Re: Joules per bang...

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

Original poster: Terry Fritz <teslalist-at-twfpowerelectronics-dot-com>

Hi,

At 06:46 PM 1/19/2004, you wrote:


>Tesla list wrote:
>
>>Original poster: DRIEBEN-at-midsouth.rr-dot-com
>>Scot,
>>
>>Apparently, your math is off. 5.625 joules = 5.625 watts, therefore,
>>if we fire 5.625 watts per bang at 120 a second, that yields a much more
>>conservative 795 watts per second.

I think this error was corrected in another post but...

5.625 joules x 120 BPS = 675 joules/second = 675 watts.

>>.........
>
>
>hmmm   now Im lost again and I thought I had this figured out...    ok as 
>I see it
>
>J= the amount of energy a capacitor of a certian size and voltage can 
>produce...

J = 1/2 x C x V^2


>W= the amount of power dissapated over a time interval...
>and 1J will produce 1W in 1 second
>am I right so far??

Yes, in our case,

W = J x BPS


>okay   if we have our 2000uF cap charged to 1000V   that should give us 
>1000J...

J = 1/2 x C x V^2  =  1/2 x 2000e-6 x 1000^2  =  1000 joules

>and if discharged thru a resistor (ummm 1000ohms?estimating here...) that 
>will drain the cap in 1 second approximately (ignoring the logorithmic curve),

RC = R x C = 2000e-6 x 1000 = 2 seconds.  The cap usually takes 5 x RC to 
discharge.  How about a 100 ohm resistor to get the time you want here.

>should produce 1000W of  power/heat/energy... in other words we need a 
>1000W  1 ohm resistor to handle the current....safely...

You have 1000 joules of energy.  "heat" is measured in joules too but, lets 
not think of heat since we have enough problems already ;-)).

Since we use 1000 J in 1 second, we used 1000 watts.  Most of it early in 
the second and just a little near the end.


>if discharged thru a near 0 ohm resistor taking about .0001 seconds that 
>should produce about 10MW of power in that time duration...peak power...

Still 1000W as measured over 1 second.  There is a very fast peak, but on 
average over our second it was still 1000W

Peak power in this case is V^2 / R = 1000^2 / 0.0001  =  10,000,000,000 
Watts peak.


>now if we were to do this 10 times a second...  there should be 10KJ of 
>energy used   ergo 10KW avg power per second
>
>am I still good here?

Yes,

Cheers,

         Terry


>Scot D
>
>
>