Figure 10.27 illustrates both the lift coefficient, C_{L} versus the angle of attack, α, and the drag coefficient, C_{D} versus the angle of attack, α for a typical cambered (nonsymmetrical) airfoil. Consider an airplane that weighs 45,000lb with cambered (nonsymmetrical) wings with a span width of 30 ft and a chord length of 25 ft, as illustrated in Figure EP 10.21. Assume that the airplane cruises at an altitude of 45,000 ft (air at standard atmosphere at an altitude of 45,000ft, \rho = 0.00046227 slug/ft^{3} ) at a steady cruising speed of 250 mph. (a) Determine the angle of attack to cruise at a steady speed at cruising altitude. (b) Determine the drag force acting on the aircraft wings during steady cruising speed. (c) Determine the thrust power required by the plane engines in order to overcome the wing drag force.
Question 10.21: Figure 10.27 illustrates both the lift coefficient, CL versu...
(a) The lift coefficient is determined by applying Equation 10.21 F_{L} = C_{L} \frac{1}{2} \rho.V^{2} . A. The lift force must equal the weight of the aircraft at steady cruising speed.
slug: = 1 lb \frac{sec^{2}}{ft} b: = 30 ft c: = 25 ft A_{plan}: = 2.b.c = 1.5 \times 10^{3} ft^{2}
\rho _{cruise} : = 0.00046227 \frac{slug}{ft^{3}} W: = 45000 lb
V_{cruise}: = 250 mph = 366.667 \frac{ft}{s} F_{Lcruise}: = W = 4.5 \times 10^{4} lb
Guess value: C_{Lcruise}: = 0.8
Given
F_{Lcruise} = C_{Lcruise} \frac{1}{2} \rho _{cruise} .V_{cruise}^{2} . A_{plan}
C_{Lcruise} =Find (C_{Lcruise}) = 0.965
Furthermore, the angle of attack to cruise at a steady speed at cruising altitude for the computed lift coefficient is determined from Figure 10.27 as follows:
α := 10 deg
(b) The drag force acting on the aircraft wings during steady cruising speed is determined by applying Equation 10.12 F_{D} = C_{D} \frac{1}{2} \rho v^{2}A . Furthermore, the drag coefficient is deter mined from Figure 10.27 for an angle of attack, \alpha = 10^{\circ } .
Guess value: F_{Dcruise}: = 100 lb
Given
F_{Dcruise} = C_{Dcruise} \frac{1}{2} \rho.V_{cruise}^{2} . A_{plan}
F_{Dcruise}: =Find (F_{Dcruise}) = 3.729 \times 10^{3} lb
(c) The thrust power required by the plane engines in order to overcome the wing drag force is determined by applying Equation 10.23 P_{engines} = F_{T} v_{cruise} \geq F_{D} v_{cruise} as follows:
F_{Tcruise}: = F_{Dcruise} P_{engines}: = F_{Tcruise}. v_{cruise} = 1.367 \times 10^{6} \frac{ft.lb}{s}
h_{P}: = 550 \frac{ft.lb}{sec} P_{engines} = 2.486 \times 10^{3} hP