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Author: Subject: Computing energetic properties and crystal density of energetic materials
Engager
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Today's research in energetic materials generates tons of new substances, but their properties are almost always missed or incomplete. Sometimes all that you know about interesting compound is it's molecular structure (e.g bond chain and ordering). In order to decide is it worth to spend time and resources on it's synthesis, abilty to predict energetic properties of new compound compound will be very helpful. In general, prediction of properties is a hard work, but can be done even on simple home PC with use of common chemical software. I my post i've tried to show it in most simple, general way, without deep going into special details. Much more detaled information about HEDM properties prediction is avialable in journal articles and quantum chemistry books. Generaly, considering new energetic material we are most interested in four main explosive properties: detonation velocity (this is proportional to substance density and heat of explosion), heat of explosion (this contributes to explosion power), volume of explosion gasses (affects explosive power ant throwing ability) and sensitivity of explosive. All this, except sensitivity can be predicted by means of some chemical calculations, without any experimental data required.

Essential software

First off all you will need to get some common chemical software, such as molecular structure modeling tools and quantum chemistry package. Some additional visualization and file editing software is usefull as well. Here is a list of programs, witch i found apropriate for our task.

1. Chemoffice Suite. This one is surely familiar to many people who use PC for chemistry. This package contains some general compound modeling tools and some more advanced features, such as rectefication of structure using data from build in compound structure database or simple optimization tools like MM2. Program can also serve as generator of input to further optimization by quantum chemical software. You can review it here: www.cambridgesoft.com/software/ChemOffice/.

2. PC Gamess quantum chemistry package. Surely other quantum chemical codes such as Gaussian, or NWChem can also be used, but Gamess is best in terms of speed and is also free for academic use. PC Gamess is a freely available ab initio and DFT computational chemistry program developed to offer high performance on Intel-compatible x86, AMD64, and EM64T processors. It was initially based on the free GAMESS(US) program sources but extends its functionality in some important areas. The PC GAMESS project is maintained by the staff of the Laboratory of Chemical Cybernetics at Moscow State University (MSU). The project coordinator and leading developer is Dr. Alex A. Granovsky. You can download this package from official server here: http://classic.chem.msu.su/gran/gamess/index.html.

3. In order do calculate substance crystal density using electron density maps provided by PC Gamess, you will need special utility witch i've written specialy for this small article. If you are using Windows XP or later they will run without any setup, if you use other outdated OS you will need VB Runtime Library installed in order to run it. Utility + source code + article from Journal Of Hazardous Materials describing density prediction principle + examples of PC Gamess input & output + Chem3D files used for calculation, can be downloaded in single archive here: http://rapidshare.de/files/40492387/Density_Calc.rar.html

4. To make your work more comfortable and enjoy visualization on the fly you can get some aditional tools such as Chemcraft (visualization tool for PC Gamess and 3D molecular structures), Ultraedit32 file editing tool, capable for fast browsing and editing of big files and FAR file mannager for easy dos style file operations. Chemcraft is shareware, but is fully functional for 30 days after first run, there is also simplified free version avialable. You can get this program here: http://www.chemcraftprog.com.

Then you got essential software, you can beging compound structure modeling. Properties witch we want to calculate are only drop in the ocean, from listed software capabilities, so this programs can be very usefull in many other fields of numerical chemical research.

Generation of initial compound structure

After MM2 optimizaton your molecule is much closer to real one, but must be further optimized to get valid structure. Final optimization and energy + properties prediction require job of quantum chemical software. Now you must export structure from Chem3D to PC Gamess input file. Chem3D can export PC Gamess input files, however it can't set correct computation options in it, so you must create this file by hand. Gamess input deck require set of all atom coordinates in your molecule, and assumes that molecule is better or less optimized, overvise calculation may not converge. This is a bit of art in providing good initial structure, but one optimized by Chem3D usualy works well. To export file with atom coordinates in format requred by PC Gamess, save file with structure from Chem3D, selecting "Gamess Input" file format. Now our initial structure is complete and we can proceed to PC Gamess quantum chemical calculations.

Predicting heat of formation and optimizing molecular structure

As you may know, compound's heat of formation is related to structure of compound and is a measure of it's energy, so positions of atoms in molecule relative to each other precisely defines structure energy and dirivated heat of formation value. This is not completely precise, because we take our molecule as single one in free space (igeal gas), but in real crystaline cell molecule is always surrounded by many others and this also affects structure energy. However energy difference done by placing molecule in crystalline cage (heat of vaporisation) is usualy much lower in value then heat of formation, and can be completely skipped if we solve our problem in gas phase aproximation. Heat of formation in solid or liquid state is always lower then one in gas phase, but heat of vaporisation is hard to be predicted due to extreme complexity of calculations needed for it. You should realize that quantum chemestry calculations require a lot of computing power, and ammount of calculations rise dramaticaly then using higher level of theory or growing number of atoms in molecule. In order to compute middle scale HEDM molecule in reasonable time while still getting good quality output property values, we must select computation method very carefully. The most suitable method for our case is semiemperical PM3, witch is fast and good in prediction of geometry and heats of formation. Now we must arrange input file for our job, this file has general structure witch is shown on left picture below.

To give you any idea about what exactly listed parameters do, i give them a short description. First parameter group ($CONTRL) are most importans, as they control the job of PC Gamess program, and designate target of calculation. Our target is specified by keyword "RUNTYP=OPTIMIZE" informing a program that we need to optimize molecule structure and get it's energy, keywords "COORD=CART" and "UNITS= ANGS" set mode of input cordinates to cartesian 3D system using angstrom (10^-10 m) as units of lenth, keywords "SCFTYP=RHF" and "MAXIT=50" forces program to use Restricted Hartree-Fock method to solve Schrodinger equation (core of quantum chemistry theory) for our molecular system and set iteration limiter for it at 50 iterations.$GUESS and $BASIS groups determine choce of initial aproximate orbital primitives, witch will be used for numerical solution of Schrodinger equation in our molecule, more detailed information about basis sets can be found in quantum chemistry literature.$STATPT group controls stationary point search mode. Leave theese all untouched.

