The jet a cone sprays sideways

The duct that works backwards, the throat that stops listening and one area, two answers built the whole theory of a nozzle on one simplification: at each station along the duct there is one velocity, pointing along the axis, and its Mach number is fixed by the area there. Everything followed from that — the throat, choking, the two roots of the area ratio, the shock that stands between them.

The simplification is very good, and it hides one thing a rocket designer cannot ignore. A divergent section is not a duct with straight parallel walls. It widens, so its walls lean outwards, and the gas next to a leaning wall cannot be moving along the axis. The simplest divergent section is a cone, and the flow it produces is simple enough to compute exactly.

A source flow leaving a cone #

Once the gas is well past the throat, a conical section behaves as though every particle had come from a single point, the apex of the cone projected backwards. The streamlines are straight lines from that point, each leaving at its own angle between the axis and the wall, and by symmetry the flow’s speed, pressure and Mach number are the same everywhere on a sphere centred on it.

That is the one-dimensional theory again, with the stations bent. The area that matters at a distance $R$ from the apex is not a flat disc but a piece of a sphere, and it grows as $R^2$; the Mach number follows from that area through the same area–Mach relation, on its supersonic branch. What changes is only the direction the gas is moving: along the axis at the centre, at the wall’s angle at the edge, and at every angle in between.

The factor is a ratio of two areas #

The thrust of a nozzle is, by the momentum theorem applied to a control volume, the momentum flux leaving it along the axis plus the pressure pushing on its exit, less the ambient pressure on the same area. Take the control volume’s exit face to be the spherical cap through the lip. On it the flow is uniform and crosses the surface at right angles, which makes both terms easy to write down.

The mass leaving per second is the mass flux $\rho V$ times the cap’s own area. The axial momentum leaving is $\rho V^2$ times the axial component of each element’s direction, integrated over the cap — and the integral of the axial component of a surface’s normal over a surface is just that surface’s shadow on a plane perpendicular to the axis: the exit disc. So the momentum term is $\dot m V$ times

$$\lambda = \frac{\text{area of the exit disc}}{\text{area of the cap}} = \frac{\pi R^2\sin^2\alpha}{2\pi R^2(1-\cos\alpha)} = \frac{1+\cos\alpha}{2}.$$

The same argument for a two-dimensional wedge, whose cap is an arc, gives $\sin\alpha/\alpha$.

The formula has nothing in it but the angle. It does not contain the gas, the Mach number, the area ratio or the temperature, because it is a statement about the directions in which the streamlines leave and nothing else. For small angles the loss is $\alpha^2/4$ for a cone and $\alpha^2/6$ for a wedge — so halving a cone’s half-angle quarters its loss. A 15° cone throws away 1.7 per cent of its jet’s momentum, a 30° cone 6.7.

The momentum is not destroyed. Every element of the jet moving outwards on one side has a partner moving outwards on the other, so the sideways momentum cancels round the axis and no sideways force appears; the gas simply carries it away, as kinetic energy that pushes nothing. A jet deflected by a plate splits in a proportion with the same $(1 + \cos\beta)/2$ in it, for a related reason: in both, what decides the push along one direction is the projection of the flow’s direction onto it.

Checked across a plane where nothing is uniform #

The cap is a convenient surface because the flow on it is uniform. The flat exit plane, the one a test stand would put a pressure rake across, is not.

The centre of the exit plane is closer to the apex than its edge is, because the plane is flat and the spheres are not. So along the axis the gas has expanded less than it has at the lip, its Mach number is lower and its pressure higher, and at every radius its direction leans outwards by a different angle. For a 15° cone the Mach number rises from 3.877 on the axis to 3.925 at the lip; for a 30° cone from 3.762 to 3.961.

Integrating the axial momentum flux, $\gamma p M^2\cos^2\theta$, and the pressure over that plane is a calculation with no factor in it anywhere: at each radius it finds the distance to the apex, the local area ratio, the Mach number from the area–Mach relation and the flow’s angle, and adds up. It returns thrust coefficients of 1.812597 and 1.725060 in vacuum for the two cones — the cap’s values with the $(1+\cos\alpha)/2$ factor, to fourteen figures.

