Advanced10 min read8 / 9

Ramjets & scramjets

Fly fast enough and the compressor becomes dead weight: simply slowing down the oncoming air compresses it more than any spinning machinery could. The ramjet is that realisation with fuel injectors — no compressor, no turbine, no moving parts. The scramjet is what you are forced to build when even slowing the air down becomes impossible to afford.

Ram compression is free — and it compounds

Bring a Mach $M_0$ flow smoothly to rest and its temperature and pressure rise by:

\tau_r = 1 + \frac{\gamma-1}{2}M_0^2, \qquad \pi_r = \tau_r^{\gamma/(\gamma-1)}

(1)

Fig. 1. Equation (1) plotted. The temperature ratio (orange) reads on the linear left axis; the pressure ratio (blue) grows so violently it needs the logarithmic right axis. At Mach 3 the free pressure ratio is already ~37 — matching a whole modern compressor. At Mach 5 it passes 500. No rotating machinery could add anything useful.

The ideal ramjet is therefore the Brayton cycle with $\tau_c = 1$ : inlet compression, constant-pressure burn to $T_{t4}$ , nozzle expansion. With no turbine to feed, the exit velocity collapses to something beautiful:

V_9 = V_0\,\sqrt{\frac{\tau_\lambda}{\tau_r}}

(2)

— the exhaust is the flight velocity, amplified by how much hotter the burner makes the gas than the inlet already did. Specific thrust is $V_0(\sqrt{\tau_\lambda/\tau_r} - 1)$ , and every property of the ramjet reads straight off it:

Zero thrust at zero speed. $V_0 = 0 \Rightarrow F = 0$ . A ramjet cannot taxi, cannot take off, cannot even idle — it must be boosted to Mach ~2 by a rocket or a turbojet before it wakes up. (The demo’s ramjet shows exactly this: drag the Mach slider to zero and watch thrust die.)
A thermal ceiling. As $M_0$ rises, $\tau_r$ chases $\tau_\lambda$ : the air arrives pre-heated, the materials-limited $T_{t4}$ stays fixed, and the burner can add less and less. When $\tau_r \to \tau_\lambda$ , thrust → 0 with perfect efficiency — the engine becomes a very hot pipe. The sweet spot sits near Mach 3–4.

Fig. 2. The family portrait, computed by the demo's own solver at 12 km: the turbofan owns subsonic, the turbojet carries into low supersonic, the ramjet takes over where its ram compression matures — each curve fading exactly where its physics says it must.

The hypersonic wall

Past Mach ~5 the ramjet’s own trick turns against it. Try the numbers:

Flight Mach

M_0

3.0

T_{t0} = T_0\left(1 + \tfrac{\gamma-1}{2}M_0^2\right) = 217 \times 2.80

T_{t0} = 607\ \text{K}

Fully-stopped air at 20 km. Harmless — any inlet can do this.

Stopping Mach 6 air makes it ~1 900 K before combustion — hotter than turbine metallurgy, hot enough that fuel added to it largely refuses to release energy (the combustion products dissociate back apart). Deceleration through shocks also destroys total pressure ruinously at these speeds. The conclusion is radical: stop stopping the air.

Supersonic combustion: the scramjet

A scramjet’s inlet decelerates the flow only partially — from Mach 6+ outside to Mach ~2–3 inside the combustor — so the static temperature stays low enough for fuel to burn usefully. The price: fuel must inject, mix, ignite and finish burning in air crossing the engine at kilometres per second — famously compared to lighting a match in a hurricane. Fuel-air residence time is of order a millisecond, which is why flight-proven scramjets (X-43, X-51) burn fast-mixing hydrogen or pre-cracked hydrocarbons, and why the demo gives the scramjet a combustor Mach control where the ramjet has none.

$M_c$: combustor-entry Mach: 0 (full stop) for a ramjet, ~2–3 for a scramjet [–]
$\tau_r'$: effective ram heating — only the slowed-down fraction of τr is felt as temperature [–]

In the demo’s model this appears as a reduced effective ram ratio: the cycle only “pays” the temperature rise of the deceleration it actually performs, which postpones the $\tau_r \to \tau_\lambda$ ceiling by several Mach numbers. That is the entire scramjet advantage, bought with the hardest combustion problem in aerospace.