An airplane has a mass of 2*10^6 kg and air flows past the power surface of the wings at 100ms¯¹. If the wings have a surface area of 1200m², how fast must the air flow over the upper surface of the wing if the plane is to stay in the air? Consider only the Bernoulli's effect.

190 m/s

Explanation:

For the plane to stay in the air, the lift force must equal the weight.

The lift force is also equal to the pressure difference across the wings (pressure at the bottom minus pressure at the top) times the area of the wings.

Therefore:

mg = (P₂ - P₁) A

P₂ - P₁ = mg / A

Using Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

½ ρ (v₁² - v₂²) = P₂ - P₁

½ ρ (v₁² - v₂²) = mg / A

v₁² - v₂² = 2mg / (Aρ)

v₁² = v₂² + 2mg / (Aρ)

Substituting values (assuming air density of 1.225 kg/m³):

v₁² = (100 m/s) ² + 2 (2*10⁶ kg) (9.8 m/s²) / (1200 m² * 1.225 kg/m³)

v₁² = 36,666.67 m²/s²

v₁ = 191 m/s

Rounding to two significant figures, the air must move at 190 m/s over the top of the wing.