Sciencemadness Discussion Board

Deuterated chemical explosives

Fusionfire - 10-8-2011 at 03:48

Hello all,

I am trying to think of a chemical explosive that has all the <sup>1</sup>H replaced by <sup>2</sup>H.

The following factors are of interest:
1) High mole % of H in comparison to other elements.

2) Easy to synthesise on a laboratory and industrial scale.

3) Minimal wastage of valuable <sup>2</sup>H, supplied as heavy water, during the conversion process to the explosive.

4) Chemically stable.

5) Relatively insensitive.

The best I could think of was an oxy-hydrogen explosive mixture. It would have a 67% mole percentage of H, can be produced by the electrolysis of heavy water (in theory with very little loss of <sup>2</sup>H) and is chemically stable + insensitive when the reactants are kept apart.

The problem is that this would be in gas or liquid phase and a solid state energetic is preferable. I am also not sure how the VoD and sensitivity vary with pressure. If you keep raising the pressure of a oxy-hydrogen mixture isothermally, does a spontaneous reaction occur at some point?

What about frozen hydrogen peroxide crystals intimately mixed with lithium hydride (all deuterated) and kept frozen?

hissingnoise - 10-8-2011 at 05:06

Before initiating a self-sustaining fusion reaction on the global scale, I think you should consider, for a while, the numerous questions that will forever remain unanswered . . .

Fusionfire - 10-8-2011 at 05:19

Oh, do calm down :D

A bit of armchair hypothesising never led to a nuclear armageddon!

Bot0nist - 10-8-2011 at 05:23


Sorry to sound ignorant, but could you all elaborate on the theoretical apocalyptic outcome of such an undertaking.

Me no understand.:(

hissingnoise - 10-8-2011 at 05:36

Quote:
Oh, do calm down.

We're doomed! Doomed, I tell you! All doooommmed . . .
Och Aye!
:D


Fusionfire - 10-8-2011 at 05:37

Well to initiate nuclear fusion you would need to subject light nuclei to very high temperatures, so that their kinetic energies exceeds the electrostatic repulsion of the protons in their nuclei. At close enough distances the strong nuclear force takes over and you get nuclear fusion + a large release of energy.

Apart from rather expensive research using pulsed laser arrays and magnetic confinement, the only way this has ever been achieved is by gravitational confinement in stars and a fission stage.

Fusionfire - 10-8-2011 at 05:38

Quote: Originally posted by hissingnoise  
Quote:
Oh, do calm down.

We're doomed! Doomed, I tell you! All doooommmed . . .
Och Aye!
:D




Yes yes ha ha that is all very funny, but now back to the original post... :P

Nicodem - 10-8-2011 at 05:51

By dissolving ammonium nitrate in D2O and rotavaping you will get deuterated ammonium nitrate. Doing this twice or trice will give you >90% deuterium enrichment. You can then add those few % of whatever needs to be added to make the thing detonable. Totally nonsensitive secondary explosive and as easy as it gets using only cheap D2O.

hissingnoise - 10-8-2011 at 05:54

OK! Go ahead - take out the entire solar system while you're at it . . .
Who really cares? :D

Fusionfire - 10-8-2011 at 06:01

Quote: Originally posted by Nicodem  
By dissolving ammonium nitrate in D2O and rotavaping you will get deuterated ammonium nitrate. Doing this twice or trice will give you >90% deuterium enrichment. You can then add those few % of whatever needs to be added to make the thing detonable. Totally nonsensitive secondary explosive and as easy as it gets using only cheap D2O.


Haha very funny, dissolving ammonium nitrate in heavy water and then evaporating it.

hissingnoise - 10-8-2011 at 06:05

But on the plus side, if you electrolyse 'heavy water' and spark the product-mixture, you'll end up with the original substrate.
Its collection, though, just might pose problems! ;)

Nicodem - 10-8-2011 at 06:16

Quote: Originally posted by Fusionfire  
Haha very funny, dissolving ammonium nitrate in heavy water and then evaporating it.

What is so funny? That is the common method for deuteration of exchangeable protons. And besides D2O is relatively cheap if you buy it from chemical suppliers. It is massively used and massively produced. Si*ma sells it for less than a $ per g (99% deuterium rich!). Of course, if you want to make it yourself it will cost you way more.

Fusionfire - 10-8-2011 at 06:31

So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.

N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <---> N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O

Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with N-<sup>2</sup>H bonds?

What determines the equilibrium position?

Nicodem - 10-8-2011 at 06:58

Quote: Originally posted by Fusionfire  
So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.

N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <---> N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O

Of course! Have you never heard of H/D exchange? That's how exchangeable protons are routinely detected in NMR spectroscopy. You add a drop of D2O or deuterated methanol to the sample solution in DMSO-d6 or CDCl3 and check which 1H NMR peaks decrease or disappear.
Quote:
Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with N-<sup>2</sup>H bonds?

