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RE: Capacitor Sizing for Power
Original poster: "Loudner, Godfrey by way of Terry Fritz <twftesla-at-qwest-dot-net>" <gloudner-at-SINTE.EDU>
Hi Matt
All your first equation says is that 120 caps with a capacitance of 0.2uF
placed in parallel will be able to store 2700 joules of energy at a
potential of 15,000 volts. This is already enough to lock in the charging
time of a cap, say 5RC = 5(15,000volt/0.240amp)(0.0000002F) = 0.0036sec. =
1/278sec. You are asking for all 120 caps to charge and discharge in 1sec.
This is the same as asking a single cap to charge and discharge in 1/120sec,
or a charging time of 1/240 sec. Now1/240sec. is about 116% of 1/278sec. So
you want to place a 0.2uF cap in a tank circuit that requires 1/278sec. to
charge, but you are asking it to forget all about that and charge in
1/240sec.
Godfrey Loudner
> -----Original Message-----
> From: Tesla list [SMTP:tesla-at-pupman-dot-com]
> Sent: Wednesday, July 18, 2001 5:54 PM
> To: tesla-at-pupman-dot-com
> Subject: Capacitor Sizing for Power
>
> Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <Mddeming-at-aol-dot-com>
>
> Hi All!
>
> I just started working up the requirements for my 15/240 NST farm
> coil and decided to try the P=rate*0.5*C*V^2 method. using a SRSG -at-120
> BPS. I
> have read that this is a "good ballpark figure" for minimum cap size
> needed
> to handle the power.
> Now 15KV*240mA= 2700 W Substituting gives us
> 2700=120*0.5*15000^2*C or
> 2700=60*15*15*1E6*C which reduces to
> 3/15*1E-6=C or
> 0.20uF=C = 200nF
>
> Using the resonance relationship to calculate the minimum capacitance:
> XL=2*pi*f*L=Xc=1/(2*pi*f*C)
> if R<<XL then Z=E/I ~XL
> XL=15000/.240=62500=377*L or L=166.8Hy
> Xc=1/(2pi*f*C)
> 62500=1/377C or 2.35625*1e7=1/C
> 4.244*1e-8=C or,
> C=42.44nF
> A difference of 470% is not close in my ballpark. Even doubling the value
> for
> LTR gives an error >230%
> Am I missing something obvious or am I just brain dead this week?
>
> Matt D.
>