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,

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?

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? We are talking about the same energy
per pulse right?


--- 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

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