The protons in ammonium ions are exchangeable (their pKa in water is 9.3) so that their exchange is nearly instantaneous. You don't need to break any bond, they continuously break themselves when the ammonium ion is solvated in protic solvents (proton exchange is a dynamic equilibrium).
Quote:
What determines the equilibrium position?

It essentially boils down to statistics as this is by far the most contributing factor, so you can calculate a rough estimate of the deuterium enrichment from the amounts of NH4NO3 vs. D2O used and the number of process repeatings (if you have some basic mathematical skills you will easily find out how to do this). The boiling point difference between D2O and HOD is too low to contribute anything and even if it would it would contribute positively. Thermodynamic factors such as the solvation difference between the ammonium ion and its deuterated counterparts are surely observable, but can also be largely ignored in practice. So at the end it is all about statistics.

Of course you can recover most of the D2O by distillation instead of rotavaping it away, but it will obviously be less deuterated than the starting one. Yet, you can use it for the first run for a fresh batch of NH4NO3 to increase efficiency in the second process with the non-recovered D2O. The rationality of such recycling depends on the amount of ND4NO3 you want to prepare. For small amounts it is not worth the fuss.

sternman318 - 10-8-2011 at 07:07

Quote: Originally posted by Fusionfire  
So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.

N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <---> N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O

Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with N-<sup>2</sup>H bonds?

What determines the equilibrium position?


Well this is merely a guess, but both water and ammonium dissasociate in solution

NH4+ + H2O <--> NH3 + H3O+

and

H20 + H20 <--> OH- + H3O+

Of course the dissasocation is small for both, which is why you might need to run the procedure a few times to ensure a high amount of deuteration. However statistics favors the hydrogen-1 being replaced by the hydrogen-2, since there is large excess deuterium/ hydrogen-2.

EDIT: Nevermind, Nicodem has this :P

[Edited on 10-8-2011 by sternman318]

The WiZard is In - 10-8-2011 at 07:30

Quote: Originally posted by Fusionfire  
Hello all,

I am trying to think of a chemical explosive that has all the <sup>1</sup>H replaced by <sup>2</sup>H.


Sorry this is done-did every day and we are all still here.


Accession Number : ADA468050
Title : Role of Thermochemical Decomposition in Energetic
Material Initiation Sensitivity and Explosive Performance
Descriptive Note : Conference paper (preprint)
Corporate Author : AIR FORCE RESEARCH LAB EDWARDS AFB CA
Personal Author(s) : Shackelford, Scott A.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA468050
Report Date : 05 FEB 2007
Pagination or Media Count : 31

Abstract : Catastrophic initiation of an energetic material
consists of a complex, interactive, sequential train of
mechanistic mechanical, physical, and chemical processes
which occur over a finite time period and proceed from
macroscopic into sub-microscopic composition levels (bulk >
crystalline > molecular > atomic). Initiation results when these
processes proceed at a rate which generates sufficient energy
(heat) to reach a threshold stage within this finite time period.
Thus, the rate at which these mechanistic processes occur
defines initiation sensitivity and affects performance.
Thermochemical decomposition processes regulate the rate at
which heat energy is released at the molecular level, and
therefore to some extent, control energetic material initiation
sensitivity and performance characteristics. Kinetic deuterium
isotope effect (KDIE) data, obtained during ambient pressure
thermochemical decomposition process, identifies the
mechanistic rate-controlling bond rupture which ultimately
regulates the energy release rate of a given energetic material.
This same rate-controlling bond rupture also appears as a
significant rate-limiting feature in higher order deflagration,
combustion, and explosion phenomena. The effect the KDIE
-determined rate-controlling bond rupture exerts on initiation
sensitivity, and its potential influence in combustion and
explosion performance is delineated.



Accession Number : ADA121652
Title : Explosive-Driven Hemispherical Implosions for Generating Fusion Plasmas
Descriptive Note : Interim rept.
Corporate Author : TORONTO UNIV DOWNSVIEW (ONTARIO) INST FOR AEROSPACE STUDIES
Personal Author(s) : Sagie, D. ; Glass, I. I.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA121652
Report Date : MAR 1982
Pagination or Media Count : 104

Abstract : The UTIAS explosive-driven-implosion facility was
used to produce stable, centered and focused hemispherical
implosions to generate neutrons from D-D reactions. A high
resolution scintillator-detection system measured the neutrons
and gamma-rays resulting from the fusion of deuterium.
Several approaches were used to initiate fusion in deuterium.
The simpliest and most direct proved to be in predetonated
stoichiometric mixture of deuterium-oxygen. The other
successful method was miniature Voitenko-type compressor
where a plane diaphragm was driven by the implosion wave
into a secondary small spherical cavity that contained pure
deuterium gas at one atmosphere. A great deal of work still
remains in order to measure accurately the neutron flux and its
velocity distribution as well as the precise interactions of the
neutrons with the steel chamber which produced the gamma
-rays. Nevertheless, this is the only known work where fusion
neutrons were produced by chemical energy only in a direct
and indirect manner.


