need a new design

[SeaMonkey]

Flash Bang,

With the Low Voltage Cell plate configuration the total current is shared by all the commonly connected plates. When the plates are interleaved, those plates which are between plates of the opposite polarity will be conducting on both sides to the opposite plates.

Therefore: Your plate size of 2.5" X 12" yields 30 square inches per side.

Assuming you have a total of 12 plates; 6 Anodes and 6 Cathodes interleaved, spaced at 1/8 inch; the total current your cell would be able to support at the 1/4 Ampere per square inch maximum is 82.5 Amperes.

Your total conductive surface area for the Anode group is 330 square inches.

Your total conductive surface area for the Cathode group is 330 square inches.

330 square inches X 0.25 Amperes = 82.5 Amperes.

That is the advantage of the low voltage cell. You take into account the total surface area of the conductive sides of your common plates.

I tried to explain the calculation process in the earlier post with an example of one plate and did not sufficiently clarify the procedure to find the total surface area of all the common plates; which is additive.

[SeaMonkey]

Flash Bang,

In answer to your question specifically: How many plates would be required for a 20 Ampere cell?

20 Amperes of current flow through a low voltage cell requires a minimum of 80 square inches of total Cathode surface area and 80 square inches of total Anode surface area.

For your 2.5" X 12" plates you would need 2 Cathode plates and 2 Anode plates. A total of 4 plates.

This would result in 90 total square inches of conductive surface area for the Cathode group; and 90 total square inches of conductive surface area for the Anode group. This surface area will support a maximum current of 22.5 Amperes.

Since DC to DC coverters are available which will provide 40 Amperes of current to the low voltage cell (requiring about 11 Amperes from the 14 Volt automotive electrical system), I'd shoot for a cell that would accomodate the 40 Amperes.

With your plate size of 2.5" X 12" you would need a total of 8 plates; 4 Cathode plates and 4 Anode plates.

Each plate group would have 210 square inches of conductive surface area and thus would be able to receive up to 52.5 Amperes maximum.

[FlashBang]

[SeaMonkey]

[SeaMonkey]

[mrgalleria]

[FlashBang]

[SlickGixxer]

[chemelec]

[a3holerman]

[chemelec]

[SeaMonkey]

[FlashBang]

[SlickGixxer]