Fluid Mechanics is 6 to 9 questions on the 110-question FE Civil exam, and it is the area where a small setup error costs the most. A manometer sign, a centroid depth measured to the wrong surface, or a velocity head dropped from the energy equation all produce answers that appear in the option list.
These free practice problems follow the same pattern as the exam: a short scenario, a figure, and four numeric choices. Solutions show the control volume or datum chosen, the equation as it appears in the handbook, and the unit conversion step, because most fluids errors on the FE are unit errors in disguise.
Exam weight: NCEES lists Fluid Mechanics at 6-9 questions of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Fluid Mechanics
NCEES splits Fluid Mechanics into four areas across its 6 to 9 questions: fluid properties, fluid statics, energy and impulse-momentum, and flow measurement. Statics questions ask for hydrostatic force on a gate or dam wall and the location of its center of pressure, manometer pressure differences, and buoyancy or stability of a floating body. These are geometry problems as much as fluids problems.
The energy side runs from continuity and the Bernoulli equation to the full energy equation with pump head, turbine head, and friction loss from Darcy-Weisbach and the Moody diagram or Hazen-Williams. Momentum questions want the force on a pipe bend or a jet striking a vane. Flow measurement covers venturi meters, orifices, pitot tubes, and weirs, each with its own discharge coefficient.
5 Free Fluid Mechanics Practice Problems
Each fluid mechanics problem below comes from the FEExamClick FE Civil question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — A. Flow measurement
A pitot-static probe in air measures a stagnation-static pressure difference of \( 50\,\mathrm{Pa} \). Take air density as \( 1.20\,\mathrm{kg/m}^{3} \). What is the flow speed?
Answer: B) 9.1 m/s
Step 1 – Identify given information:
\( \Delta p = 50 \text{ Pa} \) (stagnation minus static), \( \rho_{\text{air}} = 1.20 \text{ kg/m}^3 \).
Step 2 – Velocity from pitot tube: From the FE Handbook (Bernoulli's equation):
$$v = \sqrt{\frac{2\Delta p}{\rho}}$$
Step 3 – Substitute and solve:
$$v = \sqrt{\frac{2 \times 50}{1.20}} = \sqrt{83.33} = 9.13 \text{ m/s}$$
Step 4 – Answer: The correct answer is (B) 9.1 m/s. This formula applies to incompressible flow (low Mach number). For high-speed compressible flows, corrections are needed.
Problem 2 — B. Fluid properties
A liquid has specific weight \( \gamma = 12.0\,\mathrm{kN/m}^{3} \). What is its density \( \rho \) (kg/m\(^{3}\))? Use \( g = 9.81\,\mathrm{m/s}^{2} \).
Answer: C) 1220
Step 1 – Identify given information:
\( \gamma = 12.0 \text{ kN/m}^3 = 12{,}000 \text{ N/m}^3 \), \( g = 9.81 \text{ m/s}^2 \)
Step 2 – Density from specific weight: From the FE Handbook:
$$\gamma = \rho g \quad \Rightarrow \quad \rho = \frac{\gamma}{g}$$
Step 3 – Substitute and solve:
$$\rho = \frac{12{,}000}{9.81} = 1{,}223 \approx 1{,}220 \text{ kg/m}^3$$
Step 4 – Answer: The correct answer is (C) 1220 kg/m\(^{3}\). Since \( \rho > 1000 \) kg/m\(^{3}\), this liquid is denser than water (SG > 1). Examples include glycerin and some brines.
Problem 3 — C. Fluid statics
A steel block (\( \rho = 7850\,\mathrm{kg/m}^{3} \)) has volume \( 0.010\,\mathrm{m}^{3} \) and is fully submerged in water. What is the apparent weight (tension) needed to hold it at rest?
Answer: C) 672 N
Step 1 – Weight of steel block:
$$W = \rho_s g V = 7{,}850 \times 9.81 \times 0.01 = 770 \text{ N}$$
Step 2 – Buoyancy force:
$$F_B = \rho_w g V = 1000 \times 9.81 \times 0.01 = 98.1 \text{ N}$$
Step 3 – Cable tension:
$$T = W - F_B = 770 - 98.1 = 672 \text{ N}$$
Correct answer: (C)
Problem 4 — D. Energy, impulse, and momentum of fluids
Using pump similarity \( P \propto \rho N^3 D^5 \), a model pump has \( D_{model} = 0.5 D_{prototype} \) and is run at \( N_{model} = 2 N_{prototype} \) (to match head). What is the power ratio \( P_{model}/P_{prototype} \) (same fluid)?
Answer: A) 0.25
Step 1 – Identify given information:
\( D_m = 0.5 D_p \), \( N_m = 2 N_p \), same fluid (\( \rho_m = \rho_p \)). Power scales as \( P \propto \rho N^3 D^5 \).
Step 2 – Pump power scaling law: From the FE Handbook:
$$\frac{P_m}{P_p} = \left(\frac{\rho_m}{\rho_p}\right)\left(\frac{N_m}{N_p}\right)^3 \left(\frac{D_m}{D_p}\right)^5$$
Step 3 – Substitute and solve:
$$\frac{P_m}{P_p} = (1)(2)^3(0.5)^5 = 8 \times \frac{1}{32} = 0.25$$
Step 4 – Answer: The correct answer is (A) 0.25. Despite running at double speed, the much smaller diameter (raised to the 5th power) dominates, resulting in only 25% of prototype power.