That agreement is not a coincidence of the numbers. The cap and the flat plane both pass through the lip, and between them they enclose a lens of gas that touches nothing else. In steady flow, whatever axial momentum enters that lens through the cap must leave through the plane. The factor derived on one surface is a consequence of conservation on every surface, and the calculation on the non-uniform one is the check that the derivation on the uniform one did not cheat.

Why a wedge loses less than a cone #

The two formulas differ for a reason that is visible on the cap. A wedge’s exit cap is an arc, and each degree of it carries the same share of the flow. A cone’s cap is a piece of a sphere, and each degree of it is a ring whose circumference grows with the angle from the axis — so a cone puts more of its gas at large angles, where the axial share is smallest, than a wedge of the same half-angle does. The outermost ring of a 30° cone’s cap carries half again as much gas per degree as the ring at 20°, and that ring is leaning further.

That is why a two-dimensional nozzle with flat diverging walls keeps 95.5 per cent of its momentum at 30° where a cone keeps 93.3, and it is one of the few ways in which a rectangular nozzle is better than a round one. It is also why the cone’s factor is the one that matters in practice: nearly every nozzle that has to produce thrust efficiently is round, because a round section has the least wall for its area.

Measured across the plane, the average misleads #

The non-uniform exit plane has a consequence for anyone testing a nozzle. A rake of pitot probes across the exit, reading the Mach number at each radius, sees the variation in the figure above: lower on the axis, higher at the lip. Averaging those readings and treating the jet as uniform at that average gives the wrong thrust, because the flow at each radius also leans outwards by a different amount, and neither the Mach number nor the direction is uniform.

The correct average weights each probe by the area of its ring and by the square of the cosine of the local flow angle, and it is exactly the integral the check above performs. A meter that samples a profile has the same difficulty in a pipe, for a velocity that varies across it; here the variation is smaller and the direction varies as well, which is the part a probe pointed along the axis cannot see at all.

Only the momentum pays #

The pressure term in the thrust is the exit pressure times the area it acts on, projected onto the axis — and the projection of the cap is the exit disc, the same disc the pressure acts on in the one-dimensional theory. So the cone’s shape reduces the momentum term and leaves the pressure term untouched.

That makes the thrust loss a little smaller than the momentum loss, by the share of the thrust the pressure supplies. In vacuum a nozzle of area ratio 4 gets almost 11 per cent of its thrust from the pressure on its exit and a nozzle of area ratio 25 about 5 per cent, so a 15° cone loses 1.52 per cent of its thrust at the first and 1.62 at the second, against the 1.70 per cent of momentum it sprays sideways in both. A nozzle that expands further leans more on momentum, and so pays more of its divergence loss.

The loss moves the curve, not its peak #

A nozzle working against an atmosphere has a best area ratio. Expanding the gas further than the ambient pressure lets the exit pressure fall below it, and the ambient then pushes back on an exit area that has grown; expanding it less leaves pressure unconverted into speed. The peak is where the exit pressure equals the ambient — the design condition one area, two answers described — and a search over area ratio finds it at 11.87 for a chamber at a hundred times the ambient pressure, with the exit pressure there matching the ambient to four figures.

The cones’ curves lie below the ideal one by close to their divergence losses at every area ratio, and their peaks sit at almost the same place. That is the practical consequence of the loss being a factor on one term: it does not change what the best expansion is, only how much the best expansion delivers. A designer chooses the area ratio from the pressures and the cone angle from the length, and the two choices barely interact.

The pressure term is where the two regimes either side of the peak live. A nozzle with too small an area ratio for its ambient leaves its gas above the ambient pressure, and the plume goes on expanding outside the nozzle, where the expansion pushes on nothing; one with too large an area ratio leaves its gas below the ambient, and the atmosphere pushes back on the exit. A rocket rising through the atmosphere passes through both, because the ambient falls while the nozzle stays the same, and its best area ratio at launch is far smaller than its best area ratio at altitude. The divergence factor sits on top of all of that unchanged — which is one reason the cone angle is chosen once, from length and mass, while the area ratio is argued over for the whole of a vehicle’s trajectory.

