PMA power curves & efficiency

[sh123469]

I think I remember seeing something about the pma's at hydrogenappliances.com on here a while back.  In particular, I though someone had stated how to figure out power requirements (turning) versus output into a real load.

The stated power curves are open circuit voltage and short circuit amps.  hydrogenappliances.com/pmacurves.html

There has to be a way to figure the output of these in real and useable terms.

I have seen where many people have a sort of distrust for these folks.  I have tried conversing with these folks via email and have found them to be generally evasive about efficiency, input vs. output figures, and real world situations.  They won't even supply separate rpm/voltage and rpm/amps curves or any power input to power specific load requirements, won't give coil resistances/wire sizes or anything else.

If their stated outputs were correct in relation to the hp input, they would have a "perpetual motion" generator system in some cases.  We all know that with the inherent lossesdue to efficiency, this isn't going to happen.  If you could really get 20000 watts out for an input of 6000 watts...I'll buy several!

Any comments, suggestions, formulae would be very much appreciated.

Thanks

Steve

[finnsawyer]

Steve,if the voltage and current are in phase or dc, then the ratio of open circuit voltage Voc to closed circuit current Icc is equal to the effective resistance Ref of the generator.  Think of the generator as being a battery with voltage equal to Voc in series with Ref.  If you hang another resistance across the generator you can then calculate the power dissipated in both resistances.  I made a comment about this once before.

[Victor]

Hi GeoM,

 You are correct in you thinking except that technically the current is limited by the impedance not the resistance. The impedance is a vector of the resistance and the inductive reactance. There is also an effect called armature reaction which is caused by the stator currents distorting the field flux. Only the resistance portion  is dissapated as heat, so the other portion of the impedance, while having a negative effect on load voltage, does not require extra shaft power.

Make the wind fun!

Victor

[finnsawyer]

I wanted to stay away from this stuff and give the guys something to work with.  Reff is a simple model that gives an idea of what the device can do.  Ultimately, of course, you need to take phase into account (which the manufactures should discuss).  That's why I keep harping about serious experimenters getting oscilloscopes.  No, I don't have stock in any oscilloscope manufacturers.

[spacejunk]

Hmm..I looked at their site a while ago and concluded that this company,by presenting the PMA output data in this way was deliberately trying to mislead its customers.

What other reason would they have for not showing real-world data. PowerOut/PowerIn is one thing. But as far as Pout vs RPM goes, well it aint rocket science! All the issues of Winding impedance, Core losses etc can be ignored if one is to simply rectify the AC,  Shunt regulate the DC output to say 15,30 or 60V (for 12,24 or 48V system -PMA specific) and measure the Output Current Io vs RPM.

Thus Pout=VxI.

Simple! Why do they skirt around the issue and give confusing curve data. ?

Was that "nice" enough?

SpaceJunk