A loss with no viscosity in it

Water flows along a pipe and the pipe suddenly gets wider. Something is lost there — every handbook has a table of it, every pump calculation includes it, and the usual explanation is friction: extra wall, a churning eddy in the corner, and dissipation.

The usual explanation is wrong, and the way to see that it is wrong is that the loss can be written down exactly, in closed form, from a calculation in which no viscosity appears. Not approximated: derived. It has been known since Borda in 1766, it agrees with measurement to a few per cent, and its size depends on the geometry alone.

Two balances that disagree, on purpose #

The method is the field’s. Draw a box in the fluid spanning the step, add up what crosses its faces, and refuse to look inside.

What crosses the faces is momentum flux and pressure, and both have to be counted — which is exactly the accounting error the popular momentum argument for lift makes by keeping only the first. Here the pressure term is the whole difficulty, because one of the faces is an awkward shape.

Continuity is immediate, and it is the statement that a streamtube’s mass has nowhere to go applied to a tube whose walls happen to be pipe. With $\beta = A_1/A_2$ the area ratio,

$$V_2 = \beta V_1$$

Momentum needs one statement about the annular face of the step, and it is the only assumption in the essay. The fluid in the corner is nearly stationary, so the pressure acting on that annulus is taken to be the upstream pressure $p_1$. Then the pressure force on the whole upstream face is $p_1 A_2$, and the balance gives

$$p_2 - p_1 = \rho V_1^2\,\beta(1 - \beta)$$

Energy, if nothing were lost, would be Bernoulli’s:

$$p_2 - p_1 = \tfrac{1}{2}\rho\left(V_1^2 - V_2^2\right) = \tfrac{1}{2}\rho V_1^2 (1 - \beta^2)$$

The two do not agree, and they are not supposed to. Momentum is conserved across the step and energy is not, so the difference between the two answers is exactly the energy that went missing:

$$\Delta h = \tfrac{1}{2}\rho V_1^2\left(1 - \beta\right)^2 = \tfrac{1}{2}\rho\left(V_1 - V_2\right)^2$$

Half the square of the velocity change. That is Borda’s result, and it contains no material property of any kind.

Why momentum survives and energy does not #

This is the part worth being careful about, because it is the same asymmetry that decides which side of a shock a gas may be on and which way a hydraulic jump may go, and it is not obvious.

Momentum is conserved because the only things that can change it are forces on the boundary of the box, and those are pressures on the two ends and on the step — all of which are accounted. The turbulence inside the box exchanges momentum between one bit of fluid and another and cannot create or destroy any. Whatever happens in there, it happens between parties whose momenta sum to a constant.

Energy is different because there is a place for it to go that the balance does not track. Mechanical energy can become internal energy — heat — and the internal energy leaves in the fluid at a temperature rise so small nobody measures it. The turbulence in the corner is the route: the jet issuing into the wide pipe shears against the slow fluid around it, that shear feeds a cascade, the cascade ends at the smallest scales in viscous dissipation, and the energy is gone from the mechanical account.

So viscosity is genuinely responsible for the loss, and is genuinely irrelevant to its size. The cascade will dispose of whatever mechanical energy is handed to it, at whatever rate is required, and the amount handed to it is fixed upstream by the momentum balance. Halving the viscosity does not halve the loss; it makes the eddies smaller and the dissipation happen at a finer scale, and the number is unchanged.

That is a genuinely surprising thing about turbulence, and it is one of the few places on this site where the fact that turbulence cannot be computed does not matter in the least.

There is a mechanical analogy that makes the asymmetry feel inevitable rather than clever. Two railway wagons collide and couple together. Momentum is conserved — nobody needs to know anything about the buffers, the springs or the noise — and the final velocity follows in one line. Kinetic energy is not conserved, and the amount lost is exactly $\tfrac{1}{2}\mu(\Delta v)^2$ with $\mu$ the reduced mass: a formula containing no property of the buffers whatever. The sudden enlargement is that collision, between a fast jet and the slow fluid around it, run continuously. Carnot said so in 1783 and the resemblance is not an analogy; it is the same theorem.

The reason it feels surprising in a fluid and obvious in wagons is that in a fluid the mess is visible. A recirculating eddy looks like something that ought to need describing, and a smooth picture of it would prove nothing anyway.

The one assumption, examined #

A derivation with a single modelling statement in it deserves to have that statement looked at, and this one has been asserted twice above without any argument for it. The claim is that the pressure on the annular face of the step — the flat ring of wall the fluid never touches at speed — equals the upstream pressure $p_1$. Everything exact in this essay rests on it.

