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

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

$${the}\:\mathrm{velocities}\:\mathrm{of}\:\mathrm{air}\:\mathrm{particles}\: \\ $$$$\mathrm{above}\:\mathrm{and}\:\mathrm{below}\:\mathrm{the}\:\mathrm{wing}\:\mathrm{of}\:\mathrm{an}\:\mathrm{aircraft} \\ $$$$\:\mathrm{speeding}\:\mathrm{down}\:\mathrm{the}\:\mathrm{runway}\:\mathrm{at}\:\mathrm{a} \\ $$$$\:\mathrm{given}\:\mathrm{instant} \\ $$$$\mathrm{are}\:\mathrm{210m}/\mathrm{s}\:\mathrm{and}\:\mathrm{200m}/\mathrm{s}\: \\ $$$$\mathrm{respectively}.\mathrm{If}\:\mathrm{the}\:\mathrm{density}\:\mathrm{of}\:\mathrm{air}\:\mathrm{is}\: \\ $$$$\mathrm{1}.\mathrm{2kg}/\mathrm{m}^{\mathrm{3}} ,\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{pressure} \\ $$$$\:\mathrm{difference}\:\mathrm{between}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{and} \\ $$$$\mathrm{lower}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wing}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by physicstutes last updated on 13/May/21

You might have come across the concept found in Benouli′s equation P + (1/2)ρv^2 = constant. which tells us that the relationship between pressure, density and speed is P = (1/2)ρv^2 so P_(above) = (1/2)ρv_(above) ^2 = (1/2)(1.2)(210)^2 = 26,460 Pa P_(below) = (1/2)ρv_(below) ^2 = (1/2)(1.2)(200)^2 = 24,000 Pa pressure difference = ΔP = P_(above) −P_(below) = 26,460 − 24000= 2,460 Pa

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

Commented by jlewis last updated on 13/May/21

thankz alot

$${thankz}\:{alot} \\ $$

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