Evaluation of the Effects and Consequences of Major Accidents in Industrial Plants
Industrial Safety Series Vol. 1. Safety of Reactive Chemicals (T. Yoshida) Vol. 2. Individual Behaviour in the Control of Danger (A.R. Hale and A.I. Glendon) Vol. 3. Fluid Mechanics for Industrial Safety and Environmental Protection (T.K. Fannelöp) Vol. 4. Thermal Hazards of Chemical Reactions (T. Grewer) Vol. 5. Safety of Reactive Chemicals and Pyrotechnics (T. Yoshida, Y. Wada and N. Foster) Vol. 6. Risk Assessment and Management in the Context of the Seveso II Directive (C. Kirchsteiger, Editor and M. Christou and G. Papadakis, Co-editors) Vol. 7. Critical Temperatures for the Thermal Explosion of Chemicals (T. Kotoyori) Vol. 8. Evaluation of the Effects and Consequences of Major Accidents in Industrial Plants (J. Casal)
Industrial Safety Series, 8
Evaluation of the Effects and Consequences of Major Accidents in Industrial Plants Joaquim Casal Centre for Studies on Technological Risk Department of Chemical Engineering Universitat Politècnica de Catalunya Barcelona, Spain
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Preface
This book presents the basic aspects of the various kinds of major accidents that can occur in industrial plants and during the transport of dangerous goods, as well as the methods and mathematical models used to predict the effects and consequences on buildings, equipment and the population. The various chapters analyse leaks of dangerous substances, fires, various types of explosion, and atmospheric dispersion of toxic or flammable products. They also present vulnerability models that can predict the consequences of such an accident on a sensitive element (person or equipment). The chapter dedicated to quantitative risk analysis explains how to use the aforementioned models and methods to determine the individual and collective risk posed by a particular plant or activity. The study of this type of accident is a basic part of risk analysis. It is essential to improving the safety of industry and related activities. This eminently practical book covers basic topics that will help the reader understand these phenomena. The calculation models included herein are relatively simple and can be used to obtain useful, applicable results; in fact, they are often used by professionals. In order to demonstrate how these models are used, I have applied them to a series of examples and real-life cases. Although these calculations are usually performed by hermetic computer codes, strong conceptual knowledge can help us avoid the error of accepting absurd or excessively conservative or optimistic results as correct. This is only possible if we have a good understanding of the phenomena involved and the equations and hypotheses of the applied models. This book is designed for engineers working in (or who aim to work in) the risk analysis field, students finishing an undergraduate engineering degree - fortunately, such programmes have begun placing more emphasis on safety and risk - and postgraduate students. I have based this book on my professional experience and career. I would therefore like to acknowledge some colleagues with whom I have had the pleasure of working over the past twenty years. I especially want to thank my friend Norberto Piccinini, a professor at the Polytechnic University of Turin, who taught a pioneering course on risk analysis in Spain and introduced me to the subject. I am also grateful to all those who offered their comments and criticism on this book, especially my colleagues at the Centre for Technological Risk Studies (CERTEC): Professors Josep Arnaldos and Eulàlia Planas and researchers Jordi Dunjó, Mercedes Gómez-Mares, Miguel Muñoz and Adriana Palacios. I would also like to thank Professor Juan A. Vílchez for providing very interesting original material on quantitative risk analysis. The doctoral thesis of Andrea Ronza has been a very useful source of information. Finally, I would like to thank Professor Josep M. Salla, a colleague at my university, and Professor Roberto Bubbico of La Sapienza University for their comments.
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Writing this book has been a personally enriching experience. I hope that it contributes in some way to improving the safety of industrial facilities and the quality of the environment. Let me finish with a reflection: In the fields of accident modelling and risk analysis, we work with a considerable degree of uncertainty. We often make up for this by making simplifying assumptions. Even if we apply a model more or less correctly, we may still obtain erroneous results. As in other fields of engineering, experience and good judgement are essential. Joaquim Casal Barcelona
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Contents
Preface v 1. Introduction 1 1. Risk 1 2. Risk analysis 2 3. Major accidents 5 3.1 Types 5 3.2 Damage 9 4. Domino effect 12 4.1 Classification of domino effects 12 4.2 An example case 12 5. Mathematical modelling of accidents 14 Nomenclature 16 References 16 2. Source term 19 1. Introduction 19 2. Liquid release 21 2.1 Flow of liquid through a hole in a tank 21 2.2 Flow of liquid through a pipe 24 2.2.1 Liquid flow rate 24 2.2.2 Friction factor 27 3. Gas/vapour release 30 3.1 Flow of gas/vapour through a hole 30 3.1.1 Critical velocity 30 3.1.2 Mass flow rate 33 3.1.3 Discharge coefficient 33 3.2 Flow of gas/vapour through a pipe 35 3.3 Time-dependent gas release 40 4. Two-phase flow 42 4.1 Flashing liquids 42 4.2 Two-phase discharge 43 5. Safety relief valves 44 5.1 Discharge from a safety relief valve 45 6. Relief discharges 47
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6.1 Relief flow rate for vessels subject to external fire 48 6.2 Relief flow rate for vessels undergoing a runaway reaction 49 7. Evaporation of a liquid from a pool 53 7.1 Evaporation of liquids 53 7.2 Pool size 53 7.2.1 Pool on ground 53 7.2.2 Pool on water 53 7.3 Evaporation of boiling liquids 53 7.4 Evaporation of non-boiling liquids 54 8. General outflow guidelines for quantitative risk analysis 55 8.1 Loss-of-containment events in pressurized tanks and vessels 56 8.2 Loss-of-containment events in atmospheric tanks 56 8.3 Loss-of-containment events in pipes 56 8.4 Loss-of-containment events in pumps 56 8.5 Loss-of-containment events in relief devices 56 8.6 Loss-of-containment events for storage in warehouses 57 8.7 Loss-of-containment events in transport units in an establishment 57 8.8 Pool evaporation 57 8.9 Outfllow and atmospheric dispersion 58 Nomenclature 58 References 59 3. Fire accidents 61 1. Introduction 61 2. Combustion 61 2.1 Combustion reaction and combustion heat 62 2.2 Premixed flames and diffusion flames 63 3. Types of fire 63 3.1 Pool fires 64 3.2 Jet fires 65 3.3 Flash fires 65 3.4 Fireballs 66 4. Flammability 66 4.1 Flammability limits 66 4.1.1 Estimation of flammability limits 67 4.1.2 Flammability limits of gas mixtures 69 4.1.3 Flammability limits as a function of pressure 70 4.1.4 Flammability limits as a function of temperature 70 4.1.5 Inerting and flammability diagrams 71 4.2 Flash point temperature 72 4.3 Autoignition temperature 73 5. Estimation of thermal radiation from fires 74 5.1 Point source model 74 5.2 Solid flame model 77 5.2.1 View factor 78 5.2.2 Emissive power 80 6. Flame size 83 6.1 Pool fire size 84
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6.1.1 Pool diameter 84 6.1.2 Burning rate 86 6.1.3 Height and length of the flames 87 6.1.4 Influence of wind 87 6.2 Size of a jet fire 90 6.2.1 Jet flow 90 6.2.2 Shape and size of the jet fire 92 6.2.3 Influence of wind 94 6.3 Flash fire 99 7. Boilover 100 7.1 Tendency of hydrocarbons to boilover 102 7.2 Boilover effects 103 8. Fireball 104 8.1 Fireball geometry 104 8.1.1 Ground diameter 104 8.1.2 Fireball duration and diameter 104 8.1.3 Height reached by the centre of the fireball 105 8.2 Thermal features 106 8.2.1 Radiant heat fraction 106 8.2.2 Emissive power 107 8.2.3 View factor 108 8.3 Constant or variable D, H and E 108 9. Example case 109 Nomenclature 113 References 115 4. Vapour cloud explosions 119 1. Introduction 119 2. Vapour clouds 120 3. Blast and blast wave 121 3.1 Blast wave 121 3.2 Detonations 122 3.3 Deflagrations 123 3.4 Blast scaling 123 3.5 Free-air and ground explosions 124 4. Estimation of blast: TNT equivalency method 125 5. Estimation of blast: multi-energy method 129 6. Estimation of blast: Baker-Strehlow-Tang method 133 7. Comparison of the three methods 136 8. A statistical approach to the estimation of the probable number of fatalities in accidental explosions 138 9. Example case 140 Nomenclature 144 References 144 5. BLEVEs and vessel explosions 147 1. Introduction 147 2. Mechanism of BLEVE 149
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2.1 Liquid superheating 151 2.2 Superheat limit temperature 153 2.3 Superheat limit temperature from energy balance 156 2.4 When is an explosion a BLEVE? 159 3. Vessel failure 163 3.1 Mechanism 163 3.2 Pressure required for vessel failure 164 4. Estimation of explosion effects 165 4.1 Thermal radiation 165 4.2 Mechanical energy released by the explosions 165 4.2.1 Ideal gas behaviour and isentropic expansion 166 4.2.2 Real gas behaviour and irreversible expansion 168 4.3 Pressure wave 169 4.4 Using liquid superheating energy for a quick estimation of 'P 173 4.5 Estimation of 'P from characteristic curves 176 4.6 Missiles 178 4.6.1 Range 181 4.6.2 Velocity 182 5. Preventive measures 183 6. Example cases 186 Nomenclature 190 References 192 6. Atmospheric dispersion of toxic or flammable clouds 195 1. Introduction 195 2. Atmospheric variables 195 2.1 Wind 196 2.2 Lapse rates 199 2.3 Atmospheric stability 200 2.4 Relative humidity 204 2.5 Units of measurement 204 3. Dispersion models 205 3.1 Continuous and instantaneous releases 205 3.2 Effective height of emission 207 4. Dispersion models for neutral gases (Gaussian models) 208 4.1 Continuous emission 209 4.2 Instantaneous emission 215 4.3 Short-term releases 218 5. Dispersion models for heavier-than-air gases 219 5.1 Britter and McQuaid model 221 5.1.1 Continuous release 221 5.1.2 Instantaneous release 223 5.1.3 Finite duration release 225 6. Calculating concentration contour coordinates 227 6.1 The Ooms integral plume model 227 6.2 Determining concentration contour coordinates 227 7. Dispersion of dust 230 8. Atmospheric dispersion of infectious agents 231
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8.1 Emission source 231 8.2 Dispersion of airborne pathogenic agents 232 8.3 Epidemics: dispersion of airborne viruses 232 9. Escaping 236 10. Sheltering 236 10.1 Concentration indoors 236 10.1.1 Continuous release 236 10.1.2 Temporary release 237 10.1.3 Instantaneous release 239 10.1.4 A simplified approach 241 11. Example case 242 Nomenclature 244 Annex 6-1 246 References 247 7. Vulnerability 249 1. Introduction 249 2. Population response to an accident 249 3. Probit analysis 250 4. Vulnerability to thermal radiation 254 4.1 Damage to people 254 4.1.1 Probit equations 257 4.1.2 Clothing 258 4.1.3 Escape 258 4.1.4 Effect of hot air 261 4.2 Material damages 261 5. Vulnerability to explosions 263 5.1 Damage to human beings 263 5.1.1 Direct consequences 263 5.1.2 Indirect consequences 265 5.1.3 Collapse of buildings 268 5.2 Consequences of an explosion for buildings and structures 269 6. Vulnerability to toxic substances 271 6.1 Dose and probit equations 273 6.2 Substances released from a fire 275 7. Inert gases 277 8. Influence of sheltering 279 8.1 Thermal radiation 279 8.2 Blast 280 8.3 Toxic exposure 280 9. Relationship between the number of people killed and the number of people injured in major accidents 280 10. Zoning according to vulnerability 281 11. Example case 283 Nomenclature 286 Annex 7-1 287 References 288
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8. Quantitative risk analysis 291 1. Introduction 291 2. Quantitative risk analysis steps 292 3. Individual and societal risks 294 3.1 Individual and societal risks definition 294 4 Risk mapping 296 4.1 Individual risk contours 296 4.2 Procedure 296 4.3 Societal risk 298 5. Introductory examples of risk calculation 299 6. Frequencies and probabilities 306 6.1 Frequencies of most common loss-of-containment events 306 6.2 Failure of repression systems 306 6.3 Human error 306 6.4 Probabilities for ignition and explosion of flammable spills 306 6.5 Meteorological data 309 7. Example case 309 7.1 Estimation of the frequencies of initiating events 311 7.2 Event trees of the diverse initiating events 312 7.3 Effects of the different accidental scenarios 319 7.4 Calculation of the individual risk 327 Nomenclature 329 References 331 Annex 1 Constants in the Antoine equation 333 Annex 2 Flammability levels, flash temperature and heat of combustion (higher value) for different substances 335 Annex 3 Acute Exposure Guideline Levels (AEGLs) 337 Annex 4 Immediately Dangerous to Life and Health concentrations (IDLH) 345 Annex 5 Determining the damage to humans from explosions using characteristic curves 347 Index 353
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Chapter 1
Introduction 1 RISK Risk is a familiar concept in many fields and activities including economics, business, sport, industry, also in everyday life, but it is not always referred to with exactly the same meaning. A strict definition is required, however, when the term is used in a professional environment. Various definitions have been proposed, for example: “a situation which can lead to an unwanted negative consequence in a given event”; “the probability that a potential hazard occurs”; “the unwanted consequences of a given activity, in relation to their probability of occurrence”; or, more specifically, “a measure of human injury, environmental damage or economic loss in terms of both the incident likelihood and the magnitude of the loss or injury”. In order to make a thorough risk assessment it is important to first establish an accurate definition through which the risk can be quantified. In the currently accepted definition risk is calculated by multiplying the frequency with which an event occurs (or will occur) by the magnitude of its probable consequences: Risk = frequency · magnitude of consequences Therefore, if an accident occurs once every 50 years and its consequences are estimated to be one hundred fatalities, the risk is two fatalities·year-1. If, with the same frequency, it causes financial losses of 30·106 €, the risk is 6·105 €·year-1. The concept of risk can be distinguished from hazard (“a chemical or physical condition that has the potential for causing damage to people, property or the environment” [1]) in that it takes into account the frequency of occurrence. This definition of risk is very convenient, but it also creates several difficulties. The first of these is to establish the units in which risk is measured, since they cannot only be fatalities or money per unit time: the consequences can also be measured in terms of injuries to people or damage to the environment, which are more difficult to assess. It is also difficult to estimate the frequency of occurrence of a given type of accident and the magnitude of its consequences. Fortunately, these difficulties can be overcome by applying appropriate methodologies which can be used to obtain a final risk estimation. When analyzing the risk of a given accident, it is likely that exact values will not be known for certain variables, for example the conditions of the released material (temperature, pressure) and meteorological conditions (wind speed and direction). In addition, it is often difficult to make accurate predictions of some specific circumstances related to the source of the accident; for example, if the accident is caused by the loss of containment of a fluid
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through a hole in a pipe or tank where it is only possible to guess the size and location of the hole. As a result, the values obtained are often approximate and we should refer to “estimation” rather than “calculation” (which implies a higher degree of accuracy). Since there are various types of risk they can be classified according to different criteria. Generally speaking, risks can be classified into three categories: Category A risks: those that are unavoidable and accepted without any compensation (for example, the risk of death caused by lightning). Category B risks: those that are, strictly speaking, avoidable but which must be considered unavoidable in everyday life (for example, the risk of dying in a traffic accident). Category C risks: those that are clearly avoidable but to which people expose themselves because they can be rewarding (for example, climbing or canoeing). This classification constitutes a frame of reference that can be used to establish tolerability criteria for certain risks. For example, a widely accepted criterion in several countries sets the tolerability of the risk generated by a given industrial installation at 10-6 fatalities·year-1. This is ten times the typical Category A risk of death caused by lightning (10-7 year-1) and 10-2 times the risk of death due to any cause for a young person. Risks are usually classified into three further categories for industrial activities: Conventional risks: those related to activities and equipment typically found in most industries (for example, electrocution). Specific risks: those associated with handling or using substances that are considered hazardous due to their properties and nature (for example, toxic or radioactive substances). Major risks: those related to exceptional accidents and situations whose consequences can be especially severe as large amounts of energy or hazardous substances may be released during short periods of time. Conventional and specific risks usually affect on-site employees. Since these types of risk are not related to exceptional situations, they are relatively easy to predict and can be prevented or mitigated by implementing standard safety measures. However, the effects of major risks can cover much greater distances, which means that they can also affect the external population and are often more difficult to predict and evaluate. As a result, a set of methodologies has been developed to analyse and quantify such risks. These methodologies are referred to collectively as “risk analysis”. 2 RISK ANALYSIS Risk analysis is used to assess the various types of risk associated with a given industrial installation, a particular activity or the transportation of hazardous materials. Risk analysis methodologies can provide reasonably accurate estimates of potential accidents, the frequency of these accidents and the magnitude of their effects and consequences. Fig. 1-1 [2] shows a simplified outline of the different steps used to apply risk analysis to a given project, activity or plant. The first step is to identify the potential accident types. In this case, it is first necessary to analyse the external events, i.e. hazards that are external to the system being studied: these include, for example, the flooding of a nearby river or an explosion in a neighbouring process plant. There is no specific methodology for this analysis. The first step in assessing the hazards associated with the system being analysed is to apply a historical analysis. Historical analysis consists in studying previous accidents in
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similar systems to the one under analysis, i.e. in a similar plant (for example, a process plant or a storage area), in the same operation or activity (for example, loading/unloading tanks), or involving the same material. This is essentially a qualitative approach, although in cases where there are a sufficiently large number of accidents a statistical analysis can be used to obtain numerical or quantitative results (see Chapter 7, Section 9). Historical analysis is usually performed using an accident database [3, 4]. It can identify the weak points of a system or the types of failure that can be expected on the basis of past experience. It is an essential tool in establishing the basic data required in risk analysis, such as the frequencies of initiating events (see Chapter 8). It is also the only source of experimental data on large-scale accidents; this information is essential for validating and improving mathematical models of major accidents and vulnerability models. However, historical analysis is not a systematic tool for identifying the hazards that exist in a given plant.
Fig. 1-1. Risk analysis steps (taken from [2], with permission).
The HAZOP (hazard and operability) analysis is a powerful tool for identifying potential accidents. It consists of a critical, formal and systematic analysis of a process plant or an engineering project to evaluate the potential risk derived from the abnormal operation or
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failure of individual components and the resulting effects on the whole system. The procedure is based on the evidence that deviations from normal operating conditions often lead to a system failure: when an accident occurs, one or more process variables have deviated from their normal values. HAZOP is performed by a team which analyses all possible deviations of the operating variables in the various nodes of the unit being studied. A set of guide words (no, less, more, other than, etc.) is exhaustively applied to the different operating variables (flow rate, level, temperature, pressure, etc.) in order to identify the possible consequences of all deviations. The HAZOP analysis reduces the frequency of failure or accidents in a particular unit as it identifies potential accidents that can reasonably be expected to occur and the additional safety measures required. Once the hazards have been identified it is necessary to quantify their effects and consequences. Mathematical models of the different types of accident are used to establish the effects (the intensity of thermal radiation from a fire, the overpressure from an explosion, or the concentration of a material released into the atmosphere). The consequences of an accident are estimated by determining the intensity of its effects relative to the distance over which they are felt and by identifying the distribution of the vulnerable elements (population, equipment, etc.). This can be done by using vulnerability models (Fig. 1-1), which show the relationship between the intensity of an effect and the degree of damage caused to a given target. If the analysis is concluded at this point, it can be considered a deterministic approach. i.e. it establishes the potential accidents and estimates their effects and consequences. However, a further step is required to evaluate the risk more comprehensively, which is to estimate the expected frequencies of the different potential accident scenarios. This can be done by plotting and solving the appropriate fault trees and event trees. The fault tree is a schematic representation of the logical sequence of events that must occur in order to reach a top event, i.e. a given accident. By applying this technique it is possible to “descend” from a fairly unlikely event (the accident) to primary events such as the failure of a valve or control device. These are relatively frequent events in process plants and their frequency or probability of occurrence is known. Therefore, once the fault tree has been constructed it is possible to logically combine these values to provide a reasonably accurate estimate of the expected frequency with which the top event or accident will occur. Fault trees can also be obtained from the HAZOP analysis. Event trees are graphical representations of the different sequences that lead from an initiating event (for example, a loss of containment through a hole in a pipe) to diverse accident scenarios (pool fires, flammable clouds or explosions). The sequences follow the different branches of a tree depending on the success or failure of the different measures taken to prevent the emergency (for example, the activation of a water spray system or the intervention of an operator) or the various events which can occur (for example, immediate or delayed ignition of a flammable cloud). Again, the frequency of the initiating event, the probability of failure/success of the different safety measures, and the intermediate events that occur can be used to estimate the frequency of the different accident scenarios. Risk is estimated from the magnitude of the consequences and the frequency of occurrence of an accident. It can be used to calculate the individual risk and societal risk over the area of influence of a given installation or activity and to draw iso-risk lines for individual risk. This is Quantitative Risk Analysis (QRA). Once this analysis has been performed, if the risk is considered too high the plant or project must be modified to increase safety until a tolerable level of risk is reached.
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3 MAJOR ACCIDENTS Major accidents have been defined [5] as “an occurrence such as a major emission, fire, or explosion resulting from uncontrolled developments in the course of the operation of any establishment … and leading to serious danger to human health and/or the environment, immediate or delayed, inside or outside the establishment, and involving one or more dangerous substances”. Major accidents involve the release —instantaneous or over a relatively short period— of significant amounts of energy or of one or more hazardous materials. This can occur both in industrial establishments and during transportation: for example, major accidents have occurred that involved train or road tankers. Major accidents are associated with one or more of the following dangerous phenomena: thermal: thermal radiation mechanical: blast (pressure wave) and ejection of fragments chemical: release of toxic materials. These accidents can affect people, property and the environment. Human consequences can be physical (fatalities or injuries) or psychological and can affect both the employees of the establishment in which the accident occurs and the external population. The consequences on property are usually the destruction of equipment or buildings. Environmental consequences can be immediate or delayed and include the release of a hazardous material into the atmosphere, into the soil or into water. In addition, major accidents usually cause indirect losses such as loss of profits by the company involved. 3.1 Types Major accidents are associated with the occurrence of fires, explosions or atmospheric dispersions of hazardous materials. An accident can also involve more than one of these phenomena: an explosion can be followed by a fire, a fire can cause the explosion of a vessel, and an explosion can cause the dispersion of a toxic cloud. Fire accidents can be classified into the following general categories (Fig. 1-2): Pool fires. Steady state combustion of a pool of flammable liquid (usually a hydrocarbon) with a given size and shape, determined by the presence of a dike or by the ground slope; most pool fires occur in the open air. Combustion is poor and large amounts of black smoke are released. Large pool fires are turbulent with variable flame length (intermittency). A pool fire can also take place when a flammable, non-miscible liquid is spilled on water. Tank fires. Similar to pool fires but usually with a circular shape, where the diameter is determined by the tank size; the flames are located at a certain height above the ground. Jet fires. Steady state turbulent diffusion flames with a large length/diameter ratio, caused by the ignition of a turbulent jet of flammable gas or vapour. The entrainment of air into the flame improves the combustion, which is much more efficient than in pool fires. The shape and position of the jet is mainly determined by the jet velocity influence (particularly in the case of high-speed jets) and buoyancy effects are observed at the jet fire tip. Pool fires, tank fires and jet fires can produce very high heat fluxes, although this effect is limited to a relatively short distance that is much shorter than those associated with explosions or the atmospheric dispersion of pollutants. However, other equipment within this area may be severely damaged by the fire. Flash fires. These are sudden, intense fires in which flames propagate through a mixture of air and flammable gas or vapour within the flammability limits. They are associated with the atmospheric dispersion of gas/vapour under certain meteorological conditions: when the cloud meets an ignition source, the flame propagates through the flammable mixture. In
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certain conditions, mechanical effects (blast) can also occur. If the vapour comes from a liquid pool the flash fire will lead to a pool fire. Fireballs. Ignition of a mass of liquid/vapour mixture that is typically associated with the explosion of a vessel containing a superheated flammable liquid. Since there is no oxygen inside the cloud the fire only burns on the outside of the fireball. As droplets evaporate due to the strong thermal radiation, the density of the mixture decreases and the diameter of the fireball increases. Large (but short duration) fireballs can also occur in tank fires in the event of a boilover.
Fig. 1-2. Accidents involving fire.
Explosions are associated with major accidents involving mechanical phenomena. Explosions occur when there is a rapid increase in volume due to the expansion of a pressurized gas or vapour, the sudden vaporization of a liquid (physical explosions), or a fast chemical reaction (often combustion). Explosions can be classified into the following categories (Fig. 1-3): Vapour cloud explosions. Chemical explosions involving a significant amount of a flammable gas or vapour mixed with air. They are usually associated with the release of flammable liquids or vapour-liquid mixtures. A vapour cloud explosion is always accompanied by a flash fire and the severity of the mechanical effects (blast) is determined by the mass involved and the characteristics of the environment (confinement/congestion): we can consider confined, partly confined and unconfined explosions. Vessel explosions and BLEVEs. Physical explosions caused by the sudden failure of a vessel containing a pressurized gas or superheated liquid (i.e. a liquid at a temperature that is significantly higher than its boiling point at atmospheric pressure) in equilibrium with its vapour. Under certain conditions (currently under discussion) this type of explosion may be referred to as a BLEVE (Boiling Liquid Expanding Vapour Explosion). Dust explosions. When finely divided oxidizable particulate solids (such as flour, sugar, cork, aluminium, aspirin and coal) undergo very fast combustion when dispersed in air, which causes severe explosions. Dust explosions are determined by particle size and solid concentration in air and are very difficult to model. They occur in confined environments, commonly inside equipment (silos, cyclones). An initial explosion often generates strong
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turbulence which disperses a large amount of dust; it is then followed by a second, much stronger explosion.
Fig. 1-3. Types of explosion.
Finally, the release of a toxic material can produce a toxic cloud. Depending on the density of the cloud (heavier than air or with a density that is equal to or less than that of air) and on the meteorological conditions, the cloud is either dispersed quickly into the atmosphere or evolves close to the ground and moves at wind speed. The major accidents which can occur in industrial installations or during the transportation of hazardous materials are usually related to a loss of containment. The loss of containment can be caused by an impact, by the failure of a piece of equipment (a pipe or tank) due to the effects of corrosion, by human error during a loading or unloading operation, or by various other factors. The loss of containment can also be a consequence of the accident itself, for example in the case of the explosion of a pressurized tank. Once the release has taken place, the evolution will depend on the physical state of the substance spilled (Fig. 1-4) [6]. When a liquid is spilled onto the ground and no concrete layer is present, both the soil and underground waters can be contaminated. If the spill occurs on water or the substance reaches water (for example, a river or the water in a port) then the water will be polluted. If the product is less dense than and non-miscible in water (as is the case of hydrocarbons) it can evaporate into the atmosphere. If a pool is formed and the material is flammable, an ignition point will cause a pool fire: large amounts of (possibly toxic) smoke are released and intense thermal radiation can affect nearby equipment. If no immediate ignition occurs, a toxic or flammable cloud can develop.
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Fig. 1-4. Simplified schematic representation of the accidents that can occur following a loss of containment, their effects and the potential associated damage.
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A flammable cloud can be ignited, which leads to a flash fire and possibly an explosion, depending on the amount of material involved; in this case the remaining liquid in the pool will immediately produce a pool fire. The wind and meteorological conditions may favour the formation of a toxic cloud, which represents a potential threat to people within a given area. If the released material is a vapourliquid mixture (this is usually the case when a hot, pressurized liquid is released into the atmosphere) the formation of a vapour cloud is very likely, as the vaporization of liquid droplets will increase the concentration in the mixture with air. If the material is released as a gas or vapour and the exit velocity is sufficiently low, a cloud may still be formed. If the exit velocity is high, the large entrainment of air will dilute the mixture and the release will just be dispersed into the atmosphere (the build-up of a flammable cloud is unlikely). If ignition occurs it can lead to a jet fire in both cases. Dust can also create dangerous clouds when released into the atmosphere; for example, soybean dust released while unloading a ship has been known to cause serious problems due to the allergenic substances it contains. Furthermore, fine dust can produce strong explosions when it is dispersed in air. These explosions do not usually follow a loss of containment but instead occur inside equipment (in a silo, dryer or cyclone, for example), although their effects may be felt over a significant area. Finally, pressurized tanks can explode if the pressure increases above a certain value or if the vessel loses strength due to increased temperatures during a fire, to give a common example. In this case, the blast will affect a certain area and missiles can be ejected over large distances. If the material is flammable the explosion —possibly a BLEVE— can be followed by a fireball. This type of accident always leads to the release of energy (overpressure from an explosion or thermal radiation from a fire) or the release of a hazardous material which will eventually be dispersed into the atmosphere, be spilled on water or penetrate into the soil. Overpressure, thermal energy and missiles can have serious consequences on people and property (buildings and equipment), while the release of a hazardous material onto water, into the soil or into the atmosphere can represent a danger to people and the environment. Table 1-1 Distribution of major accidents in process plants and in the transportation of hazardous materials Type of accident % Fire 47 Explosion 40 Gas cloud 13
Historical analysis can be used to deduce the relative frequency with which major accidents occur. A survey of accidents in process plants and in the transportation of hazardous materials [7] produced the distribution shown in Table 1-1. The most frequent accident type was fire, followed by explosion and gas cloud. 3.2 Damage The magnitude of the accident will depend on various parameters: Inventory. The mass or energy directly involved in an accident is proportional to the amount of material present in the plant in which the accident takes place, which is
9
why it is always a positive safety measure to reduce hazardous material inventories (this was one of the lessons learnt from the accidents in Flixborough and Bhopal). Energy. The magnitude of the consequences of an accident is proportional to the amount of energy contained in a system (pressure energy, heat of combustion). Time. The magnitude of the consequences is inversely proportional to the time during which a given amount of energy or hazardous material is released: the intensity of the phenomenon at a given distance will be higher and the possibility of escape will decrease. Exposure. The degree of exposure can have a considerable effect on the consequences of an accident on people in the vicinity. For example, a building can provide very efficient protection against a toxic release and being inside a car can protect against the thermal radiation from a flash fire (if the car is not engulfed by the fire). Exposure is also related to the distance between the population and the source of the accident. If there is a reasonable distance, the intensity of the effects that reach people in the vicinity will be much lower than if the population is located close to the plant. This was a key factor in the severity of the consequences of major accidents such as those happened at San Juan Ixhuatepec Mexixo) and Bhopal (India) in 1984. The potential damage caused by major accidents can affect people, equipment and the environment. The probability vs. number of fatalities curve that represents the hazards of the substances associated with these accidents (Fig. 1-5) clearly shows that the most severe human consequences are caused by explosions, followed by fires and then gas clouds [8]. 1
EXPLOSIVE FLAMMABLE TOXIC
Probability
0.1
0.01
1E-3 1
10
100
Number of fatalities
Fig. 1-5. p-N curve as a function of the hazard category of the substance (taken from [8], with permission).
