Author Topic: axial flux stall  (Read 1243 times)

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fanman

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axial flux stall
« on: January 10, 2006, 03:34:40 AM »
hi everyone

im going to build a axial flux machine for charging 120 volt battery, i would like to know why the machines that are made on this site sometimes stall, could you please define that term, also if the machine stalls couldnt you use bigger blades to overcome the internal resistance, make sense to me,

my axial flux machine will be about 24" dia. 24 magnets, 18 poles, my magnets were going to be 2x2x1 but now i think i will a bit smaller one, i want it to produce enough voltage to charge 120 volt battery, and one comment i got was about the stall problem with this machine because of the internal resistence, i would like to wind this accordingly to not produce stall, is that possible?


thanks  fanman

« Last Edit: January 10, 2006, 03:34:40 AM by (unknown) »

Flux

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Re: axial flux stall
« Reply #1 on: January 10, 2006, 02:10:05 AM »
I thought I have flogged this to death in the last few weeks but perhaps you have come here too late to have understood the basic problem.


Stall is not peculiar to axial machines, it just happens that most people here build that type.


I don't want to go too deeply into the theory of aerofoils but for satisfactory operation the angle of attack for reasonable lift must not exceed a certain angle.

The best lift/drag ratio for the common aerofoils we use here is near 4deg and we try to run near there at cut in. Things work fairly well up to about 12 deg but much beyond that the air separates from the blade surface and becomes turbulent and lift virtually collapses and this is the stall that we are bothered about.


Now that is out of the way let's see the implications for a windmill.


To run the blades at the optimum angle of attack we need to keep the vector between blade and wind speed constant. The blade speed must follow wind speed. We normally consider this for convenience at the blade tip and it means that the blade tip speed must track wind speed. Everywhere here you will meet tsr ( tip speed ratio) and this is just simply the ratio of blade tip speed to wind speed ( in the same units )


The implication of all this is that you should operate at a constant tsr for best power out. Your rotational speed should be proportional to the wind speed.


Now the snag. If you charge a battery our alternator is effectively clamped to a constant voltage load ( battery volts). If your alternator is 100% efficient the prop speed will rise until the dc output voltage equals the battery voltage and then sit absolutely constant at the battery voltage. If the alternator has resistance then there is an internal volt drop  and the speed will rise as the thing will have to supply an internal volt drop and this drop unlike the battery is load dependent.

The speed will now rise with load, but the steepness of the speed rise depends on the internal resistance. for ideal tracking of a prop you will need to let the speed rise by a factor of 3 from cut in at 7 mph to a wind speed of about 21 mph. This implies that you will have twice as many volts lost in the internal resistance as the battery volts. ( efficiency about 33 %).


This is a pretty sorry state of affairs from the efficiency point of view and has even worse implications for large machines. A 100W machine may well handle a stator loss of 200W but a 5kW one will never get rid of 10kW without frying.


For this reason we never track wind speed exactly, we make use of the fact that we can increase the angle of attack over a fair range before we reach stall, the lift/ drag is not as goo but it still works reasonably.


It so happens that if we run above design tsr at cut in when the load is light, we can manage to run below design in higher winds. For a nominal tsr of 7 we may manage tsr 9 at cut in and we may get as low as 5 before we hit stall with a bang.


In real terms this works out at nearer 2;1 speed change in stead of 3:1 so we reach about 50 % efficiency at nominal full load. This is much better but stator heating will still decide the wind speed that you need to furl at to reduce power from the wind.


To design the alternator to operate without external resistance you choose just enough magnet material to have the internal resistance you need for 50% efficiency at full load. The losses will be negligible at light load as it depends on current squared.


Small machines are always optimised this way, it is the cheapest way to do it.


If you can't get the output you need within the constraints of the stator dissipation you will have to use bigger magnets and thicker wire in a more expensive alternator. To avoid stall you now have to keep that efficiency down to 50% but if half that loss is dissipated in a resistor outside the stator the only heating of the stator will be from the component lost within it.


The small losses from a small machine would not be worth capturing and it would make no sense to over engineer the alternator to catch perhaps 100W in a heater.


When you are looking at 5 Kw that is serious heat and if you could catch 3kW in an external heater it would make life pleasant on a cold day so it is not all doom and gloom but it may be a problem in summer.


I hope this makes sense, if it puts this problem in perspective it will have been worthwhile.

Flux

« Last Edit: January 10, 2006, 02:10:05 AM by Flux »