A bowed meniscus keeps most of its slip until the groove floods

Twice as slippery along as across computed Philip’s slip lengths for a surface of parallel stripes, solid and gas, and a post holds liquid back by its radius did the same for an array of posts. Both treated every gas surface as a flat plane in line with the solid tops. That is the shape a meniscus takes when the liquid above it and the gas below it are at the same pressure, and it is almost never the case in use. A liquid flowing through a channel walled with such a texture is at a higher pressure than the gas trapped in the grooves, by however much drives it plus whatever hydrostatic head sits on it, and the meniscus answers that difference the only way a liquid surface can: it curves.

Pinned at the two edges of its stripe, a meniscus under a pressure difference $\Delta p$ becomes a circular arc of radius $\gamma/\Delta p$, with $\gamma$ the surface tension, dipping into the groove beneath. The worry that follows is obvious. The slip came from the liquid being free to slide over gas, and a meniscus that sags into its groove has drawn the liquid down into the texture, closer to the groove’s walls and further from the plane the slip was measured against. Every increment of pressure should cost some slip, and a surface used under load should be delivering markedly less than the flat-meniscus number advertised for it.

The calculation says that the worry is right in direction and wrong in size. The slip does fall. But it falls slowly, and by the time the pressure is high enough for the arc to reach the groove’s wall at its advancing angle — the point at which the liquid floods the texture and the slip vanishes altogether — most of it is still there.

The arc and the pressure that ends it #

The geometry is fixed by two numbers. The gas stripe has a half-width $a$, and the meniscus spanning it is an arc of radius $R = \gamma/\Delta p$. Measuring the pressure in the stripe’s own units, $P = a\Delta p/\gamma = a/R$, the arc meets the solid’s edge at an angle $\beta$ below the horizontal with $\sin\beta = P$, and it dips at its middle by

$$s = a\,\tan(\beta/2).$$

A flat meniscus is $P = 0$. A semicircle is $P = 1$, and no arc pinned at those edges can hold more: past that pressure the meniscus would have to bulge sideways beyond its own pinning points.

Long before then the groove floods. The meniscus stays pinned at the edge only while its angle against the groove’s wall is below the wall material’s advancing contact angle, $\theta_a$. For a groove with vertical walls the arc’s slope at the edge is $\beta$ below the horizontal, which puts the liquid at $90^\circ + \beta$ against the wall, and the contact line moves down the wall — the texture floods, the Cassie state becomes the Wenzel state — when $90^\circ + \beta = \theta_a$. The flooding pressure is therefore $P = \sin(\theta_a - 90^\circ)$: 0.342 for a wall advancing at 110°, 0.5 at 120° and 0.866 at 150°. That is the classical result $\Delta p = -2\gamma\cos\theta_a/w$ for a groove of width $w = 2a$, and the force a contact line holds is a range explains why it is the advancing angle, not the equilibrium one, that decides it: a pinned line can carry any angle up to that limit and then gives way at once.

The dips at those floods are modest: 0.176 of the half-width at 110°, 0.268 at 120° and 0.577 at 150°. For water on a texture with an 8-micrometre half-width and walls advancing at 120° the flood comes at 4.5 kPa — under half a metre of water head — and the meniscus has dipped 2.1 micrometres by then, which is also the shallowest groove that survives to the flood without the arc touching its floor.

The flow over a bowed meniscus #

The flow is the one the stripes essay solved with the boundary moved. Shear runs along the stripes, so the velocity $w$ is along them and depends only on the position across them and the height; in the creeping limit it is harmonic. It vanishes on the solid tops, carries no shear stress on the meniscus — the normal derivative of $w$ is zero there — and far above the surface grows as $y + b$, which defines the slip length $b$. Flat, the answer is Philip’s $b = (L/\pi)\ln\sec(\pi\phi/2)$ for period $L$ and gas fraction $\phi$.

Bowed, there is no closed form, and the solution here is by a boundary-element method: the meniscus and the solid top of one period are cut into panels, crowded towards the pinned edges where the flow has a square-root singularity, and the periodic Green’s function of Laplace’s equation turns the problem into a linear system for the unknown surface velocity on the meniscus and the unknown shear on the solid. The slip length is one more unknown, tied to the rest by the condition that no net flux leaves through the surface. With the meniscus flat the method returns Philip’s slip at a gas fraction of 0.8 to four parts in a hundred thousand.

