Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss: Visual walkthrough

The symbols we will earn (read this once)

Before any picture, here is the tiny dictionary. Every one of these gets a figure below — this is just so you know the cast.

$a$ = speed of sound in the gas — how fast a tiny pressure ripple travels through it.
$V$ = flow speed — how fast the gas as a whole is streaming along.
$M$ = Mach number = $V / a$ = (how fast the gas moves) ÷ (the speed of sound in that gas). $M = 1$ means "exactly sound speed"; $M > 1$ is supersonic.
$γ$ = ratio of specific heats — a single number describing the gas (air ≈ 1.4, hot rocket exhaust ≈ 1.2–1.3). Treat it as a fixed dial for now.
$p$ = static pressure (the push the gas exerts sideways on anything touching it).
$p_{0}$ = total (stagnation) pressure — the pressure the gas would have if you smoothly brought it to a dead stop. In a frictionless (isentropic) expansion this stays constant even while the ordinary pressure $p$ drops.
$θ$ = an angle the flow direction turns through.
$ν (M)$ = the Prandtl-Meyer function — the star of the show. We will define it as an accumulated turning angle.
expansion fan = the spread-out sheaf of Mach lines that a supersonic flow bends through when it turns around a corner into more room — like the ribs of a hand-fan opening from a single pivot. We draw it in Step 5.

Step 1 — A supersonic message can only travel in a cone

WHAT. A tiny disturbance (a pebble of pressure) sitting in moving gas sends ripples outward at the speed of sound $a$ . If the gas itself streams past at flow speed $V$ that is faster than sound, those ripples get swept downstream and pile up into a single straight line called a Mach line.

WHY. This is the whole reason expansion happens through waves instead of all at once. In supersonic flow no signal can run upstream, so the gas learns "the pressure ahead dropped" only along these Mach lines. Every step below rides on this fact.

PICTURE.

The blue disturbance point emits a circle of sound in one time-tick (radius = sound speed $a$ ). In that same tick the flow carries the point forward by distance = flow speed $V = M a$ (since $M = V / a$ ). Because $M > 1$ , the flow distance is longer than the sound circle, so all the circles tuck inside a cone. The half-angle of that cone is the Mach angle $μ$ :

$μ = arcsin (\frac{1}{M})$

Term by term: $\frac{1}{M}$ = (one sound-tick, length $a$ ) ÷ (one flow-tick, length $V = M a$ ) = the ratio of the two radii in the picture; $arcsin$ = "which angle has this sine?" — it reads the angle straight off the right triangle whose opposite side is the sound circle and hypotenuse is the flow distance.

Recall

Why can't the disturbance send a wave upstream? ::: The flow sweeps every sound ripple downstream faster than the ripple can travel, so signals live only inside the Mach cone. What does $μ$ get bigger or smaller with? ::: As $M \to 1$ , $1/ M \to 1$ , $μ \to 90°$ (wave is nearly perpendicular). As $M \to \infty$ , $μ \to 0°$ (wave lies almost along the flow).

Step 2 — Turn the flow by a whisker and read off one Mach wave

WHAT. Put supersonic gas against a wall that bends away from the flow by a tiny angle $d θ$ . The gas has room to spread, so it speeds up a hair (its Mach number rises by $d M$ ) and drops pressure a hair — and it does this across exactly one Mach line.

WHY. A big smooth turn is just millions of these tiny turns stacked up. If we can find the exact bookkeeping rule for one infinitesimal turn — how much $M$ changes per degree turned — we can add them all up. This is the seed of the whole formula.

PICTURE.

Where the formula comes from — the velocity triangle. Look at the picture. The incoming velocity has magnitude $V$ pointing along the wall. The single green Mach line sits at the Mach angle $μ$ to that flow. Now split the velocity into two pieces measured relative to the Mach line: a piece along the line, $V cos μ$ , and a piece across the line (perpendicular to it), $V sin μ$ .

The key physical fact: only the across-the-line piece changes; the along-the-line piece is untouched by the wave. So the flow turns because the small perpendicular piece $V sin μ$ grows a little while $V cos μ$ stays put — tilting the total vector. Chasing this tilt through the little triangle gives, cleanly,

$d θ = M^{2} - 1 \frac{d V}{V},$

because $tan μ = 1/ M^{2} - 1$ so $cot μ = M^{2} - 1$ is exactly the lever that converts a fractional speed change $d V / V$ into a turn. That is the pure geometry — no gas physics yet.

