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Calculating Energy to Pressure

  1. Oct 29, 2009 #1
    I am a student at the University of Delaware, working for the Disaster Research Center under grant from the National Science Foundation. My project has me analyzing building collapses and rubble/debris pile structure. I have been loaned a program by the Livermore Laboratory called Cheetah which calculates explosive pressures. This program has been unbelievably useful but the output for particular data is given to me in energy (Kg/cc) and I need it in a pressure form to calculate the pressures acting on the structure.

    How can I convert the energy to pressure? Remember this is acting on a structure and is not contained in a specific volume other than the container that the device is located in (which disintegrates upon initial reaction) and is not moving the entire object (so calculating the amount of work being done is a no go). Thanks for any help you can give me.
     
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  3. Oct 29, 2009 #2

    Andrew Mason

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    Pressure is a measure of the energy density of a pressurized fluid. If you are given units of energy per unit volume (this would be KJ/cc perhaps but not Kg/cc which is simple mass density), that would be the internal energy of the gas per unit volume. The relationship between pressure and internal energy depends on the specific heat of the gas. If we are dealing with air, the Cv of air is about 5R/3 (mostly diatomic gas) where R is the gas constant. So to convert Energy density to pressure you would multiply by 3/5:

    U = nCvT = 5nRT/3

    PV = nRT (ideal gas equation)

    So: U = 5PV/3; P = 3U/5

    Note: energy is a measure of a force multiplied by the distance through which that force acts. E = Fd. Pressure is force per unit area: P = F/A = F/(V/d) = Fd/V = E/V. Pressure x volume = Energy. Pressure x change in volume represents the work done by a gas when expanding in volume.

    AM
     
    Last edited: Oct 29, 2009
  4. Oct 29, 2009 #3
    I did not see the typo, yes I have the energy in kJ/cc. When I go in tomorrow I will try to work out the calculations and see if they match up to what I expect. Thanks for your help!
     
  5. Nov 2, 2009 #4
    I was looking through and I am still not getting what I am expecting, the issue is what I am expecting was previously calculated by experts.

    What I know:

    The compound is 94.5% AN (2,057kg, Density: 0.84 kg/L) and 5.5% Fuel Oil (120kg, Density: 0.86 kg/L)

    This is stored in (13) 55-gallon drums

    Total density = 2,177,000g (components) ÷ 2,706,569 cc (capacity of drums) = 0.804
    g/cc

    The velocity of detonation (VoD) is: ~3,900 meters per second

    The software from the Livermore Laboratory has given me this based on those numbers: The mechanical energy of detonation = -3.345 kJ/cc

    Sorry for not understanding this and thanks for any help you can provide.

    (I am getting core pressures of 1.96 GPa, but that is a little more than half what experts have gotten)
     
    Last edited: Nov 2, 2009
  6. Nov 5, 2009 #5
    I found an error in the densities so I started using those supplied by the Livermore Lab and came up with numbers that are twice what was expected. Unfortunately I do not know how the original pressure was calculated and the guy who did it is not available to explain. His core pressure is 500,000psi and I'm getting around 1 million. The other inconsistency I am getting is that the pressure from ANFO is higher than ANNM (which is more explosive).

    I am consulting a former NASA physicist here at the University to see if I can figure that out. MY question now is, am I using the correct formula to calculate pressure drop off at distance x. I have this entered in an excel spreadsheet so I don't have to do hundreds of calculations.

    I am using this: =SUM(A4/(B5)^3)

    where A is the pressure and B is the distance. I know that pressure drop off is an inverse function of the distance cubed, and that after a certain distance it is to drop to an inverse function of the distance squared. I don't know where that magic number is but considering the structure is 10' away from the device I don't need the 2nd. Does this look correct so far? Thanks
     
  7. Nov 5, 2009 #6

    Andrew Mason

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    Pressure is a function of volume and temperature. Pressure will vary inversely as the volume and directly with temperature (in Kelvin - 0 K = -273 C) in an ideal gas that is in thermal equilibrium. As the gas expands, it cools because it is doing work (ie. converting its internal energy into mechanical work). The temperature will depend on how much internal energy is converted to mechanical work. The more rapidly it performs work, the more rapidly it cools and therefore the more rapidly the pressure decreases.

    For example, if the explosion takes place in open air, the exploding gases do work [itex]P_{atm}\Delta V[/itex] against the atmosphere. If the explosion is inside a building, the explosion does a lot more work in ripping apart the building. In the first case, the pressure front of the exploding gases will be higher for a given volume than in the latter case.

    To further complicate matters, the exploding gases are not in thermal equilibrium - there will be a temperature gradient within the exploding gases. This makes it difficult to analyse. I expect that the Lawrence Livermore lab people can help you do that.

    AM
     
  8. Nov 5, 2009 #7
    Thanks Andrew, you have been a great help for me in understanding how it all works. I have been working with them over the past few months but I just want to double and triple check my work and numbers before I put them in this report. Thanks again!
     
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