Exercises — Nozzle area ratio ε = A_e - A - — choosing for optimal performance

• Under-expanded: $p_e>p_a$. Gas still bursting with pressure at the exit; the nozzle is too short (ε too small). This is the left plume in figure s01, still bulging outward.
• Perfectly expanded (optimum): $p_e=p_a$. The middle plume in figure s01 — a clean straight column.
• Over-expanded: $p_e<p_a$. Gas expanded past ambient; outside air squeezes the plume inward (the pinched right plume in figure s01), pressure thrust turns negative, separation can start.

$\epsilon =A_e/A^{\ast }=0.30/0.02=15$. Since $\epsilon >1$ there is a diverging section — the flow area grows past the throat, which is exactly what accelerates already-sonic gas to supersonic speeds.

The supersonic ($M_e>1$) root. As figure s02 shows, the $1/M_e$ front factor makes the area–Mach curve U-shaped: the dashed line at $\epsilon =15$ crosses the blue curve at both the green (subsonic) dot and the red (supersonic) dot. Upstream of the throat (orange $M=1$ line) the flow is subsonic; once it passes the sonic throat it becomes supersonic, so downstream we read the right-hand red ($M>1$) dot.

Step 1 (WHAT): Perfect expansion means $p_e=p_a=0.20$ bar. So $p_0/p_e=100/0.20=500$. Step 2 (WHY this tool): We know pressures, want Mach — that is exactly the Mach–pressure relation. $500={\left(1+0.1M_e^2\right)}^6,\frac{\gamma }{\gamma -1}=\frac{1.2}{0.2}=6.$ Step 3: Take the 6th root: $1+0.1M_e^2=500^{1/6}=2.817$. $0.1M_e^2=1.817\Rightarrow M_e^2=18.17\Rightarrow M_e=\mathrm{4.26.}$

WHY this tool: We now know $M_e$ and $\gamma$ but want a geometry — the exit-to-throat area ratio. The relation that connects Mach number to area is the area–Mach relation $\epsilon =A_e/A^{\ast }$; it is the only one of our four tools with area in it, so it is the correct next step. (Reading figure s02, we are converting the red supersonic $M_e$ into its $\epsilon$ height on the blue curve.) Exponent: $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.2}{0.4}=5.5$. Front bracket factor $\frac{2}{\gamma +1}=\frac{2}{2.2}=0.909$. $\epsilon =\frac{1}{4.26}[0.909(2.817){]}^{5.5}=\frac{1}{4.26}(2.561{)}^{5.5}.$ $(2.561{)}^{5.5}=175.9$, so $\epsilon =175.9/4.26=41.3$. Read the size: a near-vacuum-ish design ($p_a$ small) needs lots of expansion → large ε (tens), consistent with the parent note.

WHY these tools: "Perfect expansion at the ground" fixes $p_e=p_a=1.0$ bar, giving the pressure ratio, so we first use the Mach–pressure relation (pressures → $M_e$), then feed $M_e$ into the area–Mach relation to turn that Mach into the geometry ε — the same two-step chain as L2.1–L2.2. $p_0/p_e=70/1.0=70$. Exponent $\gamma /(\gamma -1)=1.25/0.25=5$. $70=(1+0.125M_e^2{)}^5\Rightarrow 1+0.125M_e^2=70^{1/5}=\mathrm{2.336.}$ $0.125M_e^2=1.336\Rightarrow M_e^2=10.69\Rightarrow M_e=\mathrm{3.27.}$ Area exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.25}{0.5}=4.5$; front factor $\frac{2}{2.25}=0.889$. $\epsilon =\frac{1}{3.27}[0.889\times 2.336{]}^{4.5}=\frac{1}{3.27}(2.077{)}^{4.5}.$ $(2.077{)}^{4.5}=27.1$, so $\epsilon =27.1/3.27=8.3$. Small ε — exactly what a sea-level stage wants.

$(p_e-p_a)A_e=(0.20-1.0)\times 10^5\times 1.2=(-0.80\times 10^5)(1.2)=-9.6\times 10^4\text{N}.$ This term is −96 kN: it subtracts from thrust. The nozzle is badly over-expanded at the ground (its $p_e$ is far below the thick sea-level air) — the red region of figure s03, far to the right of the matched point. Ambient air pushes back on the exit plane; in practice this severe mismatch would also trigger flow separation (Flow Separation in Over-expanded Nozzles).

$dF/d{A}_e=p_e-p_a=(0.6-1.0)\times 10^5=-0.4\times 10^5<0$. Adding area (bigger ε) would decrease thrust — you are over-expanded. So you should shrink ε until $p_e$ rises to meet $p_a=1.0$ bar. Figure s03 shows exactly this: on the over-expanded (red) side the thrust curve slopes downward, so you slide left toward the green matched peak.

Engine A: $p_0/p_e=80/1.0=80\Rightarrow (1+0.1M_e^2{)}^6=80\Rightarrow 1+0.1M_e^2=80^{1/6}=2.076\Rightarrow M_e^2=10.76\Rightarrow M_e=\mathrm{3.28.}$ Engine B: $p_0/p_e=80/0.10=800\Rightarrow (1+0.1M_e^2{)}^6=800\Rightarrow 1+0.1M_e^2=800^{1/6}=3.06\Rightarrow M_e^2=20.6\Rightarrow M_e=\mathrm{4.54.}$ Since the area–Mach relation increases with $M_e$ (for the supersonic branch — the rising right arm of the blue curve in figure s02), and B has the larger $M_e$, ${\epsilon }_B>{\epsilon }_A$. Physically: thinner outside air ⇒ smaller target $p_e$ ⇒ more expansion ⇒ larger ε. High-altitude engines are the "bell-flared" giants.

