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View Full Version : Small, slow AC machines: surface to volume ratio?


DGoncz@aol.com
Jun25-05, 03:35 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In recent threads in sci.electronics.design, rec.crafts.metalworking,\nand sci.physics, I have written about my attempt to convert Burden\'s\nSurplus Center #10-1134 motor into a pedal powered self-excited\ninduction generator (SEIG). The consensus is it won\'t work, and I am\ngetting ready to dispose of the evidence. :)\n\nThere are many reasons it won\'t work. The one I understand the best is\nthat when R, the winding DC resistance plus certain AC resistance\ncomponents, is greater than the square root of L*C, where L is the\nwinding inductance including the effect of the rotor, and C is an\nauxiliary capacitor across the windings to achieve self-excitation,\nthen the resonant "tank" circuit has no impedance hump, and therefore Q\n= 1.\n\nI was wondering if this is a surface to volume ratio effect. That is, R\ncomes from the cross sectional *area* of the wire selected, and L comes\nfrom the turns, in a three dimensional *volume* selected, of individual\nwires. Small objects have a higher surface to volume ratio, in general,\nthan larger ones, so for a specific design at some scale factor, R &gt;\nsqrt(L*C).\n\nAlso, I speculate room temperature superconductors might allow small,\nslow self-excited induction generators with minimal R, and was\nwondering if anyone here agrees or disagrees with that speculation.\n\nThe motor is:\n\nhttp://surpluscenter.com/item.asp?UID=2005062415255748&catname=&qty=1&item=10-1134\n\nThe threads are:\n\nhttp://groups-beta.google.com/groups?q=author:DGoncz@+generator&start=0&scoring=d&hl=en&lr=&ie=ISO-8859-1&num=10&\n\nYours,\n\nDoug Goncz\nReplikon Research\nFalls Church, VA 22044-0394\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In recent threads in sci.electronics.design, rec.crafts.metalworking,
and sci.physics, I have written about my attempt to convert Burden's
Surplus Center #10-1134 motor into a pedal powered self-excited
induction generator (SEIG). The consensus is it won't work, and I am
getting ready to dispose of the evidence. :)

There are many reasons it won't work. The one I understand the best is
that when R, the winding DC resistance plus certain AC resistance
components, is greater than the square root of L*C, where L is the
winding inductance including the effect of the rotor, and C is an
auxiliary capacitor across the windings to achieve self-excitation,
then the resonant "tank" circuit has no impedance hump, and therefore Q
= 1.

I was wondering if this is a surface to volume ratio effect. That is, R
comes from the cross sectional *area* of the wire selected, and L comes
from the turns, in a three dimensional *volume* selected, of individual
wires. Small objects have a higher surface to volume ratio, in general,
than larger ones, so for a specific design at some scale factor, R >
\sqrt(L*C).

Also, I speculate room temperature superconductors might allow small,
slow self-excited induction generators with minimal R, and was
wondering if anyone here agrees or disagrees with that speculation.

The motor is:

http://surpluscenter.com/item.asp?UID=2005062415255748&catname=&qty=1&item=10-1134

The threads are:

http://groups-\beta.google.com/groups?q=author:DGoncz@+generator&start=0&scoring=d&hl=en&lr=&ie=ISO-8859-1&num=10&

