The pump that is better the more it squeezes

A wave on the wall is a pump found that a channel whose wall moves only in and out, in a wave travelling along it, delivers a steady flow — and that continuity alone, in the frame moving with the wave, fixes nearly everything about it. The momentum equation was needed for one number, the pressure the pump can push against, and it came out as a straight line: most pressure at no flow, none at the free-pumping flow $3\varphi^2/(2+\varphi^2)$.

A straight pump curve is a characteristic, not an account. It says what the pump delivers and nothing about what the wall had to do to deliver it. The same question for the mirror-image problem, a sheet that swims by waving, had a clean answer: the work per unit distance travelled contains no amplitude at all, because the sheet’s power and its speed both go as the amplitude squared.

The pump has the same mechanism running the other way — a wall wave driving fluid rather than fluid resisting a wall wave — and it would be natural to expect the same independence. It is not there, and the reason is the whole of this essay.

The account, written down #

The wall moves only across the channel in the laboratory frame. Shear stress acts along the wall, where the wall does not move, so it does no work; the wall’s power is its pressure times its normal speed and nothing else. With the pressure $p(\theta)$ along one wavelength and the half-width $h = a(1 + \varphi\sin\theta)$, the mean power the wall delivers per unit length of the half-channel, in units of $\mu c^2/a$, is

$$W = 2\pi\varphi\,\langle p\cos\theta\rangle = 3\big[I_1 + (\Theta - 2)I_2 - (\Theta - 1)I_3\big],$$

after an integration by parts that trades the pressure for its gradient, which the lubrication solution already supplies. $I_1$, $I_2$ and $I_3$ are the means of $H^{-1}$, $H^{-2}$ and $H^{-3}$ over a wavelength, all known in closed form, and $\Theta$ is the mean flow in units of $ca$.

The useful power is the mean flow times the pressure it is pushed up: $\Theta\,\Delta p$, with $\Delta p = -3[(\Theta - 1)I_3 + I_2]$. The efficiency is the ratio. Both are polynomials in $\Theta$ — the useful power a quadratic, the wall power a straight line — so the efficiency is a quadratic over a line and its maximum is the root of a quadratic. The best efficiency of a peristaltic pump has a closed form at every amplitude.

The account closed a second way #

An energy account can be wrong in a way that still looks plausible, so the one above was checked by a route that shares none of its algebra.

Whatever the wall supplies and the pump does not deliver must be dissipated by viscosity. The dissipation is $\mu(\partial u/\partial y)^2$ integrated across the channel and along the wavelength, and it can be computed directly from the velocity profile — differentiating the profile numerically and integrating the square on a grid of 1,600 stations and 200 levels, with no use of $I_1$, $I_2$, $I_3$ or the integration by parts.

At four operating points, including one pumping backwards against the wave, the wall power less the useful power agrees with the directly integrated dissipation to between 2.7 and 8.2 parts in a million, which is the grid’s own error. The account is right.

Efficiency along the pump curve #

The shape of each curve is forced at its ends. At no flow the pump holds its full pressure and delivers nothing; at free pumping it moves its full flow and lifts nothing. Between them the useful power is a hump.

What differs from one amplitude to the next is everything else. A wave closing a fifth of the channel never does better than 4.5 per cent. A wave closing half of it reaches 27 per cent. A wave closing nine-tenths reaches 82 per cent, and one closing ninety-seven hundredths reaches 94. The best efficiency is not only amplitude-dependent: it spans almost the whole of the range an efficiency can have.

The two ends of the best efficiency #

For a shallow wave the best efficiency is $9\varphi^2/8$. Expanding the three integrals for small $\varphi$, the useful power becomes $\Theta(\tfrac32\varphi^2 - \Theta)$ times a constant and the wall power becomes $\varphi^2/2$ times the same constant, to leading order. The ratio is largest at $\Theta = \tfrac34\varphi^2$ — half the free-pumping flow — where it is $\tfrac98\varphi^2$. The closed form, evaluated at $\varphi = 0.02$, agrees with that to a part in ten thousand.

For a nearly closed wave the best efficiency approaches one, with a shortfall that shrinks in proportion to the remaining gap. At $\varphi = 0.99$ the shortfall is 1.88 per cent, and at $\varphi = 0.999$ it is ten times smaller, to within half a per cent.

