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Vienna, Austria
Joined Apr 2007
1,316 Posts
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Quote:
- thrust rises with power, but only proportional to the square root (4x the power, 2x the thrust) because of the change in kinetic energy proportional to the velocity squared. - the load of a prop, and (probably also a fan) rises approximately with the square of the RPM, we all know that going from 3 to 4 cells isn't just having a load one third bigger than before (if it was, current wouldn't rise then because the higher voltage already makes for one third more power at equal amps. Amps do rise a lot more, but you don't get to see a quadratic effect in the power because the motor will not be able to provide the rpm proportional to voltage at higher loads etc... so, regarding RPM, the two effects (load = power rising with the square of the RPM, and thrust rising with the square root of the power) cancel out - approximately of course still, empirical data would be interesting! |
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Rutland, England
Joined Feb 2005
160 Posts
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It's also important to remember when considering this, that any reference to power is totally irrelevant to this particular argument, and is likely to confuse anyone who is thinking about this. We are talking here about one performance parameter (Thrust) of a simple stand-alone fan; the faster you turn it the more thrust you get out of it. The only argument is: is the relationship between Thrust and RPM linear? or does it take an exponential form? It really is as simple as that ... and if i'm proven wrong, it's now going to cost me shedloads of money ![]() (If) later we did go on to consider the amount of power needed to produce that Thrust, then things do get a bit more complicated, because the power used in a real system will be made up of two distinct elements: 1. The raw shaft power required to rotate the fan, but additionally 2. The power wasted as heat in the system driving it. Afraid i'd also have to disagree with the red bit above. From the equations i've derived for these fans, the relationship between Thrust and Power required is not so bad/exponential as Mopetista is *suggesting (ie 4 times the power required for 2 times the thrust) - even when you take into account all the energy lost as heat in the motor etc. I'll share these equations (and hopefully better still - real characteristic graphs) with everyone when i've done some more work evaluating them with some real hardware (it's really frustrating not having any kit yet!) *Out of interest, in full size gas-turbines there ARE certain times when to get twice the Thrust you have to apply 4 times the rate of energy/fuel to achieve it. This is when 'Re-heat' is used (when it gets really noisy and there's a red flamy bit showing from the back of the engine). The reason for this is that the mass flow of air (mDot in our Thrust = mDot x dU equation, from my previous post) doesn't actually increase during Reheat conditions - this being the case, the only way that thrust can be increased is to massively increase the (dU term) - ie increase the exit velocity of the air by a factor of 2. This in turn increases the Kinetic Energy of the exhaust air by a factor of 4. It's a very in-efficient/expensive process ![]() Still... since we don't have re-heat (yet) in our models, i don't think we need to worry about mathematics for that sorta stuff at the moment ![]() Now forget equations, and get on with Chrismas; have a good one. Rgs, Santa |
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Vienna, Austria
Joined Apr 2007
1,316 Posts
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power = change of energy / time; acceleration = change of velocity /time
sorry Andy, you should be more careful, the part you highlighted is VERY basic physics (and therefore not worth any beer :-)
- power is work done (equivalent to the change of the system's energy) per unit of time - acceleration is a change in velocity per unit of time now to keep things simple we talk about a duration of one second, and then you can forget it in the following calculations and concentrate on the differences in energy and the differences in velocity Your power will linearly change the energy of the air. I.e., twice the power, twice the enegry change. However, the velocity of the air will only rise with the square root of the (kinetic) energy, so you put in 4 x the power you will get 4 times the energy change (per unit of time) but only 2 x the change in velocity - and a change in velocity is just acceleration, and thrust = force = mass x acceleration Now that's hardly rocket science but a bit of juggling the units. the empirical part rather lies in the square relationship of power vs. RMP at a prop, and possibly an EDF rotor: I remember having read that the power needed to turn a prop rises with the square of the RPM. I don't know how accurate that "law" is. Merry X-mas to all power-hungry EDFs out there |
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Rutland, England
Joined Feb 2005
160 Posts
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Quote:
, and wait for someone to produce some simple test results. I'm quite imovable in my claim that for an axial flow compressor (which is what these little fans effectively are), the thrust you get out can not possibly be directly proportional to the rotational speed at which you turn it.Let's just call it quits for now, till some kind soul does a test for us ![]() Kind regards, Andy |
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Vienna, Austria
Joined Apr 2007
1,316 Posts
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why twice the RPM needs more than twice the power
One last effort on my side to explain why it gets MUCH harder to turn a fan faster - one might think the load would be proportional to RPM, at best.
There are actually two effects: Firstly, by turning faster, its blades hit more air. Strictly proportional effect, the area swept by a fan blade will rise with the rotational speed. That's not all, however. The second effect is less obvious at first but just as real: The fan blades hit the air with more speed, thereby transferring more momentum. There you have the other proportionality to RPM. There you go, load rising with the sqare of the RPM. Of course real life is messy, and it won't be precisely quadratic, plus that is the power turning the fan, and not the power put in the motor. Since the efficiency rather drops than rises at the power levels we run in EDFs (efficiency curves look like the flight paths of projectiles) when amps are increased, things look even worse regarding thrust... |
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Australia, VIC, Melbourne
Joined Nov 2006
6,493 Posts
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Huh? The area swept by a fan never changes.....
