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

KYA: Woh measured (operational) form $c^{\ast }=p_0{A}_t/\dot{m}$ use karti hai. KYUN: Ye teeno quantities seedha test stand par read hoti hain. Theoretical form ko $T_0$, $\mathcal{M}$, $\gamma$ chahiye — chemistry data jo shayad uske paas nahi hai. $c^{\ast }$ ko isi tarah define karne ki poori wajah yahi hai ki yeh sirf stand-measurable numbers use karke chamber performance isolate karta hai. Dekho Choked Flow and the Throat — kyun throat (exit nahi) reference point hai.

$\frac{[\text{Pa}][{\text{m}}^2]}{[\text{kg/s}]}=\frac{[{\text{N/m}}^2][{\text{m}}^2]}{[\text{kg/s}]}=\frac{[\text{N}]}{[\text{kg/s}]}=\frac{[{\text{kg⋅m/s}}^2]}{[\text{kg/s}]}=[\text{m/s}].$ KYA: Humne step by step $\text{Pa}={\text{N/m}}^2$ aur $\text{N}={\text{kg⋅m/s}}^2$ replace kiya. KYUN: yeh prove karne ke liye ki $c^{\ast }$ velocity units carry karta hai, chahe woh koi real gas speed nahi hai — yeh ek figure of merit hai.

Step 1 — throat area. Diameter $d_t$ wale circle ka area $A_t=\frac{\pi }{4}{d}_t^2$ hota hai. $A_t=\frac{\pi }{4}(0.050{)}^2=1.9635\times 10^{-3}{\text{m}}^2.$ Kyun: throat circular hai; area ke liye radius squared chahiye, aur $d^2/4=r^2$. Step 2 — definition apply karo. $c^{\ast }=\frac{p_0{A}_t}{\dot{m}}=\frac{6\times 10^6\times 1.9635\times 10^{-3}}{18}=\frac{11781}{18}\approx 654.5\text{m/s}.$ Kyun: koi bhi nozzle data ki zaroorat nahi — yeh raw measured chamber score hai.

Step 1 — $\Gamma$. Exponent hai $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.25}{2(0.25)}=\frac{2.25}{0.5}=4.5$. $\Gamma =\sqrt{1.25}{\left(\frac{2}{2.25}\right)}^{4.5}=1.1180\times (0.8889{)}^{4.5}.$ Ab $(0.8889{)}^{4.5}\approx 0.5890$, toh $\Gamma \approx 1.1180\times 0.5890\approx 0.6585$. Kyun: $\Gamma$ saari $\gamma$-dependence ek number mein pack kar deta hai, toh hum ise ek baar compute karke reuse karte hain; dekho Vandenkerckhove Function Γ. Step 2 — "hot & light" term. $\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\sqrt{\frac{8.314\times 3000}{0.020}}=\sqrt{1.2471\times 10^6}=1116.7\text{m/s}.$ Kyun: yeh square-root grouping hi $c^{\ast }$ ki physics hai — yeh "hot ($T_0$ upar) aur light ($\mathcal{M}$ neeche)" dependence carry karta hai, aur seedha throat sound speed $\sqrt{\gamma R_{s}T}$ se aata hai jisme $R_s=R_u/\mathcal{M}$ hai. Jo kuch bhi ek propellant ko accha banata hai woh is term mein hai. Step 3 — combine karo. $c^{\ast }=\frac{1}{\Gamma }\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\frac{1116.7}{0.6585}\approx 1696\text{m/s}.$ Kyun: $\Gamma$ se divide karna $\gamma$-dependent throat geometry ko fold in karta hai — bada $\Gamma$ matlab choked throat zyada mass pass karta hai, jo $c^{\ast }$ ko ghatata hai (yaad karo $c^{\ast }=p_0{A}_t/\dot{m}$, toh zyada $\dot{m}$ score ko chhota karta hai). Yeh assembled theoretical figure of merit hai.

