Question

Consider an airplane modeled after the twin-engine Beechcraft Queen Air executive transport. The airplane has the following characteristics: weight is 38,220 N; wing area is 27.3 m2; aspect ratio is 7.5; Oswald efficiency factor is 0.9; and zero-lift drag coefficient CD,0 is 0.03.
1) Calculate the thrust required to fly at a velocity of 725 km/h at standard sea level. Assume the value of ?? = 1.225 kg/m3 at standard sea level. (Round the final answer to the nearest whole number.) The thrust required to fly at a velocity of 725 km/h at standard sea level is _______N.
2) Calculate the thrust required to fly with a velocity of 725 km/h at an altitude of 4.5 km. Assume the value of ?? = 0.777 kg/m3 at an altitude of 4.5 km. (Round the final answer to the nearest whole number.) The thrust required to fly with a velocity of 725 km/h at an altitude of 4.5 km is _______ N.

Answer 1

Answer:

(A) 3052.7 N

(B)13,089.1 N

Explanation:

weight (w) = 38,220 N

wing area (A) = 27.3 m^{2}

Oswald efficiency factor (E) = 0.9

zero-lift drag coefficient (CD0) = 0.03

velocity (v) = 725 km/h = 201.4 m/s

density of air (p) = 1.225 kg/m^{3}

aspect ratio (AR) = 7.5

(A) at standard sea level

  weight (w) = $\frac{1}{2}.pv^{2} A.Cl$   where Cl = lift coefficient    

Cl = $\frac{2w}{pAv^{2} }$= $\frac{2x38,220}{1.225x27.3x201.4^{2} }$ = 0.5635

coefficient of drag (Cd) = CDO + $\frac{Cl^{2} }{n.E.AR}$ (take note that π is shown as $n$)

Cd = 0.03 + $\frac{0.5635^{2} }{3.142x0.9x7.5}$ = 0.045

Cl/Cd = L/D (lift to drag ratio) = 0.5635/0.045 =12.52

Thrust required = w/(l/d) = 38220 / 12.52 = 3052.7 N

(B) At altitude of 4.5 km

density (p) = 0.777 kg/m^{3}

Cl = $\frac{2w}{pAv^{2} }$= $\frac{2x38,220}{0.777x27.3x201.4^{2} }$ = 0.0888

coefficient of drag (Cd) = CDO + $\frac{Cl^{2} }{n.E.AR}$ (take note that π is shown as $n$)

Cd = 0.03 + $\frac{0.0888^{2} }{3.142x0.9x7.5}$ = 0.0304

Cl/Cd = L/D (lift to drag ratio) = 0.0888/0.0304 =2.92

Thrust required = w/(l/d) = 38220 / 2.92 = 13,089.1 N