A scraper held flat is pushed off the wall it scrapes

The eddies nobody stirs asked what a slow flow does in a sharp corner when it is driven from far away, and found that the corner fills with a geometric sequence of counter-rotating eddies, each about three times smaller and sixteen hundred times weaker than the one outside it. That question had no particular flow in it: the corner’s walls were still and whatever moved the fluid was somewhere else. The opposite question is the one that a painter’s scraper, a windscreen wiper and the doctor blade on a printing roll all pose. One wall of the corner moves. A surface carrying a viscous liquid slides under the edge of a blade that is held still, and the blade’s job is to stop the liquid going past.

That corner is driven at the corner. The motion is supplied at the apex itself, where the sliding wall meets the blade, rather than leaking in from a stirring elsewhere, and the answer changes completely. There are no eddies at any angle. Every streamline comes in along the moving wall, turns once and leaves along the blade. And the stress on the blade grows without limit as the apex is approached, so the force on it depends logarithmically on a length that the problem does not contain.

G. I. Taylor worked the problem out in a short note in 1962, under the title of scraping viscous fluid from a plane surface, and the result that a user of a scraper can feel is its direction. The liquid does not merely drag the blade along with the wall. It pushes the blade away from the wall, and the flatter the blade is held, the harder it pushes — roughly as one over the square of the angle.

The flow that the geometry permits #

The setting is the creeping-flow limit, the one in which inertia has been dropped entirely and the stream function obeys the biharmonic equation $\nabla^4\psi = 0$. For paint under a scraper that is a good approximation where it matters: a thick paint has a viscosity of order a pascal-second, a scraper’s speed is a fraction of a metre per second, and near the edge the relevant length is a fraction of a millimetre, which puts the local Reynolds number well below one. Far from the edge inertia may matter; at the edge, where the stress is largest, it does not.

Take the moving wall along $\theta = 0$, sliding towards the apex at speed $U$, and the blade along $\theta = \alpha$, still. The corner has no length in it, as Moffatt’s did, but now the wall’s speed is a velocity that is the same at every distance from the apex. A stream function that reproduces a fixed velocity at every radius must grow in proportion to the radius, so the form is

$$\psi = r\,f(\theta),$$

the exponent $\lambda = 1$ fixed by the driving instead of chosen by an eigenvalue equation. The biharmonic equation then allows

$$f(\theta) = A\sin\theta + B\cos\theta + C\,\theta\sin\theta + D\,\theta\cos\theta,$$

and four conditions decide the four constants: no flow through the wall, $f(0) = 0$; the wall’s speed on the wall, $f'(0) = -U$; no flow through the blade, $f(\alpha) = 0$; and no slip on the blade, $f'(\alpha) = 0$. The radial velocity is $u_r = f'(\theta)$, independent of the radius, so the profile across the corner is the same at every distance from the apex. The pressure is $p = -\mu(f' + f''')/r$ and the vorticity $-(f + f'')/r$, and both diverge as the apex is approached.

That single solution is the whole local flow. Moffatt’s eddies are still available — the eigensolutions with complex $\lambda$ can be added to it to match whatever the far field demands — but at 45° the slowest of them has an exponent whose real part is 6.39, so they fall off towards the apex as $r^{6.39}$ against the driven flow’s $r^1$. Near the edge of a scraper they are swamped. The eddies of an undriven corner exist because nothing else is there to dominate them; a sliding wall is something else.

Why there are no eddies #

The streamlines in the figure are drawn at five levels of the stream function, and because $\psi = r f(\theta)$ each is the same curve magnified: it arrives along the moving wall far out, sweeps round the corner, and returns along the blade. The pattern is the same at 20°, at 90° and at 135°. What would make an eddy is a change of sign of $f$ across the corner — a band of angles in which the fluid circulates the other way — and $f$ keeps one sign at every angle from the smallest to the largest. The check is direct: the function is sampled at four hundred angles across the corner for blades at 20°, 45°, 90°, 135° and 170°, and never crosses zero.

