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Old Apr 06, 2004, 03:56 PM
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Edf Faq

There have been several threads recently, which indicate that more general knowledge about EDFs would be welcome.
There also have been some answers, which indicate that this is also quite necessary.
Now, I certainly don't know all the answers, but I think that I have put together a fair amount of general available knowledge in this field which will be published shortly as a series of articles in Q&EFI.
If you find any mistakes tell me here, but I find it not necessary to shoot the pianist. After all these things are not my personal opinion but derived from accepted engineering praxis.

Lets make a start and if YOU have any questions in respect of EDFs, ask! I will do my best.

Meteor:
"I'm trying to get my head around the definition of "thrust" and "efficiency" with respect to fans that share the same swept area...
Assuming we have a Vasa 65 shroud and minifan480 and Kyosho rotors turned down to exactly the same diameter as the Vasa fan, how does the design of the rotor affect thrust produced by the unit? In the prop world, thrust is commonly defined strictly on diameter. How does it work with fans?"


Basically the efficiency of an EDF can be divided into two parts: “external” and “internal”. The external efficiencies concern the power provided by the battery and the power absorbed by the aeroplane itself in flight. The internal efficiencies are the ones which govern the EDFs performance in relation to the shaft power provided by the motor. The overall efficiency is the product of the part efficiencies.
When the plane is at rest the absorbed power is obviously zero but the electrical power probably at the most. So referring to the “external” efficiency in this case is quite useless!
For this case however it is very useful to know the thrust which the EDF can produce and the efficiency of the power conversion of the shaft power to the air flow.
The following equation is very useful in this respect and can be solved easily with a pocket calculator:
Static Thrust = 1.557 * De^2 * (eta * P / De^2)^2/3
The static thrust in this instance is calculated in N = Newton, if the power P is given in W = Watt and the exit diameter De in m (meter). The “internal” efficiency eta must be estimated in the first instance, and , being optimistic, I tend to use a figure of 0.8 for this, commonly expressed as 80%.

Using this basic equation and rearranging we can also calculate the internal efficiency:

Eta = N / (1.557 * De)^2/3 * P^2/3

Now the fun starts: if you measure the static thrust and the exit diameter and convert them into IS units and have a good guess at the shaft power of your motor (see published figures at various el. input powers) you can then calculate the internal efficiency of the EDF. Be prepared for surprises!
Now, this is pure basic physics – nothing fancy, no guess work (apart from the motor figures, but they are quite reliable these days) and no experimental and sometimes questionable data!
The factor 1.557 is derived from phi and the air density. If the air density is lower than at see level you have to multiply the factor by the density ratio for higher altitudes. This is approx. 0.82 at a height of 6000ft.

To make calculations easier I have attached an Excel spread sheet where you can input the variables.


"This is my first time playing with ducted fans. I thought that since we have no real gas combustion/expansions in a ducted fan, there's no need to reduce or expand the areas.
I've attached an image. Let's say the total annular area of a ducted fan is A1. Shouldn't the intake and exit area be equal to A1?"

This may be just ok for a plane at rest. But consider the plane flying with a certain speed through the air: In the worst case the velocity of the air entering your constant area tube is the same as the flying speed. This is also the same axial velocity through the fan. The fan can not do anything else but increase the total pressure which is converted into velocity pressure at the exit nozzle – the increase of velocity providing the thrust (remember T = delta v * M)!
If you don’t decrease the exit area to accommodate the higher exit velocity two things can happen: 1. the airflow follows the tube and contracts after it has left the exit nozzle resulting in a loss of thrust (as in a propeller) or 2. the airflow detaches from the tube wall and leaves the tube as a contracted free stream with a lot of friction and even more losses as in 1.
Ergo: it is quite important to reduce the exit nozzle area to accommodate the increased velocity produced by the fan (rotor and stator). The ratio of fan swept area (called fan blade annulus in the text books) to the exit nozzle area is in the region of 1.2 to 1.5 depending on the fan design and should be provided by the manufacturer. It can however also vary with the length of the jet pipe, if this is longer than 2 * De.

