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.

Question

Philoid · Accepted Answer

The Bernoulli’s theorem states the relation,

P_a + V_a²= P_b + V_b²

Thus we get,

P_a – P_b =(v_a² – v_b²) …..(i)

Where,

‘P_a’ is the pressure on the upper surface of the wing

‘P_b’ is the pressure on the lower surface of the wing.

‘v₁’ is the speed on the upper surface of the wing.

‘v₂’ is the speed on the lower surface of the wing.

‘ρ’ is the density of the air.

The reason for the lift of the airplane is the difference between the pressure differences between the upper and lower surface,

Thus,

The force which is responsible for the airplane lift, F = (P_b-P_a) A

Replacing the value of Pressure difference from equation (i),

⇒F = (v_a²– v_b²) A

Given,

Speed of the wind on the upper surface, v₁ = 70 m/s

Speed of the wind on the lower surface , v₂ = 63 m/s

Area of the wing, A = 2.5 m²

Density of the air, ρ = 1.3 kg / m³

⇒

⇒ F = 1512.87 N = 1.512 × 10³ N

the lift on the wing of the airplane is 1.512 × 10³ N.

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.