FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions

What is the gauge pressure at a depth of 10 m in water (density 1000 kg/m^3, g = 9.81 m/s^2)?

• a.100 kPa
• b.981 kPa
• c.9.81 kPa
• d.98.1 kPa✓

Hydrostatic pressure p = rho·g·h = 1000 x 9.81 x 10 = 98,100 Pa = 98.1 kPa. Pressure increases linearly with depth and is independent of the container's shape.

Incompressible water flows through a pipe. If the cross-sectional area is reduced to half, what happens to the flow velocity?

• a.It doubles✓
• b.It halves
• c.It quadruples
• d.It stays the same

By the continuity equation A1·V1 = A2·V2, velocity is inversely proportional to area for incompressible flow. Halving the area doubles the velocity so that volumetric flow rate is conserved.

Water flows at 3 m/s through a circular pipe of 0.1 m diameter. What is the volumetric flow rate?

• a.0.03 m^3/s
• b.0.0236 m^3/s✓
• c.0.236 m^3/s
• d.0.0079 m^3/s

Flow rate Q = A·V. The area A = (pi/4)·D^2 = (pi/4)(0.1)^2 = 0.00785 m^2. Then Q = 0.00785 x 3 = 0.0236 m^3/s. Continuity ties cross-sectional area and velocity to discharge.

In pipe flow, flow is generally considered laminar when the Reynolds number is below approximately what value?

• a.4,000
• b.2,100✓
• c.500,000
• d.10

For flow in a circular pipe, laminar flow generally persists below Re of about 2,100; transitional and turbulent flow occur at higher values. The Reynolds number Re = rho·V·D/mu compares inertial to viscous forces.

What is the specific weight of water at standard conditions (density 1000 kg/m^3, g = 9.81 m/s^2)?

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

Specific weight gamma = rho·g = 1000 x 9.81 = 9810 N/m^3 = 9.81 kN/m^3. Specific weight is weight per unit volume, distinct from density (mass per unit volume).

What is the gauge pressure at the bottom of a 5 m deep tank of oil with specific gravity 0.8 (g = 9.81 m/s^2)?

• a.4.9 kPa
• b.39.2 kPa✓
• c.392 kPa
• d.49.1 kPa

Hydrostatic pressure p = SG·rho_water·g·h = 0.8 x 1000 x 9.81 x 5 = 39,240 Pa = 39.2 kPa. Using water (SG = 1) would wrongly give 49.1 kPa; the fluid's specific gravity scales the result.

A manometer shows a mercury column height of 0.2 m (mercury SG = 13.6). What pressure does this represent (g = 9.81 m/s^2)?

• a.26.7 kPa✓
• b.1.96 kPa
• c.2.72 kPa
• d.267 kPa

p = rho·g·h = (13.6 x 1000) x 9.81 x 0.2 = 26,683 Pa = 26.7 kPa. Mercury's high density (SG 13.6) makes it compact for measuring large pressures; using water would give only 1.96 kPa.

Water discharges from a small opening 5 m below the free surface of a large tank. What is the ideal exit velocity (g = 9.81 m/s^2)?

• a.9.9 m/s✓
• b.98.1 m/s
• c.49.1 m/s
• d.19.8 m/s

By Torricelli's theorem (from Bernoulli), v = sqrt(2·g·h) = sqrt(2 x 9.81 x 5) = sqrt(98.1) = 9.9 m/s. Forgetting the square root gives the incorrect 98.1.

Water (density 1000 kg/m^3) flows at 2 m/s through a duct of cross-sectional area 0.01 m^2. What is the mass flow rate?

• a.20 kg/s✓
• b.0.02 kg/s
• c.2 kg/s
• d.200 kg/s

Mass flow rate m_dot = rho·A·V = 1000 x 0.01 x 2 = 20 kg/s. This is the continuity equation expressed in mass terms; leaving out density yields the volumetric rate 0.02 m^3/s instead.

Water (density 1000 kg/m^3, viscosity 1x10^-3 Pa·s) flows at 2 m/s in a 0.05 m diameter pipe. What is the Reynolds number?

