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PE Water Resources practice questions

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Question 1Hydraulics and Hydrology

A 10-acre watershed has a runoff coefficient of 0.60. If the design rainfall intensity is 2.0 in/hr, what is the peak discharge estimated by the rational method?

  1. 6.0 cfs
  2. 120 cfs
  3. 12 cfs Correct answer.
  4. 20 cfs

Explanation

The rational method is Q = CiA, with i in in/hr and A in acres giving Q in cfs. Here Q = 0.60 x 2.0 x 10 = 12 cfs. The coefficient accounts for infiltration and surface storage losses, so it must be applied.

Question 2Hydraulics and Hydrology

A concrete pipe with n = 0.013 and a diameter of 3.0 ft flows full on a slope of 0.005 ft/ft. What is the average flow velocity?

  1. 12.8 ft/s
  2. 4.2 ft/s
  3. 10.6 ft/s
  4. 6.7 ft/s Correct answer.

Explanation

For a circular pipe flowing full, the hydraulic radius is D/4 = 0.75 ft. Manning's equation in US customary units gives V = (1.486/n) R^(2/3) S^(1/2) = (1.486/0.013)(0.75^(2/3))(0.005^0.5) = 6.7 ft/s.

Question 3Hydraulics and Hydrology

Time of concentration is best defined as:

  1. The time required for rainfall excess to travel from the outlet upstream to the divide
  2. The rainfall duration that produces the largest peak discharge
  3. The travel time from the most remote point of the watershed to the outlet Correct answer.
  4. The lag between the rainfall-excess centroid and the runoff-hydrograph centroid

Explanation

Time of concentration is the travel time of a water particle from the hydraulically most remote point of the basin to the outlet. It is the duration at which the entire watershed begins contributing to outflow. It is a physical travel-time property, not a hydrograph timing statistic.

Question 4Hydraulics and Hydrology

Which land use and hydrologic soil group combination would produce the highest NRCS curve number?

  1. Impervious pavement over soil group D Correct answer.
  2. Open space in good condition, soil group C
  3. Woodland in hydrologic soil group A
  4. Cultivated row crops, contoured, soil group B

Explanation

Curve numbers increase with impervious cover and with poorer soil drainage. Group D soils have the lowest infiltration capacity, and pavement is essentially impervious, so this combination yields a curve number near 98. Permeable soils with vegetal cover yield much lower values.

Question 5Hydraulics and Hydrology

For a rectangular sharp-crested weir operating with fully developed crest contraction, doubling the head produces a discharge increase of approximately what factor?

  1. 1.41
  2. 2.83 Correct answer.
  3. 2.00
  4. 4.00

Explanation

A sharp-crested rectangular weir follows Q = C L H^(3/2). Discharge therefore scales with the 1.5 power of head, and 2^1.5 = 2.83. Linear or square scaling applies to other structures, not to weirs.

Question 6Hydraulics and Hydrology

A unit hydrograph for a watershed is the direct runoff hydrograph produced by:

  1. one inch per hour of rainfall excess sustained for one hour
  2. 1 inch of rainfall excess regardless of duration or distribution
  3. 1 inch of rainfall excess over a specified duration Correct answer.
  4. 1 inch of total rainfall uniformly distributed over the basin

Explanation

A unit hydrograph represents the direct runoff from one inch of rainfall excess occurring uniformly over the basin in a stated duration. Both the depth and the duration must be specified because the shape changes with storm duration. Total rainfall is not used because abstractions would not be accounted for.

Question 7Hydraulics and Hydrology

An aquifer has a hydraulic conductivity of 25 ft/day and a hydraulic gradient of 0.004. What is the specific discharge by Darcy's law?

  1. 0.0016 ft/day
  2. 6.25 ft/day
  3. 0.10 ft/day Correct answer.
  4. 0.40 ft/day

Explanation

Darcy's law gives specific discharge q = K i, the flow per unit cross-sectional area. Multiplying 25 ft/day by 0.004 yields 0.10 ft/day. The actual seepage velocity would be larger when divided by porosity.

Question 8Hydraulics and Hydrology

In open channel flow, critical flow occurs when the Froude number equals:

  1. 0.0
  2. 1.0 Correct answer.
  3. 2.0
  4. 0.5

Explanation

Critical flow is the boundary between subcritical and supercritical flow, and it occurs at Froude number equal to 1.0. At that condition specific energy is minimum for a given discharge. Values below 1 indicate subcritical flow and above 1 indicate supercritical flow.

