Question 5 of 7: Range and Endurance Speeds, Take-Off Distance, Gust Load and Ceiling

Question 5: Range and Endurance Speeds, Take-Off Distance, Gust Load and Ceiling (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The Question 4 aircraft: $C_D = 0.031 + 0.034 C_L^2$, $T_{SL} = 85$ kN, $m = 29\,500$ kg ($W = 289.4$ kN), $S = 65$ m². Additionally:

Find. The jet endurance and range speeds at 8000 m; the take-off distance from brake release to the 15 m screen height at sea level; the load factor produced by a sharp-edged 40 km/h upward gust; and the absolute ceiling.

Approach. The endurance and range speeds for a jet follow from minimising drag and drag-per-unit-speed respectively; the take-off is split into a uniformly accelerated ground run and a steady climb-out; the gust load factor follows from the incremental angle of attack the vertical gust imposes; and the absolute ceiling is where the lapsed thrust has fallen to the minimum drag, so that no excess power remains.

1. Part (a) — maximum endurance for a jet is minimum drag. A turbojet burns fuel in proportion to thrust, so the time aloft per unit fuel is greatest when the thrust required — that is, the drag — is least. From Question 4(d) that condition is $C_L = \sqrt{C_{D0}/K} = 0.9549$, so at 8000 m $$V_{endurance} = V_{md} = \boxed{133.3\ \text{m}\,\text{s}^{-1}}$$ with $D_{min} = 18.79$ kN.
2. Maximum range for a jet minimises drag per unit speed. Distance per unit fuel is proportional to $V/D$, so the range condition maximises $C_L^{1/2}/C_D$, which for a parabolic polar occurs at $$C_{L,range} = \sqrt{\frac{C_{D0}}{3K}} = \sqrt{\frac{0.031}{0.102}} = 0.5513$$ $$V_{range} = \sqrt{\frac{2W}{\rho S C_{L,range}}} = \sqrt{\frac{2 \times 289\,395}{0.5252 \times 65 \times 0.5513}} = \boxed{175.4\ \text{m}\,\text{s}^{-1}}$$ The ratio $V_{range}/V_{endurance} = 3^{1/4} = 1.316$ is exact for a parabolic polar and is the quickest check on both answers. Note the contrast with a propeller aircraft, where the relationship is inverted: propeller range is flown at minimum drag and propeller endurance at $C_L = \sqrt{3C_{D0}/K}$.
3. Part (b) — establish the stall and lift-off speeds. In the take-off configuration at sea level, $$V_{stall} = \sqrt{\frac{2W}{\rho_0 S C_{L,max}}} = \sqrt{\frac{2 \times 289\,395}{1.225 \times 65 \times 1.7}} = 65.4\ \text{m}\,\text{s}^{-1}$$ $$V_{LO} = 1.25 \times 65.4 = \boxed{81.7\ \text{m}\,\text{s}^{-1}}$$
4. Evaluate the net accelerating force at the representative ground-run speed. During the run the aeroplane is opposed by aerodynamic drag and by rolling friction on the residual weight the wheels still carry: $$F = T - D - \mu\,(W - L)$$ Both $D$ and $L$ vary as $V^2$, so evaluating $F$ once at $V_{LO}/\sqrt{2} = 57.8$ m/s (where $V^2$ equals its mean over the run) gives the correct average. There $q = \tfrac12(1.225)(57.8)^{2} = 2046\ \text{Pa}$ and, with $C_{D,run} = 0.031 + 0.034(0.25)^{2} = 0.033125$, $$L = 2046 \times 65 \times 0.25 = 33.25\ \text{kN},\qquad D = 2046 \times 65 \times 0.033125 = 4.41\ \text{kN}$$ $$F = 85.0 - 4.41 - 0.02\,(289.4 - 33.2) = 75.47\ \text{kN}$$
5. Integrate the uniformly accelerated run. $$a = \frac{F}{m} = \frac{75\,470}{29\,500} = 2.558\ \text{m}\,\text{s}^{-2} \quad\Longrightarrow\quad s_{g} = \frac{V_{LO}^{2}}{2a} = \frac{81.7^{2}}{2 \times 2.558} = \boxed{1306\ \text{m}}$$ Rolling friction supplies 5.1 kN of the 9.5 kN of retarding force, and the lift generated during the run — which unloads the wheels — trims only 0.67 kN off it, so at this low ground-run lift coefficient the friction term is set almost entirely by the weight.
