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) = [tex]\frac{1}{2}.pv^{2} A.Cl[/tex] where Cl = lift coefficient
Cl = [tex]\frac{2w}{pAv^{2} }[/tex]= [tex]\frac{2x38,220}{1.225x27.3x201.4^{2} }[/tex] = 0.5635
coefficient of drag (Cd) = CDO + [tex]\frac{Cl^{2} }{n.E.AR}[/tex] (take note that π is shown as [tex]n[/tex])
Cd = 0.03 + [tex]\frac{0.5635^{2} }{3.142x0.9x7.5}[/tex] = 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 = [tex]\frac{2w}{pAv^{2} }[/tex]= [tex]\frac{2x38,220}{0.777x27.3x201.4^{2} }[/tex] = 0.0888
coefficient of drag (Cd) = CDO + [tex]\frac{Cl^{2} }{n.E.AR}[/tex] (take note that π is shown as [tex]n[/tex])
Cd = 0.03 + [tex]\frac{0.0888^{2} }{3.142x0.9x7.5}[/tex] = 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
