Question 1 of 7: The Standard Atmosphere, Altitude Definitions and Bernoulli’s Equation

Question 1: The Standard Atmosphere, Altitude Definitions and Bernoulli’s Equation (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.

Part (a) — the assumed temperature distributions and how they generate the pressure variation. The Standard Atmosphere is not a measurement; it is an agreed model built from one assumed temperature-altitude relation per layer, plus two pieces of physics that are common to every layer. In the troposphere, from sea level to the tropopause at $11\ \text{km}$, the temperature is assumed to fall linearly with geopotential altitude at the constant lapse rate $L = 6.5\ \text{K}/\text{km}$, so that $T = T_0 - L h$ with $T_0 = 288.15\ \text{K}$; at the tropopause this gives $216.65\ \text{K}$. In the lower stratosphere, from $11\ \text{km}$ to about $20\ \text{km}$, the temperature is assumed constant at that tropopause value — the layer is isothermal. (Above $20\ \text{km}$ the model resumes a positive gradient, but that is beyond anything asked here.)

The two pieces of common physics are the hydrostatic equation and the perfect-gas equation of state. The atmosphere is assumed to be in static equilibrium, so the weight of a slab of air of thickness $dh$ is carried by the pressure difference across it:

$$dp = -\rho\, g\, dh, \qquad p = \rho R T .$$

Eliminating the density between them separates the variables and leaves a relation in which the only unknown function is the assumed temperature distribution:

$$\frac{dp}{p} = -\frac{g}{R\,T(h)}\, dh .$$

In the troposphere we substitute $T = T_0 - L h$, so $dh = -dT/L$, and the integral becomes $\int dp/p = (g/LR)\int dT/T$. Integrating from sea level gives the power law

$$\frac{p}{p_0} = \left(\frac{T}{T_0}\right)^{g/(LR)} = \left(\frac{T}{T_0}\right)^{5.2559}, \qquad \frac{\rho}{\rho_0} = \left(\frac{T}{T_0}\right)^{g/(LR)-1} = \left(\frac{T}{T_0}\right)^{4.2559},$$

the density result following at once from $p = \rho R T$. In the stratosphere $T$ is a constant, $T_1 = 216.65\ \text{K}$, so it comes straight out of the integral and the result is an exponential decay measured from the tropopause conditions $p_1, \rho_1$ at $h_1 = 11\ \text{km}$:

$$\boxed{\ \frac{p}{p_1} = \frac{\rho}{\rho_1} = \exp\!\left[-\frac{g\,(h-h_1)}{R\,T_1}\right]\ }$$

So the assumed temperature distribution is doing all of the work: a gradient layer always integrates to a power law whose exponent is fixed by the lapse rate, and an isothermal layer always integrates to a simple exponential whose scale height is $R T_1/g \approx 6.34\ \text{km}$. Pressure and density are continuous at the join, which is what allows the layers to be chained upwards.

Part (b) — pressure, temperature and density altitude. The real atmosphere is never the standard one, so each of these three terms answers the question “at what altitude in the Standard Atmosphere would I find the property I have just measured?” The pressure altitude is the altitude in the Standard Atmosphere at which the standard pressure equals the ambient static pressure actually measured. It is what a barometric altimeter set to 1013.25 hPa displays, and because every aircraft in controlled airspace uses the same setting above the transition altitude, it is the quantity that guarantees vertical separation even when it is not the true height above the ground. The temperature altitude is the altitude in the Standard Atmosphere at which the standard temperature equals the measured ambient temperature. The density altitude is the altitude in the Standard Atmosphere at which the standard density equals the actual ambient density; equivalently it is the pressure altitude corrected for the departure of temperature from standard, since $\rho = p/RT$.

Density altitude is the one that matters for performance. Lift, drag and jet thrust all scale with $\rho$, so an aircraft performs as though it were at its density altitude, not at its geometric height: on a hot day at a high-elevation aerodrome the density altitude can exceed the geometric elevation by a kilometre or more, lengthening the take-off run and flattening the climb gradient. Note that the three altitudes need not be in any fixed order — air that is warmer than standard at a given pressure is thinner than standard, which pushes the density altitude above the pressure altitude.

Part (c) — the highest speed at $M = 0.82$ and $10\,000\ \text{m}$.

Given. The aircraft is limited to a maximum Mach number $M_{max} = 0.82$ and is flying at $h = 10\,000\ \text{m}$ in the Standard Atmosphere, which is inside the troposphere.

