Laid flat, a warm plate is pushed by the pressure of its own layer

The exact vertical plate was driven by buoyancy acting along its layer: warm fluid next to the wall is lighter than the fluid beside it, and gravity pushes it up the wall directly. Turn the plate through a right angle, warm side up, and that push disappears. Gravity now acts across the layer, perpendicular to the wall, and has no component along it at all. Something still makes the warm fluid flow away from the plate’s edge, because a warm horizontal surface in still air does make a flow, and what it is changes the answer from the vertical plate’s in a way a constant cannot absorb.

Buoyancy across the layer becomes pressure along it #

Across a thin layer the pressure is hydrostatic. Warm fluid weighs less than the cold fluid above it, so the pressure at the wall, under a column that includes the warm layer, is lower than the pressure at the same height far from the plate, where the column is cold all the way up. The deficit is the layer’s buoyancy integrated across it,

$$p'(x, y) = -\rho g\beta\int_y^\infty (T - T_\infty)\,\mathrm{d}y',$$

and it is deeper where the layer is thicker. A layer that thickens away from the edge therefore has a pressure that falls away from the edge, and that gradient is the push. It is a push and not a pull, in the sense nothing sucks insists on: the fluid at the edge sits under a full cold column and the fluid further in under a partly warm one, and the higher pressure at the edge drives the fluid inward from there. The fluid is drawn in at the edge and along the plate by the pressure its own warmth has left behind it. The pressure that has no speed of an incompressible flow is here pinned to the temperature field by hydrostatics, and it does the work the along-plate buoyancy did on the vertical plate.

The push is strongest at the wall and falls to nothing at the layer’s edge, where the deficit no longer changes along the plate. It is a favourable pressure gradient, of the kind a layer climbing uphill meets in reverse, and the velocity profile it makes has the full, rounded shape of an accelerated layer: a peak at $\eta = 1.58$ and a warm layer ending near 5.9 in air.

The exponents a pressure-driven layer must have #

A push that depends on the layer’s own thickness sets its own scales. With $\delta \propto x^{a}$ the deficit goes as $g\beta\Delta T\,\delta$ and its gradient as $g\beta\Delta T\,\delta/x$; balancing that against the inertia $u^2/x$ and the inertia against viscosity, $u^2/x \sim \nu u/\delta^2$, gives $\delta \propto x^{2/5}$ and $u \propto x^{1/5}$. The similarity variable is $\eta = (y/x)\mathrm{Gr}_x^{1/5}$, with $\mathrm{Gr}_x = g\beta\Delta T x^3/\nu^2$, and with $\psi = \nu\mathrm{Gr}_x^{1/5}f(\eta)$ and $p'/\rho = (\nu/x)^2\mathrm{Gr}_x^{4/5}P(\eta)$ the boundary-layer equations become

$$5f''' + 3ff'' - f'^2 - 2(P - \eta\theta) = 0,\qquad 5\theta'' + 3\,\mathrm{Pr}\,f\theta' = 0,\qquad P' = \theta,$$

with $f(0) = f'(0) = 0$, $\theta(0) = 1$, and $f'$, $\theta$ and $P$ vanishing far away. The heat transfer is $\mathrm{Nu}_x = -\theta'(0)\,\mathrm{Gr}_x^{1/5}$: a fifth root, where the vertical plate has a fourth. The pressure appears as one more unknown with one more far condition, and the box scheme that solved the vertical plate solves this one on the same kind of grid by the same continuation in the Prandtl number.

The reduction, checked against the equations it came from #

A similarity reduction is algebra, and algebra can be wrong in ways the reduced equations never reveal: a sign in the pressure term, a factor in the convective one, and the box scheme will solve the wrong equations as accurately as the right ones. So the solution was rebuilt as dimensional fields — velocity, temperature and pressure for air twenty kelvin above the room, twenty centimetres from the edge — and the original boundary-layer equations were evaluated on them by central differences in $x$ and $y$, in a code that never sees $\eta$. The momentum, energy and hydrostatic balances close to six parts in ten thousand of their largest term. With the pressure’s sign reversed — the plate warm underneath — the same check misses by a factor of two, so it is capable of failing. The vertical plate’s reduction passes the same test at four parts in a thousand.

A fifth root against a fourth #

In air the two coefficients come out almost equal — $\mathrm{Nu}_x = 0.3568\,\mathrm{Gr}_x^{1/4}$ vertical and $0.3574\,\mathrm{Gr}_x^{1/5}$ flat — so the flat plate’s local heat transfer is very nearly the vertical one’s times $\mathrm{Gr}_x^{-1/20}$. That ratio is 0.63 at a Grashof number of $10^4$, half at $10^6$ and 0.40 at $10^8$. A twentieth is a small exponent, and over the four decades of a laminar plate it is a factor of 1.6: laying a plate down costs it between a third and three fifths of its heat transfer, more the longer the plate, because the push that drives the flat layer weakens as the layer thickens relative to the distance the push acts over.