• a.100,000✓
• b.100
• c.10,000
• d.1,000

Re = rho·V·D/mu = (1000 x 2 x 0.05)/(1x10^-3) = 100/0.001 = 100,000. This far exceeds ~2100, so the flow is turbulent. Re is the ratio of inertial to viscous forces.

An object of volume 0.02 m^3 is fully submerged in water (density 1000 kg/m^3, g = 9.81 m/s^2). What is the buoyant force on it?

• a.196.2 N✓
• b.1,962 N
• c.19.6 N
• d.98.1 N

By Archimedes' principle, buoyant force F_b = rho·g·V_displaced = 1000 x 9.81 x 0.02 = 196.2 N. It equals the weight of the displaced fluid and is independent of the object's own weight.

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 it (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.88.3 kN✓
• b.44.1 kN
• c.176.6 kN
• d.58.9 kN

Resultant force F = rho·g·h_c·A, where h_c is the depth to the centroid (1.5 m) and A = 2 x 3 = 6 m^2. F = 1000 x 9.81 x 1.5 x 6 = 88,290 N = 88.3 kN.

For a vertical rectangular gate 3 m tall with its top edge at the free surface, at what depth does the resultant hydrostatic force act (center of pressure)?

• a.1.5 m
• b.1.0 m
• c.2.0 m✓
• d.3.0 m

For a surface with its top at the free surface, the center of pressure is at 2/3 of the height = (2/3)(3) = 2.0 m. It lies below the centroid (1.5 m) because pressure increases with depth.

A fluid has dynamic viscosity 1x10^-3 Pa·s and density 1000 kg/m^3. What is its kinematic viscosity?

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

Kinematic viscosity nu = mu/rho = (1x10^-3)/1000 = 1x10^-6 m^2/s. It is the dynamic viscosity normalized by density and appears in the Reynolds number.

In a horizontal pipe of constant elevation, when the fluid speeds up as it enters a narrower section, what happens to the static pressure?

• a.Decreases✓
• b.Increases
• c.Drops to zero
• d.Stays constant

By Bernoulli's equation, along a horizontal streamline p + (1/2)rho·V^2 is constant, so higher velocity means lower static pressure. This inverse pressure-velocity relationship underlies venturi and airfoil behavior.

Water flows at 2 m/s through a 0.1 m diameter pipe 100 m long with a Darcy friction factor of 0.02. What is the head loss due to friction (g = 9.81 m/s^2)?

• a.0.41 m
• b.4.08 m✓
• c.40.8 m
• d.2.04 m

Darcy-Weisbach: h_f = f·(L/D)·(V^2/2g) = 0.02 x (100/0.1) x (2^2/(2 x 9.81)) = 0.02 x 1000 x 0.204 = 4.08 m. Head loss scales with the square of velocity.

What is the hydraulic diameter of a square duct with side length 0.2 m?

• a.0.05 m
• b.0.10 m
• c.0.40 m
• d.0.20 m✓

Hydraulic diameter D_h = 4A/P = 4(0.2^2)/(4 x 0.2) = 4(0.04)/0.8 = 0.20 m. For a square duct D_h equals the side length; it lets non-circular ducts use pipe-flow relations.

A volumetric flow rate of 0.1 m^3/s passes through a pipe of diameter 0.2 m. What is the average flow velocity?

• a.6.37 m/s
• b.1.59 m/s
• c.0.50 m/s
• d.3.18 m/s✓

V = Q/A, with A = (pi/4)D^2 = (pi/4)(0.2)^2 = 0.0314 m^2. V = 0.1/0.0314 = 3.18 m/s. Using diameter instead of area in the denominator is a common error.

A pump delivers water at 0.05 m^3/s against a head of 20 m (rho = 1000 kg/m^3, g = 9.81 m/s^2). What is the ideal hydraulic power required?

• a.9.81 kW✓
• b.4.9 kW
• c.98.1 kW
• d.0.98 kW

Hydraulic power P = rho·g·Q·H = 1000 x 9.81 x 0.05 x 20 = 9810 W = 9.81 kW. Actual shaft power is higher because pump efficiency is below 100%.

A body with drag coefficient 0.4 and frontal area 2 m^2 moves through air (density 1.2 kg/m^3) at 30 m/s. What is the drag force?

