Plate Area & amps (or watts) relationship??

[Roland Jacques]

Ive read a post, cant find it now. It suggested that there is a ideal surface area of plates to amps. Anymore amps would accelerate corrosion of SS.

Does anyone know that formula?

2nd part
I wanted to know if that plate "Area" included just one side of each electrode, and both sides of each neutral plates?

If each side of a plate is 12" of area and arranged +nnnnnn- would be 14 x 12 =168" of area.

3rd part
If i slag blast the plates for a 300% more surface area. Does this rule apply to the 12" or the larger 36" per plate area?

[Buster]

It's supposed to be 2-4 square inches per Amp, per side of each plate in a series.
So, in your example of 12 inch square plates in a +nnnnnn- configured cell, you only count one side of 1 plate, which is 12/2 to 12/4 =between 6 and 3 amps for the complete series.

I don't know exactly why this figure was arrived at, but I imagine that it's to do with efficiency/temperature/plate life etc.
It may not be a good recommendation if steam input is useful to the engine as some suggest, if you get my drift. It still needs some more investigation.

Not sure about the slag blasting. In theory it seems like it should increase the number of potential amps but in practice it probably would be adding to heat etc as it would still be the same amount of metal carrying the current???

[Roland Jacques]

That's strange that the number of plates dont factor into the equation somehow.

Also Id think the thicker the plates the more resistance equaling more heat. the thinner the less resistance.

The steam thing is a entirely different subject...

[Buster]

I can understand why the per plate figure is used, as the amps are the same moving through each of the plates no matter how many there are.

The thinner plates is an interesting point I never considered, and you're probably right about that being a factor. I imagine the equation mentioned was based on a particular plate thickness.
It's all just a guide, I suppose, and there's nothing like testing out what actually works best with what we have.

[Philldpapill]

Roland, the number of plates doesn't factor in because you have the same amount of amps passing through those surfaces as well. You need to look at it as "how many electrons per second are jumping on/off of this particular section of plate area".

If the plates were in parallel, then sure - you have more amp draw and more surface area, so the equation still holds.

[biggy boy]

What about plating?
It's claimed that plating the plates with say Nickel or titanium or Platinum,
Reduces the resistance of the plate!
But plating is just adding a layer on the surface of the plate!!
So Yes the new plate surface may have a lower Resistance, due to the plating.
But the current HAS to travel through each plate and on to the next, how is this going to help.
The current still has to travel through the Stainless steel portion of the plate!
I can't see how the plating lowers the resistance of the stainless, beneath the plating??
Current traveling across the plates surface OK I can see that, through it ??

I can see it would make it easier for the current flow to transition (transfer) from the electrolyte to the plate, due to the plating!
Maybe that is the point!!!


Glen

[Roland Jacques]

I think i partially understand this rule, but i really cant get my head around the why's.

I understand the amps going through all the plates at the same amount. But I guess its the role of the plates I'm not to sure about. One part that confuses me is the fact the amps are going through wires, bolts, & terminals with a relatively small surface area, but the area for the plates need to be many times greater.

1. I would think current density rule would have to be primarily concerned with what the electrolyte is. Or more accurately how conductive the electrolyte is. I would think that KOH at 5% would require more plate area than KOH @ 33%. Is this the case?

2. Does the current density rule only apply to the negative and positive electrodes. I cant see how this rule would apply to neutral plates.

In other words, if i used nano coated electrodes, i could push many times more amps through nano plates because they have many times more surface area. A 6x6 nano plate could handle 80 amps 14 volts no problem. (so I hear)
2.a. Couldn't we just use SS neutral plates with the Nano electrode?
2.b. And wouldn't thinner neutrals be better than thicker ones? less resistance...

[Philldpapill]

To be honest, plating plates is really hype. The IDEA is that it increases surface area... BUT that really isn't what's important. You don't really generate most of the heat at the plate boundry anyway. Most of the heat is generated by the resitance of the water itself.

Let's say you have a plate with 10" squared. If the roughness gives the total area 20" squared, you STILL only have 10" squared of water to conduct through... You may have a lower resistance at the water-plate interface, but the bulk resistance of the water/electrolyte is the same.

However, this is only theory(a pretty sound theory), so real world expirements would have to be done to verify it. I'm expecting MAYBE a 1-2% decrease in power loss........

[IM2L844]

Don't get current density (amps per cm squared) mixed up with power density (watts per cm cubed) the main reason for limiting current density is to limit the rate of plate deterioration which not only deteriorates your plates (obviously) but adds that material to the electrolyte causing all kinds of other problems. Of course the two are related in that it requires a certain minimum amount of potential (voltage) to achieve a particular minimum current density, but, as I think Phil was trying to point out, excess power density is a more reliable indicator of whether or not there will be a lot of heat produced, but it can be a little more difficult to calculate...now, I've lost track of the point I was trying to make. Oh well. Did that make any sense to anyone besides me?

[Roland Jacques]

To be honest, plating plates is really hype. The IDEA is that it increases surface area... BUT that really isn't what's important. You don't really generate most of the heat at the plate boundry anyway. Most of the heat is generated by the resitance of the water itself.

Let's say you have a plate with 10" squared. If the roughness gives the total area 20" squared, you STILL only have 10" squared of water to conduct through... You may have a lower resistance at the water-plate interface, but the bulk resistance of the water/electrolyte is the same.

So then doesn't this make the most important factor in determining Current density (or power density) the conductivity's of the electrolyte more than the plate area? So the more conductive your e-lyte the smaller the plate can be. right?

Don't get current density (amps per cm squared) mixed up with power density (watts per cm cubed) the main reason for limiting current density is to limit the rate of plate deterioration which not only deteriorates your plates (obviously) but adds that material to the electrolyte causing all kinds of other problems. ?

So would the rate of deterioration change if i slag blasted the electrodes tripling the surface area? And would thinner neutral plates be better than thicker ones.
Shouldn't they be looked at defferently, electrodes and neutral plates?