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Sep 20, 2017, 05:16 AM
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# Actual Theory of EDF's

As promised I have written out a brief description of how EDF's work. This description is only part of the whole story and I will endeavour at some later date add to it in such a way that it will benefit users. I have as yet, not worked out exactly all the consequences of a 'Real' world unit but so far I believe that the basic theory will remain both intact and applicable. Please read, I know many theories have been advanced on the pages of a good number of forums and for the reader this is just 'another' example. I do hope there will be some who read it through and digest.

ELECTRIC DUCTED FANS.
When I first looked at these little flight engines, like most, I thought they are somewhere between a propellor and an axial compressor stage, however I quickly realised they are something on their own. The nearest is a compressor blade that produces no increase in density, that is the compression should, ideally, be only a constant volume process. Having said that, like a propellor, they use the principle of forward procession - or pitch.
I will now lay down the configuration of this device, a duct that is parallel from entry to the end of the vanes, a 'concentrating nozzle'* and a fan with a 'solidity'** less than but close to one with a parallel platform continuing past the vanes. Finally this applies only to ducted fans that operate at less than 100m/s or 5000 watts/kilogram.

* ' Concentrating nozzle' : A nozzle that might appear to be converging but actually does not change the pressure of the flow or its velocity.
** 'Solidity' : where when viewed along the axial axis if the leading edge of one blade exactly matches the trailing edge of the blade preceeding it, it is a value of 1. Overlap would be greater than 1. A gap would be a value less than 1.

THE 'IDEAL' PROCESS.

I will start with an explanation of what I call the 'IDEAL' in other words if the unit was not constrained by the limits imposed by friction etc what would happen. We have to start by accepting two conditions that I will prove later are, in fact, a result of the power available and using a fan/duct combination.

THE VELOCITY, AT ANY GIVEN HEIGHT ON THE BLADE, THROUGH THE FAN DISC IS CONSTANT.

THE DENSITY OF THE AIR REMAINS CONSTANT.

I will refer to this velocity as 'Vf' and any subsequent reference to this will refer to the constant velocity through the fan or to IMPLY the power for 1kg mass I.E. (Vf^2)/2. Note Vf will be the axial value of the pitch.

The next stage is to set up four planes in the following order :

Pa1 a plane which sits sufficiently forward of the fan such that there is no effect from the fan.

Pf1 a plane which is positioned at the front of the fan at minimum presssure and a velocity of Vf.

Pf2 a plane which is positioned at the rear of the fan at maximum pressure and a velocity of Vf.

Pa2 a plane which sits sufficiently rear of the fan such that there is no effect from the fan.

Next start the fan and allow it to reach a stable condition.

The air starts to accelerate at some point between Pa1 and Pf1 moving towards the low pressure plane created by the mass of air moving from Pf1 to Pa2. It reaches Pf1 at a velocity that is equivilent to the pressure drop - Vf. The energy required for this acceleration is taken from the air. Consequently with no change in density it cools by the amount of energy required by this acceleration. THE TEMPERATURE AND PRESSURE ARE BELOW AMBIENT BY THE AMOUNT OF HEAT REQUIRED BY THE ACCELERATION. THE TOTAL ENERGY BALANCE HAS NOT CHANGED - ONLY CONVERTED FROM PRESSURE TO VELOCITY.

The air now enters the fan and the energy added although equivilent to the energy required for the acceleration is infact the extra energy required by a CONTINUOUS motion at Vf through the fan ( Note that at Pf1 the air has already reached velocity Vf). The result is that at Pf2 the velocity remains unchanged, the density remains unchanged but the temperature has risen by the energy added by the fan. This results in a pressure rise. There is a difference in pressure over the fan which we measure as thrust. THE TEMPERATURE AND PRESSURE ARE RETURNED TO AMBIENT. I initially struggled with how the energy was added by both the pressure side and the suction side of the blade, however I realised, after considerable thought, many sketches and calculations, that since the motion is considered constant both would add power more or less equally. So a single equation for power would reflect quite accurately the conversion of power.

Finally the air is returned to the atmosphere where its velocity energy is used to 'PULL' in the air entering the fan. This decays its velocity before plane Pa2.

