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SCE PE Water Resources & Environmental study materials
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SCE PE Water Resources & Environmental practice questions

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Question 1Hydrology

In the hydrologic cycle, which process returns water directly from the atmosphere to the land surface?

  1. Transpiration
  2. Infiltration
  3. Precipitation Correct answer.
  4. Percolation

Explanation

Precipitation is the atmospheric delivery of water as rain, snow, hail, or other forms to the land surface. Infiltration and percolation move water within the soil, while transpiration releases water vapor from plants.

Question 2Hydrology

Which instrument measures evaporation from an open water surface?

  1. Neutron probe
  2. Class A pan Correct answer.
  3. Orifice weir
  4. Current meter

Explanation

A Class A pan is a standard evaporation pan used to measure water loss from an open surface, with pan coefficients applied to estimate lake or reservoir evaporation. The other devices measure streamflow, velocity, or soil moisture rather than evaporation.

Question 3Hydrology

The maximum rate at which soil can absorb water under given conditions is called the:

  1. Wilting point
  2. Field capacity
  3. Storage deficit
  4. Infiltration rate Correct answer.

Explanation

Infiltration rate is the rate at which water enters the soil surface, and its maximum under given conditions is the infiltration capacity. Field capacity and wilting point describe soil-water retention states rather than an entry rate.

Question 4Hydrology

Which component of streamflow is sustained by groundwater discharge during dry periods?

  1. Interflow
  2. Storm runoff
  3. Baseflow Correct answer.
  4. Direct runoff

Explanation

Baseflow is the relatively steady portion of streamflow fed by groundwater discharge, and it maintains flow between storms. Direct runoff, interflow, and storm runoff are quicker storm-related contributions.

Question 5Hydrology

Which watershed characteristic is defined as the area drained by a stream at a specified point?

  1. Drainage area Correct answer.
  2. Drainage density
  3. Relief ratio
  4. Stream order

Explanation

Drainage area, or basin area, is the plan area contributing runoff to a specified outlet. Stream order, relief ratio, and drainage density describe network and terrain characteristics rather than the contributing area.

Question 6Hydrology

Which method estimates areal average precipitation by weighting gage values according to polygon areas?

  1. Isohyetal method
  2. Arithmetic mean
  3. Thiessen method Correct answer.
  4. Double-mass analysis

Explanation

The Thiessen method constructs polygons around rain gages and weights each gage by the fraction of basin area it represents. The isohyetal method interpolates precipitation contours, while the arithmetic mean gives equal weight to all gages.

Question 7Hydrology

A stage-discharge rating curve relates stream discharge to:

  1. Gage height Correct answer.
  2. Sediment concentration
  3. Water temperature
  4. Channel slope

Explanation

A rating curve expresses discharge as a function of stage, or gage height, at a stable channel cross section. Temperature, sediment concentration, and slope may affect measurements but are not the primary paired variable.

Question 8Hydrology

For a watershed with no groundwater flow across its boundaries, which expression represents the annual water balance?

  1. Q = P + ET + ΔS
  2. P = ET + Q + ΔS Correct answer.
  3. ΔS = P + ET + Q
  4. P = ET - Q + ΔS

Explanation

The water-balance equation sets precipitation equal to evapotranspiration plus runoff plus change in storage when no external groundwater flow occurs. A positive ΔS indicates storage gain, and a negative value indicates storage loss.

Question 9Hydrology

Using the rational method, a basin has C = 0.4, rainfall intensity i = 2.0 in/hr, and area A = 10 acres. What is the approximate peak discharge?

  1. 2.0 cfs
  2. 4.0 cfs
  3. 8.1 cfs Correct answer.
  4. 16.1 cfs

Explanation

The rational method gives Q = CiA, so Q = 0.4 × 2.0 × 10 = 8.0 acre-in/hr. Converting with about 1.008 cfs per acre-in/hr gives approximately 8.1 cfs.

Question 10Hydrology

Using the Kirpich formula, Tc = 0.0078 L^0.77 S^-0.385, with L = 1,000 ft and S = 0.01 ft/ft, what is the time of concentration?

