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*To*: tesla-at-pupman-dot-com*Subject*: Re: ballasting problems*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Fri, 11 May 2001 16:41:51 -0600*Resent-Date*: Fri, 11 May 2001 16:43:21 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <d3o97C.A.gQH.vrG_6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "BunnyKiller by way of Terry Fritz <twftesla-at-uswest-dot-net>" <bigfoo39-at-telocity-dot-com> Tesla list wrote: > Original poster: "Mark Broker by way of Terry Fritz <twftesla-at-uswest-dot-net>" <broker-at-uwplatt.edu> > > I was attempting to calculate the proper ballast for The Geek Group's Pig. We > wound the core (he posted the size in "The Mighty Inductor of Doom") according > to my calcs (actually used more turns for a Safety Factor). The result: I've > NEVER seen a circuit breaker pop that hard! Wow! There was enough resistance, > or enough inductance to prevent the circuit breaker from tripping until we drew > an arc from the pig outputs. > > Anyway, I used XL=w*L to determine the inductance needed. Then I used the > standard inductance equation L=N^2*A*mu/len modified such that len = > N*wire_diameter. A is the area of the core, from wire center to wire center. > It's approximate. mu is 4*pi*10^-7 * 5000 (approximate mu for silicon steel). > > So, the results show that I need a whopping ONE TURN to ballast 10kVA (5.8mH)! Snipperz... Hi Mark.... sounds like your inductor either saturated or the H is too small ... you should be looking for around 6 ohms of "resistance" from the inductor to get 40A -at- 240V do you have a variac of substantial size ??? ( 30+ amp rating??) hook up the variac as normal, place the inductor across the outputs of the variac. Slowly apply voltage and record at 10 volt intervals the voltage out of the variac AND the current running thru one of the wires feeding the inductor ( clamp on amp meter is great here) . using the V/A=R formula, you can average out the multiple readings and get the average resistant value (XL ) of the inductor. From there you can use XL/377 to find the inductance of the coil. for example you have V=10 A=1, V=20 A=2... etc... we have an avg of 10 ohms (R). 10/377= H.... 10/377=.0265H or 26.5mH what I have found in my playing around with scrapped cores is that with 180- 200 winds of #10THHN on various core sizes have resulted in about 2.5 amps per square inch of area ( center leg measurements) before saturation begins to become a problem. Scot D

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