One of the distractors is what you get by carrying the flow relation's D³ into the power law, or by scaling head as N² with no diameter term at all, which under-weights the diameter.
Problem 5 — A. Flow measurement
A venturi meter in a horizontal water line has \( D_1 = 0.25\,\mathrm{m} \), \( D_2 = 0.15\,\mathrm{m} \), and \( C_d = 0.98 \). If the flow rate is \( Q = 0.080\,\mathrm{m^3/s} \), what pressure drop \( \Delta p \) between inlet and throat is required?
Answer: D) 9.3 kPa
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| \( D_1 \) | \( 0.25 \text{ m} \) |
| \( D_2 \) | \( 0.15 \text{ m} \) |
| \( C_d \) | \( 0.98 \) |
| \( Q \) | \( 0.080 \text{ m}^3\text{/s} \) |
Step 2 – Venturi pressure drop formula: From the FE Handbook, solving the venturi equation for \( \Delta p \):
$$\Delta p = \frac{\rho}{2}\left(\frac{Q}{C_d A_2}\right)^2\left(1 - \beta^4\right)$$
where \( \beta = D_2/D_1 \).
Step 3 – Substitute and solve:
\( \beta = 0.15/0.25 = 0.6 \), \( \beta^4 = 0.1296 \)
$$A_2 = \frac{\pi(0.15)^2}{4} = 0.01767 \text{ m}^2$$
$$\frac{Q}{C_d A_2} = \frac{0.080}{0.98 \times 0.01767} = 4.623 \text{ m/s}$$
$$\Delta p = \frac{1000}{2}(4.623)^2(1 - 0.1296) = 500 \times 21.37 \times 0.8704 = 9{,}298 \text{ Pa} \approx 9.3 \text{ kPa}$$
Step 4 – Answer: The correct answer is (D) 9.3 kPa.
Option (C) is what you get by squaring the diameter ratio once instead of twice, using 1 − β² where the area-ratio term is 1 − β⁴.
Using the FE Reference Handbook for Fluid Mechanics
The Fluid Mechanics section is dense, so navigate by figure rather than by search term. The hydrostatic force diagram with the center of pressure offset, the Moody diagram, and the table of loss coefficients for fittings are the three pages you will open most. Note that open-channel flow and pipe networks sit in the Water Resources section, not here.
Four Mistakes That Cost Points
- Using the centroid depth where the center of pressure belongs Resultant hydrostatic force uses the depth to the centroid of the surface, but its line of action acts lower, at the center of pressure offset by I divided by y-bar times A. Moment questions about a gate hinge fail whenever the force is applied at the centroid.
- Applying Bernoulli where energy is lost Bernoulli assumes no friction, no pump, and no turbine. The moment a problem mentions pipe length, roughness, fittings, or a pump, you need the full energy equation with head loss terms. Sudden expansions and exits also remove energy that Bernoulli alone will not account for.
- Mixing gauge and absolute pressure Manometer and hydrostatic problems are almost always gauge pressure, but property questions involving gas density, vapor pressure, or cavitation need absolute. Adding or omitting one atmosphere changes the answer by a factor the distractors are built around, so decide which reference you are on before writing an equation.
- Forgetting that momentum is a vector Force on a pipe bend requires x and y components summed separately, then combined. Reversing the sign on the outlet velocity is the usual error, because the momentum flux leaves the control volume while the inlet enters it. Sketch the control volume and label positive directions before substituting numbers.
Frequently Asked Questions
How many fluid mechanics questions are on the FE Civil exam?
NCEES assigns 6 to 9 of the 110 questions to Fluid Mechanics. Water Resources and Environmental Engineering is a separate and larger area, but it leans on the same energy equation and friction loss tools, so time spent here pays off twice. Treat fluids as foundation work rather than as an isolated five-question block.
Do I have to read the Moody diagram on screen?
Yes, and it is worth practicing. On the computer-based exam you read the friction factor off the on-screen chart, which is harder than using a printed copy. Compute the Reynolds number and relative roughness first, then locate the curve. For fully turbulent flow at high roughness the factor flattens out and precise reading matters less.
What units cause the most trouble in FE fluids problems?
Pressure and density in US customary units. Pounds per square inch versus pounds per square foot costs a factor of 144, and slugs versus pounds-mass costs a factor of 32.2. Write every quantity in a consistent set before substituting, and check that your final head comes out in feet and your force in pounds.
Are the fluid mechanics questions conceptual or numerical?
Mostly numerical, with a few conceptual items on fluid properties such as viscosity, surface tension, and the difference between laminar and turbulent behavior. The numerical ones are short by design, since the whole exam averages under three minutes per question. If a fluids problem is taking five minutes, you have probably picked the wrong equation.
Keep Going
Done with fluid mechanics? Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline. Or test yourself against the clock with the free FE Civil practice exam: 20 timed questions in the CBT format.