FE Civil · Chapter 9 · 4–6 exam questions

FE Civil Fluid Mechanics

This chapter covers fluid properties, hydrostatic pressure and forces, Bernoulli's equation, pipe flow and head loss, the momentum equation, flow measurement devices, and dimensional analysis — the core fluid mechanics topics tested on the FE Civil exam.

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What the FE tests in Fluid Mechanics

Fluid Properties & Statics

As a civil engineer, you need fluid properties to size every hydraulic system. Density and specific weight let you compute hydrostatic forces on dams and retaining walls. Viscosity determines whether flow is laminar or turbulent in water mains. Manometers and pressure calculations are how you verify pressures in the field and on design drawings. Buoyancy governs whether a buried tank floats out of saturated ground.

Fluid Dynamics

As a civil engineer, continuity and Bernoulli's equation are the first tools you reach for in any pipe or channel analysis. The Darcy-Weisbach equation gives you friction head loss for pipe sizing, and the momentum equation lets you compute forces at pipe bends and nozzles — which is how you design thrust blocks and restrained joints for water mains.

Flow Analysis & Measurement

As a civil engineer, you measure flow to monitor treatment processes, calibrate hydraulic models, and verify stormwater controls. Pitot tubes, venturi meters, and orifice plates each exploit Bernoulli's principle to give you velocity or flow rate. Dimensional analysis and similitude let you build scale models of spillways and bridge decks that actually predict full-scale behavior.

Key Fluid Mechanics formulas

$\gamma = \rho g$
Specific WeightFE Handbook p. 176
$\tau = \mu \frac{dv}{dy}$
Newton's Law of ViscosityFE Handbook p. 176
$P_2 - P_1 = \gamma h$
Hydrostatic PressureFE Handbook p. 177
$F_R = \gamma h_C A$
Hydrostatic ForceFE Handbook p. 179
$A_1 v_1 = A_2 v_2$
ContinuityFE Handbook p. 180
$\frac{P_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{v_2^2}{2g} + z_2$
Bernoulli's EquationFE Handbook p. 180
$Re = \frac{\rho v D}{\mu} = \frac{vD}{\nu}$
Reynolds NumberFE Handbook p. 181
$h_f = f\frac{L}{D}\frac{v^2}{2g}$
Darcy-WeisbachFE Handbook p. 182
$\Sigma F = \rho Q(v_2 - v_1)$
Momentum EquationFE Handbook p. 186

Sample Fluid Mechanics problems

Q1. An oil has a density of $\rho = 870\,\text{kg/m}^3$ . What is its specific weight? Use $g = 9.81\,\text{m/s}^2$ .

Answer: $8{,}535\,\text{N/m}^3$

Explain it simply

Specific weight is density times gravitational acceleration: $\gamma = \rho \times g = 870 \times 9.81 = 8{,}534.7$ N/m³. Choice A confuses density with specific weight (forgot to multiply by $g$ ). Choice C divides by $g$ instead of multiplying. Choice D has the right number but wrong units — specific weight is force per volume (N/m³), not mass per volume.

Q2. A plate slides over a $0.5\,\text{mm}$ oil film at $1.2\,\text{m/s}$ . The dynamic viscosity of the oil is $\mu = 0.04\,\text{Pa}\cdot\text{s}$ . Assuming a linear velocity profile, what is the shear stress on the plate?

Answer: $96\,\text{Pa}$

Explain it simply

For a linear profile, the velocity gradient is $dv/dy = v/\delta$ . Convert the gap: $0.5$ mm $= 0.0005$ m. Then $dv/dy = 1.2/0.0005 = 2{,}400$ s⁻¹. Shear stress: $\tau = \mu \times dv/dy = 0.04 \times 2{,}400 = 96$ Pa. Choice B forgets to convert mm to m (uses $\delta = 0.5$ m). Choice A uses $\delta = 0.005$ m. Choice D uses $\delta = 0.00005$ m.

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Common Fluid Mechanics mistakes on the FE

Forgetting to convert units — pressure in kPa, velocity in m/s, diameter in m (not mm or cm).
Bernoulli only applies along a streamline for steady, incompressible, inviscid flow — adding friction requires the energy equation.
Using the wrong friction factor — Moody diagram gives Darcy f, not Fanning f (which is 4× smaller).
The resultant hydrostatic force uses the pressure at the centroid, but it acts at the center of pressure — they are NOT the same point.
Re < 2,100 = laminar, Re > 10,000 = fully turbulent — the transition zone is 2,100–10,000.
For Froude similitude, velocity scales as the square root of the length ratio — not the linear ratio (that’s Reynolds).

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