A washed filter works as long as it rests

A filter is slowed by what it has caught follows a constant-pressure filtration onto Ruth’s straight line: the particles a cloth catches form a cake, the cake is a packed bed whose resistance grows with every litre filtered, and the filtrate grows only as the square root of time. It finds that a batch filter which has to be emptied and refilled between runs should be stopped while it is still producing well, because the last litres of a long run are the most expensive in the cycle.

Most batch filters do one more thing before they are emptied. The finished cake is full of the liquid it was made from — mother liquor, in the trade — and that liquid carries whatever was dissolved in the slurry. If the solids are the product, the dissolved salts are an impurity; if the liquid is the product, what is left in the pores is a loss. Either way the cake is washed: clean liquid is pushed through it to displace what is in its pores. This essay adds that wash to the cycle and asks what it does to the best cycle, how much wash a cake needs, and whether a soft cake should be made thin because a thick one would spend its life resisting its own wash.

The cycle, and the one rule its optimum obeys #

Per square metre of cloth, Ruth’s law for a filtration at constant pressure $\Delta p$ gives the time to collect a filtrate volume $V$ as

$$t_f = aV^2 + bV, \qquad a = \frac{\mu\alpha c}{2\Delta p}, \qquad b = \frac{\mu R_m}{\Delta p},$$

with $\mu$ the liquid’s viscosity, $\alpha$ the cake’s specific resistance, $c$ the mass of solid deposited per cubic metre of filtrate and $R_m$ the cloth’s resistance. The quadratic term is the cake and the linear one the cloth. A cycle then stands idle for a downtime $t_d$ while the cake is discharged and the filter closed again, and its average output is $V/(t_f + t_d)$.

Maximising that output over $V$ takes one line. The derivative vanishes where $aV^2 = t_d$: the part of the working time the cake is responsible for equals the time the filter stands still. The cloth’s share, $bV$, drops out of the condition entirely. For the silica slurry of the earlier essay — ten-micron particles packed at a porosity of 0.4, one bar, a downtime of twenty minutes — that puts the best cycle at 21.4 minutes of filtering, of which twenty are the cake’s and 1.4 the cloth’s.

The same arithmetic has a long history somewhere quite different. In 1913 Ford Harris, an engineer at Westinghouse, asked how many parts a factory should make in one batch when every batch carries a fixed set-up cost and every part made early carries a storage cost that grows with the batch. His answer, the economic order quantity, sets the batch where the set-up cost per part equals the storage cost per part, and it is the square-root formula still taught in every operations course. A batch filter is the same problem with time in place of money: the downtime is the set-up charged once per batch, and the cake’s growing resistance is a cost per litre that rises with the size of the batch. The best filter cycle is an economic order quantity, and $V^* = \sqrt{t_d/a}$ is Harris’s formula.

A wash runs at the slowest rate the filtration ever had #

The wash goes through the finished cake. Nothing about the cake changes while it does — no solids arrive — so the wash flows at a constant rate, and that rate is the one the filtration had at its last instant, $1/(2aV + b)$.

That is the slowest rate of the whole filtration, and for a cake-dominated filter it is half of the filtration’s mean rate, $V/t_f = 1/(aV + b)$. Each cubic metre of wash therefore takes twice as long to pass as an average cubic metre of filtrate did. The factor of two is not a detail of this slurry: it is the square-root law read backwards, because a volume that grows as $\sqrt{t}$ is collected at a rate that has fallen to half its average by the time the volume is collected.

A cake built from $V$ of filtrate holds $kV$ of liquid in its pores, with $k = \varepsilon c/\rho_s(1 - \varepsilon)$ from the cake’s porosity $\varepsilon$ and the solid’s density $\rho_s$. A wash of $w$ pore volumes then takes

$$t_w = wkV\,(2aV + b) = 2awk\,V^2 + bwk\,V,$$

which is quadratic in the cake exactly as the filtration was. So washing does not change the shape of the problem: it multiplies the cake’s coefficient by $1 + 2wk$ and the cloth’s by $1 + wk$, and the rule survives unchanged. The cake’s share of filtering and washing together equals the downtime, whatever the wash ratio.

