Solid rocket Isp derivation from grain properties: Visual walkthrough

3.3.38 · D2 · Physics › Rocket Propulsion › Solid rocket Isp derivation from grain properties

Step 0 — Woh words jo hum use kar sakte hain

Kisi bhi algebra se pehle, physical pictures par agree kar lete hain. Neeche jo bhi hai woh sab cheezein hain jinhe tum point karke dikha sakte ho.

Definition Starting objects

Grain — fuel ka solid block. Ek candle socho, bas shaped.
Burning surface $A_{b}$ — woh exposed area jo abhi jal raha hai, square metres mein measure hota hai ( $m^{2}$ ). Sirf yahi ek jagah hai jahan solid gas banta hai.
Burn rate $r$ — fire solid ke andar kitni tezi se jaati hai, metres per second mein ( $m/s$ ). Surface is speed se andar ki taraf badhti hai.
Density $ρ_{p}$ — ek cubic metre mein kitne kilograms solid hain ( $kg/m^{3}$ ).
Chamber — woh sealed space jahan gas hot aur pressurised hokar ikatti hoti hai.
Chamber pressure $p_{c}$ — chamber ke andar gas ka pressure (pascals, Pa).
Nozzle exit — bilkul peeche ka wide open munh jahan gas finally nikalti hai.
Exit pressure $p_{e}$ — us munh par gas ka pressure, jab gas expand aur cool ho chuki hoti hai (Pa). Hamesha $p_{c}$ se kam hota hai.
Molar mass $M$ — exhaust gas ke ek mole ka mass ( $kg/mol$ ); "har gas molecule kitna bhaari hai." Halke molecules (jaise water vapour, $H_{2}$ ) zyada tezi se nikalte hain.
Exhaust — woh hot gas jo peeche se nikalti hai. Uski speed hi hamara asli concern hai.

Prerequisite pictures Grain Geometry and Thrust Profiles aur Saint-Robert Burn Rate Law mein hain; jo pieces chahiye woh hum yahan rebuild kar lete hain.

Step 1 — Ek shell of solid ko gas bante hue dekho

KYA. Hum poochh rahe hain: ek tiny time slice $d t$ mein, kitne kilograms solid gas ban jaata hai?

KYUN. Thrust hai "mass thrown backward per second times how fast." Toh sabse pehle humein chahiye ki grain har second kitna mass produce karta hai. Jab tak yeh number nahi hoga, rocket ke baare mein kuch bhi kaam nahi karega.

PICTURE. Figure dekho: burning surface (cyan) ek paper-thin shell of thickness $r d t$ (amber) ke andar chali gayi hai. Woh shell woh propellant hai jo $d t$ ke dauran vaporise hua.

Figure — Solid rocket Isp derivation from grain properties

Us thin shell ka volume hai area × thickness:

$d V = fire kitna wide hai A_{b} \times time d t mein kitna deep gaya (r d t)$

Volume ko density se multiply karo mass milega, phir $d t$ se divide karo rate milegi:

$d m = ρ_{p} A_{b} (r d t) ⟹ \overset{m}{˙} = ρ_{p} A_{b} r$

$\overset{m}{˙}$ (kg/s) — "dot" ka matlab hai per second: har second produce hone wala mass.
$ρ_{p} A_{b} r$ — density $\times$ area $\times$ speed. Units check: $\frac{kg}{m ^{3}} \cdot m^{2} \cdot \frac{m}{s} = \frac{kg}{s}$ . ✓

Step 2 — Burn rate kya set karta hai? Pressure fire ko push karta hai

KYA. Hum $\overset{m}{˙}$ ke andar ke $r$ ko khol rahe hain: burn rate koi fixed number nahi hai, yeh chamber pressure $p_{c}$ ke saath badhta hai (Step 0 mein define kiya).

KYUN. Pressure mein aag hotter aur surface ke against tezi se jalti hai. Rocket engineers ne yeh measure kiya aur ek curve fit kiya — Saint-Robert Burn Rate Law.

PICTURE. Figure kuch exponents $n$ ke liye $r$ ko $p_{c}$ ke against plot karta hai. Kyunki $n < 1$ hai, curve concave down hai — badhta rehta hai lekin zyada slowly, jaise pressure badha waise flatten hota jaata hai. Wahi flattening exactly hai jo motor ko blast hone se bachati hai (concave-up, steepening curve danger wali case hoti).