Question 9Hydraulics and Hydrology

A 12-in diameter ductile iron pipe (C = 130) carries 2.0 cfs. Using the Hazen-Williams equation, what is the friction head loss per 1000 ft of pipe?

  1. 2.1 ft per 1000 ft Correct answer.
  2. 1.6 ft per 1000 ft
  3. 4.2 ft per 1000 ft
  4. 3.4 ft per 1000 ft

Explanation

In US customary units h_f = 4.727 L Q^1.852 / (C^1.852 d^4.871) with Q in cfs and d in feet. Substituting L = 1000 ft, Q = 2.0 cfs, C = 130 and d = 1.0 ft gives about 2.1 ft of head loss.

Question 10Hydraulics and Hydrology

A 12-in diameter orifice with a discharge coefficient of 0.62 sits 6.0 ft below the water surface of a reservoir. What is the discharge?

  1. 12.2 cfs
  2. 4.8 cfs
  3. 9.6 cfs Correct answer.
  4. 15.4 cfs

Explanation

Orifice flow is Q = C_d A sqrt(2gh). The area is 0.785 ft² and the ideal velocity is sqrt(2 x 32.2 x 6.0) = 19.7 ft/s, so Q = 0.62 x 0.785 x 19.7 = 9.6 cfs.

Question 11Hydraulics and Hydrology

A 40-acre watershed contains 25 acres of impervious area with C = 0.90 and 15 acres of lawn on sandy soil with C = 0.20. What is the area-weighted runoff coefficient?

  1. 0.46
  2. 0.90
  3. 0.55
  4. 0.64 Correct answer.

Explanation

Composite runoff coefficients are weighted by subarea: C = sum(C_i A_i)/sum(A_i). Here (25 x 0.90 + 15 x 0.20)/40 = 25.5/40 = 0.64. Weighting by area, not simple averaging, is required.

Question 12Hydraulics and Hydrology

A watershed has a curve number of 80. Using Ia = 0.2S, what is the direct runoff depth for a 4.0-in rainfall?

  1. 3.5 in
  2. 1.1 in
  3. 2.5 in
  4. 2.0 in Correct answer.

Explanation

The retention is S = 1000/CN - 10 = 2.5 in, so the initial abstraction is 0.5 in. The NRCS runoff equation gives Q = (P - 0.2S)²/(P + 0.8S) = (3.5)²/6.0 = 2.0 in.

Question 13Hydraulics and Hydrology

A flow path is 2000 ft long with an average slope of 0.01 ft/ft. Using the Kirpich equation, what is the time of concentration?

  1. 16 min Correct answer.
  2. 60 min
  3. 8 min
  4. 35 min

Explanation

Kirpich's equation is Tc = 0.0078 L^0.77 S^-0.385 with L in feet and S in ft/ft. Substituting gives 0.0078 (2000^0.77)(0.01^-0.385) = 16 min. The negative exponent on slope means flatter slopes lengthen the time of concentration.

Question 14Hydraulics and Hydrology

In Muskingum routing, the reach travel time is K = 3 h and the weighting factor is x = 0.25. For which routing time step does the coefficient C0 equal zero?

  1. 0.75 h
  2. 3.0 h
  3. 1.5 h Correct answer.
  4. 6.0 h

Explanation

The Muskingum coefficient is C0 = (0.5Δt - Kx)/(K - Kx + 0.5Δt). Setting the numerator to zero gives Δt = 2Kx = 2(3)(0.25) = 1.5 h. At that step the routed outflow depends only on prior inflows.

Question 15Hydraulics and Hydrology

Which statement correctly describes level-pool reservoir flood routing?

  1. Peak outflow coincides in time with the peak inflow
  2. Peak outflow always exceeds peak inflow at an uncontrolled spillway
  3. Peak outflow is lower than peak inflow and lags it in time Correct answer.
  4. Peak outflow equals peak inflow when the spillway is uncontrolled

Explanation

Reservoir storage absorbs part of the flood volume, which attenuates the peak and delays it. Outflow peaks after inflow because the pool level continues rising after the inflow peak. Only a reservoir with no storage would pass the hydrograph unchanged.

Question 16Hydraulics and Hydrology

A well pumps 400 gpm at a drawdown of 20 ft. Assuming specific capacity stays constant, what drawdown would be expected at 600 gpm?

  1. 20 ft
  2. 13.3 ft
  3. 45 ft
  4. 30 ft Correct answer.

