Work out of a change of swirl

There is one equation for every rotating machine that moves a fluid. Not one per type — one, for a centrifugal pump, an axial compressor, a steam turbine, a jet engine’s fan, a torque converter and a kitchen extractor:

$$w = U_2 V_{\theta 2} - U_1 V_{\theta 1}$$

The work per unit mass equals the blade speed times the change in the swirl component of the fluid’s velocity. There is no blade shape in it, no pressure, no efficiency, no viscosity, no gas properties and no Mach number. It follows from angular momentum on a control volume drawn round the rotor, with nothing at all assumed about what is inside.

Angular momentum, on the same box as everything else #

The derivation is the field’s method for the last time.

Draw a control volume enclosing the rotor. Angular momentum crosses its faces as a flux, $\dot{m}\,r V_\theta$, exactly as linear momentum crosses as $\dot{m}\,V$. In steady operation the angular momentum inside the box is constant, so the torque applied by the blades equals the net flux out:

$$Q = \dot{m}\left(r_2 V_{\theta 2} - r_1 V_{\theta 1}\right)$$

The shaft turns at $\Omega$, so the power is $Q\Omega$, and dividing by the mass flow gives the work per unit mass with $U = \Omega r$. That is the whole derivation and it fits in three lines.

What is worth dwelling on is what never appeared. No pressure. The pressure acting on the faces of the box is radial or axial and exerts no moment about the axis, so it drops out of an angular-momentum balance entirely — which it does not do in the linear ones this field has used up to now. The blades exert a torque and that torque is the only thing the balance can see.

This is the same argument the wind-turbine rung makes, run the other way. There, angular momentum leaving in the wake was a loss the axial theory had not accounted for. Here it is the entire mechanism.

Two routes, and why one of them is not a route #

The work is computed twice, and the count matters because it is easy to overstate.

The Euler form is the angular-momentum statement above.

The kinematic identity splits the same work into three physically distinct pieces:

$$w = \frac{V_2^2 - V_1^2}{2} + \frac{U_2^2 - U_1^2}{2} + \frac{W_1^2 - W_2^2}{2}$$

— the change in the fluid’s own kinetic energy, the work of the centrifugal field, and the diffusion in the frame rotating with the blades. It is an exact algebraic consequence of the triangle geometry and it shares no arithmetic with the first, so the two agreeing is a real check. They agree to $2\times10^{-16}$.

What is not a second route, and is reported rather than asserted against, is the torque form $Q\Omega/\dot{m}$: that is the Euler expression written with radius and rotation rate instead of blade speed, and comparing them would be comparing a statement with itself. Saying so is worth the sentence, because “three independent checks” is exactly the sort of claim that inflates without anybody noticing.

There is a genuine third check, and it is a quadrature rather than an identity. It appears three sections down.

The three terms are the taxonomy of turbomachinery, and reading them off is more useful than any list of machine types.

• A centrifugal pump lives on the middle term. Its inlet is at a small radius and its outlet at a large one, so $U_2^2 - U_1^2$ is large and positive, and it can produce a large pressure rise in one stage without asking the blade passage to diffuse the flow much at all.
• An axial compressor has $U_1 = U_2$, so the middle term vanishes and the work has to come from the other two. That is why axial stages produce so little pressure rise each and why a compressor has fifteen of them.
• An impulse turbine has $W_1 = W_2$: the relative speed is unchanged and the blade merely turns the flow. All the work comes from the first term, the static pressure is the same either side of the rotor, and the machine is doing full work with no pressure drop across the moving blades at all.

That last is this essay’s refutation as a worked case. The gas is certainly pushing on the blades, and the force it exerts is certainly what the shaft feels — but the work is not the pressure drop times the volume, because in an impulse stage there is no pressure drop across the rotor and there is plenty of work.

The rotor that turns nothing #

The complementary refusal is asserted directly: a rotor with the same swirl on both sides does exactly zero work, whatever else is happening.

$$V_{\theta 2} = V_{\theta 1} \quad\Longrightarrow\quad w = 0$$

to fourteen decimal places, and the site’s gate also checks that a stage claiming work with unchanged swirl is refused. Both are worth having because the failure is a common piece of reasoning rather than a coding mistake. A duct with a pressure drop across it, a screen, a heat exchanger, a throttle — all of them have a fluid pushing hard on a solid and none of them extracts any work, because none of them changes the angular momentum. A rotor that spins freely in a stream, taking no load, is in the same category.

