EXPLOSIVES: Effects of an Explosion Classification of Explosives
EXPLOSIVES: Effects of an Explosion Classification of Explosives Low Explosives High Explosives
Primary Secondary Conclusion When an explosive is detonated, the material is instantly converted from a solid into a mass of rapidly expanding gases. Causes 3 primary
effects: Blast pressure Fragmentation Thermal effects Taken in part from a seminar by Jim Kahoe and Greg Brown 2 Effects of an Explosion: Blast Pressure At the time of detonation, the gases can rush out at velocities of up to 7,000 mph and can exert pressure of up to 700 tons per square inch. This gas travels in a outward circular
pattern like a giant wave, smashing and shattering everything in its path. 3 4 Order of Priorities Priority Composition of Products of Decomposition 1 2 3 4
5 6 7 8 9 A metal and chlorine Metallic chloride(solid) Hydrogen and chlorine HCl (gaseous) A metal and oxygen Metallic oxide (solid) Carbon and oxygen CO (gaseous) Hydrogen and oxygen H2O (gaseous) CO and oxygen CO2 (gaseous) Nitrogen N2 (elemental) Excess oxygen O2 (elemental) Excess hydrogen H2 (elemental) 5
Order of Priorities Priority Composition of Products of Decomposition A B C D E F Carbon and oxygen CO (gaseous) Hydrogen and oxygen H2O (gaseous) CO and oxygen CO2 (gaseous)
Nitrogen N2 (elemental) Excess oxygen O2 (elemental) Excess hydrogen H2 (elemental) 6 Balancing Chemical Explosion Equations * The progression is from top to bottom; you may skip steps that are not applicable, but you never back up. * At each separate step there are never more than two compositions and two products. * At the conclusion of the balancing, elemental forms, nitrogen, oxygen, and hydrogen, are always found in diatomic form. Example, TNT: C6H2(NO2)3CH3; constituents: 7C + 5H + 3N + 6O Using the order of priorities priority 4 gives the first reaction products: 7C + 6O -> 6CO with one mol of carbon remaining Next, since all the oxygen has been combined with the carbon to form CO, priority 7 results in:
3N -> 1.5N2 Finally, priority 9 results in: 5H > 2.5H2 The balanced equation, showing the products of reaction resulting from the detonation of TNT is: C6H2(NO2)3CH3 -> 6CO + 2.5H2 + 1.5N2 + C The number of moles of gas formed is 10. The product, carbon, is a solid. 7 Volume of Products of Explosion The molecular volume of any gas at 0 C and under normal atmospheric pressure is very nearly 22.4 liters or 22.4 cubic decimeters. Thus, considering the nitroglycerin reaction. C3H5(NO3)3 -> 3CO2 + 2.5H2O + 1.5N2 + .25O2 One mole of nitroglycerin produces 3 + 2.5 + 1.5 + .25 = 7.25 molecular volumes of gas; and these molecular volumes at 0 C and atmospheric pressure form an actual volume of 7.25 X 22.4 = 162.4 liters of gas. (Note that the products H2O and CO2 are in their gaseous form.) Further, by employing Charles' 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 of its volume at 0 C, for each degree of rise in temperature. Therefore, at 15 C the molecular volume of any gas is, V15 = 22.4 (1 + 15/273) = 23.63 liters per mol Thus, at 15 C the volume of gas produced by the explosive decomposition of one gram molecule of nitroglycerin becomes V = 23.63 l (7.25 mol) = 171.3 liters/mo 8 The potential of an explosive is the total work that can be performed by the gas resulting from its explosion, when expanded adiabatically from its original volume, until its pressure is reduced to atmospheric pressure and its temperature to 15 C. The potential is therefore the total quantity of heat given off at constant volume when expressed in equivalent work units and is a measure of the strength of the explosive. An explosion may occur under two general conditions: the first, unconfined, as in the open air where the pressure (atmospheric) is constant; the second, confined, as in a closed chamber where the volume is constant. The same amount of heat energy is
liberated in each case, but in the unconfined explosion, a certain amount is used as work energy in pushing back the surrounding air, and therefore is lost as heat. In a confined explosion, where the explosive volume is small (such as occurs in the powder chamber of a firearm), practically all the heat of explosion is conserved as useful energy. If the quantity of heat liberated at constant volume under adiabatic conditions is calculated and converted from heat units to equivalent work units, the potential or capacity for work results. 9 Qmp represents the total quantity of heat given off by a gram molecule of explosive of 15 C and constant pressure (atmospheric); Qmv represents the total heat given off by a gram molecule of explosive at 15 C and constant volume; and W represents the work energy expended in pushing back the surrounding air in an unconfined explosion and thus is not available as net theoretical heat; Then, because of the conversion of energy to work in the constant pressure case, Qmv = Qmp + W
