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

WHAT is $c^*$ ?

WHY define it this way? Because $p_0$ , $A_t$ , $\dot m$ are the three things you can measure on a test stand without knowing anything about the nozzle exit. It isolates chamber performance.

The thrust coefficient $C_F$ handles the nozzle; specific impulse splits cleanly: $I_{sp}\,g_0 = c_e = c^* \, C_F$ So $c^*$ = chemistry & chamber, $C_F$ = nozzle. This separation is the reason $c^*$ exists.

HOW to derive $c^*$ from first principles

We use 1-D isentropic flow of a perfect gas that is choked (Mach 1) at the throat.

Step 1 — Mass flow through a choked throat. $\dot m = \rho_t A_t v_t$ Why this step? Continuity: mass in = mass out through the throat area.

Step 2 — At the throat $M=1$ , so $v_t = a_t = \sqrt{\gamma R_s T_t}$ where $R_s = R_u/\mathcal{M}$ is the specific gas constant. Why? Choking means the flow reaches the local speed of sound at the minimum area.

Step 3 — Relate throat conditions to chamber (stagnation) conditions. For isentropic flow with $M=1$ : $\frac{T_t}{T_0} = \frac{2}{\gamma+1}, \qquad \frac{p_t}{p_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}},\qquad \frac{\rho_t}{\rho_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{1}{\gamma-1}}$ Why? These come from the isentropic energy equation $T_0/T = 1+\frac{\gamma-1}{2}M^2$ evaluated at $M=1$ .

Step 4 — Assemble $\dot m$ . Using $\rho_0 = p_0/(R_s T_0)$ : $\dot m = \underbrace{\rho_0\left(\tfrac{2}{\gamma+1}\right)^{\frac1{\gamma-1}}}_{\rho_t} A_t \underbrace{\sqrt{\gamma R_s T_t}}_{v_t}$ Substitute $T_t = T_0\frac{2}{\gamma+1}$ and $\rho_0=p_0/(R_sT_0)$ and simplify: $\dot m = \frac{p_0 A_t \,\gamma}{\sqrt{\gamma R_s T_0}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$

Step 5 — Invert into $c^* = p_0A_t/\dot m$ :

Why this final form matters: it puts the dependence naked in front of you.

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

Worked Examples

Worked example Example 1 — Compute

$c^*$ for a $H_2/O_2$ engine Given $T_0 = 3600\text{ K}$ , $\mathcal{M} = 16\text{ g/mol}=0.016\text{ kg/mol}$ , $\gamma=1.20$ , $R_u=8.314\text{ J/mol·K}$ .

Step 1 — $\Gamma$ : $\Gamma=\sqrt{1.2}\left(\frac{2}{2.2}\right)^{\frac{2.2}{0.4}} = 1.0954\times(0.9091)^{5.5}$ . Why? Plug $\gamma$ into Vandenkerckhove. $(0.9091)^{5.5}\approx0.5827$ , so $\Gamma\approx 0.6384$ .

Step 2 — $\sqrt{R_uT_0/\mathcal{M}}$ : $\sqrt{8.314\times3600/0.016}=\sqrt{1.8707\times10^{6}}=1367.7$ m/s. Why? This is the "hot & light" energy term.

Step 3 — Combine: $c^* = 1367.7/0.6384 \approx \mathbf{2143\ m/s}$ . Why? $c^*=\frac1\Gamma\sqrt{R_uT_0/\mathcal M}$ . (Real $H_2/O_2$ hits ~2350 m/s with lower $\mathcal M$ ; matches ballpark.)

Worked example Example 2 — Measure

$c^*$ on a test stand A motor shows $p_0=5\text{ MPa}$ , throat diameter $d_t=40\text{ mm}$ , $\dot m=12\text{ kg/s}$ . Find $c^*$ .

Step 1 — Throat area: $A_t=\frac{\pi}{4}(0.040)^2=1.257\times10^{-3}\text{ m}^2$ . Why? Circular throat.

Step 2 — Apply definition: $c^*=\dfrac{p_0A_t}{\dot m}=\dfrac{5\times10^{6}\times1.257\times10^{-3}}{12}$ . Why? This is the measured $c^*$ , needing no nozzle data.

Step 3: $c^*=\dfrac{6283}{12}\approx \mathbf{524\ m/s}$ (low → poor combustion / heavy products).