• a.216 N
• b.432 N✓
• c.43.2 N
• d.864 N

Drag force F_D = (1/2)·C_d·rho·V^2·A = 0.5 x 0.4 x 1.2 x 30^2 x 2 = 432 N. Drag grows with the square of velocity, so doubling speed quadruples drag.

A liquid has a density of 850 kg/m^3. What is its specific gravity (reference water = 1000 kg/m^3)?

• a.8.5
• b.0.85✓
• c.850
• d.1.18

Specific gravity SG = rho_fluid/rho_water = 850/1000 = 0.85. It is a dimensionless ratio; a value below 1 means the fluid is lighter than water and will float on it.

A pressure gauge reads 150 kPa. If atmospheric pressure is 101.3 kPa, what is the absolute pressure?

• a.150 kPa
• b.251.3 kPa✓
• c.48.7 kPa
• d.101.3 kPa

Absolute pressure = gauge pressure + atmospheric pressure = 150 + 101.3 = 251.3 kPa. Gauge pressure is measured relative to atmosphere; absolute is measured from a perfect vacuum.

In a hydraulic press, a 100 N force is applied to a small piston of area 0.01 m^2. What force is produced on the large piston of area 0.1 m^2?

• a.10,000 N
• b.10 N
• c.1,000 N✓
• d.100 N

By Pascal's principle, pressure is equal throughout: p = 100/0.01 = 10,000 Pa. Output force = p·A_large = 10,000 x 0.1 = 1,000 N. Force multiplies by the area ratio (10x).

Incompressible water flows at 2 m/s in a 0.1 m diameter pipe that narrows to 0.05 m diameter. What is the velocity in the narrow section?

• a.1 m/s
• b.8 m/s✓
• c.16 m/s
• d.4 m/s

Continuity A1·V1 = A2·V2 with area proportional to D^2 gives V2 = V1·(D1/D2)^2 = 2 x (0.1/0.05)^2 = 2 x 4 = 8 m/s. Velocity scales with the square of the diameter ratio, not linearly.

What is the velocity head of water flowing at 10 m/s (g = 9.81 m/s^2)?

• a.1.02 m
• b.51.0 m
• c.5.10 m✓
• d.10.2 m

Velocity head = V^2/(2g) = 10^2/(2 x 9.81) = 100/19.62 = 5.10 m. It is the kinetic-energy term of Bernoulli's equation expressed as an equivalent column height.

What is the pressure head corresponding to a water pressure of 98.1 kPa (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.10 m✓
• b.5 m
• c.100 m
• d.1 m

Pressure head = p/(rho·g) = 98,100/(1000 x 9.81) = 10 m. Pressure head expresses pressure as an equivalent height of the fluid column.

Water (surface tension 0.072 N/m, contact angle ~0) rises in a glass tube of radius 0.001 m. What is the approximate capillary rise (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.7.3 mm
• b.29.4 mm
• c.14.7 mm✓
• d.1.5 mm

Capillary rise h = 2·sigma·cos(theta)/(rho·g·r) = 2 x 0.072/(1000 x 9.81 x 0.001) = 0.144/9.81 = 0.0147 m = 14.7 mm. Rise is inversely proportional to tube radius.

What is the density of air modeled as an ideal gas at 101.3 kPa and 300 K (R = 287 J/kg·K)?

• a.1.00 kg/m^3
• b.2.35 kg/m^3
• c.1.18 kg/m^3✓
• d.0.85 kg/m^3

Ideal gas law rho = p/(R·T) = 101,300/(287 x 300) = 101,300/86,100 = 1.18 kg/m^3. Temperature must be absolute (kelvin).

For fully developed laminar pipe flow at a Reynolds number of 1600, what is the Darcy friction factor?

• a.0.016
• b.0.032
• c.0.080
• d.0.040✓

For laminar flow, f = 64/Re = 64/1600 = 0.040. This exact relation applies only below Re ~2100; turbulent flow requires the Moody chart or Colebrook equation.

A solid body of specific gravity 0.6 floats freely in water. What fraction of its volume is submerged?

