Over-expanded nozzle — oblique shocks in plume, efficiency loss: Exercises

This page is a self-test ladder for the parent topic. Each problem hides its full worked solution — try first, then reveal. We climb from L1 (just recognise the words) to L5 (design-level synthesis).

Before you start, own these symbols — no equation below uses them until you've read this:

The thrust equation we will reuse constantly (built in the parent):

We will also grade nozzle quality with a velocity efficiency $C_{v}$ . Here is why it takes the square-root form it does — read this before the L3 problems use it:

The figure below is the physical picture behind every problem on this page — refer back to it whenever a problem mentions "the plume," "the lip," or "oblique shocks":

Figure — Over-expanded nozzle — oblique shocks in plume, efficiency loss

Trace it left to right: the red arrow is the supersonic flow leaving the grey diverging nozzle; at the exit plane $p_{e} < p_{a}$ , so the orange arrows (ambient air) squeeze inward; that squeeze folds the flow into the two blue oblique shocks springing from the nozzle lips at angle $β$ ; they cross on the centreline and the green lines show the flow re-expanding into the next diamond. Problems 2.3, 3.x and 5.1 all live on those blue shock lines.

Level 1 — Recognition

Problem 1.1 (L1)

A nozzle exits gas at $p_{e} = 40 kPa$ into air at $p_{a} = 101 kPa$ . Is it over-expanded, under-expanded, or perfectly matched?

Worked example Solution 1.1

Compare $p_{e}$ with $p_{a}$ . Here $p_{e} = 40 < p_{a} = 101$ , so the jet pressure is below ambient. What this means physically: the outside air is winning, squeezing the plume inward. Answer: Over-expanded ( $p_{e} < p_{a}$ ).

Problem 1.2 (L1)

In the thrust equation $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ , which term becomes negative for an over-expanded nozzle, and what does a negative thrust term mean?

Worked example Solution 1.2

The pressure thrust term $(p_{e} - p_{a}) A_{e}$ . Since $p_{e} < p_{a}$ , the bracket is negative, and $A_{e} > 0$ , so the whole term is negative. Physical meaning: it is a retarding force — the ambient air pushes on the exit disc against the direction of flight, subtracting from your total thrust.

Problem 1.3 (L1)

When an over-expanded plume compresses back up to ambient pressure, does it do so through shock waves or expansion fans? One word.

Worked example Solution 1.3

Shock waves. Rising pressure in supersonic flow is compression, and compression in supersonic flow happens through shocks. (Expansion fans do the opposite — they drop pressure smoothly.)

Level 2 — Application

Problem 2.1 (L2)

A nozzle has $p_{e} = 30 kPa$ , $p_{a} = 101 kPa$ , $A_{e} = 1.5 m^{2}$ . Compute the pressure thrust term. Is it helping or hurting?

Worked example Solution 2.1

What we do: plug into $(p_{e} - p_{a}) A_{e}$ . $(30 - 101) \times 1 0^{3} \times 1.5 = (- 71 \times 1 0^{3}) (1.5) = - 106.5 \times 1 0^{3} N = - 106.5 kN$ Why: we isolate the pressure part to see the penalty alone. Answer: $- 106.5 kN$ — it hurts, subtracting 106.5 kN from the momentum thrust.

Problem 2.2 (L2)

For the nozzle in 2.1, the momentum thrust is $\overset{m}{˙} v_{e} = 480 kN$ . Find the actual total thrust, and the fractional thrust loss caused by over-expansion.

Worked example Solution 2.2

Total thrust: $F = 480 + (- 106.5) = 373.5 kN$ Fractional loss relative to the momentum thrust: $\frac{106.5}{480} = 0.2219 \approx 22.2%$ Answer: $F = 373.5 kN$ ; you lose about $22.2%$ of the jet's thrust to pressure mismatch alone.

Problem 2.3 (L2)

A shock has upstream Mach number $M_{1} = 3.0$ and shock angle $β = 3 0^{\circ}$ . Find the normal Mach component $M_{1 n} = M_{1} sin β$ . Why is this number, not $M_{1}$ itself, the one that governs the shock strength?

Worked example Solution 2.3

$M_{1 n} = 3.0 \times sin 3 0^{\circ} = 3.0 \times 0.5 = 1.5$ Why the normal component rules: a shock is a wall of compression that the flow crosses perpendicular to the shock face. Only the velocity component pointing into that wall gets slammed and slowed. The component sliding along the shock passes through untouched. So the "violence" of the shock is set by $M_{1 n}$ , and here $M_{1 n} = 1.5$ (a modest shock) even though $M_{1} = 3$ looks fierce.

