Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn: Worked examples

3.3.37 · D3 · Physics › Rocket Propulsion › Grain geometry — BATES, star, wagon wheel; neutral - progres

Shuru karne se pehle, teen symbols jo parent se laye hain, simple words mein dobara batate hain taaki koi undefined cheez kabhi touch na ho:

Definition Paanch symbols (expand karne ke liye click karo)

$w$ = web burned = kitne metres tak burning surface solid ke andar perpendicular direction mein ghus gayi. $w = 0$ ignition par.
$A_{b}$ = burning surface area (m²) = abhi kitna propellant aag par hai.
$r$ = linear burn rate (m/s) = kitni tezi se har burning surface andar ki taraf khaati hai. Ek given instant par surface ke har jagah same value hoti hai.
$ρ_{p}$ = propellant density (kg/m³).
$\overset{m}{˙}$ = mass generation rate (kg/s) = $ρ_{p} A_{b} r$ (mass = density × area × thickness-per-second).

Scenario matrix

Hume har tarika cover karna hai jisme $A_{b}$ behave kar sakta hai. Yeh puri list hai, har ek ke saath example jo use hit karta hai:

Cell	Case class	Degenerate / limit?	Example
C1	Progressive — $A_{b}$ badhta hai (BATES inner-only)	—	Ex 1
C2	Regressive — $A_{b}$ girta hai (outer-burning / shrinking annulus)	—	Ex 2
C3	Neutral — $A_{b}$ ≈ const (end-burner)	zero-slope limit	Ex 3
C4	Competing terms — port badhta hai aur ends shrink hote hain (full BATES)	$d A_{b} / d w$ ka sign	Ex 4
C5	Degenerate: web khatam hota hai — $w \to w_{ma x}$ , area $\to 0$	limiting/zero input	Ex 5
C6	Perimeter → thrust — star vs wagon wheel, high initial $A_{b}$	large-input limit	Ex 6
C7	Pressure amplification — $\frac{1}{1 - n}$ exponent kaata hai	non-linear map	Ex 7
C8	Real-world word problem — mission ke liye geometry choose karo	design choice	Ex 8
C9	Exam twist — "kaun sa neutral hai?" hidden regressive tail ke saath	trap case	Ex 9

Neeche har numeric answer machine-verified hai.

C1 — Progressive: BATES inner surface only

Inner surface = ek cylinder wall. Uska area hai circumference × length: $A_{b} (w) = 2 π (R_{i} + w) L .$ Yeh step kyun? Jab sirf inner face jalta hai, burning surface wahi cylindrical hole ki wall hai. Radius $ρ$ aur length $L$ wali cylinder wall ka area $2 π ρ L$ hota hai (ise ek rectangle mein unroll karo: width = circumference $2 π ρ$ , height = $L$ ). Radius $R_{i} + w$ ho jaati hai kyunki wall $w$ se bahar ki taraf recede karti hai.
$w = 0$ plug karo: $A_{b} (0) = 2 π (0.02) (0.30) = 0.03770 m^{2} .$ Yeh step kyun? Hum pehle ignition par evaluate karte hain taaki starting burning area fix ho — woh baseline jisse baad ki har value compare hoti hai.
$w = 0.01$ plug karo: $A_{b} (0.01) = 2 π (0.03) (0.30) = 0.05655 m^{2} .$ Yeh step kyun? Same $L$ , lekin radius $0.02$ se $0.03$ ho gayi — bada circle lambi circumference deta hai, to zyada surface aag par hai.
Trend: area $50%$ badha. Kyunki $A_{b}$ $w$ ke saath badhta hai, yeh progressive hai.

Verify: ratio $= \frac{0.03}{0.02} = 1.5$ exactly (dono mein $2 π L$ common hai), to $50%$ rise rounding artifact nahi — yeh radius ratio khud hai. Units: $m \cdot m = m^{2}$ . ✔

Figure — Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn

C2 — Regressive: outer-burning grain

Outer surface bhi ek cylinder wall hai, lekin yeh andar ki taraf recede karti hai, to uski radius kam hoti hai: $A_{b} (w) = 2 π (R_{o} - w) L .$ Minus sign kyun? Andar ki taraf jalne wali outer surface radius khoti hai jaise $w$ badhta hai. Geometry wahi cylinder wall hai; sirf offset ka sign flip hota hai.
$w = 0$ : $A_{b} = 2 π (0.05) (0.40) = 0.12566 m^{2} .$ Yeh step kyun? Ignition baseline fix karo, bilkul Ex 1 ki tarah, taaki baad ki value ke paas kuch compare karne ko ho.
$w = 0.01$ : $A_{b} = 2 π (0.04) (0.40) = 0.10053 m^{2} .$ Yeh step kyun? Radius $0.05$ se $0.04$ par aa gayi; chhote circle ki circumference choti hoti hai, to kam surface aag par hai.
Trend: area $20%$ giri. Kyunki $A_{b}$ ghatta hai, yeh regressive hai — thrust waqt ke saath droops karta hai.