Length against the square of the angle #

The reason nobody simply uses a very narrow cone is the length it takes.

To reach an exit radius five times the throat’s, a cone of half-angle $\alpha$ needs a divergent section $(\sqrt{\varepsilon}-1)/\tan\alpha$ throat radii long. The loss goes as the square of the angle and the length as one over its tangent, so the product of the loss and the square of the length is nearly constant for small angles: halving the length roughly quadruples the loss. A 10° cone loses under one per cent and is nearly twenty-three throat radii long; a 30° cone is a third of that length and loses nine times as much.

Length costs mass, which in a rocket costs everything, and it costs wall — a long divergent section grows a thick boundary layer, and friction in a nozzle does more than slow the gas. The standard compromise is not a cone at all. A contoured bell turns the wall back towards the axis before the lip, so that the gas leaving near the wall is nearly axial, and recovers most of a narrow cone’s momentum in a length close to a wider cone’s. The cone remains the reference against which a bell is measured, and $(1+\cos\alpha)/2$ remains the number it is measured with.

What one and a half per cent is worth #

A loss of 1.7 per cent of the momentum sounds like the kind of correction that matters to an engineer’s report and to nobody else. For a rocket it is not.

A rocket’s performance is its exhaust speed, and its exhaust speed enters the rocket equation linearly: the velocity change a stage can deliver is the exhaust speed times the logarithm of its mass ratio. Take away 1.7 per cent of the momentum and, to the accuracy that the pressure term’s share allows, the stage delivers 1.7 per cent less velocity change for the same propellant. Reaching low orbit takes something over nine kilometres a second, and 1.7 per cent of that is about 160 metres a second — a velocity a launch vehicle has to buy back with propellant, which it has to carry, which it has to lift.

A 30° cone would cost four times as much. The nozzles on real launch vehicles are neither: they are contoured bells whose divergence loss is a fraction of a per cent, and the difference between them and a cone of the same length is paid for in the machining of the contour. That is a price the one-dimensional theory, with its uniform axial exit, gives no reason to pay.

What the source-flow model leaves out #

The throat region. Near the throat the flow is not a source flow: it has just turned from converging to diverging, the sonic line is curved, and the gas at the wall is moving at a different angle from the gas on the axis. The source flow is the flow the cone settles into some distance downstream.

The sonic area. The cap’s area ratio is taken relative to the geometric throat, as if the throat were exactly sonic and exactly flat. With friction it is neither, and with a curved throat the discharge is slightly less than the ideal.

The boundary layer. The wall’s layer thickens along the section, displacing the flow inwards and dragging on it. Both reduce the thrust further, and both grow with the section’s length, which is the other side of the length trade.

The gas. A rocket’s exhaust is a reacting mixture whose ratio of specific heats changes as it expands and cools. The factor does not care; the Mach numbers and pressures do.

Separation. A nozzle working at a much lower ambient pressure than it was designed for is fine; one working at a much higher one can let the flow separate from the wall inside the section, behind a shock, and then the exit plane is not full of nozzle flow at all. The jump the equations allow is where that begins.

A factor from the rocket-motor literature #

The factor was in the rocket-motor literature by the early 1940s, derived exactly as here from a source flow and a cap, and it has been the first correction in every account of nozzle performance since. What makes it worth deriving again is how little it depends on. A great deal of nozzle theory changes when the gas, the chamber or the expansion changes. This number is the one piece that is pure geometry — the shadow of a sphere on a plane — and it survives the calculation that tries hardest to break it.

Still open: the throat that is not the sonic point #

Every nozzle in this essay, and in the three before it, reaches Mach one at its throat. That is exact for a nozzle with frictionless walls, because the throat is the only place where the area stops changing, and it stops being exact the moment the wall drags on the gas. Friction, on its own, drives a flow in a constant-area pipe towards Mach one — the second way a flow can choke — and in a nozzle it competes with the widening that is trying to carry the flow past it.

That is friction moves the sonic point past the throat: the condition that puts the sonic point downstream, the saddle that the only smooth solution must pass through, and what a throat that is still subsonic does to the mass flow a choked nozzle can pass.