Converging nozzle — subsonic flow, Mach 1 at exit: Worked examples

The scenario matrix

Every converging-nozzle question is one of these cells. The single decision that splits the whole tree is: is $p_{b} / p_{0}$ above or below the critical value $0.528$ ?

Cell	Case class	Input signature	What is fixed	What you solve for
A	No flow (degenerate)	$p_{b} = p_{0}$	$V = 0$ everywhere	Nothing flows; $\overset{m}{˙} = 0$
B	Subsonic exit (unchoked)	$0.528 < p_{b} / p_{0} < 1$	$p_{e} = p_{b}$	Exit $M$ , $T_{e}$ , $V_{e}$ , $\overset{m}{˙}$
C	Exactly critical	$p_{b} / p_{0} = 0.528$	$M_{e} = 1$ boundary	Confirm it just chokes
D	Choked, under-expanded	$p_{b} / p_{0} < 0.528$	$M_{e} = 1$ , $p_{e} = 0.528 p_{0}$	$p_{e}$ , $\overset{m}{˙}$ (max), why $\overset{m}{˙}$ is flat
E	Limiting: $p_{b} \to 0$ (vacuum)	$p_{b} / p_{0} \to 0$	Still $M_{e} = 1$ , $p_{e} = 0.528 p_{0}$	Show exit does NOT change
F	Real-world word problem	Given engine/tank numbers	mix of above	Full pipeline to a physical answer
G	Exam twist — sizing	Want a target $\overset{m}{˙}$	Solve for area $A$	Invert the choked mass-flow law
H	Exam twist — non-air gas	$γ \neq = 1.4$	New critical ratio	Recompute $0.528 \to$ new value

Below, each example is tagged with the cell(s) it covers. Together they fill the whole table.

Setting the sign conventions (build before use)

Definition Symbols, in plain words

Pressures & the gate

$p_{0}$ — stagnation pressure: the pressure of the still gas in the tank, before it moves.
$p_{b}$ — back pressure: the pressure of the room the nozzle dumps into.
$p_{e}$ — exit pressure: the actual pressure of the gas right at the nozzle mouth.
$r \equiv p_{b} / p_{0}$ — the one ratio that decides everything. Always between $0$ and $1$ (you can only push from high to low pressure).

Temperatures

$T_{0}$ — stagnation temperature: temperature of the still gas in the tank.
$T_{e}$ — exit temperature, the gas temperature at the nozzle mouth.

The "star" (critical) state — the value a quantity takes when the exit is exactly at $M = 1$ . The little star superscript ( $*$ ) always means "the choked value":

$p^{*}$ — critical (choked) exit pressure $= 0.528 p_{0}$ for air.
$T^{*}$ — critical exit temperature $= \frac{2}{γ + 1} T_{0}$ .
$ρ^{*}$ — critical exit density.
$a^{*}$ — the speed of sound at the choked exit; when $M = 1$ the gas moves at exactly this speed.

Motion & flow

$M_{e}$ — exit Mach number = (exit speed) ÷ (local speed of sound). $M_{e} = 1$ means the gas moves exactly at the speed of sound.
$a$ — speed of sound: how fast a small pressure signal travels through the gas, $a = γ R T$ (see Speed of Sound). We write $a_{e}$ for its value at the exit.
$V$ — the bulk flow speed of the gas ( $V_{e}$ = exit speed). Related by $V = M a$ .
$ρ$ — density: mass of gas per cubic metre ( $ρ_{e}$ = exit density, $ρ^{*}$ = choked density).
$A$ — nozzle exit (throat) area: the cross-sectional area of the narrow mouth the gas leaves through, in $m^{2}$ .
$\overset{m}{˙}$ — mass flow rate: kilograms of gas leaving per second, $\overset{m}{˙} = ρ A V$ .
$R$ — specific gas constant of the gas (air: $287 J/kg\cdotK$ ), the constant in the ideal-gas law $p = ρR T$ . Also sets the speed of sound.
$γ$ — ratio of specific heats ( $1.4$ for air, $5/3$ for a monatomic gas like helium).

The decision figure below is the flowchart every example follows.

