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Worked Examples · Example 9.7

Q.A fully loaded Boeing aircraft has a mass of 3.3×105 kg3.3 \times 10^{5}\ \text{kg}. Its total wing area is 500 m2500\ \text{m}^{2}. It is in level flight with a speed of 960 km/h960\ \text{km/h}.

(a) Estimate the pressure difference between the lower and upper surfaces of the wings.
(b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ=1.2 kg m−3\rho = 1.2\ \text{kg m}^{-3}]
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The lift on the wings must equal the aircraft's weight, giving a pressure difference of about 6.5×1036.5\times10^3 Pa between the lower and upper wing surfaces. Applying Bernoulli's equation, this pressure difference corresponds to the air over the upper surface moving about 7.6% faster than the air below.

(a) Pressure difference across the wings

In level flight, the total lift force balances the aircraft's weight: L=mgL=mg, and lift is the pressure difference times the wing area, L=ΔP⋅AL=\Delta P\cdot A. With m=3.3×105m=3.3\times10^5 kg, A=500A=500 m2^2, g=9.8g=9.8 m/s2^2:

ΔP=mgA=(3.3×105)(9.8)500=3.234×106500≈6.5×103 Pa\Delta P = \frac{mg}{A} = \frac{(3.3\times10^5)(9.8)}{500} = \frac{3.234\times10^6}{500} \approx 6.5\times10^3\text{ Pa}

(b) Fractional increase in speed on the upper surface

Convert the cruise speed to SI units:

v=960 km/h=960×10003600≈266.7 m/sv = 960\text{ km/h} = \frac{960\times1000}{3600} \approx 266.7\text{ m/s}

Bernoulli's equation for horizontal flow (no height difference between the wing surfaces) gives

ΔP=12ρ(vu2−vl2)\Delta P = \frac12\rho\left(v_u^2-v_l^2\right)

so

vu2−vl2=2ΔPρ=2×6.5×1031.2≈1.083×104 m2/s2v_u^2-v_l^2 = \frac{2\Delta P}{\rho} = \frac{2\times6.5\times10^3}{1.2} \approx 1.083\times10^4\text{ m}^2/\text{s}^2 …

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