A speed nobody imposed

Take any of the groups this collection is organised by and look for the velocity in it. A Reynolds number has the tunnel speed; a Mach number has the flight speed; a Keulegan–Carpenter number has the orbital speed of a wave. In every case somebody set a dial.

A warm plate in still air has no dial. It makes its own flow, and the only way to write a Reynolds number for it is to solve the problem first and then divide. What is left when the speed is removed is

$$\mathrm{Gr}_x = \frac{g\beta\,\Delta T\,x^3}{\nu^2},$$

which is the square of a Reynolds number built from the velocity scale $\nu/x$ — a viscous diffusion speed, not a flow speed, and not a quantity anything in the room possesses.

That distinction is the whole essay. The Grashof number is a perfectly good input: g, β, ΔT, x and ν are all set by the apparatus, so it belongs on the list of hypotheses beside Re and Ma and not on the list of results beside the Nusselt number. What is unusual is that the flow’s velocity — the thing a Reynolds number would be built from — is on the other list.

The derivation, done rather than quoted #

The exact similarity solution of this problem is Ostrach’s and is a two-point boundary-value problem with two unknowns at the wall. What is derived here instead is Squire’s integral solution, because it is closed-form, because its constants are the ones everybody quotes, and because its single assumption is exactly the thing that turns out to fail.

Assume the two profiles the boundary conditions and one interior condition force:

$$\frac{u}{u_0} = t(1-t)^2,\qquad \frac{T - T_\infty}{\Delta T} = (1-t)^2,\qquad t = y/\delta,$$

both vanishing at the outer edge, the velocity vanishing at the wall as well because of no slip and the temperature reaching ΔT there.

Put them into the integral momentum and energy equations,

$$\frac{\mathrm d}{\mathrm dx}\int u^2\,\mathrm dy = g\beta\int(T - T_\infty)\,\mathrm dy - \nu\,u'(0), \qquad \frac{\mathrm d}{\mathrm dx}\int u(T-T_\infty)\,\mathrm dy = \frac{2\alpha\Delta T}{\delta},$$

and three beta integrals are needed: $\int t^2(1-t)^4 = 1/105$, $\int(1-t)^2 = 1/3$ and $\int t(1-t)^4 = 1/30$. All three are computed here by Simpson rather than looked up, to a part in 10⁹.

The powers cancel, which is the check #

With $u_0 = C_1x^{1/2}$ and $\delta = C_2x^{1/4}$ the x-dependence cancels identically from both equations. That is not a convenience: it is the check that the assumed powers are the right ones, and it is the same argument a similarity solution makes, done with two integrals instead of a change of variable.

What is left is two algebraic equations,

$$C_1C_2^2 = 80\alpha,\qquad \frac{C_1^2C_2}{84} = \frac{g\beta\Delta T\,C_2}{3} - \frac{\nu C_1}{C_2},$$

whose solution is

$$\left(\frac{\delta}{x}\right)^4 = \frac{240\,(20/21 + \mathrm{Pr})}{\mathrm{Pr}^2\,\mathrm{Gr}_x}, \qquad \frac{\mathrm{Nu}_x}{\mathrm{Gr}_x^{1/4}} = \frac{2}{240^{1/4}}\,\mathrm{Pr}^{1/2}\left(\tfrac{20}{21} + \mathrm{Pr}\right)^{-1/4}.$$

And $2/240^{1/4} = 0.50813$ while $20/21 = 0.95238$: the 0.508 and the 0.952 every textbook prints as Squire–Eckert’s constants, arrived at from three quadratures and one page of algebra. Getting them out rather than in is worth the page, because the whole of the next section depends on the form being right and not merely the values.

An exponent that can be differentiated #

The reason the closed form matters is that the Prandtl dependence is now a function rather than a fit.

$$\frac{\mathrm d\ln \mathrm{Nu}}{\mathrm d\ln \mathrm{Pr}} = \frac12 - \frac14\cdot\frac{\mathrm{Pr}}{20/21 + \mathrm{Pr}}.$$

At small Prandtl number it is exactly 1/2 and at large Prandtl number exactly 1/4, with no numerics anywhere. That is two collapses rather than one: writing the answer on the Rayleigh number $\mathrm{Ra} = \mathrm{Gr\,Pr}$ works at the top and fails at the bottom, and writing it on $\mathrm{Gr\,Pr}^2$ works at the bottom and fails at the top, and there is no single group that does both.

And the crossover has an exact position. The slope is 3/8 at Pr = 20/21, which is a regime boundary with a closed form in a problem where nothing else has one — and 20/21 is 0.952, so the boundary sits almost exactly at the Prandtl number of air. Every correlation for natural convection in air is fitted in the one place where neither asymptote holds.

This is the same difficulty the forced flat plate has, arriving by a completely different route: there the two exponents come from which boundary layer is inside which, here they come from a ratio of two terms in an algebraic solution, and in both cases the fluid everybody works with sits between them.

