Question

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^–1and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m² ? Take the density of air to be 1.3 kg m^–3.

Answer 1

Given

• Speed over upper surface: \( V_{1} = 70 \,\text{m s}^{-1} \)
• Speed over lower surface: \( V_{2} = 63 \,\text{m s}^{-1} \)
• Area of the wing: \( A = 2.5 \,\text{m}^{2} \)
• Density of air: \( \rho = 1.3 \,\text{kg m}^{-3} \)

1. Pressure difference using Bernoulli’s theorem

Take region 1 just above the wing and region 2 just below the wing, at the same height. For steady, incompressible, non-viscous flow:

\( P_{1} + \dfrac{1}{2}\rho V_{1}^{2} = P_{2} + \dfrac{1}{2}\rho V_{2}^{2} \)

Rearrange to get the pressure difference between lower and upper surfaces:

\( P_{2} - P_{1} = \dfrac{1}{2}\rho \left( V_{1}^{2} - V_{2}^{2} \right) \)

Substitute values:

\( V_{1}^{2} = 70^{2} = 4900 \) \( V_{2}^{2} = 63^{2} = 3969 \) \( V_{1}^{2} - V_{2}^{2} = 4900 - 3969 = 931 \)

\( P_{2} - P_{1} = \dfrac{1}{2} \times 1.3 \times 931 \approx 0.65 \times 931 \approx 605.15 \,\text{Pa} \)

2. Lift on the wing

Lift is the net upward force due to this pressure difference acting over area \( A \):

\( F_{\text{lift}} = (P_{2} - P_{1}) A = 605.15 \times 2.5 \,\text{N} \approx 1512.9 \,\text{N} \)

Lift on the wing \( \approx 1.5 \times 10^{3} \,\text{N} \).