The first figure shows the budget for a concentrated slurry, 400 kilograms of silica per cubic metre of filtrate. With no wash the best cycle filters for twenty minutes and a little more. With a wash of two pore volumes it filters for 14.7 minutes and washes for 5.8. With five pore volumes the two times are nearly equal, 10.3 minutes of filtering and 10.2 of washing. Throughout, the dashed line — everything the cake is responsible for, filtering and washing together — sits on the downtime. The wash does not stretch the working part of the cycle; it takes a share of it from the filtering.

What the wash costs depends on the slurry, not on the wash #

Because washing multiplies the cake’s coefficient by $1 + 2wk$, the best output falls by very nearly the square root of that factor: $V^*$ and the output both scale as $1/\sqrt{a(1+2wk)}$ when the cloth is negligible. The number that decides whether a wash is cheap or expensive is therefore $2wk$ — the wash volume as a fraction of the filtrate volume, doubled because the wash runs at the final rate.

For a dilute slurry, 50 kilograms of silica in each cubic metre, the cake holds only 1.26 per cent of the filtrate’s volume in its pores. Two pore volumes of wash then cost 2.4 per cent of the output — one minute of washing in a cycle of forty-one, which no plant would notice. The concentrated slurry is eight times richer in solid, so its cake holds eight times the pore liquid per litre filtered: $k = 0.10$. The second figure is that slurry. Its best output without a wash is 0.455 cubic metres per square metre of cloth per hour, from a cycle that collects 0.307 cubic metres. One pore volume of wash lowers it to 0.415, two to 0.384 and five to 0.321; two pore volumes cost 15.6 per cent.

The marks move left as the wash grows, from 0.307 to 0.217 cubic metres per cycle at five pore volumes. That is the square-root law again: the best batch is $\sqrt{t_d/A}$ and washing makes $A$ larger. A washed filter should be emptied sooner than an unwashed one, and a filter washing a concentrated slurry sooner still. The engineer who has tuned a filter’s cycle by trial on an unwashed product and then adds a wash has made the cycle too long, by about a sixth in cake thickness for two pore volumes of a concentrated slurry.

The same rule settles a question that is often argued the other way, whether to wash at a lower pressure than the filtration to avoid cracking the cake. The wash then runs at a lower rate than the final filtration rate, the factor $2wk$ is multiplied by the ratio of the two pressures, and the cost of the wash rises in proportion. Cracks are a real reason to do it; the price in output can be read straight off $2wk$.

How many pore volumes: the wash front smears #

A wash of one pore volume would displace every drop of the pore liquid if the wash advanced through the cake as a flat front. It does not. Liquid in the fast channels between grains runs ahead of liquid in the slow ones, and molecular diffusion blurs the boundary between wash and mother liquor as it goes. The mixing is the same longitudinal dispersion that smears a pulse through a packed bed into a residence-time distribution, and it enters the washing problem as one number: the cake’s Péclet number, $\mathrm{Pe} = uL/D_L$, its thickness $L$ over the length $D_L/u$ on which the front smears.

With $\xi$ the depth through the cake as a fraction of its thickness and $w$ the wash passed in pore volumes, the concentration of the original liquid in the pores obeys

$$\frac{\partial c}{\partial w} + \frac{\partial c}{\partial \xi} = \frac{1}{\mathrm{Pe}}\,\frac{\partial^2 c}{\partial \xi^2},$$

starting at one everywhere, with clean wash entering the face of the cake and the effluent leaving at the cloth. It was solved here by the Crank–Nicolson method on up to thirteen thousand cells, fine enough that the numerical smearing is far below the physical. Two checks come with it. What is left in the cake plus what has left in the effluent equals what was there to begin with — solute, like mass, has nowhere else to go — to $6\times10^{-11}$, and at a Péclet number of 400 the effluent concentration agrees with the error-function front that a sharp-fronted dispersion gives to within 0.014 — the remainder being the error-function formula’s own approximation, which neglects the cake’s finite length.

The third figure is the washing curve, on a logarithmic scale so that the last few per cent — the ones that matter to anyone washing for purity — are visible. Every curve starts along the same line: for the first half pore volume the wash is only pushing out liquid that has not yet met it, and the fraction left falls as one minus the wash. They part as the front approaches the cloth. At a Péclet number of 300 the front arrives nearly intact, and the solute left drops by a factor of ten within a tenth of a pore volume. At three, the front is smeared through the whole cake before it arrives, and what remains leaves slowly by diffusion out of liquid the wash has already passed, falling by a factor of ten for every pore volume or so.

