Fluid Dynamics

Volume flow rate is given by: Q = A × v

where: A = cross-sectional area = 0.02 m² v = fluid speed = 3 m/s

Q = 0.02 m² × 3 m/s = 0.06 m³/s

The volume flow rate is 0.06 cubic meters per second (or 60 liters per second).

Use continuity equation: A₁v₁ = A₂v₂

Step 1: Calculate areas A₁ = π(0.02 m)² = 0.001257 m² A₂ = π(0.01 m)² = 0.000314 m²

Step 2: Apply continuity A₁v₁ = A₂v₂ 0.001257 m² × 2 m/s = 0.000314 m² × v₂ v₂ = (0.001257 × 2) / 0.000314 v₂ = 8 m/s

Alternatively, since area depends on r²: v₂ = v₁(r₁/r₂)² = 2 m/s × (2/1)² = 8 m/s

The water speeds up to 8 m/s in the narrow section.

Bernoulli's equation: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Given: P₁ = 200,000 Pa, v₁ = 4 m/s, h₁ = 0 m v₂ = 8 m/s, h₂ = 5 m, ρ = 1000 kg/m³

Solve for P₂: P₂ = P₁ + ½ρv₁² - ½ρv₂² + ρgh₁ - ρgh₂

P₂ = 200,000 + ½(1000)(16) - ½(1000)(64) + 0 - (1000)(10)(5) P₂ = 200,000 + 8,000 - 32,000 - 50,000 P₂ = 126,000 Pa = 126 kPa

The pressure in the elevated section is 126 kPa.

Note: Pressure decreases due to both increased speed and increased height.

Use Bernoulli's principle for pressure difference: P_bottom - P_top = ½ρ(v_top² - v_bottom²)

Step 1: Calculate pressure difference ΔP = ½(1.2 kg/m³)[(250 m/s)² - (200 m/s)²] ΔP = 0.6[62,500 - 40,000] ΔP = 0.6 × 22,500 = 13,500 Pa

Step 2: Calculate lift force F = ΔP × A F = 13,500 Pa × 20 m² F = 270,000 N = 270 kN

The lift force on the wing is 270 kilonewtons.

Physical principle: Faster air over the top creates lower pressure, resulting in net upward force.

Use Torricelli's theorem (from Bernoulli's equation): v = √(2gh)

For h = 10 m: v₁ = √(2 × 10 m/s² × 10 m) v₁ = √200 = 14.1 m/s

For h = 40 m: v₂ = √(2 × 10 m/s² × 40 m) v₂ = √800 = 28.3 m/s

Observations: • Exit speed at 10 m depth: 14.1 m/s • Exit speed at 40 m depth: 28.3 m/s • Doubling depth by factor of 4 increases speed by factor of 2 (since v ∝ √h)

This is the same speed an object would have if dropped from height h!