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

3.3.37 · D4 · Physics › Rocket Propulsion › Grain geometry — BATES, star, wagon wheel; neutral - progres

Poore time, master chain yaad rakho (parent se): $A_{b} \Rightarrow K = \frac{A _{b}}{A _{t}} \Rightarrow p_{c} \propto K^{\frac{1}{1 - n}} \Rightarrow F = C_{F} p_{c} A_{t}$

Intuition Yeh exact chain kyun hai?

$K = A_{b} / A_{t}$ ratio kyun banate hain? Chamber pressure ek tug-of-war hai gas ke banne (burning area $A_{b}$ ke proportional) aur gas ke escape karne (throat area $A_{t}$ ke proportional) ke beech. Pressure akela $A_{b}$ ya $A_{t}$ set nahi karta, balki unka ratio karta hai — dono double karo aur pressure unchanged rahega. Woh single dimensionless number $K$ (aksar $K_{n}$ bhi kehte hain) isliye natural knob hai (dekho Chamber pressure & throat area (Kn ratio)). $C_{F}$ kaise $F$ mein aata hai? Ek baar $p_{c}$ pata ho, nozzle us pressure ko thrust mein convert karta hai. Thrust coefficient $C_{F}$ woh bookkeeping factor hai jo kehta hai "is particular nozzle se ek given $p_{c} A_{t}$ se kitna thrust milta hai" — yeh nozzle ki expansion geometry ko package karta hai (dekho Thrust coefficient and nozzle). Toh $F = C_{F} p_{c} A_{t}$ : pressure times throat area ek force scale deta hai, aur $C_{F}$ use nozzle ke liye scale up karta hai.

Yahan:

$A_{b}$ = burning surface area (m², woh area jo abhi jal rahi hai).
$A_{t}$ = nozzle ka throat area (m², sabse chota hole jisse gas escape karta hai).
$w$ = web burned = kitna aage, apne aap ke perpendicular, flame front solid mein ghus gaya hai (metres).
$r$ = linear burn rate (m/s), kitni tezi se woh front recede karta hai.
$ρ_{p}$ = propellant density (kg/m³).
$n, a$ = do Saint-Robert constants (dekho Saint-Robert burn rate law).
$c^{*}$ = characteristic velocity (m/s), ek propellant-aur-combustion property jo measure karta hai ki burnt gas kitni efficiently chamber pressure build karta hai; jitna zyada $c^{*}$ , utna zyada pressure per unit mass flow (dekho Rocket equation & specific impulse).

Level 1 — Recognition

L1.1

Burn class (neutral / progressive / regressive) batao aur ek sentence mein kyun: (a) ek single-face end-burner, (b) ek BATES segment jo sirf apne inner cylinder par burn karta hai, (c) burnout ke paas late-stage rounded star.

Recall Solution

(a) Neutral. Sirf ek flat circular face jalta hai; uska area $π R^{2}$ kabhi nahi badalta jab face length ke neeche jaata hai. (b) Progressive. Port perimeter $2 π (R_{i} + w)$ badhta hai jab surface outward recede karta hai, toh $A_{b}$ rise karta hai. (c) Regressive. Jab star ke points gone ho jaate hain, outer burning boundary ek shrinking arc hai; perimeter girta hai → $A_{b}$ girta hai.

L1.2

Ek case-bonded grain (ends inhibited, toh sirf port perimeter jalta hai) length $L$ aur port perimeter $P (w)$ ke saath, $A_{b} (w)$ likho. Kaunsi ek geometric quantity ko tum isliye "design" karte ho poori thrust curve shape karne ke liye?

Recall Solution

$A_{b} (w) = P (w) \cdot L$ Kyunki $L$ ek fixed constant hai, $A_{b}$ ka poora time-shape (aur isliye thrust) ==the port perimeter $P (w)$ == carry karta hai. Burn design karna = yeh design karna ki jab port outward khulta hai to perimeter kaise change hoti hai.

