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Two identical fans, one is unducted and other is ducted,both operate at same RPM.

•Does ducted fan require more engine torque than unducted-fan ,measured both at same fan RPM?

power=torque x RPM so in my case more torque mean same as more power("induced power")...

(Engine torque is direct indicator which fan has higher drag at blades.We can make analogy with wall to wall airfoil(represent ducted fan) vs 3D wing (no side walls at wingtips )represent unducted fan.

Here is problem:

I think ducted fan requier more torque for same RPM(maybe I am wrong?),which means it has higher drag than unducted fan,but in theory wall to wall airfoil must has lower drag than 3D wing at same speed and same AoA.

•How explain this contradiction?

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Two identical fans […] I think ducted fan require more torque for same RPM

Yes, a little more. The shroud prevents the air flowing directly around the tips, allowing them to produce more thrust. And this extra thrust comes with induced power.

but in theory wall to wall airfoil must has lower drag than 3D wing at same speed and same AoA.

No, it doesn't. The comparison is always made for the same lift, never the same AoA.

For the same AoA the drag should be higher for the same reason as above.

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  • $\begingroup$ No, it doesn't. The comparison is always made for the same lift, never the same AoA. For the same AoA the drag should be higher for the same reason as above. I agree with your conlusion.But then this answers from @JZYL and Auberron are wrong.It will be good to fix this topic.aviation.stackexchange.com/questions/77573/… $\endgroup$ – user52248 Sep 5 '20 at 21:34
  • $\begingroup$ How explain increase in torque with type of drags at blades?Higher torque at ducted fan is because higher airspeed in ducted tunnel cause higher skin friction at blades or the main reason is increased induced drag at blades in ducted fan? $\endgroup$ – user52248 Sep 5 '20 at 23:31
  • $\begingroup$ @Сократ, the effect we were talking about is a pressure drag. The pressure difference near the tips is larger and the tips still have non-zero angle of attack, so the counter-rotation component of the pressure force is larger. $\endgroup$ – Jan Hudec Sep 6 '20 at 11:49
  • $\begingroup$ … the velocity is still the same, because that's how we defined the comparison. If it was an equilibrium state without the shroud, it will not be with it. The flow speed will increase, which will reduce the angle of attack (assuming fixed pitch fan), which will reduce the torque. The same engine power will probably result in higher flow speed, slightly higher RPM and slightly lower torque. $\endgroup$ – Jan Hudec Sep 6 '20 at 11:53
  • $\begingroup$ the effect we were talking about is a pressure drag. Induced drag is pressure drag.If we striktly follow drag definition total drag=zero lift drag(skin friction+pressure drag) + induced drag,then pressure drag is measured only at zero lift angle and induced drag is measured whenever wing produce lift,so when fan rotating then he produce thrust/lift so this part of increasing pressure at tips "belongs" to induced drag. Do you agree? $\endgroup$ – user52248 Sep 6 '20 at 13:22

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