Exercises — Characteristic velocity c - = P_c A - ṁ — derivation, combustion efficiency measure

WHAT: apply $c^{\ast }=P_c{A}^{\ast }/\dot{m}$ directly — no chemistry, no nozzle needed. WHY: this is the measured face; it works from stand data alone. $c^{\ast }=\frac{(4.0\times 10^6\text{Pa})(0.020{\text{m}}^2)}{40\text{kg/s}}=\frac{80000}{40}=2000\text{m/s}$ Units check: $\frac{\text{Pa}\cdot {\text{m}}^2}{\text{kg/s}}=\frac{\text{N}}{\text{kg/s}}=\text{m/s}$ ✓ Answer: $c^{\ast }=2000\text{m/s}$.

WHAT: use the throat area $A^{\ast }$, never $A_e$. WHY: the whole $c^{\ast }$ relation exists because the flow is choked (sonic) at the throat. The exit area belongs to nozzle expansion, which is $C_F$'s job — see Thrust Coefficient C_F. $c^{\ast }=\frac{(6.0\times 10^6)(0.008)}{30}=\frac{48000}{30}=1600\text{m/s}$ Answer: $c^{\ast }=1600\text{m/s}$ (using $A^{\ast }$, not $A_e$).

WHAT: evaluate $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ piece by piece. WHY first? $\Gamma$ packages all the $\gamma$-dependence of choked flow into one number, so building it once makes every later $c^{\ast }$ formula short.

• Exponent: $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.20}{2(0.20)}=\frac{2.20}{0.40}=5.5$
• Base: $\frac{2}{\gamma +1}=\frac{2}{2.20}=0.9091$
• Power: $(0.9091{)}^{5.5}=0.5921$
• Prefactor: $\sqrt{1.20}=1.0954$ $\Gamma =1.0954\times 0.5921=0.6486$ Answer: $\Gamma \approx 0.649$.

WHAT: get $R$, then plug into $c_{\text{ideal}}^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }$. WHY: the thermochemistry (from Combustion Chamber Thermochemistry) sets $T_c,\mathcal{M},\gamma$; those feed the theoretical face.

• Specific gas constant: $R=\frac{R_u}{\mathcal{M}}=\frac{8314}{22}=377.9\text{J/(kg⋅K)}$
• $R{T}_c=377.9\times 3200=1.2093\times 10^6$
• $\sqrt{R{T}_c}=1099.7$ $c_{\text{ideal}}^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }=\frac{1099.7}{0.6486}=1695\text{m/s}$ Answer: $c_{\text{ideal}}^{\ast }\approx 1695\text{m/s}$.

WHAT: ratio measured over ideal. ${\eta }_{c^{\ast }}=\frac{1610}{1695}=0.9498\approx 95.0\%$ WHY it means something: both faces ignore the nozzle, so the $\sim 5\%$ gap can only be chamber-side — incomplete combustion, heat loss to the walls, or imperfect propellant mixing (exactly the isentropic/adiabatic assumptions listed in the domain-of-validity box breaking a little). It can not be a nozzle fault (that would move $C_F$, not $c^{\ast }$). Answer: ${\eta }_{c^{\ast }}\approx 0.950$ ($95.0\%$); shortfall = incomplete burning / wall heat loss / mixing loss.

WHAT / WHY: with $\gamma$ fixed, $\Gamma$ is fixed, so $c_{\text{ideal}}^{\ast }=\sqrt{R{T}_c}/\Gamma \propto \sqrt{T_c/\mathcal{M}}$. Both $T_c$ and $\mathcal{M}$ sit under the same square root, so a $\pm 10\%$ change in either moves $c^{\ast }$ by nearly the same factor.

• (a) $T_c\to 3520$: factor $\sqrt{1.10}=1.0488\Rightarrow c^{\ast }=1695\times 1.0488=1778\text{m/s}$.
• (b) $\mathcal{M}\to 19.8$: factor $\sqrt{1/0.90}=\sqrt{1.1111}=1.0541\Rightarrow c^{\ast }=1695\times 1.0541=1787\text{m/s}$. WHAT IT MEANS — read the figure: the bar chart below plots all three $c^{\ast }$ values on the same vertical axis (m/s). The leftmost black bar is the baseline ($1695$); the middle black bar is "+10% $T_c$" ($1778$); the red bar on the right is "−10% $\mathcal{M}$" ($1787$) and is the tallest. Geometrically it wins because $\sqrt{T_c/\mathcal{M}}$ responds to $1/\sqrt{0.9}=1.054$ (dividing by a smaller number) versus $\sqrt{1.1}=1.049$ (multiplying by a bigger one) — the red bar edges just above its black neighbour. That tiny-but-real gap is why rocket designers chase light exhaust (hydrogen) as hard as they chase heat.

Answer: (b) lowering $\mathcal{M}$ wins slightly — $1787$ vs $1778\text{m/s}$.

