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Nitrate Flash Stoichiometry


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#1 alany

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Posted 06 January 2005 - 03:15 AM

Degn & Allen (et al) suggest this composition for "Nitrate Flash":

5 Potassium Nitrate
2 Aluminium (dark pyro flake)
3 Sulfur

Try as I might I just can't work out the reaction(s) this composition is designed around.

I started with the basic Potassium Nitrate + Sulfur reaction to yield Oxygen (and Potassium Sulfide and Nitrogen waste):

2(KNO3) + S => K2S + N2 + 3(O2)

Then I considered Aluminium burning in Oxygen:

4(Al) + 3(O2) => 2(Al2O3)

Nievely wacking these together you get:

2(KNO3) + S + 4(Al) => K2S + N2 + 2(Al2O3)

Which if you subst in the molar masses and divide through by the smallest gives you roughly:

63 Potassium Nitrate
34 Aluminium
10 Sulfur

(59:32:9 normalised to percent)

Only Miller has something similar with 6:3:1 and 6:1:1 nitrate flashes in "Bangor Powder" from the PFP - the only reference I can find even including rec.pyrotechnics which talks about the PFP.

What other pathways is the reaction taking to make the 5:2:3 composition work?

Aluminium and Sulfur can react directly with each other I guess:

2(Al) + 3(S) => Al2S3

That is a mass ratio of about 1.8:1 S:Al or:

18 Sulfur
10 Aluminium

That means there is way too little Aluminium in the 5:2:3 composition for it to be a simple mixture of both reactions. It must be "burning" sulfur to produce a gas rather than just using it to soak up all the Potassium from the Nitrate and relase all its oxygen.

Using the Potassium Disulfide (3.2:1) reaction instead of the Potassium Sulfide (6.3:1) one just makes things worse. In practice it would probably burn slowly producing a bubbling puddle of potassium polysulfides.

I don't get it! Degn sourced compositions are often 'odd' but so many people use this nitrate flash composition I would have imagined it was well tuned. I'd like to understand what tricks it is using.

BTW: I haven't tried this ~ 6:3:1 in practice, I wonder how it compares to the traditional 5:2:3?

Now, for Barium Nitrate flash:

Ba(NO3)2 + S + 4(Al) => BaS + N2 + 2(Al2O3)

I get (normalised to percent for comparison):

65 Barium Nitrate
27 Aluminium
8 Sulfur

While the conventional composition (Lancaster) is:

68 Barium Nitrate
23 Aluminium
9 Sulfur

Which is a pretty good match, slight over-oxidisation and additional sulfur is a common trick to speed up a composition. I can't be totally on the wrong track with 6:3:1 (potassium) nitrate flash can I?

[edited to fix typo in title]

Edited by alany, 06 January 2005 - 09:54 AM.


#2 paul

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Posted 06 January 2005 - 09:23 AM

I think it is not that easy. Different temperatures = different combustionproducts. The thermo-dynamical behaviour of mixtures with more than 2 reactants is very hard to calculate.

So, I wonder, that your calculation is THIS exact. Good work, if it works for other formulas too :D

Edited by paul, 06 January 2005 - 09:25 AM.

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#3 Guest_Daniel Scott_*

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Posted 06 January 2005 - 09:48 AM

Not meaning to sound like a smart arse or anything, your very intelligent with chemistry but what about moisture in the air? Would that not effect burning speeds? Also particle size? If you have BP thats ball milled and one thats just 100mesh charcoal but the rest is ball milled the decomposition products would be different because the not all the charcoal would be "burnt" in the mix but rather thrown out as sparks.

#4 alany

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Posted 06 January 2005 - 10:42 AM

Paul: Yes most reactions are never that simple. The Nitrate/Sulfur one in particular has many paths it can follow, producing sulfur oxides, metal polysulfides, oxygen, nitrogen, nitrogen oxides, sulfur nitride polymers, metal nitrides, etc.

Many simple compositions are based on the assumption that a single reaction occurs and minor pathways can be ignored. For example BP, its ratios are tuned only slightly (by experiment) from a stoichiometric mix that nievely assumes a simple reaction (despite huge amounts of evidence that BP is anything but simple) not too different to the one I am assuming here. ("Stoichiometric BP" is approximately 75:13:12).

What I am trying to figure out is what assumptions or rules the designer of the 5:2:3 composition was using. i.e. Try to read their mind and reverse engineer the reaction(s) they assume are happening. Using the BP example again it is easy to see that 75:15:10 BP is "pretty close" to the stoichiometric ratios you get if you assume the reaction is dominated by:

2(KNO3) + 3© + S => K2S + N2 + 3(CO2)

It is quite possible that the 5:2:3 composition was derived by trial an error, in which case it may be the empirically "fastest" you can get for "round numbers". The thing that interests me (and prompted this whole piece of research) is why it is so vastly different than the "first guess" stoichiometric composition. Does this composition make use of some chemistry that I don't understand or haven't considered? I'd love to find out!

