Exercises — Injector design — impinging, coaxial, swirl injectors

They can raise the pressure drop $\Delta p$ (area, density, $C_d$ are fixed by hardware/propellant here). Because $\dot{m}\propto \sqrt{\Delta p}$, doubling $\Delta p$ multiplies $\dot{m}$ by $\sqrt{2}\approx 1.41$, not by 2. Flow grows as a square root of pressure drop. Why square root and not linear? The pressure drop turns into kinetic energy $\frac{1}{2}\rho v^2$, and energy scales with $v^2$; so $v$ (and hence $\dot{m}$) scales with $\sqrt{\Delta p}$.

• (a) Impinging → collision of jets (two streams smash into a fan sheet).
• (b) Coaxial → shear between two streams (fast outer stream shreds the slow inner jet).
• (c) Swirl → centrifugal thinning of a hollow cone (spin flings liquid outward into a thin cone). Key distinction (see Atomization and the d-squared Law): coaxial needs a second high-speed stream; swirl can atomize a single propellant on its own.

False. It is mainly a stability requirement. A high enough $\Delta p$ acts like a stiff spring between the feed lines and the chamber: chamber pressure wobbles cannot push back up through the orifice. Drop $\Delta p$ too low and feed + chamber couple → Combustion Instability. It sets metering and protects stability.

(a) Velocity from Bernoulli, $v=\sqrt{2\Delta p/\rho }$. Why this form? Ideal jet speed depends only on pressure drop and density. $v=\sqrt{\frac{2(3\times 10^5)}{810}}=\sqrt{740.7}=27.2\text{m/s}$ (b) Area from $\dot{m}=C_{d}A\sqrt{2\rho \Delta p}$. $\sqrt{2\rho \Delta p}=\sqrt{2(810)(3\times 10^5)}=\sqrt{4.86\times 10^8}=2.204\times 10^4$ $A=\frac{0.5}{0.75\times 2.204\times 10^4}=3.02\times 10^{-5}{\text{m}}^2$ Diameter from $A=\pi d^2/4$: $d=\sqrt{4A/\pi }=\sqrt{\frac{4(3.02\times 10^{-5})}{\pi }}=6.2\times 10^{-3}\text{m}=6.2\text{mm}$

First case: $\tan \varphi =v_t/v_x=20/20=1\Rightarrow \varphi =\arctan (1)=45^{\circ }$. Why arctan? $\arctan$ answers "which angle has this tangent?" — it undoes the tangent to hand us the angle back. Doubled spin: $\tan \varphi =40/20=2\Rightarrow \varphi =\arctan (2)=63.4^{\circ }$. More spin ⇒ wider, thinner cone ⇒ (from the d²-law) finer droplets that evaporate faster.

Inner momentum flux: ${\rho }_i{v}_i^2=1140\times 12^2=1140\times 144=1.6416\times 10^5$. Outer momentum flux: ${\rho }_o{v}_o^2=1.2\times 320^2=1.2\times 1.024\times 10^5=1.2288\times 10^5$. $J=\frac{1.2288\times 10^5}{1.6416\times 10^5}=0.749$ $J<1$: the outer stream's momentum is slightly less than the inner jet's, so shearing is only moderate. To improve atomization the designer would raise $v_o$ (the light gas must go fast, since its density is tiny) or lower $v_i$.

Balanced case. Put jet 1 on the $+$ side ($+\sin 30^{\circ }$) and jet 2 on the $-$ side ($-\sin 30^{\circ }$): $\tan \alpha =\frac{P\sin 30^{\circ }-P\sin 30^{\circ }}{P\cos 30^{\circ }+P\cos 30^{\circ }}=\frac{0}{2P\cos 30^{\circ }}=0\Rightarrow \alpha =0^{\circ }.$ Transverse (sideways) momenta cancel exactly ⇒ spray straight down the axis — exactly what avoids scrubbing the wall.

