Sciencemadness Discussion Board

D2O enrichment

APO - 29-12-2012 at 13:49

I only have 20g of D2O and I would like more, but it is extremely expensive, so I want to make my own. The method I am interested in is electrolysis, H2O splits at 1.48v and D2O splits slightly higher, if I use exactly 1.48v the H2O will hopefully split leaving the D2O in soulution,then I would add more water to the slightly enriched soulution, and repeat the process. Any more ideas?

neptunium - 29-12-2012 at 16:11

are you trying to get the deuterium out of the heavy water? because unless you have another source of deuterium you cannot make D2O!
enriching heavy water from water (which always contain some ) with this process would take years or an industrial size set up!
its expensive for a reason..is that what you are trying to do?

on second consideration , you might be able to get a little bit by the electrolysis process...its worth trying!

[Edited on 30-12-2012 by neptunium]

phlogiston - 29-12-2012 at 16:23

I don't know where you got it from, but as chemicals go it is really not particularly expensive. Typical price is between 0.6 to 2 US dollar per ml, depending on the quantity you order and special requirements (sterility, impurity testing etc).

It would be awesome if you could enrich some D2O yourself and it would earn you some mad scientists respect but it will never be cost effective. Besides the sheer number of electrolysis (or distillation) steps you need to perform to reach a respectable level of enrichment, I think the first not so trivial problem you need to tackle is how to actually measure the level of enrichment in an home lab.

There have previously been threads on this. Use search.

neptunium - 29-12-2012 at 16:36

I would LOVE to own a mass spectrometer....wouldnt you?

APO - 29-12-2012 at 17:21

I can determine the enrichment by by the increase of density, so I think i'll try the electrolysis process.

neptunium - 29-12-2012 at 17:25

more power to you! what do have to lose? if anything you'll gain a new experience in physics and chemistry!
looks like a win win to me!
keep us posted !

99chemicals - 29-12-2012 at 17:26

http://unitednuclear.com/index.php?main_page=product_info&am...

1$ per ml + shipping.



[Edited on 12-30-2012 by 99chemicals]

APO - 29-12-2012 at 17:37

I want to make my own because 1$ a gram is kinda expensive compared to most of my stuff, and yes I got the D2O off of united nuclear.

neptunium - 29-12-2012 at 17:51

i agree ...i'd like a liter or 500ml of it but 500 bucks or even 675! is way out of my reach!
but what do you want to do with it? i mean there is not many interesting experiments to do with heavy water...and how much do you need?

APO - 29-12-2012 at 19:22

Are you kidding? There are plenty of experiments, like fusion studies, examining DDO vs HHO burn tempature, deuterating hydrocarbons, study of higher energy hydrogen, liquid deuterium rocket fuel, and way more!

For many experiments you need compounds with specific isotopes, especially for nuclear reactions, like bubble fusion, and the fleishman and ponds experiment.

Also I just remebered that under high energy conditions with a mix of water and lithium hydroxide that it will evolve helium and deuterium oxide.

2(H2O,LiOH) x 11(He),9(D2O)

I won't give so many details, because I heard some people on the form haven't been using proper safety precautions.

elementcollector1 - 29-12-2012 at 19:31

"High energy conditions"
As in, gas phase for both compounds, an insanely high temp, .... that's the only way I could even think of lithium hydroxide and water reacting together and *not* making a solution of lithium hydroxide.
"I won't give so many details, because I heard some people on the forum haven't been using proper safety precautions."
We give them advice on how to make such things as white phosphorus, elemental potassium, cyanide, and so much more. There's an entire subforum dedicated to the manufacture of explosives. In short, the danger factor of some k3wl killing him/herself from trying this is about as high as the danger factor for a large amount of the other substances that are popular in the forums right now. Spill ahead on those fusion details!

APO - 29-12-2012 at 20:16

Ok, I'll say a little more but I won't explain the apparatus needed because I don't wanna help people hurt them selves. What I meant by high energy conditions, was that this will only work using a super heated plasma compressed by strong magnetic fields, and being excited by a high voltage. The electrodes need to be in a grid inside another gride as the cathode and anode, also the high energy state is started and maintained by a high voltage. The chemical mixture it 2 parts lithium hydroxide monohydrate, and 9 parts water(never use heavy water). If you use heavy water it will not absorb the resulting neutrons, and will instead keep them constant by moderating them, if you did do that you would need radiation protection. This kind of fusion is unique, because if done correctly it poses no threat of a meltdown, or hazerdous by-products. The rest of the details are classified for safety and financial reasons.

kristofvagyok - 30-12-2012 at 04:18

Quote: Originally posted by APO  
Also I just remebered that under high energy conditions with a mix of water and lithium hydroxide that it will evolve helium and deuterium oxide.

