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

This page assumes you know nothing beyond "hot gas comes out fast." Every letter, ratio, and squiggle used by the parent note parent topic is built here, in the order they depend on one another. Read top to bottom once and no symbol will ever ambush you.

1. Pressure — $p$

Picture a box of air. The tiny molecules bounce off every wall. Each bounce is a small shove; billions per second add up to a steady outward push. Pack more molecules in, or heat them so they bounce harder, and $p$ rises.

Why does the topic need $p$? Because the entire story is a contest between two pressures:

• $p_e$ — pressure of the gas at the nozzle exit (subscript $e$ = "exit").
• $p_a$ — pressure of the ambient air outside (subscript $a$ = "ambient", the sky around the rocket).

Look at the red arrows: the outside air pushes inward on the exhaust jet's edge. The exhaust pushes outward with its own pressure. Whoever is stronger wins the boundary.

2. Area of the exit — $A_e$

Why do we need it? Pressure is force per area. To turn a pressure into an actual push (a force), you multiply by the area it acts on:

$\text{force}=p\times A$

So the unbalanced pressure acting on the exit disc gives a real force $(p_e-p_a)A_e$. That single product decides how many kilonewtons you gain or lose from pressure mismatch. Hold this — it returns in §7.

3. Mass flow rate — $\dot{m}$

Picture standing at the exit with a stopwatch, weighing every kilogram of gas that streams past. That per-second weight is $\dot{m}$. A big engine might spit out hundreds of kg every second.

Why needed? Thrust is fundamentally throwing mass backward fast. You need to know how much mass ($\dot{m}$) and how fast (next symbol) to compute the push.

4. Exit velocity — $v_e$

The whole purpose of the nozzle's diverging cone is to make $v_e$ as large as possible. Multiply the mass thrown per second by the speed it's thrown at, and you get the momentum thrust:

$\dot{m}{v}_e(\text{units: }\frac{\text{kg}}{\text{s}}\cdot \frac{\text{m}}{\text{s}}=\text{newtons})$

This is the "rocket kicks because it hurls exhaust" part of thrust — always positive, always helping.

5. The full thrust equation — $F$

Now that $\dot{m}$, $v_e$, $p_e$, $p_a$, $A_e$ all exist, we can assemble the parent note's master formula:

• First term: throwing mass backward. Always helps.
• Second term: leftover pressure push on the exit disc. Helps if $p_e>p_a$, hurts if $p_e<p_a$.

For an over-expanded nozzle $p_e<p_a$, so $(p_e-p_a)$ is negative — the second term subtracts. That is the first, gentlest way over-expansion steals thrust. See the full derivation in Thrust Equation.

6. Speed of sound $a$ and the Mach number — $M$

Everything violent in this topic happens because the gas moves faster than sound. First we need a symbol for the sound speed itself.

Now we can compare the flow's speed to that message speed.

Why does this matter so much? Sound (travelling at $a$) is how a gas "sends a message" that pressure is changing ahead. If the flow is faster than its own messages ($M>1$), the gas cannot be warned in advance to turn or slow down smoothly. The adjustment must happen abruptly — in a razor-thin wall called a shock. At the exit of a rocket nozzle, $M_e$ is typically $3$–$5$: deeply supersonic. Subscript $e$ = exit again, so $M_e$ = "Mach at exit". More at Gas Dynamics.

The red circle is the flow; it outruns the black sound-wave ripples it emits, so the ripples pile into a cone behind it — the physical origin of a shock.

7. Shock wave — the abrupt squeeze

Two flavours matter here:

• Normal shock — the shock surface sits perpendicular (at $90°$) to the flow. Maximum squeeze, maximum loss.
• Oblique shock — the shock sits at an angle to the flow. Only the part of the velocity crossing the shock is squeezed; the rest slides along. Gentler, less loss.

The red line is the oblique shock. Notice the velocity arrow splits into a normal component (crosses the shock, gets compressed) and a tangential component (slides along, unchanged). This split is the key trick of §8. Foundations in Shock Wave Fundamentals.

