Worked examples — Nozzle area ratio ε = A_e - A - — choosing for optimal performance

Before we compute anything, we agree on the alphabet. Every symbol below is earned — if you have not met it, read this box.

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The scenario matrix

Every rocket-nozzle question is one of these cells. The examples below hit each one — the last column tells you which example covers it.

Cell	Case class	What is special	Covered by
A	$p_e=p_a$ (matched)	optimum, pressure term $=0$	Ex 1
B	$p_e>p_a$ (under-expanded)	ε too small, positive pressure thrust remains	Ex 2
C	$p_e<p_a$ (over-expanded)	ε too large, negative pressure thrust, separation risk	Ex 3
D	$p_a\to 0$ (vacuum limit)	limiting behaviour, huge ε	Ex 4
E	$\epsilon =1$ (degenerate)	throat = exit, sonic ($M_e=1$) only	Ex 5
F	two Mach roots of one ε	sign/branch choice (sub- vs supersonic)	Ex 6
G	real-world word problem	choose ε across a flight	Ex 7
H	exam twist (thrust comparison)	rank two nozzles, catch the trap	Ex 8
I	sub-critical: never chokes	$p_0/p_a<1.772$, flow stays subsonic	Ex 9

Throughout, take $\gamma =1.20$ so $\frac{\gamma }{\gamma -1}=6$ and $\frac{\gamma +1}{2(\gamma -1)}=5.5$, matching the parent note. Sea-level $p_a=1.0$ bar; $1\text{bar}=10^5\text{Pa}$.

Figure 1 below shows the three physical cases side by side — keep glancing back at it as we sweep the sky.

In Figure 1 the left panel (blue plume, straight edges) is the matched case of Cell A; the middle (yellow, ballooning plume) is under-expanded Cell B; the right (pink, pinched plume) is over-expanded Cell C. The yellow dashed line is the throat $A^{\ast }$; the blue vertical line is the exit plane $A_e$.

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Cell A — matched (optimum) expansion

Forecast: guess before computing — is $\epsilon$ closer to 3, 15, or 150? Is the pressure thrust $+$, $0$, or $-$? Look at the left panel of Figure 1 — its plume is straight because $p_e=p_a$.

1. Pressure ratio. $p_0/p_e=70/0.5=140$. Why this step? Equation (pressure–Mach) is written in terms of $p_0/p_e$; we need that number first.
2. Solve for $M_e$. Invert: $1+0.1M_e^2=140^{1/6}=2.276$, so $M_e^2=12.76$, $M_e=3.573$. Why this step? Mach number is the bridge — the area equation only speaks Mach, not pressure.
3. Area ratio. With $M_e=3.573$: $\epsilon =\frac{1}{3.573}{\left[\frac{2}{2.2}(2.276)\right]}^{5.5}=\frac{1}{3.573}(2.069{)}^{5.5}\approx \frac{54.8}{3.573}\approx \mathrm{15.3.}$ Why this step? This converts the chosen pressure into the geometry we can machine.
4. Pressure thrust. At the design altitude $p_e=p_a=0.5$ bar, so $(p_e-p_a)A_e=0$. Why this step? This is the definition of Cell A — matched means the second thrust term dies.

Verify: Plug $M_e=3.573$ back into pressure–Mach: $(1+0.1(12.77){)}^6=2.277^6=140.0$ ✓. Magnitude sanity: an upper-stage/vacuumish design giving $\epsilon \approx 15$ is exactly the "tens" range the parent predicted. Units: ε is dimensionless ✓.

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Cell B — under-expanded ($p_e>p_a$)

Forecast: smaller ε means less expansion. Do you expect $p_e$ to be higher or lower than the matched case? Sign of $(p_e-p_a)$? The middle panel of Figure 1 (ballooning plume) is what we are testing for.

1. Find $M_e$ from $\epsilon =8$. Root-find the area equation for the supersonic root (bracket $M\in [1,10]$, bisection as in the recipe above): $M_e\approx 3.117$. Why this step? Geometry is given, not pressure — so we go area → Mach → pressure this time, and area → Mach needs root-finding because it cannot be solved by algebra.
2. Find $p_e$. $p_0/p_e=(1+0.1(3.117{)}^2{)}^6=(1.9716{)}^6=58.9$, so $p_e=70/58.9=1.188$ bar. Why this step? We need $p_e$ to compare against $p_a$ — that comparison names the cell.
3. Classify. $p_e=1.188>p_a=0.20$ → under-expanded (Cell B). The gas still bursts with pressure. Why this step? The sign of $p_e-p_a$ is the classification.
4. Pressure thrust. $(p_e-p_a)A_e=(1.188-0.20)\times 10^5\times 0.16=+1.58\times 10^4\text{N}=+15.8\text{kN}$. Why this step? Positive term ⇒ shortening was "safe" here — no separation, and it still adds push. But had we lengthened it toward the matched ε at this altitude we'd have squeezed out even more $v_e$.

