Question 5.6: The pilot of an F-16 wants to maintain a constant altitude o......
The pilot of an F-16 wants to maintain a constant altitude of 30,000~ \mathrm{ft} flying at idle power. Recall from the discussion at the start of the chapter, for flight in a horizontal plane (i.e., one of constant altitude), where the angles are small, the lift must balance the weight of the aircraft, which is 23,750 pounds. Therefore, as the vehicle slows down, the pilot must increase the angle of attack of the aircraft in order to increase the lift coefficient to compensate for the decreasing dynamic pressure. Use the lift curves for the F-16 aircraft, which were provided by the General Dynamics Staff (1976) and are presented in Fig. 5.26 for several Mach numbers. Assume that the lift curve for M=0.2 is typical of that for incompressible flow, which for the purposes of this problem will be Mach numbers of 0.35 , or less. Prepare a graph of the angle of attack as a function of the air speed in knots (nautical miles per hour) as the aircraft decelerates from a Mach number of 1.2 until it reaches its minimum flight speed. The minimum flight speed (i.e., the stall speed), is the velocity at which the vehicle must fly at its limit angle of attack in order to generate sufficient lift to balance the aircraft’s weight. The wing (reference) area S is 300 ~\mathrm{ft}^2.
Let us first calculate the lower limit for the speed range specified (i.e., the stall velocity). Since the aircraft is to fly in a horizontal plane (one of constant altitude), the lift is equal to the weight. Thus,
W=L=C_L\left(0.5 \rho_{\infty} U_{\infty}^2\right) S (5.42a)
In order to compensate for the low dynamic pressure that occurs when flying at the stall speed \left(U_{\text {stall }}\right), the lift coefficient must be the maximum attainable value, C_{L, \max }. Thus,
W=L=C_{L, \max }\left(0.5 \rho_{\infty}\left(U_{\text {stall }}\right)^2\right) S
Solving for U_{\text {stall }},
U_{\text {stall }}=\sqrt{\frac{2 W}{\rho_{\infty} C_{L, \max } S}}
Referring to Fig. 5.26, the maximum value of the lift coefficient (assuming it occurs at an incompressible flow condition) is 1.57 , which corresponds to the limit angle of attack. From Table 1.2 \mathrm{~b}, the free-stream density at 30,000~ \mathrm{ft} is 0.0008907 ~\mathrm{slugs} / \mathrm{ft}^3, or 0.0008907~ \mathrm{lbf}~ \mathrm{s} / \mathrm{ft}^4. Thus,
\begin{aligned} & U_{\text {stall }}=\sqrt{\frac{2(23750 \mathrm{lbf})}{\left(0.0008907 \mathrm{lbf} \mathrm{s}^2 / \mathrm{ft}^4\right)(1.57)\left(300~ \mathrm{ft}^2\right)}} \\ & U_{\text {stall }}=336.5 ~\mathrm{ft} / \mathrm{s}=199.2 \mathrm{knots} \end{aligned}
Thus, the minimum velocity at which the weight of the F-16 is balanced by the lift at 30,000~ \mathrm{ft} is 199.2 knots with the aircraft at an angle of attack of 27.5^{\circ}. The corresponding Mach number is
M_{\text {stall }}=\frac{U_{\text {stall }}}{a}=\frac{336.5 \mathrm{ft} / \mathrm{s}}{994.85 \mathrm{ft} / \mathrm{s}}=0.338
To calculate the lift coefficient as a function of Mach number from M_{\infty}=1.2 down to the stall value for the Mach number,
C_L=\frac{2 W}{\rho_{\infty}\left(M_{\infty} a_{\infty}\right)^2 S} (5.42b)
The velocity in knots is given by
U_{\infty}=M_{\infty} \mathrm{a}_{\infty}(0.59209)
where U_{\infty} has the units of knots. The values obtained using these equations and using the lift curves presented in Fig. 5.26 to determine the angle of attack required to produce the required lift coefficient at a given Mach number are presented in the following table.
table 1
The correlation between the angle of attack and the velocity (in knots), as determined in this example and as presented in the table, is presented in Fig. 5.27. Note how rapidly the angle of attack increases toward the stall angle attack in order to generate a lift coefficient sufficient to maintain the altitude as the speed of the F-16 decreases toward the stall velocity. The angle of attack varies much more slowly with velocity at transonic Mach numbers.
table 1
\begin{array}{lccrc} \hline \begin{array}{c} M_{\infty} \\ (-) \end{array} & \begin{array}{c} C_L \\ (-) \end{array} & \begin{array}{c} \alpha \\ \left({ }^{\circ}\right) \end{array} & {\begin{array}{c} U_{\infty} \\ (\mathrm{ft} / \mathrm{s}) \end{array}} & \begin{array}{c} U_{\infty} \\ (knots) \end{array} \\ \hline 1.2 & 0.125 & 1.9 & 1193.82 & 706.8 \\ 0.9 & 0.222 & 2.4 & 895.37 & 530.1 \\ 0.8 & 0.281 & 3.5 & 795.88 & 471.2 \\ 0.6 & 0.499 & 6.2 & 596.91 & 353.4 \\ 0.338 & 1.57 & 27.5 & 336.50 & 199.2 \\ \hline \end{array}