Over - under expanded nozzle flows: Visual walkthrough

Every symbol below is earned before it is used. If you have never seen the word "Mach" or the letter $γ$ , start reading at line one and you will be fine.

Step 1 — What is a nozzle, and what are we tracking?

WHAT. A nozzle is a pipe whose cross-section area $A$ changes along its length. Gas flows through it. We follow one thin "slice" of gas and ask: as area $A$ changes, what happens to the gas's speed $V$ and its pressure $p$ ?

WHY. Before any formula, we must agree on the four things we will measure. Pressure $p$ is how hard the gas pushes sideways on the walls (think: air in a balloon). Speed $V$ is how fast the slice moves down the pipe. Density $ρ$ (Greek letter "rho") is how tightly packed the molecules are — mass per unit volume. Area $A$ is the width of the pipe there.

PICTURE. Look at the pipe below: it narrows to a waist and widens again — this is the converging–diverging (de Laval) shape. The green slice is our tracked chunk of gas.

Step 2 — The speed of sound, and the number $M$

WHAT. Sound is a tiny pressure ripple. It travels through the gas at a fixed speed we call $a$ (the local speed of sound). We define one dimensionless number: $M = \frac{V}{a}$ Here $M$ ("Mach number") is the gas speed compared to its own sound speed. $M < 1$ means slower than sound (subsonic); $M = 1$ means exactly sound speed (sonic); $M > 1$ means faster than sound (supersonic).

$V$ — the gas speed we care about.
$a$ — how fast news (a pressure ripple) can travel inside this gas.
$M$ — the ratio; the single most important number in this whole chapter.

WHY this ratio and not just $V$ ? Because the ability of pressure to send a message upstream depends entirely on whether $V$ beats $a$ . A number like "300 m/s" tells you nothing on its own; " $M = 2$ " tells you the gas outruns its own sound — that is the fact that will lock $p_{e}$ . We need the ratio, so we build $M$ .

PICTURE. Two runners: a pressure-ripple runner (speed $a$ ) trying to go left, standing on a walkway moving right at speed $V$ . When $V > a$ ( $M > 1$ ) the ripple is swept backward — it can never reach the throat. This is the seed of "the exit is deaf."

Recall Why "deaf to downstream" already follows

If $M > 1$ ::: the gas moves faster than any pressure signal can crawl upstream, so the back pressure $p_{b}$ literally cannot send a message to the exit. The exit "hears nothing" from outside.

Step 3 — Conservation of energy → pressure tied to $M$

WHAT. We track the gas as isentropic (smooth, no shocks, no heat lost). Energy conservation says the total energy per unit mass stays constant. Write it as a "stagnation" (subscript $0$ ) quantity: imagine slowing the gas gently to rest — the pressure it would then have is the stagnation pressure $p_{0}$ (this is the chamber pressure). Energy + the perfect-gas isentropic link $p / ρ^{γ} = const$ give: $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{γ}{γ - 1}}$

Term by term:

$p_{0}$ — chamber (stagnation) pressure, the "full tank" reference, a fixed constant.
$p$ — the local static pressure we want.
$γ$ ("gamma") — the gas's stiffness ratio (for air, $1.4$ ); it says how pressure and density trade off when squeezed.
$M^{2}$ — appears because kinetic energy scales with speed squared; faster flow ⇒ more of the energy is motion ⇒ less is pressure ⇒ $p$ drops relative to $p_{0}$ .
The whole bracket raised to $γ / (γ - 1)$ — the exact exponent that the isentropic (constant-entropy) rule demands.

WHY this tool? We want a knob connecting pressure to $M$ . Energy conservation is that knob: it is the master relation of Isentropic Flow Relations. Read it as "tell me $M$ , I'll tell you how far $p$ has fallen below $p_{0}$ ."

PICTURE. As $M$ climbs left→right, the curve $p / p_{0}$ slides down toward zero. High Mach = low pressure. Memorise that slope.

Step 4 — Conservation of mass → the throat sets a reference

WHAT. Mass cannot pile up: the mass flowing past every station per second is the same, $ρ A V = const$ . Combine this with the isentropic relations and one special area falls out: the area where $M = 1$ , called the throat area $A^{*}$ (star = sonic). We measure every other area against $A^{*}$ : $\frac{A}{A ^{*}} = \frac{1}{M} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$

$A$ — area at the station of interest.
$A^{*}$ — throat area, where $M = 1$ ; the flow is choked there (see Choked Flow & the Throat (M=1)).
$\frac{1}{M}$ out front + the bracket — together they make $A / A^{*}$ have a minimum of 1 exactly at $M = 1$ , rising on both sides.

