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Unformatted text preview: AOE 3104 Homework #5 Assigned: March 22, 2010 Due: March 29, 2010 (Place your homework in the box outside my office by 5 PM.) For Problems 1 through 3, C D TO = 0 . 02 + 0 . 04 C 2 L , C L max , TO = 1 . 5 , V TO = 1 . 2 V stall , and TO = 0 . 02 . Problem 1. A certain plane weighs W = 115 , 000 lbs, has a wing loading W/S = 59 psf and has four jet engines, each of which produces 6900 pounds of thrust. Find the takeoff ground run for this aircraft, in standard sea level conditions, for the following two cases: The takeoff angle of attack is used during the entire ground run Optimum ground run conditions. (i.e., maximum acceleration) Solution. The total thrust is T SL = 4(6900) N = 27 , 600 N The stall speed in takeoff configuration is V stall = radicalBigg 2 W SL SC L max = radicalBigg 2(59 psf) (2 . 377 E 3 sl / ft 3 )(1 . 5) = 182 ft / s so V TO = 1 . 2 V stall = 218 ft / s . Part 1) The lift coefficient at takeoff is C L TO = 2 W V 2 TO S = 1 . 04 The takeoff acceleration equations is V = A BV 2 where A = g parenleftbigg T SL W parenrightbigg B = g W parenleftbigg 1 2 S ( C D C L G ) + a parenrightbigg Note that a is zero for a constantthrust aircraft. Assuming the lift coefficient for takeoff is maintained throughout the ground roll, C L G = C L TO , we have A = (32 . 2) parenleftbigg 27 , 600 115 , 000 . 02 parenrightbigg = 7 . 1 ft / s 2 and B = 32 . 2 59 parenleftbigg 1 2 (2 . 377 e 3) ( . 02 + (0 . 05)(0 . 25) 2 (0 . 02)(0 . 25) ) parenrightbigg = 2 . 759 E 5 ft 1 The takeoff distance is S TO = 1 2 B ln parenleftbigg A A BV 2 TO parenrightbigg = 3724 ft Part 2) The lift coefficient for maximum acceleration is C L G = 2 K = . 02 2(0 . 04) = 0 . 25 . In this case, B = g W parenleftbigg 1 2 S ( C D C L G ) + a parenrightbigg = 32 . 2 59 parenleftbigg 1 2 (2 . 377 e 3) ( . 02 + (0 . 05)(0 . 25) 2 (0 . 02)(0 . 25) ) parenrightbigg = 1 . 1342...
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 Fall '09

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