8. The shock angles — $\beta$ and $\theta$

Two Greek letters describe an oblique shock's geometry:

In an over-expanded plume, the ambient air bends the exhaust inward, so $\theta$ is the inward turn and $\beta$ is the angle of the shock rising from the nozzle lip. Using the upstream Mach $M_1$ from §7 (the flow entering the shock), the normal component of that Mach is

$M_{1n}=M_1\sin \beta$

Here the extra subscript $n$ means "normal component" — the part of $M_1$ pointing across the shock line. Why $\sin \beta$? Because $\sin \beta$ picks out exactly the fraction of the velocity that points across the shock line — the only part that gets compressed. The tangential part uses $\cos \beta$ and is untouched. (If $\beta =90°$, $\sin \beta =1$ and we recover a normal shock — every case is covered.)

9. The gas property — $\gamma$ (gamma)

You don't need to derive $\gamma$ here — just know it's a fixed property of the exhaust, plugged into every shock and expansion formula. It sets how much pressure rises for a given squeeze. It appears in the parent's oblique-shock relations and in the isentropic formulas of Isentropic Flow.

10. Stagnation pressure — $p_0$ and the efficiency $C_v$

A perfect (lossless) process keeps $p_0$ constant. A shock destroys some $p_0$ forever — that lost fraction is the efficiency you can never get back. Using the §7 shock subscripts, we write $p_{01}$ for the stagnation pressure upstream of a shock and $p_{02}$ for downstream, so the survival fraction across one shock is $p_{02}/p_{01}<1$.

Now the efficiency. We compare two exit velocities:

• $v_{e,\text{ideal}}$ = the velocity you would get with a perfectly lossless (no-shock) expansion.
• $v_{e,\text{actual}}$ = the velocity you really get after shocks eat some of it.

Their ratio is the velocity coefficient. And we compare two stagnation pressures:

• $p_{0,\text{chamber}}$ = the flow's full energy back in the combustion chamber (before any loss).
• $p_{0,\text{actual}}$ = the energy that survives all the way to the exit after the shocks.

Where does the square root come from? Kinetic energy per kilogram is $\frac{1}{2}{v}^2$, and the usable energy the flow carries is tracked by $p_0$. So energy $\propto v^2\propto p_0$. Taking the ratio of the actual case to the ideal case:

$\frac{v_{e,\text{actual}}^2}{v_{e,\text{ideal}}^2}=\frac{p_{0,\text{actual}}}{p_{0,\text{chamber}}}⟹\frac{v_{e,\text{actual}}}{v_{e,\text{ideal}}}=\sqrt{\frac{p_{0,\text{actual}}}{p_{0,\text{chamber}}}}$

The square root appears because we started from velocity squared and had to undo the square to recover a plain velocity ratio. Efficiency links to Nozzle Area Ratio and Altitude Compensation.

11. Area ratio — $\epsilon$

Before dividing, we need the bottom of the fraction.

Written as e.g. "$40:1$", it says the exit is 40 times wider in area than the narrowest point. Too large a $\epsilon$ for the current air pressure is exactly the definition of over-expansion, and it's why the same engine that shines in space over-expands on the launch pad. See Nozzle Area Ratio and Nozzle Flow Separation for what happens when the crush gets severe.

Prerequisite map — how these feed the topic

Read the map bottom-up: the boxes at the top are the raw symbols from §1–§11; the arrows show which ones combine into which idea, all funnelling into the final topic box.

• Left branch (§1–§5): pressure $p$, exit area $A_e$, mass flow $\dot{m}$, exit velocity $v_e$ all merge into the thrust equation $F$.
• Middle branch (§1): comparing $p_e$ against $p_a$ gives the pressure-match verdict, which becomes over-expansion when $p_e<p_a$.
• Right branch (§6–§9): Mach number $M$ (built on sound speed $a$) makes shock waves possible; with $\gamma$ they set the oblique-shock angles $\beta ,\theta$.
• Bottom join (§10): shocks cut stagnation pressure $p_0$, which sets efficiency $C_v$. Area ratio $\epsilon$ (§11) feeds over-expansion. Everything lands in the topic.

Equipment checklist

Test yourself — reveal only after answering aloud.