Over a plate in mercury the warm layer flows as if inviscid

A speed nobody imposed derived the heat transfer of a warm vertical plate by Squire’s integral method and set the answer against Ostrach’s exact similarity solution, which it quoted rather than computed. The reason it gave for quoting was a real one: the similarity equations are a boundary-value problem with two conditions at infinity, the usual way to solve them is to shoot outwards from the wall and adjust two starting values until the far conditions are met, and that shooting is unstable in exactly the place the comparison most needed it, at low Prandtl number.

Computing the exact solution properly turns out to change one of that essay’s conclusions and to explain the rest of them. The integral method does fail worst where the two layers have different thicknesses, as it said — but it fails low, by ten per cent at a Prandtl number of a hundredth, not high by twenty-seven. And the reason the exact answer behaves as it does at that end is that a liquid metal’s warm layer is very nearly inviscid: viscosity is confined to a sliver against the wall, and the rest of the layer’s speed is set by buoyancy against inertia alone.

The equations, and why a shot misses #

With $\eta = (y/x)(\mathrm{Gr}_x/4)^{1/4}$, a stream function $\psi = 4\nu(\mathrm{Gr}_x/4)^{1/4}F(\eta)$ and the temperature excess scaled to the wall’s, $\theta$, the boundary-layer equations for a vertical isothermal plate in the Boussinesq approximation reduce to

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

with $F(0) = F'(0) = 0$ and $\theta(0) = 1$ at the wall and $F'$ and $\theta$ vanishing far away. The heat transfer is $\mathrm{Nu}_x = -\theta'(0)(\mathrm{Gr}_x/4)^{1/4}$, so $\mathrm{Nu}_x/\mathrm{Gr}_x^{1/4} = -\theta'(0)/\sqrt2$; the wall shear is $F''(0)$. Shooting guesses $F''(0)$ and $\theta'(0)$, integrates outwards, and corrects the guesses until the velocity and temperature die at the far end.

The difficulty can be measured rather than described. Start an outward integration from the exact wall values with the wall shear wrong by a stated fraction, and record how far it gets before its velocity is off by a tenth of the peak or its temperature by a tenth of the wall’s excess. In air a shot one per cent off leaves the solution at $\eta = 3.2$, inside a layer whose temperature falls to one per cent of its wall value at 4.4; three thousandths off it survives to 16, past the layer’s edge. At a Prandtl number of 0.01 the warm layer reaches to $\eta = 35$, and a shot three thousandths off has left the solution by $\eta = 8$. Only within a thousandth does it reach the edge.

That is the cliff the earlier essay described, and its consequence is worse than slowness. A Newton iteration on the wall values needs the far conditions to evaluate its error; if the shot diverges before it gets there, the error is a property of the divergence rather than of the guess, and the iteration has nothing to steer by until it is already within a thousandth of the answer.

Posing it as one system instead #

The box scheme does not integrate at all. It writes the five first-order equations as differences on every interval of a grid at once — centred on each interval, so second-order accurate however the grid is stretched — adds the three wall conditions and the two far ones, and solves the whole set as a single nonlinear algebraic system by Newton’s method. The far conditions become equations at the last node rather than targets a trajectory has to hit, and an instability that grows outwards has nowhere to grow, because no value is ever computed from the values inside it alone.

The grid here crowds towards the wall in geometric progression, with the interval standing in for infinity set to twelve times the thicker layer’s natural width — $\mathrm{Pr}^{-1/2}$ in similarity units at low Prandtl number, $\mathrm{Pr}^{1/4}$ at high — and the crowding set so the first cell resolves the thinner. Every solution is reached by continuation from $\mathrm{Pr} = 1$ a quarter of a decade at a time, each the next one’s first guess.

At $\mathrm{Pr} = 0.72$ it returns $F''(0) = 0.67602$ and $-\theta'(0) = 0.50463$, against the 0.6760 and 0.5046 every table prints. Doubling the far boundary and refining the grid moves $-\theta'(0)$ by less than three parts in a million at Prandtl numbers of a thousandth, a hundredth, one and a hundred. Every entry of the table the earlier essay quoted, from a hundredth to a thousand, is reproduced to two parts in a thousand — every entry but one as it was printed.