Indirect damage is also produced during the post-accident situation. The various types of damage caused by an accident can be classified according to the general outline shown in Fig. 1-6 [6].
10
Human damage refers to both loss of human lives and injuries (light, severe or very severe) caused by the accident and the intervention and evacuation costs of containing the effects of the event and minimizing its consequences. Environmental damage refers to the environmental resources/areas affected by the accident in addition to the social aspects affected, which include cultural life, historical buildings and landscape. Water is affected by direct spills, soil pollution and when polluted fire-fighting water used to combat a large fire is not adequately controlled and is spilled into a river or into sea water. The atmosphere is polluted in almost all types of accident, although in most cases the pollutant is dispersed and diluted relatively quickly. TYPES OF DAMAGES
DAMAGE TO HUMAN LIFE/HEALTH
ENVIRONMENTAL DAMAGE
MATERIAL DAMAGE
LOSS OF PROFITS
deaths
biosphere air
storage (liquid & liquefied gases)
breakdown costs
injured people evacuation costs
water
warehouses
soil
land vehicles
indirect costs (loss of image, etc.)
loss of wages
process equipment utilities roads & railways buildings & industrial areas
Fig. 1-6. Potential damage derived from major accidents.
Material damage includes all financial losses derived from damage to equipment and replacement requirements. Loss of profits derives from the breakdown of certain installations after the accident. The scale of these losses depends on the time required to resume normal activity. This category also includes some costs attributed to the damage to the company’s image, although this is obviously difficult to quantify. Equipment (process units, buildings, vehicles, etc.) is damaged by explosions and all types of fire. Finally, many of these accidents interrupt activity across the entire affected area, which causes loss of profits during a certain period of time. Historical data obtained from various sources indicate a clear tendency towards more serious accidents, the average cost of which is growing considerably each year [9]. Windhorst and Koen (cited in [10]) suggested that risk increases as a function of plant size according to the following relationship: Risk = k · capital2 This could be partly attributed to the current trend of building larger plants (some of them in developing countries, such as bulk chemical plants) while essentially maintaining the same design. This leads to larger inventories, larger release rates in the event of an accident and more complicated piping systems (valves, flanges and welds). In general terms, the result is an increase in risk.
11
4 DOMINO EFFECT The domino effect has been defined [11] as a cascade of events in which the consequences of a previous accident are increased both spatially and temporally by following ones, thus leading to a major accident. A domino effect involves a primary event that affects a primary installation, which induces one or more secondary accidents that affect other installations. The spread of damage can be either spatial (areas not involved in the primary accident are damaged) or temporal (the same area is involved but the secondary events are delayed), or both. Installations involved in a domino effect may or may not belong to the same establishment. 4.1 Classification of domino effects Historical analysis has shown that domino effects can be classified according to two criteria: the type of primary and secondary installations involved, and the nature of the primary and secondary physical effects produced. The types of installation that are most frequently affected by domino effects are: pressure storage tanks, atmospheric or cryogenic storage tanks, process equipment, pipe networks, small conditioners and solid storage areas. The relative occurrence of these different types of unit is summarized in Table 1.2 [11]. Loading/unloading areas were not included due to the lack of available data. Table 1-2 Contribution of the different types of installation to primary or secondary accidents Type of installation Primary, % Secondary, % Pressurized storage tanks 30 33 Atmospheric or cryogenic storage tanks 28 46 Process equipment 30 12 Pipe networks 12 -Small conditioners -9
Primary effects can be either thermal or mechanical. For example, a vapour cloud explosion following a release of flammable material can cause equipment to collapse or a jet fire can produce a tank explosion. Secondary effects can be thermal, mechanical or toxic. Toxic phenomena do not cause a domino effect. The different physical effects found in primary and secondary accidents and their relative frequencies are shown in Table 1-3 [11]. Table 1-3 Effects of major accidents related with domino effect Effects in primary accidents Effects in secondary accidents Mechanical (35%) Mechanical (37%) Thermal (77%) Thermal (93%) Toxic (10%)
4.2 An example case An example case of the domino effect occurred in a petrochemical plant in Priolo (Italy) in 1985 [12]. Instrument failure in the reboiler of a distillation column caused a temperature increase that activated a safety relief valve. The chattering of the relief valve led to a release
12
of flammable gas from a flange. The gas ignited to produce a jet fire, which was increased by the gas released from a broken pipe with a diameter of 150 mm. The jet fire reached a larger pipe with a diameter of 600 mm located at a distance of 16 m, which contained ethylene at 18.2 bar and was connected to another distillation column. A large fireball was produced, followed by a very large jet fire. The secondary jet fire impinged on the bottom of a set of eight cylindrical storage tanks (diameter: 3.6 m, height: 48 m) located at a distance of 60 m and which contained LPG. One of the tanks underwent a BLEVE explosion followed by a fireball; the amount of LPG involved was approximately 50,000 kg and the centre of the fireball reached a height of 250 m above the ground. The vessel was broken into several fragments and the largest two were propelled to distances of 25 m and 125 m from the initial tank location. Smaller fragments were ejected to distances of up to 700 m. The impact of these fragments caused three other storage tanks to fall on other equipment including a pipe rack. The thermal effects on equipment were significant in a radius of 250 m.
Fig. 1-7. Example case of domino effects.
The domino effects of this accident can be summarized as follows: Release from a flange o jet fire: flame impingement on a large pressurized pipe o failure of the pipe o fireball plus large jet fire: flame impingement on pressurized LPG tank o BLEVE plus fireball from one tank: blast, radiation effects plus mechanical impact o serious damage to other equipment, various large fires. There were no fatalities (one employee was injured) because the plant was evacuated in time. However, the plant was severely damaged. A village located close to the plant was also evacuated. Total damage was estimated at 65·106 $US (1985).
13
5 MATHEMATICAL MODELLING OF ACCIDENTS Mathematical models can be used to estimate the effects of accidents; they consist of sets of equations that describe the phenomenon and provide predictions of the thermal radiation emitted by a fire, the peak overpressure from an explosion, the path followed and distance reached by the ejected fragments or the evolution of the concentration in the atmospheric dispersion of a release. The first step when trying to predict these phenomena is to estimate the amount of material involved in the accident and the rate at which it is spilled or released. This is done by applying source term models. Source term models are based on fluid dynamics and heat transfer and require the exact or estimated values of the temperature and pressure of the material involved. This often constitutes a factor of uncertainty, as these conditions may depend on the evolution of the situation: if a vessel is heated by a fire, pressure and temperature will probably increase with time and the variation will depend on the heating rate. The definition of the problem itself creates a further difficulty: in order to calculate the flow rate through a hole in a pressurized vessel it is necessary to know the shape and size of the hole, but this information is often unavailable. Consequently, models commonly apply simplifying assumptions and assume standard initiating events. A number of models have been published that describe fires, atmospheric dispersion and the effects of explosions. Their degree of complexity varies significantly: some are very simple, some are more complex and some are very complex. Overly simplistic models are easy to use but they can sometimes lead to significant errors. In theory, complex models should provide good results but in practice they often require information and data which are unavailable. Fires are sometimes modelled with the point source model, in which a fire is represented by a point that irradiates thermal energy in all directions. It is a very simple model that overestimates the intensity of the thermal radiation close to the fire and provides overly conservative results. The solid flame model is more complex but provides relatively accurate descriptions of fires. It assumes that the fire is a solid body which radiates thermal energy with a certain intensity. It requires only a small amount of data but cannot produce a very accurate prediction of the shape and size of the flames. In addition, the emissive power of the flames can be known only approximately In accidents that involve the explosion of a vessel, uncertainty arises from the lack of information about the energy available to create overpressure. This depends, first of all, on the pressure in the vessel just before the explosion, which cannot be predicted: the vessel can fail well before the maximum theoretical value is reached if its wall loses strength as a result of the increased temperature. Furthermore, the energy released by the explosion depends on the thermodynamic process responsible for the expansion of the gas. Finally, a significant – and generally unknown – proportion of the explosion energy is used in the ductile breaking of the vessel and in ejecting the fragments. The existence of directional effects (higher overpressure in certain directions) can also complicate the situation. In vapour cloud explosions, the first difficulty is to determine the mass of flammable vapour involved in the explosion (only the mixture within the flammability limits will contribute to the blast). This will depend on the amount of material released, the evolution of the cloud, the time elapsed between the start of the release and the moment of ignition, and the meteorological conditions. In this type of explosion the mechanical yield is very low –i.e. only a small fraction of the released energy is used in creating the pressure wave– but it
14
cannot be accurately predicted. Furthermore, it is strongly influenced by confinement and by the degree of congestion of the air mass through which the pressure wave moves. The atmospheric dispersion can be predicted with a reasonable degree of accuracy (for a given source term) in the case of neutral or light substances. However, the dispersion of heavier-than-air gases is not yet sufficiently well known and the various existing models are relatively complex and show significant scattering in their predictions. Once the properties of the explosion have been determined vulnerability models are applied to estimate its effects on people or property. The best way to analyse the effects of thermal radiation or a dose of a toxicant on people or property is to use probit equations. These are expressions that establish the relationship between the magnitude of an aggressive action and the degree of damage that it causes to the exposed population. They are relatively reliable for most dangerous phenomena. Tabulated reference values are commonly used to predict the damage to property (buildings and equipment). Due to the difficulties mentioned above, some authors have suggested adopting conservative criteria such as high efficiencies in vapour cloud explosions and the assumption that all the mass in the vapour cloud contributes to the blast. However, these solutions can lead to overestimates of the distances at which significant effects are observed or of the magnitude of the effects at a given distance. This is illustrated in the two examples below. ______________________________________ Example 1-1 Consider a release of cyclohexane: a cloud containing 10,000 kg of hydrocarbon is formed. Estimate the overpressure at a distance of 200 m if: a) all the cyclohexane in the cloud contributes to the build up of overpressure, with a yield of 10%; and b) only 30% of the cyclohexane in the cloud contributes to overpressure, with a yield of 3%. Solution a). As explained in Chapter 4, the equivalent mass of TNT can be calculated by considering that 1 kg of cyclohexane releases the same amount of energy as 10 kg of TNT,
M TNT
K 10 M cyclohexane
0.1 10 10,000 10,000 kg
This corresponds to a scaled distance dn = 9.3 m kg-1/3; with this value we obtain a peak overpressure of 0.18 bar. This will cause (see Table 7-14) the following damage: 50% destruction of the brickwork of houses. b). For 3,000 kg of cyclohexane and a yield of 3%, M TNT
K 10 M cyclohexane
0.03 10 3,000
900 kg
This corresponds to a scaled distance dn = 20.7 m kg-1/3 and a peak overpressure of 0.054 bar, which will cause minor damage to house structures or occasional damage to window frames. The two predictions are therefore significantly different. ______________________________________ ______________________________________ Example 1-2 The explosion of a tank containing 10,000 kg of liquefied propane creates a fireball with a diameter of 270 m and a duration of 16 s. Estimate the effects of thermal radiation on a group
15
of 83 people located at a distance of 300 m from the initial position of the tank if: a) the radiant heat fraction of the fireball is 0.4; and b) the radiant heat fraction is 0.2 (for fireballs, this variable depends on the brightness of the flames and ranges between 0.2 and 0.4, while its maximum value is 0.4). Solution Taking into account the fireball diameter, the view factor is F = 0.138 at a distance of 300 m (see Chapter 3). An atmospheric transmissivity of 0.68 is taken. a). Taking the radiant heat fraction as 0.4, we obtain a flame emissive power of E = 400 kW m-2. By applying the solid flame model, a thermal radiation intensity of I = 37.5 kW m-2 would reach the group of people. By calculating the dose and the probit variable (Chapter 7) it is found that 75 people would die. b). Assuming that the radiant heat fraction is 0.2, E = 250 kW m-2. The same model gives a radiation value of I = 23.4 kW m-2. By using the corresponding values for the dose and the probit variable it is found that 44 people would die, which is a significantly different outcome to that obtained in the previous estimation. ______________________________________ The mathematical modelling of major accidents should therefore be performed with caution by applying reasonable assumptions and establishing a certain margin of safety, but also taking into account that excessively conservative approaches lead to overpredictions of the effects and influence-areas of accidents. As in many other fields, experience plays an important role. NOMENCLATURE
dn E F I M MTNT
K
scaled distance (m kg-1/3) flame emissive power (kW m-2) view factor (-) thermal radiation intensity (kW m-2) mass of hydrocarbon in the cloud contributing to overpressure (kg) equivalent mass of TNT (kg) explosion yield factor (-)
REFERENCES [1] Center for Chemical Process Safety. Guidelines for Consequence Analysis of Chemical
Releases. AIChE. New York, 1999. [2] N. Piccinini. Affidabilitá e Sicurezza nella Industria Chimica. IEC. Barcelona, 1985. [3] MHIDAS. Major Hazard Incident Data Service. Health and Safety Executive. London,
2007. [4] FACTS. Failure and Accidents Technical Information System Base. TNO. The Hague,
2007. [5] Council Directive 96/82/EC of 9 December 1996 on the control of major-accident
hazards involving dangerous substances. OJ No. 10, 14 January 1997.
16
[6] A. Ronza. PhD thesis. UPC. Barcelona, 2007. [7] E. Planas, H. Montiel, J. Casal. Trans IChemE, 75, Part B (1997) 3. [8] S. Carol, J. A. Vílchez, J. Casal. J. Loss. Prev. Process Ind. 15 (2002) 517. [9] S. Carol, J. A. Vílchez, J. Casal. J. Loss. Prev. Process Ind. 13 (2000) 49. [10] J. D. W. Edwards. J. Loss. Prev. Process Ind. 18 (2005) 254. [11] C. Delvosalle. Domino Effects Phenomena: Definition, Overview and Classification.
First European Seminar on Domino Effects. Leuven, 1996. [12] Ministerio dell’Interno. DGPCSA. Servizio Tecnico Centrale. Rassegna comparata
incidenti di notevole entità. SDRPVVF. Roma, 1986.
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Chapter 2
Source term 1 INTRODUCTION Major accidents always start with a loss of containment. A material contained inside a piece of equipment (for example a tank, a distillation column or a pipe) exits to the atmosphere through an opening such as a hole, a crack or an open valve. The origin may be corrosion, a mechanical impact or a human error. The loss of containment itself can also be an accident, as in the case of the explosion of a pressurized tank. Once the loss of containment has started, the evolution of the event will depend on a series of circumstances such as the condition of the material (gas, liquid or a mixture of the two), its properties, the meteorological conditions and the measures taken to mitigate the leak. In order to predict the effects and consequences of a given accident, we must calculate (or, better, estimate) the velocity at which the material will be released, the size of the liquid pool that will form and the velocity at which the liquid will evaporate. This information is required in order to apply mathematical models of the various dangerous phenomena that can occur (fire, explosion, toxic cloud, etc.) and thus predict the physical effects of the accident (concentration, thermal radiation, blast, etc.) as a function of time and distance. As mentioned in Chapter 1, a number of accidents can occur in the event of a loss of containment of a hazardous material. Fig. 2-1 summarizes the various possibilities. The released material is often a fluid (a gas or a liquid). The loss of containment can be continuous over a certain time or instantaneous. Continuous releases can take place through a hole in a tank, a broken pipe or a safety valve. In these cases, we must calculate the mass flow rate (sometimes as a function of time), the total amount released or the time during which the release takes place. If the released material is a liquid, a pool will probably form (or the liquid will be retained in a dike) and evaporate at a certain velocity. Vapour can also be released if a liquid is depressurized in the loss of containment and flashes. The loss of containment can also be instantaneous, e.g. if a storage tank breaks or a pressurized tank bursts. A set of equations is used to perform these calculations. Often, when too many variables are unknown, the calculation is not direct and an iterative procedure must be applied. Furthermore, to predict the effects of a hypothetical accident, some assumptions must be made: the size of the hole, its position (if it is located at the bottom of a tank, the release is likely to be a liquid, whereas if it is located at the top, the release will probably be a gas), the time during which the release will take place, the ground temperature, etc. Here, the experience of the person performing the calculations is very important. For a very specific case or unit, a few probable sources can usually be identified. For risk analysis of industrial installations, however, some general rules are often applied (see the last section of this chapter).
19
Fig. 2-1. Simplified scheme of the various source terms that can occur when there is a loss of containment in a plant or during transportation.
This chapter discusses the source term models for the most common loss-of-containment events: - Flow of liquid through a hole in a tank. - Flow of liquid through a pipe.
20
- Flow of gas or vapour through a hole. - Flow of gas through a pipe. - Two-phase flow. - Evaporation of a liquid from a pool. Some illustrative examples have also been included. For a more extensive treatment, additional literature is recommended [1, 2, 3, 4]. 2 LIQUID RELEASE Liquids are common in industrial installations, where they are stored in tanks and flow through pipes and equipment. If a liquid is released through a hole or a broken pipe, its subsequent behaviour will depend on the conditions prior to the loss of containment (pressure and temperature): it can flash, form a pool and then evaporate. In order to foresee what will happen, it is necessary to estimate the liquid flow rate. The following sections analyse the most common situations. For a flowing fluid, the mechanical energy balance can be written as follows [1]: § u2 g 'z ' ¨¨ © 2
· dP ¸¸ ³ Ws F f U ¹
0
(2-1)
where g is the acceleration of gravity (m s-2) z is the height above an arbitrary level (m) u is the velocity of the fluid (m s-1) P is the pressure (Pa) Ws is the shaft work (kJ kg-1) Ff is the friction loss (kJ kg-1), and U is the density of the fluid (kg m-3). For incompressible fluids, the density is constant and
dP
³U
l
'P
(2-2)
Ul
Furthermore, if there is no pump or turbine in the line (see Fig. 2-2), Ws = 0. Then, the mechanical energy can be simplified to: § u2 g 'z ' ¨¨ © 2
· 'P ¸¸ Ff ¹ Ul
0
(2-3)
2.1 Flow of liquid through a hole in a tank Consider a tank containing a liquid up to a certain level, and with an absolute pressure Pcont above the liquid. This pressure can be kept constant (as, for example, when the tank is padded with nitrogen to avoid the formation of a flammable atmosphere) at a certain value or simply at atmospheric pressure. In the latter case, if the tank is emptied, a gas entry must be provided to prevent a vacuum from forming, which could lead to the collapse of the vessel.
21
Finally, if the tank contains partly liquefied gas, the vapour-liquid equilibrium will maintain constant the pressure above the liquid.
Fig. 2-2. Flow of liquid from a hole in a tank.
If there is a hole in the wall of the tank (Fig. 2-2), there will be a liquid leak. The mass flow rate can be calculated with the following expression, which is based on the mechanical energy balance and a discharge coefficient CD, in order to take into account the frictional losses in the hole: m
Aor U l C D
· § P P0 2 ¨¨ cont g hl ¸¸ Ul ¹ ©
(2-4)
where m is the liquid mass flow rate (kg s-1) Aor is the cross-sectional area of the orifice (m2) CD is a discharge coefficient (-) Pcont is the pressure above the liquid (Pa) P0 is the outside pressure (usually the atmospheric pressure) (Pa), and hl is the height of liquid above the leak (m). The discharge coefficient CD is a complicated function of the leak geometry, the ratio dor/dpipe and the Reynolds number in the hole. For turbulent flow, the following values can be used: sharp-edged orifices, CD = 0.62; straight orifices, CD = 0.82; and rounded orifices, CD = 0.97. If, due to depressurization, the liquid undergoes a flash vaporization, it can be assumed that this phenomenon occurs downstream from the hole, so it is not taken into account for the calculation of m. As the release of liquid proceeds, the liquid level in the tank decreases, and as a consequence the flow rate through the hole decreases as well. The mass discharge rate at any time t can be calculated with the following expression [2]:
22
m
Aor U l C D
§ P P0 2 ¨¨ cont g hlinitial Ul ©
· U l g C D2 Aor2 ¸¸ t At ¹
(2-5)
where At is the cross-sectional area of the tank (m2) and t is the time from the onset of the leak (s). However, in risk analysis, the conservative assumption of constant discharge rate—at the initial flow rate—is sometimes applied until all of the liquid above the level of the hole has been released. Finally, the time required for the vessel to empty to the level of the leak is, for a constant tank cross section: te
1 § At ¨ C D g ¨© Aor
· ª § Pcont P0 ¸¸ « 2¨¨ g hlinitial Ul ¹ «¬ ©
· ¸¸ ¹
2 Pcont P0 º » Ul »¼
(2-6)
For a non-constant cross-sectional area of a tank (sphere, horizontal cylinder), see [4, 5]. ______________________________________ Example 2-1 A cylindrical tank 10 m tall and 5 m in diameter contains toluene at 20 ºC. The pressure above the liquid surface is kept at essentially atmospheric pressure with nitrogen. The tank is filled to 85%. A collision creates a hole in the tank wall (dor = 50 mm) 1 m above the bottom. The leak is repaired and stopped 30 min after the onset. Calculate: a) the initial flow rate through the hole; b) the amount of toluene spilled, and c) the time during which the toluene would have been spilled if the leak had not been repaired. (Utoluene = 867 kg m-3).
Solution a) Initial height of liquid in the tank: 0.85 · 10 = 8.5 m Initial height of liquid above the leak: 8.5 -1 = 7.5 m The initial (maximum) flow rate is calculated with Eq. (2-4): m
S 0.05 2 4
867 0.62 2 9.81 7.5
12.8 kg s-1
b) As toluene is spilled, the height of liquid in the tank decreases and, consequently, the leak flow rate decreases as well. Therefore, the release must be divided into several discrete time segments, assuming constant liquid head and leak flow rate over each segment. Ten segments are taken for this case. For the first time segment (from t = 0 s to t = 180 s):
m = 12.8 kg s-1, hl = 7.5 m. During the first 180 s, the amount of toluene spilled is: 12.8 · 180 = 2304 kg 2.66 m3 After the first time segment, the height of the liquid above the leak is:
23
hl
7.5
2.66 S d2 4
7.365 m
The following table summarizes the calculation results for the various time segments. t, s
m, kg s-1
hl, m
0-180 180-360 360-540 540-720 720-900 900-1,080 1,080-1,260 1,260-1,440 1,440-1,620 1,620-1,800 1,800
12.80 12.68 12.57 12.45 12.34 12.23 12.11 11.99 11.88 11.76
7.500 7.365 7.231 7.099 6.968 6.832 6.704 6.577 6.451 6.324 6.201
Accumulated spilled mass, kg 2,304 4,586 6,850 9,092 11,313 13,514 15,694 17,853 19,991 22,107
The amount spilled is 22,107 kg. c) The time is calculated with Eq. (2-6): § S 52 · ¨ ¸ 1 4 ¨ ¸ 2 9.81 7.5 19,944 s 5.54 h. te 0.62 9.81 ¨ S 0.05 2 ¸ ¨ ¸ 4 © ¹ ______________________________________ 2.2 Flow of liquid through a pipe When a fluid flows through a pipe, there is friction between the fluid and the pipe wall (depending on the roughness of the wall) and the mechanical energy of the fluid is partially converted into thermal energy. For a given flow rate, the fluid experiences a certain pressure drop, and the pressure upstream must be high enough to overcome it. Even though the pressure changes, for incompressible fluids (liquids) the density is constant along the pipe. In risk analysis, a common problem is to calculate the liquid flow through a hole in a pipe or through a broken pipe (Fig. 2-3).
2.2.1 Liquid flow rate The relationship between pressure drop and fluid velocity for an incompressible liquid flowing through a piping system can be obtained from the Fanning equation: 'P
2 f F Ul L u 2
(2-7-a)
dp
24
or
u
'P d p
(2-7-b)
2 f F Ul L
where 'P is the pressure drop over the pipe (Pa) L is the pipe length (m) u is the fluid velocity (m s-1) dp is the diameter of the pipe (m), and fF is the Fanning friction factor (-). The Fanning friction factor is the ratio between the mechanical energy dissipated by friction and the kinetic energy of the flowing fluid. The mass flow rate in the pipe can then be calculated by: m
Ap U l
'P d p
(2-8)
2 f F Ul L
where Ap is the cross-sectional area of the pipe (m2).
Fig. 2-3. Flow of liquid from a pipe.
The Fanning factor is a function of the Reynolds number, which depends on the velocity of the liquid in the pipe; therefore, the mass flow rate in the pipe must be calculated by iteration [3]. The following procedure can be followed: 1. Guess a value for the Reynolds number. 2. Calculate the value of the Fanning friction factor. 3. Calculate the liquid velocity in the pipe. 4. Calculate a new Reynolds number.
25
5. Compare the two Reynolds numbers. If they are not equal, correct the value of Re and repeat the procedure. The pressure drop in the piping system is not only due to friction with the pipe itself. Pipe fittings, elbows, valves, contractions, etc., also play a role. This contribution is usually expressed as an equivalent length of straight pipe; thus:
L
Lstraight pipe 6 Lequivalent
(2-9)
Table 2-1 shows the equivalent length of various pipe fittings for turbulent flow. Table 2-1 Equivalent length of pipe fittings (turbulent flow only)* Pipe fitting Globe valve, wide open Angle valve, wide open Gate valve, wide open ¾ open ½ open ¼ open 90º elbow, standard long radius 45º elbow, standard Tee, used as elbow, entering the stem Tee, used as elbow, entering one of two side arms Tee, straight through 180º close return bend Ordinary entrance (pipe flush with wall of vessel) Borda entrance (pipe protruding into vessel) Rounded entrance, union, coupling Sudden enlargement from dp to D Laminar flow in dp
Lequivalent/dp
| 300 | 170 |7 | 40 | 200 | 900 30 20 15 90 60 20 75 16 30 Negligible 2 Re ª §¨ d p «1 32 «¬ ¨© D 2
·º ¸» ¸» ¹¼
ª § d p2 «1 ¨ 2 «¬ ¨© D
·º ¸» ¸» ¹¼
1 f F ,ind p 4
Turbulent flow in dp Sudden contraction from D to dp; all conditions except high-speed gas flow where P1/P2 t 2 Laminar flow in dp
2
§ d p2 Re ª «1.25 ¨ 2 ¨D 160 ¬« © 1 f F ,ind p 10
Turbulent flow in dp
2
·º ¸» ¸» ¹¼ 2 ª § d p ·º «1.25 ¨ 2 ¸» ¨ D ¸» © ¹¼ ¬«
*Taken from O. Levenspiel, Engineering Flow and Heat Exchange, p. 25, Plenum Press, New York (1984), with permission of Springer Science and Business Media.
26
2.2.2 Friction factor The Fanning friction factor (fF) can be calculated as follows [1]. For laminar flow:
16 Re
fF
(2-10)
In the transition regime, the value of fF is uncertain. For turbulent flow, the Colebrook equation can be used: 1 fF
§ 1 H 1.255 4 log ¨ ¨ 3.7 d p Re f F ©
· ¸ ¸ ¹
(2-11)
where H is the roughness of the pipe (-) (Table 2-2). Table 2-2 Roughness of clean pipes* Pipe material Riveted steel Concrete Wood stave Cast iron Galvanized iron Asphalted cast iron Commercial steel or wrought iron Drawn tubing Glass Plastic (PVC, ABS, polyethylene)
H, mm 1 - 10 0.3 - 3 0.2 - 1 0.25 - 0.26 0.15 0.12 0.043 - 0.046 0.0015 0 0
*Taken from O. Levenspiel, Engineering Flow and Heat Exchange. Plenum Press, New York, 1984.
