Question 4 of 7: Maximum Speed, Glide, Climb and Minimum Drag
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 4: Maximum Speed, Glide, Climb and Minimum Drag (20 marks)
Find. The maximum level speed at sea level and 5000 m; the flattest glide
angle at 2000 m and the speed at which it occurs; the best rate of climb at sea level and at
8000 m; and the minimum-drag speed at 8000 m with its parasite and induced drag components.
Figure 4.1 — Thrust required (total drag) against airspeed at sea
level for the Question 4 aircraft, split into its parasite and induced components, with the
85 kN thrust-available line. The minimum of the drag curve fixes $V_{md}$ and
$(L/D)_{max}$; the upper intersection with the thrust line fixes $V_{max}$. The lower
intersection, on the "back side" of the curve, is a second equilibrium that is not usable in
practice.
Approach. Every part follows from the same drag equation written in terms
of dynamic pressure, $D = q S C_{D0} + K W^2/(qS)$: the maximum speed is where thrust
available equals it, the flattest glide is at its minimum, the best climb rate maximises the
excess power $ (T-D)V$, and the minimum-drag point splits the drag exactly in half between its
two contributions.
Part (a) — write the level-flight thrust balance as a quadratic in
$q$. With $L = W$, the lift coefficient is $C_L = W/(qS)$, so
$$D = qSC_{D0} + \frac{K W^{2}}{qS}$$
Setting $D = T$ and multiplying through by $q$:
$$S C_{D0}\,q^{2} - T q + \frac{K W^{2}}{S} = 0
\quad\Longrightarrow\quad
q = \frac{T \pm \sqrt{T^{2} - 4 C_{D0} K W^{2}}}{2 S C_{D0}}$$
The larger root is the maximum speed; the smaller is the low-speed equilibrium on the back
side of the drag curve (Figure 4.1).
Evaluate at sea level. $T = 85\,000\ \text{N}$ and
$W = 289\,395\ \text{N}$, so
$$T^{2} - 4C_{D0}KW^{2} = 7.225\times10^{9} - 4(0.031)(0.034)(289\,395)^{2}
= 7.225\times10^{9} - 3.531\times10^{8}$$
$$\sqrt{\;} = 8.290\times10^{4},\qquad
q = \frac{85\,000 + 82\,897}{2 \times 65 \times 0.031} = 41\,662\ \text{Pa}$$
$$V_{max} = \sqrt{\frac{2q}{\rho_0}} = \sqrt{\frac{2 \times 41\,662}{1.225}}
= \boxed{260.8\ \text{m}\,\text{s}^{-1}}$$
Substituting back gives $D = 85.0$ kN, confirming the root. This is $M = 0.766$ at sea level,
so ignoring compressibility, as the question directs, is a real simplification.
Repeat at 5000 m with the lapsed thrust. There
$\rho = 0.7361\ \text{kg}\,\text{m}^{-3}$ and, on the assumed lapse law,
$$T = 85 \times \frac{0.7361}{1.225} = 51.08\ \text{kN}$$
The same quadratic gives $q = 40\,624\ \text{Pa}$ and
$$V_{max} = \sqrt{\frac{2 \times 40\,624}{0.7361}} = \boxed{257.8\ \text{m}\,\text{s}^{-1}}$$
The maximum speed is almost unchanged. That is not a coincidence: at high speed the drag is
dominated by the parasite term $qSC_{D0}$, so $q_{max} \approx T/(SC_{D0})$, and with
$T \propto \rho$ the dynamic pressure and hence $V^2 = 2q/\rho$ come out nearly independent of
altitude. The small reduction is the residual induced-drag term. In Mach terms, however,
$M = 0.804$ at 5000 m against 0.766 at sea level, so the aircraft is closer to its
compressibility limit up high.
