What a turning frame keeps

The rung below this one is about an equation with almost nothing in it. The work a rotor does per unit mass is $U_2V_{\theta2} - U_1V_{\theta1}$, and no blade shape, no pressure, no gas property and no Reynolds number appears anywhere in it.

Its silence about pressure is the interesting part. A compressor exists to raise pressure and Euler’s equation does not mention it — so something else has to say where the pressure comes from, and the answer turns out to have two halves that behave completely differently.

A frame that should not conserve anything #

Stand in the frame turning with the rotor. The flow in it is steady — that is the whole reason for standing there — and the price is that the frame is accelerating, so Newton’s laws acquire two extra terms: a centrifugal force pointing outward and a Coriolis force perpendicular to the relative motion.

Bernoulli’s theorem needs a steady flow along a streamline in an inertial frame, and this frame is not inertial. The expectation should be that nothing survives.

Something does, and the reasons are one line each.

The centrifugal force has a potential. It is $\Omega^2 r$ outward, which is the gradient of $\Omega^2r^2/2 = U^2/2$ — so it can be absorbed into the pressure term exactly as gravity is, and it contributes a $-U^2/2$ to the conserved quantity.

The Coriolis force does no work. It is $-2\boldsymbol{\Omega}\times\mathbf{W}$, which is perpendicular to $\mathbf{W}$ at every instant, so its dot product with the displacement is identically zero. It bends the relative streamlines and it never appears in an energy balance — which is the same reason the Earth’s rotation cannot decide which way a bath drains by doing work on it.

What is left is the rothalpy:

$$\frac{p}{\rho} + \frac{W^2}{2} - \frac{U^2}{2} = \text{constant along a relative streamline},$$

and it is Bernoulli’s theorem in a rotating frame with one extra term.

The two halves of a pressure rise #

Rearranged, the rothalpy statement says what a rotor’s static pressure rise is made of:

$$\frac{\Delta p}{\rho} = \frac{U_2^2 - U_1^2}{2} + \frac{W_1^2 - W_2^2}{2}.$$

The second term is diffusion: the blade passage is a duct that slows the relative flow, and slowing a flow raises its pressure. That is ordinary and it is severely limited, because a boundary layer asked to climb a pressure rise separates — past a relative velocity ratio of about 0.72 the passage stalls, which is the constraint the rung below identifies as the real limit on a stage.

The first term is the radius, and it is a different kind of quantity altogether. It has no velocity gradient in it, no boundary layer, no diffusion and no limit. It is available whenever the flow leaves at a larger radius than it entered, and it is available in full whatever the blade is doing.

So a designer has two sources of pressure and only one of them can be exhausted. The second is the one every diffuser in this collection is limited by, and the first is not a fluid-mechanical quantity at all.

The number that separates a fan from a turbocharger #

Put numbers on that split and the whole architecture of turbomachinery falls out of it.

Take a rotor at 1,000 radians a second with an inlet radius of 80 millimetres and a relative flow of 200 metres a second, diffused as hard as the limit allows.

• Axial — no radius change at all. The pressure rise is 11,600 pascals, every joule of it from diffusion, and there is nothing left to take.
• Radius ratio 2. 23,100 pascals, of which half came from the radius.
• Radius ratio 3. 42,300 pascals, of which 73 per cent came from the radius.

The radial machine is not diffusing any harder; the de Haller ratio is 0.72 in all three cases and the blade is doing exactly the same aerodynamic work. It is collecting a term the axial machine has no access to.

That is why a centrifugal compressor reaches a pressure ratio of four or five in one stage and an axial one manages 1.2 to 1.4. It is not that radial blades are better; it is that a radial machine has a source of pressure that is geometry rather than fluid mechanics, and the axial machine’s only source is the one with a stall in it.

And the price the axial machine pays instead #

The trade is not free and the axial machine’s answer is stages, which connects this rung directly to the staging arithmetic in the applied field.

An axial compressor cannot take more per stage, so it takes less per stage many times. Fifteen stages at 1.3 gives a pressure ratio of about 40, which no single radial stage approaches — so the axial machine wins on total ratio and loses on ratio per stage, and it wins on frontal area, which is why aircraft engines are axial and why a turbocharger, which has no frontal-area constraint, is not. Frontal area is drag, and drag on a nacelle is the same accounting as drag on anything else.

The two architectures are the same equation solved with different terms, and once the split is written down the choice between them stops being a matter of tradition.

There is a second consequence about turbines worth stating, because the sign reverses and the conclusion does not. A radial inflow turbine has the flow entering at a large radius and leaving at a small one, so its centrifugal term is negative — and it is helping, because a turbine wants the pressure to fall. The same free term, working the other way, is why a radial inflow turbine extracts so much per stage and why a turbocharger’s hot end looks the way it does.

Reading the split as a design decision #

The two-term form is more useful as a question than as a formula, and the question is: of the pressure rise this stage must produce, how much is the geometry going to give me?

