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.}

Asked by Pragya Singh | 1 year ago | 85

Given:

Speed of wind on the upper side of the wing, V_{1} = 70 m/s

Speed of wind on the lower side 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}

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, P_{1} = Pressure on the upper side of the wing

P_{2} = Pressure on the lower side of the wing

Now the lift on the wing = ( P_{2} – P_{1} ) 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 10^{3} N

Therefore the lift experienced by the wings of the air craft is 1.51 x 10^{3} N.

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