Next we have a $DATA group in witch our initial molecule is defined, by provided atom coordinates and core charges. First line below$DATA contains job title witch will be displayed in output file, C1 key defines symetry of our molecule (C1 means no symetry and can be used in any case), lines below contains atom names, core charges and coordinates. Don't worry we not need to calculate them by hand, they are already saved to file by Chem3D program we used before, we just need to copy them to our input file. Contents of file saved by Chem3D is shown in middle picture below. Then editing input file you must ensure that first character on all lines is space and there is line with keyword $END on end of atom coordinate listing. Note that file saved by Chem3D also saves Lp atoms (lone pairs) witch are junk and must be discarded. Result of copy&paste work is complete PC Gamess input file, witch is shown on right picture below: Once input file is prepared it should be executed by PC Gamess, save input file to directrory where PC Gamess executable is located with name "input", launch comand prompt, navigate to PC Gamess dirrectory and type "pcgamess >pm3.out", this will launch calculations and save results to specified "pm3.out" file. Gamess calculations can take some time, so you can be interested in viewing progress at the moment. To do that you can open "pm3.out" with FAR file viewer tool or by Chemcraft, in file viewer you can monitor calculation process in real time and watch all current actions done by program, Chemcraft will show you 3D structure of your molecule at all optimization steps taken by PC Gamess. All this shown on pictures below: After execution is complete, you will get both, optimized molecular structure and corresponding heat of formation. To get them open output file "pm3.out" and find line with statement "EQUILIBRIUM GEOMETRY LOCATED", below you will see set of atomic coordinates with optimized molecule structure, corresponding heat of formation is slightly above this line, in the begining of final optimization iteration. You can aslo visualize optimized (real) molecular structure by opening "pm3.out" by Chemcraft and selecting Optimized geometry in the step list. Results of calculations viewed as mentioned above are shown on screenshots below: As you can see in on the screenshot above, heat of formation of our structure in gas phase is calculated as 40.571 kcal/mol. If you want to convert this value to kJ/mol, you must multiply it by 4.1868, if you want heat of formation in kJ/kg value in kcal/mol must be multiplied by (4.1868*1000/molar mass). Our optimized molecule structure can now be used for calculation of crystal density, we will find it in the next step. Predicting Crystal density Density is very important for HEDM, because it dirrectly affects detonation velocity. That's why then looking for new effective energetic compounds density is one of the main aims in search, and it very important to have ability to predict it at least aproximately before spending time on synth of new compound. Modern theoretical work on computation of substane crystal density assumes that density is a function of electron density in molecule and that volume of molecule in crystaline cell is proportional to volume of 3D region in witch electron density is above 0.001 electrons per cubic bohr (bohr is unit based on hydrogen electron orbit radius). Electron density map can be computed by means of quantum chemical software, so derived density value can also be determined. PC Gamess can produce electron density map, but it can't calculate volume of reqired zone, this must be done by user or some other external program. Numericaly this task transforms in task of creating isosurface with 0.001 e/bohr3 density value, and calculation of it's inside volume, using electron density values mapped by PC Gamess in set of 3D mesh points. In order to calculate requested volume we should split 3D mesh on cubes, each of those has some volume and electron density values at it's corners. It is clear that if all this values are above 0.001 e/bohr3 our cube is inside isosurface and it's volume must be added to sum, otherwise if values at all corners is below 0.001 e/bohr3 cube is outside isosurface and it's volume is discarded. If values at some corners of cube are bigger then 0.001 and lower then 0.001 at other - this means than isosurface intersects our test cube, cutting off some volume that must be added to sum, this volume can be aproximated by deviding cube volume to 8 parts (number of cube corners) and multiplying this value by number of corners witch have value above 0.001 (e.g number of corners inside isosurface). Because electron density map produced by PC Gamess contains enumorous number of small cubes (millions), volume sum obtainted then considering all this small cubes together will be very precise, but will also require some intense calculations, fortunately we are not forced to do this by hand, this is PC live for. I've written small utility witch will do this operation using PC Gamess generated density map, so our job is launch games electron mapping job and to open resulted output file by utility. Typical electron density mapping job input file is shown on left picture below, lines with atomic cordinates are missed and should be filled with coordinates of our previous optimization run, witch we done then predicting heat of formation. After copy&paste work we will get input file like one shown on right picture below: Before processing this new input file we must clean up PC Gamess directory from output generated from previous job, otherwise this data could be overwritten and PC Gamess execution will terminate with warning about old output file existance. Make a new folder inside PC Gamess one, and move old "input" and "punch" files to this folder. Now you can save new "input" file in PC Gamess folder, and execute it as in previous run, using command prompt and typing "pcgamess >dens.out". I you examing new input file closely you will note that this time we used another basis set, and used RUNTYP=ENERGY, theese are changed because PC Gamess can not generate electron density maps using PM3 and other semiemperical basis sets, however to prevent computation of new molecular structure we used single point energy run, in witch structure of our molecule will be preserved. After execution of our mapping job is completed we can open output files to watch obtained results, electron density map, in witch we are interested is located in file called "punch" inside PC Gamess dirrectory. This file contains many other information, and our map is located under$CUBE group and it's end is marked be $END keyword, as shown on screens below: Now we can use produced electron density map to estimate crystal density of our compound. To do this just open it with utility i've provided, type path to PUNCH file and click Load File button. Remember that utility recognizes only CHNO substances, and will fail if substance contain other elements or if input file format is incorrect, so be sure to supply correct input file. Results of this operation are shown on screenshot below: As you can see on this screenshot, calculated crystal density of our compound is 1.896 g/cm3. However this value must be corrected by multiplying it with 0.955 witch is correction factor for our basis set (PM3), introduced because calculated density produced by this basis set is systematically higher then real. So for our compound we estimate density of 1.896*0.955 = 1.81 g/cm3, that is very close to experimental value 1.82 g/cm3. If you want to visualize your resulting density map by Chemcraft, or to use it for input in some other programs, you can convert it to CUB file format. This is easily done by copying all data between$CUBE and corresponding \$END group in "punch" file and saving it a new file with .CUB extention. This file can be visualized by Chemcraft program, density isosurface drawn by chemcraft and part of this exported .CUB file are shown on the screens below:

Predicting heat of explosion and explosion gas volume

Heat of explosion determines, how much energy is released then our substance detonate. As you may know heat generated from rection is sum of formation heats of explosion products, subtracted by heat of formation of initial compound. To calculate heat of explosion we must assume some detonation equation, revealing products of detonation process and their proportions against each other. Generaly detonation can be assumed as simple burning process, in witch carbon inside explosive compound is oxidized to CO2, hydrogen to H2O and nitrogen to N2, by using oxygen contained inside explosive molecule. But how can we be then oxygen contained in molecule present in larger or smaller ammount then required to get only this products? Then there is too many oxygen in molecule we must assume that excess oxygen is released in free state (O2), if there is not enough oxygen for full oxidation situation is bit more complicated and we must take to account atom oxidation "priority". Hydrogen gets oxidized on first place, and oxygen will be used to form H2O until all hydrogen atoms will be oxidized, and only then oxygen will be used to oxidize carbon backbone, free hydrogen is released in explosion only if there is not enougth oxygen in initial molecule to fully oxidize is (for example in HN3 explosion). Carbon is first oxidized to carbon monixide CO, CO2 will form only if there is enough oxygen remaining, to form CO and CO2 mixture, and only CO2 if oxygen content is enough for full oxidation. Free carbon can be released only then not enough oxygen remaining to convert all carbon atoms to CO, and will be released in mixture with CO if oxygen content of explosive is too low. This model is reasonably simpified, but can be used for getting some usefull information about properties of explosive. Four our case (RDX, C3N6H6O6) we can assume following decomposition equation:

C3N6H6O6 = 3H2O + 3CO + 3N2

Heat of formation of initial explosive compound is already known from calculation and to get heat of explosion we need to know heats of formation of explosion products. These can be calculated in same way by using PM3 method in PC Gamess, and are precalculated as (also in kcal/mol): CO = -19,74453 ; H2O = -53,42649 ; CO2 = -85,03881 ; N2 = 17,56604 ; O2 = -21,42657. Using this value and molecular mass of our compound we can calculate heat of explosion in several units:

Hexpl (kcal/mol) = [3*(-53.42) + 3*(-19.74) + 3*(17.56)] - [40.57] = -166.8 - 40.57= -207,37 kkal/mol = -868,216716 kJ/mol = -3908,949 kJ/kg

Minus sign means that heat is released (exotermic reaction). This value is lower then experimental (-5297 kJ/kG) by about 1400 kJ/kg. Our result is inaccurate because experimental heat is taken using solid compound (and we assumed gas state reaction) and also because heat of formation depends from temperature whitch is not same as room temperature used then computing our heat of formation values, simplified detonation equation also takes part in this inaccuracy. While resulted value is not very precise, it still very usefull, because it can be used to compare different energetic molecules with each other, taking to account the fact that real heats of explosion have same proportions to each other as calculated ones if compound is of the same class. For example, we know experimental and calculated heats of explosion for compound A, and for compound B of same class we know only calculated heat of explosion, we can use this data to get real heat of explosion for compound B: Acalc/Bcalc = Areal/Breal ; Breal = Areal*Bcalc/Acalc.