And a SL of other refs.

Fusionfire - 10-8-2011 at 07:52

Crikey you just smacked my head with a realisation of a physical phenomenon I had under my nose but took for granted! Sorry for a moment I thought you were taking the mickey out of me ;)
Quote: Originally posted by Nicodem  
Quote: Originally posted by Fusionfire  
So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.

N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <---> N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O

Of course! Have you never heard of H/D exchange? That's how exchangeable protons are routinely detected in NMR spectroscopy. You add a drop of D2O or deuterated methanol to the sample solution in DMSO-d6 or CDCl3 and check which 1H NMR peaks decrease or disappear.


No, unfortunately. They didn't cover that in engineering.
Quote:

Quote:
Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with N-<sup>2</sup>H bonds?

The protons in ammonium ions are exchangeable (their pKa in water is 9.3) so that their exchange is nearly instantaneous. You don't need to break any bond, they continuously break themselves when the ammonium ion is solvated in protic solvents (proton exchange is a dynamic equilibrium).


What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?

What if you use a proton exchange polymer like Nafion?
Quote:

Quote:
What determines the equilibrium position?

It essentially boils down to statistics as this is by far the most contributing factor, so you can calculate a rough estimate of the deuterium enrichment from the amounts of NH4NO3 vs. D2O used and the number of process repeatings (if you have some basic mathematical skills you will easily find out how to do this).


Since it seems I am thankfully in the company of experts, I'll risk looking like an idiot and take a dive in hope of getting a bit of education outside my usual realm...

1 mole of NH4NO3 has 4 moles of H.
2 moles of D2O has 4 moles of D.
So if you mix 1 mole of NH4NO3 with 2 moles of D2O, assuming proton exchange to be unbiased*, you'd end up with 1 mole of H2O and 1 mole of D2O, and NH2D2NO3. Probably the reaction will maximise entropy so you wouldn't end up with 1 mole of H2O and 1 mole of D2O, but 2 moles of DHO.

Quote:

Of course you can recover most of the D2O by distillation instead of rotavaping it away, but it will obviously be less deuterated than the starting one. Yet, you can use it for the first run for a fresh batch of NH4NO3 to increase efficiency in the second process with the non-recovered D2O. The rationality of such recycling depends on the amount of ND4NO3 you want to prepare. For small amounts it is not worth the fuss.


Understood. It is kind of like trying to increase the thermodynamic efficiency of a heat engine by using the waste heat to increase the temperature of the gas prior to combustion.

BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?

*Wikipedia says:
Quote:

The lone electron pair on the nitrogen atom (N) in ammonia, represented as a pair of dots, forms the bond with a proton (H+). Thereafter, all four N-H bonds are equivalent, being polar covalent bonds

So I understand that all four N-H bonds can be replaced by N-D bonds with unbiased probability

Fusionfire - 10-8-2011 at 08:00

Quote: Originally posted by The WiZard is In  
And a SL of other refs.


It seems though, that work on a chemical energy -> nuclear fusion energy device stopped in the late 20th century.

http://en.wikipedia.org/wiki/Pure_fusion_weapon
Quote:

Accession Number : ADA121652
Title : Explosive-Driven Hemispherical Implosions for Generating Fusion Plasmas
Descriptive Note : Interim rept.
Corporate Author : TORONTO UNIV DOWNSVIEW (ONTARIO) INST FOR AEROSPACE STUDIES
Personal Author(s) : Sagie, D. ; Glass, I. I.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA121652
Report Date : MAR 1982
Pagination or Media Count : 104


and

Quote:

Despite the many millions of dollars spent by the U.S. between 1952 and 1992 to produce a pure fusion weapon, no measurable success was ever achieved. In 1998, the U.S. Department of Energy (DOE) released a restricted data declassification decision stating that even if the DOE made a substantial investment in the past to develop a pure fusion weapon, "the U.S. Is not known to have and is not developing a pure fusion weapon and no credible design for a pure fusion weapon resulted from the DOE investment".


I'm not just thinking about a pure fusion weapon but, for example, imagine the enormous potential of a pulsed or continuous (chemical) detonation engine that releases clean nuclear energy by fusion as well.

niertap - 10-8-2011 at 08:21

This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires ~100,000,000k.

