Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn: Question bank

3.3.37 · D5 · Physics › Rocket Propulsion › Grain geometry — BATES, star, wagon wheel; neutral - progres

Shuru karne se pehle, teen words jo hum baar baar use karte hain, plain language mein:

True or false — justify karo

TF1. "Ek solid motor ki flame port ke neeche candle wick ki tarah travel karti hai."

False — flame har exposed surface ke perpendicular ek saath recede karti hai, port ke along nahi. Saari burning skin ek saath baahir ki taraf jaati hai, isliye thrust area se set hota hai, kisi flame-front position se nahi.

TF2. "Agar port hole visibly bada ho raha hai, toh motor zaroor progressive hai."

False — tumhe saari burning surfaces ka sum karna hoga. Free-ended BATES mein inner port bada hota hai jabki do end faces chhoti hoti jaati hain, isliye net

A_{b}

flat reh sakta hai ya girr bhi sakta hai.

TF3. "Neutral burn matlab thrust ek perfectly flat, ruler-straight line hai."

False — neutral sirf main burn ke dauran approximately constant hota hai. Real stars mein choti ripples hoti hain aur burnout par ek regressive sliver tail bhi hoti hai, isliye hum kehte hain "≈ constant".

TF4. "Burn rate

r

ko double karne se thrust double ho jaata hai lekin burn duration unchanged rehti hai."

False —

\overset{m}{˙} = ρ_{p} A_{b} r

, isliye double

r

se instantaneous mass flow roughly double hota hai aur web aadhe time mein burn ho jaati hai. Tum ek taller, shorter thrust curve paate ho, sirf taller nahi.

TF5. "Ek end-burner (cigarette burn) neutral hai kyunki uski area mein kuch nahi badalta."

True — sirf ek flat circular face jalta hai,

A_{b} = π R^{2}

constant hai, isliye thrust flat hai aur (kyunki poori length ko end-to-end burn karna hota hai) bahut long-lasting hai. Iske slow pace ke liye Saint-Robert burn rate law dekho.

TF6. "Wagon wheel ek smooth, constant-thrust sustainer ke liye choose kiya jaata hai."

False — uska huge initial perimeter high early thrust deta hai boost phase ke liye, phir thin slivers se ek regressive tail aati hai. Smooth constant profile star ka kaam hai.

TF7. "Kyunki

p_{c} \propto A_{b}

hai, area mein

20%

rise se chamber pressure mein

20%

rise hogi."

False — ye

p_{c} \propto A_{b}^{1/ (1 - n)}

hai.

n = 0.4

ke saath exponent

1.67

hai, isliye

1. 2^{1.67} \approx 1.36

— ek

36%

pressure jump. Geometry effects amplified hote hain.

TF8. "Perimeter-times-length rule

A_{b} = P (w) \cdot L

kisi bhi grain ke liye kaam karta hai."

False — ye sirf case-bonded, end-inhibited grains ke liye hold karta hai jahan ends jal nahi sakte, isliye saari area port walls se aati hai. Ek free-ended BATES ko end-annulus terms bhi chahiye. Case-bonded vs free-standing grains dekho.

Error dhundho

SE1. "Star grains neutral hain kyunki pointed fingers perimeter ko change hone se rokti hain."

Error yeh hai: aisa nahi ki kuch change hi nahi hota — yeh ek balance hai. Jaise sharp fingers burn hoti hain (area girta hai) valleys web mein khulti hain (area badhta hai); ek achha design inhe cancel kar deta hai. Koi bhi ek effect hatao aur neutrality toot jaati hai.

SE2. "Rising thrust curve banana ho toh aise propellant use karo jiska burn-rate exponent

n

bada ho."

Error yeh hai:

n

control karta hai ki

r

pressure ke saath kitna sensitive hai aur amplification

1/ (1 - n)

— yeh

A_{b}

ko time ke saath nahi badhata. Curve shape geometry

A_{b} (w)

se aati hai;

n

pressure sensitivity aur stability tune karta hai.

