Worked examples — Nozzle thermodynamics — isentropic expansion from chamber to exit

This page is the drill ground for the parent nozzle note. The parent gave you the formulas. Here we throw every kind of case at those formulas — subsonic vs supersonic, the throat, the limits, and the tricky exam twists — so no scenario surprises you later.

Before anything, let's re-pin the four workhorse equations from the parent so we never use a symbol we haven't re-anchored.

Recall The symbols we will keep reusing

• $T_0,P_0,{\rho }_0$ ::: chamber (stagnation) values — the gas brought fully to rest, all energy thermal, no motion.
• $T,P,\rho$ ::: local static values at some point where gas moves with speed $V$.
• $M=V/a$ ::: Mach number, gas speed divided by sound speed $a=\sqrt{\gamma RT}$. $M<1$ subsonic, $M=1$ sonic, $M>1$ supersonic.
• $\gamma =c_p/c_v$ ::: heat-capacity ratio, $\approx 1.2$ for hot rocket exhaust.
• $A^{\ast }$ (and any "$\ast$" quantity) ::: value at the throat, where $M=1$. $A$ is the area at any other station.

Throughout, unless told otherwise, we use the parent's gas: $\gamma =1.2$, $R=320\text{J/(kg⋅K)}$, $P_0=10\text{MPa}$, $T_0=3500\text{K}$.

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The scenario matrix

Every question this topic can ask is one of these cells. Each worked example below is tagged with the cell it fills.

Cell	What makes it distinct	Covered by
A. Chamber → throat (M=1)	the sonic point, critical ratios	Ex 1
B. Subsonic station (M<1)	converging section, area above $A^{\ast }$	Ex 2
C. Supersonic station (M>1)	diverging section, area above $A^{\ast }$ again	Ex 3
D. Zero / degenerate input (M→0)	chamber itself, ratios → 1, $A/A^{\ast }\to \infty$	Ex 4
E. Limiting behaviour (M→∞)	maximum possible exhaust speed	Ex 5
F. Inverse problem	given area ratio, find M (two roots!)	Ex 6
G. Real-world word problem	design a nozzle for a target exit velocity	Ex 7
H. Exam twist — sign/regime trap	same $A/A^{\ast }$, which branch? over/under-expanded	Ex 8

The two subtle traps are cell D (a degenerate input that breaks the naive area formula) and cell F/H (the area ratio gives two Mach numbers — you must pick the physically correct one). We hit those head-on.

Figure s01 — "one ratio, two Machs". The horizontal axis is Mach number $M$; the vertical axis is the area ratio $A/A^{\ast }$. The blue curve is the subsonic branch ($M<1$), the red curve the supersonic branch ($M>1$), and they meet at the throat minimum $A/A^{\ast }=1$ at $M=1$ (white dotted line). The dashed green horizontal line at $A/A^{\ast }=4$ crosses the curve twice — once on blue (subsonic root), once on red (supersonic root), each marked by a yellow dot. That double crossing is the visual reason cells B, C, F and H all exist: geometry alone never tells you which regime you are in.

See also Nozzle Flow Regimes and Mach Number and Sound Speed for the regime language used below.

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Example 1 — Cell A: chamber to throat

Forecast: at $M=1$ the ratios are fixed numbers (they only depend on $\gamma$). Guess: temperature drops a little, pressure roughly halves.

1. Temperature. $T^{\ast }=T_0\cdot \frac{2}{\gamma +1}=3500\cdot \frac{2}{2.2}=3181.8\text{K}$. Why this step? Set $M=1$ in $T_0/T=1+\frac{\gamma -1}{2}{M}^2=1+0.1=1.1$, so $T=T_0/1.1$ — this converts the general relation into the special throat value.
2. Pressure. $P^{\ast }=P_0{\left(\frac{2}{2.2}\right)}^6=10\times 0.9091^6=5.645\text{MPa}$. Why this step? The pressure formula's exponent is $\gamma /(\gamma -1)=1.2/0.2=6$; at $M=1$ the bracket collapses to $2/(\gamma +1)$.
3. Sound speed = exit speed. $a^{\ast }=\sqrt{\gamma R{T}^{\ast }}=\sqrt{1.2\times 320\times 3181.8}=1105.6\text{m/s}$. Why this step? At the throat $M=1$ means $V^{\ast }=a^{\ast }$ exactly, so computing the sound speed is computing the gas speed there.

Verify: temperature dropped only $9\%$ but pressure dropped $44\%$ — pressure is far more sensitive because its exponent (6) is much larger than temperature's (1). Units of $a^{\ast }$: $\sqrt{(\text{J/kg⋅K})(\text{K})}=\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✓.

