Re: DC charging reactor

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "D.C. Cox by way of Terry Fritz <twftesla-at-qwest-dot-net>" <resonance-at-jvlnet-dot-com>


The steel density is a fixed value with a value for magnetic permeability.
This is usually dependent on the silicon content of the iron you are using
(this data from mfgr).  You need to see the actual hysteresis curve for the
steel you have.

Your core would be the 4.5 x 6.5 value = 30 sq. in.

Depending on the value of your core steel you should be able to run around
15-20 kva without problems.

There is a good section in the ARRL Radio Amateur's Handbook on these
hysteresis curve factors.

Dr. Resonance




----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Saturday, September 28, 2002 10:22 PM
Subject: Re: DC charging reactor


> Original poster: "Daniel Barrett by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <dbarrett1-at-austin.rr-dot-com>
>
> > Original poster: "D.C. Cox by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <resonance-at-jvlnet-dot-com>
> >
> >
> > Imagine slicing through your core with a knife. Looking at the end view
of
> > the core you just sliced through, the will be looking at the cross
> sectional
> > area, ie, cutaway view.  If the core measures 5 inches wide x 4 inches
> high
> > then the 5 x 4 in = 20 sq. inches of cross sectional area.
> >
> > This is what they are talking about --- the rectangular or square cross
> > section area of the core material.  The larger the cross sectional area
> the
> > more lines of magnetic flux lines that can be produced, and a higher
> > magnetic flux produces more power per unit area.
> >
>
>     Hmmm. I'm still having trouble picturing this in my head, the part
about
> slicing through the core. There are about 6 dimentional ways of doing this
> and I'm not sure which way you mean... If I throw out an example can you
> elaborate?
>     For my reactor I cut a MOT core into its 'E' and 'I' sections. I sawed
> the middle leg out of the E, turning it into a 'C'. I split the C into 2
> piles of laminations and stuck these end to end forming a rectangular
ring.
> It measures 6.5 inches long by 4.5 inches wide. The ring width is about 1
> inch and the laminiations are 1 inch thick. So it's as if I took a 1"x1"
> iron bar and bent it 90 degrees four times forming a 4.5 x 6.5 rectangle.
> There is a winding on each of the 6.5" sections.
>     So is the 'cross sectional area' simply 6.5 x 4.5? Surely the
thickness
> of the laminations (read overall iron mass) has to figure into this
> somewhere. Help, I'm geometrically challenged and I cant get up ;)
>
>     Smoke test says it handles 1400KVA  without saturation...
>
> db
>
>
>
>