For fully developed turbulent flow in rough pipes, fF can be calculated with: 1 fF
dp · § ¸ 4 log ¨¨ 3.7 H ¸¹ ©
(2-12)
For smooth pipes and Re < 100,000, fF can be calculated with the Blasius equation: fF
0.079 Re 0.25
(2-13)
And for smooth pipes and Re > 100,000, the following expression can be applied [6]: fF
0.0232 Re 0.1507
(2-14)
The value of the Fanning factor can also be obtained from the classical plot of fF as a function of H/d and Re (see Fig. 2-4).
27
Fig. 2-4. The Fanning factor as a function of the Reynolds number and pipe roughness. Taken from O. Levenspiel, Engineering Flow and Heat Exchange, p.20, Plenum Press, New York (1984), with permission of Springer Science and Business Media.
______________________________________ Example 2-2 A cylindrical tank 10 m tall and 5 m in diameter contains toluene at 20 ºC. The pressure above the liquid surface is kept at essentially atmospheric pressure with nitrogen. The tank is filled to 85%. Estimate the initial liquid outflow through a commercial steel pipe (with an inside diameter of 100 mm) connected to the bottom of the tank. The pipe has been broken 65 m from the tank; it is horizontal and has a gate valve (open) and two 90º elbows. Data: Ptoluene, 20 ºC = 5.9 · 10-4 kg m-1 s-1; Utoluene, 20 ºC = 867 kg m-3; Hcommercial steel = 0.046 mm.
28
Solution For the commercial steel pipe:
H dp
0.046 100
0.00046
The pressure drop through the piping system will be the difference between the pressure at the pipe inlet and the atmospheric pressure: 'P
U l g hl
867 9.81 0.85 10
72,295 Pa
The equivalent length of the pipe, including the fittings, is: L
65 0.1 7 2 30 16 32 76.5 m
Guessed Re = 25,000 u
P Re d p Ul
5.9 10 25,000 4
0.1 867
0.17 m s-1
From Fig. 2-4, fF = 0.0064. By applying Eq. (2-7-b): u
72,295 0.1 2 0.0064 867 76.5
2.92 m s-1
Therefore, Re must be corrected. The following table summarizes the results of the trial-anderror procedure. Re 25,000 70,000 300,000 400.000 500,000 520,000 530,000
u, m s-1 0.170 0.544 2.041 2.722 3.402 3.539 3.631
fF 0.00640 0.00520 0.00450 0.00440 0.00435 0.00433 0.00430
Finally, u = 3.547 m s-1. Thus, the mass flow rate is: m
S 0.12
24.1 kg s-1. 4 ______________________________________ 3.547 867
29
u (Eq. (2-7b)), m s-1 2.920 3.240 3.480 3.520 3.540 3.544 3.560
3 GAS/VAPOUR RELEASE
When a gas or a vapour is released from a given piece of equipment (pipe, tank, etc.), the pressure energy contained in the gas is converted into kinetic energy as the gas leaves and expands through the exit. The density, pressure and temperature of the gas change during the loss of containment. In practice, if the density change of the gas is small (U1/U2 < 2, P1/P2 < 2) and the velocity is relatively low (u < 0.3 times the velocity of sound in the gas), then the flow can still be considered incompressible. However, at high pressure changes and high flow velocities, kinetic energy and compressibility effects become dominant in the mechanical energy balance and the flow is considered compressible. Therefore, the accurate analysis of such systems involves four equations: the equation of state and those of continuity, momentum and energy. This makes the analysis rather complicated. To simplify matters, it is usually assumed that the flow is reversible and adiabatic, which implies isentropic flow. Furthermore, it is often assumed that the fluid is an ideal gas with constant specific heat (average value). The models for estimating the release flow rate of a gas can also be applied to a vapour, as long as no condensation occurs. Therefore, in this chapter, both categories (gas and vapour) will be referred to as “gas”. 3.1 Flow of gas/vapour through a hole For liquids and gases with low pressure changes and low velocities (P1/P2 < 2, Ma < 0.3) (Ma is the Mach number), the flow can be considered incompressible and the expressions presented in the previous section can be applied. However, with a gas flow, if the pressure change is significant and the velocity is high, then kinetic and compressibility effects play an important role [1] and the pressure, temperature and density change significantly when the gas flows through an opening. The flow is considered compressible and different expressions must be applied to calculate it.
3.1.1 Critical velocity When a gas or vapour exits through a hole, there are two possible situations: sonic velocity and subsonic velocity. This is discussed below.
Fig. 2-5. Flow of gas or vapour through a hole.
30
Let us assume that gas is flowing from a tank at a certain pressure (Pcont) through a hole in the wall (Fig. 2-5). If the pressure downstream from the hole (Pout) decreases, the velocity of the gas through the hole increases. This velocity will increase until, at a certain value of Pout, it reaches the velocity of sound in that gas (at that temperature). Further decrease of Pout will not cause any increase in fluid velocity: the velocity of sound at Pchoked, Tchoked, is the maximum velocity at which the gas can flow through the orifice (to reach supersonic speed, specially designed converging-diverging nozzles would be required). The pressure at the hole outlet will be Pchoked, even though Pout decreases further. Pchoked is called choked or critical pressure, and the velocity at the hole in these conditions is called choked or critical velocity. Assuming isentropic expansion, the relationship between the choked pressure and the pressure inside the tank can be expressed as: Pchoked Pcont
J
ª 2 º J 1 « » ¬ J 1¼
(2-15)
where Pcont is the pressure inside the container or the pipe (Pa), and J is the ratio of heat capacities, cp/cv (-). The choked velocity is the maximum possible velocity in an accidental release. It is found in most accidental gas releases. Since Pout is usually the atmospheric pressure (essentially constant), the same conditions are reached if the pressure inside the container (a tank, a pipe) increases up to a certain value: further increases in Pcont will not produce any further increase in the gas exit velocity. Therefore, critical velocity will be reached if the following condition is fulfilled: J
Pcont ª J 1 º J 1 t« » P0 ¬ 2 ¼
(2-16)
In fact, J is the isentropic coefficient of the gas or vapour at the relieving conditions. However, for gases with properties similar to those of an ideal gas, J is the ratio of heat capacities. J is always greater than unity. For most gases, it ranges from 1.1 to 1.4; therefore, sonic velocity will be reached when Pcont/P0 t 1.9. For air (J , for example, sonic velocity is reached when Pcont/P0 t 1.893, i.e. when the downstream absolute pressure is 52.8% of the upstream absolute pressure. The speed of sound in an ideal gas at a temperature T can be calculated with the following expression:
us
J T R 10 3
(2-17)
Mv
where us is expressed in m s-1 R is the ideal gas constant (8.314 kJ kmole-1 K-1) Mv is the molecular weight of the gas (kg kmole-1) Table 2-3 shows the ratio of heat capacities, J = cp/cv, for various gases. The density of a gas increases with pressure. Therefore, once the critical velocity has been reached, if Pcont is further increased, the release velocity will still be the speed of sound, but
31
the density of the gas will be higher. Therefore, the mass flow rate will increase with Pcont. It is therefore clear that what becomes critical or choked is the velocity (m/s) of the gas rather than the flow. Thus, critical velocity is a better term than critical flow. Critical flow —i.e. both choked gas velocity and mass flow rate— can be reached when there is a given pressure upstream from the hole and vacuum conditions downstream from the hole so that the sonic velocity is reached. In this case, the inlet gas density is constant and therefore the mass flow rate is also choked. The temperature of the gas in the jet at the orifice is:
Tchoked
§P Tcont ¨¨ choked © Pcont
§ J 1 · ¨ ¸ J ¸¹
· ¨© ¸¸ ¹
§ 2 · ¸¸ Tcont ¨¨ © J 1 ¹
(2-18)
where Tcont is the temperature in the container or pipe (K). Table 2-3 Molecular weight, heat capacity ratio and sonic velocity for various gases and vapours at 298 K and 101.3 kPa. Calculated from [7, 8] Gas Molecular weight us, m s-1 J = cp/cv Acetylene 26.0 1.247 345 Acrylonitrile 53.1 1.149 232 Air 29.0 1.400 246 Ammonia 17.0 1.311 437 Benzene 78.1 1.112 188 Butane 58.1 1.091 216 Carbon dioxide 44.0 1.301 271 Carbon monoxide 28.0 1.400 352 Chlorine 70.9 1.330 216 Cyclohexane 84.2 1.085 179 Ethane 30.1 1.188 313 Ethylene 28.0 1.253 333 Ethylene oxide 44.0 1.215 261 Helium 4.0 1.660 1014 Hexane 86.2 1.062 175 Hydrogen chloride 36.5 1.399 308 Hydrogen 2.0 1.405 1314 Hydrogen sulphide 34.1 1.326 310 Methane 16.0 1.304 449 Natural gasa 18.1 1.270 419 Nitrogen 28.0 1.406 352 Oxygen 32.0 1.395 329 Propane 44.1 1.146 253 Propylene 42.1 1.148 260 Sulphur dioxide 64.1 1.264 221 Toluene 92.1 1.087 171 a
86.15% CH4, 12.68% C2H6, 0.09% C4H10, 0.68% N2
32
3.1.2 Mass flow rate The mass flow rate of gas through an orifice can be calculated with the following expression, obtained from the mechanical energy balance by assuming isentropic expansion and introducing a discharge coefficient: J 1
mhole
Aor C D Pcont\
Mv § 2 · J 1 ¸ J ¨¨ ¸ 3 © J 1 ¹ Z Tcont R 10
(2-19)
where m is the mass flow rate (kg s-1) Cd is a dimensionless discharge coefficient (-) Aor is the cross-sectional area of the orifice (m2), and Z is the gas compressibility factor at Pcont, Tcont (-) (for ideal gas behaviour, Z = 1). \is a dimensionless factor that depends on the velocity of the gas. For sonic gas velocity:
\ 1
(2-20)
and for subsonic gas velocity: J 1
\
2
2 § J 1 · J 1 § P0 ¨ ¸ ¨ J 1 © 2 ¹ ¨© Pcont
2 § · J ¨ § P0 ¸¸ ¨1 ¨¨ ¹ ¨ © Pcont ©
· ¸¸ ¹
J 1 J
· ¸ ¸ ¸ ¹
(2-21)
The value of \ has been plotted as a function of P0/Pcont, for various heat capacity ratios, in Fig. 2-6. The length of a free jet of a gas can be estimated [9] with the following expression: Lj
6 u j d or
(2-22)
uw
where uj is the velocity of the jet at the source (m s-1) dor is the diameter of the source, and uw is the average ambient wind speed (m s-1) (default value: 5 m s-1). 3.1.3 Discharge coefficient CD is a coefficient that takes into account the fact that the process is not isentropic. Its value is CD = 1.0 for a full-bore rupture in a pipe. For sharp-edged orifices in accidental releases (high Reynolds number), some authors recommend CD = 0.62 and others recommend a conservative value of 1.0. ______________________________________ Example 2-3 Due to an incorrect manoeuvre, an impact creates a hole with an approximate diameter of 2 cm in the top of a tank containing propane at 25 ºC and 10 bar. The level of the liquid is low, so gas is released through the hole. Calculate the mass flow rate.
33
Solution For propane, J = 1.15. Therefore: J
1.15
ª J 1º J 1 «¬ 2 »¼
ª1.15 1º 1.151 «¬ 2 »¼
1.74
Since Pcont P0
10 bar = 9.87 1.013 bar
the propane velocity at the orifice is critical. Therefore, by applying Eq. (2-19): J 1
Mv § 2 · J 1 Aor C D Pcont\ J ¨¨ ¸¸ Z T R 10 3 J 1 ¹ © cont
mhole
1.151
S 0.02 2 4
§ 2 · 1.151 44.1 ¸¸ C D 10 10 1 1.15 ¨¨ 1 . 15 1 8.314 10 3 1 273 25 © ¹
5
0.525 kg/s if C D 0.62 ® ¯0.847 kg/s if C D 1.0
The velocity at the hole can now be calculated. The conditions at the choked jet are: 1.15
Pchoked
§ 2 · 1.151 ¸¸ 10 ¨¨ © 1.15 1 ¹
Tchoked
§ 5.74 · 298 ¨ ¸ © 10 ¹
U choked
5.744 10 44.1 8.314 10 277.2
1.151 1.15
5.744 bar
277.2 K
5
3
11 kg m-3
The velocities at the hole corresponding to the two mass flow rates are: u
0.525 § 0.02 2 · ¨¨ S ¸ 11 4 ¸¹ ©
152 m s-1
u
0.847
245 m s-1
§ 0.02 2 ¨¨ S 4 ©
· ¸¸ 11 ¹
34
The speed of sound in propane at choked conditions is: 1.15 277.2 8.314 10 3 245 m s-1 44.1 ______________________________________ us
1,00
0,95
0,90
g = 1,10 1,15 1,20 1,25 1,30 1,35 1,40 1,45
1,10
0,85
0,80
y
0,75
0,70
0,65
0,60
0,55
0,50
0,45
0,40 1,00
0,95
0,90
0,85
0,80
0,75
0,70
0,65
0,60
0,55
0,50
P0/Pcont
Fig. 2-6.\ as a function of P0/Pcont and J.
3.2 Flow of gas/vapour through a pipe A typical case in risk analysis is the calculation of a gas flow from a pipe connected to an upstream constant pressure source (usually a vessel) (Fig. 2-7). The gas outflow can take place through a full-bore rupture of the pipe or through a hole in the pipe wall. In both cases, the pressure in the pipe must be estimated at a point just in front of the orifice. This requires knowledge of the gas flow rate which, in turn, depends on the pressure drop between the upstream constant pressure source and the aforementioned point. A trial-and-error procedure is therefore required. A relatively simple method [3] is presented below.
35
Fig. 2-7. Flow of a gas through a pipe.
The overall pressure drop between the upstream pressure source and the environment is the pressure drop in the pipe plus the pressure drop through the opening (a hole or the fully broken pipe):
P
'P Pcont P0
cont
Pp Pp P0
'Ppipe 'Phole
(2-23)
where Pp is the pressure inside the pipe just in front of the opening (Pa), and P0 is the ambient pressure (Pa). The mass flow rate through the pipe depends on the pressure drop through the pipe and the mass flow rate through the hole depends on the pressure drop through the hole. According to the law of conservation of mass, the mass flow through the pipe due to the loss of containment (mpipe) must be equal to the mass flow through the hole (mhole): m pipe
mhole
(2-24)
The mass flow rate through a pipe depends on the pressure at both ends of the pipe, and can be calculated with the following expression [3]: Pp
2 m pipe
Ap
³ U P dP
Pcont
§ L 4 fF ¨ ¨d © p
(2-25)
· ¸ ¸ ¹
where Ap is the cross-sectional area of the pipe (m2) U (P) is the density of the gas (kg m-3) fF is the Fanning friction factor (-) L is the length of the pipe (m), and dp is the diameter of the pipe (m). The following relationships apply:
36
1
U
§ P ·] constant ¨ ¸ ©Z¹
(2-26)
]
1
Z R 10 3 cv M v
(2-27)
where Z is the compressibility factor. For ideal gas behaviour: Z
1 and ] J
The integral in Eq. (2-25) may be solved analytically, assuming a constant compressibility factor and constant specific heat at a constant volume, cv; for Z = 1: Pp
³ U P dP | P
cont
U cont
Pcont
§ ] ¨¨ © 1 ]
§ · ¨ § Pp ¸¸ ¨ ¨¨ ¹ ¨ © Pcont ©
1]
· ¸¸ ¹
]
· ¸ 1¸ ¸ ¹
(2-28)
(the units are kg2 m-4 s2). This set of equations can be solved by trial and error by guessing the internal pressure inside the pipe just in front of the pipe opening, Pp. The following procedure must be followed: 1. Guess the pressure inside the pipe just in front of the pipe opening (P0 < Pp < Pcont). 2. Calculate the mass flow rate through the pipe opening with Eq. (2-19) (CD = 0.62 for a hole in the pipe wall; CD = 1 for full-bore rupture). 3. Calculate the mass flow rate through the pipe with Eqs. (2-25) and (2-28). 4. Compare the two mass flow rates. If they are not equal, correct the value of Pp and repeat the procedure. To calculate the mass flow rate through the pipe opening (Eq. (2-19)), the temperature of the gas in the pipe in front of the opening must be estimated. This requires a trial-and-error procedure. By defining a parameter Y, the following expressions can be applied [1]: Yi
1
Tp Tcont Pp Pcont
J 1 2
Mai2
Ycont Yp Ma cont Ma p
(2-30) Ycont Yp
§ Ma 2p Ycont ln ¨ 2 ¨ Ma cont 2 Yp ©
J 1
(2-29)
· § 1 1 ·¸ ¸¨ J 2 ¸ ¨ Ma cont Ma 2p ¸ ¹ © ¹
(2-31)
§ 4 fF L · ¨ ¸ 0 ¨ dp ¸ © ¹
37
(2-32)
For sonic flow at the exit, Map = 1 and Eqs. (2-29) and (2-31) become: Tp Tcont J 1 2
2 Ycont J 1 § 2 Ycont ln ¨¨ 2 J 1 Ma cont ©
(2-33)
· § 1 · ¸¨ ¸ ¸ ¨ Ma 2 1¸ J cont ¹ © ¹
§ 4 fF L · ¸ 0 ¨ ¨ dp ¸ ¹ ©
(2-34)
For the case of a hole in the wall of the pipe, to calculate the mass flow rate, the pressure in the pipe at a point on a level with the hole must be known. If the only flow is the one caused by the leak, then a trial-and-error procedure must be applied to establish the value of Pp. If there is, furthermore, a certain flow rate through the pipe, it must be taken into account in order to estimate the value of Pp. ______________________________________ Example 2-4 A constant pressure source (5 bar, 288 K) of natural gas (Mv = 17.4; J = 1.27) is connected to a smooth polyethylene pipe with a diameter of 164 mm. Calculate the release flow rate a) if the pipe is completely broken 330 m from the source, and b) if there is a hole in the pipe wall, at L = 330 m, with a diameter of 50 mm. Solution a) For full-bore rupture, CD = 1.0. For a smooth polyethylene pipe (Fig. 2-4), a value of the Fanning factor fF = 0.002 will be assumed. Guessed pressure: Pp = 4 bar. To estimate the temperature of the gas at the end of the length of pipe, just in front of the opening, Eq. (2-34) is applied: § § 1.27 1 ·· 2 Macont ¨ 2 ¨1 ¸¸ 1.27 1 ¨ © 2 ¹ ¸ §¨ 1 1·¸ 1.27 § 4 0.002 330 · ln ¨ ¸ 2 2 ¸ ¨ 2 0.164 1.27 1 Macont ¸ ¨© Macont © ¹ ¹ ¸ ¨ ¹ © i.e.: 2 § 2 0.27 Macont 1.135 ln ¨¨ 2 © 2.27 Macont
· 1 ¸ ¸ Ma 2 20.44 cont ¹
0
By trial and error, Macont = 0.2. Therefore, by applying Eq. (2-29):
Ycont
1
1.27 1 0.2 2 2
1.005
From Eq. (2-33):
38
0
Tp 288
2 1.005 1.27 1
Tp = 255 K Mass flow rate through the opening (Eq. (2-19)): 1.27 1
mhole
S 0.164 2 4
§ 2 · 1.27 1 17.4 ¸¸ 1.0 4 10 1 1.27 ¨¨ 255 8.314 10 3 © 1.27 1 ¹
5
16 kg s -1
Mass flow rate through the pipe (Eq. (2-25)) (Ucont = 3.68 kg m-3): 11.27 § § · · § 1.27 · ¨ § 4 · 1.27 ¨ ¸ ¸ 5 ¨ ¸ 2 ¨ 5 10 3.68 ¨ 1¸ ¸ ¸¨¨ 5 ¸ 1 1 . 27 ¸ ¸ ¨ © ¹¨© ¹ © ¹ ¹ © § 330 · 4 0.002 ¨ ¸ © 0.164 ¹
m pipe
S 0.164 2 4
4.33 kg s -1
Since two different results have been obtained, a new trial is required. The results of the calculation procedure are summarized in the following table: Pp, bar 4 3 2.7 2.0 1.75 1.74
mhole, kg s-1 16.00 12.00 10.01 8.00 6.99 6.96
mpipe, kg s-1 4.33 5.84 6.36 6.78 6.95 6.96
Therefore, m = 6.96 kg s-1. b) For the hole in the pipe wall, CD = 0.62. The influence of pressure decrease on gas temperature is now neglected. A trial-and-error procedure is again applied. Due to the change in the flow, a new Fanning friction factor value, fF = 0.003, is now assumed (it will be checked later). Guessed pressure: Pp = 4.5 bar. To estimate the temperature of the gas, by trial and error the value of Macont is found: Macont = 0.17. The temperature of the gas just in front of the opening is calculated with Eqs. (2-29) and (2-33): Tp = 254.7 K. The mass flow rates are now: 1.27 1
mhole
S 0.05 2 4
§ 2 · 1.27 1 17.4 ¸¸ 0.62 4.5 10 5 1 1.27 ¨¨ 254.7 8.314 10 3 © 1.27 1 ¹
39
1.04 kg s -1
11.27 § § ·· § 1.27 · ¨ § 4.5 · 1.27 ¨ ¸¸ 5 ¸¸ ¨ ¨ 2 ¨ 5 10 3.68 ¨¨ 1 ¸ ¸¸ 1 1 . 27 5 ¹ ¸¸ ¨ ¹¨© © © ¹¹ © § 330 · 4 0.003 ¨ ¸ © 0.164 ¹
m pipe
S 0.164 4
2
2.56 kg s -1
The two values are different. By trial and error, the following values are ultimately obtained: Pp = 4.9 bar, m = 1.13 kg s-1. Although the critical velocity is not reached, the use of \ is correct (see Fig. . With this mass flow rate, the average Reynolds number in the pipe is calculated and the value of fF is found. Since two different results are obtained, a new trial is required. The results of the calculation procedure are summarized in the following table: Pp, bar
m, kg s-1
Re
fF*
4,90 4,50 4,47 4,46
1,13 1,04 1,03 1,03
891,53 819,78 813,17 812,52
0,01795 0,01952 0,01968 0,01969
*fF calculated with Eq. (2-10) because there is laminar flow.
Therefore, Pp=4.46 bar and m = 1.03 kg s-1. ______________________________________ A special, relatively complex case is the calculation of accidental releases from distribution systems (gas networks). For these situations, more powerful models are required (see, for example [6]). 3.3 Time-dependent gas release In the above examples, the proposed expressions estimate the mass flow rate of a gas release to a constant pressure source. If the pressure upstream is not constant, as, for example, in the case of a vessel containing a pressurized gas, then these expressions can only be applied to the first few minutes of the release. As the release proceeds, the pressure inside the vessel decreases. Therefore, in this case, the upstream pressure must be corrected at given time intervals in order to calculate the mass flow rate as a function of time (the assumption of a constant flow rate, at the initial value, throughout the period will lead to a conservative prediction). The release decay is basically a function of two factors: the initial leak rate and the initial mass of gas inside the vessel. The decay of the leak, which follows an exponential trend, can be roughly described [10] by the following expression:
m t
§ minitial t · minitial exp ¨ ¸ ¹ © W
(2-35)
where m(t) is the mass flow rate at the time t after the onset of the leak (kg s-1)
40
minitial is the mass flow rate at the onset of the leak (kg s-1) W is the initial mass of gas in the vessel (kg), and t is the time after the onset of the leak (s). ______________________________________ Example 2-5 A vessel with a volume of 9 m3 contains nitrogen at an absolute pressure of 14 bar and 25 ºC. Calculate the mass flow rate through a hole with a diameter of 2.54 cm a) at the onset of the release, and b) as a function of time. Solution At the initial pressure, the density of nitrogen is 15.8 kg m-3. Therefore, the initial content of the vessel is: W
9 15.8 142.2 kg
Due to the pressure in the vessel, it is evident that critical velocity will be reached at the hole. The initial mass flow rate can be calculated with Eq. (2-19): 1.411
mhole
S 0.0254 2 4
§ 2 · 1.411 28 ¸¸ 0.62 14 10 1 1.41¨¨ 298 8.314 10 3 © 1.41 1 ¹
5
1.015 kg s -1
(A value of CD = 0.62 has been assumed here.) The flow as a function of time is calculated with Eq. (2-35). For example, thirty seconds after the onset of the release:
§ 1.015 30 · m 30 1.015 exp ¨ ¸ © 142.2 ¹
0.82 kg s-1.
Fig. 2-8 shows the variation of mass flow rate as a function of time.
Fig. 2-8. Flow of a gas through a hole as a function of time.
______________________________________
41
4 TWO-PHASE FLOW
In some cases, a mixture of liquid and vapour or gas can occur. Two-phase flow may occur, for example, when a hot, pressurized liquid is significantly depressurized, causing it to boil and suddenly vaporize. If liquid droplets are entrained or foam is formed, a spray will be released through the opening or the relief device. If the mixture of liquid and gas/vapour is ejected into the atmosphere, the spray droplets may be evaporated or may fall to the ground. The existence of two phases has a significant influence on the mass flow rate of the release. Two-phase flow is a complex phenomenon, not yet sufficiently well understood, so conservative design is often applied in order to estimate relief requirements. The following paragraphs explain a simple approach. Furthermore, Section 5.2 discusses a particular case for runaway reactions. For a more accurate prediction, more complex methodologies or computer programs should be used [11]. 4.1 Flashing liquids If there is a leak or a relief to the atmosphere from a vessel containing a hot pressurized liquid, upon depressurization the liquid becomes superheated, its temperature being higher than its boiling temperature at atmospheric pressure. Therefore, it undergoes a sudden, or “flash”, vaporization: a mixture of vapour and droplets exits to the atmosphere. This process is so fast that it can be assumed to be adiabatic. The vapour takes the excess energy from the remaining liquid to become vaporized, and the resulting vapour/liquid mixture reaches its atmospheric boiling temperature. Thus, the vaporization of a differential mass of liquid dwl implies a decrease in the temperature of the remaining liquid dT:
dwl
wl c pl 'H v
dT
(2-36)
This expression can be integrated between the initial temperature of the liquid before depressurization, Tcont, and the final temperature of the mixture (the atmospheric boiling point if the mixture is released into the atmosphere), Tb, in order to calculate the mass of vapour: c pl Tcont Tb § · 'H v ¸ wv wil ¨1 e ¨ ¸ © ¹
(2-37)
where wv is the mass of vapour (kg) wil is the initial mass of liquid (kg) 'Hv is the mean latent heat of vaporization between Tcont and Tb (kJ kg-1), and cpl is the mean heat capacity of the liquid between Tcont and Tb (kJ kg-1 K-1). The ratio between the mass of vapour formed and the initial mass of liquid is usually called the vaporization fraction: f
wv 1 e wil
c p Tcont Tb
'H v
(2-38)
42
In practice, a significant amount of droplets will be entrained by the vapour, incorporated into the vapour cloud (thus increasing its density) and later vaporized. Therefore, the “real” value of f will be higher than that predicted by Eq. (2-38). In some studies, it has been observed that in fact there was no rainout, and all of the liquid droplets were incorporated into the cloud. Thus, although a value of 2f has been suggested for the vaporization fraction by some authors, a conservative approach is to assume that all of the mass released is entrained in the cloud. For multicomponent mixtures, there is preferential vaporization of the more volatile components and the calculation of f becomes significantly complicated. 4.2 Two-phase discharge When a mixture of liquid and vapour or liquid and gas is discharged, the cross-sectional area required for a given relief is significantly larger than that corresponding to vapour alone. The relationship between the venting area and the discharge rate is complex. An accurate design requires the use of complex procedures. The following paragraphs present a set of equations that allow an approximate calculation. The flashing process approaches equilibrium conditions if the pipe is sufficiently long, the minimum length being approximately 0.1 m or greater than 10 diameters. For a pipe length less than 0.1 m (non-equilibrium regime), the flow rate (kg s-1 m-2) increases sharply as length decreases, approaching liquid flow as the length approaches zero. For non-equilibrium regime, flashing flow is choked and can be estimated by [12]:
Gne |
'H v vlv
(2-39)
N Tcont c p l
where Gne is the discharge rate (kg m-2 s-1) 'Hv is the latent heat of vaporization at boiling temperature of liquid (kJ kg-1) Tcont is the storage temperature (K) vlv is the change in specific volume brought about by the change from liquid to vapour (m3 kg-1), and c p l is the specific heat of the liquid (kJ kg-1 K-1). N is a non-equilibrium parameter: N
2 Pcont
'H v2 L P0 U l C D2 vlv2 Tcont c pl Le
(2-40)
where Pcont is the pressure inside the vessel (Pa) P0 is the atmospheric pressure (Pa) L is the pipe length to opening (ranging from zero to 0.1 m) (m) Ul is the liquid density (kg m-3) CD is the discharge coefficient (-), and Le = 0.1 m. Eq. (2-40) shows that the degree of non-equilibrium varies directly with the pipe length L. For L/dp = 0, there is no flashing and Eq. (2-39) reduces to the orifice equation for incompressible liquid flow. For L t 0.1 m (or greater than 10 diameters), we can assume that equilibrium flashing conditions are reached; the flow rate is a weak function of the L/dp ratio and Eq. (2-39) becomes:
43
Ge |
'H v
(2-41)
vlv Tcont c p l
If the physical properties ('Hv, vlv) are unknown, Eq. (2-41) can be expressed as: Ge |
dP dT
T cp
(2-42)
The effect of subcooling on the discharge rate is: G sub
2 Pcont Pv U l
(2-43)
where Pv is the vapour pressure at the storage temperature (Pa). The mass flow rate for a two-phase discharge of subcooled or saturated liquids can be expressed as [13] [14]: 2 G2 p C D G sub
Ge2 N
(2-44)
where G2p is the discharge mass flow rate (kg m-2 s-1), and CD is a discharge coefficient (-). As mentioned in Section 3.1, in the discharge of flashing liquids through a hole in a vessel wall, we can assume that the discharge is a liquid, and that flashing occurs downstream of the hole. 5 SAFETY RELIEF VALVES
A safety valve is a device used to prevent overpressure in a given piece of equipment (vessel or pipe). When a predetermined maximum (set) pressure is reached, the safety valve reduces the excess pressure by releasing a volume of fluid. Safety valves can act in various situations, such as exposure to plant fires, failure of a cooling system and runaway chemical reactions. A safety relief valve has been defined by the ASME as a pressure relief valve characterized by rapid opening or pop action, or by opening in proportion to the increase in pressure over the opening pressure, depending on the application. It may be used either for liquid or compressible fluid. In general, safety relief valves act as safety valves when used in compressible gas systems, but open in proportion to overpressure when used in liquid systems. The basic elements of the design of a safety valve consist of a right-angle pattern valve body with a valve inlet mounted on the pressure-containing system and an outlet connected to a discharge system or venting directly into the atmosphere. Fig. 2-9 shows a typical safety valve design [15]. Under normal operating conditions, a disc is held against the nozzle seat by a spring, which is housed in a bonnet mounted on top of the body. The amount of compression on the spring (which provides the force that closes the disc) is usually adjustable, so the pressure at which the disc is lifted off its seat to allow relief may vary. When the
44
pressure inside the equipment rises above the set pressure, the disc begins to lift off its seat. As the spring starts to compress, the spring force increases and further overpressure is required for any further lift. The additional pressure rise required before the safety valve will discharge at its rated capacity is called the overpressure [15]. The allowed overpressure depends on the standards being followed and on each specific case. For compressible fluids, it normally ranges from 3 to 10%, and for liquids, from 10 to 25%. Once normal operating conditions have been restored or the pressure inside the equipment has dropped below the original set pressure, the valve closes again.
