Burn rate r = a·P^n — Vieille's law: Visual walkthrough

3.3.36 · D2 · Physics › Rocket Propulsion › Burn rate r = a·P^n — Vieille's law

Step 1 — Burning surface ek aisi wall hai jo move karti hai

KYA: hum moving-wall speed ko $r$ naam dete hain. KYUN: solid rocket ki har cheez — thrust, pressure, burn time — isi wall ke kitni tezi se move karne ke downstream hai. Isliye pehle isko apna symbol milna chahiye. PICTURE: figure mein magenta line time $t$ par burning surface hai; violet dashed line woh jagah hai jahan woh ek instant baad hogi. Unke beech ka gap, time se divide karo, woh $r$ hai.

Step 2 — Wall ko move karne ke liye, tumhe agla layer cook karna hoga

Chalo energy demand banate hain ek second ki burning ka, surface ke unit area ke liye.

KYA: ek second mein wall $r$ distance sweep karti hai, toh thickness $r$ ka ek slab (unit area per) consume hota hai. Uska mass hai $ρ_{p} \times r$ — yeh mass flux hai, har square metre mein har second cross karne wala mass.

$c_{p} (T_{s} - T_{0})$ se multiply kyun karte hain? Har kilogram ko $T_{0}$ se $T_{s}$ tak le jaana padta hai; uske liye $c_{p} (T_{s} - T_{0})$ joules per kilogram lagte hain. Mass-per-second ko joules-per-kilogram se multiply karo → joules per second, yani power demand:

$\overset{q}{˙}_{need} = mass in a m^{3} ρ_{p} layers/s r J per kg per K c_{p} temperature climb (T_{s} - T_{0})$

PICTURE: figure mein shaded band woh layer hai jo ek second mein sweep hoti hai; arrow dikhata hai ki woh $T_{0}$ se $T_{s}$ tak temperature ladder climb kar rahi hai.

Step 3 — Flame woh heat conduction se deliver karta hai

Heat kitni tezi se conduct hoti hai, iska rule hai Fourier's law: heat flux proportional hoti hai temperature ke slope ke.

KYA: temperature $T_{f}$ (flame par) se $T_{s}$ (surface par) tak gap $δ$ ke across girta hai. Us drop ka slope hai $(T_{f} - T_{s}) / δ$ .

$δ$ se divide kyun karte hain? Gradient ek change hai distance ke upar. Usi temperature drop ko chote gap mein squeeze karo aur slope steep ho jaata hai, toh heat zyada tezi se neeche bahti hai. Yahi is law ka poora engine hai.

$\overset{q}{˙}_{supply} \approx λ \frac{T _{f} - T _{s} temp drop}{gap it drops across δ}$

PICTURE: orange gradient triangle dikhata hai temperature $T_{f}$ se $T_{s}$ tak distance $δ$ ke upar girta hua. Slope jitna steep (chota $δ$ ), utna mota downward heat arrow.

Step 4 — Steady burning mein supply exactly equals demand

KYA: dono expressions ko equal karo. KYUN: isse hum $r$ — woh quantity jis mein hume interest hai — ko solve kar sakte hain baaki sab ke terms mein.

$ρ_{p} r c_{p} (T_{s} - T_{0}) = λ \frac{T _{f} - T _{s}}{δ}$

Rearrange karo (dono sides ko $ρ_{p} c_{p} (T_{s} - T_{0})$ se divide karo):

$r = \frac{λ ( T _{f} - T _{s} )}{ρ _{p} c _{p} ( T _{s} - T _{0} ) δ}$

Is formula ka message padho: $r$ inversely proportional hai $δ$ se. Iske baaki sab kuch — conductivity, densities, temperatures — kisi given propellant ke liye roughly fixed rehta hai. Toh bacha hua ek hi knob hai standoff distance $δ$ :

$r \propto \frac{1}{δ}$

PICTURE: do panels — ek wide gap (weak, cool, slow $r$ ) versus ek hugging flame (chota gap, fierce, fast $r$ ).

Step 5 — Pressure flame ko paas kheencha hai: $δ \propto P^{- n}$

KYA: ek power $P^{- n}$ (not $- P$ , not $e^{- P}$ ). Power kyun aur kuch nahi? Gas mein reaction rates concentration ko power tak raise karne se scale karti hain (do molecules milte hain → concentration $^{2}$ ; concentration khud $\propto P$ hai). Jo bhi "rate $\propto P^{power}$ " se bana hai, woh distance ke liye bhi power law pe collapse hota hai. Power law natural shape hai jab underlying physics hai "pressure multiply karo aur effect fixed factor se multiply ho jaata hai."

PICTURE: teen flames $P$ , $2 P$ , $4 P$ par — har ek apna standoff roughly halving karta hua, scale par drawn.

Step 6 — Substitute karo aur constants ko $a$ mein collect karo

KYA: $δ \propto P^{- n}$ ko $r \propto 1/ δ$ mein daalo.

