Converging-diverging (de Laval) nozzle — subsonic, supersonic flow: Visual walkthrough

We assume nothing but arithmetic and the idea that a picture can be measured. Every new symbol gets a plain-word meaning and a place on a drawing before it is used.

Step 0 — The picture we are describing

Before any algebra, look at what we are talking about.

Figure — Converging-diverging (de Laval) nozzle — subsonic, supersonic flow

We follow the gas as it slides from a fat slice to a thin slice. Three things can change: the width $A$ , the speed $V$ , the packing $ρ$ . The whole derivation is just bookkeeping on how these three trade off.

Step 1 — Conservation of mass: the quantity that never changes

WHAT. In steady flow, whatever mass enters the fat end each second must leave the thin end each second — no gas piles up, none vanishes. The mass passing any slice per second is called the mass flow rate, written $\overset{m}{˙}$ (the dot means "per second").

$\overset{m}{˙} = kg per m^{3} ρ \times m^{2} A \times m / s V = constant$

WHY it multiplies like that. In one second the gas sweeps out a plug of length $V$ (metres travelled) and cross-section $A$ . That plug's volume is $A \cdot V$ cubic metres, and each cubic metre holds $ρ$ kilograms — so the mass is $ρ A V$ . This is continuity.

PICTURE. Look at the two plugs below: the fat plug is short-and-wide, the thin plug is long-and-thin, but they contain the same number of pink dots (same mass). That is the whole meaning of " $\overset{m}{˙}$ constant."

Step 2 — Newton's law for a gas slug (the Euler equation)

WHAT. Now ask: what makes the gas speed up at all? A force. For a frictionless gas the only push is the pressure difference across a slug. First we must name pressure itself.

WHY pressure. If the pressure behind is higher than the pressure ahead, there's a net forward shove and the slug accelerates. To turn that into an equation we carefully write Newton's second law, $force = mass \times acceleration$ , for one thin slug of gas.

Formula Building each piece of

$F = ma$ Take a thin slug of face area $A$ and tiny length $d x$ .

Mass of the slug. Its volume is $A d x$ and each cubic metre holds $ρ$ kilograms, so its mass is $ρ A d x$ .
Net force on the slug. The back face is pushed forward by $p A$ ; the front face is pushed backward by $(p + d p) A$ . Adding them, the net force is $p A - (p + d p) A = - d p A$ .
Acceleration of the slug. Here is the key kinematic move. In steady flow the speed depends on position, $V (x)$ , not on time directly. As the slug moves forward by $d x$ its speed changes by $d V$ , and it covers that $d x$ in time $d t = d x / V$ . So its acceleration is $a_{slug} = \frac{d V}{d t} = \frac{d V}{d x / V} = V \frac{d V}{d x} .$ This is why the acceleration comes out as $V d V / d x$ and not simply $d V / d t$ : a slug in steady flow speeds up by moving into a faster region, so its acceleration is speed times the speed-gradient.

Now assemble $F = ma$ : $net force - d p A = mass ρ A d x \times acceleration V \frac{d V}{d x}$ The $A$ cancels on both sides and the $d x$ cancels inside the acceleration, leaving: $- d p = ρ V d V$

PICTURE. The slug below has pressure $p$ on its back face and $p + d p$ on its front. If $d p$ is negative (pressure drops going forward), the back push wins and the blue arrow lengthens — the gas accelerates.

Step 3 — The speed of sound enters (and why it must)

WHAT. We now have two equations, but three unknowns changing ( $ρ$ , $A$ , $V$ , $p$ — four really). We need one more link. It comes from asking: how are pressure changes and density changes related in this gas?

WHY this exact tool. A small pressure ripple travelling through a gas is a sound wave. The speed of that ripple, the speed of sound $a$ , is defined by exactly the ratio "how much pressure change per unit density change":

$a^2 \;=\; \left.\frac{dp}{d\rho}\right|_{\text{isentropic}}\qquad\Longrightarrow\qquad d\rho = \frac{dp}{a^2}\tag{3}$

PICTURE. The figure shows a squeeze: push the gas (raise $p$ ) and it packs tighter (raise $ρ$ ). A stiff gas (large $a$ ) barely compresses; a soft gas (small $a$ ) compresses a lot. The number $a^{2}$ is the slope of that squeeze curve, measured along the no-heat-loss (isentropic) path.

Step 4 — Fold the three equations together

WHAT. We combine (2) and (3) to write the density change in terms of the speed change, then substitute into the mass balance (1).

Step 4a — put (2) into (3). From (2), $d p = - ρ V d V$ . Drop that into (3): $d ρ = \frac{- ρ V d V}{a ^{2}} ⟹ \frac{d ρ}{ρ} = - \frac{V ^{2}}{a ^{2}} \frac{d V}{V}$

Step 4b — name the ratio. The clump $\frac{V ^{2}}{a ^{2}}$ is (speed / sound-speed) squared. That ratio has a name:

Step 4c — substitute into the mass balance (1). $from d ρ / ρ - M^{2} \frac{d V}{V} + \frac{d A}{A} + \frac{d V}{V} = 0$

Group the two $\frac{d V}{V}$ terms: $\frac{d A}{A} + (1 - M^{2}) \frac{d V}{V} = 0$

Move it across and flip the sign:

PICTURE. The figure below plots the multiplier $(M^{2} - 1)$ against $M$ . Notice it crosses zero exactly at $M = 1$ . To the left it's negative (area and speed move opposite ways); to the right it's positive (area and speed move the same way). That single sign flip is the entire secret of the nozzle.