Yours,

Doug Goncz
Replikon Research
Falls Church, VA 22044-0394

DGoncz@aol.com
Jun26-05, 11:43 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>To droll on a bit, and expand the discussion:\n\nClearly a motor is an inductive load. Without argument we may take that\na motor is an inductive source, when used as a generator. We\'re talking\nAC motors here....\n\nWhat the hell is an inductive source?\n\nWell, a transformer could be one, but there\'s this issue of impedance\nreflection, so let\'s leave that out and *not* assume a classic linear\nAC-tranformer-AC-diode-capacitor power supply is an inductive source.\n\nI do know one motor sucks at driving another, so we\'ll just go with\nthat. An inductive source is one which "doesn\'t like" inductive loads.\nAnd so we may presume it does like resistive or capacitive loads.\n\nNow, universally, what is done with an AC induction motor to turn it\ninto an SEIG is "to slap some caps on it". This is power factor\ncorrection, making the generator system more happy with resistive\nloads, which predominate.\n\nHow does this work?\n\nWell, I understand resonance and I think I have a pretty good\nexplanation:\n\nThe stator winding inductance and slapped on capacitor bank need to\nform a *resonant tank* with frequency "somewhat less" than the\nfrequency desired.\n\nHow much less? Or, a better question, why *less* and not *more*?\n\nOh, that\'s easy....\n\nThe driving source, the prime mover, is essentially along the X axis.\nAnd output, a function of resonance, is along Y.\n\nNow, what do you *not* want a generator to do when you load it?\n\nYou *don\'t* want it to sag, that is, lower its output. And that is\nprecisely what will happen if you\'re operating on the low-frequency,\ninductive side of the Q curve:\n\nA little load, a little less speed (conservation of power or energy). A\nlittle less speed, a little *less* output. A little less output, well,\nyou get the idea. dy/dx &gt; 1.\n\nA positive feedback system, right? Remove the load, it overspeeds,\nblowing out the prime mover. Add load, it sags, stalling or lugging the\nprime mover.\n\nSo what happens on the other side?\n\nEasy. Negative feedback. dy/dx &lt; 1.\n\nMore later....\n\nDoug\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>To droll on a bit, and expand the discussion:

Clearly a motor is an inductive load. Without argument we may take that
a motor is an inductive source, when used as a generator. We're talking
AC motors here....

What the hell is an inductive source?

Well, a transformer could be one, but there's this issue of impedance
reflection, so let's leave that out and *not* assume a classic linear
AC-tranformer-AC-diode-capacitor power supply is an inductive source.

I do know one motor sucks at driving another, so we'll just go with
that. An inductive source is one which "doesn't like" inductive loads.
And so we may presume it does like resistive or capacitive loads.

Now, universally, what is done with an AC induction motor to turn it
into an SEIG is "to slap some caps on it". This is power factor
correction, making the generator system more happy with resistive
loads, which predominate.

How does this work?

Well, I understand resonance and I think I have a pretty good
explanation:

The stator winding inductance and slapped on capacitor bank need to
form a *resonant tank* with frequency "somewhat less" than the
frequency desired.

How much less? Or, a better question, why *less* and not *more*?

Oh, that's easy....

The driving source, the prime mover, is essentially along the X axis.
And output, a function of resonance, is along Y.

Now, what do you *not* want a generator to do when you load it?

You *don't* want it to sag, that is, lower its output. And that is
precisely what will happen if you're operating on the low-frequency,
inductive side of the Q curve:

A little load, a little less speed (conservation of power or energy). A
little less speed, a little *less* output. A little less output, well,
you get the idea. dy/dx > 1.

A positive feedback system, right? Remove the load, it overspeeds,
blowing out the prime mover. Add load, it sags, stalling or lugging the
prime mover.

So what happens on the other side?

Easy. Negative feedback. dy/dx < 1.

More later....

Doug

DGoncz@aol.com
Jun27-05, 10:58 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I wrote:\n\n&gt; A little load, a little less speed (conservation of power or energy). A\n&gt; little less speed, a little *less* output. A little less output, well,\n&gt; you get the idea. dy/dx &gt; 1.\n\n&gt; A positive feedback system, right? Remove the load, it overspeeds,\n&gt; blowing out the prime mover. Add load, it sags, stalling or lugging the\n&gt; prime mover.\n\n&gt; So what happens on the other side?\n\n&gt; Easy. Negative feedback. dy/dx &lt; 1.\n\nEr, dy/dx &lt; -1. That sounds right.\n\nDoug\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>I wrote:

> A little load, a little less speed (conservation of power or energy). A
> little less speed, a little *less* output. A little less output, well,
> you get the idea. dy/dx > 1.

> A positive feedback system, right? Remove the load, it overspeeds,
> blowing out the prime mover. Add load, it sags, stalling or lugging the
> prime mover.

> So what happens on the other side?

> Easy. Negative feedback. dy/dx < 1.

Er, dy/dx < -1. That sounds right.

Doug