PE Civil (NCEES Principles and Practice of Engineering) — All Questions

Using the Rational Method, what is the peak runoff for a 20-acre watershed with runoff coefficient C = 0.5 and rainfall intensity i = 3 in/hr?

• a.3 cfs
• b.60 cfs
• c.30 cfs✓
• d.10 cfs

The Rational Method is Q = C·i·A, where in U.S. units i is in in/hr and A in acres, giving Q directly in cfs (the unit conversion factor is approximately 1). Q = 0.5 x 3 x 20 = 30 cfs. The method suits small urban drainage areas.

Water flows at 2 m/s through a channel with cross-sectional flow area 0.5 m^2. What is the discharge?

• a.4 m^3/s
• b.1 m^3/s✓
• c.0.5 m^3/s
• d.2 m^3/s

The continuity equation gives discharge Q = V·A = 2 x 0.5 = 1 m^3/s. For steady incompressible flow, discharge is constant along the channel even as area and velocity vary.

By Manning's equation, if the channel slope is doubled (all else equal), the flow velocity increases by approximately what factor?

• a.4
• b.2
• c.1
• d.1.41✓

Manning's velocity is V = (1/n)·R^(2/3)·S^(1/2), so velocity is proportional to the square root of slope. Doubling S multiplies velocity by sqrt(2) = 1.41. Slope has a weaker-than-linear effect on velocity.

The resultant hydrostatic force on a vertical rectangular gate holding back water acts at what location, measured from the bottom of the gate?

• a.2H/3 above the bottom
• b.H/3 above the bottom✓
• c.H/2 (the centroid)
• d.H/4 above the bottom

Because hydrostatic pressure increases linearly with depth, the pressure distribution is triangular and its resultant acts at the centroid of that triangle, which is H/3 above the bottom (or 2H/3 below the surface). This is the center of pressure.

In the Rational Method, the design storm duration is typically taken equal to what quantity so that the entire watershed contributes to peak flow?

• a.One hour
• b.The return period
• c.The time of concentration✓
• d.The storm frequency

Peak runoff occurs when the storm duration equals the time of concentration, the time for runoff to travel from the hydraulically most distant point to the outlet. At that duration the whole watershed contributes simultaneously, maximizing the peak.

An open channel is rectangular, 3 m wide, flowing at a depth of 1.0 m, with Manning n = 0.013 and a bed slope of 0.0016. Using Manning's equation (SI), what is the mean flow velocity?

• a.3.08 m/s
• b.0.088 m/s
• c.1.10 m/s
• d.2.19 m/s✓

Manning (SI): V = (1/n)·R^(2/3)·S^(1/2). Area A = 3(1) = 3 m^2, wetted perimeter P = 3 + 2(1) = 5 m, so hydraulic radius R = A/P = 0.6 m. V = (1/0.013)(0.6^0.667)(0.0016^0.5) = 2.19 m/s. Using R = depth (1 m) gives the wrong 3.08; using S instead of sqrt(S) gives 0.088.

For the same rectangular channel (3 m wide, depth 1.0 m) flowing at a mean velocity of 2.19 m/s, what is the discharge?

• a.3.29 m^3/s
• b.2.19 m^3/s
• c.6.57 m^3/s✓
• d.10.95 m^3/s

Continuity: Q = V·A. The flow area is A = b·y = 3 x 1.0 = 3 m^2, so Q = 2.19 x 3 = 6.57 m^3/s. Discharge is velocity times cross-sectional area, not velocity alone.

A circular pipe of diameter 0.6 m flows completely full. What is its hydraulic radius?

• a.0.3 m
• b.0.075 m
• c.0.15 m✓
• d.0.6 m

For a full circular pipe, R = A/P = (piD^2/4)/(piD) = D/4 = 0.6/4 = 0.15 m. A common error is to use the diameter (0.6) or radius (0.3) directly; the hydraulic radius of a full circle is one-quarter of the diameter.

A rectangular open channel is 4 m wide and flows at a depth of 1.5 m. What is the hydraulic radius?

• a.0.857 m✓
• b.0.667 m
• c.4 m
• d.1.5 m

R = A/P. Area A = 4 x 1.5 = 6 m^2; wetted perimeter (bed plus two sides) P = 4 + 2(1.5) = 7 m; R = 6/7 = 0.857 m. The hydraulic radius uses the wetted perimeter, not the flow depth or width alone.

Using the Rational Method, what is the peak runoff from a 12-acre watershed with runoff coefficient C = 0.7 and a design rainfall intensity of 2.5 in/hr?

