Fundamentals8 min read2 / 9

The atmosphere jets fly in

A jet engine’s working fluid is whatever the sky serves at that altitude — and the sky’s menu changes drastically between the runway and cruise. Every performance number in the engine demos starts by asking the International Standard Atmosphere for the local temperature, pressure and density, so we build that model first.

Two rules make the whole model

Rule 1 — temperature. In the troposphere (0 to 11 km) air is heated from below by the ground and mixed by weather, and temperature falls linearly with altitude at the lapse rate:

T(h) = T_{sl} - \Lambda h, \qquad T_{sl}=288.15\ \text{K},\quad \Lambda = 6.5\ \text{K/km}

(1)

Above the tropopause at 11 km the mixing stops and temperature holds constant at 216.65 K (−56.5 °C) through the lower stratosphere — the demo’s envelope ends at 20 km, still inside this isothermal layer.

Rule 2 — pressure. Each layer of air must carry the weight of everything above it (hydrostatic balance, $dp = -\rho g\,dh$ ). Combine with the ideal-gas law $\rho = p/(RT)$ and integrate. With the linear $T(h)$ of the troposphere the result is a power law; in the isothermal stratosphere it becomes a pure exponential:

\frac{p}{p_{sl}} = \left(\frac{T}{T_{sl}}\right)^{\tfrac{g}{\Lambda R}} \quad (h \le 11\,\text{km}), \qquad p = p_{11}\,e^{-\tfrac{g\,(h-11\,\text{km})}{R\,T_{11}}} \quad (h > 11\,\text{km})

(2)

$R$: specific gas constant of air, 287.05 [J/(kg·K)]
$g$: gravitational acceleration, 9.81 [m/s²]
$g/(\Lambda R)$: the tropospheric exponent ≈ 5.26 [–]

Fig. 1. The profiles the demos actually use, sampled from the code. The temperature line kinks at the tropopause; pressure never kinks — it just keeps decaying, ~22% of sea level at 11 km and ~5.5% at 20 km.

The speed of sound rides on temperature

Mach number — the unit every jet flies by — is speed over the local speed of sound, and sound speed depends on temperature alone:

a = \sqrt{\gamma R T}

(3)

At sea level $a = \sqrt{1.4 \times 287 \times 288} \approx 340\ \text{m/s}$ ; at cruise it is only 295 m/s. So the same Mach number is a slower true airspeed up high — and everything aerodynamic (shock waves, the compressibility effects of the aerodynamics course) follows Mach, not m/s. This is why the demos take Mach + altitude as inputs and derive velocity, never the reverse.

Altitude

h

11000 m

T = 216.6\ \text{K}\ (-57\,^\circ\text{C})

p = 22.6\ \text{kPa},\ \ \rho = 0.364\ \tfrac{\text{kg}}{\text{m}^3},\ \ a = 295\ \tfrac{\text{m}}{\text{s}}

Stratosphere: temperature holds at −56.5 °C while pressure keeps falling exponentially. Cold air is exactly what a jet wants — the same Tt4 buys a bigger temperature ratio τλ.

Why airliners cruise at 11 km

Thin air = less drag. Drag scales with density; at the tropopause the aircraft pushes through a quarter of the sea-level air for the same true airspeed.
Cold air = a happier cycle. The engine’s key non-dimensional number is $\tau_\lambda = T_{t4}/T_0$ — turbine-inlet temperature over ambient. The metallurgy fixes $T_{t4}$ ; a colder sky raises the ratio and with it thermal efficiency and specific work. −56 °C is a gift.
But thin air also starves the engine. Thrust scales with mass flow, and mass flow with density: an engine that makes 300 kN on the runway makes ~60 kN at cruise. Aircraft climb until available thrust meets required drag — that balance point is cruise altitude.