Exercises — Solid rocket Isp derivation from grain properties

• Left grey box (geometry): $A_b$, $A_t$, aur burn law $r=a{p}_c^n$ $p_c$ mein feed hote hain. Blue arrows follow karo.
• Blue arrows throughput path carry karte hain: $p_c\to r\to \dot{m}\to F$. Jo kuch bhi grain touch karta hai.
• Green box (chemistry): $\gamma ,T_c,M,p_e/p_c$ green arrow ke saath $v_e$ feed karte hain.
• $F$ (orange) wahan hai jahan blue aur green milte hain: $F=\dot{m}{v}_e$ (matched).
• Red arrow to $I_{sp}$ sirf green $v_e$ se aata hai — koi blue arrow wahan nahi pahunchta. Yahi "grain cancels" ki poori kahaani ek picture mein hai.

Formulas ko order mein walk karo (aur s01 map par trace karo).

• $\dot{m}={\rho }_p{A}_{b}r$: $A_b$ linearly aata hai, toh $A_b$ double karne se $\dot{m}$ double hoga.
• $F=\dot{m}{v}_e$ (matched nozzle): $\dot{m}$ double, $v_e$ unchanged (chemistry fixed) → $F$ double.
• $I_{sp}=F/(\dot{m}{g}_0)$: numerator aur denominator dono double → unchanged. Answer: $\dot{m}$ aur $F$ double; $I_{sp}$ untouched. Geometry ek power knob hai, efficiency knob nahi — exactly map ka "no blue arrow reaches $I_{sp}$" wala point.

Seedha $\dot{m}={\rho }_p{A}_{b}r$ mein substitute karo: $\dot{m}=1750\times 0.40\times 0.006=4.2\text{kg/s}.$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot {\text{m}}^2\cdot \frac{\text{m}}{\text{s}}=\frac{\text{kg}}{\text{s}}$. ✓ Answer: $4.2\text{kg/s}$.

Saint-Robert: $r=a{p}_c^n$. $p_c^{0.3}=(5\times 10^6{)}^{0.3}.$ Logs lo: $\ln (5\times 10^6)=15.42$; times $0.3=4.627$; $e^{4.627}=102.2$. $r=4\times 10^{-5}\times 102.2=4.09\times 10^{-3}\text{m/s}\approx 4.1\text{mm/s}.$ Answer: $r\approx 0.0041\text{m/s}$.

Step 1 — prefactor $\frac{2\gamma }{\gamma -1}=\frac{2.5}{0.25}=10$. Kyun: yeh enthalpy-to-KE conversion factor hai. Step 2 — gas energy scale $\frac{R{T}_c}{M}=\frac{8.314\times 3200}{0.022}=1.209\times 10^6\text{J/kg}$. Step 3 — expansion bracket exponent $\frac{\gamma -1}{\gamma }=\frac{0.25}{1.25}=0.2$; $(0.008{)}^{0.2}$: $\ln 0.008=-4.828$, times $0.2=-0.9657$, $e^{-0.9657}=0.3808$. Bracket $=1-0.3808=0.6192$. Step 4 — exhaust velocity $v_e=\sqrt{10\times 1.209\times 10^6\times 0.6192}=\sqrt{7.487\times 10^6}=2736\text{m/s}$. Step 5 — seconds mein $I_{sp}=2736/9.81=278.9\mathbf{s}$. Answer: $I_{sp}\approx 279\text{s}$. Ek bhi grain number andar nahi gaya — exactly jaisa parent note ka key insight predict karta hai.

Step 1 — mass flow $\dot{m}=1800\times 0.75\times 0.007=9.45\text{kg/s}$. Step 2 — effective exhaust velocity $c=g_0{I}_{sp}=9.81\times 279=2737\text{m/s}$ (matched nozzle ke liye $v_e$ ke barabar, expected hai). Step 3 — thrust $F=\dot{m}c=9.45\times 2737=2.586\times 10^4\text{N}\approx 25.9\text{kN}$. Answer: $\dot{m}=9.45\text{kg/s}$, $F\approx 25.9\text{kN}$.