Verify: Cross-check $M_e$: area formula $\frac{1}{3.117}[\frac{2}{2.2}(1.9716){]}^{5.5}=\frac{1}{3.117}(1.7924{)}^{5.5}=\frac{24.9}{3.117}=8.0$ ✓. Pressure term positive matches "under-expanded still pushes" ✓. Units: $\text{Pa}\cdot {\text{m}}^2=\text{N}$ ✓.

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Cell C — over-expanded ($p_e<p_a$)

Forecast: which sign will the pressure term carry now? Bigger or smaller ε than Cell A? (Same nozzle — trick: ε is fixed; only the sky changed.) The right panel of Figure 1 (pinched plume) is this cell.

1. Nothing about the nozzle changed. $p_e$ is still $0.5$ bar because ε is fixed geometry. Why this step? The core insight: a rigid nozzle keeps its $p_e$; only $p_a$ moved from $0.5$ to $1.0$ bar.
2. Classify. $p_e=0.5<p_a=1.0$ → over-expanded (Cell C). Outside air pushes back into the bell. Why this step? Sign of $p_e-p_a$ is now negative — the danger cell.
3. Momentum thrust. $\dot{m}{v}_e=250\times 2650=6.625\times 10^5\text{N}=662.5\text{kN}$. Why this step? The dominant, always-positive term — we need it to see how much the penalty eats.
4. Pressure thrust. $(0.5-1.0)\times 10^5\times 0.30=-1.5\times 10^4\text{N}=-15\text{kN}$. Why this step? This is the penalty — it subtracts, exactly matching the parent's number.
5. Total. $F=662.5-15=647.5\text{kN}$; the pressure term costs $\approx 2.3\%$. Why this step? Puts the loss in perspective, and warns of possible Flow Separation in Over-expanded Nozzles once $p_e$ falls well below $p_a$.

Verify: $\dot{m}{v}_e=250\cdot 2650=662500$ N ✓. Pressure term $=(0.5-1.0)\cdot 10^5\cdot 0.30=-15000$ N ✓. Fraction $15/662.5=0.0226\approx 2.3\%$ ✓.

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Cell D — vacuum limit ($p_a\to 0$)

Forecast: does $\epsilon$ head for a fixed number, or grow without bound as $p_e\to 0$?

1. Pressure ratio. $p_0/p_e=70/0.02=3500$. Why this step? Tiny $p_e$ makes a huge ratio — this is the source of the large ε.
2. Solve $M_e$. $1+0.1M_e^2=3500^{1/6}=3.911$, $M_e^2=29.11$, $M_e=5.396$. Why this step? Higher pressure ratio ⇒ higher exit Mach (faster, colder exhaust).
3. Area ratio. $\epsilon =\frac{1}{5.396}[\frac{2}{2.2}(3.911){]}^{5.5}=\frac{1}{5.396}(3.556{)}^{5.5}\approx \frac{1069}{5.396}\approx 198$. Why this step? Converts to geometry — an ε near two hundred, exactly the deep-vacuum regime.
4. Limit reasoning. As $p_e\to 0$, $M_e\to \infty$ and $\epsilon \to \infty$. Perfect vacuum expansion needs an infinitely large bell — impossible — so real vacuum engines just make ε "as big as mass/packaging allows." See Altitude Compensation — Aerospike Nozzles for the escape from this trap. Why this step? Covers the limiting behaviour cell honestly: the optimum is unreachable, only approached.

Verify: $(1+0.1\cdot 29.11{)}^6=3.911^6=3500$ ✓. Area: $\frac{1}{5.396}(3.556{)}^{5.5}\approx 198$ ✓. Trend: larger $p_0/p_e\Rightarrow$ larger ε ✓.

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Cell E — degenerate throat ($\epsilon =1$)

Before this example, meet the master curve. Figure 2 plots the area ratio $\epsilon$ against exit Mach $M_e$ for $\gamma =1.20$. Study it now — Cells E and F both live on it.

In Figure 2 the yellow dot at the very bottom is the throat of Cell E ($\epsilon =1$ at $M=1$); the pink and blue dots on the dashed $\epsilon =2$ line are the two roots of Cell F. The curve falls as $M$ rises toward 1, hits its single lowest point $\epsilon =1$ exactly at $M=1$, then rises again for $M>1$. That single-minimum shape is the key to both cells.

Forecast: with no flare at all, is the exhaust subsonic, sonic, or supersonic?