The fourth figure extracts the one number a plant needs, the wash that leaves one per cent behind. It is 3.1 pore volumes at a Péclet number of 3, 2.0 at 10, 1.43 at 30, 1.17 at 100 and 1.02 at 1000, and it tends to 0.99 — plug flow, which removes exactly one per cent less than everything in 0.99 of a pore volume. Most of the curve’s fall happens between Pe of 3 and 100, and above a few hundred more sharpness buys nothing.

The cake is not the grains #

What Péclet number does a filter cake have? For the concentrated slurry’s best washed cake, 65 millimetres of ten-micron silica, the textbook estimate of dispersion in a packed bed — molecular diffusion through tortuous pores, $D_m/\sqrt{2}$, plus mechanical mixing on the scale of the grain, $\tfrac12 ud$ — gives a dispersion coefficient of $2.6\times10^{-9}$ square metres per second at the wash’s interstitial speed of 0.37 millimetres per second — slow enough, at a pore Reynolds number of a few thousandths, that Darcy’s law has no competition. The cake’s Péclet number is then about 9,500, the mark at the right of the fourth figure. By that estimate every filter cake washes as plug flow: a cake is thousands of grains thick, and a front that smears over a few grains arrives sharp.

Measured washing curves are not that sharp. The wash breaks through early and the last solute leaves slowly, which in the dispersion model’s terms means a Péclet number of tens rather than thousands — a smearing length of millimetres in a cake whose grains are microns, as though the cake were a bundle of tubes smearing a front the way shear smears one in a single tube. The model has not failed; it is being told something the grain-scale estimate left out. A real cake is not uniform. It is laid down layer by layer from a slurry whose particles settle and segregate, so its permeability varies with depth and across the cloth; its fine particles fill some pores and leave others open; and it may contain dead-end pores whose liquid the wash can reach only by diffusion. Each of those makes the front run unevenly over distances far larger than a grain. The washing curve therefore measures the cake’s heterogeneity, and a fitted Péclet number is a statement about the cake’s structure rather than about its particles. That is the same lesson the packed-bed reactor learns from its outlet: the spread in a bed’s residence times comes from the bed’s large-scale disorder long before it comes from the grains.

Holding the purity, and where the best cake goes #

A plant does not wash a fixed number of pore volumes; it washes until the product is clean enough. The wash ratio is then not a choice but a consequence of the cake’s thickness, because the Péclet number is the thickness over a fixed smearing length $\ell$. A thick cake washes to a given purity in fewer pore volumes than a thin one. That seems to argue for thick cakes, against the square-root law’s preference for thin ones, and the question is which wins.

The fifth figure answers it for the concentrated slurry, washing every cake to one per cent. With a smearing length of half a millimetre the best cake is built from 0.280 cubic metres of filtrate, has a Péclet number of 141, needs 1.13 pore volumes and delivers 0.410 cubic metres per square metre per hour. With two millimetres the best cake is 0.278, at a Péclet number of 35 and 1.38 pore volumes, delivering 0.402. With eight millimetres it is 0.274, at 8.6 and 2.10 pore volumes, delivering 0.380.

Both arguments are real and neither wins by much. Against a filter washed to a fixed 2.10 pore volumes, whose best cake by the rule would be 0.257, holding the purity instead moves the best cake about 6.5 per cent thicker, because a thicker cake earns a smaller wash ratio. But the gain in output from making that move is a fifth of a per cent: 0.3803 against 0.3810 from the fixed-ratio rule applied at the same wash. The peak is flat, as the peak of every square-root optimum is, and a six per cent error in the batch size costs almost nothing. What dispersion costs is not the position of the optimum but its height — 7 per cent of the output between the sharpest and the most smeared of these three cakes, all of it spent on extra pore volumes.

A soft cake washes no worse for being thick #

The earlier essay closed with a prediction: a compressible cake, which packs tighter against the cloth the harder it is pushed, should be washed thin and often, because a thick one would spend most of its life resisting its own wash. The calculation says the prediction is wrong, and the reason is worth having.