Fig. 2-9. Typical safety valve design (DIN valve).
5.1 Discharge from a safety relief valve The discharge flow from a safety relief valve depends on the pressure inside the equipment and the cross-sectional flow area of the valve. The relationship between these three variables is established by the existing methods for sizing safety valves. The following paragraphs present the standard AD-Merkblatt A2, DIN 3320, TRD 421. The minimum required orifice area for a safety valve used in air and gas applications can be calculated (for sonic flow) with the following expression: Asv
0.1791 m sv < D w Pcont
TZ Mv
(2-45)
and for liquid applications: Asv
0.6211 m sv
(2-46)
D w U l Pcont Pback
45
where Asv is the minimum cross-sectional flow area of the safety valve (mm2) msv is the discharge mass flow rate (kg h-1) Pcont is the absolute relieving pressure (bar) Pback is the absolute backpressure (bar) T is the inlet temperature (K) Ul is the liquid density (kg m-3) Mv is the molecular weight (kg kmol-1) Z is the compressibility factor (-) Dw is an outflow coefficient specified by the manufacturer (-), and < is an outflow function (see Fig. 2-10).
Fig. 2-10. The outflow function as used in AD-Merkblatt A2, DIN 3320, TRD 421. Taken from Spirax Sarco Steam Engineering Tutorials [15], by permission.
For two-phase flow, a conservative approach consists in calculating the area required to discharge the vapour fraction and, separately, that required for the liquid fraction, and then adding them together in order to establish the minimum cross-sectional area. ______________________________________ Example 2-6 In a train accident, a derailed wagon tank containing propane is heated by an external fire to 47 ºC (Pcont = 16.3 bar). A safety relief valve, now located in the liquid zone, opens and discharges into the atmosphere. Estimate the discharge rate.
46
Data: boiling temperature of the liquid at atmospheric pressure = - 42 ºC; average latent heat of vaporization between - 42 ºC and 47 ºC = 358 kJ kg-1; average cp of the liquid between these two temperatures = 2.54 kJ kg-1 K-1; Ul = 440 kg m-3; k = 1.15; molecular weight = 44.1; Av = 3.25 cm2. Solution Taking into account the liquid temperature in the tank and its boiling temperature at atmospheric pressure, flash vaporization will occur. The vaporization fraction is (Eq. (2-38)): f
wv § 2.54 320 231 · 1 exp ¨ ¸ 358 wil © ¹
0.468
Taking into account the pressure values inside and outside the tank, it can be assumed —with a certain degree of uncertainty— that sonic flow will occur in the valve. Because this is a two-phase flow (although there is some uncertainty related to this type of flow at the nozzle), vapour and liquid flow is estimated separately with Eqs. (2-45), (2-46). Vapour flow: 0.468 m sv
Ag < D w Pcont
0.1791
Mv TZ
Ag 0.45 0.7 16.3
0.1791
44.1 320
Liquid flow: 0.532 m sv
Al D w U l Pcont Pback
Al 0.7 440 16.3 1.013
0.6211
0.6211
Furthermore, Ag + Al = 325
By solving these three equations, the discharge mass flow rate is obtained: msv = 6,535 kg h-1 ______________________________________ 6 RELIEF DISCHARGES
When a vessel containing a liquid or a gas is heated (by an external fire or by an exothermic chemical reaction, for example), the internal pressure increases. If both temperature and pressure continue to increase, at a certain value, the vessel will burst. To avoid this, a relief device is usually provided to allow the discharge of fluid once a given (set) pressure is reached. If the discharge is controlled and adequately treated (by using a secondary tank, a cold water tank, a scrubber, a flare, etc.), no dangerous release into the atmosphere will occur. However, if the discharge is released directly into the atmosphere, it can lead to an accident. To foresee its eventual effects, the discharge flow rate must be estimated.
47
6.1 Relief flow rate for vessels subject to external fire If a vessel containing a liquid is subject to a certain heat flux, the liquid will evaporate and the pressure will rise. Vapour or gas will have to be released through a pressure relief valve to prevent the pressure from exceeding a given value. However, even with a relief device, an accident (the failure of the tank) can occur if the metal above the wetted surface area (i.e. the metal that is not cooled by the liquid inside the vessel) is excessively heated and loses its strength (Fig. 2-11).
Fig. 2-11. Relief in a vessel exposed to fire.
The following paragraphs describe a method for calculating the gas discharge rate from the relief device (for vessels containing stable liquids) proposed in NFPA 30 [16]. 1. The surface of the vessel exposed to fire is assumed to be: - spherical tank: 55% - horizontal tank: 75% - rectangular tank : 100% (excluding the top surface) - vertical tank: 100% (up to a height of 9 m). 2. The heat flux to the vessel is estimated as a function of the exposed vessel surface and the design pressure: Aexp 18.6 m2 o Q 63,080 Aexp
(2-45-a)
0.566 18.6 m2 < Aexp < 92.9 m2 o Q = 224,130 Aexp
(2-45-b)
0.338 92.9 m2 < Aexp < 260 m2 o Q = 630,240 Aexp
(2-45-c)
Aexp ! 260 m2 and Pdesign < 0.07 barg o Q 4,130,000
(2-45-d)
0.82 Aexp > 260 m2 and Pdesign > 0.07 barg o Q = 43,185 Aexp
(2-45-e)
where Aexp is the exposed vessel surface (m2), and Q is the heat flux (W).
48
For tanks containing LPG (pressurized tanks), NFPA58 [17] recommends the following expression: 0.82 Q = 70,945 Aexp
(2-45-f)
where Aexp is the entire surface area of the vessel (m ). 3. The gas discharge rate is calculated using the following expression: 2
m
FQ 'H v
(2-46)
where m is the discharge rate (kg s-1) F is a reduction factor (-), and 'Hv is the latent heat of vaporization at the boiling temperature of the liquid (kJ kg-1). The reduction factor F depends on the protective measures applied: - Aexp greater than 18.6 m2 and drainage with a minimum slope of 1% leading the spill to a remote (at a distance of at least 15 m) impounding area: F = 0.5. - adequately designed water spray system and drainage: F = 0.3. - adequate thermal insulation: F = 0.3. - water spray system plus thermal insulation plus drainage: F = 0.15. - none of the aforementioned protective measures: F = 1. ______________________________________ Example 2-7 Calculate the mass discharge rate required in a horizontal cylindrical tank containing nhexane that is being exposed to fire. The tank (length = 7 m, diameter = 4 m) is located inside a containing dike and drainage of the spilled liquid is provided. Data: 'Hv n-hexane = 334 kJ kg-1.
Solution The area exposed to fire is.
Aexp 0.75 S 4 7 S 4 2 / 2
2
103.7 m
Taking into account the drainage system, F = 0.5. The heat flux received by the tank is: 0.338 Q = 630,240 Aexp 630,240 103.7 0.338
3,025,790 W
And the required discharge mass flow rate is:
F Q 0.5 3,025,790 4,530 kg h-1 = 1.26 kg s-1 'H v 334 ______________________________________ m
6.2 Relief flow rate for vessels undergoing a runaway reaction Runaway reaction is the term used to define the uncontrolled development of one or more exothermic chemical reactions. Runaway reactions have been the origin of a number of
49
accidents in chemical plants, including the well-known cases of Seveso (Italy, 1976) and Bhopal (India, 1984). They may involve the loss of control of a desired chemical reaction or the development of an undesired reaction. Highly exothermic chemical reactions are potentially dangerous, and slightly exothermic reactions can cause an increase in temperature that may set off highly exothermic reactions. Uncontrolled exothermic reactions can occur not only in chemical reactors, but also in other units such as distillation columns, storage tanks, etc. If the rate at which the reacting material generates heat is higher than the rate at which the system can dissipate it, the temperature will increase up to a value at which the process is uncontrollable. The essential condition is the existence of a self-accelerated heating process: as the temperature increases, the reaction rate increases exponentially up to very high values. This process can be very slow in its first steps, but very fast in its final step. The formation of gas products or an increase in vapour pressure will raise the pressure inside the vessel. If there is a relief device, gas/vapour of two-phase flow will be released when the set pressure is reached. This will prevent explosion, but if the released stream is not adequately treated, there will be a loss of containment of some dangerous material. The potential danger involved in a runaway reaction is not always taken into account, at least in comparison with other risks involving much smaller amounts of energy. Take, for example, the case of a storage tank containing acrylonitrile [18] at 10 ºC. If the volume of the tank is 13,000 m3 and it has been filled to 90%, it will contain 725,000 kg. If no inert gas has been used, in the vapour head (100 m3) the mixture will contain 11% acrylonitrile and be within flammability limits. The danger associated with this explosive gas mixture is obvious. However, if the energy released in the combustion of the acrylonitrile contained in this gas mixture is compared with the energy released in the exothermic polymerization of the acrylonitrile, the following values are obtained: - energy released in the polymerization process: 900,000 MJ. - energy released in the combustion of the vapour phase 900 MJ. Therefore, the risk associated with the reactivity of a chemical can be higher than the — generally more evident— risk posed by some of its properties (flammability, toxicity, etc.).
Fig. 2-12. Evolution of pressure in a vessel with a runaway reaction. Taken from [19], by permission.
50
The evolution of pressure in a vessel in which a runaway reaction takes place will depend on the kinetics of the reaction and on the type of discharge vented. If the vessel is closed, the pressure will increase exponentially. If there is a relief device and it is opened, the pressure will still increase somewhat, reaching a maximum and then decreasing (Fig. 2- 12). The value of the maximum pressure reached during the loss of control will depend on the kinetics of the reaction, the temperature of the mixture when the runaway starts, the initial concentration of the reactants, the mass initially contained in the vessel, the relationship between this mass and the cross-sectional area of the relief device, and the set pressure of the device. The relief mass flow rate can be estimated from the mass and heat balances by applying some simplifying assumptions: - the discharge mass flow rate is essentially constant. - the heat of reaction per unit mass (q) is practically constant. - the physical and thermodynamic properties are constant. With these assumptions, the differential equation obtained from the mass and heat balances can be integrated. Thus, Leung [19] proposed the following expression to estimate the relief mass flow rate for the case of homogeneous-vessel venting, i.e. assuming zero disengagement of liquid and vapour within the vessel:
m GA
Wq º ª V 'H v c v 'T » « ¼» ¬« W vlv
2
(2-47)
where m is the discharge rate (kg s-1) G is the discharge rate per unit cross-sectional area of venting (kg m-2 s-1) W is the initial mass contained in the vessel (kg) q is the heat flow released by the chemical reaction (kW kg-1) V is the volume of the vessel (m3) cv is the specific heat at constant volume (kJ kg-1K-1) 'Hv is the latent heat of vaporization of the liquid (kJ kg-1) vlv is the change in specific volume when changing from liquid to vapour (m3 kg-1) 'T is the “overtemperature”, the increase in temperature corresponding to the overpressure, (K). The heat flow released by the chemical reaction can be calculated as the arithmetic mean of the values corresponding to the temperature rise rate at the moment at which the relief starts and the moment at which the maximum temperature is reached: q
1 ª§ dT · § dT · º c v «¨ ¸ ¨ ¸ » 2 ¬© dt ¹ set © dt ¹ max ¼
(2-48)
The ratio ('Hv/vlv) can be obtained from P-T data by applying the following relationship obtained from the Clapeyron relation: 'H v dP Tset vlv dT
(2-49)
51
This expression is very accurate for single-component fluids and is a good approximation for multicomponent mixtures in which composition change is minimal [19] The mass flux or mass flow rate per unit area can be calculated as follows: 'H v G 0.9 vlv
§ 1 · ¨ ¸ ¨c T ¸ p © ¹
0.5
(2-50)
where 'Hv must be expressed in J kg-1 and cp in J kg-1 K-1. The Leung method is quick and simple and, thus, widely used for approximate calculations. A classic example proposed by Huff [20] and also applied by Leung [19] is used to illustrate the application of this method. ______________________________________ Example 2-8 Estimate the discharge rate and the vent area for a tank of styrene monomer undergoing adiabatic polymerization after being heated to 70 ºC. The maximum allowable working pressure of the tank is 5 bar. System parameters: vessel volume = 13.2 m3; W = 9,500 kg; Ps = 4.5 bar; Ts = 482.5 K; (dT/dt)set = 0.493 K s-1 (sealed system); Pmax = 5.4 bar; Tm = 492.7 K; (dT/dt)max = 0.662 K s-1. Material properties: Ps = 4.5 bar 0.001388 0.08553 2.470 310.6
Specific volume, liquid, m3 kg-1 Specific volume, gas, m3 kg-1 Cpl, kJ kg-1 K-1 'Hv, kJ kg-1
Pm = 5.4 bar 0.001414 0.07278 2.514 302.3
Solution From Eq. (2-48), and assuming cv | cp for an incompressible fluid: q
1 2.47 >0.662 0.493@ 1.426 kJ kg-1 s-1 2
The mass discharge rate and the mass flux can be calculated from Eqs . (2-47) and (2-50): m
9,500 1.426 º ª 13.2 310.6 2.47 492.7 482.5 » « 9 , 500 0.08553 0.001388 ¼ ¬
§ · 310,600 1 ¨ ¸ G 0.9 0.08553 0.001388 ¨© 2,470 482.5 ¸¹
2
255 kg s-1
0.5
3040 kg m-2 s-1
And the corresponding vent area is: 255 A 0.084 m2 3040 ______________________________________
52
7 EVAPORATION OF A LIQUID FROM A POOL 7.1 Evaporation of liquids On land or on water, a spilled liquid forms a pool. On water, the pool is unconfined, but on land its size and shape are often established by the existence of a retention dike. Immediately after the spill, the liquid starts to receive heat from the surroundings (the ground, the atmosphere, solar thermal radiation; see Fig. 2-13) and is vaporized. Initially, the vaporization process is usually controlled by heat transfer from the ground, especially in the case of boiling liquids. These two cases, boiling and non-boiling liquid pools, should be considered separately.
Fig. 2-13. Evaporation of a liquid from a pool.
7.2 Pool size
7.2.1 Pool on ground If the spill creates a pool on the ground and there is (as is often the case in industrial facilities) a containment barrier, the pool diameter is fixed. If the dike is right-angled, the equivalent diameter must be used: D
4
surface area of the pool
(2-51)
S
If there is no dike, the shape and size of the pool must be determined by the features of the release (continuous or instantaneous) and of the ground (slope). An equilibrium diameter can be reached as a function of the vaporization rate (see [9] for more detailed information). 7.2.2 Pool on water For a hydrocarbon spill on water (usually seawater), the expressions applied to pool fires (Chapter 3) (prior to ignition) can be applied. 7.3 Evaporation of boiling liquids When there is a spill of, for example, a pressurized liquefied gas, the liquid will be at its boiling temperature and at a temperature lower than that of the surroundings. Therefore, heat
53
will be transferred from the atmosphere by solar thermal radiation and, mostly in the first stage, from the ground. By analysing the heat conduction from the ground to the liquid, the following expression can be obtained [2, 21]: Q pool
k s Ts T pool
(2-52)
S Ds t
where Qpool is the heat flux reaching the pool from the ground (W m-2) ks is the thermal conductivity of the soil (see Table 2-4) (W m-1 K-1) Ts is the temperature of the soil (K) Tpool is the temperature of the liquid pool (K) Ds is the thermal diffusivity of the soil (m2 s-1) (see Table 2-4), and t is the time after the spill (s). Table 2-4 Values of thermal conductivity and thermal diffusivity of various solid media [18, 19] Substance ks, W m-1 K-1 Ds, m2 s-1 Average ground 0.9 4.3 · 10-7 Sandy ground (dry) 0.3 2.0 · 10-7 Sandy ground (8% water) 0.6 3.3 · 10-7 Wood 0.2 4.5 · 10-7 Gravel 2.5 11 · 10-7 Carbon steel 45.0 127 · 10-7 Concrete 1.1 10 · 10-7
If the entire heat flux is devoted to evaporating liquid, then the mass boiling rate is: m pool
Q pool A pool
(2-53)
'H v
where mpool is the mass boiling rate (kg s-1) A is the area of the pool (m2), and 'Hv is the latent heat of vaporization (expressed in J kg-1). From Eq. (2-52) it is clear that, as time passes and the ground becomes cooler, the heat flux to the pool decreases. After a certain time, the atmospheric and solar heat fluxes become more important and control the process [3]. 7.4 Evaporation of non-boiling liquids If the spilled liquid has a boiling temperature higher than the ambient temperature and is stored at a temperature lower than its boiling point (often at or near the ambient temperature), then the pool will not boil. Evaporation will occur essentially by vapour diffusion, the driving force being the difference between the vapour pressure of the liquid and the partial pressure of the liquid in the atmosphere (Pv – Pamb). The mass transfer process will be significantly influenced by the movement of air above the pool. The evaporation rate can be estimated with the following expression [22]:
54
G pool 2 10 3 u w0.78 r 0.11
M v P0 § Pv Pamb ln¨¨1 RT P0 Pv ©
· ¸¸ ¹
(2-54)
where Gpool is the evaporation rate (kg m-2 s-1) uw is the wind speed (m s-1) r is the radius of the circular pool (m) Mv is the molecular weight of the liquid (kg kmol-1) R is the ideal gas constant (expressed in J kmol-1 K-1) T is the temperature (K) Pv is the vapour pressure of the liquid at its temperature (Pa) Pamb is the partial pressure of the liquid in the atmosphere (Pa), and P0 is the atmospheric pressure (Pa). For Pv < 2·104 Pa, Eq. (2-54) may be simplified to: G pool 2 10 3 u w0.78 r 0.11
Mv Pv Pamb RT
(2-55)
If the pool is not circular but rectangular, r must be substituted in Eqs. (2-54) and (2-55) by L, the length of the side of the pool parallel to the wind (m). ______________________________________ Example 2-9 A broken pipe creates a pool of n-hexane with a diameter of 22 m. The ambient temperature is 20 ºC and the wind speed is 3 m s-1. Calculate the evaporation rate. Data: Mv = 86; Pv n-hexane, 20 ºC = 121 mm Hg. Solution The boiling temperature of n-hexane at atmospheric pressure is 68.7 ºC. Therefore, the pool will not boil. The partial pressure of n-hexane in the atmosphere is, of course, negligible. By applying Eq. (2-54): G pool 2 10 3 3 0.78 110.11
§ · 16,130 0 86 1.0132 10 5 ¸ 0.00224 kg m-2 s-1 ln¨¨1 3 5 8.314 10 273 20 © 1.0132 10 16,130 ¸¹
And, for the whole pool: m = 0.00224 kg m-2 s-1 · 380 m2 = 0.851 kg s-1 ______________________________________ 8 GENERAL OUTFLOW GUIDELINES FOR QUANTITATIVE RISK ANALYSIS
To predict the loss of containment for a given case and a given accidental scenario, the specific conditions corresponding to that situation and equipment must be taken into account and the appropriate hypothesis concerning the outflow —a hole, a broken pipe, etc.— must be considered. However, if a generic quantitative risk analysis must be performed over a given
55
installation, a set of hypotheses concerning the various loss-of-containment events is usually applied systematically in order to cover, with a conservative approach, the most common events that can significantly influence risk estimation. The “Purple Book” [9] offers a very interesting compilation of general guidelines covering such hypotheses. The following paragraphs contain a summary with the essential aspects of these guidelines. In quantitative risk analysis, the maximum duration of a release is usually 30 min. Usually, loss-of-containment events are included only if their frequency of occurrence is equal to or greater than 10-8 per year and lethal damage occurs outside the establishment. 8.1 Loss-of-containment events in pressurized tanks and vessels The following loss-of-containment events (LOCs) are usually considered in pressure, process and reactor vessels: Instantaneous release of the complete inventory. Continuous release of the complete inventory in 10 min at a constant rate of release. Continuous release from a hole with an effective diameter of 10 mm. If the discharge is from the liquid section of the vessel, pure liquid is released. Flashing in the hole is not modelled (flashing takes place outside the vessel). 8.2 Loss-of-containment events in atmospheric tanks In atmospheric storage tanks, pressure is atmospheric or slightly higher. The following LOCs are usually considered: Instantaneous release of the complete inventory: - Directly into the atmosphere. - From the tank into an unimpaired secondary container. Continuous release of the complete inventory in 10 min at a constant release rate: - Directly into the atmosphere. - From the tank into an unimpaired secondary container. Continuous release from a hole with a diameter of 10 mm: - Directly into the atmosphere. - From the tank into an unimpaired secondary container. 8.3 Loss-of-containment events in pipes Full-bore rupture (outflow from both sides). A leak with a diameter of 10% of the nominal diameter (with a maximum of 50 mm). For a full-bore rupture in a pipe, CD = 1.0. In other cases, CD = 0.62. Assume that the pipe has no bends and a wall roughness of approximately 45 Pm. 8.4 Loss-of-containment events in pumps Full-bore rupture of the largest connecting pipeline. A leak with a diameter of 10% of the nominal diameter of the largest connecting pipe (with a maximum of 50 mm). If no pump specifications are available, assume a release rate of 1.5 times the nominal pumping rate. 8.5 Loss-of-containment events in relief devices Discharge at the maximum discharge rate.
56
8.6 Loss-of-containment events for storage in warehouses Solids: dispersion of a fraction of the packaging unit inventory as respirable powder. Solids: spill of the complete packaging unit inventory. Emission of unburned toxics and toxics produced in the fire. 8.7 Loss-of-containment events in transport units in an establishment LOCs for road tankers and tank wagons in an establishment: - Instantaneous release of the complete inventory. - Continuous release from a hole (with the size of the largest connection). If the tank is partly filled with liquid, a liquid phase release from the largest liquid connection is assumed. - Full-bore rupture of the loading/unloading hose (outflow from both sides). - Leak in the loading/unloading hose (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). - Full-bore rupture of the loading/unloading arm (outflow from both sides). - Leak in the loading/unloading arm (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). - Fire under the tank (release from the connections under the tank followed by ignition or fire in the surroundings of the tank): instantaneous release of the complete inventory of the tank. Ships in an establishment: - Full-bore rupture of the loading/unloading arm (outflow from both sides). - Leak in the loading/unloading arm (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). - External impact, large spill: o Gas tanker: continuous release of 180 m3 in 1800 s. o Semi-gas tanker (refrigerated): continuous release of 126 m3 in 1800 s. o Single-walled liquid tanker: continuous release of 75 m3 in 1800 s. o Double-walled liquid tanker: continuous release of 75 m3 in 1800 s. - External impact, small spill: o Gas tanker: continuous release of 90 m3 in 1800 s. o Semi-gas tanker (refrigerated): continuous release of 32 m3 in 1800 s. o Single-walled liquid tanker: continuous release of 30 m3 in 1800 s. o Double-walled liquid tanker: continuous release of 20 m3 in 1800 s. 8.8 Pool evaporation If the liquid spill is contained in a bund, the spreading of the liquid will be limited: the maximum pool size will be the size of the bund and the pool size and shape will be defined. The effective pool radius can then be calculated from the bund area as follows: rpool
Abund
(2-56)
S
If there is no bund, the released liquid will spread onto the soil or the water surface; the pool diameter will increase up to the moment in which the evaporation rate from the pool equals the release rate; at this time, the pool will reach its maximum diameter. The thickness of the pool will be a function of the surface roughness. A minimum value of 5 mm can be assumed.
57
As for the maximum surface for unconfined pools, the following values have been proposed: 1,500 m2 (D = 44 m) for pools on land and 10,000 m2 (D = 113 m) for pools on water. 8.9 Outflow and atmospheric dispersion If the release mass flow rate changes with time, this variation should be taken into account in applying the atmospheric dispersion model. Therefore, the release must be divided into several discrete time segments, with constant outflow over each segment. A division into five segments is suitable for most cases. The following general rules can be applied [6]: - Calculate the mass released in the first 30 min (Mrel). - Calculate the mass released in each time segment, Msegment = Mrel/number of segments. - Calculate the duration of the first time segment, tsegment = time required to release Msegment. - Calculate the release rate in the first time segment = Msegment/tsegment. - Apply the same procedure to the other time segments. As for the cloud dispersion, each time segment is treated as an independent steady-state release neglecting the dispersion downwind until the total release can be considered instantaneous. NOMENCLATURE
Abund Aexp Aor Ap Asv At CD cp c pl
bund area (m2) vessel surface exposed to external fire (m2) cross-sectional area of the orifice (m2) cross-sectional area of the pipe (m2) minimum cross-sectional flow area of a safety valve (mm2) cross-sectional area of the tank (m2) discharge coefficient (-) specific heat at constant pressure (kJ kg-1 K-1) liquid specific heat at constant pressure (kJ kg-1 K-1)
cv dp dor F f Ff fF G2p Gsub hl 'Hv L Lj Ma Mv m msv N
specific heat at constant volume (kJ kg-1 K-1) pipe diameter (m) diameter of the orifice (m) reduction factor (-) vaporization fraction (-) friction loss term (J kg-1) Fanning friction factor (-) discharge mass flow rate (kg m-2 s-1) term accounting for subcooling (kg m-2 s-1) height of liquid above leak (m) latent heat of vaporization (kJ kg-1) equivalent length of a pipe plus pipe fittings (m) length of a free jet of gas (m) Mach number (= u/us)(-) molecular weight (kg kmole-1) mass flow rate (kg s-1) mass flow rate discharged from a safety valve (kg h-1) dimensionless parameter (-)
58
P Pback Pchoked Pcont Pdesign Pmax P0 Pp Pout Ps Pv 'P Q R Re rpool T Tchoked Tcont Tj Tp u uj us uw vlv W wil wl wv Z
Dw H ] J U Ul
20 m, E | Esoot | 40 kW m-2. Calculation of atmospheric transmissivity: from Eq. (3-19), Pwa = 2027 N m-2, therefore,
Pw
2027
70 100
1419 N m-2
By applying Eq. (3-17-b):
W
2.02 1419 25
0.09
0.79
The view factor can be obtained from Table 3-4; H/(D/2) = 2.3, l/(D/2) = 1.8; interpolating, Fv = 0.265. For a horizontal surface, from Table 3-5, Fh = 0.165. Therefore,
89
Fmax
0.265 2 0.165 2
0.312
By applying the solid flame model, I
0.79 0.312 40
9.8 kW m-2
b) With a wind speed of 6 m s-1, the flame drag will be significant; it can be estimated using Eq. (3-47): D'
§ 62 · ¸¸ 60 1.5 ¨¨ © 9.81 60 ¹
0.069
75.5 m
D’ – D = 15.5 m > 14 m, the valves will be engulfed by the fire. ______________________________________ 6.2 Size of a jet fire Jet or flare fires are characterized by highly turbulent diffusion flames. They can occur due to the accidental release of a fuel gas —for example, through a broken pipe or a flange, or from a relief valve— or in process or emergency flaring. Accidental jet fires have occurred in many parts of process plants or in transportation accidents and often impinge on equipment; in this case, large heat fluxes occur due to the high convective heat transfer caused by the relatively good combustion and the high flow velocities. A number of BLEVEs or similar explosions have been caused by jet fires. Flares also release large amounts of radiant energy, although they are located in high stacks to assure safe operation. In both cases, the prediction of the jet fire size and of the thermal flux as a function of distance is required in order to determine the effects of a jet fire and to establish safety distances.