1. Put $\epsilon =1$ in the area equation. The curve $\epsilon (M)$ in Figure 2 has its unique minimum value $\epsilon =1$, and that minimum occurs at exactly $M=1$. So $\epsilon =1\iff M_e=1$ — there is nowhere else on the curve that low. Why this step? $\epsilon =1$ is the very bottom of the curve; because the minimum is unique, it can only be reached at the single Mach $M=1$ (the throat is the choke point, Choked Flow and the Throat Condition).
2. Check it. At $M_e=1$: $\frac{1}{1}[\frac{2}{2.2}(1+0.1){]}^{5.5}=(1.0{)}^{5.5}=1$ ✓. Why this step? Confirms the degenerate case is self-consistent — no diverging section, no supersonic exit.
3. Pressure at exit. $p_0/p_e=(1+0.1{)}^6=1.1^6=1.772$, so $p_e=70/1.772=39.5$ bar. Why this step? Shows the exhaust leaves still very high pressure — hugely under-expanded at any altitude; almost all expansion is wasted. This is why you need the diverging bell.

Verify: $\frac{2}{2.2}(1.1)=1.0$ exactly, so $\epsilon =1.0$ at $M_e=1$ ✓. $1.1^6=1.7716$, $70/1.7716=39.5$ bar ✓.

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Cell F — two Mach roots, one ε (branch choice)

Forecast: which root lives downstream of the throat? Trace the horizontal $\epsilon =2$ line in Figure 2 — it stabs the curve twice.

1. Why two roots exist. The area–Mach curve dips to its minimum ($\epsilon =1$) at $M=1$, then rises on both sides. So any $\epsilon >1$ is hit twice — once going subsonic, once supersonic. Why this step? This is the whole point of the cell: $\epsilon$ alone does not pin the flow; a sign/branch choice does. See the two marked dots in Figure 2.
2. Subsonic root. Root-find on the bracket $M\in [0.01,1]$ (bisection): $g(M)$ goes from large-positive at tiny $M$ down through zero. The solution is $M_e\approx 0.306$ (pink dot). Why this step? This is the branch in the converging part before the throat — not the exhaust. We quote three decimals because that is what a few bisection halvings deliver.
3. Supersonic root. Root-find on the bracket $M\in [1,10]$: the solution is $M_e\approx 2.197$ (blue dot). Why this step? Past the throat the flow is choked at $M=1$ and keeps accelerating in the diverging bell — so it is this root.
4. Choose. For the nozzle exit (downstream of throat), take $M_e=2.197$. Why this step? Physical placement, not algebra, selects the branch — the classic exam trap the parent warned about.

Verify: Supersonic: $\frac{1}{2.197}[\frac{2}{2.2}(1+0.1\cdot 4.827){]}^{5.5}=\frac{1}{2.197}(1.348{)}^{5.5}=\frac{1}{2.197}(4.41)=2.01$ ✓. Subsonic: $\frac{1}{0.306}[\frac{2}{2.2}(1+0.1\cdot 0.0936){]}^{5.5}=\frac{1}{0.306}(0.9176{)}^{5.5}=\frac{0.614}{0.306}=2.01$ ✓.

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Cell G — real-world word problem

Forecast: should the first stage have the bigger or smaller bell?

1. First-stage strategy. It spends most of its work in thick air; choose $p_e$ near the mid of its range to avoid heavy over-expansion at liftoff. Take design $p_e\approx 0.7$ bar. Why this step? Designing for exactly $1.0$ bar (liftoff) would over-expand badly higher up; a slight compromise avoids separation at $t=0$. See Flow Separation in Over-expanded Nozzles.
2. First-stage ε. $p_0/p_e=100$; $1+0.1M_e^2=100^{1/6}=2.154$, $M_e^2=11.54$, $M_e=3.397$; $\epsilon =\frac{1}{3.397}[\frac{2}{2.2}(2.154){]}^{5.5}=\frac{1}{3.397}(1.958{)}^{5.5}\approx \frac{40.7}{3.397}\approx 12.0$. Why this step? Turns the chosen $p_e$ into buildable geometry — a modest ε ≈ 12, right for a booster.
3. Second-stage strategy. It works in thin air, so match a low $p_e$. Take $p_e\approx 0.05$ bar. Why this step? Low ambient ⇒ low optimum $p_e$ ⇒ large expansion, big bell.
4. Second-stage ε. $p_0/p_e=1400$; $1+0.1M_e^2=1400^{1/6}=3.386$, $M_e^2=23.86$, $M_e=4.885$; $\epsilon =\frac{1}{4.885}[\frac{2}{2.2}(3.386){]}^{5.5}=\frac{1}{4.885}(3.078{)}^{5.5}\approx \frac{484}{4.885}\approx 99$. Why this step? Confirms the second stage needs a much larger bell (ε ≈ 99) — the "size ε to the sky" mnemonic in action; larger ε ⇒ higher Specific Impulse Isp in vacuum.

