A row is not a set of aerofoils

Everything this collection has said about lift has been about a body in an unbounded stream. The circulation is round one object, the Kutta condition applies at one trailing edge, and the free stream is the velocity a long way away, where the body’s influence has died out.

A compressor blade has an infinite row of identical blades above and below it, and their influence does not die out. It is the same everywhere along the row, it is different upstream and downstream, and it is part of what the blade in the middle is flying through.

What a row does, exactly #

The vortex row has a closed field, and it is the foundation of everything below.

A row of point vortices of strength Γ spaced $s$ apart along $y$ has

$$u - iv = -\frac{i\Gamma}{2s}\coth\frac{\pi\zeta}{s},$$

from summing $\sum 1/(\zeta - iks) = (\pi/s)\coth(\pi\zeta/s)$. Far upstream $\coth \to -1$ and far downstream $\coth \to +1$, so the row induces $v = -\Gamma/2s$ ahead of itself and $+\Gamma/2s$ behind:

The row turns the flow by Γ/s in total, half of it before the blades and half after. That is checked here to a part in 10¹⁵ against the closed form, and the kernel is checked separately to reduce to a single vortex at large spacing — which is where it turned out to be badly conditioned, and is worth a sentence.

Written as $\sinh 2a/(\cosh 2a - \cos 2b)$ the coth loses ten digits at wide spacing, because both terms in the denominator are 1 to within $10^{-10}$ and their difference is the whole answer. Rewriting the denominator as $2(\sinh^2 a + \sin^2 b)$ — the same number, and not a difference of two quantities near one — fixes it. The limit the kernel is checked against is the limit it is worst-conditioned in, which is not a coincidence: wide spacing is small argument.

The velocity that exists nowhere #

The immediate consequence is the one that makes a cascade a different problem.

Because the row induces $\mp\Gamma/2s$ symmetrically, the uniform part of the field is exactly halfway between the inlet and the outlet. A blade is not flying through the inlet velocity and it is not flying through the outlet velocity: it is flying through the vector mean of them, and no probe anywhere in the machine measures that.

So the cascade form of Kutta–Joukowski is $L = \rho|\mathbf W_m|\Gamma$ with $\mathbf W_m$ the mean, and the blade’s incidence is measured from the mean flow direction rather than from the inlet.

That is a structural fact rather than a correction, and getting it wrong is expensive. Imposing the inlet angle on the uniform stream in the linear system — the obvious thing, and the thing this essay tried first — over-turns the flow by seventeen degrees at unit solidity, with every other check in the calculation still passing.

Recovering the isolated aerofoil #

The solve is a discrete-vortex calculation: forty panels along the chord, a vortex at each quarter point, a control point at each three-quarter point, tangency enforced with the row kernel. That placement makes the Kutta condition automatic for a flat plate, and the first thing to check is that it reproduces the answer everybody already knows.

At a solidity of 0.02 the slope is 6.2809 against 2π = 6.2832: three parts in ten thousand. The isolated aerofoil is recovered, which is the check that licenses everything at higher solidity.

And then it falls. At σ = 0.5 the slope is 90 per cent of 2π, at σ = 1 it is 67 per cent, at σ = 2 it is 37 per cent. A blade in a tight row carries far less lift per unit of incidence than the same blade alone.

Which is not the same as turning less #

The obvious reading of that is that a tight row is worse, and the obvious reading is exactly backwards.

An open row barely turns the flow at all: at σ = 0.02 the blades are set at 30° in a flow arriving at 45°, and the flow leaves at 44.35°. The deviation — the angle between the outlet flow and the blade — is 14.35°, which is nearly the whole of the intended turning.

A tight row guides the flow to its own angle: at σ = 3 the flow leaves at 30.001°, a deviation of a thousandth of a degree, and the turning is 14.999° out of an intended 15.

So a tight row turns the flow completely and each blade carries less. Those are the same fact: the turning is $\Delta\tan\beta = \sigma C_L/2$, and a tight row achieves a large turning with a small $C_L$ because there are many blades doing it. The lift per blade falls and the number of blades rises faster.

What an isolated calculation predicts #

The designer’s calculation is $C_L = 2\pi(\beta_1 - \lambda)$ on the inlet incidence, and $\Delta\tan\beta = \sigma C_L/2$. At σ = 0.1 it predicts 3.8° of turning against the row’s 3.0 — right to twenty per cent, in a regime where it is predicting almost nothing. At σ = 1 it predicts 34.9° against the row’s 14.16°: a factor of two and a half.

