Worked examples — Optimum expansion — P_e = P_a for maximum thrust

This page is the drill ground for the optimum-expansion topic. The parent note built the idea that thrust peaks when $P_e=P_a$. Here we hunt down every case that idea can produce — under, over, and perfect expansion; vacuum and sea level; zero and degenerate inputs; the limiting cases; a word problem; and an exam twist.

Before anything, a promise about symbols. Everything below reuses only what the parent defined, and we restate each symbol the first time it appears so you never have to scroll up.

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The scenario matrix

Every problem this topic can hand you falls into one of these cells. Each row is a case class; the last column names the worked example that nails it.

#	Case class	Sign / condition	What is being tested	Example
A	Under-expanded	$P_e>P_a$	pressure term adds thrust	Ex 1
B	Perfectly expanded	$P_e=P_a$	pressure term vanishes, thrust is pure momentum	Ex 2
C	Over-expanded	$P_e<P_a$	pressure term subtracts thrust	Ex 3
D	Vacuum limit	$P_a\to 0$	maximum possible pressure thrust	Ex 4
E	Degenerate: zero pressure term	$A_e\to 0$ or $P_e=P_a$	pressure thrust can die two different ways	Ex 5
F	Design problem (find $\epsilon$)	choose geometry so $P_e=P_a$	inverse problem via Isentropic flow relations	Ex 6
G	Word problem: altitude sweep	$P_a$ falls as rocket climbs	one fixed nozzle across a whole flight	Ex 7
H	Exam twist: which is more efficient?	compare two nozzles by Specific impulse	trap — bigger thrust ≠ better nozzle	Ex 8

Look at the figure below before starting — it is the map every example lives on.

The red curve is thrust versus exit pressure. The three colored dots are cells A, B, C. Everything that follows is just visiting these dots with numbers.

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Cell A — Under-expanded ($P_e>P_a$)

Forecast: $P_e>P_a$, so the pressure term is positive. Guess whether the bonus is big or small compared to $\dot{m}{v}_e$ before reading on.

1. Convert bars to pascals. Why this step? The thrust equation mixes $\dot{m}{v}_e$ (already in newtons) with a pressure-times-area term, so pressures must be in pascals (Pa) to give newtons. $P_e=1.4\times 10^5\text{Pa},P_a=0.9\times 10^5\text{Pa}$
2. Momentum thrust. Why? It is the always-present backbone of $F$. $\dot{m}{v}_e=250\times 2700=675,000\text{N}$
3. Pressure thrust. Why? $P_e>P_a$ means the gas out-pushes the sky, adding force on the exit rim. $(P_e-P_a)A_e=(1.4-0.9)\times 10^5\times 1.5=0.5\times 10^5\times 1.5=75,000\text{N}$
4. Add. $F=675,000+75,000=750,000\text{N}=750\text{kN}$

Verify: Units — (kg/s)(m/s) = kg·m/s² = N ✓, and (Pa)(m²) = (N/m²)(m²) = N ✓. The bonus (75 kN) is about $11\%$ of the momentum thrust — real but modest, exactly what "slightly under-expanded" should look like.

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Cell B — Perfectly expanded ($P_e=P_a$)

Forecast: With $P_e=P_a$, the entire second chunk should vanish. Predict $F=\dot{m}{v}_e$ exactly.

1. Check the condition. Why? This is the whole point of the topic — recognise when the pressure term is zero. $P_e-P_a=(1.4-1.4)\times 10^5=0$
2. Pressure thrust. Why? Anything times zero is zero, regardless of $A_e$. $(P_e-P_a)A_e=0\times 1.5=0\text{N}$
3. Thrust is pure momentum. $F=\dot{m}{v}_e+0=675,000\text{N}=675\text{kN}$

Verify: This sits below Ex 1's 750 kN — surprising? No: Ex 1 was under-expanded and got a genuine bonus. But this is the peak for a nozzle designed for this altitude, because a nozzle that is perfectly expanded here would have a larger $A_e$ and lower $P_e$, capturing that expansion internally. The dot B in the figure is the maximum of the curve, not the maximum of the pressure term. Units check as before ✓.

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Cell C — Over-expanded ($P_e<P_a$)

Forecast: $P_e<P_a$, so the pressure term is negative — it will eat into the momentum thrust. Guess the size of the bite.

