Area-Mach number relation A - A - = f(M) — isentropic flow: Worked examples

This page is the drill deck for the Area–Mach relation. The parent note built the formula; here we push it through every kind of situation a nozzle can throw at you — subsonic branch, supersonic branch, the sonic point, tiny and huge Mach limits, choking, real word problems, and an exam twist.

Throughout we use the master formula (air, $γ = 1.4$ , so the exponent $\frac{γ + 1}{2 ( γ - 1 )} = 3$ ):

Before the examples, let us name every case so nothing sneaks up on you.

The scenario matrix

Think of the $A / A^{*}$ curve as a U-shaped valley (see below). Every problem lands in one of these cells:

#	Case class	What is special about it	Which example
C1	Sonic point $M = 1$	the exact bottom of the valley, $A / A^{*} = 1$	Ex 1
C2	Subsonic branch $M < 1$	left wall of the valley; $A / A^{*} > 1$	Ex 2
C3	Supersonic branch $M > 1$	right wall; same $A / A^{*}$ as some subsonic $M$	Ex 2
C4	Both roots for one area	geometry alone is ambiguous	Ex 3
C5	Limit $M \to 0$	$A / A^{*} \to \infty$ (huge reservoir, still air)	Ex 4
C6	Limit $M \to \infty$	$A / A^{*} \to \infty$ (hypersonic, area blows up)	Ex 4
C7	Choked mass flow	throat sits at $M = 1$ ; $\overset{m}{˙}$ frozen	Ex 5
C8	Real-world word problem	wind-tunnel / rocket design phrasing	Ex 6
C9	Exam twist — property from area	chain $A / A^{*} \to M \to p, T, ρ, V$	Ex 7
C10	Exam twist — back-pressure picks branch	same duct, two flow states	Ex 8

Figure — Area-Mach number relation A - A - = f(M) — isentropic flow

The figure is the map for everything below. Point to a dot on it as you read each example.

Worked examples

Ex 1 — Case C1: the sonic point (a sanity anchor)

Worked example The bottom of the valley

A stream is exactly at $M = 1$ . What is $A / A^{*}$ ? What would $A / A^{*}$ be at $M = 1.001$ and $M = 0.999$ ?

Forecast: Guess before reading. At $M = 1$ the flow is the sonic reference, so surely $A / A^{*} = 1$ . Just off $M = 1$ , should the ratio go up or down? (It can only go up — 1 is a minimum.)

Step 1 — Plug $M = 1$ . $\frac{A}{A ^{*}} = \frac{1}{1} [\frac{1 + 0.2 ( 1 )}{1.2}]^{3} = [\frac{1.2}{1.2}]^{3} = 1.$ Why this step? This is the definition of $A^{*}$ turned into arithmetic: at $M = 1$ the local area is the sonic area, so the ratio must be exactly 1. If your formula did not give 1 here, it is wrong.

Step 2 — Nudge to $M = 0.999$ and $M = 1.001$ . $\frac{A}{A ^{*}} (0.999) \approx 1.0000003, \frac{A}{A ^{*}} (1.001) \approx 1.0000003.$ Why this step? Both sides rise above 1 — that is the visual proof the valley has a flat minimum at $M = 1$ (the derivative is zero there). Flat bottom = the throat area changes very little for a big Mach swing near sonic, which is why nozzle throats are so sensitive.

Verify: $A / A^{*} = 1$ at $M = 1$ exactly, and the two neighbours are both $> 1$ and nearly equal — consistent with a smooth minimum. ✔

Ex 2 — Cases C2 & C3: one subsonic, one supersonic (direct evaluation)

Worked example Walk up both walls

Compute $A / A^{*}$ at (a) $M = 0.5$ (subsonic) and (b) $M = 2.0$ (supersonic).

Forecast: Both are away from $M = 1$ , so both give $A / A^{*} > 1$ . The steep supersonic wall usually climbs faster, so expect the $M = 2$ value to be a fair bit above the $M = 0.5$ value.

Step 1 — Subsonic, $M = 0.5$ . $\frac{A}{A ^{*}} = \frac{1}{0.5} [\frac{1 + 0.2 ( 0.25 )}{1.2}]^{3} = 2 [\frac{1.05}{1.2}]^{3} = 2 (0.875)^{3} \approx 1.340.$ Why this step? Direct substitution — no ambiguity when $M$ is given. The $1/ M = 2$ factor dominates the small bracket, so we sit modestly above 1 on the left wall.

