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Conversion relationship between TNT equivalent and explosion pressure

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P0=140e6[N]*(Q/1[kg])^(2/3) "Load magnitude"

Hello everyone, I am having a problem now. I don't know what kind of physical quantity 140e6 is in the formula. How did it come from? Seek answers from the great gods, thank you first!


2 Replies Last Post 2 mai 2019, 08:09 UTC−4
Henrik Sönnerlind COMSOL Employee

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Posted: 6 years ago 30 avr. 2019, 09:22 UTC−4

Hi,

The quoted expression is the definition of a parameter in the Wave Propagation in Rock Under Blast Loads example. That example is based on data in a report I wrote some 15 years ago, so I do not remember all details.

The force amplitude 140e6[N] must be considered just as a value that may be known from for example experience. In real life, there is of course some area on which the explosive acts, so the force is derived from a pressure amplitude times that area (which is not directly known in this example )

The fact that the peak pressure during an explosion is proportional to the power "2/3" of the amount of explosive is however well established in that field, if I remember correctly.

Regards,
Henrik

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Henrik Sönnerlind
COMSOL
Hi, The quoted expression is the definition of a parameter in the *Wave Propagation in Rock Under Blast Loads* example. That example is based on data in a report I wrote some 15 years ago, so I do not remember all details. The force amplitude 140e6\[N\] must be considered just as a value that may be known from for example experience. In real life, there is of course some area on which the explosive acts, so the force is derived from a pressure amplitude times that area (which is not directly known in this example ) The fact that the peak pressure during an explosion is proportional to the power "2/3" of the amount of explosive is however well established in that field, if I remember correctly. Regards, Henrik

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Posted: 6 years ago 2 mai 2019, 08:09 UTC−4

I think I understand, thank you very much for your answer.

Regards, Mr. Li

I think I understand, thank you very much for your answer. Regards, Mr. Li

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