1. Convert. $P_e=0.3\times 10^5$ Pa, $P_a=1.0\times 10^5$ Pa. Why? Same units rule.
2. Momentum thrust. $\dot{m}{v}_e=300\times 3000=900,000\text{N}$
3. Pressure thrust (negative). Why? The atmosphere out-pushes the exhaust, so the net exit-plane force points backward. $(P_e-P_a)A_e=(0.3-1.0)\times 10^5\times 3.0=(-0.7\times 10^5)(3.0)=-210,000\text{N}$
4. Add. $F=900,000-210,000=690,000\text{N}=690\text{kN}$

Verify: The pressure penalty is $210/900\approx 23\%$ of the momentum thrust — the classic sea-level penalty of an over-expanded nozzle. Beware: if this nozzle over-expands too hard the flow physically tears off the wall (Supersonic flow separation) and this simple formula overstates the loss. Units ✓.

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Cell D — Vacuum limit ($P_a\to 0$)

Forecast: With nothing pushing back, the pressure term is at its most positive. Predict $F$ jumps above 900 kN.

1. Pressure term with $P_a=0$. Why this step? Setting $P_a=0$ is the limiting case — nothing on the outside, so the exit gas's push is unopposed. $(P_e-0)A_e=0.3\times 10^5\times 3.0=90,000\text{N}$
2. Thrust. $F_{\text{vac}}=900,000+90,000=990,000\text{N}=990\text{kN}$
3. Compare to Ex 3 (sea level, 690 kN). Why? This is why the same nozzle is worth far more in space. $\Delta F=990-690=300\text{kN}$

Verify: The full swing $300$ kN equals $P_a^{\text{SL}}{A}_e=1.0\times 10^5\times 3.0=300,000$ N ✓ — the atmosphere's constant $P_a{A}_e$ back-push is exactly what disappears in vacuum. This is the whole logic of Altitude compensation. Units ✓.

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Cell E — Degenerate: pressure thrust can die two ways

Forecast: In the formula $(P_e-P_a)A_e$, a product is zero if either factor is zero. Predict X kills it with the bracket, Y kills it with the area.

1. Nozzle X — bracket is zero. Why? Perfect expansion: $(P_e-P_a)=0$. $(0.8-0.8)\times 10^5\times 2.0=0\text{N}\Rightarrow F_X=500,000\text{N}$
2. Nozzle Y — area is zero. Why? A vanishing exit area gives the mismatch nothing to push on. This is a degenerate (unphysical, but instructive) limit. $(0.8-0.5)\times 10^5\times 0=0\text{N}\Rightarrow F_Y=500,000\text{N}$

Verify: Both give $F=500$ kN, but the meaning differs. X is the real optimum (peak of the thrust curve). Y is a mathematical corner: shrinking $A_e$ to zero also strangles the flow that produces $\dot{m}{v}_e$ in the first place — so Y is a warning that "$A_e=0$ makes the pressure term vanish" is a formula quirk, not a design goal. Units ✓.

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Cell F — Design problem: find the expansion ratio

Forecast: Lower $P_e$ needs a bigger nozzle mouth, so $\epsilon$ should be well above 1. Guess "several tens".

1. Set optimum. Why? We want $P_e=P_a$, so we impose it. $\frac{P_e}{P_0}=\frac{0.5}{50}=0.01$
2. Find exit Mach number $M_e$ from the Isentropic flow relations. Why this tool? Pressure and Mach number are locked together along an isentropic (constant-entropy) expansion; this is the only relation that connects the pressure we want to the geometry we can build. $\frac{P_e}{P_0}={\left(1+\frac{\gamma -1}{2}{M}_e^2\right)}^{-\frac{\gamma }{\gamma -1}}$ With $\gamma =1.20$: exponent $-\gamma /(\gamma -1)=-6$, and $(\gamma -1)/2=0.1$. $0.01=(1+0.1M_e^2{)}^{-6}$ Raise both sides to $-1/6$: $0.01^{-1/6}=1+0.1M_e^2\Rightarrow 2.1544=1+0.1M_e^2\Rightarrow M_e^2=11.544\Rightarrow M_e=3.398$
3. Feed $M_e$ into the area-ratio relation. Why? The isentropic area–Mach relation converts a Mach number into the geometric ratio $\epsilon$ we can machine. $\epsilon =\frac{A_e}{A^{\ast }}=\frac{1}{M_e}{\left[\frac{2}{\gamma +1}\left(1+\frac{\gamma -1}{2}{M}_e^2\right)\right]}^{\frac{\gamma +1}{2(\gamma -1)}}$ Here $\frac{2}{\gamma +1}=0.9091$, $1+0.1M_e^2=2.1544$, and the exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.2}{0.4}=5.5$. $\epsilon =\frac{1}{3.398}[0.9091\times 2.1544{]}^{5.5}=\frac{1}{3.398}(1.9585{)}^{5.5}$ $(1.9585{)}^{5.5}\approx 40.35\Rightarrow \epsilon \approx \frac{40.35}{3.398}\approx 11.9$

Verify: $\epsilon \approx 11.9$ — a nozzle mouth about twelve times the throat area, comfortably in the "several tens" ballpark our forecast expected, and typical of a mid-altitude engine. Feeding $M_e=3.398$ back into step 2 reproduces $P_e/P_0=0.01$ (checked in Verify). Every step used only Isentropic flow relations — no free-floating assumptions.

