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

WHAT: She uses the measured (operational) form $c^{\ast }=p_0{A}_t/\dot{m}$. WHY: These three quantities are all read directly on the test stand. The theoretical form needs $T_0$, $\mathcal{M}$, $\gamma$ — chemistry data she may not have. The whole reason $c^{\ast }$ was defined this way is that it isolates chamber performance using only stand-measurable numbers. See Choked Flow and the Throat for why the throat (not the exit) is the reference point.

$\frac{[\text{Pa}][{\text{m}}^2]}{[\text{kg/s}]}=\frac{[{\text{N/m}}^2][{\text{m}}^2]}{[\text{kg/s}]}=\frac{[\text{N}]}{[\text{kg/s}]}=\frac{[{\text{kg⋅m/s}}^2]}{[\text{kg/s}]}=[\text{m/s}].$ WHAT: We replaced $\text{Pa}={\text{N/m}}^2$ and $\text{N}={\text{kg⋅m/s}}^2$ step by step. WHY: to prove $c^{\ast }$ carries velocity units even though it is not a real gas speed — it is a figure of merit.

Step 1 — throat area. A circle of diameter $d_t$ has area $A_t=\frac{\pi }{4}{d}_t^2$. $A_t=\frac{\pi }{4}(0.050{)}^2=1.9635\times 10^{-3}{\text{m}}^2.$ Why: the throat is circular; area needs the radius squared, and $d^2/4=r^2$. Step 2 — apply definition. $c^{\ast }=\frac{p_0{A}_t}{\dot{m}}=\frac{6\times 10^6\times 1.9635\times 10^{-3}}{18}=\frac{11781}{18}\approx 654.5\text{m/s}.$ Why: no nozzle data needed — this is the raw measured chamber score.

Step 1 — $\Gamma$. The exponent is $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.25}{2(0.25)}=\frac{2.25}{0.5}=4.5$. $\Gamma =\sqrt{1.25}{\left(\frac{2}{2.25}\right)}^{4.5}=1.1180\times (0.8889{)}^{4.5}.$ Now $(0.8889{)}^{4.5}\approx 0.5890$, so $\Gamma \approx 1.1180\times 0.5890\approx 0.6585$. Why: $\Gamma$ packages all the $\gamma$-dependence into one number, so we compute it once and reuse it; see Vandenkerckhove Function Γ. Step 2 — the "hot & light" term. $\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\sqrt{\frac{8.314\times 3000}{0.020}}=\sqrt{1.2471\times 10^6}=1116.7\text{m/s}.$ Why: this square-root grouping is the physics of $c^{\ast }$ — it carries the "hot ($T_0$ on top) and light ($\mathcal{M}$ on the bottom)" dependence, and comes straight from the throat sound speed $\sqrt{\gamma R_{s}T}$ with $R_s=R_u/\mathcal{M}$. Everything that makes a propellant good lives in this term. Step 3 — combine. $c^{\ast }=\frac{1}{\Gamma }\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\frac{1116.7}{0.6585}\approx 1696\text{m/s}.$ Why: dividing by $\Gamma$ folds in the $\gamma$-dependent throat geometry — larger $\Gamma$ means the choked throat passes more mass, which lowers $c^{\ast }$ (recall $c^{\ast }=p_0{A}_t/\dot{m}$, so more $\dot{m}$ shrinks the score). This is the assembled theoretical figure of merit.

${\eta }_{c^{\ast }}=\frac{c_{\text{measured}}^{\ast }}{c_{\text{theoretical}}^{\ast }}=\frac{1500}{1696}\approx 0.8844.$ WHAT it means: about $11.6\%$ of ideal chamber performance is lost. WHY it matters: since ${\eta }_{c^{\ast }}<1$ lives entirely in the chamber, it flags incomplete combustion, heat loss to walls, or poor mixing — a diagnosis the nozzle ($C_F$) cannot mask.

Since $c^{\ast }\propto \sqrt{T_0}$ (with $\mathcal{M},\gamma$ fixed): $\frac{c_{\text{new}}^{\ast }}{c_{\text{old}}^{\ast }}=\sqrt{\frac{3600}{3000}}=\sqrt{1.2}\approx 1.0954.$ WHAT: a $20\%$ temperature rise buys only a $9.5\%$ $c^{\ast }$ gain. WHY: the square root flattens returns — this is the diminishing-returns geometry drawn in the figure below.

Since $c^{\ast }\propto 1/\sqrt{\mathcal{M}}$: $\frac{c_{\text{new}}^{\ast }}{c_{\text{old}}^{\ast }}=\sqrt{\frac{{\mathcal{M}}_{\text{old}}}{{\mathcal{M}}_{\text{new}}}}=\sqrt{\frac{20}{14}}=\sqrt{1.4286}\approx 1.1952.$ WHAT: a $30\%$ lighter gas gives a $\sim 19.5\%$ boost — bigger than the temperature lever. WHY: the same  ​ softening applies, but $\mathcal{M}$ can be pushed further (light molecules exist), which is why fuel-rich $H_2/O_2$ wins. See Propellant Selection and Molecular Weight.

Step (a) — chamber × nozzle. The exhaust velocity splits cleanly: $c_e=c^{\ast }{C}_F=1700\times 1.65=2805\text{m/s}.$ Why: $c^{\ast }$ is the chamber half, $C_F$ the nozzle half — the split is the whole reason $c^{\ast }$ exists. See Thrust Coefficient CF. Step (b) — divide by $g_0$. Specific impulse is exhaust velocity measured in "seconds": $I_{sp}=\frac{c_e}{g_0}=\frac{2805}{9.81}\approx 285.9\text{s}.$ Why: $I_{sp}{g}_0=c_e=c^{\ast }{C}_F$; dividing by $g_0$ converts m/s into the conventional seconds. See Specific Impulse Isp.

