The Best Cell Ive seen

[BoyntonStu]

What part of that calculation do you not agree with? The shortest link of a chain does not define the total length of the chain.

My calculation of 153.86 represents the true working surface area of the cell: all negative, gas-producing surfaces.

In your calculation, you've calculated the actual length of the .5 tube, not the effective length. You've also (mis)calculated the inside of the .5 tube (.40 dia.), which does not produce gas.

To recap, gas is produced on the o.d. of the .50" tubes and on the i.d. of the 1.00" tubes.

"pi x .5 x 42" for one side"

That's all that needs to be calculated.


It is not where the gas is produced that defines the amps/area of a cell.

The limit is the smallest plate or tube area.

Larger plates spread the current out and produce no more gas than the smaller plate.

BoyntonStu

[HomeGrown]

"pi x .5 x 42" for one side"

That's all that needs to be calculated.


It is not where the gas is produced that defines the amps/area of a cell.

The limit is the smallest plate or tube area.

Larger plates spread the current out and produce no more gas than the smaller plate.

BoyntonStu

If you have your aforementioned example of a 6x6 & 1x1 cell, it is agreed that the cell area is defined by the 1x1 plate. However, if in this cell you add a 2x2 plate on the opposite side of the 6x6, is not your total cell area now 5 sq. in.? Applying your logic, the cell would still be 1x1.

You are ignoring the other half of the cell, which is the 1.00 / .75 tubes. Also, why are you calculating the full length of the .5" tube?

[BoyntonStu]

If you have your aforementioned example of a 6x6 & 1x1 cell, it is agreed that the cell area is defined by the 1x1 plate. However, if in this cell you add a 2x2 plate on the opposite side of the 6x6, is not your total cell area now 5 sq. in.? Applying your logic, the cell would still be 1x1.

You are ignoring the other half of the cell, which is the 1.00 / .75 tubes. Also, why are you calculating the full length of the .5" tube?

We have to define what we mean by cell area.

My definition is the area that limits the current.

However, if in this cell you add a 2x2 plate on the opposite side of the 6x6, is not your total cell area now 5 sq. in.?

Yes, we agree here as a simpler example below.

Place a 1x1 plate centered on each side of a 6x6 plate.

Connect - + -

I calculate 2 sq inches because you have 2 cells.

In your 3 tube cell if you do it - + - I believe that you add together the outer areas of the .5 tube and the .75 tubes.

The .5 tube 'sees' the .75 tube and the .75 tube 'sees' the 1.0 tube.

"Also, why are you calculating the full length of the .5" tube?" My bad. Should be 5.3"

So here goes: pi x .5 x 5.3 x 7 + pi x .875 x 5.3 x 7 = Your previous result Check!

Sorry, I misunderstood you before.


BoyntonStu

[HomeGrown]

Glad that's cleared up!

[HomeGrown]

Homegrown Series or parallel How do you get power to the electrodes

Sorry Boltazar, I missed your question.

The cell is parallel, and the actual power connection configuration is still in the works. I will post pics when it's done. I'm playing around with a couple different ideas.