dump load

[brianschanafelt]

Hey everybody

I just bought a c-40 charge controller/diversion control. I have looked all over to find ideas on building a dump load, but i cant find how to figure how many amps i am pulling from a element. in my case four. i have four small toaster oven elements connected, but am not sure how many amps it is pulling. Witch i think it is nearly not enough. I have the controller set for 14.0v bulk, 13.5v float. And i am going way over 14.0v. so i have hooked up a headlight also for the time being until i build a dump load that is worthy of dumping 30-35A max.

I am currently charging with a Hugh Piggott 8' axial flux witch on a windy puts out 25A easy. with 5 deep cycle 12v marine batteries

(ya i know there crappy but it is a start)

Just trying to find something that is easily available that would draw 30A. want to put a heating element in the water heater, but i am preparing to move to colorado in a month or so. if anybody could point in the right direction it would be much appreciated

thanks everyone

[wdyasq]

"if anybody could point in the right direction it would be much appreciated"

It will be best if you learn Ohm's law, V = IR , and its uses.

Ron

[richhagen]

I have seen toaster elements of different designs, some that run 120VAC to each individual run of elements, and some that are connected with two outer elements in series and those two connected in parallel with a two sided middle element.   In all cases they would be unsuited for 14V because of the low amount of power that they would draw.  I use the elements with two in series and another modified element as my dump load for my solar panels at 48V.  They draw about 580 Watts or so at 55 Volts or so. If I were to connect them to a 12V system then since the Power consumed is equal to the square of the voltage divided by the resistance, I would only draw one sixteenth as much power.  For use on a 12V system I would have to seriously modify the element, basically making for strands of the nichrome wire only one fourth as long in order to draw the same amount of power out.  Rich

[Capt Slog]

I used an old convector heater for my elements.  

To dump that much current, you will need to share it over a few resistors (wires/elements), as you're doing at the moment.

for example..  V/I = R  so 14 / 30 = 0.466 ohms.  Rather than having 420 on that single element (14 x 30 = 420 ), split it into 6 elements for example.

If we round up the 0.466 to 0.5 then you would need 6 elements of around 3 ohms each in parallel, (1/R = 1/r + 1/r + 1/r + 1/r + 1/r + 1/r)

But to find the lengths of elements you need.......

Hook up a power supply with the voltage you want with an ammeter.

Connect one end of an element to the power and then work along the element with a probe/crocodile clip untill the ammeter indicates 5 amps (30 / 6 ).

This is just an example, I'm not saying it will work but it gives you the method.  I use 6 elements to dump around HALF what you're planning, because I don't like things getting too hot. I had Ghurds help with this when I did mine, I'm passing on what he taught me. :-)

.

[jmk]

 This took the confusion out of it. Thanks for posting it.