${\eta }_{c^{\ast }}=\frac{c_{\text{measured}}^{\ast }}{c_{\text{theoretical}}^{\ast }}=\frac{1500}{1696}\approx 0.8844.$ KYA matlab hai: ideal chamber performance ka lagbhag $11.6\%$ lost ho raha hai. KYUN matter karta hai: kyunki ${\eta }_{c^{\ast }}<1$ sirf chamber mein hota hai, yeh incomplete combustion, wall mein heat loss, ya poor mixing flag karta hai — ek diagnosis jo nozzle ($C_F$) chhupa nahi sakta.

Kyunki $c^{\ast }\propto \sqrt{T_0}$ (jisme $\mathcal{M},\gamma$ fixed hain): $\frac{c_{\text{new}}^{\ast }}{c_{\text{old}}^{\ast }}=\sqrt{\frac{3600}{3000}}=\sqrt{1.2}\approx 1.0954.$ KYA: $20\%$ temperature rise se sirf $9.5\%$ $c^{\ast }$ gain milta hai. KYUN: square root returns ko flatten karta hai — yeh wahi diminishing-returns geometry hai jo neeche figure mein dikhaya gaya hai.

Kyunki $c^{\ast }\propto 1/\sqrt{\mathcal{M}}$: $\frac{c_{\text{new}}^{\ast }}{c_{\text{old}}^{\ast }}=\sqrt{\frac{{\mathcal{M}}_{\text{old}}}{{\mathcal{M}}_{\text{new}}}}=\sqrt{\frac{20}{14}}=\sqrt{1.4286}\approx 1.1952.$ KYA: $30\%$ halka gas $\sim 19.5\%$ boost deta hai — temperature lever se bada. KYUN: wahi  ​ softening apply hoti hai, lekin $\mathcal{M}$ ko zyada push kiya ja sakta hai (halke molecules exist karte hain), isliye fuel-rich $H_2/O_2$ jeet jaata hai. Dekho Propellant Selection and Molecular Weight.

Step (a) — chamber × nozzle. Exhaust velocity cleanly split hoti hai: $c_e=c^{\ast }{C}_F=1700\times 1.65=2805\text{m/s}.$ Kyun: $c^{\ast }$ chamber half hai, $C_F$ nozzle half — yeh split hi poori wajah hai ki $c^{\ast }$ exist karta hai. Dekho Thrust Coefficient CF. Step (b) — $g_0$ se divide karo. Specific impulse exhaust velocity hai "seconds" mein measure ki gayi: $I_{sp}=\frac{c_e}{g_0}=\frac{2805}{9.81}\approx 285.9\text{s}.$ Kyun: $I_{sp}{g}_0=c_e=c^{\ast }{C}_F$; $g_0$ se divide karna m/s ko conventional seconds mein convert karta hai. Dekho Specific Impulse Isp.

Step 1 — throat area. $A_t=\frac{\pi }{4}(0.045{)}^2=1.5904\times 10^{-3}{\text{m}}^2$. Kyun: circular throat ka area $\frac{\pi }{4}{d}_t^2$ hota hai (wahi $d^2/4=r^2$ logic jaise L2.1 mein); $A_t$ ${\text{m}}^2$ mein chahiye kyunki $c^{\ast }$ ki definition ise $p_0$ ke saath $\text{Pa}$ mein combine karti hai. Step 2 — definition rearrange karo. $c^{\ast }=p_0{A}_t/\dot{m}$ se $\dot{m}$ solve karo: $\dot{m}=\frac{p_0{A}_t}{c^{\ast }}=\frac{7\times 10^6\times 1.5904\times 10^{-3}}{1750}=\frac{11133}{1750}\approx 6.362\text{kg/s}.$ Kyun: $c^{\ast }$ ratio $p_0{A}_t/\dot{m}$ ko fix karta hai; jab tum ek $c^{\ast }$ demand karo aur $p_0,A_t$ set karo, mass flow forced ho jaata hai.

