Chamber-to-exit relation - all quantities as f(M_e, γ): Worked examples

Before anything, one reminder of the vocabulary, so no symbol is used before it is earned:

Definition The symbols you need on this page

$M_{e}$ ::: exit Mach number = exit speed divided by the local speed of sound. A pure number, no units.
$γ$ ::: ratio of specific heats of the exhaust gas (about $1.2$ for rocket products, $1.4$ for air). A pure number.
Subscript $0$ ::: stagnation (chamber) value — what you'd measure if the flow were brought gently to rest.
Subscript $e$ ::: static value actually present at the exit plane.
$A^{*}$ ::: throat area, the cross-section where the flow is exactly sonic ( $M = 1$ ).
$R$ ::: specific gas constant, units $J/(kg\cdotK)$ .
$V_{e}$ ::: exit velocity — the actual speed of the exhaust gas as it leaves the exit plane, units $m/s$ .
$a_{e}$ ::: local speed of sound at the exit, $a_{e} = γ R T_{e}$ , units $m/s$ . (So $M_{e} = V_{e} / a_{e}$ .)
$c_{p}$ ::: specific heat at constant pressure, $c_{p} = γ R / (γ - 1)$ , units $J/(kg\cdotK)$ . It measures how much thermal energy each kilogram of gas stores per degree.
$\overset{m}{˙}$ ::: mass flow rate — kilograms of exhaust passing any cross-section per second, units $kg/s$ .
$g_{0}$ ::: standard gravity, $9.81 m/s^{2}$ , used to convert exhaust speed into Specific Impulse in seconds.

For the whole page we lean on the one grouping that appears everywhere. Give it a name:

The scenario matrix

Everything this topic can throw at you falls into one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	What makes it distinct	Covered by
A. Subsonic exit $M_{e} < 1$	diverging section would slow flow; ratios mild	Ex 1
B. Sonic / throat $M_{e} = 1$	the reference state $A^{}$ itself; $A_{e} / A^{} = 1$	Ex 2
C. Supersonic exit $M_{e} > 1$	the real rocket case; steep drops	Ex 3
D. Degenerate $M_{e} = 0$	chamber = exit; all ratios collapse to $1$	Ex 4
E. Limiting $M_{e} \to \infty$	vacuum limit, $V_{e} \to V_{m a x}$	Ex 5
F. Two-branch trap given area ratio, find $M_{e}$	one $A_{e} / A^{*}$ ⇒ two Mach numbers	Ex 6 (figure)
G. Real-world word problem	full chamber→thrust chain	Ex 7
H. Exam twist $M_{e}$ hidden behind $P_{e} / P_{0}$	invert the pressure relation	Ex 8
I. Sensitivity to $γ$	same $M_{e}$ , different gas	Ex 9

Prerequisites if any cell feels shaky: Isentropic Flow Relations, Area Ratio and Mach Number, Converging-Diverging Nozzle.

Worked example Ex 1 — Cell A: Subsonic exit (

$M_{e} = 0.5$ , $γ = 1.4$ ) Given air, $M_{e} = 0.5$ , $T_{0} = 300 K$ , $P_{0} = 2.00 bar$ . Find $T_{e}$ , $P_{e}$ , $ρ_{e} / ρ_{0}$ .

Forecast: Will the temperature drop a lot or a little? (Guess: only a little — the flow is slow, so it steals only a little thermal energy.)