Explanation

Specific capacity is discharge divided by drawdown, here 400/20 = 20 gpm/ft. Assuming it remains constant, drawdown at 600 gpm is 600/20 = 30 ft. Drawdown grows in proportion to pumping rate under this linear assumption.

Question 17Hydraulics and Hydrology

A pump curve is H = 120 - 0.002Q² and the system curve is H = 40 + 0.001Q², with H in feet and Q in gpm. What is the operating discharge?

  1. 283 gpm
  2. 200 gpm
  3. 163 gpm Correct answer.
  4. 115 gpm

Explanation

The operating point is where pump head equals system head. Setting 120 - 0.002Q² = 40 + 0.001Q² gives 80 = 0.003Q², so Q = 163 gpm. At that flow both curves give the same head of about 67 ft.

Question 18Hydraulics and Hydrology

Water flows 1.0 ft deep at 12 ft/s in a rectangular channel. What is the sequent depth after the hydraulic jump?

  1. 1.5 ft
  2. 2.0 ft
  3. 4.0 ft
  4. 2.5 ft Correct answer.

Explanation

The approach Froude number is Fr = 12/sqrt(32.2 x 1.0) = 2.11. The sequent depth relation y2/y1 = 0.5[sqrt(1 + 8Fr²) - 1] gives y2 = 0.5(6.06 - 1)(1.0) = 2.5 ft. The jump converts supercritical flow to subcritical flow with an energy loss.

Question 19Hydraulics and Hydrology

A culvert's headwater elevation is essentially independent of barrel slope and roughness but changes markedly with entrance geometry. The culvert is operating under:

  1. Inlet control Correct answer.
  2. Friction control
  3. Tailwater control
  4. Outlet control

Explanation

In inlet control the barrel is not the limiting section, so headwater depends on the entrance shape and the headwater-to-diameter ratio. Slope and roughness of the barrel have little influence in that regime. Outlet control, by contrast, makes headwater sensitive to slope, length, and roughness.

Question 20Hydraulics and Hydrology

What is the probability that a 25-year flood will be equaled or exceeded at least once in the next 10 years?

  1. 67%
  2. 40%
  3. 4%
  4. 33% Correct answer.

Explanation

The annual exceedance probability is 1/25 = 0.04. The probability of at least one exceedance in n years is 1 - (1 - p)^n = 1 - 0.96^10 = 0.335. The risk accumulates nonlinearly because years are treated as independent trials.

Question 21Hydraulics and Hydrology

For a 50-year return period, what is the Gumbel reduced variate y = -ln[-ln(1 - 1/T)]?

  1. 1.50
  2. 2.25
  3. 3.90 Correct answer.
  4. 4.60

Explanation

With T = 50, the non-exceedance probability is 0.98. Taking -ln(0.98) gives 0.0202, and -ln(0.0202) = 3.90. The reduced variate increases without bound as the return period grows.

Question 22Hydraulics and Hydrology

Water moving at 4.0 ft/s in a steel pipe is stopped instantly by a valve closure. If the pressure wave celerity is 4000 ft/s, what is the maximum surge head rise?

  1. 124 ft
  2. 400 ft
  3. 497 ft Correct answer.
  4. 248 ft

Explanation

Instantaneous closure gives the Joukowsky surge Δh = aΔV/g. Substituting 4000 ft/s and 4.0 ft/s yields 4000 x 4.0/32.2 = 497 ft of head. Slower closure reduces the surge because part of the wave energy is dissipated during closure.

Question 23Hydraulics and Hydrology

Horton infiltration parameters are f0 = 3.0 in/hr, fc = 0.5 in/hr, and k = 2.0 hr^-1. What total depth infiltrates during the first hour?

  1. 1.08 in
  2. 3.00 in
  3. 2.50 in
  4. 1.58 in Correct answer.

Explanation

Integrating f(t) = fc + (f0 - fc)e^(-kt) over one hour gives F = fc t + (f0 - fc)(1 - e^(-kt))/k. Substituting yields 0.5 + (2.5/2)(1 - e^-2) = 0.5 + 1.08 = 1.58 in. The rate declines from the initial value toward the final constant rate.

Question 24Hydraulics and Hydrology

An NRCS graphical peak-discharge calculation uses a unit peak discharge of 400 cfs per square mile per inch of runoff. The watershed is 1.0 mi², the 24-hour rainfall is 8.0 in, and CN = 75 with Ia = 0.2S. What is the estimated peak discharge?

  1. 1,200 cfs
  2. 1,600 cfs
  3. 2,500 cfs
  4. 2,020 cfs Correct answer.