Daniel: I don't think humidity is a big factor in the nitrate flash composition design. It wasn't really the speed that worried me, wet or dry the design may be assuming something I'd like to understand. Conventional 7:3 flash is designed to make Alumina and Potassium Chloride with no regard for anything else, yet it works quite well and is empirically known to be the best "round numbers" composition. (Stoichiometric Flash is approximately 66:34 - hence why 2:1 is very popular too).

You do make a good point about mesh sizes and the practical reality of compositions. The system isn't "closed" so partially consumed reactants escape, as do intermediates from all the pathways. The temperature and pressure of the reaction also drives which pathways are more favoured than others, no doubt analytical estimates are fantastically complex and would keep professional chemists busy for years. BP is very well studied and good approximations are in the texts, but they were largely derived experimentally.

I understand that there are tables of the oxygen requirements for various fuels in the texts, but so far I haven't seen them in anything I've purchased (anyone know where I can get them?). They are experimentally derived values that take into consideration the conditions typical in stars (for example) and are used as guides for tuning the oxygen balance in compositions.

#5 cat

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Posted 06 January 2005 - 08:54 PM

Very good post Alan, I don?t see many well thought out posts like this. I?m going to express my *opinion* of the reaction. I?m only going to consider an empirical type reaction as I can?t predict the actual products produced (in other words I?m excluding the (abnormal sulfur oxides, metal polysulfides, sulfur nitride polymers, metal nitrides, etc).

Daniel is also right water will affect the products produced in this reaction but, for simplicity I?m going to assume a dry comp. We are also going to assume that this comp is contained and that outside reactants (air, moisture, etc) will play no role in the reaction.

I know this somewhat contradicts what you?ve already read but from experience I?ve learned when there is KNO3 and S in a comp they react first (let the snide comments begin). The sulfur works as a fast fire since the reaction doesn't require much energy.

Let?s pretend that the reaction will continue until the limiting reactant (KNO3/K) is used up

2KNO3 + S - > K2S + N2 + 3O2

This would consume 5mol (all) of KNO3, and 2.5mol (5/6ths) of S that will produce 7.5mol O2

The remainder of the sulfur will be consumed to create Sulfur Dioxide (max 0.5mol O2 consumed)

O2 + S -> SO2

The nitrogen will react with the oxygen creating various oxides of nitrogen (max 2.5mol N2 + 1.25mol O2)

2N2 + O2 -> N2O

That?s if the reactions proceeded in subsection, rather then all at once without producing oxides of potassium

We still have the Al and 5.75mol of Ox to consume

4Al + 3O2 -> 2Al2O3

That will consume all the Al and 1.5mol Ox (leaving 4.25mol Ox)

2.5mol K2O, could consume a max 1.25mol O2,

2K2S + O2 ?> 2K2O + 2S

The (now) free sulfur (2.5mol) can consume an equal amount of O2,

O2 + S -> SO2

Leaving .5mol O2, who knows maybe that?s lost as free nitrate ions, dissipated, consumed in the combustion of the casing. Even a binary product created by elemental reactants will contain more/less complex reaction products then the empirical formula.

OK, it?s been many, many years since chemistry class any comments?
-cat

#6 alany

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Posted 06 January 2005 - 09:24 PM

Cat: Yep, It seems reasonable to assume the excess sulfur and oxidiser will react to produce sulfur dioxide (stealing sulfur from the potassium sulfides as well asremaining free sulfur) and the excess oxygen combine with the nitrogen in the high temperature flame to form nitrogen oxides.

The question becomes which works better in practice I guess. The extra gasses might make the 5:2:3 composition a better report charge.

#7 cat

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Posted 06 January 2005 - 11:46 PM

Alan: Sorry I didn't see this question above. For a report a violent reaction counts for more then a gaseous reaction. Ever use Al that?s too fine when working with CuO or PbO thermite? Neither creates (much) gas content but both have substantial brisance. KNO3 is not known for its ability to release its Ox quickly, my first suggestion is to look for a more suitable Ox supply (yes I know not everyone has access). The earlier post ignores some basics for instance once the sulfur and KNO3 begins reacting enough energy and heat is present to start reacting Al+ S/O a considerably more energetic reaction when speaking in chem calories.

It would be nice if someone had a reference, listing some basic reactions and ccal?s released.

As a side note what did you mean by ?tables of the oxygen requirements for various fuels? ?

#8 alany

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Posted 07 January 2005 - 08:30 AM

Several people on rec.pyrotechnics have mentioned that such tables exist, I guess I should just ask the question there. They apparently contain 'real world' one mole of X needs Y moles of O2 information. They are not stoichiometric tables, which anyone could just compute, they are based on oxygen demand experiments.

I don't think I really need them, but it would be interesting to read the text.




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