Compute $ϕ$ . $ϕ = 1 + \frac{0.4}{2} (0.5)^{2} = 1 + 0.2 \cdot 0.25 = 1.05$ . Why this step? Every ratio is a power of $ϕ$ ; get it once.
Temperature. $T_{e} = T_{0} ϕ^{- 1} = 300/1.05 = 285.7 K$ . Why? $T_{e} / T_{0} = ϕ^{- 1}$ by energy conservation.
Pressure. Exponent $- γ / (γ - 1) = - 1.4/0.4 = - 3.5$ . $P_{e} = 2.00 \cdot (1.05)^{- 3.5} = 2.00 \cdot 0.84261 = 1.685 bar$ . Why this step? Pressure follows temperature along an isentrope, but with the extra exponent $- γ / (γ - 1)$ from $P V^{γ} = const$ — so pressure falls faster than temperature.
Density. Exponent $- 1/ (γ - 1) = - 2.5$ . $ρ_{e} / ρ_{0} = (1.05)^{- 2.5} = 0.88474$ . Why this step? Density is fixed once $P$ and $T$ are, via the ideal-gas law $ρ \propto P / T$ ; its exponent is exactly $γ / (γ - 1) - 1 = 1/ (γ - 1)$ .

Verify: Cross-check with the ideal-gas identity $ρ_{e} / ρ_{0} = (P_{e} / P_{0}) / (T_{e} / T_{0}) = 0.84261/0.95238 = 0.88474$ — matches step 4 exactly ✓. Units: all ratios dimensionless ✓; $T_{e} < T_{0}$ and $P_{e} < P_{0}$ as expected for accelerating flow ✓. As forecast, changes are small (< 6% in $T$ ).

Worked example Ex 2 — Cell B: Sonic throat (

$M_{e} = 1$ , $γ = 1.2$ ) Given $M_{e} = 1$ . Find every ratio and the area ratio $A_{e} / A^{*}$ .

Forecast: What must $A_{e} / A^{*}$ equal when the exit is the sonic point? (Guess: exactly 1 — the exit is the throat.)

$ϕ$ at $M = 1$ . $ϕ^{*} = 1 + \frac{0.2}{2} (1)^{2} = 1.1$ . Why? This defines the throat "starred" reference state.
Temperature ratio. $T^{*} / T_{0} = 1/1.1 = 0.9091$ . Why this step? $T / T_{0} = ϕ^{- 1}$ is our master temperature relation; here evaluated at the sonic point to get the throat temperature.
Pressure ratio. exponent $- 1.2/0.2 = - 6$ : $P^{*} / P_{0} = (1.1)^{- 6} = 0.5645$ . Why this step? This particular value (pressure at $M = 1$ ) is the "critical pressure ratio" that tells you when a nozzle chokes — worth computing on its own.
Area ratio. Plug $M_{e} = 1$ into the area–Mach formula: $\frac{A _{e}}{A ^{*}} = \frac{1}{1} [\frac{2}{2.2} \cdot 1.1]^{\frac{2.2}{0.4}} = [1]^{5.5} = 1.$ Why this step? The bracket becomes $\frac{2}{γ + 1} ϕ^{*} = \frac{2}{2.2} \cdot 1.1 = 1$ , and $1$ to any power is $1$ .

Verify: $A_{e} / A^{*} = 1$ exactly — the exit coincides with the throat, as forecast ✓. Sanity: $P^{*} / P_{0} = 0.5645$ is the classic "critical pressure ratio" for $γ = 1.2$ (choked-flow condition) ✓.

Worked example Ex 3 — Cell C: Supersonic exit (

$M_{e} = 2.5$ , $γ = 1.4$ ) Given air, $M_{e} = 2.5$ , $T_{0} = 600 K$ , $R = 287 J/(kg\cdotK)$ . Find $T_{e}$ , $P_{e} / P_{0}$ , $V_{e}$ , $A_{e} / A^{*}$ .

Forecast: Bigger $M_{e}$ than Ex 1 — expect a big temperature drop and a fat area ratio. How fat? Guess before computing.