Turning the flow is the mechanism. Pushing on the blades is what the turning feels like.

There is a version of the same confusion on the other side of this site, and putting the two together is instructive. Air must be pushed down is a popular account of lift that names the right mechanism and does the accounting badly, and the reason it goes wrong is that it counts momentum flux and forgets pressure. This one goes wrong in the mirror image: it counts pressure and forgets that pressure exerts no moment about the axis. In both cases the correct answer comes from writing down everything that crosses the box’s faces and being told by the algebra which terms survive, rather than from picking the term that feels responsible.

Where the work goes, and what actually limits a stage #

Euler’s equation says how much work a stage does. It says nothing about whether the stage will work at all, and the quantity that decides that is not in the equation.

The degree of reaction is the fraction of the stage’s static-enthalpy rise that happens in the rotor rather than the stator, and for a repeating stage at constant blade speed it is $1 - (V_{\theta 1}+V_{\theta 2})/2U$. Zero reaction is the impulse stage; fifty per cent splits the job evenly between the rotor and the stator and is what most axial compressors use, because it shares the diffusion between two blade rows instead of loading one.

The diffusion ratio $W_2/W_1$ is the real constraint. A compressor blade passage is a diffuser: it slows the relative flow in order to raise its pressure, and a boundary layer asked to climb a pressure rise separates. Past a diffusion ratio of about 0.7 the passage stalls, the stage stops producing pressure rise, and if enough stages do it at once the compressor surges — a violent flow reversal that can destroy an engine.

So the design of a stage is a negotiation between two equations that live in different essays. Euler says how much work is available from a given change of swirl; the separation criterion says how much turning a blade row can do without letting go. The first is exact and contains no fluid mechanics worth the name; the second is the entire content of a field of this site and is not exact at all.

The radial problem, and the third check #

Everything so far is a mean-line calculation: one radius, one triangle, one answer. A real annulus has a hub and a tip, the blade speed varies across it by a factor of two or more, and the swirl has to be distributed somehow.

The free vortex distribution, $rV_\theta$ constant, has a property that makes it very nearly universal in axial machines. The torque integral

$$Q = \int \rho V_x \, r V_\theta \, 2\pi r\,\mathrm{d}r$$

has $rV_\theta$ as a constant factor, so it comes straight out of the integral and what is left is just the mass flow. The mean-line formula is then exact, not approximate, and every station along the blade does the same work per unit mass.

That is the third check, and it is a quadrature rather than an identity: the integral is done numerically across the annulus and compared with $\dot{m}\,rV_\theta$, agreeing to $10^{-16}$. The same integral for a solid-body distribution misses the mean-line answer by 18.37 per cent, and that disagreement is asserted too — a mean-line formula that agreed with the integral for both distributions would mean the integral was not being done.

There is a reason the free-vortex distribution has that name, and it is on this site already. $rV_\theta$ constant is exactly an irrotational vortex: the flow goes round and the fluid parcels do not spin. So a free-vortex compressor design is one whose swirl carries no vorticity, and the reason it is used is that no vorticity means no radial variation in the work — which is the same statement twice.

The price is the twist. Because $V_\theta \propto 1/r$ while $U \propto r$, the flow angle a blade section meets changes enormously from hub to tip, and a free-vortex blade has to be twisted through tens of degrees along its span. That twist is the visible signature of this design choice on every jet engine fan, and it is the same geometric requirement a wind-turbine blade has for the same reason — a section whose relative flow direction changes continuously outward has to be set continuously differently.

The distribution has a second consequence that is easier to feel than to derive. Since the work per unit mass is the same at every radius, a free-vortex stage is uniformly loaded: no annulus is doing more than its share, so no annulus reaches its diffusion limit before the others. A design that overloaded the hub would stall there first, and a stalled hub in a multi-stage compressor is how a surge begins. Uniformity is not elegance here; it is the whole margin.

The swirl the blade asks for and the swirl it gets #

Every triangle above takes the outlet swirl as given. A designer does not have it; a designer has a blade angle, and assumes the flow leaves along it. That assumption is wrong by ten or fifteen per cent in a centrifugal machine, for a reason that is pure kinematics rather than any kind of loss.

Consider the fluid in one passage of an impeller, in the frame rotating with it. The flow arrived from a stationary inlet carrying no vorticity, and vorticity is not created by an inviscid flow being carried about — so in the absolute frame it is still irrotational. Seen from a frame turning at $\Omega$, an irrotational flow has vorticity $-2\Omega$: the fluid in each passage appears to circulate backwards relative to the impeller, as though there were a slow eddy filling it.