Detonation equation witch we assumed above can also be used to calculate explosion gas volume, that is important factor contributing to power of our explosive. As you can see in decomposition equation above 1 mol of our compound (RDX) generates 9 mol of gaseous products. Taking to account the fact that 1 mol of any ideal gas at room temperature is 22.4 liters, we can calculate volume of explosion gases:

Vgas = 22.4 * 9 = 201.6 l/mol = (1000/222.11) * 201.6 = 907 l/kg

Calculated value is very close to experimental (908 l/kg), so volume of explosion products can be predicted quite accurately, if explosion detonation equation is composed correctly. You may heard about Trauzl bomb test for explosives, it measures expansion power of explosion gases, and depends only from heat of explosion and explosion gas volume. This value can aslo be derivated using parameters we calculated above, but has no much interest for us.

In next post i will show how to calculate detonation velocities at different densities using calculated data above.

497
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This is very interesting stuff indeed. I can't wait to see more.

I have recently downloaded Hyperchem 7.5. I have not play with it much so far, I don't know how its capabilities compare to the above stated programs but it looks like it does some similar functions. Do you know anything about it?
Engager
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Predicting Velocity of detonation

Detonation velocity is one of the main factors, determining power of explosive. The detonation velocity is the rate of propagation of a detonation in an explosive; if the density of the explosive is at its maximum value, and if the explosive is charged into columns which are considerably wider than the critical diameter, the detonation velocity is a characteristic of each individual explosive and is not influenced by external factors. It decreases with decreasing density of packing in the column. The standard hydrodynamic theory used for computing VOD of an explosive is concerned only with the amount of energy liberated and the nature of end products, and is independent of chemical reactions. The density, heat of formation and atomic composition can be integrated into an empirical formula for predicting performance of a proposed explosive.

There are many different relations describing dependence of detonation velocity from explosive constitution, and they use different parameters witch are used for velocity estimation, however most common are explosive constitution, heat of formation, oxygen balance and of course density. Formulas have different degree in accuracy and often give good estimations for some class of explosive compounds, while giving prediction with less accuracy for another compound classes. In our case we are interested in ones, witch give reasonably good predictions for any compound class, and use computationaly available parameters. It should be noted that in many formulas detonation velocity is much more dependent on constitution of explosive, then from heat of formation, and inaccurate heats of formation got from calculations (gas phase value, not taking to account crystalline energy effects) are perfectly suitable for prediction of detonation velocity. In reverse to heat of formation, density greatly affects detonation velocity and is one of the main factors controlling it, good thing is that computed densities are precise enough to perform good estimations of detonation velocity.

Now, I will show you how to use two different sets of empirical formulas to estimate detonation velocity and detonation pressure. First formula, having more simple nature, was described by Aisenshtadt in Russian explosive material journal, it can be used to estimate detonation velocity from heat of formation, explosive constitution and oxygen balance:

At this formula Hf is heat of formation in kcal/mol, Mw is a molecular mass of explosive compound in g/mol, p is explosive density in g/cm3, D1.6 is detonation velocity in km/sec at 1.6 g/cm3 and Dp is detonation velocity at target density (p) it is also in km/sec. Use of this formula is pretty straight forward, just use your explosive formula, predicted heat of formation and select density you want to use in calculation, but don’t forget to convert your heat of formation into kcal/mol, before substitution to formula. Example results for our calculated molecule (RDX) are shown below:

C3H6O6N6 ; Hf= 40.571 kcal/mol ; r = 1.8 g/cm3 ; B = ((3 + 6 + 6 + 6)/222.117)*1000 = 94.544 ; OB = |16(6-6-3)/222.117|*100 = 21.610 ; Qv = - |(40.571 + 0.3(6+6+6))/222.117| * 1000 = -206.964 ; M = 3 ; D1.6 = (0.73*94.544 – 0.24*21.610 + 0.0073*206.964)^0.5 = 8.083 km/sec ; D1.8 = 8.083 + 3(1.8-1.6) = 8.683 km/sec.

Note, that in formula above B is factor dependent from substance constitution, OB is form of oxygen balance, and Qv is approximate heat of explosion. All this are multiplied by coefficients got by fitting experimental data for many explosive compounds, and all this values are finally united then completing detonation velocity. This is a philosophy of empirical formulas – select properties witch as we think are determining value being estimated, and then fit experimental data to get coefficients for estimated value formula. Formula above can also be used for mixtures of explosives, in this case explosive formula and heat of formation must be substituted by corresponding average values for explosive mixture. As you can see calculated detonation velocity is pretty close to experimental (8.750 km/sec), error is only 0.75%. Divergence can vary for different classed of compounds, but formula usually give satisfactory results.

To give an alternative to Aisenshtadt formula, i will also show usage of formulas proposed by M.H. Keshavarz (References: Propellants, Explosives, Pyrotechnics 30 (2005), No. 2 ; Predicting the Detonation Velocity of CHNO Explosives by a Simple Method ; pp 105-108 and Journal of Hazardous Materials A119 (2005) 25–29 ; Simple determination of performance of explosives without using any experimental data). This formula uses calculated gas phase heat of formation (calculated in same way as we done – by PM3 method) of explosive compound instead of experimental one, and volume of explosion gasses derivated from estimated explosive detonation equation. Method assumes that all heat generated by explosion is used to heat explosion products, that allows to estimate explosion temperature witch is then used in pair with explosion gas volume and density to estimate detonation velocity. Estimation of explosion temperature is bit of art in numerical way, and requires some math operations or numerical analysis. To estimate explosion we must use heat balance for our process, but we can’t simply use standard heats of formation of explosion products as we done then estimating heat of explosion, method requires to take account to temperature dependence of heat of formation in explosion products. Heat of formation is function of temperature, exact equation for this dependence can be obtained from isobaric heat capacity vs temperature dependency, this one is usually approximated in polynomial form as shown on illustration below. Heat of formation vs temperature function is obtained by integration of this dependency, as shown below:

Cp is heat capacity at constant pressure ; Hf are heats of formation ; T is temperature ; Nj is number of moles of product j in detonation equation ; A,B,C,D,E,F are constants for each individual compound obtained by fitting experimental heat capacity data. Equation at the bottom of illustration above is heat balance equation, showing that all heat released by explosion is used entirely to heat explosion product (adiabatic explosion), since explosion process is extremely fast, adiabatic assumption is pretty much accurate. To solve heat balance equation we can use simple numerical method called bisection.

Bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. Suppose we want to solve the equation f(x) = 0. Given two points a and b such that f(a) and f(b) have opposite signs, we know by the intermediate value theorem that f must have at least one root in the interval [a, b] as long as f is continuous on this interval. The bisection method divides the interval in two by computing c = (a+b) / 2. There are now two possibilities: either f(a) and f(c) have opposite signs, or f(c) and f(b) have opposite signs. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs (watch illustration below).

To solve our heat balance, first we must transform our heat balance equation to form suitable for bisection f(x)=0, next we reasonably assume that explosion temperature lies somewhere between room temperature and some high temperature for example 6000K. Now we are ready to solve our problem, we just need to get coefficients from explosion equation (Nj’s), coefficients got by fitting experimental heat capacity (A, B, C, D, E, F) and heat of formation of our explosive witch we calculated before. Be aware that these experimental coefficients are calculated to fit heat equation then T/1000 is used instead of simple temperature value, so to get correct results you must place T/1000 in place of T in equation below.

C3N6H6O6 = 3H2O + 3CO + 3N2

For example i will show single calculation of f(T) at T=600K, so you can see how data listed above is working in our equation. Be careful, heat balance root finding takes a lot of simple calculations, be sure to make no errors in signs, numbers and choose proper temperature intervals in table, or you will get completely confusing results. Here is an example, i’ve mentioned:

Root bracketing is now straight forward, bisection method is stable and usually converges quite fast. Method is the same as in bisection method description, but f(x) is replaced with our function f(T), results are shown in table below. As you can see method converged to temperature about 3731K witch is our explosion temperature. Amount of calculations is too big to by hand, that’s why i’ve written utility for calculation of detonation parameters, PC must pay by fair work for money you spent on it.

