EDIT: It's not a duplicate of Are ducted fans more efficient? That question and the answers doesn't address the reason for the higher theoretical efficiency, it is more about efficiency in practice (drag on the duct, weight, etc) and hence why they aren't used despite the higher theoretical efficiency

I was told shrouded propellers are more efficient becuase tip vorticies are eliminated by the wall which would imply no induced drag but apparently that is wrong Do ducted fans eliminate induced drag? therefore I've been trying to figure out why they are still more efficient than an open propeller despite still having induced drag. It must have less induced drag.

The vortex around an unshrouded propeller:

enter image description here

At first I thought the wall somehow increases the effective wingspan, moving the vorticies to the top of the wall much like winglets do and thus reducing the induced drag that way

enter image description here

However unlike winglets the walls don't have a pressure difference on either side (it's not an airfoil) therefore the vortex can't be there.

So the vortex must be around the whole wall because above the wall the pressure is low and below it the pressure is high.

enter image description here

But this doesn't change the effective wingspan so why does it have less induced drag?

My explanation is that the existence of the wall causes the inside of the vortex to "straighten out" with the flow through the duct which makes the downwash velocity constant along the length of the blades (since the flow is irrotational). This means that this condition is satisfied:

enter image description here

from page 7 of http://naca.central.cranfield.ac.uk/reports/1923/naca-report-121.pdf

Therefore induced drag is minimized. Is this correct?

  • 1
    $\begingroup$ @CrossRoads In this question I clearly say that I thought the explanation was what I wrote in that other question (lower induced drag due to increased effective wingspan) but now I think it's something else (lower induced drag due to more elliptical lift distribution) $\endgroup$ – Aviator S Oct 16 '18 at 18:10
  • $\begingroup$ @CrossRoads H O V E R $\endgroup$ – Aviator S Oct 16 '18 at 18:52
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    $\begingroup$ The approximate speed that a ducted fan is no longer efficient is about 100mph. So it depends on the specific flight conditions and aircraft configuration. $\endgroup$ – jwzumwalt Oct 16 '18 at 19:05
  • $\begingroup$ @jwzumwalt Not doubting your statement, but is there any information citing the 100mph efficiency limit? I would be interested in learning about the physics behind this. $\endgroup$ – YAHsaves Oct 16 '18 at 20:18
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    $\begingroup$ Interest peaked on ducted fans in the late 70's and early 80's. I got my information from the EAA "Sport Aviation" magazine (Aug 68, Jun 73, etc) which ran several authoritative articles that included engineers such as Molt Taylor. He considered it for the Aero Car and Mini-IMP but found it impractical. Some articles included wind tunnel testing and drag analysis. It shouldn't be to hard to find with a Google Search. Peter Kemp on this forum is one of the more prominent aerodynamist and would also be a good source - he loves math and probably has it somewhere in his library. $\endgroup$ – jwzumwalt Oct 16 '18 at 21:01

Basically, yes. The difference between shrouded and unshrouded propeller is that the shrouded one can produce uniform thrust across the diameter, while for the unshrouded one the thrust decreases near the tips.

That way a shrouded propeller accelerates more air than an unshrouded one of the same diameter. This air therefore needs to be accelerated to lower speed, and therefore carries away less kinetic energy, requiring less induced power¹.

However, diameter can be varied, so the efficiency comparison is not that straightforward. When the propeller spins relatively slowly, making it larger is better, similarly to how increasing wing span is better, aerodynamically, than adding winglets.

However increasing the speed of the tips increases parasite drag, especially if it becomes supersonic. And since increasing size while maintaining angular speed increases the orbital speed of the tips, increasing size only helps to a certain point. That's when shrouds become useful.

¹ In propellers and rotors it is called induced power rather than induced drag, because it counts directly against the engine power. It also describes the physics better, since in both cases it is the work that is done on the air by the reaction to the generated lift/thrust.

