Injector design — impinging, coaxial, swirl injectors: Worked examples

3.3.34 · D3 · Physics › Rocket Propulsion › Injector design — impinging, coaxial, swirl injectors

Koi bhi formula use karne se pehle, har symbol ko plain words mein naam dete hain taaki kuch bhi undefined na aaye.

Intuition

$Δ p$ bahut chhota kyun nahi hona chahiye (stability floor, yahan derive kiya gaya) Orifice law kehta hai $\overset{m}{˙} \propto Δ p$ . Sensitivity differentiate karo: ek chhoti si wobble $δ (Δ p)$ flow wobble produce karti hai $δ \overset{m}{˙} / \overset{m}{˙} = \frac{1}{2} δ (Δ p) /Δ p$ . Agar $Δ p$ bada hai kisi bhi chamber-pressure ripple ke comparison mein, toh woh ripple $Δ p$ ka ek chhota fraction hai, toh $δ \overset{m}{˙}$ tiny hai — feed ko chamber ki koi khabar nahi. Agar $Δ p$ chhota hai, toh ek given size ki chamber ripple $Δ p$ ka ek bada fraction hai, toh woh flow ko strongly modulate karti hai, aur modulated flow ripple ko feed back karti hai — ek self-amplifying loop (dekho Combustion Instability). Engineers isliye $Δ p$ ko kam se kam $\sim 15%$ of chamber pressure $p_{c}$ rakhte hain; hum yeh floor $Δ p ≳ 0.15 p_{c}$ likhte hain aur isko neeche use karte hain.

The scenario matrix

Ab jo symbols hain, aao har situation list karte hain jisme yeh formulas push ki ja sakti hain. Har row ek "cell" hai jisme reader ko survive karna chahiye. Neeche har example ko us cell ke saath tag kiya gaya hai jise woh kill karta hai.

Cell	Tool	Tricky part
A normal metering	orifice law	seedha plug-and-chug, area aur diameter nikalo
B inverse metering	orifice law	fixed hole diya hai, $Δ p$ solve karo
C zero / degenerate	orifice law	agar $Δ p \to 0$ toh kya? $\overset{m}{˙} \to 0$ ka physically matlab kya
D balanced impinging	momentum balance	transverse terms cancel → seedha spray ( $α = 0$ )
E unbalanced impinging	momentum balance	ek jet zyada strong → spray tilts, kaunsa sign?
F extreme impinging	momentum balance	ek jet mar jaata hai ( $\overset{m}{˙}_{2} \to 0$ ) → sheet survivor ke saath chalti hai
G coaxial normal	momentum ratio $J$	light fast gas vs heavy slow liquid, kya $J > 1$ hai?
H coaxial limit	momentum ratio $J$	$Δ v \to 0$ → koi shear nahi → atomization fail
I swirl angle	$tan ϕ = v_{t} / v_{x}$	spin vs axial, "cone kitna chauda hai" ke charon cases
J word problem	orifice + $d^{2}$ -law	ek short chamber mein burn time ke liye holes ki count chuno
K exam twist	orifice law	$Δ p$ $p_{c}$ ke percent ke roop mein diya gaya, hidden units

Cell A + K — metering, seedha aur disguised

Worked example Example 1 (Cell A) — ek single orifice size karo

$\overset{m}{˙} = 0.80 kg/s$ liquid oxygen deliver karo, $ρ = 1140 kg/m^{3}$ , saath mein $Δ p = 4 bar = 4 \times 1 0^{5} Pa$ aur $C_{d} = 0.80$ . Orifice area $A$ aur diameter $d$ nikalo.

Forecast: Zyada flow ke liye zyada area chahiye; zyada $Δ p$ jet ko faster banata hai toh hole chhota ho sakta hai. Guess: kuch square millimetres, toh kuch mm across.

