Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss: Worked examples

This page is the "no case left behind" companion to 3.3.15 Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss (Hinglish)'s English parent. We march through every kind of situation an under-expanded nozzle problem can hand you — every sign of the pressure gap, the degenerate "just barely" cases, the extreme limits, a real flight problem, and an exam trap — and solve each one fully.

Before line one, one promise: nothing gets used before it is built. Let us build the three tools we lean on.

The three tools (built from zero)

The scenario matrix

Every under-expanded problem is one (or a blend) of these cells. Each example below is tagged with the cell it fills.

Cell	What makes it special	Example
A. Baseline	$r > 1$ , ordinary supersonic exit	Ex 1
B. Degenerate $r = 1$	pressure tie — the "zero turn" boundary	Ex 2
C. Wrong-sign trap	student is given $r < 1$ — must reject the P-M method	Ex 3
D. Small mismatch limit	$r \to 1^{+}$ , tiny angle, thrust loss $\to 0$	Ex 4
E. Large mismatch limit	high altitude, big $r$ , big turn	Ex 5
F. Thrust-force accounting	uses $F = \overset{m}{˙} v_{e} + A_{e} (p_{e} - p_{\infty})$	Ex 6
G. Real-world word problem	rocket climbs, $p_{\infty}$ falls with altitude	Ex 7
H. Exam twist	different $γ$ , asks for the loss factor $1 - cos θ$	Ex 8

Worked examples

Worked example Example 1 — Cell A: the baseline under-expanded jet

Given: $M_{e} = 3.5$ , $p_{e} = 1.5$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Find: $M_{plume}$ and the turning angle $θ_{turn}$ .

Forecast: Guess — will $M_{plume}$ be bigger or smaller than $3.5$ ? Will $θ$ be a few degrees or tens of degrees?

Confirm the regime. $r = 1.5/1 = 1.5 > 1$ → under-expanded. Why this step? The whole method only applies when the gas is still over-pressured; check first.
Plume Mach. Exponent $γ / (γ - 1) = 1.3/0.3 = 4.333$ ; and $\frac{γ - 1}{2} = 0.15$ . Denominator bracket $= 1 + 0.15 (3. 5^{2}) = 1 + 0.15 (12.25) = 2.8375$ . $1.5 = (\frac{1 + 0.15 M _{plume}^{2}}{2.8375})^{4.333}$ So $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1. 5^{1/4.333} = 2.8375 (1.0964) = 3.111$ . Thus $M_{plume}^{2} = (3.111 - 1) /0.15 = 14.07$ , giving $M_{plume} \approx 3.75$ . Why this step? Pressure falling to $p_{\infty}$ forces the flow faster — the isentropic relation is the only bridge from a pressure ratio to a Mach number.
Turning angle with $(γ + 1) / (γ - 1) = 2.3/0.3 = 2.769$ : $ν (3.5) = 2.769 arctan (\frac{1}{2.769} 11.25) - arctan 11.25 = 66.3°$ $ν (3.75) = 2.769 arctan (\frac{1}{2.769} 13.06) - arctan 13.06 = 71.0°$ $θ_{turn} = ν (3.75) - ν (3.5) = 71.0° - 66.3° = 4.7°$ Why this step? The turning bank account: cost of going from $M_{e}$ to $M_{plume}$ is the difference of the two totals.

Verify: $M_{plume} > M_{e}$ ✓ (faster, as pressure dropped). $θ \approx 4.7°$ is a "few degrees" — matches the parent's $3$ – $5°$ estimate for $r = 1.5$ . Look at the fan geometry below.

Figure — Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

The red waves are the centered expansion fan; the violet arrow is the exit flow, the orange arrow the deflected plume — the small angle between them is your $θ_{turn}$ .

Worked example Example 3 — Cell C: the wrong-sign trap

Given: $M_{e} = 3.5$ , $p_{e} = 0.7$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . A student blindly applies Prandtl-Meyer. Find: What is actually happening, and why the P-M method is illegal here.

Forecast: Is this even an under-expanded nozzle?

Regime. $r = 0.7 < 1$ → $p_{e} < p_{\infty}$ → over-expanded, not under-expanded. Why this step? The sign of $r - 1$ decides which physics runs. Here ambient wins.
Why P-M fails. Expansion fans drop pressure. Here the gas must raise its pressure to reach $p_{\infty}$ . Pressure-raising in supersonic flow happens through oblique shocks, not expansion fans — see 4.1.3-Oblique-shock-theory and 3.3.14-Over-expanded-nozzle-shock-diamonds.
The plume Mach direction. If you did force the isentropic formula: $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 0. 7^{1/4.333} = 2.8375 (0.9192) = 2.608$ , giving $M_{plume}^{2} = (2.608 - 1) /0.15 = 10.72$ , $M_{plume} \approx 3.27 < M_{e}$ . Why this step? A Mach that dropped is the mathematical fingerprint of compression — the formula is quietly telling you a shock, not a fan, is needed.