Level 2 — Application

L2.1

Ek BATES segment sirf apne inner cylinder par jalta hai (ends inhibited). Given $R_{i} = 0.025$ m, $L = 0.40$ m, $ρ_{p} = 1800$ kg/m³, $r = 6 \times 1 0^{- 3}$ m/s. $A_{b}$ aur mass generation rate $\overset{m}{˙} = ρ_{p} A_{b} r$ compute karo (a) $w = 0$ par aur (b) $w = 0.015$ m par. $\overset{m}{˙}$ kitne percent rise hua?

Recall Solution

Inner surface radius $R_{i} + w$ aur length $L$ ka ek cylinder hai: $A_{b} = 2 π (R_{i} + w) L$ . (a) $w = 0$ : $A_{b} = 2 π (0.025) (0.40) = 0.06283 m^{2}$ . $\overset{m}{˙} = 1800 \times 0.06283 \times 0.006 = 0.6786 kg/s$ . (b) $w = 0.015$ : $R_{i} + w = 0.040$ , $A_{b} = 2 π (0.040) (0.40) = 0.10053 m^{2}$ . $\overset{m}{˙} = 1800 \times 0.10053 \times 0.006 = 1.0857 kg/s$ . Rise $= \frac{0.10053 - 0.06283}{0.06283} = 0.60 = 60%$ . Same $r$ , bada radius → lamba circumference → zyada surface → zyada gas. Progressive. ✔

L2.2

Amplification law $p_{c} \propto A_{b}^{1/ (1 - n)}$ use karo $n = 0.35$ ke saath. Agar $A_{b}$ upar find kiye gaye 60% se rise hua, toh $p_{c}$ kitne factor se rise karta hai? ( $A_{t}$ fixed assume karo.)

Recall Solution

Area ratio hai $\frac{A _{b 2}}{A _{b 1}} = 1.60$ . Toh $\frac{p _{c 2}}{p _{c 1}} = (\frac{A _{b 2}}{A _{b 1}})^{\frac{1}{1 - n}} = 1.6 0^{\frac{1}{0.65}} = 1.6 0^{1.5385} \approx 2.06 .$ 60% area rise ~106% pressure rise ban jaati hai — yahi hai $\frac{1}{1 - n}$ amplification kaam karta hua (dekho Chamber pressure & throat area (Kn ratio)).

Level 3 — Analysis

L3.1

Ek BATES segment with free (uninhibited) ends mein inner cylinder bhi grow kar raha hai aur do annular end-faces bhi shrink kar rahi hain: $A_{b} (w) = port 2 π (R_{i} + w) (L - 2 w) + two ends 2 π [R_{o}^{2} - (R_{i} + w)^{2}] .$ $\frac{d A _{b}}{d w}$ find karo $w = 0$ par aur $L$ ki woh condition batao jo segment ko exactly neutral at the start banati hai ( $d A_{b} / d w = 0$ ). $R_{i} = 0.02$ m, $R_{o} = 0.06$ m use karo — neutral $L$ solve karo.

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

Recall Solution

Term by term differentiate karo (figure mein red = growing port, violet = shrinking ends dekho). Port term $P = 2 π (R_{i} + w) (L - 2 w)$ : $\frac{d P}{d w} = 2 π [(L - 2 w) - 2 (R_{i} + w)] .$ Ends term $E = 2 π [R_{o}^{2} - (R_{i} + w)^{2}]$ : $\frac{d E}{d w} = 2 π [- 2 (R_{i} + w)] = - 4 π (R_{i} + w) .$ $w = 0$ par sum: $\frac{d A _{b}}{d w}_{0} = 2 π [L - 2 R_{i}] - 4 π R_{i} = 2 π L - 4 π R_{i} - 4 π R_{i} = 2 π (L - 4 R_{i}) .$ Ise zero set karne par: $L = 4 R_{i}$ . $R_{i} = 0.02$ ke saath: $L = 0.08$ m. Interpretation: agar barrel chhota hai ( $L < 4 R_{i}$ ) toh shrinking ends jeet jaate hain → regressive; agar lamba hai ( $L > 4 R_{i}$ ) toh growing port jeet jaata hai → progressive; exactly $L = 4 R_{i}$ shuru mein balance karta hai → neutral onset. ( $R_{o}$ derivative mein drop out ho gaya — yeh sirf total web fix karta hai, starting slope nahi.)

