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Re: MMC cap bank



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

Hi Gerry,

The way I was thinking about it was that the energy
lost must be the same per bang, so the RMS must be
SQRT10x.

When I went to justify it, I kept the time constant
but with a higher sum of squares. It's just another
way of looking at the same thing.

--- Tesla list <tesla-at-pupman-dot-com> wrote:
 > Original poster: "Gerry Reynolds by way of Terry
 > Fritz <teslalist-at-qwest-dot-net>"
 > <gerryreynolds-at-earthlink-dot-net>
 >
 > Hi Terry and Jimmy,
 >
 > I have to admit this issue hooked me too.  It seemed
 > that you both were
 > right and there is this apparent paradox.  I think
 > whats happening here is
 > the average current goes up 10x.  But the RMS
 > current goes up SQRT 10.  If
 > you perform the RMS integral:
 >
 >      RMS = sqrt ( (sum of the squares) / T)   where
 > T is the period the
 > integral (sum)
 >
 >                     is
 > taken over
 >
 > If the current profile per pulse doesn't change,
 > then the "sum of the
 > squares" for one pulse interval doesn't change
 > between 100BPS and 1000BPS.
 > The only thing that changes between 100BPS and
 > 1000BPS is that T gets 1/10
 > as big.  Therefore SQRT (1/T) is the SQRT of 10.
 >
 > Any comments
 >
 > Gerry
 >
 >
 > ----- Original Message -----
 > From: "Tesla list" <tesla-at-pupman-dot-com>
 > To: <tesla-at-pupman-dot-com>
 > Sent: Friday, July 04, 2003 3:03 PM
 > Subject: Re: MMC cap bank
 >
 >
 >  > 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
 >  > > >
 >  > > > At 09:07 PM 7/3/2003 -0700, you wrote:
 >  > > >
 >  > > > > > Second, keep the BPS rate LOW.  Capacitor
 >  > > > current is
 >  > > > > > directionally
 >  > > > > > proportional to BPS.  3000 BPS has 10X
 > the
 >  > > > current
 >  > > > > > of 300 BPS and 100X the
 >  > > > > > cap heating!!!
 >  > > > >
 >  > > > >Hi Terry,
 >  > > > >
 >  > > > >I "think" the heating should only be 10x as
 > much.
 >  > > > If
 >  > > > >it were 100x then each burst would have to
 > waste
 >  > > > 10x
 >  > > > >the power, and I don't see how that is
 > possible.
 >  > > > The
 >  > > > >average current will be 10x as much, but the
 > RMS
 >  > > > will
 >  > > > >only be SQRT10x as much because the duty
 > cycle is
 >  > > > >greater. Check me on this because it's
 > getting late
 >  > > > >for me ;-)).
 >  > > > >
 >  > > > >=====
 >  > > > >Jimmy
 >  > > >
 >  > > >
 >  > >
 >  > >
 >  > >
 >  > >=====
 >  > >Jimmy
 >  >
 >  >
 >
 >


=====
Jimmy

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