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Original poster: "rheidlebaugh by way of Terry Fritz <twftesla-at-qwest-dot-net>" <rheidlebaugh-at-zialink-dot-com>
I have used old variac cores just as you intend with no cutting. I think you
should have room for 80T if you overlap your turns. I never throw those
cores away. They are to good. They are made of wound iron strapping not
ceramic solid so I dont think you want to be cutting them.
> From: "Tesla list" <tesla-at-pupman-dot-com>
> Date: Fri, 15 Feb 2002 13:10:39 -0700
> To: tesla-at-pupman-dot-com
> Subject: Ballast
> Resent-From: tesla-at-pupman-dot-com
> Resent-Date: Fri, 15 Feb 2002 13:14:22 -0700
> Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> I currently use a 5,000 watt variac for ballast for my coil using a 5 kva
> 14,400 v transformer, normally running at about 30 amps. I can hear this
> variac groan sometimes and suspect I might be into saturation - and I don't
> want to cut the core.
> I recently came across a 7.5 amp Powerstat variac that has burned windings.
> I stripped them all off and am considering slicing .25" out of the core ( to
> allow ease of winding and prevent saturation) and winding it with insulated
> #8 house wire. If I have enough room to overlap the inside windings, like
> the variac manufacturers do, I could get about 68 turns of #8 wire on there.
> Reality might be 40 to 50 turns.
> The core weight about 6.5 lbs. It is 3.0" tall, 4.0" wide with a 1.8"
> diameter hole through the center - i.e. the sides are about 1.1" thick.
> After winding I would epoxy a .25" thick piece of phenolic back into the core
> to give structural integrity again.
> Does anyone think this might work as a 30 amp ballast? What might the
> inductance be? I see where Jim Lux supplied a formula of L = k * N squared.
> Is k permeability? If so, what is the k of this material likely to be? I
> would think that we would need to account for the core area and the
> permeability of the material. The core is 1.1 x 3.0 which would be 3.3
> square inches times maybe a 10.0" winding length would be 33 cubic inches.
> I do need some help here.
> Thanks, Ed Sonderman