The entry that had been converted twice #

The table’s value at $\mathrm{Pr} = 0.01$ was 0.0570. The box solution gives $-\theta'(0) = 0.08059$ there, converged to five figures, and $0.08059/\sqrt2 = 0.0570$. The entry was already the heat transfer in the form $\mathrm{Nu}_x/\mathrm{Gr}_x^{1/4}$, and the comparison divided it by $\sqrt2$ a second time on the way to the same form. Every other entry was $-\theta'(0)$ as tabulated, which is why the error showed in one place only.

Corrected, Squire’s closed form is 0.900 of the exact heat transfer at $\mathrm{Pr} = 0.01$, and 0.858 of it as the Prandtl number goes to zero. Above $\mathrm{Pr} = 0.7$ it is high by between one and eight per cent, peaking at 1.074 near a Prandtl number of seven — water — and settling to 1.01 at the viscous end. The integral method is therefore worst at low Prandtl number, as the earlier essay argued, and wrong in the other direction from the one it reported: the single-thickness profile under-predicts a liquid metal’s heat transfer. That essay has been corrected and now says so.

The sign is not a detail. A designer who believes the simple formula over-predicts sizes a liquid-metal exchanger with a margin in hand; the margin was never there, and the formula’s error and the safety factor point the same way. The direction follows from the physics the integral method leaves out, which is the subject of the rest of this essay.

A layer whose speed needs no viscosity #

The profiles say what is different. At $\mathrm{Pr} = 0.01$ the velocity rises to its peak at $\eta = 1.49$ and then stays near it across most of a warm layer that reaches to $\eta = 35$; the velocity layer and the warm layer end together. At $\mathrm{Pr} = 1000$ the warm layer is gone by $\eta = 0.44$ and the velocity reaches out to 21, fifty times as far, carried by viscosity into fluid that was never warmed. The two layers change places as the Prandtl number passes one, which is the whole of the other layer’s argument for a forced flow, now in a flow that makes its own speed.

At the low end the change is more than one of thickness. Over almost all of the warm layer the viscous term $F'''$ is negligible: the balance is buoyancy against inertia, and inertia alone carries the warmed fluid upward. Viscosity matters only in a sublayer next to the wall, where the no-slip condition has to be met, and the sublayer is thin against the warm layer by a factor that falls nearly as $\sqrt{\mathrm{Pr}}$.

The peak’s distance is 4.3 per cent of the warm layer at $\mathrm{Pr} = 0.01$ and 0.21 per cent at $10^{-5}$, a local slope of 0.45 over the lowest decade, still rising slowly towards a half. And the peak velocity approaches a value that can be named. Drop the viscous term and evaluate the momentum equation at the wall, where $F = 0$: what is left is $2F'^2 = \theta$, so an inviscid layer must slide past the wall at $F' = 1/\sqrt2$. The exact solution’s peak velocity, 0.576 at a hundredth and 0.699 at $10^{-5}$, is climbing to 0.7071. The sublayer’s job is to bring that slip speed to rest at the wall, and the more inviscid the fluid the less of the layer it needs to do it.

Why the integral method runs low there #

Squire’s derivation assumes one thickness and a velocity profile $t(1 - t)^2$ that peaks a third of the way out. In a liquid metal the true peak is a twentieth of the way out at $\mathrm{Pr} = 0.01$ and closer at lower, with the velocity nearly flat across the rest of the warm layer. The assumed profile puts too little fluid at speed near the wall, so it carries heat away from the wall too slowly, and the Nusselt number it predicts is too small. In water the error reverses: the true velocity layer is wider than the warm one, the single thickness drags the temperature profile out with the velocity, and the heat transfer comes out high. The comparison figure is a record of which way the shared thickness is wrong.

