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Sulfur-free black powder equation and burning temperature?

I am trying to calculate the pressure that builds up in a closed vessel containing deflagrating gunpowder.  The web gave me several conflicting equations for 70% KNO3 : 30% charcoal gunpowder:

1.  6 KNO3 + C8H4O --> 3 K2CO3 + 8 CO + 4 H2O + 3 N2
Oxygen is unbalanced.

2.  6 KNO3 + C7H8O --> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2
Hydrogen is unbalanced.

3.  6 KNO3 + C7H4O --> 3 K2CO3 + CO2 + 6 CO + 2 H2O + 3 N2
Carbon is unbalanced.

None of these equations have 70:30 by weight saltpeter to charcoal, and none is balanced.  To fix the weight: it is possible that the charcoal is not to be completely dry, and that there are a few water molecules hiding out, but there would have to be 8:1 water to charcoal molecules for equations 1 and 2, and 9:1 for equation 3.  To maybe fix the balance and masses: if the charcoals had a coefficient of 2, equation 1 would need 1:2 water to charcoal, equation 2 would need 2:2, and 3 would need 3:2.  This is assuming that the saltpeter is carrying no water.

Potential equations with 2 moles charcoal and carrying water to achieve the right weight ratios and balances are following:
(I offer two potential equations each because I lack a reactivity series table and know not how extra atoms would come together.  Please alert me if the following equations are just B.S.)

4. a.  6 KNO3 + 2 C8H4O . H2O --> 3 K2CO3 + 3 N2 + 11 CO + 4 H2 + H2O + 2 C ?
4. b.  6 KNO3 + 2 C8H4O . H2O --> 3 K2CO3 + 3 N2 + 7 CO + 5 H2O + 6 C ?
Excess carbon would take oxygen from water, or stay carbon, or would the charcoal not be all used up?  I assume that the reactant water molecules will remain unchanged.

5. a.  6 KNO3 + 2 C7H8O . 2H2O --> 3 K2CO3 + 3 N2 + 11 CO + 8 H2 + 2 H2O ?
5. b.  6 KNO3 + 2 C7H8O . 2H2O --> 3 K2CO3 + 3 N2 + 10 H2O + 3 CO + 8 C ?
Same problem as #4

6. a.  6 KNO3 + 2 C7H4O . 3H2O --> 3 K2CO3 + 3 N2 + 11 CO + 4 H2 + 3 H2O ?
6. b.  6 KNO3 + 2 C7H4O . 3H2O --> 3 K2CO3 + 3 N2 + 3 CO + 7 H2O + 8 C ?
Same problem as #4

If anyone can confirm any of these equations or provide a more accurate one, please do.  Also, anyone who can tell me the burning temperature of the correct equation, please do.

lemonie2 years ago
You are trying to obtain a theoretical delta-H or similar?
I think you probably really want to test this, because fast heterogeneous reactions with radicals, temperature-curves, pressure-curves, thermal-transfers to cylinder-walls etc are rather difficult to model.

-What are you doing...?

L
tincanz (author)  lemonie2 years ago
I don't think testing is very feasible, because I have no access to sulfur-free black powder and don't want to try to make it at this time. I want to calculate the pressure exerted by the gasses emitted by burning gunpowder. If I know the proper equation (maybe Jack A Lopez just gave it to me below) then I can calculate the moles of gas created from particular masses of reactants. I can then use the burning temperature (which I don't know) with PV=NRT to find what thickness and material sphere- or cylinder-walls can be to either withstand the pressure or to break.
lemonie tincanz2 years ago
Do you know what "muzzle-flash" is?
These reactions are about as far from the ideal gas law as gases get.
You would do best to look for data that already exists, e.g. http://www.frfrogspad.com/intballi.htm

L
tincanz (author)  lemonie2 years ago
This helps very much. Thank you.
Prfesser1 year ago
The reason for the conflicting equations is that there is no single simple stoichiometric equation that represents the combustion process at higher temperatures. For example, the products at about 1500 C and 1000 psi include K2CO3, N2, CO, CO2, H2O, H2, and tiny amounts of several other species.  When the chamber pressure is changed, the temperature and amounts of products will also change.

Professionals sometimes use PROPEP or GUIPEP or another modelling programme to predict results of propellant combustion.
tincanz (author)  Prfesser1 year ago
Oh, thanks. I'll look up those programmes.
rickharris2 years ago
Errr! High pressure - Closed container - Your making a bomb WHY would anyone her help you you could do anything with it.
tincanz (author)  rickharris2 years ago
Its only gunpowder. Its not a high explosive. It would need a humungous amount of reactants to be present to hold any threat. If I was a terrorist, I could just run tests to find the pressure and decide the ideal container based on the results. This is theoretical, hypothetical, and safe.
But then again if you intended to do harm you would say all that wouldn't you?
tincanz (author)  rickharris2 years ago
Yes, but if I intended to to harm I would not need these equations because I could just try different materials and see which sends shrapnel the farthest or with the most force. I am trying to learn this for the sake of knowledge. All manufacturing processes used by the "modern world" should exist in people's heads, because then they appreciate the created objects much more. Besides gunpowder, I am learning about glassblowing, bloomsmithing, blacksmithing, pottery, carpentry, spinning, weaving, tanning, rubber-making, paper-making, farming, saving seeds, maintaining soil integrity, masonry, electrical sysems, and navigation. They are all being learned solely to preserve human knowledge and make it more accessible to the common person.
For those of you just starting this thread, it looks like the empirical formula for charcoal, C7H4O, and some of the chemical equations came from this section, of the Wikipedia article on gunpowder:
http://en.wikipedia.org/wiki/Gunpowder#Sulfur-free_gunpowder

I think this equation
6 KNO3 + C7H4O --> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2
[7C,4H,6K,6N,19O]
is balanced.

And this one too:
4 KNO3 + C7H4O --> 2 K2CO3 +  5CO + 2 H2O + 2 N2
[7C,4H,4K,4N,13O]

And the mass ratios of grams KNO3 to KNO3 and charcoal together are:

606/(606+104) = 0.85 for the first reaction,

and 404/(404+104) = 0.80 for the second one,

with the main difference between those reactions being the first one has enough oxygen to make CO2, and the second one just enough to make CO.

I admit neither of those numbers is exactly equal to 0.70, but they're kind of close.  I  think for this kind of reaction, there is going to be a lot of guesswork here anyway. 

I mean you might get more meaningful results from just mixing stuff together and lighting it on fire.

Which makes me think:  you are playing with fire here.  So you should take those precautions necessary to keep from blowing yourself up.

Wearing safety glasses is probably a good idea.
tincanz (author)  Jack A Lopez2 years ago
Maybe this means that the ideal ratio is actually 8:2, because they both produce 9 moles of gasses and the latter equation uses fewer reactants and produces less smoke. Thank you for the help.