First order: a dip always costs slip #

For a shallow dip the answer can be written down. Lowering a shear-free boundary by $s\,\eta(x)$, where $\eta$ is the dip’s shape, changes the slip length by an amount that follows from the reciprocal identity between the flat flow and the perturbed one:

$$\delta b = -\,s \int_{\text{gas}} \eta(x)\left(\frac{\partial w_0}{\partial x}\right)^{2} dx,$$

where $w_0$ is the flat flow’s velocity along the meniscus. The integrand is positive wherever the meniscus dips, so a dip always costs slip, and the cost is weighted by the square of how fast the gas surface’s own velocity changes across the stripe. Philip’s solution gives that velocity exactly — $w_0 = (L/\pi)\,\mathrm{arccosh}\!\big(\cos(\pi x/L)/\cos(\pi a/L)\big)$ — so the first-order loss is a single integral.

The loss per unit dip grows with the gas fraction, from 0.43 of the slip per half-width of dip on a surface one-fifth gas to 0.78 on a surface 95 per cent gas. A gassier surface is more slippery and more of its slip is carried by the meniscus, so more of it is exposed to the meniscus’s shape. The boundary elements at a dip of one hundredth of a half-width agree with the integral to seven parts in a thousand at gas fractions of 0.5 and 0.8 — the size of the second-order term at that dip — which is the check that both the integral and the solver are right.

Where along the meniscus the slip is lost #

The integrand says something the total does not: where across the stripe the loss happens.

The loss is concentrated near the pinned edges. On a surface 0.8 gas, 89 per cent of it is incurred in the outer half of each gas stripe, and the density is largest right at the edge, where the dip itself is vanishingly small. The middle of the meniscus, where it sags furthest, contributes nothing at first order. The reason is the weighting. In the middle of a gas stripe the surface velocity is at its maximum and flat across, so moving the surface there changes nothing to first order; at the edge the surface velocity rises from zero, at the pinned solid, as the square root of the distance, and its gradient is unbounded. A small dip in a region where the surface is being sheared hard costs more than a large dip where it is not.

The same weighting explains why a soft or moving boundary is so often mistaken for slip and back again. A soft wall is read as slip only at second order found a wall that gives under pressure changing a drainage force at a different order from the one a slip length acts at; here the boundary that moves is the slip-carrying one itself, and it moves the slip at first order, but with a coefficient set almost entirely by the few tenths of the stripe nearest its edges. And how many things a flow must be told is the reason a moved boundary matters at all: the shape of the surface on which the shear-free condition is applied is part of the condition, and changing it changes the answer even when the condition’s wording does not change.

That is also why the boundary elements had to be crowded at the edges. The physics that sets the loss and the singularity that makes the flat problem hard are in the same place.

The deep arc loses less than its first-order line #

A linear estimate made at small dips and carried to the flood would overstate the loss.

At the dip a 120° wall allows, the first-order line predicts that a surface 0.8 gas keeps 0.840 of its slip, and the full solution keeps 0.879. At a 150° wall’s dip the line predicts 0.655 and the arc keeps 0.804. The curve bends away from its tangent, and the reason is again the edge. As the arc deepens its slope at the pinned edge steepens, so near the edge the meniscus no longer lies in the plane of the solid but angles down away from it, and the liquid in the corner between the solid’s top and the steepening meniscus moves less and less. The edge region is where the first-order loss was concentrated; as it slows, there is less there to lose. Most of the deepening then happens in the middle of the stripe, which was costing little to begin with.

No gas fraction loses a quarter before its groove floods #

The first figure puts the pieces together, and the conclusion is uniform across the range a working surface occupies. At a 110° wall’s flood a surface half gas keeps 0.930 of its slip, one 0.8 gas keeps 0.913 and one 0.95 gas keeps 0.888. At a 120° wall’s flood those become 0.903, 0.879 and 0.845. At a 150° wall’s — a pressure no plain vertical-walled texture of an ordinary hydrophobic coating can reach, since smooth surfaces rarely advance past 120° — they are 0.840, 0.804 and 0.752.

The flat-meniscus slip length is therefore an honest number for a striped surface under load, within ten to fifteen per cent, all the way to the pressure at which the surface stops working. The decline under pressure is a correction; the failure, when it comes, is total. A measured slip that collapses with rising pressure is evidence that grooves are flooding — the menisci depinning one by one, or gas dissolving into an undersaturated liquid until the arcs can no longer span their grooves — not that the menisci are merely bowing.

