Over-expanded nozzle — oblique shocks in plume, efficiency loss: Visual walkthrough

This page rebuilds the parent result — the over-expanded nozzle story — from absolutely nothing. No prior symbol is assumed. By the end you will have drawn every term in the thrust equation and watched the shocks form.

We only use these ideas as building blocks (each is defined the moment it appears): pressure, a supersonic jet, and one simple triangle. Everything else we build here.

Step 1 — What is "pressure pushing on a surface"?

WHAT. Before we talk about rockets, we build the single fact everything rests on: a gas presses on any wall it touches, and that push is a force.

WHY. The whole efficiency loss is one push winning over another. If you cannot see pressure as a force, none of the thrust equation will feel real.

PICTURE. In the figure, a flat plate of area $A$ (measured in square metres, $m^{2}$ ) sits in gas at pressure $p$ (measured in pascals, $Pa$ — that is newtons per square metre). Notice the amber arrows on the left: each is the gas hammering the plate, all pointing the same way. Now follow the single white arrow on the right — that is what you get when you add every little hammer-blow together: one clean total force.

Figure — Over-expanded nozzle — oblique shocks in plume, efficiency loss

Step 2 — Where the thrust comes from: two separate pushes

WHAT. We split a rocket's total thrust into two clean pieces and draw each one.

WHY. The parent's master formula has two terms. Beginners blur them together. We keep them apart so the loss later has an unambiguous home.

PICTURE. The figure shows the nozzle's circular exit plane (the white ellipse), area $A_{e}$ (the little $e$ means "at the exit"). Trace the two arrows leaving it: the fat cyan arrow pointing left is the mass being flung backward — the momentum push; the thin amber arrow just below it is the gas pressure acting on that exit disk — the pressure push. They point the same way here, but in Step 3 the amber one can flip.

This links to the full Thrust Equation.

Step 3 — The three signs of the pressure term (cover every case)

WHAT. The sign of $p_{e} - p_{a}$ has exactly three possibilities. We draw all three.

WHY. The contract demands every case. "Over-expanded" is one specific sign, and you only understand it by seeing its two siblings.

PICTURE. Three mini-panels. In each, compare the cyan bar ( $p_{e}$ , the exhaust) against the amber bar ( $p_{a}$ , the atmosphere). Left panel: cyan is taller → the push helps. Middle: equal → no push. Right panel (our villain): amber towers over cyan → the atmosphere wins and the push reverses.

Case	Sign of $p_{e} - p_{a}$	Pressure push	Name
$p_{e} > p_{a}$	$+$	helps thrust	under-expanded
$p_{e} = p_{a}$	$0$	zero — pure momentum	matched (ideal)
$p_{e} < p_{a}$	$-$	hurts thrust	over-expanded

Step 4 — Why the atmosphere doesn't just squeeze in gently: supersonic flow

WHAT. The exhaust leaves faster than sound. We show what that forbids.

WHY. If the flow were slow, the atmosphere could compress it smoothly and there would be no shocks, no loss. The entire drama exists because the flow is supersonic.

PICTURE. Follow the cyan arrow — the gas flow at speed $v_{e}$ , drawn long because it is fast. Now the shorter amber arrow below it is a sound signal travelling at speed $a$ , trying to carry the message "pressure is too high out here!" upstream (to the left). It cannot: the cyan arrow overtakes it. The message never reaches the nozzle smoothly, so the flow must find out the hard way.

At the exit of an over-expanded rocket, $M_{e} > 1$ (often around $4$ ) — supersonic. More on this in Gas Dynamics and Isentropic Flow.

Step 5 — Why the shock is oblique, not head-on

WHAT. The compression comes from the sides (the atmosphere squeezing the plume's edges), so the shock sits at a slant.

WHY. A head-on normal shock (defined in Step 4) would be the most violent, most wasteful compression. Nature uses the cheapest legal option — a tilted (oblique) shock.