The physical reason is that the driven flow has somewhere to go. The moving wall drags a sheet of liquid into the corner; the blade stops it; the liquid must return, and the only way out is back along the blade. There is a single circulation, of the sense the wall imposes, filling the whole wedge. The undriven corner has no such source and no such sink: its motion is a residue of something outside it, and each eddy is driven by the one outside it through a dividing streamline, which is what makes the sequence and what makes it decay. A corner with its own drive has no need of a sequence.

This is the first half of the contrast between a corner stirred from far away and one driven at its apex. A corner’s flow is decided not by its angle alone but by where the motion enters. Entering from afar, it produces structure all the way to the apex and almost nothing measurable there. Entering at the apex, it produces one simple circulation and the largest stresses in the problem.

The flow turns back a third of the way across #

The radial velocity across the corner has a shape that hardly depends on the blade’s angle. Next to the moving wall the liquid travels into the corner at the wall’s speed; next to the blade it is still; in between it must reverse, because the net flow into the corner through any arc is zero. The figure shows where it reverses: at 0.335 of the corner’s angle for a 30° blade, 0.339 at 60°, 0.344 at 90° and 0.355 at 150°.

The third is not a coincidence, and it is the bridge to the most useful thing about this solution. When the blade is nearly flat the corner is a thin wedge, a gap $h = \alpha r$ that closes towards the apex, and the flow in a thin closing gap is exactly what lubrication theory describes: a sheared Couette profile from the moving wall plus a parabolic Poiseuille profile driven by the pressure gradient that the gap’s narrowing produces. With zero net flux the profile is $u = U(1 - 3s)(1 - s)$ across the gap, with $s$ the fraction of the way from the moving wall, and it vanishes at $s = 1/3$. The exact solution reverses at 0.335 for a 30° blade; the thin-gap theory says a third exactly, and the difference grows only slowly with the angle.

The force, and how it grows as the blade goes flat #

On the blade the normal stress is $\mu f'''(\alpha)/r$ and the shear stress $\mu f''(\alpha)/r$. Both grow as one over the distance from the apex. That is the singularity of the driven corner, and it means the force on a blade of length $R$ is an integral of $1/r$,

$$F = \mu U\,\ln(R/\varepsilon)\,\big[\,\text{a number set by } \alpha\,\big],$$

finite only because the corner is ended at some small distance $\varepsilon$. The numbers in the first figure are those brackets, resolved along the wall and away from it.

They show the whole of Taylor’s practical conclusion. At 90°, a blade held upright, the liquid pushes it off the wall with 1.36 and drags it with 2.14, both in units of $\mu U \ln(R/\varepsilon)$. At 45° the lifting part is 8.56 and the drag 4.88. At 10° the lifting part is 196 and the drag 22.9. Laying the blade from 45° down to 10° multiplies the drag by 4.7 and the lift by 22.9. The resultant turns towards the wall’s normal as the blade goes flat: it points 32° from the wall at 90°, 60° at 45° and 83° at 10°.

The small-angle law has the same lubrication origin as the reversal at a third. In a thin wedge with zero net flux, the pressure gradient must cancel the Couette flux exactly, which needs $dp/dr = 6\mu U/h^2 = 6\mu U/(\alpha^2 r^2)$, and integrating that pressure over the blade gives a normal force of $(6/\alpha^2)\,\mu U \ln(R/\varepsilon)$. The shear on the moving wall is $4\mu U/h$, which gives an along-wall force of $(4/\alpha)\,\mu U\ln(R/\varepsilon)$. At 10° those are 197.0 and 22.92 against the exact 195.8 and 22.87: lubrication is within 0.6 and 0.2 per cent. At 45° it overstates the lift by 14 per cent, at 90° by 78 per cent, and past that the thin-gap picture has nothing to say. The scraper is lubrication theory’s wedge carried to angles where the wedge is not thin, and lubrication theory is the scraper’s small-angle limit.

