Viscosity lets a jet break, later and into bigger drops

Where a jet stops being a jet placed every liquid jet on a map of two numbers: the gas Weber number, which decides which of four break-up regimes it is in, and the Ohnesorge number, $\mathrm{Oh} = \mu/\sqrt{\rho \sigma d}$, which compares the liquid’s viscosity with its surface tension at the jet’s own size. It took the growth of the ripples that break the jet from Rayleigh’s inviscid analysis — the fastest ripple 4.51 diameters long, computed where the instability was, and drops of 1.89 jet diameters — and it noted that small nozzles are viscous because the Ohnesorge number grows as the nozzle shrinks. What it did not do was put the viscosity into the instability.

That gap matters because the commonest intuition about viscous jets is that viscosity stops them breaking. A thread of honey poured from a spoon falls a metre unbroken; a thread of water from a tap breaks into drops within a few centimetres. This essay puts the viscosity in, and finds that the intuition is right about the length and wrong about the reason.

What viscosity can and cannot do #

A round jet of radius $a$ is unstable to varicose ripples — necks and bulges — of wavenumber $k$ whenever the ripple is longer than the jet’s circumference, $ka < 1$: a rippled column of the same volume has less surface than a straight one, so surface tension pumps liquid from the necks into the bulges. That is Plateau’s geometry and Rayleigh’s mechanism, and it does not depend on the viscosity at all. Weber put the viscosity in as damping. In units of the capillary time $\sqrt{\rho a^3/\sigma}$, with $\xi = ka$ and the Ohnesorge number on the radius, the growth rate $\omega$ obeys

$$\omega^2 + 3\,\mathrm{Oh}\,\xi^2\,\omega = \xi\,\frac{I_1(\xi)}{I_0(\xi)}\,(1 - \xi^2).$$

The right-hand side is Rayleigh’s drive, exact for an inviscid jet. The second term on the left is the viscous resistance to the flow that moves liquid from neck to bulge, proportional to the square of the wavenumber because the flow’s gradients are. The equation’s structure says the whole story in advance: wherever the drive is positive, so is the growth rate, however large the damping. Viscosity changes how fast a ripple grows. It cannot make an unstable ripple stable.

Every ripple still grows #

The figure solves the relation for every ripple. At an Ohnesorge number of zero it is Rayleigh’s curve, fastest at $ka = 0.697$ with a rate of 0.343 per capillary time. At 0.1 it is lower everywhere and its peak has moved to longer ripples; at 1, much lower, fastest at 0.344; at 10, fastest at 0.123 with a rate of 0.0114 — thirty times slower than water’s. Every curve still reaches zero at exactly $ka = 1$, Plateau’s limit, and is positive below it.

The shift of the peak is the important part. The damping grows as $\xi^2$ and the drive, at long wavelength, also as $\xi^2$, so for long ripples both scale together and the damping can only slow them in proportion; for short ripples the drive weakens towards zero at $ka = 1$ while the damping keeps growing, so they lose most. The ripple that grows fastest is pushed to longer wavelengths.

The fastest ripple lengthens #

In jet diameters, the fastest ripple is 4.51 long for an inviscid jet and grows with the Ohnesorge number: 9.14 at one, 25.5 at ten. Weber’s long-wave approximation, which replaces Rayleigh’s ratio of Bessel functions with its small-argument form, gives this in closed form,

$$\frac{\lambda_{\rm fastest}}{d} = \pi\sqrt2\,\sqrt{1 + 3\,\mathrm{Oh}},$$

with $\mathrm{Oh}$ on the diameter. At zero viscosity it is 4.44 against the exact 4.51, and as the viscosity grows the two converge, to within three per cent at an Ohnesorge number of ten, because the fastest ripple is then long enough that the long-wave form is the exact one. The square root is the balance read off directly: the damping’s $\xi^2$ against the drive’s $\xi^2(1 - \xi^2)$ puts the fastest ripple where $\xi^2$ is about $1/\sqrt{\mathrm{Oh}}$ at large Ohnesorge number.

With no inertia there is no fastest ripple #

Rayleigh worked out the other extreme too, a thread so viscous that inertia can be dropped altogether, and it is worth seeing what the relation says there because it explains the square root. Without the $\omega^2$ term the growth rate is the drive divided by the damping, which at long wavelength is $(1 - \xi^2)/6\,\mathrm{Oh}$: largest for the longest ripple, and the same for every ripple long enough. In a world with no inertia a viscous thread has no preferred wavelength at all; it would break at the longest scale available to it. What selects a fastest ripple in a viscous jet is the inertia that was dropped — small, but present, and it penalises long ripples because they must move liquid far. Balancing it against the viscous damping is what puts the fastest ripple at $\xi^2 \sim 1/\sqrt{\mathrm{Oh}}$, so the square root in Weber’s formula is the trace of a term that the viscous limit says is negligible.