  1. 7.1 min
  2. 9.4 min Correct answer.
  3. 18.8 min
  4. 12.2 min

Explanation

Substituting L = 1,000 ft and S = 0.01 gives L^0.77 ≈ 204 and S^-0.385 ≈ 5.89. Multiplying by 0.0078 yields approximately 9.4 minutes.

Question 11Hydrology

A 2-hour unit hydrograph has a peak of 100 cfs. Assuming linear unit-hydrograph response, what is the peak flow from a 2-hour storm with 3 inches of excess rainfall?

  1. 600 cfs
  2. 100 cfs
  3. 150 cfs
  4. 300 cfs Correct answer.

Explanation

A unit hydrograph represents 1 inch of excess rainfall, so the response scales linearly with rainfall depth. Three inches of excess rainfall gives a peak of 3 × 100 = 300 cfs.

Question 12Hydrology

For a curve number CN = 80, what is the potential maximum retention S in inches using S = 1000/CN - 10?

  1. 2.0 in
  2. 4.0 in
  3. 2.5 in Correct answer.
  4. 3.0 in

Explanation

Substituting CN = 80 gives 1000/80 - 10 = 12.5 - 10 = 2.5 inches. The retention S is used in the NRCS curve-number runoff equation.

Question 13Hydrology

A flood has an annual exceedance probability of 2%. What is its return period?

  1. 100 years
  2. 50 years Correct answer.
  3. 2 years
  4. 20 years

Explanation

Return period is the reciprocal of annual exceedance probability, so T = 1/0.02 = 50 years. This means the flood has a 1-in-50 chance of being equaled or exceeded in any year.

Question 14Hydrology

What is the probability that at least one 100-year flood occurs during a 30-year period?

  1. 30%
  2. 10%
  3. 26% Correct answer.
  4. 74%

Explanation

The probability of no 100-year flood in one year is 0.99, so over 30 years it is 0.99^30 ≈ 0.74. Therefore, the probability of at least one event is 1 - 0.74 ≈ 0.26, or 26%.

Question 15Hydrology

The straight-line baseflow separation method connects which two points on a hydrograph?

  1. Rise and recession points Correct answer.
  2. Peak and end of recession
  3. Start of rise and peak
  4. Two successive peaks

Explanation

In the straight-line method, a line is drawn from a point on the rising limb to a point on the recession limb to separate baseflow from direct runoff. The points are selected based on the start of storm response and the end of direct runoff.

Question 16Hydrology

Using Horton's equation f(t) = fc + (f0 - fc)e^(-kt), with f0 = 2.0 in/hr, fc = 0.5 in/hr, k = 0.5/hr, and t = 2 hr, what is the infiltration rate?

  1. 1.25 in/hr
  2. 0.75 in/hr
  3. 1.50 in/hr
  4. 1.05 in/hr Correct answer.

Explanation

The exponent is -0.5 × 2 = -1, so e^-1 ≈ 0.368. Then f = 0.5 + (2.0 - 0.5)(0.368) ≈ 0.5 + 0.55 = 1.05 in/hr.

Question 17Hydrology

On a flow-duration curve, a flow that is exceeded 20% of the time is read at which position?

  1. Minimum flow only
  2. Peak flow only
  3. Exceeded 20% of the time Correct answer.
  4. Exceeded 80% of the time

Explanation

A flow-duration curve plots discharge against the percentage of time that discharge is equaled or exceeded. The 20% value means the flow is equaled or exceeded during 20% of the record.

Question 18Hydrology

In Muskingum flood routing, the parameter K primarily represents the:

  1. Reach length
  2. Storage weighting factor
  3. Peak attenuation
  4. Travel time through reach Correct answer.

Explanation

K is a storage-time constant that approximates the travel time of the flood wave through the reach. The parameter X is a weighting factor that describes the relative influence of inflow and outflow on storage.

Question 19Hydrology

A Class A pan records 8.0 mm/day of evaporation, and the pan coefficient is 0.7. What is the estimated lake evaporation?

  1. 11.4 mm/day
  2. 2.4 mm/day
  3. 8.0 mm/day
  4. 5.6 mm/day Correct answer.

