Foundations — Injector design — impinging, coaxial, swirl injectors

This page assumes you know nothing. We build every letter the parent note uses before you meet it in an equation. Read top to bottom — each idea is a brick for the next.

1. Pressure $p$ — a push spread over area

Picture a crowd pressing on a door. Ten people leaning on a wide door barely bend it; the same ten squeezed onto a tiny door panel could snap it. Same total push, smaller area → bigger pressure.

$p=\frac{\text{force}}{\text{area}}$

The parent note uses $p_1$ (pressure inside the feed manifold, before the hole), $p_2=p_c$ (chamber pressure, after the hole), and $\Delta p=p_1-p_2$.

Figure 1 — Cross-section of an injector wall. On the left the manifold sits at high pressure $p_1$ with the fluid nearly at rest; on the right the chamber sits at lower pressure $p_2=p_c$. The leftover push $\Delta p=p_1-p_2$ (black arrow) shoves fluid through the hole and out as the fast red jet of speed $v$.

WHY the topic needs it: that leftover push $\Delta p$ is the engine of the whole injector — it is what turns stored high-pressure liquid into a fast jet.

2. Density $\rho$ — how much mass is packed in

Picture a shoebox filled with feathers versus the same box filled with lead. Same box (same volume), wildly different mass. Lead has higher density.

$\rho =\frac{\text{mass}}{\text{volume}}$

WHY the topic needs it: heavy fluid is harder to accelerate. When the same $\Delta p$ pushes on kerosene (dense) versus hydrogen gas (light), the light stuff ends up moving much faster. That single fact is why coaxial injectors put fast light gas on the outside.

3. Velocity $v$ — speed in a direction

Picture a leaf carried by a stream — its velocity is an arrow: length = speed, arrowhead = direction. In the jet coming out of an injector hole, $v$ is that arrow pointing out of the hole.

WHY the topic needs it: combustion is a race against the clock. Fluid that moves faster and gets shredded into mist burns before it reaches the chamber's far wall. Velocity is the middle-man between pressure (which creates speed) and mass flow (which speed carries).

4. Mass flow rate $\dot{m}$ — kilograms per second through the hole

Picture standing at a doorway counting people walking through per second. If they move faster, or the doorway is wider, or they are packed more densely, more people pass per second. Fluid is the same:

$\dot{m}=\rho Av$

• $\rho$ = how densely packed (mass per volume),
• $A$ = how wide the opening,
• $v$ = how fast they cross.

Figure 2 — People streaming through a doorway of area $A$ at speed $v$ (red arrow); the black dots on the left show how densely packed they are (density $\rho$). More density, a wider opening, or higher speed each raise the count per second, which is exactly $\dot{m}=\rho Av$.

WHY the topic needs it: the whole point of "metering" is hitting a target $\dot{m}$. The engine's thrust is set by how much propellant burns per second, so $\dot{m}$ is the number the injector must deliver exactly.

5. From $\Delta p$ to jet speed — where the factor of 2 comes from

Now we can build the parent's key result $v=\sqrt{2\Delta p/\rho }$ step by step, so the mysterious factor of $2$ is no longer mysterious.

Derivation (the parent's Bernoulli step, spelled out):

WHAT we do — write "energy per cubic metre is conserved along a streamline" for two points on the same blob's path: point 1 inside the manifold (pressure $p_1$, essentially at rest so $v\approx 0$) and point 2 at the hole exit (pressure $p_2=p_c$, speed $v$). WHY — under the three assumptions above no energy is added or lost between them, so the total (pressure energy + motion energy) must match: $\underset{\text{point 1}}{\underbrace{p_1+\frac{1}{2}\rho (0{)}^2}}=\underset{\text{point 2}}{\underbrace{p_2+\frac{1}{2}\rho v^2}}$

WHAT next — subtract $p_2$ from both sides so all the pressure sits on the left. WHY — we want to isolate the motion-energy term: $p_1-p_2=\frac{1}{2}\rho v^2⟹\Delta p=\frac{1}{2}\rho v^2$

There it is: the pressure drop equals one-half density times speed-squared. WHAT last — undo the $\frac{1}{2}$ and the square to free $v$. WHY — multiply both sides by $2$, divide by $\rho$, then take the square root (the only tool that undoes a square, see §6): $v^2=\frac{2\Delta p}{\rho }⟹v=\sqrt{\frac{2\Delta p}{\rho }}$

See Bernoulli Equation for the fuller statement including height terms (which we drop here because an injector hole is tiny and horizontal).

6. The square root x​ — undoing a square

WHY the topic needs it — and why this tool and not another: we just found $\Delta p=\frac{1}{2}\rho v^2$, i.e. $\Delta p$ grows as speed squared. To pull $v$ back out alone, the only operation that undoes a square is the square root. That is why the flow law carries a ​: $v=\sqrt{\frac{2\Delta p}{\rho }}$ This is why doubling the pressure drop does not double the speed — it multiplies speed by only $\sqrt{2}\approx 1.41$. The square root is baked into every incompressible injector flow law.

7. Orifice area $A$ and diameter $d_{\text{h}}$ — the size of the hole

Picture a coin: measure across it to get $d_{\text{h}}$; the flat face's area is $A$. The $\pi /4$ comes from a circle being "boxed" inside a square of side $d_{\text{h}}$ — the circle fills about $78.5\%$ of that square.