Now the compressibility term. We want the relation in terms of $d M / M$ , not $d V / V$ , because $M$ is what our lookup table will use. But $M$ and $V$ are not proportional in a compressible gas: as the gas speeds up it also cools, so $a$ changes too. Starting from $M = V / a$ and using the isentropic energy balance (total temperature conserved), one gets the exact link

$\frac{d V}{V} = \frac{1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M} .$

The factor $1 + \frac{γ - 1}{2} M^{2}$ is precisely the ratio of total temperature to static temperature; it appears because part of a Mach-number increase is "used up" cooling the gas rather than pure speeding-up. Substitute this into the geometry result and you land on the master differential relation:

$d θ = \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M}$

Reading it:

$d θ$ = the tiny angle we turned (left side — the thing we control).
$\frac{d M}{M}$ = the fractional change in Mach number (turn a little, gain a little speed).
$M^{2} - 1$ = the geometry lever $cot μ$ from the velocity triangle; it is zero at $M = 1$ (a sonic flow can't be turned by a wave at all) and grows with $M$ .
$1 + \frac{γ - 1}{2} M^{2}$ = the compressibility tax = (total temperature)/(static temperature); it shrinks the turn because raising $M$ also cools the gas, and $γ$ sets how strongly.

Step 3 — Add up all the whiskers: define $ν (M)$

WHAT. Integrate (sum up) that differential turn from the moment the flow was exactly sonic ( $M = 1$ ) up to any Mach $M$ . The running total of angle is what we name $ν (M)$ .

WHY. We want a lookup table: "if the flow is at Mach $M$ , how much has it turned in total since it was sonic?" That single running total lets us handle any expansion by subtraction, which is Step 5. We anchor the count at $M = 1$ because that is the one Mach number where turning first becomes possible, and we set $ν (1) = 0$ so the counter starts at zero.

PICTURE.

The curve is $ν$ climbing as $M$ increases. Each shaded sliver is one $d θ$ from Step 2; the height of the curve at a given $M$ is the sum of all slivers to its left — the total accumulated turn.

$ν (M) = \int_{1}^{M} \frac{M ˉ ^{2} - 1}{M ˉ ( 1 + \frac{γ - 1}{2} M ˉ ^{2} )} d \overset{ˉ}{M}$

Here $\overset{ˉ}{M}$ is just the dummy "sweeping" Mach number that runs from 1 up to the target $M$ . Doing this integral with the substitution $\overset{ˉ}{M} = sec ϕ$ (which turns the ugly square root into a clean $tan ϕ$ ) collapses it into the closed form from the top of the page. You never have to integrate again — the two $arctan$ terms are the answer to this integral.

Step 4 — The ceiling: maximum turn at infinite Mach

WHAT. Push $M \to \infty$ in $ν (M)$ . Both $arctan$ terms hit $90° = π /2$ , and the whole thing settles to a finite number — the largest angle supersonic gas can ever turn through.

WHY. This is the degenerate/limiting case, and it matters physically: a plume cannot turn outward forever. If the pressure mismatch demanded more turn than $ν_{m a x}$ , the flow would peel away into vacuum and the model breaks. Knowing the ceiling tells us when to stop trusting the formula.

PICTURE.

The $ν$ curve flattens against a horizontal dashed ceiling. For $γ = 1.4$ :

$ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1) \approx 130.5°$

Term by term: the $\frac{π}{2}$ is what each $arctan$ maxes out at; the bracket is (gas-constant factor) minus 1 because the second $arctan$ subtracts its own $90°$ . For hotter exhaust ( $γ$ smaller) the ceiling is higher — the gas can turn more.

