Injector design — impinging, coaxial, swirl injectors: Exercises

3.3.34 · D4 · Physics › Rocket Propulsion › Injector design — impinging, coaxial, swirl injectors

Shuru karne se pehle, teen master formulas ka ek reminder — seedhe plain words mein, taaki koi symbol surprise na kare.

Formula Teen tools jo baar baar kaam aayenge

Metering (kitna flow hoga): $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ — $\overset{m}{˙}$ hai mass per second (kg/s), $C_{d}$ hai discharge coefficient (real-orifice losses ke liye ek fudge factor, 0.6–0.9), $A$ hai hole area (m²), $ρ$ hai liquid density (kg/m³), $Δ p$ hai hole ke across squeeze hua pressure (Pa). Yeh seedha Bernoulli se aata hai.

Impinging spray direction: $tan α = \frac{m ˙ _{1} v _{1} sin θ _{1} - m ˙ _{2} v _{2} sin θ _{2}}{m ˙ _{1} v _{1} cos θ _{1} + m ˙ _{2} v _{2} cos θ _{2}}$ — $α$ woh angle hai jis par merged sheet nikalta hai, chamber axis se measure karke (seedha neeche = 0°). $θ$ har jet ka aiming angle hai; $\overset{m}{˙} v$ ek jet ka momentum flux hai (kitni zyada push karti hai).

Coaxial atomization: $J = \frac{ρ _{o} v _{o}^{2}}{ρ _{i} v _{i}^{2}}$ — outer-over-inner momentum flux ratio. Bada $J$ ⇒ zyada fine droplets.

Swirl cone: $tan ϕ = \frac{v _{t}}{v _{x}}$ — tangential spin $v_{t}$ ko axial speed $v_{x}$ se divide karne par milta hai half-cone angle.

Level 1 — Recognition

L1.1 Ek designer chahta hai ki bina propellant badle ek single fixed orifice se zyada mass flow aaye. $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ dekhte hue, woh kaun si ek quantity badha sakta hai, aur kya $\overset{m}{˙}$ us quantity ke saath linearly badhega ya square root ki tarah?

Recall Solution

Woh pressure drop $Δ p$ badha sakta hai (area, density, $C_{d}$ yahan hardware/propellant se fix hain). Kyunki $\overset{m}{˙} \propto Δ p$ hai, $Δ p$ double karne par $\overset{m}{˙}$ sirf $2 \approx 1.41$ guna hoga, 2 guna nahi. Flow badhta hai pressure drop ke square root ki tarah. Square root kyun, linear kyun nahi? Pressure drop kinetic energy $\frac{1}{2} ρ v^{2}$ mein convert hota hai, aur energy $v^{2}$ ke saath scale karti hai; isliye $v$ (aur hence $\overset{m}{˙}$ ) $Δ p$ ke saath scale karta hai.

L1.2 Har injector type ko uske primary atomization mechanism se match karo: (a) impinging, (b) coaxial, (c) swirl. Options hain: collision of jets / shear between two streams / centrifugal thinning of a hollow cone.

Recall Solution

(a) Impinging → collision of jets (do streams ek fan sheet mein smash hoti hain).
(b) Coaxial → shear between two streams (fast outer stream slow inner jet ko shred karti hai).
(c) Swirl → centrifugal thinning of a hollow cone (spin liquid ko outward fling karke ek thin cone banata hai). Key distinction (dekho Atomization and the d-squared Law): coaxial ko ek second high-speed stream chahiye; swirl akele ek propellant ko atomize kar sakta hai.

L1.3 Sach ya jhooth: " $Δ p$ ko kam se kam ~15% chamber pressure $p_{c}$ rakhna sirf ek metering requirement hai." Explain karo.

Recall Solution

Jhooth. Yeh mainly ek stability requirement hai. Ek kaafi bada $Δ p$ feed lines aur chamber ke beech ek stiff spring ki tarah kaam karta hai: chamber pressure ki wobbles orifice ke through upar push back nahi kar saktin. $Δ p$ bahut kam karo toh feed + chamber couple ho jaate hain → Combustion Instability. Yeh metering aur stability dono set karta hai.

Level 2 — Application

L2.1 Kerosene ( $ρ = 810 kg/m^{3}$ ) ek orifice se $\overset{m}{˙} = 0.5 kg/s$ par flow kar raha hai jisme $C_{d} = 0.75$ aur $Δ p = 3 bar = 3 \times 1 0^{5} Pa$ hai. (a) jet velocity aur (b) orifice diameter nikalo.