• a.30 cfs
• b.14.7 cfs
• c.21 cfs✓
• d.4.29 cfs

Rational Method: Q = C·i·A, with i in in/hr and A in acres yielding Q in cfs (conversion factor ~1). Q = 0.7 x 2.5 x 12 = 21 cfs. The method applies to small drainage areas where the storm can cover the whole basin.

A watershed has 5 acres of pavement (C = 0.9) and 15 acres of lawn (C = 0.3). What is the area-weighted composite runoff coefficient?

• a.0.90
• b.0.60
• c.0.45✓
• d.0.30

The composite coefficient is area-weighted: C = (5(0.9) + 15(0.3))/(5 + 15) = (4.5 + 4.5)/20 = 0.45. Averaging the two C values without weighting by area (0.60) is the common mistake.

Water flows in a channel at 3.0 m/s with a depth of 0.5 m. What is the Froude number?

• a.1.35✓
• b.0.61
• c.0.45
• d.2.72

Froude number Fr = V/sqrt(g·y) = 3.0/sqrt(9.81 x 0.5) = 3.0/2.215 = 1.35. The Froude number is the ratio of inertial to gravitational forces and classifies open-channel flow regime.

For flow with a Froude number of 1.35, the flow regime is best described as:

• a.critical
• b.subcritical
• c.tranquil
• d.supercritical✓

When Fr > 1 the flow is supercritical (rapid, shallow, high velocity); Fr < 1 is subcritical (tranquil); Fr = 1 is critical. A Froude number of 1.35 exceeds 1, so the flow is supercritical.

A rectangular channel 3 m wide carries 6 m^3/s. What is the critical depth?

• a.0.204 m
• b.1.13 m
• c.0.408 m
• d.0.742 m✓

Critical depth for a rectangular channel: yc = (q^2/g)^(1/3), where unit discharge q = Q/b = 6/3 = 2 m^2/s. yc = (2^2/9.81)^(1/3) = (0.408)^(1/3) = 0.742 m. At critical depth the specific energy is minimum.

Open-channel flow has a depth of 1.2 m and a velocity of 2.0 m/s. What is the specific energy?

• a.0.204 m
• b.1.20 m
• c.2.24 m
• d.1.40 m✓

Specific energy E = y + V^2/(2g) = 1.2 + (2.0^2)/(2 x 9.81) = 1.2 + 0.204 = 1.404 m. Specific energy is the energy head measured relative to the channel bottom.

Water flows through a 0.2-m-diameter pipe, 100 m long, at 2.0 m/s. With a Darcy friction factor f = 0.02, what is the friction head loss?

• a.10.2 m
• b.2.04 m✓
• c.4.08 m
• d.0.20 m

Darcy-Weisbach: h_f = f(L/D)(V^2/2g) = 0.02(100/0.2)(2.0^2/(2 x 9.81)) = 0.02 x 500 x 0.204 = 2.04 m. Head loss grows with the square of velocity.

Water (kinematic viscosity 1x10^-6 m^2/s) flows at 2.0 m/s in a 0.2-m pipe. What is the Reynolds number?

• a.400,000✓
• b.40,000
• c.4,000,000
• d.200,000

Re = V·D/nu = (2.0 x 0.2)/(1x10^-6) = 400,000. The Reynolds number is the ratio of inertial to viscous forces and sets the flow regime.

For pipe flow with a Reynolds number of 400,000, the flow is:

• a.turbulent✓
• b.transitional
• c.laminar
• d.critical

In pipe flow, Re < 2,000 is laminar, 2,000-4,000 transitional, and Re > 4,000 turbulent. A Reynolds number of 400,000 is well into the turbulent range.

A pump lifts 0.05 m^3/s of water against a total head of 20 m. What is the water (hydraulic) power delivered to the fluid?

• a.98.1 kW
• b.1.0 kW
• c.9.81 kW✓
• d.19.6 kW

Water power P = gamma·Q·H = 9,810 N/m^3 x 0.05 m^3/s x 20 m = 9,810 W = 9.81 kW. This is the power added to the water; the required brake/shaft power is larger by the pump efficiency.

A pump must deliver 9.81 kW of water power. If the pump efficiency is 75%, what brake (shaft) power is required?

• a.7.36 kW
• b.13.1 kW✓
• c.9.81 kW
• d.12.3 kW

Brake power = water power / efficiency = 9.81/0.75 = 13.1 kW. Efficiency losses mean the motor must supply more than the useful hydraulic power; multiplying by efficiency (7.36) is the common error.