That relative eddy is superposed on the through-flow, and near the exit its backward motion subtracts from the tangential velocity the blade angle would have delivered. The flow leaves under-turned, and it leaves under-turned even with no viscosity, no boundary layer and no separation anywhere.

The ratio of what is delivered to what the blade asked for is the slip factor, and it depends chiefly on how many blades there are: more passages mean narrower ones, a weaker eddy across each, and less slip. The classical estimates put the shortfall in tangential velocity at roughly $\pi U_2 \sin\beta_2 / Z$ for $Z$ blades, giving slip factors around 0.85 to 0.92 for a typical fifteen- to twenty-bladed impeller.

And it is not a loss, which is the point most often got wrong. Slip reduces the swirl change, so by Euler’s equation it reduces the work — and it reduces the shaft power required in exactly the same proportion. The head falls and the power falls together; the efficiency is essentially untouched. A machine with a slip factor of 0.87 is not eighty-seven per cent efficient at anything. It is a machine that does eighty-seven per cent of the work its geometry promised, at very nearly the efficiency it would otherwise have had.

Which is a good closing note for an equation with no blade in it. Euler’s relation is exact given the swirl; the whole difficulty of turbomachinery is that the swirl is not the blade angle.

Where the model stops #

No losses. Euler’s equation gives the work exchanged with the fluid. How much of it survives as useful pressure rise — or how much shaft work is needed to achieve it — depends on friction, tip clearance, secondary flows and shocks, none of which is here. A real compressor stage has an isentropic efficiency around 0.9 and every point of that is somebody’s career.

No compressibility. The whole calculation is kinematic and holds for any fluid, but the useful output is a pressure rise, and relating work to pressure rise needs an equation of state and an efficiency. In a fan or a compressor operating near sonic relative velocities the blade sections develop their own shocks, which is the dominant loss in a transonic first stage.

No radial equilibrium. The swirl distribution is not free: a rotating flow needs a radial pressure gradient to balance the centrifugal acceleration, and that couples the axial velocity to the swirl. The mean-line and free-vortex arguments above hold constant axial velocity, which the radial equilibrium equation does not generally permit.

Nothing unsteady. A rotor blade passes through the wakes of the stator upstream of it once per blade passage. That unsteadiness sets the noise, the blade fatigue and a good deal of the loss, and a steady triangle knows nothing of it.

Who found it, and when #

Leonhard Euler published the equation in 1754, in a memoir on hydraulic machines written for the Berlin Academy — a paper about a proposed water turbine for the fountains at Sanssouci. It is the oldest result in this field by seventy years and one of the oldest on this site.

It has aged extraordinarily well. Euler had no gas turbines to describe, no boundary-layer theory, no concept of a Reynolds number and no thermodynamics in the modern sense; his equation predates Bernoulli’s own son’s publication of the energy theorem. It is nevertheless the equation used unaltered in every turbomachinery design office today, and it is used unaltered because it contains nothing about the machine — the property that makes it look almost content-free is exactly the property that made it survive.

That is worth putting beside Borda’s loss of 1766 and Bélanger’s jump of 1828. All three are eighteenth- or early-nineteenth-century control-volume results, all three are exact, and all three are still in current use in a subject that has otherwise been rebuilt twice since.

Where the field ends #

Nine fields, and this one has a different character from the eight before it.

The others ask what a flow does. This one asks what a flow is for, and the answers turn out to be unusually certain: Betz’s ceiling, Froude’s efficiency, the Borda–Carnot loss, the conjugate depths of a jump, Murray’s cube law and Euler’s equation are consequences of conservation and optimisation alone, and every one of them holds for machines nobody has built.

The reason is one sentence, and it is the sentence this field exists to make: a control volume does not need to know what is inside it. That is why a model containing no turbine sets a limit on every turbine, why the loss at a step can be written down without a viscosity, and why an equation from 1754 is still the one a compressor is designed with.

And it is also the field’s limit, stated in the same sentence. The box reports what crosses its faces and nothing else — so it is silent about blades stalling, wakes going unstable, layers separating, and every one of the things that decides whether a machine works. Those live in the eight fields before this one, where nothing is exact and the flow will not stay laminar. A design is the negotiation between the two, and knowing which of the two a given number came from is most of what it means to understand it.