Common Mistakes (Steel-manned)

Flashcards

Define

c^*

operationally

c^* = p_0 A_t/\dot m

— chamber pressure × throat area ÷ mass flow.

What does

c^*

isolate?

Combustion chamber & propellant quality, independent of the nozzle.

c^*

scales with

T_0

and

\mathcal M

how?

c^*\propto\sqrt{T_0/\mathcal M}

(hot & light gas wins).

Why only

\sqrt{T_0}

dependence?

Comes from

v\propto a\propto\sqrt{T}

(speed of sound

\propto\sqrt T

Split of exhaust velocity

c_e = c^* C_F

(chamber × nozzle).

Vandenkerckhove function

\Gamma

\Gamma=\sqrt\gamma\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}

c^*

in terms of

R_u,\mathcal M

c^*=\frac1\Gamma\sqrt{R_uT_0/\mathcal M}

Why fuel-rich

H_2/O_2

maximizes

c^*

Excess

H_2

lowers mean

\mathcal M

, boosting

\sqrt{1/\mathcal M}

more than the small

T_0

loss costs.

Define

c^*

efficiency

\eta_{c^*}=c^*_{measured}/c^*_{theoretical}

; <1 indicates incomplete combustion/heat loss.

Which condition holds at the throat in the derivation?

M=1

(choked flow),

v_t=a_t

Recall Feynman: explain to a 12-year-old

Imagine you're blowing up a balloon and letting it fly. $c^*$ measures how good your "puff" is — how much push-pressure you build up in the balloon before it even reaches the mouth. Two things make a great puff: the air should be really hot (energetic) and made of light, zippy little molecules (like helium instead of heavy air). Hot + light = they bounce around super fast and push hard. The shape of the balloon's mouth (the nozzle) is a separate story — $c^*$ only grades the balloon's inside.

Connections

Thrust Coefficient CF — the nozzle half; $c_e=c^*C_F$ .
Specific Impulse Isp — $I_{sp}g_0=c^*C_F$ .
Choked Flow and the Throat — the $M=1$ condition used in Step 2.
Isentropic Flow Relations — source of the $\frac{2}{\gamma+1}$ ratios.
Adiabatic Flame Temperature — where $T_0$ comes from.
Propellant Selection and Molecular Weight — why we chase low $\mathcal M$ .
Vandenkerckhove Function Γ — the $\gamma$ -only lumped factor.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, $c^*$ (c-star) ko samajhna simple hai: yeh sirf tumhare combustion chamber ka report card hai — nozzle se koi lena-dena nahi. Definition hai $c^* = p_0 A_t/\dot m$ , matlab kitna chamber pressure ban raha hai per unit mass flow jo throat se nikalta hai. Test stand pe yeh teen cheezein easily naap sakte ho, isiliye engineer log chamber quality judge karne ke liye $c^*$ use karte hain.

Ab asli physics: derive karne pe milta hai $c^* = \frac{1}{\Gamma}\sqrt{R_u T_0/\mathcal{M}}$ . Iska matlab clear hai — garam aur halka gas jeet-ta hai. Flame temperature $T_0$ zyada ho toh $c^*$ badhta hai, par sirf $\sqrt{T_0}$ ke hisaab se (double temperature = sirf 1.41 times). Aur molecular weight $\mathcal{M}$ kam ho toh $c^*$ zyada, kyunki halke molecules same energy pe fast bhaagte hain. Isiliye $H_2/O_2$ engines mein thoda extra hydrogen daala jaata hai — $\mathcal{M}$ gir jaata hai aur $c^*$ chhalang maar jaata hai.

Do bade galtiyaan avoid karo. Ek: $c^*$ ko exhaust velocity mat samajhna — actual exhaust velocity toh $c_e = c^* \times C_F$ hai, jahan $C_F$ nozzle ka contribution hai. Do: formula mein $R_s = R_u/\mathcal{M}$ (specific gas constant) use karna, plain $R_u$ nahi — yahi se toh $1/\sqrt{\mathcal{M}}$ wali dependence aati hai. Ratta maarne ke bajaye yaad rakho: "Chamber Star = Hot over Heavy, square-root," aur "Star before Force" ( $c^*$ pehle, phir $C_F$ ).

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

WHAT is $c^*$ ?

HOW to derive $c^*$ from first principles

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rocket Propulsion

Connections