Step 5 — Case A: subsonic gas ( $M < 1$ )

WHAT. When $M < 1$ , the factor $(M^{2} - 1)$ is negative. To make the gas accelerate we want $d V > 0$ , so $\frac{d V}{V} > 0$ . Then: $\frac{d A}{A} = negative (M^{2} - 1) \cdot positive \frac{d V}{V} < 0$ so $d A < 0$ : the area must shrink. Squeeze to speed up.

WHY it matches everyday life. For slow flow, density barely changes (the gas acts almost incompressible, like water). So keeping $ρ A V$ constant means: shrink $A$ , and $V$ must rise to fill the gap. This is the garden-hose intuition, and it is correct — but only below Mach 1.

PICTURE. Converging pipe, blue arrow lengthening as the walls close in.

Step 6 — Case B: supersonic gas ( $M > 1$ )

WHAT. When $M > 1$ , the factor $(M^{2} - 1)$ is positive. To accelerate ( $\frac{d V}{V} > 0$ ) we now need: $\frac{d A}{A} = positive (M^{2} - 1) \cdot positive \frac{d V}{V} > 0$ so $d A > 0$ : the area must grow. Widen to speed up. The opposite of everyday intuition.

WHY. Look back at Step 4b: $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V}$ . With $M > 1$ the density plunges faster than the speed rises. In the mass balance $ρ A V = const$ , if $ρ$ crashes then $A V$ must climb — and it turns out both $A$ and $V$ increase together. The thinning gas out-thins the widening pipe, and speed makes up the rest.

PICTURE. Diverging pipe: walls flare out, the pink dot-cloud thins dramatically, and the blue arrow grows longest of all.

Step 7 — The degenerate case: exactly sonic ( $M = 1$ )

WHAT. At $M = 1$ the factor $(M^{2} - 1) = 0$ exactly, so the law becomes: $\frac{d A}{A} = 0 \cdot \frac{d V}{V} = 0 ⟹ d A = 0$

WHY this pins the throat. $d A = 0$ means the area is neither growing nor shrinking — it is at a turning point, a local minimum. That location is the throat. Therefore the gas can hit exactly the speed of sound only at the throat, never in the middle of a converging or diverging run. This is why a de Laval nozzle must have a throat: it is the only door between the subsonic and supersonic worlds.

PICTURE. The hourglass throat: converging on the left, the pinch where $M = 1$ and $d A = 0$ , diverging on the right.

The one-picture summary

Everything above compresses into a single hourglass with the sign of $(M^{2} - 1)$ colouring each region. Read the summary figure left to right: the gas enters slow on the left in the blue converging cone (there $M < 1$ , and the factor $(M^{2} - 1)$ is negative, so shrinking the tube speeds it up); it reaches exactly the speed of sound at the yellow pinch in the middle (there $M = 1$ and the factor is zero, so the area must be at its minimum — this is the throat); and it races out supersonic on the right through the pink diverging cone (there $M > 1$ , the factor is positive, so widening the tube speeds it up). One equation governs all three regions, and only the sign of the bracket changes.

Recall Feynman retelling — the whole walkthrough in plain words

We followed one packet of gas sliding down a pipe. First we noticed the same amount of gas passes every slice each second — that's our unbreakable rule ( $ρ A V$ never changes). We wrote it as three fractional changes that must add to zero: width, speed, and packing. Then we asked what pushes the gas faster: a pressure drop. Falling downhill in pressure buys speed. Next we needed to know how much the gas packs tighter when pressure changes — and that "stiffness" number is literally the speed of sound squared. Fold those together and the packing change equals $- M^{2}$ times the speed change, where $M$ is just "how many times the speed of sound." Plug that back into the never-changing rule and out pops one clean equation: $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ . Below the speed of sound the bracket is negative, so squeezing the pipe speeds the gas up (the hose trick). Above the speed of sound the bracket flips positive, so you must widen the pipe to speed up — because the gas thins out faster than the pipe grows. And right at the speed of sound the bracket is zero, so the pipe can't be changing width there at all: that's the throat, the one and only doorway from slow to fast. An hourglass tube: squeeze in, hit Mach 1 at the pinch, then flare out to blast supersonic. That's a rocket nozzle.

Recall Quick self-check

Why must $M = 1$ occur at the throat and nowhere else? ::: Because at $M = 1$ the factor $(M^{2} - 1) = 0$ forces $\frac{d A}{A} = 0$ , i.e. the area is at a minimum (the throat). In subsonic flow, which way must area change to accelerate the gas? ::: It must shrink ( $d A < 0$ ), because $(M^{2} - 1) < 0$ . In supersonic flow, why does widening the pipe speed the gas up? ::: Because density falls faster than area grows, so to keep $ρ A V$ constant the speed must rise. What single equation links the physics of momentum, mass, and sound? ::: The Area–Velocity law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ .

Where this leads: the sign flip here is the reason for choking, the shapes of rocket and steam-turbine nozzles, and where a normal shock can stand in the diverging section.