2(H2O,LiOH) x 11(He),9(D2O)

I won't give so many details, because I heard some people on the form haven't been using proper safety precautions.

At first, nuclear reactions looks easy on paper but if you want to make them then it's a bit problematic.

Second: mixing some LiOH and some water, exposing them to a little energy will result a big nothing, you won't get helium and D2O from those.

Third: lithium is a mixture of Li6 and Li7 what are two completely different thing, let's see just their reaction with a neutron:
Li-6 + n -> T + He-4 +4.78 MeV
Li-7 + n -> T + He-4 + n -2.47 MeV

Fourth: If you do not add D2O to the mixture or something what will give off a neutron at elevated temperature/pressure/ect than what will start the reaction what will turn the LiOH to He and T/D2O? Or how will it work? Get out a particle from here, put it to there?

Fifth: we've got a few kg of LiOH, I would be really thankful if you would describe the method for turning it to D2O, because currently we can't use LiOH to anything useful.

Sixth: back to the original topic, D2O is one of the cheapest deuterated solvents. Unless you got free electricity, there is no chance to make it cheaper.

Seventh:
Quote:
This kind of fusion is unique, because if done correctly it poses no threat of a meltdown, or hazerdous by-products.

Please, tell a few more details, I'm interested in nuclear chemistry.

neptunium - 30-12-2012 at 04:20

Quote: Originally posted by APO  
Are you kidding? There are plenty of experiments, like fusion studies, examining DDO vs HHO burn tempature, deuterating hydrocarbons, study of higher energy hydrogen, liquid deuterium rocket fuel, and way more!

For many experiments you need compounds with specific isotopes, especially for nuclear reactions, like bubble fusion, and the fleishman and ponds experiment.

Also I just remebered that under high energy conditions with a mix of water and lithium hydroxide that it will evolve helium and deuterium oxide.

2(H2O,LiOH) x 11(He),9(D2O)

I won't give so many details, because I heard some people on the form haven't been using proper safety precautions.


thats exactly what i thought you might say....i just wanted to hear it from you

neptunium - 30-12-2012 at 04:28

Quote: Originally posted by APO  
Ok, I'll say a little more but I won't explain the apparatus needed because I don't wanna help people hurt them selves. What I meant by high energy conditions, was that this will only work using a super heated plasma compressed by strong magnetic fields, and being excited by a high voltage. The electrodes need to be in a grid inside another gride as the cathode and anode, also the high energy state is started and maintained by a high voltage. The chemical mixture it 2 parts lithium hydroxide monohydrate, and 9 parts water(never use heavy water). If you use heavy water it will not absorb the resulting neutrons, and will instead keep them constant by moderating them, if you did do that you would need radiation protection. This kind of fusion is unique, because if done correctly it poses no threat of a meltdown, or hazerdous by-products. The rest of the details are classified for safety and financial reasons.


i knew where this was going...you are talking about the polywell my friend.

if done correctly ? this is beyond the scope of a home scientist, you WILL get radiations out of your fusor/polywell

dont want to discourage you or anything but good luck!

APO - 30-12-2012 at 13:36

I know I haven't said much about the nuclear method, but I will say that it won't work properly inside a farnsworth fusor. Also scince this takes place in water , the neutrons will usually rip into the protium and make deuterium, scince the deuterium oxide doesn't absorb neutrons it just moderates them,then it will make more D2O out of the normal water. The chance of tritated water is very low because of this and it's decay is helium-3. The isotopes used in this method use hydrogen-1 ,lithium-7 and oxygen-16, and because there are 11 water molecules and only two lithium hydroxide molecules only the the atoms of two water molecules are conductive enough to gain enough energy to split and then fuse. When it's taking place, first at 220 volts the conductive water molecules split effectively enough to were there is a brief atomosphere around the electrodes, allowing a spark gap, which ignites the resulting HHO, and then ionizes it, so that will get it up to a pretty high tempature but no fission/fusion. I'm not sure what exact voltage it must get up to for the nuclear reaction to occur, but when it does each lithium hydroxide monohydrate molecule is split into 22 protons, 20 neutrons, and 20 electrons, so it will form 11 helium-3 atoms and 9 deuterium oxide molecules.


