Converging-diverging (de Laval) nozzle — subsonic, supersonic flow: Worked examples

Before any example, we pin down four symbols this whole page leans on, in plain words and pictures.

Definition The four symbols you must own first

Stagnation (reservoir) pressure $p_{0}$ : the pressure of the gas at rest, deep inside the tank/combustion chamber that feeds the nozzle, where the flow speed $V = 0$ . Picture a big calm tank; $p_{0}$ is the gauge on its wall. Because the gas there is not moving, all its energy is "stored"; as it accelerates down the nozzle the local pressure $p$ falls below $p_{0}$ . So $p_{0}$ is the fixed ceiling; $p$ is what's left after some energy became speed.
Stagnation temperature $T_{0}$ : same idea for temperature — the temperature in that calm tank where $V = 0$ . As the gas speeds up it cools, so local $T < T_{0}$ .
Throat / sonic area $A^{*}$ : the cross-sectional area where the flow is exactly sonic, $M = 1$ . In a running de Laval nozzle this is the physical throat (the narrowest point). We measure every other area against it as the ratio $A / A^{*}$ — that is why the master relation (AR) below is written in terms of $A / A^{*}$ and not raw area.
Back pressure $p_{b}$ : the pressure of the environment the nozzle exhausts into — the atmosphere for a rocket, or the condenser for a steam turbine. It is imposed from outside the nozzle, at the exit plane. Lowering $p_{b}$ is like sucking harder on the exit; it is the single dial that decides which of the many flow regimes Nature picks.

Figure — Converging-diverging (de Laval) nozzle — subsonic, supersonic flow

Read the picture: the calm tank on the left holds gas at $p_{0}, T_{0}$ ( $V = 0$ ); the hourglass duct accelerates it; the environment on the right imposes $p_{b}$ . Everything else on this page is bookkeeping between these four labels.

The scenario matrix

Every nozzle question is really one of these case classes. The right-hand column names the example that covers it.

#	Case class	The distinguishing feature	Covered by
A	Subsonic root of area ratio	$A / A^{*} > 1$ , pick the $M < 1$ answer	Ex 1
B	Supersonic root of area ratio	$A / A^{*} > 1$ , pick the $M > 1$ answer	Ex 2
C	Degenerate area: $A / A^{*} = 1$	the throat, $M = 1$ exactly	Ex 3
D	Limiting input: $A / A^{*} \to \infty$	vacuum / infinite expansion	Ex 3
E	Choking test (sign of back pressure)	is $p_{b}$ below the critical?	Ex 4
F	Not choked (venturi regime)	high $p_{b}$ , throat never reaches $M = 1$	Ex 5
G	Normal shock inside the diverging part	$p_{b}$ between shock-at-exit and design	Ex 6
H	Real-world word problem	rocket thrust from exit state	Ex 7
I	Exam twist / trap	converging-only nozzle "goes supersonic?"	Ex 8

The three master relations every example plugs into. They are the isentropic ratios; each says "how much has this quantity dropped below its calm-tank value once the gas reaches Mach $M$ ?"

The picture above is the key to reading (AR): for any area ratio bigger than $1$ , a horizontal line crosses the curve twice — once on the falling (subsonic) branch and once on the rising (supersonic) branch. That single fact generates cases A and B.

Example 1 — Subsonic root (cell A)

Forecast: guess before solving — will $M$ be close to $1$ or close to $0$ ? And will $p / p_{0}$ be near $1$ (barely expanded) or tiny?

Steps.

Write (AR) with $A / A^{*} = 2$ : solve $\frac{1}{M} [\frac{2}{2.4} (1 + 0.2 M^{2})]^{3} = 2$ . Why this step? (AR) is the only equation linking geometry to Mach; $2 > 1$ so two roots exist and we must choose the subsonic one because the whole flow is stated subsonic.
Numerically the subsonic root is $M \approx 0.306$ . Why this step? On the falling branch of the curve in the figure, the horizontal line at $2.0$ hits a small Mach number — matching our forecast that a wide, non-choked venturi is only mildly sped up.
Pressure from (P): $\frac{p _{0}}{p} = (1 + 0.2 \cdot 0.30 6^{2})^{3.5} \approx 1.067$ , so $p / p_{0} \approx 0.937$ . Why this step? Low Mach means little kinetic energy taken, so pressure barely drops — the gas is still nearly at reservoir pressure $p_{0}$ .