A slider bearing generates its load-carrying pressure the same way: the moving surface drags oil into a gap that narrows, and the pressure has to rise to push back the excess. The difference is that a bearing’s gap stays open at its exit and the scraper’s closes, so the bearing’s pressure is bounded and the scraper’s is not. A scraper is a slider bearing with no exit, which is why it carries load in a direction its user did not intend.

What a flat blade does #

The refuted claim is the intuition that a blade laid low sits more firmly on the surface because more of it is close to the surface. The force says the opposite. The liquid in the wedge is a bearing film, and a flatter wedge is a better bearing. A blade held at 10° with a given pressing force is lifted by a film whose lifting force per unit of $\ln(R/\varepsilon)$ is 23 times larger than at 45°; the logarithm must then be 23 times smaller for the same force, which for a centimetre blade means a gap of millimetres where there was a micrometre. The blade rides up on the paint it was meant to remove.

That is the reason the working instruction for scraping and for doctoring a roll is a steep angle, and why a blade worn to a flat heel behaves as if it had been laid down. It is also why a blade that is intended to leave a film — a coating knife, a squeegee laying down a layer of known thickness — is set at a shallow angle or given a flat land: there the lift is the mechanism rather than the nuisance.

The drag, the part along the wall, grows too, though far more slowly, as $4/\alpha$ rather than $6/\alpha^2$. The power the wall spends is the drag times its speed, and that power is the same thing as the rate at which viscosity converts the corner’s motion to heat. The model checks one against the other: the wall’s power, computed from the shear stress on the wall, matches the dissipation integrated over the corner to four parts in a hundred million at 45°.

Where the work is spent #

The dissipation is concentrated next to the moving wall. In the quarter of the angle that borders it, 66.6 per cent of the work is burnt for a 30° blade and 61.6 per cent for a 90° one; only at 150°, where the corner is so wide that the wall’s region has room to spread, does the share fall to half. The thin-gap theory gives the reason: the shear rate in the gap is $(U/h)(6s - 4)$, largest at the moving wall where the Couette and Poiseuille parts add, and the share of its square in the first quarter of the gap is 67.2 per cent. The heat is made where the paint is sheared hardest, in a layer next to the surface being cleaned rather than next to the blade.

For a scraper that matters little. For a doctor blade running continuously on a roll it decides where the heat goes — into the roll, which carries it away — and why a blade on a fast roll stays cooler than its contact stresses suggest. The present model has no temperature in it, and a film that heats itself takes up that coupling; the scraper only says where the heat is released.

The length the problem does not contain #

The force is proportional to $\ln(R/\varepsilon)$, and nothing in the creeping-flow problem sets $\varepsilon$. The stress grows as one over the distance from the apex all the way in, and its integral diverges at both ends: at the outer end the blade has a length and the similarity solution stops being the flow, and at the inner end something must stop the corner from being a corner.

The figure puts numbers on it for a blade 10 cm wide and 1 cm long, a thick paint of 1 Pa·s, and a wall moving at 0.2 m/s. At 45° the paint pushes the blade off with 1.58 N if the corner ends a micrometre from the apex and 0.788 N if it ends a tenth of a millimetre away. Each halving of that length adds 0.119 N, which is exactly what doubling the blade’s length would add: in a logarithm the inner and outer lengths enter only as their ratio. At 30° the same blade is pushed off with 3.81 N at a micrometre; at 60°, with 0.797 N.

What ends the corner is the physics the model has left out. It might be a real gap under the blade, the edge never quite reaching the surface and the liquid squeezing under it. It might be the radius to which the blade’s edge has been ground or worn. It might be slip at the wall, if the liquid is a polymer melt that really does slip or the surface is coated. It might be the paint’s own structure — pigment particles a few micrometres across, below which the paint is not a continuum. Each gives a different $\varepsilon$, and the force depends on it only weakly: a factor of a hundred in $\varepsilon$, from a micrometre to a tenth of a millimetre, changes the force on a centimetre blade by a factor of two. The logarithm forgives almost everything, which is why Taylor’s estimate is useful despite the singularity.