The gap grows with solidity, and it grows because the isolated calculation uses the inlet incidence where the blade sees the mean. As the row turns the flow more, the mean moves further from the inlet, and the error compounds.

That gap is what every turbomachinery deviation correlation exists to patch — Carter’s rule, Howell’s rule, the whole apparatus of cascade testing. They are empirical because the effect is geometric and depends on the camber and the stagger as well as the solidity, and because a real cascade has a boundary layer on each blade and a wake behind it.

What the correlations do not say #

Carter’s rule is worth writing down, because it is the empirical answer to the question this essay computes. It says the deviation is

$$\delta = m\,\theta/\sqrt{\sigma},$$

for a camber θ and a coefficient m that depends on the stagger — an inverse square root in solidity, and nothing else.

The flat-plate row does not obey it, and the disagreement is the informative part. Between σ = 0.4 and σ = 1.6 the computed deviation falls from 5.57° to 0.10°, a factor of 56 across a factor of four in solidity. Carter’s rule over the same range gives a factor of 2. The computed curve is not of that family at all: it closes like the row kernel, which is exponential in the spacing, and not like a power law.

The reason is the camber the flat plate does not have. A cambered blade’s deviation cannot go to zero however tight the row is, because the flow leaving a curved passage has been asked to keep turning right up to the trailing edge and it stops turning slightly early — that residual is proportional to the camber and survives the limit. A flat plate has no camber to fail to follow, so at high solidity it is guided exactly and its deviation vanishes.

So the two statements are about different things, and the θ in Carter’s rule is where the difference lives. The solidity dependence in the correlation is the part a real cascade shares with this calculation; the camber is the part the flat-plate model discarded, and it is the part that sets the floor.

What this means for a stage #

A compressor or turbine stage is a row of rotor blades followed by a row of stators, and the numbers above translate into the two things a designer trades.

Work per stage. The work a rotor does on the fluid is $\Omega r\,\Delta c_\theta$ by Euler’s turbomachine equation, so it is proportional to the turning. Turning more per stage means fewer stages, a shorter and lighter machine, and a higher solidity to achieve it — which is why compressor solidities sit between about 1 and 1.7 and not at 0.3.

And diffusion. A compressor row is a diffuser: it slows the flow in the passage, and a boundary layer in an adverse gradient will only take so much before it separates. The standard measure is de Haller’s ratio $W_2/W_1 > 0.72$, or Lieblein’s diffusion factor, and both are limits on how much turning a row may be asked for regardless of what the inviscid calculation says it can do.

So the design point is squeezed from two sides. The inviscid calculation says more solidity turns the flow more completely; the viscous limit says more turning is not allowed; and the answer is a row that is tight enough to guide the flow and loaded lightly enough not to separate.

A turbine is easier, and for a reason worth naming: it accelerates the flow through the passage, so its boundary layers are in a favourable gradient and it can be asked for far more turning per row. A turbine stage routinely turns the flow through 90 degrees or more; a compressor stage manages 20 to 40. The same cascade theory describes both, and the asymmetry is entirely in the boundary layer.

The other reading: a duct with blades in it #

There is a second way to see the tight-row limit, and it is the one a turbomachinery engineer uses.

At high solidity the passage between two adjacent blades is a long narrow channel, and the flow leaves it parallel to the walls because it has nowhere else to go. The blades have stopped being aerofoils and have become the walls of a duct. In that limit the outlet angle is the blade angle, the turning is whatever the geometry says, and the aerodynamics has been reduced to the shape of a channel.

At low solidity the blades are isolated aerofoils and the outlet angle is whatever the circulation produces. Between those two limits is where machines are built, at solidities of about 1 to 1.5, and the whole of blade-row design lives in the crossover.

That crossover is a collapse and a residual in the shape these essays keep finding. The isolated-aerofoil picture collapses a row onto one blade, and the solidity is what survives the collapse.

The same argument in three other places #

The structure — a body whose free stream includes the influence of its own periodic images — turns up elsewhere in this collection under other names, and the family resemblance is worth drawing.

A biplane. Two wings one above the other, each flying in the other’s downwash, is the same calculation with two members instead of infinitely many. What comes out is Munk’s stagger theorem — the total induced drag does not depend on the fore-and-aft spacing — which the biplane essay computes, and it is the same statement that the interference depends on the wake trace rather than on the arrangement.

