Review of the vulnerability of chemical equipment subjected to blast wave

doi:10.16085/j.issn.1000-6613.2018-0756

Abstract

Abstract:

Blast wave is one of the important escalation factors of Domino effect in chemical industry parks. Based on the literature analysis, the inadequacies of current research on vulnerability of chemical equipment subjected to blast wave were pointed out, from the four aspects of the intensity characterization of blast wave, the dynamic response of equipment damage, the damage probability of equipment and the vulnerability analysis. It was suggested that the characteristic of overpressure history curve and energy spectrum about different explosion types, quantitative study on spatial distribution parameter model or lumped parameter model, damage probability methods based on the structural reliability and the vulnerability analysis of blast wave intensity parameters and equipment resistance parameters should be further studied. Based on these, the flow chart of the vulnerability of chemical equipment subjected to blast wave was built. The critical sections of equipment vulnerability analysis were the reliability method of lumped parameter and stochastic finite element method. The review can be considered as the reference for enhancing the reliability and accuracy of quantitative risk assessment of domino effect.

Key words: blast wave, Domino effect, equipment vulnerability, reliability method of lumped parameter, stochastic finite element method

摘要：

爆炸冲击波是导致化工园区多米诺效应事故升级的重要原因之一。在文献分析的基础上，本文从爆炸冲击波强度表征、化工设备破坏失效的动力响应、破坏失效概率计算方法、易损性分析等4个方面，指出目前爆炸冲击波作用下化工设备易损性研究存在的不足，进而提出了进一步研究方向，包括不同爆炸类型的超压时程曲线特征与能量频谱规律研究、空间分布参数模型或集中参数模型的定量研究、基于结构可靠性的破坏失效概率计算方法、爆炸冲击波强度参数与设备抗性参数的易损性分析等，在此基础上系统性建立了爆炸冲击波作用下化工设备易损性研究的流程图，指出集中参数的可靠性方法与随机有限元法是爆炸冲击波作用下化工设备易损性研究的关键性方向，为提高爆炸冲击波多米诺效应定量风险评估的可靠度与精确度提供参考。

关键词: 爆炸冲击波, 多米诺效应, 设备易损性, 集中参数的可靠性方法, 随机有限元法

CLC Number:

TQ 086
X 937

Kun HU, Guohua CHEN, Zhihang ZHOU, Kongxing HUANG. Review of the vulnerability of chemical equipment subjected to blast wave[J]. Chemical Industry and Engineering Progress, 2019, 38(04): 1634-1645.

胡昆, 陈国华, 周志航, 黄孔星. 爆炸冲击波作用下化工设备易损性研究评述[J]. 化工进展, 2019, 38(04): 1634-1645.

Figures/Tables 8

初始事故场景	事故升级因素	可能产生的二级事故场景
VCE	火灾热辐射，爆炸冲击波	池火灾，喷射火，火球，闪火，机械爆炸、约束爆炸、BLEVE、VCE、有害介质泄漏
BLEVE	碎片，爆炸冲击波
约束爆炸	爆炸冲击波
机械爆炸	碎片，爆炸冲击波
凝聚相爆炸	爆炸冲击波

