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

Before we use any formula, let us name every symbol in plain words so nothing sneaks in undefined.

The scenario matrix

Now that the symbols exist, let us list every situation these formulas can be pushed into. Each row is a "cell" the reader must be able to survive. Every example below is tagged with the cell it kills.

Cell	Tool	The tricky part
A normal metering	orifice law	plain plug-and-chug, get area & diameter
B inverse metering	orifice law	solve for $Δ p$ given fixed hole
C zero / degenerate	orifice law	what if $Δ p \to 0$ ? What does $\overset{m}{˙} \to 0$ mean physically
D balanced impinging	momentum balance	transverse terms cancel → straight spray ( $α = 0$ )
E unbalanced impinging	momentum balance	one jet stronger → spray tilts, which sign?
F extreme impinging	momentum balance	one jet dies ( $\overset{m}{˙}_{2} \to 0$ ) → sheet follows the survivor
G coaxial normal	momentum ratio $J$	light fast gas vs heavy slow liquid, is $J > 1$ ?
H coaxial limit	momentum ratio $J$	$Δ v \to 0$ → no shear → atomization fails
I swirl angle	$tan ϕ = v_{t} / v_{x}$	spin vs axial, all four "how wide is the cone" cases
J word problem	orifice + $d^{2}$ -law	pick number of holes for burn time in a short chamber
K exam twist	orifice law	$Δ p$ given as percent of $p_{c}$ , hidden units

Cell A + K — metering, plain and disguised

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

Deliver $\overset{m}{˙} = 0.80 kg/s$ of liquid oxygen, $ρ = 1140 kg/m^{3}$ , with $Δ p = 4 bar = 4 \times 1 0^{5} Pa$ and $C_{d} = 0.80$ . Find the orifice area $A$ and diameter $d$ .

Forecast: More flow needs more area; higher $Δ p$ makes the jet faster so we can shrink the hole. Guess: a few square millimetres, so a few mm across.

Rearrange the orifice law for $A$ . Why this step? Everything is known except $A$ , so isolate it. $A = \frac{m ˙}{C _{d} 2 ρ Δ p}$
Compute the root $2 ρ Δ p$ . Why? This radicand is the product of density and the ideal kinetic-energy term; it is not the jet speed. (The true ideal jet speed is $v = 2 Δ p / ρ$ ; here we deliberately keep $ρ$ inside the root so that $C_{d} A 2 ρ Δ p$ comes out directly as a mass flow. The radicand's job is simply to convert pressure-and-density into the quantity $ρ v$ .) $2 (1140) (4 \times 1 0^{5}) = 9.12 \times 1 0^{8} = 3.020 \times 1 0^{4}$
Divide. Why? Now plug the pieces in. $A = \frac{0.80}{0.80 \times 3.020 \times 1 0 ^{4}} = 3.311 \times 1 0^{- 5} m^{2}$
Diameter from $A = π d^{2} /4$ . Why? Real orifices are round holes. $d = \frac{4 A}{π} = \frac{4 ( 3.311 \times 1 0 ^{- 5} )}{π} = 6.49 \times 1 0^{- 3} m \approx 6.5 mm$

Verify: Units of $2 ρ Δ p$ are $(kg/m^{3}) (Pa) = kg^{2} / (m^{4} s^{2}) = kg / (m^{2} s)$ , so $A \cdot$ that $=$ kg/s ✓. Cross-check the true jet speed $v = 2 Δ p / ρ = 2 (4 \times 1 0^{5}) /1140 = 26.5 m/s$ , and indeed $ρ v = 1140 \times 26.5 = 3.02 \times 1 0^{4}$ matches the radicand ✓. Plug back: $0.80 \times 3.311 \times 1 0^{- 5} \times 3.020 \times 1 0^{4} = 0.80 kg/s$ ✓.

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

$Δ p$ hidden as a percentage Same LOX ( $ρ = 1140$ ), same $\overset{m}{˙} = 0.80 kg/s$ , $C_{d} = 0.80$ , but now you are told chamber pressure $p_{c} = 70 bar$ and $Δ p = 15%$ of $p_{c}$ . Find $A$ .

Forecast: $15%$ of 70 bar is about 10 bar — bigger $Δ p$ than Example 1, so the hole should be smaller than 6.5 mm.

