Characteristic velocity c - and its relation to flame temperature, MW: Worked examples

Before touching numbers, let us name every symbol on this page so nothing is used blind:

Definition The symbols we will use (physical meaning of each)

$c^{*}$ — characteristic velocity (m/s): a quality score for the chamber + propellant alone.
$p_{0}$ — chamber (stagnation) pressure (Pa): the pressure built up inside the combustion chamber before the gas escapes.
$A_{t}$ — throat area (m²): the cross-section of the narrowest point of the nozzle, where the flow chokes.
$\overset{m}{˙}$ — mass flow rate (kg/s): how many kilograms of propellant leave per second.
$M$ — molecular weight of the exhaust (kg/mol): the mass of one mole of gas; small = light, zippy molecules.
$γ$ — specific heat ratio (dimensionless): how "springy" the gas is; ~1.1–1.67 for combustion gases.
$R_{u}$ — universal gas constant $= 8.314 J/mol\cdotK$ : the same for every gas.
$T_{0}$ — chamber (flame) temperature (K): how hot the burned gas is.
$Γ$ — Vandenkerckhove function (dimensionless): a bundle of all the $γ$ -dependence.
$C_{F}$ — thrust coefficient (dimensionless): the nozzle's amplification of chamber performance.
$c_{e}$ — effective exhaust velocity (m/s): the real average speed of the exhaust jet, given by $c_{e} = c^{*} C_{F}$ (chamber quality × nozzle amplification).
$I_{s p}$ — specific impulse (s): thrust per unit weight-flow of propellant; a fuel-economy score, $I_{s p} g_{0} = c_{e}$ .

Now recall the two faces of the same quantity:

The scenario matrix

Every problem this topic can pose sits in one of these cells. The examples below each carry a (Cell …) tag so you can see the whole grid is covered.

Cell	Case class	What's given → what's asked	Which form
A	Forward (chemistry)	$T_{0}, M, γ$ → $c^{*}$	theoretical
B	Measured (test stand)	$p_{0}, A_{t}, \overset{m}{˙}$ → $c^{*}$	measured
C	Backward for $\overset{m}{˙}$	$c^{*}, p_{0}, A_{t}$ → $\overset{m}{˙}$	measured, inverted
D	Backward for $T_{0}$	$c^{*}, M, γ$ → $T_{0}$	theoretical, inverted
E	Efficiency / two-form clash	measured vs theoretical → $η_{c^{*}}$	both
F	Limiting $γ$	$γ \to 1$ and $γ$ large → $Γ$ behaviour	$Γ$ alone
G	Degenerate input	$\overset{m}{˙} \to 0$ , $A_{t} \to 0$ , $T_{0} \to 0$	sanity of formula
H	Real-world word problem	mixture ratio shift → $c^{*}$ trade	theoretical, ratio
I	Exam twist (chain to $I_{s p}$ )	$c^{*}, C_{F}$ → $c_{e}, I_{s p}$	link forms

We now walk every cell.

Example 1 — Forward from chemistry (Cell A)

Worked example Kerosene/LOX chamber

Given $T_{0} = 3500 K$ (flame temperature), $M = 22 g/mol = 0.022 kg/mol$ (molecular weight), $γ = 1.22$ (specific heat ratio), $R_{u} = 8.314 J/mol\cdotK$ (universal gas constant). Find $c^{*}$ .

Forecast: heavier gas than $H_{2} / O_{2}$ (22 vs 16) → expect $c^{*}$ lower than the parent's 2143 m/s. Guess ~1800 m/s.

Step 1 — Compute $Γ$ . $Γ = 1.22 (\frac{2}{2.22})^{\frac{2.22}{0.44}} = 1.1045 \times (0.9009)^{5.045}$ . Why this step? $Γ$ packages all the $γ$ -dependence; compute it once, reuse it. $(0.9009)^{5.045} \approx 0.5905$ , so $Γ \approx 0.6522$ .