Verify: ratio $= 0.04/0.05 = 0.8$ , yani $20%$ drop, purely radius ratio se. $d A_{b} / d w = - 2 π L < 0$ ka sign "decreasing" confirm karta hai. ✔

C3 — Neutral limit: the end-burner

Burning face ek flat disc hai, area $A_{b} = π R^{2}$ , aur yeh $w$ par depend nahi karta: $A_{b} (w) = π R^{2} = π (0.03)^{2} = 2.827 \times 1 0^{- 3} m^{2} .$ Yeh step kyun? Disc sirf rod ke andar translate karti hai. Uski radius $R$ kabhi nahi badlti (walls inhibited), to area frozen hai. Yahi exact zero-slope limit hai: $d A_{b} / d w = 0$ ⇒ perfectly neutral.
Duration = web to burn / burn rate. Yahan "web" poora length $L$ hai: $t = \frac{L}{r} = \frac{0.50}{2 \times 1 0 ^{- 3}} = 250 s .$ Yeh step kyun? Face $L$ distance $r$ speed se travel karta hai; time = distance / speed. Constant chota area ⇒ kam thrust lekin bahut lamba burn — classic sustainer.

Verify: $d A_{b} / d w = 0$ ( $π R^{2}$ mein koi $w$ nahi) → exactly neutral. $t$ units: $m / (m/s) = s$ . $250$ s ≈ 4 min, sustainer ke liye sensible. ✔

C4 — Competing terms: full BATES segment

Do competing surfaces likho (parent note se): $A_{b} (w) = port 2 π (R_{i} + w) (L - 2 w) + do ends 2 π [R_{o}^{2} - (R_{i} + w)^{2}] .$ Yeh step kyun? Har burning surface ko sum karna hoga — inner wall (badhti hai) plus do ring-shaped ends (shrink hote hain jaise port unhe khaata hai). Koi bhi term miss karna classic galti hai.
Slope paane ke liye differentiate karo (woh tool jo "rising or falling?" ka jawab deta hai): $\frac{d A _{b}}{d w} = 2 π [(L - 2 w) - 2 (R_{i} + w)] + 2 π [- 2 (R_{i} + w)] .$ Derivative kyun? "Progressive vs regressive" literally woh sawaal hai jo derivative answer karta hai: kya $A_{b}$ abhi upar ja raha hai ( $> 0$ ) ya neeche ( $< 0$ )? Koi aur tool instantaneous trend nahi deta.
$w = 0$ par evaluate karo: $\frac{d A _{b}}{d w}_{0} = 2 π [L - 2 R_{i} - 2 R_{i}] = 2 π (L - 4 R_{i}) .$ Numbers plug karo: $L - 4 R_{i} = 0.20 - 4 (0.02) = 0.20 - 0.08 = 0.12 > 0.$ To $d A_{b} / d w ∣_{0} = 2 π (0.12) = 0.7540 m^{2} / m > 0.$ Yeh step kyun? $w = 0$ set karne se ignition trend isolate hota hai — ekdum pehla instant, jahan designers sabse zyada care karte hain ki thrust rise ho raha hai ya fall.
Sign padho: positive ⇒ area ignition par badh raha hai ⇒ shuru mein progressive. Port ki growth ends ke shrink ko beat karti hai kyunki $L$ $R_{i}$ ke relative kaafi lamba hai.