Now, then temperature of explosion and volume of explosive gasses is known, we can simply calculate corresponding detonation velocity, by means of simple formula shown below. In formula (r) is density at witch detonation velocity is being predicted (g/cm3), Texp is temperature of explosion in K found in calculations above, (n) is a amount of gas generated by 1 gram of explosive (simple sum of mole numbers of explosion products in right side of detonation equation, divided by molecular weight of our explosive, value in mol/g), resulting detonation velocity will be in km/sec.

Substitution of our data (T=3731K, p = 1.8 g/cm3; n = 9/222.11 mol/g) we get detonation velocity of 8899 m/sec, experimental value at 1.8 g/cm3 is 8750 m/sec. So found detonation velocity is slightly overestimated, error is 1.67%, so accuracy is also very good, especialy taking to account the fact that when we started we known absolutely nothing about our compound! Formula works well on many classes of explosive, but requires some intense calculations.

Now we have predicted velocity witch is quite precise, but one may also want to predict detonation pressure, fortunately it’s very easy as it can be predicted using only heat of formation and explosive composition, as shown in another M.H Keshavarz article. Formula for calculation of detonation pressure is shown below:

Here Hf is explosive heat of formation (we already calculated it way before), r is density in g/cm3, Mw is molecular mass of our explosive, resulted detonation pressure will be in kbar. Using data for our explosive (a=3, b=6, c=6, d=6, Hf=169.86 kJ/mol, r=1.8 g/cm3) we can calculate detonation pressure for our explosive, witch is 414 kbar (experimental is 341 kbar) with 18% error, however with no surprise due to gas phase calculated heat of formation, instead of crystalline experimental one. On experimental heat of formation formula gives much more precise result (355 kbar calculated vs 341 kbar measured).

In total, from zero knowledge starting point we have predicted: heat of formation, crystal density, volume of explosion gasses, detonation temperature, detonation pressure, detonation velocity. Of course there are many other estimation methods, but this ones are simple and can be made at home. For convenience i have written small utility witch calculates estimated parameters using methods described above and in references mentioned. Utility requires VB runtime libraries, but if you use windows XP it should run without any setup. Utility with the source code + references for this article in small archive file are attached to this post, screenshot is shown below:

That’s all for now, any opinions & comments about this work will be gladly appreciated. If you found some bugs in utilities i provided please report them to me by personal message. Questions on property calculation topic can be asked in this topic.

[Edited on 20-9-2008 by Engager]

[Edited on 20-9-2008 by Engager]

Attachment: Detonation2.rar (345kB)

Engager
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 Quote: Originally posted by 497 This is very interesting stuff indeed. I can't wait to see more. I have recently downloaded Hyperchem 7.5. I have not play with it much so far, I don't know how its capabilities compare to the above stated programs but it looks like it does some similar functions. Do you know anything about it?

Thanks for your inerest, it's good to see that i'm not alone here. Regretably i haven't got Hyperchem to watch for it's capabilities, but if you need any of software i've mentioned in my post i can upload it.

[Edited on 20-9-2008 by Engager]

497
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That is a pretty awesome little program! And an amazing feat of mathematics too. It works perfectly so far.

I think a useful addition would be a table of actual values (det vel, gas vol, etc) of various explosives. That would give one something to compare their outputs to see accuracy, and just to get a practical idea of its power. I know that information is scattered around, just I've never seen it condensed into a single table.

Also having some kind of scale to get an idea of the extrapolated test values (sand crush, lead block expansion, etc.) based on the output values from the program would be useful.

A heat of formation calculator would be really nice too.

Another idea: similar programs except for fuel-air detonations, thermobaric enhanced explosives, metalized explosives, etc. I suppose it might be harder to find computational methods for those though.

I would love it if you could upload those programs if its not too much trouble.

Here are a couple resources I ran into recently. They have some interesting computational and expiremental research and going on at Caltech. The military site has a lot of nice old scanned in stuff from the 80s, on all kinds of subjects.

http://www.galcit.caltech.edu/EDL/publications/references/ed...
http://www.dtic.mil/dtic/search/tr/index.html

I for one think you are doing a great service for all of us interested in energetics. Keep up the good work!
Engager
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I've uploaded software i've metioned to: http://rapidshare.com/files/146893683/Essentials.rar.html

Package is ~100 Mb size and contains the following stuff: Chemoffice8 & Serial ; PC Gamess 7.1 with samples, help files, and some additional extentions ; Chemcraft 2.63 + keygen ; UltraEdit32 File Editor + crack ; Density and Det.Velocity utilities for this thread with source codes on VB6 ; files used for calculations in topic above ; references on computational methods used.

Formatik
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Thanks for the programs. It's also possible to estimate the relative strength equivalent, which results most resemble the trauzl test. They are formulas most likely built on deriatives, and likley most applicable to CHNO compounds, but I've seen favorable results even with some inorganics (see below). Among one of these formulas is one seen here: http://www.fas.org/man/dod-101/navy/docs/es310/chemstry/chem... Here are some estimates: Method 1 is the one in the site above. For the second method I'd have to find it again.

urea nitrate..............91.................44................90
EtONO2...................122...............147..............140
trinitromethane.........84................120..............137
HNAB.........................--.................133...............122
MeONO2..................242..............197..............203
NH4NO3....................--..................61...............60
NH4ClO4....................--..................51...............65
isoPrONO2.................56..................42..............56
N2H4.HClO4...............--...................120............121
TEGDN.......................92.................79...............106
Hexanitrobiphenyl.....111...............114..............120
DEGDN......................136................133.............136
TeNMe........................46..................69...............21

For heat of formation values, see here: http://cobweb.ecn.purdue.edu/~propulsi/propulsion/comb/prope...
Microtek
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@ Engager:

I have tried your method with the supplied programs and am very impressed with the amount of work that you have done.

I have a few comments however; I tried it with TNT and got a density of ca 1.76 (after the 0.955 correction factor). Needless to say, this gave quite a high VOD. Do you know if this method gives poor predictions for nitroaromatics, or if maybe I made an error?

Secondly (and more importantly) I tried with HNIW but couldn't get the PM3 calculation to converge. It gave an error message saying that it made too many iterations, so I increased the MAXIT parameter to 1000 which didn't resolve the issue.
Could you try to run the program for HNIW and report if you can make it reach a conclusion? I am doing some work with HNIW and related compounds at the moment, so it would be quite valuable with a predictive tool.
Engager
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 Quote: Originally posted by Microtek @ Engager: I have tried your method with the supplied programs and am very impressed with the amount of work that you have done. I have a few comments however; I tried it with TNT and got a density of ca 1.76 (after the 0.955 correction factor). Needless to say, this gave quite a high VOD. Do you know if this method gives poor predictions for nitroaromatics, or if maybe I made an error? Secondly (and more importantly) I tried with HNIW but couldn't get the PM3 calculation to converge. It gave an error message saying that it made too many iterations, so I increased the MAXIT parameter to 1000 which didn't resolve the issue. Could you try to run the program for HNIW and report if you can make it reach a conclusion? I am doing some work with HNIW and related compounds at the moment, so it would be quite valuable with a predictive tool.

Well, 0.955 factor works very fine on caged nitramines and tetrazoles, witch i've tried to calculate. Benzene aromatics like TNT can have different value because PM3 computaion method produces results of different quality on differented molecule classes. This is not a error of calculation by my utilities. In order to get good correction factor for benzene nitroaromatics, take a set of molecules with known density, compute them to get calculated densities without 0.955 correction and select correction factor witch suits computed vs experimental data.