W T N T = α W f Q f Q T N T

；②

Z = R W T N T 1 / 3

；③TNT曲线（峰值超压p _s与比例距离Z之间的关系图）、峰值超压及正相持续时间经验公式

①爆源形状：点源

②需要确定的参数：W _f、α、Q _f、R

③结果：峰值超压p _s、冲量i _s

④适用条件：仅适用于远场预测

TNO多能法

①

R ˉ = R p 0 E 1 / 3

；②TNO曲线（量纲为1峰值超压

p -

、量纲为1正相持续时间

t + -

与量纲为1距离

R -

之间的关系图）；③p _s=

p -

p ₀，

t + = t ˉ + E / p 0 1 / 3 c 0

①爆源形状：半球形对称蒸气云

②需要确定的参数：E、R、爆源体积、爆源强度^[17]（通过点火能、爆源内阻塞度、边界约束程度确定）

③结果：峰值超压p _s、正相持续时间t ₊

④以爆源强度等级选取TNO曲线

⑤适用条件：近场、远场预测均可

Baker-Strehlow法

①

R ˉ = R p 0 E 1 / 3

；②Baker-Strehlow曲线（量纲为1峰值超压

p -

、量纲为1冲量

i -

与量纲为1距离

R ˉ

之间的关系图）；③p _s=

p -

p ₀，

i s = i ˉ E 1 / 3 p 0 2 / 3 c 0

①爆源形状：球形对称蒸气云

②需要确定的参数：E、R、爆源体积、最大火焰传播速度-马赫数Ma ^[26]（通过火焰传播方向、燃料活性、障碍物密度确定）

③结果：峰值超压p _s、冲量i _s

④以马赫数Ma等级选取Baker-Strehlow曲线

⑤适用条件：近场、远场预测均可

方法

计算过程

特征

TNT当量法

①

W T N T = α W f Q f Q T N T

；②

Z = R W T N T 1 / 3

；③TNT曲线（峰值超压p _s与比例距离Z之间的关系图）、峰值超压及正相持续时间经验公式

①爆源形状：点源

②需要确定的参数：W _f、α、Q _f、R

③结果：峰值超压p _s、冲量i _s

④适用条件：仅适用于远场预测

TNO多能法

①

R ˉ = R p 0 E 1 / 3

；②TNO曲线（量纲为1峰值超压

p -

、量纲为1正相持续时间

t + -

与量纲为1距离

R -

之间的关系图）；③p _s=

p -

p ₀，

t + = t ˉ + E / p 0 1 / 3 c 0

①爆源形状：半球形对称蒸气云

②需要确定的参数：E、R、爆源体积、爆源强度^[17]（通过点火能、爆源内阻塞度、边界约束程度确定）

③结果：峰值超压p _s、正相持续时间t ₊

④以爆源强度等级选取TNO曲线

⑤适用条件：近场、远场预测均可

Baker-Strehlow法

①

R ˉ = R p 0 E 1 / 3

；②Baker-Strehlow曲线（量纲为1峰值超压

p -

、量纲为1冲量

i -

与量纲为1距离

R ˉ

之间的关系图）；③p _s=

p -

p ₀，

i s = i ˉ E 1 / 3 p 0 2 / 3 c 0

①爆源形状：球形对称蒸气云

②需要确定的参数：E、R、爆源体积、最大火焰传播速度-马赫数Ma ^[26]（通过火焰传播方向、燃料活性、障碍物密度确定）

③结果：峰值超压p _s、冲量i _s

④以马赫数Ma等级选取Baker-Strehlow曲线

⑤适用条件：近场、远场预测均可

学者	TNT当量法的当量系数α	TNO多能法的爆源强度	Baker-Strehlow法的马赫数Ma	超压预测结果
Lobato等^[20]	0.1	10	0.59	当距离爆源较近时，预测结果大小顺序为：TNT当量法> TNO多能法>Baker-Strehlow法；当距离爆源较远时，3种方法预测结果相近
Sari^[27]	—	10	5.2	当距离爆源较近时，预测结果大小顺序为：TNO多能法>Baker-Strehlow法；当距离爆源较远时，两种方法预测结果相近
张网等^[26]	0.1	5	0.25	当距离爆源较近时，预测结果大小顺序为：TNT当量法>Baker-Strehlow法>TNO多能法；当距离爆源较远时，3种方法预测结果相近

热力学和物理假设	学者	备注
恒定体积能量增量	Brode^[29]	爆炸冲击波由气相与闪蒸的液相恒体积膨胀产生的
理想气体行为与等熵膨胀	Prugh^[30]	①爆炸冲击波由气相与闪蒸的液相膨胀产生； ②β=0.4； ③TNT曲线
热力学的有效能量	Crowl^[31,32]	爆炸冲击波由闪蒸的过热液相膨胀产生
等温膨胀	Smith等^[33]	爆炸冲击波由气相与闪蒸的液相的快速恒温膨胀产生
实际气体行为与绝热不可逆膨胀	Planas-Cuchi等^[34]	①爆炸冲击波由气相膨胀产生； ②β=0.4； ③TNT曲线
	Casal等^[35]	①爆炸冲击波由液相过热能产生； ②β=0.035-0.05；③TNT曲线
	Hemmatian等^[28]	①爆炸冲击波由气相与闪蒸的液相膨胀产生； ②β=0.4；③TNT曲线
实际气体行为与等熵膨胀	Roberts^[36]； CCPS^[37]	①爆炸冲击波由气相与闪蒸的液相膨胀产生； ② β=1（未使用）； ③TNO的Sachs 曲线
	Casal等^[35]	①爆炸冲击波由液相过热能产生； ② β=0.07-0.14； ③ TNT曲线
	Genova^[38]	①爆炸冲击波由液相过热能产生；② β=0.07； ③ TNO的Sachs曲线
等熵膨胀	Birk^[39]	①爆炸冲击波由气相膨胀产生；② β=2； ③ Kinney & Graham经验公式

准则	适用范围
超压准则	t ₊≥10T
冲量准则	t ₊≤T/4
超压-冲量准则	T/4≤t ₊≤10T

作者	模型
路胜卓^[51]	$a = - k i s m 1 + m 2 e - k t v = i s m 1 + m 2 e - k t w = i s (m 1 + m 2) k (1 - e - k t)$
Salzano等^[75]	$J = i s 2 ρ t σ D L θ 3$

作者	模型
路胜卓^[51]	$a = - k i s m 1 + m 2 e - k t v = i s m 1 + m 2 e - k t w = i s (m 1 + m 2) k (1 - e - k t)$
Salzano等^[75]	$J = i s 2 ρ t σ D L θ 3$

