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Exercises · 9.14

Q.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−170\ \text{m s}^{-1} and 63 m s−163\ \text{m s}^{-1} respectively. What is the lift on the wing if its area is 2.5 m22.5\ \text{m}^{2}? Take the density of air to be 1.3 kg m−31.3\ \text{kg m}^{-3}.

Yanam CbseNCERTSubjective· 3mImportance★★★★★est
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Using Bernoulli's equation, the pressure difference between the upper and lower wing surfaces is found from the difference of the squares of the flow speeds; multiplying by the wing area gives a lift of about 1.51×1031.51\times10^3 N.

Setting up Bernoulli's equation

For horizontal flow (no height change between the two surfaces), Bernoulli's equation gives

Pl−Pu=12ρ(vu2−vl2)P_l - P_u = \frac12\rho\left(v_u^2-v_l^2\right)

where vu=70v_u=70 m/s (upper surface) and vl=63v_l=63 m/s (lower surface), and ρ=1.3\rho=1.3 kg/m3^3.

Computing the pressure difference

vu2=702=4900,vl2=632=3969v_u^2 = 70^2 = 4900, \qquad v_l^2 = 63^2 = 3969

vu2−vl2=4900−3969=931v_u^2-v_l^2 = 4900-3969 = 931

Pl−Pu=12×1.3×931=0.65×931≈605.2 PaP_l-P_u = \frac12\times1.3\times931 = 0.65\times931 \approx 605.2\text{ Pa}

Computing the lift …

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