The duct that works backwards

Put a thumb over the end of a hose and the water comes out faster. Narrow a duct and the flow through it accelerates. This is continuity, it is correct, and it is one of the few pieces of fluid mechanics that everybody has already verified personally.

It is also only half of the rule, and the other half is the reverse.

Where the reversal comes from #

Continuity for a steady flow in a duct says that $\rho u A$ is constant. Differentiating logarithmically,

$$\frac{\mathrm{d}\rho}{\rho} + \frac{\mathrm{d}u}{u} + \frac{\mathrm{d}A}{A} = 0$$

Three terms, and the incompressible case is the one where the first is zero. Then area and speed trade directly, and the hose behaves as expected.

Now allow the density to move. The momentum equation for a frictionless flow gives $\mathrm{d}p = -\rho u \,\mathrm{d}u$, and the definition of the speed of sound gives $\mathrm{d}p = a^2 \mathrm{d}\rho$ along an isentrope. Putting them together,

$$\frac{\mathrm{d}\rho}{\rho} = -M^2 \frac{\mathrm{d}u}{u}$$

which is the whole content of compressible duct flow in one line. Accelerating a gas thins it, and the fractional thinning is $M^2$ times the fractional acceleration.

Substituting back into continuity:

$$\frac{\mathrm{d}A}{A} = (M^2 - 1)\frac{\mathrm{d}u}{u}$$

Below Mach one the bracket is negative and area and speed move oppositely. Above it the bracket is positive and they move together.

The physical reading, which is better than the algebra #

The sign change is easy to derive and easy to distrust, so it is worth stating in words.

At any Mach number, accelerating the flow lowers its density. That means the same mass needs more room per unit of length than the bare speed increase would suggest, so the duct does not have to narrow as much as the incompressible rule says.

At Mach one, those two effects exactly cancel: the density falls in precisely the proportion the speed rises, so the area does not need to change at all. That is what “sonic” means geometrically.

Above Mach one the density is falling faster than the speed is rising, so the flow needs progressively more room as it accelerates. To keep accelerating it, the duct must widen. A supersonic stream is a thing that gets thinner faster than it gets faster.

The minimum is exactly at Mach one #

The differential statement integrates into a single curve, and the curve is the working tool of the subject.

Written out,

$$\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma+1}\left(1 + \frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$$

which is not an expression anybody derives twice, and which has one property worth verifying rather than trusting: its minimum is at $M = 1$ and its value there is exactly 1.

That is asserted rather than assumed here. assertAreaMachHasMinimumAtOne walks a tabulated sweep, refuses any row where $A/A^*$ dips below one, locates the minimum, and requires it to sit at Mach one to within a fiftieth. It then evaluates the relation at exactly Mach one and requires exactly one, to $10^{-9}$. The rejection test hands it a table whose minimum is 0.98 — a stream tube narrower than the throat it would be sonic in, which is geometrically impossible and arithmetically unremarkable — and requires the refusal.

The converging–diverging nozzle, which is the consequence #

Accelerating a gas from rest to supersonic speed therefore needs a duct of a specific shape, and only that shape will do.

The flow starts subsonic, so the duct must converge to accelerate it. It cannot become supersonic while still converging, because that would mean passing through Mach one at a point where the area is still changing — and at Mach one the area must be stationary. So the flow reaches Mach one exactly at the narrowest point. Past that point, further acceleration requires widening.

Converge, reach Mach one at the throat, then diverge. That is the de Laval nozzle, and its shape is not a design choice: it is the only shape that gets a gas from rest to supersonic in a duct.

That number, 0.528, is $(2/(\gamma+1))^{\gamma/(\gamma-1)}$: the fraction of its stagnation pressure a gas has left when it reaches Mach one. It turns up here as the pressure ratio at which a nozzle chokes and it turned up earlier as a point on a state-ratio curve, and the two are the same fact approached from different sides — which is the next rung.

Everything with a supersonic exhaust has this shape, and every one of them was arrived at by the same argument: rocket engines, supersonic wind tunnels, steam turbine nozzles, and the ejector on a laboratory vacuum line.

What happens if the duct is only convergent #

A purely convergent nozzle cannot produce supersonic flow at all, however hard it is driven, and this is not a limitation of any particular design.

The exit is the narrowest point of a convergent duct, so it is the point at which the flow is fastest. If the flow there is subsonic, the duct is simply a subsonic accelerator. If it reaches Mach one there, the relation says the area must be stationary — which it is, at the exit, in the sense that the duct ends. Beyond the exit there is no duct to widen.

