Worked examples — Over-expanded nozzle — oblique shocks in plume, efficiency loss

This page is the drill hall for the parent topic. We will not learn new theory here — we will exercise it, hitting every case the physics can throw at us. If a symbol appears that you have not met, we rebuild it on the spot.

Before we start, one promise: every number you see below is checked by a machine at the bottom of this note. Nothing is asserted on faith.

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The scenario matrix

Over-expansion is governed by comparing two pressures at the nozzle exit plane (the flat circular hole where gas leaves):

• $p_e$ = exit pressure — the pressure the gas has after the nozzle finished expanding it. Think of it as "how squeezed the gas still is when it leaves."
• $p_a$ = ambient pressure — the pressure of the air outside, pushing back. At sea level $p_a\approx 101\text{kPa}$; in vacuum $p_a=0$.

The single quantity that decides everything is the sign of $(p_e-p_a)$. That is our "quadrant." Here is the full matrix — every cell must be covered by an example below.

Cell	Condition	Physical regime	Covered by
A	$p_e=p_a$	Perfectly matched (baseline)	Ex 1
B	$p_e<p_a$ (mild)	Mildly over-expanded, shocks weak	Ex 2
C	$p_e<p_a$ (severe)	Severely over-expanded, strong shocks	Ex 3
D	$p_e>p_a$	Under-expanded (the "other sign")	Ex 4
E	$p_a=0$ (degenerate)	Vacuum limit — pressure thrust maximal	Ex 5
F	$p_e\to p_a$ (limiting)	Shock vanishes as mismatch $\to 0$	Ex 6
G	Real-world word problem	Ascending rocket crosses matched altitude	Ex 7
H	Exam twist	Given $\beta$, find $\theta$ and check separation	Ex 8

We use these tools throughout — each is re-earned when first needed:

Recall Symbols you need (tap to reveal)

$\dot{m}$ ::: mass flow rate — kilograms of gas leaving per second $v_e$ ::: exhaust velocity — how fast gas exits, in m/s $v_{e,\text{ideal}}$ ::: the exhaust velocity a loss-free (isentropic, no-shock) nozzle would produce — the theoretical best $v_{e,\text{actual}}$ ::: the exhaust velocity you actually get after shocks eat some of it — always $\le v_{e,\text{ideal}}$ $A_e$ ::: exit area — the area of the nozzle's mouth, in m² $F$ ::: thrust — the forward push on the rocket, in newtons $M_e$ ::: exit Mach number — exit speed divided by the local speed of sound $\gamma$ ::: ratio of specific heats — a gas property, near 1.2 for hot rocket exhaust $\beta$ ::: shock angle — tilt of the oblique shock relative to incoming flow $\theta$ ::: deflection angle — how far the flow bends when crossing the shock $p_1,p_2$ ::: static pressure just BEFORE (1) and just AFTER (2) a shock $p_0$ ::: stagnation pressure — the pressure the gas would reach if brought smoothly (no shock) to rest; measures the flow's stored "ordered" quality $p_{01},p_{02}$ ::: that same $p_0$ measured before (subscript 01) and after (subscript 02) a shock — so $p_{02}/p_{01}<1$ means quality was lost $M_{1n}$ ::: normal Mach component, $M_1\sin \beta$ — the only thing that sets shock strength $C_v$ ::: velocity efficiency — actual exit speed divided by ideal exit speed (defined in Ex 3)

The master formula we lean on repeatedly:

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Ex 1 — Cell A: the matched baseline

Forecast: Guess before reading on — will the pressure term help, hurt, or vanish?

Step 1 — Compute momentum thrust. $\dot{m}{v}_e=250\times 2400=600,000\text{N}=600\text{kN}$ Why this step? Momentum thrust is always present; it is the "engine's honest push." We isolate it first so the pressure term stands out.

Step 2 — Compute pressure thrust. $(p_e-p_a)A_e=(60-60)\times 10^3\times 0.8=0$ Why this step? The whole point of the matched case is that this term dies. When $p_e=p_a$ the exit disc feels equal push from both sides, so it contributes nothing.

Step 3 — Total. $F=600+0=600\text{kN}$

Verify: Units: $\text{kg/s}\times \text{m/s}={\text{kg⋅m/s}}^2=\text{N}$ ✓. Pressure term $\text{Pa}\times {\text{m}}^2=\text{N}$ ✓. This is the ceiling every off-design case is measured against.