In between, at half closure, the numbers become exact. With $\varphi = \tfrac12$ the three integrals are $2/\sqrt3$, $8/(3\sqrt3)$ and $4/\sqrt3$, the efficiency reduces to $2\Theta(1 - 3\Theta)/(1 - 2\Theta)$, and its maximum is at $\Theta = (3 - \sqrt3)/6$ with a value of exactly $2 - \sqrt3$, or 26.8 per cent. That is a curiosity rather than a principle, but it is a useful fixed point to check any computation against.

So the independence the swimming sheet showed is simply not there. Its reason was that the sheet’s power and its output — its speed — scaled together. Here the useful power and the wall power scale together only for a shallow wave, where both go as $\varphi^2$ and the ratio is still small, and they separate as the wave deepens because the losses and the output depend on the narrowest gap in different ways.

Where on the curve the pump is best #

Half of the free flow is the answer for any pump whose pressure falls linearly with flow and whose losses do not care where on the curve it runs: useful power is then a symmetric hump and the input is constant, so the peak is in the middle. A shallow peristaltic pump is that kind of machine.

A deep one is not, and the reason is visible in the wall power’s formula. The wall power falls as the flow rises, because the coefficient of $\Theta$ in it, $I_2 - I_3$, is negative. At high flow the pressure the wall pushes against is lower, so the wall does less work — and for a deep wave the reduction is large enough that the losses fall faster than the output as the flow approaches free pumping. The best point therefore moves towards free pumping, to 91 per cent of it at $\varphi = 0.9$. A peristaltic pump with a deep wave is most efficient when asked for little pressure.

Where the wall’s work goes #

There is nowhere else for the power to go: no inertia to store it, no free surface to radiate it, no heat conducted away that is not already the dissipation. The shallow pump turns nearly all of its wall’s work into heat, and the useful ten per cent is almost incidental. The deep pump turns most of it into pressure-driven flow.

The deep pump’s losses gather at the throat. At $\varphi = 0.9$ the fifth of the wavelength where the gap is less than 0.3 of the mean half-width carries two-thirds of all the dissipation. That is where the lubrication film is thinnest, the shear across it is greatest, and — for a pump nearly closing its tube — where the fluid that leaks back past the wave is squeezed through.

The two zeros are a structural feature rather than a numerical one. In the wave’s frame the flux between the centreline and the wall is the same at every section, and where the gap happens to equal that flux plus the wall’s own speed times the gap, the parabolic profile degenerates into a uniform plug with no shear at all. Every operating point with a flow between zero and the free-pumping flow has two such sections per wavelength.

The pump that becomes a piston #

The approach to occlusion explains the efficiency’s approach to one. The pump curve is a straight line whose intercept on the flow axis — free pumping — tends to the full volume the wave displaces, and whose slope, the flow lost per unit of pressure, is $1/(3I_3)$. That slope is the leakage back past the throat, and $I_3$ grows as the gap to the minus five-halves because the flow through a lubrication gap goes as the gap cubed and the gap’s length along the wave goes as its square root. With a gap of a hundredth, the leakage per unit pressure is some 2,600 times smaller than with a gap of a quarter.

A pump that closes its tube completely has no leakage and no lubrication film at the throat to dissipate in, and it moves the wave’s displaced volume against any pressure. That is a positive-displacement pump — a roller pump, a hose squeezed by a travelling clamp — and its efficiency as a fluid machine is limited only by what the lubrication model has thrown away: the deformation of the tube, the friction of the rollers, the fluid’s inertia in a real pulsing flow. The calculation here says how a peristaltic pump passes continuously from a wave that barely pumps to a piston that pumps perfectly, and that the passage takes the efficiency from nine-eighths of a small square to one.

Against a stated pressure, the squeeze has an optimum #

Everything so far has let each amplitude run at its own best point. A pump is rarely chosen that way. It is asked to deliver a pressure — the head of a ureter, the back-pressure of a dialysis circuit, the resistance of whatever is downstream — and the question is how hard its wave should squeeze to do that job with the least work.

At a fixed pressure there is a best squeeze, and squeezing harder than it costs efficiency. The reason is the position figure read at a fixed duty. A wave that barely holds the pressure is working at the shut-off end of its curve, delivering almost no flow for a great deal of wall work. A wave that closes the tube far more than the pressure needs is working near its own free-pumping end, where the flow is large but the pressure it lifts is a small fraction of what it could, and the losses in its throat are paid on the whole of that flow. Between the two lies an amplitude at which the stated pressure is the right fraction of the wave’s own capacity.

