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^{–1}and 63 m s^{–1} respectively. What is the lift on the wing if its area is 2.5 m^{2}? Take the density of air to be 1.3 kg m^{–3}.

#### Solution 1

Speed of wind on the upper surface of the wing, *V*_{1} = 70 m/s

Speed of wind on the lower surface of the wing, *V*_{2} = 63 m/s

Area of the wing, *A* = 2.5 m^{2}

Density of air, *ρ* = 1.3 kg m^{–3}

According to Bernoulli’s theorem, we have the relation:

`P_1 + 1/2rhoV_1^2 = P_2 + 1/2 rhoV_2^2`

`P_2 - P_1 = 1/2 rho(V_1^2 - V_2^2)`

Where,

*P*_{1} = Pressure on the upper surface of the wing

*P*_{2} = Pressure on the lower surface of the wing

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

Lift on the wing = `(P_2 - P_1)A`

`=1/2 rho(V_1^2 - V_2^2)A`

`= 1/2 1.3((70)^2-(63)^2)xx 2.5`

= 1512.87

= 1.51 × 10^{3} N

Therefore, the lift on the wing of the aeroplane is 1.51 × 10^{3} N.

#### Solution 2

Let v_{1}, v_{2} be the speeds on the upper and lower surfaces of the wing of the aeroplane, and P_{1} and P_{2} be the pressure on upper and lower surfaces of the wing respectively.

Then `v_1 = 70 ms^(-1), v_2 = 63 ms^(-1); rho = 1.3 kg m^(-3)`

From Berniulli's theorem

`P_1/rho + gh + 1/2v_1^2 = P_2/rho + gh + 1/2 v_2^2`

`:.P_1/rho - P_2/rho = 1/2 (v_2^2 -v_1^2)`

or `P_1 -P_2 = 1/2 rho(v_2^2 - v_1^2) = 1/2 xx 1.3[(70)^2 - (63)^2] Pa = 605.15 Pa`

This difference of pressure provides the lift to the aeroplane. So lift on the aeroplane = pressiure difference x area of wing

= 605.15 x 2.5 N = 1512.875 N

= 1.15 x 10^{3} N