Exercises — Burn rate r = a·P^n — Vieille's law

Kya: hum $P=1$ plug in karte hain. Kyun: kisi bhi cheez ko power mein raise karo, agar base $1$ ho, toh result $1$ hi hoga: $1^{0.4}=1$. $r=a\cdot 1^{0.4}=a=3\text{mm/s}.$ Yahi toh $a$ ko "$1\text{MPa}$ par" define karne ka poora point hai: $a$ hi unit pressure par burn rate hai. Exponent ko kaam karne ka mauka nahi milta kyunki $1$ kisi bhi power mein $1$ hi rehta hai.

(i) exponent $n=0.35$ — yeh sensitivity hai; bada $n$ matlab pressure changes rate ko zyada swing karti hain. (ii) coefficient $a$ — rate jab $P=1$ ho. (iii) $r$ khud — woh speed jisme burning surface solid ke andar eat karti hai, exhaust speed nahi.

Kya: direct substitution. Kyun exponential aata hai: $8^{0.3}$ sabse aasaan $8^{0.3}=e^{0.3\ln 8}$ ke through hai — log, power ko ek multiply mein badal deta hai, phir $e$ log ko undo kar deta hai. $8^{0.3}=e^{0.3\times 2.0794}=e^{0.6238}=\mathrm{1.866.}$ $r=5\times 1.866=9.33\text{mm/s}.$ Sense check: pressure $8\times$ badhi lekin $r$ muskil se double hua — yahi exactly ek chote $n$ ko karna chahiye.

Kya: $12=6P^{0.5}$ ko $P$ ke liye solve karo. Kyun: pehle divide karo taaki power isolate ho, phir $0.5$ power ko $1/0.5=2$ tak raise karke undo karo. $P^{0.5}=\frac{12}{6}=2\Rightarrow P=2^{1/0.5}=2^2=4\text{MPa}.$ Kyunki $n=0.5$ ka matlab hai "$P$ ka square root", double burn rate paane ke liye tumhe chaar guna pressure chahiye.

Kyun ratio trick: hum $a$ nahi jaante, lekin do pressures par law ko divide karne se yeh completely cancel ho jaata hai: $\frac{r_2}{r_1}={\left(\frac{P_2}{P_1}\right)}^n={\left(\frac{9}{3}\right)}^{0.45}=3^{0.45}.$ $3^{0.45}=e^{0.45\times 1.0986}=e^{0.4944}=\mathrm{1.6396.}$ $r_2=7\times 1.6396=11.48\text{mm/s}.$

Step 1 (kyun divide karein): ratio $a$ ko khatam kar deta hai taaki hum $n$ isolate kar sakein: $\frac{6}{4}={\left(\frac{5}{2}\right)}^n\Rightarrow 1.5=2.5^n.$ Step 2 (kyun logs): logs $n$ ko exponent se neeche kheench laate hain: $n=\frac{\ln 1.5}{\ln 2.5}=\frac{0.4055}{0.9163}=\mathrm{0.4426.}$ Step 3 — $a$ recover karo ek point mein $n$ wapas daal kar: $a=\frac{r_1}{P_1^n}=\frac{4}{2^{0.4426}}=\frac{4}{1.359}=2.943\text{mm/s}.$ Sense check: $n<1$ → yeh propellant stable hai. ✔

Kyun seedhi line hoti hai: $r=a{P}^n$ ka ${\log }_{10}$ lene par $\log r=\log a+n\log P$ milta hai — yeh ek line ki equation hai jisme slope $n$ aur intercept $\log a$ hai. Slope (=$n$): do marked points ke beech rise over run: $n=\frac{\log 12.6-\log 5}{\log 10-\log 1}=\frac{1.1004-0.6990}{1-0}=0.4014\approx \mathrm{0.40.}$ Intercept (=$\log a$): $\log P=0$ par (yaani $P=1$), $\log r=\log 5$ hai, isliye $a=5\text{mm/s}$. Yeh exactly woh extraction method hai jo engineers real strand-burner runs par use karte hain.

Kyun exponents compare karein: equilibrium par generation = exhaust. $P$ ko thoda bump up karo; jo term zyada tez grow karegi woh jeet jaayegi.

• A ($n=0.85<1$): exhaust ($P^1$) generation ($P^{0.85}$) se zyada tez grow karta hai. Extra gas banne se zyada tez nikalta hai → pressure wapas girta hai → stable, survive karta hai. ✅
• B ($n=1.15>1$): generation ($P^{1.15}$), exhaust ($P^1$) se aage nikal jaati hai. Bump khud ko feed karta hai → runaway → blast ho jaata hai. ❌ Ek akelaa dividing line $n=1$ hai. Yeh is poore topic ka 80/20 fact hai.

