The number that is an answer

There is a distinction between the groups this collection is organised by that is almost never drawn, and drawing it changes what a regime diagram can be.

A Reynolds number is built from a speed, a size and a fluid, all three of which somebody chooses before the experiment starts. So is a Mach number, a Froude number, a Deborah number and every other group here. Each of them is a hypothesis: it licenses a model, it can be set, and it can be held fixed while something else is varied.

$$\mathrm{Nu} = \frac{hL}{k}$$

cannot. The thermal conductivity and the length are chosen; $h$, the heat-transfer coefficient, is not. It is defined as the wall heat flux divided by the temperature difference that drove it, and it is what comes out of the apparatus. The Nusselt number is on the left-hand side.

Two of three, and the third is determined #

The way to make that concrete is to solve the problem and see how many things had to be supplied.

The flat plate at zero incidence gives the Blasius layer,

$$f''' + \tfrac12 f f'' = 0,\qquad f(0) = f'(0) = 0,\ f'(\infty) = 1,$$

and its thermal partner,

$$\theta'' + \tfrac{\mathrm{Pr}}{2} f\,\theta' = 0,\qquad \theta(0) = 0,\ \theta(\infty) = 1,$$

with $\mathrm{Nu}_x/\sqrt{\mathrm{Re}_x} = \theta'(0)$ and $C_f/2 = f''(0)/\sqrt{\mathrm{Re}_x}$. Two inputs go in — a Reynolds number and a Prandtl number — and a Nusselt number comes out. Nothing anywhere in the system allows Nu to be specified.

The shot returns $f''(0) = 0.332057337$ against the published 0.332057336, and the same solve gives a displacement thickness of 1.7207876 — a second number it was not fitted to, and the check that the momentum half is right before the thermal half is asked anything.

Two routes to the same number #

The energy equation is first order in $\theta'$, which makes a second and completely different route available:

$$\theta'(0) = \left[\int_0^\infty \exp\!\left(-\tfrac{\mathrm{Pr}}{2}\int_0^\eta f\,\mathrm{d}\eta'\right)\mathrm{d}\eta\right]^{-1}.$$

That is a quadrature. The other route is a Runge–Kutta shot at the second-order equation. Neither was tuned to the other, and across Prandtl numbers from 10⁻⁵ to 10³ the worst disagreement between them is 1.2 × 10⁻¹³.

A number computed one way is a formula. A number computed two ways is a result — and the point of insisting on it here is precisely the essay’s subject: the Nusselt number is the thing on the other side of the equals sign, so it had better be determined to more digits than anybody needs.

One thing had to be fixed to get there and it is worth recording. The thermal layer at low Prandtl number is far thicker than the viscous one, so an integration stopped at η = 10 returns $\theta'(0) \approx 1/\eta_{\max}$ — which is 0.1 whatever the fluid is, and produces a Prandtl exponent of 0.019 where the physics has a half. The fix costs nothing, because out there $f$ is exactly linear: the tail is built rather than solved, and reaches twelve thermal thicknesses.

The whole of a flat plate, in one curve #

The curve everybody quotes as $\mathrm{Nu} = 0.332\,\mathrm{Re}^{1/2}\mathrm{Pr}^{1/3}$ is the top half of that plot. At the bottom the exponent is a half, not a third, and the asymptote there is $\sqrt{\mathrm{Pr}/\pi} = 0.5642\sqrt{\mathrm{Pr}}$ in closed form — because when the thermal layer is far thicker than the viscous one, $f \approx \eta - \beta^*$ everywhere it reaches and the integral becomes a Gaussian.

At Pr = 10⁻⁵ the solve is within 0.3 per cent of that limit, which is a third opinion nothing was fitted to.

Why the exponent changes, and where #

The two exponents are not arbitrary and the reason for each is a statement about which layer is inside which.

At large Prandtl number the thermal layer is buried deep inside the viscous one, where the velocity profile is still nearly linear: $f' \approx f''(0)\eta$. Heat is being carried across a region in which the fluid moves in proportion to its distance from the wall, and balancing conduction against that linear convection gives a thermal thickness going as $\mathrm{Pr}^{-1/3}$ — hence the cube root. This is the regime of oils and of water, and it is the one the textbook correlation was written for.

At small Prandtl number the thermal layer is far outside the viscous one, in a region where the fluid is moving at essentially the free-stream speed. Heat is being conducted sideways out of a slab of uniformly moving fluid, which is a much simpler problem — it is the Graetz problem’s outer limit — and the balance gives $\mathrm{Pr}^{-1/2}$. This is the regime of liquid metals, where a sodium-cooled reactor has a Prandtl number near 0.005 and the cube-root correlation is out by a factor of three.

The crossover is not sharp because there is no third layer to separate them; the two limits meet where the two thicknesses are comparable, which is Pr of order one, and the transition between them takes about four decades because the exponents differ by only a sixth.

And the exponent has nothing to say about the coefficient. A correlation with the right exponent and a coefficient fitted in air is wrong by tens of per cent in water, because the prefactor moves through the same crossover.

