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

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.