Explanation

First compute runoff depth: S = 1000/75 - 10 = 3.33 in, Ia = 0.67 in, and Q = (8.0 - 0.67)²/(8.0 + 2.67) = 5.04 in. Then qp = qu A Q = 400 x 1.0 x 5.04 = about 2,020 cfs. The unit peak discharge already reflects the time of concentration and Ia/P ratio.

Question 25Hydraulics and Hydrology

A pump draws from an open tank with a static suction lift of 10 ft, suction-line friction loss of 4 ft, and water at 60 F (vapor pressure head 0.6 ft). If atmospheric pressure head is 33.9 ft, what is the available net positive suction head?

  1. 19.3 ft Correct answer.
  2. 29.3 ft
  3. 39.3 ft
  4. 23.3 ft

Explanation

NPSH available is the absolute suction head at the pump flange minus the vapor pressure head: NPSHA = h_atm - h_static - h_friction - h_vapor. Substituting gives 33.9 - 10 - 4 - 0.6 = 19.3 ft. This value must exceed the pump's required NPSH to avoid cavitation.

Question 26Hydraulics and Hydrology

In the Manning equation, the roughness coefficient n accounts for:

  1. the longitudinal slope of the channel bed
  2. the kinematic viscosity of the water
  3. boundary roughness that retards flow Correct answer.
  4. the depth of flow relative to critical depth

Explanation

Manning's n is an empirical resistance coefficient that represents the roughness of the wetted boundary. It reflects channel material, vegetation, and irregularities that retard flow.

Question 27Hydraulics and Hydrology

The Froude number for open-channel flow is defined as:

  1. Fr = Q / (A sqrt(g))
  2. Fr = V / sqrt(gD) Correct answer.
  3. Fr = sqrt(gD) / V
  4. Fr = V^2 / (2g)

Explanation

The Froude number is the ratio of inertial forces to gravity forces in open-channel flow. It is calculated as velocity divided by the square root of gravitational acceleration times hydraulic depth.

Question 28Hydraulics and Hydrology

The rational method is generally considered appropriate for drainage areas no larger than about:

  1. 1,000 acres
  2. 100 acres
  3. 25 acres
  4. 200 acres Correct answer.

Explanation

The rational method assumes uniform rainfall over the entire drainage area and a constant runoff coefficient. It is commonly limited to small watersheds of about 200 acres or less; larger areas generally require hydrograph or routing methods.

Question 29Hydraulics and Hydrology

The Theis equation is used to:

  1. route a flood wave through a gated reservoir spillway
  2. estimate transient drawdown in a confined aquifer Correct answer.
  3. compute peak discharge from a small rural watershed
  4. determine normal depth in a uniform grassed channel

Explanation

The Theis equation is the transient well-hydraulics solution for a confined aquifer. It relates drawdown to pumping rate, transmissivity, storativity, and time since pumping began.

Question 30Hydraulics and Hydrology

The time of concentration of a watershed is the time required for:

  1. runoff to travel from the most remote point to the outlet Correct answer.
  2. the recession limb to return to base flow
  3. a flood wave to pass through a reservoir storage pool
  4. rainfall to infiltrate completely into the soil profile

Explanation

Time of concentration is the travel time for runoff from the hydraulically most remote point of the watershed to reach the outlet. It is a key input to many peak-discharge and hydrograph methods.

Question 31Hydraulics and Hydrology

For a sharp-crested rectangular weir with crest length L and head H, the Francis formula in U.S. customary units is:

  1. Q = 0.62 A sqrt(2gH)
  2. Q = 3.33 L H^1.5 Correct answer.
  3. Q = 1.49/n A R^(2/3) S^0.5
  4. Q = 3.33 L H^0.5

Explanation

The Francis formula for a sharp-crested rectangular weir in cfs is Q = 3.33 L H^1.5, with L and H in feet. The discharge varies with the three-halves power of head.

Question 32Hydraulics and Hydrology

Darcy's law states that specific discharge through a porous medium is directly proportional to:

  1. the square of porosity
  2. the capillary fringe height
  3. the reciprocal of conductivity
  4. the hydraulic gradient Correct answer.

Explanation

Darcy's law is q = -K(dh/dl), so specific discharge is proportional to the hydraulic gradient. The constant of proportionality is hydraulic conductivity.

Question 33Hydraulics and Hydrology

The annual exceedance probability of a 100-year flood is:

  1. 0.01 Correct answer.
  2. 0.10
  3. 1.00
  4. 0.001

Explanation

Annual exceedance probability is the reciprocal of return period, so for T = 100 years the probability is 1/100 = 0.01, or 1 percent in any year.

