Understanding the relationship between vawt and pma

[wellingtonex]

Hello to all, new member here, stand back! I've built a few prototype vertical axis wind turbines and have been on a major learning curve trying to find the best pma( permanent magnet alternator) to match my mill. At first the reasoning seemed simple. Find a pma and adapt the mill to drive it at the desired rpm. 80 watts later my belts are squealing and pma won't turn any faster. At a ratio of 1:9 I believe I've spent all available torque. So my thoughts are now directed at a low speed pma. My mill is exposed to roughly 6000w of wind energy, keeping betz in mind, I factor that my mill is 10% efficient leaving me with about 600 watts of extracted energy. Heres where I struggle, the low speed pma requires 554nm at 2500watts. I'm sure i won't be producing this much torque but i know my mill is quite strong. Is there a formula or a way to calculate torque with only primative information? Does my reasoning toward this matter seem adequate? Can anyone give me their insights on this matter?
Thank you

 

[Baluncore]

Welcome to PF.
What type of vertical axis mill have you built ? If it is a Savonius rotor then that would explain your problems.

How do you remove the energy from the PMA? If it charges a battery directly then RPM will be limited by battery voltage as energy will only start to flow when the RPM of the PMA exceeds the battery voltage. Any higher voltage will stall the aerodynamics of the windmill.

You should not be using belts that slip. Use bicycle roller chain or stepped belts. V-belts are inefficient.

 

[wellingtonex]

Baluncore, the mill is a savonous/darreus h-rotor with variable pitch. My first prototype was reaching tsr's of 1.68, proving the mill is generating lift, operating at 100rpm. My newest build has a much larger radius which operates at a slower speed but quite strong.
Because of the mills unique design, I've been bound to a v belt drive system, which could be changed, but quite difficult. I should rephase squealing to squeaking belts... tthe belts are capable of stopping the mill, but there is slippage.
My results of testing show that changing the ratio between the mill and pma only change the rpm of the mill with slight differences in pma wattage output. As i reduce the ratio, the mill speeds up accordingly with small increases in amperage from the pma.
As you mentioned, the chosen voltage will restrict the output of the pma. At this point, I have tested the following... 12v, 24v, 36v, 48v, and 120v inverter. My best test results always seem to come from 36v which in some way must match the power curves of the mill.
At this point I'm not comfortable winding up this mill to higher sppeed . Its quite large and is very sensitive under load, when i tested 120v, the freewheeling point before power production was quite scary, once it started producing she settled down?
My target mill rpms are 15-30 Which explains the 1:9 gear induction however, I'm using high speed pma's. I am currently considering a low speed pma which boasts 2.5kw at 50rpm. This could be Direct driven but I struggle with the torque requirements to do this. 554nm or 408 ft lbs seems quite high and I don't want to overestimate my mills capability. I realize I won't be utilizing the full potential of pma, I just want to be certain I'm choosing one that will work for my situation.
Now youve got me wondering what my tsr is on my mill at 30rpm... as I haven't tested it. I may not be generating lift at 30rpm. A worthy test.

I really apreciate your input... thank you, Rod

 

[Baluncore]

I have seen too many people fail because they liked the elegance of the Savonius rotor. Unfortunately, the Savonius rotor is the kiss of death for power generation. Once it infects the brain it can be considered a terminal case. It is probably the most inefficient design, but it needs the strongest tower.

I guess you only use the Savonius to start the Darius, but if the Savonius remains engaged it will waste much of the available energy since the Savonius and Darius have different optimum RPM to wind-speed curves. While the Darius is generating power, the Savonius will be consuming power. If the Savonius is built inside the envelope of the Darius then you will have problems because the dirty air downstream of the Savonius will upset the Darius. That airflow interference will also cause tower oscillation at some speeds. If you have an anemometer that indicates sufficient wind, and if you sense the Darius not rotating, then you can start the Darius with a motor / generator. That is a more efficient strategy than employing a Savonius rotor.

If you know the wind speed then you should be able to measure or calculate the optimum RPM for the windmill to extract maximum power. As you load the mill with the alternator the mill will slow down until it passes the optimum RPM for that wind-speed. That will be when the angle of attack is optimum. If you draw more power you will begin to stall the foils.

If you used a car alternator with a field winding, then you could regulate energy extraction and transfer to the battery by control of the field current. Unfortunately you cannot do that with a PMA.
The windmill will operate best at different RPM in different wind speeds. A PMA will need to generate into a capacitor at the voltage generated by the PMA at the optimum RPM for the wind speed. A switching converter can then transfer power from the capacitor into the fixed voltage battery. Drawing the right amount of power from the mill via the capacitor is critical to getting maximum energy from the windmill.

