Injector design — impinging, coaxial, swirl injectors: Visual walkthrough

3.3.34 · D2 · Physics › Rocket Propulsion › Injector design — impinging, coaxial, swirl injectors

Hum vault mein pehle se maujood ideas pe lean karenge: Bernoulli Equation (spinning liquid speed kaise gain karta hai), Atomization and the d-squared Law (kyun ek thin cone matter karta hai), aur Combustion Instability (kyun hum spray ki parwah karte hain).

Step 1 — Do motions, alag alag draw kiye

KYUN. Space mein koi bhi motion, chahe kitni bhi curved ho, hamesha chosen directions ke saath simple straight-line pieces mein split ki ja sakti hai. Physics ka poora trick yahi hai: decompose karo, phir recombine karo. Hum do directions choose karte hain jo injector ki apni geometry se match karti hain.

Axial direction — seedha chamber axis ke neeche, jis taraf rocket apna exhaust push karega. Blob ki speed is direction mein $v_{x}$ kaho (chota $x$ "axis" ke liye).
Tangential direction — sideways, us circle ke saath jis par blob vortex chamber ke andar spin kar raha tha. Is speed ko $v_{t}$ kaho ( $t$ "tangential" ke liye, matlab "circle ke kinare ke saath").

Figure — Injector design — impinging, coaxial, swirl injectors

Amber blob ko dekho: cyan arrow aage point karta hai ( $v_{x}$ ), white arrow sideways point karta hai ( $v_{t}$ ). Akela koi bhi arrow real motion nahi hai — real motion abhi aani hai.

Step 2 — Sideways motion kahan se aati hai: vortex chamber

KYUN yeh design. Agar tum paani seedha drain par squirt karo toh woh bas seedha ghus jaata hai. Agar tum use ek round bowl ki wall ke saath squirt karo, toh woh swirl karta hai — bilkul waisi tarah jaise paani plughole se neeche jaata hai. Swirl injector jaanboojhkar plughole vortex banata hai taaki nikalta liquid pehle se spin kar raha ho.

Amber inlet arrow wall ke saath enter karta hai, center ki taraf nahi. Cyan spiral follow karo: blob wall par ride karta hai, chote exit ke paas aate aate tezi se spin karta hai (yeh speed-up ek Bernoulli effect hai — narrow path, faster flow). Jab tak woh exit tak pahunchta hai tab tak usne ek real tangential speed $v_{t}$ carry kar li hoti hai.

Step 3 — Do arrows combine karo: velocity triangle

KYUN tip-to-tail. Velocities displacements ki tarah add hoti hain: agar ek second mein tum aage $v_{x}$ slide karo aur sideways $v_{t}$ slide karo, to tera net move start se finish tak ki straight line hai — un dono ke banaye rectangle ka diagonal. Woh diagonal resultant velocity hai.

Blue right triangle mein ek cyan leg hai ( $v_{x}$ , forward), ek white leg hai ( $v_{t}$ , sideways), aur ek amber diagonal hai ( $v$ , true motion). Corner mein chota square forward aur sideways ke beech $9 0^{\circ}$ mark karta hai — woh construction se perpendicular hain, aur yehi wajah hai ki Pythagoras apply hota hai.

Step 4 — Angle $ϕ$ paida hota hai, aur yeh ek tangent hai

KYUN tangent, sine ya cosine nahi? Hum ek aisa number chahte hain jo "kitna sideways per unit forward" — tilt ki steepness — capture kare. Step 3 ke right triangle par:

$ϕ$ ki opposite side (woh leg jo angle ka samna kare) sideways leg $v_{t}$ hai.
Adjacent side (woh leg jo angle ko touch kare, axis ke saath) forward leg $v_{x}$ hai.

Ratio opposite ÷ adjacent tangent ki definition hai — "har ek step forward ke liye tum kitna sideways utha jaate ho". Woh ratio precisely tilt ki steepness hai, isliye tangent natural tool hai. Sine ke liye diagonal length chahiye; cosine ke liye bhi; lekin hum tilt ko do legs se akele padh sakte hain, aur woh ratio $tan$ hai.

Corner par amber wedge $ϕ$ hai. Notice karo kaise woh axis (adjacent, cyan) aur true-velocity diagonal (amber) ke beech baith ta hai. Yeh bilkul wahi right-triangle reasoning hai jo tum $θ = arctan (A_{y} / A_{x})$ ke liye kisi bhi vector par use karte — yahan $v_{t}$ "up" component ka role play karta hai aur $v_{x}$ "along" component ka.

Step 5 — Ek tilted blob se poora hollow cone

KYUN hollow (filled nahi). Liquid vortex chamber ki wall par rida, isliye woh sab exit hole ki rim se nikalte hain, center se nahi. Center empty hai (air core). Rim + constant tilt = ek thin conical shell, beech mein khula — classic hollow-cone spray.

Har white arrow ek blob hai jo half-angle $ϕ$ se nikalta hai; us arrow ko axis ke around rotate karo aur woh cyan cone sweep karta hai. Amber dashed circle dikhata hai kahan sheet land karti hai — ek ring, dot nahi, yeh prove karta hai ki cone hollow hai.

