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Consider this system, rotor with shroud: enter image description here

From this research, we see that: at the same rpm, the rotor thrust decreases,the shroud thrust increases as tip clearance decreases. But why does the rotor thrust decrease at smaller tip clearance? You can see author said at page 11:

"Note that the increased inflow in the plane of the rotor at smaller tip clearances causes a small reduction in the effective angle of attack seen by the rotor blade, which results in a small decrease in the rotor thrust."

I understand that the inflow increases that will causes a small reduction in the effective angle of attack seen by the rotor blade. But final question: why do we have increased inflow at smaller tip clearance?

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  • $\begingroup$ "final question: why do we have increased inflow": When I read the excerpt and see your diagram, I understand that if you have a longer blade (resulting from a reduced clearance with constant shroud diameter), the rotor disk diameter is increased, hence the air flow mass (assuming the angular speed is kept constant). Am I wrong in my reading? $\endgroup$ – mins Jan 16 at 9:29
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    $\begingroup$ @mins actually, the blade diameter is kept consant and the shroud diameter changes (the author used CFD and easily changed shroud diameter ) $\endgroup$ – Dat Jan 16 at 10:21
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This is the reason why a jet engine needs a shroud around its core: to reduce blade tip loss. When you decrease the wingtip clearance, you are curbing the wingtip vortices or the wingtip pressure bleed, and in doing so you are making each blade more and more effective at approximating its ideal i.e. infinite span high-low pressure spatial distribution under and above it, and this more ideal pressure difference accelerates the airflow better, which results in more inflow, which reduces the angle of attack of each blade a little, but increases the overall thrust produced. Think of the shroud as a big end plate for each and every blade. Think of this setup as the fan in a high bypass ratio turbofan: less fan tip clearance = more thrust.

EDIT: I'm not sure the author understood it as I do in his paper. He never clearly explained it in his writing. Plus, just why is there so much lift distributed at the wingtip? I've never seen anyone design a propeller with so much bending moment. He is making a propeller much heavier than it really is because the propeller could be less rigid and lighter and still providing the same thrust. If you really want an efficient prop, put the majority of lift near the midsection, and let the lift taper when approaching the tip.

EDIT: See how shroud works on a bona fide jet engine compressor: compressor shroud

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  • $\begingroup$ Could you please explain these two points in your answer? 1) "which reduces the angle of attack of each blade a little" 2) "If you really want an efficient prop, put the majority of lift near the midsection, and let the lift taper when approaching the tip." $\endgroup$ – Hans Feb 9 at 1:52
  • $\begingroup$ @Hans 1) the angle of attack of each blade is determined by the linear velocity of the blade and incoming speed of air when the latter increases the angle of attack decreases 2) Google the lifting line theory. $\endgroup$ – Meatball Princess Feb 9 at 10:52
  • $\begingroup$ 1) If I understand you correctly, you are saying that in order to get the same amount of inflow at higher incoming air speed the blade is designed to have its angle of attack reduced, rather than that for a fixed design, when the air speed increases the blade angle of attack adapts and increases. Is it right? $\endgroup$ – Hans Feb 10 at 22:56
  • $\begingroup$ @Hans no, no blade is designed for any fixed AoA. Blades are rotating wings, and wings naturally tolerate a certain range of AoA before separation and stalling. How can the AoA not change when inflow increases if RPM is held constant? Unless the air is compressed, and that is unlikely, as air is decompressed a bit before the prop. $\endgroup$ – Meatball Princess Feb 11 at 6:29
  • $\begingroup$ You are saying the following: The inflow increases the downward air speed through the blade while the rotor speed is held constant, the net direction of the air relative to the blade is turned more downwards compared to that at lower downward airflow speed, therefore the angle of attack of the blade relative to the net air flow direction decreases. Is it right? $\endgroup$ – Hans Feb 11 at 7:45

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