Also I accidentally messed up the equation so here is the right one: 1(H2O,LiOH),9(H2O) x E = 11(He),9(D2O)

It makes Helium-3, and Deuterium Oxide, so some pretty useful stuff is obtained. Of course the method is still being refined. Feel free to ask questions, just note that the aparatus to get this to occur is currently classified.

APO - 30-12-2012 at 13:49

Also it will not work in a polywell, it needs a very special reactor.

kristofvagyok - 30-12-2012 at 14:09

Quote: Originally posted by APO  
Also I accidentally messed up the equation so here is the right one: 1(H2O,LiOH),9(H2O) x E = 11(He),9(D2O)


We've got some uranyl nitrate... According to this I could make some DIY plutonium-nitrate? AWESOME!

APO - 30-12-2012 at 14:27

I see no way that this would make a nitrate, also plutonium production would require way too much electricity. You might be able to make plutonium from having a rod of uranium-238 dioxide in D2O with tungsten carbide walls though. Also plutonium is illegal.

phlogiston - 30-12-2012 at 15:05

So, in summary you've got a prototype of a tabletop fusion device up and running but just need a bit of D2O to fuel it, but that's too expensive.

So, you are clearly about to solve the worlds energy and natural resources crisises. You'll be easily able to afford a few $ worth of heavy water in the near future so why not find an investor, unless you simply enjoy the challenge ofcourse, which I -DO- totally understand.

APO - 30-12-2012 at 15:48

No, I have a reactor that makes D2O for other experiments, but I'm interested in other D2O enrichment processes.

IrC - 30-12-2012 at 16:52

So that hydroxide, would that be Li 6 or 7? I have long dreamed of exceeding the Lawson criterion with water in a liquid state using a tabletop device I built for 50 bucks with materials from Home Depot. I would then tell a public forum how classified it is while I kept bringing it up. From reading many alternative energy sites over time often I read classified as code for "how the hell would I know".

I also wonder how many gallons of water you process per each liter of D2O, or is this yet another fusion device where you are inserting the neutrons into each Hydrogen atom.

[Edited on 12-31-2012 by IrC]

neptunium - 30-12-2012 at 18:06

come on guys ! tone the sarcasm down a little ! this is a science forum ! everyone with a made up system and no proof should be able to share its tails!

APO - 30-12-2012 at 19:48

All you care about is that it can't be partly classified? I don't make the rules but I gotta follow them. I just thought you guys would like to know the reaction, that's all.

[Edited on 31-12-2012 by APO]

IrC - 30-12-2012 at 20:25

Classified means you do not discuss it period. No admitting any existence. You are merely showing the classic symptoms of coming here to blow your own horn with nothing to back it up. Problem is I don't give a damn. I doubt many do. You said it, this is a science forum. Where we DISCUSS science. Not where we claim to know more than God but are not allowed to talk about it. Don Lancaster has many articles on the certain way in general the hocus pocus science crowd behaves and this thread is classic. And irritating.

Took Ford little time to get people driving. Edison little time to light homes. Tesla even less time to light the entire planet. They made something, made no secret maybe someday we will see it claims, got it out there so we can go buy it fast. I have been listening to claims such as yours since the 60's and so far not one of tens of thousands has appeared at Walmart. Or the gas station. I still burn dead dinosaurs to get to Walmart by the way.

You complain to me about being sarcastic yet read my post carefully. I asked two simple questions. You answered neither. Typical. As you said this is a science forum where we discuss things. I recall few saying I have a mysterious black box, believe what I say, but you are not allowed to see the box. This does not constitute a discussion in any way. Accomplishes no discourse among members other that to convey an impression you know things with fantastic claims only you are allowed to know. Where I must ask is the discussion in that? You merely worked there. Did you sign a non disclosure agreement and if so why are you talking about it here in public? No employer would tolerate this. If you did not sign one I can promise all here nothing you are talking about has any super secret repercussions concerning the near future for anyone. I spent many years in billion dollar R&D facilities, I know the drill quite well.