L3.2

Ek short paragraph mein explain karo kyun ek star grain neutral ki taraf tend karta hai, jabki uska port clearly burn hone ke saath open up hota hai.

Recall Solution

Ek star mein do tarah ke features hote hain: andar ki taraf nikle propellant ke sharp points, aur unke beech valleys (slots). Jab front recede karta hai: points kha jaate hain, jo perimeter drop karta; usi waqt valleys chaudi aur gehri hoti hain, jo perimeter raise karti hain. Ek achhe se cut star ka proportion aisa hota hai ki yeh do opposite trends main burn ke dauran roughly cancel ho jaate hain, $P (w) \approx$ const dete hain, isliye $A_{b} \approx$ const → neutral. Sirf burnout ke paas, jab points gone ho jaate hain, perimeter shrink shuru hoti hai (regressive sliver tail).

Level 4 — Synthesis

L4.1

Ek designer flat thrust curve chahta hai aur (i) free-ended BATES with $L = 4 R_{i}$ , aur (ii) case-bonded neutral star ke beech choose karna hai. Motor head-end par structural buzzing prone hai. Kaunsa choice combustion instability ko better resist karta hai, aur kyun? (Ek paragraph.)

Recall Solution

Case-bonded star generally safer hai. Ek free-standing BATES mein grain case mein loosely baitha hota hai, toh uske ends aur annular gaps open surfaces present karte hain jo acoustic oscillations drive aur sustain kar sakte hain; grain ek free body ki tarah vibrate bhi kar sakta hai (dekho Case-bonded vs free-standing grains). Ek case-bonded star wall se glued hota hai, structural modes damp karta hai, aur uski port shape (bahut saare slots) clean acoustic standing waves break up karti hai jo Combustion instability feed karti hain. Dono ≈neutral thrust dete hain, lekin bonded star us flat curve ko better damping ke saath pair karta hai. Trade-off: case-bonded grains bond line par thermal-stress cracking suffer karte hain, jo separately check karna hoga.

L4.2

L2.1 ka progressive BATES lo ( $R_{i} = 0.025$ , $L = 0.40$ ) lekin ab demand karo ki chamber pressure web ke $w = 0$ se $w = 0.015$ tak jaate hue apne mean ke $\pm 10%$ ke andar rahe. $p_{c} \propto A_{b}^{1/ (1 - n)}$ use karo $n = 0.35$ ke saath, show karo ki yeh inner-only BATES requirement fail kyun karta hai, aur ek geometry naam do jo pass hoti.

Recall Solution

L2.2 se pressure ratio end-to-start $1.6 0^{1.5385} \approx 2.06$ hai — pressure double se zyada ho jaata hai. Mean ke baare mein measure karne par bhi, woh roughly $\pm 34%$ swing hai, $\pm 10%$ se kaafi bahar. Numerically: $p_{c} (0) \propto 0.0628 3^{1.5385}$ aur $p_{c} (w) \propto 0.1005 3^{1.5385}$ ; ratio $2.06$ . Ek $\pm 10%$ band factor $1.10/0.90 = 1.22$ span karta hai, aur $2.06 ≫ 1.22$ → fails. Ek neutral star (case-bonded, $P (w) \approx$ const) $A_{b}$ ko nearly flat rakhta hai, toh $p_{c}$ kuch percent ke andar rehta hai — woh pass karta hai. Yahi exact reason hai kyun constant-thrust missiles stars use karti hain, plain inner-burning cylinders nahi.