I have attached a picture which shows the pressure development in an EDF with real figures

Have fun with fans

Klaus
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Old Apr 06, 2004, 04:04 PM
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And here is the promised Excel Spresd sheet
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Old Apr 07, 2004, 04:58 PM
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no.2

Hi all

here is some more, but don't forget, if you don't put any more questions I have to ask myself.

Meteor
"Now, the actual aerodynamic efficiency of that rotor to turn those watts into thrust becomes the factor with distinguishes the "good" from the "bad". But is there a -really- significant difference?
It always puzzles me when people throw out statements like: "The minifan 480 is 20% more efficient than the Kyosho fan...", but what is the measuring criteria, and are all other variables locked in that comparison?"

This is where the “internal" efficiency of an EDF comes into play.

To explain this we first have to define a few parameter which are important:
There is first the ratio Vax/U. Vax is the axial air velocity through the fan, which is, in the first instance, constant over the radius and also constant through the fan and stator unless we change the hub or shroud diameter. U is the circumferential blade speed, dependent on the motor’s rpm. U changes with the radius of the fan blade. It is least at the hub and greatest at the tip for the same revs. The ratio Vax/U can be anything from let’s say 0.2 to 1, the ratio is also called flow coefficient. Since U changes over the radius the flow coefficient also changes over the radius at constant rpm and constant Vax. For a first approximation a medium value for the flow coeff. is used, based on the mean radius of the blade. Low powered fans (GWS’s for example) have a low mean value of flow coefficient of perhaps 0.3, high powered fans can have up to 0.6 or even 0.7 depending on the blade loading.
Only in connection with a particular design can we deduct the fan efficiency from the flow coefficient but it is also evident that very high values, particularly when encountered at the hub, where they can be larger than 1, equate to excessive blade loads which may lead to local stalled air flow called blockage. On the other hand a too low value of the flow coefficient can result in high aerodynamic losses due to high blade friction and very low or even negative angles of attack of the blade, which then operates at low L/D ratios.
Compared to propellers low flow coefficients are similar to low pitch and high flow coeff. to high pitch.

Depending also on the ratio Vax/U is the ratio Wm/U. Wm is the mean velocity of the airflow in respect of the fan blade. This again varies with the blade radius under consideration. If we take the mean radius as above with the correct suffixes it becomes Wmm. The angle of this velocity in respect to the peripheral direction may be called beta.
From geometric considerations in line with airfoil calculations one can deduct an efficiency:
eta(blade) = 1 – Wm/U * tg eta(section) / sin (beta + eta(section))
Eta(blade) would be the total for that mean radius and needs to be repeated for other stations of the blade and then averaged. Eta(section) is the glide angle of the blade section used at this station. Typical values for eta(section) is 1deg, or tg eta = 0.0175 and beta = 37deg. Wm/U can be (as in one of my sample calculations) 1.04, then the above becomes:
Eta(blade) = 1 – 1.04* 0.0175/0.62 = 1 – 0.0296 = 0.97 or 97%!
But please, that is only at this particular station, over the blade length it calculates to a total of approx. 0.94. To this we have to add the tip losses and the friction losses and the diffusion losses and the losses of the stator. All together the efficiency of the complete fan stage will be in the region of 85% or even 90% if we are lucky.
However to show the point lets consider a different example where we have increased U but decreased Vax. Then Wm/U may become 1.0, beta = 12deg and eta(section) 4deg. With these the blade efficiency is down to 0.75 or 75%! And the stage efficiency may be as low as 0.5 = 50%.
It's the designers job to get it right.

Have fun with fans

Klaus
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Old Apr 07, 2004, 05:17 PM
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Probably too general but......

Winmodels, this is probably much too general, but what I would like to see would be a series of articles something on the order of: "I have a CD ROM conversion brushless motor and I would like to construct my own version of a Ducted Fan unit for use in my Fan-Tastic Mig 15 and this is how I designed it."
Best,
Ed
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Old Apr 07, 2004, 07:28 PM
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Ed

at the moment I have no CD ROM conversion BL motor available, but I'm building a 80mm EDF for standard 600 size motor. After Easter I will be able to post first pictures of parts.
I only started this to answer those more general questions mentioned.
Now if you would tell me that you have this CD ROM motor and want to design an EDF around it to fit into XY model , then I will gladly answer the questions which this design will pose.