$ϕ$ . $ϕ = 1 + 0.2 \cdot (2.5)^{2} = 1 + 0.2 \cdot 6.25 = 2.25$ . Why this step? As always, get the master group first so every later ratio is a one-line power of it.
Temperature. $T_{e} = 600/2.25 = 266.7 K$ . Why? Nearly $2.25 \times$ cooler — kinetic energy dominates now.
Pressure. exponent $- 3.5$ : $P_{e} / P_{0} = (2.25)^{- 3.5} = 0.0585$ . Why this step? We need the pressure ratio both to feel the "50-to-1 drop that makes thrust" and to check the flow is fully expanded; it's $ϕ$ raised to $- γ / (γ - 1)$ .
Velocity. $V_{e} = M_{e} \frac{γ R T _{0}}{ϕ} = 2.5 \frac{1.4 \cdot 287 \cdot 600}{2.25}$ . Inside: $1.4 \cdot 287 \cdot 600 = 241, 080$ ; $/2.25 = 107, 147$ ; $= 327.3$ . So $V_{e} = 2.5 \cdot 327.3 = 818.4 m/s$ . Why? $V_{e} = M_{e} a_{e}$ with $a_{e} = γ R T_{e}$ ; the $ϕ$ in the denominator turns chamber temperature into exit temperature.
Area ratio. bracket $= \frac{2}{2.4} \cdot 2.25 = 1.875$ ; exponent $\frac{2.4}{0.8} = 3$ : $A_{e} / A^{*} = \frac{1}{2.5} (1.875)^{3} = \frac{6.592}{2.5} = 2.637$ . Why this step? The flow relations set the thermodynamics; the area ratio translates them into the geometry you must actually machine — for $M_{e} > 1$ the diverging cone has to widen by this factor.

Verify: Standard isentropic tables give $A / A^{*} = 2.637$ at $M = 2.5$ , $γ = 1.4$ ✓. Speed check via $a_{e}$ : $a_{e} = 1.4 \cdot 287 \cdot 266.7 = 327.2$ , and $V_{e} = 2.5 \cdot 327.2 = 818$ ✓ (same as step 4). Units of $V_{e}$ : $J/kg = m^{2} / s^{2} = m/s$ ✓.

Worked example Ex 4 — Cell D: Degenerate zero-Mach (

$M_{e} = 0$ ) Given $M_{e} = 0$ , any $γ$ . What are all the ratios?

Forecast: If the exit flow isn't moving, how does the exit differ from the chamber? (Guess: not at all.)

$ϕ$ . $ϕ = 1 + \frac{γ - 1}{2} (0)^{2} = 1$ . Why? No motion ⇒ no kinetic-energy term.
All ratios. $ϕ^{- 1} = 1$ , $ϕ^{- γ / (γ - 1)} = 1$ , $ϕ^{- 1/ (γ - 1)} = 1$ . So $T_{e} = T_{0}, P_{e} = P_{0}, ρ_{e} = ρ_{0}$ . Why this step? Every ratio is a power of $ϕ$ , and any power of $1$ is $1$ — so with no flow the exit state simply is the chamber state.
Velocity. $V_{e} = 0 \cdot (\dots) = 0$ . Why this step? $V_{e} = M_{e} a_{e}$ ; with $M_{e} = 0$ the exhaust isn't moving, confirming "chamber = exit" physically.
Area ratio. $A_{e} / A^{*} = \frac{1}{0} [\dots]$ → diverges to infinity. Why? To carry the same mass flow at zero speed you'd need an infinitely wide passage. Physically, $M_{e} = 0$ can't be reached in a real supersonic nozzle — it's the mathematical edge.

Verify: At $ϕ = 1$ every power of $1$ is $1$ , so exit = chamber, consistent with "no expansion happened" ✓. The $1/ M_{e}$ blow-up correctly flags $M_{e} = 0$ as unphysical for a flowing nozzle ✓.

Worked example Ex 5 — Cell E: Vacuum limit (

$M_{e} \to \infty$ ) Given $γ = 1.2$ , $T_{0} = 3500 K$ , $R = 520 J/(kg\cdotK)$ . What is the maximum possible exit velocity?

Forecast: As $M_{e}$ grows, $V_{e}$ keeps rising — but does it rise forever, or level off? (Guess: it levels off, because temperature can't drop below absolute zero.)