At a fixed pressure a compressible cake’s solid stress rises from nothing at its face to the full pressure drop at the cloth, built up grain by grain from the drag of the liquid passing through, as the stress in a bed about to fluidise is, and its porosity and specific resistance follow the stress. The stress profile obeys $dp_s/dW = \mu q\,\alpha(p_s)$ through the cake’s mass $W$ per unit area, and the flux $q$ through a cake of mass $W$ is fixed by requiring the stress to reach $\Delta p$ at the cloth. Measured as a fraction of the cake’s mass, that profile is identical for every thickness: a cake twice as heavy has the same stresses at the same fractional depths, with half the flux. Its mean resistance and its mean porosity are then independent of thickness, which makes its filtration time and its pore volume scale exactly as an incompressible cake’s do. The wash time over the filtration time comes out as $2wcm$, with $m$ the pore volume per kilogram of solid, whatever the cake’s thickness.

The sixth figure shows it. For cakes whose resistance grows with stress to the powers 0, 0.5, 1 and 1.3, washed with two pore volumes of a slurry of 50 kilograms per cubic metre, the ratio of washing to filtration time is 0.130, 0.153, 0.186 and 0.208 at two bar, and it does not change by one part in $10^{15}$ between a cake of one kilogram per square metre and one of sixty. Compressibility raises the ratio — in a softer cake the stress climbs later, close to the cloth, so more of its mass sits at low stress and open porosity — but it does not make it depend on thickness, and so it does not move the best cycle away from the rule. The right panel adds what compressibility does do: pressing a soft cake harder squeezes liquid out of its pores, and at ten bar the ratios fall to 0.079, 0.102, 0.146 and 0.181. A soft cake is cheaper to wash at high pressure even when high pressure has stopped buying filtration rate, which the earlier essay found it does for compressibilities above one.

What the picture cannot show #

A wash that displaces and nothing else. The wash is assumed to have the density and viscosity of the liquid it displaces. A wash that is less viscous fingers through the cake ahead of its own front, which no one-dimensional dispersion describes; a denser one run downwards is stable, and upwards is not.

A cake that does not crack or shrink. Cakes crack as they are washed and, above all, as they are blown dry, and a crack is a channel that short-circuits both the wash and its purity. The drying step that follows most washes is not in the cycle here.

One dispersion length for a whole cake. The smearing length stands in for all the cake’s heterogeneity at once, and the washing curve of a real cake has a longer tail than the advection–dispersion equation gives, from liquid held in dead-end pores and inside porous particles, which leaves by diffusion on its own timescale. Those tails decide the last fraction of a per cent, and a wash specified to a part per thousand needs them measured.

A batch filter. A rotary drum filter makes and washes its cake continuously, and its cycle is set by the drum’s speed rather than by a batch volume. The same square-root arithmetic applies to it zone by zone, but the optimum is a speed and not a batch.

The convention the numbers depend on #

Volumes are per square metre of cloth. The wash ratio $w$ counts pore volumes of the finished cake, the liquid its pores hold at the end of filtration, and not volumes of the cake or of the filtrate. The Péclet number is the cake’s thickness over the smearing length $D_L/u$, with $u$ the interstitial speed, the speed in the pores rather than the superficial one. “One per cent left” is the fraction of the solute that was in the pores at the start of the wash, averaged through the cake. The slurries are the earlier essay’s silica, with a specific resistance of $6.4\times10^9$ metres per kilogram from the Kozeny–Carman formula, a cloth of $10^{10}$ per metre, one bar, and water.

Who found it, and when #

The cycle arithmetic follows from Bernard Ruth’s constant-pressure law of the 1930s, and the rule that the best cycle spends as long on the cake as on the downtime is a standard result of filtration textbooks from the middle of the twentieth century. Ford Harris published the economic order quantity in 1913, and it is usually credited to R. H. Wilson, who popularised it in the 1930s. The model of washing as axial dispersion through the cake was developed in the 1960s and 1970s, notably by Richard Wakeman, whose fits of the dispersion model to measured washing curves are still the usual way to specify a wash. Danckwerts gave the inlet condition used here in 1953, for the residence-time distributions of reactors.

Still open: drying the cake the wash has left #

The washed cake is still full of liquid — the wash, now, rather than the mother liquor — and most batch filters end the cycle by blowing gas through the cake to drain it before it is discharged. Drainage is a two-phase flow through the pores, governed by the capillary pressure needed to empty a pore of a given size, and it stops at a residual saturation the gas cannot remove. The next calculation adds that stage to the cycle: the threshold pressure below which no liquid leaves at all, set by the cake’s largest pores and the capillary length of its liquid, the rate at which it drains above it, and whether the drying time grows as the square of the cake’s thickness, as filtering and washing do, or as some other power — which is what decides whether the rule that the cake’s time equals the downtime survives a third stage.