Recall Solution

(a) Velocity Bernoulli se, $v = 2Δ p / ρ$ . Yeh form kyun? Ideal jet speed sirf pressure drop aur density par depend karti hai. $v = \frac{2 ( 3 \times 1 0 ^{5} )}{810} = 740.7 = 27.2 m/s$ (b) Area $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ se. $2 ρ Δ p = 2 (810) (3 \times 1 0^{5}) = 4.86 \times 1 0^{8} = 2.204 \times 1 0^{4}$ $A = \frac{0.5}{0.75 \times 2.204 \times 1 0 ^{4}} = 3.02 \times 1 0^{- 5} m^{2}$ Diameter $A = π d^{2} /4$ se: $d = 4 A / π = \frac{4 ( 3.02 \times 1 0 ^{- 5} )}{π} = 6.2 \times 1 0^{- 3} m = 6.2 mm$

L2.2 Ek swirl injector element mein axial speed $v_{x} = 20 m/s$ aur tangential (spin) speed $v_{t} = 20 m/s$ hai. Half-cone angle $ϕ$ nikalo. Phir, agar designer $v_{t}$ ko double karke $40 m/s$ kar de (wahi $v_{x}$ ), toh naya $ϕ$ nikalo.

Pehle geometry dekho — figure mein label kiya gaya hai ki kaun sa arrow axial speed $v_{x}$ hai, kaun sa tangential spin $v_{t}$ hai, aur dono milke kaisa resultant banate hain jo cone angle set karta hai:

Figure — Injector design — impinging, coaxial, swirl injectors

Recall Solution

Pehla case: $tan ϕ = v_{t} / v_{x} = 20/20 = 1 \Rightarrow ϕ = arctan (1) = 4 5^{\circ}$ . Arctan kyun? $arctan$ jawaab deta hai "is tangent wala angle kaun sa hai?" — yeh tangent ko undo karke angle wapas deta hai. Doubled spin: $tan ϕ = 40/20 = 2 \Rightarrow ϕ = arctan (2) = 63. 4^{\circ}$ . Zyada spin ⇒ wider, thinner cone ⇒ (d²-law se) zyada fine droplets jo tezi se evaporate hote hain.

L2.3 Ek coaxial element: inner LOX $ρ_{i} = 1140 kg/m^{3}$ , $v_{i} = 12 m/s$ ; outer gaseous hydrogen $ρ_{o} = 1.2 kg/m^{3}$ , $v_{o} = 320 m/s$ . Momentum flux ratio $J$ compute karo aur batao ki atomization acchi hogi ya nahi.

Recall Solution

Inner momentum flux: $ρ_{i} v_{i}^{2} = 1140 \times 1 2^{2} = 1140 \times 144 = 1.6416 \times 1 0^{5}$ . Outer momentum flux: $ρ_{o} v_{o}^{2} = 1.2 \times 32 0^{2} = 1.2 \times 1.024 \times 1 0^{5} = 1.2288 \times 1 0^{5}$ . $J = \frac{1.2288 \times 1 0 ^{5}}{1.6416 \times 1 0 ^{5}} = 0.749$ $J < 1$ : outer stream ka momentum inner jet se thoda kam hai, isliye shearing sirf moderate hai. Atomization improve karne ke liye designer $v_{o}$ badhayega (light gas ko fast jaana padega, kyunki uski density bahut kam hai) ya $v_{i}$ ghataayega.

Level 3 — Analysis

L3.1 Ek balanced doublet: do equal jets ( $\overset{m}{˙}_{1} v_{1} = \overset{m}{˙}_{2} v_{2} = P$ ), dono axis se $θ = 3 0^{\circ}$ par lekin opposite sides par. Dikhao ki resultant sheet seedha neeche jaata hai ( $α = 0$ ). Phir dikhao kya hoga agar ek jet 20% kamzor ho jaye ( $\overset{m}{˙}_{2} v_{2} = 0.8 P$ , same angles).

Momentum vectors dekho:

Recall Solution

Balanced case. Jet 1 ko $+$ side par rakho ( $+ sin 3 0^{\circ}$ ) aur jet 2 ko $-$ side par ( $- sin 3 0^{\circ}$ ): $tan α = \frac{P s i n 3 0 ^{\circ} - P s i n 3 0 ^{\circ}}{P c o s 3 0 ^{\circ} + P c o s 3 0 ^{\circ}} = \frac{0}{2 P c o s 3 0 ^{\circ}} = 0 \Rightarrow α = 0^{\circ} .$ Transverse (sideways) momenta exactly cancel ho jaate hain ⇒ spray seedha axis ke neeche jaata hai — exactly wahi jo wall scrubbing se bachata hai.

