Converging nozzle — subsonic flow, Mach 1 at exit: Visual walkthrough

Before step 1 we need a few plain-language ideas. Everything else is built on them.

Definition The words we start from

==Speed of sound $a$ ==: how fast a tiny pressure "tap" (a whisper) travels through the gas. Think of it as the top speed at which one gas molecule can tell the next one "something changed upstream." We take as known (from Speed of Sound) that $a = γ R T$ (units: metres per second) — bigger when the gas is hotter.
==Mach number $M$ ==: the flow's own speed $V$ measured in units of $a$ , i.e. $M = V / a$ . $M = 0.5$ means "moving at half the whisper-speed." $M = 1$ means "moving exactly at the whisper-speed." That last one is the whole story.
==Stagnation state (subscript $0$ )==: the gas sitting still in the big reservoir before the pipe, with pressure $p_{0}$ , temperature $T_{0}$ . "Stagnation" = "brought to rest." (See Stagnation Properties.)
==Specific heat at constant pressure $c_{p}$ ==: the amount of energy (in joules) needed to warm one kilogram of the gas by one kelvin while its pressure is held fixed. Units: $J / (kg \cdot K)$ . So $c_{p} T$ is simply "how much thermal energy one kilogram is carrying at temperature $T$ ." For air, $c_{p} \approx 1005 J / (kg \cdot K)$ .

The symbols $γ$ (a fixed number, $1.4$ for air, describing how the gas stores heat) and $R$ (the gas constant, units $J / (kg \cdot K)$ ) are just labels for known numbers — we never need to open them up.

Step 1 — Draw the corridor, and why narrower means faster

WHAT. A gas reservoir feeds a pipe that only gets narrower. Mass conservation for steady, one-dimensional flow says the same amount of gas must pass every slice each second: $ρ A V = constant$ where $ρ$ is density (kg per cubic metre), $A$ is the slice's cross-sectional area (square metres), and $V$ is the flow speed. If $A$ shrinks and $ρ$ barely changes (true while the flow is slow), then $V$ must rise to keep the product fixed.

WHY. This is the engine of the whole chapter: it is why squeezing the pipe speeds the gas up. But it also warns us — $ρ$ stops holding still once the gas moves fast, and that is exactly where the $M = 1$ ceiling comes from. So we plot the resulting speed and watch for the ceiling.

PICTURE.

Figure — Converging nozzle — subsonic flow, Mach 1 at exit

The blue curve climbs left-to-right (narrower ⇒ faster, straight from $ρ A V =$ const). The red dashed line is $M = 1$ , the whisper-speed. Notice the curve can touch the red line at the exit but never rise above it while still inside. That "touch but never cross" is what "choked" means, and Steps 2–6 prove it must be so.

Step 2 — Energy can't vanish: the stagnation-temperature relation

WHAT. Follow one packet of gas from the reservoir into the fast stream. Nothing heats it from outside (adiabatic), no paddle-wheel does work on it, and height changes are ignored. So the sum "thermal energy stored + energy of motion" is the same everywhere:

$stored (thermal) c_{p} T + of motion \frac{1}{2} V^{2} = all stored, at rest c_{p} T_{0}$

Term by term: $c_{p} T$ is the thermal energy per kilogram at temperature $T$ (defined above); $\frac{1}{2} V^{2}$ is its kinetic energy per kilogram at speed $V$ ; $c_{p} T_{0}$ is the reservoir value where $V = 0$ so all the energy is thermal.

WHY. This is just "energy is conserved," but written for a moving gas. It tells us the gas pays for speed by cooling down — motion energy is bought with thermal energy. That trade is the seed of the speed limit.

PICTURE.

The green thermal bar shrinks and the yellow motion bar grows as we move down the pipe, but the two bars always stack to the same total height (the dashed line). That fixed ceiling is $c_{p} T_{0}$ .

Now the key algebraic move — and why we choose it. The equation mixes three quantities that all change down the pipe ( $T$ , $V$ , and implicitly $a$ ). We want a single clean variable that already "knows" about the sound-speed ceiling, and that variable is $M = V / a$ . So we engineer $M$ into the equation:

Divide every term by $c_{p} T$ . This makes the left side dimensionless and turns the reservoir term into the ratio $T_{0} / T$ — the very thing we're hunting.
Rewrite the motion term. After dividing, it reads $\frac{V ^{2}}{2 c _{p} T}$ . Substitute $V^{2} = M^{2} a^{2}$ and $a^{2} = γ R T$ , so it becomes $\frac{M ^{2} γ R T}{2 c _{p} T}$ — and the $T$ cancels.
Substitute $c_{p} = \frac{γ R}{γ - 1}$ . Now the $γ R$ cancels too, leaving only the pure number $\frac{γ - 1}{2} M^{2}$ .