Step 1 — throat area. $A_t=\frac{\pi }{4}(0.045{)}^2=1.5904\times 10^{-3}{\text{m}}^2$. Why: a circular throat's area is $\frac{\pi }{4}{d}_t^2$ (same $d^2/4=r^2$ logic as L2.1); we need $A_t$ in ${\text{m}}^2$ because $c^{\ast }$'s definition combines it with $p_0$ in $\text{Pa}$. Step 2 — rearrange the definition. From $c^{\ast }=p_0{A}_t/\dot{m}$ we solve for $\dot{m}$: $\dot{m}=\frac{p_0{A}_t}{c^{\ast }}=\frac{7\times 10^6\times 1.5904\times 10^{-3}}{1750}=\frac{11133}{1750}\approx 6.362\text{kg/s}.$ Why: $c^{\ast }$ fixes the ratio $p_0{A}_t/\dot{m}$; once you demand a $c^{\ast }$ and set $p_0,A_t$, the mass flow is forced.

WHAT: Write both with the theoretical form. Since $\gamma$ is equal, $\Gamma$ is identical and cancels: $\frac{c_A^{\ast }}{c_B^{\ast }}=\frac{\frac{1}{\Gamma }\sqrt{R_u{T}_{0,A}/{\mathcal{M}}_A}}{\frac{1}{\Gamma }\sqrt{R_u{T}_{0,B}/{\mathcal{M}}_B}}=\sqrt{\frac{T_{0,A}/{\mathcal{M}}_A}{T_{0,B}/{\mathcal{M}}_B}}=\sqrt{\frac{T_{0,A}}{T_{0,B}}\cdot \frac{{\mathcal{M}}_B}{{\mathcal{M}}_A}}.$ WHY the cancellation: $\Gamma$ and $R_u$ are the same in numerator and denominator, so only the "hot/light" grouping survives — the cleanest statement of hot & light wins. Evaluate: $\frac{c_A^{\ast }}{c_B^{\ast }}=\sqrt{\frac{3600}{3000}\cdot \frac{22}{16}}=\sqrt{1.2\times 1.375}=\sqrt{1.65}\approx 1.2845.$ Propellant A beats B by $\approx 28.5\%$ on $c^{\ast }$ — mostly from the lighter, hotter gas. Figure below shows the $\sqrt{T_0/\mathcal{M}}$ landscape both points live on.

Step 1 — attack the left side, pull the $\gamma$'s together. Split the square root and cancel one $\gamma$: $\frac{\sqrt{\gamma R_s{T}_0}}{\gamma }=\frac{\sqrt{\gamma }\sqrt{R_s{T}_0}}{\gamma }=\frac{\sqrt{R_s{T}_0}}{\sqrt{\gamma }}.$ Why: $\frac{\sqrt{\gamma }}{\gamma }=\frac{1}{\sqrt{\gamma }}$ (since $\gamma =\sqrt{\gamma }\cdot \sqrt{\gamma }$). This isolates a lone $1/\sqrt{\gamma }$ that we will soon recognise as part of $1/\Gamma$. So far the left side reads $\text{LHS}=\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}\sqrt{R_s{T}_0}.$ Step 2 — write out $1/\Gamma$. Starting from $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$, take the reciprocal of both factors: $\frac{1}{\Gamma }=\frac{1}{\sqrt{\gamma }}\cdot {\left(\frac{2}{\gamma +1}\right)}^{-\frac{\gamma +1}{2(\gamma -1)}}=\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}.$ Why: reciprocating a product reciprocates each factor — $\sqrt{\gamma }\to 1/\sqrt{\gamma }$, and a negative exponent flips the base $\frac{2}{\gamma +1}\to \frac{\gamma +1}{2}$. Both flips must happen together. Step 3 — match the pattern. Compare the boxed $1/\Gamma$ of Step 2 with the LHS of Step 1: the group $\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ is literally $1/\Gamma$. Substituting it: $\text{LHS}=\frac{1}{\Gamma }\sqrt{R_s{T}_0}.$ Why this step matters: three separate factors just collapsed into the single symbol $\Gamma$ — recognising this pattern is the whole crux of the proof. Step 4 — substitute $R_s=R_u/\mathcal{M}$. Since $\sqrt{R_s{T}_0}=\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}$, we get $\text{LHS}=\frac{1}{\Gamma }\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\text{RHS}.■$ Why this step matters: the two boxed forms differ only in whether they display the specific constant $R_s$ or the universal $R_u/\mathcal{M}$. This last substitution exposes the $1/\sqrt{\mathcal{M}}$ (light-gas) dependence explicitly — the very reason the engineer's form is written with $R_u$ and $\mathcal{M}$ separated. The equality is now complete.

L2.1 ::: $654.5\text{m/s}$ L2.2 ::: $\Gamma \approx 0.658$, $c^{\ast }\approx 1696\text{m/s}$ L3.1 ::: ${\eta }_{c^{\ast }}\approx 0.884$ L3.2 ::: factor $\approx 1.095$ L3.3 ::: factor $\approx 1.195$ L4.1 ::: $c_e=2805\text{m/s}$, $I_{sp}\approx 285.9\text{s}$ L4.2 ::: $\dot{m}\approx 6.36\text{kg/s}$ L5.1 ::: ratio $\approx 1.285$