Fluid dynamics applications

Fluid dynamics applications examine how moving fluids behave in real-world systems, applying principles of continuity, Bernoulli's equation, and viscosity.

Key Definitions:

Continuity Equation describes mass conservation in fluid flow: A₁v₁ = A₂v₂ where A is cross-sectional area and v is fluid velocity. For incompressible fluids, this ensures constant volume flow rate (Q = Av).

Bernoulli's Equation represents energy conservation along a streamline: P + ½ρv² + ρgh = constant. Here P is pressure, ρ is fluid density, v is velocity, g is gravitational field strength, and h is height. Each term represents pressure energy, kinetic energy per unit volume, and gravitational potential energy per unit volume respectively.

Viscosity (η) measures a fluid's internal resistance to flow. Poiseuille's Law governs laminar flow through pipes: Q = πr⁴ΔP/8ηL where r is radius, ΔP is pressure difference, and L is pipe length.

Stokes' Law describes drag force on spherical objects: F = 6πηrv where r is sphere radius and v is velocity.

Reynolds Number (Re) predicts flow patterns: Re = ρvL/η where L is characteristic length. Re < 2000 indicates laminar flow; Re > 4000 suggests turbulent flow.

Cambridge Note: Always state assumptions (incompressible fluid, steady flow, no friction) when applying these equations.

Applications include Venturi meters (flow measurement), airfoil lift (Bernoulli effect), blood circulation (viscous flow), and sedimentation (Stokes' Law).

Understanding fluid dynamics through practical applications makes abstract concepts concrete.

Venturi Meter Application: Consider water flowing through a pipe that narrows (like a garden hose with your thumb partially covering it). By continuity, velocity increases in the narrow section. Bernoulli's principle reveals that faster-moving fluid exerts lower pressure. Engineers exploit this in carburettors where fast-moving air creates low pressure, drawing fuel into the airstream. Think of it as fluid being "too busy speeding up to push sideways as hard."

Aircraft Wing Lift: Airfoil shapes force air traveling over the curved top surface to move faster than air beneath (longer path in same time). This velocity difference creates pressure difference - lower pressure above, higher below - generating upward lift force. Modern Formula 1 cars use inverted airfoils to create downforce, pressing cars onto tracks at high speeds.

Blood Flow in Arteries: Your cardiovascular system demonstrates Poiseuille's Law beautifully. When arteries narrow (atherosclerosis), the r⁴ relationship means flow rate decreases dramatically - halving radius reduces flow by 16 times! Your heart must work exponentially harder to maintain circulation. This explains why doctors monitor arterial diameter closely.

Terminal Velocity in Fluids: Raindrops, sediment particles, and parachutists all reach terminal velocity when drag force (Stokes' Law) balances gravitational force. Small spherical particles in liquids demonstrate this perfectly - cream rising in milk or cells settling in centrifuges follow predictable patterns based on size and density differences.

Memory Aid: "VCAP" - Venturi (pressure drops), Continuity (area affects speed), Airfoils (lift from speed), Poiseuille (resistance from radius).

Example 1: Venturi Meter

Water flows through a horizontal pipe of diameter 6.0 cm that narrows to 2.0 cm. If velocity in the wide section is 2.0 m/s and pressure is 150 kPa, calculate pressure in the narrow section. (ρ_water = 1000 kg/m³)

Solution:

Step 1: Apply continuity equation

• A₁v₁ = A₂v₂
• π(0.03)²(2.0) = π(0.01)²v₂
• v₂ = 2.0 × (0.03/0.01)² = 18 m/s

Step 2: Apply Bernoulli's equation (horizontal pipe: h₁ = h₂)

• P₁ + ½ρv₁² = P₂ + ½ρv₂²
• 150,000 + ½(1000)(2.0)² = P₂ + ½(1000)(18)²
• 150,000 + 2,000 = P₂ + 162,000
• P₂ = −10,000 Pa = −10 kPa

Examiner Note: Negative gauge pressure indicates pressure below atmospheric. State this explicitly.

Example 2: Terminal Velocity

A spherical oil droplet (radius 2.0 μm, density 900 kg/m³) falls through air (η = 1.8×10⁻⁵ Pa·s, ρ = 1.2 kg/m³). Calculate terminal velocity.

Solution:

At terminal velocity: Weight = Drag + Buoyancy

• (4/3)πr³ρ_oil·g = 6πηrv + (4/3)πr³ρ_air·g
• v = [(4/3)r²g(ρ_oil − ρ_air)] / (6η)
• v = [4(2.0×10⁻⁶)²(9.8)(900−1.2)] / [18(1.8×10⁻⁵)]
• v = 0.043 m/s = 4.3 cm/s

Cambridge Tip: Always show buoyancy term even if negligible; omitting it loses method marks.