Re: Ok, what is an LTR cap.

[Terry Fritz <twftesla-at-uswest-dot-net>]

O0ps!  You may have wanted teh "resonant" size instead of the full power LTR
size I gave....

C = I / (2 x pi x F x V)

Where:

C = the resonant capacitance value in Farads
I = The transformers current rating in amps RMS
pi = 3.14159...
F = The line frequency (50 or 60 Hz)
V = The transformers rated voltage (volts RMS)

So....

C = 0.694 / (2 x pi x 60 x 14400)  ==  127.84nF

Cheers,

        Terry



At 08:17 PM 8/24/00 -0500, you wrote: 
>
> Hi Terry, Mark, 
>
> Tesla list wrote: 
>>
>> Original poster: Terry Fritz <twftesla-at-uswest-dot-net> 
>>
>> Hi Bart, 
>>
>>         A 14.4kV 694mA pig should deliver 10000 watts.  If we divide 10000
>> watts 
>> by 120 BPS we get the energy per bang as 83 joules.  The peak voltage is 
>> 14.4 x SQRT(2) = 20.4 kV peak.  So what size cap when charge to 20.4 kV 
>> will hold 83 joules?
>
> 14.4kV x .694A = 9,994W (10kva). The peak is 20.4kV. But, 83 joules? How long
> do you think it's going to take to charge a .401uF cap? Well, 41.8ms. At a
> BPS of 120 (8.33ms), it would take more than one revolution using an 1800 RPM
> sync gap with 1 eletrode. Well, slow the RPM's down and yes, but this isn't
> realistic nor is this "impedance matched" to the transfomer. To "impedance
> match" the cap to the rated transformer, I've been taught 1/(2*pi*Z*F), or in
> this case, 0.128uF. I don't understand how .401uF (or .351uF) sounds perfect?
> Getting really confused with this concept, and that's how I see this, as more
> of a concept that's unrealisitic. Of course, I guess if we ran at 120 BPS
> anyway, we would be down to 33 joules due to the cap only charging to 1 time
> constant or so, but I'm not sure of the point here. How many coils can handle
> 33 joules? 
>
> (confused with this in sunny CA), 
> Bart 
>>
>>   
>> E = 1/2 x C * V^2  or  C = 2 x E / V^2 
>>
>> C = 0.401uF  Assuming losses eat some of that up the 0.351 number sounds 
>> perfectly correct. 
>>
>> However, pole pigs can be ballasted in all kinds of ways so you would have 
>> to look at the total system to be sure the the equations would apply to a 
>> pig system. 
>>
>> Cheers, 
>>
>>         Terry 
>>
>> At 06:35 PM 8/24/00 -0500, you wrote: 
>> Hi Mark, Terry, 
>>
>> Tesla list wrote: 
>> Original poster: "Mark Broker" <broker-at-uwplatt.edu> 
>>
>> > Actually, PIG's are STR's (smaller than resonant). Typically, coilers use
>> a 
>> > 10kva, 14.4kv pig's. Resonant caps are 0.128uF (from memory). Obvisouly 
>> HUGE! 
>> > We typically use smaller than resonant caps (STR's) on our systems. NST's 
>> > however (and OBIT's), can use LTR's. Could you imagine the joules for a
>> pig 
>> > using an LTR? I see smoke, fire, wooosh!!! Well, for most of us, not Greg,
>>
>> > Bill, Hull, etc... or you Chris (yet?). 
>>
>> Actually, Terry's MMC Calc gives me .351uF. 
>> For a 240/14.4kv, 694mA, 20,749 ohm pig? Terry is .351uF correct? I though 
>> a matching reosnant cap size should be 0.128uF. Maybe my math is wrong 
>> here. Someone please clarify? 
>>
>> Bart 
>>  Chris and I have a design, and are 
>> attempting to gather enough caps (912 of Terry's, or any other combo that
>> will 
>> work).  Did you know that there's a world-wide tantulum shortage?  I even 
>> have the 
>> PC board mask already laid out (I think). 
>>
>> I thought that it was really 13.8kV nominal? 
>>
>> Mark
>