Part (b) — the flattest glide is the best lift-to-drag ratio. In an
unpowered steady glide at angle $\gamma$ below the horizontal, resolving along and normal to
the flight path gives $D = W\sin\gamma$ and $L = W\cos\gamma$, so
$\tan\gamma = D/L = 1/(L/D)$. The angle is therefore minimised when $L/D$ is greatest, which
for a parabolic polar occurs at
$$C_L = \sqrt{\frac{C_{D0}}{K}} = \sqrt{\frac{0.031}{0.034}} = 0.9549,\qquad
\left(\frac{L}{D}\right)_{max} = \frac{1}{2\sqrt{C_{D0}K}}
= \frac{1}{2\sqrt{0.031 \times 0.034}} = 15.40$$
$$\gamma_{min} = \arctan\!\left(\frac{1}{15.40}\right) = \boxed{3.72^{\circ}}$$
Both $C_L$ and the glide angle are independent of altitude and of weight — only the
speed at which they occur depends on those.
Find the speed at which the flattest glide occurs at 2000 m. At 2000 m,
$T = 275.15\ \text{K}$ and $\rho = 1.0065\ \text{kg}\,\text{m}^{-3}$. Using
$L = W\cos\gamma$ with $\cos 3.72^{\circ} = 0.99789$:
$$V = \sqrt{\frac{2W\cos\gamma}{\rho S C_L}}
= \sqrt{\frac{2 \times 289\,395 \times 0.99789}{1.0065 \times 65 \times 0.9549}}
= \boxed{96.2\ \text{m}\,\text{s}^{-1}}$$
From 2000 m the aeroplane would therefore glide $15.40 \times 2000 = 30.8$ km in still air.
Part (c) — set up the maximum-rate-of-climb condition. In a shallow
steady climb the rate of climb is the excess power divided by the weight:
$$R\!/\!C = \frac{(T - D)V}{W},\qquad
D = \tfrac12\rho V^{2} S C_{D0} + \frac{2KW^{2}}{\rho V^{2} S}$$
Differentiating the excess power $P = TV - D V$ with respect to $V$ and setting
$\mathrm{d}P/\mathrm{d}V = 0$ gives, with $x = V^{2}$,
$$1.5\,\rho S C_{D0}\,x^{2} - T x - \frac{2KW^{2}}{\rho S} = 0$$
Only the positive root is physical.
Repeat at 8000 m. There $T_{8} = 236.15\ \text{K}$,
$\rho = 0.5252\ \text{kg}\,\text{m}^{-3}$ and the lapsed thrust is
$85 \times 0.5252/1.225 = 36.44$ kN. The same cubic-in-$V$ condition gives
$V = 163.9\ \text{m}\,\text{s}^{-1}$ and
$$R\!/\!C = \boxed{9.07\ \text{m}\,\text{s}^{-1}}\ (544\ \text{m/min})$$
The best-climb speed rises with altitude while the climb rate collapses to about one third of
its sea-level value — the behaviour that, extrapolated, defines the ceiling computed in
Question 5(d). Note that the best-climb speed is 1.77 times the sea-level minimum-drag speed (154.2 against 87.2 m/s) but only 1.23 times it at 8000 m (163.9 against 133.3 m/s, part (d)),
because the ratio falls towards one as the lapsed thrust approaches the minimum drag; a best-climb speed below $V_{md}$ would mean the wrong root was taken.
Part (d) — minimum-drag speed at 8000 m. Minimum drag occurs at
the same $C_L = \sqrt{C_{D0}/K} = 0.9549$ found in part (b), so
$$V_{md} = \sqrt{\frac{2W}{\rho S C_L}}
= \sqrt{\frac{2 \times 289\,395}{0.5252 \times 65 \times 0.9549}}
= \boxed{133.3\ \text{m}\,\text{s}^{-1}}$$
The corresponding dynamic pressure is
$q = 0.5 \times 0.5252 \times 133.3^{2} = 4665\ \text{Pa}$.
Split the drag into its two components.
$$D_{parasite} = q S C_{D0} = 4665 \times 65 \times 0.031 = 9.40\ \text{kN}$$
$$D_{induced} = \frac{KW^{2}}{qS} = \frac{0.034 \times (289\,395)^{2}}{4665 \times 65}
= 9.40\ \text{kN}$$
They are equal, which is the defining property of the minimum-drag point, and their
sum
$$D_{min} = 2W\sqrt{C_{D0}K} = 2 \times 289\,395 \times 0.032465
= \boxed{18.79\ \text{kN}}$$
recovers $W/D = 289.4/18.79 = 15.40 = (L/D)_{max}$, closing the loop with part (b).