Verify: neutral crossover $L = 4 R_{i}$ hai, yani $L = 0.08$ m. Humara $L = 0.20 > 0.08$ , to progressive — consistent. Agar $L$ ko $0.08$ m se neeche shrink karein to sign flip ho jaega regressive mein. ✔

C5 — Degenerate: web runs out

Radial web tab khatam hota hai jab port outer wall tak pahunche: $R_{i} + w_{ma x} = R_{o} \Rightarrow w_{ma x} = R_{o} - R_{i} = 0.05 - 0.02 = 0.03 m .$ Yeh step kyun? Us moment port aur casing ke beech koi propellant nahi bachta — wall ke aage recede karne ki jagah nahi.
$w = w_{ma x} = 0.03$ par $A_{b}$ evaluate karo:
- Port term: $2 π (R_{i} + w) (L - 2 w) = 2 π (0.05) (0.20 - 0.06) = 2 π (0.05) (0.14) = 0.04398$ m².
- End term: $2 π [R_{o}^{2} - (R_{i} + w)^{2}] = 2 π [0.0025 - 0.0025] = 0$ . $A_{b} (w_{ma x}) = 0.04398 + 0 = 0.04398 m^{2} .$ Yeh step kyun? Ends exactly burnout par vanish ho jaate hain kyunki $(R_{i} + w) = R_{o}$ , to annulus $R_{o}^{2} - (R_{i} + w)^{2}$ zero ho jaata hai. Yahi degenerate/zero case hai: ek poori surface disappear ho jaati hai.
Thrust behaviour: reality mein yahan burn clean nahi hoti — thin leftover slivers of propellant toot jaate hain, area zero ki taraf collapse karta hai, aur thrust sharply girta hai — regressive tail. Model ka $A_{b}$ finite rehta hai kyunki woh fragmentation ignore karta hai.

Verify: end term $= 2 π [R_{o}^{2} - R_{o}^{2}] = 0$ identically $w_{ma x}$ par — annulus genuinely gone hai. $w_{ma x} = 0.03$ m poori radial thickness $R_{o} - R_{i}$ hai. ✔

C6 — Perimeter → thrust: star vs wagon wheel

Case-bonded grain ke liye, burning area = perimeter × length (parent ka rule, ends inhibited): $A_{b} = P \cdot L .$ Yeh step kyun? Ends sealed hone se, sirf port wall burning surface hai, jiska area hai (outline kitna lamba hai) × (grain kitna deep hai) = $P \cdot L$ .
Star: $A_{b, A} = 1.2 \times 0.60 = 0.72$ m². Wheel: $A_{b, B} = 3.0 \times 0.60 = 1.80$ m². Yeh step kyun? Har perimeter ko burning area mein convert karte hain taaki chemistry laane se pehle dono grains ko same footing par compare kar sakein.
Initial mass flow $\overset{m}{˙} = ρ_{p} A_{b} r$ :
- Star: $1800 \times 0.72 \times 0.006 = 7.776$ kg/s.
- Wheel: $1800 \times 1.80 \times 0.006 = 19.44$ kg/s. Yeh step kyun? Same chemistry ( $ρ_{p}, r$ ), to mass flow directly burning area ke saath scale karta hai, jo perimeter ke saath scale karta hai. Wheel ka high perimeter boost phase ke liye high initial thrust deta hai.
Ratio: $19.44/7.776 = 2.5$ — exactly perimeter ratio. Wagon wheel $2.5 \times$ launch thrust deta hai — thin spokes ke regressive sliver tail ki cost par.

Verify: $\overset{m}{˙}_{B} / \overset{m}{˙}_{A} = P_{B} / P_{A} = 3.0/1.2 = 2.5$ (baaki sab factors cancel ho jaate hain). ✔

C7 — $\frac{1}{1 - n}$ amplification kaata hai

Numbers se pehle, do symbols enter hote hain jo is page par define karne chahiye:

Pressure law yaad karo (Chamber pressure & throat area (Kn ratio)): $p_{c} \propto A_{b}^{\frac{1}{1 - n}} .$ Yeh exponent kyun, sirf $A_{b}^{1}$ nahi? Kyunki $p_{c}$ mass balance ke dono sides par hai — yeh burn rate $r = a p_{c}^{n}$ ke zariye feedback deta hai (dekho Saint-Robert burn rate law). $p_{c}$ isolate karne par factor $(1 - n)$ invert hota hai, to geometric change exponent $\frac{1}{1 - n}$ se amplify hoti hai.
Exponent compute karo: $\frac{1}{1 - n} = \frac{1}{1 - 0.4} = \frac{1}{0.6} = 1.6 \overline{6} .$ Yeh step kyun? Humein exact power chahiye jo area ratio ko pressure ratio mein map kare, pehle kuch usmein raise karein.
$20%$ area rise (factor $1.20$ ) ko exponent se map karo: $\frac{p _{c}^{new}}{p _{c}^{old}} = 1.2 0^{1.6667} = 1.3551.$ To chamber pressure lagbhag $35.5%$ badhta hai, $20%$ nahi. Yeh step kyun? Ratio ko $1.667$ par raise karna use magnify karta hai — non-linear feedback ka poora point yahi hai.
Thrust $F = C_{F} p_{c} A_{t}$ mein $C_{F}, A_{t}$ fixed hain, to thrust $p_{c}$ ko track karta hai: bhi $\approx 35.5%$ upar. Ek "chhota" area drift ek bada thrust drift ban jaata hai — isliye designers true neutrality dhundhte hain.