A wing in a wind tunnel. The walls are equivalent to an infinite array of image wings, so a tunnel model is a cascade in two directions, and the correction for it is exactly the induced velocity of that array. The walls are in the answer makes the same point about the same algebra: the free stream a model flies in is not the tunnel speed.

And a formation of aircraft. The lift beside a wing is the two-member version again, with the members side by side rather than stacked, and its result — a pair tip to tip costs exactly half what the two cost apart — is the low-solidity limit of the same arithmetic.

In each case the isolated-body theory is exact and is being asked a question about a configuration, and the correction is not a small perturbation: it is a change in what the free stream is.

What the model contains #

The calculation above is flat plates, and that is a real restriction, so it is worth separating what survives from what does not.

The kernel is exact and holds for any row, whatever the blades are. It is a statement about an infinite periodic array of point vortices, and nothing in its derivation asks what the vortices are attached to, so it survives every refinement of the blade below.

The mean-velocity theorem is exact and follows from the kernel alone: it does not depend on the blade shape at all, only on the fact that the row induces symmetric far fields.

The lift slope’s fall with solidity is qualitatively right and quantitatively a flat plate’s. A cambered blade at the same solidity has a different curve, and the direction and rough magnitude are the same.

And the deviation is a flat plate’s deviation, which is the least transferable of the four. A real cascade’s deviation depends on the camber, the stagger, the thickness distribution and the boundary layer, and correlations for it are the reason cascade tunnels were built.

The other velocity nobody measures at the blade #

The mean velocity is not the only quantity in the triangle that is assumed rather than known. Every figure here holds the axial component fixed between inlet and outlet, which is what a two-dimensional row of constant spacing in an incompressible fluid means. A machine does not hold it fixed, and the way it drifts is the reason multistage compressors have moving parts they would otherwise not need.

Three things change the axial velocity through a row. The annulus is contoured, deliberately. The density rises, in a compressor, which shrinks the volume flow. And the endwall boundary layers thicken, so a growing fraction of the passage is carrying almost nothing — the blockage — and the flow that does get through has to move faster to fit.

Individually those are a few per cent per stage. The trouble is that they accumulate. Each row’s outlet triangle is the next row’s inlet triangle, so an axial velocity a few per cent high hands the following blade an incidence that is wrong, which changes its turning, which hands the row after that a worse one. By the back of a fifteen-stage compressor the accumulated error is not a correction.

Which is what the machinery is for. A multistage compressor carries variable stator vanes in its front stages and bleed valves in its middle, and neither exists to improve any blade’s aerodynamics. They exist to re-aim the velocity triangles when the accumulated axial-velocity error has pushed the front stages towards stall and the rear ones towards choke — the classic off-design mismatch, which appears whenever the machine is run away from the speed its triangles were drawn at.

So the essay’s residual has a companion. The blade’s incidence is measured from a mean nobody probes, through an axial velocity nobody measures at the blade, and both are recovered by correlation.

What is not in it at all #

No viscosity. A real blade row’s loss is almost entirely boundary layers and wakes: profile loss on the blade, endwall loss in the corners, and — in a compressor — the possibility that the diffusion is more than the boundary layer will take. That last is what limits how much turning a row can do, and this model has no opinion about it.

No three-dimensionality. A real row has endwalls, and the secondary flows in the corners between blade and endwall are a large fraction of the loss in a modern machine.

No compressibility, which is what actually limits a transonic compressor stage.

And no unsteadiness. A rotor row and a stator row pass each other, so every blade sees a wake once per blade-pass, and the resulting unsteady loading is a fatigue problem and a noise source. It is also a gust problem of exactly the kind the unsteady-lift essays are about, arriving at a few thousand times a second.

Where the row came from #

Cascade theory begins with the recognition that a blade row is a two-dimensional periodic problem rather than a set of one-body problems, which is Weinig’s in the 1930s. The conformal mappings that solve a flat plate cascade exactly are his, and Howell’s cascade tunnel measurements in the 1940s are what turned the theory into a design method by supplying the deviation.

The discrete-vortex method used here is the same 1/4–3/4 rule that a panel method rests on, with one kernel replaced. That the whole of the periodicity fits into a single hyperbolic function is the reason a cascade is tractable at all: an infinite sum that happens to have a closed form.

What this leaves #

An isolated aerofoil’s velocity is measured a long way away and a cascade blade’s is measured halfway between two things a probe can reach. The residual the collapse discards is the solidity, and it is worth a factor of two and a half at the values machines are built at.

The next essay takes the Kutta condition itself and asks what happens when the sharp edge it depends on is taken away: the condition that can be bought.