Take the limit of $V_{e}$ . $V_{e} = M_{e} \frac{γ R T _{0}}{1 + \frac{γ - 1}{2} M _{e}^{2}}$ . For large $M_{e}$ the " $1 +$ " is negligible: $V_{e} \approx M_{e} \frac{γ R T _{0}}{\frac{γ - 1}{2} M _{e}^{2}} = \frac{2 γ R T _{0}}{γ - 1}$ . Why this step? The $M_{e}^{2}$ inside cancels the outer $M_{e}$ , leaving a finite ceiling $V_{m a x}$ .
Number. $V_{m a x} = \frac{2 \cdot 1.2 \cdot 520 \cdot 3500}{0.2} = \frac{4 , 368 , 000}{0.2} = 21, 840, 000 = 4673 m/s$ . Why this step? Plugging the actual chamber numbers turns the abstract ceiling into a concrete "no exhaust can beat 4673 m/s from this gas" — the hard physical limit on performance.
Interpretation. All thermal energy has become directed kinetic energy; $T_{e} \to 0$ . Why this step? It explains why the ceiling exists: you can't extract more speed than the total stored thermal energy $c_{p} T_{0}$ allows, and that pool is finite.

Verify: Energy check: max KE per unit mass $= \frac{1}{2} V_{m a x}^{2} = c_{p} T_{0}$ with $c_{p} = γ R / (γ - 1) = 1.2 \cdot 520/0.2 = 3120$ . Then $\frac{1}{2} (4673)^{2} = 1.092 \times 1 0^{7}$ and $c_{p} T_{0} = 3120 \cdot 3500 = 1.092 \times 1 0^{7}$ ✓ — perfect match. Real nozzles reach maybe $70%$ of this; a finite ceiling exists as forecast ✓.

Worked example Ex 6 — Cell F: The two-branch trap (given

$A_{e} / A^{*}$ , find $M_{e}$ ) Given $γ = 1.4$ and area ratio $A_{e} / A^{*} = 2.0$ . Find $M_{e}$ .

Forecast: One area ratio — how many Mach numbers satisfy it? (Guess before looking at the figure below.)

The figure below plots the area ratio $A_{e} / A^{*}$ (red curve) against the exit Mach number $M_{e}$ (horizontal axis). Notice the curve dips to a single lowest point of exactly $1$ at $M_{e} = 1$ (the black square) and climbs on both sides. The dashed black line is our target height $A_{e} / A^{*} = 2.0$ ; where it crosses the red curve (the two black dots) are the answers.

See the shape. In the figure, the red curve $A_{e} / A^{*}$ vs $M_{e}$ has a minimum of 1 at $M = 1$ and rises on both sides. So a horizontal line at height $2.0$ cuts it twice. Why? A converging-diverging nozzle passes the same mass through the same area either subsonically ( $M < 1$ ) or supersonically ( $M > 1$ ).
Subsonic root. Solving numerically, $M_{e} \approx 0.306$ . Why this step? The left crossing is the subsonic solution — a nozzle that stays below Mach 1 throughout can also produce this area ratio, so we must report it too.
Supersonic root. Solving numerically, $M_{e} \approx 2.197$ . Why? The right crossing is the supersonic solution; the formula is genuinely two-valued, so the geometry alone cannot pick a branch.
Pick physically. A rocket nozzle designed for thrust runs the supersonic branch: $M_{e} = 2.197$ . Why this step? Extra physics (the nozzle is "running full" / chamber pressure is high) selects the supersonic root; without that context both are mathematically valid.

Verify: Plug $M_{e} = 2.197$ : $ϕ = 1 + 0.2 (2.197)^{2} = 1.9654$ ; bracket $= \frac{2}{2.4} (1.9654) = 1.638$ ; $^{3} = 4.394$ ; $/2.197 = 2.000$ ✓. And $M_{e} = 0.306$ : $ϕ = 1.01874$ ; bracket $= 0.8489$ ; $^{3} = 0.6117$ ; $/0.306 = 1.999$ ✓. Both branches confirmed.

