Asci 309: Aerodynamics Exercize 1

ASCI 309: Aerodynamics

Exercise I: Rectilinear Motion

Uniformly Accelerated Rectilinear Motion and Newton’s Law of Momentum

Equations to use (remember to keep track of units):

[pic]F =ma F-Force g-gravitational acceleration

[pic]F =(T-D-f)= (W/g)*a T-Thrust a-acceleration

Takeoff distance (s) = V2/2a D-Drag V-Velocity

s = ½ at2 f- friction t-time

KE = ½ mV2 W- Weight m-mass

KE-Kinetic Energy

s- distance

GIVEN INFORMATION: (Questions 1 to 5)

Gross Weight = 300,000 pounds

Average Drag = 15,000 pounds

Average Friction Force = 3,000 pounds

Average Thrust = 90,000 pounds

Lift Off Speed = 140 Knots True Airspeed (KTAS)

1. Compute the acceleration on the aircraft during the takeoff roll (ft / s 2).

w = mg

m = 300000 / 32

m = 9375slugs

a = F / m

a = (T – d – F) / m

a = (9000 – 15000 – 3000) / 9375

a = 7.68ft/s²

2. What would be the length of the takeoff run (ft)?

S = (V2 – V0²) / 2a

S = [[(140)(1.69)]² - 0] / 2 (7.68)

S = 3643ft

3. How long would it take until liftoff once the takeoff roll is started (s)?

V = V0 + at

140 = 0 + 7.68t

140 x 1.69 = 7.68t

t = (140 x 1.69) / 7.68

t = 30.8sec

4. Given the information shown above, determine how fast this airplane should be going when it passes the 2000-foot runway marker (2000 feet from the start of the takeoff roll). Express your answer in knots.

S = (V² – V0²) / 2a

2000 = (V² – 0) / 2 (7.68)

V² = (2000) (2) (7.68)

V = 175.27ft/s

V = 103.85knots

5. What is the Kinetic Energy of the aircraft after climbing out to 10,000 ft MSL and 250 KTAS?

V = 250 * 1.69

V = 422.5f/s

KE = ½mv²

KE = ½ * 9375 * (422.5)²

KE = 836748046.9ftlbs

KE = 8.3674 * 10^8ftlbs