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

Question 1 of 7: Standard-atmosphere altitudes, Pitot-static reading and stagnation pressure

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 1: Standard-atmosphere altitudes, Pitot-static reading and stagnation pressure (20 marks — a 8, b 6, c 6)

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) geometric cruise altitude	$10{,}000\ \text{m}$
(a) measured ambient pressure	$p=24.5\ \text{kPa}$
(a) measured ambient temperature	$T=-40\,{}^{\circ}\text{C}=233.15\ \text{K}$
(b) true airspeed and altitude	$240\ \text{km/h}=66.667\ \text{m}\,\text{s}^{-1}$ at $2500\ \text{m}$
(c) sea-level free-stream speed	$70\ \text{m}\,\text{s}^{-1}$
ISA sea-level datum	$p_{0}=101.325\ \text{kPa}$, $T_{0}=288.15\ \text{K}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$
troposphere lapse rate and gas constant	$L=0.0065\ \text{K}\,\text{m}^{-1}$, $R=287.05\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$

Find. The pressure, temperature and density altitudes corresponding to the measured cruise state, the ambient density that produces the density altitude, the Pitot-minus-static pressure difference at the second condition, and the maximum (stagnation) pressure on the blunt body at the third.

The four altitudes compared. Only the geometric height is a geometric fact; the other three are ISA look-ups — the height at which the standard atmosphere reproduces the measured temperature, pressure and density respectively.

Approach. Each of the three altitudes is the height at which the standard atmosphere alone reproduces one measured property, so each is found by writing the ISA relation for that property and inverting it for height; parts (b) and (c) are then two applications of the same incompressible Bernoulli statement, once to read dynamic pressure off a probe and once to add it to the ambient static pressure at a stagnation point.

Part (a) — write the troposphere model and get the temperature altitude directly. Below $11\ \text{km}$ the ISA temperature falls linearly, $T=T_{0}-Lh$, so the height at which the standard atmosphere is as cold as the measurement follows immediately: $$\begin{aligned} h_{T} &= \frac{T_{0}-T}{L} \\ &= \frac{288.15-233.15}{0.0065} \\ &= 8461.5\ \text{m} \end{aligned}$$ The aeroplane is at $10{,}000\ \text{m}$ but the air around it is as warm as standard air at only $8460\ \text{m}$: this day is warmer than standard aloft.
Invert the ISA pressure law for the pressure altitude. Integrating the hydrostatic equation with the linear lapse rate gives the pressure ratio as a power of the temperature ratio, $$\begin{aligned} \frac{p}{p_{0}} &= \left(1-\frac{Lh}{T_{0}}\right)^{g/(LR)}, \\ \frac{g}{LR} &= \frac{9.80665}{0.0065\times287.05}=5.2559 \end{aligned}$$ Solving that relation for the bracket and then for the height, $$\begin{aligned} \theta_{p}&=\left(\frac{24.5}{101.325}\right)^{1/5.2559}=(0.24180)^{0.19026}=0.76325\\ h_{p}&=\frac{T_{0}}{L}\,(1-\theta_{p})=\frac{288.15}{0.0065}\times0.23675=10{,}495\ \text{m} \end{aligned}$$ so a pressure altimeter set to $1013.25\ \text{hPa}$ would indicate about $10{,}495\ \text{m}$, roughly $500\ \text{m}$ above the true height.
Get the ambient density from the perfect-gas equation. The question asks for the density built from the two measured values rather than read from a table, which is what makes the density altitude a genuinely independent third answer: $$\begin{aligned} \rho &= \frac{p}{RT} \\ &= \frac{24\,500}{287.05\times233.15} \\ &= 0.3661\ \text{kg}\,\text{m}^{-3} \end{aligned}$$
Invert the ISA density law for the density altitude. Dividing the pressure law by the temperature law lowers the exponent by one: $$\frac{\rho}{\rho_{0}}=\left(1-\frac{Lh}{T_{0}}\right)^{g/(LR)-1} =\left(1-\frac{Lh}{T_{0}}\right)^{4.2559}$$ With $\rho/\rho_{0}=0.3661/1.225=0.29883$ the bracket is $\theta_{\rho}=(0.29883)^{1/4.2559}=0.75292$, and therefore $$\boxed{\;h_{T}=8462\ \text{m},\quad h_{p}=10{,}495\ \text{m},\quad \rho=0.366\ \text{kg}\,\text{m}^{-3},\quad h_{\rho}=10{,}953\ \text{m}\;}$$ The density altitude is the highest of the three, which is the physically important statement: engine and wing behave as though the aeroplane were nearly a kilometre higher than it is.
Part (b) — evaluate the standard density at 2500 m. The question directs that ISA density be used, so the same power law is applied forwards rather than inverted: $$\begin{aligned} \theta &= 1-\frac{0.0065\times2500}{288.15}=0.94361, \\ \rho_{2500} &= 1.225\times(0.94361)^{4.2559}=0.9569\ \text{kg}\,\text{m}^{-3} \end{aligned}$$
Read the probe difference as dynamic pressure. For incompressible flow the Pitot hole senses $p_{0}=p+\tfrac{1}{2}\rho V^{2}$ while the static ports sense $p$, so the cell between them reads exactly the dynamic pressure. With $V=240/3.6=66.667\ \text{m}\,\text{s}^{-1}$, $$\boxed{\;p_{0}-p=\tfrac{1}{2}\rho V^{2} =\tfrac{1}{2}\times0.9569\times(66.667)^{2}=2126\ \text{Pa}=2.13\ \text{kPa}\;}$$ The Mach number here is only about $0.20$, so the incompressible reading is accurate to well under one per cent and no compressibility correction is needed.