• a.60%✓
• b.6%
• c.100%
• d.40%

For a floating body, the submerged fraction equals the ratio of densities: rho_body/rho_fluid = 0.6, so 60% is submerged. This follows from equating weight and buoyant force.

Open-channel flow moves at 3 m/s at a depth of 1 m. What is the Froude number (g = 9.81 m/s^2)?

• a.0.31
• b.0.96✓
• c.9.4
• d.3.0

Froude number Fr = V/sqrt(g·y) = 3/sqrt(9.81 x 1) = 3/3.13 = 0.96. Since Fr < 1 the flow is subcritical. Fr compares inertial to gravitational forces in free-surface flow.

Water (density 1000 kg/m^3) flows at 4 m/s with a static pressure of 100 kPa. What is the stagnation (total) pressure?

• a.100 kPa
• b.116 kPa
• c.108 kPa✓
• d.104 kPa

Stagnation pressure = static + dynamic = p + (1/2)rho·V^2 = 100,000 + 0.5 x 1000 x 4^2 = 100,000 + 8,000 = 108 kPa. The dynamic term is the pressure recovered when flow is brought to rest.

Pipe flow at a Reynolds number of about 3000 is best described as:

• a.Supersonic
• b.Laminar
• c.Turbulent
• d.Transitional✓

Between roughly Re 2100 and 4000, pipe flow is transitional, intermittently switching between laminar and turbulent behavior. Below ~2100 it is laminar; above ~4000 it is fully turbulent.

What height of a water column produces a pressure of 50 kPa (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.0.51 m
• b.5.1 m✓
• c.10.2 m
• d.2.5 m

From p = rho·g·h, h = p/(rho·g) = 50,000/(1000 x 9.81) = 5.1 m. This converts a pressure into an equivalent water-column height, as read on a manometer.

By how much does hydrostatic pressure in water increase per meter of depth (rho = 1000 kg/m^3, g = 9.81 m/s^2)?

• a.0.98 kPa/m
• b.1.0 kPa/m
• c.9.81 kPa/m✓
• d.98.1 kPa/m

The pressure gradient in a static fluid is dp/dz = rho·g = 1000 x 9.81 = 9810 Pa/m = 9.81 kPa/m. Pressure grows linearly with depth regardless of container shape.

Water flows from an orifice 2 m below the free surface of a large open tank. What is the ideal jet velocity (g = 9.81 m/s^2)?

• a.19.6 m/s
• b.39.2 m/s
• c.3.13 m/s
• d.6.26 m/s✓

Torricelli's theorem: v = sqrt(2·g·h) = sqrt(2 x 9.81 x 2) = sqrt(39.24) = 6.26 m/s. It comes from applying Bernoulli's equation between the surface and the orifice.

Pipe flow has a Reynolds number of 1500. The flow regime is:

• a.Choked
• b.Turbulent
• c.Laminar✓
• d.Transitional

Re = 1500 is below the ~2100 threshold, so the flow is laminar, dominated by viscous forces with smooth, orderly streamlines and no cross-mixing.

A Newtonian fluid with dynamic viscosity 1x10^-3 Pa·s experiences a velocity gradient of 100 s^-1 near a wall. What is the shear stress?

• a.0.01 Pa
• b.0.1 Pa✓
• c.10 Pa
• d.1 Pa

Newton's law of viscosity: tau = mu·(du/dy) = 1x10^-3 x 100 = 0.1 Pa. Shear stress is proportional to the velocity gradient, with viscosity as the proportionality constant.

A gauge pressure of -20 kPa indicates that the absolute pressure is:

• a.Above atmospheric
• b.Equal to absolute zero
• c.Below atmospheric✓
• d.Equal to atmospheric

A negative gauge pressure means the absolute pressure is below atmospheric, i.e., a partial vacuum. Absolute pressure = atmospheric + gauge = 101.3 + (-20) = 81.3 kPa, still positive.

For steady incompressible flow through a pipe with a single inlet and outlet, the continuity principle requires that:

• a.Mass in exceeds out
• b.Mass out exceeds in
• c.Mass in equals mass out✓
• d.Flow depends only on pressure

Conservation of mass for steady flow requires mass flow in to equal mass flow out. For incompressible flow this also means the volumetric flow rate Q is constant along the pipe.