火场条件下相邻汽油罐油蒸汽泄漏及爆炸规律

Gasoline vapor leakage and explosion law of an oil tank adjacent to fire

油库发生火灾时,临近的油罐内汽油受热形成油蒸汽从呼吸阀泄出,泄漏的油蒸汽与空气混合易点燃,易引起燃烧爆炸事故。本文中以容积为5 000 m^3(22 m×13 m)的汽油罐泄漏油蒸汽为研究对象,以数值模拟为研究方法,研究了汽油蒸汽泄漏及爆炸规律。研究发现:在距地面高1 m的平面上,当无风且呼吸阀泄漏油蒸汽速率为0.25 m/s时,距该罐中心50 m以外就可视作安全区域;当呼吸阀泄漏的油蒸汽速率为0.25 m/s时,风速达到5.0 m/s及以上,不易积聚成爆炸油蒸汽;当无风时,呼吸阀泄漏油蒸汽的速率增大1个量级,油蒸汽积聚达到爆炸下限1/2所需时间缩短2个量级;当风速为3.0 m/s、呼吸阀泄漏油蒸汽速率为0.25 m/s、泄漏时间为200 s、着火点距罐壁1 m时,距点火源距离增大1个量级,超压峰值下降1~2个量级。

When the oil depot is in fire,a large amount of gasoline vapor is formed by the heat absorption of oil in an adjacent gasoline tank with a fixed top.The gasoline vapor is ignited after mixing with air,which is likely to cause combustion and explosion accidents.In this paper,the gasoline vapor leaked from a tank of 5 000 m^3(22 m×13 m)is taken as the research object,the law of gas vapor leakage and explosion is researched by numerical simulation.It is found that the area beyond 50 m away from the tank center is safe at 1 m above the ground if there is no wind and the gasoline vapor leakage velocity is 0.25 m/s.It is not easy to accumulate into the flammable gasoline vapor as the gasoline vapor leakage velocity from the breathing valve is 0.25 m/s,and the wind speed reaches 5.0 m/s and above.As there is no wind and the gasoline vapor leakage velocity from the breathing valve is increased by 1 order of magnitude,the time to half of the lower flammability limit is reduced by 2 orders of magnitude.When the wind speed is 3.0 m/s,the gasoline vapor leaking velocity is 0.25 m/s,and the leakage time is 200 s,the peak overpressure is reduced by 1 2 orders of magnitude if the distance to the ignition source is increased by 1 order of magnitude.