Step 2 — Compute the "hot & light" term. $R_{u} T_{0} / M = 8.314 \times 3500/0.022 = 1.3227 \times 1 0^{6} = 1150.1 m/s$ . Why this step? This square-root term is the raw energy content per unit mass; $Γ$ only scales it.

Step 3 — Divide. $c^{*} = 1150.1/0.6522 \approx 1763 m/s$ . Why this step? $c^{*} = \frac{1}{Γ} R_{u} T_{0} / M$ — the right face of our master formula.

Verify: Ratio to the parent's $H_{2} / O_{2}$ value: $1763/2143 = 0.82$ . The only big change was $M$ up from 16→22, i.e. $16/22 = 0.853$ ; the small $T_{0}$ drop ( $3500/3600 = 0.986$ ) and $Γ$ shift explain the rest. Units: $J/kg = m^{2} / s^{2} =$ m/s. ✓

Example 2 — Measured on the stand (Cell B)

Worked example Read

$c^{*}$ off gauges A firing shows $p_{0} = 6.5 MPa$ (chamber pressure), throat diameter $d_{t} = 50 mm$ (so throat area $A_{t}$ follows), $\overset{m}{˙} = 18 kg/s$ (mass flow). Find $c^{*}$ .

Forecast: big throat, high pressure — chamber looks healthy. Guess ~700 m/s (this is the measured face; it can be far below theoretical if combustion is poor).

Step 1 — Throat area. $A_{t} = \frac{π}{4} d_{t}^{2} = \frac{π}{4} (0.050)^{2} = 1.9635 \times 1 0^{- 3} m^{2}$ . Why this step? The definition needs area, not diameter; a round throat gives $\frac{π}{4} d^{2}$ .

Step 2 — Apply the left face. $c^{*} = \frac{p _{0} A _{t}}{m ˙} = \frac{6.5 \times 1 0 ^{6} \times 1.9635 \times 1 0 ^{- 3}}{18}$ . Why this step? No nozzle data needed — this is exactly why $c^{*}$ was defined this way (see Choked Flow and the Throat: the throat is where the flow chokes and $\overset{m}{˙}$ locks in).

Step 3 — Arithmetic. Numerator $= 12763 N$ ; $c^{*} = 12763/18 \approx 709 m/s$ . Why this step? Just the final divide.

Verify: Units: $\frac{Pa \cdot m ^{2}}{kg/s} = \frac{N}{kg/s} = \frac{kg\cdotm/s ^{2}}{kg/s} =$ m/s. ✓ (This low value would flag a diagnostic — heavy products or bad mixing.)

Example 3 — Backward for mass flow (Cell C)

Worked example Size the propellant feed

You want $c^{*} = 1750 m/s$ at $p_{0} = 5 MPa$ (chamber pressure) through a throat of diameter $30 mm$ . What mass flow $\overset{m}{˙}$ does the chamber demand?

Forecast: rearranging $c^{*} = p_{0} A_{t} / \overset{m}{˙}$ gives $\overset{m}{˙} = p_{0} A_{t} / c^{*}$ . Higher $c^{*}$ (good propellant) → less mass flow for the same push. Guess a few kg/s.

Step 1 — Throat area. $A_{t} = \frac{π}{4} (0.030)^{2} = 7.069 \times 1 0^{- 4} m^{2}$ . Why this step? Same geometry rule as Example 2.

Step 2 — Invert the definition. $\overset{m}{˙} = \frac{p _{0} A _{t}}{c ^{*}} = \frac{5 \times 1 0 ^{6} \times 7.069 \times 1 0 ^{- 4}}{1750}$ . Why this step? $c^{*}$ is a ratio; solving for any one of its three variables is pure algebra.

Step 3 — Arithmetic. Numerator $= 3534.3 N$ ; $\overset{m}{˙} = 3534.3/1750 \approx 2.02 kg/s$ .

