Isentropic Temperature Ratio

Stop a moving gas and it gets hotter. Nothing was added — the kinetic energy the flow was carrying has simply turned back into thermal energy, and the temperature it reaches is called the stagnation or total temperature. The relation is $T_0/T = 1 + \frac{\gamma-1}{2}M^{2}$, and it is nothing more than the steady-flow energy equation written for a perfect gas.

Read it in the nozzle direction and it explains what a nozzle is for. The combustion chamber is essentially at rest, so the chamber temperature IS the stagnation temperature — 3500 K for a good kerolox mixture. As the gas accelerates out through the bell, its static temperature falls by exactly this factor: at Mach 3 with $\gamma = 1.2$ the ratio is 1.9, so 3500 K becomes 1842 K at the exit. Every kelvin of that drop has been converted into exhaust velocity, which is the whole transaction the nozzle exists to perform.

Read it in the flight direction and it explains why fast aircraft get hot. At Mach 3 in air, $T_0/T = 2.8$: a stratospheric ambient of 220 K becomes 616 K at a stagnation point, which is why the SR-71 was built of titanium and why re-entry is a thermal problem rather than a mechanical one. Note that a real surface does not quite reach the full stagnation temperature — boundary-layer effects give a recovery factor around 0.85 to 0.9 for turbulent flow — but the equation sets the ceiling.

Two properties make this relation unusually well behaved. Stagnation temperature is CONSERVED along an adiabatic duct even when the static temperature is changing wildly, and it is conserved even ACROSS A SHOCK, which the pressure ratio is not. That makes total temperature the natural bookkeeping variable in compressible flow: it stays put while everything else moves. It works in kelvin only, and the classic slip is feeding it Celsius or Fahrenheit — a ratio taken between two non-absolute temperatures is not a ratio of anything.