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

Intuition How to read Figure s01 — trace it with your finger

Put your finger on the top black box: this is where you always start, by computing $r = p_{b} / p_{0}$ . The single arrow drops straight down into the red diamond in the middle — the only decision on the whole page, the yes/no question " $r > 0.528$ ?".

Slide left along the "yes" arrow → you land in the black unchoked box: the exit is subsonic ( $M_{e} < 1$ ) and the exit pressure equals the room, $p_{e} = p_{b}$ . This is Cells A and B.
Slide right along the "no" arrow → you land in the red-outlined choked box: the exit is locked at $M_{e} = 1$ and the exit pressure is frozen at $p_{e} = 0.528 p_{0}$ , deaf to the room. This is Cells C, D, E, F.

Notice there is exactly one diamond: no example ever needs a second decision. Every worked example below is literally you dropping a number into that top box and reading off which branch it falls to.

Where the $0.528$ comes from (so we can change it)

Cell H uses a different gas, so I cannot keep quoting $0.528$ as a magic number — I must show the machine that produces it. It comes straight from the master relation.

Cell A — No flow (degenerate input)

Cell B — Subsonic exit (the everyday case)

Worked example Cell B ·

$0.528 < r < 1$ Air reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , exit area $A = 10 cm^{2}$ . Back pressure $p_{b} = 350 kPa$ . Find exit Mach, exit temperature, exit speed, and mass flow.

Forecast: Will the exit be subsonic or sonic? Guess $M_{e}$ to one decimal.

Gate: $r = 350/500 = 0.70 > 0.528$ → unchoked, so $p_{e} = p_{b} = 350 kPa$ . Why this step? Above critical, the jet still "feels" the room and matches it.
Invert the master relation to get $M_{e}$ : $\frac{p _{0}}{p _{e}} = \frac{500}{350} = 1.4286 = (1 + 0.2 M_{e}^{2})^{3.5}$ Take the $3.5$ -th root: $1.428 6^{1/3.5} = 1.1073$ , so $0.2 M_{e}^{2} = 0.1073 \Rightarrow M_{e} = 0.536 \approx 0.732$ . Why this step? The master relation ties a measurable pressure ratio to $M$ . Inverting extracts the speed.
Exit temperature from $T_{0} / T_{e} = 1 + 0.2 M_{e}^{2} = 1.1073$ : $T_{e} = 300/1.1073 = 270.9 K$ Why this step? We need the actual gas temperature at the exit because the speed of sound $a = γ R T$ depends on it — without $T_{e}$ we cannot turn $M_{e}$ into a real m/s speed.
Local speed of sound and exit speed (see Speed of Sound): $a_{e} = γ R T_{e} = 1.4 \cdot 287 \cdot 270.9 = 330.1 m/s, V_{e} = M_{e} a_{e} = 0.732 \cdot 330.1 = 241.6 m/s$ Why this step? Mach number is a ratio; to get a physical speed in m/s we must multiply it by the local sound speed $a_{e}$ , which we can only compute now that we know $T_{e}$ .
Exit density from ideal gas $p_{e} = ρ_{e} R T_{e}$ : $ρ_{e} = \frac{350000}{287 \cdot 270.9} = 4.502 kg/m^{3}$ Why this step? Mass flow counts kilograms, so we need how densely packed the gas is; the ideal-gas law converts the known $p_{e}, T_{e}$ into density $ρ_{e}$ .
Mass flow $\overset{m}{˙} = ρ_{e} A V_{e} = 4.502 \cdot (10 \times 1 0^{- 4}) \cdot 241.6 = 1.088 kg/s$ . Why this step? $\overset{m}{˙} = ρ A V$ is literally (mass per volume)×(area)×(speed) = mass leaving per second — the final physical answer.

Verify: $M_{e} = 0.732 < 1$ ✓ (subsonic, as a converging nozzle must be here). Units: $kg/m^{3} \cdot m^{2} \cdot m/s = kg/s$ ✓. All checked numerically below.

Cell C — Exactly critical (the knife-edge)

Cell D — Choked and under-expanded

Worked example Cell D ·

$r < 0.528$ Air reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , exit area $A = 10 cm^{2}$ . Now $p_{b} = 101 kPa$ (atmosphere). Find $M_{e}$ , $p_{e}$ , $T_{e}$ , $V_{e}$ , and $\overset{m}{˙}$ .