Orifice law ko $A$ ke liye rearrange karo. Kyun? Sab kuch $A$ ko chhodkar known hai, toh use isolate karo. $A = \frac{m ˙}{C _{d} 2 ρ Δ p}$
Root $2 ρ Δ p$ compute karo. Kyun? Yeh radicand density aur ideal kinetic-energy term ka product hai; yeh jet speed nahi hai. (True ideal jet speed hai $v = 2 Δ p / ρ$ ; yahan hum deliberately $ρ$ ko root andar rakhte hain taaki $C_{d} A 2 ρ Δ p$ directly mass flow ke roop mein aaye. Radicand ka kaam sirf pressure-and-density ko quantity $ρ v$ mein convert karna hai.) $2 (1140) (4 \times 1 0^{5}) = 9.12 \times 1 0^{8} = 3.020 \times 1 0^{4}$
Divide karo. Kyun? Ab pieces plug in karo. $A = \frac{0.80}{0.80 \times 3.020 \times 1 0 ^{4}} = 3.311 \times 1 0^{- 5} m^{2}$
$A = π d^{2} /4$ se diameter. Kyun? Real orifices round holes hote hain. $d = \frac{4 A}{π} = \frac{4 ( 3.311 \times 1 0 ^{- 5} )}{π} = 6.49 \times 1 0^{- 3} m \approx 6.5 mm$

Verify: $2 ρ Δ p$ ke units hain $(kg/m^{3}) (Pa) = kg^{2} / (m^{4} s^{2}) = kg / (m^{2} s)$ , toh $A \cdot$ woh $=$ kg/s ✓. True jet speed cross-check karo $v = 2 Δ p / ρ = 2 (4 \times 1 0^{5}) /1140 = 26.5 m/s$ , aur sach mein $ρ v = 1140 \times 26.5 = 3.02 \times 1 0^{4}$ radicand se match karta hai ✓. Wapas plug karo: $0.80 \times 3.311 \times 1 0^{- 5} \times 3.020 \times 1 0^{4} = 0.80 kg/s$ ✓.

Worked example Example 2 (Cell K, exam twist) —

$Δ p$ percentage ke roop mein chhupa hua Same LOX ( $ρ = 1140$ ), same $\overset{m}{˙} = 0.80 kg/s$ , $C_{d} = 0.80$ , lekin ab tumhe bataya gaya hai chamber pressure $p_{c} = 70 bar$ aur $Δ p = 15%$ of $p_{c}$ . $A$ nikalo.

Forecast: 70 bar ka $15%$ lagbhag 10 bar hai — Example 1 se bada $Δ p$ , toh hole chhota hona chahiye 6.5 mm se.

Percent ko pascals mein pressure mein badlo. Kyun? Formula ko Pa chahiye, "percent" nahi. Yahi woh jagah hai jahan exam students marks kho dete hain. $Δ p = 0.15 \times 70 bar = 10.5 bar = 1.05 \times 1 0^{6} Pa$
Root. Kyun? Pehle jaisa converter. $2 (1140) (1.05 \times 1 0^{6}) = 2.394 \times 1 0^{9} = 4.893 \times 1 0^{4}$
Area. Kyun? $A$ phir se isolate karo. $A = \frac{0.80}{0.80 \times 4.893 \times 1 0 ^{4}} = 2.044 \times 1 0^{- 5} m^{2} \Rightarrow d = 5.10 mm$

Verify: Zyada $Δ p$ → chhota hole, jaise forecast tha (5.1 mm < 6.5 mm) ✓. Aur $Δ p / p_{c} = 15%$ exactly us stability floor par baitha hai $Δ p ≳ 0.15 p_{c}$ jo humne matrix ke upar derive ki thi, toh yeh design minimum safe margin par hai.

Cell B + C — inverse metering aur zero limit

Worked example Example 3 (Cell B) — hole fixed hai, woh

$Δ p$ nikalo jo chahiye Ek finished injector ka total effective area $A = 3.0 \times 1 0^{- 5} m^{2}$ hai, $C_{d} = 0.80$ , kerosene $ρ = 810 kg/m^{3}$ flow kar raha hai. Tumhe $\overset{m}{˙} = 0.60 kg/s$ push karna hai. Kaun sa $Δ p$ required hai?