$r = a p_{c}^{n}$

$a$ — ek coefficient jo chemistry aur temperature se bana hota hai (curve ki height).
$p_{c}$ — chamber pressure (pascals).
$n$ — pressure exponent, response ki "steepness," typically $0.2$ – $0.5$ .

Step 3 — Gas kitni tezi se nikalta hai? Heat ko speed mein trade karo

KYA. Hum ab "kitna gas" se "gas kitna fast" ki taraf switch kar rahe hain. Nozzle chamber gas ki heat energy ko directed speed $v_{e}$ mein convert karta hai.

KYUN. Exhaust ki speed hi actually rocket ko push karti hai. Jo picture yaad rakhni hai: chamber mein ek hot, slow, crowded gas ek throat se squeeze hoti hai aur fan out hoti hai, cold aur fast nikal ke aati hai.

PICTURE. Figure dikhata hai chamber (hot, high $T_{c}$ , high $p_{c}$ , slow) → throat → nozzle exit (cold, low $p_{e}$ , fast $v_{e}$ ). Energy conserved hai: thermometer girta hai jaise speedometer badhta hai.

Algebra se pehle, do naye symbols. Jab tum ek kilogram gas ko ek degree warm karte ho, jo energy lagti hai woh is baat par depend karti hai ki usse expand hone dete ho ya nahi:

Ab per kilogram gas ka energy bookkeeping ("enthalpy in = enthalpy out + kinetic energy gained"):

$chamber mein heat energy c_{p} T_{c} = bachhi hui heat c_{p} T_{e} + jet ki kinetic energy \frac{1}{2} v_{e}^{2}$

Sirf ise solve karne se $v_{e} = 2 c_{p} (T_{c} - T_{e})$ milta hai — lekin $T_{e}$ directly measure nahi kar sakte. Hum sirf pressures $p_{c}$ aur $p_{e}$ jaante hain. Toh humein temperatures se pressures tak ek bridge chahiye.

The bridge — isentropic relation. Ek gas ke liye jo bina heat leak hue smoothly expand hoti hai (called isentropic), temperature aur pressure ek doosre se locked hote hain:

$\frac{T _{e}}{T _{c}} = (\frac{p _{e}}{p _{c}})^{\frac{γ - 1}{γ}}$

$γ = c_{p} / c_{v}$ — abhi define kiya gaya heat-capacity ratio.
Yeh shape kyun? Jab pressure drop hota hai, gas apne aap ko bahar push karke kaam karta hai, aur woh kaam apni khud ki heat se pay hota hai — isliye temperature ko pressure ke saath girna hi padta hai. Exponent $\frac{γ - 1}{γ}$ exact exchange rate hai, Nozzle Isentropic Expansion mein derive kiya gaya hai.

$T_{e} = T_{c} (\frac{p _{e}}{p _{c}})^{(γ - 1) / γ}$ ko $v_{e} = 2 c_{p} (T_{c} - T_{e})$ mein substitute karo, phir $c_{p} = \frac{γ}{γ - 1} \frac{R}{M}$ use karo (specific heat gas constant aur molar mass ke terms mein likha). Yeh deta hai:

$v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M} [1 - (\frac{p _{e}}{p _{c}})^{\frac{γ - 1}{γ}}]$

Ise term by term padho:

$\frac{2 γ}{γ - 1}$ — gas type se ek pure number.
$\frac{R}{M}$ — yeh gas ka specific gas constant hai (units $J/(kg\cdotK)$ ): universal constant $R = 8.314 J/(mol\cdotK)$ ko molar mass $M$ se divide karna "per mole" ko "per kilogram" mein badal deta hai. Toh $\frac{R T _{c}}{M}$ per kilogram energy ka scale hai: hotter chamber ( $T_{c}$ ↑) ya lighter molecules ( $M$ ↓, Step 0 mein define) ka matlab hai per kilogram zyada energy convert hogi.
Bracket $[1 - (\frac{p _{e}}{p _{c}})^{(γ - 1) / γ}]$ — actually cash ki gayi energy ka fraction. Zyada expand karo ( $p_{e}$ ko $p_{c}$ se bahut neeche drop karo) → bracket → 1 → maximum speed.

Step 4 — Control volume se thrust assemble karo

KYA. "Mass per second" (Step 1) aur "speed" (Step 3) ko combine karke force $F$ banao.

KYUN. Force momentum leave hone ki rate hai. Har second, $\overset{m}{˙}$ kilograms $v_{e}$ speed se jaate hain, $\overset{m}{˙} v_{e}$ momentum backward carry karte hain — rocket equal-and-opposite forward kick feel karta hai.