The same logarithm appears where a liquid’s edge moves over a solid: the drop a no-slip wall would never let spread is the same corner turned the other way, a wall sliding under a liquid wedge whose upper face is free rather than solid, and its force too diverges as the logarithm of a length that only physics outside the continuum can set. Taylor’s scraper was the first worked case of that singularity and its moving-contact-line descendant is the famous one.

The convention: what is a force here #

The forces quoted are those the liquid exerts on the blade, per unit span, in units of $\mu U \ln(R/\varepsilon)$, resolved along the moving wall — positive in the direction the wall moves — and normal to it — positive pushing the blade away. The blade is held still and its edge touches the wall at the apex; the wall slides towards the apex. Angles are measured from the wall through the liquid to the blade, so a blade at 90° stands upright and one at 170° is folded back almost flat on the side the liquid leaves by. Distances in the numerical example are a blade 1 cm long and 10 cm wide.

What the solution leaves out #

It is two-dimensional, steady and inertialess. A real scraper ends at its corners, where liquid escapes sideways, and over a blade whose width is ten times its length that leak is felt over roughly a tenth of the span. Inertia enters at a distance from the edge of order $\nu/U$, beyond which the flow is no longer a similarity solution: five millimetres for the 1 Pa·s paint at 0.2 m/s, so the corner is inertialess over half the blade, but two micrometres for water at half a metre per second, so a wiper on a wet windscreen is a creeping corner only at its very edge. Paint is shear-thinning, and since the shear rate in the corner varies as one over the distance from the apex, its viscosity varies across the corner and the $1/r$ law is replaced by a weaker power; the logarithm becomes finite without any $\varepsilon$ at all, at the cost of a parameter the Newtonian solution does not need. A liquid with a free surface upstream — paint is not a full bath — has a meniscus in front of the blade, and that is where a fibre or a plate pulled from a bath takes its film from.

None of those changes the two structural claims: that the driven corner has no eddies, and that its stress grows into the apex so the force depends on what ends it.

Who found it #

Taylor’s note of 1962 is three pages long. It sets up the corner with one sliding wall, writes the same four-term solution, and derives the force on the scraper including its logarithmic dependence on the inner cut-off, which he discusses in a sentence. Moffatt’s paper on corner eddies two years later took it as the driven counterpart of the undriven problem and drew the contrast between them, and Batchelor’s textbook carried the scraper into the standard teaching of slow flow, where it remains the cleanest example of a stress singularity at a moving contact. Huh and Scriven’s paper of 1971 on the moving contact line is the same calculation with one face of the wedge free.

How it was checked #

The four boundary conditions are met to rounding error, $9 \times 10^{-16}$. The radial momentum equation, with the pressure derived from the solution and the velocity’s Laplacian taken by central differences, is satisfied to $2 \times 10^{-8}$, the size of the differencing error. The wall’s power equals the dissipation integrated over the corner to $4 \times 10^{-8}$ of itself, which checks the stresses and the velocity against each other independently of the boundary conditions. The along-wall force at 45° from the blade’s stresses, 4.8849, matches the power divided by the wall’s speed to six figures. At 5° the lift is 0.15 per cent below lubrication theory’s $6/\alpha^2$ and the drag 0.05 per cent below $4/\alpha$. The stream function was sampled at four hundred angles for five blade angles from 20° to 170° and keeps one sign throughout. And the solver refuses a blade at zero, at 180° and at a negative angle, where the corner is not a corner.

Still open: a blade that rides on what it lets through #

Every number here is for a blade whose edge is held on the wall, with the corner ended by an $\varepsilon$ chosen from outside. A real scraper or doctor blade is pressed onto the surface with a load, and the film under its edge is not given but found: the blade rises until the lift of the liquid it lets through balances the load. That makes $\varepsilon$ an output — the thickness of the layer the blade leaves behind — and turns Taylor’s corner into a coating problem. The next calculation would add a flat land of a given length under the blade’s edge, solve the lubrication flow through it matched to the corner’s flow above, and ask how the film left on the surface depends on the blade’s angle and load: whether the steep blade that the force argues for also leaves the thinnest film, and how much load a flat blade needs to clean as well.