Step 1 — base ${\rho }_{p}a{c}^{\ast }(A_b/A_t)=1800\times 3\times 10^{-5}\times 1550\times 250$. $1800\times 3\times 10^{-5}=0.054$; $\times 1550=83.7$; $\times 250=2.0925\times 10^4$. Step 2 — exponent $\frac{1}{1-n}=\frac{1}{0.6}=1.6667$. Step 3 — power raise karo $\ln (2.0925\times 10^4)=9.949$; $\times 1.6667=16.581$; $e^{16.581}=1.588\times 10^7\text{Pa}$. Answer: $p_c\approx 15.9\text{MPa}$.

$M$-dependence isolate karo. Root ke andar sirf $\frac{R{T}_c}{M}$ mein $M$ hai; baaki sab (prefactor, bracket) fixed hai. Toh $v_e\propto \sqrt{\frac{1}{M}}⟹I_{sp}\propto \frac{1}{\sqrt{M}}.$ $M$ aadha karne se $I_{sp}$ ko $\sqrt{2}=1.414$ se multiply hota hai. Answer: $I_{sp}$ factor $\sqrt{2}\approx 1.41$ se badhta hai ($+41\%$ gain). Yahi reason hai ki hydrogen-rich exhaust prized hai: same energy ke liye light molecules tezi se move karti hain.

Full thrust: $F=\dot{m}{v}_e+(p_e-p_a)A_e$. Vacuum mein $p_a=0$: Step 1 — momentum thrust $\dot{m}{v}_e=9.45\times 2737=25,865\text{N}$. Step 2 — pressure thrust $(p_e-0)A_e=1.0\times 10^5\times 0.20=20,000\text{N}$. Step 3 — total $F=45,865\text{N}$. Step 4 — vacuum $I_{sp}$ $=\frac{F}{\dot{m}{g}_0}=\frac{45,865}{9.45\times 9.81}=494.7\text{s}$. Matched (sea-level) $I_{sp}=\frac{25,865}{9.45\times 9.81}=278.9\text{s}$ se compare karo. Answer: pressure term vacuum mein $\approx 216\text{s}$ add karta hai ($279\to 495\text{s}$). Isliye $I_{sp}$ hamesha altitude/vacuum tag ke saath quote kiya jaata hai.

Axes padhna: horizontal axis burn-rate exponent $n$ hai (dimensionless, $0$ se just under $1$ tak); vertical axis equilibrium chamber pressure $p_c$ MPa mein hai. Red curve hai $p_c=(\text{base}{)}^{1/(1-n)}$ — base fixed hai, sirf exponent $\frac{1}{1-n}$ badalta hai jab hum $n$ slide karte hain.

• Teen coloured dots $n=0.3$ (green), $0.7$ (orange), $0.9$ (red) mark karte hain. Labels se unke exponents padho: $1.43$, $3.33$, $10$. Jaise $n$ chadhta hai, $p_c$ super-linearly shoot up karta hai — dots $n$ se kahin zyada tezi se chadhte hain.
• Grey dashed vertical line at $n=1$: curve uski taraf race karti hai aur vertical ho jaati hai — exponent $\frac{1}{1-n}\to \infty$. Base mein infinitesimal change phir $p_c$ ko infinity tak bhej deta hai. Physically: ek chhota pressure bump $r=a{p}_c^n$ badhata hai, jo zyada gas banata hai, jo phir pressure badhata hai — throat itni tezi se vent nahi kar sakta → runaway burst.
• Blue shaded "stable design zone" left mein: yahan curve almost flat hai, loop settle ho jaata hai, $p_c$ well-behaved hai. Real motors yahan rehte hain.
• $n=0$ edge (green marker far left par): exponent $\frac{1}{1-0}=1$, toh $p_c$ base ke barabar hai — lekin zyada important yeh hai ki $r=a{p}_c^0=a$, chamber pressure se independent constant burn rate. Controller loop khul jaata hai: pressure ab burn rate mein feedback nahi karta. Aisa "plateau" propellant maximally stable hota hai (koi runaway possible nahi) lekin woh self-regulating flexibility chodta hai jo mild $n$ provide karta hai. Answer: $n<1$ ke neeche pressure loop self-settle hota hai; $n=1$ par diverge karta hai (explosion); $n=0$ par loop fully decoupled hai — $r$ flat aur pressure-blind hai. Safe designs shaded left zone mein rehte hain, $0\le n<1$.