Verify: $1. 2^{1/0.6} \approx 1.3551$ ; amplification factor $1.3551/1.20 \approx 1.13 > 1$ confirm karta hai ki pressure area se zyada move karta hai. $n < 1$ chahiye taaki exponent finite aur positive rahe (self-regulating). ✔

C8 — Real-world design word problem

Boost phase → wagon wheel. Kyun? Uska enormous initial port perimeter sabse bada initial $A_{b}$ deta hai, isliye sabse bada initial $\overset{m}{˙}$ aur thrust (Ex 6). Uska regressive sliver tail acceptable hai kyunki boost brief aur front-loaded hona chahiye.
Cruise phase → end-burner (ya star). Kyun? End-burner ka disc area frozen hai (Ex 3): neutral aur long-burning, poore cruise ke liye low, flat, steady thrust deta hai. Well-designed star alternative hai jab thoda zyada (lekin phir bhi flat) thrust chahiye, kyunki woh $A_{b}$ ≈ constant rakhta hai.
Unhe swap kyun nahi karein? Wagon wheel as cruiser bahut jaldi burn out ho jaata aur badly droop karta (regressive). End-burner as booster bahut kam thrust deta. Class role dictate karta hai: progressive/high-perimeter boost ke liye, neutral cruise ke liye.

Verify: matrix ke saath consistent — wagon wheel = high initial $A_{b}$ (C6), end-burner = neutral (C3). Check karne ke liye koi numeric nahi; logic har phase ko uske burn class se map karta hai. ✔

C9 — Exam twist: "kaun sa neutral hai?" (trap)

Batao neutral ka matlab actually kya hai: $A_{b}$ ≈ constant main burn par — ek approximation, identity nahi. Yeh step kyun? "Neutral" label bulk behaviour describe karta hai, har instant nahi; uska asli matlab pin karna hi trap ko disarm karta hai.
Hidden non-neutral parts dhundho. Jaise star ke points aur valleys round out hote hain, chhote ripples aate hain; aur burnout ke paas, spokes ke beech propellant ke thin slivers port chodte hain aur $A_{b}$ collapse karta hai — ek regressive tail. Yeh step kyun? Koi real geometry $A_{b}$ ko ignition se burnout tak literally frozen nahi rakh sakti; tail unavoidable hai jab web exhaust hoti hai (C5 echoing).
Correct answer: "Nahi — thrust main burn par ≈ constant hai, ek mild ripple aur ant mein ek regressive sliver tail ke saath. Neutral ek approximation hai." Student ka "exactly flat" trap hai.

Verify: parent ke mistake callout ke saath consistent ("neutral approximate hai; real stars mein regressive sliver tail hoti hai"). ✔

Recall Quick self-test (reveal karne ke liye click karo)

Progressive matlab $A_{b}$ waqt ke saath kya karta hai? ::: Badhta hai → rising thrust. End-burner exactly neutral kyun hai? ::: Uski flat disc face ka area $π R^{2}$ hai jisme koi $w$ -dependence nahi, to $d A_{b} / d w = 0$ . Full BATES: $R_{i}$ ke terms mein neutral crossover length? ::: $L = 4 R_{i}$ ( $d A_{b} / d w ∣_{0} = 2 π (L - 4 R_{i}) = 0$ se). Area $20%$ upar $n = 0.4$ ke saath → pressure kitna upar? ::: Lagbhag $35.5%$ , kyunki $p_{c} \propto A_{b}^{1/ (1 - n)} = A_{b}^{1.667}$ . BATES mein $w_{ma x}$ par end-annulus term ka kya hota hai? ::: Woh zero ho jaata hai — annulus vanish ho jaati hai ( $R_{i} + w = R_{o}$ ).

Dekho bhi: Saint-Robert burn rate law · Chamber pressure & throat area (Kn ratio) · Thrust coefficient and nozzle · Rocket equation & specific impulse · Combustion instability · Case-bonded vs free-standing grains