The probe senses stagnation pressure at the nose hole and ambient static pressure at the flush side ports; the cell between them reads the dynamic pressure directly.

Part (c) — add the sea-level dynamic pressure to ambient. The highest pressure anywhere on a body in a steady incompressible stream is the stagnation pressure at the forward stagnation point, where the flow has been brought to rest isentropically along the dividing streamline: $$\boxed{\;p_{0}=p_{\infty}+\tfrac{1}{2}\rho_{0}V^{2} =101\,325+\tfrac{1}{2}\times1.225\times70^{2}=101\,325+3001=104{,}326\ \text{Pa}\;}$$ that is $104.3\ \text{kPa}$, about three per cent above ambient. Nowhere on the body can the pressure exceed this value, because any other point still carries some kinetic energy.

Quantity	Result
(a) temperature altitude	$8462\ \text{m}$
(a) pressure altitude	$10{,}495\ \text{m}$
(a) ambient density from $p$ and $T$	$0.366\ \text{kg}\,\text{m}^{-3}$
(a) density altitude	$10{,}953\ \text{m}$
(b) Pitot minus static pressure	$2126\ \text{Pa}=2.13\ \text{kPa}$
(c) highest pressure on the blunt body	$104{,}326\ \text{Pa}=104.3\ \text{kPa}$

Topic. The three equivalent altitudes of the International Standard Atmosphere, and incompressible stagnation pressure.

Key relations. Troposphere temperature $T=T_{0}-Lh$; pressure $p/p_{0}=(1-Lh/T_{0})^{g/LR}$ with $g/LR=5.2559$; density $\rho/\rho_{0}=(1-Lh/T_{0})^{g/LR-1}$; perfect gas $p=\rho RT$; Bernoulli $p+\tfrac{1}{2}\rho V^{2}=\text{constant}$ along a streamline.

Why this works. The standard atmosphere is a one-parameter family: fix the height and every property follows. Turn that statement around and each property, measured on a real non-standard day, can be quoted as the height at which standard air would have shown the same value. Because the three properties are linked by the gas law but the real day is not standard, the three inverted heights disagree, and the disagreement is exactly the information a pilot needs. Temperature altitude says how warm the day is; pressure altitude is what the altimeter reads and what air-traffic separation is based on; density altitude governs lift, thrust and propeller performance, and is therefore the number quoted in take-off charts. In part (c) the same Bernoulli constant that lets a probe measure speed also bounds the surface pressure, because the stagnation point is the only place on the body where the whole dynamic pressure has been recovered.

Common pitfall. Two errors dominate. The first is using the pressure-altitude exponent $5.2559$ for the density inversion; the density exponent is one less, $4.2559$, and using the wrong one shifts the density altitude by several hundred metres. The second is treating the Pitot-static reading as an absolute pressure rather than a difference — the probe reads $\tfrac{1}{2}\rho V^{2}$, not $p_{0}$, and an airspeed indicator converts that difference using sea-level density, which is why indicated airspeed falls below true airspeed as height increases.

Source. Anderson, Introduction to Flight, 9th ed., §3.4 (the standard atmosphere and its three altitudes) and §4.11 (the Pitot-static tube and stagnation pressure).

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