Magnetic Field Equations Predict Overunity...

[tim123]

Hi guys. I'd like to discuss the equation for magnetic field strength. It looks to me like it predicts overunity:

H = Amps * Turns / Length

Note, there's nothing in there which allows you to determine power. You've only got amps, not volts.

The 2 things that ultimately determine how much power it takes to create a given field are:
- Size of the coil
- Resistivity of the conductor material

So, if we increase the physical size of a coil - we get more field for the same input power. There's a practical limit to this where each turn takes so much wire that the law of diminishing returns applies, but there's still scope for a vast range of field strengths for a given input.

If we reduce the temperature of the coil, things get really interesting... The resistivity drops with temperature and there's a point around -230C, even before superconductivity, where the resistance of the coil would be negligible - meaning you can get your desired magnetic field for much less input - even just a few microwatts of power.

Now, assuming that we can turn a varying magnetic field into useful work - we have, at some point on the scale - overunity... The question then becomes - where?

From what I can tell - OU is just outside the physical range we commonly use. A standard sized solenoid is underunity. Make it 10-20 times bigger, attach it to a crank, and my calcs say it's well OU.

Here's some examples - of standard solenoids with a mobile iron core. All these coils use 48 watts of input power:

Length     CoreDiam    CoilDiam     Hfield     OutputPower
100mm       10mm         20mm      48,7561         4w
100mm       20mm         30mm      37,756         23w
100mm       20mm         50mm      55.358         34w
100mm       40mm         60mm      37,801         93w
100mm     100mm        200mm     48,815       755w
100mm     100mm        300mm     59,792       925w
100mm     100mm      3000mm     81,751    1,265w
1000mm    500mm      1000mm     15,436   59,752w (80Hp!)

The output power values are determined from:
- Max possible force (from book: http://archive.org/details/solenoidselectr01undegoog)
  F = AMPTURNS * AREA / LENGTH

- average pull throughout power stroke
  PULL = MAXPULL * sin(0.77 * PI * (CoilLength / InsertedDistance))

- Throw = Coil Length.
- Crank size is half throw.
- Duty cycle is 50%.

[gyulasun]

Done.

[gyulasun]

Hi Tim,

Maybe there is a typo in the last raw of CoilDiam column: is it not 5000 mm instead of the 500? You can use the Modify feature for 12 hours from the time of posting, after that you cannot edit.

[tim123]

Thanks  . Can you delete the superfluous posts too?

[gyulasun]

Hi Tim,

My questions:

1) When considering the coil current needed for the AmperTurns in the calculations, have you thought of the coil's self inductance which will greatly modify the coil impedance (think of a series R-L circuit)? This means that you need to use higher input voltage for insuring the needed current within a certain ON time for the coil and this involves eventually a higher than calculated input power. Did you include this in the calculations shown or you think it is not an issue? (Because I assume you will switch ON and OFF the coil current with 50% duty?)

2) I have not seen any loss calculation occuring in the shorted coil and its effect on input power? How many turns are involved approximately in the shorted coil with respect to the main input coil? Have you considered this question?

Greetings,  Gyula