Re: power factor correction for NST

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

Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Mddeming-at-aol-dot-com>

In a message dated 12/31/01 1:46:48 PM Eastern Standard Time, tesla-at-pupman-dot-com
writes:





>
> Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <evp-at-pacbell-dot-net>
>
> Tesla list wrote:
> > 
> > Original poster: "Neil Richardson by way of Terry Fritz
> <twftesla-at-qwest-dot-net>" <neil-at-opticalrealities-dot-com>
> > 
> > NSTs are 50% efficient, and thats the reason. I believe it is the same with
> > all older NSTs.
>
>     Woops!  No way - if this were true they'd all burn up; the "lost poper"
> would show up in the form of heat!  There seems to be some confusion due
> to the use of magnetic shunts to increase the leakage reactance and
> provide current limiting.  I have measured the total effective
> resistance of a typical 12 kV, 60 ma transformer and find it to be about
> 6000 ohms.  At a load current of 60 ma that corresponds to about 22
> watts loss, to which must be added the core loss, which may be another
> 50 watts at most, giving a total loss of perhaps 75 watts.  The power
> output will depend on the type of load, but with a resistive load which
> gives maximum output power (resistance = leakage reactance) the power
> output would be 360 watts, for a new efficiency of about 83%.
>
> Ed



Hi Ed,
         I think he's defining efficiency as USEFUL power out/ RATED power out.
In this case, 360/(12KV X 60 ma)= 360/720=50% as stated.
Matt D.