Consider the twin-turbofan-engine supersonic, combat aircraft McDonnell Douglas (now Boeing) F/A-18 Hornet (Figure 8.12a) with the following [10] characteristics: m = 16,770 kg, S = 38 m², b = 12.3 m, Tmax = 2 × 79.2 kN Assume the maximum allowable load factor is 10 and αo = 0; CLα = 5 1/rad;

Question

Consider the twin-turbofan-engine supersonic, combat aircraft McDonnell Douglas (now Boeing) F/A-18 Hornet (Figure 8.12a) with the following [10] characteristics:

m = 16,770 kg, S = 38 m², b = 12.3 m, [latex]T_{\max}[/latex] = 2 × 79.2 kN

Assume the maximum allowable load factor is 10 and [latex]α_o [/latex]= 0; [latex]C_{L_α}[/latex] = 5 1/rad; e = 0.83, [latex]C_{D_o}[/latex] = 0.022 (subsonic), [latex]C_{D_o}[/latex]= 0.032 (transonic); [latex]C_{D_o}[/latex]= 0.042 (supersonic). By the way, the mass is the aircraft loaded mass and the engine thrust is with the afterburner.
The maximum lift coefficient of the aircraft without using flap is 1.4. Will this fighter be able to turn (coordinated level) at a bank angle of 75° and a 1000 m of turn radius at 10,000 ft? If so, determine the time that it takes to have a completed 180° turn (a half circle).

Accepted Answer

To investigate the capability of the aircraft for this turn, we need to inspect the following four items:

1. Is the desired load factor less than the maximum allowable load factor?
2. Is the desired load factor less than the maximum producible load factor?
3. Is the turn airspeed more than the turn stall speed?
4. Does the aircraft generate a sufficient thrust for this turn?

We first need to find the following two parameters:

[latex]AR=\frac{b^2}{S}=\frac{12.3^2}{38}=3.98 \quad \quad \quad \quad (3.9)\ \space \ K=\frac{1}{\pi eAR}=\frac{1}{3.14 imes 0.83 imes 3.98}=0.096 \quad \quad \quad \quad (3.8)[/latex]

At 10,000 ft, the air relative density is 0.738 and the air density is 0.904 kg/m³. Maximum engine thrust at 10,000 ft is

[latex]T=T_o\Big[\frac{ ho}{ ho_o}\Big]^1=2 imes 79.2 imes (0.738)^1=116.9 \space kN \quad \quad \quad \quad (4.24)[/latex]

1. Desired load factor is

[latex]n=\frac{1}{\cos\phi}=\frac{1}{\cos(75)}=3.86 \quad \quad \quad \quad (9.12)[/latex]

The desired load factor is less than the maximum allowable load factor (i.e., 10).
2. Maximum producible load factor is

[latex]n_{\max_{\max}}=\frac{T_{\max}}{2W\sqrt{KC_{D_o}}}=\frac{116.9}{2 imes 16,770 imes 9.81\sqrt{0.096 imes 0.022}}=7.72 \quad \quad \quad \quad (9.55)[/latex]

The desired load factor is less than the maximum producible load factor.
3. Turn airspeed

[latex]R=\frac{V^2}{g\sqrt{n^2-1}}\Rightarrow V=\sqrt{Rg\sqrt{n^2-1}}=\sqrt{1000 imes 9.81 imes \sqrt{3.86^2-1}}\Rightarrow V=191.3 \space m/s \quad \quad \quad \quad (9.24)[/latex]

This is a subsonic velocity, so [latex]C_{D_o}[/latex] of 0.022 is used in the calculation of drag.
The stall speed for this turn is

[latex]V_{s_{t}}=\sqrt{\frac{2mg}{ ho SC_{L_{\max}}\cos(\phi)}}=\sqrt{\frac{2 imes 16770 imes 9.81}{0.738 imes 1.225 imes 38 imes 1.4 imes \cos(75)}}\Rightarrow V_{s_t}=162.5 \space m/s \quad \quad \quad \quad (9.15)[/latex]

The turn airspeed (191.3 m/s) is greater than the stall speed for this turn (162.5 m/s).
4. Required engine thrust

[latex]D=\frac{1}{2} ho V^2S\Big(C_{Do}+K\Big(\frac{2nW}{ ho V^2S}\Big)^2\Big) \quad \quad \quad \quad (9.39) \ \space \ D=\frac{1}{2} imes 0.904 imes 191.3^2 imes 38\Big(0.022+0.096\Big(\frac{2 imes 3.86 imes 16,770 imes 9.81}{0.904 imes 191.3^2 imes 38}\Big)^2\Big)=75,696 \space N=75.7 \space kN[/latex]

The lift coefficient

[latex]C_L=\frac{2nW}{ ho V^2S}=\frac{2 imes 3.86 imes 16,770 imes 9.81}{0.904 imes 191.3^2 imes 38}=1.011\quad \quad \quad \quad (9.14)[/latex]

The aircraft angle of attack

[latex]C_{L_\alpha}=\frac{dC_L}{d\alpha}\Rightarrow \alpha=\frac{C_L}{C_{L_{\alpha}}}=\frac{1.011}{5}=0.202 \space rad=11.6^{\circ}\quad \quad \quad \quad (2.10)[/latex]

In a level coordinated turn, the horizontal component of the thrust is equal to the drag

T cosα = D (9.3)

The horizontal component of the thrust is

T cosα = 116.9 × cos(11.6) = 114.5 kN

The horizontal component of the available engine thrust is more than the required thrust (114.5 > 75.7) for this maneuver.
All four requirements are met. Thus, the fighter aircraft F/A-18 is capable of performing such a maneuver. The time to complete a half circle is

[latex]t_{180}=\frac{\pi R}{V}=\frac{3.14 imes 1000}{191.3}=16.4 \space s \quad \quad \quad \quad (9.27)[/latex]

(3.9): [latex]AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}[/latex]

(4.24): [latex]T=T_o\Big(\frac{ ho}{ ho_o}\Big)^k[/latex] (troposphere) where
[latex]T_o[/latex] is the sea-level thrust
k is a factor that varies from 0.9 to 1.2

(9.15): [latex]V_{s_{t}}=\sqrt{\frac{2mg}{ ho SC_{L_{\max}}\cos(\phi)}}=\sqrt{\frac{2nmg}{ ho SC_{L_{\max}}}}[/latex]

(9.39): [latex]T= D=\frac{1}{2} ho V^2S\Big(C_{Do}+K\Big(\frac{2nW}{ ho V^2S}\Big)^2\Big)[/latex]

(9.14): [latex]C_L=\frac{2nW}{ ho V^2S}=\frac{2W}{ ho V^2Scos(\phi)}[/latex]

(2.10): [latex]C_{L_{\alpha}}=a=\frac{dC_L}{d\alpha}[/latex] (aircraft)

(9.3): [latex] \sum{F_x}=0 \Rightarrow T \cos\alpha =D[/latex]

(9.27): [latex]t_{circle}=\frac{2\pi R}{V}[/latex]

Question 9.6: Consider the twin-turbofan-engine supersonic, combat aircraf...

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