A compressor needs one actuator for each way it can fail

A throttle can hold the peak only in a small plenum put a fast throttle on the Moore–Greitzer compressor and asked it to do two jobs. Driven by the plenum’s pressure it damped surge, the whole flow’s oscillation through machine and plenum. Driven by the stall cell’s amplitude it removed the jump into deep stall, but what it held was a small stalled cell rather than clean flow, because a throttle acts on the whole annulus and a stall cell lives at one angle; and above a B of about a third the held cell coupled to the surge mode through the throttle and the machine cycled. One actuator, two modes, and the coupling between them lived in the actuator.

The essay ended by naming the actuator that can reach the cell: air blown in at the rotor tips with an amplitude and a phase set by the cell’s own measured position. This essay adds it, and finds that it settles the stall question and exposes a second limit that had been hidden behind the first. The throttle’s surge damping has a ceiling as well as a floor, the ceiling does not move with B, and at B = 2 the two have crossed: the throttle alone cannot damp surge at ninety per cent of the peak flow at any gain. A second signal for the throttle reopens the window. With an injector for the cell and a two-signal throttle for surge, the machine is held.

Injection that leans against the cell #

A rotating stall cell is a first circumferential harmonic of the flow: on one side of the annulus the axial flow is low, on the other high, and the pattern turns at a fraction of the rotor’s speed. Where the flow is low the blades are at a higher incidence and the cell is deepened; that is why it grows where the compressor’s characteristic rises, left of its peak. Injecting air at the blade tips on the low-flow side energises the end-wall flow there and raises the local pressure rise, which is the opposite of what the cell does. Modelled as a shift of the local characteristic in proportion to the injected flow, at the characteristic’s steepest slope, the injection enters the cell’s amplitude equation as a term against the cell:

$$\frac{da}{d\xi} = \frac{\sigma}{2}\Big[\big(1 - u^2 - \tfrac14 a^2\big)\,a \;-\; \frac{i_1}{W}\cos\beta\Big], \qquad i_1 = \min(G W a,\ i_{\max}).$$

Here $a$ is the cell’s flow amplitude over the characteristic’s half-width $W$, $u$ the flow’s position on the characteristic, zero at the middle of the rising branch and one at the peak, and $i_1$ the injector’s first-harmonic flow: proportional to the cell with a gain $G$, limited by the injector’s capacity, and applied with a phase lag $\beta$ behind the cell, so that only its component $\cos\beta$ opposes it. Squared, with no injection, the equation is Moore and Greitzer’s for the cell’s energy $J = a^2$, to rounding.

Linearised, the cell grows only where $1 - u^2 > G\cos\beta$. With no lag, a gain of 0.4 moves the stall line from the peak to 88.7 per cent of the peak flow, a gain of 0.7 to 77.4 per cent, and a gain of one to half the peak flow, the bottom of the rising branch. That is the one thing a throttle could not do: the cell is stopped from starting, so the controlled state has no cell in it at all.

The throttle’s ceiling #

The injector does nothing to surge. Surge has no cell in it; it is the flow and the plenum pressure trading energy, and at an unstalled operating point the two modes are independent to first order. So the throttle keeps its surge job, and the throttle essay’s result applies: a throttle that opens in proportion to the plenum pressure’s rise damps the oscillation once its gain exceeds a floor that grows as $B^2$.

That essay’s result also has a second side, which only shows once the machine is asked to run well left of its peak. A throttle that opens with the pressure makes the throttle line the compressor sees flatter, and past a certain gain it is flatter than the compressor’s own rising characteristic. The operating point is then a saddle — the same static instability a compressor left of its peak on an ordinary, too-open throttle line would have — and the mode that grows is a single real root, not an oscillation. The determinant of the surge mode’s matrix sets the ceiling,

$$K_\Psi < \frac{1/S - \gamma/2\sqrt\Psi}{\sqrt\Psi},$$

with $S$ the compressor’s slope and $\gamma$ the throttle’s coefficient, and B does not appear in it. The floor rises as $B^2$, the ceiling stays put, and they meet at $B = 1/2S$.

At 95 per cent of the peak flow, where the characteristic is nearly flat, the window is open to B = 2.44. At 90 per cent it closes at 1.29; at 80 per cent at 0.72. The two actuators therefore limit the machine in different directions. The injector’s limit is a vertical line, set by its gain and independent of the plenum. The throttle’s is a curve that falls steeply as the operating point moves left, set by the plenum and independent of the injector. A large plenum leaves the injector with nothing to do: the machine surges before the cell matters.