WHY this tool? Because the geometry we can actually build is area, not Mach. Mass conservation is the bridge from "shape" ( $A / A^{*}$ ) to "speed" ( $M$ ). This is the Area-Mach Number Relation.

PICTURE. The U-shaped curve: $A / A^{*}$ dips to $1$ at $M = 1$ and climbs on both sides. So one area ratio touches the curve at two Mach numbers — one subsonic, one supersonic.

Step 5 — Picking the correct root (the edge case!)

WHAT. A given $A_{e} / A^{*} > 1$ meets the U-curve at two points: a subsonic $M < 1$ and a supersonic $M > 1$ . In the diverging part of a supersonic nozzle we take the supersonic root $M_{e} > 1$ .

WHY. In converging flow, widening slows gas (subsonic branch). Once past the throat where $M = 1$ , widening now speeds gas up — the supersonic branch. Physics chose the branch for us at the throat; we just follow it.

Degenerate cases — cover them all:

$A_{e} / A^{*} = 1$ : the only root is $M = 1$ . Exit is exactly sonic; no supersonic flow at all.
$A_{e} / A^{*} < 1$ : impossible — no area is smaller than the throat by definition. The curve never dips below 1.
$A_{e} / A^{*} \to \infty$ (huge bell): $M_{e} \to \infty$ , and from Step 3, $p_{e} \to 0$ . Infinite expansion, vacuum-like exit.
If the throat is not choked (chamber pressure too low): $M$ never reaches 1, the flow stays fully subsonic, and this whole supersonic story does not apply.

PICTURE. Same U-curve, one horizontal line at $A_{e} / A^{*} = 4$ , two intersection dots highlighted — we circle the supersonic (right) one.

Step 6 — Combine: exit pressure from geometry alone

WHAT. Feed the supersonic $M_{e}$ (found from $A_{e} / A^{*}$ in Step 5) into the Step 3 pressure law: $p_{e} = \frac{p _{0}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$

$M_{e}$ — comes only from $A_{e} / A^{*}$ (Step 4–5).
$p_{0}$ — the chamber pressure (a chosen constant).
Result $p_{e}$ — therefore depends on geometry and $p_{0}$ only. The back pressure $p_{b}$ appears nowhere.

WHY this is the punchline. Notice what is missing: $p_{b}$ . There is no line in this whole chain where the outside air enters. That is the mathematical proof that a supersonic exit cannot match itself to the atmosphere — the mismatch must spill outside as shocks or fans.

PICTURE. A flow-chart of the logic: shape $\to$ $M_{e}$ $\to$ $p_{e}$ , with $p_{b}$ shown knocking on a locked door.

Step 7 — The consequence: three regimes

WHAT. Now compare the geometry-locked $p_{e}$ against the atmosphere $p_{b}$ :

$p_{e} = p_{b}$ → perfectly expanded, clean parallel jet.
$p_{e} < p_{b}$ → over-expanded, atmosphere squeezes in with Oblique Shocks (or a Mach disk / separation if severe).
$p_{e} > p_{b}$ → under-expanded, jet keeps expanding outward through Prandtl-Meyer Expansion Fans.

WHY. Since $p_{e}$ is fixed, the only variable is $p_{b}$ . Lower the atmosphere (climb in altitude) and a nozzle can slide from over- to perfectly- to under-expanded — the physics of Rocket Nozzle Design & Thrust Optimization.

PICTURE. Three jets side by side: shocks pinching in (over), clean column (perfect), fans bulging out (under).

The one-picture summary

Everything above compresses into one arrow of logic: geometry → Mach → pressure → regime, with the outside world (atmosphere) permanently locked out of the pressure calculation.

Recall Feynman: retell the whole walkthrough

Picture a water slide sealed in a magic tube. We watched one bucket of gas ride down it. First we learned to measure it: how wide the tube is ( $A$ ), how fast the gas goes ( $V$ ), how hard it pushes ( $p$ ). Then we invented a smart number, $M$ — the gas's speed divided by the speed of a shout inside it. The big surprise: once the gas is faster than a shout ( $M > 1$ ), no shout from outside can travel back up the slide. Next, energy bookkeeping told us "faster gas = lower pressure," and mass bookkeeping told us "the tube's widest-vs-throat ratio picks the speed." Put them together and out pops the exit pressure — built entirely from the tube's shape and the tank pressure. The outside air never got a vote. So when the gas finally squirts out and finds the air at a different pressure, it has to fix the mismatch right there in the open air: squishing together into shocks if the air is too pushy, or fanning out if the air is too weak. That's the whole story — the slide can't change shape mid-ride, so the world outside does the adjusting.