Forecast: Guess $p_{e}$ . Trap: is it $101$ kPa?

Gate: $r = 101/500 = 0.202 < 0.528$ → choked. Why this step? Below critical, the exit locks at $M_{e} = 1$ regardless of $p_{b}$ .
Exit pressure is fixed by the nozzle, NOT by $p_{b}$ : $p_{e} = p^{*} = 0.528 \cdot 500 = 264 kPa (not 101!)$ Why this step? At $M = 1$ pressure signals can't travel upstream, so the room's $101$ kPa never reaches inside. Remaining drop $264 \to 101$ happens outside as expansion waves.
Exit temperature: $T_{e} = T^{*} = \frac{2}{2.4} \cdot 300 = 250 K$ . Why this step? At the choked exit $M = 1$ , so the stagnation-to-static relation collapses to the fixed factor $T^{*} / T_{0} = 2/ (γ + 1)$ — we need $T^{*}$ to get the sound speed in the next step.
Exit speed = local sound speed (because $M = 1$ ): $V_{e} = a^{*} = 1.4 \cdot 287 \cdot 250 = 316.9 m/s$
Density and mass flow (the ceiling value): $ρ^{*} = \frac{264000}{287 \cdot 250} = 3.679 kg/m^{3}, \overset{m}{˙} = ρ^{*} A V_{e} = 3.679 \cdot 1 0^{- 3} \cdot 316.9 = 1.166 kg/s$ Why this step? Same $\overset{m}{˙} = ρ A V$ recipe as Cell B, now using the choked star-state values; this is the maximum mass flow the nozzle can pass.

Verify: $\overset{m}{˙} = 1.166 kg/s$ here vs $1.088 kg/s$ in Cell B — choked flow is larger, and it is the maximum (next example proves it won't grow). ✓

Cell E — Limiting case: exhaust into vacuum

Worked example Cell E ·

$r \to 0$ Same tank, but now $p_{b} = 0$ (a vacuum chamber). What changes at the exit compared to Cell D?

Forecast: Does the exit go supersonic now that the room is empty?

Gate: $r = 0/500 = 0 < 0.528$ → still choked. Why this step? The gate only cares whether $r$ is below $0.528$ . Zero is below it, so nothing about the exit changes.
Exit stays locked: $M_{e} = 1$ , $p_{e} = 264 kPa$ , $T_{e} = 250 K$ , $\overset{m}{˙} = 1.166 kg/s$ — identical to Cell D. Why this step? This is the whole point of choking: the exit is deaf to $p_{b}$ once sonic. The extra expansion from $264 \to 0$ happens entirely outside.

Verify: The mass-flow formula's $M$ -factor $\frac{M}{( 1 + 0.2 M ^{2} ) ^{3}}$ is maximised at $M = 1$ (derivative zero there). Since the exit is pinned at $M = 1$ , $\overset{m}{˙}$ cannot rise. Even a perfect vacuum gives the same $1.166 kg/s$ . ✓ (proven numerically below)

The mass-flow "ceiling" is the shape to burn into memory:

Intuition How to read Figure s02 — walk the curve right-to-left

The horizontal axis is the gate ratio $r = p_{b} / p_{0}$ , and I have drawn it reversed (1 on the left, 0 on the right) so that the physical act of lowering the back pressure moves you steadily to the right — follow the black arrow along the bottom.

Start at the far left ( $r = 1$ ): flat on the floor, $\overset{m}{˙} = 0$ (this is Cell A).
Walk right (drop $p_{b}$ ): the black curve climbs — the unchoked regime, Cells B. More pressure drop, more flow, just like a garden hose.
You reach the red dashed vertical line at $r = 0.528$ : the exit has just hit $M = 1$ .
Keep walking right past it: the curve turns flat and red — a dead-level plateau. That red plateau is choking. Drop $p_{b}$ all the way to a vacuum on the far right and the height never changes.

