Exercises — Over - under expanded nozzle flows

WHAT we compare: $p_e=0.6$ vs $p_b=0.9$. Since $0.6<0.9$, we have $p_e<p_b$ → over-expanded. WHY: the exit pressure has fallen below ambient, so the surroundings must push the jet inward — that compression is done by oblique shocks anchored at the nozzle lip. This is the left (red) panel of the figure: the jet is squeezed inward and crossing shock lines form. (Mnemonic check: OLE = Over → Low → Exit. Over-expanded ⇒ Low exit pressure. ✓)

WHAT we compare: $p_e=0.6$ vs $p_b=1.0$. $0.6<1.0$ → $p_e<p_b$ → over-expanded, more strongly than in 1.1. WHY "more strongly": the mismatch the outside must correct is the gap $p_b-p_e$. In 1.1 it was $0.9-0.6=0.3$ bar; here it is $1.0-0.6=0.4$ bar — a bigger gap means a stronger required compression, hence stronger shocks. WHY this matters physically: if the gap grows large enough the gentle oblique shocks cannot do the job alone and a normal shock / Mach disk forms (see Normal Shock & Mach Disk). Same red panel of the figure.

WHAT we compare: $p_e=0.4$ vs $p_b\approx 0$. $0.4>0$ → $p_e>p_b$ → under-expanded — the right (yellow) panel of the figure, jet bulging outward through expansion fans. WHY the plume spreads so wide: the jet is enormously over-pressured relative to a near-vacuum, so it keeps expanding outside the nozzle in every outward direction. That is why rocket plumes fan out dramatically in space — the smaller $p_b$ is, the larger the outward turning the fans must produce.

WHAT: use tool (A) — it is the only formula linking $p_0$, $M_e$, $p_e$. WHY (A) and not (B): we are given $M_e$ directly, so we skip the area step (tool B is only needed when we must find $M_e$ from geometry). $\frac{p_0}{p_e}={\left(1+0.2\cdot 2.5^2\right)}^{3.5}=(1+1.25{)}^{3.5}=2.25^{3.5}$ Now $2.25^{3.5}\approx 17.09$. So $p_e=\frac{20}{17.09}\approx 1.17\text{bar}.$

WHAT we compare: $p_e\approx 1.17$ vs $p_b=1.0$. $1.17>1.0$ → $p_e>p_b$ → under-expanded (mildly). WHY only mildly: the gap $p_e-p_b\approx 0.17$ bar is small, so only weak expansion fans are needed — the yellow panel of the figure, but with a gently-flaring boundary rather than a dramatic bulge.

WHAT: tool (B) is the only equation relating area ratio to $M$. WHY the supersonic root: the area–Mach curve is U-shaped — every area ratio above 1 has TWO roots, one subsonic ($M<1$) and one supersonic ($M>1$). In the diverging section past a choked throat the flow is supersonic, so we take the $M>1$ root. Plugging $\gamma =1.4$: $2.0=\frac{1}{M_e}{\left[\frac{1}{1.2}\left(1+0.2M_e^2\right)\right]}^3.$ Solving numerically gives $M_e\approx 2.197$.

Step 1 (tool B) — geometry to Mach. Solve $6.0=\frac{1}{M_e}{\left[\frac{1}{1.2}(1+0.2M_e^2)\right]}^3$ for the supersonic root: $M_e\approx \mathrm{3.37.}$ WHY tool B first: we are given geometry ($A_e/A^{\ast }$), not Mach, so we must convert area ratio → Mach before touching pressure. WHY the answer is large: a bigger area ratio means the gas expands through more area, accelerating more, so $M_e$ is large (6 is a large ratio ⇒ $M_e>3$). Step 2 (tool A) — Mach to pressure. $\frac{p_0}{p_e}=(1+0.2\cdot 3.37^2{)}^{3.5}=(1+2.272{)}^{3.5}=3.272^{3.5}\approx \mathrm{63.1.}$ $p_e=\frac{50}{63.1}\approx 0.792\text{bar}.$ WHY tool A second: now that $M_e$ is known, tool (A) is the only relation that turns it into a pressure. WHY $p_e$ is so far below $p_0$: high Mach means the gas gave up nearly all its pressure to gain speed, so $p_e$ is a tiny fraction of the 50-bar chamber. Step 3 — classify. $p_e=0.792<p_b=1.0$ → over-expanded (the red panel of the figure). WHY shocks: exit pressure sits below ambient, so the outside must compress it back up — done by oblique shocks at the lip.

WHAT: perfect expansion means $p_e=p_b$, and $p_e$ is fixed by geometry. So $p_b^{\star }=p_e\approx 0.792$ bar. WHY: that is the one ambient pressure that exactly matches the geometry-locked exit pressure, so the jet leaves cleanly (the green centre panel of the figure). Reasoning about altitude: as a rocket climbs, $p_b$ falls.