PICTURE. Find the amber shock line slanting up across the figure — that is the shock, tilted by angle $β$ from the incoming flow. The cyan arrow is the incoming flow $M_{1}$ (the upstream Mach number from Step 4). Watch how the two white arrows split that cyan arrow into a right triangle: one white arrow points straight across the shock line ( $M_{1 n}$ ), the other slides along it ( $M_{1 t}$ ). Only the across-piece gets squeezed.

Step 6 — The turn: flow deflects inward by $θ$ , and when no shock can attach

WHAT. Passing through the oblique shock, the flow bends toward the centreline by an angle $θ$ . We also show that for each $θ$ there are usually two possible shock angles $β$ , and a maximum $θ$ beyond which no attached oblique shock exists.

WHY. The plume boundary has to fold inward to fit the squeeze — that is what tilts your thrust off the axis. And the "two-branch / maximum-angle" behaviour is exactly what decides whether you get a clean oblique shock or a detached, even more wasteful, normal-like shock.

PICTURE (left). The streamline kinks at the shock: it enters horizontal, leaves tilted down by $θ$ . PICTURE (right). The $θ$ – $β$ – $M$ curve for a fixed $M_{1}$ : read $θ$ across the bottom, $β$ up the side. For any $θ$ below the peak the curve is hit at two $β$ values — the lower one is the weak shock (small $β$ , flow stays supersonic, what the plume actually picks), the upper one is the strong shock (large $β$ , flow goes subsonic). At the peak sits $θ_{m a x}$ ; ask for a turn steeper than that and the curve has no solution — the shock detaches and bows out as a curved, normal-like front.

Step 7 — The real cost: stagnation pressure falls, and that IS the lost exit velocity

WHAT. Every shock converts some ordered kinetic energy into random heat. We measure this with the drop in stagnation pressure $p_{0}$ , write the exact $p_{02} / p_{01}$ formula, and show precisely how it caps the exhaust speed.

WHY. Lost stagnation pressure is lost ability to make exhaust speed. This is the number that finally answers "how much thrust did we throw away?"

PICTURE. Before the shock, a tall cyan bar of $p_{01}$ (usable "push potential"); after, a shorter amber bar $p_{02}$ . The dotted white line connects their tops so you can see the missing slice — the arrow points to it: that slice is gone forever, turned to disorder.

Definition Velocity coefficient — every symbol

$C_{v} = \frac{v _{e, actual}}{v _{e, ideal}} = \frac{p _{0, actual}}{p _{0, chamber}}$

$v_{e, actual}$ — the exhaust speed you really get after the shocks have eaten some $p_{0}$ ( $m/s$ ).
$v_{e, ideal}$ — the exhaust speed you would get with a loss-free (isentropic) expansion, no shocks ( $m/s$ ).
$p_{0, chamber}$ — chamber stagnation pressure: the full push-potential the gas started with, deep inside the combustion chamber (this is the $p_{01}$ of the first shock).
$p_{0, actual}$ — the push-potential that survives to the exhaust after every shock has taken its cut.
$C_{v}$ — velocity coefficient: the fraction of ideal exhaust speed you keep.

Why the square root: speed $\propto energy \propto p_{0}$ , from the velocity formula above. This model assumes shock $p_{0}$ -loss is the only spoiler (friction and heat leakage neglected — standard first-order).

Step 8 — Shocks repeat: the diamond pattern (edge case: over-correction)

WHAT. One shock overshoots or undershoots the pressure match, so the plume expands again, then re-compresses — again and again. This makes the visible shock diamonds.

WHY. The single shock of Steps 5–7 is not the end; losses compound. The contract wants this limiting/degenerate behaviour shown.

PICTURE. Trace the plume left to right. The cyan envelope is the plume boundary; the amber diamonds are the crossing oblique shocks. On the first crossing, look for the short white vertical line on the axis — that is the Mach disk (a normal shock, defined in Step 4). Past it the flow over-compresses, then fans back out below $p_{a}$ , and the cycle repeats. Watch the diamonds shrink to the right — each one is weaker because $p_{0}$ (and $M_{1}$ ) has fallen.

