Calculate the horsepower that the Wright-Taylor engine would have needed to prov

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Calculate the horsepower that the Wright-Taylor engine would have needed to provide if Smeaton's coefficient of air pressure w as indeed equal to 0.005, instead of the Wright Brothers' 0.0033 value. In other words, would MORE or LESS power be required if Smeaton's coefficient was 0.005 compared to the power generated when Smeaton's coefficient was 0.0033? Use the calculations starting on page 41 of Chapter 1 as a guide and make sure you answer the question: would more or less power be required if Smeaton's coefficient was 0.005? Hints: first calculate a new velocity when k=0.005 and then use your new velocity in the drag equations. Complete your calculations with only one angle of attack: 2.5 degree ! Lift L required = Total weight of aircraft + pilot + engine (in lbs) = 285 + 140 + 200 = 625 lbs A = Surface area of wings = 500 ft2 AF = Frontal area (parasitic drag) = 20 ft2 CL = Coefficient of lift = 0.311 (for a maximum angle of attack = 2.5 degree ) = 0.706 (for a maximum angle of attack = 7.5 degree ) CD/CL = Drag-to-lift ratio = 0.138 (for a maximum angle of attack = 2.5 degree ) = 0.108 (for a maximum angle of attack = 7.5 degree ) k = Smeaton coefficient of air pressure = 0.0033 Now recall that21: P = D * V = horsepower needed to overcome drag D Therefore, one needs to calculate both the total drag D and the velocity V The velocity can be obtained from the following expression: L = k V2 A GL rightarrow V = [L/k A CL]1/2

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