That is also why a very viscous thread’s break-up looks disordered: its growth-rate curve is nearly flat over a wide band of long ripples, so whichever of them the disturbance seeded most strongly wins, and the necks fall at irregular spacings. A thin jet’s sharp peak selects a spacing; a thick one’s broad plateau barely does.

The viscous time, not the capillary one #

The growth time tells why honey falls so far. An inviscid jet’s fastest ripple grows by a factor $e$ in 2.91 capillary times. Past an Ohnesorge number of about one the time grows in proportion to it — 88 capillary times at ten, 850 at a hundred — and a time proportional to $\mathrm{Oh}\sqrt{\rho a^3/\sigma}$ is $\mu a/\sigma$: the viscous time, in which surface tension drives liquid through a viscous resistance. The capillary time, in which surface tension accelerates liquid against its inertia, has dropped out. A viscous thread breaks on the time it takes surface tension to squeeze treacle, and that is a long time.

So the length of a falling thread is long because the time is long, and the thread is still breaking at the rate the relation gives — slowly, and at a long wavelength. A honey thread, given enough fall, beads and breaks; it usually lands first. A fibre coated with oil showed the same thing for a film: it beads after a number of its own viscous times, and whether it is smooth over a length depends on how fast it moves through that length, not on whether it is unstable.

Bigger drops #

If each fastest wavelength of jet becomes one drop, the drop’s volume is the wavelength’s: $\pi a^2\lambda = \pi D^3/6$, and its diameter in jet diameters is $(12\pi/ka)^{1/3}/2$. Inviscid that is Rayleigh’s 1.89. At an Ohnesorge number of one, 2.39; at ten, 3.37; at a hundred, 4.92. The longer wavelength puts more liquid in each drop, and the drop’s diameter grows as the cube root of the wavelength — so as the sixth root of the Ohnesorge number at large values. A hundredfold increase in viscosity buys only a doubling of drop size.

That weak dependence is useful. Ink-jet and spray-drying nozzles work with liquids whose viscosity varies by factors of ten between batches, and the drop size, which is what the process cares about, moves by under fifty per cent. The break-up length, which is set by the time and moves in proportion to the viscosity, is the quantity that changes a great deal — and it is the one an operator sees first.

Six liquids, one nozzle #

From a one-millimetre nozzle, the thin liquids are indistinguishable. Mercury, water and ethanol have Ohnesorge numbers of a few thousandths, and all three break at Rayleigh’s 4.5 millimetres — mercury’s much higher surface tension and density cancel in a number that has neither alone. A half-and-half mixture of glycerol and water, six times more viscous than water, is still effectively inviscid at this size. Pure glycerol, at an Ohnesorge number of five, breaks at 18.3 millimetres, and honey at about forty, at 48.8.

The Ohnesorge number also says how size changes a liquid. It grows as one over the square root of the diameter, so the same glycerol through a ten-micrometre ink-jet nozzle has an Ohnesorge number of fifty and behaves like honey from a spoon. The thin-liquid end of the figure is a statement about millimetre jets, not about thin liquids. A liquid is thin or thick only at a stated size, which is what a dimensionless number is for.

An ink-jet printer, tuned to the fastest ripple #

The relation’s most precise everyday use is in continuous ink-jet printing, where a jet is deliberately broken into a stream of equal drops by vibrating the nozzle at the frequency of the fastest ripple. A twenty-micrometre jet of an ink three times as viscous as water, with a surface tension of 0.035 newtons a metre, has an Ohnesorge number of 0.11: thin, but not negligibly so. Its fastest ripple is 5.25 jet diameters long — longer than Rayleigh’s 4.51 by the viscosity’s share — and at ten metres a second that wavelength passes the nozzle 95,000 times a second, which is where such printers are driven. Each drop is 1.99 jet diameters across and the ripple grows by a factor $e$ in 21 microseconds, so the jet breaks two and a half millimetres from the nozzle. Driving at Rayleigh’s inviscid wavelength instead would ask for a ripple 14 per cent shorter than the one the ink prefers, and the drops would come out uneven — a drop is not a tear, but whether a stream of them is uniform is decided by this number.

At the far end of the scale the instability becomes slow enough to work with. A millimetre thread of molten glass at a thousand pascal-seconds has an Ohnesorge number above a thousand; its fastest ripple is 270 diameters long and grows by a factor $e$ in ten seconds. A glass-blower or a fibre-drawing tower has all the time it needs to shape such a thread before surface tension has done anything visible — the thread is unstable, and the instability is simply too slow to matter on the time of the process. The same thread, left hanging, beads into drops more than seven of its diameters across.