WHY the topic needs it: designers choose $A$ (and $d_{\text{h}}$) to tune the flow. Example 1 in the parent solves for exactly this: given the flow you want, how big must the hole be?

8. The discharge coefficient $C_d$ — the "real hole" penalty

Picture water squeezing through a doorway: it doesn't fill the whole doorway — the stream necks inward (the "vena contracta") and rubs the edges (friction). So the effective opening is smaller than the drilled one. $C_d=0.75$ means you actually get $75\%$ of the ideal flow. This is where the friction we ignored in the inviscid Bernoulli step of §5 is quietly put back.

WHY the topic needs it: without $C_d$ your predicted $\dot{m}$ would be too high and the engine would run lean or rich. It is the bridge from clean theory to a machined part. Combining §4, §5 and $C_d$ gives the parent's metering law $\dot{m}=C_{d}A\sqrt{2\rho \Delta p}$.

9. When the fluid is a gas — compressible and choked flow

Everything above assumed the fluid is a liquid (roughly constant density $\rho$ — the third assumption of §5). But coaxial injectors push gas (like gaseous hydrogen) through their outer holes, and gases are springy — they can be squashed. At high pressure ratios this changes the rules, and the topic must not pretend otherwise.

Figure 4 — Gas mass flow $\dot{m}$ versus pressure ratio across a hole. As you lower the downstream pressure (moving right), flow rises — then flattens. Past the red critical point the throat has reached the speed of sound and the flow is choked: $\dot{m}$ stays fixed no matter how much lower the chamber pressure goes.

WHY the topic needs it: leaving this out would make a novice apply the liquid $\sqrt{\Delta p}$ law to a choked gas stream and get the wrong $\dot{m}$ — a critical regime for real rocket coaxial injectors.

10. Angles $\theta$, $\alpha$, $\varphi$ and the tangent — direction from a right triangle

The parent note builds three spray directions from angles. All of them rest on one right triangle, and all three angles are measured from the chamber axis — the straight-down centre-line of the chamber.

Figure 3 — A velocity arrow (red) drawn as the slanted side of a right triangle. Its along-axis piece $v_x$ is the horizontal (adjacent) side; its across piece $v_t$ is the vertical (opposite) side. The angle between the arrow and the axis is read as $\tan (\text{angle})=v_t/v_x$ = across ÷ along. Every one of $\theta$, $\alpha$, $\varphi$ is this same angle, just for a different arrow.

WHY this tool: the injector cares about which way the spray goes. The tangent turns two velocities (sideways $v_t$ and forward $v_x$) into one angle. That is exactly the swirl cone law $\tan \varphi =\frac{v_t}{v_x},$ and the same triangle logic gives the impinging spray direction $\alpha$ (there the "sides" are sideways and forward momenta instead of velocities).

Now the three specific angles, each defined and pictured below in Figure 3b:

Figure 3b — The three angles side by side, each measured from the chamber axis (dashed). Left: two impinging jets each at aim angle $\theta$; their collision throws off a sheet at resultant angle $\alpha$ (red) — here balanced, so $\alpha =0$ straight down. Right: a swirl injector's hollow cone opening at half-angle $\varphi$ (red).

11. Momentum $\dot{m}v$ and the momentum flux ratio $J$

Picture a fire hose knocking you backward: the water's mass times its speed is the shove you feel. The parent's impinging law adds up the sideways oomph of two jets and demands the total be conserved when they collide — the sheet flies off along the leftover oomph (this is where $\alpha$ from §10 comes from).

WHY the topic needs it: coaxial atomization is a fight between two streams. $J$ is the scoreboard.

12. Droplet diameter (droplet $d$) and the $d^2$ label

Picture an ice cube versus crushed ice: crushed ice melts far faster because tiny pieces have huge surface-to-volume. Halving droplet size cuts burn time to a quarter. See Atomization and the d-squared Law for the full derivation.

WHY the topic needs it: it is the entire reason atomization matters — small droplets burn inside the chamber; big ones leave unburned and waste performance (measured by Characteristic Velocity c-star).

13. Chamber pressure $p_c$ and the stability rule of thumb

WHY the topic needs it: the injector's own drop must stay a healthy fraction of $p_c$ (rule of thumb $\Delta p\gtrsim 0.15p_c$) so the roaring chamber can't push oscillations back through the holes. If it does, feed and chamber lock into a feedback loop — see Combustion Instability. (A choked gas orifice, from §9, does this decoupling automatically.)

How the foundations feed the topic

The map below reads top to bottom. Each box is one foundation from above; the cryptic-looking node names are just short tags, expanded here so you can trace them back: p = pressure, pc = chamber pressure, dp = pressure drop $\Delta p$, rho = density, A = orifice area, v = velocity, mdot = mass flow $\dot{m}$, energy = Bernoulli energy balance, sqrt = square root, Cd = discharge coefficient, meter = metering law, gas = compressible/choked flow, mom = momentum, J = momentum flux ratio, tri = velocity right triangle & tan, imp/coax/swirl = the three injector spray laws, drop = droplet diameter, dsq = $d^2$ burn law.

Equipment checklist

Cover the right side and test yourself. If any answer surprises you, re-read that section.