s01 map ko left-to-right follow karo. Step 1 — chamber pressure (L2.3 se, same numbers): $p_c=1.588\times 10^7\text{Pa}=15.9\text{MPa}$. Step 2 — burn rate $r=a{p}_c^n=3\times 10^{-5}\times (1.588\times 10^7{)}^{0.4}$. $\ln (1.588\times 10^7)=16.581$; $\times 0.4=6.632$; $e^{6.632}=759.6$. $r=3\times 10^{-5}\times 759.6=0.02279\text{m/s}$. Step 3 — mass flow $\dot{m}={\rho }_p{A}_{b}r=1800\times 1.2\times 0.02279=49.2\text{kg/s}$. Step 4 — specific impulse (chemistry L2.1 ke identical hai): $I_{sp}=278.9\text{s}$, toh $c=g_0{I}_{sp}=2736\text{m/s}$. Step 5 — thrust $F=\dot{m}c=49.2\times 2736=1.346\times 10^5\text{N}\approx 135\text{kN}$. Answers: $p_c\approx 15.9\text{MPa}$, $r\approx 0.0228\text{m/s}$, $\dot{m}\approx 49.2\text{kg/s}$, $I_{sp}\approx 279\text{s}$, $F\approx 135\text{kN}$. Notice karo causal chain: grain+throat $p_c$ set karte hain → $p_c$ $r$ set karta hai → $r$ $\dot{m}$ set karta hai → chemistry $I_{sp}$ set karti hai → dono $F$ mein combine hote hain.

Left side (gas made): ${\rho }_p{A}_{b}r=\dot{m}=49.2\text{kg/s}$ (L4.1 Step 3 se). Right side (gas vented): $\frac{p_c{A}_t}{c^{\ast }}=\frac{1.588\times 10^7\times 4.8\times 10^{-3}}{1550}=\frac{7.622\times 10^4}{1550}=49.2\text{kg/s}$. Dono sides $49.2\text{kg/s}$ ke barabar hain. ✓ Operating point self-consistent hai — grain bilkul usi rate par gas manufacture karta hai jis rate par throat expel kar sakta hai, yahi "equilibrium pressure" ka matlab hai.

$A_t$ chota karne se $p_c$ badhta hai ($p_c\propto (A_b/A_t{)}^{1/(1-n)}$ se), jo $r$ badhata hai, jo $\dot{m}$ badhata hai — toh $A_t$ certainly $\dot{m}$ aur $F$ mein rehta hai. Kya woh $I_{sp}$ mein bachta hai? $I_{sp}=\frac{v_e}{g_0},v_e=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}\left[1-{\left(\frac{p_e}{p_c}\right)}^{\frac{\gamma -1}{\gamma }}\right]}.$ Sirf jagah jahan $A_t$ $v_e$ mein enter ho sakta hai woh ratio $p_e/p_c$ ke through hai. Ab Nozzle Isentropic Expansion ka key fact: isentropic flow ke liye exit-to-throat area ratio $A_e/A_t$ pressure ratio $p_e/p_c$ fix karta hai isentropic area–Mach–pressure relation ke through $\frac{A_e}{A_t}=\frac{1}{M_e}{\left[\frac{2}{\gamma +1}\left(1+\frac{\gamma -1}{2}{M}_e^2\right)\right]}^{\frac{\gamma +1}{2(\gamma -1)}},\frac{p_e}{p_c}={\left(1+\frac{\gamma -1}{2}{M}_e^2\right)}^{-\frac{\gamma }{\gamma -1}},$ jahan $M_e$ exit Mach number hai. Inhe saath padho: geometric ratio $A_e/A_t$ pin karo aur $M_e$ fixed hai, jo $p_e/p_c$ pin karta hai — $p_c$ ki absolute value se independent. Chamber aur exit pressure dono saath scale up hote hain jab $p_c$ badhta hai, toh unka ratio sirf geometry se frozen hai. $p_e/p_c$ constant rakhe hue, $v_e$ mein absolute $A_t$ ka koi dependence nahi hai. Isliye $I_{sp}$ $A_t$ se independent hai. Answer: $A_t$ throughput ($\dot{m}$, $F$) scale karta hai lekin efficiency ($I_{sp}$) se cancel out ho jaata hai — same 80/20 split jaisa $A_b$ mein hai. Reason yeh hai ki sirf ratio $p_e/p_c$ (fixed area ratio $A_e/A_t$ se set), absolute pressure nahi, $v_e$ tak pahunchta hai.