Verify: Plug back: $c^{*} = 5 \times 1 0^{6} \times 7.069 \times 1 0^{- 4} /2.02 = 1750$ m/s. ✓ Units: $\frac{Pa\cdotm ^{2}}{m/s} = \frac{N}{m/s} =$ kg/s. ✓

Example 4 — Backward for flame temperature (Cell D)

Worked example Infer

$T_{0}$ from measured $c^{*}$ A monopropellant gives measured $c^{*} = 1200 m/s$ , with $M = 0.020 kg/mol$ (molecular weight), $γ = 1.25$ (specific heat ratio). Estimate the chamber temperature $T_{0}$ (assume perfect combustion, so measured $=$ theoretical).

Forecast: $c^{*} \propto T_{0}$ so $T_{0} \propto (c^{*})^{2}$ — squaring a ~1200 m/s value hints at a few thousand K.

Step 1 — $Γ$ . $Γ = 1.25 (\frac{2}{2.25})^{\frac{2.25}{0.5}} = 1.1180 \times (0.8889)^{4.5} = 1.1180 \times 0.6288 = 0.7030$ . Why this step? We need $Γ$ before we can isolate $T_{0}$ ; it depends only on $γ$ .

Step 2 — Rearrange for $T_{0}$ . From $c^{*} = \frac{1}{Γ} R_{u} T_{0} / M$ , square both sides: $T_{0} = \frac{( c ^{*} Γ ) ^{2} M}{R _{u}} .$ Why this step? The square-root is the only obstacle; squaring undoes it and lets us solve linearly for $T_{0}$ .

Step 3 — Plug in. $c^{*} Γ = 1200 \times 0.7030 = 843.6$ ; squared $= 7.117 \times 1 0^{5}$ ; times $M / R_{u}$ : $T_{0} = \frac{7.117 \times 1 0 ^{5} \times 0.020}{8.314} \approx 1712 K$ .

Verify: Forward-check with $T_{0} = 1712$ : $8.314 \times 1712/0.020 = 7.117 \times 1 0^{5} = 843.6$ ; divide by $Γ = 0.7030$ → $1200$ m/s. ✓ (See Adiabatic Flame Temperature — this is how a stand estimate cross-checks the thermochemistry.)

Example 5 — Efficiency, the two forms clash (Cell E)

Example 6 — Limiting behaviour of $Γ$ (Cell F)

Here the geometry of the function matters, so we look at a picture.

Figure — Characteristic velocity c - and its relation to flame temperature, MW

Figure 1 — $Γ$ versus $γ$ . The horizontal axis is the specific heat ratio $γ$ (1.05 to 1.67); the vertical axis is the Vandenkerckhove function $Γ$ . The blue curve is $Γ (γ)$ , rising gently from about 0.62 to 0.73. Three labelled dots sit on the curve — red at $γ = 1.10, Γ = 0.6255$ ; yellow at $γ = 1.40, Γ = 0.6847$ ; green at $γ = 1.67, Γ = 0.7262$ . A faint white horizontal band marks the real-gas range $Γ \approx 0.63$ – $0.73$ , and the curve stays inside it across the whole plot — the visual proof that $γ$ affects $c^{*} \propto 1/Γ$ only weakly.

Worked example What happens to

$Γ$ as $γ \to 1$ and $γ$ grows? Evaluate $Γ$ at $γ = 1.10, 1.40, 1.67$ and describe the trend.

Forecast: the figure shows $Γ$ rising gently with $γ$ . Real gases sit in a narrow band $Γ \approx 0.63$ – $0.73$ .

Step 1 — $γ = 1.10$ (red dot). $Γ = 1.10 (\frac{2}{2.10})^{\frac{2.10}{0.20}} = 1.0488 \times (0.95238)^{10.5} = 1.0488 \times 0.5964 = 0.6255$ . Why this step? The exponent $\frac{γ + 1}{2 ( γ - 1 )}$ blows up as $γ \to 1$ (denominator $\to 0$ ), so we must evaluate carefully rather than "take a limit in your head."

