Re: MMC cap bank

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

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
 >
 >