Have fun with fans

Klaus
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Old Apr 07, 2004, 07:52 PM
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Hi Klaus, Tom in Milwaukee WI here. I have a question about converting an OS 91 type of fan into electric. Can or has this been done?
I see a lot of older Ducted fan Jets (Bryon F-16's and so on, being sold on Ebay, EXAMPLE: http://cgi.ebay.com/ws/eBayISAPI.dll...EBWA%3AIT&rd=1)

and I am wondering how difficult it would be to convert this to Electric power.

Also, I am considering building a multi-motor (4 micro-fans) foam type airliner and I'd like to know the best power and wiring methods, i.e. number of battery packs, esc's, series or parallel hook-up's, etc.

Thanks for a great Topic, now I wish I had gone to engineering school so I could understand it all.
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Old Apr 08, 2004, 12:09 AM
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A .91 fan engine makes about 5 HP. This is about 3730W. If you find a power system that can do this it can be done. It has been done.

Greg
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Old Apr 08, 2004, 12:41 AM
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5 HP?? Dang I'm going out to get one for my pressure washer!! The current 5 hp honda motor on there is about 50 lbs!

Justin
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Old Apr 08, 2004, 09:36 AM
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A .91 ducted fan engine makes 5HP by turing 26,000RPM with little torque output. The honda turn 3000RPM with many times the torque.

Greg
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Old Apr 08, 2004, 09:37 AM
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A .91 ducted fan engine makes 5HP by turing 26,000RPM with little torque output. The Honda turns 3000RPM with many times the torque.

Greg
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Old Apr 08, 2004, 11:02 AM
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I should have stated in my question that I want to consider converting the Byrojet Fan over to electric power
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Old Apr 08, 2004, 11:10 AM
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http://www.rcgroups.com/forums/showt...ght=750+gordon

Greg
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Old Apr 08, 2004, 06:33 PM
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[QUOTE]Originally posted by RCParkflyer
[B]Hi Klaus, Tom in Milwaukee WI here. I have a question about converting an OS 91 type of fan into electric. Can or has this been done?
I see a lot of older Ducted fan Jets (Bryon F-16's and so on, being sold on Ebay, EXAMPLE: http://cgi.ebay.com/ws/eBayISAPI.dll...EBWA%3AIT&rd=1)[unquote]
Hi Tom

I think Greg has answered that question sufficiently. One could perhaps add that it is possible - but does this sort of conversion make sense? From an engineering point of view times have changed, and by todays standard the ICDFs built in the 1980s are rather outdated i.e. inefficient. They are also too heavy for the conversion to electric - same applies to the planes themselves.
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Old Apr 08, 2004, 06:38 PM
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Quote:
Originally posted by RCParkflyer

Also, I am considering building a multi-motor (4 micro-fans) foam type airliner and I'd like to know the best power and wiring methods, i.e. number of battery packs, esc's, series or parallel hook-up's, etc.

Hi

depends what motors you are going to use, brushed or brushless

Have fun with fans

Klaus
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Old Apr 08, 2004, 08:26 PM
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Hi Klaus,

Thanks for Your's and Greg's answers, I've read about Gordon's twin HW750 project before, I guess what I was looking for is some type of electric replacement for the OS .91 Engine, still using the Byrojet Fan.

I think that Gordon's, and work by StuMax and others will usher in a new style of EDF systems that will help us close the gap with our Fuel (DF and Turbine) flying friends.

As for my other question the 4 motor combination, this will be small brushless motors, possibly the CD-ROM type Motor's, or simular powered 300-400 size brushless with Mini-fans, (I'm attempting to stay within a Budget ) I'm mainly inerested in the wiring possibilities, and not specific of pack size, etc. just a rough idea of what I am going to be looking to spend.
(hope that helps)

Thanks
Tom - Milwaukee
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