Question 9.5: A civil subsonic jet (turbojet engine) transport aircraft ha...

We first need to find a few parameters:

$\Big(\frac{L}{D}\Big)_{\max}=\frac{1}{2\sqrt{KC_{D_o}}}=\frac{1}{2\sqrt{0.06\times 0.02}}=14.4 \quad \quad \quad \quad (5.28) \\ \space \\ n=\frac{1}{\cos\phi}=\frac{1}{cos(30)}=1.155 \quad \quad \quad \quad (9.12)$

• At the sea level (ρ = 1.225 kg/m³)
a. Airspeeds
The quadratic (9.65) equation has the following solutions for velocities:

$V=\sqrt{\frac{T}{\rho S C_{D_o}}\Big[1\pm \sqrt{1-\Big(\frac{nW}{T(L/D)_{max}}\Big)^2}\Big]} \quad \quad \quad \quad (9.70) \\ \space \\ V=\sqrt{\frac{110\times 1,000}{1.225\times 245\times 0.02}\times \Big[1\pm \sqrt{1-\frac{1.155\times (120,000\times 9.8)^2}{(110\times 1,000\times 14.4)^2}}\Big]}$

This results in two answers:

$V_1=166.7 \space m/s \quad and \quad V_2=94 \space m/s$

We need to inspect and ensure that these airspeeds are greater than the stall speed at this turning flight:

$V_{s_{t}}=\sqrt{\frac{2mg}{\rho SC_{L_{\max}}\cos(\phi)}}=\sqrt{\frac{2\times 120,000\times 9.81}{1.225\times 245\times 1.8\times \cos(30)}}\Rightarrow V_{s_t}=70.83 \space m/s=182.8 \space knot \quad \quad \quad \quad (9.15)$

Both airspeeds ($V_1  and  V_2$) are greater than the stall speed, so both values are acceptable.
b. Turn rate

$\omega=\frac{g\sqrt{n^2-1}}{V} \quad \quad \quad \quad (9.32) \\ \space \\ \omega_1=\frac{g\sqrt{n^2-1}}{V_1}=\frac{9.18\sqrt{1.155^2-1}}{166.7}=0.034\space rad/s=1.945 \space deg/s \quad \quad \quad \quad (9.32) \\ \space \\ \omega_2=\frac{g\sqrt{n^2-1}}{V_2}=\frac{9.18\sqrt{1.155^2-1}}{94}=0.06\space rad/s=3.45 \space deg/s \quad \quad \quad \quad (9.32)$

Thus, a turn with a lower speed yields the higher turn rate.
c. Turn radius

$V=R\omega \Rightarrow R=\frac{V}{\omega} \quad \quad \quad \quad (9.31) \\ \space \\ R_1=\frac{V_1}{\omega_1}=\frac{166.7}{0.034}=4910.7\space m=4.9\space km \quad \quad \quad \quad (9.31) \\ \space \\ R_2=\frac{V_2}{\omega_2}=\frac{94}{0.06}=1562.6\space m=1.5\space km \quad \quad \quad \quad (9.31)$

It is observed that the turn with the lower airspeed yields a lower turn radius. By comparing the results, we can conclude that at the sea level, it is recommended to turn with the lower speed (94 m/s or 182.8 knot), since it is more efficient.

• At 30,000 ft (ρ = 0.458 kg/m³)
Maximum engine thrust at 30,000 ft is

$T=T_o\Big[\frac{\rho}{\rho_o}\Big]^{0.9}=300\times \Big(\frac{0.458}{1.225}\Big)^{0.9}=123.7\space kN \quad \quad \quad \quad (4.21)$

The aircraft engine is producing enough thrust (more than 110 kN).
a. Airspeed
Two new airspeeds at 30,000 ft altitude are

$V=\sqrt{\frac{110\times 1,000}{0.458\times 245\times 0.02}\times \Big[1\pm \sqrt{1-\frac{1.155\times (120,000\times 9.8)^2}{(110\times 1,000\times 14.4)^2}}\Big]} \quad \quad \quad \quad (9.70)$

This results in two answers:

$V_1=272.7 \space m/s \quad and \quad V_2=153.8 \space m/s$

Again, we need to inspect and make sure that these speeds are greater than the stall speed at this turning flight and 30,000 ft altitude.

$V_{s_{t}}=\sqrt{\frac{2mg}{\rho SC_{L_{\max}}\cos(\phi)}}=\sqrt{\frac{2\times 120,000\times 9.81}{0.458\times 245\times 1.8\times \cos(30)}}\Rightarrow V_{s_t}=116 \space m/s=225.5 \space knot \quad \quad \quad \quad (9.15)$

Both airspeeds ($V_1\space and \space V_2$) are greater than the stall speed, so both values are acceptable.
b. Turn rate
The turn rates for these airspeed are

$\omega_1=\frac{g\sqrt{n^2-1}}{V_1}=\frac{9.18\sqrt{1.155^2-1}}{272.7}=0.021\space rad/s=1.19 \space deg/s \quad \quad \quad \quad (9.32) \\ \space \\ \omega_2=\frac{g\sqrt{n^2-1}}{V_2}=\frac{9.18\sqrt{1.155^2-1}}{153.8}=0.037\space rad/s=2.11 \space deg/s \quad \quad \quad \quad (9.32)$

Thus, a turn with a lower speed yields the higher turn rate.
c. Turn radius

$R_1=\frac{V_1}{\omega_1}=\frac{272.7}{0.021}=13,134\space m=13.1\space km \quad \quad \quad \quad (9.31) \\ \space \\ R_2=\frac{V_2}{\omega_2}=\frac{153.8}{0.037}=4179\space m=4.2\space km \quad \quad \quad \quad (9.31)$

It is again observed at 30,000 ft altitude that the turn with the lower airspeed yields a lower turn radius. Please note that at the sea level, the thrust is about 1/3 of the maximum available engine thrust, while at 30,000, the thrust is about 90% of the maximum available engine thrust.
By comparing the results for the sea level and 30,000 ft, we can have a general conclusion applicable for every aircraft. At a high altitude, the turn performance will be reduced; that is, the turn rate is reduced and the turn radius is increased. Hence, an aircraft has a lower maneuvering capability at high altitudes. It is worth mentioning that at 30,000 ft, the first airspeed is in high subsonic region (M = 0.8). However, at such speeds, $C_{Do}$ is much higher. This point is neglected in this example. In real calculation, you need to use a higher $C_{Do}$ value for high subsonic speeds.

(5.28):      $\Big(\frac{C_L}{C_D}\Big)_{\max}=\frac{1}{2\sqrt{KC_{D_o}}}$

(9.15):       $V_{s_{t}}=\sqrt{\frac{2mg}{\rho SC_{L_{\max}}\cos(\phi)}}=\sqrt{\frac{2nmg}{\rho SC_{L_{\max}}}}$

(9.65):     $S^2C_{D_o}q^2-TSq+Kn^2W^2=0$

(4.21):      $T=T_o\Big(\frac{\rho}{\rho_o}\Big)^c$    (troposphere) where
$T_o$ is the sea-level thrust
T is the engine thrust at altitude