The argument for it is not about the corner at all. It is about the jet. At the plane of the step, the fluid coming out of the small pipe is still moving straight: it has not yet begun to spread, because spreading is what the shear layer does over the following few diameters. Streamlines that are straight and parallel have no centripetal acceleration to supply, so there is no pressure gradient across them —

$$\frac{\partial p}{\partial n} = \frac{\rho V^2}{R}, \qquad R \to \infty$$

— and the pressure is therefore uniform right across the plane, from the jet’s axis out to the pipe wall, at whatever value the jet carries. That value is $p_1$. The annulus does not need to be stagnant for this; it needs only to lie on a plane the jet crosses without curving.

Which is the same argument that lets one write the pressure as constant through the thickness of a boundary layer, and it fails in the same circumstances: wherever the streamlines are curved. Here they are, slightly. The corner fluid is not motionless but slowly recirculating, and a rotating body of fluid has a pressure that varies across it, so the true annular pressure is a little away from $p_1$ and varies over the ring.

That error is small, and its smallness is a measurement rather than a deduction. Borda’s result agrees with experiments on sudden enlargements in turbulent pipe flow to within a few per cent over most of the useful range of area ratios, which is what bounds the assumption. It is not what proves it — nothing in the algebra could — and the honest description of the loss coefficient is therefore “exact given one statement about a wall pressure, and that statement is good to a few per cent”.

There is something clarifying about where the uncertainty ends up. It is not in the turbulence, not in the shear layer, and not in the eddy that every textbook figure draws; it is in the pressure on a ring of stationary metal, which is the easiest quantity in the whole problem to go and measure. Drill the annulus and tap it, and the assumption becomes a reading.

And the failure is confined. The assumption is made once, on one face, in one balance, and every other line follows from conservation. So a better number for the annular pressure would not require the derivation to be redone — it would shift one term in the momentum balance and carry through to the loss unchanged in form.

What was computed, and what the assertion catches #

The two balances are computed separately, the loss is taken as the difference, and it is compared with the closed form at four hundred area ratios from 0.0025 to 1. The worst disagreement across the whole range is $1.1\times10^{-16}$ — one bit of double precision.

That looks like a check on arithmetic and it is, but on arithmetic with a specific trap in it. The pressure force on the upstream face is $p_1 A_2$, not $p_1 A_1$: the step’s annulus is part of the upstream face and carries $p_1$ by assumption. Using $A_1$ is the natural mistake, it changes the momentum balance, and it produces a loss that is still positive, still zero at $\beta = 1$, still plausible in a table — and not Borda’s. Only the comparison with the closed form catches it.

Two further refusals are checked in the site’s gate:

The loss is never negative. Asserted at every one of the four hundred ratios, not at one. A negative Borda–Carnot loss is an enlargement that creates energy, and it is the second law wearing a plumbing hat — the sign is the only thing separating this result from a perpetual motion machine.

A pipe that does not change area loses nothing. At $\beta = 1$ the loss is zero to machine precision, which is the boundary condition any wrong algebra is most likely to violate.

And the routine refuses an area ratio above one. A sudden contraction is not this problem: the assumption that makes the momentum balance closed — the step face at upstream pressure — has no counterpart there, because the corresponding face is downstream of the contraction where the fluid is fast and the pressure is anything but ambient. A contraction’s loss coefficient is a measurement, and that fact is worth having stated by a refusal rather than buried in a caveat.

Recovery and loss are different questions #

An enlargement is usually installed to recover pressure, not to avoid loss, and those are two different optimisations that a designer has to choose between explicitly.

The recovery $C_p = 2\beta(1-\beta)$ has a maximum, found by golden section at $\beta = 0.5000000$ with $C_p = 0.5000000$, and it is a pleasing coincidence of the algebra that both come out at a half. Meanwhile the loss $(1-\beta)^2$ falls monotonically towards $\beta = 1$, where nothing is lost because nothing has happened.

So “minimise the loss” and “maximise the recovery” give opposite answers, and a designer who asks the wrong one gets told to leave the pipe alone. What is actually wanted is usually neither: it is the best recovery achievable for a required area change, and that is a question about gradual diffusers rather than sudden steps.