The physical scales follow. For air twenty kelvin above the room, twenty centimetres from the edge, the flat layer’s warmth reaches about four centimetres up, to its one-per-cent edge, and its peak speed is about five centimetres a second; the vertical plate’s layer at the same distance is under two centimetres thick and moves at twenty. The flat layer thickens as $x^{2/5}$ and speeds up only as $x^{1/5}$, so it is a slow, deep layer, and close to the edge it is deep enough to break the approximation it was derived under: a centimetre from the edge it is already about as thick as its distance from the edge, and the boundary-layer equations are not to be trusted there.

Only the upper face #

The construction works only if the pressure falls away from the edge. That needs warm fluid above the plate, so that the deficit is under it, or, equally, cold fluid below a cold plate. Under a plate warm on its lower face the warm fluid is trapped against the plate, the hydrostatic anomaly has the other sign, and the pressure rises away from the edge: a push towards the edge rather than away from it. A layer cannot start at an edge and be pushed back into it, and the similarity form from a leading edge has no solution; the flow under a warm downward-facing plate goes the other way, from the middle out to the edges, and is a different problem with a different solution.

The same symmetry decides which ceilings work. A chilled ceiling panel, cold face down, is the mirror image of a warm plate facing up — cold fluid under warm, a pressure excess under the panel that grows away from its edges — and it carries this solution with the signs reversed: a layer drawn in from the edges, its heat transfer going as the fifth root. A warm ceiling, a radiant heating panel facing down, is the mirror image of the other face, and it has no edge layer at all; its warm air pools against it, and it is known among building engineers as a poor convector for exactly that reason. The physics of the two ceilings is the physics of the two faces of one plate.

That sign has a history. Stewartson’s 1958 reduction, the first, attached the solution to the wrong face; Gill, Zeh and del Casal showed in 1965 that it belongs to the plate warm on top. The dimensional check above fails by a factor of two with the pressure’s sign reversed, which is the same statement made by arithmetic: the equations for the other face are not these equations.

The Prandtl number decides how much flatness costs #

The ratio is not a property of the geometry alone. At a Grashof number of a million the flat plate keeps 0.76 of the vertical plate’s heat transfer at a Prandtl number of a hundredth, half in air, and 0.32 at a thousand. In a liquid metal both layers are nearly inviscid, as the vertical one is, and an inviscid layer pushed by a pressure behaves much like one pushed by a body force. In an oil the warm layer is a thin skin inside a far thicker velocity layer, and the pressure push, which acts only where the fluid is warm, has to drag the whole of the velocity layer with it; the vertical plate’s buoyancy acts on the same thin skin but has no pressure to build first. The more viscous the fluid, the more it matters which way the plate faces.

The two ends of the range have the same character as the vertical plate’s. At a Prandtl number of $10^{-5}$ the flat plate’s heat transfer in the form $\mathrm{Nu}_x/(\mathrm{Gr}_x\mathrm{Pr}^2)^{1/5}$ reads 0.575, with no viscosity in the group, and at $10^5$ the form $\mathrm{Nu}_x/(\mathrm{Gr}_x\mathrm{Pr})^{1/5}$ reads 0.456, with no inertia in it. Between them the two layers trade places exactly as a forced layer’s do, and the push, which lives in the warm layer, finds itself driving either the whole of the flow or a small part of it.

A tilt of one degree #

Real horizontal surfaces are rarely level, and a small tilt restores a component of buoyancy along the plate, $g\beta\Delta T\sin\psi$ for a tilt $\psi$ from horizontal. The flat layer’s own push at the wall is $(2/5)|P(0)|\,g\beta\Delta T\,\mathrm{Gr}_x^{-1/5}$, which falls as the layer grows, so at some distance from the edge the tilt’s push must win. The two are equal at

$$\mathrm{Gr}_x = \left(\frac{(2/5)|P(0)|}{\sin\psi}\right)^5,$$

with $|P(0)| = 1.735$ in air.

The fifth power makes the crossover extraordinarily sensitive to the tilt. At ten degrees it comes at a Grashof number of about a thousand, within a few millimetres of the edge, and the layer is a vertical plate’s with $g\sin\psi$ almost everywhere. At one degree it comes at $10^8$, about a third of a metre from the edge of a plate twenty kelvin above the air — most of a roof panel or a solar collector’s absorber. A surface tilted by a degree is still a flat plate for the first third of a metre, and one tilted by a tenth of a degree is flat throughout its laminar life. The comparison is of forces at the wall, not a solution, and the change from one kind of layer to the other is gradual; but the fifth power is in the physics, not in the estimate, and it says a builder’s level is a finer instrument than the angle a designer would think worth stating.

The correlation in the handbooks #

The heat-transfer coefficient engineers use for the upper face of a warm plate is $\mathrm{Nu} = 0.54\,\mathrm{Ra}^{1/4}$, with the length taken as the plate’s area over its perimeter, from Lloyd and Moran’s measurements of 1974. Its exponent is the vertical plate’s, not this one’s.