Level 5 — Mastery

L5.1

Design task. Tumhe do-phase thrust curve chahiye: pehle third of burn ke liye ek high boost, phir ek lower sustain. Parent note ki geometries use karke ek grain concept sketch karo (words + master chain mein), aur har phase mein burn class batao. Phir qualitative $A_{b} (w)$ shape do.

Recall Solution

Concept: ek wagon-wheel (deep thin spokes) forward section jo ek plain cylindrical (BATES inner) ya thoda regressive aft section mein transition karta hai.

Boost phase: wagon wheel ki huge initial perimeter → huge initial $A_{b}$ → (chain $A_{b} \Rightarrow K \Rightarrow p_{c} \propto K^{1/ (1 - n)} \Rightarrow F$ se guzar ke) ek bada early thrust. Jab thin spokes burn through hote hain aur slivers bante hain, $A_{b}$ sharply girta hai — regressive — thrust ko sustain level par drop karta hai.
Sustain phase: remaining cylindrical bore $A_{b}$ ko roughly flat rakhta hai (near-neutral) ek lower value par, ek lamba steady tail deta hai. $A_{b} (w)$ shape: $w = 0$ par high, jab spokes vanish hote hain toh steep drop, phir ek low plateau — classic "boost-sustain" staircase (figure mein violet curve). Yeh standard tarika hai ek propellant grain se do thrust levels paane ka, chemistry change kiye bina.

L5.2

Mass balance se stability requirement $n < 1$ prove karo, aur physically explain karo agar $n \geq 1$ ho toh kya galat hota hai.

Recall Solution

Steady mass balance: gas generated = gas exhausted. Gas burning area par rate $ρ_{p} A_{b} r = ρ_{p} A_{b} a p_{c}^{n}$ se banaya jaata hai, aur throat se rate $p_{c} A_{t} / c^{*}$ se escape karta hai, jahan $c^{*}$ (characteristic velocity, upar symbol list mein define) set karta hai ki ek given pressure kitna mass flow drive karta hai. Unhe equate karo: $ρ_{p} A_{b} a p_{c}^{n} = \frac{p _{c} A _{t}}{c ^{*}} .$ $p_{c}$ ki powers collect karo: dono sides ko $p_{c}^{n}$ se divide karo, $ρ_{p} A_{b} a c^{*} = A_{t} p_{c}^{1 - n} \Rightarrow p_{c} = (\frac{ρ _{p} a c ^{*} A _{b}}{A _{t}})^{\frac{1}{1 - n}} .$ Iske liye ek finite, single positive pressure mile, exponent finite hona chahiye aur map self-correcting hona chahiye: $1 - n > 0$ , yaani $n < 1$ . Physical meaning: equilibrium se shuru karo aur $p_{c}$ ko thoda sa upar nudge karo. Generation $p_{c}^{n}$ ki tarah badhti hai; exhaust $p_{c}^{1}$ ki tarah badhta hai. Agar $n < 1$ , exhaust generation se aage nikal jaata hai — extra pressure generation se zyada tezi se bleed off ho jaata hai, toh system equilibrium par wapas aata hai (self-regulating). Agar $n \geq 1$ , generation kam se kam utni hi tezi se badhti hai jitni exhaust: koi bhi upward nudge khud ko feed karta hai → runaway pressure aur burst. (Yahi $\frac{1}{1 - n}$ hai jo Chamber pressure & throat area (Kn ratio) mein bhi aata hai.)

Recall Quick self-check clozes

Ek free-ended BATES neutral-at-onset hota hai jab ::: $L = 4 R_{i}$ . $n = 0.35$ ke saath $A_{b}$ mein 60% rise $p_{c}$ ko ek factor ::: $1.6 0^{1/0.65} \approx 2.06$ se raise karta hai. Wagon-wheel forward + cylindrical aft ek ::: boost-then-sustain (regressive→neutral) thrust curve deta hai. Stability require karti hai ::: $n < 1$ , taki nozzle exhaust ( $\propto p_{c}$ ) generation ( $\propto p_{c}^{n}$ ) se aage nikal sake.