Burn rate r = a·P^n — Vieille's law: Worked examples

3.3.36 · D3 · Physics › Rocket Propulsion › Burn rate r = a·P^n — Vieille's law

Yeh page ek complete drill hai. Hum ek chhota sa map banate hain har tarah ke questions ka jo Vieille's law $r = a P^{n}$ throw kar sakta hai, phir har box mein ek example karte hain taaki tumhara kabhi aisa case se na mila ho jo tumne pehle dekha na ho.

Yahan sab kuch sirf parent law se aata hai — Vieille's law. Agar koi symbol dikhe, toh woh wahan define kiya gaya tha ya neeche rebuild kiya gaya hai.

Recall Teen symbols jinpe hum rely karte hain (rebuild kiye, koi assumptions nahi)

$r$ = linear burn rate: kitni tezi se solid propellant ki surface andar ki taraf khaati hai, $mm/s$ mein. Ek mombatti socho jiska flame neeche wax mein ghusta jaata hai.
$P$ = chamber pressure: hot gas us surface pe kitna zor se dabaata hai, $MPa$ (mega-pascals) mein.
$a$ = burn-rate coefficient (ek number jo chemistry aur grain ka starting temperature dono carry karta hai); $n$ = pressure exponent (kitni strongly $r$ , $P$ pe react karta hai).

Law: $r = a P^{n}$ .

Scenario matrix

Vieille's law mein koi negative angles ya quadrants nahi hain — lekin inputs $P$ aur $n$ ke alag regimes zaroor hain, saath mein degenerate aur limiting cases bhi. Har exam question neeche exactly ek cell mein fit hota hai.

Cell	Case class	Kya special hai isme	Example
C1	Forward evaluate ( $P > 1$ )	Plug in karo, $P^{n}$ badhta hai	Ex 1
C2	Forward evaluate ( $P < 1$ )	$P^{n} < 1$ , burn rate $a$ se neeche girta hai	Ex 2
C3	Degenerate $P = 1$	$P^{n} = 1$ toh $r = a$ exactly	Ex 3
C4	Degenerate $n = 0$	pressure-independent burn	Ex 3
C5	Do points se $n$ nikalna	ratio $a$ ko cancel karta hai, logs use karo	Ex 4
C6	Inverse — target $r$ ke liye $P$ nikalna	power ko root se undo karo	Ex 5
C7	Stability limit $n \to 1$ aur $n > 1$	knife-edge / runaway	Ex 6
C8	Real-world word problem (web thickness / burn time)	$r$ ko time se connect karta hai	Ex 7
C9	Exam twist — temperature $a$ ko shift karta hai	winter vs summer launch	Ex 8

Woh ek tool jo hum baar baar use karte hain: logarithm

Vieille's law unknown ko exponent ke andar chhupa deta hai ( $P^{n}$ ). Exponent ko neeche kheenchne ke liye jahan hum solve kar sakein, exactly ek tool hai: logarithm, kyunki definition ke hisaab se $lo g (x^{n}) = n lo g x$ — yeh ek power ko ek multiplication mein badal deta hai. Yahi ek reason hai ki logs is page pe kahin bhi aate hain. Jab bhi unknown $n$ ho ya $P^{n}$ ke roop mein fasa ho, $lo g$ pakad lo; jab unknown $P$ khud ho, toh power ko root se undo karo. Neeche ka figure dikhata hai kyun: ordinary axes pe law muda hua hota hai, lekin log–log axes pe yeh ek straight line ban jaata hai jiska slope hi $n$ hota hai.

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

Worked examples

Worked example Example 1 — Forward evaluate,

$P > 1$ · Cell C1 $a = 5 mm/s$ , $n = 0.35$ . $P = 7 MPa$ pe $r$ nikalo.

Forecast: pressure $7 \times$ badh gayi. Kya tumhe lagta hai $r$ bhi $7 \times$ badhega? Padhne se pehle guess karo.

Law likho: $r = a P^{n} = 5 \times 7^{0.35}$ . Yeh step kyun? Direct substitution — yeh base case hai jis par baaki har cell bani hai.
$x^{n} = e^{n l n x}$ use karke power evaluate karo: $7^{0.35} = e^{0.35 l n 7} = e^{0.35 (1.9459)} = e^{0.6811} = 1.976$ . Yeh step kyun? $7^{0.35}$ koi clean number nahi hai; exponential–log identity woh tarika hai jisse ek calculator (ya tum) actually fractional power compute karta hai.
Multiply karo: $r = 5 \times 1.976 = 9.88 \approx 9.9 mm/s$ .

Verify: Units: $mm/s \times (dimensionless) = mm/s$ . ✔ Aur sense check forecast ka jawab deta hai: $7 \times$ pressure se sirf $\sim 2 \times$ burn rate mila, kyunki chhota $n$ matlab weak sensitivity. Yahi toh Vieille's law ka poora point hai.

Worked example Example 2 — Forward evaluate,

$P < 1$ · Cell C2 Wahi propellant $a = 5 mm/s$ , $n = 0.35$ . $P = 0.25 MPa$ pe (ek low-pressure ignition transient) $r$ kya hoga?

Forecast: $P$ $1$ se neeche hai, toh kya $r$ , $a = 5$ se bada hoga ya chhota?

Substitute karo: $r = 5 \times 0.2 5^{0.35}$ . Yeh step kyun? Wahi law — lekin ab base 1 se kam hai, aur 1 se kam number ki positive power ka result 1 se kam aata hai.
Evaluate karo: $0.2 5^{0.35} = e^{0.35 l n 0.25} = e^{0.35 (- 1.3863)} = e^{- 0.4852} = 0.6156$ . Yeh step kyun? 1 se kam number ka $ln$ negative hota hai, toh exponent $P^{n}$ ko 1 se neeche le jaata hai — yeh woh case hai jo log bhool jaate hain.
Multiply karo: $r = 5 \times 0.6156 = 3.08 \approx 3.1 mm/s$ .

Verify: $r = 3.1 < a = 5$ , exactly jaisa forecast mein guess hona chahiye tha: reference pressure se neeche, surface coefficient $a$ se dheere recede karti hai. ✔

Worked example Example 3 — Dono degenerate inputs ·

Cells C3 & C4 (a) Ek test rig exactly $P = 1 MPa$ pe hold karta hai. $a = 5, n = 0.35$ ke liye $r$ nikalo. (b) Ek hypothetical "plateau propellant" ka $n = 0$ hai. $r$ kya hai, aur yeh special kyun hai?

Forecast: (a) ke liye kya $n$ ka koi fark padega? (b) ke liye kya pressure ka koi fark padega?

(a) $r = 5 \times 1^{0.35}$ . Koi bhi number kisi bhi power tak, agar base $1$ hai, toh result $1$ hota hai: $1^{0.35} = 1$ . Yeh step kyun? Yeh definition anchor hai — isi liye hum kehte hain " $a$ burn rate hai $P = 1$ pe." Toh $r = 5 mm/s$ chahе $n$ kuch bhi ho.
(b) $r = a P^{0}$ . Kuch bhi (nonzero) power $0$ pe $1$ hota hai: $P^{0} = 1$ , toh $r = a = 5 mm/s$ har pressure pe. Yeh step kyun? $n = 0$ $P$ ko equation se bilkul hata deta hai — burn rate pressure-independent ho jaata hai. Real "plateau" propellants ek pressure band pe iske kareeb aate hain taaki thrust predictable rahe.