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

Along a horizontal pipe the velocity increases from 2 m/s to 8 m/s. By Bernoulli's equation (neglecting losses), the pressure:

• a.is unchanged
• b.decreases by 60 kPa
• c.decreases by 30 kPa✓
• d.increases by 30 kPa

Bernoulli (horizontal): p1 + rho·V1^2/2 = p2 + rho·V2^2/2. dp = rho(V1^2 - V2^2)/2 = 1000(4 - 64)/2 = -30,000 Pa. Pressure drops 30 kPa as the fluid speeds up.

In the Hazen-Williams formula for pipe flow, a higher C coefficient indicates:

• a.a rougher pipe with higher friction
• b.a smoother pipe with lower friction✓
• c.a higher fluid viscosity
• d.a smaller pipe diameter

The Hazen-Williams C is a roughness coefficient: higher C means a smoother interior and less friction loss (new PVC ~150, old cast iron ~100). Unlike Darcy's f, the Hazen-Williams C increases as the pipe gets smoother.

Water at 3.0 m/s passes through a fitting with a minor-loss coefficient K = 0.5. What is the minor head loss?

• a.0.459 m
• b.1.5 m
• c.0.229 m✓
• d.0.918 m

Minor loss h_L = K·V^2/(2g) = 0.5 x (3.0^2)/(2 x 9.81) = 0.5 x 0.459 = 0.229 m. Minor losses are proportional to the velocity head.

A 0.5-m-diameter pipe flows full at 2.5 m/s. What is the discharge?

• a.0.196 m^3/s
• b.1.25 m^3/s
• c.0.491 m^3/s✓
• d.0.982 m^3/s

Q = V·A = V(piD^2/4) = 2.5 x (pi x 0.5^2/4) = 2.5 x 0.1963 = 0.491 m^3/s. The full-pipe area is piD^2/4.

Normal depth in an open channel is the depth at which:

• a.the flow is critical
• b.the velocity is maximum
• c.the velocity is zero
• d.flow is uniform, with gravity balancing friction and the water surface parallel to the bed✓

Normal (uniform) depth occurs when the gravitational driving force exactly balances boundary friction, so the depth, velocity, and energy grade stay constant and the water surface parallels the channel bed. Manning's equation solves for this depth.

If Manning's roughness coefficient n doubles while slope and geometry are unchanged, the flow velocity:

• a.is quadrupled
• b.is unchanged
• c.is halved✓
• d.doubles

In Manning's equation V = (1/n)R^(2/3)S^(1/2), velocity is inversely proportional to n. Doubling n halves the velocity (and thus the discharge).

For a given flow area, which open-channel cross-section is the most hydraulically efficient (smallest wetted perimeter)?

• a.a semicircle✓
• b.a triangle
• c.a square with vertical sides
• d.a wide shallow rectangle

The most efficient section minimizes wetted perimeter for a given area, maximizing hydraulic radius and thus capacity. Among all shapes the semicircle is optimal; among trapezoids the half-hexagon is best.

A trapezoidal channel has a bottom width of 3 m, side slopes of 2H:1V, and a flow depth of 1.5 m. What is the flow area?

• a.4.5 m^2
• b.6.75 m^2
• c.12 m^2
• d.9 m^2✓

Trapezoidal area A = (b + z·y)·y, with z = 2. A = (3 + 2 x 1.5)(1.5) = (3 + 3)(1.5) = 9 m^2. The term z·y accounts for the sloped sides on each side of the base.

For that trapezoidal channel (bottom width 3 m, side slopes 2H:1V, depth 1.5 m), what is the wetted perimeter?

• a.6 m
• b.12 m
• c.7.24 m
• d.9.71 m✓

Wetted perimeter P = b + 2y·sqrt(1 + z^2) = 3 + 2(1.5)·sqrt(1 + 2^2) = 3 + 3·sqrt(5) = 3 + 6.708 = 9.71 m. Each sloped side has length y·sqrt(1 + z^2).

For the same trapezoidal channel (area 9 m^2, wetted perimeter 9.71 m), what is the hydraulic radius?