For a long strip the area over the perimeter is half its width, and the edge solution averaged over a half-width is $5/3$ of its local value at the centre, because the local coefficient falls as $x^{-2/5}$. On that common length the correlation is 1.32 times the laminar layer at a Grashof number of $10^4$ and 2.1 times at $10^8$, the gap growing as $\mathrm{Gr}^{1/20}$ — which is the whole difference between the two exponents. In water the factor is similar. The correlation is a good measurement of something, and that something is not the laminar layer this essay computes.

A worked case puts the gap in units a designer uses. A long strip forty centimetres wide, twenty kelvin above still air, has a Grashof number of $2.4\times10^7$ on its half-width. The laminar edge layers from its two long sides give an average heat-transfer coefficient of about 2.3 watts per square metre per kelvin. The handbook correlation gives 4.5 — almost exactly what the same strip would give stood on edge, where the vertical plate’s average over twenty centimetres is 4.3. The correlation, in other words, treats the flat strip as transferring heat about as well as a vertical one, and the laminar layer says it should transfer half as much. A factor of two in a coefficient that small is easily lost beside radiation, which from a dull surface at those temperatures is worth five or six of the same units, and that may be part of why the discrepancy has been lived with.

What the correlation measures is the next question, and the layer itself offers a reason. The fluid on a warm horizontal plate is warm fluid under cold, the arrangement that convects if it is still.

Taking the layer’s thickness where its temperature excess has fallen to one per cent, its own Rayleigh number passes 1,708 — the threshold of a still layer between rigid walls — at a local Grashof number of about 440 in air, a few millimetres from the edge. Beyond that the flat layer is thick enough to convect on its own, and what holds it is its own flow: the shear sweeps disturbances downstream before they grow. Measured flat layers develop rolls aligned with the flow some way downstream, and a finite plate’s two layers meet at its middle and leave it as a rising plume. Both carry heat the laminar edge solution does not, and either could account for a correlation that sits above it by a growing factor. Which one does, and over what range, the similarity solution cannot say; what it can say is that the fourth-root law is not its continuation.

The convention behind every number #

All the heat-transfer numbers are local, on the distance from one edge, with the plate at one temperature and the Grashof number on the same distance. The flat plate’s average over a length $L$ is $5/3$ of its local value at $L$, where the vertical plate’s is $4/3$; a square plate has four edges, and a strip two, and the correlation’s area-over-perimeter length is a device for collapsing different shapes onto one curve rather than a length the flow uses. The Boussinesq approximation is assumed throughout, so $\beta\Delta T$ must be small. Thicknesses are where a profile has fallen to one per cent of its wall or peak value, the same convention a boundary layer always needs, and the layer Rayleigh number inherits it: on a five per cent thickness the crossing moves by a factor of a few in the Grashof number.

What the picture cannot show #

The edge itself. The similarity form starts at a sharp leading edge in fluid at rest, where the boundary-layer approximation fails; a real plate’s edge is thick, and the fluid arriving round it has its own history. The middle. A finite plate’s layers collide at its centre and turn upward, and the pressure that drove them is then set by the plume as well; the semi-infinite plate has no middle. The rolls. The layer’s own instability is estimated here by a still layer’s threshold, which is only a scale; the shear’s stabilising effect, and the distance the rolls need to grow, are a stability problem the solution does not pose. Large temperature differences, where the Boussinesq approximation fails, and a tilted plate’s actual layer, which is not similar and would have to be marched.

Who worked it out #

Stewartson set up the similarity form for a horizontal plate in 1958. Gill, Zeh and del Casal corrected the face it applies to in 1965; Rotem and Claassen solved it across Prandtl numbers in 1969 and photographed the flow; Pera and Gebhart extended the analysis to slightly inclined plates and to the stability of the layer in 1973. Lloyd and Moran’s measurements, made by an electrochemical technique on plates of several shapes, are the source of the fourth-root correlation, and Goldstein, Sparrow and Jones’s naphthalene measurements of the same year found a sixth-root law at the smallest Rayleigh numbers — a third exponent for the same plate, and a reminder that a correlation’s exponent records the range it was fitted over. The fifth root is the one the laminar layer from the edge has, and the exponents a Grashof number licenses were never only a quarter.

Still open: the plume the middle sends up #

Two flat layers drawn in from opposite edges of a strip meet at its centre and have nowhere to go but up, and they leave as a plane plume carrying all the heat the plate gave them. A plume above a line source has no wall and no edge: its only input is the heat it carries, which stays constant with height. The next calculation asks what that one invariant fixes, and finds a centreline speed that rises with height, as the fifth root, a temperature excess falling as the three-fifths power, and a closed-form solution at one Prandtl number against which the box scheme can be checked exactly. Whether the plume above a real strip forgets the plate that fed it within a few widths, and so takes the line source’s form, is the harder question beyond that, and it is the one that would say how much of the handbook correlation is the plume rather than the layer.