5.4.1 Analysis of vapor cloud explosion model Vapor cloud explosion can produce a variety of destructive effects, such as shock wave overpressure, thermal radiation, debris effect, etc., but the most dangerous and destructive force is the destructive effect of shock wave.Common shock damage damage criteria include overpressure criterion, impulse criterion, pressure impulse criterion, etc.Overpressure criterion is adopted in this evaluation. The overpressure of vapor cloud explosion is calculated by TNT equivalent method.TNT equivalent of vapor cloud explosion can be estimated by the following formula: Where: 1.8: ground explosion coefficient; α: TNT equivalence coefficient of vapor cloud, 0.04; WF: mass of fuel involved in explosion in vapor cloud formed by LPG, kg; QF: combustion heat of fuel, kJ / kg; Qtnt: TNT explosion heat, 4520kj / kg; Wtnt: TNT equivalent of vapor cloud, kg; According to the data provided by the project unit, the composition of butane.According LPG is to the 50% propane material and 50% coefficient and characteristic table, the combustion heat HC / (103btu. LB1) of propane is 19.9, and the combustion heat HC / (103btu. LB-1) of butane is 19.4, then: Combustion heat of LPG QF = 19.9 × 103 × 0.5 + 19.4 × 103 × 0.5 = 19.7 × 103 (BTU / lb) = 19.7 × 103 × 1.055 ÷ 0.454 = 45779 (kJ / kg) The density of liquefied petroleum gas is 0.51t/m3, the filling coefficient is 0.9, and the total volume percentage involved in the explosion in the vapor cloud formed by the leaked liquefied petroleum gas is 30%, assuming that the 6m3 of liquefied petroleum gas in this grade II supply station is completely leaked (in fact, it is impossible to completely leak).Then: When 6m3 LPG leaks, the mass of fuel in the vapor cloud formed by LPG is WF = 6 × 0.51 × 103 × 0.9 × 30% = 826 (kg) WTNT=1.8×0.04×826×45779/4520=602.3(kg) Death zone If there is a lack of protection for the personnel in the area, it is considered that no one will suffer serious injury or death outside the area. The inner diameter is zero and the outer diameter is R0, indicating that the probability of death of the personnel around the outer circle due to the shock wave is 50%. The relationship between it and the explosion volume is determined by the following formula: Where: wtnt is TNT equivalent of explosion source, kg. Substitute wtnt = 602.3 (kg, TNT) The death radius r0 = 11.3m It can be considered that the number of people who do not die in the circle is exactly equal to the number of people who die outside the circle, that is to say, all people in the death zone will die, and no one outside the death zone will die.This assumption is approximately true when the damage effect decays rapidly with distance. Seriously injured area In case of lack of protection, the vast majority of people in the area will suffer serious injury, and a few people may die or be injured.The inner diameter is the death radius r0, and the outer diameter is recorded as R1, which means that the probability of rupture of eardrum caused by shock wave is 50%. It requires that the peak overpressure of shock wave is 44000pa.Shock overpressure △ PS can be calculated as follows: △Ps=0.137Z-3+0.119Z-2-0.019 E=1.8 alpha WfQf △Ps =44000/P0 =0.4344 Formula: Is the horizontal distance from the target to the explosion source, m; Is the ambient pressure, PA; Is the total energy of explosion source, J / kg. The radius of serious injury can be obtained by substituting the peak overpressure 44000pa of shock wave: R1=1.082(E/101300)=32.4m1/3 Minor injury area In case of lack of protection, the vast majority of the people in the area will suffer slight injury, a few will be seriously injured or safe, and the possibility of death is very small.The inner diameter of this area is the outer diameter of the serious injury areaR1, the outer diameter is R2, which represents that the probability of eardrum rupture due to shock wave at the outer boundary is 1%, and the peak overpressure of shock wave required is 17000pa.According to the calculation method of serious injury radius, it can be concluded that: R2=1.956(E/101300)=58.6m1/3 Safety zone Even if there is no protection, the vast majority of people in the area will not be injured, and the probability of death is almost zero.The inner diameter of the area is R2, and the outer diameter is infinite. ⑤ Damage zone of buildings The degree of damage to the building can be determined by the following methods. Table 5.

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