Because the answer is known before any blade exists. The inlet and outlet radii are set by the machine’s layout — where the flow comes from and where it has to go — and the shaft speed is set by whatever drives it. Those three numbers give the centrifugal term immediately, with no aerodynamics at all. The blade is then asked to make up the difference by diffusion, and whether it can is a single comparison against the de Haller limit.

So the design order is the opposite of the intuitive one. A designer does not shape a blade and then discover the pressure rise. A designer computes what the radius supplies, subtracts it from what is required, converts the remainder into a required relative-velocity ratio, and checks whether that ratio is achievable. Only then is there a blade to draw.

That order also explains a fact about real machines that looks arbitrary. A centrifugal impeller’s blades are often radial — straight, with no backward sweep at all — in high-pressure-ratio applications, which looks like a crude design. It is not: a radial-bladed impeller does no diffusion in the relative frame worth the name, takes essentially all of its pressure rise from the centrifugal term, and is therefore immune to the stall that limits everything else. It buys robustness by taking only the free term, and it pays in efficiency, which is exactly the trade the split makes visible.

The term that is not in either half #

There is a quantity conspicuously missing from the pressure accounting and its absence is worth a paragraph, because a reader who has come from Euler’s equation will be looking for it.

The work the rotor does is $U_2V_{\theta2} - U_1V_{\theta1}$, and that appears nowhere above. The static pressure rise is not the work: the difference between them is the change in the flow’s own absolute kinetic energy, which the rotor has also imparted and which is still in the flow when it leaves.

A stage recovers some of that in its stator, which is why a stator exists and why the degree of reaction — the fraction of the static rise that happens in the rotor — is a design variable at all. A fifty per cent reaction stage takes half the static rise in the rotor and half in the stator, and the arithmetic above is the rotor’s half.

Which is the cleanest way to see what rothalpy is for. Euler’s equation is about the whole machine and knows nothing about pressure. Rothalpy is about the rotor, in the frame where it is steady, and knows nothing about work. The two together are the stage, and neither alone is.

Why the frame is worth standing in at all #

It is worth being explicit about what the rotating frame buys, because a reader could reasonably ask why a calculation is being done in an accelerating frame when an inertial one is available.

In the ground’s frame, a compressor rotor is a set of blades sweeping past. The flow at any fixed point is violently unsteady — the velocity there swings every time a blade goes by, at a frequency of thousands per second — and no steady analysis of any kind applies. There is no Bernoulli, no streamline that stays put, and no station where a quantity may be evaluated.

In the blade’s frame the same flow is steady. The blades are stationary, the passages are ducts, and the flow through them settles into a pattern that does not change. Every tool this collection has for steady flow becomes available at once, at the price of two fictitious forces — and both of them turn out to be harmless, one because it has a potential and one because it does no work.

That is a very good trade and it is the reason turbomachinery is a subject rather than a collection of measurements. The whole of blade design happens in a frame nobody occupies.

There is a caution attached and it is where the trade stops being free. The frame is steady only if the flow really is periodic with the blade passing — one passage identical to the next, repeating for ever. A rotor with a distorted inlet, an upstream strut wake, or a rotating stall cell does not satisfy that, and in those cases the relative flow is unsteady too and the analysis above applies to none of it. The frame removes the unsteadiness the blades themselves cause, and no other.

Where the free term is not free #

Three qualifications, because a term with no limit in the equation always has a limit somewhere else.

The radius costs blade speed, and blade speed costs stress. Doubling the outlet radius at a fixed shaft speed doubles the tip speed, and the centrifugal load on the blade grows as its square. That is not in the rothalpy statement at all and it is what actually caps a radial machine — the subject of the rung above this one.

The flow has to be turned back. A radial machine discharges outward and something must collect it, which is a diffuser or a volute, and a radial diffuser at high Mach number is one of the hardest components in the subject. The rothalpy statement is about the rotor and hands the problem to the next part of the machine.

And the relative flow is not free of losses. The statement above is inviscid. Real losses appear as a shortfall in the pressure rise achieved for a given change in the kinematic terms, and the rothalpy form is exactly where they are measured — a rotor’s loss coefficient is defined as the departure from this equation.

One more thing the equation does not contain #

A quantity that appears in every real machine and in none of the arithmetic above deserves naming, since its absence is the same kind of fact as the chord’s absence from the stress in the rung above.

The number of blades is not in the rothalpy statement. The relation holds along a relative streamline and every passage carries one, so a rotor with twelve blades and one with thirty obey it identically at the same velocities. What the blade count changes is whether those velocities are the ones the blade angles asked for — which is slip, and the rung below prices it — and the pressure accounting then runs on whatever the flow actually did.

That is a useful separation to keep. Rothalpy is an exact relation between the velocities at two stations; the blade count is one of the things that decides what those velocities are. Confusing them is how a designer ends up believing that adding blades raises the pressure rise, when what it does is bring the achieved swirl closer to the intended one and raise the work — and the pressure follows the work only through the accounting above.

What happens to rothalpy when the machine is not adiabatic #

The conserved quantity above assumes the rotor neither gains nor loses heat, and it is worth asking what survives when it does — because one important class of machine is deliberately cooled.

