Gas Problem Dealing with Explosives

  • Thread starter hankw
  • Start date
  • #26
Danger
Gold Member
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I afraid that you're way out of my league now. My "Blasters' Handbook" doesn't contain those sort of formulae. It's restricted to tables of explosives' properties, along with how many sticks of what go where, and electrical formulae for designing the detonation circuit. Sounds like you have some heavy experts helping you already, though.
 
  • #27
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Pervect, Danger, and all,

I am going to have to work through the ideal gas law and adiabatic process. I have never dealt with gases before. I will be back in a few days.

Hank
 
  • #28
Clausius2
Science Advisor
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I Hope we are not collaborating to another London's bombing or with an AlQaeda member....(joking).
 
  • #29
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Sorry for not returning as i promised. I was out of town for a construction job. I am back for a while.

I took a long look at those two analysis i posted earlier. I agree with Pervect. They are backdoor-ish. Almost misinformation. They both throw jargon around and then go through a laborious effort of determining energy, pressure, velocity. Who needs this? After i studied the Gas laws some, it looks like all one needs is the amount of explosive gas and a temperature to determine how much air volume an explosive creates. That is a two-step problem! Since the explosion is close to a adiabatic process, one does not need to consider heating of the surrounding air. How simple can it get? Am i right?

Below is information i collected on TNT.

C6H2 (NO2) 3CH3 -> 6CO + 2.5H2 + 1.5N2 + C
Gas: 1 mol TNT/10 mol gas
Weight: 227.1 g/mol
Temp: 3,000K (trying to confirm this)
Velocity: 7,197 m/sec
Pressure: 187,807 atm

I don't see how i need to consider things such as velocity and pressure. I just want to find out the final air volume. I think i can just use the ideal gas law. PV = nRT. I am going solve for kg so i update the gas and weight numbers to reflect a kg quantity- 4.40 mol/kg and 44 mol gas/kg.

PV = nRT
T = 273 + 3,000 + 21 = 3293 K
1 atm * V = 44 moles * 0.08 * 3,293 K
V = 11,591 L

So (1) kg of TNT produces about 11,591 liters of air volume. Is this right? It seems that the hardest part about all this was deciphering all the encrypted information in those two analysis. Sheesh. I didn't think science was that hard.

Since i am self-taught, i hope someone can confirm that i am doing things correctly so i don't learn things wrong. Thanks for all the help.

Hank
 
  • #30
ZA
13
0
explosives

Although my gas laws education is basic, I can say that you are dealing with too many variables to solve the problem here. The thing is, as already mentioned, you don't know so many things about the room itself. for example, how would you know at which pressure the room would be destroyed? Maybe more explosives than were needed were used. Maybe a window shattered and let the gasses out quickly, giving you a variable air pressure and maybe as i already mentioned, there was a blast crater in the floor, weakening it and causing a structural faliure. Another thing is that you would need to use (as already stated) Cv gas values, not Cp. I have tried and tried to calculate certain aspects of rocket fuel (no luck so far, but I'm getting there) and even an idealized and simplified version of the reactions is VERY complicated. Another question. how do you know that the explosives were tnt or c3 and not c4 or something (I don't know the formulas for c3 or c4, so I'm just assuming)? Finally, you have to account for Vander Waals forces (I probably misspelled it) because you're dealing with extreme heats and temperatures and Ideal gas laws are designed for very low pressure and non polar molecules (and probably other things, I forgot). By the way, real gas laws are a major problem in rocket engine calculations, (ie: gamma values). If you could actually solve for the exact mass and identity of the explosives used under certain conditions (withinn 20%) I would be truly amazed. good luck!
 

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