Figure 10.28 illustrates the lift-to-drag ratio, FL/FD = CL/CD versus different angles of attack, α for a typical cambered (nonsymmetrical) airfoil. Consider an airplane that weighs 50,000 lb with cambered (nonsymmetrical) wings with a span width of 40 ft and a chord length of 35 ft, as illustrated

Question

Figure 10.28 illustrates the lift-to-drag ratio, [latex] F_{L}/F_{D} = C_{L}/C_{D} [/latex] versus different angles of attack, α for a typical cambered (nonsymmetrical) airfoil. Consider an airplane that weighs 50,000lb with cambered (nonsymmetrical) wings with a span width of 40 ft and a chord length of 35 ft, as illustrated in Figure EP 10.22. Assume that the airplane cruises at an altitude of 45,000 ft (air at standard atmosphere at an altitude of 45,000 ft, [latex] \rho = 0.00046227 slugs/ft^{3} [/latex]) at a steady cruising speed of 300 mph. (a) Determine the lift force and the drag force acting on the aircraft wings during steady cruising speed at the target operating point for the airfoil.

Accepted Answer

(a) The drag coefficient is determined by applying Equation 10.12 [latex] F_{D} = C_{D} \frac{1}{2} ho v^{2}A [/latex]. The lift force must equal the weight of the aircraft at steady cruising speed. Furthermore, the angle of attack, α that produces the maximum value of [latex] F_{L}/F_{D} = C_{L}/C_{D} [/latex], which is the target operating point for an airfoil, is determined from Figure 10.28.

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] b: = 40 ft [/latex] [latex] c: = 35 ft [/latex] [latex] A_{plan}: = 2.b.c = 2.8 imes 10^{3} ft^{2} [/latex]

[latex] ho _{cruise} : = 0.00046227 \frac{slug}{ft^{3}} [/latex] [latex] W: = 50000 lb [/latex]

[latex] V_{cruise}: = 300 mph = 440 \frac{ft}{s} [/latex] [latex] F_{Lcruise}: = W = 5 imes 10^{4} lb [/latex] [latex] \alpha : = 6.5 deg [/latex]

Guess value: [latex] C_{Lcruise}: = 0.01 [/latex] [latex] C_{Dcruise}: = 0.01 [/latex] [latex] F_{Dcruise}: = 100 lb [/latex]

Given

[latex] \frac{F_{Lcruise}}{F_{Dcruise}} = 16 [/latex] [latex] \frac{C_{Lcruise}}{C_{Dcruise}} = 16 [/latex]

[latex] F_{Dcruise} = C_{Dcruise} \frac{1}{2} ho _{cruise} .V_{cruise}^{2} . A_{plan} [/latex]
[latex] \left ( \begin{matrix} C_{Lcruise} \ C_{Dcruise} \ F_{Dcruise} \end{matrix} ight ) : =Find (C_{Lcruise}, C_{Dcruise}, F_{Dcruise} ) [/latex]

[latex] C_{Lcruise} = 0.399 [/latex] [latex] C_{Dcruise} = 0.025 [/latex] [latex] F_{Dcruise} = 3.125 imes 10^{3} lb [/latex]

As a check, the lift force is computed by applying Equation 10.21 [latex] F_{L} = C_{L} \frac{1}{2} ho.V^{2} . A[/latex] as follows:

[latex] F_{Lcruise} = C_{Lcruise} \frac{1}{2} ho _{cruise} .V_{cruise}^{2} . A_{plan} = 5 imes 10^{4} lb [/latex]

Question 10.22: Figure 10.28 illustrates the lift-to-drag ratio, FL/FD = CL/...

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