Question 1 of 7: Level Flight, Terminal Dive, Pitot-Static Measurement and the Absolute Ceiling

Question 1: Level Flight, Terminal Dive, Pitot-Static Measurement and the Absolute Ceiling (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.

(a) Lift coefficient and drag in level flight

Given. The aircraft cruises straight and level in the standard atmosphere, so its lift must equal its weight and the drag polar then fixes the drag.

Find. The lift coefficient $C_L$ required for level flight and the resulting drag force $D$.

Approach. Evaluate the standard-atmosphere density at 2500 m, obtain the dynamic pressure, set lift equal to weight to get $C_L$, substitute into the polar for $C_D$, and multiply by $qS$ to get the drag.

1. Standard-atmosphere density at 2500 m. In the troposphere the temperature falls linearly and the density follows a power law, $$T = T_0 - Lh = 288.15 - 0.0065(2500) = 271.90\ \text{K}, \qquad \rho = \rho_0\left(\frac{T}{T_0}\right)^{4.2559}$$ so that $\rho = 1.225\,(271.90/288.15)^{4.2559} = 0.9569\ \text{kg}/\text{m}^3$.
2. Dynamic pressure and weight. With $V = 600/3.6 = 166.67\ \text{m}/\text{s}$, $$q = \tfrac{1}{2}\rho V^2 = \tfrac{1}{2}(0.9569)(166.67)^2 = 13\,290\ \text{Pa}, \qquad W = mg = 6500(9.81) = 63\,765\ \text{N}.$$
3. Lift coefficient from the level-flight condition. Steady level flight requires $L = W$, hence $$C_L = \frac{W}{qS} = \frac{63\,765}{13\,290 \times 33} = \boxed{0.1454}$$ which is a very low lift coefficient — the aircraft is flying far faster than its minimum-drag speed, so only a small angle of attack is needed.
4. Drag coefficient and drag. Substituting into the given polar, $$C_D = 0.028 + 0.038(0.1454)^2 = 0.028 + 0.000803 = 0.02880,$$ and therefore $$D = q S C_D = 13\,290 \times 33 \times 0.02880 = \boxed{12\,630\ \text{N} = 12.63\ \text{kN}}$$

The induced contribution is only $0.0008$ of a total $0.0288$, i.e. under 3 % of the drag; at this speed the aeroplane is almost entirely parasite-drag limited. The corresponding lift-to-drag ratio is $L/D = 63\,765/12\,630 = 5.05$, well below the maximum of $1/(2\sqrt{C_{D0}K}) = 15.3$ available to this polar, which confirms that the aircraft is flying well to the fast side of its minimum-drag point.

(b) Maximum speed in a vertical dive

Given. The same aircraft ($m = 6500\ \text{kg}$, $S = 33\ \text{m}^2$, $C_{D0} = 0.028$) is put into a vertical dive at 1500 m with the engine throttled back, so thrust is zero and the lift coefficient is zero.

Find. The maximum (terminal) velocity attainable in the dive.

Approach. In a vertical dive the flight path is along the weight vector, so the equation of motion along the path is $m\,dV/dt = W - D$. The maximum speed is the terminal condition $D = W$, evaluated with $C_L = 0$ so that $C_D = C_{D0}$ only.

1. Density at 1500 m. $T = 288.15 - 0.0065(1500) = 278.40\ \text{K}$ and $$\rho = 1.225\,(278.40/288.15)^{4.2559} = 1.0581\ \text{kg}/\text{m}^3.$$
2. Terminal-velocity condition. Setting drag equal to weight with $C_D = C_{D0} = 0.028$, $$W = \tfrac{1}{2}\rho V^2 S\,C_{D0} \quad\Longrightarrow\quad V = \sqrt{\frac{2W}{\rho S\,C_{D0}}}.$$
3. Substitution. $$V = \sqrt{\frac{2(63\,765)}{1.0581 \times 33 \times 0.028}} = \sqrt{130\,440} = \boxed{361\ \text{m}/\text{s} = 1300\ \text{km}/\text{h}}$$

This answer must be closed with a physical check, because it is the point of the question. The speed of sound at 1500 m is $a = \sqrt{\gamma R T} = \sqrt{1.4(287.05)(278.40)} = 334.5\ \text{m}/\text{s}$, so the predicted dive speed corresponds to $M = 361/334.5 = 1.08$. The drag polar supplied in part (a) is a low-speed, incompressible polar, and at $M \approx 1$ the zero-lift drag coefficient of an aeroplane of this class rises by a factor of three or more. The genuine terminal speed of the real aircraft is therefore substantially lower than 361 m/s, and in practice such an aeroplane would be limited by a placarded $V_{NE}$ or by Mach buffet long before it reached this figure. The number above is the correct answer to the question as posed, with the given constant $C_{D0}$; the compressibility caveat should be stated alongside it.