Your problem with HNIW is based on fact that you used incorrect starting struture for input of Gamess, remember it will never converge unless structure you give is correct and at least slightly optimized by implementing MM2 in Chem3D as i described. HNIW computation works very fine for me, i got density 2.043 vs 2.04 experimental with usual 0.955 (2.141 * 0.955 = 2.044 g/cm3) correction. Input files for PC Gamess jobs, and results of successful jobs on HNIW using method i've described in my article are placed in archive here:

http://rapidshare.com/files/148417525/CL-20.rar.html

To give you any idea how good initial Chem3D - MM2 structure looks like, watch the picture below. Left structure is produced by Chem3D+MM2 and right one is optimized by gamess and is taken from converged PC Gamess PM3 run.

[Edited on 26-9-2008 by Engager]

Microtek
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OK I tried it again and I noticed that the nitrogroups that I draw in chemoffice look somewhat different from yours. Mine are drawn with a delocalized doublebond while yours look like N-O single bonds. Also, if I use the MM2 operation after attaching the NO2's, then the nitro groups are changed to sp3 hybridization rather than sp2 (which is how they start out).
Anyway, by running the MM2 operation on just the skeleton without any NO2 groups and then attaching these afterwards gave me the correct result. I suspect I'm not using chemoffice quite correctly.
Engager
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Due to the report of Microtec i found error in detonation velocity utility, problem is in Visual Basic representation of numbers, this is now fixed, and work correctly then HNIW data is entered (Formula - C 6 H 6 N 12 O 12 ; Hf - 451,12 kJ/mol ; Density - 2,04 g/cc), new version is attached to this post.

[Edited on 27-9-2008 by Engager]

Attachment: Detonation3.rar (27kB)

The_Davster
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Some silly questions:

How do I use the chemoffice keygen? My virus scanner picks up keygen.nfo because whenever I click it it attempts to contact the internet....

I have no idea how to install the PCGamess7.1...where is the setup.exe file?

Microtek
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I couldn't get chemoffice activated either, so I found another version on avaxhome. As for PCGAMESS, it is ready to go when unpacked; no installation required.
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Maybe it's possible to predict some of the properties of the explosive, as long as the bondings and reaction-products are known, thereby the temperature, pressure, molecular velocity etc ..

But I'm not gonna believe properties of materials could be predicted (today) unless somebody shows that he can deliberately invent eg. new superconducting materials, alloys with specifyable thermal hardening properties or something else. Science is far from that today.

The method also seems to be limited to the organic chemistry, also the empirical formulae are adjusted on the known properties of the materials: No wonder, if some of the results comply with the lead-block test ... .

But anyhow, a useful thing to have is a structural database: From there on you can investigate and go onto adventure. Unfortunately I never found any rapidshare-link to a version of the CCSD (CAmbridge crystal structure database, http://www.ccdc.cam.ac.uk/), maybe someone could help with that ?
Engager
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 Quote: Originally posted by The_Davster Some silly questions: How do I use the chemoffice keygen? My virus scanner picks up keygen.nfo because whenever I click it it attempts to contact the internet.... I have no idea how to install the PCGamess7.1...where is the setup.exe file?

This is funny =) nfo files are not executables but are simple text files opened by notepad, in one of theese following numbers vere given for Chemoffice registration: serial: 7654321 reg-key: VLPG-CG5O-5K22-VKVK-RF . Microtek told me that negative heats of formation can not be entered in detonation utility, this is now fixed, fixed version is attached to this post. And one more thing if some admin reads this post, please remove attached previous versions of this file, so people can download only fixed one.

Attachment: Detonation4.rar (27kB)

franklyn
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PROJECTING AN ESTIMATE OF PERFORMANCE

I believe old tricks are the best especially if there is no apparent improvement with newer methods.
To estimate a hypothetical explosive's performance one need only know it's structure , be able to
do arithmetic , and have a calculator that does square roots.
- - here you must enter fractional exponents within parenthesis ^ ( 1/2 )
http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl
- - this does not accept fractional exponents, one must do power and root separately
http://firstyear.chem.usyd.edu.au/calculators/chemical_calcu...

The two references immediately following and previously cited by me outline exactly how to proceed.

Honestly I don't know why I bother to post references when apparently no one takes notice

Estimation of the Density of Organic Explosives from Their Structural Formulas
http://fas.org/sgp/othergov/doe/lanl/lib-www/la-pubs/0032140...
also available in the Science Madness library
It's not necessary to re-invent the wheel if it gets you to the same place
the method is very simply stated by example on pages 3 , 4 , 18

and in an earlier post
A Simple Method for Calculating Detonation Properties of C-H-N-O Explosives
the relevant formulas and procedure is concisely expressed on pages 9 , 13 , 25
- - Another reprint with sequels is attached at the bottom of this post - -
- - the same procedure and formulas are on pages 26 , 28 , 33
A demonstration by steps is in the following post in this thread
If anyone has any quarrel with the preceding , know that the method has also been cited
by the luminary Gerald Hurst. That's good enough for me.
http://yarchive.net/explosives/detonation_pressure.html

Out of academic interest only , the following linked from my second cited post above
and made available again by engager in an attachment to this current thread here.

Simple determination of performance of explosives without using any experimental data

- This is essentially the same paper in a different journal -
Predicting the Detonation Velocity of CHNO Explosives by a Simple Method

Predicting Heats of Formation of Energetic Materials Using Quantum Mechanical Calculations
http://www.arl.army.mil/arlreports/2001/ARL-RP-15.pdf

.

Attachment: Calc Det Prop of CHNO HE.djvu (681kB)

franklyn
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HOW TO APPRAISE PERFORMANCE OF HYPOTHETICAL - C H N O - EXPLOSIVE

- In 4 easy steps

The following example demonstrates the steps to estimate how one can project the
probable performance of an explosive.
Pg 33 Organic Chemistry of Explosives - Agrawal
Addition and condensation reactions employing nitroform or its methylol derivative,
2,2,2-trinitroethanol, have been used to synthesize a huge number of compounds
containing the trinitromethyl group. Such compounds often have a very favorable
oxygen balance which contributes to explosive performance. However, compounds
containing the trinitromethyl group often exhibit an unacceptably high sensitivity to
shock and impact.
[ Only testing of the actual compound can determine if it merits
further consideration ]

I have been reviewing literature on a notion I have for applying nitroform as a functional
group. The idea is to attach four trinitromethyl groups to pentaerythrite. I had planned
to post how this hypothetical molecule might be synthesized and possible approaches
to producing the precursors. Nitroalkanes are derived from this trinitromethyl group by
the well established and long ago documented method of nitro-dehalogenation. The
pathway indicated would use a tetra(halo)pentaerythrite with a metal nitroformate
salt to form Pentaerythrite Tetranitroformate - PETNF
This yields 20 mols of gas and has a slight positive oxygen balance.

. . . . . . . . C[ CH2.C( NO2 )3 ]4 => 9 CO2 + 4 H2O + 6 N2 + O2

1 )
The first thing then is to draw a structure
Arguslab a feature rich , superb molecular modeling program is what I use to resolve
structure , just click the " pliers " icon in the tool bar until converged. This entire part
can be skipped , it only serves as a visual aid and to confirm that the object can exist.
The 3D Argus *.agl file of PETNF is in the zip file attached below.

[url][/url]

2 )
To derive a value for the density I have applied the method of H. H. Cady cited above in
Estimation of the Density of Organic Explosives from Their Structural Formulas

To establish the density , the molecule must be dismantled into component pieces
for which the volume is known. Density is then derived by adding up all the pieces
and dividing the molar weight by the sum of these volumes.
Tables of these, are given in the cited reference along with their values.
C[ CH2.C( NO2 )3 ]4 , breaks up into the following functional groups or free radicals.