  • $\begingroup$ This air therefore needs to be accelerated to lower speed, and therefore carries away less kinetic energy, requiring less induced power Is engine torque smaller at ducted fan compare to unducted fan, if both fans has same RPM? $\endgroup$ – user52248 Sep 5 '20 at 16:03
  • $\begingroup$ @Сократ, please, be so kind and ask a question. Comments are not suitable for this kind of discussion. $\endgroup$ – Jan Hudec Sep 5 '20 at 20:54
  • $\begingroup$ My question is already there...aviation.stackexchange.com/questions/80839/… $\endgroup$ – user52248 Sep 5 '20 at 21:00

I suppose the confusion may be due to different sources using different definitions of induced drag. As used most commonly in my experience induced drag is the energy wasted as non-useful work in the creation of turbulence directly attributable to the generation of lift. Most noticeably in the wingtip vortex. In other words imparting flow in directions other than the ideal direction; in keeping with the photo-quote, an infinitely long completely uniform wing would have on lateral pressure gradient and thus no sidewards flow, nor the vortex formed from sideways flow.(It is not the vortex that causes drag, the vortex is just a symptom of the lateral pressure gradient.)

A close shroud, possibly even attached to propeller tips, of sufficient width will stop these vortex. However it may not stop all lateral flow because of the helical flow of the propwash. The helical flow can be reduced with static blades much like those used in axial flow compressors on gas-turbine engines. The shroud does increase add parasitic and form drag but it most definitely does reduce induced drag.

The first drawing in the Question with the un-shrouded prop, has the wrong perspective or axis so the vortex is in the wrong spacial plane; the second drawing has the vortex flowing through a solid wall and the prop again has the wrong axis. The third drawing is not a vortex it is the bulk flow encountered with a stationary fan in an enclosed room.(and the prop has the wrong rotational axis)

  • $\begingroup$ Induced drag has only one definition: the work done on the air to produce lift. Most of this work is imparting the flow the desired direction, and is therefore useful. You correctly note that the vortex is just a symptom, not cause, but it is mainly result of the lift generation itself, not the sideways flow. Induced drag is mainly proportional to lift to span ratio. An infinite wing only has no induced drag if you only let it produce finite total drag, which means zero lift per unit of span. If you let it produce non-zero lift per unit of span, it will have induced drag too. $\endgroup$ – Jan Hudec Feb 7 '19 at 20:03
  • $\begingroup$ Induced drag is a term used well beyond aircraft design, so I'm only using "lift" as an example familiar to the aviation field but other forms of dynamic fluid reaction could be applied. A wing with no sideways flow produces no vortex yet still produces lift. Wingspan ratio reduces induced drag by reducing the lateral pressure gradient and by extension induces less lateral/sideways flow. $\endgroup$ – Max Power Feb 8 '19 at 23:44
  • $\begingroup$ I suppose I could narrow the term to three possible definitions: 1, total drag created from the creation of useful lift plus the energy wasted on non-ideal flow as a result of lift; 2, just the portion of drag attributed to the energy wasted on non-ideal flow as a result of lift; 3, only the portion of drag attributable to lift produced by an ideal-flow. $\endgroup$ – Max Power Feb 8 '19 at 23:44
  • $\begingroup$ Everybody except you agrees that induced drag is what you have under option 1. $\endgroup$ – Jan Hudec Feb 9 '19 at 19:18

No, the vortices are trapped in the tip clearance.

What if there's no "outside" at all to your setup? Imagine a theoretical scenario where the entire space outside the duct is solid to $\infty$. Where are the vortices now? They could only be within the tip clearance.

I just realized this paper from another question is the perfect answer to this question.

And your assumption of shroud somehow making the downwash uniform is also wrong. Note that although the drawing in the statement of the problem or this paper is for a two-blade shrouded fan, even a fan with a very high solidity factor, e.g. 0.8~0.9, as used in high bypass-ratio turbofans, does not equalize fan wake, and that equalization happens only due to shear friction between the infinitesimal pockets of air themselves.

enter image description here enter image description here

Found this CFD of a turbofan's fan.

enter image description here

  • $\begingroup$ What if there is zero tip clearance, because the shroud is attached to the blades and rotates with them? And you can combine both cases and consider a pipe with rotating section with blades in it... $\endgroup$ – Jan Hudec Feb 7 '19 at 20:05
  • $\begingroup$ @JanHudec then the wing and shroud should be seen as a wing. $\endgroup$ – Meatball Princess Feb 8 '19 at 5:30
  • $\begingroup$ There is atmospheric pressure outside the wall and higher pressure inside the wall, of course there is going to be flow around the wall. If it were only within the tip clearance then induced drag would limit to zero as tip clearance limits to zero but it doesn't, it limits to the induced drag on an elliptical wing. As to your second claim that downwash velocity isn't constant, we are talking about a perfect irrotational fluid going through an actuator disk. $\endgroup$ – Aviator S Feb 13 '19 at 16:09

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