PICTURE. Figure motor ke around ek dashed box (control volume) draw karta hai. Gas momentum right se bahar flow karta hai; exit par ek leftover pressure mismatch ek doosra, chota push add karta hai.

$F = momentum thrust \overset{m}{˙} v_{e} + pressure thrust (p_{e} - p_{a}) A_{e}$

$\overset{m}{˙} v_{e}$ — main event: mass flow $\times$ exhaust speed.
$p_{e}$ — exit pressure (Step 0); $p_{a}$ — outside air pressure; $A_{e}$ — exit area.
$(p_{e} - p_{a}) A_{e}$ — ek bonus (ya penalty) agar jet surroundings se alag pressure par leave kare.

Step 5 — $I_{s p}$ define karo aur grain ko cancel hote dekho

KYA. Specific impulse banao — "jale hue weight ke har kilogram per kick."

KYUN. Hume ek fairness number chahiye jo ek tiny motor aur ek giant motor ko — same stuff se bane — compare kare. Thrust ko mass flow se divide karne se "kitna bada" remove ho jaata hai aur "kitna acha" bachta hai.

PICTURE. Figure do motors dikhata hai — ek bada $A_{b}$ , ek chota — gas same speed se spitting karte. Bada wala zyada gas banata hai (taller arrow bundle), lekin har kilogram equally good hai.

Effective exhaust velocity $c = F / \overset{m}{˙}$ define karo, phir

$I_{s p} = \frac{c}{g _{0}} = \frac{F}{m ˙ g _{0}} = \frac{1}{g _{0}} [v_{e} + \frac{( p _{e} - p _{a} ) A _{e}}{m ˙}]$

$g_{0} = 9.81 m/s^{2}$ — ek bookkeeping constant, taaki answer seconds mein aaye. Local gravity nahi.
Notice karo $\overset{m}{˙}$ denominator mein hai: numerator ke $F$ mein $A_{b}$ ne jo boost kiya tha woh wापस divide ho jaata hai.

Matched nozzle ke liye ( $p_{e} = p_{a}$ ) pressure term khatam ho jaata hai:

$I_{s p} = \frac{v _{e}}{g _{0}} = \frac{1}{g _{0}} \frac{2 γ}{γ - 1} \frac{R T _{c}}{M} [1 - (\frac{p _{e}}{p _{c}})^{\frac{γ - 1}{γ}}]$

Har symbol chemistry ya nozzle ka hai ( $γ, T_{c}, M, p_{e} / p_{c}$ ). Grain quantities $ρ_{p}, A_{b}, a, n$ chale gaye. Yahi wahi conclusion hai jo Tsiolkovsky Rocket Equation view se milta hai: efficiency exhaust speed mein hai, is baat mein nahi ki tum kitna mass carry karte ho.

Step 6 — Grain KAHAAN matter karta hai: chamber pressure

KYA. Grain geometry ek cheez zaroor control karti hai jo wapas loop hoti hai: equilibrium chamber pressure $p_{c}$ .

KYUN. Grain se bana gas throat se vent hone wale gas ke equal hona chahiye, warna pressure change hoga. Made = vented set karna $p_{c}$ pin karta hai.

"Gas vented" side derive karna. Narrow throat itna hi gas pass kar sakta hai. Choked-flow theory (Characteristic Velocity c-star se) kehti hai ki area $A_{t}$ ke throat se nikalne wala mass chamber pressure aur ek chemistry-only speed constant $c^{*}$ se set hota hai:

$c^{*} \equiv \frac{p _{c} A _{t}}{m ˙ _{vent}} ⟹ \overset{m}{˙}_{vent} = \frac{p _{c} A _{t}}{c ^{*}}$

$c^{*}$ — characteristic velocity, defined as chamber pressure times throat area divided by mass flow jo woh pass karne deta hai. Yeh measure karta hai "vented mass ke per unit kitna pressure chamber build karta hai," aur sirf $T_{c}$ , $M$ , $γ$ par depend karta hai.
Bada $p_{c}$ ya bada throat $A_{t}$ → zyada gas vented. Makes sense.

Made = vented set karo. Gas made hai $\overset{m}{˙} = ρ_{p} A_{b} r$ (Step 1); gas vented hai $p_{c} A_{t} / c^{*}$ . Steady operation mein ye equal hote hain:

$ρ_{p} A_{b} r = \frac{p _{c} A _{t}}{c ^{*}}$

PICTURE. Figure ek balance scale hai: left par "gas made" $= ρ_{p} A_{b} r$ ; right par "gas vented" $= p_{c} A_{t} / c^{*}$ . Yeh ek pressure par settle hote hain.

