Characteristic velocity c - = P_c A - ṁ — derivation, combustion efficiency measure: Worked examples

This page is the worked-example workshop for Characteristic velocity $c^*$. We do not re-derive the theory here — we use it, and we deliberately walk into every corner: normal numbers, zero inputs, limiting values of $γ$ , a word problem, and an exam trap. If you have not seen the boxed formula yet, read the parent first.

Everything below rests on the two faces of $c^{*}$ :

Intuition Why the two theoretical forms are the same

They look different but are algebraically identical. Start from the right form $\frac{1}{Γ} γ R T_{c}$ and pull $\frac{1}{Γ}$ apart. Flipping the fraction inside the $Γ$ bracket flips the sign of its exponent, turning $\frac{2}{γ + 1}$ into $\frac{γ + 1}{2}$ : $\frac{1}{Γ} = \frac{1}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ Now multiply by $γ R T_{c}$ : $\frac{1}{Γ} γ R T_{c} = \frac{1}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} \cdot γ R T_{c} = (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} R T_{c} .$ The two $γ$ factors cancel cleanly (one from $1/Γ$ , one from inside $γ R T_{c}$ ), leaving $(\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} R T_{c}$ . To recover the left form $\frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\dots}$ , note $\frac{γ R T _{c}}{γ} = \frac{γ R T _{c}}{γ} = \frac{R T _{c}}{γ}$ — so the left form equals $\frac{1}{γ} (\frac{γ + 1}{2})^{\dots} R T_{c}$ , i.e. the left form is $\frac{1}{γ}$ times the right-hand expression above. In other words the truly-equal pair is $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ No stray $γ$ appears or disappears anywhere — every one is accounted for. Left form is handy when you already know $Γ$ ; right form is handy when you don't.

Two symbols you must never mix up:

$A^{*}$ = throat area (the pinch, where the flow is choked). This is the "star" area.
$\overset{m}{˙}$ = mass flow rate in kg/s (the "m-dot" — the little dot means "per second").

The scenario matrix

Every question this topic can throw at you falls into one of these cells. Each example below is tagged with the cell it fills, and every cell carries its own unit sanity-check inside the worked example.

Cell	What makes it tricky	Example
A. Plain measured	just plug into $P_{c} A^{*} / \overset{m}{˙}$	Ex 1
B. Plain theoretical	build $Γ$ first, then $γ R T_{c}$	Ex 2
C. Efficiency	ratio measured / ideal	Ex 3
D. Solve backwards	given $c^{*}$ , find $\overset{m}{˙}$ or $P_{c}$	Ex 4
E. Degenerate: $\overset{m}{˙} \to 0$	limiting behaviour, why it blows up	Ex 5
F. Limiting $γ$	$γ \to 1^{+}$ vs $γ \approx 1.67$ — sanity of $Γ$	Ex 6
G. Design mantra: change $M$	swap propellant, why $H_{2}$ wins	Ex 7
H. Word problem	real test stand, unit soup	Ex 8
I. Exam trap	someone hands you $A_{e}$ or a nozzle detail	Ex 9

We hit every cell A–I. Units matter as much as the arithmetic — most exam blood is lost to units.

The examples

Worked example Example 2 — Cell B: theoretical

$c^{*}$ from chemistry $γ = 1.20$ , $T_{c} = 3200 K$ , $M = 22 g/mol$ . Find $c_{ideal}^{*}$ .

Forecast: Theoretical $c^{*}$ is usually a bit higher than what a real stand measures (real engines lose a few percent). Expect ~ $1500$ – $1600$ m/s.

Gas constant: $R = R_{u} / M$ . Fix the units first: $R_{u} = 8314 J/(kmol\cdotK)$ and $M = 22 g/mol = 22 kg/kmol$ , so $R = 8314/22 = 378 J/(kg\cdotK)$ . Why this step? The formula needs the specific gas constant $R$ (per kg). Matching $R_{u}$ in J/(kmol·K) with $M$ in kg/kmol keeps the prefixes consistent.
Exponent: $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.2}{0.4} = 5.5$ . Why this step? This single power packages the density ratio ( $\frac{1}{γ - 1}$ ) plus the sound-speed from the isentropic relations.
Build $Γ = 1.2 (\frac{2}{2.2})^{5.5} = 1.0954 \times (0.9091)^{5.5} = 1.0954 \times 0.5875 = 0.6436$ . Why this step? $Γ$ collects all the $γ$ -dependence into one factor, so the rest of the formula is just thermochemistry.
Thermochemical part: $γ R T_{c} = 1.2 \times 378 \times 3200 = 1.452 \times 1 0^{6} = 1205 m/s$ . Why this step? This is the "burn hot, make light gas" core — it carries $γ R T_{c}$ so we can divide by $Γ$ directly.
Divide: $c_{ideal}^{*} = \frac{γ R T _{c}}{Γ} = \frac{1205}{0.6436} = 1872 m/s$ . Why this step? The clean identity $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c}$ assembles the two pieces in one division.