Step 2 — Supersonic, $M = 2.0$ . $\frac{A}{A ^{*}} = \frac{1}{2} [\frac{1 + 0.2 ( 4 )}{1.2}]^{3} = \frac{1}{2} [\frac{1.8}{1.2}]^{3} = \frac{1}{2} (1.5)^{3} = \frac{3.375}{2} = 1.6875.$ Why this step? Now the bracket $(1.5)^{3}$ grows fast and beats the $1/2$ — we are on the right wall, higher than the subsonic case, as forecast.

Verify: $1.340$ (sub) $< 1.6875$ (super), both $> 1$ . Locate both dots on the s01 map: same height rule fails here because these are different $M$ — but each sits above the sonic minimum. ✔

Ex 3 — Case C4: one area, TWO Mach numbers (the inverse problem)

Worked example Geometry is ambiguous

A duct station has $A / A^{*} = 2.0$ , $γ = 1.4$ . Find all Mach numbers.

Forecast: $2 > 1$ , so the valley is cut by a horizontal line at height 2 — it crosses twice. One root well below 1 (near $0.3$ ), one well above 1 (near $2.2$ ).

Step 1 — Set up the equation. $2 = \frac{1}{M} [\frac{1 + 0.2 M ^{2}}{1.2}]^{3} .$ Why this step? We are inverting $f (M)$ . There is no closed-form inverse (it mixes $1/ M$ with a cubed polynomial), so we iterate or read a table — this is exactly why compressible-flow tables exist.

Step 2 — Subsonic root by iteration. Trying $M = 0.3$ gives $A / A^{*} \approx 2.035$ ; $M = 0.31$ gives $\approx 1.978$ . Interpolating: $M_{sub} \approx 0.306.$ Why this step? Bracket the answer between two guesses whose outputs straddle 2, then squeeze — a hand-version of Newton's method.

Step 3 — Supersonic root by iteration. $M = 2.2$ gives $A / A^{*} \approx 2.005$ ; $M = 2.19$ gives $\approx 1.985$ . So $M_{super} \approx 2.197.$ Why this step? The same height line hits the right wall too. Which one is physical depends on back-pressure (see Ex 8) — the formula alone will not choose.

Verify: Plug both back: $f (0.306) \approx 2.00$ and $f (2.197) \approx 2.00$ . Two valid roots for one area — exactly the U-shape prediction from the parent note. ✔

Ex 4 — Cases C5 & C6: the two limits (why the valley walls fly to infinity)

Worked example What happens at the extremes?

Estimate $A / A^{*}$ as $M \to 0$ and as $M \to \infty$ . Use $M = 0.05$ and $M = 5$ as stand-ins.

Forecast: Both limits blow up. Near-still air ( $M \to 0$ ) needs a huge pipe to carry the same mass slowly; hypersonic flow ( $M \to \infty$ ) needs a huge pipe because density collapses. So both walls climb to $\infty$ .

Step 1 — $M \to 0$ tail. For tiny $M$ , the bracket $\to (1/1.2)^{3} \approx 0.5787$ constant, while $1/ M \to \infty$ : $\frac{A}{A ^{*}} \approx \frac{0.5787}{M} M \to 0 \infty.$ Check $M = 0.05$ : $\frac{1}{0.05} (0.5787) \approx 11.57.$ Why this step? The $1/ M$ factor is the culprit — it says "to squeeze a fixed mass flow through nearly-stationary gas you need enormous area." Physically: $V \to 0$ in $\overset{m}{˙} = ρ A V$ forces $A \to \infty$ .

Step 2 — $M \to \infty$ tail. The bracket $\sim (\frac{0.2}{1.2} M^{2})^{3} = (0.1 \overline{6} M^{2})^{3}$ grows like $M^{6}$ , divided by $M$ gives $\sim M^{5} \to \infty$ . $\frac{A}{A ^{*}} (5) = \frac{1}{5} [\frac{1 + 0.2 ( 25 )}{1.2}]^{3} = \frac{1}{5} (5)^{3} = \frac{125}{5} = 25.0.$ Why this step? At high Mach, density $ρ$ crashes toward zero, so continuity again demands large area. The right wall wins the race to infinity much faster (power $M^{5}$ vs. $1/ M$ ).