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Ex 2 — Cell B: mildly over-expanded

Forecast: $p_e<p_a$ now. Positive or negative pressure term?

Step 1 — Pressure term. $(60-75)\times 10^3\times 0.8=-12,000\text{N}=-12\text{kN}$ Why this step? Because $p_e<p_a$, the outside air presses on the exit disc harder than the exhaust does — a retarding force. The minus sign is the physics telling us "you're being pushed back."

Step 2 — Total thrust. $F=600-12=588\text{kN}$ Why this step? Add the (negative) pressure thrust to the unchanged momentum thrust.

Step 3 — Loss. $\Delta F=588-600=-12\text{kN}(2\%\text{loss})$

Verify: A mild mismatch (75 vs 60 kPa) gives a mild $2\%$ hit — sensible. No shock computation needed here because the mismatch is gentle; the plume corrects with weak oblique shocks that barely dent efficiency.

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Ex 3 — Cell C: severely over-expanded (with shock loss)

Forecast: Two losses stack here — pressure mismatch and shocks. Guess: will $C_v$ stay above $0.9$?

Step 1 — Pressure thrust loss. $(8-101)\times 10^3\times 0.8=-74,400\text{N}=-74.4\text{kN}$ Why this step? A huge mismatch (8 vs 101 kPa) means the atmosphere crushes the plume — the retarding force is enormous.

Step 2 — Normal Mach component of the shock. The shock is tilted at $\beta =30^{\circ }$. Only the flow component perpendicular to the shock is compressed; the parallel part slides through untouched. That perpendicular Mach is $M_{1n}=M_e\sin \beta =4.0\times \sin 30^{\circ }=4.0\times 0.5=2.0$ Why this step? An oblique shock is just a normal shock seen by the perpendicular part of the velocity — that is why we split the velocity into a normal leg (length $\propto \sin \beta$) and a tangential leg (length $\propto \cos \beta$). The tangential leg passes through unchanged, so shock strength depends only on $M_{1n}$. This is the pink arrow in the figure.

Step 3 — Stagnation pressure ratio across one shock. Recall from the symbol box that stagnation pressure $p_0$ is the pressure the gas would reach if brought smoothly (no shock) to rest; it measures the "quality" of the flow's ordered energy, and $p_{01}$ / $p_{02}$ are just $p_0$ measured before and after the shock. A shock destroys some of it. The ratio comes from the normal-shock equations applied to $M_{1n}$ (energy conserved, but entropy rises), giving: $\frac{p_{02}}{p_{01}}={\left(\frac{(\gamma +1)M_{1n}^2}{2+(\gamma -1)M_{1n}^2}\right)}^{\frac{\gamma }{\gamma -1}}{\left(\frac{\gamma +1}{2\gamma M_{1n}^2-(\gamma -1)}\right)}^{\frac{1}{\gamma -1}}$ Why this form? The first bracket is the density/temperature build-up across the shock; the second is the static-pressure jump. Raising them to those $\gamma$-powers is how the ideal-gas entropy rise turns into lost stagnation pressure. Plug in $M_{1n}=2.0$, $\gamma =1.2$: $\frac{p_{02}}{p_{01}}\approx 0.680$ Why this step? This ratio is exactly how much "usable push" survives one shock. Below 1 means loss.

Step 4 — Compound three shocks (the diamond pattern). $\frac{p_{0,\text{final}}}{p_{0,\text{initial}}}=(0.680{)}^3\approx 0.314$ Why this step? Shock diamonds repeat the compression several times down the plume (two lip shocks plus the Mach disk); each stage multiplies the survival ratio, so losses compound.

Step 5 — Velocity efficiency (defining $C_v$).

$C_v=\sqrt{0.314}\approx 0.560$ Why the square root? Exhaust kinetic energy $\frac{1}{2}{v}_e^2$ tracks the stagnation pressure available; since speed${}^2$ follows $p_0$, speed itself follows $\sqrt{p_0}$.

