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Unformatted text preview: AOE 3104 Homework #6 Solutions Problem 1. A constantpower aircraft has the following parameter values: S = 1000 ft 2 , T , SL = 15 , 000 lb , P shaft , SL = 5000 hp , η p = 0 . 8 C D L = 0 . 04 + 0 . 08 C 2 L C L max , L = 3 . 2 μ L = 0 . 25 W L = 45 , 000 lb (The subscript “L” denotes “landing configuration.”) • Given that the aircraft lands at 1.2 V stall , compute the ISA sealevel landing distance in feet assuming that spoilers are deployed ( C L → 0) and thrust is zeroed at the instant of touchdown. • Repeat the previous analysis, assuming that the aircraft lands in a 25 mph headwind. Solution. Part A) The stall speed in landing configuration is V stall , L = radicalBigg 2 W L ρ SL SC L max , L = radicalBigg 2(45 , 000 lb) (2 . 377 × 10 − 3 sl / ft 3 )(1000 ft 2 )(3 . 2) = 108 . 8 ft / s so the touchdown speed is V TD = 1 . 2 V stall , L = 130 . 5 ft / s . The landing acceleration is ˙ V = A BV 2 where A = μ L g = (0 . 25)(32 . 2 ft / s 2 ) = 8 . 05 ft / s 2 and B = g W parenleftbigg 1 2 ρSC D parenrightbigg = 32 . 2 ft / s 2 45 , 000 lb parenleftbigg 1 2 (2 . 377 × 10 − 3 sl / ft 3 )(1000 ft 2 )0 . 04 parenrightbigg = 3 . 40 × 10 − 5 ft − 1 The landing distance is s = 1 2 B ln parenleftbigg 1 B A V 2 TD parenrightbigg = 1022 ft ....
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 Fall '09
 Aerodynamics, Aviation terminology, CD0 WTO /S, Pdyn CD0 WTO, WTO /S +K

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