An ideal gradual diffuser recovers the full Bernoulli value $1 - \beta^2$ — the dashed curve on the figure — which at $\beta = 0.5$ is 0.75 against the sudden step’s 0.5. Real ones get most of the way there, and then stop, for a reason this site has spent a whole field on: a diffuser is an adverse pressure gradient by construction, and a boundary layer climbing a pressure rise separates. Past a total included angle of about seven degrees the flow lets go of one wall, the effective area ratio collapses, and the recovery falls off a cliff. The limit on a diffuser is not the loss equation in this essay; it is the separation criterion in another one.

That limiting case is worth naming, because it is the most common one in a real system. A pipe discharging into a tank has $\beta \to 0$, and its exit loss is exactly one velocity head — the whole of the kinetic energy the pump paid for. It is very often the largest single loss in a pipework calculation, it is exact, and it has nothing to do with friction.

The table of minor losses, sorted by what kind of thing each entry is #

Every pipework handbook has a table of loss coefficients $K$, defined by $\Delta h = K \tfrac{1}{2}\rho V^2$, running to a page or two: sudden enlargement, sudden contraction, entrance, exit, elbow, tee, gate valve, globe valve. They are printed in one list, in one typeface, to the same number of significant figures.

They are not one kind of object.

• Sudden enlargement, $K = (1-\beta)^2$ — derived, exact, this essay.
• Exit into a reservoir, $K = 1$ — derived, exact, the limiting case above.
• Sudden contraction, $K \approx 0.42(1-\beta)$ — measured. The flow separates at the sharp corner and forms a vena contracta, a waist narrower than the downstream pipe, and the loss is the Borda–Carnot loss of the enlargement from that waist back out to the full bore. So the physics is exact and the contraction ratio is not: it depends on the sharpness of the corner and comes from experiment.
• Elbows, valves, tees — measured, every one of them, and the numbers vary between manufacturers by more than the precision they are quoted to.

That taxonomy is more useful than the table. The first two entries will not change when somebody does a better experiment. The rest might.

The reason this essay’s calculation is allowed to use Bernoulli at all is worth stating in that light. The energy balance is applied twice — once on each side — and never across the discontinuity. Applied across it, it would be the second row of that figure without the excuse: a theorem carried into a region where its hypotheses fail. Every quantity on the left of the step and every quantity on the right is related by an exact statement; the two exact statements simply have different constants, and the difference between them is the point.

Where the box’s method reaches and where it does not #

The two internal-flow rungs make a matched pair, and the contrast between them is the most useful thing in this field.

The friction along a straight pipe cannot be computed by a control volume, because there the wall is the boundary: what crosses it is exactly what is being asked about, and drawing a box does not help. So that quantity is a correlation, fitted to somebody else’s pipes in the 1930s, with an equivalent sand roughness standing in for a surface nobody measured.

The loss at a step can be computed by a control volume, because the step is inside the box. The messiest-looking part of the flow — the recirculating corner, the shear layer, the reattachment — is precisely the part that never has to be described.

The general lesson is not that control volumes are powerful — they are, but that is a technique. It is that the difficulty of a flow and the difficulty of a question about it are unrelated. A turbulent separated recirculating corner is as hard a flow as this site contains, and the question “how much energy does it destroy” has a one-line exact answer. Meanwhile a perfectly attached laminar boundary layer on a flat plate is easy to picture and its friction took Blasius a thesis.

Who found it, and when #

Jean-Charles de Borda published the result in 1766, in a memoir to the Académie des Sciences on the discharge of fluids. Lazare Carnot rederived it in 1783 in a more general form — as the kinetic energy lost when two masses of fluid at different velocities are suddenly mixed, which is the mechanical analogue of an inelastic collision and is where the modern name comes from.

Both dates deserve attention. This is a turbulence result, obtained in closed form, seventy years before Hagen and Poiseuille measured laminar pipe flow, a century before Reynolds distinguished the two regimes, and a hundred and eighty years before anybody could compute a turbulent flow at all. Carnot’s framing is the more revealing: he did not model the mixing, he treated it as an inelastic collision and used the conservation law that survives one. That is the whole method of this field, stated in 1783.

Where the ladder goes #

The field’s next pair takes the same argument to a free surface, and the result is one of the most satisfying correspondences in fluid mechanics.

Shallow water behaves like a gas of ratio of specific heats two — the depth plays the density’s part, $\sqrt{gh}$ plays the speed of sound’s, and the Froude number plays the Mach number’s. A hydraulic jump is then the exact counterpart of a shock: momentum conserved across it, energy destroyed, and one direction forbidden. It is Borda’s argument with a river in it, the loss is again a difference between two balances, and the white water below a weir is the dissipation made visible.