Figure 10.26 illustrates how the lift coefficient, CL varies as a function of the angle of attack, α for a symmetrical airfoil and a nonsymmetrical airfoil. Consider an airplane that weighs 30,000 lb with nonsymmetrical wings (airfoils) with a span width of 25 ft and a chord length of 20 ft, as

Question

Figure 10.26 illustrates how the lift coefficient, [latex] C_{L} [/latex] varies as a function of the angle of attack, α for a symmetrical airfoil and a nonsymmetrical airfoil. Consider an airplane that weighs 30,000 lb with nonsymmetrical wings (airfoils) with a span width of 25 ft and a chord length of 20 ft, as illustrated in Figure EP 10.20. Assume the airplane takes off and lands at sea level (air at standard atmosphere at sea level, [latex] \rho = 0.0023768 slugs/ft^{3} [/latex]) at a maximum angle of attack. Also assume that the airplane cruises at an altitude of 35,000 ft (air at standard atmosphere at an altitude of 35,000 ft,[latex] \rho = 0.0007319 slugs/ft^{3} [/latex]) at an angle of attack of [latex] 5^{\circ } [/latex]. (a) Determine the minimum takeoff and landing speed. (b) Determine the steady cruising speed at the cruising altitude.

Accepted Answer

(a) The minimum takeoff and landing speed is determined by applying Equation 10.25 [latex] F_{L,mim} = W = C_{ L,max} \frac{1}{2} ho.V^{2}_{min} . A[/latex]. The lift force must equal the weight of the aircraft at takeoff and landing. Furthermore, the maximum angle of attack at takeoff and landing is read from Figure 10.26, where the lift coefficient is also a maximum.

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] b: = 25 ft [/latex] [latex] c: = 20 ft [/latex] [latex] A_{plan}: = 2.b.c = 1 imes 10^{3} ft^{2} [/latex]

[latex] ho : = 0.0023768 \frac{slug}{ft^{3}} [/latex] [latex] \alpha _{max} : = 16 deg[/latex] [latex] C_{Lmax}: = 1.6 [/latex]

[latex] W: = 30000 lb [/latex] [latex] F_{Lmin}: = W = 3 imes 10^{4} lb [/latex]

Guess value: [latex] V_{min}: = 100 \frac{ft}{sec} [/latex]

Given

[latex] F_{Lmin} = C_{Lmax} \frac{1}{2} ho.V^{2} . A_{plan} [/latex]

[latex] V_{min} =Find (V_{min}) = 125.609 \frac{ft}{s} [/latex] [latex] V_{min} =85.642 mph [/latex]

which is the stall speed. However, the FAA requires that the minimum safe speed for aircraft takeoff and landing be at least 1.2 times the stall speed, [latex] v_{min} [/latex] which is determined as follows:

[latex] V_{minsafe}: = 1.2 V_{min} = 150.73 \frac{ft}{s} [/latex] [latex] V_{minsafe}= 102.771 mph [/latex]

(b) The steady cruising speed at the cruising altitude is determined by applying Equation 10.21 [latex] F_{L} = C_{L} \frac{1}{2} ho.V^{2} . A[/latex]. The lift force must equal to the weight of the aircraft at steady cruising speed. Furthermore, the lift coefficient for the cruising angle of attack is read from Figure 10.26.

[latex] ho _{cruise} : = 0.00073819 \frac{slug}{ft^{3}} [/latex] [latex] \alpha _{cruise} : = 5 deg[/latex] [latex] C_{Lcruise}: = 0.75 [/latex]

[latex] F_{Lcruise}: = W = 3 imes 10^{4} lb [/latex]

Guess value: [latex] V_{cruise}: = 1000 \frac{ft}{sec} [/latex]

Given

[latex] F_{Lcruise} = C_{Lcruise} \frac{1}{2} ho _{cruise}. V^{2}_{cruise} . A_{plan} [/latex]

[latex] V_{cruise}: =Find (V_{cruise}) = 329.201 \frac{ft}{s} [/latex] [latex] V_{cruise} = 224.455 mph [/latex]

Question 10.20: Figure 10.26 illustrates how the lift coefficient, CL varies...

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