Re: DC charging reactor

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

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