Choked flow — condition M = 1 at throat, maximum mass flow: Worked examples

Everything below uses tools built in the parent and its prerequisites: Isentropic Flow Relations, Stagnation (Total) Properties, Speed of Sound & Mach Number, and — for the last twist — Converging–Diverging (de Laval) Nozzle.

The scenario matrix

Before any arithmetic, list every kind of question this topic can ask. One symbol first, so the table reads cleanly:

Cell	Case class	What decides it	Example
A	Not choked (subsonic exit)	$p_{b} / p_{0} > r^{*}$	Ex. 1
B	Exactly at threshold	$p_{b} / p_{0} = r^{*}$	Ex. 2
C	Choked (deep)	$p_{b} / p_{0} < r^{*}$	Ex. 3
D	Max mass flow number	plug $M = 1$ , $A = A^{*}$	Ex. 4
E	Limiting input: $T_{0}$ up	$\overset{m}{˙}_{m a x} \propto 1/ T_{0}$	Ex. 5
F	Limiting input: $p_{0}$ up	$\overset{m}{˙}_{m a x} \propto p_{0}$	Ex. 5
G	Degenerate: $p_{b} = p_{0}$ (no drop)	$\overset{m}{˙} = 0$	Ex. 6
H	Different gas ( $γ$ change)	recompute $r^{*}$	Ex. 7
I	Real-world word problem	pick the right cell	Ex. 8
J	Exam twist: C-D nozzle back pressure	throat still chokes	Ex. 9

The figures below anchor the map from $p_{b} / p_{0}$ to behaviour (Cells A–C) and the $\overset{m}{˙}$ -vs- $p_{b}$ saturation curve (Cells D–G).

Figure — Choked flow — condition M = 1 at throat, maximum mass flow

Constants used throughout unless stated: air, $γ = 1.4$ , $R = 287 J kg^{- 1} K^{- 1}$ , so $r^{*} = (2/2.4)^{3.5} \approx 0.528$ and the bracket $(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (0.8333)^{3} \approx 0.5787$ .

Worked example Example 1 — Cell A:

Not choked Air reservoir $p_{0} = 120$ kPa exhausts to atmosphere $p_{b} = 101.3$ kPa through a converging nozzle. Choked? Forecast: The pressure drop is gentle (only ~16%). Guess: not choked, subsonic exit. Step 1 — Form the back-pressure ratio. $p_{b} / p_{0} = 101.3/120 = 0.844$ . Why this step? Choking is decided by comparing this single number to $r^{*}$ — nothing else. Step 2 — Compare to $r^{*} = 0.528$ . $0.844 > 0.528$ . Why this step? Above threshold means the exit pressure the nozzle needs ( $= p_{b}$ ) is still higher than $p^{*}$ , so the throat never has to reach $M = 1$ . Answer: Not choked. Exit pressure $= p_{b} = 101.3$ kPa, exit is subsonic ( $M_{exit} < 1$ ), and lowering $p_{b}$ would raise $\overset{m}{˙}$ . Verify: Sanity — a mere 16% drop cannot produce sonic flow; the whole point of $r^{*} \approx 0.528$ is you need ~47% drop first. Consistent. ✔

Worked example Example 2 — Cell B:

Exactly at threshold $p_{0} = 192$ kPa, $p_{b} = 101.3$ kPa, converging nozzle, air. Choked, borderline, or not? Forecast: $101.3/192 \approx 0.53$ — suspiciously close to 0.528. Guess: right on the edge. Step 1 — Ratio. $p_{b} / p_{0} = 101.3/192 = 0.5276$ . Why this step? Same single-number test as Cell A. Step 2 — Compare. $0.5276 \approx 0.528 = r^{*}$ . Why this step? At exactly $r^{*}$ the exit reaches $M = 1$ but only just; the exit pressure equals ambient AND equals $p^{*}$ simultaneously — the last subsonic state and the first choked state coincide. Answer: Just choked. $M_{exit} = 1$ , $p_{exit} = p^{*} = p_{b}$ . Any further drop in $p_{b}$ locks it into full choke. Verify: $0.528 \times 192 = 101.4 kPa = p_{b}$ (to rounding). The exit pressure matches ambient exactly, the defining feature of the threshold. ✔

Worked example Example 3 — Cell C:

Choked (deep) $p_{0} = 500$ kPa, $p_{b} = 101.3$ kPa, air, converging nozzle. Choked? What is the exit pressure? Forecast: Big drop (80%). Guess: choked, exit locked above ambient. Step 1 — Ratio. $p_{b} / p_{0} = 101.3/500 = 0.2026$ . Why this step? The choke test. Step 2 — Compare. $0.2026 < 0.528$ → choked. Why this step? A converging exit cannot fall below $p^{*}$ while remaining a smooth subsonic-to-sonic flow; it clamps at $M = 1$ . Step 3 — Locked exit pressure. $p_{exit} = p^{*} = 0.528 \times 500 = 264 kPa$ . Why this step? Once choked, exit pressure freezes at $p^{*}$ , independent of how low $p_{b}$ is. Answer: Choked. Exit pressure $= 264$ kPa (well above ambient 101.3 kPa → under-expanded jet finishes expanding outside). Verify: $264 > 101.3$ , so the jet is under-expanded — exactly what "choked with low $p_{b}$ " predicts. ✔