The flow signal the throttle essay dismissed #

The throttle essay showed that a throttle driven by the flow cannot damp surge at all, because a flow gain does not enter the surge matrix’s trace. That is still true, and it is beside the point here, because the ceiling is not the trace’s — it is the determinant’s, and the flow gain is in the determinant. A throttle that opens as the flow falls, $\gamma = \gamma_0 + K_\Psi(\Psi - \Psi_0) + K_\Phi(\Phi - \Phi_0)$ with $K_\Phi$ negative, makes the throttle line steeper again in exactly the sense the pressure gain made it flatter, and lifts the ceiling without touching the damping.

At B = 2 and ninety per cent of the peak flow, the pressure alone fails at every gain: below 2.76 the oscillation grows, above it the static root does, and the two regions overlap. With a flow gain of −3.8 the static root is pushed back and every pressure gain from 7.3 to 12.5 is stable. The two signals have two jobs: the pressure damps the oscillation through the trace and the flow keeps the operating point a node through the determinant. That is a structural statement of the same kind as the throttle essay’s, one level further: a single signal on the throttle reaches one of the surge matrix’s two invariants, and holding a large plenum left of the peak needs both.

The instinctive bleed valve — one that opens as the measured flow falls — is the flow gain with exactly the sign used here. It was right for a reason its designers did not give: not because it damps the surge, which it cannot, but because it stops the pressure controller from turning surge into a static collapse.

Why the two actuators do not fight #

The throttle essay’s stall controller failed because the throttle was asked to hold a stalled state, and in a stalled state the two modes are joined. In the Moore–Greitzer equations the cell acts on the mean flow through a term proportional to its energy, $J$: a cell of amplitude $a$ lowers the annulus-averaged pressure rise in proportion to $a^2$. About a state with a cell already in it — $J$ of order one — a change in the cell changes the mean flow at first order, the mean flow changes the plenum, and the plenum is where surge lives. Any actuator that moves one moves the other, and the throttle, which can only move the mean flow, was moving both at once.

About a state with no cell, the same term is second order. A small cell of amplitude $a$ moves the mean flow by an amount of order $a^2$, too small to matter while the cell is small, so the linearised system splits into two blocks: the surge pair, which the throttle governs, and the cell, which the injector governs, with nothing in between. That is why the injector changes the calculation’s character and not only its numbers. It does not have to be stronger than the throttle’s stall feedback was; it has to act on a state in which the modes are separate, and the only way to get such a state left of the peak is to stop the cell from starting, which the throttle could not do and the injector can.

The separation lasts only while the cell is small. A kick large enough to make $a^2$ comparable to the distance from the stall line joins the blocks again, and the basin computed below is where that happens: the injector’s capacity bounds the cell from one side, the throttle’s design bounds the coupling from the other, and whichever is weaker sets the largest disturbance the machine survives.

Kicked, and held #

The opening figure puts all four controllers through the same test: the machine at ninety per cent of its peak flow and B = 2, a stall cell of amplitude 0.05 — a flow disturbance of two and a half per cent of the peak — started at time zero, and an injector of gain 0.7 whose capacity is two per cent of the peak flow. With the throttle fixed, the injector kills the cell, and the machine surges anyway: the flow swings from backflow at 45 per cent of the peak to 150 per cent and back. With the pressure throttle added, the static root takes over and the flow runs away to zero. With the two-signal throttle and no injector, surge is held and the cell is not: the machine falls into deep stall at 39 per cent of the peak flow with a cell of amplitude two. Only with both does the cell die and the flow return to where it was.

Dialled down to B = 1, the pressure throttle alone is enough for surge and the injector with it recovers; at B = 0.5 the same. The number of actuators a compressor needs depends on its installation as much as on its blades — which is the point the plenum decides whether a compressor surges made about the instability, now made about its cure.

Sizing the injector #

An injector delivers a bounded flow. While the cell is small the injection grows with it, and the linear argument holds. Once the cell is large enough that $GWa$ exceeds the capacity, the injection stops growing, the cell’s own growth is limited only by the cubic term, and a large enough disturbance escapes into deep stall. The capacity therefore sets how large a disturbance the controlled machine survives.

At small capacities the answer is a ratio: the controlled machine survives a cell of about 2.75 times the injector’s capacity, at either B. The injector does not have to cancel the cell’s flow deficit; it has to stop it growing, and the cell’s growth rate is small near the stall line, so a modest opposing push suffices. At larger capacities the injector stops being the limit. At B = 2 the survivable cell levels off near twelve per cent of the peak flow, because a cell that large reduces the mean flow enough to disturb the surge mode, and the throttle’s pair, designed for small disturbances, is what fails. More injected air then buys nothing; a stiffer throttle would.