Forecast: Kyunki $\overset{m}{˙} \propto Δ p$ , flow badhane ke liye $Δ p$ quadratically badhana padta hai — bada pressure expect karo.

$Δ p$ free karne ke liye orifice law ko square karo. Kyun? $Δ p$ square root ke andar hai; dono sides square karne se woh bahar aa jaata hai. $\overset{m}{˙}^{2} = C_{d}^{2} A^{2} (2 ρ Δ p)$
$Δ p$ ke liye solve karo. Kyun? Unknown isolate karo. $Δ p = \frac{m ˙ ^{2}}{2 ρ C _{d}^{2} A ^{2}} = \frac{0.6 0 ^{2}}{2 ( 810 ) ( 0.80 ) ^{2} ( 3.0 \times 1 0 ^{- 5} ) ^{2}}$
Calculate karo. Kyun? Number nikalo. $Δ p = \frac{0.36}{2 ( 810 ) ( 0.64 ) ( 9.0 \times 1 0 ^{- 10} )} = \frac{0.36}{9.331 \times 1 0 ^{- 7}} = 3.86 \times 1 0^{5} Pa \approx 3.86 bar$

Verify: Ise forward law mein feed karo: $0.80 \times 3.0 \times 1 0^{- 5} \times 2 (810) (3.86 \times 1 0^{5}) = 0.80 \times 3.0 \times 1 0^{- 5} \times 2.500 \times 1 0^{4} = 0.600 kg/s$ ✓.

Worked example Example 4 (Cell C) — degenerate zero case

Example 3 ke injector mein, $Δ p \to 0$ hone par $\overset{m}{˙}$ ka kya hota hai? Aur agar pumps fail ho jaayein toh $Δ p = 0$ exactly ho jaata hai?

Forecast: Koi push nahi → koi jet nahi → koi flow nahi. Lekin kitni tezi se yeh khatam hota hai?

Law ki dependence padho. Kyun? $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ mein $Δ p$ root ke andar hai, toh $\overset{m}{˙} \propto Δ p$ .
Limit lo. Kyun? Hum behavior chahte hain, sirf ek point nahi. $lim_{Δ p \to 0^{+}} \overset{m}{˙} = C_{d} A 2 ρ \cdot 0 = 0$
Approach ki shape note karo. Kyun? Square root ka matlab hai curve zero mein infinite slope ke saath aata hai — $Δ p$ ko half karne se flow sirf $2 \approx 1.41$ factor se cut hoti hai, 2 se nahi. Toh flow pehle dheere girta hai, phir zero ke paas plunge karta hai.

Verify: $Δ p = 3.86 bar$ par, $\overset{m}{˙} = 0.60$ . Half karke $1.93 bar$ : $\overset{m}{˙} = 0.60/ 2 = 0.424 kg/s$ ✓ — $0.30$ nahi. Yahi reason hai chhota $Δ p$ dangerous hai: flow barely metered hoti hai aur chamber pressure peeche dhakka de sakta hai → instability.

Cells D, E, F — impinging jets, har collision geometry

Us convention ke saath, transverse-momentum balance resultant sheet angle $α$ axis se deta hai: $tan α = \frac{m ˙ _{1} v _{1} sin θ _{1} + m ˙ _{2} v _{2} sin θ _{2} net sideways push}{total downward push m ˙ _{1} v _{1} cos θ _{1} + m ˙ _{2} v _{2} cos θ _{2}}$

Neeche ke figure ko dekho jab tum Examples 5–7 padho. Blue arrow jet 1 hai (right ki taraf jhuka, $θ_{1} = + 3 5^{\circ}$ ), pink arrow jet 2 hai (left ki taraf jhuka, $θ_{2} = - 3 5^{\circ}$ ), aur yellow arrow resultant sheet hai jiska dashed axis se tilt woh angle $α$ hai jo hum solve karte hain. Neeche right mein chhota horizontal "+ transverse" arrow notice karo: woh mark karta hai kaun si sideways direction positive count hoti hai, toh positive $α$ matlab sheet right ki taraf jhukti hai.