The single message of the picture: the red flat top is a ceiling. Lowering $p_{b}$ below $0.528$ buys you zero extra mass flow — which is exactly why Cell D and Cell E give the identical $1.166 kg/s$ .

Cell F — Real-world word problem

Worked example Cell F · applied

A pressurised nitrogen tank ( $γ = 1.4$ , $R = 297 J/kg\cdotK$ ) feeds a small thruster through a converging nozzle of exit area $A = 2 cm^{2}$ . Tank: $p_{0} = 800 kPa$ , $T_{0} = 290 K$ . It fires into space ( $p_{b} \approx 0$ ). How much propellant leaves per second?

Forecast: Choked or not in space? And is $\overset{m}{˙}$ small or huge?

Gate: $r = 0/800 = 0 < 0.528$ → choked (space is well below critical). Why this step? Same universal gate; nitrogen has $γ = 1.4$ so the $0.528$ number still applies.
Exit conditions: $p_{e} = 0.528 \cdot 800 = 422.5 kPa$ , $T_{e} = \frac{2}{2.4} \cdot 290 = 241.7 K$ . Why this step? Choked means $M = 1$ , so the exit pressure and temperature are locked to the fixed star-state fractions of $p_{0}, T_{0}$ — we need both before we can find speed and density.
Exit speed: $V_{e} = a^{*} = 1.4 \cdot 297 \cdot 241.7 = 316.9 m/s$ . Why this step? $M_{e} = 1$ so exit speed is the local sound speed — for Stagnation Properties this is the "throat" state.
Density and mass flow: $ρ^{*} = \frac{422500}{297 \cdot 241.7} = 5.885 kg/m^{3}, \overset{m}{˙} = 5.885 \cdot (2 \times 1 0^{- 4}) \cdot 316.9 = 0.373 kg/s$ Why this step? $\overset{m}{˙} = ρ A V$ turns the exit state into the actual kilograms-per-second of propellant expelled — the physical answer the thruster designer wants.

Verify: Units chain to kg/s ✓. Sensible for a small thruster (hundreds of grams per second). Checked below.

Cell G — Exam twist: size the nozzle for a target flow

Worked example Cell G · invert the choked mass-flow law for area

You need a choked air nozzle passing $\overset{m}{˙} = 0.5 kg/s$ from $p_{0} = 600 kPa$ , $T_{0} = 320 K$ . What exit area $A$ ?

Forecast: Bigger or smaller than a $cm^{2}$ ? Guess first.

Start from the general mass-flow law written in stagnation terms (from Mass Flow Rate & Choking): $\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} \frac{M}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$ Because we are told it is choked, set $M = 1$ . The $M$ -factor collapses to the fixed choked constant: $\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ Why this step? Choked flow fixes $M = 1$ , so $\overset{m}{˙}$ now depends on only $A, p_{0}, T_{0}, γ$ — every quantity except the one unknown $A$ . That makes it a one-line solve.
Rearrange to isolate the unknown area: $A = \frac{m ˙}{p _{0} \frac{γ}{R T _{0}} ( \frac{2}{γ + 1} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$ Why this step? We want $A$ , so we divide the target $\overset{m}{˙}$ by everything else. This is the literal "invert the choked law" the matrix promised.
Evaluate the two constant factors for air. First the exponent: $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.4}{0.8} = 3$ . Then $\frac{1.4}{287 \cdot 320} = 1.524 \times 1 0^{- 5} = 3.904 \times 1 0^{- 3}, (\frac{2}{2.4})^{3} = 0.5787$ Why this step? Pin down every gas/reservoir constant before touching $A$ , so the final division is pure arithmetic.
Per unit area the nozzle passes: $\frac{m ˙}{A} = p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{3} = 600000 \cdot 3.904 \times 1 0^{- 3} \cdot 0.5787 = 1355 kg/(s\cdotm^{2})$ Why this step? This "flow per square metre" is exactly what we divide the target into.
Solve: $A = 0.5/1355 = 3.69 \times 1 0^{- 4} m^{2} = 3.69 cm^{2}$ . Why this step? Dividing the demanded $\overset{m}{˙}$ by the per-square-metre flow gives the required throat area — the design answer.