The only way is to make an explosive flux power generator and focus multiple explosions into 1.

Fusionfire - 10-8-2011 at 10:29

Quote: Originally posted by niertap  
This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires ~100,000,000k.

The only way is to make an explosive flux power generator and focus multiple explosions into 1.


I didn't say I could make it happen. It was just armchair speculation, which was why I tried to focus on the chemistry and not the engineering (as is the remit of posts in this sub-forum).

The WiZard is In - 10-8-2011 at 10:30

Quote: Originally posted by niertap  
This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires ~100,000,000k.

The only way is to make an explosive flux power generator and focus multiple explosions into 1.



Or Star Trek .... Matter - antimatter.


djh
----
Many scientist, most notable Carl
Sagan, have believed the sheer
number of environments in the
universe makes it possible that life
had developed elsewhere. But as
Clancy points out, "It's one thing to
believe that life might exist on other
planets, and quite another to believe
that it is secretly examining your
private parts."

Stuart Vyse's review of :—
Susan A Clancy
Abducted : How People Come to Believe
They Were Kidnapped By Aliens
Science 310 [5752] 1280-81

Sickman - 10-8-2011 at 10:51

Quote: Originally posted by Fusionfire  
Hello all,

I am trying to think of a chemical explosive that has all the <sup>1</sup>H replaced by <sup>2</sup>H.

The following factors are of interest:
1) High mole % of H in comparison to other elements.

2) Easy to synthesise on a laboratory and industrial scale.

3) Minimal wastage of valuable <sup>2</sup>H, supplied as heavy water, during the conversion process to the explosive.

4) Chemically stable.

5) Relatively insensitive.

The best I could think of was an oxy-hydrogen explosive mixture. It would have a 67% mole percentage of H, can be produced by the electrolysis of heavy water (in theory with very little loss of <sup>2</sup>H) and is chemically stable + insensitive when the reactants are kept apart.

The problem is that this would be in gas or liquid phase and a solid state energetic is preferable. I am also not sure how the VoD and sensitivity vary with pressure. If you keep raising the pressure of a oxy-hydrogen mixture isothermally, does a spontaneous reaction occur at some point?

What about frozen hydrogen peroxide crystals intimately mixed with lithium hydride (all deuterated) and kept frozen?


Deuterated Ammonium picrate is likely going to be one of the better answers, and it will still have one light hydrogen if the picric acid hasn't been deuterated as well.

The ND3 is easily made from Mg3N2 (magnesium nitride) and heavy water.

As far as it igniting a fusion reaction: maybe, when hell freezes over!;)

Dr.Bob - 10-8-2011 at 11:48

Quote: Originally posted by Fusionfire  

What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?

What if you use a proton exchange polymer like Nafion?

BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?


The C-H bonds in an alkane are too strong to break easily without some strong reagent or strong base. They do not exchange with solvent protons in normal conditions or unless they are activated by certain substituents. The proton exchange resins/polymers will not affect that.

I can't imagine any practical way to make any mixture of LiH and H2O2 that would not react to generate H2 and LiOOH or LiOH and O2. Even frozen, once a few molecules got close enough to react, they would generate an exotherm which would lead to melting and reaction/explosion. Please don't try this at home...

Bob

Fusionfire - 10-8-2011 at 14:03

Quote: Originally posted by Dr.Bob  
Quote: Originally posted by Fusionfire  

What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?

What if you use a proton exchange polymer like Nafion?

BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?


The C-H bonds in an alkane are too strong to break easily without some strong reagent or strong base. They do not exchange with solvent protons in normal conditions or unless they are activated by certain substituents. The proton exchange resins/polymers will not affect that.


Presumably as the C-C bond energy (346 kJ/mol) is less than the C-H one (411 kJ/mol) if you heat up an alkane in the presence of deuterium it will break up into shorter alkanes or alkenes or alkynes, and then the C-H bonds will get substituted for C-D. C=C bond energies are 602 kJ/mol and C≡C are 835 kJ/mol. Please correct me if I am wrong but I guess this must be how deuterated methane is made.

http://www.caslab.com/Methane-D4_deuterated_methane_CAS_558-...

How would you go from here to get deuterated paraffin? (see below why you'd be interested in this...)

My reference to proton exchange polymers was separate. If you have a block of Nafion and soak it in heavy water, will you start to find that the H+ (technically sulphonic acid groups) start picking up deuterium so as to maximise the system's entropy?
Quote: Originally posted by Dr.Bob  

I can't imagine any practical way to make any mixture of LiH and H2O2 that would not react to generate H2 and LiOOH or LiOH and O2. Even frozen, once a few molecules got close enough to react, they would generate an exotherm which would lead to melting and reaction/explosion. Please don't try this at home...
Bob


You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy coating to passivate your LiH from runaway exothermic oxidation by the peroxide.

Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.

Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source of external heat would set this off!

[Edited on 10-8-2011 by Fusionfire]

Sickman - 10-8-2011 at 15:20

Quote: Originally posted by Fusionfire  

You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy coating to passivate your LiH from runaway exothermic oxidation by the peroxide.

Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.

Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source of external heat would set this off!

[Edited on 10-8-2011 by Fusionfire]


External heat would only achieve activation energy for a chemical reaction! Your hopes of fusion are kaput without a source of neutron bombardment. Please don't clutter the forum with your "cold fusion mumbo jumbo"!:o

http://grooveshark.com/#/s/Tell+No+Tales/3mu0wU?src=5

[Edited on 10-8-2011 by Sickman]

Fusionfire - 10-8-2011 at 23:12

Quote: Originally posted by Sickman  
Quote: Originally posted by Fusionfire  

You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy coating to passivate your LiH from runaway exothermic oxidation by the peroxide.

Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.

Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source of external heat would set this off!

[Edited on 10-8-2011 by Fusionfire]


External heat would only achieve activation energy for a chemical reaction! Your hopes of fusion are kaput without a source of neutron bombardment. Please don't clutter the forum with your "cold fusion mumbo jumbo"!:o

http://grooveshark.com/#/s/Tell+No+Tales/3mu0wU?src=5

[Edited on 10-8-2011 by Sickman]


I never said the external heat would set off anything but the chemical reaction.

Sickman - 11-8-2011 at 01:28

Quote: Originally posted by Fusionfire  
Quote: Originally posted by Sickman  
Quote: Originally posted by Fusionfire  

You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy coating to passivate your LiH from runaway exothermic oxidation by the peroxide.

Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.

Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source of external heat would set this off!

[Edited on 10-8-2011 by Fusionfire]


External heat would only achieve activation energy for a chemical reaction! Your hopes of fusion are kaput without a source of neutron bombardment. Please don't clutter the forum with your "cold fusion mumbo jumbo"!:o

http://grooveshark.com/#/s/Tell+No+Tales/3mu0wU?src=5

[Edited on 10-8-2011 by Sickman]


I never said the external heat would set off anything but the chemical reaction.


When you say "A slightest source of external heat would set this off!" a normal person would ask "set what off"? The question goes back to your use of the anaphor "this", which seemed, to me, to point back to your use of the antecedent in the preceding sentence: "nuclear fusion"! If however the antecedent you intended was "chemical explosive" then I apologize for my rush to judgement. And yet your use of the phrase "set off" is ambiguous given the context of the preceding sentence.

I know, I know, enough with the grammar already. My point: words do matter, alot!
:P

http://www.youtube.com/watch?v=BkM05jjHMNw

[Edited on 11-8-2011 by Sickman]

KemiRockarFett - 11-8-2011 at 05:34

Stupid idea related to the "cold fussion" project runned by Mr Rosso that I am sure you all already have heard about.

Mr Rosso claims that he, in his device, runs this reaction:

62Ni + p --> 63Cu E equals -> 0.006687 mass units.

The device is some kind of steal tube with fine Ni powder, rheney nickel?, and H2 gas and some secrets. Some experts in physics, bought?, thinks that this could be serous and for real.

Now to my idea based on if this above is true:

Deuterised explosive or ordinary HE mixed with superfine Al or similar to get out H2 in big ammounts. The mixtures VoD will go down as the mix become fueled with metal ( for H2/D2 production). Dont se this as a problem. Produce an explosive lens, think of a Pu-fission bomb, there this HE-mixture is the sperhe. The Ni-powder or NiD3,NiH3 in the center of this sphere. As ususally use bridgewire dets to get uniform implosion of the spherically HE/metal mix. Surrounding the explosive-H2/D2 sphere should be a sphere of Uranium238 or similar neutron eating materialand around this a sphere with neutron reflector material and after this the outer layer should be steel. The spere is evacuated and filled upp with 200 bar hydrogen/deuterium gas. This will never work somebody claims, the energi will be to low. Please calculate the effectdenstity in the center för one meter diameter sphere. How much work will be focused into an centervolume and for how long time?


Check this, and compare with above, Canada cold fussion reactor with pistons compressing the reactants to the center of the sphere:

http://www.technologyreview.com/business/23102/

If experiments are made with explosively driven magnetic field compressors use demagnitisation of neurudynium permanentmagnets to get the input current. Its the easiset.
How to do this? Just detonatate PETN/RDX located inside the magnet to get rid of the magnetization. The change in magnetic flux induces the starting current and off cource the PETN/RDX also detonates the explosive filled metal tube that short circuits and compress the magnetic field.


