22-Mec-B7 Aero and Space Flight · May 2016

Question 2 of 7: Steady Level Flight, Minimum Speed and the Drag Breakdown

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Mec-B7 Aero and Space Flight, National Examinations, May 2016. Three hours, open book; any non-communicating calculator permitted. Seven questions of equal value; any six constitute a complete paper and only the first six answered are marked, so full marks are 120 and the percentage grade is (mark obtained / 120) × 100. Several questions call for an essay-format answer, where clarity and organisation carry marks. All seven questions are solved here.

Reference texts. Solutions follow the conventions of the texts recommended for this examination code:

J. D. Anderson Jr., Introduction to Flight, 9th ed. — the standard atmosphere and the four altitudes (Ch. 3), incompressible and compressible flow with the Pitot-static tube (§3.4, §4.11, §8.7), airplane performance (Ch. 6), stability and control (Ch. 7), propulsion (Ch. 9), space flight and atmospheric entry (Ch. 8). This is the primary reference throughout.
J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed. — finite-wing theory and induced drag (Ch. 5), transonic flow, drag divergence and the area rule (Ch. 11).
B. N. Pamadi, Performance, Stability, Dynamics and Control of Airplanes, 3rd ed. — take-off and landing distances and gust load factors (Ch. 2, Ch. 5).
G. P. Sutton and O. Biblarz, Rocket Propulsion Elements, 9th ed. — the ideal rocket equation and propellant-system comparison (Ch. 4, Ch. 11–12).
H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed. — the orbit equation, vis-viva and orbital elements (Ch. 2–3).

Check — assumptions carried through the performance questions. The paper's page-1 note invites the candidate to "submit with their answer paper a clear statement of any assumptions made", and three quantities the performance questions need are never stated:

Standard atmosphere. ISA sea-level values $T_0 = 288.15\ \text{K}$, $p_0 = 101\,325\ \text{Pa}$, $\rho_0 = 1.225\ \text{kg}\,\text{m}^{-3}$, troposphere lapse rate $L = 0.0065\ \text{K}\,\text{m}^{-1}$ to 11 km, then isothermal at $216.65\ \text{K}$; $R = 287.05\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$, $\gamma = 1.4$. Inside the atmosphere model $g = 9.80665\ \text{m}\,\text{s}^{-2}$, giving the exponents $g/(LR) = 5.2559$ for pressure and $4.2559$ for density; aircraft weights use $g = 9.81\ \text{m}\,\text{s}^{-2}$.
Thrust lapse (Questions 4 and 5). For a fixed-geometry turbojet the thrust is taken proportional to density, $T = T_{SL}\,(\rho/\rho_0)$. Nothing in the paper states a lapse law, and this is the conventional first approximation.
Ground-run averaging (Question 5b). The net accelerating force is evaluated once at $V_{LO}/\sqrt{2}$ — the speed at which $V^2$ equals its mean over the run — and the run is then taken as uniformly accelerated.

Compressibility corrections to lift and drag are ignored wherever the question says so (Question 4a) and elsewhere in Questions 4 and 5, consistent with the parabolic drag polar supplied.

Question 2: Steady Level Flight, Minimum Speed and the Drag Breakdown (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.

Quantity	Part (a)	Part (b)
Wing area $S$	30 m²	23 m²
Flight condition	$V = 280$ km/h at 1700 m	$h = 500$ m
Mass / coefficients	$C_L = 1.1$, $C_D = 0.08$	$m = 5000$ kg, $C_{L,max} = 1.5$ and 2.8

Find. In (a) the thrust required and the aircraft weight in steady level flight; in (b) the stalling (minimum level-flight) speeds clean and with high-lift devices deployed; in (c) definitions of the four named drag contributions.

Approach. Steady level flight requires lift to balance weight and thrust to balance drag, so both parts (a) and (b) reduce to evaluating $\tfrac12 \rho V^2 S$ at the ISA density for the stated altitude and multiplying by the appropriate coefficient; the minimum speed is the speed at which the maximum available lift coefficient is just enough to carry the weight. Part (c) is descriptive.

Part (a) — evaluate the ISA density and dynamic pressure at 1700 m. $T = 288.15 - 0.0065 \times 1700 = 277.10\ \text{K}$, so $$\rho = 1.225\left(\frac{277.10}{288.15}\right)^{4.2559} = 1.0372\ \text{kg}\,\text{m}^{-3}$$ With $V = 280/3.6 = 77.78\ \text{m}\,\text{s}^{-1}$, $$q = \tfrac12 \rho V^{2} = 0.5 \times 1.0372 \times 77.78^{2} = 3137\ \text{Pa}$$
Balance lift against weight. In steady level flight $L = W$, so $$W = q S C_L = 3137 \times 30 \times 1.1 = \boxed{103.5\ \text{kN}}$$ which corresponds to a mass of $103\,527/9.81 = 10\,553\ \text{kg}$.
Balance thrust against drag. With the flight path horizontal and the thrust line aligned with it, $T = D$: $$T = q S C_D = 3137 \times 30 \times 0.08 = \boxed{7.53\ \text{kN}}$$ The ratio provides an immediate audit: $T/W = C_D/C_L = 0.08/1.1 = 0.0727$, and $7.53/103.5 = 0.0727$. The aeroplane is operating at a lift-to-drag ratio of 13.75, which is plausible for a propeller transport at this speed.
Part (b) — find the density at 500 m. $T = 288.15 - 3.25 = 284.90\ \text{K}$ and $$\rho = 1.225\left(\frac{284.90}{288.15}\right)^{4.2559} = 1.1673\ \text{kg}\,\text{m}^{-3}$$ The weight is $W = 5000 \times 9.81 = 49.05\ \text{kN}$.
Set the lift equation at maximum lift coefficient. The minimum level-flight speed is the speed at which the wing, working at its maximum lift coefficient, just carries the weight: $$V_{min} = \sqrt{\frac{2W}{\rho S C_{L,max}}}$$ Clean, with $C_{L,max} = 1.5$: $$V_{min} = \sqrt{\frac{2 \times 49\,050}{1.1673 \times 23 \times 1.5}} = \sqrt{2436.4} = \boxed{49.4\ \text{m}\,\text{s}^{-1}}\ (177.7\ \text{km/h})$$
Repeat with the high-lift devices deployed. With $C_{L,max} = 2.8$ the same expression gives $$V_{min} = \sqrt{\frac{2 \times 49\,050}{1.1673 \times 23 \times 2.8}} = \sqrt{1305.0} = \boxed{36.1\ \text{m}\,\text{s}^{-1}}\ (130.0\ \text{km/h})$$ Because $V_{min} \propto C_{L,max}^{-1/2}$, raising the maximum lift coefficient by the factor 2.8/1.5 = 1.87 reduces the minimum speed only by $\sqrt{1.87} = 1.37$ — a 26.8 per cent reduction. Landing distance, which scales with the square of the approach speed, falls by nearly half, which is why the modest-looking coefficient gain is worth the mechanical complexity.

