
Post by matterbeam on May 17, 2017 14:35:29 GMT
Hello! I'm having trouble with a set of equations meant to describe the velocity of an Explosively Formed penetrator, called the Gurney equations. Specifically, the kinetic energy delivered by the explosive charge, and the kinetic energy contained in the EFP, do not add up. Here is the setup:M is the metal plate, or the 'flyer'. C is the explosive charge. N is the tamper or backplate. In our scenario, M is the projectile, C is the beryllium filler and N is the nuclear warhead of mass about 10 times greater than explosive charge. It allows us to consider the metal plate as infinitely tamped. Here is the equation:Vm: Velocity of the metal plate E: yield energy converted into thermal energy within the filler. (2E)^0.5 is the specific velocity of our device. M: mass of metal plate C: mass of filler Let's use a 1 kiloton yield warhead, massing 100kg. It is configured like a pulse propulsion unit for an Orion driver, delivering about 85% of its energy into heating a beryllium filler. This is 3.56 TJ. The beryllium masses 10kg. The metal plate is 10kg. Using the equation, we get an EFP flying out at 2311km/s. The problem:10kg at 2311km/s contains 26TJ of kinetic energy. This is higher than the energy delivered by the warhead. What could be the problem?



Post by leerooooooy on May 17, 2017 17:30:05 GMT
The formula is almost certainly an approximation that only works for far lower speeds and energies. Just like how kinetic energy = 1/2mv^2 breaks down completely for v near the speed of light



Post by matterbeam on May 17, 2017 17:33:26 GMT
The formula is almost certainly an approximation that only works for far lower speeds and energies. Just like how kinetic energy = 1/2mv^2 breaks down completely for v near the speed of light So which model is appropriate?



Post by Enderminion on May 17, 2017 22:13:28 GMT
4Tj of heat but how much radiation and plasma/light?



Post by n2maniac on May 18, 2017 7:52:29 GMT
Units do not match up in the equation. Right side is unitless. Left numerator is velocity. Denominator is sqrt(energy), which is a factor of sqrt(mass) away from matching up. I think it is probably meant to be sqrt(2E/C), but the C got omitted somewhere and now people just look at a table for the value of sqrt(2E). Taking the limit as C>>M, the value of V should be a bounded constant.
Relativity can still be ignored here.



Post by underwhelmed on May 19, 2017 13:36:39 GMT
Wiki says sqrt(2E) is the Gurney constant and has dimensions of velocity.



Post by leerooooooy on May 19, 2017 19:41:14 GMT
Wiki says sqrt(2E) is the Gurney constant and has dimensions of velocity. Then the formula cannot be applied here because metal is not an explosive and there is no way to calculate said constant for it



Post by n2maniac on May 20, 2017 0:16:09 GMT
Wiki says sqrt(2E) is the Gurney constant and has dimensions of velocity. Okay, but inputting total energy applied to the Be will need mass divided somewhere for units to be correct.



Post by matterbeam on May 20, 2017 11:59:16 GMT
Wiki says sqrt(2E) is the Gurney constant and has dimensions of velocity. Okay, but inputting total energy applied to the Be will need mass divided somewhere for units to be correct. Further research into conventional explosives (which probably put me on a list somewhere) shows that the Gurney constant is an energy density. However, there is not simple relationship between energy density and detonation velocity because the explosive gasses have different molecular weights. So, for most hydrocarbons at 23000K temperatures, detonation velocity is a third of the gurney constant. It cannot be scaled up accurately to nuclear levels of energy or with any propellant.



Post by n2maniac on May 21, 2017 6:25:44 GMT
Sanity check: over how long does the nuclear blast last?



Post by matterbeam on May 21, 2017 10:27:06 GMT
Sanity check: over how long does the nuclear blast last? 10 nanoseconds.



Post by Enderminion on May 21, 2017 14:44:12 GMT
Sanity check: over how long does the nuclear blast last? 10 nanoseconds. thats how long a fission reaction takes, since its a CHAIN reaction I would say <200 nanoseconds



Post by n2maniac on May 21, 2017 15:47:43 GMT
Okay, so I would be suspicious of seeing a <200mm thick piece of Be absorb that and tries to push something much faster than 200mm/200ns = 1000km/s (dissassembles itself more efficiently as this ratio is approached). Not itself necessarily a limitation. Echoing leerooooooy , the piece of Be will not act like an explosive. It is going to be heated more uniformly by the blast. I think all this really changes is that less energy is forced to go into the Be. This may require a different model.