Level 3 — Analysis

Problem 3.1 (L3)

Across an oblique shock the static pressure ratio is $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1 n}^{2} - 1) .$ For $γ = 1.4$ and $M_{1 n} = 1.5$ , compute $p_{2} / p_{1}$ .

Worked example Solution 3.1

What we do: substitute $γ = 1.4$ , $M_{1 n} = 1.5$ . First the front coefficient: $\frac{2 γ}{γ + 1} = \frac{2 ( 1.4 )}{2.4} = \frac{2.8}{2.4} = 1.1667$ . Then $M_{1 n}^{2} - 1 = 2.25 - 1 = 1.25$ . $\frac{p _{2}}{p _{1}} = 1 + 1.1667 \times 1.25 = 1 + 1.4583 = 2.4583$ Answer: $p_{2} / p_{1} \approx 2.46$ . A modest $M_{1 n} = 1.5$ shock raises pressure by a factor 2.46, confirming that a shallow oblique shock is a gentle compressor.

Problem 3.2 (L3)

The stagnation-pressure ratio (the "energy quality budget" that survives) across a shock is $\frac{p _{02}}{p _{01}} = (\frac{( γ + 1 ) M _{1 n}^{2}}{2 + ( γ - 1 ) M _{1 n}^{2}})^{\frac{γ}{γ - 1}} (\frac{γ + 1}{2 γ M _{1 n}^{2} - ( γ - 1 )})^{\frac{1}{γ - 1}} .$ With $γ = 1.4$ , $M_{1 n} = 1.5$ , compute it. What fraction of stagnation pressure is lost?

Worked example Solution 3.2

Where the formula comes from (so it isn't a black box): stagnation pressure is set by two things a shock changes — the density-like squeeze and the temperature-like heating. The formula is literally the product of those two effects.

Bracket 1 is the density ratio raised to $\frac{γ}{γ - 1}$ . That exponent is the isentropic link between density and pressure: for a smooth (reversible) part of the change, pressure scales as density $^{γ / (γ - 1)}$ when written in stagnation terms.
Bracket 2 carries the irreversible piece — the entropy jump that only a shock produces — through the exponent $\frac{1}{γ - 1}$ . This is the bracket that makes the product drop below 1; without the shock's entropy rise it would be exactly 1. Multiplying them gives "how much of the total-pressure budget survives." The two exponents are just the two ways $γ$ enters compressible flow (reversible scaling vs. entropy accounting).

Now the numbers ( $γ = 1.4$ so $\frac{γ}{γ - 1} = 3.5$ , $\frac{1}{γ - 1} = 2.5$ ):

$M_{1 n}^{2} = 2.25$ .
Bracket 1: $\frac{( 2.4 ) ( 2.25 )}{2 + ( 0.4 ) ( 2.25 )} = \frac{5.4}{2.9} = 1.8621$ . Raise to $3.5$ : $1.862 1^{3.5} = 8.798$ .
Bracket 2: $\frac{2.4}{2 ( 1.4 ) ( 2.25 ) - 0.4} = \frac{2.4}{6.3 - 0.4} = \frac{2.4}{5.9} = 0.40678$ . Raise to $2.5$ : $0.4067 8^{2.5} = 0.10565$ .
Product: $8.798 \times 0.10565 = 0.9297$ . Answer: $p_{02} / p_{01} \approx 0.930$ , so about $7.0%$ of the stagnation-pressure budget is destroyed by this one shock. Interpretation: even a gentle shock ( $M_{1 n} = 1.5$ ) already burns 7% of your usable energy quality — and diamonds stack several of these.

Problem 3.3 (L3)

A plume passes through three identical shocks, each with $p_{02} / p_{01} = 0.930$ . Find the overall stagnation-pressure survival, then the velocity efficiency $C_{v} = \frac{p _{0, final}}{p _{0, initial}} .$

Worked example Solution 3.3

What we do: losses multiply, because each shock acts on the survivor of the previous one. $\frac{p _{0, final}}{p _{0, initial}} = 0.93 0^{3} = 0.8044$ Velocity efficiency (recall: velocity goes like the square-root of the surviving energy budget): $C_{v} = 0.8044 = 0.8969 \approx 0.897$ Answer: about $80.4%$ of stagnation pressure survives; $C_{v} \approx 0.90$ . You've lost ~10% of exhaust velocity to the diamond stack alone.

