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

Question 1 of 7: Altitudes, Mach-Limited Speed and Stagnation Pressure

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 1: Altitudes, Mach-Limited Speed and Stagnation Pressure (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	Symbol	Value
Measured ambient pressure (a)	$p$	26 kPa
Measured ambient temperature (a)	$T$	−39°C = 234.15 K
Limiting Mach number (b)	$M$	0.86
Altitudes of interest (b)	$h$	sea level and 9000 m
Flight speed (c)	$V$	210 m/s
Altitude (c)	$h$	5000 m

Find. The pressure, temperature and density altitudes corresponding to the measured ambient conditions; the true airspeed at the Mach limit at two altitudes; and the largest pressure appearing anywhere on the aircraft surface in the third flight condition.

Approach. Each "altitude" is the height in the standard atmosphere at which one property matches the measured value, so each is obtained by inverting the ISA relation for that property; the Mach limit converts to true airspeed through the local speed of sound; and the highest surface pressure is the stagnation pressure at the forward stagnation point, which at $M = 0.655$ must be computed with the compressible isentropic relation rather than Bernoulli's equation.

Part (a) — write the three ISA relations to be inverted. In the troposphere the temperature falls linearly and the pressure and density follow as powers of the temperature ratio: $$T = T_0 - Lh,\qquad \frac{p}{p_0} = \left(\frac{T}{T_0}\right)^{g/(LR)},\qquad \frac{\rho}{\rho_0} = \left(\frac{T}{T_0}\right)^{g/(LR)-1}$$ with $g/(LR) = 9.80665/(0.0065 \times 287.05) = 5.2559$, so the density exponent is 4.2559. Each altitude is found by setting the relevant property equal to the measured one and solving for $h$.
Temperature altitude — invert the lapse law. This is the direct one, since temperature appears linearly: $$h_T = \frac{T_0 - T}{L} = \frac{288.15 - 234.15}{0.0065} = \boxed{8308\ \text{m}}$$ The measured air is as cold as standard air at about 8.3 km.
Pressure altitude — invert the pressure ratio. With $p/p_0 = 26\,000/101\,325 = 0.25660$, $$\frac{T}{T_0} = (0.25660)^{1/5.2559} = 0.77198 \quad\Longrightarrow\quad h_p = \frac{T_0}{L}\left(1 - 0.77198\right)$$ $$h_p = \frac{288.15}{0.0065}\,(0.22802) = \boxed{10\,108\ \text{m}}$$ The aircraft's altimeter, which senses pressure, would read about 10.1 km.
Ambient density from the perfect gas equation, as instructed. $$\rho = \frac{p}{RT} = \frac{26\,000}{287.05 \times 234.15} = 0.3868\ \text{kg}\,\text{m}^{-3}$$ This is the actual density; the density altitude follows by asking where the standard atmosphere is this thin.
Density altitude — invert the density ratio. With $\rho/\rho_0 = 0.3868/1.225 = 0.31577$, $$\frac{T}{T_0} = (0.31577)^{1/4.2559} = 0.76274 \quad\Longrightarrow\quad h_\rho = \frac{288.15}{0.0065}\,(0.23726) = \boxed{10\,518\ \text{m}}$$ The three answers order themselves as $h_T \lt h_p \lt h_\rho$: at the measured pressure the air is warmer than standard (234.15 K against the 222.4 K standard value at 10.1 km), so it is thinner than standard and the density altitude sits above the pressure altitude. That ordering is the physical check on the arithmetic.
Part (b) — convert the Mach limit at sea level. True airspeed follows the local speed of sound, $V = M a$ with $a = \sqrt{\gamma R T}$: $$a_0 = \sqrt{1.4 \times 287.05 \times 288.15} = 340.3\ \text{m}\,\text{s}^{-1} \quad\Longrightarrow\quad V_0 = 0.86 \times 340.3 = \boxed{292.7\ \text{m}\,\text{s}^{-1}}$$ that is 1054 km/h.
Repeat at 9000 m, where the air is colder. $T = 288.15 - 0.0065 \times 9000 = 229.65\ \text{K}$, so $$a = \sqrt{1.4 \times 287.05 \times 229.65} = 303.8\ \text{m}\,\text{s}^{-1} \quad\Longrightarrow\quad V = 0.86 \times 303.8 = \boxed{261.3\ \text{m}\,\text{s}^{-1}}$$ that is 940.5 km/h. The Mach-limited true airspeed falls with altitude by about 11 per cent, purely because the speed of sound depends on temperature alone.
Part (c) — identify where the highest pressure acts. The largest pressure anywhere on the surface occurs where the flow is brought to rest relative to the aircraft, at the forward stagnation point on the nose; its value is the stagnation (total) pressure $p_{0}$. At 5000 m the standard atmosphere gives $T = 255.65\ \text{K}$, $p = 54.02\ \text{kPa}$, $\rho = 0.7361\ \text{kg}\,\text{m}^{-3}$, and $$M = \frac{V}{\sqrt{\gamma R T}} = \frac{210}{\sqrt{1.4 \times 287.05 \times 255.65}} = \frac{210}{320.5} = 0.655$$
Apply the compressible isentropic stagnation relation. At $M = 0.655$ the density change through the stagnation process is not negligible, so $$p_{0} = p\left(1 + \frac{\gamma-1}{2}M^{2}\right)^{\gamma/(\gamma-1)} = 54.02\,\bigl(1 + 0.2 \times 0.655^{2}\bigr)^{3.5}$$ $$p_{0} = 54.02 \times 1.3341 = \boxed{72.1\ \text{kPa}}$$ For comparison, the incompressible estimate $p + \tfrac12\rho V^2 = 54.02 + 16.23 = 70.25\ \text{kPa}$ is 2.5 per cent low — a useful reminder that above roughly $M = 0.3$ the Bernoulli form of the dynamic pressure understates the stagnation rise.

