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 m2 ? Take the density of air to be 1.3 kg m–3.

Asked by Pragya Singh | 1 year ago |  85

Solution :-

Given:
Speed of wind on the upper side of the wing, V1 = 70 m/s
Speed of wind on the lower side of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m -3

Using Bernoulli’s theorem, we get :
$$P_{ 1 } + \dfrac{ 1 }{ 2 }\rho V_{ 1 }^{ 2 } = P_{ 2 } + \dfrac{ 1 }{ 2 }\rho V_{ 2 }^{ 2 }$$

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

Where, P1 = Pressure on the upper side of the wing
P2 = Pressure on the lower side of the wing

Now the lift on the wing = ( P2 – P1 ) x Area
$$\dfrac{ 1 }{ 2 }\rho \left ( V_{ 1 }^{ 2 } – V_{ 2 }^{ 2 } \right ) \times A$$

$$\dfrac{ 1 }{ 2 }\times 1.3 \left ( \left ( 70 \right )^{ 2 } – \left ( 63 \right )^{ 2 } \right ) \times 2.5$$

= 1.51 x 103 N

Therefore the lift experienced by the wings of the air craft is 1.51 x 103 N.

Answered by Abhisek | 1 year ago

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