A condensing film throttles itself by a fourth root

The number that is an answer set out what kind of number a Nusselt number is. It is not a property of a fluid or a parameter of a problem. It is the result, the heat transfer written without dimensions, and the job of the theory is to find the function that gives it. In nearly every practical case the function is a correlation, measured and then fitted. So it is worth having one case in which the number can be derived entirely, on one page, with nothing fitted. Wilhelm Nusselt himself provided it, and the essay that bears his name in most heat-transfer texts has this at its centre.

Hold a cold vertical wall in steam. Steam condenses on it. On a clean metal the condensate wets the surface and runs down as a thin continuous film, and that film is the only thing between the steam and the wall: the steam side has very little resistance to heat, because steam is at its saturation temperature right up to the water’s surface, and it condenses on arrival. So all the temperature difference falls across the film. The heat crosses the film by conduction, since the film is thin and it is slow.

So the wall makes its own insulation out of what it is condensing, and there’s a neat feedback, in which the film at each height depends on all the condensation above it.

Three lines #

Take a film of thickness $\delta(x)$ at distance $x$ down the wall. It is thin and slow enough to be viscous, with gravity balanced by shear:

$$\mu\frac{d^2u}{dy^2}=-\rho g,\qquad \Gamma=\int_0^\delta\rho u\,dy=\frac{\rho^2g\delta^3}{3\mu}.$$

That’s the same falling film as anywhere, and its flow per unit width goes as the cube of its thickness. A lubrication film has the same cube, for the same reason.

Heat crosses by conduction, at a rate $k\Delta T/\delta$ per unit area, and each joule condenses $1/L$ of a kilogram:

$$\frac{d\Gamma}{dx}=\frac{k\Delta T}{\delta L}.$$

Substituting the first in the second gives $\delta^3\,d\delta=\frac{\mu k\Delta T}{\rho^2gL}\,dx$, which integrates at once:

$$\delta^4=\frac{4\mu k\Delta T\,x}{\rho^2gL}.$$

There isn’t anything else to do. That is the whole theory. The local heat-transfer coefficient is $h=k/\delta$, and because it goes as $x^{-1/4}$ the mean over a height $H$ is exactly 4/3 of its value at the bottom:

$$\bar h=0.943\left(\frac{\rho^2gLk^3}{\mu\,\Delta T\,H}\right)^{1/4}.$$

The coefficient 0.943 isn’t fitted: it’s $\frac43/\sqrt[4]{4}$. Everything in it can be measured with a table of the properties of water.

The fourth root is the signature of a film that supplies itself, and the identical power appears in a completely different setting, in a Leidenfrost drop’s vapour film. There vapour is produced beneath a drop by conduction across the film, and it has to flow out sideways. Here liquid is produced beside a wall by conduction across the film, and it has to flow out downward. In each, flow goes as a cube of the thickness and the supply as one over it. One is a film of vapour keeping a liquid off a hot wall, the other a film of liquid keeping a vapour off a cold wall. A coincidence of form, with a real reason.

The number, and what it is a function of #

Written the way the earlier essay on this number would want it, the mean over a wall is

$$\overline{Nu}=\frac{\bar hH}{k}=0.943\left(\frac{\rho^2gLH^3}{\mu k\,\Delta T}\right)^{1/4},$$

and the group in the bracket is a Galilei number, $gH^3/\nu^2$, times a Prandtl number, over a Jakob number, $c_p\Delta T/L$. The Jakob number is what is unusual in that list. It’s the ratio of the heat a film has to shed by cooling to the heat it sheds by condensing, and because latent heats are large it’s small: 0.019 for a drop of 10 K in steam. It sits in the denominator, and that is the reason condensing heat transfer is so strong. On a half-metre wall in steam at that drop, the Nusselt number is 5700, which the same wall could only reach in a turbulent single-phase stream at a substantial speed. And nothing is flowing there but a film 117 micrometres thick, moving at fifteen centimetres a second.

Steam is at one end of what’s possible, mainly because of water’s conductivity and latent heat. Put a refrigerant such as R134a on the same wall at the same drop. Its film is almost as thin, 104 micrometres. But its conductivity is less than an eighth of water’s, its latent heat is about a thirteenth, and the mean coefficient comes out at 1030 $\mathrm{W/m^2K}$, against 7750. That’s the reason a domestic refrigerator’s condenser and an air conditioner’s are built with such careful attention to the outside and to the condensing side’s geometry, while a power station’s steam side can be left nearly alone, with the worst resistance in the cooling water’s boundary layer instead. Comparing the two figures takes one line and no tests, and that’s the value of having a formula with no fitted constant: it transfers.

