When starting a 4 litre 6 cylinder gas engine..using math, how long will an 80 farad supercap array deliver
sufficient volts(above 12.4) before it falls below that??The caps would start at 13 v and deliver up to 250 amps..??

 

Energy stored in a cap is 1/2 x C x V x V

@ 13 V ................... = 1/2 x 80 x 13 x 13 = 6760 J

@ 12.4 ..................= 1/2 x 80 x 12.4 x 12.4 = 6150 J

You have to subtract one figure from the other to get 610 Joules .... this is the energy released when discharged from 13 to 12,4 ....

@ 250 Amps and 12.7 V energy is consumed at the rate of 250 x 12.7 = 3175 Joules per second

So the cap alone would only crank the engine for less than 0.2 secs !!!

In practice the battery is usually helping , and cranking still occurs down to about 10 V releasing more energy from the cap.

 

The cap voltage drop is ΔV= (i*t)/C so for a 0.6V drop the time is:
t = ΔV*C/i = 0.6V*80F/250A = 0.192 sec
(as oz93666 calculated above using energy).

If you allow it to drop to 10V, then the time would be:
3V*80F/250A = 0.96 sec.

 

The cap voltage drop is ΔV= (i*t)/C so for a 0.6V drop the time is:
t = ΔV*C/i = 0.6V*80F/250A = 0.192 sec
(as oz93666 calculated above using energy).

If you allow it to drop to 10V, then the time would be:
3V*80F/250A = 0.96 sec.

That seems a lot simpler your way , crutschow ... I'm just trying to get my head around how you did it!

 

thats not much time. definitely need to double that..

 

I'm just trying to get my head around how you did it!

From basic definitions.
The definition for capacitance is C = Q/V.
Q (charge) is current times time or i*t, thus C = (i*t)/V.
Rearranging gives t = (C*V)/i