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Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...
Ash Small wrote:
These two equations then become exactly the same, which implies that the peripheral losses at the interface between the moving column of air and the surrounding, stationary, air completely dominate all of the losses from the system.
These equations describe the kinetic energy that is contained in a moving cylinder of air. For a propeller, the air has to be accelerated in order to generate thrust, so it is not a "lossy" mechanism, but the intent of the prop. Not all of the power put into the prop is converted into air energy, so one usually defines a props efficiency by the ratio of air power to prop power.
In the case of air drag on a car, the idea behind the equation is, that a column of air in front of the car is accelerated to the speed of e.g. the wind shield. So the power lost by the car is the same as the power needed to give the air column its speed. From this a drag force can be calculated.
All of this has nothing to do with vortices around the prop tips. The equations quoted describe an ideal behaviour, where such effects are not taken into account. They can accounted for by adding some fudge factor to the equations.
This it what one would presumably call a 'first order approximation'. I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.
I've drawn a quick graph. I assume the x axis (A) is m^2 and the y axis (v^3) is m/s.
I'm not mathematically minded, but I'll try and answer any further questions.
EDIT: Actually I'm still trying to work out what the graph means
(if in doubt, draw a picture!!)
(x axis is 0-10, y axis is 0-100)
I'll sort out the PPT thing later, thanks for the tips.
I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.
All of the props energy ends up as kinetic energy of air. Some part of it is useful, i.e. the downward air velocity, which results in an upward thrust. Other energies, e.g. the swirl around the blades and a helical component of the downward stream don't contribute to the upward thrust and are therefore regarded as losses. Being picky: the resistance of air being accelerated is due to its mass, not its viscosity.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...
Ash Small wrote:
I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.
All of the props energy ends up as kinetic energy of air. Some part of it is useful, i.e. the downward air velocity, which results in an upward thrust. Other energies, e.g. the swirl around the blades and a helical component of the downward stream don't contribute to the upward thrust and are therefore regarded as losses. Being picky: the resistance of air being accelerated is due to its mass, not its viscosity.
All of these losses pretty much follow the same law, ie the 1/2A x v^3 thing, which seems to be common through all of the propeller efficiency/drag stuff.
I think it is something to do with viscosity, especially as the drag forces increase. I seem to remember getting a similar graph when dropping different sized ball bearings into a tube of oil as a student.
There is certainly more than one way to analyse any problem.
All of these losses pretty much follow the same law, ie the 1/2A x v^3 thing, which seems to be common through all of the propeller efficiency/drag stuff.
I think that we have a terminology problem. What in terms of the v^3 law the drag is for e.g. a car, is the thrust for a propeller. For the case of the car I would call it loss, but not for the prop.
Registered Member #2529
Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600
It is somewhat relevant, depending on the Reynolds number.
The Reynolds number for a small drone is going to be a lot lower than a full size helicopter.
The lower Reynolds number means that viscosity is more significant than normal; model aircraft are designed slightly differently to full size aircraft for this reason.
Ash Small wrote ...
BigBad wrote ...
I'm not sure the disadvantage is quite as big as you think it is, otherwise twin rotor helicopters like the Chinook wouldn't exist.
Tip losses are treated fairly well here:
Personally, I think the Chinook is a compromise for a number of reasons.
Well, yeah, all aircraft are a bunch of compromises flying in close proximity, but I think you're completely overstressing the tip loss thing.
Ash Small wrote ...
EDIT: Can't open PPT files, I boycott Microsoft software as much as I'm able to. Open Office doesn't seem to open them.
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