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EDITED addition : The velocity is converted back to heat and returns to the ambient. I added this because someone pointed out that to a new reader this was missing. I felt it was implicit, but I believe the critism was correct.
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In other words the energy added to the exit stream is required to make the process 'CONTINUOUS' and cannot be used for anything else.
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EDITED addition : I wrote the above to elicit some form of re-action, it is based on some work I did in the late 90's and early 2000's and is almost certainly an incorrect interpretation. I didn't receive the replies I'd hoped for and consequently since I will leave this still available on my blog, feel that it is important to add what is the most likely explanation.
PLEASE NOTE that this will make no difference to the power or thrust calculations and is just a difference in how the power is converted. The general principle remains exactly the same. I will ammend the analogy to indicate the difference.
Here is an analogy which may help you to understand. Imagine a line of men stretching from plane Pa1 to Pf2 each holding the hand of both the man ahead and the man behind. Those between Pf2 and Pa2 are not holding each others hands.You are the man at Pa1, you are accelerated towards the fan by the man ahead pulling you, as you enter the fan you find that strong forces envelope you body pushing towards the exit of the fan, you pull the man behind with increased strength, reaching its maximum as you exit the fan. As the man ahead exits the fan he lets go of your hand. You pass out of the fan letting go of the man behind you. As you exit you are moving with a slightly higher velocity than when you entered. You move away from the fan gradually going slower and slower. Finally you find yourself at Pa2 suspended exactly the same as you were at Pa1.
In other words the energy added to the stream through the fan makes the process 'CONTINUOUS'.
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It is a beautiful example of a device obeying Newtons Laws, Impulse and Thermodynamics. However when you analyse it in a little more detail you realise it produces thrust with NO POWER. How can that be? - as it is its just moving air around, ALL the fan power is used up just moving the air. The Thrust? This is a reaction force, much the same as leaning up against a wall, the wall exerts a reaction force that stops you from falling over but cannot instigate any motion. (It doesn't push you the other way and make you fall away from the wall and if it did you'd really give it a telling off for contravening Newtons laws and impulse.) This reaction or 'thrust' is the necessary reaction to MAINTAIN the operation of the unit.

So far so good, (as a man jumping from a ten story building says as he passes the fourth floor.). We now need to examine what happens when the fan is moving forward at Vf. Take care to remember that it is the Fan Unit that is moving forward and NOT the air moving backwards. O.K. so now the air is stationary and the fan has a forward velocity of Vf. From the point of view of the fan nothing has changed, air is arriving at Pf1 at the same velocity as when it was stationary, the only difference is that the pressure and temperature of the air is ambient. Why ? Well, there doesn't need to be any reduction in pressure to accelerate the air - after all its just stationary, sitting around waiting for something to happen. The fan does its job and adds exactly the same amount of energy to the air as when it was stationary, it would appear that this would increase its temperature and consequently pressure to a level that raises them to above ambient by Vf. Hmmm in a sense it does but how does it convert this change into power for forward motion.

That dogged me for ages, I knew I was on the right track but how is this power converted into forward motion of the fan. Gradually I realised that in this 'IDEAL' state with no friction and therefore drag, (hard bit!) because of a change in direction of the resultant force applied by the motor to the fan, the power was converted almost instantly to forward motion. With no friction the blades are literaly sliding along a tube of air in a spiral motion exactly following the path of the pitch. Remember in this model the air is incompressible and acts like a fluid. Of course in the real thing we could never reach this stage, there must be some 'slip'.

Now imagine whats happening from the point of view of the fan. Create a tube of air Vf long but only as it is seen by the fan. Not the entry or exit. This tube is stationary and the blades of the fan are sliding along the spiral that is the pitch much like a propeller. Its very difficult to get your head round because we all instinctively imagine a reactive force but there isn't one because we've reached a limit where there is no force/thrust only power, all acceleration has ceased. It's much easier to comprehend from a point just before the limit, when we still have a little thrust and therefore reaction in the air, that is, where this tube still has some motion with relationship to the ambient reference frame and the fan has not yet reached Vf. Now continue this process until your tube of air is moving with a velocity of Vf and the fan has ceased motion. If you can do this then it becomes very obvious that from the fans perspective very little changes, the reaction either powers the air or the forward motion and has very little implications for the fan.

Before I continue, I will give a little background of how I arrived at this theory. The first thing I needed to know was how the power is added, I will start with a brief description of Euler's turbine equation. This tells us that there is a maximum amount of power that can be added or absorbed from the fluid. I will be working 'per kilogram' so we don't need mass in the equations.

At at a radius 'R' on the fan calculate the velocity in metres per second. We will call this 'U'. I don't have the symbol for pi so I'm afraid it will be pi!