Verify: plug $M = 0.306$ back into (AR): $\frac{1}{0.306} [0.833 (1 + 0.2 \cdot 0.0936)]^{3} = 3.27 \cdot 0.611 \approx 2.0$ ✅. Pressure ratio $0.937$ is close to $1$ , consistent with "subsonic and barely expanded." Units: all ratios, dimensionless ✅.

Example 2 — Supersonic root (cell B)

Forecast: the same area ratio — will you get the same Mach as Example 1, or something bigger than 1?

Steps.

Use (AR) with $A / A^{*} = 2$ again, but take the supersonic root ( $M > 1$ ). Why this step? Past a choked throat the diverging section obeys $d A / A = (M^{2} - 1) d V / V$ with $M > 1$ , so widening area accelerates the flow — Nature is on the rising branch of the curve.
Numerically $M \approx 2.197$ . Why this step? The horizontal line at $2.0$ in the figure also strikes the rising branch — same area, wildly different speed. This is the "two-solution fact" made concrete.
Pressure: $\frac{p _{0}}{p} = (1 + 0.2 \cdot 2.19 7^{2})^{3.5} = (1.966)^{3.5} \approx 10.7$ , so $p / p_{0} \approx 0.0935$ . Why this step? Supersonic flow has drained huge kinetic energy from the gas, so static pressure has collapsed to roughly one-tenth of reservoir — utterly unlike the $0.937$ of Example 1.

Verify: re-plug $M = 2.197$ : $\frac{1}{2.197} [0.833 (1 + 0.2 \cdot 4.827)]^{3} = 0.455 \cdot 4.40 \approx 2.0$ ✅. Note both examples solve the same equation to $2.0$ from opposite branches — that is exactly cases A and B being twins.

Example 3 — Degenerate + limiting inputs (cells C & D)

Forecast: for (a), only one answer survives — which? For (b), does $M$ keep climbing forever, or hit a ceiling?

Steps.

(a) Set $A / A^{*} = 1$ in (AR). The curve's minimum value is $1$ , reached only at $M = 1$ . Why this step? At the minimum the two roots merge into one — the subsonic and supersonic branches touch. This is the throat, the place we defined as the sonic area $A^{*}$ ; the master law forces $d A = 0$ when $M = 1$ .
So the only solution is $M = 1$ . There is no other Mach with area ratio $1$ . Why this step? This kills the "two-solution" ambiguity precisely at the throat — a degenerate, single-root case.
(b) Take (AR) as $M \to \infty$ : the bracket grows like $M^{2}$ , raised to power $3$ , gives $M^{6}$ , times $1/ M$ → $A / A^{*} \sim M^{5}$ . So $A / A^{*} \to \infty$ needs $M \to \infty$ . Why this step? An infinitely large exit would give infinite Mach — but temperature from (T), $T / T_{0} = 1/ (1 + 0.2 M^{2}) \to 0$ , so the gas freezes; real gases liquefy and this is only a limit, never reached.
The maximum exit velocity is finite though: $V_{ma x} = 2 c_{p} T_{0}$ — the speed reached when all the tank's thermal energy $c_{p} T_{0}$ has become kinetic energy $\frac{1}{2} V^{2}$ (set $T \to 0$ in the energy line $c_{p} T_{0} = c_{p} T + \frac{1}{2} V^{2}$ ). Even at $M \to \infty$ , $V$ is capped because the local sound speed $a = γ R T \to 0$ . Why this step? Distinguishes "Mach $\to \infty$ " (because sound speed vanishes) from "speed $\to \infty$ " (which does not happen). Here $c_{p}$ is the same specific heat introduced above, tied to $R$ by $c_{p} = γ R / (γ - 1)$ .