• a.0.75 m
• b.1.5 m
• c.0.927 m✓
• d.1.08 m

R = A/P = 9/9.71 = 0.927 m. The hydraulic radius combines area and wetted perimeter and feeds directly into Manning's velocity.

At critical flow in an open channel, the Froude number equals:

• a.0.5
• b.1.0✓
• c.2.0
• d.0

Critical flow is defined by Fr = 1.0, where specific energy is minimum for a given discharge. Below 1 the flow is subcritical; above 1 it is supercritical.

A hydraulic jump occurs where open-channel flow transitions from:

• a.laminar to turbulent
• b.supercritical to subcritical✓
• c.steady to unsteady
• d.subcritical to supercritical

A hydraulic jump forms where rapid (supercritical) flow abruptly slows to tranquil (subcritical) flow, dissipating energy through turbulence. It is often used deliberately below spillways as an energy dissipator.

In a flow system, the energy grade line (EGL) lies above the hydraulic grade line (HGL) by an amount equal to:

• a.the elevation head
• b.the friction slope
• c.the pressure head
• d.the velocity head, V^2/(2g)✓

The EGL is total head; the HGL is the piezometric head (elevation + pressure head). Their difference is the velocity head V^2/(2g), which is why the two lines converge where velocity is low.

What is the specific weight (unit weight) of water at standard conditions?

• a.981 kN/m^3
• b.9.81 kN/m^3✓
• c.98.1 kN/m^3
• d.1000 kN/m^3

Specific weight gamma = rho·g = 1,000 kg/m^3 x 9.81 m/s^2 = 9,810 N/m^3 = 9.81 kN/m^3 (about 62.4 lb/ft^3). Density is mass per volume; specific weight is force per volume.

Water has a dynamic viscosity of 1x10^-3 Pa·s and a density of 1,000 kg/m^3. What is its kinematic viscosity?

• a.1x10^3 m^2/s
• b.1x10^-3 m^2/s
• c.1000 m^2/s
• d.1x10^-6 m^2/s✓

Kinematic viscosity nu = mu/rho = (1x10^-3)/(1,000) = 1x10^-6 m^2/s. Kinematic viscosity is dynamic viscosity normalized by density and appears in the Reynolds number.

In the NRCS (SCS) curve-number method, a higher curve number (CN) indicates:

• a.lower rainfall
• b.more infiltration and less runoff
• c.more evaporation
• d.more runoff and less infiltration✓

The curve number reflects the runoff potential of soil and land cover: high CN (impervious or wet clay soils) produces more runoff and less infiltration, while low CN (sandy, well-drained, vegetated) infiltrates more.

The primary hydraulic purpose of a stormwater detention basin is to:

• a.reduce and delay the peak runoff discharge leaving the site✓
• b.increase the peak flow to the outlet
• c.provide sewage treatment
• d.remove dissolved nitrogen

A detention basin temporarily stores runoff and releases it slowly, attenuating (lowering and delaying) the peak discharge so downstream channels are not overwhelmed. It manages flow rate, not primarily water quality.

The Rational Method (Q = C·i·A) is best suited for estimating peak runoff from:

• a.only agricultural watersheds
• b.groundwater aquifers
• c.small drainage areas such as urban catchments✓
• d.large river basins

The Rational Method assumes uniform rainfall and a steady-state peak, valid only for small areas (commonly under about 200 acres) where the time of concentration is short. Larger basins require unit-hydrograph or continuous simulation methods.

A storm produces 2 inches of rainfall excess over a 10-acre watershed. What is the runoff volume in acre-feet?

• a.20 ac-ft
• b.0.167 ac-ft
• c.1.67 ac-ft✓
• d.16.7 ac-ft

Volume = depth x area = (2 in / 12 in/ft) x 10 ac = 0.1667 ft x 10 ac = 1.67 acre-feet. Converting inches of depth to feet before multiplying by acres gives acre-feet directly.

A city of 10,000 people uses 150 gallons per capita per day. What is the average daily water demand?

