Author Topic: Initial calculations for tests?  (Read 1533 times)

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copilador

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Initial calculations for tests?
« on: February 20, 2017, 01:39:51 AM »
They are correct these calculations? (To start testings)


electrondady1

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Re: Initial calculations for tests?
« Reply #1 on: February 20, 2017, 05:33:29 AM »
D is greater than E
in a single phase machine D = E+(E/2)

copilador

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Re: Initial calculations for tests?
« Reply #2 on: February 20, 2017, 09:11:44 AM »
Hi, thanks electrondady1,

Magnet measures:  30mm x E->20mm x B->10mm (N50 Neodymium)

D = (20+20*0.66)/2 = 13.3 mm three phases?

D= 20+(20/2) = 30mm single phase

it is right?

joestue

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Re: Initial calculations for tests?
« Reply #3 on: February 20, 2017, 09:49:30 AM »
To find the optimum coil thickness you can cut aluminum washers out, the same size of the coil, and measure the drag exerted on the stack of washers by the rotor turning at a fixed rpm. Change the air gap and the number of washers and you should find an optimum thickness of aluminum. You may want to make the washer the size of three coils, so that the force is constant.


a potentially more accurate and easier way to perform this test is to set the rotor up, with a shaft, and a rope coiled around the shaft with a weight on it. the weight will apply a constant torque, and the eddy current in the aluminum washers (or test coils) will resist the motion. the RPM of the rotor should very quickly stabilize at some speed.

the more efficient the coil, the slower that rpm will be. once you get to the test coils, you could use this method to find the optimal inside diameter of the coil. as you probably know, the inside diameter of the coil is of diminishing returns. adding more turns only marginally increases the volts, at a significant increase in resistance. you can also vary the outside diameter of your generator, moving the magnets closer together or further away.

the constant torque applied, shorted coil experiment i described will quickly identify the most efficient coil structure.. the rpm will be the lowest.

its ok to use a solid block of aluminum or copper of various thicknesses as well, but there are some differences between a coil of x number of turns of uniform amps, and a solid block. the results should be similar though. i would try both and see if you arrive at the same optimal thickness of copper or aluminum. you will have to change the air gap for each datapoint of course.
« Last Edit: February 20, 2017, 10:02:17 AM by joestue »

Ungrounded Lightning Rod

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Re: Initial calculations for tests?
« Reply #4 on: February 20, 2017, 08:39:04 PM »
Could others check me on this, please?

I thought it was:

A ~= 2B

The sweet spot for maximum generation from a given set of magnets is for the gap to be the same thickness as the total thickness of the magnetic material, stuffed solid with the thickness of the coil.
 - Widening the gap beyond that means loss from weaker field goes up more than gain from more turns.
 - Narrowing the gap beyond that means loss from fewer turns goes up more than gain from stronger field.
Approximate both because the air gap means the coil doesn't completely fill the space (make the space a tad bigger) and because the curve is bending over at the maximum, so it changes little with minor deviations from the idea and thus isn't too critical.  (I'd make the coil as thick as the total magnetic material and the gap larger by the desired clearance.  That should be close to optimum, though I didn't do the math to be sure.)

For THREE phase:

D ~= (E / 2)  (Maybe slightly more because the field isn't straight across but bows out slightly.)

The voltage waveform in this geometry is more a trapezoid (sawtooth with the peaks flattened) than a sine.  Three of these waveforms 120 degrees apart, like three sine waves, add to zero.  So you not only have a piecewise linear approximation to a sine but you can connect your machine in delta and not have substantial circulating currents.

Gap ~= E  (Can't get it exact in this geometry because the gap is a wedge and the magnets are usually rectangular.  Again don't forget the slight field spread-out.)

Because:
 - Magnet over both sides cancels, so why waste copper and add resistance losses?
 - Magnet narrower than the gap makes the waveform flatten out temporarily as it crosses zero - both wasting an opportunity to generate and fouling up the delta-connection cancellation.

« Last Edit: February 20, 2017, 08:44:18 PM by Ungrounded Lightning Rod »

copilador

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Re: Initial calculations for tests?
« Reply #5 on: February 20, 2017, 11:09:35 PM »
Hi,
With your help I made the first design.
The current doubt A:  A ~= 2B or  A ~= B

electrondady1

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Re: Initial calculations for tests?
« Reply #6 on: February 21, 2017, 07:39:17 AM »
hold on, did you have that illustration of a coil  showing D as the coil leg width yesterday ?
if so, i am withdrawing my previous answer.
A=2B
D=E  is ok, but  is dictated by the next coil, that is dictated by the space to the next mag.
C=2mm or 1/16" if your stator is perfectly flat but this being the world you may have to open up A to get more C
 

copilador

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Re: Initial calculations for tests?
« Reply #7 on: February 21, 2017, 09:57:41 PM »
Thank, great help for me  :)

Ungrounded Lightning Rod

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Re: Initial calculations for tests?
« Reply #8 on: February 23, 2017, 02:55:52 PM »
With your help I made the first design.

I see that's a three-phase design.  The geometry can be substantially improved, as follows:

 - Change from rectangular coils to trapezoidal coils.
 - Butt them up tightly to each other so they touch.  The lines where they touch will be radial.
 - D ~= 1/2 E, so a magnet would just cover the TWO coil-sides that are touching.
 - The hole is ~E at the center of the magnet, so a magnet over the hole just (pretty much) covers the hole.
 - Space the magnets ~ 1/2 E, so that the layout is like this, working clockwise:
    a) The first magnet (N) is exactly over the combination of the right side of coil 1 with the left side of coil 2
    b) The second magnet (S) has its right edge over the boundary between the right side of coil two with the left side of coil 3
    c) The third magnet (N) is over the hole in coil 3
    d) The fourth magnet (S) has its left edge is over the boundary between the right side of coil 3 with the left side of coil 4
    e) (like a) The fifth magnet (N) is over the combination of the right side of coil 4 with the left side of coil 5
   and so on around the device.

This makes the waveform of each phase work in six equal steps:
 1 Ramp up from zero to max +.
 2 Hold at max +
 3 Ramp down to zero.
 4 Continue ramping down to max -
 5 Hold at max -
 6 ramp up to zero.

At the position drawn (assuming clockwise magnet motion and the handedness of the coils causes N over left side to generate +):
 - Coils 1, 4, 7 are at the boundary between steps 5 and 6
 - Coils 2, 5, 8 are at the boundary between steps 1 and 2
 - Coils 3, 6, 9 are at the boundary between steps 3 and 4
« Last Edit: February 23, 2017, 03:13:24 PM by Ungrounded Lightning Rod »

copilador

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Re: Initial calculations for tests?
« Reply #9 on: February 24, 2017, 11:37:43 AM »
Hi, Thanks, i modified , is it correct? (12 coils  - 16 magnets) MOD.

Ungrounded Lightning Rod

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Re: Initial calculations for tests?
« Reply #10 on: February 24, 2017, 01:55:53 PM »
Yes, that's it.