6.2.1 Jet flow In an accidental release, the sonic velocity (velocity of sound in the gas in exit gas conditions) is reached if the following relationship is fulfilled: J
P0 ª 2 º J 1 d« Pcont ¬ J 1»¼
(3-48)
where P0 is the atmospheric pressure (N m-2) and Pcont is the pressure inside the container or the pipe (N m-2). The sonic velocity is the maximum possible velocity in an accidental release. It is also called choked velocity. Once the speed of sound has been reached, further increases in Pcont will not produce any further increase in the gas exit velocity. However, as the density of gas increases with pressure, the mass flow rate will increase linearly with pressure. For most gases, the sonic velocity is reached if the pressure at the source is greater than 1.7-1.9 bar. This is usually the situation in accidental releases. The following expressions concerning the gas jet are of interest. The speed of sound in a given gas at a temperature T is:
90
J T R 10 3
us
(3-49)
Mv
where us is expressed in m s-1 and R is the ideal gas constant (8.314 kJ kmol-1 K-1). The temperature of the gas in the expanding jet at the orifice outlet is:
Tj
§ P Tcont ¨¨ 0 © Pcont
§ J 1 · ¨ ¸ J ¸¹
· ¨© ¸¸ ¹
(3-50)
where P0 is the atmospheric pressure (N m-2) Pcont is the initial pressure of the gas in the container or pipe (N m-2) and Tcont is the temperature in the container or pipe (K). The effective orifice diameter, ds, is the diameter of an imaginary orifice that would release air with a density Ua at the same flow rate at which gas is being emitted. It can be calculated with the following expression: 4m' S U au j
ds
(3-51-a)
where m’ is the mass flow rate of gas (kg s-1) and uj is the velocity in the expanding jet at the gas outlet (m s-1). ds can also be calculated as follows: ds
d or
Uj Ua
(3-52-a)
where Uair is the density of ambient air (kg m-3) and Uj is the density of the gas at the outlet. For unchoked flow, Uj = U g0 (273/Tj); for choked flow, the jet expands to atmospheric pressure downstream of the exit hole. Then: ds
dj
Uj Ua
(3-52-b)
where dj
4m
(3-53)
S uj Uj
where Uj is the density of the gas in the expanded jet (kg m-3). uj can be calculated as follows:
91
uj
Mj
J R Tj
(3-54)
Mv J 1
Mj
§P · J J 1 ¨¨ or ¸¸ 2 © P0 ¹ J 1
(3-55)
where Mj is the Mach number for choked flow of an expanding jet (-) P0 is the atmospheric pressure (N m-2) and Por is the static pressure at the orifice exit plane (N m-2): J
Por
Pcont
§ 2 · J 1 ¨¨ ¸¸ © J 1¹
(3-56)
6.2.2 Shape and size of the jet fire There are various sets of equations proposed by different authors for predicting the shape and size of a jet fire, with significant scattering in the results. Two classical treatments, for calm situations and the presence of wind, respectively, have been selected here. In a calm wind situation, the length of the flames in a jet fire can be estimated in a simple way using the expression proposed by Hawthorne et al. [34]: L d or
5.3 ª Tad « c st vol ¬D st Tcont
§ M ¨¨ c st 1 c st a Mv ©
·º ¸¸» ¹¼
1/ 2
(3-57)
where L is the length of the visible flame, from the lift-off distance to the tip (m) cst-vol is the mole fraction of fuel in the stoichiometric fuel-air mixture (-) Tad is the adiabatic flame temperature (K) Ma is the molecular weight of air (kg kmole-1) and Dst is the ratio of the number of moles of reactants to moles of product for a stoichiometric fuel-air mixture (-). Eq. (3.57) can be simplified for hydrocarbon gases to [6] [34]: L d or
15 § M a ¨ c st vol ¨© M v
· ¸¸ ¹
1/ 2
(3-58)
The lift-off distance s can be estimated using the following expression [35]: s
6.4 S d or u j
(3-59)
4 u av
where dor is the diameter of the orifice (m), and uav is the average jet velocity (m s-1) (uav | 0.4 uj).
92
Finally, the diameter of the jet fire can be estimated as a function of its length using the following expression: Dj
ª L sº 0.29 x «ln x »¼ ¬
1/ 2
(3-60)
where x is the axial distance from the orifice (m), and s is the lift-off distance (m). ____________________________________ Example 3-6 A cylindrical tank containing butane has been heated to 51 ºC. Gas is vented upwards from a release device (outlet internal diameter: 0.025 m) located on the top of the tank, 4 m above ground (Fig. 3-11). There is no wind. Estimate the maximum thermal radiation on the wall of a tank located at a horizontal distance of 9 m from the jet axis, at a height of 4.5 m above the ground. 'Hc = 45700 kJ kg-1. J = 1.11. Constants in the Antoine equation for butane: A = 4.35576, B = 1175.58, C = -2.071. Ambient temperature = 18 ºC. Relative humidity = 50%.
Fig. 3-11. Jet fire in a calm situation.
Solution The combustion reaction is: C4H10 +
c st vol
13 O2 o 4CO2 + 5H2O 2
1 13 1 1 2 0.21
0.0313
93
Estimation of the length of the flame using Eq. (3-58): L 0.025
15 § 29 · ¨ ¸ 0.0313 © 58 ¹
1/ 2
8 .4 m
Estimation of the lift-off distance using Eq. (3-59): 6.4 S 0.025 u j
s
4 0 .4 u j
0 .3 m
Pressure inside the vessel: log P
4.35576
1175.58 ; P = 5 bar. 324 2.071
Calculation of the mass flow rate of fuel using Eq. (2-19): 1.111
m
'
0.025 2 58 § 2 · 1.111 S 0.62 5 10 5 1.11¨ ¸ 4 1 . 11 1 324 8 . 314 10 3 © ¹
0.447 kg s-1
For butane jet fires, Brzustowski [35] obtained the following value for the radiant heat fraction: Krad = 0.3. If the jet fire is assumed to be a cylinder, from Eq. (3-60) an average diameter D | 1 m is obtained. Estimation of the average emissive power using Eq. (3-27): E
0.3 0.447 45,700 12 S 1 8.4 2 S 4
215 kW m-2
Estimation of the view factor from Table (3-4): Fv = 0.0238. For a relative humidity of 50% and l = 9 m, W = 0.88. Therefore, the thermal radiation intensity (Eq. (3-20)) is:
I 0.0238 215 0.89 4.5 kW m-2 ______________________________________ 6.2.3 Influence of wind The wind can have a significant influence on the jet fire. The model proposed by Chamberlain [36, 37], relatively complex, describing the jet flames by the frustrum of a cone (Fig. 3-12) has been selected here. First of all, the auxiliary parameter Y must be calculated by iteration: §g d 0.024 ¨ 2 s ¨ u © j
· ¸ ¸ ¹
1/ 3
Y 5 / 3 0 .2 Y 2 / 3 c c
0
(3-59)
94
For parafins, cc = (2.85/cst-mass)2/3 where cst-mass is the stoichiometric mass fraction of fuel (-). In still air, the length of the flame measured from the centre of the exit orifice to the tip of the flame can be calculated with the following expression: Lb0
Y ds
(3-60)
where Lbo is expressed in m. W 2
uw L
Lb Lbv
D
W
Db
1
S
dor
Fig. 3-12. Influence of wind on a jet fire [36, 39].
Under the influence of wind:
Lb
Lb0 0.51 e 0.4u w 0.49 1 6.07 10 3 T jv 90
(3-61)
where Tjv is the angle between the hole axis and the wind vector (º). The lift-off distance is: s
Lb
sin 0.185 e 20 Rw 0.015 D sin D
(3-62-a)
where Rw is the ratio of wind speed to jet velocity: Rw = uw/uj. In still air, s
0.2 Lb
(3-62-b)
The length of the flames (length of frustrum) is:
95
L
L2b s 2 sin 2 D s cos D
(3-63)
If Rw d 0.05, the tilt angle can be calculated as follows:
D
T
jv
90 1 e 25.6 Rw 8000
Rw Ri Lbo
(3-64-a)
and if Rw > 0.05,
D
T
jv
134 1726Ri R
90 1 e 25.6 Rw
0.026
1/ 2
w
(3-64-b)
Lbo
where RiLbo is the Richardson number based on Lbo,
§ g Lb0 ¨ 2 2 ¨d u © s j
Ri Lbo
· ¸ ¸ ¹
1/ 3
and cos T jv
cos : cos T j
: is the angle between the wind direction and the normal perpendicular to the pipe in the horizontal plane; Tj is the angle between the hole axis and the horizontal in the vertical plane. Finally, the width of frustrum (base and tip, respectively)can be calculated with the following expressions:
6 Rw
ª ª §U 1.5 « 1 « 1 ¨ air ¨ U « « © j ¬ ¬
· ¸ ¸ ¹
1/ 2
W1
d s 13.5 e
W2
Lb 0.18 e 1.5 Rw 0.31 1 0.47 e 25 Rw
º 1 º» 70 Rids C ' Rw » e » 15 » ¼ ¼
(3-65)
(3-66)
where Rids is the Richardson number based on the source diameter and C’ is a function of Rw:
Rids C'
§ g ds ¨ 2 2 ¨d u © s j
· ¸ ¸ ¹
1/ 3
1000 e 100 Rw 0.8
The diameter of the jet fire can be estimated as a function of its length using Eq. (3-58). The surface of the flames (A) can also be approximated by considering a cylinder with an average width:
96
A
S § W1 W2 · ¨ 2©
2
2
§ W W2 · ¸ LS ¨ 1 ¸ 2 ¹ © ¹
(3-67)
The value of E can be estimated using Eq. (3-27), with [36]
K rad
0.21e
0.00323u j
0.11
(3-68)
For relatively large distances, for example in the case of flares, the point source model can be applied. ______________________________________ Example 3-7 Based on Example 3-6, with a wind speed of 6 m s-1. Determine the size and shape of the jet fire. Estimate the radiation (point source model) on a target located at ground level at 15 m downwind. Solution Calculation of the diverse jet parameters: 1.09
Por
§ 2 · 1.091 5¨ ¸ © 1.09 1 ¹
Tj
§ 1.01 · © 324 ¨ ¸ © 5 ¹
Uj
2 .6
§ 1.09 1 · ¸ ¨ 1.09 ¹
273 284
2.9 bar
284 K
2.5 kg m-3 1.09 1
Mj
uj
dj
ds
1.09 1.09 1 §¨ 2.9 ·¸ 2 1 . 01 © ¹ 1.09 1
1.76
1.09 8314 284 58.1
4 0.447
S 370 2.6 0.0245
2.6 1.2
1.76
370 m s-1
0.0245 m
0.036 m
Calculation of Y by trial and error:
97
58.1 13 29 58.1 2 0.21
c st vol
§ 9.81 0.036 · 0.024 ¨ ¸ 2 © 370 ¹
Y
2/3
0
295 0.036 10.6 m
10.6 0.51 e 0.4 6 0.49 1 6.07 10 3 90 90 5.67 m
Lb Ri Lbo
D
§ 2.85 · Y 5 / 3 0 .2 Y 2 / 3 ¨ ¸ © 0.067 ¹
295
Lb 0
Rw
1/ 3
0.0607
§ 9.81 10.6 ¨¨ 2 2 © 0.036 370 6 370
8000
· ¸¸ ¹
1/ 3
4.04
0.0162 0.0162 4.04
32º
Lift-off distance: s
5.67
sin 0.185 e 20 0.0162 0.015 32 sin 32
0.88 m
Flame length: L
5.67 2 12 sin 2 32 1 cos 32
4.8 m
Width of frustrum base and tip: Ri ds
§ 9.81 0.036 ¨¨ 2 2 © 0.036 370
· ¸¸ ¹
1/ 3
0.0137
C'
1000 e 100 0.0137 0.8
W1
1/ 2 ª ª 1 º 70 0.0137 254.5 0.0162 º § 1.2 · 0.036 13.5 e 6 0.0162 1.5 « 1 « 1 ¨ » ¸ »e © 2.6 ¹ 15 ¼» ¬« ¬« ¼»
254.5
98
0.5 m
W2
5.67 0.18 e 1.5 0.0162 0.31 1 0.47 e 25 0.0162 1.9 m
Taking into account the flames tilt and size, the distance from the centre of the flames to the target is 15.4 m. From Eqs. (3-17-b) and (3-19), W = 0.84. Therefore, the intensity of thermal radiation at the target is: 0.3 0.447 45700 0.84 1.7 kW m-2. 4 S 15.4 2 ______________________________________ I
6.3 Flash fire In a flash fire two completely different situations must be considered in terms of the thermal flux: the targets that are engulfed by the fire and those that are beyond the area covered by the flames. For natural gas, heat fluxes of 160-300 kW m-2 were measured within the fire contour [40], while outside it —although relatively near the contour— the heat flux was approximately 5 kW m-2. For natural gas and propane, the average flame speed measured was in the range of 12 m s-1, although transient values of up to 30 m s-1 were detected. For these substances, an average surface emissive power of 173 kW m-2 was recorded. The size and position of a flammable cloud (i.e. the volume of the cloud that is within the flammability limits) can be predicted for a given case by applying atmospheric dispersion models. However, from the point of view of the mathematical modelling of fire features, flash fires are practically unknown. The only method for estimating the size of the flames in a flash fire was proposed by Raj and Emmons [40], who gave the following semiempirical expression for the visible flame height:
H
ª S2 § U f a ¨¨ 20 h « «¬ g h © U a
2 · w r2 º ¸¸ » 3 ¹ 1 w »¼
1/ 3
(3-70)
where h is the cloud height (m) S is the flame speed (m s-1) Uf-a is the density of the fuel-air mixture (kg m-3) Ua is the density of air (kg m-3) r is the stoichiometric air-fuel mass ratio (-), and
I I st for I ! I st D ' 1 I st w 0 for I d I st w
(3-71)
where D’ is the constant pressure expansion ratio for stoichiometric combustion (typically 8 for hydrocarbons) Iis the fuel volume ratio in the fuel-air mixture and Ist is the stoichiometric fuel volume ratio.
99
Predicting the thermal radiation from a flash fire requires a series of simplified assumptions to be made: the composition of the cloud is fixed and homogeneous, and the surface of the flame is similar to a vertical plane moving through the stationary cloud. Overall, the prediction will be only a rough approximation. However, as stated before, the consequences of a flash fire are very serious inside the flame contour, while outside it they are far less severe and often negligible. 7 BOILOVER
Fires in large fuel storage tanks are relatively frequent. These large fires are comparatively difficult to extinguish, requiring large amounts of water/foam, and often last several hours. In this case, a particular phenomenon can occur during the fire, which increases the size of the flames and the area covered by the thermal radiation and thus increasing potential serious consequences of the event. This phenomenon is known as a boilover. A boilover can occur essentially in tanks containing mixtures of different hydrocarbons with a wide range of boiling temperatures, for example with crude oil. A typical scenario would consist in an initial explosion blowing out the tank roof, followed by a fire. During the fire, in the top boiling fuel layer the most volatile compounds are preferentially vaporized. There is in fact a distillation process, and this layer is progressively enriched in the heaviest (higher boiling point) components: its temperature also increases progressively. As the fire burns, the thickness of this layer —rich in high boiling temperature components— increases and progresses in depth. The expansion of the hot zone is caused by convective motions induced by vapour bubble formation during the vaporization of the lighter fuel components [41].The speed at which the thickness of the layer increases is greater than the speed at which the surface of the fuel descends. Thus, a heat wave propagates towards the bottom of the tank. If the tank contains a water layer that is denser than the fuel, or an oilwater emulsion layer suspended in the fuel, at a certain point the heat wave (at a temperature higher than the boiling temperature of water) will reach this aqueous layer. This will cause the initial vaporization of some of the water. The turbulence of the phenomenon will cause both layers to mix, causing the extensive vaporization of water. The practically instantaneous generation of a large amount of steam —with a specific volume 1600 times that of liquid water— will cause a violent eruption, ejecting ignited fuel out of the tank and dramatically increasing the size of the flames (Fig. 3-13). The boiling temperature of the water layer will be higher than 100 ºC due to the hydrostatic pressure. For a high tank, this temperature may reach 120 ºC. In a boilover, the upper fuel layer may reach temperatures of up to 430 K. The speed at which the heat wave progresses usually ranges between 0.3 m h-1 and 1 m h-1, although in some cases it has reached values of 1.2 m h-1. The presence of water inside a fuel tank may be explained in several ways: a) it may enter with the fuel; b) it may be rain water, which enters through holes in the roof; c) it may be due to atmospheric humidity, as a result of the tank breathing and condensing the humidity inside; and d) it may be water used by firemen to extinguish the fire. There are also two special cases associated with this situation, known as frothover and slopover. Frothover occurs when the vaporization of water is smoother and the steam bubbles cause the tank to overflow due to the continuous frothing of burning fuel. Slopover can occur when water is applied to the burning surface of the fuel and sinks into the hot oil. The vaporization of the water causes the ignited fuel to overflow.
100
Fig. 3-13. Boilover in a fuel tank.
In special circumstances, a boilover could also occur without the presence of water, although this is fairly infrequent. In this case, the density of the upper hot layer increases with the distillation process. The temperature of the lower fuel layers increases slowly due to heat conduction, while at the same time its density decreases. The upper layer has a high content of components with high boiling points, while the lower layer is still rich in volatile components. At a certain point, this situation —the existence of a dense, very hot layer above another with a lower density— can lead to the turbulent mixing of the two layers, resulting in the instantaneous vaporization of the volatile liquid and the occurrence of boilover. The phenomenon described in the previous paragraphs corresponds to the so-called hot zone boilover. A somewhat different phenomenon is the thin layer boilover, which occurs when a thin layer of fuel burns over a layer of water. This can happen when there is a spillage of fuel on the ground. If the fuel is ignited, after a short time (about one minute) the water starts to boil and the bubbles eject fuel upwards, significantly increasing the size of the flames. This phenomenon is characterized by a strong crackling sound. An important aspect when dealing with boilover is the ability to predict the moment at which it will occur. The highest value (tboilover) can be estimated from a simple heat balance of
101
the mass of fuel contained in the tank: the fuel is heated by the fire from its surface until all of the fuel has reached the heat wave temperature:
U l c p hHC Thw Ta
t boilover
(3-72)
Q f m 'hv c p T0 av Ta
where Ul is the density of fuel at Ta (kg m-3) cp is the specific heat of fuel at Ta (kJ kg-1 K-1) hHC is the initial height of fuel in the tank before the fire starts (m) m is the burning rate (kg m-2 s-1) 'hv is the vaporization heat at T0av (kJ kg-1) Ta is the ambient temperature (K) Thw is the temperature of the heat wave when the boilover occurs (K) T0av is the average boiling temperature of the fuel (K), and Qf is the thermal flow entering the fuel from its surface ( | 60 kW m-2). Thw can be estimated from the distillation curve of the fuel by an iterative procedure. However, Eq. (3-72) assumes that the layer of water is at the bottom of the tank, and does not take into account the possibility of a layer of water-hydrocarbon emulsion in a higher position, which would shorten the time needed for boilover to occur. In the case of a tank fire, the progression of the heat wave can be followed if vertical bands of intumescent paint have been applied to the tank wall. Another possibility is to throw water against the tank wall and observe its behaviour (i.e. whether it boils or not). Recently, the use of thermographic (IR) cameras has been proposed. However, although both procedures may indicate the progression of the heat wave, the possible presence of water at a certain height inside the tank again creates uncertainty about the moment at which the boilover may occur. From Eq. (3-72), a theoretical expression for the speed at which the heat wave progresses can be obtained: Q f m 'hv c p T0 av Ta
u wave
(3-73)
U l c p Thw Ta
7.1 Tendency of hydrocarbons to boilover For a hot zone boilover to occur, the following conditions are required: the presence of water inside the tank; the existence of a mixture of components with a wide range of boiling temperatures; a fuel with a relatively high viscosity. The conditions for a boilover were quantified by Michaelis et al. [42] as follows. The average boiling temperature of the fuel (T0av) must be higher than that of water at the pressure at the water-fuel interface. T0av can be determined using the following expression:
T0av
T
0 min
T0max
1/ 2
(3-74)
The pressure at the water-fuel interface is: Pinterface
P0 hHC U l g
(3-75)
For common fuel storage conditions, this criterion generally reduces to:
102
T0av ! 393K
The range of boiling temperatures in the fuel must be sufficiently wide to generate the heat wave. 'T0 =T0max - T0min must be higher than 60 ºC if T0min is higher than the boiling temperature of water at the pressure at the water-fuel interface (a temperature of 393 K is assumed); if T0min < 393 K, then (393 - T0max) should be greater than 60 ºC [43]. Finally, these authors state that the kinematic viscosity of the fuel must be higher than that of kerosene at the boiling temperature of water at the water-fuel interface, QHC t 0.73 cSt. These three criteria were combined in an empirical parameter called the factor of propensity to boilover [42], which indicates the tendency of a hydrocarbon to generate a boilover during a fire: PBO
§ 393 ·§ 'T0 · 2 § Q HC ·1 / 3 ¸ ¨1 ¨ T0 ¸¨© 60 ¸¹ ¨© 0.73 ¸¹ av ¹ ©
(3-76)
According to this criterion, hydrocarbons with a value of PBO t 0.6 could generate a boilover. However, this expression should be applied with caution. 7.2 Boilover effects The effects of the boilover are essentially the generation of a fireball, which is the most serious effect, and, to a minor extent, the ejection of ignited fuel around the tank. If a fireball is created, the value of E for liquid hydrocarbons can be assumed to be approximately 150 kW m-2. The heat radiation over a given target can be calculated using the solid flame model (see the next section in this chapter). For the threshold values of 1000 (kW·m-2)4/3s for 1% lethality and 600 (kW·m-2)4/3s for irreversible consequences (serious burns), applying the conservative assumption that the entire contents of the tank at the moment of the boilover participate in the fireball, INERIS [44] proposed the following expression for calculating the distances corresponding to lethality and irreversible consequences: d lethality
d irreversible
k lethalityW 0.45
(3-77-a)
k irreversibleW 0.45
(3-77-b)
where W is the mass of fuel in the tank at the beginning of the fire. The exponent 0.45 is in fact an average value and the other constants depend on the type of fuel (Table 3-9). Table 3-9 Values of constants in Eqs. (3-68) and (3-69) [44] Fuel klethality kirreversible Fuel oil N. 2 0.420 0.573 Kerosene 0.387 0.525 Domestic fuel 0.317 0.439 Diesel oil 0.319 0.439 Crude oil 0.267 0.363
103
8 FIREBALL
When a BLEVE explosion involves a flammable substance, it is usually followed by a fireball, which releases intense thermal radiation. Fireballs can also occur during boilover. The thermal energy is released rapidly, which is a function of the mass in the tank. The phenomenon is characterized from the beginning by strong radiation, eliminating the possibility of escape for individuals nearby (who will also have suffered the effects of the blast). To estimate the radiation received by a surface located at a given distance, the solid flame model can be applied (Eq. (3-20). It is necessary, therefore, to know the value of the emissive power (E), the view factor (F), the atmospheric transmissivity Wand the distance between the flame and the target. To know this distance, it is necessary to estimate the diameter of the fireball as well as the height at which its centre is located. The shape of the fireball can vary according to the type of tank failure. Rapid failures produce approximately spherical fireballs, whilst slower BLEVEs tend to produce cylindrical fireballs with high lift-offs. However, to estimate their effects, a spherical shape is usually assumed. The parameters that must be evaluated to predict the effects of a fireball are the diameter, the duration and the height at which the fireball is located; this will allow the thermal radiation to be estimated at any given distance. In this section, a methodology is described with which to estimate these values. 8.1 Fireball geometry
8.1.1 Ground diameter The zone on the ground that can be engulfed by flames during the initial development of the fireball can be approximated by the following expression [45]: D groundflas h
1.3 Dmax
(3-78)
where Dmax is the maximum diameter achieved by the fireball (see Eq. (3-83)). 8.1.2 Fireball duration and diameter Various authors have proposed correlations for the prediction of the diameter and duration of a fireball generated by a given mass M of fuel [40]. Most of them have the following general expression: D t
aMb
(3-79)
cMe
(3-80)
where a, b, c and e are empirical or semi-empirical constants. A comparative study of 16 of these expressions was made by Satyanarayana et al. [46]. Although it is rather difficult to establish which is really the best equation, due to the lack of experimental data at large scale, taking into account this and other studies, the duration and diameter of the fireball can be estimated using the following expressions.
104
For the duration (time): t
0.9 M 0.25
(3-81)
where the units are kg (M) and s (t). According to a model proposed by Martinsen and Marx [47, 48], the fireball main features (D, H, E) change as a function of time. The fireball is considered to reach its maximum diameter during the first third of the fireball duration. Thus, while the fireball is growing the following equation applies:
D Dmax
8.664 M 0.25 t i
1/ 3
for 0 d t i d t/3
(3-82)
5.8 M 1 / 3 for t/3 t i d t
(3-83)
where Dmax is the maximum value achieved by D and ti is the time at any instant i. It is worth noting, however, that there is very little experimental data available to support this type of comparative analysis. Furthermore, these data —obtained from real accidents in the case of large fireballs— are not always accurate, as often the films are incomplete or of poor quality. In fact, the lack of accuracy is not only due to differences in the predictions arising from diverse correlations. Another factor influencing it is the estimation of the fraction of the overall mass of fuel that is really involved in the fireball. As happens in many cases of risk analysis, the inaccuracy arises from the definition of the problem itself. It should be taken into account that some fuel has been leaving the vessel through the safety valves from the moment at which they opened; the amount released will depend on the time elapsed between this moment and that of the explosion. Furthermore, more fuel is sucked into the wake of the propelled fragments. Consequently, it is essentially impossible to accurately establish the mass of fuel that will contribute to the fireball. If more accurate information is not available, 90% of the maximum capacity of the vessel should be used. The fact that in Eqs. (3-82) and (3-83) the mass of fuel is affected by an exponent equal to approximately 1/3 considerably reduces its influence on the value of D. Finally, the lack of accuracy is also due to the fire wake left by the fireball, the size of which can be significant [49]; this modifies the flame surface and, consequently, the radiation that will reach a given point. The correlations mentioned in the previous paragraphs nevertheless allow an estimation of the size of the fireball. It should be taken into account that, as its size and position change continuously, the thermal radiation is not constant. The available films of BLEVE accidents show that the fireball grows quickly to its maximum diameter, remaining at this diameter for a short time and then dissipating. Sometimes, calculation of the radiation received by a given target is performed by supposing that the fireball achieves its maximum size immediately after reaching a certain height. Some guidelines suggest calculating the hazard distances for the fireball located just over the ground (i. e., H = D) [50]. 8.1.3 Height reached by the centre of the fireball This height is a function of the specific volume and the latent heat of vaporization of the fuel; therefore, strictly speaking, it varies with the substance. This is not usually taken into account. Usually, the fireball rises at a constant rate from its lift-off position to three times
105
this height during the last two-thirds of its duration. The following equations [47] can be used to estimate this height: H
0.5 D for 0 d t i d t/ 3
(3-84)
H
3Dmax t i 2t
(3-85)
for t/ 3 t i d t
where H is the height at which the centre of the fireball is located (m). If an average value must be taken, the following expression can be used:
H
0.75 D
(3-86)
The values obtained with these expressions have been compared with those corresponding to four real cases (Table 3-10). The heights correspond to the top of the fireball (h = H+D/2). The diameter has been calculated with Eqs. (3-82) and (3-83). The height has been calculated with Eq. (3-85) and (3-86); with Eq. (3-85) a value of ti = 2t/3 has been taken: note that if ti = t/2 is assumed, the results are the same as those obtained with Eq. (3-86). Table 3-10 Predicting the height (top of the flame) of the fireball Accident Fuel M, kg H, m (observed) Crescent City Priolo Priolo Paese
Propane Ethylene Propylene Propane
35,000 80,000 50,000 800
230 225 250 95
H, m [Eq. (3-86)] 237 312 267 67
H, m [Eq. (3-85)] (at ti = 2t/3) 284 375 320 81
The results from both expressions are relatively good, taking into account the accuracy of the data; therefore, Eqs. (3-84) and (3-85) can be used if a variable height is assumed, and Eq. (3-86) can be used if a constant value is assumed. 8.2 Thermal features
8.2.1 Radiant heat fraction Once more, we do not know for certain what fraction of the energy released is emitted as thermal radiation. In fact, this is one of the most important uncertainties in the calculation of the thermal radiation from a fireball. The following correlation has been proposed [51] to estimate this value:
K rad
0.00325 P 0.32
(3-87)
where P is the pressure in the vessel just before the explosion, in N·m-2 Typically, this pressure can be supposed to be the relief pressure (when calculated for fire). The value of K rad usually ranges between 0.2 and 0.4, its maximum value being limited to 0.4. From this radiation coefficient and the heat released from the fireball, the radiated energy can be
106
deduced. If P is not known, a thumb rule value of K rad = 1/3 can be assumed according to some well known guidelines for offsite consequence analysis [52].