 

[wellingtonex]

Wow, youre just the guy I need to talk to. I couldn't agree more with what youve said. Dont get me wrong, I'm no genius, and I have fallen into the arms of the savonous. I've done much research on prior work with the vertical and tried to learn from others mistakes. With this research I've developed a savonous rotor that doesn't spend extracted energy to force downwind blades thru drag zones. My design has its draw backs no doubt, but it has huge upsides aswell. This towerless design works extremely well in very poor conditions. It targets the urban market eliminating high speeds and the need for massive towers and expensive transmission lines. I am not trying to oversell my mill as I am having trouble just trying to rate its performance and capability. I'm probably in over my head but I'm having too much fun and I'm too stubborn to give up! Lol My design is patent pending at this point, but I'm not anxious to advertise it online yet.
With help from people like Yourself, I'll be able to bring out the full potential of my design. I've read some very interesting articles and many well educated people will disagree with me, " there's NO power in dirty wind!", they say. I disagree, as a sheet of plywood and a light gust will prove otherwise! We just haven't found an acceptable means of harnessing it yet.
I would be very interested in your thoughts on my design, concept, etc.
I don't want to assume i have massive amounts of torque, and I don't want to order the wrong lowspeed pma( as its coming from China) I can rig up a scale to measure ftlbs at start up but I'm sure the torque will increase as the mill reaches its optimum operating range. I've recorded the maximum amps for different gear ratios which I believe I can do a bench torque test to relate the relative torque. I've heard of the pma feeding a capacitor but I guess At this point I'm more interested in capability more so than efficiency. ( hoping I understood your post)
The capacitor makes perfect sense, now that I think more about it. It alllows the pma to operate at a much wider range, annd maintain full effiency.. Would the switching converter cause surging on the mill when it dumps A full capacitor? At higher speeds I'm sure it would switch very quickly( depending on capacitor size) but what about low. Speeds?

I can't thank you enough for your inut, I look forward to more!

Rod

 

[Baluncore]

Would the switching converter cause surging on the mill when it dumps A full capacitor?

The switching converter would only take small bites of energy from the capacitor. The cap would be large enough that it had only small voltage changes. When the windmill is generating more energy, the converter will bite more often. Knowing the wind speed and your windmill characteristics you can know the optimum RPM and hence the PMA output voltage. Whenever V_cap rises above that voltage the converter takes another bite and efficiently puts that small bite of energy into the battery.

Power; W = V * I. Charge; Q = I * t. Capacitance; C = Q / V
If your optimum PMA output voltage is V at some RPM, and it is generating W watt then the average current will be I = W / V.
If the time between bites is dt, and the voltage step you can tolerate is dV volt, then C = I * dt / dV.
That gives you the capacitor value needed. It will need a bigger capacitor for lower RPM when the voltage steps will be bigger because C_energy = 0.5 * C * V². There will be some speed below which it is better not to extract the little power available but to keep the blades spinning ready for the next gust.

 

[wellingtonex]

The switching converter would only take small bites of energy from the capacitor. The cap would be large enough that it had only small voltage changes. When the windmill is generating more energy, the converter will bite more often. Knowing the wind speed and your windmill characteristics you can know the optimum RPM and hence the PMA output voltage. Whenever V_cap rises above that voltage the converter takes another bite and efficiently puts that small bite of energy into the battery.

Power; W = V * I. Charge; Q = I * t. Capacitance; C = Q / V
If your optimum PMA output voltage is V at some RPM, and it is generating W watt then the average current will be I = W / V.
If the time between bites is dt, and the voltage step you can tolerate is dV volt, then C = I * dt / dV.
That gives you the capacitor value needed. It will need a bigger capacitor for lower RPM when the voltage steps will be bigger because C_energy = 0.5 * C * V². There will be some speed below which it is better not to extract the little power available but to keep the blades spinning ready for the next gust.

 

[wellingtonex]

I think I'm still with you, but for clarification... And a better understanding, is the voltage measured off the 3 phase or is it optimum voltage rectified? if it were off the three phase I would assume I'd need three capacitors.
Will the windmill freewheel if the capacitor is full and converter fails? Sorry if that's a dumb question, I am curius on how the capactor behaves and if its a load for pma at all times or just when discharged.
To better understand the converter... if the pma is at entry level charging voltages, does the converter dump on a voltage difference? In other words, the cap doesn't need to be full for converter to dump?
Does the cap accumulate voltage or does it just store the peak voltage sent to it?
I know more voltage means more potential, in this case is more voltage the best? Should I pick the highest voltage pma that will suit the rpms and torque of my mill?
I am still running a high speed pma at this point and only have 23volts at my optimum rpm. I think I'd be better off to order a lowspeed pma before I start entering in values to those equations.
I'm not sure about this voltage step tolerance... when the converter dumps to load I'll assume the mill will respond accordingly with load. If the converter dumps longer... the mill could stall... therefore the tolerance could be adjusted by the length of time each dump takes? Am I on the right page with this?
Once again I thank you for your input! I really appreciate it
Rod

 

[Baluncore]

is the voltage measured off the 3 phase or is it optimum voltage rectified? if it were off the three phase I would assume I'd need three capacitors.