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Example 2 — Cell B: a subsonic station

Forecast: subsonic, so barely any acceleration yet — expect $T,P$ close to chamber values, and area larger than the throat.

1. Bracket factor. $1+\frac{\gamma -1}{2}{M}^2=1+0.1(0.25)=1.025$. Why this step? This one number feeds all four formulas — compute it once and reuse, so a single arithmetic slip can't hide in three places.
2. Temperature. $T=T_0/1.025=3500/1.025=3414.6\text{K}$. Why this step? Temperature is just $T_0$ divided by the bracket — the gas has given up only a sliver of thermal energy because it is barely moving.
3. Pressure. $P=P_0/1.025^6=10/1.1596=8.624\text{MPa}$. Why this step? Same bracket, now raised to the pressure exponent $6$; pressure falls faster than temperature, as always.
4. Area ratio. $\frac{A}{A^{\ast }}=\frac{1}{0.5}[\frac{2}{2.2}(1.025){]}^{5.5}=2\times (0.9318{)}^{5.5}=2\times 0.6807=1.361$. Why this step? The inside term $\frac{2}{2.2}(1.025)=0.9318$ is below 1, so raising to $5.5$ shrinks it — the $1/M=2$ out front then lifts the ratio back above 1, giving a duct wider than the throat.

Verify: $A/A^{\ast }=1.36>1$ ✓ — the converging section is wider than the throat, correct for subsonic. Compare with Ex 3: the same value of area ratio can also appear supersonically, which is the whole point of the matrix.

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Example 3 — Cell C: a supersonic station

Forecast: big acceleration — expect temperature roughly halved, pressure down by two orders of magnitude, velocity near $2500\text{m/s}$, and a wide exit bell.

1. Bracket factor. $1+0.1(9)=1.9$. Why this step? Same master number as before; $M^2=9$ makes it much larger than the subsonic case, signalling a big energy conversion.
2. Temperature. $T_e=3500/1.9=1842.1\text{K}$. Why this step? Divide $T_0$ by the bracket — half the thermal energy has already become motion.
3. Pressure. $P_e=10/1.9^6=10/47.05=0.2125\text{MPa}=212.5\text{kPa}$. Why this step? Bracket to the $6$th power; the huge drop is what drives the acceleration.
4. Velocity. $V_e=3\sqrt{1.2\times 320\times 1842.1}=3\times 841.4=2524\text{m/s}$. Why this step? $V=Ma$ with $a=\sqrt{\gamma R{T}_e}$ — the exhaust speed that ultimately makes thrust.
5. Area ratio. First the inside term: $\frac{2}{2.2}(1.9)=0.9091\times 1.9=1.7273$. Raise to $5.5$: $1.7273^{5.5}=20.22$. Then $\frac{A_e}{A^{\ast }}=\frac{1}{3}\times 20.22=6.74$. Why this step? Here the inside term is $>1$, so the $5.5$ power amplifies it strongly (to $20.22$); dividing by $M=3$ leaves a wide expansion ratio of $6.74$. This is the exponent everyone mis-evaluates — note $1.7273^{5.5}\ne 14.14$.

Verify: $P_e=212\text{kPa}>P_a=101\text{kPa}$, so the jet leaves above ambient → the nozzle is under-expanded (it could have expanded more). Note $A_e/A^{\ast }=6.74>1$ again, exactly like the subsonic Ex 2 — proving one area ratio, two regimes. Links: Thrust Equation Derivation, Real Nozzle Losses.

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Example 4 — Cell D: degenerate input, M → 0 (the chamber itself)

Forecast: with no motion, static = stagnation, so all three thermodynamic ratios → 1. But the area ratio has a $1/M$ out front — what happens?

1. Thermodynamic ratios. $1+\frac{\gamma -1}{2}(0{)}^2=1$, so $T/T_0=1$, $P/P_0=1$, and $\rho /{\rho }_0=1^{-1/(\gamma -1)}=1$. Why this step? No kinetic energy means no conversion — the gas is at stagnation. The density ratio matters because it is the $\rho$ that appears in mass flow $\rho VA=\text{const}$; at $M=0$ it equals the chamber density ${\rho }_0$ exactly, which we use in the very next step.
2. Area ratio limit. $\frac{A}{A^{\ast }}=\frac{1}{M}[\dots {]}^{5.5}\to \infty$ as $M\to 0$. Why this step? The $1/M$ blows up while the bracket term stays finite (it tends to $(2/2.2{)}^{5.5}$). Physically: mass flow $\rho VA=\text{const}$ with $\rho \to {\rho }_0$ (from step 1) and $V\to 0$ forces area $\to \infty$. A real chamber is approximately this — huge area compared to the throat.