Verify: (a) plug $M = 1$ into (AR): $\frac{1}{1} [0.833 \cdot 1.2]^{3} = [1.0]^{3} = 1.0$ ✅ minimum confirmed. (b) at $M = 5$ , (AR) $= \frac{1}{5} [0.833 (1 + 0.2 \cdot 25)]^{3} = 0.2 \cdot 125 = 25$ — already huge, confirming $A / A^{*}$ blows up with $M$ ✅.

Example 4 — Choking test: is the back pressure low enough? (cell E)

Forecast: choking happens once throat pressure drops to the critical fraction of $p_{0}$ . Guess: is $3/8$ above or below that threshold?

Steps.

Critical pressure ratio at $M = 1$ : $\frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{3.5} = (0.8333)^{3.5} \approx 0.528$ . Why this step? This is (P) evaluated at $M = 1$ ; it is the pressure the throat must fall to before it can go sonic. See Choked Flow & Mass Flow Limit.
So $p^{*} = 0.528 \times 8 = 4.23 bar$ . Why this step? Converts the dimensionless threshold into a real pressure to compare against $p_{b}$ .
Compare: $p_{b} = 3 bar < p^{*} = 4.23 bar$ . Why this step? If the environment demands an exit pressure below the critical throat pressure, the throat has already hit $M = 1$ — flow is choked, and $\overset{m}{˙}$ is locked at $\overset{m}{˙}_{ma x}$ .
Conclusion: choked. Lowering $p_{b}$ from $3$ to $2$ or $1$ bar would not raise the mass flow. Why this step? This is the sign-based decision at the heart of cell E — everything hinges on which side of $0.528 p_{0}$ the back pressure lies.

Verify: $(2/2.4)^{3.5}$ : $2/2.4 = 0.83333$ ; $0.8333 3^{3.5} = e^{3.5 l n 0.83333} = e^{3.5 (- 0.18232)} = e^{- 0.6381} = 0.5283$ ✅. And $0.528 \cdot 8 = 4.23$ bar, and $3 < 4.23$ ✅ → choked.

Example 5 — Not choked: the venturi regime (cell F)

Forecast: with $p_{b}$ so close to $p_{0}$ , is the throat sonic? Where in the nozzle is the fastest flow?

Steps.

Check choke: $p_{b} / p_{0} = 7/8 = 0.875 > 0.528$ . Why this step? Above the critical ratio the throat cannot reach $M = 1$ — the nozzle acts as a subsonic venturi, not a supersonic accelerator.
Since it's all subsonic, the geometric throat never reaches $M = 1$ , so no station is sonic and the label $A^{*}$ (the sonic area) has no physical location inside this flow — it exists only as a bookkeeping reference for (AR). The flow speeds up in the converging part, peaks (subsonic) at the narrowest point, then slows down in the diverging part. Why this step? Subsonic flow obeys $d A < 0 \Rightarrow d V > 0$ ; the diverging section has $d A > 0$ so it decelerates — recovery of pressure like a venturi.
Exit pressure equals back pressure, $p_{e} = p_{b} = 7$ bar. From (P): $\frac{p _{0}}{p _{e}} = \frac{8}{7} = 1.143 = (1 + 0.2 M_{e}^{2})^{3.5}$ . Why this step? In fully subsonic flow the exit pressure is set by the back pressure the environment imposes; invert (P) to read off $M_{e}$ .
Solve: $1 + 0.2 M_{e}^{2} = 1.14 3^{1/3.5} = 1.0388$ , so $M_{e}^{2} = 0.194$ , $M_{e} \approx 0.441$ . Why this step? Confirms a modest subsonic exit — the highest speed sits back at the narrowest point, not the exit, the opposite of the supersonic case.