Explanation

Lake evaporation is estimated by multiplying pan evaporation by the pan coefficient: 8.0 × 0.7 = 5.6 mm/day. The coefficient adjusts for differences between the pan and a natural water body.

Question 20Hydrology

For the Gumbel reduced variate y = -ln[ln(T/(T-1))], what is y for a return period T = 100 years?

  1. 2.25
  2. 4.00
  3. 4.60 Correct answer.
  4. 3.14

Explanation

For T = 100, T/(T-1) = 100/99 ≈ 1.0101, and ln(1.0101) ≈ 0.01005. Then ln(0.01005) ≈ -4.60, so the negative of that value gives y ≈ 4.60.

Question 21Hydrology

In reservoir flood routing, why is the routed outflow peak usually lower than the inflow peak?

  1. Storage attenuates the peak Correct answer.
  2. Outflow volume exceeds inflow volume
  3. Outflow peak occurs first
  4. Storage eliminates all outflow

Explanation

Reservoir storage temporarily detains part of the flood volume, which reduces and delays the downstream peak. The outflow volume still approximately equals the inflow volume over the complete flood event.

Question 22Hydrology

A 1-hour unit hydrograph has ordinates of 0, 100, 50, and 0 cfs at 0, 1, 2, and 3 hours. A storm has 1 inch of excess rainfall in the first hour and 2 inches in the second hour. What is the peak direct runoff?

  1. 300 cfs
  2. 100 cfs
  3. 150 cfs
  4. 250 cfs Correct answer.

Explanation

Convolution sums the lagged, scaled unit-hydrograph responses. At 2 hours, the runoff is 1 × 50 + 2 × 100 = 250 cfs, which is the peak.

Question 23Hydrology

To convert a 1-hour S-hydrograph to a 3-hour unit hydrograph, which operation is used?

  1. Lag, subtract, divide by 3 Correct answer.
  2. Multiply the S-curve ordinates by 3
  3. Divide S-curve by 3 without lagging
  4. Add S-curve and its 3-hour lag

Explanation

The difference between the S-curve and its lagged version represents the runoff volume from 3 hours of continuous unit inflow. Dividing that difference by 3 converts the volume to a 3-hour unit hydrograph.

Question 24Hydrology

In low-flow frequency analysis, the statistic 7Q10 denotes the:

  1. 7-day low flow with 10-year recurrence Correct answer.
  2. 7-month low flow exceeded 10% of time
  3. 7-day low flow exceeded 10 times yearly
  4. 7-day peak flow with 10-year recurrence

Explanation

7Q10 is the lowest 7-day average flow expected with a 10-year recurrence interval, equivalent to a 10% annual exceedance probability. It is commonly used as a low-flow design statistic.

Question 25Hydrology

A sudden break in the slope of a double-mass curve of station precipitation versus reference precipitation indicates:

  1. Random error in precipitation measurement
  2. Station record change from relocation Correct answer.
  3. A normal seasonal cycle in precipitation
  4. Increase in contributing watershed area

Explanation

A double-mass curve compares cumulative station catch with a consistent reference group to detect changes in record consistency. A persistent slope change suggests a relocation, exposure change, or instrument shift at the station.

Question 26Hydrology

On an intensity-duration-frequency curve for a fixed return period, how does the average rainfall intensity change as storm duration increases?

  1. It increases as duration increases
  2. It stays the same for all durations
  3. It changes with duration only for durations shorter than one hour
  4. It decreases as duration increases Correct answer.

Explanation

A longer storm averages its most intense bursts together with much lower-intensity periods, so the mean intensity over the longer duration is smaller. IDF curves therefore slope downward with duration, with each return period plotted as a separate curve.

Question 27Hydrology

According to Darcy's law, the specific discharge through a saturated porous medium is equal to which quantity?

  1. Hydraulic conductivity divided by the hydraulic gradient measured across the sample
  2. Hydraulic conductivity multiplied by the hydraulic gradient Correct answer.
  3. Hydraulic gradient divided by porosity
  4. Porosity multiplied by the hydraulic gradient

Explanation

Darcy's law is written q = K i, where q is specific discharge, K is hydraulic conductivity, and i is the hydraulic gradient. Porosity influences the pore velocity of the water but does not enter the Darcy flux expression directly.