D/2
d
x
Fig. 3-14. Position of fireball and target.
8.2.2 Emissive power An average value of the emissive power can be calculated as the radiant heat emitted divided by the surface of the fireball:
E
K rad M 'H c S D2 t
(3-88)
where t is the time corresponding to the duration of the fireball (s) M is the molecular weight of the fuel, and 'Hc is the heat of combustion (lower value) of the fuel (kJ kg-1). Experimental work shows that the emissive power varies with time, reaching a maximum very quickly at the end of the fireball expansion and then decreasing slowly until extinction. Again, the value of E can be calculated [47] separately for the fireball growth phase and for the last two thirds of the duration. For the growth phase, Emax, this expression can be used: E max
0.0133 K rad 'H c M 1 / 12 for 0 d t i d t/3
(3-89)
If Eq. (3-89) gives a value higher than 400 kW·m-2, then 400 kW·m-2 must be taken. During the last two-thirds of the duration of the fireball, E can be calculated using this expression:
E
ª 3 § t ·º E max « ¨1 i ¸» for t/3 t i d t t ¹¼ ¬2 ©
(3-90)
The average value of E ranges between 200 and 350 kW·m-2 [53] and for an LPG fireball usually ranges between 250 and 400 kW·m-2.
107
8.2.3 View factor The maximum view factor is that corresponding to a sphere and a surface perpendicular to its radius. Due to the geometrical simplicity of this system, this factor can be calculated using a very simple equation:
4 S ( D 2 / 4)
Fmax
ªD º 4S « d » 2 ¬ ¼
2
D2 ªD º 4 « d» 2 ¬ ¼
2
(3-91)
where (D/2+d) is the distance between the surface receiving the radiation and the centre of the fireball (Fig. 3-14). For a given source, this is the situation corresponding to the maximum radiation intensity on the target. For other positions of the surface of the target, the value of F must be corrected by using the angle formed between this surface and the surface perpendicular to the radius of the fireball. Fig. 3-14 clearly shows that a given radiation bunch falls over a surface whose area varies with its inclination, the minimum area (and maximum radiation flux) corresponding to a surface perpendicular to the radiation. Thus,
Fvertical
F max cosD
Fhorizontal
(3-92-a)
F max sin D
(3-92-b)
8.3 Constant or variable D, H and E Fireball height, diameter and emissive power can thus be calculated using Eqs. (3-86), (385) and (3-88), assuming that they are constant with time, or, when it is assumed that they are not constant, using Eqs. (3-84), (3-85), (3-82), (3-83), (3-89) and (3-90) with different values for the growth phase and for the last two-thirds of the duration, respectively. Fig. 3-15 shows how these variables change as a function of time (variable D, H and E model), for a given case (100,000 kg of propane; see Example 3-8). The trend of these three variables is quite different during the first five seconds and during the rest of the duration of the fireball. For accurate results, this alternative is probably better. For a rapid, estimative calculation, constant values of D, H and E can be taken; in this way, more conservative results are obtained. Fig. 3-16 shows the variation in the thermal radiation reaching a vertical surface located at a certain distance as a function of time, for a given case (see Example 3-8). It can be observed that in the variable D model, Iv increases quickly with time, reaching a sharp maximum at ti = t/3, subsequently decreasing abruptly. In the constant D model, the thermal radiation intensity has a constant value. This different behaviour has some influence on the calculation of the dose received by a target located at a given distance, dose
t I 4/3
(3-93)
If the dose is calculated using the two methods (see Fig. 3-17), more conservative values are obtained for the constant D model, although higher values of the thermal radiation intensity are calculated for he variable D model.
108
Fig. 3-15. Emissive power, fireball height and fireball diameter as a function of time, for a BLEVE of 100,000 kg of propane (see Example 3-8).
Recently [54], it has been emphasized that a good estimation of D, Ep and W is very important for the calculation of thermal radiation from a fireball, whilst the influence of H is not so important. 9 EXAMPLE CASE
______________________________________ Example 3-8 3 A tank with a volume of 250 m , 80% filled with propane (stored as a pressurized liquid at room temperature), is heated by a fire to 55 ºC (~19 bar) and bursts. The thermal radiation, as well as the consequences on people, must be estimated at a distance of 180 m from the initial location of the tank. Data: Room temperature = 20°C; HR = 50% (partial pressure of water vapour, 1155 Pa); J =1.14; Hc -1
-3
= 46,000 kJ·kg ; Tc = 369.8 K; Tboil. atm. pres. = 231.1 K; Uliquid, 20 °C = 500 kg·m , Uliquid, 55 °C = -3 3 -1 -1 444 kg·m-3; U = 37 kg·m ; cp = 2.4·10 J · kg ·K . vapour, 55 °C
liquid
Solution: First of all, the mass of propane involved is calculated:
M
Vl U l , 20ºC
(0.8 250 m 3 ) 500 kg m -3
100,000 kg
The thermal radiation will be calculated using the two models discussed previously. Estimation of thermal radiation (D, H and E constant) By using Eq. (3-83), the fireball diameter is estimated:
109
D
5 .8 M 1 / 3
5.8 100000 1 / 3
269 m
Its duration is estimated with Eq. (3-81): t
0.9 M 0.25
0.9 100000 0.25
16 s
and the height reached by the fireball is estimated by Eq. (3-86):
H
0.75 D
0.75 269
202 m
The distance between the flame and the target, according to Fig. 3-14, can be calculated as follows: d
H 2 x2
D 2
202 2 180 2 134.5 136 m
The atmospheric transmissivity will be:
W
2.85 (1155 136) 0.12
0.68
The view factor is calculated with Eq. (3-91): F
D2 · §D 4¨ d ¸ ¹ ©2
269 2 2
· § 269 4¨ 136 ¸ ¹ © 2
2
0.25
The fraction of heat radiated is:
K rad
0.00325 (1.9 10 6 ) 0.32
0.33
The emissive power is (Eq. (3-88)): E
0.33 100,000 46,000 S 269 2 16
417 kW m -2
A value of E = 400 kW m-2 will be taken. The radiation intensity on a surface perpendicular to the radiation will be:
I
W F Ep
0.68 0.25 417
70.9 kW m -2
on a vertical surface, Iv
I cos D
70.9 0.67
47.5 kW m -2
and on a horizontal surface,
110
Ih
I sin D
70.9 0.75
53.2 kW m -2
The dose received by a person exposed to a radiation intensity Iv for the entire duration of the fireball is: dose 16 47.5 10 3
4/3
2.8 10 7 s ( W m -2 ) 4 / 3
Estimation of thermal radiation (variable D, H and E) Growth phase (first 5.3 s; see Fig. 3-15): The diameter increases up to Dmax = 269 m. H increases up to 135 m. The emissive power has a constant value of Emax = 400 kW·m-2 (Eq. (3-90) gives a value of 527 kW·m-2). Last two thirds (from t5.3 to t16): The diameter is practically constant at Dmax = 269 m. The average height of the fireball centre, H, increases steadily. The emissive power decreases steadily. The thermal radiation received by a vertical surface located at 180 m varies as a function of time, as shown in Fig. 3-16: it increases up to a maximum value of approximately 82 kW m-2 during the growth phase (first 5.3 s) and afterwards it decreases significantly during the second third and more smoothly during the last third. In this figure, the value corresponding to the constant D, H and E model has also been plotted.
Fig. 3-16. Variation of the thermal radiation intensity from a fireball as a function of the time, according to the two models, for a given case (see Example 3-8).
The dose received by a person exposed to this radiation can be calculated (Fig. 3-17) with the following expression:
111
dose
³I
4/3 v ,t
dt
The dose received by a person exposed to the thermal radiation intensity Iv for the entire duration of the fireball is 2·107 s (W·m-2)4/3. It can be observed that the values of the dose calculated by the two methods are similar, with a lightly higher dose obtained with the constant D, H and E model. Consequences on people Thermal radiation: For a dose of 2.8·107 s (W m-2)4/3 (constant D, H and E model), the probit function for lethality (unprotected people) is (see chapter 7, section 4): Y = -36.38 + 2.56 ln 2.8·107 = 7.52
This value implies (chapter 7, Table 7-1) 99.4 % mortality. By applying the same expression, for a dose of 2·107 s (W m-2)4/3 (variable D, H and E model): Y = -36.38 + 2.56 ln 2·107 = 6.65 This value implies 95 % mortality.
Fig. 3-17. Variation of the dose received by a person located at 180 m as a function of time, according to both models.
Taking into account the accuracy of the probit function, both values can be considered similar. ______________________________________
112
NOMENCLATURE
A surface of the solid flame through which heat is radiated (m2) Aor cross sectional area of the orifice (m2) a constant in Eq. (3-12) (K); constant in Eq. (3-79) (-) b constant in Eq. (3-12) (K); constant in Eq. (3-79) (-) c constant in Eq. (3-12) (K); constant in Eq. (3-80) (-) CD discharge coefficient (-) ci concentration of component i on a fuel basis (% volume) cp specific heat at constant pressure (kJ kg-1 K-1) cst-vol mole fraction of fuel in the stoichiometric fuel-air mixture (-) cst-mass stoichiometric mass fraction of fuel in Eq. (3-59) (-) cv specific heat at constant volume (kJ kg-1 K-1) D pool or fireball diameter (m) D’ pseudo pool diameter in the wind direction (m) Deq pool equilibrium diameter (m) Dgroundf initial fireball diameter (m) Dj diameter of the jet fire (m) Dmax fireball maximum diameter (m); maximum pool diameter for an instantaneous spill on water (m) Dpool pool diameter at time t (m) d distance between the surface of the flames and the target (m) dor orifice or outlet diameter (m) ds effective orifice diameter (m) E emissive power of the flames (kW m-2) Elum value of E for the luminous zone of the flames (kW m-2) Esoot value of E for the non-luminous zone of the flames (kW m-2) e constant in Eq. (3-80) (-) F view factor (-) Fh view factor, horizontal surface (-) fr interface tension (N m-1) Fv view factor, vertical surface (-) g acceleration of gravity (m s-2) H average height of the fire (m); height at which the fireball centre is located (m) HR relative humidity of the atmosphere (%) h height at which the fireball top is located (m); cloud height (m) hHC initial height of fuel in the tank before the fire starts (m) 'hv vaporization heat at boiling temperature (kJ kg-1) 'Hc’ combustion heat (kJ mole-1 or kJ kg-1) 'Hc net combustion heat (kJ mole-1 or kJ kg-1) I intensity of the thermal radiation reaching a given target (kW m-2) L length of the visible flame (m) l distance between the centre of the cylindrical fire and the target (m) LBv vertical distance between the gas outlet and the flame tip (m) lp distance between the point source and the target (m) M mass of substance (kg) m fuel mass burning rate per unit surface and per unit time (kg m-2 s-1) m’ mass flowrate in the jet (kg s-1); mass burning rate per unit time (kg s-1)
113
mf Ma Mv P Pi Pinterf Pcont Po Por Pw Pwa QF QL Qr qv R Rids RiLbo Rw r Re S s T t Ta Tad Tcont Tf Tfl Thw T0 'T0 T0av ti Tj Tspill Tv uav uj usound uw uw* uwave V Vi Vl vl
burning velocity of an infinite diameter pool (kg m-2 s-1) molecular weight of air (kg kmole-1) molecular weight of fuel (kg kmole-1) pressure in the vessel just before the explosion (N m-2) pressure (N m-2) pressure at the water-fuel interface (N m-2) pressure inside the container or pipe (N m-2) atmospheric pressure (N m-2) static pressure at the orifice (N m-2) partial pressure of water in the atmosphere (N m-2) saturated water vapour pressure at the atmospheric temperature (N m-2) heat flux from the flame (kW m-2) heat lost from the fuel surface (kW m-2) heat released as thermal radiation (kW) heat required to produce the gas or vapour (kJ kg-1) ideal gas constant (8.314 kJ kmole-1 K-1) Richardson number based on ds (-) Richardson number based on Lbo (-) ratio between wind velocity and jet velocity at gas outlet (-) mass-ratio in the stoichiometric air/fuel mixture (-) Reynolds number (-) flame speed (m s-1) lift-off distance (m) temperature (K) fireball duration (s) ambient temperature (K) adiabatic flame temperature (K) temperature inside the container (K) flash point temperature (K) radiation temperature of the flame (K) temperature of the heat wave when boilover occurs (K) boiling temperature at atmospheric pressure (K) range of boiling temperatures in the fuel (K) average boiling temperature of the fuel (K) time at instant i (s) jet temperature at the gas outlet (K) duration of the spill (s) temperature of the fuel before it is released (K) average jet velocity (m s-1) velocity in the jet at the gas outlet (m s-1) velocity of sound in a given gas (m s-1) wind speed (m s-1) dimensionless wind speed (-) velocity at which the heat wave progresses (m s-1) flow rate of liquid release (m3 s-1) volume of liquid spilled instantaneously (m3) volume of liquid in the vessel (m3); total volume of spilled liquid (m3) liquid leak rate (m3 s-1)
114
W w x xlum y
D D’ DE Dst Tj Tjv H M
I Ist J
Krad QHC :
Ul Ua Uf-a Uj Uw V W
mass of fuel in the tank at the beginning of the fire (kg) width of the flame front (m) horizontal distance from the flames to the target (m) (Table 3-6); axial distance from the orifice in Eq. (3-60) (m); horizontal distance between the centre of the fireball and the target (m) fraction of fire surface covered by luminous flame (-) burning rate (m s-1) tilt angle of a pool fire or a jet fire (º) constant pressure expansion ratio for stoichiometric combustion (-) angle between the axis of the orifice and the line joining the centre of the orifice and the tip of the flame (º) moles of reactant per mole of product in a stoichiometric fuel-air mixture (-) angle between the hole axis and the horizontal in the vertical plane (º) angle between the hole axis and the horizontal in the wind direction (º) emissivity (-) angle between the plane perpendicular to the receiving surface and the line joining the source point and the target (º) fuel volume ratio in the fuel-air mixture (-) stoichiometric fuel volume ratio (-) ratio of specific heats of the gas, Cp/Cv (-) radiant heat fraction (-) kinematic viscosity of the fuel (cSt) angle between the wind direction and the normal to the pipe in the horizontal plane (º) liquid fuel density (kg m-3) air density (kg m-3) density of the fuel-air mixture (kg m-3) density of gas fuel in the outlet (kg m-3) water density (kg m-3) Stefan-Boltzmann constant (5.67·10-8 W m-2 K-4) atmospheric transmissivity (-)
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[30] Thomas, P. H. The Size of Flames from Natural Fires. 9 Symposium on Combustion.
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3rd International Combustion Symposium. The Combustion Institute. Pittsburgh, 1949. [35] Brzustowski, T. A. Progress in Energy and Comb. Sci., 2 (1976) 129. [36] Chamberlain, G. A. Chem. Eng. Res. Dev. 65 (1987) 299. [37] Committee for the Prevention of Disasters. Methods for the Calculation of Physical Effects (the “Yellow Book”), 3rd ed. The Hague, SDU, 1997. [38] Kern, G. R. Hydrocarbon Proc. 43 (1964) 121-125. [39] Kalghatgi, G. T. Comb. and Flame 52 (1983) 91. [40] CCPS. Guidelines for Evaluating the Characteristics of Vapor Cloud Explosions, Flash Fires and BLEVEs. AIChE. New York, 1994. [41] Broeckmann, B., Schecker, H. G. J. Loss Prev. Process Ind. 8 (1995) 137. [42] Michaelis, P., Mavrothalassitis, G., Hodin, A. Boilover. Rapport Interne. TOTAL-EDFINERIS. 1995.
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Chapter 4
Vapour cloud explosions 1 INTRODUCTION Accidental explosions are associated with a very fast release of energy that produces large quantities of expanding gas. This gas may be compressed gas that undergoes a sudden loss of containment (note that pressure is measured in energy per unit volume) or high-temperature combustion products released as a result of a very rapid combustion process. In both cases, the rapid expansion of the gases gives rise to a blast or overpressure wave that can have a significant effect on the surroundings. This chapter deals with vapour cloud explosions. Explosions associated with the sudden loss of containment of compressed gases or superheated liquids (bursting of pressurized vessels) are covered in Chapter 5. Of the major accidents that occur in the process industry and in the transportation of hazardous substances, explosions are relatively frequent. In a historical analysis performed on 5,325 accidents [1], 36% were explosions (fires were more frequent, constituting 44%). In a more recent survey on transport accidents, one in every 9.5 accidents led to an explosion and one in every 15 accidents led to a fire-explosion sequence. A survey of port areas showed that one in every six release-fire events produces an explosion [2]. Explosions are important because they can have destructive effects over relatively large areas. Several circumstances, such as the generally short lapse between the start of the emergency and the occurrence of the explosion, can increase the severity of explosions in terms of their effects on people in the vicinity. In fact, the p-N curves corresponding to the different types of accident —explosion, fire and gas cloud— clearly show that explosions have the most severe consequences, followed by fires and lastly by gas clouds (Fig. 4-1) [3]. In the figure, the abscissae represent the severity of the accident, expressed as the number of fatalities; the values of the ordinate axis are the probability that an accident causes a number of fatalities equal to or higher than N (for N = 0, p = 1). In all types of accidents, this probability decreases as the severity of the accident (number of fatalities) increases, but for a given probability the number of fatalities is generally higher for an explosion. Explosions must be modelled to predict the potential destructive power of the blast that can be produced in a given installation. In the process industry, the substances that can cause an explosion are essentially hydrocarbons such as LPG, gasoline or cyclohexane. A typical scenario would be the loss of containment of a hot and pressurized liquid fuel; the depressurization causes flash vaporization, which forms a cloud containing a mixture of fuel and air. After a relatively short period, the cloud —a large part of which is within the flammability limits— is ignited and an explosion occurs. Hydrocarbons are fairly poor explosives, but the quantities involved can be very large. Furthermore, the partial confinement
119
of the cloud and the eventual turbulence inside it (due to a jet, for example) increase the blast, so these explosions can ultimately be very destructive. A similar effect is observed with dust: apparently inoffensive substances such as sugar, flour or aspirin have produced very serious explosions. 1
EXPLOSION FIRE GAS CLOUD
Probability
0.1
0.01
0.001 1
10
100
Number of fatalities
Fig. 4-1. p-N curve as a function of accident type: explosion, fire and gas cloud. Taken from [3], by permission.
Therefore, an important aspect of risk analysis is the determination of explosion hazards and the effects of these explosions on people and equipment. This chapter explains the prediction of blast damage from vapour cloud explosions and discusses representative practical cases. 2 VAPOUR CLOUDS A vapour cloud is formed due to the loss of containment of a certain mass of a flammable vaporizing liquid or gas. This can be a spill of liquid that is then evaporated from a pool, a release of gas or vapour, or a loss of containment of a superheated liquid which, under depressurization, undergoes a flash vaporization that produces a biphasic (spray) release. Under certain meteorological conditions, a flammable cloud may form. If the cloud is ignited, the substance will burn and a flash fire will occur. It is possible that, in addition to the fire, a mechanical explosion will also take place. For an explosion to occur, several conditions must be fulfilled. Firstly, the substance released must be flammable, and there must be a certain delay in ignition, because if ignition occurs immediately, it is a different phenomenon (a jet fire). If there is a significant delay, it is possible that a sufficiently large cloud of a fuel-air mixture will develop. Additionally, part of the fuel-air cloud must be within the flammability limits, i.e. it must be flammable, and the vapour cloud must be of a minimum size. Several values have been suggested, but it is not clear whether a threshold value really exists, as this probably depends on the circumstances (for example, the confinement). A minimum amount of 1,000 kg has been suggested. Finally, the presence of turbulences is required; these turbulences can be produced either by the release mode (a jet) or by the interaction with obstacles that create a partial confinement.
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If these conditions are fulfilled and the cloud meets an ignition source, an explosion will occur and a blast wave will affect a certain area. The overall energy released will be a function of the amount of flammable substance involved in the explosion and its explosion energy, although only a relatively small part of this energy will be used to create the blast. For practical purposes, the energy of explosion can be taken as the lower combustion energy, as for many substances these two quantities differ only by a small percentage [4]. 3 BLAST AND BLAST WAVE From a practical point of view, the most important feature of an explosion is the blast. The explosion energy of a high explosive or the heat of combustion of a fuel are partly converted into mechanical energy (blast) due to the expansion of the combustion gas products, which is caused by the stoichiometry of the reaction (higher number of moles) and, essentially, by thermal expansion. Under atmospheric conditions, the maximum theoretical efficiency of this conversion in a hydrocarbon-air explosion is approximately 40%, although in practice it is always much lower. With conventional explosives, the efficiency is much higher. 3.1 Blast wave The mechanical energy of the explosion constitutes a blast wave that moves at a certain velocity through the atmosphere. Overpressure is the result of two competing phenomena: the pressure build-up due to combustion and the pressure decrease due to the expansion of gases. The shape of this wave depends on the type of explosion (Fig. 4-2). Before the blast wave arrival, the pressure is ambient pressure Po. In an ideal blast wave, the overpressure increases almost instantaneously to a value Po+'P, then decreases less rapidly to negative values, reaches a minimum and finally returns to the ambient value.
Fig. 4-2. The shape of the blast wave for a) detonations and b) deflagrations.
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This type of blast wave, called a shock wave (instantaneous pressure increase), would correspond to a detonation (Fig. 4-2 a). 'P is usually called peak side-on overpressure. In a deflagration, the increase in pressure is less rapid (Fig. 4-2 b) and, for a given amount of explosive, the maximum (peak) overpressure is less. Therefore, the blast wave has a positive phase followed by a negative or suction phase. If a blast wave encounters a rigid object, a transient pressure is induced over its surface during the positive phase. On the front wall, the overpressure is reflected and a local reflected wave is produced; this reflected overpressure ranges between two and several times the blast wave overpressure [5]. The effects on wide objects are essentially caused by the positive phase. Although the destructive effects of a blast wave are usually associated with the peak overpressure, another significant variable is the dynamic pressure [6]. The dynamic pressure is created by the air movement (blast wind) induced by the blast wave and is proportional to the square of the air speed and to the density of the air behind the blast wave. Once the positive phase has passed, the object undergoes the effects (drag force) of the blast wind that follows the positive phase, which is associated with the negative phase; the effects on narrow objects are essentially caused by this drag force. Duration and impulse are also important parameters of a blast wave. The duration is the time for which the overpressure persists; in Fig. 4-2, the positive phase duration is t2-t1 and the negative phase duration is t3-t2. The impulse is the area under the overpressure/time curve. Positive and negative impulses can be defined as [7]: t2
i
³ P t P dt
i
³ P
t3
t2
(4-1)
o
t1
o
P t dt
(4-2)
where P is the overpressure (Pa) and P0 is the atmospheric pressure (Pa). Explosions can be divided into detonations and deflagrations. These have different features and levels of severity, which are briefly commented on in the following sections. 3.2 Detonations In a detonation, the blast wave propagates through the unreacted mixture at supersonic velocity. As mentioned above, the blast wave has a characteristic shape and an almost instantaneous increase in pressure (Fig. 4-2 a). The overpressure produced by the fast combustion is propagated through the mixture as a shock wave, which compresses the mixture and causes it to ignite. Therefore, a flame front and a shock wave propagate together at supersonic velocity. Because of these features, detonations are, for a given quantity of explosive, more destructive than deflagrations. Blast waves from high explosives (for example, TNT) are close to the ideal wave, due essentially to the relatively small volume of explosive material and the rapid rate of energy release [6] that is associated with a very fast chemical reaction. However, the probability of a detonation with an unconfined flammable cloud is in practice negligible, essentially because of the lack of homogeneity of the fuel-air mixture, which prevents any eventual detonation from being propagated. In fact, there is only one case in which this may have occurred and it was caused by very specific circumstances: the explosion of a vapour cloud involving approximately 23,000 kg of propane in Port Hudson (Missouri, 1970) [8].
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3.3 Deflagrations In vapour cloud explosions, because the volume of the vapour-air mixture is large and the rate of energy release is relatively slow (the chemical reaction is slower than in the case of a detonation), the explosion is a deflagration. Deflagrations are characterized by a smoother blast wave (Fig. 4-2 b) that propagates at subsonic velocity. The flame propagation is initially associated with heat conduction and molecular diffusion and in the first steps of a gas cloud explosion the flame propagation is laminar and overpressures are very small. As the explosion proceeds, turbulence is generated and, as a result, the combustion rate increases in a feedback process. The blast wave from a vapour cloud explosion approaches the ideal blast wave shape the further it travels away from the edge of the vapour cloud [6]. Furthermore, in vapour cloud explosions the negative impulse is no longer small compared to the positive impulse. Large negative overpressures cause damage to structures by suction; therefore, it is more difficult to predict the blast wave from a vapour cloud explosion because it is produced by a high explosive. 3.4 Blast scaling There are scaling laws that relate the properties of blast waves from different explosions (amount of explosive, distance) and that therefore allow the blast wave properties from an explosion to be extrapolated from data obtained under different conditions. There is experimental evidence that if a spherical explosive charge of diameter D produces a peak overpressure 'P at a distance d from its centre, as well as a positive-phase duration t+ and an impulse i (the integral of the pressure-time history), then a charge of diameter kD of a similar explosive in the same atmosphere will produce an overpressure wave with a similar form and the same peak overpressure 'P (as well as a positive-phase duration kt+ and an impulse ki) at a distance kd from the charge centre. This is called the Hopkinson or “cube root” scaling law.
Fig. 4-3. Hopkinson scaling law. Taken from [5] (copyright 1994 by the American Institute of Chemical Engineers), by permission of AIChE.
If we consider that the mass of the charge is proportional to its volume, i.e. to the third power of its diameter, a “scaled distance” is defined as the ratio between the real distance and the cube root of the charge mass:
123
dn
d M 1/ 3
(4-3)
where dn is the scaled distance (m kg-1/3) d is the real distance from the centre of the explosion to the point at which the overpressure must be estimated (m), and M is the charge mass (kg). Therefore, the cube root of the charge mass is often used as a scaling parameter; when two charges of the same explosive, with similar geometry but of different sizes, explode in the same atmosphere, similar peak overpressures are produced at the same scaled distance. This is the most simple and common form of blast scaling. Another approach, which will also be used in the following sections, is that proposed by Sachs [7, 9]. The blast wave can be expressed as a function of scaled overpressure (or Sachs scaled overpressure),
'P s
'P P0
(4-4)
combustion energy-scaled distance (or Sachs scaled distance),
R
d
(4-5)
E / P0 1 / 3
and Sachs scaled impulse,
is
i us P E1/ 3
(4-6)
2/3 0
where i is the incident impulse (Pa s) E is the energy involved in the explosion (J) P0 is the atmospheric pressure (Pa) 'P is the side-on peak overpressure (Pa) and us is the speed of sound in air (m s-1). 3.5 Free-air and ground explosions If either the explosive mass or the bursting vessel is located far from surfaces that might reflect the overpressure wave, the blast wave will be spherical; these are called free-air explosions and are not usually observed in accidental explosions (although these conditions can be found in certain military activities). However, if the explosion takes place near a surface —for example, the ground— the blast wave will be reflected by this surface and will act over a hemispherical volume instead of a spherical one. Furthermore, the reflected wave will overtake the first wave and thereby increase its strength. These are called surface or ground explosions. If the reflection from the ground is perfect, the reflection factor is 2, as the energy contained in the blast wave in the lower hemisphere of a free-air explosion is used to generate the blast wave in the hemisphere above the ground. However, in practice, the reflection factor for high-energy explosives is often lower (approximately 1.8), due to the fact that some of the
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energy released in the explosion is dissipated in the production of ground shock or in cratering [7]. For low energy-density sources such as gas cloud explosions, the amount of energy dissipated into the ground is very small and a factor of 2 can be applied. 4 ESTIMATION OF BLAST: TNT-EQUIVALENCY METHOD A common approach for determining the damage caused by a given explosion consists in estimating the “TNT equivalency”, i.e. the mass of TNT that would produce the same degree of damage. The main features of TNT and other high explosives have been extensively studied and are therefore reliable references. In vapour cloud explosions, the equivalent mass of TNT can be calculated using the following expression: WTNT
K
M 'H c 'H TNT
(4-7)
where M is the mass of fuel in the cloud (kg) 'Hc is the lower heat of combustion of the fuel (kJ kg-1) K is the explosion yield factor (-) and 'HTNT is the blast energy of TNT (4680 kJ kg-1). Using Eq. (4-7) incurs a number of difficulties. Firstly, the value of the explosion yield factor Kmust be established. As hydrocarbons —the most common substances involved in vapour cloud explosions— are in fact fairly poor explosives, only a small part of the energy released is used to create the blast wave: values between 1% and 10% have been proposed by different authors. Although the value of K is probably influenced by the reactivity of the fuel involved and the eventual partial confinement of the cloud, the currently accepted value is 3% (K = 0.03). Regarding the mass of fuel involved in the explosions, in a worst-case scenario it should be assumed that the total amount of flammable substance in the cloud contributes to the generation of the blast. However, other authors suggest that only the mass within the lower flammability limit contour should be considered. Once the value of the equivalent mass of TNT, WTNT, has been determined, the scaled distance must be calculated (Eq. (4-3)). From the scaled distance, the blast peak overpressure can be found in a standard blast chart (Fig. 4-4): for any given scaled distance there is a corresponding value of peak overpressure. Different plots are available in the literature and give slightly different values of peak overpressure (N.B. blast charts for free-air explosions are also found in the literature). The peak overpressure can also be estimated from the scaled distance by using the following expression:
'P P0
1 4 12 d n d n2 d n3
(4-8)
As mentioned above, the main features of TNT and vapour cloud explosions are different. TNT is a high energy-density explosive and vapour clouds are a low energy-density source. TNT explosions are detonations, while vapour cloud explosions are deflagrations: the shape
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and velocity of their respective blast waves are different. This is why the TNT-equivalency method is not an exact procedure. The error would be higher in the near field (up to approximately 3 cloud diameters from the centre of the explosion) and lower in the far field (which begins at approximately 10 cloud diameters from the centre of the explosion), as the blast wave from a vapour cloud explosion tends to develop the characteristics of a TNT blast wave as it travels away from the centre of the explosion [6]. Thus, the TNT-equivalency method may be useful in the far field, for example, in estimating the blast damage to the areas surrounding a chemical plant.