It just covers the area faster, as per what your second part of the 'fransferring the momentum' bit is. Covering the area faster transfers more energy into the air... that is all that changes. I don't believe it is a squared law, but it is not a linear law either. I think it is a form of exponential, and I would expect there are two reasons for that: 1) The blades are most effificent at a certain speed, and as you go higher above that they lose efficiency, thus you need more power than linear to gain more thrust.at a linear rate 2) Because the fan is inside a "pipe" there is a point where you go from just pushing air through it easily (eg lower RPM) - like water just running through a pipe and not filling the pipes cross-sectional area - and then out to a point where you are 'forcing' it through, under pressure then, and from the point onwards where this pressure becomes a factor, things get non-linear too. A fan blade setup actually has X % of 'open area' - no blade covering that area. This means that you will not cross the 'pipe too full' point, where pressures then builds, for a fair while. For eg, if the blades cover 50% (even less for 5 blade fans) it has X amount of RPM increase before it hits that "walls" of pressure factor. But the CS10/12, that have more like 70% blade area coverage, hit that much sooner. Total pitch/volume pumping maths comes into that, but I am pretty sure the CS10/12 have higher numbers for that then 5 blade or 6 blade (6 blade generally having more than 5 blade). You don't need to know the maths, or even have the numbers, because acoross those 5,6,10,12 blade fans you can see that direct relationship just from results. Lower blade count give more thrust, for less watts. This is just a 'fluke' of how it is. That the 5 and 6 blades seem to use the same blade/pitch approx - thus a 5 blade has higher RPM before hitting that pressure wall than the 6 blade. The CS10/12 hit that wall sooner, but those two end up being extremely close to the same total result of their blade areas and pitches, thus their results come out pretty much dead equal. These are just made up numbers to demonstrate: - Where they hit the pressure wall 5-blade 40k RPM 6-blade 37k RPM 10/12 blade 30k RPM (they are probably fractionally different really, but unseen in testing so far) Now all of those above, are typically run at RPM ABOVE that pressure wall anyway. X amount above for each type... and likely that they are all being used ONLY to an area that is Y amount above the pressure wall, in total energy wastage, because people see it is futile driving them harder, and/.or by then the power levels required are unsustainable anyway (or that the fans explode!) I made the 10/12 blade a LOT lower RPM... to demonstrate the other spin-off they have in how they HATE exhaust restriction. Which is part because of their greater "blade to air gap" ratio (they have more total blade area than 5/6 blades), and that this causes there 'pressure wall' point to occur much sooner in RPM. The sooner their 'pressuried airflow' can get out of the plane (fan housing/ducting) the sooner energy wastage is halted. (friction per distance of 'pipe') |
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Australia, VIC, Melbourne
Joined Nov 2006
6,493 Posts
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But I do ponder over the factors of airflow speed, versus airflow pressure, and where those came from (blade shape, RPM, exhaust ducting back pressure etc) so that makes for some 'overlapping'/grey possibility areas. And we can't measure those ourselves.
So I can see there can be some other factors, and interactions, that cloud exactly what is going on. |
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Lots of theory guys, but here is more testing....
Since I couldnt find the cheaper Tacon 2860 2550kv motor anywhere, I decided to go with the more available, but more expensive version, the red Leopard 2860 2550kv, which is the exact same motor. I just got through with some bench running and overall I am happy with the results. Asthetically the motor looks really cool, and came in some spiffy packaging. Lots of boat guys run these motors, but I havent seen many EDF'ers, so I thought I would give it a shot. Back from the bench run and the results look good, here is my setup and the numbers: Leopard 2860 4D 2550kv motor Changesun 70 rotor, dynamically balanced by uncle tam Eflite Delta V-15 70mm housing Gens Ace 5s 3300mah batt On the bench this setup pulled a peak of 1269 watts @ 69 amps, and settled to around 1200 watts @ 61 amps. The motor was nice and cool after a 4 min run and the fan sounds really smooth. I still had about 50% left in the pack after a 4 min bench run with lots of WOT burts. I plan on installing this in my Habu 2 over the next week or so. Hopefully the numbers will improve when its installed in the jet. Thoughts? |
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They do boats, but they carry the version of the motor I got with the cooling holes. I thought it was going to be the sealed version, but even from the description, they clearly state it's the version with the cooling holes, which is perfect. This is the only place I found the motor not in china so I recieved iit in just a few days. |
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Australia, WA, Perth
Joined May 2007
3,129 Posts
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Well, it's not this particular fan, but given they all work basically the same:
![]() The math should describe, not prescribe. If the math and experiment disagree, the math is wrong. These look like curves to me - nothing liner, especially then you look at where the X axis begins. |
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sorry for more sidetracking. I didnt beleive it myself (that Thrust ~ rpm^2), so after a little searching on the web the MIT page is helpful in showing that Thrust is (amongst other things) proportional to rpm^2
http://web.mit.edu/16.unified/www/SP...es/node86.html if you look further down the Power coefficient equation also confirms Andy's statement that Power is proportional to rpm^3 If you combine the two equations (substituting out rpm) Thrust is propotional to Power^2/3. |
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