Two laws, and the fluids between them #

The two limits have clean forms. In the inviscid limit the heat transfer cannot depend on the viscosity at all, and the only combination of the remaining quantities is

$$\mathrm{Nu}_x = 0.600\,(\mathrm{Gr}_x\mathrm{Pr}^2)^{1/4},\qquad \mathrm{Gr}_x\mathrm{Pr}^2 = \frac{g\beta\Delta T\,x^3}{\alpha^2},$$

the Boussinesq number, which contains the thermal diffusivity twice and the viscosity not at all. In the opposite limit inertia drops out instead, and $\mathrm{Nu}_x = 0.503\,(\mathrm{Gr}_x\mathrm{Pr})^{1/4}$, the Rayleigh number’s form. The exact solution computed here climbs to both from below: 0.5994 at $\mathrm{Pr} = 10^{-5}$ against 0.600, and 0.5023 at $10^{5}$ against 0.503.

The surprising connection is that the low end is the same statement the forced-convection layer makes. The number that is an answer found the Blasius plate’s Prandtl exponent tending to a half at the bottom, where the heat transfer depends on the Péclet number alone: the velocity the heat meets is the free stream’s, not the boundary layer’s, and the viscosity has no say. Here there is no free stream, and the flow’s own inviscid speed takes its place; the group that results, $\mathrm{Gr}_x\mathrm{Pr}^2$, is the natural-convection Péclet number squared in effect, with the same viscosity-free character. Both flows lose their viscosity in the same limit for the same reason, and neither boundary layer is the one a sketch of a boundary layer would draw.

The two laws cross where they are equal, at $\mathrm{Pr} = (0.503/0.600)^4 = 0.494$. There the exact solution is 0.736 of either. Air, at 0.72, reads 0.421 against 0.600 in the first form and 0.387 against 0.503 in the second — each law over-predicts air’s heat transfer by thirty per cent or more. Air sits next to the crossover, where neither limit is within a fifth, which is why every correlation for air is a fit with a Prandtl function in it rather than either law, and why the integral method’s single Prandtl function did as well as it did there: a smooth interpolation between two powers is the right kind of object for a fluid that sits between them.

The Rayleigh number’s reputation as the natural-convection group comes from a problem where inertia genuinely has no say. A layer heated from below between rigid walls starts to convect at a Rayleigh number of 1,708 whatever its Prandtl number, because the question is whether a fluid at rest can be destabilised, and a fluid at rest has no inertia to weigh. A plate’s layer is never at rest; it moves from its leading edge upward, and how fast it can move against its own inertia is exactly what the Prandtl number decides. The same group that sets the threshold of a stationary layer is therefore the right collapse for a moving one only where the moving one is too viscous for its inertia to matter — at the top of the Prandtl range, and nowhere near a liquid metal.

What a liquid-metal designer reads off it #

Mercury at room temperature has a Prandtl number of about 0.025, and sodium near 500 °C about 0.005. At those values the inviscid law over-predicts by eight per cent and four per cent: close enough to use with a stated correction. The Rayleigh-number law, which is what a correlation fitted to air, water and oils becomes when its Prandtl function is dropped, over-predicts mercury’s heat transfer by a factor of 2.3. That is not a refinement; it is a different physics applied to the wrong fluid, and the one number that decides which physics applies here is the Prandtl number, before any Grashof number is consulted.

The same fluids misbehave the same way in a pipe. The heat the eddies do not carry found a liquid metal’s temperature spread across a turbulent pipe by conduction while its velocity was flattened by eddies, so that correlations built on air and water’s analogy between the two fail by a factor of three. The natural-convection version of that failure is quieter, because a laminar layer has no eddies to disagree with, but its origin is the same: the heat in a liquid metal goes where conduction takes it, and a correlation that assumed the heat follows the momentum has assumed the one thing these fluids do not do.