Result	Value
(a) Temperature altitude	8308 m
(a) Pressure altitude	10 108 m
(a) Ambient density (perfect gas)	0.3868 kg·m⁻³
(a) Density altitude	10 518 m
(b) Maximum speed at sea level	292.7 m/s (1054 km/h)
(b) Maximum speed at 9000 m	261.3 m/s (940.5 km/h)
(c) Highest surface pressure (stagnation)	72.1 kPa

Topic. The four altitudes of the standard atmosphere, and stagnation conditions in compressible flow.

Key relations. Troposphere lapse $T = T_0 - Lh$; pressure and density ratios $(T/T_0)^{g/LR}$ and $(T/T_0)^{g/LR-1}$; perfect gas $p = \rho R T$; speed of sound $a = \sqrt{\gamma R T}$; isentropic stagnation $p_0/p = (1 + \tfrac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)}$.

Why this works. Geometric altitude is what a tape measure would give; pressure, temperature and density altitudes are defined quantities — each is the height in the standard atmosphere at which that one property matches what the aircraft actually measures. They coincide only when the real atmosphere happens to be standard. The distinction matters operationally: the altimeter reports pressure altitude, while aircraft performance (lift, thrust, propeller efficiency, take-off distance) depends on density, so it is density altitude that determines whether the aeroplane will clear the obstacle at the end of a hot-and-high runway. The stagnation calculation rests on a different idea: along a streamline that stagnates on the nose, the flow is decelerated isentropically, and the recovered pressure exceeds the static pressure by a factor that grows faster than $\tfrac12\rho V^2$ once the density itself starts to rise appreciably.

Common pitfall. Two recur. First, using the density exponent 5.2559 instead of 4.2559 (or vice versa) — the two differ by exactly one because $\rho \propto p/T$. Second, computing the highest surface pressure from $p + \tfrac12 \rho V^2$ at Mach numbers well above 0.3; here that costs 1.8 kPa. A third, subtler error is computing the density altitude from the ISA density at the pressure altitude rather than from the measured $p$ and $T$ — that would return the pressure altitude again and hide the very effect the question is testing.

Source. Anderson, Introduction to Flight, 9th ed., §3.4–3.5 (standard atmosphere and the four altitudes) and §8.7 (isentropic stagnation relations).

← Paper overview Next: Question 2 →