Turn the percent into a pressure in pascals. Why? The formula needs Pa, not "percent". This is where exam students lose marks. $Δ p = 0.15 \times 70 bar = 10.5 bar = 1.05 \times 1 0^{6} Pa$
Root. Why? Same converter as before. $2 (1140) (1.05 \times 1 0^{6}) = 2.394 \times 1 0^{9} = 4.893 \times 1 0^{4}$
Area. Why? Isolate $A$ again. $A = \frac{0.80}{0.80 \times 4.893 \times 1 0 ^{4}} = 2.044 \times 1 0^{- 5} m^{2} \Rightarrow d = 5.10 mm$

Verify: Higher $Δ p$ → smaller hole, as forecast (5.1 mm < 6.5 mm) ✓. And $Δ p / p_{c} = 15%$ sits exactly on the stability floor $Δ p ≳ 0.15 p_{c}$ that we derived above the matrix, so this design is at the minimum safe margin.

Cell B + C — inverse metering and the zero limit

Worked example Example 3 (Cell B) — the hole is fixed, find the

$Δ p$ it demands A finished injector has total effective area $A = 3.0 \times 1 0^{- 5} m^{2}$ , $C_{d} = 0.80$ , flowing kerosene $ρ = 810 kg/m^{3}$ . You must push $\overset{m}{˙} = 0.60 kg/s$ . What $Δ p$ is required?

Forecast: Since $\overset{m}{˙} \propto Δ p$ , to raise flow you raise $Δ p$ quadratically — expect a large pressure.

Square the orifice law to free $Δ p$ . Why? $Δ p$ lives inside a square root; squaring both sides lifts it out. $\overset{m}{˙}^{2} = C_{d}^{2} A^{2} (2 ρ Δ p)$
Solve for $Δ p$ . Why? Isolate the unknown. $Δ p = \frac{m ˙ ^{2}}{2 ρ C _{d}^{2} A ^{2}} = \frac{0.6 0 ^{2}}{2 ( 810 ) ( 0.80 ) ^{2} ( 3.0 \times 1 0 ^{- 5} ) ^{2}}$
Crunch it. Why? Get the number. $Δ p = \frac{0.36}{2 ( 810 ) ( 0.64 ) ( 9.0 \times 1 0 ^{- 10} )} = \frac{0.36}{9.331 \times 1 0 ^{- 7}} = 3.86 \times 1 0^{5} Pa \approx 3.86 bar$

Verify: Feed it back into the forward law: $0.80 \times 3.0 \times 1 0^{- 5} \times 2 (810) (3.86 \times 1 0^{5}) = 0.80 \times 3.0 \times 1 0^{- 5} \times 2.500 \times 1 0^{4} = 0.600 kg/s$ ✓.

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

In Example 3's injector, what happens to $\overset{m}{˙}$ as $Δ p \to 0$ ? And what if the pumps fail so $Δ p = 0$ exactly?

Forecast: No push → no jet → no flow. But how fast does it die?

Read the law's dependence. Why? $\overset{m}{˙} = C_{d} A 2 ρ Δ p$ has $Δ p$ under a root, so $\overset{m}{˙} \propto Δ p$ .
Take the limit. Why? We want the behaviour, not just one point. $lim_{Δ p \to 0^{+}} \overset{m}{˙} = C_{d} A 2 ρ \cdot 0 = 0$
Note the shape of the approach. Why? The square root means the curve comes into zero with infinite slope — halving $Δ p$ only cuts flow by a factor $2 \approx 1.41$ , not by 2. So flow drops slowly at first, then plunges near zero.

Verify: At $Δ p = 3.86 bar$ , $\overset{m}{˙} = 0.60$ . Halve to $1.93 bar$ : $\overset{m}{˙} = 0.60/ 2 = 0.424 kg/s$ ✓ — not $0.30$ . This is exactly why a tiny $Δ p$ is dangerous: flow is barely metered and chamber pressure can shove backwards → instability.

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

With that convention, the transverse-momentum balance gives the resultant sheet angle $α$ from the axis: $tan α = \frac{m ˙ _{1} v _{1} sin θ _{1} + m ˙ _{2} v _{2} sin θ _{2} net sideways push}{total downward push m ˙ _{1} v _{1} cos θ _{1} + m ˙ _{2} v _{2} cos θ _{2}}$

Look at the figure below as you read Examples 5–7. The blue arrow is jet 1 (leaning right, $θ_{1} = + 3 5^{\circ}$ ), the pink arrow is jet 2 (leaning left, $θ_{2} = - 3 5^{\circ}$ ), and the yellow arrow is the resultant sheet whose tilt from the dashed axis is the angle $α$ we solve for. Notice the small horizontal "+ transverse" arrow at the bottom right: that marks which sideways direction counts as positive, so a positive $α$ means the sheet leans right.