Kyun distance ko rate se divide karein: $r$ recession ki speed hai, isliye time = distance ÷ speed, exactly jaise walking distance ko walking speed se divide karte hain: $t_b=\frac{w}{r}=\frac{30\text{mm}}{9\text{mm/s}}=3.33\text{s}.$ Link: web $w$ Solid Rocket Motor Grain Geometry se aata hai; $r$ ki value Vieille's law se operating $P$ diye jaane par milti hai. Yahan burn-rate law actual hardware se milti hai.

Kyun ratio: percent change ke liye sirf factor chahiye, aur factor $a$ ko drop kar deta hai: $\frac{r_2}{r_1}={\left(\frac{6}{4}\right)}^{0.4}=1.5^{0.4}=e^{0.4\times 0.4055}=e^{0.1622}=\mathrm{1.176.}$ Toh $r$ $1.176-1=0.176=17.6\%$ badhti hai. Sense check: pressure $50\%$ badhi lekin rate sirf $\sim 18\%$ — chota $n$ response ko soften karta hai, jaise hona chahiye.

Kyun $a$ temperature ke saath move karta hai: ek garam grain apni ignition temperature ke zyada paas se shuru hoti hai, isliye agle layer ko jalaane ke liye kam heat chahiye → zyada tez recession → bada $a$. Yeh Temperature Sensitivity of Propellants action mein hai. $5^{0.35}=e^{0.35\times 1.6094}=e^{0.5633}=\mathrm{1.7565.}$ $r_{\text{cold}}=4.6\times 1.7565=8.080\text{mm/s},r_{\text{hot}}=5.4\times 1.7565=9.485\text{mm/s}.$ Kyunki $P$ aur $n$ shared hain, ratio sirf $a_{\text{hot}}/a_{\text{cold}}$ hai: $\frac{r_{\text{hot}}}{r_{\text{cold}}}=\frac{5.4}{4.6}=1.1739\Rightarrow +17.4\%.$ Consequence: wahi rocket ek garmi ke din $\sim 17\%$ zyada tez burn karta hai — zyada peak $P$, chota burn. Isliye motors ko fire karne se pehle temperature-condition kiya jaata hai.

Pehle stability: dono mein $n<1$ hai, isliye dono stable hain — woh filter akela decide nahi kar sakta. Doubling test: jab $P\to 2P$, rate factor $=2^n$: $2^{0.35}=1.274(+27\%),2^{0.95}=1.932(+93\%).$ Y ($2^{0.95}=1.93\approx 2$) almost double ho jaata hai — yeh spec (ii) meet karta hai. X sirf $+27\%$ deta hai. Tension: Y, $n=1$ knife-edge ke zyada paas hai, isliye yeh stable hai lekin kam forgiving hai; ek real designer "responsiveness" (bada $n$) ko "safety margin" ($n=1$ se neeche ki doori) ke against weigh karta hai. Sirf spec-matching yahan Y ki taraf point karta hai.

Kyun ratios check karein: agar ek power law fit hota hai, toh $\log P$ ki har doubling par same slope aana chahiye. Do ratios use karo: $n_1=\frac{\ln (4.9/3)}{\ln (4/1)}=\frac{\ln 1.6333}{\ln 4}=\frac{0.4907}{1.3863}=0.3540,$ $n_2=\frac{\ln (8.0/4.9)}{\ln (16/4)}=\frac{\ln 1.6327}{\ln 4}=\frac{0.4903}{1.3863}=\mathrm{0.3537.}$ Dono slopes agree karte hain ($\approx 0.354$) → ek law fit karta hai. Pehle point $P=1$ se: $a=r=3\text{mm/s}$. $\boxed{r=3P^{0.354}}.$ Verify middle point: $3\times 4^{0.354}=3\times 1.633=4.90$ ✔.

Yahan har problem ek law hai, $r=a{P}^n$, teen directions mein padha gaya: forward ($P$ plug in karo), backward ($P$ ke liye solve karo), aur sideways (ratios aur logs se $n,a$ pana bina $a$ jaane). Stability ek simple inequality hai $n<1$; temperature $a$ ke andar rehti hai; web-over-rate burn time deta hai. Woh chaar moves master karo aur D4 khatam.