A crevice keeps the nucleus a free bubble loses rests on the same pinned arc for a different purpose: there the curvature lets gas survive in a crack that would dissolve as a free bubble. Here the arc’s pinning is what keeps the slip, and its depinning is what ends it.

Scale changes the pressure, not the share #

Every number so far is in the stripe’s own units, so the share of slip kept at a flood depends only on the gas fraction and the wall’s advancing angle. The texture’s size enters in two places, and they pull in opposite directions. The slip length is proportional to the period. The flooding pressure is proportional to one over the half-width.

Their product is independent of scale. For water on a surface 0.8 gas with walls advancing at 120°, the slip kept at the flood times the flooding pressure is 0.41 times the surface tension: 6.57 micrometres to 4.5 kilopascals with a 20-micrometre period, 0.657 micrometres to 45 kilopascals with a 2-micrometre one. Half gas, the product is 0.20 times the surface tension; 0.95 gas, 0.72. A designer can buy pressure resistance by making the texture finer and pays for it in slip one for one, at a rate set by surface tension and nothing else. The bowing of the meniscus — the subject of this essay — enters the product only through the kept share, and moves it by twelve per cent.

The convention: what is held fixed #

Shear is along the stripes. Period, half-width and dip are lengths; the dip is quoted as a share of the half-width, the pressure as $a\Delta p/\gamma$, and the slip as a share of the flat meniscus’s at the same gas fraction. The meniscus is clean, so it carries no shear stress, and pinned at the solid’s edges, so the gas fraction is the same at every pressure. The grooves have vertical walls, so flooding is decided by the advancing angle alone; a groove deep enough that the arc never touches its floor is assumed. The liquid is in creeping flow and the gas is taken as inviscid.

What the model leaves out #

The direction across the stripes. Shear across them makes the liquid flow over and under each arc, a biharmonic problem with the same boundary moved, and the across slip — half the along slip when flat — need not lose the same share. Posts. A post array’s menisci are saddle-shaped rather than cylindrical arcs, and the post essay’s larger slip may be more or less fragile. The gas’s own viscosity, which at the deepest dips begins to matter in the lower part of the groove. Contamination. A meniscus carrying a trace of surfactant resists being sheared, and a thousandth of a monolayer holds a bubble still — for slip, that loss is likely larger than anything the bowing does. Dissolution. The gas in a pressurised groove dissolves into an undersaturated liquid; the arc then deepens at constant pressure until it floods, which is a slow failure this model describes only as a sequence of arcs.

Who worked it out #

Philip gave the flat-meniscus slip lengths in 1972. The effect of meniscus curvature on longitudinal slip was taken up in the 2000s, first to first order in the protrusion angle by Sbragaglia and Prosperetti and then numerically and analytically by Teo and Khoo, Davis and Lauga, and Crowdy, mostly for menisci bulging out into the liquid rather than sagging into the groove. The classical flooding pressure for a groove is the capillary-pressure balance of the Cassie–Wenzel transition, written in terms of the advancing angle.

How it was checked #

The flat meniscus returns Philip’s slip to $4 \times 10^{-5}$. Doubling the panels at the 120° flood’s dip moves the slip by $3 \times 10^{-4}$ of itself, and raising them from 240 to 400 at the 150° flood’s dip by $5 \times 10^{-4}$. A dip of a two-hundredth of the half-width loses the first-order integral’s slip to $7 \times 10^{-3}$ at gas fractions of 0.5 and 0.8, a difference of the size of the second-order term and the panels’ own error together. The slip falls monotonically with pressure at every gas fraction tried, and every deep arc sits above its first-order line. At the semicircle, which no real wall reaches, the panels converge slowly — the arc meets the solid at a right angle and the corner’s singularity strengthens — and the solver’s value there is quoted nowhere. The solver refuses a gas fraction of zero, a dip past a semicircle, a pressure no arc can hold and a wall that wets.

Still open: across the stripes, where the liquid must climb #

Every number here is for shear along the stripes, where the liquid slides over each arc without crossing it and the flow is harmonic. Shear across the stripes carries the liquid down into each dip and up out of it, so a bowed meniscus there is an obstacle as well as a free surface. The next calculation would solve the biharmonic problem across the stripes with the same pinned arcs, ask whether the across slip — exactly half the along slip when the menisci are flat — keeps its factor of two as the arcs deepen, and so whether the anisotropy a striped surface is designed for survives the pressure it is used at.