That makes the number the throttle essay asked for — how much injected air, as a fraction of the machine’s flow — into a statement about disturbances. An injector of one or two per cent of the flow holds the line against disturbances a few per cent of the flow in size, which is the scale of a gust at the inlet or a distortion from a crosswind. Laboratory demonstrations of tip injection, on low-speed and transonic rigs in the 1990s, used injectors of the order of a few per cent of the compressor’s flow, which is where this model puts the useful range.

The injector’s lag #

An injector is a valve and a jet, and both take time. A lag of $\tau$ seconds is a phase $\beta = \omega\tau$ behind a cell turning at $\omega$, and stall cells turn at between a fifth and a half of the rotor’s speed, so a lag of a third of a revolution is a radian. Only the part of the injection in phase with the cell’s deficit opposes it.

The line is robust and the margin is not. With no lag the injector holds the cell down to 77.4 per cent of the peak flow; lagging by a radian it still holds to 89.4 per cent, just covering an operating point at ninety. But the disturbance it survives there falls from nine per cent of the peak flow to two, and at 69 degrees to nothing: the operating point is left of the line and the cell grows from any size. The effective gain is $G\cos\beta$, so a lagged injector is a weaker one, and a weaker injector’s line moves towards the operating point until the margin between them is the only thing absorbing a disturbance. That is a slow sensor making a bearing cycle in another machine: a correction applied to a rotating quantity has to be applied where the quantity is now, not where it was.

What the injector cannot be asked to do #

Every number here is for the throttle essay’s compressor: a cubic characteristic with shut-off pressure rise 0.3, height 0.18 and half-width 0.25, a duct eight rotor radii long, time in rotor radii travelled. The injector is modelled by what it does to the cell’s equation — a shift of the local characteristic proportional to the injected flow at the characteristic’s steepest slope — not by the jets’ fluid mechanics, and the factor that converts an injected mass flow into a pressure-rise shift is the model’s largest assumption; a jet that energises the end-wall flow less effectively needs a proportionally larger gain for the same line. The injector’s bias — jets can blow but not suck, so a modulated injector runs about a mean — is assumed to raise the characteristic uniformly and is left out, though steady injection alone is known to move the stall line a little. Only the first circumferential harmonic is kept; a real cell has a shape, and higher harmonics can grow where the first is held. The compressor is incompressible and its blades respond at once, which is the low-speed limit; in a high-speed machine the blade rows’ own lag and the compressibility of the flow through them add phase of the kind the last figure prices. Underneath it all is the work a blade row does, which is what the characteristic’s peak is the most of and what the controllers are protecting.

Who blew air at the tips #

Epstein, Ffowcs Williams and Greitzer proposed active control of compressor instabilities in 1989, and Paduano and colleagues damped the first circumferential harmonic of rotating stall with an array of moving guide vanes at MIT in 1993. Day demonstrated stall suppression by air injection on a low-speed compressor at Cambridge in the same year. Haynes, Hendricks and Epstein carried the guide-vane control to a three-stage machine, and Weigl and colleagues took modulated tip jets, driven by the measured cell, to a transonic stage. The amplitude equation is Moore and Greitzer’s of 1986 with the forcing term added in the way Mansoux and colleagues’ first-harmonic models did. The static ceiling on the pressure gain follows from the surge mode’s determinant, which the throttle essay’s trace condition did not examine.

How it was checked #

With no injection, twice the amplitude times its rate of change equals Moore and Greitzer’s rate for $J$ to $10^{-17}$. The cell’s linear growth rate, read off the full right-hand side by a difference quotient, matches $(\sigma/2)(1 - u^2 - G\cos\beta)$ to $10^{-15}$ at three combinations of flow, gain and lag. The two-signal throttle’s design, checked linearly, is confirmed by integration at three operating points: the nudged machine returns in each. The pressure throttle’s floor and ceiling at B = 2 and ninety per cent of the peak flow are 7.29 and 2.76, and the growth rate computed from the eigenvalues is positive at every pressure gain between. The model refuses a negative gain, a B of zero, a negative flow and an injector of no capacity.

Still open: a cell the sensors see late #

Every controller here reads the cell perfectly. In a machine the cell is inferred from a ring of wall-pressure sensors, a handful of them, and when the cell is small — exactly when the injector must act — its first harmonic is buried in the blade-passing signal and in turbulence. The next calculation would read the cell through a ring of six or eight sensors with stated noise, estimate its phase and amplitude as a real controller must, and ask how the noise in the estimated phase spends the margin the last figure drew: whether an injector of a given capacity, fed by a noisy estimate, still survives a cell three times its size, or whether the line itself moves because a noisy phase averages to a weaker gain.