Jaise $p_e/p_c\to 0$, bracket $\left[1-(p_e/p_c{)}^{(\gamma -1)/\gamma }\right]\to 1-0=1$ (uski ceiling). Toh $v_{e,\max }=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}}=\sqrt{10\times 1.209\times 10^6\times 1}=\sqrt{1.209\times 10^7}=3477\text{m/s}.$ $I_{sp,\max }=3477/9.81=354.4\text{s}.$ Answer: ceiling $v_e\approx 3477\text{m/s}$, $I_{sp}\approx 354\text{s}$. Physically yeh saari chamber enthalpy kinetic energy mein convert hai — isse hotter ya lighter gas ke bina beat nahi kar sakte. Ek real finite nozzle uska sirf $0.62$ fraction recover karta hai (L2.1 mein), $279\text{s}$ deta hai.

• Mass flow: $\dot{m}={\rho }_p{A}_{b}r\to 0$ — grain gas banana band karta hai.
• Chamber pressure: $p_c=({\rho }_{p}a{c}^{\ast }{A}_b/A_t{)}^{1/(1-n)}\to 0$ jaise $A_b\to 0$ (base $\to 0$, positive power) — chamber depressurise hota hai (tail-off).
• Specific impulse: $I_{sp}=v_e/g_0$. Lekin $v_e$ sirf $\gamma ,T_c,M,p_e/p_c$ par depend karta hai. Jaise $p_c\to 0$, ratio $p_e/p_c\to \infty$, bracket $\to 1-(\text{large}{)}^{+}<0$ — formula toot jaata hai kyunki expansion ab physical nahi hai (nozzle dead chamber se expand nahi kar sakta). Valid regime mein, usse pehle, $v_e$ aur isliye $I_{sp}$ apni design value ke paas rehte hain jabki $F=\dot{m}{v}_e\to 0$. Answer: thrust aur $\dot{m}$ smoothly zero par fade ho jaate hain burnout par; $I_{sp}$ apni chemistry-set value hold karta hai jab tak chamber pressure us point se neeche nahi gir jaata jahan nozzle still expand kar sake. Efficiency aakhir mein marti hai, aur suddenly, gradually nahi — classic thrust "tail-off."

Bracket constant rakh ke, $v_e\propto \sqrt{T_c/M}$.

• $T_c$ $20\%$ badhao: factor $\sqrt{1.20}=1.0954$ → $+9.54\%$.
• $M$ $20\%$ ghatao (matlab $M\to 0.8M$): factor $\sqrt{1/0.8}=\sqrt{1.25}=1.1180$ → $+11.80\%$. Answer: $M$ $20\%$ ghatana jeetta hai ($+11.8\%$ vs $+9.5\%$). Asymmetry: $T_c$ aur $M$ $\sqrt{T_c/M}$ ke andar aate hain, lekin denominator mein $20\%$ reduction ($\times 1/0.8$) numerator mein $20\%$ increase ($\times 1.2$) se bada multiplier hai. Light exhaust hot exhaust ko beat karta hai, dollar for dollar — aur chamber walls par bhi zyada aasaan hai.

Causal chain, ek line ::: geometry $(A_b,A_t)$ + burn law $\to p_c\to r\to \dot{m}\to$ (chemistry $v_e$ ke saath) $\to F$; jabki $I_{sp}=v_e/g_0$ sirf chemistry hai. Kyun $A_b$ aur $A_t$ $I_{sp}$ se cancel ho jaate hain ::: dono $\dot{m}$ aur $F$ ko saath scale karte hain, aur $I_{sp}=F/(\dot{m}{g}_0)$ unhe divide kar deta hai; sirf ratio $p_e/p_c$ (area ratio $A_e/A_t$ se fixed) $v_e$ tak pahunchta hai. Fixed chemistry ke liye $I_{sp}$ ki ceiling ::: $v_{e,\max }=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}}$ at $p_e/p_c\to 0$ (full enthalpy conversion). $n=0$ par kya hota hai? ::: burn rate $r=a$ pressure-independent ban jaata hai (plateau propellant); pressure feedback loop decoupled ho jaata hai — maximally stable, koi runaway nahi. Better lever: $+20\%$ $T_c$ ya $-20\%$ $M$? ::: $-20\%M$ ($\times 1.118$) $+20\%T_c$ ($\times 1.095$) ko beat karta hai.