Unbalanced case ($P_1=P$, $P_2=0.8P$): $\tan \alpha =\frac{P\sin 30^{\circ }-0.8P\sin 30^{\circ }}{P\cos 30^{\circ }+0.8P\cos 30^{\circ }}=\frac{0.2P(0.5)}{1.8P(0.8660)}=\frac{0.1}{1.5588}=\mathrm{0.06415.}$ $\alpha =\arctan (0.06415)=3.67^{\circ }.$ The sheet now tilts $3.67^{\circ }$ toward the stronger jet's original side — that residual transverse momentum can push the spray at the wall, a real burn-through risk.

From $\dot{m}=C_{d}A\sqrt{2\rho \Delta p}$, holding $\dot{m}$ fixed means $A\sqrt{\Delta p}=\text{const}$, so $A\propto 1/\sqrt{\Delta p}$. New $\Delta p=0.15p_c$, old $\Delta p=0.10p_c$, ratio $=1.5$. $\frac{A_{\text{new}}}{A_{\text{old}}}=\sqrt{\frac{\Delta p_{\text{old}}}{\Delta p_{\text{new}}}}=\sqrt{\frac{0.10}{0.15}}=\sqrt{0.6667}=\mathrm{0.8165.}$ The orifice area must shrink to 81.6 % of its old value (about an 18.4 % reduction) to keep the same $\dot{m}$ at the higher, safer $\Delta p$. Smaller holes, higher $\Delta p$ — the classic stability trade.

$\frac{t_B}{t_A}={\left(\frac{d_B}{d_A}\right)}^2={\left(\frac{80}{50}\right)}^2=(1.6{)}^2=\mathrm{2.56.}$ B's droplets take 2.56× as long to burn. If A only just finished within the available residence time, B needs 2.56× that time and will not finish — unburned propellant leaves the chamber, cutting c* efficiency. B is not safe. (See Atomization and the d-squared Law.)

(a) $O/F={\dot{m}}_{ox}/{\dot{m}}_{fuel}=2.5$ and ${\dot{m}}_{ox}+{\dot{m}}_{fuel}=3.5$. So ${\dot{m}}_{fuel}(2.5+1)=3.5\Rightarrow {\dot{m}}_{fuel}=3.5/3.5=1.0\text{kg/s}$, and ${\dot{m}}_{ox}=2.5\text{kg/s}$. (See O/F Ratio and Mixture Ratio.) (b) First recompute the metering constant $\sqrt{2\rho \Delta p}$ from scratch for these fuel values ($\rho =810$, $\Delta p=3\times 10^5\text{Pa}$) — it happens to match L2.1 because the fuel and pressure drop are the same, but we show its origin so nothing is borrowed blindly: $\sqrt{2\rho \Delta p}=\sqrt{2(810)(3\times 10^5)}=\sqrt{4.86\times 10^8}=2.204\times 10^4\text{(units: }\sqrt{\text{Pa}\cdot {\text{kg/m}}^3}={\text{kg/(m}}^2\text{s)}).$ One 1.0 mm hole area: $A_1=\pi d^2/4=\pi (1.0\times 10^{-3}{)}^2/4=7.854\times 10^{-7}{\text{m}}^2$. Flow per hole: ${\dot{m}}_1=C_d{A}_1\sqrt{2\rho \Delta p}=0.75(7.854\times 10^{-7})(2.204\times 10^4)=1.298\times 10^{-2}\text{kg/s}$. Holes needed: $N={\dot{m}}_{fuel}/{\dot{m}}_1=1.0/1.298\times 10^{-2}=77.0\Rightarrow$ round up to 78 orifices (you never under-provide flow area).

$J=\frac{{\rho }_o{v}_o^2}{{\rho }_i{v}_i^2}=3\Rightarrow v_o^2=\frac{3{\rho }_i{v}_i^2}{{\rho }_o}=\frac{3(1140)(15^2)}{1.5}=\frac{3(1140)(225)}{1.5}=513000.$ $v_o=\sqrt{513000}=716.2\text{m/s}.$ The number is huge because hydrogen is ~760× lighter than LOX; the only way its momentum flux ${\rho }_o{v}_o^2$ can exceed the liquid's is to make $v_o$ enormous. This is why hydrogen-oxygen coaxial injectors run the gas near sonic speeds — the light propellant must go very fast to shear the heavy one. It also ties to Regenerative Cooling: that hot, fast H₂ often first cooled the chamber walls.