Unbalanced case ( $P_{1} = P$ , $P_{2} = 0.8 P$ ): $tan α = \frac{P s i n 3 0 ^{\circ} - 0.8 P s i n 3 0 ^{\circ}}{P c o s 3 0 ^{\circ} + 0.8 P c o s 3 0 ^{\circ}} = \frac{0.2 P ( 0.5 )}{1.8 P ( 0.8660 )} = \frac{0.1}{1.5588} = 0.06415.$ $α = arctan (0.06415) = 3.6 7^{\circ} .$ Sheet ab stronger jet ki original side ki taraf $3.6 7^{\circ}$ tilt ho jaati hai — woh residual transverse momentum spray ko wall ki taraf push kar sakta hai, jo ek real burn-through ka risk hai.

L3.2 Chamber $p_{c} = 70 bar$ par run kar raha hai. Current injector mein $Δ p = 7 bar$ hai (yaani $0.10 p_{c}$ ). Stability rule of thumb $Δ p \geq 0.15 p_{c}$ use karke, $\overset{m}{˙}$ -fixed orifice area ko kis factor se change karna hoga agar designer $Δ p$ ko exactly $0.15 p_{c}$ tak raise kare aur $\overset{m}{˙}$ , $ρ$ , $C_{d}$ constant rakhe?

Recall Solution

$\overset{m}{˙} = C_{d} A 2 ρ Δ p$ se, $\overset{m}{˙}$ fixed rakhne par $A Δ p = const$ hoga, isliye $A \propto 1/ Δ p$ . Naya $Δ p = 0.15 p_{c}$ , purana $Δ p = 0.10 p_{c}$ , ratio $= 1.5$ . $\frac{A _{new}}{A _{old}} = \frac{Δ p _{old}}{Δ p _{new}} = \frac{0.10}{0.15} = 0.6667 = 0.8165.$ Orifice area ko apni purani value ka 81.6% tak shrink karna hoga (lagbhag 18.4% reduction) taaki zyada safe $Δ p$ par wahi $\overset{m}{˙}$ mile. Chhote holes, zyada $Δ p$ — classic stability trade.

L3.3 Do droplet populations do injectors se nikalte hain: injector A $d_{A} = 50 μ m$ diameter ke droplets banata hai, injector B $d_{B} = 80 μ m$ ke. $d^{2}$ -law (evaporation/burn time $t \propto d^{2}$ ) use karke, burn times ka ratio $t_{B} / t_{A}$ nikalo. Agar injector A chamber ke andar just fully burn ho jaata hai, toh kya B safe hai?

Recall Solution

$\frac{t _{B}}{t _{A}} = (\frac{d _{B}}{d _{A}})^{2} = (\frac{80}{50})^{2} = (1.6)^{2} = 2.56.$ B ke droplets ko burn hone mein 2.56× zyada time lagta hai. Agar A sirf just available residence time mein finish hua, toh B ko 2.56× woh time chahiye aur woh finish nahi hoga — unburned propellant chamber se bahar nikalta hai, jisse c* efficiency cut hoti hai. B safe nahi hai. (Dekho Atomization and the d-squared Law.)

Level 4 — Synthesis

L4.1 Ek engine LOX + kerosene ko oxidizer-to-fuel ratio $O / F = 2.5$ (mass basis) par burn karta hai, total propellant flow $\overset{m}{˙}_{tot} = 3.5 kg/s$ . (a) $\overset{m}{˙}_{o x}$ aur $\overset{m}{˙}_{f u e l}$ mein split karo. (b) Kerosene ( $ρ = 810$ ) ko $C_{d} = 0.75$ , $Δ p = 3 \times 1 0^{5} Pa$ se meter kiya jaata hai. Kitne 1.0 mm-diameter fuel orifices chahiye?