Question 28Hydrology

Using the Green-Ampt infiltration equation f = K(1 + ψΔθ/F), a soil has K = 0.5 in/hr, wetting-front suction ψ = 8.0 in, moisture deficit Δθ = 0.30, and cumulative infiltration F = 2.4 in. What is the infiltration rate at that moment?

  1. 0.75 in/hr
  2. 1.5 in/hr
  3. 1.0 in/hr Correct answer.
  4. 0.5 in/hr

Explanation

The capillary term ψΔθ/F equals 2.4/2.4 = 1.0, so the bracketed factor is 2.0 and f = 0.5 × 2.0 = 1.0 in/hr. Infiltration rate declines toward K as F grows because the suction contribution diminishes.

Question 29Hydrology

The Penman combination equation estimates open-water evaporation from which pair of driving terms?

  1. Air temperature and atmospheric pressure
  2. Soil moisture content and rooting depth
  3. Wind speed alone and surface albedo alone
  4. Net radiation and aerodynamic vapor transport Correct answer.

Explanation

Penman combined the energy budget with a mass-transfer approach, so evaporation depends on available radiant energy and on the ability of the air to remove vapor through wind and vapor-pressure deficit. The aerodynamic term dominates under dry, windy, advective conditions.

Question 30Hydrology

Orographic precipitation is produced primarily by which mechanism?

  1. Surface heating causing buoyant updrafts over flat terrain
  2. Convergence of air near the center of a low-pressure system
  3. Moist air rising over a topographic barrier and cooling Correct answer.
  4. Two air masses of contrasting temperature meeting along a front

Explanation

When moist air is forced up the windward slope of a mountain range it expands and cools, raising relative humidity until condensation and precipitation occur. The leeward side typically receives much less rain because the air has lost moisture and warms as it descends.

Question 31Hydrology

Baseflow at a stream gage falls from 80 cfs to 60 cfs over 5 consecutive days of no rainfall. Using the exponential depletion relation Q = Q0 K^t, what is the daily recession constant K?

  1. 0.750
  2. 0.988
  3. 0.944 Correct answer.
  4. 0.600

Explanation

Rearranging the depletion relation gives K = (Qt/Q0)^(1/t) = (60/80)^(1/5) = 0.944. The constant represents the fraction of baseflow remaining after each day and is applied with the number of days since recession began as the exponent.

Question 32Hydrology

How do depression storage and interception chiefly affect the rainfall-runoff process on a small watershed?

  1. They delay runoff and reduce total runoff volume Correct answer.
  2. They raise infiltration rates by sealing the soil surface
  3. They increase peak discharge by concentrating flow into channels
  4. They increase the time of concentration by lengthening the flow path

Explanation

Water caught on foliage or held in surface hollows must be satisfied before overland flow can begin, creating an initial abstraction. This both delays the start of the rising limb and removes water that would otherwise appear as runoff.

Question 33Hydrology

In Snyder's synthetic unit hydrograph method, the coefficient Ct is used primarily to estimate which basin characteristic?

  1. The basin lag, which is related to length and slope Correct answer.
  2. The recession shape, which is controlled by channel storage
  3. The peak discharge, which is controlled by the coefficient Cp
  4. The base time, which is controlled by the drainage area

Explanation

Snyder related basin lag to the product of main channel length and length to the centroid through tp = Ct (L Lca)^0.3. The companion coefficient Cp governs the peak discharge of the resulting unit hydrograph.

Question 34Hydrology

In the log-Pearson Type III method used for flood-frequency analysis, what role does the skewness coefficient play?

  1. It is computed from untransformed peaks and shifts the mean of the distribution
  2. It is computed from log peaks and shapes the fitted curve Correct answer.
  3. It is assumed to equal zero at every station regardless of record length
  4. It is calculated only from the largest ten percent of the recorded peaks

Explanation

Log-Pearson Type III fits a three-parameter distribution to the base-10 logarithms of annual peak flows, with mean, standard deviation, and skewness estimated from those logarithms. The skewness controls the asymmetry of the fitted curve and therefore the magnitude of extreme quantiles.