6. Add the airborne segment to the 15 m screen. Taking the climb-out as steady at $V_{LO}$, the lift equals the weight, so $$q_{LO} = \tfrac12 (1.225)(81.7)^{2} = 4091\ \text{Pa},\qquad C_L = \frac{W}{q_{LO}S} = \frac{289\,395}{4091 \times 65} = 1.088$$ $$D = q_{LO} S \left(C_{D0} + K C_L^{2}\right) = 4091 \times 65 \times 0.0713 = 18.95\ \text{kN}$$ $$\sin\gamma = \frac{T - D}{W} = \frac{85.0 - 18.95}{289.4} = 0.2282 \;\Rightarrow\; \gamma = 13.2^{\circ},\qquad s_{a} = \frac{15}{0.2282} = 66\ \text{m}$$
7. Total the take-off distance. $$s_{TO} = s_{g} + s_{a} = 1306 + 66 = \boxed{1371\ \text{m}}$$ The ground run is 95 per cent of the total, which is typical for a lightly loaded jet with a thrust-to-weight ratio of 0.29; the airborne segment matters far more for aircraft with poorer climb gradients.
8. Part (c) — the sharp-edged gust changes the angle of attack. An aircraft flying at $V$ that instantaneously enters an upward gust of speed $U$ sees its relative wind rotated by $\Delta\alpha = U/V$ (small angles), so the lift increases by $$\Delta L = \tfrac12\rho V^{2} S\,a\,\frac{U}{V} = \tfrac12 \rho V S a U$$ Dividing by the weight and adding the 1 g of level flight, $$n = 1 + \frac{\rho V a U}{2\,(W/S)}$$ This is the classical sharp-edged gust formula; it assumes the gust is encountered instantaneously and that the aircraft has no time to respond in pitch or heave, so it is conservative — certification practice multiplies it by a gust-alleviation factor $K_g$ of about 0.7–0.9.
9. Substitute the numbers. At 500 m, $\rho = 1.1673\ \text{kg}\,\text{m}^{-3}$; $V = 350/3.6 = 97.22$ m/s; $U = 40/3.6 = 11.11$ m/s; the wing loading is $W/S = 289\,395/65 = 4452\ \text{Pa}$. Then $$\Delta n = \frac{1.1673 \times 97.22 \times 5.5 \times 11.11}{2 \times 4452} = \frac{6935}{8904} = 0.779$$ $$n = 1 + 0.779 = \boxed{1.78}$$ A 40 km/h gust therefore adds 78 per cent to the wing loading — well inside a transport aircraft's typical 2.5 g manoeuvre envelope, but a useful illustration of why high wing loading (which appears in the denominator) makes an aeroplane ride more smoothly through turbulence.
10. Part (d) — the absolute ceiling is where the excess power vanishes. Climb ceases when the maximum thrust available has fallen to the minimum drag, so that the best rate of climb is zero: $$T_{SL}\frac{\rho}{\rho_0} = D_{min} = 2W\sqrt{C_{D0}K} = 18.79\ \text{kN}$$ $$\rho = 1.225 \times \frac{18\,790}{85\,000} = 0.2708\ \text{kg}\,\text{m}^{-3}$$
11. Convert that density to an altitude. Since $0.2708\ \text{kg}\,\text{m}^{-3}$ is below the tropopause value $\rho_{11} = 0.3639\ \text{kg}\,\text{m}^{-3}$, the ceiling lies in the isothermal stratosphere, where $\rho = \rho_{11}\exp[-g(h - 11\,000)/(R\,T_{11})]$ with $T_{11} = 216.65\ \text{K}$: $$h = 11\,000 + \frac{287.05 \times 216.65}{9.80665}\ln\!\left(\frac{0.3639}{0.2708}\right) = 11\,000 + 6342 \times 0.2956$$ $$h_{ceiling} = \boxed{12\,874\ \text{m}}$$ At the ceiling the aircraft is constrained to fly at the single speed $V = \sqrt{2W/(\rho S C_{L,md})} = 185.6\ \text{m}\,\text{s}^{-1}$, which is $M = 0.63$ — still comfortably subsonic, so ignoring compressibility remains defensible. The service ceiling, defined by a residual climb rate of 0.5 m/s (100 ft/min), would lie a few hundred metres lower.