Recall Solution

(a) $O / F = \overset{m}{˙}_{o x} / \overset{m}{˙}_{f u e l} = 2.5$ aur $\overset{m}{˙}_{o x} + \overset{m}{˙}_{f u e l} = 3.5$ . Toh $\overset{m}{˙}_{f u e l} (2.5 + 1) = 3.5 \Rightarrow \overset{m}{˙}_{f u e l} = 3.5/3.5 = 1.0 kg/s$ , aur $\overset{m}{˙}_{o x} = 2.5 kg/s$ . (Dekho O/F Ratio and Mixture Ratio.) (b) Pehle metering constant $2 ρ Δ p$ ko scratch se recompute karo in fuel values ke liye ( $ρ = 810$ , $Δ p = 3 \times 1 0^{5} Pa$ ) — yeh L2.1 se match hota hai kyunki fuel aur pressure drop same hain, lekin hum iska origin dikhate hain taaki kuch blindly borrow na ho: $2 ρ Δ p = 2 (810) (3 \times 1 0^{5}) = 4.86 \times 1 0^{8} = 2.204 \times 1 0^{4} (units: Pa \cdot kg/m^{3} = kg/(m^{2} s)) .$ Ek 1.0 mm hole area: $A_{1} = π d^{2} /4 = π (1.0 \times 1 0^{- 3})^{2} /4 = 7.854 \times 1 0^{- 7} m^{2}$ . Flow per hole: $\overset{m}{˙}_{1} = C_{d} A_{1} 2 ρ Δ p = 0.75 (7.854 \times 1 0^{- 7}) (2.204 \times 1 0^{4}) = 1.298 \times 1 0^{- 2} kg/s$ . Holes chahiye: $N = \overset{m}{˙}_{f u e l} / \overset{m}{˙}_{1} = 1.0/1.298 \times 1 0^{- 2} = 77.0 \Rightarrow$ round up karke 78 orifices (flow area kabhi kam nahi provide karte).

L4.2 Ek coaxial element ko achhi atomization ke liye $J \geq 3$ chahiye. Inner LOX: $ρ_{i} = 1140$ , $v_{i} = 15 m/s$ . Outer gaseous hydrogen $ρ_{o} = 1.5$ . $J = 3$ achieve karne ke liye minimum outer velocity $v_{o}$ kya chahiye? Comment karo ki number itna bada kyun hai aur incompressible formula ko yahan caution ke saath kyun use karna chahiye.

Recall Solution

$J = \frac{ρ _{o} v _{o}^{2}}{ρ _{i} v _{i}^{2}} = 3 \Rightarrow v_{o}^{2} = \frac{3 ρ _{i} v _{i}^{2}}{ρ _{o}} = \frac{3 ( 1140 ) ( 1 5 ^{2} )}{1.5} = \frac{3 ( 1140 ) ( 225 )}{1.5} = 513000.$ $v_{o} = 513000 = 716.2 m/s .$ Number itna bada hai kyunki hydrogen LOX se ~760× halka hai; uska momentum flux $ρ_{o} v_{o}^{2}$ liquid ke momentum flux se exceed karne ka ek hi rasta hai — $v_{o}$ ko enormous banana. Isliye hydrogen-oxygen coaxial injectors gas ko near sonic speeds par run karte hain — light propellant ko heavy one ko shear karne ke liye bahut fast jaana padta hai. Yeh Regenerative Cooling se bhi juda hai: woh hot, fast H₂ pehle chamber walls ko cool karta tha.

Caution — compressibility. Formula $J = ρ_{o} v_{o}^{2} / ρ_{i} v_{i}^{2}$ incompressible momentum bookkeeping se aaya hai, jo quietly assume karta hai ki gas density $ρ_{o}$ stream speed up hone par fixed rehti hai. Lekin cold hydrogen mein speed of sound sirf ~ $1250 m/s$ hai, isliye $v_{o} = 716 m/s$ ka Mach number roughly $716/1250 \approx 0.57$ hai — yeh compressible regime mein kaafi andar hai, jahan $ρ_{o}$ actually drop karta hai jab gas injector ke through accelerate hoti hai. Real design mein tum manifold value ki jagah actual injector-exit density (gas ke pressure aur temperature se) use karte, aur is $J$ ko ek first-cut estimate maante; Mach numbers ke paas 1 ke, true momentum flux incompressible prediction se meaningfully alag hota hai.

Level 5 — Mastery

L5.1 Tumhe ek chamber diya gaya hai jisme $p_{c} = 100 bar$ , total flow $\overset{m}{˙} = 8 kg/s$ , kerosene fuel fraction $O / F = 2.6$ se. Fuel side design karo: (a) minimum stable value $0.15 p_{c}$ par $Δ p$ pick karo; (b) $C_{d} = 0.8$ , $ρ = 810$ ke saath, total fuel orifice area nikalo; (c) agar mission finer atomization bhi chahta hai, quantitatively argue karo (empirical trend $d \propto (Δ p)^{- 0.4}$ use karke) ki $Δ p = 0.25 p_{c}$ par run karne se burn time kitna improve hoga. Baaki sab fixed maano.