Verify (the other form, step by step): Use the left form $c_{ideal}^{*} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{5.5}$ .

$\frac{γ R T _{c}}{γ} = \frac{1205}{1.2} = 1004$ .
$(\frac{γ + 1}{2})^{5.5} = (1.1)^{5.5} = 1.8642$ .
Product: $1004 \times 1.8642 = 1872 m/s$ . ✓

Both forms give $1872$ m/s exactly (to rounding). Note this is a touch above the $1500$ – $1600$ forecast because $γ = 1.2$ with this $R$ is a slightly higher-energy case than the mental estimate — the forecast was a rough guess, and the math is the arbiter.

Worked example Example 5 — Cell E: degenerate input

$\overset{m}{˙} \to 0$ What does $c^{*} = P_{c} A^{*} / \overset{m}{˙}$ do as you throttle the flow toward zero (with $P_{c}$ , $A^{*}$ held)? Is that physical?

Forecast: The formula divides by $\overset{m}{˙}$ . Guess what dividing by "almost nothing" does.

Take the limit: as $\overset{m}{˙} \to 0^{+}$ , $c^{*} = \frac{P _{c} A ^{*}}{m ˙} \to + \infty$ . Why this step? This is the mathematical behaviour of a $1/ \overset{m}{˙}$ curve — it explodes.
Ask if the premise survives: to hold a fixed $P_{c}$ with no propellant flowing is impossible. Real chambers need $\overset{m}{˙}$ and the choking condition ( $M = 1$ , i.e. gas moving at the speed of sound, at the throat — see the $M$ definition above) for the tidy formula to hold at all. Why this step? Cell E teaches that a formula's math and its physical validity are different questions — the limit is real math but forbidden physics.
The honest physical limit: for a fixed propellant, $c^{*}$ is a constant ( $\propto T_{c} / M$ stuff). If $\overset{m}{˙}$ genuinely drops, $P_{c}$ drops in lockstep so the ratio stays put. You cannot move one without the other while choked.

Verify: Numerically with the Ex-1 numbers, halving $\overset{m}{˙}$ from $20$ to $10$ while forcing $P_{c}$ fixed doubles the apparent $c^{*}$ from $2400$ to $4800$ m/s — physically flagging that you have violated the choked coupling. ✓ The figure below makes this concrete.

Reading the figure: the horizontal axis is mass flow $\overset{m}{˙}$ (kg/s), the vertical axis is the apparent $c^{*}$ (km/s), both explicitly labelled on the plot. The coral curve is $c^{*} = P_{c} A^{*} / \overset{m}{˙}$ with $P_{c}$ artificially held fixed. Follow it leftward: as $\overset{m}{˙}$ shrinks into the buttery-shaded danger zone it rockets toward the sky — that vertical wall is the $\overset{m}{˙} \to 0$ blow-up of step 1, annotated on the figure. The mint dot (also labelled) is the only honest reading: the real operating point where $P_{c}$ and $\overset{m}{˙}$ are locked together by choking, giving the sensible $2400$ m/s of Ex 1. The picture's lesson: the blow-up is an artefact of pretending $P_{c}$ stays put — nature never lets you sit on the left of that dot.

Worked example Example 6 — Cell F: limiting values of

$γ$ Probe the theoretical formula across the whole physical range of $γ$ : the soft limit $γ \to 1^{+}$ (heavy, many-atom exhaust) and the hard limit $γ \approx 1.67$ (monatomic gas like helium). Hold $R T_{c} = 1.21 \times 1 0^{6} m^{2} / s^{2}$ (the Ex-2 value) and use the left form $c_{ideal}^{*} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} = \frac{R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}}$ .

Forecast: Which $γ$ gives a bigger $c^{*}$ — the soft low- $γ$ gas or the stiff monatomic one? Guess before computing.