Level 4 — Synthesis

Problem 4.1 (L4)

A vacuum-optimised engine is test-fired at sea level. Data: $p_{e} = 6 kPa$ , $p_{a} = 101 kPa$ , $A_{e} = 0.6 m^{2}$ , $\overset{m}{˙} v_{e} = 520 kN$ . Additionally the plume passes through shocks giving overall $p_{0, final} / p_{0, initial} = 0.62$ . (a) Pressure thrust loss. (b) Thrust after pressure mismatch only. (c) Velocity efficiency $C_{v}$ . (d) One sentence: which loss dominates and why.

Worked example Solution 4.1

(a) Pressure thrust: $(p_{e} - p_{a}) A_{e} = (6 - 101) \times 1 0^{3} \times 0.6 = (- 95 \times 1 0^{3}) (0.6) = - 57 kN$ (b) Thrust after pressure mismatch: $F = 520 + (- 57) = 463 kN$ (c) Velocity efficiency: $C_{v} = 0.62 = 0.7874 \approx 0.787$ (d) Which dominates: The pressure mismatch costs $57/520 = 11.0%$ of momentum thrust, but the shock losses cut exhaust velocity by ~ $21%$ ( $1 - 0.787$ ). Shock losses dominate, because they destroy the quality of the energy (stagnation pressure), which sets the achievable $v_{e}$ , whereas the pressure term is a one-time subtraction at the exit disc.

Problem 4.2 (L4)

Same engine, but you add a diffuser (a suppressor that raises effective back-pressure smoothly) so the plume no longer needs shocks: $p_{0, final} / p_{0, initial} = 0.97$ and the effective ambient at the exit is lowered so $(p_{e} - p_{a}) A_{e} = - 12 kN$ . Recompute $F$ and $C_{v}$ , and state the total thrust recovered versus 4.1.

Worked example Solution 4.2

New pressure thrust: $- 12 kN$ (given). New total thrust: $F = 520 + (- 12) = 508 kN$ . New efficiency: $C_{v} = 0.97 = 0.9849 \approx 0.985$ . Thrust recovered: $508 - 463 = 45 kN$ more usable thrust from the pressure term alone, and velocity efficiency jumps from $0.787$ to $0.985$ . Why it works: the diffuser lets the plume compress gradually (near-isentropic) instead of through violent shocks, sparing the stagnation-pressure budget.

Problem 4.3 (L4) — the matched edge case

The same engine is now flown to the altitude where its nozzle is exactly matched: $p_{e} = p_{a} = 40 kPa$ , $A_{e} = 0.6 m^{2}$ , $\overset{m}{˙} v_{e} = 520 kN$ , and with no over-expansion there are no plume shocks ( $p_{0, final} / p_{0, initial} = 1$ ). (a) Pressure thrust term. (b) Total thrust. (c) Velocity efficiency. (d) In one line, why this is the design-optimal condition.

Worked example Solution 4.3

(a) Pressure thrust: $(p_{e} - p_{a}) A_{e} = (40 - 40) \times 1 0^{3} \times 0.6 = 0 N$ . The term vanishes. (b) Total thrust: $F = 520 + 0 = 520 kN$ — pure momentum thrust, nothing subtracted. (c) Velocity efficiency: $C_{v} = 1 = 1.00$ (no shock losses, so the entire stagnation-pressure budget survives). (d) Why optimal: with $p_{e} = p_{a}$ the pressure term is neither a penalty (over-expanded) nor a missed opportunity (under-expanded), and there are no shocks to destroy stagnation pressure — every joule of thermal energy that could become ordered jet motion did. This is the target every altitude-compensation scheme chases; see Altitude Compensation.

Level 5 — Mastery

Problem 5.1 (L5)

Derive the pressure ratio $p_{2} / p_{1}$ for a normal shock from the oblique formula, and use it as a sanity check. The oblique relation is $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} sin^{2} β - 1) .$ (a) What value of $β$ turns an oblique shock into a normal shock, and why? (b) For $M_{1} = 4.2$ , $γ = 1.2$ , compute the normal-shock pressure ratio. (c) Compare to a weak oblique shock at $β = 2 5^{\circ}$ (compute $p_{2} / p_{1}$ ) and explain in one line why nature prefers the oblique one.