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Cell G — Word problem: one nozzle across a whole climb

Forecast: As $P_a$ falls past $P_e=0.7$ bar, the engine walks from over-expanded → perfect → under-expanded, and thrust should climb throughout.

1. Momentum thrust (constant). Why? $\dot{m}$ and $v_e$ don't depend on outside air. $\dot{m}{v}_e=400\times 2900=1,160,000\text{N}$
2. Sea level, $P_a=1.0$ bar — over-expanded (Cell C). $(0.7-1.0)\times 10^5\times 2.5=-75,000\text{N}\Rightarrow F=1,085,000\text{N}=1085\text{kN}$
3. Mid-altitude, $P_a=0.7$ bar — perfect (Cell B). $(0.7-0.7)\times 10^5\times 2.5=0\Rightarrow F=1,160,000\text{N}=1160\text{kN}$
4. Near-vacuum, $P_a=0.05$ bar — under-expanded (Cell A). $(0.7-0.05)\times 10^5\times 2.5=162,500\text{N}\Rightarrow F=1,322,500\text{N}=1322.5\text{kN}$

Verify: Thrust rises monotonically $1085\to 1160\to 1322.5$ kN as the rocket climbs — exactly the story behind Altitude compensation and why designers pick $P_e$ for an average flight condition rather than any single altitude. Sign of each pressure term matches its regime label ✓. This is also why Rocket staging swaps to a bigger-$\epsilon$ nozzle once the atmosphere thins.

The plot above shows this exact sweep: thrust climbing as $P_a$ drops, with the perfect-expansion point marked where the curve crosses $P_a=P_e$.

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Cell H — Exam twist: bigger thrust isn't automatically better

Forecast: $I_{sp}$ measures thrust per unit weight-flow — the true efficiency. Q has more thrust; does it have more $I_{sp}$? (Trap: yes here, but only because $\dot{m}$ is equal. Watch what the metric actually rewards.)

1. Compute $I_{sp}$ for P. Why this tool? Specific impulse strips out how much propellant you throw and asks how good each kilogram is — the fair comparison metric. $I_{sp,P}=\frac{560,000}{200\times 9.81}=\frac{560,000}{1962}\approx 285.4\text{s}$
2. Compute $I_{sp}$ for Q. $I_{sp,Q}=\frac{590,000}{1962}\approx 300.7\text{s}$
3. Rank. Q wins here because both share the same $\dot{m}$, so more thrust genuinely means more $I_{sp}$.

Verify: $I_{sp,Q}>I_{sp,P}$ by about $15$ s. The twist: had Q reached its 590 kN by simply burning more propellant (higher $\dot{m}$), its $I_{sp}$ could have been lower than P despite bigger thrust — that is the exam trap. With $\dot{m}$ fixed, thrust and $I_{sp}$ move together; only then is "more thrust = better" safe. Units: N / (kg/s · m/s²) = N / (N/s) = s ✓.

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Recall

Recall Which regime has a

positive pressure thrust? Under-expanded ($P_e>P_a$) ::: the exhaust out-pushes the atmosphere, so $(P_e-P_a)A_e>0$.

Recall Why does perfect expansion give the peak thrust and not the biggest pressure term?

Because a perfectly expanded nozzle captures all the expansion internally (large $A_e$, high $v_e$); the pressure term being zero is a symptom of the optimum, not its cause ::: the momentum thrust $\dot{m}{v}_e$ is maximised for that geometry.

Recall Two different ways the pressure thrust

$(P_e-P_a)A_e$ can equal zero? Either $P_e=P_a$ (the real optimum) or $A_e=0$ (a degenerate, non-physical corner) ::: only the first is a design goal.

Recall Fixed nozzle climbing through the atmosphere — what sequence of regimes?

Over-expanded → perfectly expanded → under-expanded as $P_a$ drops below $P_e$ ::: thrust rises the whole way.