An exponent that is two numbers #

Differentiating the curve numerically gives 0.497 below Pr = 10⁻⁴ and 0.336 above Pr = 1, and a continuous drift between them that takes about four decades. There is a decade of Prandtl number for which almost any exponent between 0.33 and 0.5 can be fitted honestly, and which one a correlation reports is a statement about the fluids its author had access to.

That is the same difficulty the slender-body expansion has and the same one a single relaxation time has: an exponent measured over a finite range is a property of the range unless the function is a power law, and this one is not.

The case where the analogy is exact #

At Pr = 1 something happens that is worth more than the number it produces.

The equation satisfied by $f''$ is $(f'')' = -\tfrac12 f f''$, and the energy equation at Pr = 1 is $\theta'' = -\tfrac12 f\theta'$. They are the same equation. With the same boundary conditions $\theta = f'$ identically, so $\theta'(0) = f''(0)$ and

$$\mathrm{St} = \frac{\mathrm{Nu}}{\mathrm{Re}\,\mathrm{Pr}} = \frac{f''(0)}{\sqrt{\mathrm{Re}}} = \frac{C_f}{2}.$$

The Reynolds analogy, exact, with no coefficient in it. Computed here at $\mathrm{Re} = 10^5$ it gives $\mathrm{St} = 1.050057499\times10^{-3}$ and $C_f/2 = 1.050057499\times10^{-3}$, agreeing to fifteen digits.

Momentum and heat are carried by the same mechanism when they diffuse at the same rate, so a measurement of the friction is a measurement of the heat transfer. Away from Pr = 1 it is not: the ratio is 16.6 at Pr = 0.02 and 0.047 at Pr = 100. A factor of three hundred and fifty between the two ends, from an identity that holds exactly in the middle.

What the analogy is for, and where it breaks #

The analogy matters because friction is easy to measure and heat transfer is not. A skin-friction gauge is a strain measurement; a heat-transfer coefficient needs a controlled wall temperature, a known flux, and a correction for every conduction path out of the model. Where Pr ≈ 1 — which is every gas — Colburn’s modification $\mathrm{St}\,\mathrm{Pr}^{2/3} = C_f/2$ extends it usefully.

It breaks in a way worth naming, because the break is not about Prandtl number. In a flow with a pressure gradient the friction and the heat transfer stop tracking each other: the friction goes to zero at a separation point and the heat transfer does not, because the fluid is still exchanging thermal energy with the wall while it is exchanging no momentum. A separated region has zero skin friction and a heat-transfer coefficient that may be higher than the attached value.

So the analogy holds where the two transport processes see the same mean flow and fails where the momentum equation has a term the energy equation does not. That is a sharper statement than “it holds for Pr near one” and it is the one that predicts where it will let somebody down.

The turbulent case, which is where the correlations live #

Everything above is laminar, and almost every heat exchanger in the world is turbulent, so it is worth saying which parts survive and which do not.

The Reynolds analogy survives and gets better. In a turbulent boundary layer the transport of heat and of momentum is dominated by the same eddies, and the molecular Prandtl number matters only in the viscous sublayer — so $\mathrm{St} \approx C_f/2$ is a good approximation for gases over a wide range, and the Colburn form with $\mathrm{Pr}^{2/3}$ extends it to liquids. The identity’s reason changes completely: laminar it is because two differential equations coincide, turbulent it is because one mixing process carries both quantities.

The exponents do not survive. The turbulent flat plate gives $\mathrm{Nu} \propto \mathrm{Re}^{0.8}$ rather than $\mathrm{Re}^{0.5}$, because the wall gradient is set by the log layer rather than by a similarity solution, and the Prandtl exponent settles nearer 0.4 without a clean asymptotic argument on either side.

And the two-layer structure survives in a different form. The thermal sublayer sits inside the viscous sublayer for Pr > 1 and outside it for Pr < 1, which is the same geometry as the laminar case with the layer thicknesses replaced by wall units. That is why liquid metals remain the hard case: the conduction reaches out past the region where the turbulence is doing the carrying.

None of that changes the essay’s point. Nu is still what comes out; Re and Pr are still what go in; and the geometry is still in the constant.

What a correlation throws away #

The universal form of a forced-convection correlation is $\mathrm{Nu} = C\,\mathrm{Re}^m\,\mathrm{Pr}^n$, and the exponents are what gets discussed. At $\mathrm{Re} = 10^4$ and $\mathrm{Pr} = 0.7$ a flat plate gives Nu = 29.3, a cylinder in cross-flow 53.3 and a sphere 60.8 — a spread of 2.08, all of it in the constant $C$.

The geometry is worth more than the exponents, and the correlation puts the geometry into a number it calls a constant. That is what a collapse onto two groups does: it is not that the answer is wrong, it is that the part of the answer nobody argues about is the larger part.

Where regime diagrams go wrong #

The consequence for how the subject is drawn is a real one.