Worked example Example 4 — Cell D:

The maximum mass-flow number Same nozzle as Ex. 3: throat area $A^{*} = 2 \times 1 0^{- 4} m^{2}$ , $T_{0} = 300$ K, $p_{0} = 500$ kPa, air. Find $\overset{m}{˙}_{m a x}$ . Forecast: Order of magnitude? Small hole, half-MPa reservoir — guess a fraction of a kg/s. Step 1 — Bracket factor. $(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (0.8333)^{3} = 0.5787$ . Why this step? This dimensionless factor is the "efficiency penalty" for the sonic throat; exponent $\frac{γ + 1}{2 ( γ - 1 )} = 3$ for air. Step 2 — Root factor. $γ / (R T_{0}) = 1.4/ (287 \times 300) = 1.626 \times 1 0^{- 5} = 4.032 \times 1 0^{- 3}$ . Why this step? This carries the units $kg / J = s/m$ that make the final answer come out in kg/s. Step 3 — Assemble. $\overset{m}{˙}_{m a x} = A^{*} p_{0} γ / (R T_{0}) (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (2 \times 1 0^{- 4}) (5 \times 1 0^{5}) (4.032 \times 1 0^{- 3}) (0.5787)$ . Why this step? This is the boxed choked-flow formula from the parent with $M = 1$ , $A = A^{*}$ substituted. Answer: $\overset{m}{˙}_{m a x} \approx 0.233 kg/s$ . Verify: Units: $m^{2} \cdot Pa \cdot (s/m) = m^{2} \cdot \frac{kg}{m s ^{2}} \cdot \frac{s}{m} = kg/s$ . ✔ Magnitude ~0.2 kg/s matches the "fraction of a kg/s" forecast. ✔

Worked example Example 5 — Cells E & F:

Turn the input knobs Starting from Ex. 4 ( $\overset{m}{˙}_{m a x} = 0.233$ kg/s): (a) double $T_{0}$ to 600 K; (b) instead double $p_{0}$ to 1000 kPa. New $\overset{m}{˙}_{m a x}$ each? Forecast: Hotter gas → guess less flow (density loss). Higher $p_{0}$ → guess more flow, linearly. Step 1 — (a) Temperature scaling. $\overset{m}{˙}_{m a x} \propto 1/ T_{0}$ , so doubling $T_{0}$ multiplies by $1/ 2$ : $0.233/ 2 = 0.165 kg/s$ . Why this step? Only the $T_{0}$ in the denominator changes; all other factors are held. Hotter gas is less dense, and sound speed rises only as $T$ — density loss wins. Step 2 — (b) Pressure scaling. $\overset{m}{˙}_{m a x} \propto p_{0}$ , so doubling $p_{0}$ doubles it: $2 \times 0.233 = 0.466 kg/s$ . Why this step? $p_{0}$ appears to the first power; more upstream pressure means proportionally denser gas at the throat. Answer: (a) $0.165$ kg/s (less); (b) $0.466$ kg/s (more). Verify: $0.233/1.4142 = 0.1648$ ✔ and $0.233 \times 2 = 0.466$ ✔. Direction matches the forecast: heat hurts, pressure helps. ✔

Worked example Example 6 — Cell G:

Degenerate — no pressure drop $p_{0} = 500$ kPa, $p_{b} = 500$ kPa (valve shut on the same reservoir), converging nozzle. Find $\overset{m}{˙}$ . Forecast: No drop → guess zero flow. Step 1 — Ratio. $p_{b} / p_{0} = 500/500 = 1$ . Why this step? The choke test still applies but at its ceiling. Step 2 — Interpret. Ratio $= 1 > 0.528$ , so not choked; the exit Mach solves the isentropic relation with $p_{exit} / p_{0} = 1$ , giving $M_{exit} = 0$ . Why this step? $\frac{p _{0}}{p} = 1 \Rightarrow 1 + \frac{γ - 1}{2} M^{2} = 1 \Rightarrow M = 0$ : with no pressure difference there is nothing to accelerate the gas. Step 3 — Mass flow. $\overset{m}{˙} = ρ A V$ with $V = M a = 0 \Rightarrow \overset{m}{˙} = 0$ . Why this step? The master $\overset{m}{˙}$ formula has a factor $M$ ; at $M = 0$ it vanishes. Answer: $\overset{m}{˙} = 0$ . This is the far-left end of the saturation curve in figure s02. Verify: In the parent's master formula the $M$ -factor is $\frac{M}{( \dots )}$ , which is exactly 0 at $M = 0$ . ✔

Worked example Example 7 — Cell H:

