This page hunts down every case a nozzle-pressure problem can throw at you: too-low exit pressure, too-high exit pressure, the exact match, the degenerate "no expansion" nozzle, the limiting "vacuum" case, and the real-world climbing rocket. First we map the territory, then we conquer each cell with a full worked example.
Intuition The single question every example answers
==Given the chamber pressure p 0 , the shape (area ratio A e / A ∗ ), and the outside pressure p b , is the jet over-expanded, under-expanded, or perfect — and by how much?== Everything below is that one question in different disguises.
For the machinery (the two master equations) see the parent Over - under expanded nozzle flows , and the building blocks Isentropic Flow Relations , Area-Mach Number Relation , and Choked Flow & the Throat (M=1) .
Before any example, let me re-state — in plain words — the only two formulas we need, so no symbol appears unearned.
Definition The symbols, in words
p 0 = pressure deep inside the chamber where the gas is nearly still (the "stagnation" pressure).
p e = the pressure of the gas right at the exit plane of the nozzle mouth.
p b = the pressure of the air outside , the "back" pressure.
M e = the Mach number at the exit = (exit gas speed) ÷ (local speed of sound). M e > 1 means supersonic.
A e / A ∗ = exit area divided by throat area. The throat is the narrowest point, where M = 1 . This ratio is a pure number ≥ 1.
γ (gamma) = the "stiffness" of the gas; for air/combustion products γ = 1.4 .
The verdict rule, stated once:
Every nozzle-pressure problem is one of these cells. The worked examples below are tagged [Cell n] .
#
Case class
What is special
Example
1
p e < p b (over-expanded, mild)
oblique shocks at lip
Ex 1
2
p e < p b strongly
Mach disk / separation risk
Ex 2
3
p e = p b (perfect)
pressure term vanishes, max thrust
Ex 3
4
p e > p b (under-expanded)
expansion fans, jet bulges
Ex 4
5
Degenerate A e / A ∗ = 1
throat = exit, M e = 1 (sonic)
Ex 5
6
Limiting p b → 0 (vacuum)
always under-expanded
Ex 6
7
Same nozzle, changing p b (climb)
regime flips mid-flight
Ex 7
8
Thrust sign across regimes
( p e − p b ) A e can be + or −
Ex 8
9
Exam twist: solve for design altitude
invert the logic
Ex 9
We fix γ = 1.4 throughout unless stated, so 2 γ − 1 = 0.2 and γ − 1 γ = 3.5 .
Worked example Ex 1 — Mild over-expansion
[Cell 1]
p 0 = 10 bar, A e / A ∗ = 4.0 , p b = 1 bar.
Forecast: A fat area ratio (4×) means the gas expanded a lot, so p e is probably small — guess over-expanded .
Find M e from Tool 1. Area ratio 4.0 ⇒ supersonic root M e ≈ 2.94 . Why this step? Geometry alone fixes the exit Mach; the throat is choked so we only need the shape.
Find p e from Tool 2. p e p 0 = ( 1 + 0.2 ⋅ 2.9 4 2 ) 3.5 = ( 2.729 ) 3.5 ≈ 34.0 . Why? This converts the Mach number into the pressure it corresponds to.
Compute p e . p e = 10/34.0 ≈ 0.294 bar. Why? Just rearrange.
Verdict. p e = 0.294 < p b = 1.0 ⇒ PR = 0.294 < 1 ⇒ over-expanded . Oblique shocks at the lip. See Oblique Shocks .
Verify: Pressure ratio p 0 / p e = 10/0.294 = 34.0 , matching step 2. Units: bar/bar dimensionless ✓. Forecast correct ✓.
Worked example Ex 2 — Strong over-expansion, separation risk
[Cell 2]
p 0 = 10 bar, A e / A ∗ = 8.0 , p b = 1 bar.
Forecast: Even bigger area ratio ⇒ even lower p e ⇒ strongly over-expanded. The pressure gap is huge, so watch for a Mach disk or the shock creeping inside.
M e from Tool 1. A e / A ∗ = 8.0 ⇒ M e ≈ 3.65 . Why? Doubling the area ratio pushes the supersonic root higher.
p e from Tool 2. p 0 / p e = ( 1 + 0.2 ⋅ 3.6 5 2 ) 3.5 = ( 3.665 ) 3.5 ≈ 92.5 , so p e = 10/92.5 ≈ 0.108 bar. Why? Same isentropic conversion.