Figure — Injector design — impinging, coaxial, swirl injectors

Worked example Example 5 (Cell D) — balanced doublet

Equal push $P_{1} = P_{2} = P$ (jahan $P = \overset{m}{˙} v$ ) ki do jets, symmetrically aimed: jet 1 right par $θ_{1} = + 3 5^{\circ}$ jhuka, jet 2 left par $θ_{2} = - 3 5^{\circ}$ jhuka. $α$ nikalo.

Forecast: Bilkul symmetric → do sideways pushes equal aur opposite hain → cancel ho jaate hain → seedha neeche, $α = 0$ .

Signed angles use karke har transverse component likho. Kyun? Sign convention bookkeeping karta hai: jet 1 ka sideways push $+ P sin 3 5^{\circ}$ hai, jet 2 ka $P sin (- 3 5^{\circ}) = - P sin 3 5^{\circ}$ hai.
Unhe add karo (numerator = net sideways push). Kyun? Formula signed components sum karta hai. $num = P sin (+ 3 5^{\circ}) + P sin (- 3 5^{\circ}) = P sin 3 5^{\circ} - P sin 3 5^{\circ} = 0$
Denominator (dono neeche push karte hain). Kyun? $cos (+ 3 5^{\circ}) = cos (- 3 5^{\circ})$ , toh downward parts add hote hain. $den = P cos 3 5^{\circ} + P cos 3 5^{\circ} = 2 P cos 3 5^{\circ}$
Ratio. Kyun? $tan α = 0/ (2 P cos 3 5^{\circ}) = 0 \Rightarrow α = 0^{\circ}$ .

Verify: $tan^{- 1} (0) = 0^{\circ}$ ✓. Ek balanced doublet axis ke neeche spray karta hai, wall se door — exactly design goal.

Worked example Example 6 (Cell E) — unbalanced doublet, kis taraf jhukta hai?

Same symmetric aiming — jet 1 $θ_{1} = + 3 5^{\circ}$ par, jet 2 $θ_{2} = - 3 5^{\circ}$ par — lekin jet 1 zyada strong hai: $P_{1} = 120 N$ , $P_{2} = 80 N$ . $α$ aur uski direction nikalo.

Forecast: Zyada strong right-leaning jet ka sideways push ab fully cancel nahi hoga, toh ek net rightward push bachega → sheet right ki taraf jhukegi (positive $α$ ) kuch degrees se.

Signed angles ke saath transverse components. Kyun? Same convention: jet 1 deta hai $+ 120 sin 3 5^{\circ}$ , jet 2 deta hai $80 sin (- 3 5^{\circ}) = - 80 sin 3 5^{\circ}$ . $num = 120 sin 3 5^{\circ} + 80 sin (- 3 5^{\circ}) = (120 - 80) sin 3 5^{\circ} = (40) (0.5736) = 22.94 N$
Denominator = total downward push. Kyun? $cos (- 3 5^{\circ}) = cos 3 5^{\circ}$ , toh dono add hote hain. $den = 120 cos 3 5^{\circ} + 80 cos 3 5^{\circ} = (200) (0.8192) = 163.8 N$
Arctan lo. Kyun? $tan α$ = (sideways)/(down); arctan jawab deta hai "kaun sa angle yeh tan rakhta hai?" $α = tan^{- 1} (\frac{22.94}{163.8}) = tan^{- 1} (0.1400) = 7.9 7^{\circ}$

Verify: Equal jets ke saath num→0 aur $α \to 0$ (Example 5) ✓. Positive num → positive $α$ : sheet $\approx 8^{\circ}$ zyada strong jet ki side ki taraf tilts karti hai — yeh wall scrub kar sakti hai, toh ek designer momenta re-balance karega.