The rothalpy statement in its general form is $I = h + W^2/2 - U^2/2$, and it is conserved along a relative streamline in steady adiabatic flow in the rotating frame. Add a heat flux and it is not: $I$ rises where heat is added and falls where it is removed, exactly as a total enthalpy does in a stationary duct.

A cooled turbine rotor is therefore a machine whose conserved quantity is not conserved, and the practical consequence is that a rotor’s performance cannot be assessed by the departure of $I$ from constancy in the way an uncooled one’s can. The loss and the cooling are mixed in the same number, and separating them needs the cooling flow’s own energy accounted for — which is where a great deal of turbine analysis’s complexity lives.

There is a second and subtler failure that is worth naming because it is not about heat at all. The quantity is conserved along a relative streamline, and that requires the relative flow to be steady — which is the whole reason for standing in the rotating frame. A rotor with an unsteady relative flow has no such conserved quantity, and the cases in which the relative flow is unsteady are exactly the ones a designer cares about: rotating stall, a distorted inlet, and the wakes of an upstream blade row chopping through the passage.

So the tool works where the machine is behaving and fails where it is not, which is the standing condition of every steady analysis in this collection and is worth stating rather than discovering.

The stage, and the other half of the accounting #

This essay is about a rotor and a stage is a rotor and a stator, so the accounting is worth completing in a paragraph because the stator’s half is where the rotor’s leftovers go.

The rotor does work $U_2V_{\theta2} - U_1V_{\theta1}$ per unit mass. Some of that appears as static pressure in the rotor — which is what rothalpy computes — and the rest is still in the flow as absolute kinetic energy, largely swirl.

The stator has no work to do, so its rothalpy is an ordinary Bernoulli: it is stationary, $U = 0$, and $p/\rho + V^2/2$ is constant through it. Its job is to remove the swirl and turn that kinetic energy into pressure, and it does so by diffusing — which puts it under exactly the same de Haller limit as the rotor, in the absolute frame instead of the relative one.

So a stage has two diffusions and one free term. The rotor’s diffusion is limited, the stator’s diffusion is limited, and only the rotor has access to the centrifugal term. That is why a fifty per cent reaction stage is the common choice: it shares the diffusion evenly between the two rows, so neither is run closer to its limit than it needs to be.

And it explains the axial machine’s shape one level further down. An axial compressor stage is two blade rows each diffusing as hard as they safely can, twice per stage, with no free term anywhere — which is why its pressure ratio is what it is and why it takes fifteen of them.

What the picture cannot show #

Incompressible throughout. Every figure treats the density as constant, which is honest for a fan and a fair approximation for one stage of a compressor at moderate pressure ratio, and it is wrong for the four-to-one radial stage the essay uses as its illustration. The compressible form replaces $p/\rho$ with the enthalpy and the argument survives unchanged; the numbers do not.

The de Haller limit is borrowed. Nothing in this collection derives 0.72. It is a summary of when blade passages have been observed to stall, and every figure here draws it accordingly.

No blade shape anywhere. The relative velocities are inputs. What blade produces them, at what incidence, with what deviation, is the design problem and this is the accounting that constrains it.

And nothing here is a stage. A rotor is half a stage; the stator that follows it recovers more pressure and removes the swirl, and the stage’s pressure ratio is the two together. The split computed here is the rotor’s alone.

The assertion behind these figures is the one that could reject: the kinematic part of the rothalpy must fall by exactly the static pressure rise divided by the density, checked to machine precision and computed by two routes that share no arithmetic. Checking that the two rothalpies are equal would be a tautology, since the pressure term is what makes them so; checking that their difference is the pressure rise is the statement.

Who found it, and when #

The rotating-frame energy equation is a nineteenth-century result in its mechanics and a twentieth-century one in its application. The name rothalpy is a portmanteau of rotational and enthalpy and dates from the mid-twentieth century, from the compressor literature; the quantity itself had been in use without a name for considerably longer.

The historically interesting point is which machine came first. The centrifugal compressor predates the axial one in practical use — the earliest jet engines, Whittle’s included, were centrifugal — precisely because one stage did enough. The axial compressor won on frontal area and on efficiency once enough stages could be made to work together, which took a decade of understanding about matching stages that had nothing to do with this equation.

So the free term was exploited first and the constrained one took longer to master, which is the order one would expect from the split above and is not the order the modern textbook’s emphasis suggests.

Where the ladder goes next #

The free term has a cost and the cost is not in this equation. The rung above is the stress at the blade root, and it is the number that decides an annulus area before any aerodynamicist sees it: integrating a blade’s own weight outward gives a root stress proportional to the annulus area times the square of the shaft speed, with the chord cancelled, the blade count cancelled and every property of the gas cancelled. It is a solid-mechanics constraint on an aerodynamic variable, and it is where the radius the rothalpy statement wants is finally capped.

The one beside it is the stage-matching problem the history above points at. A multistage machine’s stages must each be at a workable operating point simultaneously, at every speed the machine runs at, and the fact that they cannot be is why compressors have bleed valves and variable stators. That is a question about a series of the control volumes this field is built on, and it is the one place where drawing a box round each part separately genuinely fails.