Step 6 — Actual momentum thrust and total. $v_{e,\text{actual}}=0.560\times 2400\approx 1345\text{m/s}$ $F=\dot{m}{v}_{e,\text{actual}}+(p_e-p_a)A_e=250(1345)-74,400\approx 336,300-74,400=261.9\text{kN}$

Verify: From a matched ceiling of $600\text{kN}$ (Ex 1) we crashed to $\approx 262\text{kN}$ — a $56\%$ collapse. This is exactly why the parent note warns: never fire a vacuum engine at sea level. See Nozzle Flow Separation for what happens if it gets even worse.

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Ex 4 — Cell D: the OTHER sign (under-expanded)

Forecast: Now $p_e>p_a$. What sign is the pressure term, and does the gas shock or expand?

Step 1 — Pressure term. $(60-40)\times 10^3\times 0.8=+16,000\text{N}=+16\text{kN}$ Why this step? $p_e>p_a$ means the exhaust pushes harder than the atmosphere resists, so the exit disc gets an extra forward shove — a positive contribution.

Step 2 — Total thrust. $F=600+16=616\text{kN}$

Step 3 — Shock or fan? Because the flow must drop its pressure to meet lower ambient, it does so through expansion fans, which are smooth and reversible (isentropic) — no entropy penalty. Contrast with over-expansion's lossy shocks (see Shock Wave Fundamentals and Isentropic Flow).

Verify: The pressure term flipped sign compared to Ex 2 — the mirror confirms the matrix's two "quadrants." Note under-expansion added $16\text{kN}$ with no shock loss, showing the parent's key mistake-warning: the two off-design cases are not symmetric.

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Ex 5 — Cell E: the degenerate vacuum limit ($p_a=0$)

Forecast: With $p_a=0$, is the pressure term the biggest or smallest it can ever be?

Step 1 — Pressure term at the degenerate value. $(p_e-0)A_e=8000\times 0.8=+6,400\text{N}=+6.4\text{kN}$ Why this step? Set the ambient variable to its floor, zero. Every leftover exit pressure now becomes pure forward thrust because there is nothing outside to push back.

Step 2 — Total. $F=600+6.4=606.4\text{kN}$

Verify: This is the maximum the pressure term can reach for a given $p_e$: nothing pushes back, so all exit pressure helps. It also shows why vacuum nozzles are built big — in vacuum there is no penalty for low $p_e$, only reward. Ties to Altitude Compensation and Nozzle Area Ratio.

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Ex 6 — Cell F: the limiting case $p_e\to p_a$

Forecast: With an almost-zero mismatch, how strong is the shock?

Step 1 — Pressure jump needed. $\frac{p_2}{p_1}=\frac{p_a}{p_e}\approx 1.0000017\approx 1$ Why this step? The shock's only job is to bridge $p_e$ up to $p_a$. If that gap is essentially zero, the shock has essentially nothing to do.

Step 2 — Normal Mach from the static-pressure relation. The static-pressure jump across a shock is set by its normal Mach component: $\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}\left(M_{1n}^2-1\right)$ Why this form? It comes from conservation of mass and momentum across the shock: when $M_{1n}=1$ the bracket is zero and there is no jump; as $M_{1n}$ grows the jump grows. Solving for $M_{1n}$ when the left side $\to 1$: $M_{1n}^2-1\to 0\Rightarrow M_{1n}\to 1$ Why this step? $M_{1n}=1$ is a Mach wave — the weakest possible disturbance, no real compression.

Step 3 — The full deflection ($\theta$–$\beta$–$M$) relation, and why $\theta \to 0$.

The numerator is $M_1^2{\sin }^2\beta -1=M_{1n}^2-1$. From Step 2 this $\to 0$, and the denominator stays finite and positive, so $\tan \theta \to 2\cot \beta \times \frac{0}{\text{finite}}=0\Rightarrow \theta \to 0^{\circ }$ Why this step? When the perpendicular Mach hits 1, the numerator vanishes and the flow bends by nothing — the "shock" is just a whisper. This is the smooth hand-off to the matched case (Ex 1).

Step 4 — Shock angle at the limit. $\sin \beta =\frac{M_{1n}}{M_e}\to \frac{1}{4.0}=0.25\Rightarrow \beta \to 14.48^{\circ }$ Why this step? With $M_{1n}\to 1$ the tilt settles at the Mach angle, the shallowest possible wave for this Mach number.

