Original poster: "Dmitry (father dest)" <dest@xxxxxxxxxxx>
Hallo Barton,
very quick reply (no time):
imo you have confused power factor and efficiency somehow - your
"power factor of .1" is the power factor with shorted output, but if
you connect an appropriate load, you would get that "something between
.3 and .5", so i still think that there is only about 50w and 447var
(not 400 as i said before - sorry ^_^).
> Original poster: "Barton B. Anderson" <bartb@xxxxxxxxxxxxxxxx>
> Hi Dmitry,
> Yes, it won't be 450 watts. I would guess higher than 50 as that
> would indicate a power factor of .1. I would expect it to be
> something between .3 and .5, so in the neighborhood of 135 to 225
> watts based on nameplate data. Given that actual Rs and Rp values are
> unknown, reality might be somewhere between 100 and 200 watts.
> Anyway, certainly not 450.
> Take care,
> Bart
> Tesla list wrote:
>>Original poster: "Dmitry (father dest)" <dest@xxxxxxxxxxx>
>>
>>Hallo guys,
>>
>>do you really think that there would be 450 watts of heat? : )))
>>i think that there would be about 450 va-mps, about 400 var-s, and
>>maybe only 50 watts of pure heat : D
>>
>> > Original poster: Yurtle Turtle <yurtle_t@xxxxxxxxx>
>>
>> > I think he meant that it would loosen the frozen tar
>> > closest to the windings. A 15/30 is roughly 450 watts.
>> > Though I've never tried that method, it should be
>> > enough heat to soften the tar closest to the windings.
>>
>> > Adam
>>
>> > --- Tesla list <tesla@xxxxxxxxxx> wrote:
>>
>> > > Original poster: "Lau, Gary" <Gary.Lau@xxxxxx>
>> > >
>> > > Since it's current-limited, running an NST with the
>> > > secondary shorted
>> > > won't draw any more power than its faceplate rating.
>> > > So I don't know
>> > > how that could result in melting of the tar.
-----
Let the bass kick! =:-D