Group. . . . . . . . . . . . . . Carbon . . . . . . . Methylene . . . . . . . N i t r o
. . . . . . . . . . . . . . . . . . . . . . C . . . . . . . . . . . C H 2 . . . . . . . . . . N O 2
Volume of each . . . . . . .3.3 . . . . . . . . . . .10.16 . . . . . . . . . . 17.02
How many of each . . . .X 5 . . . . . . . . . . . . .X 4 . . . . . . . . . . . X 1 2
Totals for each . . . . . . 16.5 . . . . . . . . . . 40.69 . . . . . . . . . 204.24
Table II ( C - ? ) + ( ? - NO2 ) => C - NO2. . . Added correc. .0.54 . . .Total of Volumes = 2 6 7 .9 1 = ( V )

One also needs to estimate the packing factor or coefficient ( k ) figured by formula 4
on page 18. k = 0.7686 - 0.1280 (a) where (a) is the fraction of
( number of hydrogen atoms bonded to carbon ) divided by ( total atoms in molecule )
in our case 8 / 53 = 0.01932 and so , k = 0.7686 - 0.1280( 0.01932 ) = 0.7493

Density ( ρ ) is weight divided by volume , according to formula 6 page 18
factor ( k ) is included thus ( mol weight )( k ) divided by ( V )
The molar weight of the substance is 668.23 and so times 0.7493
and divided by 2 6 7 .9 1 comes to ρ = 1.87 gm /cm³

* Note : Density of a crystal does not necessarily reflect actual loading density of a charge

comparing to ( k ) values of related known explosives we may assess this further
explosive . . . .Hexanitroethane . . . . PETN . . . . ( average )
calculated . . . . . . 0.77 . . . . . . . . . . . . 0.73 . . . . . . 0.75 . .
observed . . . . . . . 0.68 . . . . . . . . . . . . 0.76 . . . . . . 0.72 . .

3 )
It's now necessary to determine the heat of formation of the compound to be analyzed.
Enthalpy changes for reactions can be obtained by simply subtracting the heats of
formation of the reactants from the heats of formation of the products. Thus if the
heats of formation of reactants that form the compound are known , the ∆Hf
Heat of formation of the compound will also be known.
It is important to know that for the purposes of determining enthalpies , the reaction
pathway need not be realizable in practice , and merely serves to ascertain the
∆Hf Heat of formation. http://pages.prodigy.net/anderhan/ch11thermo.pdf
For simplicity the hypothetical condensation of 1 mol of Pentaerythritol with
4 mols Trinitromethane as a base-acid esterfication is assumed.
The following values in kilocalories per mol are applied for thermodynamic calculations
- 57.8 for H2O ( g ), - 68.3 for H2O ( lg ) , - 94 for CO2 , - 220 for Pentaerythritol , - 16.25 (liq) Trinitromethane
Sources of thermodynamic properties
http://cobweb.ecn.purdue.edu/~propulsi/propulsion/comb/prope...
http://webbook.nist.gov/chemistry
http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf
http://www.arl.army.mil/arlreports/2001/ARL-RP-15.pdf

In accepting the condensation of Pentaerythritol with Nitroform , you see the
advantage for calculation of showing only water as a co-product of synthesis
. . . . C(CH2OH)4 + 4 HC(NO2)3 => C[ CH2.C( NO2 )3 ]4 + 4 H2O
Heats of formation of the reactants is subtracted from that of the products to
infer the ∆Hf Heat of formation : 4 H2O ( lq ) is - 273.2 , minus ( - 220 + 4 (- 16.3 )
Arithmetically subtracting a negative is the same as adding a positive so ∆Hf is + 11.8 kcal /mol

Now the heat of explosion can be calculated simply
∆Hf Heat of formation of Pentaerythrite Tetranitroform is subtracted from the detonation
products to infer the ∆He Heat of explosion: C9H8(NO2)12 -> 9 CO2 + 4 H2O + 6 N2 + O2
(9 CO2 is - 846) + (4 H2O ( g ) is - 231.2) + (N2 & O2 are zero) minus (+ 11.8) = - 1089 kcal /mol

Dividing the ∆He Heat of explosion of 1 mol of the compound by it's molar weight , - 1089 / 668.23
obtains ∆He Heat of explosion = - 1 6 3 0 Kcal per kilogram of compound
This will be assigned to the constant Q in step 4

4 )
To estimate the Detonation pressure generated and Velocity of Detonation ,
I have applied the method of M.J. Kamlett, S.J. Jacobs outlined in
The Journal of Chemical Physics , vol 48 , num 1 , cited and also attached above
A Simple method for Calculating Detonation Properties of C-H-N-O Explosives
Also in LLNL Handbook of Explosives section 8.1.2 Estimation on page 8.9
what follows does not referer to this LLNL Handbook , but to the Journal of Chem. Phy.

The formulas presented amount to generalized shortcut calculations for determining
values that according to its author will typically occur around 2 % of real values.

M is grams of gas per mol of gas , equation 14 page 27
N is mols of gas per gram , equation 13 page 27
P is detonation pressure expressed in kilobars ( 1 Bar = 1 atmosphere )
ρ density given as grams per cubic centimeter 1.87
Q is ∆He Heat of explosion in calories per gram 1 6 3 0
So this can be easily understood ,
the constants that count each type of atom , given in the article as a , b , c , d
in the formulas for M and N , are changed here to the atom each represents a = C (Carbon)
b = H (Hydrogen) , c = N (Nitrogen) , d = O (Oxygen)

M = . . .56 N - 88 O - 8 H . . . = . . . 56( 12 ) - 88( 24 ) - 8( 8 ) . . . = 35.6
. . . . . . . . 2 N + 2 O + H . . . . . . . . . . 2( 12 ) + 2( 24 ) + ( 8 )

N = ... 2 N + 2 O + H . . . . . . . . . . = . . . .2( 12 ) + 2( 24 ) + ( 8 ) . . . . . . . . . . . . . .≈ 0.03
. . . . 48 C + 4 H + 56 N + 64 O . . . . . . 48( 9 ) + 4( 8 ) + 56( 12 ) + 64( 24 )

The formulas given for M and N intend to acount for what carbon remains unreacted
in explosives with negative oxygen balance. This does not apply in this case and the
value of M X N = G = calculates to 1.068 , indicating a corrective subtraction of - 6 %
specified on the bottom of page 33 and 35

We now have all the necessary values to calculate this : ( N √ M √ Q )
( Q ( 1 6 3 0 ) is substituteded here without regard for sign )
( * Note this expression is given as ф ( phi ) in the article ) . . . ( 0.03 ( 5.97 ) ( 40.37 ) ) = 7.23

Detonation pressure = P = 15.58 ρ ² ( N √ M √ Q )
. . . . . . . . . . . . . . . . . . . . . . . . . 15.58 ( 1.87 ) ² ( 7.23 ) = 3 9 4 kilobars
minus - 24 , 6% correction factor comes to 3 7 0 kilobars

This is comparable to HMX in performance.

Velocity of Detonation = VOD = 1.01 ( 1 + 1.3 ρ ) √ ( N √M √Q)
. . . . . . . . . . . . . . . . .1.01 ( 1 + 1.3 ( 1.87 ) ) √ ( 7.23 ) = 9.318 millimeters per microsecond
multiplied by one million to obtain meters per second ,
minus - 559 , 6% correction factor comes to ≈ 8 7 6 0 Meters / sec

.

Attachment: Pentaerythitol tetranitroformate.zip (6kB)

Microtek
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@ Franklyn

In step 3 where you calculate heat of formation, you would also need to know the temperature change in the reaction; if the hypothetical condensation reaction is endo- or exothermic you would under- or overestimate the heat of formation respectively.
franklyn
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 Quote: Originally posted by Microtek @ Franklyn In step 3 where you calculate heat of formation, you would also need to know the temperature change in the reaction; if the hypothetical condensation reaction is endo- or exothermic you would under- or overestimate the heat of formation respectively.