• When $p_b>0.792$ bar (low altitude, e.g. sea level) → $p_e<p_b$ → over-expanded.
• When $p_b<0.792$ bar (high altitude) → $p_e>p_b$ → under-expanded. WHY it flips: $p_e$ stays put (geometry never changes), so as $p_b$ slides down through $0.792$ bar the sign of $p_e-p_b$ reverses. The nozzle flips regime as it climbs through that altitude — the core design tension in Rocket Nozzle Design & Thrust Optimization.

Step 1 — exit pressure. From 3.1's geometry ($A_e/A^{\ast }=6$), $M_e\approx 3.37$ and $p_0/p_e\approx 63.1$. WHY reuse 3.1: the area ratio is identical, so the Mach and pressure ratio are identical — only $p_0$ changed. With $p_0=70$ bar $=7.0\times 10^6$ Pa, $p_e=\frac{7.0\times 10^6}{63.1}\approx 1.109\times 10^5\text{Pa}(\approx 1.11\text{bar}).$ Step 2 — momentum thrust (same in both cases). $\dot{m}{V}_e=250\times 2900=7.25\times 10^5\text{N}.$ WHY this piece is altitude-independent: $\dot{m}$ and $V_e$ are set by the internal supersonic flow, which cannot hear $p_b$; only the pressure term feels the atmosphere. Step 3 — pressure thrust, sea level. $(p_e-p_b)A_e=(1.109\times 10^5-1.01325\times 10^5)(0.90)=(9.575\times 10^3)(0.90)\approx 8.62\times 10^3\text{N}.$ WHY it is positive here: $p_e>p_b$ (under-expanded), so the exit pushes out harder than the atmosphere pushes in — the term adds to thrust. $F_{\text{sea}}=7.25\times 10^5+8.62\times 10^3\approx 7.336\times 10^5\text{N}.$ Step 4 — pressure thrust, vacuum ($p_b=0$): $(p_e-0)A_e=1.109\times 10^5\times 0.90\approx 9.981\times 10^4\text{N}.$ WHY it grows: removing the atmosphere removes the inward $-p_b{A}_e$ push, so the full exit pressure now contributes. $F_{\text{vac}}=7.25\times 10^5+9.981\times 10^4\approx 8.248\times 10^5\text{N}.$ Step 5 — gain. $\Delta F=F_{\text{vac}}-F_{\text{sea}}\approx 8.248\times 10^5-7.336\times 10^5\approx 9.12\times 10^4\text{N}.$ WHY exactly this: the difference is precisely $p_b{A}_e=1.01325\times 10^5\times 0.90\approx 9.12\times 10^4$ N — the atmospheric back-push that disappears. Thrust rises by about $91$ kN climbing to vacuum.

WHAT we compare: $p_e\approx 1.11$ bar vs $p_b=1.013$ bar. $1.11>1.013$ → $p_e>p_b$ → under-expanded at sea level (barely). WHY it is nearly perfect at sea level: the gap is tiny, so the sea-level penalty is small and it only improves with altitude (as $p_b$ falls, the positive pressure-thrust term grows). WHY a large area ratio is still risky at sea level: push the ratio higher and $p_e$ drops below $p_b$, giving over-expansion and, if strong enough, flow separation and side-loads. Big bells are efficient in vacuum but hazardous at launch.

Step 1 — required $p_e$. Perfect expansion ⇒ $p_e=p_b=0.30$ bar. So $p_0/p_e=40/0.30=133.3$. WHY: perfect expansion is defined by $p_e=p_b$, which pins the pressure ratio we must engineer. Step 2 — invert tool (A) for $M_e$. We undo the exponent by raising to $1/3.5$: $1+0.2M_e^2=(133.3{)}^{1/3.5}=133.3^{0.2857}\approx \mathrm{4.05.}$ $M_e^2=\frac{4.05-1}{0.2}=15.25\Rightarrow M_e\approx \mathrm{3.91.}$ WHY invert: we know the pressure ratio and want the Mach that produces it — the reverse of Problem 2.1. Step 3 — tool (B) for area ratio. Plug $M_e=3.91$: $\frac{A_e}{A^{\ast }}=\frac{1}{3.91}{\left[\frac{1}{1.2}\left(1+0.2\cdot 3.91^2\right)\right]}^3=\frac{1}{3.91}{\left[\frac{4.057}{1.2}\right]}^3=\frac{1}{3.91}(3.381{)}^3.$ $(3.381{)}^3\approx 38.64$, so $A_e/A^{\ast }\approx 38.64/3.91\approx 9.88$. Answer: $M_e\approx 3.91$, $A_e/A^{\ast }\approx 9.9$. A big, high-altitude nozzle — which is exactly why upper-stage engines have huge bells.