A rectangular sharp-crested weir 3 m long has a head of 0.4 m over the crest. Using Q = 1.84·L·H^(3/2) (SI), what is the discharge?

• a.0.88 m^3/s
• b.1.40 m^3/s✓
• c.3.49 m^3/s
• d.2.21 m^3/s

Rectangular weir: Q = 1.84·L·H^(3/2) = 1.84 x 3 x 0.4^1.5 = 1.84 x 3 x 0.253 = 1.40 m^3/s. Head is raised to the 3/2 power, so weir flow is quite sensitive to head.

A 90-degree V-notch weir has a head of 0.3 m. Using Q = 1.38·H^(5/2) (SI), what is the discharge?

• a.0.068 m^3/s✓
• b.0.124 m^3/s
• c.0.152 m^3/s
• d.0.041 m^3/s

For a 90-degree V-notch weir, Q = 1.38·H^(5/2) = 1.38 x 0.3^2.5 = 1.38 x 0.0493 = 0.068 m^3/s. The triangular notch gives good accuracy at low flows because the exponent on head is 5/2.

Water discharges through a sharp-edged orifice of area 0.01 m^2 under a head of 3 m, with discharge coefficient Cd = 0.6. What is the discharge?

• a.0.077 m^3/s
• b.0.023 m^3/s
• c.0.046 m^3/s✓
• d.0.092 m^3/s

Orifice discharge: Q = Cd·A·sqrt(2gh) = 0.6 x 0.01 x sqrt(2 x 9.81 x 3) = 0.6 x 0.01 x 7.67 = 0.046 m^3/s. Omitting Cd (using 1.0) overestimates the flow at 0.077.

Water flows freely from an orifice under a head of 5 m. Neglecting losses, what is the theoretical jet velocity?

• a.98.1 m/s
• b.9.90 m/s✓
• c.7.00 m/s
• d.49.1 m/s

Torricelli's theorem: V = sqrt(2gh) = sqrt(2 x 9.81 x 5) = sqrt(98.1) = 9.90 m/s. The jet velocity equals that of a body freely falling through the same head.

What is the gauge pressure at a depth of 8 m below the surface of still water?

• a.157 kPa
• b.78.5 kPa✓
• c.784 kPa
• d.8 kPa

Hydrostatic pressure p = gamma·h = 9.81 kN/m^3 x 8 m = 78.5 kPa. Pressure increases linearly with depth in a static fluid.

A vertical rectangular gate 2 m wide and 3 m tall has its top edge at the water surface. What is the total hydrostatic force on the gate?

• a.58.9 kN
• b.29.4 kN
• c.176.6 kN
• d.88.3 kN✓

F = gamma·h_c·A, where h_c is the depth to the centroid. h_c = 3/2 = 1.5 m, A = 2 x 3 = 6 m^2. F = 9.81 x 1.5 x 6 = 88.3 kN. The pressure at the centroid times the area gives the resultant force.

For that same gate (2 m wide, 3 m tall, top at the surface), at what depth below the water surface does the resultant hydrostatic force act?

• a.2.0 m✓
• b.2.5 m
• c.1.5 m
• d.1.0 m

For a surface-piercing vertical plane, the pressure distribution is triangular, so the resultant (center of pressure) acts at two-thirds of the depth: 2/3 x 3 = 2.0 m below the surface. It lies below the centroid (1.5 m) because pressure is greater lower down.

A submerged object displaces 0.02 m^3 of water. What is the buoyant force on it?

• a.98.1 N
• b.196 N✓
• c.392 N
• d.20 N

Archimedes' principle: buoyant force = weight of displaced fluid = gamma·V = 9,810 N/m^3 x 0.02 m^3 = 196 N. Buoyancy depends only on the displaced volume and fluid unit weight.

Groundwater flows through an aquifer with hydraulic conductivity k = 0.001 m/s under a hydraulic gradient of 0.02 through a cross-sectional area of 10 m^2. What is the discharge (Darcy's law)?

• a.0.02 m^3/s
• b.0.0002 m^3/s✓
• c.0.001 m^3/s
• d.0.002 m^3/s

Darcy's law: Q = k·i·A = 0.001 x 0.02 x 10 = 0.0002 m^3/s. Flow through porous media is proportional to hydraulic conductivity, gradient, and area.

For the aquifer above (Darcy velocity = k·i = 2x10^-5 m/s) with a porosity of 0.25, what is the seepage (pore) velocity?