Verify: Stage 1: $2.154^6=100.0$ ✓, $\epsilon \approx 12.0$. Stage 2: $3.386^6=1400$ ✓, $\epsilon \approx 99$. Ordering ${\epsilon }_1<{\epsilon }_2$ (12 < 99) matches "higher altitude ⇒ bigger ε" ✓.

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Cell H — exam twist (rank two nozzles)

Forecast: N1 has the faster exhaust — does that settle it?

1. N1 thrust. Momentum $=200\times 2650=530,000$ N. Pressure $=(0.5-0.4)\times 10^5\times 0.30=+3000$ N. Total $=533.0\text{kN}$. ($p_e>p_a$, slightly under-expanded here — good.) Why this step? At $p_a=0.4$, the big nozzle is near matched, so its pressure term is small and positive.
2. N2 thrust. Momentum $=200\times 2500=500,000$ N. Pressure $=(1.188-0.4)\times 10^5\times 0.16=+12,608$ N. Total $=512.6\text{kN}$. Why this step? The small nozzle is more under-expanded (bigger positive pressure term) but its slower $v_e$ costs more.
3. Compare. $533.0>512.6\text{kN}$ ⇒ N1 wins here — but because it is near-matched, not merely "bigger." Why this step? Defuses the trap: if we instead fired at sea level ($p_a=1.0$), N1's term would be $(0.5-1.0)\cdot 10^5\cdot 0.30=-15$ kN and N2's would be $(1.188-1.0)\cdot 10^5\cdot 0.16=+3008$ N — the gap narrows and the ranking can flip. Optimality is about matching, not size.

Verify: N1 $=530000+3000=533000$ N ✓. N2 $=500000+12608=512608$ N ✓. N1 > N2 ✓. Sea-level check: N1 $=530000-15000=515000$; N2 $=500000+3008=503008$ — N1 still wins there but by less; the general warning stands (a more over-expanded N1 could lose).

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Cell I — sub-critical: the flow never chokes

Forecast: with only a $1.5\times$ pressure ratio, do you think the gas can reach Mach 1 anywhere?

1. Find the choke threshold. From Ex 5, reaching $M=1$ at the throat needs a pressure ratio of at least $(1+\frac{\gamma -1}{2}{)}^{\gamma /(\gamma -1)}=1.1^6=1.772$. Why this step? This single number is the gatekeeper: below it the throat cannot go sonic.
2. Compare. $p_0/p_a=1.5<1.772$ → the flow is sub-critical. The throat stays below Mach 1; nowhere in the nozzle does the gas reach the speed of sound. Why this step? $1.5<1.772$ decides the whole case — no choking.
3. What the nozzle does instead. With no sonic throat, the diverging "bell" no longer accelerates the flow — for subsonic gas a widening duct slows it down (like a river spreading out). So the diverging section acts as a diffuser, the exit stays subsonic, and $p_e=p_a=1.0$ bar automatically (a subsonic jet always matches ambient pressure). Why this step? This is the crucial reversal: the same geometry that accelerates choked supersonic flow decelerates un-choked subsonic flow. The area–Mach curve's subsonic branch (left of the yellow dot in Figure 2) governs here.
4. Thrust consequence. Since $M_e<1$ and $p_e=p_a$, exit velocity is modest and the pressure term is zero — this rig produces feeble thrust. Real rockets always run $p_0/p_a$ far above $1.772$ precisely to avoid this dud regime. Why this step? Closes the case: sub-critical operation is a failure mode to design away from, not a useful cell.

Verify: Threshold $1.1^6=1.7716$ ✓. Since $1.5<1.7716$, sub-critical (no choke) ✓. A choked run would need $p_0/p_a\ge 1.772$, e.g. the $p_0=70$ bar engines above all clear it by a wide margin ✓.

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Recall One-line recap of every cell

Matched ::: pressure term $=0$, optimum (Ex 1). Under-expanded ::: $p_e>p_a$, term $+$, ε too small (Ex 2). Over-expanded ::: $p_e<p_a$, term $-$, ε too large, separation risk (Ex 3). Vacuum limit ::: $p_a\to 0\Rightarrow \epsilon \to \infty$, unreachable (Ex 4). Degenerate $\epsilon =1$ ::: only $M_e=1$, sonic, hugely under-expanded (Ex 5). Two roots ::: pick supersonic downstream of throat (Ex 6). Multi-stage ::: small ε low, big ε high (Ex 7). Ranking trap ::: bigger ε $\ne$ always more thrust; matching wins (Ex 8). Sub-critical ::: $p_0/p_a<1.772$, never chokes, diverging section acts as diffuser (Ex 9).

Connections

• Parent topic
• Thrust Equation and Effective Exhaust Velocity
• De Laval Converging-Diverging Nozzle
• Isentropic Flow Relations
• Choked Flow and the Throat Condition
• Flow Separation in Over-expanded Nozzles
• Altitude Compensation — Aerospike Nozzles
• Specific Impulse Isp