Worked example Ex 7 — Cell G: Real-world word problem (full chain to thrust)

Given an upper-stage engine: $γ = 1.2$ , $T_{0} = 3200 K$ , $P_{0} = 50 bar$ , $R = 390 J/(kg\cdotK)$ , design exit $M_{e} = 3.5$ , mass flow $\overset{m}{˙} = 8.0 kg/s$ , expanded to matched exit pressure (ambient $= P_{e}$ ). Find $V_{e}$ and the ideal thrust $F = \overset{m}{˙} V_{e}$ .

Forecast: Roughly how fast does the exhaust leave — hundreds or thousands of m/s? (This drives Specific Impulse.)

$ϕ$ . $ϕ = 1 + 0.1 (3.5)^{2} = 1 + 1.225 = 2.225$ . Why this step? Every downstream quantity ( $T_{e}$ , then $V_{e}$ , then thrust) is built on $ϕ$ , so compute it once up front.
Exit temperature. $T_{e} = 3200/2.225 = 1438 K$ . Why this step? We need $T_{e}$ because exit speed is set by the local speed of sound $a_{e} = γ R T_{e}$ — velocity can't be found without the exit temperature.
Exit velocity. $V_{e} = 3.5 \frac{1.2 \cdot 390 \cdot 3200}{2.225}$ . Inside: $1.2 \cdot 390 \cdot 3200 = 1, 497, 600$ ; $/2.225 = 673, 079$ ; $= 820.4$ . $V_{e} = 3.5 \cdot 820.4 = 2871 m/s$ . Why this step? Thrust is momentum flow $\overset{m}{˙} V_{e}$ , so $V_{e}$ is the pivotal number; here we combine $V_{e} = M_{e} a_{e}$ with the temperature from step 2.
Thrust (pressure-matched). $F = \overset{m}{˙} V_{e} = 8.0 \cdot 2871 = 22, 970 N \approx 23.0 kN$ . Why this step? When exit pressure equals ambient, the Thrust Equation reduces to the momentum term $\overset{m}{˙} V_{e}$ alone.

Verify: Units: $kg/s \cdot m/s = kg\cdotm/s^{2} = N$ ✓. Cross-check $V_{e}$ via $a_{e} = 1.2 \cdot 390 \cdot 1438 = 820.3$ , $V_{e} = 3.5 \cdot 820.3 = 2871$ ✓. Specific impulse sanity: $I_{s p} = V_{e} / g_{0} = 2871/9.81 = 293 s$ , a believable upper-stage value ✓.

Worked example Ex 8 — Cell H: Exam twist (

$M_{e}$ hidden behind $P_{e} / P_{0}$ ) Given $γ = 1.3$ , and you're told only $P_{e} / P_{0} = 0.02$ . Find $M_{e}$ and $T_{e} / T_{0}$ .

Forecast: The Mach number isn't handed to you — you must invert the pressure relation. Do you expect $M_{e}$ above or below 1? (Guess: well above 1, since the pressure fell 50-fold.)

