Fluid Statics and Dynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Fluid statics describes fluids at rest; the key relation is the hydrostatic pressure equation P = ρ g h, where ρ is fluid density (kg/m³), g is gravitational acceleration (≈9.81 m/s²), and h is depth below the surface (m). A submerged body experiences an upward buoyant force F = ρ g V, where V is the displaced volume (Archimedes’ principle).

Fluid dynamics describes fluids in motion under three governing rules:

Continuity (mass conservation): A₁ v₁ = A₂ v₂
Flow rate: Q = A v
Bernoulli (energy conservation along a streamline): P₁ + ½ ρ v₁² + ρ g h₁ = P₂ + ½ ρ v₂² + ρ g h₂

High-yield ECAT pointers:

Distinguish absolute pressure (P_abs = P_gauge + P_atm) from gauge pressure.
In a Venturi tube, the narrower section has higher speed and lower static pressure.
Bernoulli’s equation only applies to steady, incompressible, non-viscous (inviscid) flow along a single streamline.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Hydrostatic Pressure and Pascal’s Principle

In a static fluid, pressure increases linearly with depth because each layer must support the weight of the fluid column above it: P(h) = P₀ + ρ g h, where P₀ is the pressure at the reference surface (often atmospheric, ≈101.3 kPa). Because pressure at a given depth acts equally in all directions, a force applied to an enclosed incompressible fluid transmits undiminished throughout the fluid — this is Pascal’s principle, the operating basis of hydraulic lifts and brakes. A small force F₁ on piston area A₁ produces F₂ = (A₂/A₁) F₁ at the output piston.

Buoyancy (Archimedes’ Principle)

Any body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced: F_b = ρ_fluid · g · V_displaced. The net force determines whether the body floats (F_b > mg), is neutrally buoyant (F_b = mg), or sinks (F_b < mg). An object floats when its average density is less than the fluid’s density; this is why steel ships float despite steel being denser than water.

Continuity Equation

For a steady flow of an incompressible fluid, the mass entering a pipe per second must equal the mass leaving. With constant density, the continuity equation reduces to A₁ v₁ = A₂ v₂ = Q, where Q is the volumetric flow rate (m³/s). Doubling the cross-section halves the flow speed.

Bernoulli’s Equation

Along a streamline for steady, incompressible, inviscid flow, mechanical-energy density is conserved:

P + ½ ρ v² + ρ g h = constant.

This combines pressure energy (P), kinetic energy density (½ ρ v²), and gravitational potential energy density (ρ g h). The Venturi effect follows directly: when cross-section narrows, v rises, so P must drop to keep the sum constant — used in carburettors, atomisers, and air-flow sensors.

Quantity	Symbol	Typical Unit
Pressure	P	Pa (N/m²)
Density	ρ	kg/m³
Flow speed	v	m/s
Depth/height	h	m
Flow rate	Q	m³/s

Exam Pattern in ECAT

Expect 3–5 short conceptual or computational questions (≈3% weight). Typical formats: (a) compute pressure at a given depth given density; (b) apply the continuity equation to a pipe with two cross-sections; (c) use Bernoulli to find speed or pressure difference in a Venturi setup; (d) verify floating condition for a body of given mass and volume.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Limits of the Equations

Bernoulli’s equation fails for viscous flow, for turbulent flow (where energy dissipates into heat), and across shocks or sudden expansions where streamlines break. It also assumes a single fluid of constant ρ — across an interface (e.g., water–air), apply it on each side separately. For real pipe flow, the Hagen–Poiseuille law replaces Bernoulli’s idealisation: ΔP = (8 μ L Q)/(π r⁴), where μ is dynamic viscosity (Pa·s) — pressure drop scales with the fourth power of pipe radius, explaining why arterial constrictions cause large pressure changes.

Surface Tension and Capillarity

At fluid interfaces, surface tension γ (N/m) arises from cohesive forces. Capillary rise in a tube of radius r is h = 2 γ cos θ / (ρ g r), where θ is the contact angle. Water (θ ≈ 0°) rises in glass; mercury (θ ≈ 140°) is depressed. ECAT rarely tests this numerically but may ask the qualitative direction.

Worked Micro-Example

Water (ρ = 1000 kg/m³) flows through a horizontal pipe that narrows from A₁ = 0.04 m² to A₂ = 0.01 m². The pressure in the wider section is 150 kPa. Find the pressure in the narrower section when v₁ = 2 m/s.

Continuity: v₂ = (A₁/A₂) v₁ = (0.04/0.01)(2) = 8 m/s.
Bernoulli (horizontal, h₁ = h₂):

P₂ = P₁ + ½ ρ (v₁² − v₂²) = 150 000 + ½ (1000)(4 − 64) = 150 000 − 30 000 = 120 kPa.

The narrower section shows lower static pressure despite higher speed — the classic Venturi signature.

Common Mistakes ECAT Exploits

Treating gauge and absolute pressures interchangeably in tank/pipe problems — always read the stem for which one P = ρ g h gives.
Assuming Bernoulli applies in viscous pipes (it does not; use Poiseuille if asked).
Using A v = constant where the fluid is compressible (e.g., gas above Mach 0.3) — continuity in mass form A ρ v = constant is required.
Forgetting that buoyancy depends on displaced fluid, not on the body’s own density alone — a hollow object’s average density matters.

Exam Strategy for 3% Weight Topics

ECAT questions on fluids are short and formula-driven. Memorise the four governing equations and the conditions under which each holds. In MCQs, the discriminator is usually a sign error or a unit slip (Pa vs kPa). Allocate ~45 seconds per question on fluid topics.

Practice Prompts

A 2 cm cube of aluminium (ρ = 2700 kg/m³) is fully submerged in water. Find the buoyant force and the apparent weight.
A pipe of diameter 10 cm carries water at 0.5 m/s and branches into four identical smaller pipes. Assuming incompressible steady flow, find the speed in each branch.

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Fluid Statics and Dynamics