Step 2 — $γ = 1.40$ (yellow dot, air). $Γ = 1.40 (\frac{2}{2.40})^{\frac{2.40}{0.80}} = 1.1832 \times (0.83333)^{3} = 1.1832 \times 0.5787 = 0.6847$ . Why this step? $γ = 1.40$ is the everyday value for diatomic air; checking it anchors the curve at a value most readers already know, and confirms $Γ$ has climbed from the $γ = 1.10$ result.

Step 3 — $γ = 1.67$ (green dot, monatomic, e.g. He). $Γ = 1.67 (\frac{2}{2.67})^{\frac{2.67}{1.34}} = 1.2923 \times (0.74906)^{1.9925} = 1.2923 \times 0.5619 = 0.7262$ . Why these three? They bracket every real combustion gas. The trend: $Γ$ increases with $γ$ , but slowly — the whole span is 0.63→0.73.

Verify: Because $c^{*} \propto 1/Γ$ , a larger $γ$ (larger $Γ$ ) gives a smaller $c^{*}$ for the same $T_{0} / M$ . The parent note said " $γ$ enters weakly" — the figure and these three numbers prove the band is tight. ✓ (See Vandenkerckhove Function Γ and Isentropic Flow Relations.)

Example 7 — Degenerate & zero inputs (Cell G)

Worked example What do the edge cases say?

Discuss $\overset{m}{˙} \to 0$ , $A_{t} \to 0$ , and $T_{0} \to 0$ using the two forms.

Forecast: dividing by zero should either blow up or give a physically-meaningful "nothing".

Step 1 — $\overset{m}{˙} \to 0$ at fixed $p_{0}, A_{t}$ . $c^{*} = p_{0} A_{t} / \overset{m}{˙} \to + \infty$ . Why this step? Physically impossible with real combustion — it says "infinite quality from a whisper of mass," a warning that you cannot hold $p_{0}$ up while starving the flow. A real chamber's $p_{0}$ collapses if $\overset{m}{˙} \to 0$ , so the ratio stays finite. Lesson: the three variables are not independent.

Step 2 — $A_{t} \to 0$ . For fixed $\overset{m}{˙}$ , choking requires the pressure to climb without bound; $c^{*} = p_{0} A_{t} / \overset{m}{˙}$ stays finite because $p_{0} \propto 1/ A_{t}$ . The formula is scale-invariant in throat size. Why this step? Confirms $c^{*}$ is a material property, not a geometry property — shrinking the hole doesn't change propellant quality.

Step 3 — $T_{0} \to 0$ . Theoretical form: $c^{*} = \frac{1}{Γ} R_{u} T_{0} / M \to 0$ . Why this step? Cold gas has no thermal energy to push with — $c^{*}$ correctly vanishes. Sanity confirmed.

Verify: All three limits are consistent between the two faces once you remember the physical coupling $p_{0} (T_{0}, \overset{m}{˙})$ . No contradiction. ✓

Example 8 — Real-world word problem: mixture-ratio trade (Cell H)

Worked example Should we run fuel-rich?

An $H_{2} / O_{2}$ engine at stoichiometric burns to $T_{0} = 3600 K$ , $M = 18 g/mol$ (pure $H_{2} O$ ). Running fuel-rich drops $T_{0}$ to $3300 K$ but lowers $M$ to $14 g/mol$ (excess $H_{2}$ ). Take $γ = 1.22$ (so $Γ = 0.6522$ from Example 1's $γ$ ) for both. Which gives higher $c^{*}$ ?

Forecast: the parent note claims fuel-rich wins because $1/ M$ beats the $T_{0}$ loss. Let's confirm with numbers.

Step 1 — Stoichiometric $c^{*}$ . $8.314 \times 3600/0.018 = 1.6628 \times 1 0^{6} = 1289.5$ ; $c^{*} = 1289.5/0.6522 = 1977 m/s$ . Why this step? Baseline case, right face of the formula.