方法	作者	公式
静态峰值超压阈值方法	Gledhill和Lines^[80]	$p s < Δ p t h : F d = 0 p s < Δ p t h : F d = 1$ Δp _th为静态峰值超压阈值，常压容器Δp _th=7kPa，压力容器Δp _th=38kPa
比例方法	Bagster和Pitblado^[81]	$F d = 1 - R R t h 2$ 式中，r为到爆源中心的距离，m；R _th为静态超压等于Δp _th时到爆源中心的距离，m；Δp _th为静态峰值超压阈值，对所有设备Δp _th=36kPa
比例方法	Cozzani^[7]	$F d = 1, R < 12 R t h 32 - R R t h 2, 12 R t h ≤ R ≤ 32 R t h 0, R > 32 R t h$ 式中，R _th为25%破坏失效概率的静态超压时的爆心距，m；Δp _th为静态峰值超压阈值，对所有设备Δp _th=36kPa
Probit模型方法	Khan等^[82]	$Y = - 23.8 + 2.92 l n p s$
	Khan和Abbasi^[5]	$F d = 0, p s < 70 k P a Y = - 23.8 + 2.92 l n p s, p s ≥ 70 k P a$
	Cozzani和Salzano^[7]	$Y = 18.96 + 2.44 l n p s, 常压容器 - 42.44 + 4.33 l n p s, 压力容器 28.07 + 3.16 l n p s, 塔设备 - 17.79 + 2.18 l n p s, 辅助设备$
	Zhang等^[83]	$Y = - 9.36 + 1.43 l n p s, 常压容器 - 14.44 + 1.82 l n p s, 压力容器 - 12.22 + 1.65 l n p s, 塔设备 - 12.42 + 1.64 l n p s, 辅助设备$
	Mukhim和Abbasi等^[84]	$Y = - 88.88 + 8.79 l n p s, 卧式压力容器 - 49.16 + 4.93 l n p s, 球形压力容器 - 248.00 + 22.33 l n p s, 立式压力容器 - 13.31 + 2.02 l n p s, 锥顶式常压容器 - 22.74 + 3.00 l n p s, 其他常压容器 - 15.79 + 2.02 l n p s, 浮顶式常压容器 - 6.56 + 1.24 l n p s, 冷却塔 - 35.10 + 3.95 l n p s, 分馏塔 - 55.89 + 5.63 l n p s, 萃取塔 - 22.67 + 2.67 l n p s, 分解反应釜 - 26.76 + 3.08 l n p s, 其他化学反应器 - 201.20 + 18.98 l n p s, 热交换器 - 17.42 + 2.19 l n p s, 过滤装置$

方法	作者	公式
静态峰值超压阈值方法	Gledhill和Lines^[80]	$p s < Δ p t h : F d = 0 p s < Δ p t h : F d = 1$ Δp _th为静态峰值超压阈值，常压容器Δp _th=7kPa，压力容器Δp _th=38kPa
比例方法	Bagster和Pitblado^[81]	$F d = 1 - R R t h 2$ 式中，r为到爆源中心的距离，m；R _th为静态超压等于Δp _th时到爆源中心的距离，m；Δp _th为静态峰值超压阈值，对所有设备Δp _th=36kPa
比例方法	Cozzani^[7]	$F d = 1, R < 12 R t h 32 - R R t h 2, 12 R t h ≤ R ≤ 32 R t h 0, R > 32 R t h$ 式中，R _th为25%破坏失效概率的静态超压时的爆心距，m；Δp _th为静态峰值超压阈值，对所有设备Δp _th=36kPa
Probit模型方法	Khan等^[82]	$Y = - 23.8 + 2.92 l n p s$
	Khan和Abbasi^[5]	$F d = 0, p s < 70 k P a Y = - 23.8 + 2.92 l n p s, p s ≥ 70 k P a$
	Cozzani和Salzano^[7]	$Y = 18.96 + 2.44 l n p s, 常压容器 - 42.44 + 4.33 l n p s, 压力容器 28.07 + 3.16 l n p s, 塔设备 - 17.79 + 2.18 l n p s, 辅助设备$
	Zhang等^[83]	$Y = - 9.36 + 1.43 l n p s, 常压容器 - 14.44 + 1.82 l n p s, 压力容器 - 12.22 + 1.65 l n p s, 塔设备 - 12.42 + 1.64 l n p s, 辅助设备$
	Mukhim和Abbasi等^[84]	$Y = - 88.88 + 8.79 l n p s, 卧式压力容器 - 49.16 + 4.93 l n p s, 球形压力容器 - 248.00 + 22.33 l n p s, 立式压力容器 - 13.31 + 2.02 l n p s, 锥顶式常压容器 - 22.74 + 3.00 l n p s, 其他常压容器 - 15.79 + 2.02 l n p s, 浮顶式常压容器 - 6.56 + 1.24 l n p s, 冷却塔 - 35.10 + 3.95 l n p s, 分馏塔 - 55.89 + 5.63 l n p s, 萃取塔 - 22.67 + 2.67 l n p s, 分解反应釜 - 26.76 + 3.08 l n p s, 其他化学反应器 - 201.20 + 18.98 l n p s, 热交换器 - 17.42 + 2.19 l n p s, 过滤装置$

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