• a.15 MGD
• b.0.15 MGD
• c.1.5 MGD✓
• d.150 MGD

Average demand = population x per-capita use = 10,000 x 150 = 1,500,000 gpd = 1.5 million gallons per day (MGD). Per-capita demand times population gives total average flow.

A system with an average demand of 1.5 MGD has a maximum-day peaking factor of 1.8. What is the maximum-day demand?

• a.2.7 MGD✓
• b.1.5 MGD
• c.3.3 MGD
• d.0.83 MGD

Maximum-day demand = average x peaking factor = 1.5 x 1.8 = 2.7 MGD. Peaking factors scale average demand up to size treatment and supply for high-use days.

In disinfection design, the CT value is the product of:

• a.chlorine and temperature
• b.cost and time
• c.disinfectant concentration and contact time✓
• d.clarity and turbidity

CT is disinfectant residual concentration (mg/L) multiplied by contact time (min). Regulations set required CT values for target pathogen log-removal; higher concentration or longer contact both raise CT.

The settling of discrete, non-flocculating particles in a sedimentation basin is described by:

• a.Stokes' law✓
• b.Manning's equation
• c.Bernoulli's equation
• d.Darcy's law

Type I (discrete particle) settling velocity follows Stokes' law, in which velocity is proportional to particle diameter squared and the density difference between particle and fluid. It underlies sedimentation-basin design.

Which of the following is a suspended-growth biological wastewater treatment process?

• a.rotating biological contactor
• b.trickling filter
• c.activated sludge✓
• d.slow sand filter

In activated sludge, microorganisms grow suspended (as flocs) in an aerated tank. Trickling filters and rotating biological contactors are attached-growth (biofilm) processes where the biomass grows on a fixed medium.

Primary treatment in a conventional wastewater plant mainly removes:

• a.dissolved nitrogen by denitrification
• b.soluble BOD by aeration
• c.pathogens by disinfection
• d.settleable solids by sedimentation✓

Primary treatment is physical sedimentation that removes settleable and floatable solids (and some associated BOD). Soluble BOD and nutrients are addressed in secondary and tertiary (biological/chemical) stages.

A horizontal curve has a degree of curve of 2 degrees (arc definition). What is its radius?

• a.11,459 ft
• b.1,432.4 ft
• c.2,864.79 ft✓
• d.5,730 ft

By the arc definition, R = 5,729.58 / D = 5,729.58 / 2 = 2,864.79 ft. The degree of curve is the central angle subtending a 100-ft arc; radius is inversely proportional to it.

A horizontal curve has a deflection (intersection) angle of 30 degrees and a degree of curve of 5 degrees (arc definition). What is the length of the curve?

• a.150 ft
• b.600 ft✓
• c.6 ft
• d.6,000 ft

For the arc definition, L = 100·(Delta/D) = 100 x (30/5) = 600 ft. The curve length is proportional to the deflection angle and inversely proportional to the degree of curve.

A horizontal curve has a radius of 500 ft and an intersection angle of 40 degrees. What is the tangent length?

• a.364 ft
• b.420 ft
• c.182 ft✓
• d.146 ft

Tangent length T = R·tan(Delta/2) = 500 x tan(20 deg) = 500 x 0.3640 = 182 ft. The tangent runs from the PC (or PT) to the PI.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the external distance?

• a.64.2 ft
• b.32.1 ft✓
• c.100 ft
• d.16.0 ft

External distance E = R[sec(Delta/2) - 1] = 500 x (1/cos 20 deg - 1) = 500 x 0.0642 = 32.1 ft. It is the distance from the PI to the middle of the curve.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the middle ordinate?

• a.60.3 ft
• b.30.2 ft✓
• c.15.1 ft
• d.100 ft

Middle ordinate M = R[1 - cos(Delta/2)] = 500 x (1 - cos 20 deg) = 500 x 0.0603 = 30.2 ft. It is the distance from the midpoint of the long chord to the midpoint of the curve.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the length of the long chord?

• a.684 ft
• b.342 ft✓
• c.383 ft
• d.171 ft

Long chord LC = 2R·sin(Delta/2) = 2 x 500 x sin 20 deg = 1,000 x 0.3420 = 342 ft. It connects the PC to the PT directly.

