rho_o (density, sea level) = 0.002377 (slug/ft^3) or (1.225 kg/m^3) R (ideal gas constant) = 1716 (ft^2/R deg. sec.^2) or 287 (m^2/K deg sec^2) mu (viscosity air) = 1.7894 times 10^-5 (kg/m sec.) gamma (specific heat ratio air) = 1.4 (dimensionless); Jet-A Weight: 6 lbf per Gallon g (acceleration of gravity) = 32.1 (ft/sec.^2) or 9.8 (m/sec.^2) Jetman, Yves Rossy is seen* flying formation with an AirBus 380 commercial transport. The wings are of carbon composite material 6 5 ft span and with a plan-form as shown below in the Figure 1, the wings weigh 121 pounds with 8 gal of fuel included (10 min. flight time at full power). Power is provided by 4 small jet engines of app, 50 pounds of thrust (maximum) each. FAA has waivered the seat belt requirement for Yves who weighs 170 pounds. Determine the Aspect ratio for the strap-on wing system. Is a correction warranted for Aspect Ratio? If so, modify the airfoil coefficient of lift slope to correct for Aspect Ratio. Which correction is appropriate? Calculate a new Lift coefficient function using the result of A.2. Assume that the airfoil is the NACA 2412 (Data Sheet Attached). What is the Reynolds Number at a Velocity 125 MPH? What is the required angle of attack for the modified Lift Coefficient curve? Based on the Drag Coefficient at this angle of attack plus 5% to account for body drag of Mr. Rossy what is the total drag of the configuration?
SOLUTION:-
(1) Aspect ratio AR = wing span^2 / wing area = 6.5^2 / [(6.5x2) + 2(0.5x3.25x2)] = 2.2
Aspect ratios AR = 2.2 (ANSWER) So chord c = b / AR = 6.5 / 2.2 = 2.95
(2) By formula CL = L / (qxS) Where L lift force = total plane weight = 291 lb, S =19.2 and q is determined by formula q = 1/2xxu2 (where u = 125MPH = 183.32ft/s)
So q = 0.5 x 0.076 x 183.32^2 = 1277.04 lb/ft2
So CL = 291 x 32.1 / (1277.04 x 19.2) = 0.38 For this CL the angle of attack is -ve, so we need to modify the wing area S from 19.2 to 9.75 to achieve a CL = 0.75 (angle of attack = 2degree)
Aspect Ratio AR is warranted to be modified and is appropriate. (ANSWER)
(3) For AR = 4.33, CL = 0.75 and angle of attack = 2 degrees
By formula reynolds number Re for u = 125 MPH = 183.32 ft/s is given by Re = u x c / v
where c = span / AR = 6.5 / 4.33 = 1.50, v = viscosity = 1.08 x 10-5 lb/ft-s
So Re = 183.32 x 1.5 / 1.08 x 10^-5 = 254.6 x 10^5 = 25460000 (ANSWER)
(4) Cd = CL^2 / (3.142x4.33x1) = 0.75^2 / 13.6 = 0.041
By formula Drag D = Cd x x S x u^2 / 2 = 0.041 x 0.076 x 9.75 x 183.32^2 /2 = 510.5 lb
Total drag = 510.5 + 5%(510.5) = 510.5 + 25.5 = 536 lb (ANSWER)
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