A regime map is a plot of what happens against the quantities somebody sets: Reynolds against Mach, Reynolds against Prandtl, Keulegan–Carpenter against β. Every axis of such a map is an input, and reading it consists of finding the point that describes the apparatus and looking at what is written there.

A “Nusselt number regime” cannot be found on such a map, because there is no apparatus at which the Nusselt number is a setting. The correct picture is a response surface: Nu drawn above the Re–Pr plane, with contours. It looks similar and it means something different, and the difference shows up the moment somebody tries to hold one of the axes fixed.

The same is true of the friction factor, of the discharge coefficient, of a drag coefficient and of a span efficiency. All of them are dimensionless, all of them are outputs, and none of them is a hypothesis about anything.

The list of which is which #

It is worth writing the two classes out, because the collection contains both and does not usually separate them.

Inputs, which license a model: Reynolds, Mach, Prandtl, Froude, Knudsen, Weber, Bond, Deborah, Keulegan–Carpenter, Rossby, Richardson, Womersley, reduced frequency, aspect ratio, viscosity ratio. All built from things somebody sets.

Outputs, which report a result: Nusselt, Stanton, Sherwood, drag coefficient, lift coefficient, skin-friction coefficient, pressure coefficient, discharge coefficient, friction factor, span efficiency, power coefficient.

And two that pretend to be one and are the other. The Rayleigh number of a convecting layer looks like an input and is one — it is built from the imposed temperature difference. The Grashof number of a natural-convection boundary layer also looks like an input, and the velocity it implies is an output, which is what the next essay is about. And the Reynolds number of a buoyant plume is straightforwardly an output: nobody set the speed.

What this does to a measurement #

The input–output distinction is not only tidy-mindedness. It changes what an experiment can be asked for.

If a quantity is an input, an experiment can be designed around it: hold it, vary the others, and the result is a controlled comparison. The whole practice of dynamic similarity rests on that — a model test is worth something because Re and Ma can be matched, or their mismatch reasoned about.

If a quantity is an output, none of that is available. Nobody can build a rig at Nu = 50 and see what happens; a rig runs at a Reynolds number and a Prandtl number and reports whatever Nusselt number it reports. So a measurement campaign in heat transfer is a sweep of the inputs and its result is a surface, and the correlation fitted to it is a summary of that surface with the geometry folded into a constant.

The practical consequence is that a heat-transfer correlation is only as good as the range it was fitted over, and the range is a rectangle in the Re–Pr plane rather than a range in Nu. Quoting the range of validity in Nusselt number, which handbooks occasionally do, is not merely unhelpful — it is not a statement about anything, because two different geometries at the same Nu are at different points of the plane and there is no experiment that visits them both.

Three conventions, where the Reynolds number has one #

The length buried in a Reynolds number makes a value meaningless without its convention. The Nusselt number carries the same defect three times over, and being an output does not excuse it.

The coefficient is a flux divided by a temperature difference. The flux is measured. The difference is a choice. On a flat plate it is unambiguous — wall minus free stream — which is why the solve above could be quoted to nine digits. Inside a duct there is no free stream: the fluid heats up as it goes, so the reference is the bulk mean temperature, which is a mixing-cup average that varies with position. A heat exchanger has to use the logarithmic mean of its two terminal differences, and a Nusselt number computed against an arithmetic mean and one computed against a log mean are different numbers describing the same apparatus.

The second convention is worse because it is silent. Conductivity, viscosity and the Prandtl number itself all depend on temperature, and a boundary layer spans the whole range from wall to bulk — so a correlation has to say where its properties were evaluated. The usual answer is the film temperature, halfway between; some correlations use the bulk, a few the wall. Applying a correlation with the wrong convention is an error that no amount of care with exponents recovers.

The third is the admission that the second is not enough. For an oil the viscosity can change by a factor across the layer, so a single evaluation temperature cannot represent it, and correlations carry an explicit ratio of bulk to wall viscosity raised to a small power. That correction exists precisely because the property convention is doing real work rather than tidying.

So: a length, a temperature difference, and an evaluation state. A Nusselt number quoted without all three is a number without coordinates.

Where the numbers came from #

Wilhelm Nusselt gave the group its modern form in 1915, in the paper that established dimensional analysis as the organising method of heat transfer. Osborne Reynolds proposed the analogy in 1874 — forty years before the boundary layer existed as a concept, and on an argument about eddies carrying both momentum and heat that is, in its own terms, correct. Blasius solved the flat plate in 1908 and Pohlhausen the thermal problem in 1921.

The historical order is the interesting part. The analogy came first, as a physical argument; the similarity solutions came later and showed exactly when it is an identity rather than an analogy. That is the usual direction in this subject and it is worth noticing whenever a plausible argument is available before a calculation is.

What this leaves #

The Nusselt number collapses nothing, because it is not a collapse: it is the answer, and what a correlation for it discards is the geometry, which is worth a factor of two where the exponents are worth per cents.

The next essay takes a group that is genuinely an input and whose velocity is an output — a speed nobody imposed — where a Reynolds number of six thousand appears in a problem in which nothing was ever set moving.