A worked case shows the scales. A plate five centimetres tall in mercury, ten kelvin above it, has a local Grashof number at its top of about $1.7\times10^8$, inside the laminar range. The exact solution gives a local Nusselt number of 10.0 there, a heat-transfer coefficient near 1,700 watts per square metre per kelvin, and a peak upward speed of about three centimetres a second. The viscous sublayer is 0.85 millimetres thick and the warm layer 14 millimetres: the velocity has reached its peak within the first seventeenth of the warm layer and stays near it across the rest.

How much fluid the layer draws in #

The far value $F(\infty)$ is the layer’s entrainment: a volume flux of $4\nu(\mathrm{Gr}_x/4)^{1/4}F(\infty)$ per unit width drawn sideways into the layer. It falls as $\mathrm{Pr}^{-1/2}$ at the low end and as $\mathrm{Pr}^{-1/4}$ at the high: 4.85 at a hundredth, 0.599 in air, 0.275 in water and 0.0765 in an oil of Prandtl number a thousand. In units of the thermal diffusivity rather than the viscosity, the low end’s entrainment is independent of the viscosity too, which is the same inviscid statement read on the mass budget rather than the heat.

The convention every number here rests on #

All of it is local: a Nusselt number on the distance $x$ from the leading edge, the wall-to-ambient temperature difference, and the Grashof number on the same $x$, for a plate held at one temperature. A plate’s average Nusselt number over a height $L$ is $4/3$ of the local one at $L$, and a plate heated at constant flux has different constants; a figure from one convention applied to another is a third of the way wrong before any physics has been missed. The fluid properties are constant and the density varies only in the buoyancy term, the Boussinesq approximation, which needs $\beta\Delta T$ small. The thicknesses quoted are where the temperature excess, or the velocity, has fallen to one per cent of its wall or peak value — conventions, as how thick is thin insists every boundary-layer thickness is.

What the picture cannot show #

The leading edge. Similarity holds only where the layer is thin against $x$; at the very bottom of the plate the boundary-layer equations themselves fail, and the local Nusselt number’s divergence as $x^{-1/4}$ is the model’s, not the plate’s. Transition. A vertical plate’s layer goes unstable and then turbulent near a Grashof number of $10^9$ in air, and earlier in a liquid metal whose inviscid layer has the inflected profile that instability favours; nothing here is valid past it. Variable properties. A liquid metal’s conductivity and a gas’s viscosity both change with temperature, and the Boussinesq approximation’s error grows with the temperature difference. Three dimensions. A real plate has side edges and a finite width, and a real liquid-metal pool has walls that the layer’s large entrainment reaches. And the sublayer’s own asymptotics. The slope of 0.45 over the lowest decade is still rising, and whether it reaches a half or carries a slowly varying correction is a question for a matched expansion this solution only approaches numerically.

Who solved it, and the tables it left #

Schmidt and Beckmann measured the air layer on a heated plate in 1930, and Pohlhausen reduced it to the similarity equations in the same paper and integrated them for air. Ostrach extended the solution across Prandtl numbers from a hundredth to a thousand in a NACA report of 1953, by numerical integration with the hand-adjusted shooting this essay measured, and his table has been reprinted in heat-transfer texts ever since. Le Fevre gave the two limiting coefficients in 1956, and Sparrow and Gregg the low-Prandtl behaviour the same year; Kuiken’s matched expansions of the late 1960s put the inviscid layer and its viscous sublayer on a formal footing. The box scheme is Keller’s, from 1970, written for exactly this class of boundary-layer problem. What is new here is only the error: a table entry copied after conversion, and a comparison that converted it again, in a number nobody had recomputed because the table it came from was trusted.

Still open: the plate laid flat #

A vertical plate drives its layer with buoyancy acting along it. Lay the plate flat, warm side up, and buoyancy acts across the layer instead; the only thing that can push the fluid along the plate is the pressure deficit the warm layer itself leaves beneath the colder fluid above, which grows with the layer’s thickness. That changes the exponents — the next calculation finds a heat transfer going as the fifth root of the Grashof number rather than the fourth — and it raises a question with a practical edge, since the correlations engineers use for a warm horizontal plate carry a fourth root anyway: whether the plate they were measured on had a boundary layer at all, or a plume rising from its centre.