The 3 phase alternator is rectified by a 6 diode bridge to produce DC with a ripple voltage. I expect that the peak rectified voltage will be proportional to RPM. The capacitor will be charged to close to that peak DC voltage.

Will the windmill freewheel if the capacitor is full and converter fails?

Yes. The capacitor must be rated for the maximum voltage that the PMA can generate when it freewheels in a windstorm. You should attach some form of speedbrake, something like rubber flaps to your Darius blades that will flip out and become air-brakes at some large RPM. That will protect the PMA and electronics, it will also prevent destruction of your windmill.

Im curius on how the capactor behaves and if its a load for pma at all times or just when discharged.

The capacitor will accept charge whenever PMA voltage exceeds the capacitor voltage. The capacitor smooths the voltage from the PMA. It should not be operated full or empty. Capacitor voltage will average just below the PMA output voltage. The PMA will deliver most current, (charge), at phase peaks, less current in the valleys, so the capacitor is always there as a charge reservoir for the PMA.

Does the cap accumulate voltage or does it just store the peak voltage sent to it?

A capacitor accumulates and stores charge, Q = I * t. A current of one amp for one second delivers one coulomb of charge. The voltage on a capacitor is proportional to the charge stored in the capacitor. V = Q / C.
The switching converter is designed to keep the capacitor voltage close to the optimum for that wind speed.

Should I pick the highest voltage pma that will suit the rpms and torque of my mill?

It is best to have a higher voltage so you get V_cap = V_bat at the lowest sensible RPM for generation. A switching converter can step Vcap up to Vbat if that is needed, but it is more efficient to step down from a higher voltage.

I'm not sure about this voltage step tolerance... when the converter dumps to load I'll assume the mill will respond accordingly with load. If the converter dumps longer... the mill could stall... therefore the tolerance could be adjusted by the length of time each dump takes?

The converter switches at up to 100,000 times per second. The voltage dV due to the bites of energy taken from the capacitor will be less than the natural ripple on the 3 phase rectified supply. There may be several bites taken during the time that one phase is peaking.

The critical thing is to take energy from the capacitor when the mill is spinning faster than is optimum for the particular wind speed. The converter should always wait until V_cap has recovered and is high enough before it takes the next bite. It will not stall. I expect dV to be maybe a couple of volts.

You will need to know the optimum RPM for maximum sustained power at different wind speeds. That will decide the optimum angle of attack of the Darius' blade.
You might gather that information automatically by sweeping power extraction at different wind speeds to build up a map of windspeed to RPM_opt for maximum power over time.

 

[wellingtonex]

With all this information I find myself approaching old problems with new thinking. Of course it arouses more questions. When I first hooked up my high speed 3.5kw to my mill the lowest ratio I used was 1:3 and upto 1:9.

Today I changed the ratio to 1:1.10 with much better results. Before I switched it I was at 1:3 getting 4 amps at 15 rpm (mill). Does that mean at 1:1 ratio I would produce 3 times the torque/amps?

The 4 amps I was generating was a dead short on the rectifier. Is this a practical means of testing? I usually test into a battery bank.

I measured open end volts off rectifier as well as voltage under load, when calculating pma wattage do I use open ended or loaded volts* amps?

The 1:10 ratio works well as long as I restrict it with a 12volt supply. As wind power cubes with every mile per hour I wonder if torque does as well? As the rpm increases I'm aware the drag increases on a vertical buT I would imagine the torque increases until the power curve starts diminishing.

Thanks again, Rod

 

[Baluncore]

The 4 amps I was generating was a dead short on the rectifier. Is this a practical means of testing? I usually test into a battery bank.

Power is the product of voltage and current. If you short the rectifier bridge with an amp meter then you have no volts and therefore no external power. The windings in the alternator may get quite hot.

I measured open end volts off rectifier as well as voltage under load, when calculating pma wattage do I use open ended or loaded volts* amps?

If you measure OC voltage, then there is no current flow and no voltage drop in the alternator windings. Without current your power is zero. You must generate into a reasonable load and measure both current and voltage at the same time to find the power.

The optimum gear ratio you use will be determined by wind speed and the voltage of the battery being charged. With a switching converter that ratio becomes unimportant as the effective ratio is handled by the converter.

Always remember that Power is the product of torque and RPM.