URGENT! WATER AS FUEL DISCOVERY FOR EVERYONE TO SHARE

[alpeko]

For demios:

      Tc =0.5×Cload ×Vcharge×Vrated/Ppeak

Calculating Charge Time (Tc)
Using the peak power rating of the power supply, charge time Tc can be calculated using equation 1 below.
Where;
Tc - load charge time in seconds
Ppeak - unit peak power rating in J/sec
Cload - load capacitance in Farads
Vcharge - load charge voltage in volts
Vrated - power supply rating in volts
In many applications the load charge voltage (Vcharge) and the power supply rated voltage (Vrated) will be the same, but it is important
to use equation 1 if the load is charged to a voltage less than the power supply rating, otherwise an incorrect charge time will be calculated.

Best reg.

Alpeko

[alpeko]

Hi all,    Down is litle theory from me.  For explanation only.


Which is the best equation to use?
Want to know: load capacitor charge time.
Already know: load capacitance, the charge voltage, the power supply rated voltage, power supply peak power rating.
Equation to use: 1
Want to know: supply peak power required for my application.
Already know: load capacitance, the charge voltage, the power supply rated voltage, desired charge time.
Equation to use: 2
Want to know: supply average power required for my application.
Already know: load capacitance, the charge voltage, the power supply rated voltage, and the repetition rate.
Equation to use: 3
Want to know: maximum repetition rate possible in my application*.
Already know: load capacitance, the charge voltage, the power supply rated voltage, and the supply average power rating.
Equation to use: 4
Want to know: peak charge current for a given supply.
Already know: the power supply rated voltage and peak power rating.
Equation to use: 5
Want to know: load capacitor charge time.
Already know: the load capacitor charge voltage, the load capacitance,
peak charging current.
Equation to use: 6



Tc =0.5×Cload ×Vcharge×Vrated /  Ppeak         . . . . . . . . . . . equation 1     




Ppeak =0.5×Cload ×Vcharge×Vrated/Tc              . . . . . . . . . . equation 2             



Pav =0.5×Cload ×Vcharge×Vrated×Reprate                   . . . . . equation 3



Reprate= Pav  /  0.5×Cload ×Vcharge×Vrated          . . . . . . . . equation 4



Icharge= 2×Ppeak /Vrated                              . . . . . . . . . . . . . equation 5




Tc=Cload×Vcharge/Icharge                             . . . . . . . . . . . . equation 6


            Alpeko            

[demios]

Thanks again Alpeko !!! 

[gmeast]

Hvala Alpeko, sada mi je jasnije. Usput, da li cu brze napuniti kondenzator sa niskom voltazom i visokom amperazom ili obrnuto? Tj. kako cu brze da ga napunim?

Hvala na odgovoru!

Thank you for explanation Alpeko! It makes much more sense now. By the way, do I need higher current in order to fill-up the capacitor or higher voltage? Which way do I fill the cap faster?

Thank you for the answer!

Hi demios,

Charging a capacitor is a POWER issue.  You can have all the current in the world but a cap won't charge without a voltage potential to drive the current.   You need a voltage potential and current generating capacity to keep up with the charge rate.  They go together.  Capacitor characteristics are expressed in "Joules"  and defined as 1/2C X V^2 or ' the voltage squared times the capacitance divided by two.  It is one of electrical power's mechanical counterparts.  Kinetic energy is defined similarly ... 1/2 M X V^2 or 'the velocity squared times the mass divided by two.  A very similar statement goes for energy in the famous E= M C^2 or 'energy equals Mass time the speed of light squared' ... everything is the same as everything else the closer you look.

Greg 

[alpeko]

Hi Greg,

Yes it is theory.Please give us more practical experiences about increasig mpg, or how much you improve mpg.Which method is practical for water injection ?Did you test which point is better for conect protecting diodes.On side spark plugs,or on distributor.And how many diodes need.In my test on desk I conect only 5-8 pcs  1N 5408 without problem ?? I think that one diode per kilovolt is enough,but working(on test desk),on couple diodes ? 
If you conect 300V on distributor side did you change HV cables? If you use solid wire HV cables,RFI is much higher ?Did you think about noise supresion circuit?And my last question:How much plasma in practical improve mpg,in percent ??

Thank you again.

Alpeko.