$r = a p_{c}^{n}$ (Step 2) daalo aur $p_{c}$ ke liye solve karo:

$ρ_{p} A_{b} a p_{c}^{n} = \frac{p _{c} A _{t}}{c ^{*}} ⟹ p_{c}^{1 - n} = \frac{ρ _{p} a c ^{*} A _{b}}{A _{t}} ⟹ p_{c} = (\frac{ρ _{p} a c ^{*} A _{b}}{A _{t}})^{\frac{1}{1 - n}}$

Yahan grain terms ( $ρ_{p}, a, A_{b}$ ) finally appear hote hain — lekin sirf pressure ke liye, jo thrust aur burn time feed karta hai, $I_{s p}$ formula ko nahi.

Step 7 — Degenerate case: $n \to 1$ blow up kar deta hai

KYA. Examine karo ki $p_{c}$ ka kya hota hai jab exponent $n$ 1 ke paas (aur 1 se aage) jaata hai.

KYUN. Yahi edge case hai jo parent ne flag kiya tha. Humein reader ko exactly dikhana hai ki stability kahan khatam hoti hai.

PICTURE. Figure exponent $\frac{1}{1 - n}$ ko $n$ ke against plot karta hai: $n = 0.3$ ke paas flat, phir $n \to 1$ ke saath infinity ki taraf shoot karta hai. Ek vertical amber asymptote cliff ko mark karta hai.

$p_{c} \propto (\dots)^{\frac{1}{1 - n}}$

$n = 0.35$ par: exponent $= 1/0.65 = 1.54$ — mild, self-correcting.
$n \to 1^{-}$ ke saath: exponent $\to + \infty$ — base mein ek tiny sa change $p_{c}$ ko explode kar deta hai.
$n \geq 1$ par: koi stable equilibrium nahi — ek pressure bump ek bade bump ko feed karta hai. Thermal runaway.

Ek-picture summary

Yeh single diagram poore walkthrough ko compress karta hai: grain mass flow feed karta hai (left, geometry yahan rehti hai), chemistry+nozzle exhaust speed set karte hain (right), aur $I_{s p} = F / (\overset{m}{˙} g_{0})$ geometry ko wapas divide kar deta hai — efficiency ko ek pure-chemistry number bana ke chodta hai.

Recall Feynman retelling — poora page simple words mein

Ek shaped candle socho jo ek can ke andar jal rahi hai. Flame ki wall (uska area) decide karti hai ki har second kitna smoke nikalta hai — yahi Step 1 hai. Can ko zyada squeeze karo aur flame tezi se khaati hai, lekin sirf thodi si tezi se, yahi cheez use blow up hone se bachati hai — Steps 2 aur 7. Saara woh smoke back nozzle se rush karta hai, apni heat ko speed mein trade karta hua: hotter aur lighter smoke tezi se nikalta hai — Step 3. Forward kick hai smoke-per-second times smoke-speed — Step 4. Ab yahan magic trick hai: quality judge karne ke liye hum kick ko smoke-per-second se divide karte hain, aur "flame kitna bada hai" wali part khud ko cancel kar leti hai — Step 5. Toh same wax ki ek giant candle aur ek tiny candle smoke ko same speed se shoot karti hain aur unka quality score same hota hai; giant sirf zyada smoke banata hai. Flame ka size sirf andar ka pressure set karne ke liye wapas aata hai — Step 6 — quality nahi. Quality smoke speed ke baare mein hai, aur smoke speed chemistry hai.

Quick self-check

\overset{m}{˙}

formula aur har factor ka matlab

\overset{m}{˙} = ρ_{p} A_{b} r

: density × burning area × burn rate = produced kg/s.

A_{b}

I_{s p}

se kyun vanish ho jaata hai?

A_{b}

F

aur

\overset{m}{˙}

dono ko equally multiply karta hai;

I_{s p} = F / (\overset{m}{˙} g_{0})

usse cancel kar deta hai.

Woh ek cheez kya hai jo grain geometry set karta hai aur wapas loop hoti hai?

Equilibrium chamber pressure

p_{c} = (ρ_{p} a c^{*} A_{b} / A_{t})^{1/ (1 - n)}

n < 1

kyun hona chahiye?

Exponent

1/ (1 - n)

n \to 1

ke saath diverge kar jaata hai, thermal runaway deta hai.