Instead of complaining to or about me answer my two simple questions. Li 6 or 7, how many gallons of water do you waste per liter of D2O? And why is it you only have a few ml when you have access to amazing new ways to create, or concentrate it, whichever. I will not bother to ask how you survive the neutron flux. If there is any merit in your claims then alter the way in which you are conducting this thread in a quasi secretive way. Talk about it don't talk about it. If you do, say something of use which can be openly discussed, the very purpose behind this board.

I should add this: look at the thread beginning.

D2O enrichment:

"I only have 20g of D2O and I would like more, but it is extremely expensive, so I want to make my own. The method I am interested in is electrolysis, H2O splits at 1.48v and D2O splits slightly higher, if I use exactly 1.48v the H2O will hopefully split leaving the D2O in soulution,then I would add more water to the slightly enriched soulution, and repeat the process. Any more ideas?"

OK I'll bite. If in your further posts you try to convince people you worked where this amazing new method has been achieved, then start acting secretive, then why ask "Any more ideas?"

How are we supposed to know your the one who worked there, and are not allowed to tell. This being the case, i.e., you are privy to this new dubious sounding process, why on earth would you need to ask anyone here if they have any ideas?

I am not trying to lay waste to you or your claims, who knows maybe you do have some new thing.

I am merely reacting to the lack of logic being exercised in a most annoying way. Read again your beginning post. You are not creating new D2O. Merely getting the vanishingly small amount already there. This requires the processing of an ocean of water (put right back in the ocean after D2O depletion by the way).



[Edited on 12-31-2012 by IrC]

APO - 30-12-2012 at 20:39

Sorry I meant the apparatus was classified, I didn't want people to ask about that. Sorry for the miscommunication. The method uses Li-7 hydroxide monohydrate, Li-6 will be better for helium yield but worse D2O yield and vice versa. The resulting neutrons from the reaction, usually hit the protium and turn it into deuterium. So it's a bit more like D2O synthesis, than enrichment. I was just mentioning that particular reaction, I'm looking for more info on the enrichment of D2O.

IrC - 30-12-2012 at 20:51

That is something all can discuss. Yet I fear you are still going to hit the wall of the incredible amount of water in need of processing not to mention the insane electric bills. Info on enrichment along these lines is plentiful, info on new high yield ways is not. Nor on methods of synthesis. I buy D2O to use with Li metal to create both the Deuterides as well as Deuterium gas for Fusor experiments. I have to say considering the expense of concentrating your own best to bite the bullet and just buy it. U.N. is an expensive way to go however. As far as I know the cheapest place on earth to obtain it is in Canada, although China may now be out doing them in production. Start investigating industrial suppliers, preferably ones that sell research quantities. Believe me I wish I knew of a cheap route to owning large quantities of a very interesting chemical for amateur science. Almost as fun for me as my Iridium Carbide.

APO - 30-12-2012 at 21:32

What if I use solar power?

APO - 30-12-2012 at 21:42

Also sorry IrC ,I just metioned a new method I was working for making D2O because that is similiar to enriching it. I'm also just registered on this forum a couple days ago. Please tell me how to talk, I don't mean to piss people off.

IrC - 30-12-2012 at 23:37

You are not really pissing anyone off, I'm sure it was more me being too zealous than you doing anything incorrect. What you were describing sounded so much like yet another amazing but never to be seen in the real world invention. I cannot tell anyone how to talk here I'm just another member like anyone else. From being interested in many of the outside of conventional areas and watching how they operated I imagine one develops an attitude. Or at least I did. Virtually all of them claim so much yet never allow you to see the wizard behind the curtain. Gets so irritating after a while.

To give an example, I studied the Testatika Machine for a while hoping there may be some new discovery involved. This colony claims it provides the power for their retreat or whatever it is they call it. Then I started hearing their claim that the world was not ready for free energy so they must keep it secret. Right there is all the proof of a sham required. If one assumes fossil fuels are killing the earth, would they then refuse to divulge or at least market wholesale a thing which could save the planet, including their own future generations. Think about it. Because man is not ready for their so called new technology they would condemn themselves along with everyone else to a hot polluted slow demise. In other words this is code for the fact that they are completely full of crap and pay for their gas and electricity just like everyone else. This illustrates what I mean by lack of logic in this area.