[Edited on 11-8-2011 by KemiRockarFett]

phlogiston - 11-8-2011 at 13:32

Quote:
I never said the external heat would set off anything but the chemical reaction.


Then what was the purpose of a deuterated explosive?

Explosive compression of deuterium gas using shaped charges has been used with success to initiate fusion. Look up Voitenko compressor if you like to read more.

[Edited on 11-8-2011 by phlogiston]

gregxy - 11-8-2011 at 16:01

Here is an old study on explosive driven fusion.
They did get some fusion, but the process does not scale up.


http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA121652&Locati...

Its almost amazing how many things have been tried over the years.

KemiRockarFett - 13-8-2011 at 14:28

Quote: Originally posted by gregxy  
Here is an old study on explosive driven fusion.
They did get some fusion, but the process does not scale up.


http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA121652&Locati...

Its almost amazing how many things have been tried over the years.


10^7 neutrons for a steel-hemisphere with 20 cm diameter there the hemisphere surface was charged with 200 gram PETN at 0.5-06 g/cm^3. Hemispere was placed normal to a steel plate and inside this arrangement up to 70 bar of stochimoetrical mix of D2 + O2 was initiated at the explosive lens focus point by a bridgewire. The D2+ O2 mixture initated the PETN almost perfectly but not for higher PETN densities.

My suggestion to get this going is the following "simple" steps:

1) Use a spere not a hemisphere. Increase the diameter.
2) Charge it with an explosive their all hydrogen atoms are substitutet with Detuerium ones.
3) Mix the deuterised explosive with some chemically reactive metal preferable with a large atomic mass. Balance for D2 output. (Most of the partikles that first reaches the center will be D2 ). If its possible to get fission from the fussion created neturons the reactive metal mixed in the explosive could be a fissible material. (Some one maybee thinks that its a bad idea to decrease VoD in this way but here we could use a layered explosive, without metal inmix, on top of the D2 balanced explosive. )
4) Evacuate the steelsphere totally.
5) Set the device of with a set of bridgewire detonators in a classic manner.


PHILOU Zrealone - 14-8-2011 at 10:46

What about deuteriated, C13, N15, O17 HMX?
Must be way denser than common HMX...I wonder if the increase of density will favourize higher VOD??
Because denser output gases will decrease specific impulse...but for sure not the detonic process...

Anyone has datas of higher isotopic explosives vs normal explosives...

asilentbob - 17-8-2011 at 20:08

There is a big difference between fusing a few atoms and fusing enough to get a self sustaining reaction for a time.

The first is easy... People do it in their garage all the time. You keep dumping in power and get out much less than you put in. Technology required is not much more than a neon sign transformer and high vacuum pump.
https://secure.wikimedia.org/wikipedia/en/wiki/Inertial_elec...

The second is hard.
H-Bombs. Massive magnetic confinement experiments which try to be self sustaining but are not.

Fusionfire - 18-8-2011 at 03:35

Well, in terms of the practicality of a fusion device anything but billions of funding can achieve, we can exclude laser array inertial confinement and magnetic confinement.

Assuming you could construct a spherical explosive assembly, high synchronicity detonators, wave shapers, etc. - and use deuterated chemical explosives but no fissionable core - would it be possible to trigger nuclear fusion at the core?

Temperatures in imploding bubbles at room temperature is apparently 100,000K, with some simulations (probably assuming no design imperfections) predicting 10 megakelvins.

http://en.wikipedia.org/wiki/Bubble_fusion

The actual temperature you get obviously depends crucially on the sphericity of the imploding shock. Small deviations from a spherical convergent shock would lead to jetting, lower temperatures and premature loss of confinement.

What temperatures could you expect from an axisymmetric two-point initiation device?

KemiRockarFett - 18-8-2011 at 06:27

Quote: Originally posted by Fusionfire  
Well, in terms of the practicality of a fusion device anything but billions of funding can achieve, we can exclude laser array inertial confinement and magnetic confinement.

Assuming you could construct a spherical explosive assembly, high synchronicity detonators, wave shapers, etc. - and use deuterated chemical explosives but no fissionable core - would it be possible to trigger nuclear fusion at the core?

Temperatures in imploding bubbles at room temperature is apparently 100,000K, with some simulations (probably assuming no design imperfections) predicting 10 megakelvins.

http://en.wikipedia.org/wiki/Bubble_fusion

The actual temperature you get obviously depends crucially on the sphericity of the imploding shock. Small deviations from a spherical convergent shock would lead to jetting, lower temperatures and premature loss of confinement.

What temperatures could you expect from an axisymmetric two-point initiation device?


I think its very possible as the Canadian fusion project using rammers on a steel sphere with liqud lead/lithium in it that is spinning around a hollow part in the center with D2/T2 gas.