Verify: $M_{plume} = 3.27 < 3.5$ ✓ — deceleration = compression = shock regime. Do not use $ν (M)$ for turning angle here. Correct tool: oblique shock relations.

Worked example Example 4 — Cell D: the vanishing-mismatch limit

Given: $M_{e} = 3.5$ , $p_{e} = 1.02$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Find: $θ_{turn}$ and confirm the thrust loss $\to 0$ .

Forecast: As $r \to 1^{+}$ , does $θ$ shrink smoothly to zero?

Plume Mach. $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1.0 2^{1/4.333} = 2.8375 (1.00458) = 2.8505$ . $M_{plume}^{2} = (2.8505 - 1) /0.15 = 12.337$ , $M_{plume} \approx 3.513$ . Why this step? Tiny pressure drop → tiny Mach rise.
Turn. $ν (3.513) - ν (3.5)$ . Since the change is small, $θ \approx 0.24°$ . Why this step? Nearly-equal bank balances → nearly-zero withdrawal.
Loss factor. $1 - cos (0.24°) = 1 - 0.9999912 \approx 8.8 \times 1 0^{- 6}$ → about $0.0009%$ . Why this step? Thrust loss scales with $(1 - cos θ)$ , which is $\sim θ^{2} /2$ for small $θ$ — it dies quadratically.

Verify: $θ \to 0$ and loss $\to 0$ as $r \to 1$ ✓ — continuously matching Example 2's exact zero. The physics has no jump at the boundary.

Worked example Example 5 — Cell E: the big-mismatch (high-altitude) limit

Given: $M_{e} = 3.5$ , $p_{e} = 1.0$ atm, $p_{\infty} = 0.1$ atm (high up), $γ = 1.3$ . So $r = 10$ . Find: $M_{plume}$ and $θ_{turn}$ .

Forecast: Big pressure ratio — will the turn be large (tens of degrees)?

Plume Mach. $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1 0^{1/4.333} = 2.8375 (1.6885) = 4.7912$ . $M_{plume}^{2} = (4.7912 - 1) /0.15 = 25.27$ , $M_{plume} \approx 5.03$ . Why this step? A tenfold pressure drop demands a much faster plume.
Turn. $ν (5.03) \approx 88.9°$ , $ν (3.5) = 66.3°$ , so $θ = 88.9 - 66.3 \approx 22.6°$ . Why this step? Larger $M_{plume}$ → far bigger withdrawal from the bank account.
Loss factor. $1 - cos (22.6°) = 1 - 0.9232 = 0.0768$ → up to $\sim 7.7%$ of the extra expansion momentum thrown sideways. Why this step? Now $θ$ is large; the small-angle shortcut is invalid, so use full cosine.

Verify: $θ$ jumped from $4.7°$ (Ex 1) to $22.6°$ as $r$ went $1.5 \to 10$ ✓ — bigger mismatch, bigger loss, exactly the trend the parent warns of. This is why altitude-compensation nozzles exist (3.3.16-Altitude-compensation-nozzles).

Worked example Example 6 — Cell F: full thrust-force accounting

Given: $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2800$ m/s, $A_{e} = 0.5 m^{2}$ , $p_{e} = 1.5 \times 1 0^{5}$ Pa, $p_{\infty} = 1.0 \times 1 0^{5}$ Pa. Find: Thrust $F$ , and the split between momentum and pressure terms.

Forecast: Does the over-pressure add to or subtract from thrust here?

Momentum term. $\overset{m}{˙} v_{e} = 200 \times 2800 = 560, 000$ N. Why this step? Bulk of thrust is momentum flux leaving the nozzle.
Pressure term. $A_{e} (p_{e} - p_{\infty}) = 0.5 \times (1.5 \times 1 0^{5} - 1.0 \times 1 0^{5}) = 0.5 \times 5 \times 1 0^{4} = 25, 000$ N. Why this step? Under-expanded means $p_{e} > p_{\infty}$ , so this term is positive — the over-pressure pushes forward on the exit plane.
Total. $F = 560, 000 + 25, 000 = 585, 000$ N $= 585$ kN. Why this step? Momentum theorem: sum the two contributions.

Verify: Units: $[\overset{m}{˙} v_{e}] = \frac{kg}{s} \cdot \frac{m}{s} = \frac{kg\cdotm}{s ^{2}} =$ N ✓; $[A_{e} p] = m^{2} \cdot Pa = m^{2} \cdot \frac{N}{m ^{2}} =$ N ✓. Pressure term is positive ✓ (correct sign for under-expansion). Note: the gain $25$ kN is real but is not free lunch — $v_{e}$ here is smaller than a perfectly-expanded nozzle would deliver. See 2.5.12-Thrust-coefficient-definition.

