Fluid Properties and Statics
Key Takeaways
- Fluid mechanics is the highest-weighted topic on the FE Other Disciplines exam: 12–18 questions (~14%).
- Density ρ = m/V; specific weight γ = ρg; specific gravity SG = ρ/ρ_water = γ/γ_water (dimensionless).
- Dynamic viscosity μ (Pa·s) measures resistance to shear via τ = μ(du/dy); kinematic viscosity ν = μ/ρ (m²/s).
- Hydrostatic pressure rises linearly with depth: p = p₀ + γh = p₀ + ρgh; in water γ ≈ 9,810 N/m³.
- Hydrostatic force on a plane submerged surface is F = γh̄A (h̄ = depth of the centroid); the center of pressure lies below the centroid.
- Buoyancy (Archimedes): an upward force equal to the weight of displaced fluid, F_B = ρ_fluid·g·V_displaced; a floating body submerges a fraction equal to SG.
FE Exam Weight: Fluid Mechanics is the single highest-weighted topic on the FE Other Disciplines exam — 12–18 of 110 questions (~14%). The exam is open to the searchable NCEES FE Reference Handbook, so master which formula applies and keep units consistent.
Fluid Properties
| Property | Symbol | Definition | SI units |
|---|---|---|---|
| Density | ρ | mass per volume | kg/m³ |
| Specific weight | γ | weight per volume = ρg | N/m³ |
| Specific gravity | SG | ρ/ρ_water | dimensionless |
| Dynamic viscosity | μ | resistance to shear | Pa·s (N·s/m²) |
| Kinematic viscosity | ν | μ/ρ | m²/s |
| Surface tension | σ | force per length at a surface | N/m |
| Bulk modulus | E_v | resistance to compression | Pa |
Water at 20°C (memorize)
- ρ ≈ 998 kg/m³ ≈ 1,000 kg/m³
- γ ≈ 9,790 N/m³ ≈ 9,810 N/m³ (use 9.81 kN/m³)
- μ ≈ 1.0 × 10⁻³ Pa·s; ν ≈ 1.0 × 10⁻⁶ m²/s
Specific gravity is the fast bridge: SG = 0.8 means ρ = 800 kg/m³ and γ = 7,848 N/m³.
Viscosity and Newton's Law
Newton's law of viscosity relates shear stress to the velocity gradient:
τ = μ (du/dy)
| Fluid type | Behavior | Examples |
|---|---|---|
| Newtonian | τ ∝ du/dy (constant μ) | water, air, light oils |
| Shear-thinning | μ falls as shear rate rises | ketchup, blood, paint |
| Shear-thickening | μ rises as shear rate rises | cornstarch + water |
| Bingham plastic | needs a yield stress to flow | toothpaste, drilling mud |
Worked example — viscous shear stress
Oil with μ = 0.1 Pa·s fills a 2 mm gap; the top plate moves at 0.5 m/s while the bottom is fixed. The velocity gradient is linear: du/dy = 0.5/0.002 = 250 s⁻¹. So τ = μ(du/dy) = 0.1 × 250 = 25 Pa. This is the drag stress on the moving plate.
Hydrostatic Pressure
In a fluid at rest, pressure increases linearly with depth:
p = p₀ + ρgh = p₀ + γh
Key facts the FE tests:
- Pressure acts equally in all directions at a point (Pascal's law).
- Pressure is the same at every point on a horizontal plane in a connected fluid.
- Gauge pressure = absolute pressure − atmospheric (p_atm ≈ 101.3 kPa).
Worked example — gauge pressure with depth
Gauge pressure 10 m down in water: p = ρgh = 1,000 × 9.81 × 10 = 98,100 Pa = 98.1 kPa — almost exactly 1 atmosphere. So every ~10 m of water adds one atmosphere of gauge pressure.
Manometers
For a U-tube manometer, walk from one known end to the other, adding γh going down and subtracting γh going up:
p_A + γ₁h₁ − γ₂h₂ − γ₃h₃ = p_B
The manometer fluid (often mercury, SG = 13.6) amplifies small pressure differences into readable height changes.
Hydrostatic Force on a Plane Surface
The resultant force on a submerged flat plate is:
F = γ h̄ A
where h̄ is the vertical depth of the centroid of the surface and A is its area. The force acts at the center of pressure, located below the centroid:
y_cp = ȳ + I_x̄ / (ȳ A)
The extra term I_x̄/(ȳA) pushes the resultant deeper than the centroid because pressure grows with depth — the lower part of the plate feels more pressure.
Worked example — force on a gate
A rectangular gate 2 m wide × 3 m tall has its top edge 1 m below the surface. Centroid depth h̄ = 1 + 3/2 = 2.5 m; area A = 2 × 3 = 6 m².
F = γh̄A = 9,810 × 2.5 × 6 = 147,150 N ≈ 147 kN.
Curved Surfaces
Resolve into components: the horizontal force equals γh̄A on the surface's vertical projection; the vertical force equals the weight of fluid (real or virtual) directly above the curved surface.
Buoyancy — Archimedes' Principle
A fully or partly submerged body feels an upward buoyant force equal to the weight of fluid it displaces:
F_B = ρ_fluid · g · V_displaced = γ_fluid · V_displaced
Floating bodies
A floating object displaces exactly its own weight of fluid, so:
fraction submerged = ρ_object / ρ_fluid = SG (in water)
An iceberg (SG ≈ 0.92) floats with ~92% of its volume underwater; an SG = 0.8 block floats 80% submerged. If ρ_object > ρ_fluid the body sinks; if equal it is neutrally buoyant.
Worked example — apparent weight
A 0.01 m³ steel block (ρ = 7,850 kg/m³) is fully submerged in water. Its weight in air = ρgV = 7,850 × 9.81 × 0.01 = 770 N. Buoyant force = 1,000 × 9.81 × 0.01 = 98.1 N. Apparent (submerged) weight = 770 − 98.1 = 672 N.
Stability of floating bodies
A floating body is stable when its metacenter M lies above the center of gravity G. When it tilts, the center of buoyancy shifts, creating a righting moment if M is above G. A low, wide hull (ballast low) is stable; a top-heavy one capsizes.
Common Traps
- γ vs. ρ. Use ρgh when you have density; γh when you have specific weight. Don't multiply by g twice.
- h̄ is the centroid depth, not the depth of the top edge.
- SI consistency. kPa = kN/m²; keep pressures, areas, and forces in compatible units.
- Gauge vs. absolute. Manometers and gate problems use gauge pressure unless atmospheric acts on only one side.
Quick reference
- γ = ρg; SG = ρ/ρ_water; ν = μ/ρ.
- p = p₀ + γh; F = γh̄A on a plane surface.
- F_B = γ_fluid·V_displaced; floating fraction submerged = SG.
- Water: ρ ≈ 1,000 kg/m³, γ ≈ 9,810 N/m³.
What is the gauge pressure at a depth of 10 m in water (ρ = 1,000 kg/m³)?
A rectangular gate 2 m wide and 3 m tall has its top edge 1 m below the water surface. What is the total hydrostatic force on the gate?
An object with specific gravity 0.8 floats in water. What fraction of its volume is submerged?
Oil (μ = 0.1 Pa·s) fills a 2 mm gap between two plates; the top plate moves at 0.5 m/s while the bottom is fixed. The shear stress on the plate is: