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

Question 4 of 7: Maximum speed, minimum-drag speed and best rate of climb from a parabolic drag polar

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 4: Maximum speed, minimum-drag speed and best rate of climb from a parabolic drag polar (20 marks — a 6, b 6, c 4, d 4)

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
drag polar	$C_{D}=C_{D,0}+KC_{L}^{2}$ with $C_{D,0}=0.028$, $K=0.032$
maximum thrust at sea level	$T_{0}=45\ \text{kN}$
mass, hence weight	$m=11{,}500\ \text{kg}$, $W=112{,}815\ \text{N}$
wing area	$S=43\ \text{m}^{2}$
ISA densities used	$\rho_{0}=1.225$, $\rho_{6000}=0.6599$, $\rho_{10000}=0.4127\ \text{kg}\,\text{m}^{-3}$

Find. The maximum level speed at sea level and at $6000\ \text{m}$ with the parasite and induced drag at each of those conditions, the minimum-drag speed at sea level with its drag split, the minimum-drag speed at $10{,}000\ \text{m}$, and the greatest rate of climb available at sea level.

Check: assumed thrust lapse. The paper gives the maximum thrust only at sea level, and part (a) asks for performance at $6000\ \text{m}$. Following the instruction printed on page 1 — that a candidate should assume and state any value that appears to have been omitted — the thrust of this turbojet is taken to fall in direct proportion to the ambient density, $T=T_{0}\,(\rho/\rho_{0})$, which is the standard first approximation for a fixed-geometry jet at constant Mach number. At $6000\ \text{m}$ this gives $T=45\times0.5387=24.24\ \text{kN}$. A candidate who instead assumes constant thrust with altitude would leave the dynamic pressure unchanged at its sea-level value and so obtain a maximum speed of about $336\ \text{m}\,\text{s}^{-1}$ there — comfortably supersonic, and therefore self-evidently outside the incompressible reading the question asks for; the assumption must be stated either way. Part (d) additionally ignores the $\cos\gamma$ correction to lift during the climb, which is worth under $1\%$ at the climb angles obtained here.

Approach. Every part is a statement about the same parabolic drag curve. Writing the drag in terms of dynamic pressure turns the thrust-equals-drag condition into a quadratic whose larger root is the maximum speed; differentiating the same expression gives the minimum-drag condition; and multiplying it by speed gives the power required, whose greatest shortfall below power available fixes the best climb.

Part (a) — write drag as a function of dynamic pressure alone. In level flight $L=W$, so $C_{L}=W/(qS)$ with $q=\tfrac{1}{2}\rho V^{2}$, and the drag separates into a parasite term that grows with $q$ and an induced term that falls with it: $$D=qSC_{D,0}+\frac{KW^{2}}{qS}$$ This single expression carries the whole question.
Turn thrust equals drag into a quadratic in $q$. Setting $D=T$ and multiplying through by $q$, $$\begin{aligned} SC_{D,0}\,q^{2}-Tq+\frac{KW^{2}}{S} &= 0 \\ \Longrightarrow\quad q &= \frac{T+\sqrt{T^{2}-4C_{D,0}KW^{2}}}{2SC_{D,0}} \end{aligned}$$ The larger root is taken because it is the fast, parasite-dominated intersection; the smaller root is the slow "back-side" intersection where the aircraft is again drag-limited. The discriminant is positive only if $T \ge 2W\sqrt{C_{D,0}K}$, which is exactly the statement that thrust must exceed the minimum drag for level flight to be possible at all.
Evaluate at sea level. With $T=45\,000\ \text{N}$, $4C_{D,0}KW^{2}=4(0.028)(0.032)(112\,815)^{2}=4.561\times10^{7}$ and $T^{2}=2.025\times10^{9}$, so $$\begin{aligned} q&=\frac{45\,000+\sqrt{2.025\times10^{9}-4.561\times10^{7}}}{2\times43\times0.028} =\frac{45\,000+44\,490}{2.408}=37\,164\ \text{Pa}\\ V_{\max}&=\sqrt{\frac{2q}{\rho_{0}}}=\sqrt{\frac{2\times37\,164}{1.225}} =246.3\ \text{m}\,\text{s}^{-1} \end{aligned}$$ that is $886.7\ \text{km/h}$. At this speed $C_{L}=W/(qS)=0.0706$, so the drag splits as $$\begin{aligned} D_{\text{par}} &= qSC_{D,0}=44.75\ \text{kN}, \\ D_{\text{ind}} &= qSKC_{L}^{2}=0.25\ \text{kN} \end{aligned}$$ and the two together return the thrust of $45.0\ \text{kN}$, which is the arithmetic check.
Repeat at 6000 m with the lapsed thrust. The ISA density ratio is $\sigma=\rho_{6000}/\rho_{0}=0.5387$, so $T=24.24\ \text{kN}$ and the same quadratic gives $q=19\,733\ \text{Pa}$. Hence $$\boxed{\begin{gathered} V_{\max}=246.3\ \text{m}\,\text{s}^{-1}\ \text{at sea level} \\ V_{\max}=244.6\ \text{m}\,\text{s}^{-1}\ \text{at }6000\ \text{m} \end{gathered}}$$ with the drag split $D_{\text{par}}=23.76\ \text{kN}$ and $D_{\text{ind}}=0.48\ \text{kN}$ at altitude (summing to the available $24.24\ \text{kN}$). The two maximum speeds are almost identical: thrust and parasite drag both fall in proportion to density, so their intersection moves very little in true airspeed. This is precisely why the real limit at these speeds is compressibility, which the question tells us to ignore — at $246\ \text{m}\,\text{s}^{-1}$ the sea-level Mach number is already $0.72$.
Part (b) — differentiate the drag expression. Minimum drag occurs where $\mathrm{d}D/\mathrm{d}q=0$: $$\begin{aligned} SC_{D,0}-\frac{KW^{2}}{q^{2}S} &= 0 \\ \Longrightarrow\quad q_{md} &= \frac{W}{S}\sqrt{\frac{K}{C_{D,0}}} \\ \Longrightarrow\quad C_{L,md} &= \sqrt{\frac{C_{D,0}}{K}} \end{aligned}$$ so at minimum drag the two contributions are exactly equal, each being half the total. Numerically $C_{L,md}=\sqrt{0.028/0.032}=0.9354$.
Evaluate the minimum-drag condition at sea level. Solving the lift equation at that lift coefficient, $$\begin{aligned} V_{md} &= \sqrt{\frac{2W}{\rho_{0}SC_{L,md}}} \\ &= \sqrt{\frac{2\times112\,815}{1.225\times43\times0.9354}} \\ &= 67.7\ \text{m}\,\text{s}^{-1} \end{aligned}$$ and the total drag takes its floor value $$\boxed{\begin{gathered} V_{md}=67.7\ \text{m}\,\text{s}^{-1}\ (243.6\ \text{km/h})\\ D_{\min}=2W\sqrt{C_{D,0}K}=6.754\ \text{kN}\\ D_{\text{par}}=D_{\text{ind}}=3.377\ \text{kN} \end{gathered}}$$ The lift-to-drag ratio there is $W/D_{\min}=16.7$, the best this airframe can achieve.
Part (c) — move the same condition to 10 000 m. The minimum-drag lift coefficient is a property of the polar and does not change with height; only the density in the lift equation does. With $\rho_{10000}=0.4127\ \text{kg}\,\text{m}^{-3}$, $$V_{md}=\sqrt{\frac{2\times112\,815}{0.4127\times43\times0.9354}} =116.6\ \text{m}\,\text{s}^{-1}\ (419.7\ \text{km/h})$$ The drag itself is unchanged at $6.754\ \text{kN}$, because the aircraft is flying at the same lift and drag coefficients — only the speed needed to do so has risen by the factor $\sqrt{\rho_{0}/\rho}=1.72$.
Part (d) — maximise the excess power. The rate of climb in a shallow climb is the excess power divided by the weight, $$\text{RC}=\frac{V(T-D)}{W} =\frac{1}{W}\left(TV-\tfrac{1}{2}\rho V^{3}SC_{D,0}-\frac{2KW^{2}}{\rho VS}\right)$$ Setting the derivative with respect to $V$ to zero and multiplying by $V^{2}$ produces a quadratic in $x=V^{2}$: $$\tfrac{3}{2}\rho SC_{D,0}\,x^{2}-Tx-\frac{2KW^{2}}{\rho S}=0$$
Solve the quadratic and evaluate the climb. With $\tfrac{3}{2}\rho_{0}SC_{D,0}=2.2124$ and $2KW^{2}/(\rho_{0}S)=1.5464\times10^{7}$, $$x=\frac{45\,000+\sqrt{(45\,000)^{2}+4(2.2124)(1.5464\times10^{7})}}{2\times2.2124} =20\,679\ \text{m}^{2}\,\text{s}^{-2}$$ so $V=143.8\ \text{m}\,\text{s}^{-1}$. At that speed $q=12\,666\ \text{Pa}$, $C_{L}=0.2071$, and the drag is $D=15.25+0.75=16.00\ \text{kN}$. The best climb rate is therefore $$\boxed{\begin{gathered} V_{\text{RC,max}}=143.8\ \text{m}\,\text{s}^{-1}\\ \text{RC}_{\max}=\frac{143.8\times(45\,000-16\,000)}{112\,815} =37.0\ \text{m}\,\text{s}^{-1}=2218\ \text{m}\,\text{min}^{-1} \end{gathered}}$$ Note that the best-climb speed is more than twice the minimum-drag speed: climb is a power problem, not a drag problem, and the extra speed more than pays for the extra drag.

Drag breakdown in steady level flight. The two components cross at the minimum-drag speed; the high-speed intersection of total drag with the thrust-available line fixes the maximum level speed.

Best rate of climb occurs where the vertical gap between power available and power required is widest, not where drag is least.

Quantity	Result
(a) maximum speed, sea level	$246.3\ \text{m}\,\text{s}^{-1}$ ($886.7\ \text{km/h}$)
(a) parasite / induced drag at that point	$44.75\ \text{kN}$ / $0.25\ \text{kN}$
(a) maximum speed, $6000\ \text{m}$	$244.6\ \text{m}\,\text{s}^{-1}$ ($880.4\ \text{km/h}$)
(a) parasite / induced drag at that point	$23.76\ \text{kN}$ / $0.48\ \text{kN}$
(b) minimum-drag speed, sea level	$67.7\ \text{m}\,\text{s}^{-1}$ ($243.6\ \text{km/h}$)
(b) parasite = induced drag there	$3.377\ \text{kN}$ each, $D_{\min}=6.754\ \text{kN}$
(c) minimum-drag speed, $10{,}000\ \text{m}$	$116.6\ \text{m}\,\text{s}^{-1}$ ($419.7\ \text{km/h}$)
(d) speed for best climb, sea level	$143.8\ \text{m}\,\text{s}^{-1}$
(d) maximum rate of climb, sea level	$37.0\ \text{m}\,\text{s}^{-1}$ ($2218\ \text{m}\,\text{min}^{-1}$)

Topic. Level-flight and climb performance from a parabolic drag polar.

Key relations. $D=qSC_{D,0}+KW^{2}/(qS)$; $V_{\max}$ from the larger root of $SC_{D,0}q^{2}-Tq+KW^{2}/S=0$; $C_{L,md}=\sqrt{C_{D,0}/K}$ and $D_{\min}=2W\sqrt{C_{D,0}K}$; $\text{RC}=V(T-D)/W$ maximised by $\tfrac{3}{2}\rho SC_{D,0}V^{4}-TV^{2}-2KW^{2}/(\rho S)=0$.

Why this works. Once lift is set equal to weight, the drag of an aeroplane in level flight is a function of a single variable. Writing it against dynamic pressure rather than speed makes that structure obvious: one term proportional to $q$, one inversely proportional to $q$. Every performance question in this paper is then a question about that hyperbola-plus-line. Where it crosses the thrust available, the aircraft can go no faster; where its two terms are equal, the drag is least and the lift-to-drag ratio greatest, which is the condition for maximum range in a propeller aircraft and maximum endurance in a jet; where the vertical gap between $TV$ and $DV$ is widest, the climb is best. The minimum-drag coefficient is altitude-independent because it is a property of the polar alone, which is why part (c) needed only a new density and not a new derivation.

Common pitfall. Taking the smaller root of the maximum-speed quadratic, which gives a low speed on the back of the drag curve where the aeroplane is also in equilibrium but speed-unstable. A second is assuming the best-climb speed coincides with the minimum-drag speed; minimum drag maximises the climb angle for a jet, but the best rate of climb needs excess power and always occurs faster. A third is forgetting to lapse the thrust with altitude, or lapsing it while forgetting to lapse the density as well.

Source. Anderson, Introduction to Flight, 9th ed., §6.5–6.8 (thrust required, maximum velocity, rate of climb) and §6.13; Phillips, Mechanics of Flight, 2nd ed., §3.3–3.7 for the closed-form minimum-drag and best-climb results used here.

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