Forensic Science Communications - April 2004
![]() |
![]() |
April 2004 - Volume 6 - Number 2 |
Case Report
Explosion from Nonexplosive Material: Reconstruction of the Sequence of Events
Dilip Kumar Kuila
Junior Scientific Officer (Explosives)
Ashutosh Chakrabortty
Senior Scientific Assistant (Explosives)
Central Forensic Science Laboratory
Kolkata, India
S. C. Lahiri
Department of Chemistry Professor
University of Kalyani
Kalyani, Nadia, West Bengal, India
Abstract | Introduction | Case History
Results and Discussion | Conclusions | References
Abstract
A case study of an explosion causing appreciable damage to a building in Durgapur, West Bengal, India, is presented. Systematic assessment of the damage and the physical evidence confirm it was a fuel-air aerosol explosion arising from liquid petroleum gas leakage and not an improvised explosive device. Attempts were made to reconstruct the sequence of events leading to the explosion. Different parameters and mathematical probabilities were examined to suggest the possibility of attaining the optimum condition for an aerosol-type explosion. The volume of the leaked gases was found to be close to the lower explosion limit.
Introduction
Explosions may originate from two sources—an improvised explosive device or a fuel-air explosion. The presence of explosives residue, initiators, or accelerators give sufficient evidence for an improvised explosive device. However, fuel-air aerosol, particularly liquid petroleum gas, explosion cases are difficult to investigate because of the lack of evidence (Beveridge 1998; Narayanan 1996). Fuel-air aerosol may result from mixing combustible gases, vapors, dusts, and mists or from mixing combustible liquids with air (aerosol) in appropriate proportions. Aerosols are formed by diffusion of gases and liquids in air (Sharma 1992). Fuel-air aerosol or aerosol explosions cause heavy damage to buildings resulting in damages of property or loss of lives.
The use of liquid petroleum gas for cooking and the corresponding explosions due to leakage are increasing. Also increasing are terrorist and other criminal activities in which improvised explosive devices are used. The apparent causes of explosions in many cases are difficult to ascertain because of an absence of adequate evidence. Sehgal et al. (1999) described an explosion in Punjab, India, and from the evidence and mathematical calculations, they confirmed it was an aerosol explosion.
An explosion in a building in Durgapur, West Bengal, India, is reported. Findings from examining the explosion site and the evidence excluded the possibility of an improvised explosive device. The explosion was due to liquid petroleum gas leakage. Attempts were made to quantify and estimate the amount of gas leaked and to assess the damage to the building.
To understand and reconstruct the sequence of events, different physical, chemical, and mathematical possibilities for attaining the critical concentration of fuel and air (aerosol) mixture were considered to substantiate the case for a fuel-air explosion.
Case History
In Durgapur, West Bengal, a powerful explosion occurred in a second floor apartment at 1:30 a.m. on October 10, 2002. Five people were found unconscious in the poorly ventilated room. The apartment was badly damaged, and one resident was superficially burned. Investigators originally thought an improvised explosive device caused the explosion; however, there was a strong liquid petroleum gas smell present.
Description of the Apartment
Figure 1 is a diagram of the apartment.
There are eight doors (D1, D2, D3, D4, D5, D6, D7, D8) and three windows (W1, W2, W3). D9 is the front door of the neighboring apartment.
The room dimensions are as follows:
Laboratory Tests and Other Observations
Forensic experts visited the scene and examined the gas cylinder, oven, gas pipe, and regulator. Cotton swabs from different blackened and burned areas of the house were examined in the laboratory. The results of the tests and other observations are as follows.
Observed Visible Damages
The sequence of events and available information were arranged to derive a reasonable explanation of the cause of the explosion.
Results and Discussion
The evidence (i.e., lack of epicenter, power sources, circuitry) and forensic examinations exclude the possibility of an improvised explosive device. It is, instead, a case of fuel-air (aerosol) explosion from liquid petroleum gas leakage in a congested and inadequately ventilated apartment.
The deficiency of oxygen resulting from the partial replacement of air by liquid petroleum gas from the floor level and possible narcotic effect due to the hydrocarbons and accumulated carbon dioxide caused the unconscious state of the residents.
The exact amount and the total time of the gas leakage are unknown. However, the total amount of gas consumed in 15 days (as per evidence) and gas leakage amounted to 10.4kg. According to a resident, the usual consumption of a full cylinder of gas (14.2kg) was 30 days. The amount of leakage was about 3.3kg. As per the specification of manufacturers, liquid petroleum gas contains about 70 percent propane (molecular weight 44) and 30 percent butane (molecular weight 58), and the leaked gas consisted of 2.31kg of propane (52.5gm moles) and 0.99kg of butane (17.06gm moles).
Calculation
TNT equivalency of the blast (Beveridge 1998; Noon 1995).
The total heat of combustion = (52.5 X 530.0 + 17.06 X 690) X 4.18k joules = 1655512.95k joules.
(530.0 X 4.18k joules and 690.0 X 4.18k joules are the total heat of combustion of propane and butane respectively) (Weast 1986-1987).
The energy is equivalent to ~ 35.3kg of TNT (trinitrotoluene). The energy content of TNT is 4.68MJ/kg (Beveridge 1998). Assuming 40-50 percent efficiency, the energy equivalent of the blast would be nearly 14-17kg TNT. Approximate side-on shock wave overpressure (k Pa) appears to be in the region (as per damages encountered) of 10-21 (Beveridge 1998).
Stoichiometric Ratio Between the Liquid Petroleum Gas and the Oxygen in the Room
The exact pressure of the gas and the time lag between the gas leakage and the explosion are not known. However, due to congestion and poor ventilation, leaked gas may be concentrated in Room A, Space C, and the kitchen.
