windgat wrote:I think single phase will cause quite a vibration - it means there are points in the rotation where there is no resistance. Perhaps you can use three phases to three separate heating elements?
I did think about that, and I was hoping to mount the coils very close together so that the magnet field would overlap two coils at the change-over point. Hope this doesn't become a serious factor, cause I don't know how to mount two (or three) heaters in a standard geyser.
windgat wrote:In fact the other day in a storm I saw 19amps coming into the batteries at one point. Question is at what current the wire would be damaged or melt... for that test I would need a more powerful motor. Also, in a high wind there is a lot of cooling that is hard to duplicate on the test bench.
I can work-out the resistance of the coils and thereby the anticipated heat loss. I'm trying to keep it to 2 - 2.5 Watt per coil. So far. 0.8mm wire remains a posibility.
windgat wrote:You want to use 0.8mm wire, which makes the melting point that much sooner... there again (I just read yr post again) if you operate at 240V, then 1kW is only 4 amps.
I assumed that 240V would be the peakvoltage. The effective voltage (used as an AC power source) would then be 167V. Therefore, 6A is correct.
windgat wrote:Overall I like yr idea! No more 'cutoff voltage' below which no power is extracted from the wind.
But there is another matter to consider. Something you made me think about doing that "average windspeed" exercise.
Looking at the windspeed forcasts, there are plenty of days where the max speed is 4 - 5m/s. At those speeds I still want a useful power output.
windgat wrote:I think you may get 1kW out of the design just by doubling the blade length (to 2m).
WOW
To heat-up a 100L geyser 30 degrees (ie: from 45° to 75°) over a twelve hour period, you need about 300 Watt. That level should be about a useful minimum. So, the turbine should - on relatively calm days - still produce ± 300Watt.
It follows with:P = r^2 x v^3. and v=5m/s that r= 1.55m. And with v=4m/s, r=2.12. So your suggestion of 2m blades is spot-on.
windgat wrote:So for example, for 50x25x12.5mm magnets of grade N35,90rpm gives 0.302 x 88 x 90 = 23.91V (measured value = 24V) (with std deviation of 0.00109).
Thanks for those figures. It will help a lot.
stupid Q:Earlier you indicated that the figures were per
100rpm. Now I see you are multiplying with 90rpm.