Question 35Hydrology

Which statement defines the specific yield of an unconfined aquifer?

  1. The total volume of voids divided by the total volume of the aquifer
  2. The volume of water retained against gravity per unit volume of aquifer
  3. The hydraulic conductivity multiplied by the saturated thickness
  4. The gravity-drained water released per unit area per unit water-table decline Correct answer.

Explanation

Specific yield is the drainable porosity, that is, the fraction of aquifer volume that yields water by gravity when the water table drops. Water held by capillary and molecular attraction is specific retention, and specific yield plus specific retention equals total porosity.

Question 36Hydrology

A 100-acre watershed receives 2.4 inches of rainfall and produces 0.6 inch of direct runoff. What is the total volume of direct runoff in acre-feet?

  1. 2.5 acre-feet
  2. 5.0 acre-feet Correct answer.
  3. 20 acre-feet
  4. 3.0 acre-feet

Explanation

Multiplying runoff depth by area gives 0.6 in × 100 acres = 60 inch-acres, and dividing by 12 inches per foot yields 5.0 acre-feet. Only the runoff depth contributes to the volume; the full rainfall depth includes abstractions.

Question 37Hydrology

A snowpack is 30 inches deep with an average density of 0.35 (35 percent of the density of water). What is its snow water equivalent?

  1. 8.6 inches
  2. 10.5 inches Correct answer.
  3. 3.5 inches
  4. 10.0 inches

Explanation

Snow water equivalent equals snow depth multiplied by the density expressed as a fraction of water density, so 30 in × 0.35 = 10.5 in of liquid water. This conversion allows snowmelt to be added to precipitation records for water-balance and runoff forecasting.

Question 38Hydrology

Under which combination of conditions would annual interception loss be largest?

  1. A bare soil surface with frequent dewfall
  2. A paved parking area with high albedo
  3. A sparse shrub canopy with a few intense storms
  4. A dense evergreen canopy with many small storms Correct answer.

Explanation

Interception depends on canopy storage capacity and on how often that storage is refilled and evaporated. Dense canopies hold the most water per event, and many small storms are largely intercepted and evaporated between events, whereas large storms quickly exceed storage and produce throughfall and stemflow.

Question 39Hydrology

Using the NRCS triangular unit hydrograph relation qp = 484 A / Tp, what is the peak discharge in cfs for a 2-square-mile watershed with a time to peak of 4 hours under 1 inch of direct runoff?

  1. 242 cfs Correct answer.
  2. 968 cfs
  3. 121 cfs
  4. 484 cfs

Explanation

Substituting A = 2 mi² and Tp = 4 hr gives qp = 484 × 2 / 4 = 242 cfs. The constant 484 already accounts for the unit runoff depth and the triangular shape assumption, so the runoff depth does not appear separately in the calculation.

Question 40Hydrology

A gaging station has 40 years of annual peak-flow record. Using the Weibull plotting-position formula T = (n + 1)/m, what return period is assigned to the fourth-largest peak in the record?

  1. 9.8 years
  2. 4.0 years
  3. 40 years
  4. 10.25 years Correct answer.

Explanation

With n = 40 years of record and rank m = 4 for the fourth-largest peak, the formula gives T = 41/4 = 10.25 years. Plotting positions are used to display the observed data for probability plotting rather than to extrapolate beyond the record length.

Question 41Hydraulics - Closed Conduit

Water flows at 0.020 m³/s through a 150 mm diameter pipe. Taking the kinematic viscosity as 1.0 × 10⁻⁶ m²/s, what is the Reynolds number?

  1. 1.13 × 10⁵
  2. 1.70 × 10⁵ Correct answer.
  3. 8.50 × 10⁴
  4. 2.55 × 10⁵

Explanation

The mean velocity is V = Q/A = 0.020/(π × 0.15²/4) = 1.13 m/s. The Reynolds number is Re = VD/ν = 1.13 × 0.15/(1.0 × 10⁻⁶) ≈ 1.70 × 10⁵.