A curve is designed for 60 mph on a road with a maximum side-friction factor of 0.12 and a radius of 1,500 ft. What superelevation rate is required? (Use e + f = V^2/(15R), V in mph, R in ft.)

• a.0.12
• b.0.28
• c.0.16
• d.0.04✓

e + f = V^2/(15R) = 60^2/(15 x 1,500) = 3,600/22,500 = 0.16. Then e = 0.16 - f = 0.16 - 0.12 = 0.04 (4%). Side friction carries part of the lateral demand; the rest is met by banking.

What is the minimum radius for a 70-mph curve with maximum superelevation e = 0.08 and side-friction factor f = 0.10? (Use R = V^2/[15(e + f)].)

• a.907 ft
• b.2,722 ft
• c.272 ft
• d.1,815 ft✓

R_min = V^2/[15(e + f)] = 70^2/[15(0.08 + 0.10)] = 4,900/2.70 = 1,815 ft. The minimum radius drops as the allowable superelevation and side friction increase.

A driver travels at 60 mph with a perception-reaction time of 2.5 s and a braking friction factor of 0.35 on a level road. What is the stopping sight distance? (Use SSD = 1.47Vt + V^2/[30f].)

• a.784 ft
• b.220 ft
• c.343 ft
• d.563 ft✓

SSD = brake-reaction distance + braking distance = 1.47 x 60 x 2.5 + 60^2/(30 x 0.35) = 220.5 + 342.9 = 563 ft. Using only one component (220 or 343) is the common error.

At 50 mph with a perception-reaction time of 2.5 s, what is the brake-reaction (perception-reaction) distance?

• a.367 ft
• b.184 ft✓
• c.110 ft
• d.73.5 ft

Reaction distance d = 1.47·V·t = 1.47 x 50 x 2.5 = 183.75 ft (the 1.47 converts mph to ft/s). The vehicle travels this far before braking even begins.

A vehicle at 55 mph brakes on a 3% downgrade with a friction factor of 0.35. What is the braking distance? (Use d = V^2/[30(f + G)], G negative downhill.)

• a.275 ft
• b.288 ft
• c.350 ft
• d.315 ft✓

d = V^2/[30(f + G)] = 55^2/[30(0.35 - 0.03)] = 3,025/9.6 = 315 ft. A downgrade reduces the effective deceleration, lengthening the braking distance relative to the level value (288 ft).

A crest vertical curve must provide a stopping sight distance of 500 ft with an algebraic grade change of A = 5%. Assuming S < L, what curve length is required? (Use L = A·S^2/2158.)

• a.579 ft✓
• b.290 ft
• c.158 ft
• d.1,158 ft

For S < L on a crest curve, L = A·S^2/2158 = 5 x 500^2/2158 = 1,250,000/2158 = 579 ft. The 2158 constant embeds the standard driver eye height and object height.

A sag vertical curve must provide 400 ft of headlight sight distance with A = 4%. Assuming S < L, what curve length is required? (Use L = A·S^2/[400 + 3.5S].)

• a.200 ft
• b.178 ft
• c.356 ft✓
• d.711 ft

For a sag curve controlled by headlight distance (S < L), L = A·S^2/(400 + 3.5S) = 4 x 400^2/(400 + 1,400) = 640,000/1,800 = 356 ft. The 400 and 3.5S terms model the headlight height and 1-degree upward beam.

A vertical curve is designed with a rate of vertical curvature K = 158 for an algebraic grade change of A = 5%. What is the required curve length?

• a.395 ft
• b.163 ft
• c.31.6 ft
• d.790 ft✓

L = K·A = 158 x 5 = 790 ft. K is the horizontal distance needed to change the grade by 1 percent, so multiplying by the total grade change gives the length.

A crest vertical curve 500 ft long joins a +3% grade to a -2% grade. How far from the beginning of the curve (BVC) is the high point?