I can imagine two guys out cutting wood, one who invented a new artery patch. "Hey my chainsaw broke and I'm bleeding out my neck!" "Go get that new patch you just invented it can save you!" "No I'm afraid the world will learn how it works see you on the other side" "Don't you mean in the next dimension?"

Or something like that. No one tells you how to talk here or anywhere, no one is the boss of you. Unless your married of course, but I digress.

Solar? A really big setup if you plan to make much, again think of the many gallons of water you would have to go through. Each quantity is only going to yield so much, requiring a new quantity of water over and over. Unless you know how to build this device you described yourself. In which case make two, how much for the other?


[Edited on 12-31-2012 by IrC]

watson.fawkes - 31-12-2012 at 05:55

Quote: Originally posted by APO  
The method I am interested in is electrolysis, H2O splits at 1.48v and D2O splits slightly higher, if I use exactly 1.48v the H2O will hopefully split leaving the D2O in soulution,then I would add more water to the slightly enriched soulution, and repeat the process. Any more ideas?
Yeah. Learn some statistical mechanics. Dissociation voltages are averages, not exact figures for each molecule. Exact voltages depend upon the specific thermal excitation of the molecule, and there's a distribution of these thermal energies. The effect of this is to broaden what's otherwise a cliff. The result for isotope separation is that for every voltage and temperature there a ratio of isotopes that will split. Per-pass separation isn't what you think it is.

And it's a complete newbie mistake to assume that you're going to find much D2O in nature. Almost all of it is DHO.

Chemical exchange is more energy efficient, and it's still very energy-intensive. The best known is the GS process, which uses D-H exchange between H2S and H2O. Note that the cited page mentions a 340:1 input water to output water ratio for one such plant. There's also an ammonia exchange process, but I don't know much about it.

ScienceSquirrel - 31-12-2012 at 09:44

The Girdler sulfide process is probably the cheapest and most efficient way but you are going to need some serious engineering plus a lot of power and water.
Really your ideal set up could be small water mill driving a turbine for power and the stream providing the water in the middle of nowhere so no need to worry about upsetting the neighbours if your hydrgen sulfide leaks.
Even a tiny Girdler plant will need tonnes of water and hundreds of kW hours to produce a useful yield so if you have to take them from a metered public supply your utility bill will be huge and it will excite interest from all sorts of parties :(

APO - 31-12-2012 at 11:10

Hmm... Norsk Hydro made it sound easy. I'm a pretty good engineer though. Maybe I could use a solar setup, and a rain collector, so that I could leave it for several months to get a slight yield. Anyone know the exact voltage that water splits at?

Metacelsus - 1-1-2013 at 16:52

1.48 is the theoretical minimum. However, your electrode overvoltage will cause the practical voltage to be higher.

phlogiston - 1-1-2013 at 17:27

It's good to see we are back to a real discussion on the experimental details of some mad scientist experiment, in this case deuterium enrichment.

Behind that simple question 'what is the exact voltage' lies a world of complexity you appear to have no notion of.
It is dependend on pressure, temperature, electrode material, etc. Actual cell voltage drop will also include, for instance, the drop across the ohmic resistance of the cell and the reactions taking place at the cathode.
Your questions suggest that your train of thought is that if you could hit the exact voltage where H2O splits and D2O doesn't yet, you'll end up with a much enriched residue of D2O / DHO.
Dissapointingly, it won't work to the extent that you may be hoping for. A very small fraction of enrichment in each step of electrolysis is realistic. Perhaps you can sell the vast quantities of hydrogen and oxygen to recover some of the cost, or use a fuel cell to recover the increasingly enriched fractions as D2O for input to subequent steps while cutting down on electricity costs.

Deuterium enrichment has been studied extensively and it is unlikely (but never impossible) that you will find a better way than the established methods. In a small lab setting, I think you should go for whatever process yields the best enrichment per step, regardless of energy/materials costs, which, as has been mentioned, seems to be the Girdler sulfide process (at least, of the methods published so far).

kristofvagyok - 1-1-2013 at 19:54

Long ago I also thought about deuterium purification and I ended up that there is no such cost effective way than buying it, but there is always someone who want to make it free, so here are my ideas:

As we know heavy isotopes form stronger covalent bonds than their lighter counterparts; for example, a carbon-deuterium bond is stronger than a carbon-hydrogen bond, this is why CDCl3 forms from CHCl3 in D2O.

