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Ash Small

Sun Jun 29 2014, 03:46PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Uspring wrote ...

BigBad wrote:

If you use 17 inch props, you'll get 17^2/10^2 = 2.9 times the hover time than with a 10 inch prop assuming optimal pitch and motor stuff in each case.

Are you sure? I get a linear dependency of flight time on prop diameter.
The graph you quoted is interesting. I can't derive a relationship between disk loading and efficiency, which does not take into account the mass of the aircraft explicitly.

I also have some reservations here, but I do think we're certainly on the right track.

The relationship between disc loading and efficiency is based on the 'drag losses' at the edge of the accelerated column of air. It is independant of the aircraft mass, and is dependant solely on air velocity, which is PROPORTIONAL to disc loading.

EDIT: F/A=v^2

V=SqRt(F/A)

BigBad

Sun Jun 29 2014, 05:08PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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Posts: 600

Uspring wrote ...

BigBad wrote:

If you use 17 inch props, you'll get 17^2/10^2 = 2.9 times the hover time than with a 10 inch prop assuming optimal pitch and motor stuff in each case.

edit: yes, my bad, it's linear on rotor length.

Wikipedia gives:

P = T sqrt(T/2pA)

where T is the thrust

Patrick

Sun Jun 29 2014, 08:30PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

im thinking 14 to 17 inches might be the right diameter. i certainly cant keep going bigger...

Ash Small

Sun Jun 29 2014, 09:04PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick wrote ...

im thinking 14 to 17 inches might be the right diameter. i certainly cant keep going bigger...

I'm thinking twice this size for an 'order of magnitude' improvement.

We've also 'accounted' for the mass of motors and batteries. I'm assuming 'payload' will be a 'small percentage' of this.

This equation: V=SqRt(F/A) is a tramsposition of the original thrust equation provided by Udo, and illustrates the point quite clearly, I think. I've left out Rho, for simplicity. Losses are dependant on V. When V is small, losses tend to zero, when V is large, losses tend to infinity.

Patrick

Sun Jun 29 2014, 10:24PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
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for a 1.7kg machine with 14" props, i was planning on 500 grams at least of payload. (keep in mind i couldnt get fuel cells anywhere near the energy density to achieve this.) (total AUW = 2.2kg)

my PEM fuel cell maker wanted 18,000 US$ for one prototype, at 1.5kg or there abouts. thats with a sodium borohydride reactor attcahed. at this point i know i could build a lipo hex copter with 24" props to fly 30+ mins. for a lot less tha 18k$'s.

Dr. Slack

Mon Jun 30 2014, 08:04AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Now there's general agreement that bigger props are better, what happens in the limit of an infinite length blade? I'm a bit nervous when equations seem to say 'longer is better, without limit' that some other effect has been ignored.

Obviously there are practical details, like how big your car is for transporting it, or how strong is unobtanium from which to hew an infinite length blade, or whether that geometry will be stable in flutter.

There are more physics-based real effects though, as what is being sought is best lift to drag ratio. Blade power dissipation will be a function of many variables, obviously the most important for small blades are the speed and lift dependent terms like air-column energy and tip vortices. However, as the blade gets longer, and the disc loading goes down, the viscous drag will increase in relative importance.

So what I'm asking is, at what disc loading does the drag of just hauling the blade surfaces through viscous air become a significant term in the power loss, and so at what loading is there no point in reducing it further?

Ash Small

Mon Jun 30 2014, 08:38AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

When prop diameter approaches infinity, losses approach zero, which means that all power put into the system goes into accelerating the aircraft.

Obviously there will be drag losses from the blade, which can be minimized.

The reason the 'thrust equation' (F=A x v^2) doesn't account for these is because, in a 'first order' approximation, these losses are negligible compared to 'drag losses', in most 'real world' scenarios.

All I've done is taken the thrust equation and re-arranged it, but the 'thrust equation', as written, does imply that an infinite length prop will have zero losses.

We need 'different maths' to answer your question, Neil.

EDIT: Obviously, all losses from other sources should be minimized, for example, a long, thin blade with little pitch won't produce 'a lot' of drag, but will still require 'optimization'.

Can I just point out that we would have 'sufficient lift, flight time, etc' long before we reach an 'infinite length' prop. We're only looking for one order of magnitude improvement.

EDIT: Actually, a larger prop has to do less work, as there are less losses to overcome, so it should be possible to design it with 'less drag'.

This graph covers nearly 5 orders of magnitude. If it's possible to identify where we are on the graph, we should be able to deduce how practical a single order of magnitude increase in efficiency actually is

Dr. Slack

Mon Jun 30 2014, 01:45PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

What surprises me about that graph is that there are three orders of magnitude between the disc loading of the jump jet and a helicopter, but barely one order of magnitude for hover efficiency. That suggests you will need to work very hard with disc area to bring about much improvement in flight time. The disc area goes as blade length squared, so the flight time might increase as blade length ^2/3. It's a pity the graph wasn't plotted as log-log rather than semi-log, as with log-log it's very close to a straight line.

The no-lift rotor power, from which the minimum practical disc loading can be inferred, can be measured by spinning a prop/rotor of zero pitch in a dynamometer. There will be no lift related drag terms, only viscous air, friction and non-lift turbulence.

This is exactly analogous to the motor no-load condition, where a certain minimum power is required to spin it against the viscous air between the rotor and stator, and a few other miscellaneous losses. It's well known that there is no point using a big motor to provide a very small output, the efficiency will be horrible. It's better to use a motor with lower no-load losses, even if its high load efficiency is poorer.

Similarly, there will be little point using a large swept area to produce a small thrust.

Uspring

Mon Jun 30 2014, 02:02PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Dr. Slack wrote:

Now there's general agreement that bigger props are better, what happens in the limit of an infinite length blade? I'm a bit nervous when equations seem to say 'longer is better, without limit' that some other effect has been ignored.

The efficiency of conversion of motor power to air speed seems to be remarkably independent of the prop size. In this (german) paper

is a table for props efficiencies in terms of motor energy utilisation. They use the equation (Î¶ for power conversion efficiency)

F = (2 * rho * A * (Î¶*P)^2)^(1/3)

instead of the one I quoted

F = eff * (2 * rho * A * P^2)^(1/3)

They are really the same except for the definition of efficiency. For their definition here's a translation of the table:

Î¶ = 0.7 - 0.75 can be reached for man-carrying helis
Î¶ = 0.7 dream value for models
Î¶ = 0.65 can be reached by very good model props
Î¶ = 0.55 - 0.6 average to good model props
Î¶ = 0.5 model props shouldn't be worse than this
Î¶ < 0.5 sadly happens occasionally

Ash Small

Mon Jun 30 2014, 02:03PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

But where are we in terms of Lb/Hp and Lb/ft^2?

We need some figures from Patrick.

EDIT: Udo, as I understand it, the power conversion of the prop isn't what's relevant. What's relevant is the subsequent losses due to V (air velocity).

If V is lower, then subsequent losses are lower, so the prop doesn't have to do so much work in the first place.

The way I see the losses here is analagous to current flowing in a wire of a certain cross-section.

No matter how much you increase the voltage, the losses just keep increasing. The way to reduce losses is to use a larger cross-section of wire.

Same as water down a pipe.

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