Presentation on theme: "EXPLOSIVES: Taken in part from a seminar by Jim Kahoe and Greg Brown Effects of an Explosion Classification of Explosives Low Explosives High Explosives."— Presentation transcript:
EXPLOSIVES: Taken in part from a seminar by Jim Kahoe and Greg Brown 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
3 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.
5 PriorityComposition of Products of Decomposition 1A metal and chlorine Metallic chloride(solid) 2Hydrogen and chlorine HCl (gaseous) 3A metal and oxygen Metallic oxide (solid) 4Carbon and oxygen CO (gaseous) 5Hydrogen and oxygen H2O (gaseous) 6CO and oxygen CO2 (gaseous) 7Nitrogen N2 (elemental) 8Excess oxygen O2 (elemental) 9Excess hydrogen H2 (elemental) Order of Priorities
6 PriorityComposition of Products of Decomposition ACarbon and oxygen CO (gaseous) BHydrogen and oxygen H2O (gaseous) CCO and oxygen CO2 (gaseous) DNitrogen N2 (elemental) EExcess oxygen O2 (elemental) FExcess hydrogen H2 (elemental) Order of Priorities
7 * 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 + 2.5H N2 + C The number of moles of gas formed is 10. The product, carbon, is a solid. Balancing Chemical Explosion Equations
8 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 -> 3CO H2O + 1.5N2 +.25O2 One mole of nitroglycerin produces = 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 = 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) = liters per mol Thus, at 15 °C the volume of gas produced by the explosive decomposition of one gram molecule of nitroglycerin becomes V = l (7.25 mol) = liters/mo Volume of Products of Explosion
9 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.
10 Q mp represents the total quantity of heat given off by a gram molecule of explosive of 15 °C and constant pressure (atmospheric); Q mv 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, Q mv = Q mp + W Q mp = ( viQ fi - vkQ fk ) where: Q fi = heat of formation of product i at constant pressure Q fk = 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 = ( x 10 4 N)(23.63 l)(Nmol)(10-3m3) W = (0.572)(Nmol) kcal mol
11 For TNT: C6H2(NO2)3CH3 -> with Nmol =
12 For TNT: C6H2(NO2)3CH3 -> 6CO + 2.5H N2 + C with Nmol = 10 mol Then: Qmp = 6(26.43) = 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 = (10) = kcal / mol
13 MW for TNT = g / mol Explosive Potential is defined as heat per kg of explosive Qkv = (kcal/mol) x 1000 (g/kg) / (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
14 Example The PETN reaction will be examined as an example of thermo-chemical calculations. PETN: C(CH 2 ONO 2 ) 4 MW = Heat of Formation = 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 -> 4H 2 O + 3O Then the remaining oxygen will combine with the CO to form CO and CO 2. 5CO + 3O -> 2CO + 3CO 2 Finally the remaining nitrogen forms in its natural state (N 2 ). 4N -> 2N 2 The balanced reaction equation is: C(CH 2 ONO 2 ) 4 -> 2CO + 4H 2 O + 3CO 2 + 2N 2
15 (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 re-sulting number of molecular volumes of gas (Nmol) is: Nmol = = 11 mol-volume/mol (3) Determine the potential (capacity for doing work). If the total heat liberated by an explosive under constant volume conditions (Q m ) is converted to the equivalent work units, the result is the potential of that explosive. The heat liberated at constant volume (Q mv ) is equivalent to the liberated at constant pressure (Q mp ) plus that heat converted to work in expanding the surrounding medium. Hence, Q mv = Q mp + Work (converted). a. Q mp = Q fi (products) - Q fk (reactants) where: Q f = Heat of Formation For the PETN reaction: Q mp = 2(26.43) + 4(57.81) + 3(94.39) - (119.4) = kcal/mol (If the compound produced a metallic oxide, that heat of formation would be included in Q mp.
16 b. Work = 0.572(Nm) = 0.572(11) = kcal/mol As previously stated, Qmv converted to equivalent work units is taken as the potential of the explosive. c. Qmv = (11) = kcal / mol Explosive Potential is defined as heat per kg of explosive Qkv = (kcal/mol) x 1000 (g/kg) / (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)