A last reading of the group concerns scale. The whole fourth-root family goes as $H^{-1/4}$, so the coefficient hardly cares about the size of the apparatus. A wall sixteen times as high has half the coefficient. Any correlation that was fitted on a laboratory tube may therefore be carried up to an industrial one with unusual confidence, as long as the film stays in the same regime, and the same dimensional argument about what a flow can know puts a limit on it only when viscosity or the wave pattern changes.

A larger drive and a worse coefficient #

The formula has one consequence that surprises almost anyone reading it for the first time.

The coefficient falls as the temperature difference rises. That is the opposite of what happens at a boiling wall, in which a greater drop drives more bubbles and improves the coefficient very steeply, and of natural convection, where a greater drop drives a faster flow. Here, more drop condenses more liquid, the film thickens and the conductance falls. The flux still rises, because the difference is multiplied in, but only as $\Delta T^{3/4}$.

That sets how a condenser should be run. Doubling a temperature difference doesn’t double the duty. It gives $2^{3/4}=1.68$ times, so a power station’s condenser, where every kelvin of temperature difference is taken directly from the turbine’s expansion and so from the output, is designed for a small difference on a large area. The arithmetic is unforgiving. Since the coefficient improves as the difference falls, the penalty for adding area is partly offset, and it’s a rare case of a design getting a little better at the same time as it gets more economical.

The dependence on $H$ is the same fourth root, and it says the top of the wall does most of the work. The first tenth of the wall’s height has a thin film, and it provides $0.1^{3/4}=18$ per cent of the heat; the top half provides $0.5^{3/4}=59$ per cent. If only the top of a tall wall matters so much, it would be better to strip the film off every so often. That’s exactly what a vertical fluted tube does: its flutes draw the film into the grooves by surface tension, leaving the crests almost bare, with a much smaller thickness and a coefficient several times greater, and it’s why that surface is so often found in desalination plants.

A tube is a short wall #

The most practically useful comparison the formula gives is the shape of the surface.

A horizontal tube has the same balance with a varying component of gravity along the film, and an integral around the half-circle changes 0.943 to 0.729. So comparing the two forms, a tube of diameter $D$ is the same as a wall $(0.943/0.729)^4=2.8$ diameters tall. A 20 millimetre tube therefore has the coefficient of a wall 56 millimetres high. That’s why condensers are built of horizontal tubes rather than vertical plates, and a half-metre wall has only 58 per cent of the coefficient of the tube.

Nusselt also considered a stack of $N$ tubes one above the other, with each one’s condensate falling on to the next. Its mean coefficient is then that of a single tube multiplied by $N^{-1/4}$: a rule so simple that designers have used it for a century, though real tubes drip and splash, and so do better than this. The bottom of a stack isn’t as badly off as the rule makes out, since a falling drop disturbs the film it lands on.

The fifth Nusselt could not see #

Compare the law to measurements and it turns out to be too low.

For about the whole of the working range of a condenser, measured coefficients on a clean wall lie about a fifth above the theory, even where the film is plainly laminar. Nothing about the balance has been done wrongly. What’s missing is that the film isn’t smooth. A falling liquid film at a very low Reynolds number is unstable, and it develops waves. Pyotr Kapitsa measured them in 1948, with his son Sergei. Benjamin and Yih showed in 1957 that an unstretched film on a vertical wall has no critical Reynolds number at all. It’s unstable at any flow. A film on a wall tilted from vertical has a threshold, but one on a vertical wall doesn’t, and waves arise at once with a long wavelength, which steepen and are shaped by surface tension.

Waves make the mean film thinner for the same flow. Since flux per width goes as $\delta^3$, a wavy film has a larger flow in its crests and a thinner substrate between them. That’s the reason for the enhancement, in two parts. A thinner film between crests conducts better, and the flow carried by the crests is more than their share of thickness. So a film of the same flow rate has a lower mean thickness: Kapitsa’s own calculation gave a 7 per cent reduction in mean thickness, and more elaborate treatments get higher. It’s rather a good example of a picture lying by being too tidy: a flat film would look right, and it would be wrong by a fifth.

Designers have long multiplied Nusselt’s result by 1.2 to account for that, and it was McAdams who put that into the textbooks. A Reynolds number near 1800 marks the onset of turbulence in a film, and from then on the coefficient rises instead of falling, as turbulent mixing makes the thickness less of a barrier. So a long enough wall has a minimum of the coefficient at the point of transition, with both ends better.

Waves against a calm wall #

The other large departure from the law is a mode the theory doesn’t include. If a wall’s surface isn’t wetted by the condensate, water gathers in drops rather than in a film, and each drop grows, joins its neighbours and rolls off, sweeping the wall bare for fresh nucleation. Dropwise condensation, first described by Schmidt and his colleagues in 1930, has a coefficient five to ten times that of filmwise condensation at the same conditions, because the wall between the drops has nearly no film at all. It’s among the highest heat-transfer rates available in any ordinary process. Getting it to last is the difficulty. A coating thin enough not to add a resistance of its own wears off in months in steam, so most condensers run filmwise and pay for the fourth root.