(c) Pitot–static pressure difference

Given. $V = 340\ \text{km/h} = 94.44\ \text{m}/\text{s}$, ambient static pressure $p = 74\ \text{kPa}$, ambient temperature $T = 271\ \text{K}$.

Find. The difference $p_0 - p$ between the Pitot (stagnation) pressure and the static pressure, which is the quantity the airspeed indicator actually senses.

Approach. A Pitot tube brings the flow isentropically to rest. The Mach number is well below 0.3 only marginally here, so the compressible isentropic relation is used and then compared with the incompressible result $\tfrac{1}{2}\rho V^2$ to show the size of the compressibility correction.

1. Local speed of sound and Mach number. $$a = \sqrt{\gamma R T} = \sqrt{1.4(287.05)(271)} = 330.0\ \text{m}/\text{s}, \qquad M = \frac{94.44}{330.0} = 0.2862.$$
2. Isentropic stagnation pressure. For a perfect gas brought to rest without loss, $$\frac{p_0}{p} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)} = \left(1 + 0.2 M^2\right)^{3.5}.$$ With $M = 0.2862$, $\;0.2M^2 = 0.016382$ and $p_0/p = (1.016382)^{3.5} = 1.05852$.
3. Pitot–static difference. $$p_0 - p = p\left[(1 + 0.2M^2)^{3.5} - 1\right] = 74\,000(0.05852) = \boxed{4330\ \text{Pa} = 4.33\ \text{kPa}}$$
4. Incompressible comparison. The ambient density from the perfect-gas law is $\rho = p/(RT) = 74\,000/(287.05 \times 271) = 0.9513\ \text{kg}/\text{m}^3$, so Bernoulli's incompressible form gives $\tfrac{1}{2}\rho V^2 = \tfrac{1}{2}(0.9513)(94.44)^2 = 4243\ \text{Pa}$.

The incompressible estimate is 2.0 % low. That is the practical message of the calculation: at $M \approx 0.29$ the compressibility correction is already at the level of a normal instrument tolerance, and above about $M = 0.3$ an airspeed system calibrated on $\tfrac{1}{2}\rho V^2$ must be corrected, which is exactly why an air data computer works in terms of impact pressure and Mach number rather than dynamic pressure.

(d) The absolute ceiling

The absolute ceiling is the altitude at which an aircraft's maximum rate of climb falls to zero — the greatest height at which it can sustain steady level flight. Above it, no throttle setting and no airspeed exist for which thrust available equals drag required, so level flight is impossible. Because the last few hundred metres are approached asymptotically (the climb rate decays towards zero, so the time to reach the absolute ceiling is theoretically infinite), certification uses instead the service ceiling, the altitude at which the best rate of climb has fallen to a small specified value, typically 0.5 m/s (100 ft/min) for jet transports.

The estimate follows from the fact that thrust available falls with altitude while the minimum drag the aircraft can fly at does not. For a parabolic polar the minimum drag is $D_{\min} = 2W\sqrt{C_{D0}K}$, and it is independent of density: as the air thins the aeroplane simply flies faster at the same lift coefficient $C_{L,md} = \sqrt{C_{D0}/K}$. Meanwhile the thrust of a fixed-geometry jet falls roughly in proportion to density, $T = T_0(\rho/\rho_0)$. Equating the two at the ceiling gives the density there directly,

after which the ISA is inverted to convert that density into an altitude. Two practical cautions apply. First, one must check which layer the answer falls in: if $\rho_{ceiling}$ is below the 11 km tropopause value of $0.3639\ \text{kg}/\text{m}^3$ the tropospheric power law is invalid and the isothermal stratospheric relation must be used instead — this is precisely the situation in Question 5(e) below. Second, the graphical method is equivalent and often more instructive: plot maximum rate of climb against altitude (the curve is very nearly linear for a jet) and extrapolate to $R/C = 0$. For a real aeroplane the estimate must also be checked against buffet and Mach limits, because a high-altitude aeroplane frequently reaches its aerodynamic "coffin corner" — the point where the stall speed and the buffet-onset Mach number converge — before it reaches the thrust-limited ceiling.