Step 1 — pick $p_e$: match the representative ambient, $p_e=0.55$ bar. $p_0/p_e=90/0.55=163.6$. Step 2 — $M_e$: $\gamma /(\gamma -1)=1.22/0.22=5.545$. $1+\frac{0.22}{2}{M}_e^2=163.6^{1/5.545}=163.6^{0.1803}=\mathrm{2.508.}$ $0.11M_e^2=1.508\Rightarrow M_e^2=13.71\Rightarrow M_e=\mathrm{3.70.}$ Step 3 — ε: area exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.22}{0.44}=5.045$; front factor $\frac{2}{2.22}=0.9009$. $\epsilon =\frac{1}{3.70}[0.9009\times 2.508{]}^{5.045}=\frac{1}{3.70}(2.259{)}^{5.045}.$ $(2.259{)}^{5.045}=60.1$, so $\epsilon =60.1/3.70=16.2$. Design comment: at liftoff ($p_a=1.0$) this nozzle is over-expanded ($p_e=0.55<1.0$); at burnout ($p_a=0.30$) it is under-expanded ($p_e=0.55>0.30$). By matching the middle, we minimise the total penalty across the burn — the classic single-nozzle compromise, and the reason Altitude Compensation — Aerospike Nozzles were invented.

Specific impulse is $I_{sp}=F/(\dot{m}{g}_0)$, where $g_0=9.81{\text{m/s}}^2$ is standard gravity (a fixed constant used to convert weight-flow to force, not the local gravity). With $\dot{m}$ and $g_0$ fixed, higher $F$ ⇒ higher $I_{sp}$. The perfectly expanded nozzle has $p_e=p_a$, so its pressure-thrust term is zero and nothing subtracts from momentum thrust; the over-expanded one carries a negative $(p_e-p_a)A_e$. Hence the perfectly expanded nozzle produces more thrust at the same $\dot{m}$ and therefore higher $I_{sp}$. The route matters: $I_{sp}$ is not "the exhaust speed alone" — it is total thrust per unit weight flow, and total thrust includes the pressure term.

Start from the parent's incremental result for fixed chamber and $\dot{m}$: adding a thin ring of exit area $d{A}_e$ changes thrust by $dF=p_{e}d{A}_e-p_{a}d{A}_e=(p_e-p_a)d{A}_e.$ (The internal momentum and pressure terms combine into the local wall pressure $p_e$ acting over the added area; the ambient pushes back over the same added area.) The extremum is where $dF/d{A}_e=0$: $p_e-p_a=0\Rightarrow \boxed{p_e=p_a}.$ Checking the sign: $dF/d{A}_e>0$ when $p_e>p_a$ (add area → still gaining → under-expanded), $<0$ when $p_e<p_a$ (over-expanded). So $p_e=p_a$ is a maximum, not a minimum. Altitude link: the geometry fixes $p_e$, but $p_a$ falls with altitude, so the single ε that satisfies $p_e=p_a$ is perfect at exactly one height.

(a) Perfect expansion ⇒ $p_e=p_a=0.12$ bar $=1.2\times 10^4$ Pa. $p_0/p_e=120/0.12=1000$. $\gamma /(\gamma -1)=1.18/0.18=6.556$. So $1+0.09M_e^2=1000^{1/6.556}=1000^{0.1526}=2.858$. $0.09M_e^2=1.858\Rightarrow M_e^2=20.65\Rightarrow M_e=\mathrm{4.544.}$ Area exponent $\frac{2.18}{0.36}=6.056$; front factor $\frac{2}{2.18}=0.9174$. $\epsilon =\frac{1}{4.544}[0.9174\times 2.858{]}^{6.056}=\frac{1}{4.544}(2.622{)}^{6.056}=\frac{1}{4.544}(346.6)=\mathrm{76.3.}$ (b) Design thrust ($p_e=p_a$, pressure term zero): $F=\dot{m}{v}_e+(p_e-p_a)A_e=480\times 3100+0=1.488\times 10^6\text{N}=1488\text{kN}.$ (c) At $p_a=0.40$ bar $=4.0\times 10^4$ Pa with $p_e=1.2\times 10^4$ Pa still: $(p_e-p_a)A_e=(1.2\times 10^4-4.0\times 10^4)(2.0)=(-2.8\times 10^4)(2.0)=-5.6\times 10^4\text{N}.$ $F=1.488\times 10^6-5.6\times 10^4=1.432\times 10^6\text{N}=1432\text{kN}.$ Comment: thicker air ($p_a$ up) over-expands this vacuum-optimised engine and shaves 56 kN off thrust — about 3.8% — a concrete price of firing a high-ε bell too low. This is why upper-stage engines (huge ε like 76) are never used as first stages.

Optimum thrust condition ::: $p_e=p_a$ (perfect expansion). Sign of $dF/d{A}_e$ when over-expanded ::: Negative — shrink ε. Why high altitude needs big ε ::: Small $p_a$ ⇒ small target $p_e$ ⇒ more expansion ⇒ larger $A_e/A^{\ast }$. Which Mach root past the throat ::: The supersonic $M_e>1$ root.