Verify: at $M=0.05$, the inside term is $\frac{2}{2.2}(1+0.1\times 0.0025)=0.9091\times 1.00025=0.9093$, raised to $5.5$ gives $0.6318$, times $1/0.05=20$ gives $A/A^{\ast }=12.64$ — already large and climbing as $M$ shrinks, confirming the blow-up. This is the degenerate-input case the naive user forgets: you cannot invert the area formula at the chamber, only near the throat and beyond.

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Example 5 — Cell E: limiting behaviour, M → ∞

Forecast: you can't get infinite speed — all thermal energy is finite. Guess: a fixed number set by $c_p{T}_0$.

1. Energy equation. The steady-flow energy equation from the parent, $h+\frac{V^2}{2}=h_0$, becomes for an ideal gas ($h=c_{p}T$, $h_0=c_p{T}_0$): $c_{p}T+\frac{V^2}{2}=c_p{T}_0$. This is stagnation-enthalpy conservation — it holds because the flow is adiabatic (no heat crosses the wall) and steady with no shaft work, exactly the isentropic-flow assumptions of the parent. Then as $T\to 0$ (all thermal energy spent), $V\to V_{\max }=\sqrt{2c_p{T}_0}$. Why this step? This is the physical ceiling — you cannot extract more kinetic energy than the total thermal energy $c_p{T}_0$ you started with, because total enthalpy is conserved along the streamline.
2. Compute $c_p$. $c_p=\frac{\gamma R}{\gamma -1}=\frac{1.2\times 320}{0.2}=1920\text{J/(kg⋅K)}$. Why this step? We need $c_p$ in absolute units to turn a temperature into a velocity.
3. Ceiling. $V_{\max }=\sqrt{2\times 1920\times 3500}=\sqrt{1.344\times 10^7}=3666\text{m/s}$. Why this step? Plug the numbers straight into $\sqrt{2c_p{T}_0}$.

Verify: our $M=3$ exit gave $2524\text{m/s}$ — comfortably below the $3666\text{m/s}$ ceiling ✓. No finite nozzle reaches $V_{\max }$ (would need infinite area), so real exhaust speeds always sit under this bar. See Specific Impulse Optimization and Entropy and Reversibility.

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Example 6 — Cell F: inverse problem, area ratio → Mach (two roots)

Forecast: from figure s01, a horizontal line at $4.0$ cuts the curve twice — one subsonic root, one supersonic root. Both are mathematically valid.

1. Set up the equation. Solve $4.0=\frac{1}{M}[\frac{2}{2.2}(1+0.1M^2){]}^{5.5}$ numerically. Why this step? No closed-form inverse exists — this is inherently a root-finding problem, exactly why cell F is its own case.
2. Supersonic root (bisection). Try $M=2.6$: inside $=\frac{2}{2.2}(1+0.1\times 6.76)=0.9091\times 1.676=1.5236$, $1.5236^{5.5}=10.30$, $/2.6=3.96$ (a hair low). Try $M=2.62$: inside $=0.9091\times 1.6864=1.5331$, ${}^{5.5}=10.60$, $/2.62=4.046$ (a hair high). The value $4.0$ sits between, at $M\approx 2.611$. Why this step? Bisection brackets the root: one trial gives just under $4.0$, the next just over, so the answer is trapped between $2.60$ and $2.62$.
3. Subsonic root (bisection). Try $M=0.15$: inside $=0.9091\times 1.00225=0.9111$, ${}^{5.5}=0.6024$, $/0.15=4.016$ (a hair high). Try $M=0.151$: inside $=0.9091\times 1.00228=0.9112$, ${}^{5.5}=0.6026$, $/0.151=3.991$ (a hair low). So the subsonic root is $M\approx 0.1506$. Why this step? Same bracketing on the blue branch, where the $1/M$ term dominates so a small $M$ produces a large ratio.

Verify: plug $M=2.611$ back: inside $=0.9091\times (1+0.1\times 6.817)=0.9091\times 1.6817=1.5288$; $1.5288^{5.5}=10.44$; $/2.611=3.999\approx 4.0$ ✓. Plug $M=0.1506$: inside $=0.9091\times 1.00227=0.9112$; ${}^{5.5}=0.6025$; $/0.1506=4.001\approx 4.0$ ✓. Two Mach numbers share one area ratio, and only physics (next example) picks the real one.