The flow’s own Reynolds number #

Now the velocity, which the derivation produced without being asked for it.

$$\mathrm{Re}_x = \frac{u_0 x}{\nu} = \frac{80}{\mathrm{Pr}}\left(\frac{x}{\delta}\right)^2 = \frac{80\sqrt{\mathrm{Gr}_x}}{\sqrt{240\,(20/21 + \mathrm{Pr})}},$$

exactly proportional to the square root of the Grashof number — checked here to twelve digits by doubling the Grashof number two decades at a time — and the profile’s own maximum is 4/27 of $u_0$.

A vertical plate a metre high, ten degrees warmer than the room, has $\mathrm{Gr} \approx 10^9$. That is a Reynolds number of about 126,000 on the scale velocity and 18,700 on the profile’s peak, with a peak speed of about a quarter of a metre per second — a number anybody can feel with a hand held beside a radiator, and a number that appears nowhere in the specification of the problem.

The consequence for transition is the one worth naming. A natural-convection layer goes turbulent at $\mathrm{Gr\,Pr} \approx 10^9$, and it is often said that this is a much higher threshold than a forced layer’s $\mathrm{Re} \approx 5\times10^5$. Converted, it is not: $10^9$ in air corresponds to a flow Reynolds number of about twenty thousand on the peak velocity, which is lower than the forced threshold. The comparison people make is between a number and its own square.

What the number is worth in a room #

The abstract statement deserves a few concrete evaluations, because the Grashof number’s range in ordinary life is enormous and its dependence on height is brutal.

The $x^3$ is the thing. A vertical surface ten degrees above ambient in air has $\mathrm{Gr}_x \approx 10^9\,(x/\text{m})^3$, so a 10 mm chip package is at $10^6$, a 100 mm electronics enclosure at $10^6$, a metre-high radiator at $10^9$ and a three-storey glass façade at $2\times10^{10}$. Transition sits around $10^9$, which means the same physical arrangement is laminar on a small object and turbulent on a large one, with the crossover at the scale of ordinary furniture.

That is a much sharper size dependence than a forced flow has. Doubling the length of a forced plate doubles its Reynolds number; doubling the height of a warm wall multiplies its Grashof number by eight. It is why natural-convection correlations are always written per unit height in the laminar range and for a whole surface in the turbulent one, and why a laboratory model of a building’s façade cannot be made small.

The heat flux itself is gentler than any of that suggests. With $\mathrm{Nu} \propto \mathrm{Gr}^{1/4}$ the coefficient goes as $x^{-1/4}$, so a plate ten times taller has a local coefficient about half as large — which is the familiar fact that the bottom of a radiator does most of the work.

Where the assumption fails, and it announces itself #

Squire’s derivation gives both layers one thickness. That is right when the Prandtl number is of order one and badly wrong when it is not.

From Pr = 0.7 to Pr = 1,000 the answer is between 1.9 and 8.2 per cent high — a systematic offset that leaves the shape of the Prandtl dependence intact, which is why the exponents above are worth having even though the coefficient is not exact.

At Pr = 0.01 it is 27 per cent high. And that is exactly where the thermal layer is about ten times the momentum layer, so the single δ is describing neither. The failure is not scattered across the range: it is concentrated in the one place the scaling says the assumption should break, which is the strongest evidence available that the assumption is the thing that broke.

The exact solution, and why it is not here #

The similarity reduction of the same problem is

$$F''' + 3FF'' - 2F'^2 + \theta = 0,\qquad \theta'' + 3\mathrm{Pr}\,F\theta' = 0,$$

with $F(0) = F'(0) = 0$, $\theta(0) = 1$ and both derivatives vanishing at infinity. It has two unknowns at the wall and two conditions at infinity, so it is a two-dimensional shooting problem, and its tabulated values at Pr = 0.72 are $F''(0) = 0.6760$ and $-\theta'(0) = 0.5046$.

Those are quoted here rather than recomputed, and the reason is worth recording rather than hiding. The far field of that system is nonlinearly unstable in a specific way: past the layer, any solution whose $F'$ overshoots into negative values drives $F$ negative, and then $F''' = -3FF''$ amplifies rather than damps. Integrating far enough out to resolve the low-Prandtl thermal layer puts that cliff close enough to the true trajectory that a two-dimensional Newton falls off it, and the shot finds a spurious root with a perfectly plausible-looking profile.

That is a limitation of the calculation here and it is stated as one. The closed form covers the whole range; the exact solution is imported for the comparison it is needed for; and a shorter integration that would have made the shot converge would also have returned the wrong value at Pr = 0.72, which is a worse failure than not doing it.

Two more flows with no imposed speed #

The plate is the tractable case and it is not the only one, and naming the others shows what is general in the argument.

A plume. Hot gas above a fire has no imposed velocity, no imposed length and — in the far field — not even an imposed temperature difference, because the plume entrains and cools as it rises. What is imposed is a buoyancy flux, with dimensions of length⁴ per time³, and the similarity solution built from it gives a centreline velocity going as $z^{-1/3}$ and a width growing linearly. The Reynolds number of the result grows as $z^{2/3}$, so a plume is laminar near the source and turbulent above it whatever anybody does.

