Extract heavy water?

I can't get hold of any heavy water so I want to try extract it from normal water, I only need a drop or two. I was gonna use electrolysis to separate the atoms and increase the concentration if heavy water till it gets to around 70% (min.) but how much of power do I need to pass though the wires to only separate H2O and leave the D2O? (Need the deuterium for a fusor I'm building)

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rickharris1 year ago

If your hoping to use this in a reactor then your going to want rather more than a few drops.

The Germans built a plant to make/extract heavy water during the war at Vemork, Norway,


Looking at the Wikipedia article on "Heavy Water",

the History section mentions electrolysis of water, saying Gilbert N Lewis used this method in 1933, and in the references, I find a link to this paper, here,


and if you want a copy of the full paper, the easiest way to do that is to feed the above URL to Sci-Hub,


Anyway, if I can believe what is written in this abstract, they (Gilbert N Lewis and Ronald T Macdonald) started with 20 liters of water...

Starting with twenty liters of water from an old commercial electrolytic
cell and electrolyzing in four stages until only one‐half of a cubic
centimeter of water remained, this water had the specific gravity 1.073.
Having shown that no large accumulation of the heavy isotopes of oxygen
occurs and assuming that the density varies linearly with the fraction
of H2, 2/3 of the hydrogen in this water is H2.

Note that when I cut-and-pasted that, all the superscript 2s turned into ordinary 2s. Everywhere in that quote where it says H2, it should say H<sup>2</sup>

So, at this point you are maybe wondering, how much electricity does it take to electrolyze 20 liters of water?

Well, every mol of H2O requires 2 mol of electrons.

(20000 g H2O) / (18g/ mol) = 1111 mol H2O

So that requires 2222 mol e-. The next step is to figure out what that is in electrical units, like coulombs, or ampere*hours

(2222 mol e-)*(96485 C/mol e-) = 2.144e8 = 214 400 000 coulombs

(2.144e8 C)/(3600 A*h/C) = 5.955e4 = 59 550 ampere*hours

I dunno. This looks like a lot of ampere*hours. I mean one cell, at 160 A (or 10 cells at 16 A) would take about a year of time, running the cell (or cells) continuously.

Regarding energy and power, I think a reasonable guess for the potential across the cell is about 4 volts. Multiply that by the number of ampere*hours, to get watt*hours.

I mean electricity is cheap, but the amount of time involved looks kind of serious.

That is if I did the math right.

Oops. Wait. I forgot a factor of 24 (hours in a day)

(59 500 A*h)*(day/24 h)*(year/365.25) = 6.79 ampere*year

So it would take a year at like 7 A, or like 2 weeks at 160 A.

rickharris1 year ago
rickharris1 year ago


no personal experience so use information with caution!

65Kw is a LOT of power

and 512 N is a lot of force.