I think their sphere is 3 m in diameter and around 200 rammers. So think of HE lens instead offcourse it must be a chance to work. I suggest some physisist on this phorum to do a simluation of this.

If its already done practically and succeded I think its top secret.

franklyn - 18-8-2011 at 15:05

The Neutron bomb or Enhanced Radiation bomb was conceived as a means to
greatly reduce collateral damage from the blast and heat of ordinary atomic
weapons , by relying on an increased radiation output upon detonation as the
killing mechanism. The small yield is mainly ~ 80 % neutrons the balance being
heat and blast.
Investigation into producing this effect on a smaller scale without fission never
amounted to much since the effects of the chemical explosion far exceed any
radiation that can be produced by such means , something approximate to
the effect of medical x-rays. Lightly covered in the paper cited above _
Explosive-Driven Hemispherical Implosions for Generating Fusion Plasmas
direct download => www.dtic.mil/dtic/tr/fulltext/u2/a121652.pdf

Producing neutrons is achieved by numerous means. Lithium metal and Polonium
210 , explosively driven together emits a shower of neutrons. Deuterium bound
with Tritium ( DT ) gas passing an electric discharge spark , emits a shower of
neutrons. An auxiliary device for initiating the chain reaction of an atomic bomb's
critical mass assembly called the neutron gun , consists of a tube with an electrode
at each end , a 100 kilovolt cathode shoots a tritium plasma at the anode of a
metal hydride ( scandium deuteride ) resulting in a neutron emission.

Because the output from a nuclear process is several million times greater than
the yield of chemical energy , nuclear reactions occuring in very small amounts
can optimally add significantly to the total energy that can be produced. The
conditions of pressure and temperature in detonating explosives are conducive
to processes as described. Lithium 6 fissioned by a neutron yields considerable
energy. A possible enhancement of conventional explosives may result with an
induced emission of neutrons within an explosive composition made in some
manner with Lithium 6.

.

woelen - 18-8-2011 at 22:49

I hardly see the point of deuterated explosives or explosives using heavier isotopes like C-13, N-15 and so on. Chemically, they are VERY similar to normal isotopes and so I expect the energetics of these 'enhanced' compounds to be very close to their normal counterparts. Surely there might be a few percents of difference in energy output per mole of compound and there also might be some differences in sensitivity, but do you really think that there will be some spectacular new thing?

Fusionfire - 18-8-2011 at 23:12

Quote: Originally posted by woelen  
I hardly see the point of deuterated explosives or explosives using heavier isotopes like C-13, N-15 and so on. Chemically, they are VERY similar to normal isotopes and so I expect the energetics of these 'enhanced' compounds to be very close to their normal counterparts. Surely there might be a few percents of difference in energy output per mole of compound and there also might be some differences in sensitivity, but do you really think that there will be some spectacular new thing?


VoD for deuterated explosives may even decrease contrary to usual VoD/density trends. The reason for this is that the chemical energy evolved from <sup>2</sup>H and <sup>1</sup>H chemical explosives is the same, being related to electronic re-arrangements. From the chemical energy evolved, a proportion goes to radiant/thermal energy (which we measure as temperature) and a proportion goes to bulk kinetic energy (which we measure as shock velocity/strength). For a given proportion of CE that goes to KE, increasing the mass of the constituent products decreases the velocity to conserve energy. The only way VoD for an isotopically heavier (hence denser) explosive could be higher is if the proportion of chemical energy becoming KE is higher.

The main reason for deuterated explosives is not alteration of VoD but because it is comparatively easier to initiate nuclear fusion in deuterium (D) nuclei than in protium (H).

Reason being, for a given mean particle velocity (determined by the temperature), D has about double the kinetic energy than H. As two positively charged nuclei are on a collision course with one another, their electric potential energy must be supplied by their kinetic energies until the nuclear force can take over. Therefore nuclei with double the KE but the same number of mutually repulsive protons have a higher likelihood of fusing.

If nuclear energy can be evolved in addition to chemical energy, then all bets are off and of course the deuterated system will have a much higher VoD due to greater total evolved energies.

http://en.wikipedia.org/wiki/Lawson_criterion

[Edited on 19-8-2011 by Fusionfire]

franklyn - 19-8-2011 at 03:24

The Question of Pure Fusion Explosions Under the CTBT
www.princeton.edu/sgs/faculty-staff/frank-von-hippel/Questio...

" Although U.S. progress in this area is classified , in early1992 the Russian weapon
laboratories reported neutron yields of 10^(13) to 10^(14) neutrons , corresponding to
the fusion of 10^(-10) to 10^(-9) grams of DT gas. The production of 10^(14)
neutrons would be accompanied by the release of an amount of fusion energy
equivalent to roughly 60 mg of TNT.

