In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m s - 1 and 65 m s - 1 respectively. If the wing area is 2 m 2 , the lift

Question

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m s - 1 and 65 m s - 1 respectively. If the wing area is 2 m 2 , the lift of the wing is _______ N . (Given density of air = 1 . 2 kg m - 3 )

TestHub · Accepted Answer

Applying Bernoulli's principle on upper & lower surfaces, we get

$P_{1} + \frac{1}{2} ρ {(v_{1})}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ {(v_{2})}^{2} + ρ g h_{2}$

Since, height difference between upper and lower surface is negligible.

$\Rightarrow P_{2} - P_{1} = \frac{1}{2} ρ {(v_{1})}^{2} - \frac{1}{2} ρ {(v_{2})}^{2}$

Therefore, upthrust acting due to pressure difference:

$\Rightarrow \frac{F}{A} = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) \Rightarrow F = \frac{1}{2} \times 1.2 \times (70^{2} - 65^{2}) \times 2$

$= 810 N$

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