We evaluate $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c} = \frac{R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}}$ at four points: $γ = 1.10$ , $γ = 1.13$ , $γ = 1.30$ , and the monatomic extreme $γ = 1.67$ . Here $R T_{c} = 1.21 \times 1 0^{6} = 1100$ . Why this step? Four points bracket the entire physical range and let us actually see how $Γ$ behaves.
At $γ = 1.10$ : exponent $= \frac{2.1}{0.2} = 10.5$ ; $Γ = 1.1 (2/2.1)^{10.5} = 1.0488 \times 0.6035 = 0.6329$ ; then $c^{*} = \frac{γ R T _{c}}{Γ} = \frac{1.1 \cdot 1100}{0.6329} = \frac{1153.7}{0.6329} = 1823 m/s$ . Why this step? The "soft gas" edge, computed explicitly.
At $γ = 1.13$ : exponent $= \frac{2.13}{0.26} = 8.192$ ; $Γ = 1.13 (2/2.13)^{8.192} = 1.0630 \times 0.6062 = 0.6444$ . At $γ = 1.30$ : exponent $= \frac{2.3}{0.6} = 3.833$ ; $Γ = 1.3 (2/2.3)^{3.833} = 1.1402 \times 0.5254 = 0.5991$ ; $c^{*} = \frac{1.3 \cdot 1100}{0.5991} = \frac{1254.2}{0.5991} = 2094 m/s$ . Why this step? $Γ$ rises from $0.6329$ ( $γ = 1.10$ ) to $0.6444$ ( $γ = 1.13$ ) then falls to $0.5991$ ( $γ = 1.30$ ) — a genuine hump. Note that $c^{*}$ itself keeps rising because the $γ$ factor grows; the hump is in $Γ$ , not in $c^{*}$ .
Monatomic extreme $γ = 1.67$ : exponent $= \frac{2.67}{1.34} = 1.993$ ; $Γ = 1.67 (2/2.67)^{1.993} = 1.2923 \times 0.5619 = 0.7262$ ; $c^{*} = \frac{1.67 \cdot 1100}{0.7262} = \frac{1421.5}{0.7262} = 1958 m/s$ . Why this step? Closes the high- $γ$ end: the stiff monatomic gas gives a lower $c^{*}$ than the $γ = 1.30$ peak but still above the softest gas — the interplay of $γ$ (rising) and the $Γ$ -bracket (humped) produces a broad maximum near $γ \approx 1.3$ .
Sanity of $Γ$ itself across the whole sweep: $Γ$ stays bounded in $\approx [0.60, 0.73]$ and never touches $0$ or blows up for any $γ \in (1, 1.67]$ . Why this step? A formula is only trustworthy if it is well-behaved on its whole domain — checking that $Γ$ has no poles or zeros confirms $c_{ideal}^{*}$ is finite and positive for every real propellant.

Verify: The four $Γ$ values $0.6329, 0.6444, 0.5991, 0.7262$ (all checked in the appendix) confirm the hump near $γ \approx 1.13$ in $Γ$ and the bounded range. Resulting $c^{*}$ : $1823, 2094, 1958$ m/s for $γ = 1.10, 1.30, 1.67$ — a broad peak, not a monotone slide. ✓

Worked example Example 7 — Cell G: swap the propellant (design mantra)

Keep $T_{c} = 3200 K$ , $γ = 1.20$ but switch exhaust from $M = 22$ g/mol to a hydrogen-rich $M = 10$ g/mol. What happens to $c^{*}$ ?

Forecast: The mantra says "make light gases." How much does more-than-halving $M$ buy you?

$c_{ideal}^{*} \propto γ R T_{c} = γ R_{u} T_{c} / M$ , so $c^{*} \propto 1/ M$ at fixed $T_{c}, γ$ . Why this step? Everything else ( $Γ$ , $T_{c}$ , $γ$ ) is unchanged, so only the $1/ M$ factor moves.
Ratio: $\frac{c _{10}^{*}}{c _{22}^{*}} = \frac{22}{10} = 2.2 = 1.483$ . Why this step? Direct proportionality lets us skip re-computing $Γ$ .
New value: $c_{10}^{*} = 1872 \times 1.483 = 2776 m/s$ . Why this step? A $48%$ jump in combustion score just by making the exhaust lighter — this is exactly why $H_{2}$ propellants dominate high-performance stages.

Verify: Recompute from scratch. $R = 8314/10 = 831.4$ ; $γ R T_{c} = 1.2 \times 831.4 \times 3200 = 3.193 \times 1 0^{6} = 1787$ ; $Γ = 0.6436$ (unchanged); $c^{*} = 1787/0.6436 = 2776 m/s$ . ✓ Matches the ratio method. Units of the ratio: dimensionless (g/mol over g/mol). ✓

Worked example Example 8 — Cell H: full word problem (real test stand)

An engineer logs a $12$ -second burn. The chamber held $P_{c} = 3.5 MPa$ steady, throat diameter $d^{*} = 90 mm$ , and the propellant tanks lost $216 kg$ of mass over the burn. The predicted (ideal) $c^{*}$ for this propellant is $1720 m/s$ . Report $c_{measured}^{*}$ and the combustion efficiency.