Part (c) — the four drag contributions. Skin-friction drag is the streamwise resultant of the shear stress the boundary layer exerts on the wetted surface. It is set by the wetted area, the Reynolds number and above all by whether the boundary layer is laminar or turbulent, a turbulent layer producing roughly five to ten times the local shear of a laminar one at the same Reynolds number. It exists even in an inviscid-pressure sense "drag-free" configuration and dominates the drag of slender, well-streamlined bodies at low speed.

Induced drag, or drag due to lift, is the price of generating lift with a wing of finite span. The pressure difference between lower and upper surfaces drives a spanwise flow that rolls up into trailing vortices; the downwash these induce at the wing tilts the local lift vector rearward, and the component of that tilted force along the flight direction is the induced drag. It varies as $C_L^2/(\pi A\!R\, e)$, so it is largest at low speed and high angle of attack — the opposite trend to skin friction — and it is reduced by high aspect ratio, by good spanwise load distribution and by winglets.

Parasite drag is everything that is not induced drag: skin friction plus pressure (form) drag from boundary-layer displacement and separation, plus interference drag where components meet, plus the drag of items that produce no lift at all (undercarriage, antennae, cooling flows). In the parabolic polar $C_D = C_{D0} + K C_L^2$ used throughout Questions 4 and 5, $C_{D0}$ is the parasite (zero-lift) drag coefficient and it scales with $V^2$ in force terms.

Compressibility drag, or wave drag, appears only as the flight Mach number approaches and exceeds the critical value. Local flow over the upper surface accelerates past the speed of sound, terminates in a shock wave, and the entropy rise across that shock, along with the shock-induced boundary-layer thickening or separation behind it, adds a drag increment that is absent at low speed. It rises very steeply through the transonic range — the "drag divergence" or historical "sound barrier" — and is the reason for swept wings, thin sections and area ruling. The total is $C_D = C_{D,\text{friction}} + C_{D,\text{form}} + C_{D,\text{induced}} + C_{D,\text{wave}}$, with the first two customarily grouped as parasite drag.

Result	Value
(a) Aircraft weight	103.5 kN (mass 10 553 kg)
(a) Thrust required	7.53 kN
(b) Minimum speed, clean ($C_{L,max}=1.5$)	49.4 m/s (177.7 km/h)
(b) Minimum speed, high-lift devices ($C_{L,max}=2.8$)	36.1 m/s (130.0 km/h)

Topic. The steady level-flight force balance, the stall speed, and the decomposition of aircraft drag.

Key relations. $L = W = qSC_L$ and $T = D = qSC_D$ with $q = \tfrac12\rho V^2$; $V_{min} = \sqrt{2W/(\rho S C_{L,max})}$; $C_D = C_{D0} + C_L^2/(\pi A\!R\,e)$.

Why this works. Steady, unaccelerated, level flight means the resultant force vanishes, and with small angles of attack and the thrust line nearly along the flight path the vertical and horizontal equations uncouple into $L = W$ and $T = D$. Everything else in part (a) is bookkeeping on $q$. Part (b) exploits the fact that, for a given weight and wing area, the product $q C_L$ is fixed; since $C_L$ cannot exceed $C_{L,max}$, $q$ cannot fall below a floor, and that floor is the stall speed. High-lift devices raise $C_{L,max}$ by increasing camber (flaps) and by re-energising the upper-surface boundary layer through slots so that separation is delayed to a higher angle of attack. The drag decomposition matters because the two main contributions move in opposite directions with speed: parasite drag grows as $V^2$, induced drag falls as $1/V^2$, and their crossing point is the minimum-drag speed exploited in Questions 4 and 5.

Common pitfall. Forgetting to convert 280 km/h to 77.78 m/s (using 280 directly inflates $q$ by a factor of 12.96); using sea-level density when an altitude is stated; and quoting the mass 10 553 kg as the "weight" when the question asks for a force. In part (b) a common slip is to assume the high-lift case scales linearly with $C_{L,max}$, which overstates the speed reduction by a factor of nearly two.

Source. Anderson, Introduction to Flight, 9th ed., §6.3 (thrust required), §5.14 and §6.15 (stall and high-lift devices), §5.13 and Ch. 11 of Fundamentals of Aerodynamics (drag breakdown and wave drag).

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