Verdict + severity. PR = 0.108/1 = 0.108 . Extremely below 1: strongly over-expanded. Why care? A large required pressure rise (p b / p e ≈ 9.3 ) can exceed what an attached oblique shock provides, so a normal shock / Mach disk forms and the flow may separate inside the diverging wall. See Normal Shock & Mach Disk and Flow Separation in Nozzles .
Verify: p b / p e = 1/0.108 ≈ 9.26 — a near-order-of-magnitude jump, consistent with "strong." Compare to Ex 1 where p b / p e ≈ 3.4 (milder) ✓.
Worked example Ex 3 — Perfect expansion
[Cell 3]
A nozzle with A e / A ∗ = 4.0 , p 0 = 10 bar is fired where p b = 0.294 bar.
Forecast: We just found this geometry gives p e = 0.294 bar. If ambient is also 0.294 bar, then p e = p b exactly — perfect .
p e (already known). p e = 0.294 bar from Ex 1. Why reuse? The nozzle didn't change, so p e is identical.
Verdict. PR = 0.294/0.294 = 1.000 ⇒ perfectly expanded , flow leaves parallel and clean, no waves.
Thrust consequence. Pressure term ( p e − p b ) A e = 0 ⇒ all thrust is momentum thrust m ˙ V e , the maximum for this V e . See Rocket Nozzle Design & Thrust Optimization .
Verify: PR = 1 exactly; pressure-thrust term = 0 ✓.
Worked example Ex 4 — Under-expansion
[Cell 4]
Same nozzle (p e = 0.294 bar) fired at high altitude, p b = 0.20 bar.
Forecast: Nothing about the nozzle changed, but the sky is thinner. Since p e = 0.294 > 0.20 , guess under-expanded .
p e unchanged. Still 0.294 bar (geometry-locked). Why? The exit is supersonic and "deaf" to p b .
Verdict. PR = 0.294/0.20 = 1.47 > 1 ⇒ under-expanded . The jet keeps expanding outside through Prandtl–Meyer fans and bulges out. See Prandtl-Meyer Expansion Fans .
Verify: PR = 1.47 > 1 ✓, opposite side of 1 from Ex 1 (0.294) — same p e , different sky, flipped regime ✓.
Worked example Ex 5 — Degenerate nozzle, throat = exit
[Cell 5]
p 0 = 10 bar, A e / A ∗ = 1.0 , p b = 1 bar.
Forecast: Area ratio 1 means the exit is the throat — no diverging section. The throat sits at M = 1 when choked, so guess M e = 1 , sonic (not supersonic).
M e from Tool 1. At A e / A ∗ = 1 , the only root is M e = 1 . Why? Tool 1 equals 1 exactly at M = 1 (the throat is the minimum of the area function). See Choked Flow & the Throat (M=1) .
p e from Tool 2. p 0 / p e = ( 1 + 0.2 ⋅ 1 2 ) 3.5 = ( 1.2 ) 3.5 ≈ 1.893 , so p e = 10/1.893 ≈ 5.28 bar. Why? Sonic exit has a fixed pressure ratio of about 1.89.
Verdict. p e = 5.28 > p b = 1 ⇒ under-expanded — but note the flow is only sonic , so the external adjustment is subsonic-to-transonic in character, not a classic supersonic fan. The over/under machinery still labels it by pressure.
Verify: ( 1.2 ) 3.5 ≈ 1.893 ; 10/1.893 ≈ 5.28 bar ✓. Critical pressure ratio for γ = 1.4 is the textbook 0.528 (i.e. p e / p 0 = 0.528 ) ✓.
Worked example Ex 6 — Limiting case: firing into vacuum
[Cell 6]
Same A e / A ∗ = 4.0 nozzle, but p b → 0 (deep space).
Forecast: p e = 0.294 bar is a fixed positive number; p b = 0 . Since any positive number > 0, it must be under-expanded — and maximally so.
PR . PR = p e / p b = 0.294/0 → + ∞ . Why this step? As p b → 0 + , the ratio diverges. Every real nozzle is under-expanded in vacuum.
Physical reading. The jet expands as far as physically possible; fans open to their limit (Prandtl–Meyer maximum turning). No shocks ever form. Why? You cannot need compression to reach zero pressure.
Verify: For any finite p e > 0 and p b → 0 , PR → ∞ > 1 ⇒ under-expanded for all such nozzles ✓ (checked as a limit inequality).