Question 34Hydraulics and Hydrology

A 12-in diameter pipe carries water at an average velocity of 4.0 ft/s. What is the discharge?

  1. 0.785 cfs
  2. 12.6 cfs
  3. 3.14 cfs Correct answer.
  4. 1.57 cfs

Explanation

Continuity gives Q = AV. For a 12-in pipe, the diameter is 1.0 ft and the area is 0.785 ft²; multiplying by 4.0 ft/s gives about 3.14 cfs.

Question 35Hydraulics and Hydrology

A uniform rectangular channel has hydraulic radius R = 2.0 ft, bed slope S = 0.0009, and Manning n = 0.015. What is the average velocity?

  1. 9.44 ft/s
  2. 2.36 ft/s
  3. 3.15 ft/s
  4. 4.72 ft/s Correct answer.

Explanation

Using Manning's equation V = (1.486/n) R^(2/3) S^1/2, substitute n = 0.015, R = 2.0 ft, and S = 0.0009. This gives V ≈ 4.72 ft/s.

Question 36Hydraulics and Hydrology

A 30-acre watershed has a runoff coefficient C = 0.40 and a design rainfall intensity of 2.5 in/hr. What is the peak discharge by the rational method?

  1. 75 cfs
  2. 30 cfs Correct answer.
  3. 12 cfs
  4. 7.5 cfs

Explanation

The rational method is Q = CiA, with i in inches per hour and A in acres giving Q in cfs. Thus Q = 0.40 × 2.5 × 30 = 30 cfs.

Question 37Hydraulics and Hydrology

An aquifer has hydraulic conductivity K = 40 ft/day and saturated thickness b = 50 ft. What is its transmissivity in ft²/day?

  1. 90 ft²/day
  2. 200,000 ft²/day
  3. 2,000 ft²/day Correct answer.
  4. 0.8 ft²/day

Explanation

Transmissivity is the product of hydraulic conductivity and saturated thickness, T = Kb. Here T = 40 × 50 = 2,000 ft²/day.

Question 38Hydraulics and Hydrology

A sharp-crested rectangular weir has a crest length of 10 ft and a head of 1.5 ft. Using Q = 3.33 L H^1.5, what is the discharge?

  1. 122 cfs
  2. 36.8 cfs
  3. 24.5 cfs
  4. 61.2 cfs Correct answer.

Explanation

Substitute L = 10 ft and H = 1.5 ft into the Francis formula. Since 1.5^1.5 ≈ 1.837, Q = 3.33 × 10 × 1.837 ≈ 61.2 cfs.

Question 39Hydraulics and Hydrology

A rectangular channel 20 ft wide carries 500 cfs. What is the critical depth?

  1. 2.69 ft Correct answer.
  2. 1.35 ft
  3. 19.4 ft
  4. 5.38 ft

Explanation

For a rectangular channel, critical depth is yc = (q²/g)^(1/3), where q = Q/b = 25 cfs/ft. Thus yc = (25²/32.2)^(1/3) ≈ 2.69 ft.

Question 40Hydraulics and Hydrology

A 24-in diameter storm sewer has n = 0.013 and slope S = 0.005. What is the full-flow capacity?

  1. 4.0 cfs
  2. 16.0 cfs Correct answer.
  3. 12.5 cfs
  4. 8.0 cfs

Explanation

For full flow in a circular pipe, hydraulic radius is D/4 = 0.5 ft and area is π ft². Manning's equation gives V ≈ 5.09 ft/s, so Q = VA ≈ 16.0 cfs.

Question 41Hydraulics and Hydrology

Steady radial flow to a well in a confined aquifer has Q = 0.20 cfs, T = 5,000 ft²/day, r1 = 1 ft with h1 = 50 ft, and r2 = 500 ft. Using the Thiem equation, what is the drawdown at r1 relative to r2?

  1. 34 ft
  2. 0.34 ft
  3. 1.7 ft
  4. 3.4 ft Correct answer.

Explanation

The Thiem equation gives h2 - h1 = Q ln(r2/r1)/(2πT). Converting Q to 17,280 ft³/day and using ln(500) ≈ 6.215 gives a head difference of about 3.4 ft.

Question 42Hydraulics and Hydrology

A storm has a constant intensity of 3.0 in/hr for 40 minutes. What total depth of rainfall occurs?

