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.