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Q 10.14 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 ms-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

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The speed of air above and below the wings are given to be v1 = 70 m s-1 and v2 = 63 m s-1 respectively. 

Let the pressure above and below the wings be p1 and p2 and let the model aeroplane be flying at a height h from the ground.

Applying Bernoulli's Principle on two points above and below the wings we get

\\P_{1}+\rho gh+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\rho gh+\frac{1}{2}\rho v_{2}^{2}\\ P_{2}-P_{1}=\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})\\ \Delta P=\frac{1}{2}\times 1.3\times \left ( 70^{2}-63^{2} \right )\\ \Delta P=605.15\ Pa

The pressure difference between  the regions below and above the wing is 605.15 Pa

The lift on the wing is F

\\F=\Delta P\times Area\ of\ wing\\ F=605.15\times 2.5\\ F=1512.875\ N

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