U = 2 * pi * R * Revolutions per Second. The norm these days seems to 'D' or diameter so U = pi*D*Rps

Example. Hub diameter of a fan is 35mm ( 0.035m) and it rotates at 24,000 RPMinute or 400 RPSecond

U = 2 * 3.14159 * 0.0175 * 400 = 44m/s ( Rounded up )

Eulers turbine equation tells us that the MAXIMUM amount of power that can be added/absorbed from the fluid is -

Max power in watts = U^2 ( Remember U is in metres per second so watts). per kilogram

So now we have the maximum amount of power that our little fan can add to the air flow - but in what way? Although I am 99.9999% certain it will be in the form of an acceleration I.E. SQR ( 2 * U ^2 ) I thought I'd better check, I found a number of details concerning construction ,(on the web), and also looked at drawings and actual fans and although not all, a large number have an angle between the aerofoil root section and the axial axis of about 35 degrees and this confirmed my belief. How? Well there is a simple geometric relationship between pitch angle and U. Remember that we used Vf, well this is how I arrived at this value in an 'Ideal' world. The optimum angle is actually 35.264 degrees.

Vf = U / tan(Pitch Angle) Vf = U / 0.7071

I could just do this mathematically but for many an example is easier - so we will use our 44m/s

Vf= 44 / tan(35.264)

Vf= 44 / 0.7071 = 62.226m/s

if the fan is an acceleration machine then (Vf^2)/2 = U^2

( 62.226^2 ) / 2 = 44^2 or 1936 = 1936 so they are equal.

The next point to consider, is the constant velocity within the fan, a difficult idea for many to get their head around, but look at these points :

a). The available 'Area Effective' for a given fan will be about 90% of the swept area, there are tip losses, and blade thickness losses not quite enough for the 90%. However in the real engine we would, I believe, have a slightly reduced velocity through the fan, but with an increased area, some acceleration takes place in the form of 'swirl'* which is removed by the stator vanes resulting in an increased axial velocity and therefore appears as a reduction in effective area.

b). Constant cross sectional area from the front of the fan to the end of the vanes.

c). Empirically (by experiment) arrived at nozzles with an exit cross sectional area about 90% of the swept area.

d). Incompresible flow.

e). Temperature variations are so small that variation due to expansion would be less than 1-2%

These points lead to the fact that any other explanation is almost impossible, and the clincher is the 90% cross sectional area of the nozzles. Why? well simply, the entry effective area of the fan, is the almost the same as the nozzle exit area, the flow is incompressible and since the mass flow must be the same, the velocity cannot be much different. Even at the upper limit we could not expect the entry to the fan to be more than about 95% of the swept area which would indicate a maximum variation in velocity of less than 6%.

* Swirl : A rotational movement of the air, when viewed along the axial axis, and is due to the velocity through the fan being less than the actual physical displacement of the pitch and therefore a radial acceleration takes place.

Back to the 'IDEAL' cycle. Let us see what happens as the unit moves from stationary to a forward velocity at some fraction of Vf - which I will define as Vmotion or Vm for short. A proportion of the thrust and power is liberated from 'just' moving the air and is available for flight. The velocity is constant but its RELATIVE VELOCITY is now reduced by Vm. That is, when you examine the air stream flowing through the fan unit, its velocity relative to the exterior has reduced by Vm. From the fans point of view it only has to provide the power to accelerate the air from Vm to Vf. The available power is Vf, so we have liberated Vm for propulsion. So the applied vector swings a little so that we have a force that increases the pressure/temperature of the air to provide the power for a Continuous process at this reduced velocity and a force and power that is used for flight, both acting through the fan.

This takes the form: Flight power = (Vf^2-(Vf-Vm)^2) / 2

= (Vf^2-vf^2+2VfVm-Vm^2)/2

= (2VfVm-Vm^2)/2 watts per kilogram

(For those of you who understand its a little more complicated, the velocity Vf varies at different section heights and the maths for this is much more complex requiring integration, and even then exact velocities do not follow a fixed rule but depend on the specific unit and consequently the function that will represent the velocities at a given section height is individual, so for normal purposes if you know the 'mean' velocity the difference is minimal - sufficiently accurate for most applications.)

I'm not a great user of examples but I will now give an example of a unit that has a mean efflux velocity of 60m/s with a mass flow of 0.25Kg moving through the air at 25m/s

For where it is applicable, I apologise for a lack of a dot above the 'm' (mass flow).