Worked example — RL-10 fired at sea level (every number checked)

Pressure thrust loss. $Δ F = (p_{e} - p_{a}) A_{e} = (5.5 - 101) \times 1 0^{3} \times 0.5 = - 47.75 kN$ A retarding $47.75 kN$ — the atmosphere winning the tug-of-war of Step 1.

Normal Mach at the shock (Step 5). For the first shock $M_{1} = M_{e} = 4.2$ , so $M_{1 n} = 4.2 sin 2 5^{\circ} \approx 1.775$

Stagnation loss of the first shock (Step 7). $\frac{p _{02}}{p _{01}} \approx 0.84$

After three diamonds (Step 8). $(0.84)^{3} \approx 0.60 \Rightarrow C_{v} = 0.60 \approx 0.77$

You keep only about $77%$ of ideal exhaust speed. Terrible — exactly the parent's warning.

The one-picture summary

This is the composite promised at the top: the whole over-expanded plume in one blueprint. Start at the nozzle on the left — the amber inward arrows are the atmosphere ( $p_{a} > p_{e}$ ) squeezing the plume edges. Those edges throw off the slanted amber edge shocks at angle $β$ . The shocks cross on the axis at the short white vertical Mach disk, and the cyan envelope then repeats as fading diamonds. Read left-to-right and you have replayed every step: squeeze → oblique shock → only $M_{1 n} = M_{1} sin β$ compresses → turn by $θ$ → lost $p_{0}$ → smaller $C_{v}$ and smaller thrust.

Recall Feynman retelling — say it back in plain words

A rocket pushes two ways: by throwing gas backward, and by the gas pressing on the exit disk. If the nozzle over-expands, its exhaust comes out at a lower pressure than the surrounding air, so the air presses inward harder than the exhaust presses out — that mismatch subtracts thrust. Because the gas is flying faster than sound, it can't smoothly adjust; instead it slams into slanted walls of compression called oblique shocks. Only the part of the motion pointing straight into the shock gets squeezed, which is why a tilted shock is cheaper than a head-on one — but "cheaper" is not "free." For any required turn there are two shock angles (a gentle weak one and a violent strong one), and if the turn demanded is steeper than the flow can manage, the shock gives up and detaches into a bowed front — the worst case. Each shock obeys the same three bookkeeping laws — conserve mass, momentum, and energy — and those laws force both a pressure jump and an unavoidable rise in disorder, which shows up as lost push-potential (stagnation pressure). Since exhaust speed grows like the square root of that surviving push-potential, lost $p_{0}$ is directly lost speed. The pattern repeats down the plume as glowing diamonds, and the losses multiply, so a vacuum engine tested at sea level can throw away a fifth of its exhaust speed on top of the pressure penalty.

Recall Quick self-test

What is the difference between a normal shock and an oblique shock? ::: A normal shock sits perpendicular to the flow (head-on, maximum loss); an oblique shock sits at a slant so only part of the motion crosses it head-on (gentler). What do $M$ , $M_{e}$ and $M_{1}$ each mean? ::: $M = v / a$ is the generic Mach number; $M_{e}$ is its value at the nozzle exit; $M_{1}$ is its value just upstream of a given shock (equal to $M_{e}$ for the first shock). For a fixed $M_{1}$ and turn $θ$ , how many shock angles $β$ solve the relation? ::: Usually two — a weak branch (small $β$ , stays supersonic) and a strong branch (large $β$ , goes subsonic); if $θ > θ_{m a x}$ , none — the shock detaches. Which three conservation laws produce the shock relations? ::: Conservation of mass, momentum and energy across the thin shock sheet. Why does only $M_{1 n} = M_{1} sin β$ set the shock strength? ::: Only motion perpendicular to the shock is compressed; motion parallel slides along untouched. What sign is $(p_{e} - p_{a})$ for an over-expanded nozzle, and what does it do to thrust? ::: Negative — it subtracts from thrust. Why does lost stagnation pressure mean lost thrust, and why the square root in $C_{v}$ ? ::: A rise in entropy shows up as a drop in $p_{0}$ ; since kinetic energy $\propto v^{2}$ , speed $\propto p_{0}$ , so $C_{v} = p_{0, actual} / p_{0, chamber}$ .