Recall Solution

(a) $Δ p = 0.15 \times 100 bar = 15 bar = 1.5 \times 1 0^{6} Pa$ . (b) Fuel flow: $\overset{m}{˙}_{f u e l} = \overset{m}{˙} / (O / F + 1) = 8/3.6 = 2.222 kg/s$ . $2 ρ Δ p = 2 (810) (1.5 \times 1 0^{6}) = 2.43 \times 1 0^{9} = 4.930 \times 1 0^{4} .$ $A = \frac{m ˙ _{f u e l}}{C _{d} 2 ρ Δ p} = \frac{2.222}{0.8 \times 4.930 \times 1 0 ^{4}} = 5.635 \times 1 0^{- 5} m^{2} .$ (c) Empirical trend hai $d \propto (Δ p)^{- 0.4}$ (zyada pressure drop → tezi se, thinner sheet → chhote droplets). $Δ p$ ko $0.15 p_{c}$ se $0.25 p_{c}$ tak raise karna ek factor hai $r = \frac{0.25 p _{c}}{0.15 p _{c}} = \frac{0.25}{0.15} = 1.667.$ Purane ke relative naya droplet size: $\frac{d _{new}}{d _{old}} = r^{- 0.4} = 1.66 7^{- 0.4} .$ Evaluate karne ke liye logs lo: $ln 1.667 = 0.5108$ , times $- 0.4$ deta hai $- 0.2043$ , aur $e^{- 0.2043} = 0.8152$ . Toh $d_{new} / d_{old} = 0.8152$ (droplets ~18% chhote). Burn time $d^{2}$ -law follow karta hai, $t \propto d^{2}$ , isliye $\frac{t _{new}}{t _{old}} = (\frac{d _{new}}{d _{old}})^{2} = (0.8152)^{2} = 0.6646.$ Result: burn time original ka ~66.5% ho jaata hai — lagbhag 33.5% faster burn. Physical significance: chhota burn time matlab propellant chamber ke residence time ke andar acchi tarah react kar leta hai, isliye c* efficiency badhti hai aur bahut-chhote chamber ke against margin milta hai. Lekin yeh improvement khareedni padti hai — $0.15 p_{c}$ se $0.25 p_{c}$ jaana 67% zyada pressure drop hai, isliye turbopump ko substantially zyada kaam karna padega. Yahi central injector trade hai: better atomization aur stability versus pump power.

L5.2 Ek designer propose karta hai ki unbalanced impinging doublet (residual tilt $α = 5^{\circ}$ ) ko ek swirl element se replace karein taaki wall-scrubbing risk hat sake. Do alag physics reasons batao ki yeh kaise help kar sakta hai, aur swirl ka ek naya risk — har ek ek sentence mein ek mechanism se tied.

Recall Solution

Reason 1 (symmetry): ek swirl element ek axisymmetric hollow cone produce karta hai, isliye koi net transverse momentum nahi hota jo spray ko wall ki taraf tilt kare — imbalance problem simply khatam ho jaati hai. Reason 2 (low- $Δ p$ atomization): swirl ek self-supported sheet ki centrifugal thinning se atomize karta hai, isliye yeh modest $Δ p$ par bhi fine droplets reach kar leta hai, unlike shear-based schemes jinhein high relative velocity chahiye. Naya risk: wide, low-pressure hollow cone aur uska central recirculation zone pressure waves ke liye ek receptive feedback path ka kaam kar sakta hai, jisse combustion instability ka chance badh sakta hai agar cone angle aur $Δ p$ tune na kiye jaayein.

Recall Jaane se pehle ek-line self-check

Metering law? ::: $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ , aur $\overset{m}{˙} \propto Δ p$ . $Δ p ≳ 0.15 p_{c}$ kyun rakhein? ::: feed ko chamber se decouple karne aur combustion instability se bachne ke liye. Coaxial atomization number? ::: momentum flux ratio $J = ρ_{o} v_{o}^{2} / ρ_{i} v_{i}^{2}$ ; velocity squared hoti hai (aur compressibility watch karo jab $v_{o}$ sonic ke paas ho). Swirl cone angle? ::: $tan ϕ = v_{t} / v_{x}$ ; zyada spin ⇒ wider, thinner cone ⇒ finer droplets.