Figure — Injector design — impinging, coaxial, swirl injectors

Worked example Example 5 (Cell D) — balanced doublet

Two jets of equal push $P_{1} = P_{2} = P$ (where $P = \overset{m}{˙} v$ ), aimed symmetrically: jet 1 leans right at $θ_{1} = + 3 5^{\circ}$ , jet 2 leans left at $θ_{2} = - 3 5^{\circ}$ . Find $α$ .

Forecast: Perfectly symmetric → the two sideways pushes are equal and opposite → they cancel → straight down, $α = 0$ .

Write each transverse component using the signed angles. Why? The sign convention does the bookkeeping: jet 1's sideways push is $+ P sin 3 5^{\circ}$ , jet 2's is $P sin (- 3 5^{\circ}) = - P sin 3 5^{\circ}$ .
Add them (numerator = net sideways push). Why? The formula sums signed components. $num = P sin (+ 3 5^{\circ}) + P sin (- 3 5^{\circ}) = P sin 3 5^{\circ} - P sin 3 5^{\circ} = 0$
Denominator (both push down). Why? $cos (+ 3 5^{\circ}) = cos (- 3 5^{\circ})$ , so downward parts add. $den = P cos 3 5^{\circ} + P cos 3 5^{\circ} = 2 P cos 3 5^{\circ}$
Ratio. Why? $tan α = 0/ (2 P cos 3 5^{\circ}) = 0 \Rightarrow α = 0^{\circ}$ .

Verify: $tan^{- 1} (0) = 0^{\circ}$ ✓. A balanced doublet sprays down the axis, away from the wall — exactly the design goal.

Worked example Example 6 (Cell E) — unbalanced doublet, which way does it tilt?

Same symmetric aiming — jet 1 at $θ_{1} = + 3 5^{\circ}$ , jet 2 at $θ_{2} = - 3 5^{\circ}$ — but jet 1 is stronger: $P_{1} = 120 N$ , $P_{2} = 80 N$ . Find $α$ and its direction.

Forecast: The stronger right-leaning jet's sideways push no longer fully cancels, so a net rightward push survives → the sheet leans right (positive $α$ ) by a few degrees.

Transverse components with signed angles. Why? Same convention: jet 1 gives $+ 120 sin 3 5^{\circ}$ , jet 2 gives $80 sin (- 3 5^{\circ}) = - 80 sin 3 5^{\circ}$ . $num = 120 sin 3 5^{\circ} + 80 sin (- 3 5^{\circ}) = (120 - 80) sin 3 5^{\circ} = (40) (0.5736) = 22.94 N$
Denominator = total downward push. Why? $cos (- 3 5^{\circ}) = cos 3 5^{\circ}$ , so both add. $den = 120 cos 3 5^{\circ} + 80 cos 3 5^{\circ} = (200) (0.8192) = 163.8 N$
Take arctan. Why? $tan α$ = (sideways)/(down); arctan answers "which angle has this tan?" $α = tan^{- 1} (\frac{22.94}{163.8}) = tan^{- 1} (0.1400) = 7.9 7^{\circ}$

Verify: With equal jets num→0 and $α \to 0$ (Example 5) ✓. Positive num → positive $α$ : the sheet tilts $\approx 8^{\circ}$ toward the stronger jet's side — this could scrub the wall, so a designer would re-balance the momenta.

Cells G, H — coaxial shear, and when shear dies

Cell I — swirl-cone angle, all four "how wide" cases

In the figure below, the yellow arrow is the axial velocity $v_{x}$ (straight down the dashed injector axis), and the three coloured arrows show the element's total velocity for increasing spin. As the sideways (tangential) part grows, each arrow tilts further from the axis — that tilt is the half-cone angle $ϕ$ . Watch the labelled angles climb $0^{\circ} \to 4 5^{\circ} \to 6 0^{\circ}$ as spin increases.

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

For fixed $v_{x} = 20 m/s$ , compute $ϕ$ for four tangential speeds and read the trend.

Forecast: As $v_{t}$ climbs from 0 upward, $ϕ$ grows from $0^{\circ}$ toward $9 0^{\circ}$ . At $v_{t} = 0$ the "cone" is a pencil-thin axial jet; as $v_{t} \to \infty$ it flattens toward a flat disc.

