Re: MMC cap bank

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

Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>

Hi again Jimmy,

Fortunately, the equations in Power4 and MMCcalc are correct and based off of:

""""""""""
Power Dissipation per Small Capacitor

Wo = KSYNC BPS ECAP RCAP / (RCAP + RPRI)

Where:
Wo = Power Dissipation per Small Capacitor (Watts)
KSYNC = 1 for Synchronous Gap or 0.5 for Non-Synchronous Gap
BPS = Bangs per Second
ECAP = Energy Stored in Single Capacitor per Bang (Joules)
RCAP = Internal Real Resistance of a Single Capacitor (Ohms)
RPRI = Primary Circuit Equivalent Resistance (~3 Ohms)
""""""""""""""""
You will note the Wo is indeed directly proportional to BPS ;-))  So all 
the tables and charts and theory are correct.  Just a little mind goof on 
my part 0:-))

Thanks!!  I learned something today *:-))

Cheers,

         Terry

----------------------------------
Hi Jimmy,

Oh gee, your right ;-))  It's a pulsed application...  The current and 
power are linear...

I think I built the squared thing into power4 and MMCcalc...  Time to go 
back and fix things...

Thanks for pointing this error of mine out.  It is a big one!!

Cheers,

         Terry

At 06:21 PM 7/4/2003 -0700, you wrote:
>Hmmm... but how do it know ;-)?
>The current flowing *in each burst* is completely
>independent (assuming fixed firing voltage) of the
>break rate, so how does one burst know to lose more or
>less energy? It seems to me that identical
>circumstances would lead to identical loss ;-)).
>
>Does the first burst waste a different amount of
>energy because it doesn't know the real break rate
>;-))?
>
>Why do you say that the RMS current is 10x? I agree
>that the RMS is  the "equivalent" AC current to a
>fixed DC current. To find the RMS you square, average,
>and then take the square root. After squaring and
>averaging, the result for 1000 BPS will be 10 times
>the answer for 100 BPS, because the squaring affects
>them both the same. After taking the square root, the
>ratio droppes to SQRT10 right???
>
>BTW- Thanks for being such a cool moderator guy ;-))
>
>--- Tesla list <tesla-at-pupman-dot-com> wrote:
> > Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>
> >
> > Hi Jimmy,
> >
> > At 01:12 PM 7/4/2003 -0700, you wrote:
> > >Hi Terry,
> > >
> > >I understand the I^2R thing, but isn't the RMS
> > current
> > >only square root of 10 times as much? After you
> > square
> > >and average it, it would be 10 times, but you still
> > >have to take the square root, right?
> >
> > Nope.  RMS is just the "equivalent" AC current to a
> > fixed DC current.  If I
> > run 10X the BPS rate, I draw 10X the current and get
> > 100X the heat loss
> > across a resistor.
> >
> >
> > >If it really is 500 watts lost with 1000 BPS, then
> > the
> > >amount of energy lost per pulse is 0.5 joules. If
> > you
> > >lose 5 watts at 100 BPS, then the energy lost per
> > >pulse is 0.05 joules. What would cause the increase
> > in
> > >loss per break?
> >
> > It's not linear.  "I^2"  The graph is a steep ski
> > slope.
> >
> > >We are talking about the same energy
> > >per pulse right?
> >
> > Yes.  Power = current squared times resistance.
> > Double the current, four
> > times the power.  Triple the current, nine times the
> > power...
> >
> >
> >
> > >--- Tesla list <tesla-at-pupman-dot-com> wrote:
> > > > Original poster: "Terry Fritz"
> > <teslalist-at-qwest-dot-net>
> > > >
> > > > Hi Jimmy,
> > > >
> > > > Suppose we have 10 amps RMS at 100 BPS and our
> > caps
> > > > are 0.05 ohm of
> > > > internal resistance (typecal for a 15/60).  From
> > > > P=I^2R the power lost as
> > > > heat in the caps is 5 watts.  Now lets hook it
> > to a
> > > > pole transformer and
> > > > run it at 1000 BPS for 100 amps RMS:
> > >Use my signature  Allow HTML tags [Preview]
> > >
> > > >
> > > > P = I^2 x R  == 100^2 x 0.05 = 500 watts.
> > > >
> > > >   The array can run all day at 5 watts.  But 500
> > > > watts it will die
> > > > fast!!  Probably like 15 seconds.
> > > >
> > > > Cheers,
> > > >
> > > >          Terry
> > > >