Verify: Plug $A$ back into the choked law: $1355 \cdot 3.69 \times 1 0^{- 4} = 0.500 kg/s$ ✓. Units: $Pa \cdot \frac{1}{J/kg} \cdot m^{2} = kg/s$ (since $J/kg = m^{2} / s^{2}$ ) ✓. Checked numerically below.

Cell H — Exam twist: a different gas ( $γ \neq = 1.4$ )

Worked example Cell H · monatomic gas, recompute the gate

Helium ( $γ = 5/3$ ) flows from $p_{0} = 500 kPa$ into a room at $p_{b} = 250 kPa$ . Is it choked? What is the critical ratio for helium?

Forecast: Is helium's magic number bigger or smaller than air's $0.528$ ? Guess before computing.

Do not reuse $0.528$ — that was derived with $γ = 1.4$ . Recompute the gate from the boxed critical-ratio formula with $γ = 5/3$ : $\frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}} = (\frac{2}{8/3})^{(5/3) / (2/3)} = (0.75)^{2.5} = 0.4871$ Why this step? The gate value is a pure function of $γ$ . Change the gas, change the number — this is the whole point of Cell H.
Apply the gate for helium: $r = 250/500 = 0.50$ . Compare to $0.487$ . Since $0.50 > 0.487$ → unchoked. Why this step? A monatomic gas chokes at a lower ratio, so the same $r$ that would choke air ( $0.50 < 0.528$ ) leaves helium subsonic.
Since unchoked, $p_{e} = p_{b} = 250 kPa$ ; invert the master relation with helium's $γ$ to get exit Mach. Here $\frac{γ - 1}{2} = \frac{1}{3}$ and $\frac{γ}{γ - 1} = 2.5$ : $\frac{p _{0}}{p _{e}} = \frac{500}{250} = 2 = (1 + \frac{1}{3} M_{e}^{2})^{2.5}$ Take the $2.5$ -th root: $2^{1/2.5} = 1.3195$ , so $\frac{1}{3} M_{e}^{2} = 0.3195 \Rightarrow M_{e} = 0.9585 = 0.979$ . Why this step? Same inversion machinery as Cell B, but every $γ$ -dependent exponent changed. Note $M_{e} = 0.979 < 1$ — barely subsonic, consistent with $r$ sitting just above the helium gate.

Verify: $0.487 < 0.50$ ⇒ unchoked, and inversion gives $M_{e} < 1$ ✓. The same setup in air ( $r = 0.50 < 0.528$ ) would be choked — proof that the gas matters and you must recompute the gate. Checked below.

Recall Which cell am I in? (self-test)

Compute $r = p_{b} / p_{0}$ , then answer. If $r = 1$ what flows? ::: Nothing; $\overset{m}{˙} = 0$ (Cell A). If $r$ is just below $0.528$ for air, exit Mach? ::: $M_{e} = 1$ , choked (Cell D). Choked exit pressure for air? ::: $p_{e} = 0.528 p_{0}$ , independent of $p_{b}$ . Does lowering $p_{b}$ from $0.2$ to $0$ change $\overset{m}{˙}$ ? ::: No — exit is pinned at $M = 1$ (Cell E). General critical-ratio formula? ::: $p^{*} / p_{0} = (2/ (γ + 1))^{γ / (γ - 1)}$ . Critical ratio for a monatomic gas $γ = 5/3$ ? ::: $(0.75)^{2.5} \approx 0.487$ (Cell H). To pass more $\overset{m}{˙}$ when already choked, what must change? ::: Bigger area $A$ or higher $p_{0}$ — not lower $p_{b}$ (Cell G).

Connections

Converging-Diverging (de Laval) Nozzle — parent; to break the $M = 1$ ceiling you add a diverging section.
Isentropic Flow Relations — the master relation inverted in Cells B, C, H and used to derive the critical ratio.
Mass Flow Rate & Choking — the choked $\overset{m}{˙}$ formula inverted in Cells D–G.
Speed of Sound — why exit speed equals $a^{*}$ at $M = 1$ .
Stagnation Properties — the $p_{0}, T_{0}$ reference state feeding every example.
Normal Shock Waves — what happens if you do attach a diverging section and over-compress.