Some D2O, cheers :D


So the only effective separation could work with a method what uses something else also than water. The Gilder process works, the deuterium forms a stronger bond with the sulfur so it could be enriched.

A version could be that acidic water is boiled with an organic substance in it what will get fully deuterated with time... E.G.: deuterobenzene-d6 is made simply by adding benzene to D2O and D2SO4, simply boil it a bit and distill down the pure benzene-d6. Sound easy? Yes!
The problem: a lot energy is needed to boil that amount of water and getting out the deuterium from an organic compound is also not a really easy thing .



Another idea could be (this was my long ago thought idea) that as we know hydrogen and deuterium (and tritium) is a special thing, it could be passed into palladium forming an unstable interstitial hydride. The only problem is if we use a palladium "window" for the separation of hydrogen, it will work perfectly, but it will also let through deuterium (not as well as hydrogen, but it will let it pass also). If someone is really good in material science than it would worth to look up an alloy what readily adsorbs hydrogen, but not as good the deuterium.

And a third idea is also here: as we know UH3 is a really special hydride, because it adsorbs hydrogen at low temperatures, but is releases it at elevated temperatures. I just looked up that what is known about UD3: almost nothing. It is mentioned (according to reaxys) in 3 article and they did not checked that what temperature it gives off the deuterium (or just I didn't find it).
The idea would be that that pass hydrogen/deuterium in a heated steel tube what contains uranium granulates at that temperature where the UH3 fully decomposes, but the UD3 not yet.

[Edited on 2-1-2013 by kristofvagyok]

IrC - 2-1-2013 at 02:26

Quote: Originally posted by Cheddite Cheese  
1.48 is the theoretical minimum. However, your electrode overvoltage will cause the practical voltage to be higher.


Just a thought but edge effects on the electrodes could be reduced by using spheres or partial spheres not completely immersed. This I think would allow for a more controlled potential if combined with the proper regulation scheme in the associated power source. If there is a critical yet narrow window for the proper voltage it seems this might be a useful approach. Of course they must be extremely smooth.

watson.fawkes - 2-1-2013 at 05:12

Quote: Originally posted by IrC  
Just a thought but edge effects on the electrodes could be reduced by using spheres or partial spheres not completely immersed. This I think would allow for a more controlled potential if combined with the proper regulation scheme in the associated power source.
This is in the right spirit of how to get consistency, but I don't think it's the right geometry. Any good analysis of this issue needs an account of the electric field within the electrolyte. Edge effects in electrodes come from fringing fields, and eliminating them is what's wanted. The most important aspect of this is to minimize the gap between electrodes. This leads to looking at parallel plates. I would consider putting a plate electrode within a non-conducting frame, so that the edge of the conductor was not at the edge of the plate. This would eliminate most of the field component in directions along the plane of the plate. In addition, with a small gap, you'd likely need active pumping of the electrolyte, no small complication if you have high pH electrolyte (typically using KOH) to increase conductivity.

The other thing necessary to get this kind of thing right is to use a potentiostat with a sense electrode. This kind of supply changes the voltage drop across the main, high-current electrodes in order to keep the voltage between one main electrode and the sense electrode constant. The current required to sense voltage can be minuscule, down in the nA range, in order to minimize resistance error in the sense electrode, required if you need to accurate sense in single mV so that you can regulate to 10 mV.

AJKOER - 2-1-2013 at 15:10

Here is a general idea that you may find useful.

1. Separate out (via electrolysis) the oxygen. Left with H2 and D2.

2. Combine with compound X, so you now have, say, HX and DX.

3. Select element/compound X so as to maximize a difference in a physical property that can be used in separating HX from DX. For example, you started this thread focusing on the voltage at which the H2O and D2O split occurs.

The logic here is that the element Oxygen may not be the best choice.

The fact may be it doesn't materially matter what the element X is, or you do not have the necessary data to know.

[Edited on 2-1-2013 by AJKOER]

kristofvagyok - 2-1-2013 at 15:22

Just and idea, but HF has a boiling point: 19.4 °C and DF has a boiling point 18.66 °C.

Who is enough brave to try it out? :D