Question 42Hydraulics - Closed Conduit

The hydraulic radius of a circular pipe flowing full with diameter D is:

  1. D
  2. D/2
  3. D/4 Correct answer.
  4. 2D

Explanation

Hydraulic radius is flow area divided by wetted perimeter. For a full circular pipe, R = (πD²/4)/(πD) = D/4.

Question 43Hydraulics - Closed Conduit

Which expression gives the Darcy–Weisbach friction head loss in a pipe of length L and diameter D?

  1. h_f = f (L/D)(2g/V²)
  2. h_f = f (L/D)(V²/g)
  3. h_f = f (L/D)(V²/2g) Correct answer.
  4. h_f = f (D/L)(V²/2g)

Explanation

The Darcy–Weisbach equation is h_f = f(L/D)(V²/2g), where f is the Darcy friction factor. The length-to-diameter ratio multiplies the velocity head.

Question 44Hydraulics - Closed Conduit

A valve with a loss coefficient K = 5.0 is installed in a 200 mm pipe carrying water at 2.0 m/s. What is the minor head loss?

  1. 1.02 m Correct answer.
  2. 0.51 m
  3. 0.20 m
  4. 2.04 m

Explanation

Minor loss is h_m = K V²/(2g). Substituting gives 5.0 × 2.0²/(2 × 9.81) = 1.02 m.

Question 45Hydraulics - Closed Conduit

For fully developed laminar flow in a circular pipe, the Darcy friction factor is given by:

  1. f = 0.316/Re^0.25
  2. f = 0.25/[log10(ε/3.7D)]²
  3. f = 16/Re
  4. f = 64/Re Correct answer.

Explanation

For laminar pipe flow, the Darcy friction factor is f = 64/Re. This follows from the Hagen–Poiseuille relation and applies while Re is below the transitional range.

Question 46Hydraulics - Closed Conduit

Pipe flow is conventionally treated as turbulent when the Reynolds number exceeds approximately:

  1. 10,000
  2. 500
  3. 100,000
  4. 2,000 to 4,000 Correct answer.

Explanation

The laminar-to-turbulent transition in pipe flow occurs near Re ≈ 2,000 to 4,000. Above this range, the flow is normally treated as turbulent.

Question 47Hydraulics - Closed Conduit

On the Moody diagram, the relative roughness of a pipe is expressed as:

  1. ε/D Correct answer.
  2. f/Re
  3. D/ε
  4. ε/L

Explanation

Relative roughness is the absolute roughness height ε divided by the pipe diameter D. It appears as ε/D on the Moody diagram.

Question 48Hydraulics - Closed Conduit

In the Hazen–Williams formula, a higher C value indicates:

  1. a rougher pipe with more head loss
  2. a dimensionless friction factor similar to f
  3. a pipe with higher kinematic viscosity
  4. a smoother pipe with lower loss Correct answer.

Explanation

The Hazen–Williams C is an empirical roughness coefficient. A larger C represents a smoother pipe and produces a smaller head loss for the same discharge.

Question 49Hydraulics - Closed Conduit

Two pipes are connected in series. Pipe 1 is 100 m long with D = 0.20 m and f = 0.020; pipe 2 is 50 m long with D = 0.10 m and f = 0.020. For a steady discharge of 0.030 m³/s, what is the total friction head loss, neglecting minor losses?

  1. 11.6 m
  2. 4.65 m
  3. 3.72 m
  4. 7.90 m Correct answer.

Explanation

Continuity gives V₁ = 0.955 m/s and V₂ = 3.82 m/s. Then h_f1 = 0.465 m and h_f2 = 7.44 m, so the total is about 7.90 m.

Question 50Hydraulics - Closed Conduit

Two pipes of different diameters are connected in parallel between the same two nodes. Which statement is correct for the flow between the nodes?

  1. The friction head loss is the same in both pipes. Correct answer.
  2. The velocity is the same in both pipes.
  3. The discharge divides equally between the two pipes.
  4. The longer pipe always carries the larger discharge.

Explanation

Parallel pipes have the same pressure difference between their common end nodes, so they must have the same friction head loss. The discharge divides so that each pipe satisfies that common loss.