WHAT we compare: $p_e=0.30$ vs $p_b=1.0$. $0.30<1.0$ → $p_e<p_b$ → strongly over-expanded (the red panel of the figure, pushed to its extreme). The required pressure jump the outside must impose is $p_b/p_e=1.0/0.30\approx 3.33$. WHY this is dangerous: a compression this large cannot be handled by a gentle oblique shock alone; a Mach disk (a normal shock across the jet, see Normal Shock & Mach Disk) forms, and the adverse pressure can push a shock into the diverging section, tearing the boundary layer off the walls — flow separation (see Flow Separation in Nozzles). This causes side-loads that can physically damage the nozzle at ignition — the launchpad-vs-altitude dilemma solved by dual-bell and altitude-compensating nozzles.

Step 1 — check choking. Compare the driving ratio to the critical ratio $p_0/p^{\ast }\approx 1.893$: $\frac{p_0}{p_b}=\frac{1.5}{1.0}=1.5<\mathrm{1.893.}$ WHY this test: to sonic-choke the throat the chamber must be able to drop the pressure to $p^{\ast }$ (the just-choked pressure). Since $p_b$ sits above $p^{\ast }$ would require $p_0/p_b\ge 1.893$, and we are below it, the throat never reaches $M=1$ — the nozzle is unchoked. Step 2 — consequence. With no sonic throat, the whole nozzle stays subsonic. WHY that fixes the exit pressure: subsonic flow can transmit pressure upstream, so the exit is forced to match ambient: $p_e=p_b=1.0$ bar. The diverging section acts as a diffuser (decelerates the subsonic flow), not as a supersonic accelerator. Step 3 — classification. Over-/under-expanded is defined only for a supersonic exit with $p_e$ locked by geometry. WHY it does not apply here: $p_e$ is not locked — it obeys $p_b$. So there are no external shocks or fans, just smooth subsonic flow. This is the regime the parent note's [!mistake] warns about — the naïve intuition "back pressure controls exit pressure" is correct here precisely because the flow is subsonic.

Part (a) — subsonic exit pressure. WHY tool (A) still works: tool (A) is just the isentropic energy relation (from Isentropic Flow Relations); it is valid at any Mach number, subsonic or supersonic. Only tools that assume a sonic throat ($A^{\ast }$, choking) are off-limits when unchoked. Here we are simply given $M_e$, so we plug in: $\frac{p_0}{p_e}=(1+0.2\cdot 0.5^2{)}^{3.5}=(1+0.05{)}^{3.5}=1.05^{3.5}\approx \mathrm{1.186.}$ $p_e=\frac{1.6}{1.186}\approx 1.349\text{bar}.$ WHY $p_e$ is close to $p_0$: at low Mach the gas barely accelerated, so it kept almost all its pressure — the exact opposite of the supersonic cases where $p_e\ll p_0$. Part (b) — the exact critical case $p_0/p_b=1.893$. This is the razor's edge between unchoked and choked. WHY: at exactly the critical ratio the throat just reaches $M=1$, so the exit (if the throat is the exit) has $M_e=1$ and $p_e=p^{\ast }=p_0/1.893$. With, say, $p_0=1.893$ bar this gives $p_e=1.0$ bar $=p_b$ — the flow is choked and perfectly matched, sitting at sonic ($M_e=1$) exactly. It is the unique boundary point: any higher $p_0/p_b$ pushes the throat sonic and (with a diverging section) supersonic; any lower leaves it subsonic. So the critical case is the single value where "just choked," "$M=1$ at the exit," and "$p_e=p_b$" all coincide.

First question before any classification ::: Is $p_0/p_b$ above the critical ratio ($\approx 1.893$ for $\gamma =1.4$)? If not, the flow is unchoked/subsonic and $p_e=p_b$ — no over/under-expansion. What is $p^{\ast }$? ::: The just-choked exit pressure, i.e. $p_0$ divided by the critical ratio; setting $M=1$ in tool (A) gives $p_0/p^{\ast }=((\gamma +1)/2{)}^{\gamma /(\gamma -1)}\approx 1.893$. Regime rule ::: Once choked & supersonic, compute $p_e$ from geometry (tools B then A), compare to $p_b$: $p_e<p_b$ over-expanded (shocks), $p_e>p_b$ under-expanded (fans), $p_e=p_b$ perfect. Does $p_b$ change $p_e$ in supersonic flow? ::: No — $p_e$ is locked by $A_e/A^{\ast }$ and $p_0$; the regime flips only because $p_b$ crosses the fixed $p_e$. Sign of the pressure-thrust term when over-expanded? ::: Negative — it subtracts from thrust.