A fully loaded Boeing aircraft has a mass of $5.4\times {10}^5\mathrm{kg}$. Its total wing area is $500{\mathrm{m}}^2.$It is in level flight with a speed of 1080 km/h. If the density of air $\mathrm{\rho }\text{ is }1.2{\mathrm{kgm}}^{-3}$, the fractional increases in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be $\left(\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)$

Complete Solution:

Given Data:

Mass of aircraft (m): 5.4 × 10⁵ kg

Total wing area (A): 500 m²

Aircraft speed (V): 1080 km/h = 300 m/s

Density of air (ρ): 1.2 kg/m³

Acceleration due to gravity (g): 10 m/s²

The lift force (F_L) must balance the weight of the aircraft:

F_L = m g = (5.4 × 10⁵) × 10 = 5.4 × 10⁶ N

The lift force can also be expressed in terms of pressure difference (ΔP) across the wing:

F_L = ΔP × A

ΔP = F_L / A = 5.4 × 10⁶ / 500 = 1.08 × 10⁴ N/m²

According to Bernoulli’s principle, the pressure difference is given by:

Where:

v₁ = speed of air below the wing = 300 m/s

v₂ = speed of air above the wing (to be determined)

Rearranging for v₂:

1.08 × 10⁴ = 0.6 × (v₂² - 300²)

v₂² - 90000 = 1.8 × 10⁴

v₂² = 90000 + 18000 = 108000

v₂ = √108000 ≈ 328.49 m/s

The fractional increase in speed (Δv/v₁) is:

(Δv/v₁) = (v₂ - v₁) / v₁ = (328.49 - 300) / 300 ≈ 0.10

Fractional increase in speed (percentage) = 0.10 × 100 = 10%

Final Answer:

The fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface is 10%.