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

Speed of wind on the upper surface of the wing, V1 = 70 m / s
Speed of wind on the lower surface of the wing, V2 = 63 m / s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation :

$P_1 + \frac{1 }{ 2} ρV_1^2 = P_2 + \frac{1 }{ 2} ρV_2^2$
$P_2 - P_1 = \frac{1 }{ 2} ρ(V_1^2 - V_2^2)$

Where,

P1 = Pressure on the upper surface of the wing
P2 = Pressure on the lower surface of the wing

The pressure difference between the the upper and lower surfaces of the wing provides lift to the aeroplane.

Lift on the wing = (P2 - P1)A

= $\frac{1 }{ 2} ρ(V_1^2 - V_2^2)A$

$= \frac{1 }{ 2} 1.3((70)^2 - (63)^2) × 2.5$

= 1512.87

= 1.51 × 103 N

Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.