Vf=60 Vm=25 m=0.25 Flight power (Fp) = ((2VfVm-Vm^2)/2 )*m = ((2*60*25)-25^2)/2)*.25 = ((3000-625)/2)*m =1187.5*.25 = 296.88 watts

Finally the available thrust. We can only obtain a reaction to the air exiting the fan so the thrust is the reaction to this air flow. The relative velocity of the air times the mass flow per second. When I first looked at this I thought it might be different in the respect that the mass flow would be the amount of air passing a plane relative to ambient refence plane and therefore would reduce mass flow in the form ((Vf-Vm)/Vf)*m but the mass flow relative to the fan is still at Vf and therefore this is relative thrust LEFT after the power for flight is extracted. Thus mass flow is still the same. There is a further, more mathematical check, and that is to calculate the 'thrust used' for the power equation and this bears out the following equation.

Thrust = (Vf-Vm)*m = ( 60 - 25 ) * 0.25 = 35 * 0.25 = 8.75N

The following chart is for a mass flow of 1kg, read the values at a particular velocity and multiply by the actual mass flow.

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See attachment. 60mspower.png

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NOTE : The intersection of the thrust and power curve has no bearing on flght speed, I could have altered the scale and thus moved the intersection point.

What I will now do is examine the implications of this theory when applied to the real world. I have not as yet worked out exactly the effect that friction etc has on this model but am fairly sure it only creates a shift, which will actually aid its use as a flight engine although causing an increase in power losses. That is to say that once the velocity is obtained with a 'concentrating only' nozzle. Then from cross sectional exit area or 'Area Effective' of the fan, calculate the mass flow and produce a power/thrust diagram as above. For these calculations we can ignore the actual shaft power required to produce this flow.

1). Take off, poor initial acceleration, although since these units are to a degree load sensitive there will be a reduction in air flow and and an increase in pressure produced by the fan, releasing extra power for flight. This will be noticable as a reduction in R.P.M. However this reduction will be dependant on how well the electric motor is matched to the fan, a well matched unit will only produce a small almost un-noticiable reduction. Aircraft have a higher drag during initial acceleration due to the increased angle of attack required by the lift surfaces which increases the power required. After reaching a flight speed where this drag starts reducing this effect will decline.

2). Level flight duration for a particular throttle setting will be similar to a static run at the same throttle setting. I say similar since the unit has some load sensitivity and the time can vary by a small amount both longer and shorter.

3). Probably the most important for the modeller is the limit velocity. Most EDF's are built to the 50/75 ratio, that is the hub is 50% of the fan overall diameter which results in the swept area being 75% of the full area of the fan. Additionally to reduce intake losses there is a radius entry to the duct further increasing the air collection diameter. This all helps to obtain maximum air flow in the static or low speed situations, however consider the point at which the incoming volume of air exceeds the amount that can flow through the fan. An example is much easier to follow.

An edf unit that has an O.D. of 70mm and a hub of 35mm with a 1.5mm radiused lip and a Vf of 44m/s

The volume flow through the fan at 100% of the swept area (infact it's less than 100%!) . Where 'R' is radius o.d. of the fan

1) R^2 * pi * Vf * 0.75 = .035^2 * 3.14159 * 44 * 0.75 = 0.127 m^3

The total Air entry area. R1 is R+lip = 0.0365

2) R1^2 * pi *Vf = 0.0365^2 * pi * 44 = 0.184m^3

Now divide 1 by 2 and multiply by Vf to get maximum airspeed where the air supply equals the air moved by the fan.
(0.127 / 0.184) * 44 = 30.27 m/s

That is not all, at this velocity we have no space for the acceleration of the air to 44m/s. I don't have the necessary data to specifically say at what velocity below this that drag starts to increase dramatically, but suggest its at least 10% lower than this velocity giving a maximum air speed of about 27m/s or 60% of Vf. Its a very difficult problem to resolve, since reduction in intake area will reduce the thrust and power output when at low speed and static, so making initial acceleration to flight speed even more precarius. The duct form for fans needs a considerable amount of thought and experiment and that is for the future.

Finally I could say much more about how this theory affects development of the blade forms, how the upper limit on flight speed can be improved and many many more things concerning motor selection etc. However there is sufficient here for a good start and quite honestly I could write a small book on all of the aspects of a complete system. I will sign off for now.

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Last edited by gryhrdoldfool; Sep 27, 2017 at 04:54 AM.