Worked example Ex 7 — Same rocket, whole ascent
[Cell 7]
A booster with p 0 = 10 bar, A e / A ∗ = 4.0 (p e = 0.294 bar) climbs. Classify at three altitudes: sea level p b = 1.0 , mid p b = 0.294 , high p b = 0.10 bar.
Forecast: As we climb, p b drops while p e stays put, so the regime should march over-expanded → perfect → under-expanded.
Sea level. PR = 0.294/1.0 = 0.294 < 1 ⇒ over-expanded (shocks). Why? Thick air pushes the jet in.
Mid. PR = 0.294/0.294 = 1 ⇒ perfect. Why? We hit the design altitude.
High. PR = 0.294/0.10 = 2.94 > 1 ⇒ under-expanded (fans). Why? Thin air lets the jet keep expanding.
Verify: Ratios 0.294 , 1.000 , 2.94 straddle 1 in order ✓ — one nozzle, three regimes, driven only by p b .
Worked example Ex 8 — Thrust sign across regimes
[Cell 8]
Using F = m ˙ V e + ( p e − p b ) A e . Take A e = 0.10 m 2 , p e = 0.294 bar = 29.4 kPa. Compute the pressure-thrust term only at p b = 1 bar (100 kPa) and at p b = 0.10 bar (10 kPa).
Forecast: Over-expanded should give a negative (drag-like) pressure term; under-expanded a positive one.
Sea level (over-expanded). ( p e − p b ) A e = ( 29.4 − 100 ) × 1 0 3 × 0.10 = − 7.06 × 1 0 3 N = − 7.06 kN. Why negative? Ambient outpushes the jet; pressure thrust subtracts. Units: Pa·m² = N ✓.
High altitude (under-expanded). ( 29.4 − 10 ) × 1 0 3 × 0.10 = + 1.94 × 1 0 3 N = + 1.94 kN. Why positive? Jet outpushes ambient; pressure thrust adds.
Verify: ( 29.4 − 100 ) ⋅ 1 0 3 ⋅ 0.10 = − 7060 N and ( 29.4 − 10 ) ⋅ 1 0 3 ⋅ 0.10 = + 1940 N ✓; sign flips with regime ✓.
Worked example Ex 9 — Exam twist: design for a target altitude
[Cell 9]
"You need p e = 0.30 bar (to be perfect at a chosen altitude). p 0 = 10 bar, γ = 1.4 . What M e and roughly what A e / A ∗ ?"
Forecast: We run the tools backwards : start from a pressure, get M e , then get area ratio.
Invert Tool 2 for M e . p 0 / p e = 10/0.30 = 33.33 = ( 1 + 0.2 M e 2 ) 3.5 . So 1 + 0.2 M e 2 = 33.3 3 1/3.5 = 33.3 3 0.2857 ≈ 2.717 , giving M e 2 = ( 2.717 − 1 ) /0.2 = 8.585 , M e ≈ 2.930 . Why? Take the 1/3.5 power to undo the exponent, then solve the linear-in-M 2 relation.
Tool 1 forward for area ratio. Plug M e = 2.93 : A e / A ∗ ≈ 3.97 ≈ 4.0 . Why? This tells the designer how big to make the exit relative to the throat.
Verdict at design. By construction p e = p b = 0.30 bar there ⇒ perfect.
Verify: 33.3 3 1/3.5 ≈ 2.717 ; M e ≈ 2.93 ; forward Tool 1 gives A e / A ∗ ≈ 3.97 — consistent with the A e / A ∗ = 4 nozzle used throughout ✓.
Recall Quick self-test
A nozzle gives p e = 0.30 bar. It is fired at p b = 0.30 bar. Regime? ::: Perfectly expanded (PR = 1 ), no waves, max thrust for that V e .
Same nozzle, p b = 0.05 bar. Regime and wave type? ::: Under-expanded (PR = 6 > 1 ); Prandtl–Meyer expansion fans, jet bulges out.
Same nozzle, p b = 2 bar. Regime and wave type? ::: Over-expanded (PR = 0.15 < 1 ); oblique shocks, possibly Mach disk / separation.
Firing into vacuum — over or under? ::: Always under-expanded, PR → ∞ .
Mnemonic The one-line procedure
Shape → Mach → Pressure → Compare. (Tool 1 gives M e , Tool 2 gives p e , then p e vs p b .) p e never listens to p b .