil with the properties listed below. Using the Green-Ampt model, determine the time from the start of rainfall until surface ponding occurs.
Parameter
Value
Hydraulic conductivity (K)
1.0 cm/h
Suction head (ψ)
15.0 cm
Porosity (η)
0.45
Initial moisture content (θi)
0.25
Given
Rainfall intensity i = 3.0 cm/h
Hydraulic conductivity K = 1.0 cm/h
Suction head ψ = 15.0 cm
Porosity η = 0.45
Initial moisture θi = 0.25
Find
Time to ponding (min)
Solution
Green–Ampt time to ponding under constant rainfall (Hydrology: infiltration).
Δθ = η − θi = 0.45 − 0.25 = 0.20
tp = KψsΔθ/[i(i − K)] = (1.0 × 15.0 × 0.20)/(3.0 × 2.0) = 0.50 h = 30.0 min
37.5 min uses the initial moisture 0.25 as the deficit; Δθ = η − θi = 0.20.
60.0 min divides by i alone and leaves out the (i − K) term.
67.5 min uses the porosity 0.45 as the deficit, as if the soil started dry.
90.0 min drops i from the denominator, KψsΔθ/(i − K) = 3.0/2.0 = 1.5 h.
Question 17 of 17 · Surface Water Resources and Hydrology · 9–14 items
Month
Inflow (cfs)
Demand (MGD)
Net Evaporation (in)
1
12
10
6
2
8
10
8
3
20
8
4
A proposed 120-acre reservoir is sized to meet a demand schedule. Using the provided inflow, demand, and net evaporation data, and assuming 30-day months, calculate the required active storage capacity via the sequential mass diagram method.
Given
Reservoir area = 120 acres
Month length = 30 days
Month 1: inflow 12 cfs, demand 10 MGD, net evaporation 6 in
Month 2: inflow 8 cfs, demand 10 MGD, net evaporation 8 in
Month 3: inflow 20 cfs, demand 8 MGD, net evaporation 4 in
Find
Required active storage (ac-ft)
Solution
Sequential (mass-diagram) analysis: active storage = maximum cumulative deficit (Water resources: reservoir sizing).
267 ac-ft is month 1's deficit only; month 2 is also short, so the deficits accumulate.
378 ac-ft is the cumulative balance after month 3's surplus; storage must cover the largest running deficit, reached at the end of month 2.
525 ac-ft is month 2's deficit alone; the reservoir is already 267 ac-ft down from month 1.
652 ac-ft leaves out net evaporation from the 120-acre surface (60 + 80 ac-ft).
Month 1: 714 − 921 − 60 = −267; month 2: 476 − 921 − 80 = −525; the running deficit peaks at 791 ac-ft before month 3's surplus.
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Common traps
Curve-number units. The NRCS (SCS) method works in inches: the storage term S = 1000/CN − 10 is in inches, and runoff appears only once rainfall exceeds the initial abstraction, taken as 0.2S:
Q = (P − 0.2S)2 / (0.8S + P)
Rational method: its units and the storm duration it needs are covered once, with site drainage, on FE Civil water resources.
Two rate constants in the DO sag. k1 (deoxygenation) and k2 (reaeration) are easy to swap, and both need the temperature correction from 20 °C:
kT = k20 θT − 20
Water budget signs. Change in storage = inflow − outflow, with evaporation and seepage on the outflow side. Convert depths over an area to volume (acre-feet or m3) before you add them to flows.