Verify: $1.14 3^{1/3.5} = e^{(1/3.5) l n 1.143} = e^{0.2857 \cdot 0.1335} = e^{0.03816} = 1.0389$ ; $(1.0389 - 1) /0.2 = 0.1944$ ; $0.1944 = 0.441$ ✅. Sanity: $M_{e} < 1$ , and $0.875 > 0.528$ ⇒ genuinely un-choked ✅.

Example 6 — Normal shock in the diverging section (cell G)

Forecast: across a shock, does the flow become subsonic or stay supersonic? Does pressure rise or fall?

Steps.

Post-shock Mach from the normal-shock relation: $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}} .$ Why this step? A shock is a discontinuous, non-isentropic jump — the smooth (T)/(P)/(AR) relations don't apply across it; we need the dedicated conservation-across-shock formula.
Plug $M_{1} = 2$ : numerator $= 1 + 0.2 \cdot 4 = 1.8$ ; denominator $= 1.4 \cdot 4 - 0.2 = 5.4$ . So $M_{2}^{2} = 1.8/5.4 = 0.3333$ , $M_{2} \approx 0.577$ . Why this step? Confirms the flow drops to subsonic — a shock always throws supersonic flow below $M = 1$ .
Static pressure ratio: $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) = 1 + \frac{2.8}{2.4} (4 - 1) = 1 + 1.1667 \cdot 3 = 4.5$ . Why this step? Downstream of a shock, pressure jumps up sharply — this is why the flow past the shock is a decelerating, higher-pressure subsonic stream that matches the back pressure $p_{b}$ .
After the shock the remaining diverging section acts as a subsonic diffuser (decelerates further), meeting $p_{b}$ at the exit. Why this step? Ties the shock case to the venturi behaviour of Ex 5 — same subsonic rule, just reached the hard way.

Read the figure carefully: the white walls pinch to the throat (dotted line, $M = 1$ ), the flow accelerates supersonically (amber region, $M_{1} = 2.0$ ), then the vertical amber bar is the normal shock. Look at how the labels flip across that bar — from supersonic $M_{1} = 2.0$ on the left to subsonic $M_{2} = 0.577$ on the right, with the static pressure multiplied by $4.5$ . Everything downstream of the amber bar behaves like the subsonic venturi of Example 5.

Verify: $M_{2}^{2} = 1.8/5.4 = 0.3333$ ✅, $0.3333 = 0.5774$ ✅. $p_{2} / p_{1} = 1 + (2.8/2.4) \cdot 3 = 4.5$ ✅. Sanity: $M_{2} < 1 < M_{1}$ and $p_{2} > p_{1}$ — both signatures of a compression shock ✅.

Example 7 — Real-world word problem: rocket exit state (cell H)

Forecast: with $A_{e} / A^{*} = 4$ (bigger than Example 1's $2$ ), will $M_{e}$ be above or below $2.94$ ? Will exit be hot or cold?

Steps.

Solve (AR) with $A / A^{*} = 4$ , supersonic root: $M_{e} \approx 2.94$ . Why this step? A rocket is designed to run supersonic (thrust demands high exit speed), so we take the rising branch — larger area ratio than Ex 1's $2.0$ ⇒ larger Mach, matching forecast.
Exit temperature from relation (T): $T_{0} / T_{e} = 1 + 0.2 M_{e}^{2} = 1 + 0.2 \cdot 8.644 = 2.729$ . So $T_{e} = 3000/2.729 \approx 1099 K$ . Why this step? Kinetic energy is bought with thermal energy; the gas cools dramatically as it accelerates — from $3000$ K in the calm chamber ( $T_{0}$ ) down to about $1100$ K.
Exit pressure from (P): $p_{0} / p_{e} = (2.729)^{3.5} \approx 34.1$ , so $p_{e} = 50/34.1 \approx 1.47 bar$ . Why this step? Design condition matches this to ambient for maximum thrust; it's the (P) counterpart of the temperature ratio.
Exit velocity: $a_{e} = γ R T_{e} = 1.4 \cdot 320 \cdot 1099 = 492, 300 \approx 701.6 m/s$ , then $V_{e} = M_{e} a_{e} = 2.94 \cdot 701.6 \approx 2063 m/s$ . Why this step? $V = M a$ converts the abstract Mach into the physical exhaust speed that produces thrust — the whole point of the nozzle.