WHAT: use the choked-flow result from the parent's Step 5, $\dot{m}=\frac{P_c{A}^{\ast }}{\sqrt{R{T}_c}}\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}=\frac{P_c{A}^{\ast }\Gamma }{\sqrt{R{T}_c}}$ since $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ is exactly that bundle. WHY — unpack the "same statement" claim in full: the measured face defines $c_{\text{meas}}^{\ast }=P_c{A}^{\ast }/\dot{m}$, i.e. $\dot{m}=P_c{A}^{\ast }/c_{\text{meas}}^{\ast }$. The theoretical choked-flow law says $\dot{m}=P_c{A}^{\ast }\Gamma /\sqrt{R{T}_c}$. Set these two expressions for the same $\dot{m}$ equal: $\frac{P_c{A}^{\ast }}{c_{\text{meas}}^{\ast }}=\frac{P_c{A}^{\ast }\Gamma }{\sqrt{R{T}_c}}⟹\frac{1}{c_{\text{meas}}^{\ast }}=\frac{\Gamma }{\sqrt{R{T}_c}}⟹c^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }.$ The $P_c{A}^{\ast }$ cancels on both sides — so the measured definition and the theoretical formula are literally one equation rearranged, connected by the choked-flow law. That cancellation is why $c^{\ast }$ is blind to $P_c$ and $A^{\ast }$ magnitudes and depends only on the chamber chemistry $(T_c,\mathcal{M},\gamma )$. Now the numbers:

• $R=8314/20=415.7\text{J/(kg⋅K)}$, so $\sqrt{R{T}_c}=\sqrt{415.7\times 3000}=\sqrt{1.247\times 10^6}=1116.7$.
• $\Gamma$: exponent $\frac{2.25}{0.50}=4.5$; base $\frac{2}{2.25}=0.8889$; $(0.8889{)}^{4.5}=0.5891$; $\sqrt{1.25}=1.1180$; $\Gamma =1.1180\times 0.5891=0.6587$.
• $\dot{m}=\frac{(5.0\times 10^6)(0.010)(0.6587)}{1116.7}=\frac{32935}{1116.7}=29.5\text{kg/s}$.
• Then $c_{\text{meas}}^{\ast }=\frac{P_c{A}^{\ast }}{\dot{m}}=\frac{50000}{29.5}=1695\text{m/s}$, which equals $\frac{\sqrt{R{T}_c}}{\Gamma }=\frac{1116.7}{0.6587}=1695\text{m/s}$ ✓ Answer: $\dot{m}\approx 29.5\text{kg/s}$, $c_{\text{ideal}}^{\ast }\approx 1695\text{m/s}$; the two faces agree exactly.

WHAT: invert $c=c^{\ast }{C}_F$ (the split defined in the quick reference). $C_F=\frac{c}{c^{\ast }}=\frac{2900}{1750}=1.657$ WHY the split matters: if $c^{\ast }$ hits its theoretical target but the total $c$ is low, the chamber chemistry is fine — the loss lives in $C_F$, i.e. the nozzle (poor expansion, over/under-expansion, divergence loss). See Effective Exhaust Velocity and Specific Impulse. Answer: $C_F=1.66$; blame the nozzle ($C_F$), not the combustion.

WHAT / WHY: ${\eta }_{c^{\ast }}$ isolates the chamber; $C_F=c/c_{\text{meas}}^{\ast }$ isolates the nozzle. Split the symptom. Unit A:

• ${\eta }_{c^{\ast }}=1550/1600=0.9688\approx 96.9\%$ — chamber is healthy.
• $C_F=2790/1550=1.800$ — strong nozzle. Unit B:
• ${\eta }_{c^{\ast }}=1590/1600=0.99375\approx 99.4\%$ — chamber is excellent.
• $C_F=2380/1590=1.497$ — weak nozzle. Diagnosis: Unit A burns slightly less completely (lower ${\eta }_{c^{\ast }}$) but its nozzle is fine. Unit B burns almost perfectly yet loses badly in the nozzle (much lower $C_F$) — the problem is downstream of the throat. Answer: A → mild combustion loss (${\eta }_{c^{\ast }}=96.9\%$, $C_F=1.80$); B → nozzle loss (${\eta }_{c^{\ast }}=99.4\%$, $C_F=1.50$).

Read the figure: the two black bars (left axis) are the chamber scores ${\eta }_{c^{\ast }}$ — both are tall and nearly equal, so both chambers burn well. The two red bars (right axis) are the nozzle scores $C_F$ — and here Unit B's red bar is visibly short. The eye instantly sees that B's problem is red (nozzle), not black (chamber): same diagnosis, read off in one glance.

WHAT: invert $c_{\text{ideal}}^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }$ for $R$, then $\mathcal{M}=R_u/R$. WHY: $\gamma$ fixes $\Gamma$; everything else pins $R$, hence $\mathcal{M}$ — this is exactly how a chemist is handed a target.

• $\Gamma (\gamma =1.22)$: exponent $\frac{2.22}{0.44}=5.0455$; base $\frac{2}{2.22}=0.9009$; $(0.9009{)}^{5.0455}=0.6032$; $\sqrt{1.22}=1.1045$; $\Gamma =1.1045\times 0.6032=0.6662$.
• Rearrange the target: $\sqrt{R{T}_c}=c_{\text{ideal}}^{\ast }\cdot \Gamma =1800\times 0.6662=1199.2$.
• Square: $R{T}_c=1199.2^2=1.4381\times 10^6$, so $R=\frac{1.4381\times 10^6}{3400}=422.9\text{J/(kg⋅K)}$.
• Molecular weight: $\mathcal{M}=\frac{R_u}{R}=\frac{8314}{422.9}=19.66\text{g/mol}$. Answer: $\mathcal{M}\approx 19.7\text{g/mol}$ — a moderately light exhaust, consistent with the "burn hot, make light gas" mantra. (If you needed it lighter still, you would push $\mathcal{M}$ down toward hydrogen-rich exhaust.)

1.1 → 2000 m/s · 1.2 → 1600 m/s · 2.1 → $\Gamma =0.649$ · 2.2 → 1695 m/s · 3.1 → 0.950 · 3.2 → (b) wins, 1787 vs 1778 · 4.1 → $\dot{m}=29.5$ kg/s, $c^{\ast }=1695$ · 4.2 → $C_F=1.66$ · 5.1 → A: 0.969/1.80, B: 0.994/1.50 · 5.2 → $\mathcal{M}=19.7$ g/mol.