Exercises — Solid propellants — fuel + oxidizer in polymer matrix

• ${\text{NH}}_4{\text{ClO}}_4$ = oxidizer — yeh oxygen supply karta hai taaki mixture bina atmosphere ke jale.
• Aluminium = fuel/metal — teezi se jalta hai, zyaatar energy release karta hai.
• HTPB = polymer binder — powder ko ek solid, castable grain mein chipkata hai aur khud bhi ek hydrocarbon fuel hai jo jalta hai. Yeh mnemonic hai O-F-B.

Tum $A_b$, yaani burning surface area change karte ho. Kyunki $\dot{m}\propto A_b$ aur thrust $F=\dot{m}{v}_e$, isliye burning surface ka samay ke saath shape hi thrust curve set karta hai. ${\rho }_p$ aur $r$ (fixed pressure par) port shape par depend nahi karte.

$r=6\text{mm/s}=0.006\text{m/s}$ convert karo (SI quantities multiply karne se pehle hamesha millimetres ko metres mein badlo). (a) $\dot{m}={\rho }_p{A}_{b}r=1750\times 0.040\times 0.006=0.42\text{kg/s}$. Kyun: volume eaten per second $=A_{b}r$; mass ke liye density se multiply karo. (b) Kyunki $p_e=p_a$, pressure term zero hai, isliye $F=\dot{m}{v}_e=0.42\times 2300=966\text{N}$.

Full formula: $F=\dot{m}{v}_e+(p_e-p_a)A_e$. Momentum part unchanged: $\dot{m}{v}_e=966\text{N}$. Pressure part: $(p_e-p_a)A_e=(20000-0)\times 0.012=240\text{N}$. $F=966+240=1206\text{N}$. Kyun: vacuum mein ($p_a=0$) exit pressure ek bonus push deta hai — isliye rockets space mein sea level se zyada powerful hote hain.

Ratio lo taaki constant $a$ cancel ho jaaye: $\frac{r_2}{r_1}={\left(\frac{p_2}{p_1}\right)}^n=(7/5{)}^{0.35}=1.4^{0.35}.$ $\ln 1.4=0.33647$; $\times 0.35=0.11777$; $e^{0.11777}=1.1250$. $r_2=8\times 1.1250=9.0\text{mm/s}$. Exponent kyun, simple multiply kyun nahi? Vieille's law ek power law hai: pressure double karne se burn rate double nahi hota. 40% pressure rise se sirf ≈13% zyada burn rate milta hai kyunki $n<1$.

Gas ${\dot{m}}_{\text{gen}}={\rho }_p{A}_{b}a{p}_c^n$ rate par produce hoti hai — $p_c^n$ ki tarah badhti hai. Gas throat se ek rate par nikti hai jo (choked flow ke liye) $p_c^1$ ki tarah badhti hai — linearly. Agar pressure thoda $\delta$ se badhe:

• generation ek factor $\propto p_c^n$ se badhta hai → slope $n$,
• discharge ek factor $\propto p_c^1$ se badhta hai → slope $1$. Agar $n<1$ discharge generation se aage nikal jaata hai → pressure wapas girta hai → stable. Agar $n\ge 1$ generation discharge se aage nikal jaata hai → pressure aur badhta hai → runaway / detonation. Figure dekho: jahan "make gas" curve operating point ke baad "dump gas" line ke neeche hai, motor khud correct ho jaata hai.

(a) $\dot{m}=1800\times 0.10\times 0.007=1.26\text{kg/s}$; $F=\dot{m}{v}_e=1.26\times 2500=3150\text{N}$. (b) Neutral (constant) thrust, isliye $I=Ft=3150\times 6.0=18900\text{N⋅s}$. $I=Ft$ yahan kyun kaam karta hai: sirf isliye ki $A_b$ (hence $F$) constant hai — star port ka poora yahi point hai. (c) Propellant burned $=\dot{m}\times t=1.26\times 6.0=7.56\text{kg}$.

$I_{sp}=v_e/g_0=2500/9.81=254.8\text{s}$. (Dekho Specific Impulse.) Check: $I_{sp}{g}_0{m}_{\text{prop}}=254.8\times 9.81\times 7.56=2500\times 7.56=18900\text{N⋅s}$ — Q7(b) se match karta hai. Kyun dono definitions agree karti hain: $I=Ft=\dot{m}{v}_{e}t=v_e(\dot{m}t)=v_e{m}_{\text{prop}}$, aur $v_e=I_{sp}{g}_0$.

Final mass $m_f=40-7.56=32.44\text{kg}$. $\Delta v=2500\ln (40/32.44)=2500\times \ln (1.23305)=2500\times 0.20951=523.8\text{m/s}$. Logarithm kyun? Kyunki jaise fuel jalta hai vehicle halka hota jaata hai, isliye har baad ka kilogram exhaust ek chhotey ship ko zyada accelerate karta hai — running total ek log hai, seedhi line nahi.

(a) $A_b\propto r_{\text{port}}$ ($L$ fixed ke saath), isliye factor hai $30/20=1.5$. (b) Thrust $1.5-1=0.5=50\%$ badhta hai. Burn ke dauran thrust badhna ek progressive curve hai. (c) Star port adds surface (points wapas burn karte hain) bilkul wahan jahan plain tube ise kho deta, $A_b$ ≈ constant rakhta hai → neutral (flat) thrust. Neeche geometry dekho.

Burn time $I=Ft$ se: $t=30000/5000=6.0\text{s}$. $\dot{m}=F/v_e=5000/2500=2.0\text{kg/s}$. Mass burned $=\dot{m}t=2.0\times 6.0=12.0\text{kg}$. Area $\dot{m}={\rho }_p{A}_{b}r$ se: $A_b=\frac{\dot{m}}{{\rho }_{p}r}=\frac{2.0}{1800\times 0.007}=\frac{2.0}{12.6}=0.1587{\text{m}}^2$. Backwards kyun kaam karte hain? Requirements $F$ aur $I$ fix karti hain; tum phir geometry ke liye solve karte ho — exactly aisa hi real solid motor design hota hai: grain shape target thrust curve hit karne ke liye choose ki jaati hai.

$\frac{r_2}{r_1}=(6.6/6{)}^{0.40}=1.1^{0.40}$. $\ln 1.1=0.09531$; $\times 0.40=0.038124$; $e^{0.038124}=1.03886$. Toh $r$ (aur $\dot{m}$, aur thrust) ≈ $3.9\%$ badhta hai 10% pressure rise ke liye. Kyunki $n=0.40<1$, 10% disturbance ek chhotey 3.9% response produce karta hai — motor self-limiting aur stable hai. Accha design.

Burn data se thrust ::: $F=\dot{m}{v}_e$, jahan $\dot{m}={\rho }_p{A}_{b}r$ Pressure term kab add karo? ::: jab bhi $p_e\ne p_a$: $F=\dot{m}{v}_e+(p_e-p_a)A_e$ Neutral motor ka total impulse ::: $I=Ft$ (flat thrust curve ke neeche ka area) Chhotey pressure change ⇒ burn-rate change ::: fractional pressure change ko $n$ se multiply karo Round port ki jagah star port kyun ::: $A_b$ constant rakhta hai ⇒ neutral (flat) thrust