22-Mec-B7 Aero and Space Flight · December 2013

Question 2 of 7: Minimum flight speed with and without high-lift devices, and wing surface pressures

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

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Mec-B7 Aero and Space Flight. Three hours, OPEN BOOK, any non-communicating calculator permitted. Seven questions are printed and all carry equal value; the paper states that any six constitute a complete examination and that only the first six appearing in the answer book will be marked, so each question is worth 20 of the 120 marks a candidate can attempt. The printed marking key splits every question into its sub-parts, and those weights are reproduced in the headings below. All seven questions are solved here, because this set is a study resource rather than a timed sitting. The paper also instructs the candidate to state any assumption made where a required quantity has been omitted; that instruction is used explicitly in Questions 4 and 5, where the thrust lapse with altitude and the ground-run averaging method are not given.

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 aerodynamics and the Pitot-static tube (§3.4, §4.11), airplane performance (Ch. 6), atmospheric entry (§8.11). 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 and drag divergence (Ch. 11).
B. N. Pamadi, Performance, Stability, Dynamics and Control of Airplanes, 3rd ed. — take-off and landing field lengths (Ch. 2), turning flight and the load factor (Ch. 2), static stability (Ch. 3).
W. F. Phillips, Mechanics of Flight, 2nd ed. — the parabolic drag polar, minimum-drag and maximum-climb speeds (Ch. 3).
P. H. Oosthuizen & W. E. Carscallen, Introduction to Compressible Fluid Flow, 2nd ed. — critical Mach number and compressibility effects (Ch. 1, Ch. 8).
H. J. Allen & A. J. Eggers, A Study of the Motion and Aerodynamic Heating of Ballistic Missiles Entering the Earth's Atmosphere at High Supersonic Speeds, NACA Report 1381 (1958) — the closed-form ballistic-entry solution used in Question 7.

SI units are used throughout, matching the paper. The International Standard Atmosphere constants are taken as $p_{0}=101.325\ \text{kPa}$, $T_{0}=288.15\ \text{K}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$, lapse rate $L=0.0065\ \text{K}\,\text{m}^{-1}$ through the troposphere, $R=287.05\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ and $g=9.80665\ \text{m}\,\text{s}^{-2}$ inside the atmosphere model. Aircraft weights use the rounded $g=9.81\ \text{m}\,\text{s}^{-2}$ that Canadian examination practice expects. All pressures are absolute.

Question 2: Minimum flight speed with and without high-lift devices, and wing surface pressures (20 marks — a 10, b 10)

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	Value
(a) mass and wing area	$m=4000\ \text{kg}$, $S=19\ \text{m}^{2}$
(a) maximum lift coefficient, clean	$C_{L,\max}=1.6$
(a) maximum lift coefficient, high-lift devices out	$C_{L,\max}=2.8$
(b) flight speed and altitude	$350\ \text{km/h}$ at $2500\ \text{m}$
(b) mean upper- and lower-surface speeds	$410\ \text{km/h}$ and $305\ \text{km/h}$
ISA at $2500\ \text{m}$ (from Question 1)	$\rho=0.9569\ \text{kg}\,\text{m}^{-3}$, $p_{\infty}=74.68\ \text{kPa}$
sea-level density for part (a)	$\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$

Find. The two minimum (stalling) speeds at sea level, and then the mean static pressures on the two wing surfaces together with the lift force developed per square metre of wing plan area.

Approach. Part (a) is the lift equation solved for speed at the largest lift coefficient the wing can reach; part (b) applies Bernoulli twice from the same free-stream condition, once to each surface, and subtracts the two results.