For a pressure rise of 10, in units of $\mu c\lambda/a^2$, that amplitude is 0.908, and the pump there runs at 97 per cent of its free-pumping flow with an efficiency of 78 per cent. Squeezing to 0.988 instead drops the efficiency to 60 per cent, while giving barely any more flow.

The heavier the duty, the deeper the best squeeze and the higher the efficiency reached: 38, 78 and 95 per cent for pressure rises of 1, 10 and 100. That is the continuous version of the roller pump’s design logic. A pump built for a high pressure has to nearly close its tube to hold that pressure at all, and once it does, the same near-closure that holds the pressure makes the pump efficient. A pump built for a low pressure gains nothing by closing its tube — it pays the throat’s losses on flow it did not need to force through it — and its best efficiency is modest because its best squeeze is modest.

There is a physiological reading, stated here as a consistency and no more. Ducts that must pump against substantial pressure, such as the ureter delivering into a filling bladder, contract nearly to occlusion; ducts that mainly transport contents against little resistance do not need to. This calculation has nothing in it about muscle, tissue or control, so it cannot say that is why. It says that nearly occlusive peristalsis is where a pressure-delivering pump is efficient, and a low-pressure one is not.

The swimmer and the pump, set side by side #

The comparison with Taylor’s swimming sheet is worth making precise, because the two problems are the same equations with the roles of wall and fluid swapped.

The swimming sheet’s dissipation per unit area is $\mu k^3 b^2 c^2$ and its speed is $\tfrac12 k^2 b^2 c$; both are the amplitude squared times constants, so their ratio — the work per unit distance, $2\mu k c$ — has no amplitude in it. The sheet has no second length for the amplitude to be compared with: the fluid above it is unbounded.

The pump has one: the mean half-width $a$. Its amplitude enters as $\varphi = b/a$, and every integral over the wavelength depends on how close $\varphi$ is to one. For a shallow wave the channel is effectively unbounded as far as the wave is concerned, and the pump behaves like the sheet: efficiency and flow both as $\varphi^2$. For a deep wave the second length takes over, and the gap it leaves controls everything. What gave the sheet its amplitude-free cost was the absence of a second length, and a pump is defined by having one.

A ledger of the checks #

The optimum from the quadratic and the optimum from a search that knows nothing about calculus agree to about a part in ten million of the free-pumping flow, which is the precision a golden section can reach on a smooth maximum. The shallow limit is met to a part in ten thousand at $\varphi = 0.02$, and the occlusion limit’s tenfold ratio is 9.96.

What the long-wavelength channel cannot show #

The throat at occlusion. The lubrication solution assumes the gap is thin compared with the wavelength everywhere, and it does not fail as the gap closes — it becomes more accurate. What it cannot represent is the gap reaching zero, where the fluid is not squeezed through a film but cut off by contact, and the mechanics of the wall’s contact takes over.

Inertia. A real peristaltic flow at a finite Reynolds number stores kinetic energy through each cycle and releases it, and the wave’s frame is then not quite a frame of steady flow. The account above has no kinetic energy term, which is correct in the creeping limit and an approximation outside it.

The wall’s own losses. The efficiency is hydraulic: pressure-driven flow out against pressure work in. The muscle or the roller supplying that work has losses of its own, and in a biological duct those are larger than the fluid’s.

A two-dimensional channel. A tube’s integrals are different functions of the amplitude ratio, with the gap’s cube replaced by its fourth power in the flow through the throat. The limits change their coefficients and not their character; the numbers above are for the channel.

Non-Newtonian contents. Food, urine with particles, blood and slurries all have viscosities that depend on the shear rate, and the throat is where the shear rate is highest.

Still open: the tube that deforms to make the wave #

Every calculation here prescribes the wave. In a gut, a ureter or a roller pump’s hose, the wall is an elastic tube and the wave is what results from squeezing it: the shape near the throat depends on the pressure the fluid exerts back on the wall, and the gap at the throat is decided by the balance between the squeeze and that pressure rather than chosen.

The next calculation couples the lubrication film to a wall that deflects under it — a tube whose local radius responds to the pressure through its own stiffness. The question it would answer is the one this essay’s limit raises and cannot settle: whether a pump driven by a fixed squeezing force, rather than a fixed wall shape, ever reaches occlusion at all, or whether the fluid’s own pressure at the throat holds the gap open by an amount that sets a ceiling on its efficiency below one.