How far a jet gets #

The break-up length is the distance the jet travels while its fastest ripple grows from whatever disturbance seeded it to the jet’s own radius. Taking that growth as a factor $e^{12}$ — the usual estimate, from about six millionths of the radius — the length in diameters is the growth time times the speed: at two metres a second a water jet breaks after 93 diameters, nine centimetres, and a glycerol jet after 1,720, nearly two metres. Both grow in proportion to the speed in this model, because the growth time is a property of the liquid and the jet simply travels further in it.

That linear growth is what the first essay’s break-up length found for a water jet in the Rayleigh regime, and it fails at higher speed for a reason this calculation has left out on purpose: the air. The next essay puts it in.

How the rate has been measured #

The growth rate is not only a prediction; it has been measured ripple by ripple. Donnelly and Glaberson in 1966 drove jets of water and of water–glycerol mixtures at chosen frequencies, photographed the ripples as they grew, and read the rate off the photographs: the measured rates followed the inviscid curve for water and fell below it, by the viscous damping, for the glycerol mixtures, with the peak moving to longer wavelengths as this relation says. The forced jet is the cleanest test there is of a linear instability, because the experimenter chooses the wavenumber and the jet only reports how fast it grows.

The same measurement also says where the relation stops being enough. Once the ripple’s amplitude is a sizeable fraction of the radius the necks thin faster than the bulges swell, the growth is no longer exponential, and the drop that forms is not exactly one wavelength’s volume — a satellite takes a few per cent of it. The drop sizes above are the main drops, and the satellites are what a printer’s designer works hardest to suppress, by driving the ripple’s amplitude and shape rather than only its wavelength.

What was checked #

The modified Bessel functions are computed from their power series and, for $K$, from its integral representation, and agree with tabulated values at one to two parts in $10^{15}$. With no viscosity the relation returns Rayleigh’s fastest ripple, $ka = 0.69702$, and his drop of 1.8909 diameters. With the drive replaced by its long-wave form, a direct search over two hundred thousand wavenumbers returns Weber’s closed-form fastest wavelength to five parts in a million at Ohnesorge numbers of 0, 0.1, 1 and 10 — the relation’s numerics checked against the one closed form it has.

The convention: Ohnesorge on the diameter, rates on the capillary time #

The Ohnesorge number in the figures is on the jet’s diameter, $\mu/\sqrt{\rho\sigma d}$, as the regime map uses; the relation is written with it on the radius, which is larger by $\sqrt2$. Rates are in units of the capillary time $\sqrt{\rho a^3/\sigma}$, which is a quarter of a millisecond for a millimetre water jet. Drops are diameters of a sphere holding one fastest wavelength of jet. The surrounding gas is absent.

What the picture cannot show #

The relation is linear: it gives the rate at which small ripples grow, not what happens when a neck is thin. A viscous jet does not pinch where the linear theory says; its necks stretch into long threads that pinch at their ends and leave satellite drops between the main ones, and for very viscous liquids those threads can be as large as the drops. One drop per wavelength is the linear theory’s estimate and the main drops’ size; the satellites are extra. Weber’s damping term is exact only for long ripples, which is why the full relation and the closed form differ by a few per cent at small Ohnesorge numbers; Chandrasekhar’s exact viscous relation removes that and changes no conclusion here. And the jet is Newtonian: a viscosity that depends on the question does something quite different, and polymer solutions form beads on a string that this relation does not describe, because an elastic liquid resists the thinning of a neck in a way no viscosity does.

Who found it, and when #

Plateau measured the instability’s geometry in the 1870s and Rayleigh explained it in 1878, for an inviscid jet and for a very viscous one separately. Weber’s paper of 1931 joined the two with the damping term used here and gave the closed form for the fastest wavelength, and it is still the formula engineers use. Chandrasekhar gave the exact viscous dispersion relation in 1961. The drop-size rule — one drop per fastest wavelength — is Rayleigh’s, and the sixth-root dependence on viscosity follows from Weber’s wavelength.

Still open: what the air does to it #

Every number here is for a jet in a vacuum, or in a gas so thin or slow that it does nothing. A real jet moves through air, and the air flowing over a rippled surface is faster over the crests and its pressure lower there, which pulls them out. That pull grows with the gas’s density and the square of the speed, and it acts on short ripples as well as long ones, which surface tension alone would hold flat. The next calculation adds it to the same relation and asks how far it moves the fastest ripple, whether it explains where the measured Rayleigh regime ends, and what it does to the drops.