Part (a) — equate lift to weight at the stall. In steady level flight $L=W$, and the smallest speed at which that can be achieved is the speed at which the wing is working at its greatest available lift coefficient: $$\begin{aligned} W &= \tfrac{1}{2}\rho V^{2}SC_{L,\max} \\ \Longrightarrow\quad V_{\min} &= \sqrt{\frac{2W}{\rho SC_{L,\max}}} \end{aligned}$$ The weight is $W=4000\times9.81=39{,}240\ \text{N}$.
Evaluate both configurations at sea level. Only $C_{L,\max}$ changes between the two cases, so the speeds are in the inverse ratio of the square roots of the two coefficients: $$\begin{aligned} V_{\min,\text{clean}}&=\sqrt{\frac{2\times39\,240}{1.225\times19\times1.6}} =\sqrt{2107.4}=45.9\ \text{m}\,\text{s}^{-1}\\ V_{\min,\text{high-lift}}&=\sqrt{\frac{2\times39\,240}{1.225\times19\times2.8}} =\sqrt{1204.2}=34.7\ \text{m}\,\text{s}^{-1} \end{aligned}$$ so that $$\boxed{\begin{gathered} V_{\min}=45.9\ \text{m}\,\text{s}^{-1}\ (165.3\ \text{km/h})\ \text{clean}\\ V_{\min}=34.7\ \text{m}\,\text{s}^{-1}\ (124.9\ \text{km/h})\ \text{with devices out} \end{gathered}}$$ Deploying the devices buys a $24\%$ reduction in stalling speed, and because field length scales roughly with the square of that speed it is worth some $42\%$ of the ground roll.
Part (b) — fix the free-stream state at 2500 m. From the standard atmosphere at $2500\ \text{m}$, using the same power laws as Question 1, $$\begin{aligned} \theta &= 0.94361, \\ p_{\infty} &= 101\,325\times(0.94361)^{5.2559}=74\,682\ \text{Pa}, \\ \rho &= 0.9569\ \text{kg}\,\text{m}^{-3} \end{aligned}$$ and the three speeds convert to $V_{\infty}=97.22\ \text{m}\,\text{s}^{-1}$, $V_{u}=113.89\ \text{m}\,\text{s}^{-1}$, $V_{l}=84.72\ \text{m}\,\text{s}^{-1}$.
Apply Bernoulli from the free stream to each surface. Both surfaces lie on streamlines that originate in the undisturbed flow ahead of the wing, so they share one Bernoulli constant: $$\begin{aligned} p_{u} &= p_{\infty}+\tfrac{1}{2}\rho\left(V_{\infty}^{2}-V_{u}^{2}\right), \\ p_{l} &= p_{\infty}+\tfrac{1}{2}\rho\left(V_{\infty}^{2}-V_{l}^{2}\right) \end{aligned}$$ Substituting the numbers, $$\begin{aligned} p_{u}&=74\,682+\tfrac{1}{2}(0.9569)(9452.2-12\,970.7)=74\,682-1683=73.0\ \text{kPa}\\ p_{l}&=74\,682+\tfrac{1}{2}(0.9569)(9452.2-7177.9)=74\,682+1088=75.8\ \text{kPa} \end{aligned}$$ The faster upper flow sits below ambient and the slower lower flow above it, which is the whole of the elementary explanation of lift.
Subtract to get the lift per unit area. The free-stream terms cancel identically, leaving a result that depends only on the two surface speeds: $$\begin{aligned} \frac{L}{S} &= p_{l}-p_{u} \\ &= \tfrac{1}{2}\rho\left(V_{u}^{2}-V_{l}^{2}\right) \\ &= \tfrac{1}{2}\times0.9569\times(12\,970.7-7177.9) \end{aligned}$$ $$\boxed{\begin{gathered} p_{u}=73.0\ \text{kPa} \\ p_{l}=75.8\ \text{kPa} \\ \frac{L}{S}=2771\ \text{N}\,\text{m}^{-2}=2.77\ \text{kPa} \end{gathered}}$$ As a sanity check, $2771\ \text{N}\,\text{m}^{-2}$ is a plausible wing loading for a light transport, and the implied lift coefficient $C_{L}=2771/(\tfrac{1}{2}\times0.9569\times97.22^{2})=0.613$ sits comfortably inside the linear part of a normal lift curve.

Bernoulli applied along a streamline from the free stream to each surface: the faster upper flow sits at lower static pressure, the slower lower flow at higher pressure, and the difference is the lift per unit plan area.

Quantity	Result
(a) minimum speed, clean wing	$45.9\ \text{m}\,\text{s}^{-1}$ ($165.3\ \text{km/h}$)
(a) minimum speed, high-lift devices extended	$34.7\ \text{m}\,\text{s}^{-1}$ ($124.9\ \text{km/h}$)
(b) mean upper-surface pressure	$73.0\ \text{kPa}$
(b) mean lower-surface pressure	$75.8\ \text{kPa}$
(b) lift per unit wing area	$2771\ \text{N}\,\text{m}^{-2}$

Topic. Stalling speed as the lift equation at $C_{L,\max}$, and surface pressures from Bernoulli.

Key relations. $L=\tfrac{1}{2}\rho V^{2}SC_{L}$; $V_{\min}=\sqrt{2W/(\rho SC_{L,\max})}$; $p+\tfrac{1}{2}\rho V^{2}=p_{\infty}+\tfrac{1}{2}\rho V_{\infty}^{2}$; $L/S=\tfrac{1}{2}\rho(V_{u}^{2}-V_{l}^{2})$.

Why this works. The lift equation contains three quantities the pilot cannot change in the moment — weight, wing area and air density — and two that can be traded against each other, speed and lift coefficient. Since the coefficient has a hard ceiling set by boundary-layer separation, the trade runs out at one particular speed, and that speed is the stall. High-lift devices raise the ceiling by adding camber, by adding area in the case of a Fowler flap, and by re-energising the boundary layer in the case of a slat, so the same weight can be carried more slowly. In part (b) the two surfaces are not independent systems: they are fed by the same undisturbed air, so a single Bernoulli constant serves both and the pressure difference reduces to a difference of two squared speeds. That is why the free-stream pressure disappears from the lift answer even though it is essential to the individual surface pressures.

Common pitfall. Using sea-level density in part (b) because part (a) used it. The stalling speeds are asked for at sea level and the pressures at $2500\ \text{m}$; carrying $1.225$ into part (b) inflates the lift by $28\%$. A second frequent slip is forgetting to convert km/h to m/s before squaring, which scales every pressure by $3.6^{2}\approx13$.

Source. Anderson, Introduction to Flight, 9th ed., §5.4 (the lift curve and $C_{L,\max}$), §5.17 (high-lift devices) and §6.4 (stalling speed); Anderson, Fundamentals of Aerodynamics, 6th ed., §3.2 for the Bernoulli statement used here.

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