Different gas (change $γ$ ) Helium ( $γ = 5/3 \approx 1.667$ ) reservoir $p_{0} = 500$ kPa, $p_{b} = 101.3$ kPa, converging nozzle. Choked? Give helium's critical ratio. Forecast: Same big drop as Ex. 3; guess still choked, but the threshold number differs from 0.528. Step 1 — Helium's critical ratio. $r^{*} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}}$ with $γ = 1.667$ : $\frac{2}{2.667} = 0.75$ , exponent $\frac{1.667}{0.667} = 2.5$ , so $r^{*} = 0.7 5^{2.5} = 0.487$ . Why this step? $r^{*}$ is gas-specific through $γ$ ; you cannot reuse air's 0.528. Step 2 — Compare. $p_{b} / p_{0} = 101.3/500 = 0.2026 < 0.487$ → choked. Why this step? Same threshold logic; only the number moved. Answer: Choked. Helium's critical ratio is $\approx 0.487$ (lower than air's 0.528, because a higher $γ$ pushes the sonic point to a deeper pressure drop). Verify: $0.7 5^{2.5} = 0.4871$ and $0.2026 < 0.4871$ . ✔ Also $r^{*}$ decreases as $γ$ increases (air 0.528 → helium 0.487). ✔

Worked example Example 8 — Cell I:

Real-world word problem A compressed-air tank at $p_{0} = 800$ kPa, $T_{0} = 290$ K feeds a small pneumatic tool through a converging orifice of throat area $A^{*} = 1.5 \times 1 0^{- 5} m^{2}$ . Room is at 101 kPa. What mass flow does the tank deliver? Forecast: Big drop → choked → pick the $\overset{m}{˙}_{m a x}$ formula. Guess ~0.03 kg/s (tiny hole). Step 1 — Choke check. $p_{b} / p_{0} = 101/800 = 0.126 < 0.528$ → choked, use $\overset{m}{˙}_{m a x}$ . Why this step? You must classify the case before choosing a formula; here it lands in Cell C/D. Step 2 — Root factor. $γ / (R T_{0}) = 1.4/ (287 \times 290) = 1.682 \times 1 0^{- 5} = 4.101 \times 1 0^{- 3}$ . Why this step? Same units-carrying factor as Ex. 4, now with $T_{0} = 290$ . Step 3 — Assemble. $\overset{m}{˙}_{m a x} = (1.5 \times 1 0^{- 5}) (8 \times 1 0^{5}) (4.101 \times 1 0^{- 3}) (0.5787)$ . Why this step? Choked formula with this tank's numbers. Answer: $\overset{m}{˙}_{m a x} \approx 0.0285 kg/s$ (≈ 28.5 g/s). Verify: Compute $(1.5 \times 1 0^{- 5}) (8 \times 1 0^{5}) = 12$ ; $12 \times 4.101 \times 1 0^{- 3} = 0.04921$ ; $\times 0.5787 = 0.02849$ kg/s. ✔ Matches the ~0.03 kg/s forecast. ✔

Worked example Example 9 — Cell J:

Exam twist — C-D nozzle, throat still chokes A Converging–Diverging (de Laval) Nozzle has throat $A^{*} = 2 \times 1 0^{- 4} m^{2}$ , $p_{0} = 500$ kPa, $T_{0} = 300$ K, air. Its back pressure is set so the flow is fully choked (throat sonic). Does $\overset{m}{˙}$ differ from the plain converging nozzle of Ex. 4? Forecast: The diverging part accelerates flow to supersonic — surely more mass gets through? Guess: no, mass flow is set at the throat. Step 1 — Where is $\overset{m}{˙}$ fixed? By continuity $\overset{m}{˙} = ρ A V$ is the same at every section; the choking condition ( $M = 1$ ) is imposed at the minimum area = throat. Why this step? Mass flow is a single number for the whole steady nozzle; it is capped by the sonic throat, not by the exit. Step 2 — Compute. Identical inputs to Ex. 4 → $\overset{m}{˙}_{m a x} = 0.233 kg/s$ . Why this step? The $\overset{m}{˙}_{m a x}$ formula uses only throat area and stagnation state — the diverging section does not appear. Answer: Same $\overset{m}{˙}_{m a x} = 0.233$ kg/s. The diverging section changes the exit speed and pressure (supersonic exit), not the mass flow. If the back pressure is wrong, a normal shock appears in the diverging part — but $\overset{m}{˙}$ still stays $0.233$ kg/s because the throat is unchanged. Verify: Same numbers as Ex. 4 → $0.233$ kg/s, independent of the diverging geometry or downstream shock. ✔

Recall Quick self-test across the matrix

Threshold ratio for air ::: $r^{*} \approx 0.528$ ; choked when $p_{b} / p_{0} < r^{*}$ . Does $\overset{m}{˙}_{m a x}$ depend on $p_{b}$ once choked? ::: No — it depends only on $p_{0}, T_{0}, A^{*}, γ, R$ . Effect of doubling $T_{0}$ on $\overset{m}{˙}_{m a x}$ ? ::: Multiply by $1/ 2 \approx 0.707$ (less flow). Effect of doubling $p_{0}$ ? ::: Doubles $\overset{m}{˙}_{m a x}$ . Helium's critical ratio? ::: $\approx 0.487$ (lower than air because $γ$ is higher). Does a diverging section raise $\overset{m}{˙}$ ? ::: No — the sonic throat caps it.