Jump to content

Recommended Posts

#1 A small rotor is operating at high rpm and or pitch with a high downwash velocity, #2 or a larger rotor is operating at lower rpm with lower downwash velocity.

post-57080-0-69697000-1499057082_thumb.jpg

So the answer from my book is the rotor with greater induced velocity will require the most horsepower.

since the larger rotor blade creates less downwash velocity does that mean it creates less induced velocity?

So in practical terms what's more efficient a UH 1 high aspect ratio or small narrow rotors with greater numbers like an MD500.

Note that this is not stuff required for the FAA exam, just additional knowledge. The FAA stuff doesn't get too nitty gritty on aerodynamics.

The FAA book doesn't cover this :D

  • Like 2

Share this post


Link to post
Share on other sites

Larger diameter is more efficient.

 

For pure aerodynamic efficiency the rotor would be very long, very narrow, and spin slowly, with the smallest number of blades possible. Not unlike the wings of a glider plane, for the same reasons.

 

Obviously with a helicopter there are other things that need to be taken into account, such as weight, structural strength, difference in airspeed from root to tip, practical size limitations, controllability etc, which is why we see so many different solutions for what seems like the same problem

  • Like 2

Share this post


Link to post
Share on other sites

The helicopter must follow the basic laws of physics, so a good place to start is with the physics. The rotor takes a stagnant mass of air from above the rotor and accelerated it to a given velocity below the rotor. in this sense it’s a force-mass-acceleration, that is one of Newton’s laws (i.e. Force = Mass x acceleration).

 

Therefore, we can work with either the former or latter part of that equation in order to obtain the force i.e. thrust/lift that we need. In the former, we can lower the disc load by increasing the rotor diameter, thereby working on more air (larger air mass) per second or the latter by working to increase the velocity through a given rotor diameter.

 

Rotor thrust = (Mass flow per second) x (total change in flow velocity)

 

Low Disk Loads:

Low induced velocities

Low autorotative rate of descent

Low power required to hover

Higher power loading

 

High Disk Load:

Compact rotor size

Low Empty weight

Low hub drag

 

It’s a series of trade-offs. Rotor systems optimized for high speed forward flight may be far less efficient in hover performance. Moreover, a given rotor with excellent hover performance may be very inefficient at high forward flight speeds. Hence, there is a balance between being efficient and effective. Most helicopters are designed with the scales tilted more toward effectiveness for a given task. Compared with other aircraft, like the airplane, helicopters in general are not as efficient.

 

The relationship between Disk Load (DL) and power required is given in the following equation. Also note the Figure of Merit (F.M.) and its importance in the power required. Figure of Merit (F.M.) – measure of efficiency of a hovering rotor. The geometry of the rotor basically determines the figure of merit. Things like, tip speed (function of rotor rpm and rotor radius), blade area, number of blades, taper, tip shape, twist, and airfoil section. In a perfect world, the F.M. would be 1.0; however, 0.75 - 0.80 is the general average range.

 

T = thrust or helicopter’s weight, since in any steady state thrust/lift must equal weight.

 

p = air density in slugs per cubic foot (@ sea-level p = 0.002377).

 

BcQZsFD.png

Edited by iChris
  • Like 2

Share this post


Link to post
Share on other sites

The helicopter must follow the basic laws of physics, so a good place to start is with the physics. The rotor takes a stagnant mass of air from above the rotor and accelerated it to a given velocity below the rotor. in this sense it’s a force-mass-acceleration, that is one of Newton’s laws (i.e. Force = Mass x acceleration).

 

Therefore, we can work with either the former or latter part of that equation in order to obtain the force i.e. thrust/lift that we need. In the former, we can lower the disc load by increasing the rotor diameter, thereby working on more air (larger air mass) per second or the latter by working to increase the velocity through the road for a given rotor diameter.

 

Rotor thrust = (Mass flow per second) x (total change in flow velocity)

 

Low Disk Loads:

Low induced velocities

Low autorotative rate of descent

Low power required to hover

Higher power loading

 

High Disk Load:

Compact rotor size

Low Empty weight

Low hub drag

 

It’s a series of trade-offs. Rotor systems optimized for high speed forward flight may be far less efficient in hover performance. Moreover, a given rotor with excellent hover performance may be very inefficient at high forward flight speeds. Hence, there is a balance between being efficient and effective. Most helicopters are designed with the scales tilted more toward effectiveness for a given task. Compared with other aircraft, like the airplane, helicopters in general are not as efficient.

 

The relationship between Disk Load (DL) and power required is given in the following equation. Also note the Figure of Merit (F.M.) and its importance in the power required. Figure of Merit (F.M.) – measure of efficiency of a hovering rotor. The geometry of the rotor basically determines the figure of merit. Things like, tip speed, blade area, number of blades, taper, tip shape, twist, and airfoil section. In a perfect world, the F.M. would be 1.0; however, 0.75 - 0.80 is the general average range.

 

T = thrust or helicopter’s weight, since in any steady state thrust/lift must equal weight.

 

p = air density in slugs per cubic foot (@ sea-level p = 0.002377).

 

BcQZsFD.png

Thank you for your effort, very helpful

  • Like 1

Share this post


Link to post
Share on other sites

The lower the disk loading (lb per sq.ft), the more efficient the design? I'm a novice to the subject. What struck me is that the Robinson R44 hovers around 2, for which it uses a massive 200 kg weighing 8.9 liter 6-cylinder ICE with the 'oomph' to go with that. The more complicated the design, for instance VTOL craft, the higher the DL, the lower the efficiency and the higher the downwash. You wouldn't want to be close to the landing of an Osprey.

 

623px-VTOL_DiscLoad-LiftEfficiency.svg.p

  • Like 2

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...

×
×
  • Create New...