Given the following aircraft information -Maximum Takeoff Weight (MTOw) is 408,0

Question

Given the following aircraft information -Maximum Takeoff Weight (MTOw) is 408,000 Pounds (185,066 kg) -Twin GE Engines -Engine number CF6-80C2B7F -62,100 pounds of thrust per engine Total available thrust -124,200 pounds Takeoff distance = 2.900 meters (9514 feet) -Takeoff speed = 160 kts Explain how to determine the following: «Total Acceleration = af/s^2] .Sum of Retarding Forces FR [Ub] (drag + friction) -Total Takeoff Time = t [s Power P[HP] -Potential Energy EPot [ft-Ibl *Kinetic Energy -EKin [t-Ibl -Total Energy ETot fh-Ib]) «Minimum Required Climb Angle-AOC [deal E (Ctrl) -

Explanation / Answer

given

MTOW = 185,066 kg = 408,000 pounds

Thrust, T = 124,200 pounds

Takeoff distance, d = 2900 m = 9514 ft

take off speed = 160 kts = 270.05 fps

a. total acceleration = a

2*d*a = 270.05^2

a = 3.83261 ft/s/s

b. drag force = F

T - F = MTOW*a/g

F = 75645.46583850 pounds

c. total take off time = t

270.05 = a*t

t = 70.46130564 s

d. Power = T*d/t = 16770111.12530 pounds ft/s

P = 30491.109104 HP

e. PE = 0 ( as no energy is used for ascending)

f. KE = 0.5(MTOW)270.05^2 = 14877108510 pounds ft^2/s^2

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