Cells G, H — coaxial shear, aur jab shear khatam ho jaata hai

Worked example Example 9 (Cell H) — velocity-matched limit (shear fail ho jaata hai)

Same densities rakho, lekin ek throttling fault hydrogen ko $v_{o} = 12 m/s$ par drop kar deta hai, LOX $v_{i} = 12 m/s$ ke equal. $J$ aur relative velocity $Δ v = v_{o} - v_{i}$ nikalo.

Forecast: Equal speeds → koi relative motion nahi → koi shredding nahi. $J$ 1 se bahut neeche collapse hona chahiye.

Relative velocity. Kyun? Shear atomization $Δ v$ par run karta hai, streams ke beech velocity difference. $Δ v = 12 - 12 = 0 m/s$
Momentum ratio. Kyun? Dikhao $J$ crash karta hai. $J = \frac{2.0 \times 1 2 ^{2}}{1140 \times 1 2 ^{2}} = \frac{2.0}{1140} = 1.75 \times 1 0^{- 3}$

Verify: $Δ v = 0$ matlab zero shear; $J \approx 0.0018 ≪ 1$ ✓. Koi relative velocity nahi hone par outer gas inner jet ko strip nahi kar sakti — bade droplets, poor combustion, exactly woh failure mode jo ek coaxial injector ko avoid karna chahiye.

Cell I — swirl-cone angle, "kitna chauda" ke charon cases

Neeche ke figure mein, yellow arrow axial velocity $v_{x}$ hai (dashed injector axis ke seedha neeche), aur teen coloured arrows increasing spin ke liye element ki total velocity dikhate hain. Jaise jaise sideways (tangential) part badhta hai, har arrow axis se aur door jhukta hai — woh jhukav hi half-cone angle $ϕ$ hai. Labelled angles $0^{\circ} \to 4 5^{\circ} \to 6 0^{\circ}$ ko spin badhne ke saath dekhte rehna.

Worked example Example 10 (Cell I) — chaar spin/axial combinations

Fixed $v_{x} = 20 m/s$ ke liye, chaar tangential speeds par $ϕ$ compute karo aur trend padho.

Forecast: Jaise $v_{t}$ 0 se upar badhta hai, $ϕ$ $0^{\circ}$ se $9 0^{\circ}$ ki taraf badhta hai. $v_{t} = 0$ par "cone" ek pencil-thin axial jet hai; jaise $v_{t} \to \infty$ yeh flat disc ki taraf flatten hota hai.

$v_{t} = 0$ (koi swirl nahi, degenerate case). Kyun? Zero end check karo. $tan ϕ = 0/20 = 0 \Rightarrow ϕ = 0^{\circ} (seedha jet, sabse kharab atomization)$
$v_{t} = 20 m/s$ (equal spin aur axial). Kyun? Symmetric middle case. $tan ϕ = 20/20 = 1 \Rightarrow ϕ = 4 5^{\circ}$
$v_{t} = 34.6 m/s$ (ek wide practical cone). Kyun? Ek typical design value. $tan ϕ = 34.6/20 = 1.732 \Rightarrow ϕ = 6 0^{\circ}$
$v_{t} \to \infty$ (huge spin, limit). Kyun? Doosra extreme check karo. $lim_{v_{t} \to \infty} tan ϕ = \infty \Rightarrow ϕ \to 9 0^{\circ} (flat sheet, sideways spray karta hai)$

Verify: $tan^{- 1} (1) = 4 5^{\circ}$ , $tan^{- 1} (1.732) = 60. 0^{\circ}$ ✓. Trend: zyada spin → wider, thinner cone → finer atomization — lekin ~90° se aage sheet wall se takraayegi, toh real cones lagbhag 45–60° par hote hain.

Cell J — word problem: ek short chamber ke liye kitne holes chahiye?