Yes point taken , but then I'm not a chemist.
Applicable enthalpies can be found in references I'm sure.
But that depth of involvement in thermodynamics is beyond me.
It appears to be endothermic as related compounds.
so won't be too far off.

.
franklyn
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Now that you have designed the mother of all energetic compounds,
what can be done with It , and how would you know it ?

That is a very complicated question
Blast radius is not a line in the sand where the explosion stops. The effect of
the explosion proceeds outward to an arbitrary distance of your choosing. If
you are at some distance away and the sound damages your hearing, does
this figure into your definition of "Blast Radius" ? All explosions are qualitatively
different. The hyperbaric effect of fuel air ordinance is not the same as the
underwater explosion of Torpex.
Upon detonation, a military high explosive expands the metal housing of the
ordinance to 1.5 times it's original diameter before breaking into fragments.
Typically 40 % of the weight is the explosive, the remaining 60 % is the casing.
Of the total energy liberated, half may be used up expanding the casing prior to
rupture and imparting velocity to the fragments. The remainder of the available
energy is expended in compression of the surrounding air forming an expanding
shock front that moves out supersonically. After a short distance the speed and
pressure of this front decreases much faster than the decrease in speed of the
casing fragments. This is due to the fragments having constant cross sectional
area and aerodynamic drag, while the surface area of the expanding shock front
increases as a function of the surface area of a sphere, Area = 4 ∏ ( r *squared )
( where r is the distance from the center of explosion to the shock front ).
distributing its energy at a decreasing intensity.

There are two measurements of the quality of the explosion, its peak pressure
and it's impulse or duration , this being the area under the graphed positive
pressure curve , which decreases by the reciprocal of the radius 1/ r
from the center of explosion. Peak pressure is carried by the shock front, an
extreme form of sound wave that transmits only energy. The pressure impulse
follows with the cyclonic surge of displaced air. Shock front and impulse are
disassociated and act independently. The momentum of the gas moving outward
behind the shock front expands it's volume until the pressure decreases below
atmospheric producing a rarefaction and reversal of the motion back toward the
center of explosion , while the shock front continues it's outward travel.
Remember that the energy is dissipated kinetically as a percussive wave into
the surroundings. See Pg 4 here ->
http://www.dsto.defence.gov.au/publications/2641/DSTO-TN-052...

[url][/url]

One can begin by finding the volume of gas produced by the explosive.
Consider the glycol dinitrate reaction : C2H4(NO3)2 -> 2CO2 + 2H20 + N2
the explosion of one mole of glycol dinitrate produces in the gaseous state:
2 moles of CO2; 2 moles of H2O; One mole of N2. A molar volume of gas at 0 °C
and atmospheric pressure occupies 22.4 liters. Since each mole produces one
molar volume of gas, a mole of glycol dinitrate produces 2 + 2 + 1 = 5
molar volumes of gas; and these molar volumes at 0 °C and atmospheric
pressure form an actual volume of 5 × 22.4 = 112 liters of gas. ( Note that the
products H2O and CO2 are in their gaseous form.) Based on this it can be seen
that the volume of the products of explosion can be predicted for any quantity
of an explosive.
Further, by employing Charle's Law for perfect gases, the volume of the products
of explosion may also be calculated for any given temperature. This law states
that at a constant pressure a perfect gas expands 1/273.15 of its volume at 0 °C,
for each degree Celsius of rise in temperature.
For example, at 15 °C the molar volume of an ideal gas is equal to :
The expansion by the increase in temperature ( 15 °C X 1 /273.15 X 22.414 )
plus + the original volume at 0 °C ( 22.414 ) = 23.645 liters per mole.
Thus, at 15 °C the volume of gas produced by the explosive decomposition of
one mole of glycol dinitrate becomes : V = (23.64 ltr./mol)(5 mol) = 118.2 ltr.
for the final adiabatically expanded quiescent volume. For a shallow underwater
explosion this defines the volume of the bubble that will be formed.

* Note :
For the radius volume V must be expressed in cubic meters M, there are 1000 liters
in a cubic meter so , 118.2 / 1000 , V = .1182 cubic meters

Setting this amount equal to the volume of a sphere : 4 ∏ ( r *cubed ) divided by 3

and solving for the radius : r = cube root of ( 3 V divided by 4 ∏ )

r = 30 cm , ( a ball of gas two feet across )

this gives you a naive working value for blast radius without regard to the proximity
with the ground and resulting Mach reflection. Nor the kinetic force of the shock
front and fragmentation. This later effect of fragmentation can be approximately
estimated by finding the cube root of the weight of explosive in kilograms, and
multiplying this by 130, giving the radius for the inclusion area in meters. Most,
but not all, of the fragments hurled out from an explosion will lie within this area.
Pieces moving with an azimuthal elevation between 45 and 50 degrees to the
horizontal will attain the greatest range, and a few of these will reach beyond
the inclusion area.
Without the use of calculus estimates for explosive effects are only approximate.

None of the above takes into account the energy transfer to the surrounding air.
An estimate for this can be made by applying the ideal gas law PV = nRT
which is an equation of state that describes the behavior of gases derived from
the combined gas laws used above. Pressure in atmospheres ( P ) times
volume in liters ( V ) equals = the number of moles ( n ) times
this constant .0821 ( R ) times the temperature in degrees Kelvin ( T ) ,
( which is degrees Celsius plus 273.15 )

This equation is much simpler to use than the sequence of steps applied in the
example above. Solving for the volume V = nRT/ P ,
V = ( 5 moles )(.0821 )( 15 °C + 273.15 ) / 1 atmosphere
is 118.2 Liters just as before.

The ideal gas law is really an energy equation which defines the potential
energy in a gas , the left hand side of PV = nRT
gives the value in Liter atmosphere ( L atm ) a unit of energy used for gases.
41.3106 Liter atmosphere is equal to one Kilogram Calorie.

To obtain useful and practical information about the explosion requires knowing
the initial temperature of the explosion which is itself usually only inferred by
calculation. One does however know the quantity of explosive used and it's known
caloric energy value. In the example used above, Glycol,dinitrate is rated at
approximately 1500 Kilogram Calories per Kilogram. Multiplying this value by the
amount of Nitroglycol used expressed in kilograms ( one mole ) .152 Kg., gives us
228 Kcal/ Kg. Using what we know and have established , 228 Kcal/ Kg ,
times 41.3106 Liter atmosphere , is 9419 Liter atmosphere.

9419 Liter atmosphere has a double meaning in that it can be seen as one liter
at 9419 atmospheres, as well as 9419 liters at one atmosphere. This is because
equation PV = nRT is in the form representing Boyle's law for gases PV = (constant)
This law states that for constant temperature and quantity ( moles ) of gas, it's
pressure and volume vary inversely to each other thus : ( P1 )( V1 ) = ( P2 )( V2 )
Solving for V2 = ( P1 V2 )/ P2 , since ( P1 V1 ) is 9419 , by dividing 9419 Liter atms
by a different amount of pressure ( P2 ) the volume for any arbitrarily chosen
pressure can be easily determined and with it the radius at that pressure.
Since any small elevation of ambient pressure will cause damage, the value of
one atmosphere will be used.

To calculate a blast radius, just as before, the volume is first converted into
cubic meters by dividing 9419 by 1000. Thus 9.42 cubic meters is set equal to the
volume of a sphere and solving for the radius, gives the value of 1.31 meter, a ball
more than 8 1/2 ft !
The difference between this and the 2 foot size given before, represents the
energy transmitted as a blast wave into the surroundings. Rarefaction will have
already occurred at this size having collapsed the detonation products into something
close to a 2 foot ball of gas, and the blast wave will have slowed and decayed into a
sound wave moving out at the speed of sound.