Verify: $A / A^{*} (0.05) \approx 11.57$ and $A / A^{*} (5) = 25.0$ , both large and rising — matches "both limits $\to \infty$ ". ✔

Ex 5 — Case C7: choked mass flow (the throat is frozen)

Worked example How much air can this nozzle pass?

Reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , throat $A^{*} = 10 cm^{2} = 1 0^{- 3} m^{2}$ , air ( $R = 287 J/kg\cdotK$ , $γ = 1.4$ ). Find the choked mass flow.

Forecast: With a half-megapascal reservoir and a $10 cm^{2}$ throat, expect order $\sim 1 kg/s$ of air — a lot for so small a hole, because the throat is running at Mach 1.

Step 1 — Use the choked-flow formula. See Choked Flow & Maximum Mass Flow: $\overset{m}{˙} = \frac{p _{0} A ^{*}}{T _{0}} \frac{γ}{R} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ Why this step? When the throat hits $M = 1$ ( $A = A^{*}$ ), the mass flow is set purely by reservoir state — dropping the exit pressure cannot pull more air through. This is the physical meaning of $A^{*}$ being a constant yardstick.

Step 2 — Bracket factor. $\frac{2}{γ + 1} = \frac{2}{2.4} = 0.8333$ , exponent $= 3$ , so $(0.8333)^{3} = 0.5787$ . Why this step? It is the same sonic constant we met in the $M \to 0$ tail — no coincidence: both describe the sonic reference state.

Step 3 — Numbers. $\overset{m}{˙} = \frac{500000 \cdot 1 0 ^{- 3}}{300} \frac{1.4}{287} (0.5787) \approx 1.157 kg/s .$ Why this step? Assemble the pieces. $300 \approx 17.32$ , $1.4/287 \approx 0.06983$ .

Verify (units): $\frac{Pa \cdot m ^{2}}{K} \frac{1}{J/kg\cdotK} = \frac{N}{K} \frac{kg\cdotK}{J} = \frac{N}{K} \cdot \frac{K kg}{N\cdotm} = N \frac{kg}{N\cdotm} = N\cdotkg/m = kg^{2} / s^{2} = kg/s .$ ✔ Value $\approx 1.16 kg/s$ matches parent Example 3. ✔

Ex 6 — Case C8: real-world word problem (wind-tunnel design)

Worked example Size the tunnel

You want a supersonic wind tunnel test section at $M = 3.0$ for air. Your throat is $A^{*} = 25 cm^{2}$ . What test-section area $A_{e}$ must you machine?

Forecast: $M = 3$ is deep on the supersonic wall — expect $A / A^{*}$ around 4, so $A_{e}$ several times the throat, roughly $100 cm^{2}$ .

Step 1 — Evaluate $f (3)$ . $\frac{A _{e}}{A ^{*}} = \frac{1}{3} [\frac{1 + 0.2 ( 9 )}{1.2}]^{3} = \frac{1}{3} [\frac{2.8}{1.2}]^{3} = \frac{1}{3} (2.3333)^{3} = \frac{12.7037}{3} \approx 4.235.$ Why this step? The task gives $M$ , so this is a forward evaluation — no ambiguity. We are firmly on the supersonic branch by design intent (flow is choked at the throat then expands in the diverging section — see Converging-Diverging (de Laval) Nozzle).

Step 2 — Scale by the throat. $A_{e} = 4.235 \times 25 cm^{2} \approx 105.9 cm^{2} .$ Why this step? $A / A^{*}$ is dimensionless; multiply by the real throat area to get real millimetres of hardware.

Verify: $105.9/25 = 4.235 = f (3)$ . Larger than the throat, consistent with "supersonic accelerates by diverging." ✔

Ex 7 — Case C9 exam twist: from area straight to $p, T, ρ, V$

Worked example Chain the whole toolkit

Same tunnel as Ex 6: $M_{e} = 3.0$ , $T_{0} = 300 K$ , $p_{0} = 500 kPa$ , air. Find test-section $T_{e}$ , $p_{e}$ , and speed $V_{e}$ .

Forecast: At $M = 3$ the temperature drops hard (kinetic energy stolen from thermal), so $T_{e}$ well below 300 K — maybe ~100 K; pressure crashes to a few percent of $p_{0}$ .