Step 2 — Fuel-rich $c^{*}$ . $8.314 \times 3300/0.014 = 1.9597 \times 1 0^{6} = 1399.9$ ; $c^{*} = 1399.9/0.6522 = 2147 m/s$ . Why this step? Same formula, new $T_{0}, M$ .

Step 3 — Compare. Fuel-rich is higher: $2147 > 1977$ , a $+ 8.6 %$ gain. Why this step? The ratio test: $\frac{c _{r i c h}^{*}}{c _{s t o i c h}^{*}} = \frac{3300/14}{3600/18} = \frac{235.7}{200} = 1.179 = 1.086$ — exactly the $+ 8.6 %$ , and $Γ$ cancels.

Verify: $1.179 = 1.0857$ ; $1977 \times 1.0857 = 2146.5 \approx 2147$ ✓. Conclusion: run fuel-rich — the molecular-weight lever beats the temperature penalty, precisely as Propellant Selection and Molecular Weight predicts.

Example 9 — Exam twist: chain $c^{*}$ into $I_{s p}$ (Cell I)

Worked example From chamber grade to delivered performance

A chamber has $c^{*} = 1763 m/s$ (Example 1). The nozzle delivers $C_{F} = 1.65$ (thrust coefficient). Take $g_{0} = 9.81 m/s^{2}$ . Find the effective exhaust velocity $c_{e}$ and specific impulse $I_{s p}$ .

Forecast: parent says $c_{e} = c^{*} C_{F}$ and $c_{e} \approx 1.5$ – $1.9 \times c^{*}$ . With $C_{F} = 1.65$ expect $c_{e} \approx 2900$ m/s and $I_{s p} \approx 300$ s.

Step 1 — Effective exhaust velocity. $c_{e} = c^{*} C_{F} = 1763 \times 1.65 = 2909 m/s$ . Why this step? The clean split: chamber ( $c^{*}$ ) times nozzle (Thrust Coefficient CF). This is the whole reason $c^{*}$ was invented.

Step 2 — Specific impulse. $I_{s p} = c_{e} / g_{0} = 2909/9.81 = 296.5 s$ . Why this step? $I_{s p} g_{0} = c_{e}$ is the definition (see Specific Impulse Isp); dividing by $g_{0}$ converts velocity to seconds.

Verify: $296.5 \times 9.81 = 2909$ ✓; and $c_{e} / c^{*} = 2909/1763 = 1.65 = C_{F}$ ✓. Units: m/s ÷ (m/s²) = s ✓. All three links consistent.

Recall Self-test: match each to a matrix cell

Which cell asks you to invert for $\overset{m}{˙}$ ? ::: Cell C (Example 3). Which cell shows $c^{*} \to 0$ makes physical sense? ::: Cell G, the $T_{0} \to 0$ limit (Example 7). Which cell proves fuel-rich wins numerically? ::: Cell H (Example 8). Which cell links $c^{*}$ to $I_{s p}$ ? ::: Cell I (Example 9). In which cell does $Γ$ cancel out of the answer? ::: Cell H — a ratio of two $c^{*}$ at equal $γ$ .

Connections

3.3.21 Characteristic velocity c - and its relation to flame temperature, MW (Hinglish) — Hinglish parent.
Thrust Coefficient CF — used in Example 9's $c_{e} = c^{*} C_{F}$ .
Specific Impulse Isp — Example 9's $I_{s p} g_{0} = c_{e}$ .
Choked Flow and the Throat — why the measured face (Cell B) needs no nozzle.
Isentropic Flow Relations — origin of $Γ$ (Cell F).
Adiabatic Flame Temperature — the $T_{0}$ we inferred in Cell D.
Propellant Selection and Molecular Weight — the mixture-ratio trade of Cell H.
Vandenkerckhove Function Γ — plotted in Cell F.