Thread: Firearm science question... View Single Post
 October 29, 2012, 09:58 PM #11 Jimro Senior Member   Join Date: October 18, 2006 Posts: 7,089 Unclenick, can you crunch some numbers and see if any of my calculations are remotely valid? I was going over my math and assuming that we 2.7 liters of gas is the max produced, we divide back down by 2's until we reach "approximately" our starting point for the solid. 2.3 liters/(2 to the 8th) = 0.07, so we are looking at 32 atmospheres of compression (inversely related, volume and pressure). 1 atmosphere is roughly 15 psi (slightly less, I'm rounding for easy math). 15x32 is 480 psi. So how do we turn 480 psi into 50,000 psi? Heat. P1/T1 = P2/T2, so 0.0027m3/293k = 0.00008m3/X solve for x... 0.00008(293)/0.0027 = X = 8.61 degrees Kelvin. Which means to maintain 1 atmosphere of pressure there is a minimum difference of 285 degrees kelvin (or Celcius) between the solid and gaseous states, that's a 545 degree Fahrenheit difference, which seems about right to me based on metal spalling of projectiles. So we start with a solid at normal temp, it starts turning to gas, we get 32 atmospheres of pressure from gas alone, but also a 285 degree C increase in temperature, which would need to account for the remaining pressure, and once again I'm too tired to go back to the equation and figure out what an increase of 285 K would do, especially since I calculated for the volume of powder, and not the volume of a cartridge case. And this is assuming that there isn't "waste heat" somewhere in the system. __________________ Machine guns are awesome until you have to carry one.
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