$v_{t} = 0$ (no swirl, degenerate case). Why? Check the zero end. $tan ϕ = 0/20 = 0 \Rightarrow ϕ = 0^{\circ} (straight jet, worst atomization)$
$v_{t} = 20 m/s$ (equal spin and axial). Why? The symmetric middle case. $tan ϕ = 20/20 = 1 \Rightarrow ϕ = 4 5^{\circ}$
$v_{t} = 34.6 m/s$ (a wide practical cone). Why? A typical design value. $tan ϕ = 34.6/20 = 1.732 \Rightarrow ϕ = 6 0^{\circ}$
$v_{t} \to \infty$ (huge spin, limit). Why? Check the other extreme. $lim_{v_{t} \to \infty} tan ϕ = \infty \Rightarrow ϕ \to 9 0^{\circ} (flat sheet, sprays sideways)$

Verify: $tan^{- 1} (1) = 4 5^{\circ}$ , $tan^{- 1} (1.732) = 60. 0^{\circ}$ ✓. Trend: more spin → wider, thinner cone → finer atomization — but past ~90° the sheet would hit the wall, so real cones sit around 45–60°.

Cell J — word problem: how many holes for a short chamber?

Worked example Example 11 (Cell J) — hole count from the

$d^{2}$ -law An injector must flow $\overset{m}{˙} = 2.0 kg/s$ kerosene ( $ρ = 810$ ) at $Δ p = 5 bar = 5 \times 1 0^{5} Pa$ , $C_{d} = 0.75$ . The chamber is short, so droplets must burn in the available time — that caps droplet diameter at $d_{drop} \leq 50 μ m$ . Finer sprays come from smaller orifices, and empirically halving the orifice diameter roughly halves droplet size. If a single orifice of the required total area gives $d_{drop} = 100 μ m$ , how many equal holes do you need, and what is each hole's diameter?

Forecast: Total area is fixed by metering; splitting it into more holes shrinks each hole and hence each droplet. To halve droplet size we need smaller orifices → several holes.

Total area from the orifice law. Why? Metering fixes total $A$ regardless of hole count. $A_{tot} = \frac{m ˙}{C _{d} 2 ρ Δ p} = \frac{2.0}{0.75 2 ( 810 ) ( 5 \times 1 0 ^{5} )} = \frac{2.0}{0.75 \times 2.846 \times 1 0 ^{4}} = 9.371 \times 1 0^{- 5} m^{2}$
Single-hole diameter. Why? Sets the baseline that gives 100 µm droplets. $d_{1} = 4 A_{tot} / π = 4 (9.371 \times 1 0^{- 5}) / π = 1.092 \times 1 0^{- 2} m = 10.92 mm$
Required orifice diameter for 50 µm droplets. Why? Halving droplet size needs half the orifice diameter (given rule). $d_{hole} = d_{1} /2 = 5.462 \times 1 0^{- 3} m = 5.46 mm$
Number of holes (total area conserved). Why? $N$ holes each of area $π d_{hole}^{2} /4$ must sum to $A_{tot}$ ; since each hole has $(1/2)^{2} = 1/4$ the single-hole area, $N = 4$ . $N = \frac{A _{tot}}{π d _{hole}^{2} /4} = \frac{9.371 \times 1 0 ^{- 5}}{π ( 5.462 \times 1 0 ^{- 3} ) ^{2} /4} = 4$

Verify: 4 holes × area each $= 4 \times 2.343 \times 1 0^{- 5} = 9.371 \times 1 0^{- 5} m^{2} = A_{tot}$ ✓, and by the $d^{2}$ -law halving droplet diameter cuts burn time to a quarter — safe for the short chamber. Links to Atomization and the d-squared Law and Characteristic Velocity c-star (unburned fuel would lower $c^{*}$ ).

Recall

Recall Quick self-test (reveal after guessing)

If $Δ p$ is halved, mass flow changes by what factor? ::: By $1/ 2 \approx 0.71$ — because $\overset{m}{˙} \propto Δ p$ . A balanced impinging doublet sprays at what angle from the axis? ::: $0^{\circ}$ (straight down; the two signed transverse pushes cancel). One of two impinging jets clogs — where does the surviving sheet aim? ::: Along the survivor's own injection angle (35° in Example 7) — a wall-burn hazard. Why can light hydrogen still atomize heavy LOX in a coaxial injector? ::: Momentum flux goes as $ρ v^{2}$ ; the fast gas wins on the squared velocity, giving $J > 1$ . A swirl injector with zero tangential velocity produces what? ::: $ϕ = 0^{\circ}$ — a straight pencil jet, the worst possible atomization.

Related: Bernoulli Equation (source of the root law), Combustion Instability (why $Δ p$ must stay high), O/F Ratio and Mixture Ratio (mixing sets the local ratio), Regenerative Cooling (wall-scrub from tilted sprays).