Verify: $M_{e}^{2} = 8.644$ , $1 + 0.2 \cdot 8.644 = 2.729$ ; $T_{e} = 3000/2.729 = 1099.3$ K ✅. $2.72 9^{3.5} = e^{3.5 l n 2.729} = e^{3.5 \cdot 1.0040} = e^{3.514} = 33.6$ (≈34) ✅. $a_{e} = 1.4 \cdot 320 \cdot 1099.3 = 701.6$ m/s; $V_{e} = 2.94 \cdot 701.6 = 2062.7$ m/s ✅. Units: $(J/kg) = m^{2} / s^{2} = m/s$ ✅.

Example 8 — Exam trap: "make a converging nozzle go supersonic" (cell I)

Forecast: guess the verdict — does more suction eventually break the sound barrier here?

Steps.

In a purely converging nozzle the smallest area is the exit itself; $d A < 0$ all the way, never $d A = 0$ except at the very exit. Why this step? The master law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ forbids crossing $M = 1$ anywhere except where $d A = 0$ , so the only place $M = 1$ could occur is the exit.
The best the exit can reach is $M = 1$ (choked). Set $M_{e} = 1$ in (P): $p_{e} / p_{0} = (1 + 0.2 \cdot 1)^{3.5}^{- 1} = (1.2)^{- 3.5} = 0.528$ , so $p_{e} = 0.528 \times 10 = 5.28 bar$ . Why this step? Once the exit is at $M = 1$ the nozzle is choked; that critical exit pressure $5.28$ bar is the ceiling — see Choked Flow & Mass Flow Limit.
Now pull $p_{b}$ below $5.28 bar$ . The nozzle cannot respond internally — the exit stays at $M = 1$ and $p_{e} = 5.28 bar$ ; the extra expansion down to $p_{b}$ happens outside the nozzle as expansion fans (an under-expanded jet). Why this step? No diverging section exists to satisfy $d A > 0$ for $M > 1$ , so supersonic flow inside is geometrically impossible; the gas can only finish expanding in the open air.
Verdict: disproved. A converging-only nozzle maxes out at exactly sonic exit ( $M_{e} = 1$ , $p_{e} = 5.28$ bar); any $p_{b}$ below that just moves the expansion outside. Supersonic flow requires the diverging section of a de Laval nozzle. (Contrast Steam Turbine Nozzles, which add the divergence for exactly this reason.) Why this step? This is the whole reason the de Laval shape exists — cell I is the topic's signature trap.

Verify: ceiling pressure $(1/1.2)^{3.5} = (0.8333)^{3.5} = 0.528$ , times $10 = 5.28$ bar ✅ (identical ratio to the critical value of Ex 4, as it must be — both are " $M = 1$ "). No formula produces $M_{e} > 1$ from an all-shrinking area, confirming the impossibility.

Recall Quick self-test on the matrix

Same area ratio gives two Mach numbers — which do you pick and why? ::: The subsonic root if the whole flow is subsonic (high $p_{b}$ , venturi); the supersonic root if the throat is choked and the diverging section accelerates the gas (low $p_{b}$ , design). Back pressure $p_{b} = 0.6 p_{0}$ with $γ = 1.4$ — choked or not? ::: Critical ratio is $0.528$ ; $0.6 > 0.528$ , so not choked (still subsonic at throat). After a normal shock at $M_{1} = 2$ , is the flow faster or slower than sound? ::: Slower — $M_{2} \approx 0.577 < 1$ ; every shock jumps supersonic flow to subsonic and raises pressure. Why can't suction alone make a converging nozzle supersonic? ::: The area only shrinks, so $M = 1$ can occur only at the exit (choked); with no $d A > 0$ region, $M > 1$ is geometrically impossible.