Fig. 4-4. Side-on peak overpressure for a surface TNT explosion. Taken from [5] (copyright 1994 by the American Institute of Chemical Engineers), by permission of AIChE.
Furthermore, it is important to consider that if another alternative method is used, it will also be subject to errors arising from uncertainties regarding the size, shape and composition of the vapour cloud and the influence of the confinement and congestion of the areas covered by the flammable mixture. Therefore, as the TNT-equivalency method is very simple, it remains widely used.
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A plot of characteristic curves has recently been published [11] from which it is possible to determine the approximate values of overpressure and impulse as a function of distance for an explosion whose TNT equivalent mass is known. The authors fitted power equations to the well-known relationships 'P (peak side-on overpressure) vs. dn and i’ (scaled impulse, 1 / 3 i ' i WTNT ) vs. dn for the TNT equivalent model over two different intervals of the scaled distance ( 1 d d ' < 10 m·kg-1/3 and 10 d d ' < 200 m·kg-1/3) and obtained the plot shown in Fig. 4-5. In this plot, the points corresponding to the same distance on different characteristic lines are joined by the iso-distance lines. Characteristic curves have been plotted for values of WTNT between 150 kg and 500·103 kg. For a given equivalent mass of TNT the side-on peak overpressure and the impulse at different distances can be obtained by graphical interpolation. 100000
10 3 kg TNT
500 10000
200
Imp ulse (Pa·s)
400
10 5
1000 1500 1000
50 20
600
Distances (m )
150
2
2500
1
4000
0.5
6000 0.15 25 50
100 100
Distances (m)
10 100
1000
10000
100000
1000000
10000000
Side-on overpressure (Pa)
Fig. 4-5. Characteristic curves corresponding to the explosion of different amounts of TNT at varying distances from the blast centre. Taken from [11], by permission.
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______________________________________ Example 4-1 On 1 June 1974, at a process plant located in Flixborough (UK), the rupture of a pipe led to the creation of a vapour cloud containing approximately 30,000 kg of cyclohexane ('Hcyclohexane = 43,930 kJ kg-1). The time elapsed between the rupture and the explosion was 45 s. It has been estimated that at the moment of the explosion the cloud had a volume of 400,000 m3, with an average concentration of 2%. Calculate: a) the peak overpressure at a distance of 500 m from the centre of the cloud, and b) the actual overall explosion efficiency if glass was broken at a distance of 1,950 m. Solution a) Calculate the equivalent mass of TNT (Eq. 4-7). Assume an explosion yield of 3%:
WTNT
K
M 'H c 'H TNT
0.03
30,000 43,930 4,680
8,448 kg
Calculate the scaled distance: dn
d 1/ 3 WTNT
500 3
8,448
24.55 m kg-1/3
From the chart (Fig. 4-4), the peak overpressure obtained is 'P = 0.041 bar. The effects of this blast would be the shattering of windows and minor structural damage to houses. If Eq. (4-8) is applied, 'P 1.013
1 4 12 ; 'P 0.049 bar 2 24.55 24.55 24.55 3
If the characteristic curves are used, interpolation in Fig. 4-5 for WTNT = 8,448 kg and d = 500 m gives a value of 'P = 4,500 Pa (i.e. 0.045 bar). b) The threshold overpressure for the breakage of glass is approximately 0.01 bar (Table 714). This implies (Fig. 4-4) a scaled distance dn = 80 m kg-1/3. For a distance of 1,950 m, this corresponds to an equivalent mass of TNT of dn
80
1950 3
WTNT
; WTNT = 14,480 kg
This mass implies an overall explosion yield of WTNT 'H TNT 14,480 4680 0.05 M 'H f 30,000 43,930 In fact, a yield value of 5% was also obtained by the researchers who analyzed this explosion. This relatively high value is probably due to the partial confinement and the congestion created by the plant equipment and buildings. ______________________________________
K
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5 ESTIMATION OF BLAST: MULTI-ENERGY METHOD
Several analyses of major vapour cloud explosions seem to indicate that the blast damage is not related to the overall amount of fuel present in the cloud; in fact, the analysis of hydrocarbon cloud explosions reveals a wide range of TNT equivalencies. Instead, it has been suggested that the explosion effects depend on the size, shape and nature of those portions of the flammable cloud that are partially confined or obstructed, while the unconfined parts of the cloud simply burn out and make no significant contribution to overpressure. This theory, which is increasingly accepted, is the basis of the multi-energy method for vapour cloud explosion blast modelling [12, 13]. Blast is generated in vapour cloud explosions only where the flammable mixture is partially confined and/or obstructed. In this model, the volume of the cloud within the partially confined space is converted into a hemisphere of equal volume. The model considers this hemispherical cloud to be a homogeneous mixture of hydrocarbon and air at the stoichiometric concentration. An average value of 0.1 kg m-3 for hydrocarbon-air mixtures is assumed. This corresponds to an average combustion energy of 3.5 x 106 J m-3. A numerical simulation of the blast from an explosion of this type of hemispherical charge on the Earth’s surface gave the set of curves plotted in Figs. 4-6 and 4-7.
Fig. 4-6. Multi-energy method blast chart: dimensionless positive phase duration. Taken from [15], by permission.
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Fig. 4-6 allows the positive phase duration of the vapour cloud explosion to be predicted. The figure contains 10 curves, which correspond to the 10 initial strengths or severities of the explosion considered by the model, from low strength (category 1) to detonation (category 10). A graphical indication of the corresponding blast wave shape is also provided [12].
Fig. 4-7 shows the dimensionless side-on peak overpressure 'Ps as a function of the combustion energy-scaled distance and for the 10 initial strengths or severities of the explosion. A high-strength blast corresponds to a shock wave and is represented by solid lines; low-strength pressure waves are indicated by dashed lines that may steepen into shock waves in the far field. The blast can therefore be predicted in the far field with a significant degree of accuracy for high initial strengths whenever the assumption of this initial strength is justified [12].
Fig. 4-7. Multi-energy method blast chart: dimensionless peak side-on overpressure. Taken from [15], by permission.
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The multi-energy method is applied by characterizing the vapour cloud and the congested zones and then determining their contribution to blast generation. This is achieved by the following steps [14]. a). Determine the cloud size. A dispersion calculation will be required in order to predict the concentration field in the area affected by the cloud. If no dispersion calculation is made, the overall mass must be estimated. Calculate the volume of a cloud containing the mass of fuel at the stoichiometric concentration. b). Identify the congested areas that constitute the potential sources of blast in the area affected by the vapour cloud. Potential sources of strong blast are process equipment in chemical plants, stacks of crates or pallets, the volumes between parallel planes (beneath closely parked cars, open multi-story car parks), tube-like structures (tunnels, corridors), highly turbulent fuel jets releasing into the atmosphere, etc. c). Determine the free volume of each congested area. Determine the volume of the unobstructed part of the vapour cloud. d). Estimate the energy of each area. Each area must be considered an independent blast source. Their respective combustion energies are obtained by multiplying the volume of the mixture in each area by 3.5 x 106 J m-3; the volume of the equipment must be taken into account if it is relatively large. However, if two or more areas are so close together that their combustion processes may interact, and it cannot be ruled out that their waves may overlap, their respective energies can be added. e). Determine the source strength for each area. Choose a source strength of 1 for unobstructed areas. If a certain degree of turbulence is expected in these areas due, for example, to a jet release, choose 3. A source strength of 7 is recommended for congested areas, although in a worst case scenario a value of 10 should be chosen. However, note that for overpressures below an approximate value of 0.5 bar (Fig. 4-7) there is no difference between source strengths in the range 7-10. f). Determine the location of the centre of each area. If two or more areas have been added, determine their common centre taking into account their respective centres and energies. Also, determine the location of the centre of the unobstructed part of the vapour cloud. g). Calculate the combustion energy-scaled distance at a given distance d from the centre of each area (Eq. (4-5)) and obtain the dimensionless positive phase duration and the dimensionless peak side-on overpressure from Figs. 4-6 and 4-7 respectively. The peak overpressure is calculated from 'P
'Ps P0
and the positive phase duration from
t
§E· ¨ ¸ ¨P ¸ t © 0 ¹ us
1/ 3
.
Two main difficulties are faced when the multi-energy method is applied to a given case: the selection of the source strength, which depends on the congestion in the area, and the eventual addition of the effects of two charges.
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______________________________________ Example 4-2 There is a release of propane in an LPG storage area. The flammable cloud is located in an area that contains twelve cylindrical tanks of 36 m3 (diameter: 2 m, length: 13 m) arranged in two rows. The distance between two adjacent parallel cylinders is 2 m; the distance between the two rows of cylinders is 4 m. The minimum height below the tanks is 1.5 m. The pipework on top of the tanks has a height of 0.5 m. It is estimated that, at the moment of ignition, the volume of the cloud within the flammability limits is 1,750 m3. Using the multienergy method, calculate the overpressure at a distance of 200 m from the centre of the cloud.
Fig. 4-8. LPG storage area.
Solution The volume of the congested area is the overall volume of the storage units minus the volume of the tanks:
> 2 11 13 2 4 @ 4 12 36 = 2,208 m3
V
The whole cloud is contained within the congested zone; therefore, the volume of the cloud within the flammability limits will be used to calculate the explosion energy: E = 1,750 m3 · 3.5 · 106 J m-3 = 6.125 · 109 J
d
R
200
5.1 1/ 3 § 6.125 10 9 · ¸¸ ¨¨ © 101,325 ¹ Assuming a source strength of 7, the dimensionless overpressure is obtained from Fig. 4-7:
E / P0
1/ 3
'Ps = 0.05. Then, the peak overpressure is
'P = 'Ps · P0 = 0.05 · 101,325 = 5,070 Pa This overpressure causes minor damage to house structures. ______________________________________
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6 ESTIMATION OF BLAST: BAKER-STREHLOW-TANG METHOD
This method [16] was developed to provide estimations of the overpressure (for both the positive and negative phases of the pressure wave) and impulse from vapour cloud explosions. The procedure has recently been updated. Similarly to the multi-energy method, the Baker-Strehlow-Tang (B-S-T) method is based on the assumption that only those parts of a flammable vapour cloud that are congested or partially confined contribute to the build-up of overpressure. Both methods use the same procedure to estimate the explosion energy (E), based on an average stoichiometric fuel-air mixture. The B-S-T method also uses a family of curves to determine 'Ps as a function of the combustion energy-scaled distance, although the procedure for constructing the graphical relationship between these two variables is different. The B-S-T method uses a continuum of numerically determined pressure and impulse curves (Figs. 4-9, 4-10 and 4-11) that take the flame Mach number as the parameter. The strength of the blast wave is proportional to the maximum flame speed reached within the cloud. The flame Mach number is the apparent flame speed (the vapour cloud explosion flame front relative to a stationary point of reference) divided by the ambient speed of sound. The appropriate Mach number, Mf, for each specific situation being modelled can be taken from Table 4-1 [16].
Fig. 4-9. B-S-T method: dimensionless peak side-on overpressure vs. combustion energy-scaled distance for various flame speed Mach numbers. Reprinted from [16] with permission of John Wiley & Sons, Inc.
In this table, no confining plane to flame expansion is considered to be 3D. The existence of a single confining plane implies 2D flame expansion. The confinement category 2.5D corresponds to those cases in which the confinement is made of either a frangible panel (which might be expected to fail quickly and provide ventilation) or by a nearly solid confining plane (for example, a pipe rack in which the pipes are almost touching). Congestion is considered to be low if the area blockage ratio is below 10%, medium if the ratio is between 10% and 40%, and high if the ratio is above 40%.
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Fig. 4-10. B-S-T method: negative overpressure vs. combustion energy-scaled distance. Reprinted from [16] with permission of John Wiley & Sons, Inc.
Fig. 4-11. B-S-T method: positive impulse vs. combustion energy-scaled distance. Reprinted from [16] with permission of John Wiley & Sons, Inc. Table 4-1 Flame speed Mach numbers (Mf) to be used in the Baker-Strehlow-Tang method Congestion Flame expansion Reactivity Low Medium High 2D high 0.59 DDT DDT medium 0.47 0.66 1.6 low 0.079 0.47 0.66 2.5D high 0.47 DDT DDT medium 0.29 0.55 1.0 low 0.053 0.35 0.50 3D high 0.36 DDT DDT medium 0.11 0.44 0.50 low 0.026 0.23 0.34 DDT: deflagration-to-detonation transition
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Baker et al. [17] suggest considering three different categories for the reactivity of fuels: - high reactivity fuels: hydrogen, acetylene, ethylene oxide and propylene oxide - low reactivity fuels: methane and carbon monoxide - medium reactivity fuels: all other gases and vapours. The B-S-T method is applied using the following steps: a). Determine the cloud size. Calculate the volume of a cloud containing the mass of fuel at the stoichiometric concentration. b). Identify the volume of the congested or partially confined portion of the flammable vapour cloud. c). Estimate the explosion energy (E) by multiplying the volume of the congested or partially confined portion of the flammable vapour cloud by 3.5 x 106 J m-3.
d). Calculate the combustion-energy scaled distance ( R ) for any given distance from the centre of the explosion. e). Select the appropriate flame speed (Mach number) from the values listed in Table 4-1. f). Obtain the corresponding value of the dimensionless peak side-on overpressure from Fig. 4-9. g). Multiply the dimensionless side-on peak overpressure by atmospheric pressure to obtain the peak side-on overpressure. e). Use the same steps to obtain the negative overpressure or the impulse. ______________________________________ Example 4-3 Due to a human error, a mixture of hot liquid hydrocarbons is released in a refinery, which produces a vapour cloud. The cloud drifts towards a cracker, one of the most congested units in the refinery. It is estimated that approximately 9,000 m3 of the cloud within the flammability limits covers the congested areas of the cracker; the rest of the cloud is over an open area. Use the Baker-Strehlow-Tang method to calculate the blast at a distance of 500 m from the centre of the explosion if the cloud is ignited. Solution Assume medium reactivity, 3D and high congestion. From Table 4-1, Mf = 0.50. The explosion energy is E = 9,000 m3 · (3.5 · 106) J m-3 = 3.15 · 1010 J The combustion energy-scaled distance at d = 500 m is d
R
500
E / P0
1/ 3
§ 3.15 1010 ¨¨ © 101,325
· ¸¸ ¹
1/ 3
7.38
From Fig. 4-9, 'Ps = 0.017 Therefore,
'P = 'Ps · P0 = 0.017 · 101,325 = 1,720 Pa This is essentially the same value as that calculated in [18]. It would cause glass breakage. ______________________________________
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7 COMPARISON OF THE THREE METHODS
The TNT-equivalency method assumes that the blast propagates in an idealized environment constituted by a horizontal surface and does not consider the presence of any obstacles. The multi-energy method and the B-S-T method take into account the contribution of congested zones to the generation of the blast. There is a degree of difficulty in the application of all three methods. In the case of the TNT-equivalency method, it is necessary to specify an explosion yield and the result will change depending on the value selected. In the multi-energy method, the initial blast strength must be chosen according to the degree of congestion in the area or areas covered by the flammable cloud. Finally, in the Baker-Strehlow-Tang method, the Mach number of the flame speed —a function of the congestion— must be specified. It is difficult to compare the results of the three methods, as they will be a function of the aforementioned hypothesis. However, several authors have attempted to make a comparison by applying the methods to a given case. The TNT-equivalency and multi-energy methods were compared [19] with the following hypothesis: 10% yield for the TNT-equivalency method and an initial blast of 5 for the multi-energy method. They were applied to the following scenario: a stoichiometric mixture of propane and air in a semi-confined space measuring 35 m x 35 m x 10.8 m. The results are plotted in Fig. 4-12 as side-on peak overpressure versus distance and are shown in Table 4-2.
Fig. 4-12. Comparison of the overpressure in a given scenario predicted by two different methods. Taken from [19], by permission.
It can be seen that the predicted distances to the mid-range overpressures (20.7 and 6.9 kPa) are quite similar for both models, while the distance predicted by the multi-energy method is approximately 20% greater than that predicted by the TNT-based model. In the
136
higher range of overpressures, the multi-energy method predicts maximum peak overpressures that are below 34.5 kPa. Table 4-2 Comparison of model prediction [19] VCE model Distance to specified overpressure (m) 34.5 kPa 20.7 kPa 6.9 kPa 1.0 kPa TNT-equivalency 45 61 130 670 Multi-energy -38 120 800
Another comparison has recently been published by Lobato et al. [20]. These authors applied the TNT-equivalency method, the multi-energy method and the B-S-T method to the explosion of a small cloud (264 m3) of a mixture of hydrogen and air containing 1.08 kg of hydrogen, in a low congestion environment, with the following hypothesis: an explosion yield of 10% for the TNT method (a conservative value); a blast strength of 10 due to the hydrogen explosion features for the multi-energy method; for the B-S-T method, the flame expansion was assumed to be 2D and the obstacle density lower than 10%, so Mf = 0.59. The results are plotted in Fig. 4-13 as overpressure versus distance. The TNT model predicts higher overpressures, while the multi-energy and the B-S-T methods predict similar values; the values obtained at very short distances are not significant.
Overpressure (KPa)
1000
100
10
1
TNT equivalency model Multi-energy model BST model
0.1 1
10
100
R (m)
Fig. 4-13. Overpressure as a function of distance for the three methods (TNT-equivalency: K = 10%; multi-energy method: blast strength = 10; B-S-T method: Mf = 0.59). Taken from [20], by permission.
Generally, therefore, it seems that the multi-energy and the B-S-T methods give similar results over the whole range of overpressures and distances, and that the agreement between the three methods is low for short distances and increases with distance (i.e. as overpressure decreases). No directional effects —which actually exist in many explosions and are fairly difficult to predict— are considered by these methods. For specific cases in which local overpressures must be predicted, more complex models based on computational fluid dynamics are required [15].
137
8 A STATISTICAL APPROACH TO THE ESTIMATION OF THE PROBABLE NUMBER OF FATALITIES IN ACCIDENTAL EXPLOSIONS
As seen above, the overpressure caused by an explosion is a function of the cube root of the mass M of material involved in the explosion. Therefore, the maximum distance at which the value of the overpressure is still lethal is also a function of the cube root of M. Supposing a uniform population density, the number of expected fatalities, F, is therefore proportional to the surface area affected, which in turn depends on the square of the distance from the centre of the explosion [5, 21]. The following expression has been proposed [21]: F
K M m
n
K M q
(4-9)
where K is a constant and m = 1/3, n = 2 and q = 2/3. This expression could be fairly unrealistic when applied to a real case, since: a) Population density can vary a great deal from one accident to another. Two accidents with the same characteristics can cause a very different number of fatalities depending on the number of people present in the area of influence at the time of the explosion. b) Population density in the area surrounding the source of the explosion is not homogeneously distributed, so the population affected is not necessarily proportional to the surface area. c) The fatalities resulting from an accidental explosion may not be exclusively due to the effects of the overpressure wave since other factors are also involved, such as missiles, thermal radiation and poisoning due to smoke or toxic gases. d) The overpressure attained depends not only on the amount of material involved, but also on the eventual effects of partial confinement and congestion. Eq. (4-9) has been proposed as an initial approximation for estimating the consequences of accidental explosions. Based on a series of data from 162 explosions, Marshall [21] established that the best adjustment for Eq. (4-9) would be obtained with the values K = 4 and q = 0.5. More recently, Eq. (4-9) was modified so that it could be applied differently [22]. The expression was applied to 352 explosions and new values for the constants k and p were found. However, although the fit obtained with the modified equation appeared to be very good, the correlation had little predictive value, due to the reasons mentioned above. The dispersion of data was so large that the predictive value of the correlation was practically nil. A different approach was then applied. By applying restrictive criteria, a set of 63 accidents from the period 1975-1999 were selected. The following expression (see Fig. 4-14) was proposed. It can be used to estimate the maximum number of fatalities (Fmax) that could be expected in an accidental explosion at a fixed installation, in which M kg of explosive material is involved: Fmax
3.53 M 0.23 if M d 7,000 kg
Fmax
27 if M ! 7,000 kg
(4-10)
It should be pointed out that, according to the information gathered from historical data, in accidents involving more than 7,000 kg of material, the number of fatalities rarely exceeds 27. Although this may seem illogical, it should be considered that, in most cases, these are accidents that occur in warehouses or storage areas, i.e. in areas where the population density
138
of the surroundings is rather low, so it is unlikely that a large number of fatalities will be caused.
Number of fatalities
1000
100
10
1 0,01
0,1
1
10
100
1000
10000
100000
Size of accident (tonnes of explosive)
Fig. 4-14. Maximum probable number of fatalities (Fmax) as a function of the amount of material involved in the explosion (accidents in transport or involving conventional explosives are not included). Taken from [22], by permission.
As well as knowing the maximum probable number of fatalities that would occur in an accidental explosion as a function of the amount of material involved, it is equally important to know the frequency distribution of the accidents or, in other words, the probability of reaching a specific percentage of Fmax in a given case. Thus, an intermediate variable Z is defined as Z
F Fmax
(4-11)
For each accident, Z indicates the ratio between the number of fatalities (real number of fatalities in an accident or estimated number of fatalities in a hypothetical accident) and the maximum number estimated by Eq. (4-10). It is then possible to determine the accumulated percentage of cases (Ca%) in which a given value of Z is attained. The relationship between the accumulated percentage of cases and Z is given by the following expressions [22]: Ca%
57.4 if Z d 0.022
Ca%
2.12 ln 2 Z 3.1 ln Z 100 if Z ! 0.022
(4-12)
The value 57.4 represents the percentage of cases in which no fatalities occurred. If we consider, as experience has shown, that accidents with no victims can be considered to be underrepresented in the databases, this value is in fact a conservative limit and lower than the real value.
139
______________________________________ Example 4-4 In order to calculate the maximum limit of cover for a policy in a process plant, the risk manager must analyze the possible maximum loss (PML) associated with an accident involving the explosion of a cloud containing 30,000 kg of hydrocarbon. In addition to the material losses, it is necessary to predict the number of fatalities. Estimate the maximum number of probable fatalities that could be expected in 95% of cases. Solution According to Eq. (4-12), the maximum number of probable fatalities that could be expected in this accident is Fmax = 27 The calculations (percentage of accumulated cases) corresponding to the different number of fatalities can be seen in Table 4-3. Table 4-3 Number of fatalities as a function of Z Number of % cases Z = F/Fmax fatalities (F) accumulated 0 0.00 57.4 1 0.04 68.1 3 0.11 82.8 10 0.37 94.8 15 0.56 97.5 20 0.74 98.9 27 1.00 100
The first column of the table shows the different numbers of fatalities (selected arbitrarily) for those who wish to know the probability of an accident with an equal or lesser number of fatalities. The second column shows the corresponding values of Z and the third the number of cases accumulated in which the number of fatalities (Eq. (4-12)) is the same or lower. The results indicate that in this explosion the probable number of fatalities in 95% of cases is 10 or less. ______________________________________ 9 EXAMPLE CASE
______________________________________ Example 4-5 At 1:20 pm on 23 March 2005, an explosion occurred in a refinery in Texas City (Texas) during the startup of a hydrocarbon isomerization unit. The explosion occurred when a distillation tower was overfilled with flammable liquid hydrocarbons. The liquid flowed into a blowdown drum, which was also flooded. As the drum stack vented directly outside, this produced a geyser-like release of flammable hydrocarbon that ran down and pooled on the ground during a period of six minutes. A vapour cloud covered the area of the isomerization unit and drifted towards the south. The cloud then ignited. The explosion seriously damaged
140
the plant and there were a number of secondary hydrocarbon releases and fires. Forty trailers located nearby were destroyed or damaged; 15 workers were killed and approximately 170 others were injured. An investigation of the accident [18] gave the following information. The amount of hydrocarbon between the upper flammability limit and the lower flammability limit in the cloud ranged from 3,000 to 10,000 kg, depending on the assumed condition of the discharged material (liquid/vapour fractions). The approximate plan view of the vapour cloud (corresponding to 10,000 kg between UFL and LFL) can be seen in Fig. 4-15. The vapour cloud within the UFL and LFL was located in the area surrounding the trailers, a pipe rack to the west of the isomerization unit, the area surrounding the base of the blowdown stack and an open area to the south of the unit. The volume of the cloud between the flammability levels in the congested zones was estimated to be approximately 9.000 m3. The fuel outside the congested zones was consumed in a flash fire. The vapour mixed with air below the LFL was dispersed in the atmospere. Finally, some of the fuel formed a pool on the ground. a). Using the multi-energy method, estimate the overpressure at a distance of 45 m from the centre of the cloud and the distances at which the overpressure reached values of 700 Pa and 1750 Pa, respectively. b). Using the Baker-Strehlow-Tang method, estimate the overpressure at 660 m and 300 m from the centre of the cloud, respectively. Solution According to the information published [18], the analysis of the effects of the explosion showed that the volume of the cloud located in a relatively congested area was approximately 9,000 m3. However, as a whole, this area was not as congested as is common in refinery process units. Forensic investigation data indicate that a source strength of 4 should be assumed (if more detailed information on the volumes corresponding to the different zones was available, some of them —for example, the trailer site— could be considered to correspond to a source strength of 7).
Fig. 4-15. Plan view of the vapour cloud (contours correspond to UFL and LFL) for an amount of fuel within flammability limits of 10,000 kg. Taken from [18], by permission.
141
The energy involved in the explosion can be estimated as: E = 9,000 m3 · (3.5 · 106) J m-3 = 3.15 · 1010 J a) Multi-energy method The combustion energy-scaled distance at d = 45 m is d
R
45
E / P0 1 / 3
§ 3.15 1010 ¨¨ © 101,325
· ¸¸ ¹
1/ 3
0.66
From Fig. 4-7, 'Ps = 0.096. Therefore, the peak overpressure is 'P 0.096 101,325 = 9,730 Pa. This was approximately the distance [18] at which the trailer was located and in which most of the fatalities occurred. The forensic value obtained was significantly higher (17,500 Pa), probably due to the greater local congestion in the trailer site. At these close distances, the 'P contours are influenced considerably by the arrangement of congested/confined zones; therefore, they cannot be approximated by a circle.
An overpressure of 700 Pa implies a value of 'Ps of
700 = 'Ps · 101,325
'Ps = 0.0069
This implies a value of combustion energy-scaled distance (Fig. 4-6) R = 10. Therefore, the corresponding distance is d
R 10
§ 3.15 1010 ¨¨ © 101.325
· ¸¸ ¹
1/ 3
d = 677 m And for a peak overpressure of 1,750 Pa, the same procedure gives
'Ps = 0.017, R = 4 and d = 270 m
The contours corresponding to these values are plotted in Fig. 4-13. They are essentially the same as those calculated [18] from the analysis of the diverse explosion effects (660 m and 295 m, respectively).