Worked example Solution 5.1

(a) A normal shock stands perpendicular to the flow, i.e. its face is at $9 0^{\circ}$ to the velocity, which means $β = 9 0^{\circ}$ . Then $sin β = 1$ and $M_{1} sin β = M_{1}$ — the whole velocity is normal to the shock. This is the maximum-strength, maximum-entropy case. (b) With $β = 9 0^{\circ}$ , $M_{1 n} = M_{1} = 4.2$ , $γ = 1.2$ : $\frac{2 γ}{γ + 1} = \frac{2.4}{2.2} = 1.0909, M_{1}^{2} - 1 = 17.64 - 1 = 16.64$ $\frac{p _{2}}{p _{1}} = 1 + 1.0909 \times 16.64 = 1 + 18.153 = 19.153$ So a normal shock at $M = 4.2$ jumps pressure by ~19.2× in one violent step. (c) Oblique at $β = 2 5^{\circ}$ : $M_{1 n} = 4.2 sin 2 5^{\circ} = 4.2 \times 0.42262 = 1.775$ , so $M_{1 n}^{2} = 3.151$ . $\frac{p _{2}}{p _{1}} = 1 + 1.0909 (3.151 - 1) = 1 + 1.0909 \times 2.151 = 1 + 2.346 = 3.346$ Why nature prefers oblique: the same required pressure rise can be reached in several gentle oblique steps ( $3.3 \times$ each) instead of one $19 \times$ normal shock, and because stagnation-pressure loss grows sharply with $M_{1 n}$ , many weak shocks destroy far less energy than one strong one. Nature takes the cheapest compression.

Problem 5.2 (L5)

Design task. A first-stage engine must fly from sea level ( $p_{a} = 101 kPa$ ) to $30 km$ ( $p_{a} \approx 1.2 kPa$ ). Its fixed nozzle is matched ( $p_{e} = p_{a}$ ) at $10 km$ where $p_{a} = 26 kPa$ , giving $p_{e} = 26 kPa$ . Exit area $A_{e} = 1.0 m^{2}$ . (a) At sea level, is it over- or under-expanded, and what is the pressure thrust term? (b) At $30 km$ , same questions. (c) In one sentence, connect this to why boosters are staged.

Worked example Solution 5.2

(a) Sea level: $p_{e} = 26 < p_{a} = 101$ , so over-expanded. $(p_{e} - p_{a}) A_{e} = (26 - 101) \times 1 0^{3} \times 1.0 = - 75 \times 1 0^{3} N = - 75 kN$ A 75 kN penalty low down (plus shock losses on top). (b) 30 km: $p_{e} = 26 > p_{a} = 1.2$ , so under-expanded. $(p_{e} - p_{a}) A_{e} = (26 - 1.2) \times 1 0^{3} \times 1.0 = + 24.8 \times 1 0^{3} N = + 24.8 kN$ Now the pressure term adds 24.8 kN — but the gas hasn't finished expanding, so it's a lost opportunity, not a destroyed one (expansion fans are near-isentropic, no shock losses). (c) Staging link: because no single fixed nozzle is matched across a 100:1 ambient swing, we split the ascent into stages, each carrying a nozzle sized for its altitude band — see Rocket Staging and Altitude Compensation.

Recap Ladder

Question

What single inequality defines an over-expanded nozzle?

Answer

p_{e} < p_{a}

— exit pressure below ambient.

Question

Which Mach component sets a shock's strength, and why?

Answer

The normal component

M_{1 n} = M_{1} sin β

, because only the velocity crossing the shock face perpendicularly gets compressed.

Question

Why does

C_{v}

take the square-root of the stagnation-pressure ratio?

Answer

Because kinetic energy scales as

v^{2}

and the extractable energy scales as

p_{0}

, so

v_{e}^{2} \propto p_{0}

and

C_{v} = v_{e, actual} / v_{e, ideal} = p_{0, final} / p_{0, initial}

Question

Do stagnation-pressure losses across a chain of shocks add or multiply?

Answer

They multiply:

p_{0, final} / p_{0, initial} = \prod (p_{02} / p_{01})_{i}

Question

What happens to the pressure thrust term and efficiency in the perfectly matched case

p_{e} = p_{a}

Answer

The pressure term is exactly zero and (no shocks)

C_{v} = 1

— the design-optimal condition.

Question

Why is over-expansion worse than under-expansion?

Answer

Over-expansion forces irreversible shocks (compression, entropy rise); under-expansion allows reversible expansion fans (no quality lost).

Prerequisite webs touched here: Shock Wave Fundamentals, Isentropic Flow, Gas Dynamics, Thrust Equation, Nozzle Flow Separation, Nozzle Area Ratio.