Worked example Example 7 — Cell G: a climbing rocket (word problem)

Given: A rocket keeps $p_{e} = 0.8$ atm fixed (fixed nozzle, fixed chamber). At launch $p_{\infty} = 1.0$ atm; at $11$ km altitude $p_{\infty} = 0.22$ atm. Exit $M_{e} = 4$ , $γ = 1.2$ . Find: At which altitude is the nozzle under-expanded, and roughly how the turn changes.

Forecast: A fixed nozzle can't be perfect everywhere — where does it flip from over- to under-expanded?

Launch. $r = 0.8/1.0 = 0.8 < 1$ → over-expanded (shocks, Ex 3 regime). Not our page's job here. Why this step? At sea level ambient is strong; classify by $r$ .
Altitude. $r = 0.8/0.22 = 3.64 > 1$ → under-expanded (fan forms). Why this step? As the rocket climbs, $p_{\infty}$ collapses; eventually the exhaust over-powers it.
Crossover. Perfect expansion happens when $p_{\infty} = p_{e} = 0.8$ atm. Using the barometric-style drop, that is somewhere below $11$ km. Above it, plume opens up: with $γ = 1.2$ , $\frac{γ - 1}{2} = 0.1$ , exponent $= 6$ , $(γ + 1) / (γ - 1) = 11 = 3.317$ . Denominator bracket $= 1 + 0.1 (16) = 2.6$ . $1 + 0.1 M_{plume}^{2} = 2.6 \cdot 3.6 4^{1/6} = 2.6 (1.2405) = 3.225$ , $M_{plume}^{2} = 22.25$ , $M_{plume} \approx 4.72$ . Why this step? Track the same isentropic bridge with the new $γ$ and $r$ .

Verify: Regime flips from over- ( $r < 1$ ) to under-expanded ( $r > 1$ ) as it climbs ✓ — exactly why altitude-compensating designs are pursued (3.3.16-Altitude-compensation-nozzles). $M_{plume} = 4.72 > 4$ ✓ (faster aloft).

Worked example Example 8 — Cell H: exam twist, the loss factor with air-like

$γ$ Given: $M_{e} = 2.5$ , $p_{e} / p_{\infty} = 2.0$ , $γ = 1.4$ . Find $θ_{turn}$ and the loss factor $1 - cos θ_{turn}$ .

Forecast: With springier gas ( $γ = 1.4$ ) does the same $r$ give a bigger or smaller turn than the $γ = 1.3$ cases?

Constants. $\frac{γ - 1}{2} = 0.2$ , exponent $= 1.4/0.4 = 3.5$ , $(γ + 1) / (γ - 1) = 2.4/0.4 = 6 = 2.449$ .
Plume Mach. Denominator bracket $= 1 + 0.2 (6.25) = 2.25$ . $1 + 0.2 M_{plume}^{2} = 2.25 \cdot 2^{1/3.5} = 2.25 (1.2190) = 2.7428$ , $M_{plume}^{2} = 8.714$ , $M_{plume} \approx 2.952$ . Why this step? Same isentropic bridge, air's $γ$ .
Turn. $ν (2.5) = 2.449 arctan (\frac{1}{2.449} 5.25) - arctan 5.25 = 39.12°$ . $ν (2.952) = 2.449 arctan (\frac{1}{2.449} 7.714) - arctan 7.714 = 48.78°$ . $θ_{turn} = 48.78 - 39.12 = 9.66°$ .
Loss factor. $1 - cos (9.66°) = 1 - 0.98583 = 0.01417$ → about $1.4%$ . Why this step? The exam wants the fraction of extra momentum lost sideways, not just the angle.

Verify: $M_{plume} = 2.95 > 2.5$ ✓; loss $\approx 1.4%$ is modest for $r = 2$ ✓, consistent with "few-percent" losses at moderate mismatch. Springier air turns a bit more per unit pressure drop than the $γ = 1.3$ exhaust — a subtle but examinable point.

Recall Quick self-test

Which regime is $p_{e} / p_{\infty} = 1.0$ ? ::: Perfectly expanded — zero turn, zero loss (boundary case). Which tool for $p_{e} < p_{\infty}$ ? ::: Oblique shocks, NOT Prandtl-Meyer. Turning angle formula in one line? ::: $θ_{turn} = ν (M_{plume}) - ν (M_{e})$ . Thrust loss scales like? ::: $(1 - cos θ_{turn}) \approx θ^{2} /2$ for small $θ$ . Sign of the pressure thrust term when under-expanded? ::: Positive, $A_{e} (p_{e} - p_{\infty}) > 0$ .