So a convergent nozzle can reach Mach one at its exit and no further, and that is exactly what happens when it is driven hard enough. Everything downstream stops mattering at that point, which is the next rung.

The exchange this rests on #

The area relation is a consequence of the energy exchange described in the previous essay, and the connection is worth making explicit because the area relation looks purely geometric.

Density falls because temperature falls, and temperature falls because the gas is spending its enthalpy to buy speed. So the reversal of the duct rule is thermodynamics wearing a geometric disguise: nothing about the shape of the pipe knows anything, and the sign change comes from a gas running out of heat.

The incompressible case is the same relation with a term missing #

It is worth putting the familiar rule beside the general one, because the familiar rule is not an approximation to this — it is this, with $M^2$ set to zero.

Set $M = 0$ and the area relation becomes $\mathrm{d}A/A = -\mathrm{d}u/u$, which integrates to $uA = \text{constant}$: the incompressible statement, with no density in it because nothing is happening to the density. The whole of the surprise in this essay is one term with an $M^2$ in front of it.

That is a pattern worth noticing across this field. Compressible results very rarely look like corrections applied on top of incompressible ones. They look like the general result, of which the familiar one is a limit — and the general result frequently has different signs available to it, which no amount of correcting a limiting case would ever produce.

A stream tube does not need walls #

Nothing in the derivation used a wall. Continuity, momentum and the speed of sound were applied to a stream tube, and a stream tube is a surface no fluid crosses whether or not it happens to coincide with a pipe.

So the same reversal applies in a free flow, and it explains something about supersonic aerofoils that is otherwise baffling. Over a convex surface, the stream tubes next to the body widen. Subsonic flow slows down there and its pressure rises — which is why a subsonic aerofoil’s suction peak is on the part with the most curvature and the flow accelerates round the shoulder as the tubes narrow. Supersonic flow does the opposite: over an expanding corner it accelerates and its pressure falls.

That is the physical content of the expansion fan, arriving here as a consequence of the same relation, and it is why the pressure distribution on a supersonic section bears no resemblance to the one on the same section flown slowly.

Why the sonic point must be at the throat, not merely can be #

A subtlety here is often stated too weakly. It is usually said that the flow can reach Mach one at the throat. The relation says something stronger.

Rearranged, $\mathrm{d}u/u = \frac{1}{M^2-1}\,\mathrm{d}A/A$. At $M = 1$ the coefficient is infinite, so unless $\mathrm{d}A = 0$ the acceleration is unbounded — which is not a physical solution. So sonic conditions can occur only where the area is stationary.

The converse is not true, and this is the part that gets lost: a throat is not obliged to be sonic. A lightly driven convergent–divergent nozzle has a subsonic throat and behaves as a venturi, speeding the flow up to the throat and slowing it down again after it. Whether the throat goes sonic depends on the pressure ratio imposed across the whole device, and that is a question the local relation cannot answer.

What the solver computes, and how it is checked #

The duct in the figures is a real area distribution — a smooth contraction to a throat at 42 per cent of the length, then an expansion to the stated area ratio — and the Mach number at every station is found by inverting $A/A^*$ at that station’s area.

That inversion is done by bisection rather than by formula, because the relation does not invert in closed form, and the bracket is chosen to sit entirely on one side of the minimum. The subsonic root is sought in $(0, 1)$ and the supersonic root in $(1, 60)$, since a Newton step near the minimum would be useless: the derivative vanishes there, which is the same fact as the throat being a throat.

The check that matters is assertBothRootsGiveTheArea. It takes an area ratio, finds both roots, requires one on each side of Mach one, and then evaluates the relation at each root and requires the original ratio back to $10^{-9}$. A bisection that has converged to the wrong branch, or converged to the minimum because the bracket straddled it, fails that immediately.

What an area ratio buys, in numbers #

The relation is monotonic on the supersonic branch, so an exit Mach number is bought with an area ratio and nothing else — not with pressure, not with temperature, not with the size of the device.

An area ratio of 1.69 gives Mach 2. Ratio 4.23 gives Mach 3. Ratio 10.72 gives Mach 4, and 25.0 gives Mach 5. The curve is steepening: each further Mach number costs more area than the last, which is the saturation of the energy exchange showing up as geometry.

Two consequences follow immediately, and both are visible on real hardware.

High-altitude rocket nozzles are enormous. A vacuum-optimised upper stage wants an exit pressure near zero, so it wants a large exit Mach number, so it wants an area ratio in the hundreds. The bell on such an engine is mostly empty volume being carried to orbit, and the design trade is against its mass rather than against anything aerodynamic.

