Re: MMC cap bank

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

Original poster: "jimmy hynes by way of Terry Fritz <teslalist-at-qwest-dot-net>" <chunkyboy86-at-yahoo-dot-com>

Hi Terry,

One of the reasons it's taken me so long to get my
DRSSTC working is that I make a lot of mind goofs like
that. I wasted a day of working on it because I
thought the bridge was shorted. It was just taking the
DMM along time to charge 10,000 uf ;-)).

It looks to me like that equation assumes 0%
efficiency. Wouldn't the real dissipation be around
half that?
--- Tesla list <tesla-at-pupman-dot-com> wrote:
 > 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
 > > > > >
 >
 >


=====
Jimmy

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