The difference of one meter , between being 1.31 meters away , and just 30 cm.
from a third of a pound of detonated Nitroglycol with fragmentation, is the difference
For very large detonations the margin is not so clear. For example, 100 Kg of AN + Al
is 2350 Kcal/Kg times 100 = 235000 , divided by 228 in the previous example
is 1031 times greater. The radius enlarges to 13.23 meters, or 10 times the size.
At this range you may be killed by the concussion just from the shock front.
Significant energy is still carried by the shock front out beyond the vicinity of the blast.
It's magnitude depends on the size of the initial explosion.
Note also that the zone affected and at risk for fragmentation damage
extends out 2/5 of a mile.

To review :
Determine the caloric value of the amount of explosive used , multiply this by
41.3106 to obtain V in Liter atmospheres, divide by 1000 to convert this to
cubic Meter atmospheres and solve r = cube root of ( 3 V divided by 4 ∏ )

Simplified to : ( HE wt.)( HE cal/kg )(.009862 )^1/3

As an excercise , calculate a blast radius for the BLU82 depicted here ->
http://en.wikipedia.org/wiki/BLU-82
and post your results. I am curious to see who needs to ask
how to calculate a blast radius , if all that's required is pushing a pencil.

There is a very useful calculator online that will express all the sperical values

here -> http://www.cleavebooks.co.uk/scol/calsph.htm

Further reference
http://www.fas.org/man/dod-101/navy/docs/fun/part12.htm

___________________________________________________

For F.A.E explosions _

Based on the known properties of flammable substances and explosives, it is
possible to use conservative assumptions and calculate the maximum distance
at which an overpressure or heat effect of concern can be detected. Distances
for potential impacts could be derived using the following calculation method
[described in Flammable Gases and Liquids and Their Hazards]:

D = C(nE)1/3
where D is the distance in meters to a 1 psi overpressure; C is a constant for
damages associated with 1 psi overpressures or 0.15, n is a yield factor of the
vapor cloud explosion derived from the mechanical yield of the combustion and is
assumed to be 10 percent (or 0.1) and E is the energy content of the explosive
part of the cloud in joules. E can be calculated from the mass m of substance in
kilograms times the heat of combustion Qc in joules per kilogram as follows:

E = m Qc
Combining these two equations gives:

D = 0.15 x (0.1 m Qc)1/3

.
chief
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Putting data into black-box-programs is _not_ science ! Besides commercial and half-commarcial software is a NO anyhow. A self-respecting thinker writes his own code and maybe uses some some tools, eg. for visualization ... .
Why not do the real science: Where are the quantum-mechanics ?? In the qantum-mechanical software ??? Better if the formulae are here, so anyone can follow the physics/chemistry in detail !

Then it also becomes quite interesting, and more satisfying than to handle (not so) _cool_ software-packages ....
After all: A human is not a trained monkey ...
What one really needs:
==> structural database (eg. CCSD: http://www.ccdc.cam.ac.uk/ , also ICSD: http://icsdweb.fiz-karlsruhe.de/)
==> knowledge
==> some ODE-solver-routines etc. for the language of choice might be helpful
==> a good visualization-tool: eg. http://www.geomview.org/

After all: The whole "prediction" here seems to come mainly from
==> calculating the pressure from
--> the initial density of the chemical bondings
--> the heat generated (specific to the bondings, heat gives temperature, that gives pressure)
--> the molar amount of gas produced (which is heated to the temperature(calculated from the heat))

It's simple physics, so no reason to demean oneself to black-box-operating any software, except if the aim is to impress some old grandmothers and the like ...

But anyhow it's interesting stuff above, but a real man does the calculatons himself !!
Microtek
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OK, so could you do a write-up of the quantum-chemical calculations to predict the density of RDX? Just so that those of us who don't feel so at home with quantum mechanics can see how it is done....
franklyn
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Recall that forum member enhzflep began this thread here

High Explosives Calculator

and submitted this H.E.Calc v 1.0

forum member not also provided a utility to average mixtures

forum member sylla provided code to run a program

the references posted by forum member Axt are expired now

This reference posted by enhzflep there is a good start for anyone
wanting to improve the art.

http://gltrs.grc.nasa.gov/reports/1994/RP-1311.pdf

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chief
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RDX-density-prediction:
First:
==> You just need the structure of the RDX;
==> this you get from the above mentioned CCSD
You _cannot_ relyably predict the structure, because science is just not far enough for that ! All you can do is take assumptions,
==> like how specific atoms arrange their bondings _usually_ ,
==> but this works best for compounds that are far from the unstability of explosive compounds, eg. with proteins you can toy around using the 3D-puzzle-software and try to find a place where some other molecule of the same shape might match.

Often this is sold afterwards the other way around, like: We calculated ... and experiment gave the proof; but the truth is: First by empirical means all the compounds are tested, and then the calculations are done ... .
All that can be done quantum-mechanically is to be found in the average textbook about it, like:
==> What wave-functions do exist on what boundary-conditions
==> This does not go beyond the most simple compounds, like NH3 or H2O, and even there loads of parallel theories do exist

Only _few_ scientists have an idea of how to some-sort-of predict properties of a compound, and those are also the ones who _successfuly_ make a breakthrough on some field, like eg. high-temperature-superconductors. Even Those few men (no women amongst them to my knowledge) do _lots_ of experimenting, before they find a material, and then they find just one sort of a compound, and do some substitutions on it (similar elements on similar places), like eg. the YBaCU-superconductors.

That why I say: Only thing that one can do is
==> get a quantum-mechanical textbook (maybe some internet-ressource exists too)
==> _and_ et a structural database, sor organic compounds the CCSD,
==> for inorganic structures the ICSD
==> those databases cost , no web-downloads known to me
==>but the average student might gain access via the university (data may be exported as csv-files; table-structure like mysql or other known databases, ca then be re-imported into any other database)

Now that you have the nice CCSD or ICSD (each with around 100000 compounds stored in it) you just lookup the structure, get the measurements of the unit-cell, count the atoms within it and calculate the density. But thats not prediction.

If anyone were able to relieably predict the properties of compounds, then he just could let the computer run and get himself a secretary to do the patenting-stuff. Intuition one must have for predicting chemical properties.
Today still whole departments of universitary institutions are thrown at even simple tasks, and oftentimes find nothing, like eg. solar-cell-materials and the like.

Then there of course do exist some packages, like the (GA)MESS, but these are just loads of theories put into code, just for checking around what might be. If it's true or false always the experiment decides, it's like lottery for even the most of the professionals, and computers didn't change that in the last 20 years (but are very useful for more simple tasks, like from the pressure calculate the dynamics of the pressure-wave-expansion, using ode-solvers and the like, _where_ sufficiently simple laws hold).

Thats why I say: The best thing one can do is to get structural data, write himself some code, get knowledge and see how far it gets ! This way at least the theory may be checked upon real existing evidence, the structural data within the databases !! These are what to get started from !!!
Engager
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You are talking not about computing, if you can get experimental structure you can also get experimental properties. Stuff i was posted above can give ZERO KNOWLEDGE aproximations, surely theese are not precise, but they are good then comparing theoretical structures and getting crude values to decide is there any reasons to spend money and time to produce proposed compounds. If you are a wordeater and neglect everything computed, you may continue to do so and stop loading topic with your complains. Your conversation style shows that you are not a scientist or not educated one, clever people always try to watch at problem from all sides, trying to separate good seeds from any aproach, while stupid complacent people only complain and neglect everething that not suits their oppinion.

[Edited on 1-10-2008 by Engager]

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