• a.5.0x10^-6 m/s
• b.3.2x10^-4 m/s
• c.8.0x10^-5 m/s✓
• d.2.0x10^-5 m/s

Seepage velocity vs = v_Darcy / porosity = (2x10^-5)/0.25 = 8x10^-5 m/s. Water moves only through the pore space, so the actual particle velocity exceeds the Darcy (bulk) velocity.

A tank with a volume of 500 m^3 receives a steady flow of 100 m^3/hr. What is the hydraulic detention (retention) time?

• a.0.2 hr
• b.50 hr
• c.5 hr✓
• d.2.5 hr

Hydraulic detention time = Volume / Flow rate = 500/100 = 5 hr. It represents the average time a parcel of water resides in the tank, a key parameter for sedimentation and disinfection.

A circular clarifier with a surface area of 100 m^2 treats 2,000 m^3/day. What is the surface overflow rate?

• a.2000 m/day
• b.5 m/day
• c.200 m/day
• d.20 m/day✓

Surface overflow rate = Q / surface area = 2,000/100 = 20 m/day (equivalently m^3/m^2/day). It is the design settling velocity: particles with a settling velocity above this rate are removed.

A sedimentation basin discharges 4,000 m^3/day over an effluent weir 50 m long. What is the weir loading rate?

• a.50 m^3/m/day
• b.80 m^3/m/day✓
• c.200 m^3/m/day
• d.4000 m^3/m/day

Weir loading rate = Q / weir length = 4,000/50 = 80 m^3/m/day. Keeping this rate low limits the approach velocity near the weir and prevents carryover of settled solids.

A water plant treats 5,000 m^3/day and must apply a chlorine dose of 3 mg/L. What is the chlorine feed rate?

• a.15 kg/day✓
• b.150 kg/day
• c.3 kg/day
• d.1.5 kg/day

Mass rate = dose x flow = 3 g/m^3 x 5,000 m^3/day = 15,000 g/day = 15 kg/day (using 1 mg/L = 1 g/m^3). Chemical feed is the product of concentration and volumetric flow.

A treatment plant reduces BOD from 250 mg/L to 25 mg/L. What is the BOD removal efficiency?

• a.90%✓
• b.80%
• c.10%
• d.75%

Removal efficiency = (influent - effluent)/influent x 100 = (250 - 25)/250 = 90%. This is a standard measure of process performance for BOD, TSS, and similar constituents.

Wastewater flows at 1,000 m^3/day with a BOD of 200 mg/L. What is the BOD mass loading?

• a.200 g/day
• b.2000 kg/day
• c.200 kg/day✓
• d.20 kg/day

Mass load = concentration x flow = 200 g/m^3 x 1,000 m^3/day = 200,000 g/day = 200 kg/day. Loading (mass/time) drives process sizing, whereas concentration alone does not.

100 m^3 of water at 200 mg/L chloride is blended with 300 m^3 at 40 mg/L. What is the resulting concentration?

• a.100 mg/L
• b.80 mg/L✓
• c.240 mg/L
• d.120 mg/L

Mass balance: C = (100(200) + 300(40))/(100 + 300) = (20,000 + 12,000)/400 = 80 mg/L. The mixed concentration is the flow-weighted average, not the simple average (120).

A reservoir has an inflow of 10 m^3/s and an outflow of 6 m^3/s held steady for 1 hour. What is the change in storage?

• a.14,400 m^3✓
• b.57,600 m^3
• c.4 m^3
• d.3,600 m^3

Continuity (storage) equation: dS = (I - O)·t = (10 - 6) m^3/s x 3,600 s = 14,400 m^3. Storage rises when inflow exceeds outflow; this is the basis of reservoir/detention routing.

In hydrograph analysis, a unit hydrograph represents the direct runoff resulting from:

• a.the peak discharge of any storm
• b.the baseflow component of streamflow
• c.one unit depth of excess (effective) rainfall applied uniformly over the watershed✓
• d.the total rainfall depth of a storm

A unit hydrograph is the direct-runoff response to one unit (1 in or 1 cm) of excess rainfall distributed uniformly over the basin during a specified duration. It is a linear, time-invariant transfer function used to build storm hydrographs by superposition.

Overland runoff travels 300 m to the outlet at an average velocity of 0.5 m/s. What is the travel (time of concentration) for this flow path?

• a.5 min
• b.10 min✓
• c.600 min
• d.150 min

Travel time = length / velocity = 300/0.5 = 600 s = 10 min. The time of concentration is the sum of such travel times along the longest hydraulic path and sets the design storm duration in the Rational Method.