Step 6 — Edge cases: spin bilkul band karo, phir poori tarah se badha do

Case A — no swirl, $v_{t} = 0$ . Fraction ka top zero hai, isliye $tan ϕ = 0$ , isliye $ϕ = 0^{\circ}$ . Picture: blob seedha axis ke neeche fly karta hai — ek pencil-thin jet, bilkul koi cone nahi. Yahi karta hai ek plain (non-swirl) orifice. ✓ samajh aata hai.

Case B — huge swirl, $v_{t} ≫ v_{x}$ . Fraction infinity ki taraf blow up karta hai, aur $tan ϕ \to \infty$ ka matlab hai $ϕ \to 9 0^{\circ}$ . Picture: cone almost flat disc mein flatten ho jaata hai sideways spray — great thinning, lekin chamber walls ko scrub kar dega. Designers isse pehle ruk jaate hain. ✓ limit behave karti hai.

Case C — danger zone, $ϕ \to 9 0^{\circ}$ . Jab $ϕ$ $9 0^{\circ}$ ki taraf jaata hai, $tan ϕ$ bina bound ke badhta hai — spin ka thoda sa extra bit ab cone ko enormously khol deta hai. Isliye near- $9 0^{\circ}$ designs twitchy hote hain aur wall wetting ke prone hote hain. Yehi wajah hai ki real swirl injectors comfortably $ϕ \approx 4 5^{\circ}$ – $6 0^{\circ}$ ke around baithte hain, jahan tangent gently change hota hai.

Teen sprays side by side: no-spin pencil (left), balanced cone (middle, useful design), flat disc (right). Dekho amber $ϕ$ wedge $0^{\circ}$ se almost $9 0^{\circ}$ tak kholta hai jab $v_{t} / v_{x}$ badhta hai.

Step 7 — Paas rakhne ke liye ek number

Worked example Worked value — cone angle nikalo

Ek swirl injector ek blob ko $v_{x} = 12 m/s$ forward aur $v_{t} = 20 m/s$ sideways deta hai. Half-cone angle $ϕ$ aur true exit speed $v$ nikalo.

Step 1 — ratio banao. Kyun? Formula sideways over forward maangta hai. $tan ϕ = \frac{v _{t}}{v _{x}} = \frac{20}{12} = 1.667$ Step 2 — arctan se tangent undo karo. Kyun? Hamare paas ratio hai; hum woh angle chahte hain jiska woh tangent hai. Woh "kaun sa angle?" ka sawaal exactly wahi hai jo $arctan$ answer karta hai. $ϕ = arctan (1.667) = 59. 0^{\circ}$ Step 3 — Pythagoras se true speed (Step 3 ka triangle). Kyun? Diagonal ki length woh real speed hai jo droplets carry karte hain. $v = 1 2^{2} + 2 0^{2} = 144 + 400 = 544 = 23.3 m/s$ Toh spray $\approx 5 9^{\circ}$ half-cone (wide, fine spray) ke roop mein khulta hai aur droplets $23.3 m/s$ par fly karte hain.

Ek-picture summary

Sab kuch ek frame mein: tangential inlet blob ko spin karta hai ( $v_{t}$ deta hai), blob aage bhi move karta hai ( $v_{x}$ ), dono tip-to-tail combine hoke true velocity $v$ banate hain jo $ϕ = arctan (v_{t} / v_{x})$ se tilted hai, aur us tilt ko axis ke around sweep karne se hollow cone paint ho jaata hai.

Recall Feynman: plain words mein poora walkthrough

Socho tum paani ek round sink mein aise daalte ho ki woh plughole se jaane se pehle swirl kare. Woh hole se sideways spinning aur aage drop hota hua ek saath nikalta hai. Un do motions ko do arrows ki tarah draw karo — ek forward, ek sideways — aur real path un dono ke beech ka slanted diagonal hai. Kitna slanted? "Sideways-per-forward" — exactly yahi hai jo tangent measure karta hai, isliye tilt angle $ϕ$ satisfy karta hai $tan ϕ = v_{t} / v_{x}$ . Kyunki paani hole ki rim se nikalta hai (woh wall par rida tha), aur har drop same $ϕ$ se tilts karta hai, unhe ghuma do aur ek thin hollow cone ban jaata hai. Zyada spin → wide flat cone → bahut fine mist. Koi spin nahi → ek boring straight pencil. Ek dum mein yahi hai swirl injector.

Recall Quick self-test

$tan$ kyun hai, $sin$ kyun nahi? ::: Kyunki hum tilt ko do legs se akele padhte hain (opposite $v_{t}$ , adjacent $v_{x}$ ), aur opposite/adjacent hi tangent hai. Jab $v_{t} \to 0$ toh cone ka kya hoga? ::: Woh straight axial jet mein collapse ho jaata hai, $ϕ \to 0^{\circ}$ . Spray hollow kyun hai, solid kyun nahi? ::: Liquid vortex-chamber wall par ride karta hai aur rim se nikalta hai, beech mein ek empty air core chhodta hai.