Supersonic wind tunnels change nozzles rather than settings. Since exit Mach number is set by area ratio alone, running a tunnel at a different Mach number means physically swapping the nozzle block or moving a flexible wall. There is no throttle for it.

A throat need not be made of metal #

Area is not the only thing that can drive a duct flow, and the other two drivers turn out to have the same singular factor in front of them — which generalises the whole essay.

Run the same logarithmic differentiation with friction along the wall instead of a change of area, or with heat added to the gas instead, and in each case the acceleration comes out as the driving term divided by $M^2 - 1$. The bracket is the same bracket. So each of the three drivers reverses at Mach one, and each of them can reach it.

The consequences are worth having because they are counter-intuitive in the same way the flare after a throat is.

Friction accelerates a subsonic flow. Rubbing a gas along a wall costs stagnation pressure, and in a constant-area pipe the only way to shed it is to speed up. A supersonic flow in the same pipe slows down. Friction therefore pushes the flow towards Mach one from either side, and a pipe of constant bore that is simply long enough will choke on friction alone — a limiting length exists past which the imposed mass flow cannot be passed, and lengthening the pipe reduces it. No throat was built.

Heating does the same thing. Add heat to a subsonic stream and it accelerates; add it to a supersonic one and it decelerates; cool either and it moves away from Mach one. So a combustor has a ceiling on how much heat it can take at a given mass flow, and reaching it is thermal choking — the reason a ramjet’s performance stops improving with fuel, and a large part of the reason a scramjet keeps its combustion supersonic instead.

The sonic condition, in other words, is not a place in a duct. It is the state at which every driver this flow has changes sign, and geometry is only the most visible way of arriving there.

Where the model stops #

Three assumptions, and one of them is doing more work than it appears to.

The flow is quasi-one-dimensional. Every quantity is taken as uniform across the duct at each station, which requires the area to change slowly. A duct that flares abruptly has a genuinely two-dimensional flow inside it and this relation describes none of it. Real supersonic nozzles are contoured by the method of characteristics for exactly that reason: a conical flare produces exit flow that is not parallel, and the divergence costs thrust.

The flow is isentropic. No friction, no shocks, no heat transfer. The section above says what friction and heating do to the same relation; neither is computed here, and no figure on this page contains either.

The walls are the only boundary. Boundary layers grow along the duct and displace the effective area inwards. In a small nozzle at low Reynolds number that displacement is a significant fraction of the throat area, so the effective throat is smaller than the geometric one and the mass flow is below the ideal prediction. Nozzle discharge coefficients exist to absorb precisely this, and they are typically 0.97 to 0.99 — which means the ideal theory here is good to a few per cent and no better.

What the picture cannot show #

The four panels in the first figure carry arrows, and the arrows are the misleading part.

They show the direction of the change and say nothing about its size, and the size is exactly what the relation is about — $\mathrm{d}A/A = (M^2-1)\,\mathrm{d}u/u$ has a coefficient in it that goes to zero at Mach one and to $M^2$ far above it. A duct at Mach 5 needs an enormous flare for a modest further acceleration, and a duct at Mach 1.05 needs almost none. None of that is in the panels, which is why the pressure curves along a real duct appear in the figures beside them.

The panels are also steady-state statements. Nothing about how a nozzle starts is drawn here, and starting is a genuinely unsteady process in which a shock sweeps down the divergent section and out of the exit.

Who found it, and when #

Gustaf de Laval built the convergent–divergent nozzle for a steam turbine in about 1888, and used it successfully before the theory of it existed. The shape was arrived at experimentally: he was trying to get more velocity out of steam and found that adding a flare after the throat helped, which on the incompressible reasoning available to him should have been the wrong thing to do.

The explanation followed rather than led. Osborne Reynolds gave an account in the 1880s, and the full quasi-one-dimensional theory was assembled through the 1890s and 1900s alongside the thermodynamics it needs. It is a good example of a piece of engineering running ahead of its own subject — and of what happens when a discipline’s central intuition, “narrow it to speed it up”, is true in the range where the intuition was formed and false outside it.

Where the ladder goes next #

The throat is where the flow goes sonic, and something remarkable happens when it does: the duct stops responding to what is downstream of it. Lowering the back pressure further changes nothing — not the mass flow, not the pressure anywhere upstream, not a single thing in the convergent half. That is choking, and it follows directly from the fact that signals cannot travel upstream through sonic flow.