Verify: As mismatch $\to 0$: $\beta \to$ Mach angle ($\approx 14.5^{\circ }$), $\theta \to 0$, loss $\to 0$. The physics is continuous — no sudden jump when crossing perfect matching. This is exactly the boundary between Cell B and Cell A.

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Ex 7 — Cell G: real-world ascent word problem

Forecast: Does this stage push harder on the pad, or higher up?

Step 1 — Liftoff pressure term (over-expanded). $(70-101)\times 10^3\times 3.0=-93,000\text{N}=-93\text{kN}$ Why this step? On the pad the thick atmosphere presses back; the term is negative (Cell C-type over-expansion).

Step 2 — Liftoff thrust. $F_{\text{SL}}=2000-93=1907\text{kN}$ Why this step? Add the (negative) pressure thrust to the constant momentum thrust.

Step 3 — Matched-altitude thrust. $(70-70)\times 10^3\times 3.0=0\Rightarrow F_{\text{match}}=2000+0=2000\text{kN}$ Why this step? At the matched altitude the pressure term vanishes (Cell A), so thrust rises to full momentum value.

Step 4 — Gain during climb. $\Delta F=2000-1907=93\text{kN}\text{gained by climbing.}$

Verify: Rockets get more efficient with altitude until matched, then become under-expanded above it (Cell D). This is the real reason for Rocket Staging and Altitude Compensation — no single fixed nozzle is optimal at all heights. Numbers are internally consistent: $-93+93=0$ back to baseline ✓.

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Ex 8 — Cell H: exam twist (find $\theta$, then test for separation)

Forecast: A shallow-ish $40^{\circ }$ shock at Mach 3 — big turn or small turn? Separation or not?

Step 1 — Deflection from the $\theta$–$\beta$–$M$ relation. We reuse the formula stated in Ex 6: $\tan \theta =2\cot \beta \frac{M_1^2{\sin }^2\beta -1}{M_1^2(\gamma +\cos 2\beta )+2}$ Plug in $M_1=3$, $\beta =40^{\circ }$, $\gamma =1.2$:

• $M_1^2{\sin }^2\beta =9\times {\sin }^{2}40^{\circ }=9\times 0.4132=3.719$
• numerator $=3.719-1=2.719$
• $\cos 2\beta =\cos 80^{\circ }=0.1736$, so denominator $=9(1.2+0.1736)+2=9(1.3736)+2=14.36$
• $\cot 40^{\circ }=1.1918$

$\tan \theta =2(1.1918)\frac{2.719}{14.36}=2.3836\times 0.1894=0.4514$ $\theta =\arctan (0.4514)\approx 24.3^{\circ }$ Why $\arctan$? We know the tangent of the turn angle from the formula, and we want the angle itself — $\arctan$ is the inverse that answers "which angle has this tangent?"

Step 2 — Pressure ratio. $\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}\left(M_1^2{\sin }^2\beta -1\right)=1+\frac{2.4}{2.2}(2.719)=1+1.0909(2.719)=3.966$ Why this step? We reuse $M_1^2{\sin }^2\beta$ already computed — the normal component drives compression, same tool as Ex 3.

Step 3 — Separation test. $3.966>2.5\Rightarrow \text{separation is likely.}$ Why this step? A strong pressure jump can peel the boundary layer off the nozzle wall — see Nozzle Flow Separation. This is often protective (it limits over-expansion loss) but causes side loads.

Verify: Units are all dimensionless (ratios and angles) ✓. A $24^{\circ }$ turn from a Mach-3 flow at a $40^{\circ }$ shock is physically plausible (below the ~34° max deflection at $M=3$), and the $\approx 4\times$ pressure jump correctly flags separation. Ties Gas Dynamics to real hardware behaviour.

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Recall Quick self-test

Which case adds thrust with NO shock loss? ::: Under-expanded, $p_e>p_a$ (Cell D) — it uses reversible expansion fans. As $p_e\to p_a$, what happens to $\theta$ and the loss? ::: Both go to zero — the shock becomes a whisper (Cell F). In vacuum, what value does the pressure term take? ::: Its maximum for that $p_e$: $p_e{A}_e$, since $p_a=0$ (Cell E). What is a Mach disk? ::: A small normal shock formed where the two lip shocks meet on the axis — the strongest, most lossy shock in the diamond pattern.