NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersFluids at rest and in motion, pipe flow and open channels: the base of irrigation, drainage, pumping and conservation structures.
In the exams: GATE AG Section 4 (hydraulics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit III; ICAR AICE subject 43 Unit 2; NCEES FE 12–18 questions (largest FE area)
NCEES FE Other Disciplines
Fluid mechanics is the heaviest area (12–18 questions): properties, statics and manometers, buoyancy, continuity, energy and momentum equations, pipe friction with the Moody diagram, flow measurement and similitude, often in US customary units, answered with the NCEES Reference Handbook.
How our questions for NCEES FE Other Disciplines spread across the units of Fluid Mechanics & Open Channel Hydraulics. Difficulty: 13 easy, 27 medium, 8 hard.
| Hydrostatics and manometry | 6 | |
| Forces on submerged surfaces and buoyancy | 5 | |
| Bernoulli's equation: venturimeter, orifice meter and pitot tube | 5 | |
| Properties of fluids | 4 | |
| Minor losses | 4 | |
| Continuity equation | 3 | |
| Laminar and turbulent flow in pipes | 3 | |
| Darcy-Weisbach equation and Moody diagram | 3 | |
| Dimensional analysis and similitude | 3 | |
| Momentum equation | 2 | |
| Pipes in series and parallel | 2 | |
| Orifices, mouthpieces and tank emptying | 2 | |
| Uniform flow in open channels: Manning and Chezy | 2 | |
| Specific energy and critical depth | 2 | |
| Hazen-Williams equation | 1 | |
| Weirs and notches | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A U-tube mercury manometer (relative density 13.6) is connected to a pipe carrying water; the other limb is open to air. The pipe centre is 0.2 m above the mercury surface in the pipe-side limb, and the mercury in the open limb stands 0.3 m above that surface. What is the gauge pressure at the pipe centre?
C. 38.1 kPa
Start at the pipe centre and move to the open surface: p_A + ρ_w g(0.2) − ρ_Hg g(0.3) = 0, so p_A = 13 600 × 9.81 × 0.3 − 1000 × 9.81 × 0.2 = 40025 − 1962 = 38063 Pa ≈ 38.1 kPa. Adding the water column gives 42.0 kPa; ignoring it gives 40.0 kPa. 35.1 kPa counts the whole 0.5 m (0.2 + 0.3 m) as a water column on the pipe side.
Further reading: Munson et al., Fundamentals of Fluid Mechanics
A vertical rectangular gate 1.5 m wide and 2 m high has its top edge 1 m below the water surface. At what depth below the water surface does the resultant hydrostatic force act?
A. 2.17 m
h_cp = h̄ + I_G/(A h̄). h̄ = 2 m, A = 3 m², I_G = bd³/12 = 1.5 × 8/12 = 1.00 m⁴. h_cp = 2 + 1/(3 × 2) = 2.17 m. The centre of pressure is always below the centroid (2.00 m); forgetting to divide by h̄ gives 2.33 m. 1.83 m subtracts I_G/(A h̄) and so puts the centre of pressure above the centroid.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A horizontal venturimeter 150 mm × 75 mm carries water. A mercury-water differential manometer (mercury relative density 13.6) connected between inlet and throat reads 200 mm. With a discharge coefficient of 0.97, what is the discharge?
A. 31.1 L/s
h = 0.2 × (13.6 − 1) = 2.52 m of water. a₁ = 0.0177 m², a₂ = 0.00442 m². Q = C_d a₁a₂√(2gh)/√(a₁² − a₂²) = 0.97 × 0.0177 × 0.00442 × √(2 × 9.81 × 2.52)/√(0.0177² − 0.00442²) = 0.0311 m³/s = 31.1 L/s. Using 13.6x for h gives 32.3 L/s; omitting C_d gives 32.1 L/s. 8.77 L/s takes the 0.2 m mercury reading as the water head.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Water issues from a small orifice 5 m below the free surface of a large open tank. Neglecting losses, what is the jet velocity?
D. 9.90 m/s
Torricelli's theorem: V = √(2gH) = √(2 × 9.81 × 5) = 9.90 m/s. 7.00 m/s drops the 2, and 98.1 forgets the square root. 49.1 m/s drops both the 2 and the square root (9.81 × 5).
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Oil with kinematic viscosity 1 × 10⁻⁴ m²/s flows at 1.5 m/s in a 100 mm pipe. What is the Reynolds number and the flow regime?
C. 1500, laminar
Re = VD/ν = 1.5 × 0.1/10⁻⁴ = 1500, below about 2000, so laminar. Using water's ν (10⁻⁶ m²/s) gives 150000; using the radius gives 750.0.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Oil (density 900 kg/m³, dynamic viscosity 0.2 Pa·s) flows at a mean velocity of 1 m/s through 100 m of 100 mm pipe. What is the friction head loss?
D. 7.25 m of oil
Re = ρVD/μ = 900 × 1 × 0.1/0.2 = 450, laminar. Hagen-Poiseuille: h_f = 32μVL/(ρgD²) = 32 × 0.2 × 1 × 100/(900 × 9.81 × 0.01) = 7.25 m. The same comes from f = 64/Re = 0.142 in Darcy-Weisbach. Assuming a turbulent f = 0.02 gives 1.02 m, far too low. 3.62 m uses 16 instead of 32 in Hagen-Poiseuille; 14.5 m uses 64.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Use the Swamee-Jain equation to estimate the Darcy friction factor for turbulent flow at Re = 2 × 10⁵ in a pipe of relative roughness ε/D = 0.001.
A. 0.0212
f = 0.25/[log₁₀(ε/3.7D + 5.74/Re^0.9)]² = 0.25/[log₁₀(0.000270 + 0.0000973)]² = 0.0212. The Blasius smooth-pipe value, 0.316/Re^0.25 = 0.0149, ignores roughness; 64/Re applies only to laminar flow. 0.000320 is 64/Re, valid only for laminar flow; 0.00530 is the Fanning value (one quarter of Darcy).
Further reading: White, Fluid Mechanics
For pipes connected in series, which quantity is the same in every pipe?
D. The discharge
In series the same flow passes through each pipe in turn and the losses add. In parallel the head loss is the same in each branch and the flows add. Velocity and Re change with diameter.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A vertical cylindrical tank 2 m in diameter drains through a 50 mm orifice in its base (C_d = 0.6). How long does the water level take to fall from 3 m to 1 m above the orifice?
A. 14.7 min
T = 2A(√H₁ − √H₂)/(C_d a √2g). A = π × 2²/4 = 3.14 m², a = 0.00196 m². T = 2 × 3.14 × (√3 − 1)/(0.6 × 0.00196 × 4.429) = 881 s = 14.7 min. Assuming a constant outflow at the initial head gives 11.6 min (too short); 34.8 min is the time to empty completely; omitting the factor 2 halves the answer. That halved value is 7.35 min.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A 90° V-notch with C_d = 0.6 is used to measure a tubewell's discharge. The head over the vertex is 0.2 m. What is the discharge?
D. 25.4 L/s
Q = (8/15) C_d √(2g) tan(θ/2) H^2.5 = (8/15) × 0.6 × 4.429 × tan 45° × 0.2^2.5 = 1.417 × 0.0179 = 0.0254 m³/s = 25.4 L/s. Using H^1.5 gives 127 L/s, a large overestimate. 56.7 L/s uses H² instead of H^2.5; 31.7 L/s uses the rectangular 2/3 instead of 8/15.
Further reading: USBR Water Measurement Manual; Modi and Seth
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