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

WHAT: $c^{\ast }=P_c{A}^{\ast }/\dot{m}$ directly apply karo — koi chemistry nahi, koi nozzle nahi. WHY: yeh measured face hai; yeh sirf stand data se kaam karta hai. $c^{\ast }=\frac{(4.0\times 10^6\text{Pa})(0.020{\text{m}}^2)}{40\text{kg/s}}=\frac{80000}{40}=2000\text{m/s}$ 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: throat area $A^{\ast }$ use karo, kabhi $A_e$ nahi. WHY: poora $c^{\ast }$ relation isliye exist karta hai kyunki flow throat par choked (sonic) hai. Exit area nozzle expansion ka part hai, jo $C_F$ ka kaam hai — Thrust Coefficient C_F dekho. $c^{\ast }=\frac{(6.0\times 10^6)(0.008)}{30}=\frac{48000}{30}=1600\text{m/s}$ Answer: $c^{\ast }=1600\text{m/s}$ ($A^{\ast }$ use karke, $A_e$ nahi).

WHAT: $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ ko piece by piece evaluate karo. WHY first? $\Gamma$ choked flow ki saari $\gamma$-dependence ko ek number mein package karta hai, toh isse ek baar build karne se har baad ka $c^{\ast }$ formula short ho jaata hai.

• 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: $R$ nikalo, phir $c_{\text{ideal}}^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }$ mein plug karo. WHY: thermochemistry (Combustion Chamber Thermochemistry se) $T_c,\mathcal{M},\gamma$ set karta hai; woh theoretical face ko feed karte hain.

• 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: measured over ideal ka ratio. ${\eta }_{c^{\ast }}=\frac{1610}{1695}=0.9498\approx 95.0\%$ WHY it means something: dono faces nozzle ko ignore karte hain, toh $\sim 5\%$ ka gap sirf chamber-side ho sakta hai — incomplete combustion, walls ko heat loss, ya imperfect propellant mixing (exactly woh isentropic/adiabatic assumptions jo domain-of-validity box mein listed hain, thoda tootte hain). Yeh nozzle ki galti nahi ho sakti (woh $C_F$ ko move karta, $c^{\ast }$ ko nahi). Answer: ${\eta }_{c^{\ast }}\approx 0.950$ ($95.0\%$); shortfall = incomplete burning / wall heat loss / mixing loss.

WHAT / WHY: $\gamma$ fixed hone par, $\Gamma$ fixed hai, toh $c_{\text{ideal}}^{\ast }=\sqrt{R{T}_c}/\Gamma \propto \sqrt{T_c/\mathcal{M}}$. Dono $T_c$ aur $\mathcal{M}$ same square root ke andar baithte hain, toh kisi mein bhi $\pm 10\%$ change $c^{\ast }$ ko lagbhag same factor se move karta hai.

• (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 — figure padho: neeche ka bar chart teeno $c^{\ast }$ values ko same vertical axis (m/s) par plot karta hai. Sabse left wala black bar baseline hai ($1695$); middle black bar "+10% $T_c$" hai ($1778$); red bar right mein "−10% $\mathcal{M}$" hai ($1787$) aur sabse tall hai. Geometrically yeh isliye jeetता hai kyunki $\sqrt{T_c/\mathcal{M}}$ $1/\sqrt{0.9}=1.054$ (ek chhote number se divide karna) ke response mein versus $\sqrt{1.1}=1.049$ (ek bade number se multiply karna) — red bar apne black neighbour se thoda upar hai. Yeh chhota-magar-real gap isliye hai ki rocket designers light exhaust (hydrogen) ko utni hi shiddat se dhundhte hain jitna heat ko.

Answer: (b) $\mathcal{M}$ ghataana thoda jeet jaata hai — $1787$ vs $1778\text{m/s}$.