  1. 1.50 in
  2. 2.00 in Correct answer.
  3. 0.75 in
  4. 1.20 in

Explanation

Rainfall depth equals intensity multiplied by duration, with duration expressed in hours. Since 40 minutes is 2/3 hour, depth = 3.0 × 2/3 = 2.00 in.

Question 43Hydraulics and Hydrology

Runoff travels 2,400 ft at an average velocity of 4.0 ft/s. What is the travel time in minutes?

  1. 6 min
  2. 24 min
  3. 16 min
  4. 10 min Correct answer.

Explanation

Travel time is length divided by velocity, with a conversion from seconds to minutes. Thus t = 2,400/(4.0 × 60) = 10 minutes.

Question 44Hydraulics and Hydrology

Using the pump affinity laws, if a pump's speed is doubled while the impeller diameter is unchanged, the head at a corresponding point will:

  1. decrease to one-half of the original head
  2. increase to four times the original head Correct answer.
  3. remain unchanged from the original head
  4. double the original head

Explanation

For geometrically similar pump operation, head varies with the square of rotational speed. Doubling the speed therefore increases head by a factor of four.

Question 45Hydraulics and Hydrology

Water flows at 5.0 ft/s through 1,000 ft of 12-in diameter pipe with f = 0.020. What is the friction head loss?

  1. 15.5 ft
  2. 3.88 ft
  3. 1.94 ft
  4. 7.76 ft Correct answer.

Explanation

The Darcy-Weisbach equation is hf = f(L/D)(V²/2g). With L/D = 1,000 and V²/2g ≈ 0.388 ft, hf = 0.020 × 1,000 × 0.388 ≈ 7.76 ft.

Question 46Hydraulics and Hydrology

Flood peaks at a site have a mean of 10,000 cfs and a standard deviation of 2,500 cfs. Using the Gumbel distribution with y = 2.25 for the 100-year event and KT = (y - 0.5772)/1.2825, what is the estimated 100-year peak?

  1. 15,770 cfs
  2. 13,260 cfs Correct answer.
  3. 10,000 cfs
  4. 2,500 cfs

Explanation

The Gumbel quantile estimate is Q_T = mean + KT × standard deviation. Here KT ≈ (2.25 - 0.5772)/1.2825 ≈ 1.304, so Q_T ≈ 10,000 + 1.304 × 2,500 ≈ 13,260 cfs.

Question 47Hydraulics and Hydrology

For Muskingum routing, K = 4 h, x = 0.20, and Δt = 2 h. What is the coefficient C0?

  1. 0.238
  2. 0.190
  3. 0.143
  4. 0.048 Correct answer.

Explanation

Muskingum C0 = (0.5Δt - Kx)/(K(1 - x) + 0.5Δt). Substituting gives (1.0 - 0.8)/(3.2 + 1.0) = 0.2/4.2 ≈ 0.048.

Question 48Hydraulics and Hydrology

A 1-hour unit hydrograph has ordinates of 0, 100, 200, and 0 cfs at times 0, 1, 2, and 3 h. A storm produces 1.0 in of excess rainfall in the first hour and 1.0 in in the second hour. Using discrete convolution, what is the direct runoff at time 2 h?

  1. 100 cfs
  2. 200 cfs
  3. 400 cfs
  4. 300 cfs Correct answer.

Explanation

By linear superposition, runoff at time 2 h equals the first-hour rainfall times the unit-hydrograph ordinate at 2 h plus the second-hour rainfall times the ordinate at 1 h. Thus Q2 = 1.0 × 200 + 1.0 × 100 = 300 cfs.

Question 49Hydraulics and Hydrology

For a rectangular channel, the most hydraulically efficient cross section occurs when the bottom width b is:

  1. four times the flow depth y
  2. half the flow depth y
  3. twice the flow depth y Correct answer.
  4. equal to the flow depth y

Explanation

The best hydraulic rectangular section minimizes wetted perimeter for a given area. This occurs at b = 2y, for which the hydraulic radius equals y/2.

Question 50Hydraulics and Hydrology

A pumping well is located 100 ft from a fully penetrating stream. In the method of images, the stream is represented as:

  1. a no-flow boundary 200 ft from the pumping well
  2. a pumping well of twice the strength at the stream
  3. an impermeable boundary located at the stream edge
  4. a recharge well of equal strength beyond the stream Correct answer.

Explanation

A constant-head boundary such as a fully penetrating stream is modeled by an image recharge well of equal strength placed symmetrically on the opposite side of the boundary. This maintains constant head at the stream.