I'm not sure if this has ever appeared here before but can anyone provide information on the quantity of gas generated by one gramme of gunpowder at ambient temperature and pressure?
Jump to content
Posted 29 July 2016 - 05:17 PM
Posted 30 July 2016 - 12:16 AM
I will look it up in the morning, it is in the RSC book
Edited by digger, 30 July 2016 - 12:16 AM.
Posted 30 July 2016 - 07:49 PM
Shidlovskiy P70 Principles of Pyrotechnics lists a value of 280 cm3 g-1 @ V0 for "Smoke Powder" [ KNO3 75%, Wood Charcoal 15%, Sulfur 10% ] , gaseous combustion products given as CO CO2 H2O N2 So that's quite good as it includes the water vapour produced.
why he calls it "smoke powder" I have no idea.
V0 is 00C and 760mmHg
If you want to know what volume the gas occupies at a different temperature (Vt) you use the following formula ( it's at the bottom of page 67 ) :
Vt = 280 * (1+0.00366 * T)
so 25 degrees it's :
280 * (1+0.00366 * 25) = 305.62 cm3
so you could equally find out the volume at the reaction temperature by substituting the combustion temperature of BP
Exactly how accurate it is I don't know, someone else might have a different answer.
Posted 02 August 2016 - 05:05 AM
Two things are infinite: the universe and human stupidity; and I'm not sure about the universe. ― Albert Einstein ― Insanity is doing the same thing, over and over again, but expecting different results.
0 members, 0 guests, 0 anonymous users