Step 1 — Temperature via stagnation relation (Isentropic Stagnation Relations): $\frac{T _{e}}{T _{0}} = \frac{1}{1 + 0.2 ( 9 )} = \frac{1}{2.8} = 0.3571 \Rightarrow T_{e} = 107.1 K .$ Why this step? $M$ known ⇒ every thermodynamic ratio follows. Temperature comes first because pressure and density are built from it.

Step 2 — Pressure. $\frac{p _{e}}{p _{0}} = (\frac{T _{e}}{T _{0}})^{γ / (γ - 1)} = (0.3571)^{3.5} = 0.02722 \Rightarrow p_{e} = 13.6 kPa .$ Why this step? Isentropic flow links $p$ and $T$ by the exponent $γ / (γ - 1) = 3.5$ — no heat added, so entropy fixes the relation.

Step 3 — Speed. Local sound speed (Speed of Sound a = sqrt(gamma R T)): $a_{e} = γ R T_{e} = 1.4 \cdot 287 \cdot 107.1 = 207.5 m/s, V_{e} = M_{e} a_{e} = 3 (207.5) = 622.5 m/s .$ Why this step? $V = M a$ — Mach is a ratio, so you always need the local sound speed (which depends on the local, colder, $T_{e}$ ) to recover a real velocity in m/s.

Verify: $T_{e} \approx 107$ K (cold, as forecast), $p_{e} \approx 13.6$ kPa ( $2.7%$ of $p_{0}$ ), $V_{e} \approx 623$ m/s. All internally consistent: $V_{e} / a_{e} = 622.5/207.5 = 3.0 = M_{e}$ . ✔

Ex 8 — Case C10 exam twist: back-pressure picks the branch

Worked example Same duct, two different flows

A CD nozzle has $A_{e} / A^{*} = 2.0$ . (a) If it runs fully subsonic (high back-pressure, throat NOT sonic), the exit sits on the subsonic root. (b) If it runs choked & fully expanded supersonic, the exit sits on the supersonic root. State the exit Mach in each case and explain why the number differs even though the shape is identical.

Forecast: Same $A_{e} / A^{*} = 2$ from Ex 3, so the two roots are already known: $M \approx 0.306$ and $M \approx 2.197$ . The only difference is the pressure boundary condition, not the metal.

Step 1 — Case (a) subsonic exit. Back-pressure is high; flow decelerates in the diverging part (subsonic branch), never reaching $M = 1$ at the throat. Exit root: $M_{e} = 0.306.$ Why this step? On the subsonic wall, a diverging area slows the flow ( $d A / A = (M^{2} - 1) d V / V$ with $M < 1$ ⇒ $d A$ and $d V$ opposite signs) — see Area-Velocity Relation dA-A = (M^2-1) dV-V.

Step 2 — Case (b) supersonic exit. Back-pressure low enough to choke the throat ( $M = 1$ there); the diverging section now accelerates the flow. Exit root: $M_{e} = 2.197.$ Why this step? Once $M > 1$ past the throat, the same diverging area speeds the flow up — the sign flip in the area–velocity relation. This is why geometry alone (Ex 3) could not decide.

Step 3 — The tie-break. The physical selector is the exit pressure ratio $p_{e} / p_{0}$ demanded by the downstream reservoir. High $p_{b}$ → subsonic; sufficiently low $p_{b}$ → supersonic (possibly with a normal shock inside for in-between pressures). Why this step? It closes the "which root?" question the U-shape leaves open.

Verify: Both $M = 0.306$ and $M = 2.197$ satisfy $f (M) = 2.0$ (checked in Ex 3). Same area, two legal flows, chosen by back-pressure — not by geometry. ✔

Recall Quick self-test

$A / A^{*}$ at $M = 1$ ? ::: Exactly 1 (Ex 1, the valley floor). $A / A^{*}$ at $M = 0.5$ and $M = 2.0$ ? ::: $1.340$ and $1.6875$ (Ex 2). How many Mach numbers give $A / A^{*} = 2$ ? ::: Two — $M \approx 0.306$ and $M \approx 2.197$ (Ex 3). What decides which root is physical? ::: The downstream back-pressure, not the shape (Ex 8). Choked $\overset{m}{˙}$ for $p_{0} = 500$ kPa, $T_{0} = 300$ K, $A^{*} = 10 cm^{2}$ ? ::: $\approx 1.16$ kg/s (Ex 5). Test-section area for $M = 3$ if $A^{*} = 25 cm^{2}$ ? ::: $\approx 106 cm^{2}$ (Ex 6).