The point of steps 1–3: every appearance of $T$ , $R$ and $c_{p}$ deliberately cancels, isolating $M$ as the sole survivor. That is why this particular substitution and not another.

Step 3 — Isentropy trades temperature for pressure

WHAT. The flow is also reversible (no friction, no shocks). For a reversible adiabatic (isentropic) perfect gas, temperature and pressure move together by a fixed rule:

$\frac{T _{0}}{T} = (\frac{p _{0}}{p})^{\frac{γ - 1}{γ}}$

Here $p_{0} / p$ is how much higher the reservoir pressure is than the local pressure, and the exponent $\frac{γ - 1}{γ}$ (for air, $0.286$ ) is just the gas's fixed "conversion rate" between temperature ratio and pressure ratio.

WHY. Gauges read pressure, not temperature. Step 2 gave us temperature in terms of $M$ ; this step lets us restate everything in the quantity we can actually measure and control at the exit — pressure.

PICTURE.

The curve shows: as the gas cools (moves right), its pressure drops even faster, because the exponent bends the relationship. Reading off the axes converts a temperature drop into a pressure drop.

Now combine. Step 2 gives $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ ; Step 3 gives $T_{0} / T = (p_{0} / p)^{(γ - 1) / γ}$ . Setting the two right-hand sides equal: $(\frac{p _{0}}{p})^{\frac{γ - 1}{γ}} = 1 + \frac{γ - 1}{2} M^{2}$ To free $p_{0} / p$ from its fractional exponent $\frac{γ - 1}{γ}$ , apply the inverse power $\frac{γ}{γ - 1}$ to both sides — the two exponents multiply to $1$ on the left, leaving $p_{0} / p$ bare, while the right side inherits that exponent:

Step 4 — Turn the dial to $M = 1$ : the number 0.528 appears

WHAT. We want the exact pressure ratio at which the exit just reaches the whisper-speed. So set $M = 1$ in the master relation:

$\frac{p _{0}}{p ^{*}} = (1 + \frac{γ - 1}{2})^{\frac{γ}{γ - 1}} = (\frac{γ + 1}{2})^{\frac{γ}{γ - 1}} \Rightarrow \frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}}$

The starred $p^{*}$ is the "critical" exit pressure — the pressure the gas has at the instant it hits $M = 1$ .

WHY. $M = 1$ is the ceiling from Step 1. Plugging it in tells us the one pressure ratio the nozzle will lock onto when choked. For air ( $γ = 1.4$ ):

$\frac{p ^{*}}{p _{0}} = (\frac{2}{2.4})^{3.5} \approx 0.528$

PICTURE.

The master curve $M \mapsto p / p_{0}$ is drawn once. We drop a vertical line at $M = 1$ ; where it meets the curve is the height $0.528$ . Everything left of that line is subsonic (the exit feels the back pressure); the point at the line is the locked, choked state.

Step 5 — Why lowering back pressure stops helping: the flow rate peaks at $M = 1$

WHAT. Mass flow is $\overset{m}{˙} = ρ A V$ — density × area × speed. Rewriting it in reservoir terms (using Steps 2–3 for density and temperature) gives:

$\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} \cdot call this f (M) \frac{M}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$

Everything in front of $f (M)$ is fixed by the reservoir and the fixed exit area. So the shape of $\overset{m}{˙}$ versus $M$ is the shape of $f (M)$ .

WHY. The parent claim was " $\overset{m}{˙}$ hits a ceiling." That is exactly the claim " $f (M)$ has a maximum." Where? Take $f^{'} (M) = 0$ — the top of a hump has zero slope — and it lands precisely at $M = 1$ .

PICTURE.

$f (M)$ rises from $0$ , arches over, and its peak sits exactly on the $M = 1$ line. Past $M = 1$ the curve would fall — but a converging pipe can't reach that region anyway (Step 6). So the reachable maximum flow is the peak, at $M = 1$ . This is the mathematical face of choking: once you're on the peak, pulling the back pressure lower cannot lift you higher. (More in Mass Flow Rate & Choking.)