SE3. "BATES segment ke liye,

A_{b} = 2 π (R_{i} + w) L

har time par sahi hai."

Error yeh hai: length bhi shrink hoti hai — free ends recede karti hain, isliye yeh

2 π (R_{i} + w) (L - 2 w)

hai plus do end annuli.

L

ko fixed rakhna pretend karna hai ki ends kabhi burn nahi hote.

SE4. "Kyunki burn rate uniform hai, hum

r

ko surface ke along vary karke thrust sculpt kar sakte hain."

Error yeh hai:

r = a p_{c}^{n}

chemistry aur (almost uniform) chamber pressure se set hota hai, isliye yeh ~everywhere same hai aur tum isse space mein freely program nahi kar sakte. Shaping geometry ke through hoti hai, hand-tuned

r

field ke through nahi.

SE5. "Ek stable motor ko sirf

n

koi positive number hona chahiye."

Error yeh hai: usse

n < 1

chahiye. Agar

n \geq 1

toh exponent

1/ (1 - n)

undefined ya negative ho jaata hai aur

p_{c}

runaway ho jaata hai — ek pressure feedback loop jiska koi ceiling nahi. Combustion instability se bhi relevant hai.

SE6. "

K = A_{b} / A_{t}

ko ignore kiya ja sakta hai jab humein

A_{b}

pata ho; sirf

A_{b}

maayine rakhta hai."

Error yeh hai: pressure ratio

K = A_{b} / A_{t}

par depend karta hai,

A_{b}

akele par nahi. Wahi

A_{b}

bade throat

A_{t}

ke saath lower pressure deta hai. Chamber pressure & throat area (Kn ratio) dekho.

SE7. "Thrust equals chamber pressure, bas."

Error yeh hai:

F = C_{F} p_{c} A_{t}

— pressure ko throat area aur thrust coefficient

C_{F}

se multiply kiya jaata hai.

p_{c}

thrust ko track karta hai kyunki

A_{t}

aur

C_{F}

roughly fixed hain, lekin yeh genuinely alag factors hain (Thrust coefficient and nozzle).

Why questions

WHY1. Thrust burning area ko track kyun karta hai, fixed nozzle par?

Zyada burning skin zyada gas per second banati hai (

\overset{m}{˙} = ρ_{p} A_{b} r

); zyada gas chamber pressure raise karta hai (

p_{c} \propto A_{b}^{1/ (1 - n)}

); fixed throat ke through higher pressure zyada thrust deta hai (

F = C_{F} p_{c} A_{t}

). Area us chain ke head par hoti hai.

WHY2. Ek well-chosen BATES

L / D

ratio near-neutral burn kyun produce kar sakta hai?

Badhta inner port (area rising) aur shrink hote free ends (area falling) do opposite trends hain. Length-to-diameter ratio itna choose karo ki unke rates cancel ho jaayein aur sum flat rahe.

WHY3. Ek inner-surface-only (end-inhibited, long) BATES purely progressive kyun ho jaata hai?

Ends par inhibitor paint hone ki wajah se woh jal nahi sakte, isliye sirf inner cylinder ki surface bachti hai jiska circumference

2 π (R_{i} + w)

steadily badhta hai jaise woh recede karta hai —

A_{b}

mein pure increase.

WHY4. Pressure formula mein

1

ki jagah exponent

\frac{1}{1 - n}

kyun aata hai?

Pressure mass balance ke dono sides par hota hai (ek baar directly, ek baar

r = a p_{c}^{n}

ke andar).

p_{c}

ko isolate karne par uska exponent

(1 - n)

se divide ho jaata hai, isliye ek modest area change pressure ko itna zyada swing karta hai.

WHY5. Wagon wheel boost phase ke liye achha lekin smooth cruise ke liye bura kyun hai?