WHAT: parent ke Step 5 se choked-flow result use karo, $\dot{m}=\frac{P_c{A}^{\ast }}{\sqrt{R{T}_c}}\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}=\frac{P_c{A}^{\ast }\Gamma }{\sqrt{R{T}_c}}$ kyunki $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ exactly wahi bundle hai. WHY — "same statement" claim ko poori tarah unpack karo: measured face define karta hai $c_{\text{meas}}^{\ast }=P_c{A}^{\ast }/\dot{m}$, yani $\dot{m}=P_c{A}^{\ast }/c_{\text{meas}}^{\ast }$. Theoretical choked-flow law kehta hai $\dot{m}=P_c{A}^{\ast }\Gamma /\sqrt{R{T}_c}$. Same $\dot{m}$ ke liye dono expressions barabar karo: $\frac{P_c{A}^{\ast }}{c_{\text{meas}}^{\ast }}=\frac{P_c{A}^{\ast }\Gamma }{\sqrt{R{T}_c}}⟹\frac{1}{c_{\text{meas}}^{\ast }}=\frac{\Gamma }{\sqrt{R{T}_c}}⟹c^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }.$ $P_c{A}^{\ast }$ dono sides cancel ho jaata hai — isliye measured definition aur theoretical formula literally ek hi equation hai rearranged, choked-flow law se connected. Woh cancellation isliye hai ki $c^{\ast }$ $P_c$ aur $A^{\ast }$ magnitudes se blind hai aur sirf chamber chemistry $(T_c,\mathcal{M},\gamma )$ par depend karta hai. Ab numbers:

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

WHAT: $c=c^{\ast }{C}_F$ (quick reference mein define kiya split) ko invert karo. $C_F=\frac{c}{c^{\ast }}=\frac{2900}{1750}=1.657$ WHY the split matters: agar $c^{\ast }$ apna theoretical target hit karta hai lekin total $c$ low hai, toh chamber chemistry theek hai — loss $C_F$ mein hai, yani nozzle mein (poor expansion, over/under-expansion, divergence loss). Effective Exhaust Velocity and Specific Impulse dekho. Answer: $C_F=1.66$; nozzle ($C_F$) ko blame karo, combustion ko nahi.

WHAT / WHY: ${\eta }_{c^{\ast }}$ chamber isolate karta hai; $C_F=c/c_{\text{meas}}^{\ast }$ nozzle isolate karta hai. Symptom split karo. Unit A:

• ${\eta }_{c^{\ast }}=1550/1600=0.9688\approx 96.9\%$ — chamber healthy hai.
• $C_F=2790/1550=1.800$ — strong nozzle. Unit B:
• ${\eta }_{c^{\ast }}=1590/1600=0.99375\approx 99.4\%$ — chamber excellent hai.
• $C_F=2380/1590=1.497$ — weak nozzle. Diagnosis: Unit A thoda kam completely burn karta hai (lower ${\eta }_{c^{\ast }}$) lekin uska nozzle theek hai. Unit B almost perfectly burn karta hai lekin nozzle mein buri tarah lose karta hai (much lower $C_F$) — problem throat ke downstream hai. Answer: A → mild combustion loss (${\eta }_{c^{\ast }}=96.9\%$, $C_F=1.80$); B → nozzle loss (${\eta }_{c^{\ast }}=99.4\%$, $C_F=1.50$).

Figure padho: dono black bars (left axis) chamber scores ${\eta }_{c^{\ast }}$ hain — dono tall aur nearly equal hain, toh dono chambers achha burn karte hain. Dono red bars (right axis) nozzle scores $C_F$ hain — aur yahan Unit B ka red bar visibly short hai. Aankhon ko turant dikhta hai ki B ki problem red (nozzle) hai, black (chamber) nahi: same diagnosis, ek glance mein padh lo.

WHAT: $c_{\text{ideal}}^{\ast }=\frac{\sqrt{R{T}_c}}{\Gamma }$ ko $R$ ke liye invert karo, phir $\mathcal{M}=R_u/R$. WHY: $\gamma$ se $\Gamma$ fix hota hai; baaki sab $R$ pin karta hai, isliye $\mathcal{M}$ — exactly aise hi ek chemist ko target diya jaata hai.

• $\Gamma (\gamma =1.22)$: exponent $\frac{2.22}{0.44}=5.0455$; base $\frac{2}{2.22}=0.9009$; $(0.9009{)}^{5.0455}=0.6032$; $\sqrt{1.22}=1.1045$; $\Gamma =1.1045\times 0.6032=0.6662$.
• Target rearrange karo: $\sqrt{R{T}_c}=c_{\text{ideal}}^{\ast }\cdot \Gamma =1800\times 0.6662=1199.2$.
• Square karo: $R{T}_c=1199.2^2=1.4381\times 10^6$, toh $R=\frac{1.4381\times 10^6}{3400}=422.9\text{J/(kg⋅K)}$.
• Molecular weight: $\mathcal{M}=\frac{R_u}{R}=\frac{8314}{422.9}=19.66\text{g/mol}$. Answer: $\mathcal{M}\approx 19.7\text{g/mol}$ — moderately light exhaust, "burn hot, make light gas" mantra ke consistent. (Agar aur bhi lighter chahiye, toh $\mathcal{M}$ ko hydrogen-rich exhaust ki taraf push karo.)

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.