Qmp = ( viQfi - vkQfk ) where: Qfi = heat of formation of product i at constant pressure Qfk = heat of formation of reactant k at constant pressure v = number of mols of each product/reactants (m is the number of products and n the number of reactants) The work energy expended by the gaseous products of detonation is expressed by: W = Pdv W = (10.132 x 104 N)(23.63 l)(Nmol)(10-3m3) W = (0.572)(Nmol) kcal mol 10 For TNT: C6H2(NO2)3CH3 -> with Nmol =
11 For TNT: C6H2(NO2)3CH3 -> 6CO + 2.5H2 + 1.5N2 + C with Nmol = 10 mol Then: Qmp = 6(26.43) -16.5 = 142.08 kca /l mol Note: Elements in their natural state (H2, O2, N2, C, et,.) are used as the basis for heat of formation tables and are assigned a value of zero. Qmv = 142.08 + 0.572(10) = 147.8 kcal / mol 12 MW for TNT = 227.1 g / mol Explosive Potential is defined as heat per kg of explosive Qkv = 147.8 (kcal/mol) x 1000 (g/kg) / 227.1 (g/mol)= 651 kcal/kg
Rather than tabulate such large numbers, in the field of explosives, TNT is taken as the standard explosive, and others are assigned strengths relative to that of TNT. The potential of TNT has been calculated above to be 651 kcal/kg (2.72 x 106 J/kg since 1kcal=4185 J ). Relative strength (RS) may be expressed as R.S. = Potential of Explosive/ 2.72 x 106 13 Example The PETN reaction will be examined as an example of thermo-chemical calculations. PETN: C(CH2ONO2)4 MW = 316.15 Heat of Formation = 119.4 kcal/mol (1) Balance the chemical reaction equation. Using priorities in order decide reaction products: 5C + 12O -> 5CO + 7O Next, the hydrogen combines with remaining oxygen:
8H + 7O -> 4H2O + 3O Then the remaining oxygen will combine with the CO to form CO and CO2. 5CO + 3O -> 2CO + 3CO2 Finally the remaining nitrogen forms in its natural state (N2). 4N -> 2N2 The balanced reaction equation is: C(CH2ONO2)4 -> 2CO + 4H2O + 3CO2 + 2N2 14 (2) Determine the number of molecular volumes of gas per gram molecule. Since the molecular volume of one gas is equal to the molecular volume of any other gas, and since all the products of the PETN reaction are gaseous, the resulting number of molecular volumes of gas (Nmol) is: Nmol = 2 + 4 + 3 + 2 = 11 mol-volume/mol (3) Determine the potential (capacity for doing work). If the total heat liberated by an explosive under constant volume conditions (Qm) is converted to the equivalent work units, the result is the potential of that explosive. The heat liberated at constant volume (Qmv) is equivalent to the liberated at constant pressure (Q mp) plus that heat
converted to work in expanding the surrounding medium. Hence, Qmv = Qmp + Work (converted). a. Qmp = Qfi (products) - Qfk (reactants) where: Qf = Heat of Formation For the PETN reaction: Qmp = 2(26.43) + 4(57.81) + 3(94.39) - (119.4) = 447.87 kcal/mol (If the compound produced a metallic oxide, that heat of formation would be included in Qmp. 15 b. Work = 0.572(Nm) = 0.572(11) = 6.292 kcal/mol As previously stated, Qmv converted to equivalent work units is taken as the potential of the explosive. c. Qmv = 447.87 + 0.572(11) = 454.16 kcal / mol Explosive Potential is defined as heat per kg of explosive Qkv =454.16 (kcal/mol) x 1000 (g/kg) / 316.1 (g/mol)= 1,436.8 kcal/kg = 1,436.8 x 4185 = 6.01 x 106 J / kg
This product may then be used to find the relative strength of PETN, which is e. RS = Pot (PETN = 6.01 x 106 / (2.72 x 106 ) = 2.21 Pot (TNT) 16
MS Word 2000 Mail Merge MS Word 2000's mail merge feature allows you to use a form letter or document, insert names, addresses, or other information into the letter or form automatically by inserting record fields from a data source.
Other: surrealist art. Being able to specify searches within a certain category, I.e. to choose paitning and then choose a more specific category within that to narrow the search. images that should have come up didn't. confident if I'd had...
Neonatal Parenteral Nutrition-'There's a hole in my 'tummy' dear Liza, dear Liza!' Deirdre Hartigan 13th November 2011 * * Intestinal failure occurs when the gastrointestinal tract is unable to ingest, digest and absorb sufficient macronutrients and/or water and electrolytes to...
Memorial - Norman MacCaig. Context. MacCaig is immersed in the "intolerable distance" of death, painfully conscious of its "ugliness", and painfully conscious too of the all pervading absence of his dear one. MacCaig was an atheist. As such, in the...
Flatter or negative sloping term yield curves seem to correspond to lower PE ratios and upwards sloping yield curves to higher PE ratios. Using the Regression to gauge the market… We can use the regression to predict E/P ratio in...
"The soft skills piece is something we've been hearing from business and industry a good bit." "…understanding you need to show up on time, knowing how to dress, knowing how to work in teams, knowing how to get along with...
10 Things to Remember for your Debate. 3. Simplify & be direct. w. hen you plan. Use . point form/mind map notes. 2. Know the other side's argument. Look at it from their perspective. 1 . Listen, adapt and think...