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Question Number 140869 by jlewis last updated on 13/May/21

$$the\mathrm{velocities}\mathrm{of}\mathrm{air}\mathrm{particles}$$$$\mathrm{above}\mathrm{and}\mathrm{below}\mathrm{the}\mathrm{wing}\mathrm{of}\mathrm{an}\mathrm{aircraft}$$$$\mathrm{speeding}\mathrm{down}\mathrm{the}\mathrm{runway}\mathrm{at}\mathrm{a}$$$$\mathrm{given}\mathrm{instant}$$$$\mathrm{are}210\mathrm{m}/\mathrm{s}\mathrm{and}200\mathrm{m}/\mathrm{s}$$$$\mathrm{respectively}.\mathrm{If}\mathrm{the}\mathrm{density}\mathrm{of}\mathrm{air}\mathrm{is}$$$$1.2\mathrm{kg}/{\mathrm{m}}^3,\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{pressure}$$$$\mathrm{difference}\mathrm{between}\mathrm{the}\mathrm{upper}\mathrm{and}$$$$\mathrm{lower}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{wing}?$$$$$$$$$$$$$$

Answered by physicstutes last updated on 13/May/21

$$\mathrm{You}\mathrm{might}\mathrm{have}\mathrm{come}\mathrm{across}\mathrm{the}\mathrm{concept}\mathrm{found}\mathrm{in}{\mathrm{Benouli}}^{\prime }\mathrm{s}\mathrm{equation}$$$$P+\frac{1}{2}\rho v^2=\mathrm{constant}.$$$$\mathrm{which}\mathrm{tells}\mathrm{us}\mathrm{that}\mathrm{the}\mathrm{relationship}\mathrm{between}\mathrm{pressure},\mathrm{density}\mathrm{and}\mathrm{speed}\mathrm{is}$$$$P=\frac{1}{2}\rho v^2$$$$\mathrm{so}{P}_{\mathrm{above}}=\frac{1}{2}\rho v_{\mathrm{above}}^2=\frac{1}{2}(1.2){\left(210\right)}^2=26,460\mathrm{Pa}$$$$P_{\mathrm{below}}=\frac{1}{2}\rho v_{\mathrm{below}}^2=\frac{1}{2}(1.2){\left(200\right)}^2=24,000\mathrm{Pa}$$$$\mathrm{pressure}\mathrm{difference}=\mathrm{\Delta }P=P_{\mathrm{above}}-P_{\mathrm{below}}=26,460-24000=2,460\mathrm{Pa}$$

Commented by jlewis last updated on 13/May/21

$$thankzalot$$

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