Caution — compressibility. The formula $J={\rho }_o{v}_o^2/{\rho }_i{v}_i^2$ came from incompressible momentum bookkeeping, which quietly assumes gas density ${\rho }_o$ stays fixed as the stream speeds up. But the speed of sound in cold hydrogen is only ~$1250\text{m/s}$, so $v_o=716\text{m/s}$ is a Mach number of roughly $716/1250\approx 0.57$ — well into the compressible regime, where ${\rho }_o$ actually drops as the gas accelerates through the injector. In a real design you would use the actual injector-exit density (from the gas's pressure and temperature there), not the manifold value, and treat this $J$ as a first-cut estimate; at Mach numbers approaching 1 the true momentum flux differs meaningfully from the incompressible prediction.

(a) $\Delta p=0.15\times 100\text{bar}=15\text{bar}=1.5\times 10^6\text{Pa}$. (b) Fuel flow: ${\dot{m}}_{fuel}=\dot{m}/(O/F+1)=8/3.6=2.222\text{kg/s}$. $\sqrt{2\rho \Delta p}=\sqrt{2(810)(1.5\times 10^6)}=\sqrt{2.43\times 10^9}=4.930\times 10^4.$ $A=\frac{{\dot{m}}_{fuel}}{C_d\sqrt{2\rho \Delta p}}=\frac{2.222}{0.8\times 4.930\times 10^4}=5.635\times 10^{-5}{\text{m}}^2.$ (c) The empirical trend is $d\propto (\Delta p{)}^{-0.4}$ (higher pressure drop → faster, thinner sheet → smaller droplets). Raising $\Delta p$ from $0.15p_c$ to $0.25p_c$ is a factor $r=\frac{0.25p_c}{0.15p_c}=\frac{0.25}{0.15}=\mathrm{1.667.}$ New droplet size relative to old: $\frac{d_{\text{new}}}{d_{\text{old}}}=r^{-0.4}=1.667^{-0.4}.$ Take logs to evaluate: $\ln 1.667=0.5108$, times $-0.4$ gives $-0.2043$, and $e^{-0.2043}=0.8152$. So $d_{\text{new}}/d_{\text{old}}=0.8152$ (droplets ~18 % smaller). Burn time follows the $d^2$-law, $t\propto d^2$, so $\frac{t_{\text{new}}}{t_{\text{old}}}={\left(\frac{d_{\text{new}}}{d_{\text{old}}}\right)}^2=(0.8152{)}^2=\mathrm{0.6646.}$ Result: burn time drops to ~66.5 % of the original — about a 33.5 % faster burn. Physical significance: shorter burn time means the propellant finishes reacting well within the chamber's residence time, so c* efficiency rises and you gain margin against a too-short chamber. But this improvement is bought — going from $0.15p_c$ to $0.25p_c$ is a 67 % higher pressure drop, so the turbopump must do substantially more work. That is the central injector trade: better atomization and stability versus pump power.

Reason 1 (symmetry): a swirl element produces an axisymmetric hollow cone, so there is no net transverse momentum to tilt the spray into the wall — the imbalance problem simply disappears. Reason 2 (low-$\Delta p$ atomization): swirl atomizes by centrifugal thinning of a self-supported sheet, so it reaches fine droplets even at modest $\Delta p$, unlike shear-based schemes that need high relative velocity. New risk: the wide, low-pressure hollow cone and its central recirculation zone can act as a receptive feedback path for pressure waves, raising the chance of combustion instability if the cone angle and $\Delta p$ aren't tuned.

Metering law? ::: $\dot{m}=C_{d}A\sqrt{2\rho \Delta p}$, and $\dot{m}\propto \sqrt{\Delta p}$. Why keep $\Delta p\gtrsim 0.15p_c$? ::: to decouple feed from chamber and avoid combustion instability. Coaxial atomization number? ::: momentum flux ratio $J={\rho }_o{v}_o^2/{\rho }_i{v}_i^2$; velocity is squared (and watch compressibility when $v_o$ nears sonic). Swirl cone angle? ::: $\tan \varphi =v_t/v_x$; more spin ⇒ wider, thinner cone ⇒ finer droplets.