KYA: Dono ko theoretical form se likho. Kyunki $\gamma$ equal hai, $\Gamma$ identical hai aur cancel ho jaata hai: $\frac{c_A^{\ast }}{c_B^{\ast }}=\frac{\frac{1}{\Gamma }\sqrt{R_u{T}_{0,A}/{\mathcal{M}}_A}}{\frac{1}{\Gamma }\sqrt{R_u{T}_{0,B}/{\mathcal{M}}_B}}=\sqrt{\frac{T_{0,A}/{\mathcal{M}}_A}{T_{0,B}/{\mathcal{M}}_B}}=\sqrt{\frac{T_{0,A}}{T_{0,B}}\cdot \frac{{\mathcal{M}}_B}{{\mathcal{M}}_A}}.$ KYUN cancellation hoti hai: $\Gamma$ aur $R_u$ numerator aur denominator mein same hain, toh sirf "hot/light" grouping bachti hai — hot & light wins ka sabse clean statement. Evaluate karo: $\frac{c_A^{\ast }}{c_B^{\ast }}=\sqrt{\frac{3600}{3000}\cdot \frac{22}{16}}=\sqrt{1.2\times 1.375}=\sqrt{1.65}\approx 1.2845.$ Propellant A, B se $\approx 28.5\%$ $c^{\ast }$ mein aage hai — zyaatar halke, garam gas ki wajah se. Neeche figure woh $\sqrt{T_0/\mathcal{M}}$ landscape dikhata hai jis par dono points hain.

Step 1 — left side attack karo, $\gamma$'s ko ek saath khiincho. Square root split karo aur ek $\gamma$ cancel karo: $\frac{\sqrt{\gamma R_s{T}_0}}{\gamma }=\frac{\sqrt{\gamma }\sqrt{R_s{T}_0}}{\gamma }=\frac{\sqrt{R_s{T}_0}}{\sqrt{\gamma }}.$ Kyun: $\frac{\sqrt{\gamma }}{\gamma }=\frac{1}{\sqrt{\gamma }}$ (kyunki $\gamma =\sqrt{\gamma }\cdot \sqrt{\gamma }$). Yeh ek akela $1/\sqrt{\gamma }$ isolate karta hai jise hum jald hi $1/\Gamma$ ke hisse ke roop mein pahchaanenge. Abhi tak left side padhta hai: $\text{LHS}=\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}\sqrt{R_s{T}_0}.$ Step 2 — $1/\Gamma$ likho. $\Gamma =\sqrt{\gamma }{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ se shuru karke, dono factors ka reciprocal lo: $\frac{1}{\Gamma }=\frac{1}{\sqrt{\gamma }}\cdot {\left(\frac{2}{\gamma +1}\right)}^{-\frac{\gamma +1}{2(\gamma -1)}}=\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}.$ Kyun: kisi product ka reciprocal lene par har factor ka reciprocal liya jaata hai — $\sqrt{\gamma }\to 1/\sqrt{\gamma }$, aur negative exponent base ko flip karta hai $\frac{2}{\gamma +1}\to \frac{\gamma +1}{2}$. Dono flips saath hone chahiye. Step 3 — pattern match karo. Step 2 ke boxed $1/\Gamma$ ko Step 1 ke LHS se compare karo: group $\frac{1}{\sqrt{\gamma }}{\left(\frac{\gamma +1}{2}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$ literally $1/\Gamma$ hai. Ise substitute karo: $\text{LHS}=\frac{1}{\Gamma }\sqrt{R_s{T}_0}.$ Kyun yeh step matter karta hai: teen alag factors abhi single symbol $\Gamma$ mein collapse ho gaye — is pattern ko pahchaanna hi proof ka poora crux hai. Step 4 — $R_s=R_u/\mathcal{M}$ substitute karo. Kyunki $\sqrt{R_s{T}_0}=\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}$, hume milta hai: $\text{LHS}=\frac{1}{\Gamma }\sqrt{\frac{R_u{T}_0}{\mathcal{M}}}=\text{RHS}.■$ Kyun yeh step matter karta hai: do boxed forms mein sirf itna fark hai ki woh specific constant $R_s$ dikhate hain ya universal $R_u/\mathcal{M}$. Yeh last substitution $1/\sqrt{\mathcal{M}}$ (light-gas) dependence explicitly expose karta hai — wahi wajah ki engineer's form $R_u$ aur $\mathcal{M}$ alag-alag likhi jaati hai. Equality ab complete hai.

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 $\approx 1.285$