Uske deep thin spokes bahut bada initial perimeter dete hain → bada starting

A_{b}

→ high launch thrust. Lekin thin spokes jaldi burn ho jaate hain aur slivers chhor jaate hain, jo steady cruise ke liye unsuited ek ragged regressive tail produce karta hai.

WHY6. Thrust shape ke liye engineer ka lever geometry kyun hai, chemistry nahi?

Chemistry

r

a

n

ρ_{p}

fix karti hai — timescale aur pressure sensitivity — lekin yeh space aur time mein near-uniform hain. Sirf port shape decide karti hai ki jaise web jalta hai

A_{b}

kaise evolve hota hai, jo exactly curve shape hai.

WHY7. End-burner itne kam thrust ke liye itna lamba kyun chalta hai?

Uski area ek single chhotha face

π R^{2}

hai, isliye mass flow (aur thrust) tiny hai — lekin aag ko poori length

L

ko ek ek thin layer karke chaba jana hota hai, jo bahut lamba burn deta hai. Sustainers ke liye ideal.

Edge cases

EC1. Star grain mein bilkul burnout par, aakhri moments mein kya hota hai?

Valleys saare case tak khul chuki hain aur unke beech sirf thin slivers of propellant bache hain, isliye

A_{b}

collapse ho jaata hai — ek short regressive tail even ek "neutral" grain par bhi.

EC2.

w = 0

par end-burner ka

A_{b}

kya hai versus burn ke baad mein?

Constant:

A_{b} = π R^{2}

throughout, kyunki flat face wahi radius rakhta hai jaise woh recede karta hai. Sirf remaining length shrink hoti hai, area nahi.

EC3. Degenerate case: kya hoga agar

A_{b}

kahi mid-burn mein zero ho jaaye (jaise slivers detach ho jaayein)?

Gas generation zero ki taraf drop ho jaayegi, isliye

p_{c}

aur thrust sharply fall off kar jaayenge — tail-off / burnout region. Detached slivers erratically bhi jal sakte hain aur Combustion instability ko feed kar sakte hain.

EC4. Limiting case: jaise

n \to 1

, pressure prediction ka kya hoga?

Exponent

1/ (1 - n) \to \infty

, isliye koi bhi tiny area ya pressure perturbation bina bound ke amplify ho jaati hai — us runaway ka physical meaning jo design rule

n < 1

force karta hai.

EC5. Kya hoga agar real motor mein

n \geq 1

try kiya?

Chamber pressure ka koi stable operating point nahi hoga: ek small pressure rise

r

ko itna raise karta hai ki pressure aur badhe, feedback hota rahe jab tak case fail na ho jaaye. Isliye propellants ko

n

safely

1

se below rakhke formulate kiya jaata hai.

EC6. Edge case: do grains jinka

A_{b}

identical hai lekin throat areas

A_{t}

alag hain — same thrust?

Nahi — pressure

K = A_{b} / A_{t}

par depend karta hai, isliye chhote throat wala higher

p_{c}

aur higher thrust par run karega. Same area, alag

K

, alag motor (Chamber pressure & throat area (Kn ratio)).

EC7. Boundary: bilkul pehle instant par (

w = 0

) equal outer size ke liye kaun si geometry sabse bada thrust deti hai?

Wagon wheel — uske bahut saare deep thin spokes sabse bada initial port perimeter pack karte hain, isliye sabse bada starting

A_{b}

aur sabse bada launch thrust.

Recall Quick self-check

Fixed nozzle par, instantaneous thrust kaunsi single quantity set karti hai? ::: Current burning surface area $A_{b}$ . "Web burned $w$ " physically kya measure karta hai? ::: Woh perpendicular distance jitni aag propellant ke andar recede kar chuki hai. Wagon wheel ki tail regressive kyun hoti hai? ::: Uske thin spokes slivers chhorte hain jinki area burnout ke paas collapse ho jaati hai.