Step 5 — Use it on the plume: the exit expansion

WHAT. At the nozzle lip the exhaust is at Mach $M_{e}$ with pressure $p_{e}$ . Outside sits ambient $p_{\infty}$ . If $p_{e} > p_{\infty}$ the gas expands through an expansion fan (the sheaf of Mach lines from our dictionary), speeding up to $M_{plume}$ and turning outward by $θ_{turn}$ . The turn is just a subtraction on the $ν$ counter:

$θ_{turn} = ν (M_{plume}) - ν (M_{e})$

WHY. This is exactly why we built a cumulative counter in Step 3. We don't re-integrate — we read the counter at the two Mach numbers and subtract. The bigger $M_{plume}$ is, the higher its counter, so the more the flow had to turn.

WHERE does $M_{plume}$ come from? The gas expands until its static pressure equals $p_{\infty}$ . Since the expansion is isentropic, the total pressure $p_{0}$ (defined in the dictionary — the pressure the gas would have if smoothly stopped) stays constant, which links the ordinary pressure $p$ to Mach number:

$\frac{p _{e}}{p _{\infty}} = \frac{( 1 + \frac{γ - 1}{2} M _{plume}^{2} ) ^{γ / (γ - 1)}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$

(This is the standard isentropic pressure-Mach link from 3.2.7-Isentropic-flow-area-Mach-relation; both sides are just $p_{0} / p$ ratios, so the shared $p_{0}$ cancels.) Solve it for $M_{plume}$ , feed both Machs into $ν$ , subtract — done.

PICTURE.

The straight nozzle wall ends; the green expansion fan of Mach lines opens from the lip like the ribs of a hand-fan; the flow bends down (outward) by $θ_{turn}$ ; Mach rises from $M_{e}$ to $M_{plume}$ while pressure falls from $p_{e}$ to $p_{\infty}$ .

Step 6 — Why the turn costs thrust

WHAT. Only the axial slice of the plume's momentum pushes the rocket forward. Turning the flow outward by $θ_{turn}$ multiplies the useful part by $cos θ_{turn}$ , so the fraction lost is $(1 - cos θ_{turn})$ .

WHY. Thrust is reaction to backward momentum. Momentum thrown sideways cancels itself across the axisymmetric plume and does nothing. This is the efficiency loss the parent note warned about, and it grows with $θ_{turn}$ — hence with the pressure mismatch $p_{e} / p_{\infty}$ .

PICTURE.

The plume velocity vector is split into an axial (forward, useful) leg $v cos θ_{turn}$ and a radial (sideways, wasted) leg $v sin θ_{turn}$ .

For our example, $θ_{turn} \approx 4.7°$ , so the extra-expansion momentum lost is $1 - cos (4.7°) = 1 - 0.99664 \approx 0.0034 \Rightarrow about 0.3% of the extra push.$ Small here, but it climbs fast as $p_{e} / p_{\infty}$ grows — which is exactly why 3.3.16-Altitude-compensation-nozzles exist and why the thrust coefficient $C_{F}$ falls below its optimum. (The mirror-image loss for $p_{e} < p_{\infty}$ , through shocks instead of fans, is 3.3.14-Over-expanded-nozzle-shock-diamonds; the sweet spot $p_{e} = p_{\infty}$ is 3.3.13-Optimal-expansion-ratio.)

The one-picture summary

Everything on one canvas: the Mach cone (Step 1) → one tiny turn (Step 2) → the accumulated $ν$ curve (Step 3) with its ceiling (Step 4) → the plume fan reading a $ν$ -difference (Step 5) → the velocity split that costs thrust (Step 6).

Recall Feynman retelling — say it in plain words

Supersonic gas can only whisper to itself along slanted Mach lines — that's the cone. Bend a wall away by a hair and the gas gains a hair of speed across one line; the exact exchange rate is the master differential relation — pure velocity-triangle geometry ( $cot μ = M^{2} - 1$ ) times a compressibility tax because speeding up also cools the gas. Add up every hair of turn starting from the speed of sound and you get a running counter called $ν$ — a lookup table of "how much has this flow turned in total?". That counter tops out at about $130°$ : there's a limit to how far gas can bend. Now aim it at a rocket: exhaust leaves the nozzle still over-pressured, so it expands outside through an expansion fan, its Mach counter climbs from $ν (M_{e})$ to $ν (M_{plume})$ , and the difference is the outward turn angle. Turning the flow sideways means only $cos θ_{turn}$ of the new momentum still points backward — the rest is wasted, and that's your efficiency loss.