Figure s02 — where each cell lives in the hardware. The blue outline is the nozzle wall: it converges from a wide chamber (left) down to the throat (yellow dashed line, $M=1$, area $A^{\ast }$) and then diverges to a wide exit (right). The red arrow along the axis is the flow direction. Green text marks the subsonic converging region (cell B, Ex 2), red text the supersonic diverging region (cells C, G, H), and the far-left label marks the near-stagnant chamber ($M\to 0$, cell D, Ex 4). Reading the picture left-to-right is reading the gas accelerate from rest, through Mach 1 at the throat, to supersonic at the exit.

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Example 7 — Cell G: real-world design

Forecast: $2900$ is close to Ex 5's ceiling of $3666$, so we'll need a fairly high Mach (well above 3) and a large area ratio.

1. Relate $V_e$ to $M$. $V_e=M\sqrt{\gamma RT}$ with $T=T_0/(1+0.1M^2)$. So $V_e=M\sqrt{\frac{\gamma R{T}_0}{1+0.1M^2}}$. Why this step? One equation, one unknown $M$ — velocity fixes Mach.
2. Solve. $\gamma R{T}_0=1.2\times 320\times 3500=1.344\times 10^6$. Numerically $M\approx 4.20$. Why this step? Iterating the velocity equation converges here; the high value confirms the forecast.
3. Area ratio. inside term $\frac{2}{2.2}(1+0.1\times 17.64)=0.9091\times 2.764=2.5127$; raise to $5.5$: $2.5127^{5.5}=158.8$; divide by $M=4.20$: $\frac{A_e}{A^{\ast }}=\frac{158.8}{4.20}=37.8$. Why this step? Same careful exponent handling as Ex 3 — the inside term is now well above 1, so the $5.5$ power inflates it enormously, giving the big bell of a vacuum engine.

Verify: back-substitute $M=4.20$: $T=3500/2.764=1266.3\text{K}$, $V_e=4.20\sqrt{1.2\times 320\times 1266.3}=4.20\times 697.5=2930\text{m/s}\approx 2900$ ✓ (rounding). A ratio of $\sim 38$ is exactly the kind of huge bell you see on upper-stage vacuum engines. Compare Nozzle Flow Regimes on why big ratios suit vacuum.

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Example 8 — Cell H: the exam twist (regime trap)

Forecast: the geometry forces the supersonic root (once flow chokes at the throat and stays attached), so $M_e$ and $P_e$ are fixed by geometry alone — the back-pressure only decides the shock/plume behaviour, not $M_e$.

1. Pick the branch. Attached supersonic flow → use $M_e=3.0$, not the subsonic root of the same area ratio. Why this step? This is the trap: the subsonic root only occurs in a purely converging duct or an unchoked nozzle. A supersonic diverging nozzle running full sits on the upper branch.
2. Exit pressure. bracket $=1+0.1(9)=1.9$; $P_e=10/1.9^6=0.2125\text{MPa}=212.5\text{kPa}$. Why this step? $P_e$ is set purely by $M_e$, which the geometry fixed — reuse Ex 3's number.
3. Case (a) $P_a=250$ kPa. $P_e=212.5<P_a=250$ → over-expanded, ambient squeezes the jet, oblique shocks form at the lip. Why this step? Over-expanded means the nozzle dropped the pressure below ambient.
4. Case (b) $P_a=20$ kPa. $P_e=212.5>P_a=20$ → under-expanded, the plume keeps expanding outside. Why this step? Same fixed $P_e$, but now it exceeds ambient — the label flips purely because $P_a$ changed.

Verify: $P_e$ depends only on the fixed $M_e$, so it is the same $212.5$ kPa in both cases — the back-pressure never changed $P_e$ ✓. Only the comparison $P_e$ vs $P_a$ flips the label (over-expanded at $250$ kPa, under-expanded at $20$ kPa). See Real Nozzle Losses for what those shocks do next.

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Recall Quick self-test

One area ratio gives how many Mach numbers, and how do you choose? ::: Two (one subsonic, one supersonic). Choose by the flow regime: converging/unchoked → subsonic root; diverging full-running → supersonic root. What sets the absolute ceiling on exhaust velocity? ::: $V_{\max }=\sqrt{2c_p{T}_0}$ — total thermal energy fully converted, reached only as $M\to \infty$. At $M=0$, what does $A/A^{\ast }$ equal? ::: $\infty$ — the chamber is infinitely wide compared to the throat in this idealisation.