(10^(14))(18 MeV)(1.6 x 10^(-13) J / MeV) ÷ (4.6 x 10^(8) J / g of TNT) = 0.06 g TNT.

The associated radiation dose at one meter would be about 0.2 Gy (20 rads)
significant but not great enough to cause death in the short term."


The generated neutron flux can act on fissionable elements notably lithium 6 ,
doubling the yield from nuclear energy. Current yields of this scheme is slight ,
much less than that of the explosive itself. Applied research continues.

The Physical Principles of Thermonuclear Explosives, Inertial Confinement Fusion,
and the Quest for Fourth Generation Nuclear Weapons
http://www.e-ipi.net/isri/_media/publications:ag-09-01.pdf

Transparency Measures for $ubcritical Experiments Under the CTBT
www.princeton.edu/sgs/publications/sgs/pdf/6_3jones.pdf

Project Crystal: Lithium 6 for thermonuclear weapons
www.mcis.soton.ac.uk/Site_Files/pdf/nuclear_history/Working_...

.

PHILOU Zrealone - 23-9-2011 at 14:28

Quote: Originally posted by woelen  
I hardly see the point of deuterated explosives or explosives using heavier isotopes like C-13, N-15 and so on. Chemically, they are VERY similar to normal isotopes and so I expect the energetics of these 'enhanced' compounds to be very close to their normal counterparts. Surely there might be a few percents of difference in energy output per mole of compound and there also might be some differences in sensitivity, but do you really think that there will be some spectacular new thing?

Simple examples might give a hint...
1°)Normal CH3-NO2 has a density d= 1,137 g/ccm and a molecular mass of 61 uma.
If "isotopic CH3-NO2" (deuteriated; C13, O18 and N15) was made... it would display a Molecular Mass of 70 uma.
This would take the very same volume as the normal molecule but it would be 70/61= 1,14754 times denser.
The density of the isotopic would be in the range of 1,305 g/ccm and for CHNO explosives 0,1g/ccm increase in density corresponds to an increase of VOD of about 350 m/s...resulting here in an increase of 588 m/s.
The VOD goes then from 6300m/s to 6888 m/s.
2°)Normal N2H5NO3 has a density of 1,64 g/ccm and a VOD of 8900m/s and a MW of 95 uma.
"Isotopic N2H5NO3" would have a MW of 109 uma and thus a density of 1,8817 g/ccm. The resulting VOD would be 9746 m/s...

The higher the density of the normal compound the better the VOD increase for the "isotopic explosive"...

Fusionfire - 23-9-2011 at 14:37

Quote: Originally posted by PHILOU Zrealone  
Quote: Originally posted by woelen  
I hardly see the point of deuterated explosives or explosives using heavier isotopes like C-13, N-15 and so on. Chemically, they are VERY similar to normal isotopes and so I expect the energetics of these 'enhanced' compounds to be very close to their normal counterparts. Surely there might be a few percents of difference in energy output per mole of compound and there also might be some differences in sensitivity, but do you really think that there will be some spectacular new thing?

Simple examples might give a hint...
1°)Normal CH3-NO2 has a density d= 1,137 g/ccm and a molecular mass of 61 uma.
If "isotopic CH3-NO2" (deuteriated; C13, O18 and N15) was made... it would display a Molecular Mass of 70 uma.
This would take the very same volume as the normal molecule but it would be 70/61= 1,14754 times denser.
The density of the isotopic would be in the range of 1,305 g/ccm and for CHNO explosives 0,1g/ccm increase in density corresponds to an increase of VOD of about 350 m/s...resulting here in an increase of 588 m/s.
The VOD goes then from 6300m/s to 6888 m/s.
2°)Normal N2H5NO3 has a density of 1,64 g/ccm and a VOD of 8900m/s and a MW of 95 uma.
"Isotopic N2H5NO3" would have a MW of 109 uma and thus a density of 1,8817 g/ccm. The resulting VOD would be 9746 m/s...

The higher the density of the normal compound the better the VOD increase for the "isotopic explosive"...


But the same amount of chemical energy is being released in deuterated/non-deuterated chemical explosives of the same chemical formula.

A certain percent of the chemical energy goes to kinetic energy of the expanding products. If the mass of the products increases due to replacing protium with deuterium, assuming this proportion of CE -> KE is the same then the VoD of heavier isotopes will drop.

White Yeti - 1-10-2011 at 09:33

You could incorporate some lithium while you're at it. Lithium absorbs neutrons to form tritium, yet another fuel for fusion. Boron is another one. Boron will "fission" to form helium under extreme temperature and pressure and when bombarded with protons. So I think if you really want to try this out, you should try to use as many reactions as possible, not just d-d.