Find. The corresponding true airspeed.

Approach. Get the standard temperature at $10\ \text{km}$ from the tropospheric lapse rate, convert it to the local speed of sound, and multiply by the limiting Mach number.

1. Standard temperature at 10 km. The altitude is below the tropopause, so the linear law applies: $T = T_0 - L h = 288.15 - 0.0065(10\,000) = 223.15\ \text{K}$ (that is $-50.0\ ^\circ\text{C}$).
2. Local speed of sound. For a perfect gas $a = \sqrt{\gamma R T}$, which depends on temperature alone: $a = \sqrt{1.4 \times 287.05 \times 223.15} = 299.5\ \text{m}/\text{s}$.
3. Convert the Mach limit to a true airspeed. By definition $M = V/a$, so $V = M_{max}\, a = 0.82 \times 299.5$ and $$\boxed{\ V_{max} = 245.6\ \text{m}/\text{s} = 884\ \text{km}/\text{hr}\ }$$

The limit is a Mach limit, not a speed limit, so the answer falls as the aircraft climbs through the troposphere: the same $M = 0.82$ at sea level, where $a = 340.3\ \text{m}/\text{s}$, would correspond to $279\ \text{m}/\text{s}$. This is why a jet cruising at constant Mach number appears to slow down on the true-airspeed readout as it steps up to a higher flight level.

Part (d) — mean pressures on the upper and lower wing surfaces.

Given. Level flight at $V_\infty = 88\ \text{m}/\text{s}$ and $h = 2500\ \text{m}$ in the Standard Atmosphere, with mean surface velocities $V_u = 99\ \text{m}/\text{s}$ over the upper surface and $V_l = 71\ \text{m}/\text{s}$ over the lower surface.

Find. The mean static pressure acting on each surface.

Approach. Evaluate the standard pressure and density at $2500\ \text{m}$, then apply Bernoulli’s equation along a streamline from the free stream to each surface; the free-stream Mach number is far below 0.3, so the incompressible form is legitimate.

1. Standard temperature at 2500 m. $T = 288.15 - 0.0065(2500) = 271.90\ \text{K}$.
2. Standard pressure from the tropospheric power law. $p_\infty = p_0 (T/T_0)^{5.2559} = 101\,325 \times (271.90/288.15)^{5.2559} = 74\,682\ \text{Pa}$.
3. Standard density. Either from the companion exponent or directly from the equation of state: $\rho_\infty = p_\infty/(R T) = 74\,682/(287.05 \times 271.90) = 0.9569\ \text{kg}/\text{m}^3$.
4. Check that the incompressible form is admissible. The local speed of sound is $a = \sqrt{1.4 \times 287.05 \times 271.90} = 330.6\ \text{m}/\text{s}$, so $M_\infty = 88/330.6 = 0.266$ and the fastest surface flow is at $M = 0.299$. Both are at or below 0.3, where the density change is under 5 per cent and Bernoulli’s incompressible equation is the standard tool.
5. Upper surface. Bernoulli along a streamline gives $p_u = p_\infty + \tfrac{1}{2}\rho_\infty (V_\infty^2 - V_u^2)$, so $p_u = 74\,682 + 0.4784\,(88^2 - 99^2) = 74\,682 - 984$ and $$\boxed{\ p_u = 73\,698\ \text{Pa} = 73.70\ \text{kPa}\ }$$
6. Lower surface. The same equation with $V_l$ in place of $V_u$: $p_l = 74\,682 + 0.4784\,(88^2 - 71^2) = 74\,682 + 1293$ and $$\boxed{\ p_l = 75\,975\ \text{Pa} = 75.98\ \text{kPa}\ }$$

The upper surface is below free-stream pressure and the lower surface above it, which is the correct sense: the wing lifts because the flow is accelerated over the top and retarded underneath. The mean pressure difference is $2277\ \text{Pa}$, so a wing of, say, $30\ \text{m}^2$ carrying this loading would generate roughly $68\ \text{kN}$ of lift — the right order for a light transport. Note also that the suction on the upper surface ($-984\ \text{Pa}$) and the compression underneath ($+1293\ \text{Pa}$) are of comparable size here only because the two surface velocities happen to straddle the free stream nearly symmetrically; on a real cambered wing at cruise the upper-surface suction usually supplies about two-thirds of the lift.