$r \propto \frac{1}{δ} \propto \frac{1}{P ^{- n}} = P^{n}$

Constant collect kyun karte hain? Jo kuch bhi $P$ nahi hai — conductivity $λ$ , densities, heat capacities, temperatures, $δ \propto P^{- n}$ mein proportionality factor — woh pressure ke fluctuate hone par kabhi nahi badalta. Inhe sab ek number mein bundle karo aur $a$ bol do:

$r = a P^{n}$

PICTURE: collapse — physical constants ka ek cloud single box $a$ mein funnel hota hua, peeche saaf law chhod ke.

Step 7 — Edge cases (kabhi gap mat chhodo)

$n = 0$ : toh $P^{0} = 1$ , toh $r = a$ — burn rate bilkul pressure se deaf hai. Ek flat horizontal line. (Koi real propellant exactly aisa nahi hota, lekin yeh "no sensitivity" limit hai.)
$0 < n < 1$ (useful zone): $r$ $P$ ke saath badhta hai lekin $P$ se dheere — ek gently rising curve. Yahan usable propellants rehte hain aur motor stable hota hai (generation $\propto P^{n}$ nozzle exhaust $\propto P^{1}$ se peeche reh jaata hai).
$n = 1$ : $r = a P$ , origin se ek straight line — woh knife-edge jahan generation aur exhaust ek hi pace se badhte hain. Koi bhi nudge na dampen hogi na amplify.
$n > 1$ : $r$ $P$ se tezi se upar jaata hai. Ek pressure bump generation ko exhaust se aage kar deta hai → runaway → combustion instability ka khatra.
$P \to 0$ : $r \to 0$ — koi push nahi, koi flame wapas nahi pahuncha, kuch nahi jalta. Formula floor par sensibly behave karta hai.

Chaaron $n$ shapes kyun dikhate hain? Kyunki ek akela number $n$ chupke se decide karta hai ki tumhara motor ek controllable engine hai ya bomb, aur reader ko chaaron regimes side by side dikhne chahiye.

PICTURE: chaaron curves $r$ vs $P$ ek hi axis par, colour-coded, har ek par stability verdict labelled.

Ek picture summary

Poori derivation ek single chain hai: heat needed = heat supplied → $r$ solve karo → $r \propto 1/ δ$ → $δ \propto P^{- n}$ → $r = a P^{n}$ .

Step 2: demand ρ_{p} r c_{p} (T_{s} - T_{0}) = Step 3: supply λ \frac{T _{f} - T _{s}}{δ} \Rightarrow Step 4 r \propto \frac{1}{δ}, Step 5 δ \propto P^{- n} \Rightarrow Steps 6 r = a P^{n}

Recall Feynman retelling — poora walkthrough plain words mein

Ek candle imagine karo jo neeche apne aap mein jalti hai. Fiery face kitni tezi se sink karta hai woh burn rate $r$ hai (Step 1). Ek aur layer neeche jaane ke liye, woh thanda layer cold ( $T_{0}$ ) se gas mein melt ( $T_{s}$ ) tak cook hona chahiye — aur cooking har second ek fixed amount heat khaati hai (Step 2). Woh heat deliver karta hai flame jo bilkul upar float kar raha hai, ek tiny gap $δ$ ke across neeche conduct karta hua; gap jitna chota, temperature drop ka slope utna steep aur heat utni tezi se neeche bahti hai (Step 3). Steady burning mein needed heat equals delivered heat, aur jab tum $r$ solve karte ho toh pata chalta hai woh bas $1/ δ$ hai constants ke pile times (Step 4). Ab poori cheez ko pressure se squeeze karo: zyada pressure gas ko zyada tight pack karta hai, reactions jaldi khatam hoti hain, toh flame paas aa jaata hai — $δ$ $P^{- n}$ ki tarah shrink karta hai (Step 5). Ulta karo: $r \propto 1/ δ \propto P^{n}$ , har unchanging constant ko ek letter $a$ mein sweep karo, aur $r = a P^{n}$ nikalti hai (Step 6). Aakhir mein, exponent $n$ propellant ki personality hai: 1 se neeche woh ek calm, controllable engine hai; 1 se upar woh ek bomb hai jo pressure hiccup ka intezaar kar raha hai (Step 7).

Recall

$r$ kis single quantity se inversely proportional hai? ::: Flame standoff distance $δ$ se. $P$ badhane se $δ$ kyun shrink karta hai? ::: Dense gas → faster reactions → flame front surface ke zyada paas baith jaata hai. Final law mein saare physical constants kahan jaate hain? ::: Woh single coefficient $a$ mein collect ho jaate hain. $P$ -dependence power law kyun hai exponential ki jagah? ::: Reaction rates concentration (isliye $P$ ) ke power se scale karti hain, toh standoff distance bhi $P$ ke power ko follow karti hai. $n > 1$ ka verdict? ::: Unstable — generation nozzle exhaust se aage nikal jaata hai, thermal runaway.