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The Martin Fleischmann Memorial Project is a group dedicated to researching Low Energy Nuclear Reactions (often referred to as LENR) while sharing all procedures, data, and results openly online. We rely on comments from online contributors to aid us in developing our experiments and contemplating the results. We invite everyone to participate in our discussions, which take place in the comments of our experiment posts. These links can be seen along the right-hand side of this page. Please browse around and give us your feedback. We look forward to seeing you around Quantum Heat.

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To address a lot of questions about hydrogen safety or leaks in the cell, let's talk about the amount of hydrogen we are filling in the cell.

The glass tube dimensions are:

- Outer diameter: 40.0 mm
- Inner diameter: 35.0 mm
- Length: 300.0 mm

So, the internal volume of the cell is: π⋅r2⋅L = 3.14159 * 0.1752 * 3.0 = 0.2886 liters

Then the mica boards are: 3.2 mm x 11.5 mm x 300.0 mm
So the volume of each mica part is: 0.032 x 0.115 x 3.0 = 0.01104 liters
The total volume of the 3 parts of mica is : 3 x 0.01104 = 0.03312 liters

Finally the stainless steel rod in the center of the cell is: ø6.0 mm x 300.0 mm
The volume of the rod is: π⋅r2⋅L = 3.14159 * 0.032 * 3.0 = 0.0085 liters

Thus the available volume inside the cell is: 0.2886 - 0.033120.0085 = 0.2470 liter = 2.4710-4 m3

The ideal gas law will give us the number of mole of hydrogen in the cell:

PV = nRT  =>  n = PV / RT

When we load the cell with pure hydrogen at ambient temperature (295 °K) at 3.5 bars we have:

n = 3.5105 Pa * 2.4710-4 m3 / (8.3144621 J·K−1mol-1 * 295 °K) = 0.0352 mol

The atomic mass of dihydrogen is: 2.0159 gram / mol

So at maximum, we are filling the cell with: 0.0352 * 2.0159 = 71 milligrams of H2


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0 #5 mimarob 2013-02-04 11:29
Actually a mechanic would probably use this formula:


0 #4 Ddd 2012-12-25 07:03
Even a mechanic should know a cylinder volume is a length times SQUARED radius times Pi, of course...
0 #3 Dale G. Basgall 2012-11-21 14:31
Should be 10500CC not mm, anyway it is confusing when you have been taught in shops to bore cylinders and we used to just go bore times stroke to figure volume.

I apologize for being a little questionable on this one it's just I had always used this formula and thought it would be appropriate for this one here. I would appreciate being told if I am wrong about this because I am also working with reactor volumes.
0 #2 Dale G. Basgall 2012-11-21 14:25
I guess math is done alot different in science than in general mechanics when calculating volume of a cylinder. As a mechanic I would have simply gone bore ID = 35 mm and length is 300mm so 35 x 300 = 10500 cubic mm or 1.05 litre.

Like I say I am just a mechanic so if you guys do things different in science cool but am I way wrong here on this one?
0 #1 Pekka Janhunen 2012-11-20 15:33
Two questions:
- Does daylight have access to the room, I guess not?
- How quickly was vacuuming/press ure reduction done, in comparison to Celani? If pressure is reduced too fast, maybe hydrogen outgasses violently, perhaps it kills the wire.

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