Forecast: Efficiency should land in the healthy $0.9$ – $0.99$ range if the engine is any good.

Mass flow: $\overset{m}{˙} = \frac{216 kg}{12 s} = 18 kg/s$ . Why this step? $\overset{m}{˙}$ is total mass burned over burn time — the tanks tell you this directly.
Throat area from diameter: $A^{*} = π (d^{*} /2)^{2} = π (0.045)^{2} = 6.362 \times 1 0^{- 3} m^{2}$ . Why this step? You were handed a diameter in mm — must convert to m and turn into area, or units collapse.
Measured $c^{*} = \frac{P _{c} A ^{*}}{m ˙} = \frac{( 3.5 \times 1 0 ^{6} ) ( 6.362 \times 1 0 ^{- 3} )}{18} = \frac{22267}{18} = 1237 m/s$ . Why this step? The measured definition — every quantity here is chamber/throat, nothing from the nozzle.
Efficiency: $η_{c^{*}} = 1237/1720 = 0.719 = 72%$ . Why this step? Real-over-ideal. This is low — a red flag.

Verify: $72%$ is far below the healthy $0.92$ – $0.99$ band → this engine has serious incomplete combustion or a mis-measured throat/flow. The math checks ( $0.719 \times 1720 = 1237$ ✓), but the engineering verdict is "investigate." ✓ Units all resolve to m/s and dimensionless.

Worked example Example 9 — Cell I: the exam trap (wrong area)

Exam gives: $P_{c} = 5.0 MPa$ , throat area $A^{*} = 0.010 m^{2}$ , exit area $A_{e} = 0.080 m^{2}$ , $\overset{m}{˙} = 25 kg/s$ . Compute $c^{*}$ . What is the trap?

Forecast: Two areas are dangled. Which one belongs in $c^{*}$ — and what wrong answer does the trap produce?

Identify the trap: $c^{*}$ uses the throat area $A^{*}$ , because the flow is choked there ( $M = 1$ ). The exit area $A_{e}$ belongs to the nozzle and only affects the thrust coefficient $C_F$. Why this step? Grabbing the bigger "obvious" area is the single most common $c^{*}$ error.
Correct: $c^{*} = \frac{( 5.0 \times 1 0 ^{6} ) ( 0.010 )}{25} = 2000 m/s$ . Why this step? Same as parent Ex 1 — throat only.
The trap answer (using $A_{e}$ ): $\frac{( 5.0 \times 1 0 ^{6} ) ( 0.080 )}{25} = 16000 m/s$ — physically absurd (eight times too big; no chemistry gives that). Why this step? Seeing the wrong number is what trains you to reject it on sight.

Verify: Correct $c^{*} = 2000$ m/s (matches parent). The trap gives $16000$ m/s = $8 \times$ too large, exactly the ratio $A_{e} / A^{*} = 8$ . ✓ Sanity: any $c^{*}$ above ~ $3000$ m/s for chemical propellants should trigger suspicion.

Recall Which area, and why?

In $c^{*} = P_{c} A^{*} / \overset{m}{˙}$ , why must the area be the throat, not the exit? ::: Because the flow is choked (sonic, $M = 1$ ) at the throat — that is where $\overset{m}{˙}$ locks to $(P_{c}, T_{c}, A^{*})$ . The exit area only governs $C_{F}$ (nozzle expansion).

Recall Degenerate

$\overset{m}{˙}$ As $\overset{m}{˙} \to 0$ the formula $\to \infty$ — is that a real prediction? ::: No. Holding $P_{c}$ fixed while $\overset{m}{˙} \to 0$ violates the choked coupling; physically $P_{c}$ falls with $\overset{m}{˙}$ so $c^{*}$ stays a constant set by $T_{c}, M, γ$ .

Recall Design mantra check

Halving exhaust molecular mass $M$ multiplies $c^{*}$ by what factor? ::: $2 \approx 1.41$ , since $c^{*} \propto 1/ M$ .

Recall Which

$γ$ wins? Across $γ \in (1, 1.67]$ at fixed $R T_{c}$ , is the peak $c^{*}$ at the soft end, the stiff end, or in the middle? ::: In the middle — $c^{*}$ rose to a broad maximum near $γ \approx 1.3$ ( $2094$ m/s) and was lower at both $γ = 1.10$ ( $1823$ ) and $γ = 1.67$ ( $1958$ ).