That also explains why film condensation has been studied so thoroughly. Whichever the surface, the same balance holds at the first appearance of any continuous film. Non-condensable gas is a separate hazard, and a large one: a per cent of air in a condenser gathers at the water’s surface, since steam carries it there and only the steam condenses, and it forms a diffusion barrier that can halve the coefficient. Hence the air ejectors on every power station condenser. Nothing in the formula contains that, and yet in practice it can matter more than anything else in it.

How much a thin layer is carrying #

It’s worth a moment to see the size of the thing being described, since it shows why a boundary-layer intuition is misleading here. At a metre down a wall in steam at a drop of 10 K, each square metre passes about 65 kilowatts through a film thinner than two sheets of paper. The film’s own Reynolds number, a few hundred, means it would be regarded as a slow viscous flow in any other setting. It has no boundary layer in the sense of an outer stream that a wall slows: the wall drags on a film that’s driven by gravity throughout, with the fastest fluid at its free surface, where the vapour is. Temperature falls linearly across it, because the film’s Péclet number across its own thickness is tiny. That’s the whole reason for the simple form: the temperature doesn’t need to be solved for, since at that thickness conduction wins outright.

The surface that moves with the flow records the Marangoni stresses a free surface with a temperature gradient can carry, and in the case of a mixture they’d matter here too. A surface of a pure vapour’s condensate has a constant temperature along it, so there’s no such stress, and the falling film’s surface is shear-free. That’s a piece of luck in Nusselt’s calculation, and a condensing mixture doesn’t have it.

What the picture cannot show #

A real film’s shape. The figures show a smooth film; a real one is rippled, as above. The thickness in the first figure is the smooth theory’s value, and a measured mean thickness would be somewhat smaller.

Subcooling and inertia. Rohsenow’s correction to the latent heat, 0.68 times $c_p\Delta T$, allows for the extra heat in cooling the film below saturation, and it’s in the numbers here. At a drop of 10 K in steam it adds only 1.3 per cent. Inertia is left out, along with the shear the moving vapour exerts on the surface. High-speed vapour in a condenser’s inlet drags the film along and thins it, which can raise the coefficient by much more than the waves do.

Properties. Viscosity varies across the film by a noticeable factor, and the formula uses one value at a mean temperature. For water at a modest drop it’s a small correction; for oils with a large variation it isn’t. Liquid metals are another exception: their film resistance is so small that the interface itself becomes the limit.

The wall. The wall is taken as isothermal, whereas a real one has a coolant behind it whose own resistance is usually the larger. On a real power-station tube the water side may have more resistance than the condensate, and so the fourth root is partly swamped. The formula gives the steam side alone.

The convention #

Saturated water at atmospheric pressure: density 958 kilograms a cubic metre, viscosity 0.282 millipascal-seconds, conductivity 0.679 watts a metre per kelvin, latent heat 2.257 megajoules a kilogram. The film Reynolds number is $4\Gamma/\mu$, with $\Gamma$ the mass flow per unit width. The Nusselt number in the wave figure is on the scale $(\nu^2/g)^{1/3}$, the film’s own length, so that it’s a function of its Reynolds number alone, plus a Prandtl number at high flow. The wavy and turbulent forms are Kutateladze’s and Labuntsov’s fits to measurements, not derivations.

How it was checked #

The march adds the wall’s heat step by step to the film’s flow, computing the thickness from the flow’s cube law at each height, and it has no fourth power in it anywhere. It arrives within two parts in ten thousand of the closed form. It needs a grid crowded near the top, because the film starts from nothing, and a uniform grid with a first step from nothing overestimates the film by a tenth, which is a useful reminder of how much heat is transferred in the first centimetres. The model refuses a wall of no height, a drop of nothing and a tube of no diameter.

Who found it, and when #

Nusselt’s paper, Die Oberflächenkondensation des Wasserdampfes, is from 1916. Kapitsa’s film waves date from 1948, Benjamin’s and Yih’s stability results from 1957 and 1963, and Rohsenow’s correction from 1956. McAdams’s factor of 1.2 appeared in his book of 1954. Schmidt’s report on dropwise condensation is from 1930.

Still open: how much of the fifth a wave accounts for #

The figures quote the enhancement from fits. The next calculation would march a falling film with a long-wave evolution equation, Benney’s or the Shkadov integral model, at the Reynolds numbers in this essay, with condensation at the surface as a source that depends on the local thickness, and it would ask how large the coefficient’s enhancement is when the waves are allowed to develop freely. The answer would say whether the rule of a fifth comes from the waves’ mean thinning alone, whether the local dependence of the supply on the thickness makes it larger than for a heated film, and whether there’s a Reynolds number at which it stops growing.