Question bank — Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

True. Expansion drops the static pressure toward $p_{\infty }$, and for isentropic supersonic flow a pressure drop means the gas accelerates, so $M_{\text{plume}}>M_e$.

False. The pressure term is positive, but $v_e$ is smaller (gas hasn't finished expanding inside), and the two effects don't cancel — the perfectly-expanded case $p_e=p_{\infty }$ still gives the maximum total thrust.

True. The turn happens through a continuous family of infinitesimal Mach waves, none of which are shocks, so entropy (and hence $p_0$) stays constant — unlike the over-expanded case with its lossy shocks.

False. Expansion waves cool the gas as it accelerates; the loss is geometric — momentum gets turned off-axis, so less of it points backward along the thrust axis.

True. Rising altitude lowers $p_{\infty }$ while $p_e$ (set by the nozzle geometry and chamber) barely changes, so $p_e/p_{\infty }$ grows and the plume fans out more.

True. In supersonic flow disturbances can't travel upstream, so pressure can't equalize gradually inside — it must adjust through Mach waves fanning from the lip. Subsonic jets just spread smoothly with no fan.

True. A bigger ratio forces the plume to expand to a higher $M_{\text{plume}}$, and since $\nu (M_{\text{plume}})=\nu (M_e)+{\theta }_{\text{turn}}$ with $\nu$ increasing in $M$, the required turn grows.

False. $\nu (M)$ approaches a finite ceiling ${\nu }_{\text{max}}$ (about $130.5^{\circ }$ for $\gamma =1.4$); a stream can only turn so far before it would need infinite Mach.

The inequality is backwards. Under-expanded means the nozzle expanded the gas too little, leaving $p_e>p_{\infty }$; the $p_e<p_{\infty }$ case is the over-expanded nozzle.

The dependency runs the other way. The plume expands until its static pressure equals $p_{\infty }$, so you use the isentropic pressure relation to get $M_{\text{plume}}$ first, then the Prandtl-Meyer functions give ${\theta }_{\text{turn}}$.

No — it is applied at the physical exit plane, using $p_e$ and $A_e$ there. Its whole point is to fold the still-uneven exit pressure into the thrust bookkeeping.

That describes over-expansion. Here the exhaust is at higher pressure, so it pushes outward into the air and fans out; nothing pushes back into the nozzle.

It's $\mu =\arcsin (1/M)$, and it shrinks as $M$ grows — faster flow makes the Mach cone thinner, hugging the flow direction more tightly.

Isentropic means no entropy loss, but there is still a thrust loss: the expansion happens outside with no wall to react against, so momentum is redirected off-axis instead of adding pure axial push.

They're both losses, but the physics differs: over-expanded flow can suffer shocks and even flow separation inside the nozzle (potentially damaging), while under-expansion is a smoother, often gentler external loss.

The nozzle physically ends before the gas has expanded down to $p_{\infty }$; there's no more wall to guide further expansion, so the leftover expanding happens in open air past the lip.

In supersonic flow pressure signals travel only along Mach waves and can't move upstream, so the pressure adjustment is stitched together from many infinitesimal Mach waves emanating from the lip — a centered fan.

Only the axial component $v\cos {\theta }_{\text{turn}}$ pushes the rocket forward; the sideways part is real momentum that simply doesn't point along the thrust axis, and by symmetry the sideways bits cancel and just vanish from the useful thrust.

At $p_e=p_{\infty }$ the pressure term is zero but the gas has done all its expanding inside against the walls, giving maximum $v_e$; mathematically $\partial F/\partial A_e=0$ there, so it's the peak of the thrust curve. See 3.3.13-Optimal-expansion-ratio.

A nozzle sized for maximum thrust in near-vacuum would be huge; a shorter nozzle is under-expanded low down (small loss) but performs well up high, so it's a design compromise over the whole trajectory. 3.3.16-Altitude-compensation-nozzles tackles this directly.

The useful axial part of the turned momentum is $\cos {\theta }_{\text{turn}}$ of the total, so the fraction missing from the axial direction is $1-\cos {\theta }_{\text{turn}}$ — small for tiny angles, growing fast as the turn gets large.

$C_F$ normalizes thrust against the ideal momentum flux, so it isolates how efficiently the nozzle converts flow into axial push, independent of mass flow or scale. See 2.5.12-Thrust-coefficient-definition.

No pressure imbalance at the lip, so no expansion fan (and no shock) forms — the plume leaves straight and axial. This is the perfectly-expanded boundary between the under- and over-expanded regimes.

No. Subsonic flow lets pressure signals travel upstream, so the jet just adjusts smoothly; Prandtl-Meyer expansion is a strictly supersonic phenomenon. (Real rocket nozzles run supersonic at the exit, so this is a "why it can't happen inside a choked nozzle" reminder.)

No. $M_{\text{plume}}$ and hence ${\theta }_{\text{turn}}$ are capped by ${\nu }_{\text{max}}$; once the required turn would exceed ${\nu }_{\text{max}}-\nu (M_e)$, the flow physically cannot expand further and the model breaks down into a highly under-expanded, barrel-shaped plume.

$\nu (1)=0$ by definition — the Prandtl-Meyer function is measured from Mach 1, so a sonic exit has "used up" zero turning and has the full ${\nu }_{\text{max}}$ available.

Zero — the first Mach wave marks $\theta =0$; the turning accumulates continuously across the fan up to the total ${\theta }_{\text{turn}}$ at the last wave.

$\gamma$ sets both the isentropic pressure–Mach link and the shape of $\nu (M)$, so a different propellant (different $\gamma$) reaches a different $M_{\text{plume}}$ and turning angle for the same $p_e/p_{\infty }$ — it changes the ceiling ${\nu }_{\text{max}}$ too. The area–Mach relation behind this is 3.2.7-Isentropic-flow-area-Mach-relation.