Worked example Example 11 (Cell J) —

$d^{2}$ -law se hole count Ek injector ko $\overset{m}{˙} = 2.0 kg/s$ kerosene ( $ρ = 810$ ) flow karna hai $Δ p = 5 bar = 5 \times 1 0^{5} Pa$ , $C_{d} = 0.75$ par. Chamber short hai, toh droplets ko available time mein burn karna chahiye — yeh droplet diameter ko $d_{drop} \leq 50 μ m$ par cap karta hai. Finer sprays chhote orifices se aate hain, aur empirically orifice diameter ko half karne se droplet size roughly half ho jaati hai. Agar required total area ka ek single orifice $d_{drop} = 100 μ m$ deta hai, toh kitne equal holes chahiye, aur har hole ki diameter kya hai?

Forecast: Total area metering se fixed hai; ise zyada holes mein split karne se har hole chhota hota hai aur isliye har droplet bhi. Droplet size half karne ke liye chhote orifices chahiye → kuch holes.

Orifice law se total area. Kyun? Metering total $A$ fix karta hai, hole count se koi fark nahi. $A_{tot} = \frac{m ˙}{C _{d} 2 ρ Δ p} = \frac{2.0}{0.75 2 ( 810 ) ( 5 \times 1 0 ^{5} )} = \frac{2.0}{0.75 \times 2.846 \times 1 0 ^{4}} = 9.371 \times 1 0^{- 5} m^{2}$
Single-hole diameter. Kyun? Woh baseline set karta hai jo 100 µm droplets deta hai. $d_{1} = 4 A_{tot} / π = 4 (9.371 \times 1 0^{- 5}) / π = 1.092 \times 1 0^{- 2} m = 10.92 mm$
50 µm droplets ke liye required orifice diameter. Kyun? Droplet size half karne ke liye orifice diameter half chahiye (given rule). $d_{hole} = d_{1} /2 = 5.462 \times 1 0^{- 3} m = 5.46 mm$
Holes ki count (total area conserved). Kyun? $N$ holes, har ek area $π d_{hole}^{2} /4$ ka, $A_{tot}$ sum karna chahiye; kyunki har hole ka $(1/2)^{2} = 1/4$ single-hole area hai, $N = 4$ . $N = \frac{A _{tot}}{π d _{hole}^{2} /4} = \frac{9.371 \times 1 0 ^{- 5}}{π ( 5.462 \times 1 0 ^{- 3} ) ^{2} /4} = 4$

Verify: 4 holes × har ek ka area $= 4 \times 2.343 \times 1 0^{- 5} = 9.371 \times 1 0^{- 5} m^{2} = A_{tot}$ ✓, aur $d^{2}$ -law se droplet diameter half karne se burn time quarter ho jaata hai — short chamber ke liye safe. Links to Atomization and the d-squared Law aur Characteristic Velocity c-star (unburned fuel $c^{*}$ lower karega).

Recall

Recall Quick self-test (guess karne ke baad reveal karo)

Agar $Δ p$ half ho jaaye, toh mass flow kis factor se badalta hai? ::: $1/ 2 \approx 0.71$ factor se — kyunki $\overset{m}{˙} \propto Δ p$ . Ek balanced impinging doublet axis se kitne angle par spray karta hai? ::: $0^{\circ}$ (seedha neeche; do signed transverse pushes cancel ho jaate hain). Do impinging jets mein se ek clog ho jaaye — surviving sheet kahan aim karti hai? ::: Survivor ke apne injection angle ke along (Example 7 mein 35°) — ek wall-burn hazard. Light hydrogen heavy LOX ko coaxial injector mein kyun atomize kar sakta hai? ::: Momentum flux $ρ v^{2}$ jaata hai; fast gas squared velocity par jeetta hai, $J > 1$ deta hai. Zero tangential velocity wala swirl injector kya produce karta hai? ::: $ϕ = 0^{\circ}$ — ek seedha pencil jet, sabse kharab possible atomization.

Related: Bernoulli Equation (root law ka source), Combustion Instability (kyun $Δ p$ high rehna chahiye), O/F Ratio and Mixture Ratio (mixing local ratio set karta hai), Regenerative Cooling (tilted sprays se wall-scrub).