Total volume of Room A, Space C, and the kitchen = (28490069.54 + 8480010.65 + 3927243.75) = 40897323.94cm3.
This is approximately equivalent to 365 moles of oxygen (assuming 4 : 1 ratio of N2 and O2 in air).
The equations for combustion of C3H8 and C4H10 to CO2 and H2O are
C3H8 + 5O2 = 3CO2 + 4H2O
and C4H10 + 6.5 O2 = 4CO2 + 5H2O.
Complete combustion of the leaked liquid petroleum gas would require (52.5 x 5 + 17.06 x 6.5) or 373.39 moles of oxygen. The stoichiometric ratio between the leaked liquid petroleum gas and oxygen in the room is almost one.
The main parameters for the formation of a fuel air (aerosol)-type of mixture from the accumulation of leaked gas into a compartment and inflammation leading to explosion are as follows:
The exact stoichiometry between air, the leaked liquid petroleum gas, and the proper model of the explosion are difficult to ascertain. A simple model, however, would be to consider that the leaked gas occupied the total space of Room A, Space C, and the kitchen.
Thus, there were two possibilities.
Possibility 1
Volume of propane in liquid petroleum gas (for 2.31kg) = 1176000cm3 and that of butane in liquid petroleum gas (for 0.99kg) = 382144cm3
percent of propane in the space = 1176000/40897323.94 x 100 = 2.87% (2.99%)
and percent of butane in the space = 382144/40897323.94 x 100 = 0.93 % (0.97%)
(ratio of air : propane : butane will be roughly 96 : 3 : 1)
The gas leaked is above the lower explosion limit. The values in parenthesis are calculated assuming the displacement of equal volume of air by propane and butane. However, to attain the upper explosion limit, the amount of leaked gas should be about 10 to 11kg. Therefore, attaining the upper explosion limit appears unlikely.
Possibility 2
The gas diffusion from the open end of the pipe displaces air from the ground level to the upper level because the density of air is less than the densities of propane and butane. The density profile is inverted so the proportion of butane and propane are greater on the floor and middle portion. W3, D2, and D6 were open, and other outlets were closed. Thus, due to lack of proper ventilation and congestion of the room, homogeneous mixing of the gases was not possible, but local concentrations of leaked gases may have distinct possibilities. Due to leakage of gases, displacement of an equivalent amount of air from the space would likely occur to maintain ambient pressure. Most of the gas burning was in Room A and Space C. The ignition point was the refrigerator.
The ceiling fan, light bulbs, and tube lights were intact. The cracks in the wall AB were between 128.01 to 219.45cm (concentration of liquid petroleum gas above 228.6cm was negligible). Thus the total amount of leaked gas must be in the volume between 0-228.6cm. Moreover, the gas on the floor was not connected with the blast. Therefore, it is rational to consider that the volume of leaked gases in the area between the heights 114.3 to 228.6cm was responsible for the initiation of the explosion.
The velocity of gases (taking the density)
Air Cair = {(3 x 76 x 981 x 13.6)/(14 x 0.000089)}½ = 4.9 x 104cm (O2 = 4.6 x 104cm)
Propane CProp. = 3.9 x 104cm
Butane CBut. = 3.4 x 104cm
The rate of diffusion of the gases
![]() |
|
| = 0.27 : 0.21 : 0.18 | |
The rate of diffusion shows that air will diffuse at a rate 1.5 times greater than that of butane, and 1.3 times greater than that of propane. Therefore, the middle portion (portion between 114.3 to 228.6cm) will be slightly rich in propane but less in butane and deficient in air. Moreover, the portion near W3 of the kitchen was unaffected, and the cylinder was placed near D6 (Figure 1). Therefore, half of the volume of the kitchen between heights 114.3 to 228.6cm was considered in the calculation.
Volume of air between heights of 114.3 to 228.6cm in Room A + Space C=(358.41 + 106.68cm) X 281.94 X 114.3cm = 14987870.3cm3
Half of the volume of the kitchen (between 114.3 to 228.6cm) = ½ (152.4 x 91.4 x 114.3) = 796062.9cm3
(i.e., total volume of the space = 15783933.2cm3)
percent of propane in the space = 1176000/15783933.2 . 1/2 x 100 = 3.72% (3.92%)
and percent of butane in the space = 382344/15783933.2 . 1/2 x 100 = 1.21% (1.27%)
The exact volume of the leaked gas is difficult to determine, but the gas leak did not exceed four to five percent (by volume) of the total air; however, the local concentration of the leaked gases at the ignition point may be higher. The minimum ignition energy for propane in air is only 0.00026 joules, and the volume of leaked gas was close to the lower explosion limit.
Conclusions
Assessing the damage and determining the mathematical possibilities of the fuel-air mixture (aerosol) formation led to the following conclusions.
References
Akhavan, J. Chemistry of Explosives. Royal Society of Chemistry Information Services, Cambridge, England, 1998.
Beveridge, A., ed. Forensic Investigation of Explosions. Taylor and Francis, London, 1998.
Narayanan, T. V. Modern Techniques of Bomb Detection and Disposal, R. A. Security System, New Delhi, India, 1996.
Noon, R. K. Engineering Analysis of Fires and Explosions. CRC, Boca Raton, Florida, 1995.
Sehgal, D. P. S., Sharma, S. N., and Misra, G. J. Aerosol or IED explosion: A case report, Forensic Science International (1999) 102(1): 67-72.
Sharma, B. R. Aerosol (nonexplosives) explosion, Journal of Police Research Development (1992) 26-30.
Weast, R. C., ed. Handbook of Chemistry and Physics. 67th ed., CRC, Boca Raton, Florida, 1986-1987.
Yallop, H. J. Explosion Investigation. Scottish Academic Press, Edinburgh, Scotland, 1980.