A gravity current. A dense fluid released into a lighter one runs along the floor at a speed set by $\sqrt{g'h}$ with $g'$ the reduced gravity — again a speed that comes out of the problem, and again a Froude number formed from it that turns out to be about 1/2 for reasons of energy rather than of choice. This collection meets the same structure in the number that really is one.

And the common feature is that the dimensional analysis has one group fewer than it looks. With a velocity among the inputs there would be a Reynolds number and a Froude number; without one, the two collapse into a single group and the velocity is recovered from it. That is why the Grashof number is the square of a Reynolds number rather than a Reynolds number: it is what a Reynolds number becomes when the speed it contains has been eliminated in favour of the force that produced it.

What the square is for #

The essay opens by noting that the Grashof number is the square of a Reynolds number, and treats that as an oddity. It is a design decision, and it becomes obvious the moment a forced flow is present as well.

Put a fan in the room. Now there are two mechanisms driving the same layer, and the question is which one matters. Buoyancy contributes a force of order $\rho g\beta\Delta T$ per unit volume and inertia one of order $\rho U^2/x$, and their ratio is

$$\frac{\mathrm{Gr}}{\mathrm{Re}^2},$$

which is a Richardson number. Gr is the square of a Reynolds number precisely so that this comparison is a ratio of two things of the same kind. A group built as a first power of a velocity scale could not be set against Re without a square root appearing, and the whole point of the construction is that it can.

The reading is the usual one. Below about a tenth the flow is forced and the buoyancy is a correction; above about ten it is natural and the fan might as well be off; and between them is mixed convection, a band two decades wide in which neither correlation applies and both are quoted with confidence.

Two things happen in that band that neither limit prepares anyone for.

Aiding flow beats both. A warm vertical plate with the forced flow going upwards has the buoyancy adding to the inertia near the wall, the profile becomes fuller, and the heat transfer exceeds what either mechanism gives alone.

And opposing flow can be worse than either. Turn the forced flow downwards past the same warm plate and the buoyancy is pushing the near-wall fluid the other way. At a Richardson number near one the fluid closest to the surface can be brought to rest and then reversed, so a recirculation forms inside the layer, the flow can become unsteady, and the heat transfer falls below the pure forced-convection value at the same fan speed.

That last is the practical reason the band matters. A cooling arrangement designed on a forced correlation, run at reduced flow, can pass through a condition where slowing the fan further improves the cooling — which is not a behaviour any monotone correlation can express.

The Rayleigh number of a layer, and the Grashof number of a plate #

It is worth separating this problem from the one next door, because both are called natural convection and only one of them has an instability in it.

A layer heated from below has no flow at all until a threshold is crossed: the conducting state is a solution, it is stable, and at Ra = 1707.762 it stops being. That is a bifurcation, the number is an eigenvalue, and nothing happens below it.

A vertical plate has a flow at every Grashof number. There is no conducting state to be unstable, because the buoyancy force is parallel to the wall rather than across the layer, and the fluid starts moving the instant the plate is warm. So the Grashof number is not a threshold of anything; it is a strength, in the same sense a Reynolds number is.

The two problems share a name, a fluid and a body force, and their numbers mean different things. That distinction is not usually drawn and it explains why one of them has an exact critical value quoted to seven figures and the other has a transition range quoted as “about 10⁹”.

What is left out #

Everything here is laminar and two-dimensional. Above $\mathrm{Ra} \approx 10^9$ the layer transitions, the profiles stop being self-similar, and the Nusselt number goes as $\mathrm{Ra}^{1/3}$ rather than $\mathrm{Ra}^{1/4}$ — a change of exponent that also makes the local heat-transfer coefficient independent of height, which is why turbulent natural convection correlations are written for a whole plate and laminar ones per unit height.

The Boussinesq approximation is everywhere. Density variation enters only in the buoyancy term, and the temperature difference is assumed small enough for β to be constant. For a plate a hundred degrees above ambient in air that is already questionable.

And there is no leading edge. The similarity solution and the integral solution both have $\delta \to 0$ at $x = 0$, where the boundary-layer approximation itself fails, and a real plate has a starting length over which the flow accelerates from rest. It is short, and it is the same entrance problem every developing layer has.

Where the work came from #

Ernst Schmidt and Wilhelm Beckmann measured the profiles in 1930, with a Mach–Zehnder interferometer and a thermocouple traverse, which for its date is a remarkable piece of experimental work. Eric Pohlhausen did the similarity reduction for their paper. Squire’s integral treatment is from 1938, Eckert’s version and the constants that carry both names from the 1940s, and Ostrach’s numerical solution — the table this essay compares against — from 1953.

The order is the reverse of the usual one in this collection. Here the measurement came first, the approximate closed form second, and the exact solution twenty years later, which is what happens when the exact solution is a two-point boundary-value problem and the tools are a slide rule.

What this leaves #

The Grashof number is an input like any other, and it is the only one whose flow speed is an output. The consequences are that a comparison with a forced-flow threshold is a comparison between a number and its square, and that the group which collapses the answer is Ra at one end of the Prandtl range and Gr·Pr² at the other, with air sitting exactly at the crossover.

The next essay leaves the fluid entirely and takes a group with no physics in it at all — the ratio of two lengths — which turns out to give three different exponents in three problems that look alike.