Write the pressure relation. $P_{e} / P_{0} = ϕ^{- γ / (γ - 1)}$ , exponent $- 1.3/0.3 = - 4.333$ . Why this step? This is the only master relation that connects the given ( $P_{e} / P_{0}$ ) to $ϕ$ , which in turn hides $M_{e}$ ; write it before rearranging.
Isolate $ϕ$ . $ϕ = (P_{e} / P_{0})^{- (γ - 1) / γ} = (0.02)^{- 0.3/1.3} = (0.02)^{- 0.2308}$ . Compute: $ln 0.02 = - 3.912$ ; $\times (- 0.2308) = 0.9029$ ; $e^{0.9029} = 2.467$ . So $ϕ = 2.467$ . Why this step? Raising both sides to the reciprocal power undoes the exponent, exposing $ϕ$ .
Solve for $M_{e}$ . $ϕ = 1 + \frac{γ - 1}{2} M_{e}^{2} \Rightarrow M_{e} = \frac{2 ( ϕ - 1 )}{γ - 1} = \frac{2 ( 1.467 )}{0.3} = 9.78 = 3.127$ . Why this step? $ϕ$ is just a rearranged quadratic in $M_{e}$ ; solving it recovers the Mach number the problem tried to hide.
Temperature ratio. $T_{e} / T_{0} = ϕ^{- 1} = 1/2.467 = 0.4054$ . Why this step? Once $ϕ$ is known, the temperature ratio is free — it's the simplest power of $ϕ$ ( $ϕ^{- 1}$ ).

Verify: Forward-check: $ϕ = 2.467$ , $ϕ^{- 4.333}$ : $ln 2.467 = 0.9029$ ; $\times (- 4.333) = - 3.912$ ; $e^{- 3.912} = 0.0200$ ✓ recovers the given $P_{e} / P_{0}$ . $M_{e} = 3.13 > 1$ as forecast ✓.

Worked example Ex 9 — Cell I: Sensitivity to

$γ$ (same $M_{e}$ , two gases) Given $M_{e} = 3.0$ fixed. Compare $γ = 1.2$ vs $γ = 1.4$ for $T_{e} / T_{0}$ and $P_{e} / P_{0}$ .

Forecast: Which gas keeps its exhaust hotter at the same Mach number? (Guess: the lighter one, low $γ$ — it stores energy in more molecular modes.)

$ϕ$ for each. $γ = 1.2$ : $ϕ = 1 + 0.1 \cdot 9 = 1.9$ . $γ = 1.4$ : $ϕ = 1 + 0.2 \cdot 9 = 2.8$ . Why this step? The whole comparison is really "how does $ϕ$ change with $γ$ ?" — everything else is a power of it, so isolate the difference here first.
Temperature ratio. $γ = 1.2$ : $1/1.9 = 0.5263$ . $γ = 1.4$ : $1/2.8 = 0.3571$ . Why? Higher $γ$ ⇒ bigger coefficient $\frac{γ - 1}{2}$ ⇒ bigger $ϕ$ ⇒ colder exit.
Pressure ratio. $γ = 1.2$ : exponent $- 6$ , $(1.9)^{- 6} = 0.02128$ . $γ = 1.4$ : exponent $- 3.5$ , $(2.8)^{- 3.5} = 0.02722$ . Why this step? Both the base $ϕ$ and the exponent $- γ / (γ - 1)$ move with $γ$ , so pressure sensitivity is a genuinely separate calculation from temperature — worth its own line.
Interpretation. Lower $γ$ keeps more thermal energy at the same $M_{e}$ — one reason H₂-rich exhaust (low $γ$ ) earns high Specific Impulse and good Characteristic Velocity c-star. Why this step? It turns two abstract ratios into a design rule: pick the low- $γ$ propellant when you want a hot, energetic exhaust and high specific impulse.

Verify: Matches the parent note's Worked Example 2 table exactly: $0.526, 0.357, 0.0213, 0.0272$ ✓. As forecast, low- $γ$ gas stays hotter ( $0.526 > 0.357$ ) ✓.

Recall Which cell did each example hit?

A subsonic ::: Ex 1 B sonic throat ( $A_{e} / A^{*} = 1$ ) ::: Ex 2 C supersonic ::: Ex 3 D degenerate $M_{e} = 0$ ::: Ex 4 E limit $M_{e} \to \infty$ (vacuum $V_{m a x}$ ) ::: Ex 5 F two-branch area-ratio inversion ::: Ex 6 G word problem → thrust ::: Ex 7 H invert $P_{e} / P_{0}$ to find hidden $M_{e}$ ::: Ex 8 I sensitivity to $γ$ ::: Ex 9