This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: AOE 3104 Homework #5 Solutions Problem 1. A jet-propelled airplane has a parabolic drag polar with the following parameter values: C D = 0 . 024 , AR = 7 , e = 0 . 85 , S = 120 m 2 , m = 75 , 000 kg , T SL = 150 , 000 N . Assume that T ( h ) = T SL ( h ) = T SL ( h ) SL . Compute the minimum and maximum equivalent airspeed (in m/s) for level equilibrium flight at 3000, 6000, and 9000 meters. Present your results in a table. Develop a Matlab script to generate the aircrafts flight envelope, as a plot of altitude versus equivalent airspeed (for wings level, equilibrium flight). Use the resulting plot to determine the aircrafts absolute ceiling. (Include your Matlab script and the plot with your submission.) Solution. To compute the equilibrium speeds, we must compute the conditions for wings-level equilibrium flight at constant altitude. In this condition, thrust equals drag. The parabolic drag polar is C D ( C L ) = C D + 1 eAR bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright K C 2 L . Since we are interested in equivalent airspeed, re-dimensionalize by multiplying by sea-level dynamic pres- sure times wing area: D = parenleftbigg 1 2 SL V 2 eq parenrightbigg SC D = parenleftbigg 1 2 SL V 2 eq parenrightbigg S ( C D + KC 2 L ) Noting that W = C L parenleftbigg 1 2 SL V 2 eq parenrightbigg S in wings-level, constant-altitude equilibrium flight, we find D = C D parenleftbigg 1 2 SL V 2 eq parenrightbigg S + KW 2 bracketleftbiggparenleftbigg 1 2 SL V 2 eq parenrightbigg S bracketrightbigg 1 We assume that thrust exactly balances drag in equilibrium flight and that thrust scales directly with altitude T ( h ) = ( h ) T SL . Setting T ( h ) = D and multiplying through by dynamic pressure times area, we obtain the following quadratic equation in z = V 2 eq : C D parenleftbigg 1 2 SL S parenrightbigg 2 z 2- ( h ) T SL parenleftbigg 1 2 SL S parenrightbigg z + KW 2 = 0 . The quadratic equation has real solutions if and only if the discriminant b 2- 4 ac is non-negative: bracketleftbigg- ( h ) T SL parenleftbigg 1 2 SL S parenrightbiggbracketrightbigg 2- 4 bracketleftBigg C D...
View Full Document
This homework help was uploaded on 04/06/2008 for the course AOE 3104 taught by Professor Dr.craigwoolsey during the Spring '08 term at Virginia Tech.
- Spring '08