Boeing 747 — Lift Coefficient, Drag Coefficient and Thrust Power at Cruise: 04-BS-7, December 2019, Q9

Question 9: Boeing 747 — Lift Coefficient, Drag Coefficient and Thrust Power at Cruise (5 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.

Check: the wing area as 170 m² would give C_L ≈ 1.59 — far above the chart's own plotted curves (C_L = 0.3, 0.4, 0.5). A wing area of ~511 m², the real Boeing 747-400's published value, gives C_L ≈ 0.53 — landing almost exactly on the chart's own 0.5 curve. S = 511 m² is therefore used below, checking the constant against physical fact.

Given.

Quantity	Value
Altitude / temperature / pressure	10 km / −50°C / 26 kPa
Aircraft weight (fully loaded, extended cab)	320 Mg = 3139.2 kN
Cruise Mach number	0.8
Wing area, S (see the check note)	≈511 m²
Ratio of specific heats, k / R_air	1.4 / 287 J/kg·K

Find. (a) C_L at the cruise condition; (b) the corresponding C_D from the chart; (c) the thrust power required.

Approach. Find the true airspeed from the Mach number and the local speed of sound (using the given altitude temperature), compute the air density from the ideal gas law at the given altitude pressure/temperature, then solve the lift equation for C_L. Read C_D off the referenced chart at that C_L and Mach number (extended-cab curve), and use it to find drag and thrust power.

Part (a) — speed of sound, true airspeed, air density, and C_L. $$a=\sqrt{kRT}=\sqrt{(1.4)(287)(223.15)}=\boxed{299.4\ \text{m/s}}\qquad V=\text{Ma}\cdot a=(0.8)(299.4)=\boxed{239.5\ \text{m/s}}$$ $$\rho=\frac{p}{RT}=\frac{26{,}000}{(287)(223.15)}=\boxed{0.406\ \text{kg/m}^3}$$ $$C_L=\frac{W}{\tfrac12\rho V^2S}=\frac{3{,}139{,}200}{\tfrac12(0.406)(239.5)^2(511)}=\boxed{0.528}$$
Part (b) — drag coefficient from the chart. At Mach 0.8 on the extended-cab C_L=0.5 curve (the closest plotted curve to C_L=0.528), before the transonic drag-rise becomes steep: $$\boxed{C_D\approx0.024}$$
Part (c) — drag force and thrust power. $$D=C_D\left(\tfrac12\rho V^2\right)S=(0.024)\big[\tfrac12(0.406)(239.5)^2\big](511)=142.9\ \text{kN}$$ $$P_{thrust}=DV=(142.9)(239.5)=\boxed{34.2\ \text{MW}}$$

Quantity	Value
Speed of sound, a	299.4 m/s
True airspeed, V	239.5 m/s
Air density at altitude	0.406 kg/m³
C_L	0.528
C_D (chart read)	≈0.024
Thrust power required	34.2 MW

Concept & Theory

Topic. Cruise lift/drag balance and thrust power at high subsonic Mach number.

Key relations. Speed of sound $a=\sqrt{kRT}$; lift equation $L=C_L\tfrac12\rho V^2S=W$ in level cruise; drag $D=C_D\tfrac12\rho V^2S$; thrust power $P=DV$ (thrust must equal drag in steady level flight).

Why this works. In steady, level cruise the aircraft's weight is exactly supported by lift, which pins C_L to a single value determined entirely by weight, altitude (through ρ) and speed — the chart's job is then only to supply the corresponding C_D at that C_L and Mach number, since drag (unlike lift) is not fixed by a simple equilibrium condition and must come from wind-tunnel data.

Common pitfall. Accepting a doubtful constant (here, the wing area) at face value when it produces a physically implausible result (C_L>1.5 is far beyond any subsonic transport's cruise condition) — cross-checking against the chart's own plotted range, or against a known real-world value for the same aircraft, catches the error immediately.

Source. White, Fluid Mechanics, Ch. 7 (Lift/drag of airfoils); J. D. Anderson, Fundamentals of Aerodynamics, Ch. 5 (transonic drag divergence).

Question 9 of 13: Boeing 747 — Lift Coefficient, Drag Coefficient and Thrust Power at Cruise

Question 9: Boeing 747 — Lift Coefficient, Drag Coefficient and Thrust Power at Cruise (5 marks)