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I've read in a lot of places that stators in an axial compressor are responsible for diffusing air and removing the radial velocity to make the flow parallel to the shaft. What I don't understand is how this is accomplished. One way I've heard it explained is this:

With these fundamental points in mind, let us consider the energy transfer between the moving blades and the fluid. For this discussion, consider an axial-flow compressor with such a large ratio of hub diameter to tip diameter that the blades may be approximated by a linear cascade, that is, an infinite row of blades moving in a straight line, as at the top of figure 5.1. If we were to follow an element of fluid as it passed through the moving rotor, we would note changes in both its velocity and its pressure. The character of the changes is more easily seen by transforming to a coordinate system stationary in the rotor, where the flow seems steady, but from an angle as shown by the dashed vectors. If the blades are shaped to turn the flow toward the axis in this coordinate system (in the direction of blade motion), they form diverging passages which (for subsonic flow) result in rising pressure as the fluid passes through the cascade. When this pressure increase is carried back to the stationary coordinate system, as we see from equation 5.5, energy is added to the flow.

Source: Aircraft Engines and Gas Turbines (2nd Edition) by Jack L. Kerrebrock)

Can someone explain in more detail how a stator is like a diverging passage (diffuser?) as Kerrebrock proposes?

An explanation I would propose (but would appreciate confirmation for) is that there is a pressure gradient along the axial flow due to the rotor/stator's generating lift giving them a suction side upstream and a pressure side downstream. This pressure gradient would cause the flow to slow down (and hence increase pressure). Right?

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2 Answers 2

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If you take a look at the geometry of the path between two stator blades and compare the cross section of the path(relative to the direction of the traveling gas), you can see, that they form a divergent path.

Stator geometry

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  • $\begingroup$ Thanks for the answer and sorry for the late reply, but I think the logic here is flawed. The green line $A_1$ should be drawn parallel to $A_2$. Then we can make the assumption that all the flow that enters the "diverging nozzle" though $A_1$ and exits through $A_2$. I can't think of any obvious reason why there would be no flow through the dashed green line, which needs to be shown somehow to have a well-defined control volume to apply integral forms of the conservation laws to. $\endgroup$
    – eball
    Dec 19, 2019 at 0:41
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Air needs to move through the engine for it to work. Turbine engines usually spin above 20k RPM. If there were not structures to keep the air moving through the motor then it would essentially spin in place and the motor would stall. If you cut away the side of the engine the compressor rotors would be look similar to \ and the stators would look like /. The rotor blades and stator blades are at opposing angles so the spin of the air induced by the rotor would be straightened out by the stator blades before hitting the next set of compressor blades. The goal is to have straight flowing air through the combustion chamber and the reverse is done through the turbine with varying efficiencies depending on how the thrust is used.
If a drive shaft is coming out of the motor the turbine will scavenge as much power as it can to drive the shaft and the compressor leaving very little as thrust.

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  • $\begingroup$ Thanks for the answer and again sorry for the late reply. It makes sense that the stators will straighten out the air. It even makes sense that the stators will reduce the angular velocity of the air. What doesn't make sense to me is that they will reduce the axial velocity of the air. $\endgroup$
    – eball
    Dec 19, 2019 at 11:56
  • $\begingroup$ I was trying to understand where you are going with this but I just saw your other question about the ideal fluid. You have piqued my interest as well. $\endgroup$
    – Jeff A
    Dec 19, 2019 at 19:08

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