Step 6 — Every case, including the degenerate ones

WHAT & WHY. We must be sure no scenario was skipped. Walk the back pressure $p_{b}$ from "equal to reservoir" all the way to "vacuum."

The figure plots exit Mach and $\overset{m}{˙}$ against $p_{b} / p_{0}$ from $1$ down to $0$ .

$p_{b} / p_{0} = 1$ (degenerate: no push). No pressure difference, so $V = 0$ , $M = 0$ , $\overset{m}{˙} = 0$ . The flat left end of both curves.
$0.528 < p_{b} / p_{0} < 1$ (subsonic branch). Exit unchoked: $p_{e} = p_{b}$ , and $M$ comes from inverting the master relation. As $p_{b}$ falls, $M$ rises and $\overset{m}{˙}$ rises — the "intuitive" straw-sucking region.
$p_{b} / p_{0} = 0.528$ (the critical point). Exit just reaches $M = 1$ . Both curves reach their corner.
$0 \leq p_{b} / p_{0} < 0.528$ (choked branch). Exit frozen at $M = 1$ , $p_{e}$ frozen at $0.528 p_{0}$ , $\overset{m}{˙}$ frozen at its peak. The extra pressure drop from $0.528 p_{0}$ down to $p_{b}$ happens outside the nozzle as expansion waves. Flat right ends.
$p_{b} / p_{0} = 0$ (degenerate: perfect vacuum). Still exactly $M = 1$ at the exit — the nozzle can't tell the difference between "a little below critical" and "total vacuum," because signals can't swim upstream against sonic flow.

The one-picture summary

How to read the whole derivation in one clean frame: one curve, three labelled milestones.

The blue curve is the master relation $p / p_{0}$ versus exit Mach number $M$ (Steps 2–4).
The green stretch ( $M < 1$ ) is the subsonic branch, where the exit obeys the back pressure ( $p_{e} = p_{b}$ ).
The yellow dot at $M = 1$ is the critical point: height $0.528$ — the choking pressure ratio for air.
The red dashed line marks the sonic ceiling the flow can touch but not cross inside a converging pipe.

Trace the curve left-to-right: energy conservation and isentropy build it (Steps 2–3), and the single dot where it meets $M = 1$ is the whole punchline — air chokes at $0.528$ , locked at Mach 1.

Recall Feynman: the whole walkthrough in plain words

Push still gas from a big tank into a pipe that gets narrower. Because the same gas must pass every slice each second, tightening the pipe forces the gas to hurry (Step 1). As it hurries, it trades its warmth for speed — it cools as it accelerates (Step 2). Because the flow is clean and reversible, that cooling comes with a matching pressure drop, and we can write one tidy rule connecting how fast the gas moves to how much its pressure has fallen (Step 3). Now ask: what pressure has it fallen to at the exact moment it moves as fast as a whisper travels — Mach 1? Plug $M = 1$ into the rule and out pops $0.528$ for air (Step 4). Why does the flow refuse to speed up past that? Because the amount of gas squeezing through per second is a hump that peaks right at Mach 1 (Step 5) — you're standing on the summit, and lowering the pressure outside can't lift you higher. And whether the outside is slightly low or a total vacuum, the exit stays pinned at Mach 1, dumping its leftover pressure only after it leaves the pipe (Step 6). To go faster you'd need the corridor to widen again after the pinch — but that's a different nozzle.

Recall

The single dial that decides everything ::: the back-pressure ratio $p_{b} / p_{0}$ compared to $0.528$ . Where the mass-flow hump peaks ::: exactly at $M = 1$ (its slope is zero there). What "trades" for speed as the gas accelerates ::: its thermal energy (it cools), keeping the total fixed. What makes narrower mean faster ::: mass conservation $ρ A V =$ const with $ρ$ nearly fixed at low speed. Two laws the whole derivation rests on ::: energy conservation (Step 2) and isentropy (Step 3).

Connections

Converging nozzle — subsonic flow, Mach 1 at exit — the parent result this page rebuilds picture by picture.
Isentropic Flow Relations — supplies $c_{p} = \frac{γ R}{γ - 1}$ and the $T$ – $p$ link used in Steps 2–3.
Speed of Sound — defines $a = γ R T$ and why $M = 1$ is the ceiling.
Stagnation Properties — the reservoir subscript- $0$ state.
Mass Flow Rate & Choking — the hump-peaks-at-one argument of Step 5.
Converging-Diverging (de Laval) Nozzle — the widening section needed to go past $M = 1$ .
Normal Shock Waves — what can appear downstream once supersonic.