• a.500 ft
• b.250 ft
• c.200 ft
• d.300 ft✓

The turning point is at x = g1·L/(g1 - g2) = 3 x 500/(3 - (-2)) = 1,500/5 = 300 ft from the BVC. The high (or low) point is where the curve's slope reaches zero.

In the fundamental traffic-flow relationship q = k·v, the speed term v refers to:

• a.the space-mean speed✓
• b.the maximum speed
• c.the posted speed limit
• d.the time-mean speed

The flow-density-speed identity q = k·v uses the space-mean speed (harmonic mean over a length of road). The time-mean speed (measured at a point) is always slightly higher and does not satisfy the identity exactly.

A roadway carries a flow of 1,800 veh/hr at a space-mean speed of 45 mph. What is the traffic density?

• a.25 veh/mi
• b.40 veh/mi✓
• c.81,000 veh/mi
• d.400 veh/mi

From q = k·v, density k = q/v = 1,800/45 = 40 veh/mi. Density is vehicles per unit length; multiplying flow by speed (81,000) is the classic mistake.

At a density of 40 veh/mi, what is the average spacing between vehicles?

• a.66 ft
• b.264 ft
• c.40 ft
• d.132 ft✓

Average spacing = 5,280 ft/mi / density = 5,280/40 = 132 ft per vehicle. Spacing (center-to-center distance) is the reciprocal of density expressed in feet.

A lane carries 1,200 veh/hr. What is the average time headway between vehicles?

• a.0.33 s
• b.1,200 s
• c.30 s
• d.3 s✓

Average headway = 3,600 s/hr / flow = 3,600/1,200 = 3 s. Time headway is the reciprocal of flow rate; capacity conditions correspond to the minimum safe headway.

Under the Greenshields linear model with a free-flow speed of 60 mph and a jam density of 200 veh/mi, what is the maximum flow (capacity)?

• a.1,500 veh/hr
• b.6,000 veh/hr
• c.12,000 veh/hr
• d.3,000 veh/hr✓

For Greenshields, q_max = v_f·k_j/4 = 60 x 200/4 = 3,000 veh/hr. Capacity occurs at half the free-flow speed and half the jam density.

Under the Greenshields model with a free-flow speed of 60 mph, at what speed does maximum flow (capacity) occur?

• a.45 mph
• b.30 mph✓
• c.60 mph
• d.15 mph

In the Greenshields model, capacity occurs at the optimum speed v_o = v_f/2 = 60/2 = 30 mph (and density k_j/2). Flow falls off on either side of this point.

At the jam density on a highway, the speed and flow are respectively:

• a.maximum and maximum
• b.both zero✓
• c.zero and maximum
• d.maximum and zero

At jam density vehicles are bumper-to-bumper and stopped, so speed is zero; with zero speed the flow q = k·v is also zero. Maximum flow occurs at an intermediate density, not at jam.

A location has an hourly volume of 1,800 veh and a peak 15-minute count of 500 veh. What is the peak-hour factor?

• a.0.90✓
• b.0.45
• c.1.11
• d.0.72

PHF = hourly volume / (4 x peak 15-min volume) = 1,800/(4 x 500) = 1,800/2,000 = 0.90. A PHF near 1.0 indicates uniform flow; lower values indicate sharper peaking.

An intersection approach has an hourly volume of 1,800 veh and a peak-hour factor of 0.90. What is the equivalent peak flow rate?

• a.500 veh/hr
• b.1,620 veh/hr
• c.1,800 veh/hr
• d.2,000 veh/hr✓

Flow rate = hourly volume / PHF = 1,800/0.90 = 2,000 veh/hr. Dividing by the PHF converts a full-hour count into the flow rate sustained during the peak 15 minutes, which controls capacity analysis.

A highway has an AADT of 20,000 veh/day and a K-factor of 0.10. What is the design hourly volume?

• a.20,000 veh/hr
• b.2,000 veh/hr✓
• c.1,000 veh/hr
• d.200,000 veh/hr

DHV = AADT x K = 20,000 x 0.10 = 2,000 veh/hr. The K-factor is the proportion of daily traffic occurring in the design (typically 30th-highest) hour.