142
Fig. 4-13. Estimated'P contours.
b) Baker-Strehlow-Tang method The value of Mf (Mf = 0.29) is taken from Table 4-1 for 2.5D (pipe rack, trailers), medium reactivity and low congestion. For a distance d = 660 m, the combustion energy-scaled distance can be calculated with Eq. (4-5):
R
660 § 3.15 1010 ¨¨ © 101.325
· ¸¸ ¹
1/ 3
= 9.74
From Fig. 4-9, 'Ps = 0.007. Therefore, the side-on peak overpressure is: 'P 0.007 101,325 = 710 Pa And for a distance of 300 m:
R = 4.427, 'Ps = 0.016, 'P = 1,620 Pa These results are similar to those found with the multi-energy method. ______________________________________
143
NOMENCLATURE
Ca% D d dn E F Fmax 'Hc 'HTNT i i
accumulated percentage of cases (-) diameter of the explosive charge (m) distance from the centre of the explosion (m) scaled distance (m kg-1/3) explosion energy contributing to overpressure (J) number of fatalities (-) maximum number of probable fatalities (-) lower heat of combustion of the fuel (kJ kg-1) blast energy of TNT (4,680 kJ kg-1) incident impulse (N m-2 s) positive impulse (N m-2 s)
i
negative impulse (N m-2 s)
is K M m Mf N n P p
Sachs-scaled impulse (-) constant in Eq. (4-9) (kg-q) mass of explosive or fuel (kg) constant in Eq. (4-9) (-) Mach number (flame speed divided by us) (-) number of fatalities (-) constant in Eq. (4-9) (-) overpressure (N·m-2 or bar) probability of a certain accident (-)
'P P0 'P q
scaled or dimensionless side-on peak overpressure (-) ambient pressure (N·m-2 or bar) side-on peak overpressure (N·m-2 or bar) constant in Eq. (4-9) (-)
R t t+
combustion energy-scaled distance (-) time (s) positive phase duration (s)
t us V WTNT Z
K
dimensionless positive phase duration (-) ambient speed of sound (m s-1) volume of obstacles (m3) equivalent mass of TNT (kg) ratio defined in Eq. (4-1) explosion yield factor (-)
REFERENCES
[1] J. A. Vílchez, S. Sevilla, H. Montiel, J. Casal. J. Loss Prev. Process Ind. 8 (1995) 87. [2] A. Ronza, S. Félez, R. M. Darbra, S. Carol, J. A. Vílchez, J. Casal. Loss Prev. Process Ind. 16 (2003) 551-560. [3] S. Carol, J. A.Vílchez, J. Casal. Loss Prev. Process Ind. 15 (2002) 517.
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[4] D. A. Crowl, J. F. Louvar. Chemical Process Safety. Fundamentals with Applications. Prentice Hall PTR. Upper Saddle River, 2002. [5] CCPS. Guidelines for Evaluating the Characteristics of Vapor Cloud Explosions, Flash Fires and BLEVEs. AIChE. New York, 1994. [6] D. K. J. Pritchard. Loss Prev. Process Ind., 3 (1989) 187. [7] W. E. Baker, P. A. Cox, P. S. Westine, J. J. Kulesz, R. A. Strehlow. Explosions Hazards and Evaluation. Elsevier S. P. C. Amsterdam, 1983. [8] D. S. Burgess, M. G. Zabetakis. Detonation of a flammable cloud following a propane pipeline break. The December 9, 1970, Explosion in Port Hudson, Mo. Bureau of Mines Report of Investigations No. 7752, 1973. [9] R. G. Sachs. The Dependence of Blast on Ambient Pressure and Temperature. BRL Report 466, Aberdeen Proving Ground, Maryland (1944). [10] V. C. Marshall. The Siting and Construction of Control Buildings. A Strategic Approach. I. Chem. E. Symp. Series, No. 47, 1976. [11] F. Díaz.Alonso, E. González-Farradás, J. F. Sánchez Pérez, A. Miñana Aznar, J. Ruíz Gimeno, J. Martínez Alonso J. Loss Prev. Process Ind. 19 (2006) 724. [12] A. C. van den Berg. J. Hazardous Mater. 12 (1985) 1. [13] A. C. van den Berg, A. Lannoy. J. Hazardous Mater. 34 (1993) 151. [14] CPR-E14, Methods for the Calculation of the Physical Effects of the Accidental Release of Dangerous Goods (Liquids and Gases). Committee for the Prevention of Disasters. The Director-General for Social Affair and Employment. Voorburg, 1997. [15] W. P. M. Mercx, A. C. van den Berg, C. J. Hayhurst, N. J. Robertson, K. C. Moran. Hazardous Mater. 71 (2000) 301. [16] A. J. Pierorazio, J. K. Thomas, Q. A. Baker, D. E. Ketchum. Process Safety Progr. 24 (2005) 59. [17] Q. A. Baker. Process Safety Progr. 15 (1996) 106. [18] Q. A. Baker, D. B. Olson, R. H. Bennett. Explosion Dynamics Analysis of the March 23, 2005 Explosion at the BP, Texas City Refinery Isomerization Unit. 2005. Investigation Final Report, Appendix 18. Available at: http://www.bp.com/genericarticle.do?categoryId=9005029&contentId=7015905 [19] W. E. Martinsen. The Quest Quarterly, 4 (1999) 1-4. Available at: http://www.questconsult.com/99-summer.pdf [20] J. Lobato, P. Cañizares, M. A. Rodrigo, C. Sáez, J. L. Linares. Int. J. Hydrogen Energy (in press). [21] Marshall, V. C. The Chemical Engineer, August (1977) 573. [22] Carol, S., Vílchez, J. A., Casal, J. Safety Science 39 (2001) 205.
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Chapter 5
BLEVEs and vessel explosions 1 INTRODUCTION Pressurized vessels can have a high energy content which, if suddenly released, originates a mechanical explosion. Examples of vessel explosion are the sudden rupture of a boiler or the explosion of a compressed air tank. A specific case of vessel explosions are BLEVEs. Boiling Liquid Expanding Vapour Explosions were defined by Walls [1], one of the first to propose the acronym BLEVE, as “a failure of a major container into two or more pieces occurring at a moment when the container liquid is at a temperature above its boiling point at normal atmospheric pressure”. Subsequently, Reid [2] defined BLEVEs as “the sudden loss of containment of a liquid that is at a superheated temperature for atmospheric conditions”. More recently [3] they have been defined as “an explosion resulting from the failure of a vessel containing a liquid at a temperature significantly above its boiling point at normal atmospheric pressure”. As all these definitions only refer to the explosion of a vessel, strictly speaking BLEVEs do not necessarily imply thermal effects. However, if as usually happens the material contained in the vessel is flammable, the explosion is followed by rapid combustion of the fuel, giving rise to a fireball immediately after the explosion. Therefore, in practice a BLEVE is usually associated with a fireball, making it an accident that combines both the mechanical effects of an explosion and the thermal effects of a fire. Due to this special feature, it is one of the most severe accidents that can happen in the process industry or in the transportation of hazardous materials. The substances that can lead to a BLEVE (propane, butane, water, etc.) are relatively common in the industry and are usually transported by car or rail, meaning that BLEVEs occur with a certain frequency. A historical survey performed by the authors using information contained in several databases showed the occurrence of 29 accidents of this type between 1980 and 2004 (Table 5-1). As can be seen, in half of these accidents people were killed (Mexico City 1984 being an exceptional case). A survey of 77 accidents that occurred between 1941 and 1990 [4] mentions 900 fatalities and 9,000 injured. In fact, as awareness of BLEVEs has increased, their consequences on the population (and on firefighters) have gradually decreased. An analysis of 70 accidents that occurred between 1970 and 2004 led to the determination of the most frequent causes of BLEVEs (Table 5-2). Train derailment, immediately followed by leaks caused by an impact and a fire, was the most frequent cause. Road tanker accidents and external fire had approximately the same incidence, followed by loading/unloading operations, overfilling and runaway reactions. Finally, various other factors (ruptured hose,
147
overpressure, mechanical failure, instrument failure, ship collision, etc.) were also responsible, but at a low frequency. Table 5-1 BLEVE accidents occurring between 1980 and 2004 Date Place Material 1980 Los Angeles, Gasoline USA 1980 Rotterdam, The LPG Netherlands 1981 Montonas, Chlorine México 1982 Spencer, USA Water 1982 Louisiana, USA Vinyl chloride 1982 Taft, USA Acroleine 1982 Tyne and Wear, LPG UK 1983 Reserve, USA Chlorobutadiene 1983 Houston, USA Methyl bromide 1983 Murdock, USA Propane 1984 Romeoville, USA Propane 1984 Cleveland, USA LPG 1984 Mexico City LPG 1985 1985
Priolo, Italy Pine Bluff, USA
1986 1987
Kennedy S C, USA Cairns, Australia
1988 1988
Philadelphia, USA Kings Ripton, UK
Gasoline LPG
1989
Alma Ata, Mongolia St. Peters, Australia Lyon, France
1990 1991 1995 1996
1998 1998 1999 1999 2000
Ethylene Ethylene, ethylene oxide Hydrogen
Cause Tanker, road accident
Fatalities 2
External fire in a bus station Derailment, impact, fire
29
Overheating Derailment, impact, fire Runaway reaction Tank, external fire
7 0 0 -
Runaway reaction Overfilling Derailment, impact, fire Weld failure External fire on vessel Leak and fire in storage park Leak, jet fire on tanks Derailment, impact, fire
3 2 0 15 0 500 0 0
Fire
7 0
LPG?
Human error (hose disconnected), fire Tanker, road accident Human error (hose): leak while filling vessel, fire Train collision, fire
LPG
Tank, external fire
-
Propane
-
LPG
La Plata, Argentina Paese, Italy
Propane
Albert City, USA Xian, China Dortyol, Turkey KamenaVourla, Greece Downey, USA
Propane LPG LPG LPG
External fire on small vessel Overfilling (human error) Human error while unloading a tanker, release, fire Car breakes pipes, fire Leak, fire on storage park Human error Road accident, leak, fire
Propane
Leak, fire
Propane
148
0 0
5
2 0
2 11 4 -
Some of these causes occurred simultaneously (external fire can occur immediately after derailment). Human error accounted for a number of accidents (road and train accidents, overfilling, loading/unloading, etc.). 48% of accidents occurred during transportation. An important characteristic of BLEVE is that given the right conditions —road accident involving a tanker, fire in a process plant etc.— it can happen at any moment, without warning. Thus, under these circumstances, it is best to quickly evacuate the population from within an appropriate distance; all other mitigating measures —e.g., cooling of vessels— must have been implemented before and should be applied automatically, without any presence of firemen within the hazardous zone. Table 5-2 The Most Frequent Causes of BLEVEs Cause % Train derailment 33 External fire 17 Loading/unloading 16 Road tanker accident 14 Overfilling 4 Runaway reaction 4 Other 12
In this chapter, the main features of BLEVEs and vessel explosions are discussed, as well as the methodology to evaluate their effects. In fact, the methodologies used to estimate the overpressure from a BLEVE can also be applied to the explosion of any pressurized vessel. 2 MECHANISM OF BLEVE If a tank containing a pressurized liquid is heated — for example, due to a fire — the pressure inside it will increase. At a certain moment, as heating is not homogeneous, a crack can be initiated at a hot location. This is most likely to occur in the top section of the container, where the walls are not in contact with the liquid and therefore not cooled by it; the temperature of the walls will increase and their mechanical resistance will decrease. In contrast, the wall in contact with the liquid will transfer heat to the liquid, thus maintaining a much lower temperature. Also, if a safety valve opens, the boiling liquid will have a stronger cooling action, due to the heat of evaporation: usually the safety valve is designed taking into account the fire action, and its release capacity should be able to keep the pressure inside the vessel below a given value. Once the crack is initiated, it will probably progress (although in exceptional cases this does not happen) leading to the catastrophic failure of the tank. Upon failure, due to the instantaneous depressurisation, the temperature of the liquid will be greater than that corresponding to the new pressure according to the saturation curve in the P-T diagram. In this unstable condition, it is called a ‘superheated’ liquid. Liquids can normally withstand a small amount of superheating, which under certain experimental conditions can be extended far above the atmospheric pressure boiling point [5]. Thus, at the moment of depressurisation a sudden flash of a fraction of the liquid will occur; a two-phase liquid/vapour mixture will then be released. This phenomenon occurs within a very short period of time. The significant increase in the liquid’s volume when it vaporizes —1750 times in the case of water and 250 times in the case of propane— plus the expansion of the preexisting vapour, will give rise to a strong pressure wave (explosion, bursting of the container),
149
as well as causing the container to break into several pieces that will be propelled over considerable distances. Experimental work performed with small 1 litre vessels [6] has shown that, when there is a break in the vessel, the pressure drops slightly and then rises up to a maximum; the initial depressurisation brings the fluid near the break to a superheated state, thus causing a local explosion. If the substance involved is not combustible, the pressure wave and the missiles will be the only effects of the explosion. This would happen if a steam boiler (water steam) exploded. If the substance is a fuel, however, as often happens in the process industry (for example, liquefied petroleum gas such as butane or propane), the mixture of liquid/gas released by the explosion will probably ignite, giving rise to a fireball approximately hemispherical in shape, initially at ground level. The effect of the thermal radiation in this first stage, which is usually only a couple of seconds, is very important. The whole mass of fuel can only burn at its periphery, as there is no air inside the mass (the mixture is outside the flammability limits). In the area under the fireball there can be some rain-out, causing additional fire effects. For unprotected people, this area should be considered lethal. Later on, the turbulence of the fire entrains air into the fireball. Simultaneously, the thermal radiation vaporizes the liquid droplets and heats the mixture. As a result of these processes, the entire mass increases in volume turbulently, evolving towards an approximately spherical shape that rises, leaving a wake of variable diameter. Such fireballs can be very large, resulting in very strong thermal radiation. In fact, not all the fuel initially contained in the tank is involved in the fire. Some of the fuel is sucked into the wake formed by the flying fragments. In one case (Mexico City, 1984), it has been suggested that a portion of the liquid was thrown significant distances without being ignited, which caused local fires (this effect has not been mentioned in any other case). This decreases the amount of fuel contained within the fireball and also affects its dimensions and the duration of the fire. The combined action of a BLEVE-fireball can be summarized, therefore, by the following effects: í thermal radiation í pressure wave í flying fragments. The way in which these effects are realised varies: punctual or directional in the case of projectiles, and zonal (covering a given surface) in the case of thermal radiation and blast. It is worth noting that it is practically impossible to establish the exact instant at which the explosion will take place. In fact, it can happen at any moment from the beginning of the emergency; in the San Juan Ixhuatepec accident in Mexico City [7], the time elapsed between the first explosion (which caused the fire) and the first BLEVE was only 69 seconds. The instant at which a BLEVE can occur in a tank exposed to fire depends on the following factors: í thermal flux from the fire, which will be a function of the distance from the flame to the tank and will depend on whether there is flame impingement and the type of flame (pool-fire, torching, etc.) í diameter of the tank í tank fill level í release capacity of safety valves í existence of a layer of fireproof material (passive protection). Theoretically, a thermally insulated container should resist the effect of the flames from a pool-fire (thermal flux of approximately 100 kW·m-2) for 2 hours. In the case of a jet fire, the
150
thermal flux increases significantly (up to 350 kW·m-2). Under these conditions, some BLEVEs have occurred within the first few minutes. For the generation of this type of accident, the following times have been suggested [8]: flame impingement from a jet fire, 5 min; flame impingement with turbulent flames, 30 min (this value agrees with that proposed by ASTM [9], 20 to 30 min). Although this time can vary according to the features of the installation (insulating layer, cooling devices, etc), it is clear that other factors can decrease it significantly (partial destruction due to impact, a pressure wave or local heating, for example). Clearly the most prudent response is to take into account that the explosion can occur at any moment from the beginning of the emergency. Protection measures for the people in the exclusion area should be taken and, eventually, this area should be rapidly evacuated. Several theories have been proposed to explain the BLEVE phenomenon; however, to date, no single hypothesis has been fully accepted.
Pressure
2.1 Liquid superheating While the explosion of a tank containing a pressurized flammable liquid will almost always lead to severe mechanical effects and to a fireball, according to some authors the explosion cannot always be considered strictly a BLEVE. To qualify as this type of explosion, the following conditions should be met. í Significant superheating of the liquid. Most liquefied gases under fire attack (LPG, ammonia, chlorine) fulfil this condition; it can also be fulfilled by other liquids held in closed containers which undergo anomalous heating, for example due to a fire. Also, as stated before, water can be in this condition upon instantaneous depressurisation. í Instantaneous depressurisation. This phenomenon is usually related to the type of failure of the vessel. The sudden pressure drop in the container upon failure causes the liquid to superheat. If the superheating is significant, the flash may be explosive. Isotherm Saturated liquid line CP
A
Saturated vapor line
E F
B D
C
G
Spinodal lines
H
I
Specific volume
Fig. 5-1. Liquid-vapour equilibrium under depressurization at a constant temperature. Taken from [10], by permission.
151
Equilibrium thermodynamics states that the temperature and the pressure establishing the coexistence of a liquid phase with a vapour phase are not independent. Fig. 5-1 shows the saturated liquid and vapour curves, separating the two-phase state of liquid-vapour equilibrium from the liquid and vapour single phases [10]. The curves join at the critical point (CP). In this figure an isothermal line (ABDFG) has also been plotted. The AB and FG strokes show equilibrium conditions for liquid and vapour respectively. The BDF stroke shows all the possible liquid-vapour equilibrium conditions (liquid-vapour equilibrium mixtures) at a given temperature. This figure also illustrates what happens when a compressed liquid (A) undergoes a depressurisation. The liquid evolves at a constant temperature until B without modifying its condition (liquid). When point B is reached the liquid is no longer stable and changes to a two-phase configuration (indicated by a point on the BDF line), in which the liquid and vapour phases coexist. If the two-phase configuration remains the same and the temperature is constant the pressure will also be constant. However, the real process appears to be more complex. If we acknowledge that the Redlich-Kwong (RK) equation of state adequately describes the behaviour of the thermodynamic system and can be used to calculate the isothermal, then in the two-phase region of the P-v diagram the BDF stroke assumes quite a different shape (BCDEF), with a maximum and a minimum. The equilibrium and stability criteria force the isothermal process to comply with the bound (wP/wv)T < 0. Fig. 5-1 shows that the liquid phase can be present up to point C for pressures lower than the saturation pressure. On the other side of the two-phase region, the vapour phase can exist till point E for pressures higher than the saturation pressure. The BC stroke represents metastable states of superheated liquid. These are in fact unstable states which under certain perturbations evolve towards a stable two-phase configuration. The EF line represents metastable states of supercooled vapour. Finally, the CDE line represents completely unstable states corresponding to two-phase liquid-vapour systems. Metastable states are configurations close to an equilibrium state. They are stable for small perturbations but unstable for large ones. Thus, after a certain time and given a metastable liquid a two-phase condition must be expected. Metastable conditions are often a result of fast thermodynamic processes. Thus, a superheated liquid can be described as a metastable state found at a higher temperature than the saturation temperature corresponding to the actual pressure. This metastable state is expected to evolve towards a liquid-vapour equilibrium state. The RK isothermal curve in Fig. 5-1 shows that the thermodynamic criteria for equilibrium and stability imply that the metastable liquid states are only represented by the stroke BC. Point C, the minimum in the isothermal curve, represents the limit pressure at which a metastable liquid can be found for that specific temperature. The locus of the isothermals’ minima is called the liquid spinodal curve (CP-H in Fig. 5-1). The locus of the isothermals’ maxima is the spinodal curve of the vapour phase (CP-I in Fig. 5-1). They meet at the critical point. In Figure 5-3 the liquid spinodal curve has been plotted in a P-T diagram together with the liquid-vapour saturation curve. It is clear that, from a certain point, a metastable condition can be reached by abruptly reducing the pressure from the initial pressure to atmospheric pressure (SH line). In the same figure, the tangent to the saturation curve at the critical point has also been plotted.
152
2.2 Superheat limit temperature In Fig. 5-2-a a vessel containing a liquid in equilibrium with its vapour phase is shown [10]. It is at temperature T and pressure P, which is higher than atmospheric pressure Po. Fig. 5-2-b shows what can happen when the vessel undergoes rapid depressurization to Po. The fluid remains at temperature T with a pressure lower than the equilibrium pressure, so the superheated liquid reaches a metastable state. Fig. 5-2-c shows the —immediate— final condition of thermodynamic equilibrium, where a part of the liquid vaporises in order to make the system temperature the same as the saturation temperature, To. Vapor
Vapor
Liquid
Liquid
P
Po
101.3 kN m 2
Po
101.3 kN m 2
T
Ts ( P )
T
hl
hl (T )
To
Ts ( Po )
hg
hg (T )
hl (T , Po ) | hl (T )
hlo
hlo (To )
hgo
hgo (To )
a
b
c
Fig. 5-2. A hot liquid undergoing sudden depressurization in a tank.
When the two aforementioned conditions —significant superheating and instantaneous depressurization— are met, a practically instantaneous evaporation of the contents takes place, with the formation of a large number of boiling nuclei throughout the liquid mass (homogeneous nucleation). Under these conditions, the velocity at which the volume increases is extremely high and the explosion is therefore very violent. Various authors have suggested procedures to establish a superheat limit temperature for each substance that, according to this theory, would determine the minimum temperature/pressure conditions under which a BLEVE can occur. Reid [2, 11] made a significant contribution to this field. Consider a vessel containing a pressurized liquid (for example, liquefied petroleum gas) at room temperature, in which liquid and vapour are at equilibrium at a certain pressure (Fig. 53). If, due to the thermal radiation from a fire, the temperature increases, the pressure inside the vessel will increase up to a certain pressure P’ ranging between P0 and P. If the vessel bursts under these conditions (due to the failure of the material or an impact, for example), there will be an instantaneous depressurisation from P’ to the atmospheric pressure. The depressurisation process corresponds to a drop in the pressure value equivalent to the length of the vertical line between P’ and P0. The superheating limit temperature theory states that strictly speaking there will be no BLEVE: although there will be a strong instantaneous vaporization and even an explosion, nucleation throughout the liquid mass will not occur (nevertheless, after the vessel failure probably there will be a fireball.
153
CP Pressure
Pc
S
P
Liquid-vapor saturation line Po
R
Liquid spinodal line
Tangent at CP
O
T
H Tls
To
Temperature
Fig. 5-3. Liquid spinodal curve and tangent to the saturation line at the critical point. Taken from [10], by permission.
Instead, if during the heating process the pressure evolves up to point S and the liquid temperature reaches a given value (temperature Tls in Fig. 5-3), during the depressurization the liquid spinodal line will be reached and the conditions required (superheating) for the aforementioned spontaneous homogeneous nucleation will be met: according to this theory, a BLEVE explosion will occur. This limit value depends on the substance, and is not exactly known. Diverse criteria have been proposed to establish it. From the RK equation, in accordance with the law of the corresponding states, various authors have proposed several simplified equations to assess the superheat limit temperature. The following one [11] is widely used:
Tsl Tc
0.895 Tc
(5-1)
where Tc is the critical temperature (K). The values of Tsl-Tc for a set of substances have been included in Table 5-3. Some authors have suggested estimating the value of the superheat limit temperature from the tangent line to the vapour pressure-temperature curve at the critical point (Figs. 5-3 and 54). This criterion, although not rigorous at all (it does not have any thermodynamic basis), implies a safety margin with respect to other criteria such as, for example, the one based on the use of the liquid spinodal line (Tsl-RK)). To apply it, the relationship between vapour pressure and temperature can be established by the Clausius-Clapeyron equation (however, it should be noted that, in fact, this equation should only be applied far from the critical point; thus, this method implies a certain error):
ln P
-
A B T
(5-2)
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The tangent to the saturation curve at the critical point is obtained by calculating the derivative of pressure with respect to temperature: dP dT
A
P T2
(5-3)
By applying this expression to the critical point, dPc dTc
Pc A Tc2
tgD
(5-4)
This expression gives the slope of the line tangent to the saturation curve at the critical point. The equation of this straight line is: P
tgD T b
(5-5)
An example will show how the superheat limit temperature can be calculated by two different methods. ______________________________________ Example 5-1 The superheat limit temperature will be calculated for butane. The equilibrium data corresponding to the critical point and to atmospheric pressure are: Pc = 38 bar (37.5 atm) Tc = 425.8 K P = 1.013 bar (1 atm) T = 272.5 K Solution By applying Eq. (5-1):
Tsl Tc
0.895 Tc
0.895 425.1 380.5 K
Tsl can also be obtained —with a safety margin— from Eq. (5-5). By introducing the equilibrium data into the Clausius-Clapeyron equation, the values of constants A and B for butane are calculated (the pressure expressed in bar and the temperature in K): A = 2751, B = 10.11 The slope of the tangent to the saturation curve at the critical point is therefore: tgD
§ 2751 · 38 ¨ 2 ¸ © 425.1 ¹
0.578
The value of the ordinate at the origin, b, can be found by again introducing the values corresponding to the critical point, thus obtaining b = -208. In this way, the equation of the tangent line is obtained:
155
P
0.578 T 208
Its intersection with the horizontal line at P = 1.013 bar gives a temperature of Tsl-t = 88.6 °C (Fig. 5-4).
Fig. 5-4. Saturation curve for butane and tangent at the critical point
Therefore, for a vessel containing liquid butane the minimum temperature required to achieve a degree of superheating that will cause spontaneous nucleation (and therefore BLEVE, according to the superheat limit temperature criterion) upon vessel failure ranges between 88.6 ºC and 107.5 ºC; above this range of temperatures —according to certain authors— spontaneous nucleation would occur under depressurisation and, thus, the explosiuon would be more severe. ______________________________________ In fact, the use of the tangent line to the saturation curve at the critical point as the limiting value for the occurrence of a BLEVE implies a margin of safety. The experimental data seem to indicate that, for most substances, the difference between the superheat limit temperature (Tsl) required to generate a BLEVE and the value obtained in this way (Tsl-t) is in the range of 15 to 20 ºC. Table 5-3 shows the values of Tsl-RK and Tsl-t for a set of substances. 2.3 Superheat limit temperature from energy balance A new approach to the superheat limit temperature has been proposed recently [10], based on the energy balance in the initial liquid mass just before the explosion. In Fig. 5-2 it can be seen that when the liquid is in equilibrium with its vapour it is possible to assign a specific enthalpy hl to the liquid and a specific enthalpy hg to the vapour.
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When the vessel suddenly depressurises (Fig. 5-2 b), the liquid reaches pressure Po and the temperature stays at T, and taking into account that the enthalpy of a liquid only varies slightly with the pressure, the liquid will still have an enthalpy that is basically equal to hl. In the end, when the liquid vaporises, the enthalpy associated with the vapour phase will be hgo, corresponding to the vapour enthalpy at atmospheric pressure and at the corresponding saturation temperature To. Table 5-3
Calculation of the superheat limit temperature for a variety of substances Substance To , K Tc , K Pc , atm Tsl-t , K Tsl-Tc , K Tsl-RK , K Tsl-E , K tgD Water CO2 Ammonia Methane Ethane Ethylene Propane Propylene n-Butane n-Pentane n-Hexane n-Heptane n-Octane Chlorine
373.2 194.7 239.8 111.6 184.6 169.3 231.1 225.5 272.7 309.2 341.9 371.6 398.8 238.4
647.0 304.0 406.0 191.0 305.0 282.7 369.8 365.3 425.8 470.2 507.8 539.8 569.2 419.0
217.7 73.0 112.3 45.8 48.8 50.9 43.0 45.0 37.5 33.0 29.5 26.8 24.7 93.5
2.4656 1.8429 1.8878 1.2856 0.9499 1.0534 0.7286 0.7520 0.5906 0.4717 0.4056 0.3597 0.3257 1.3368
559 265 347 156 255 235 312 307 361 403 436 467 497 350
579.0 272.0 363.4 170.9 273.0 253.0 331.0 327.0 381.1 420.8 454.5 483.1 509.4 375.0
573.0 272.5 363.1 170.9 273.8 253.3 332.0 327.8 381.5 421.5 456.0 482.0 511.5 372.2
606.4 280.2 375.2 176.8 278.9 260.8 326.8 325.4 363.0 393.7 421.6 444.4 466.8 378.3
In an adiabatic vaporization process, the fraction of liquid which is vaporised can obtain the required energy only from another liquid mass which is cooled. If qv is the required vaporisation energy per unit mass (kJ/kg), it can be expressed as a function of the enthalpy according to the following expression: qv
hgo hl
(5-6)
If ql is the heat (also per unit mass, kJ/kg) which can be released by the remaining liquid fraction when it is cooled from the initial temperature to the boiling temperature at atmospheric pressure (To), it can be expressed as: ql
hl hlo
(5-7)
where hlo is the enthalpy of the liquid at temperature To. ql will increase with the difference T-To (the superheating degree of the liquid), while qv will decrease as T-To increases. Therefore, there will be a temperature Tsl-E at which the following expression will be true: qv
ql
(5-8)
i. e.:
157
hgo hl
hl hlo
(5-9)
Eqs. (5-8) or (5-9) imply the condition that the liquid mass which cools is equal to the liquid mass which is vaporised (remember that in these expressions the enthalpies are expressed in kJ·kg-1). This means that Tsl-E defines the situation in which 50% of the liquid mass can be vaporised thanks to the energy released by the other 50% of the liquid which cools from Tsl-E to To. This situation is represented in the P vs. h diagram in Figure 5-5. Below temperature Tsl-E (for example, at a temperature T), the fraction of the liquid that cools from T to To will not release the energy required to vaporise the same amount of liquid. Instead, with superheating at temperatures higher than Tsl-E when less than 50 % of the liquid mass would cool down, more energy would be released than that required to vaporise the rest of the liquid. Pc
CP
Pressure
Saturated liquid line
Saturated vapor line T>Tsl-E
hl-hlo
hgo-hl
P
T=Tsl-E
Isotherm lines
T
(5-24)
@
By substituting the value of x in Eq. (5-22) or (5-23), 'U is found. The equivalent mass of TNT is therefore: WTNT
0.214 'U
(5-25)
4.3 Pressure wave The evolution of the overpressure from the explosion of any vessel containing superheated liquid and vapour exhibits usually two peaks separated by a very short time [25, 26]. The first peak corresponds to the expansion of the vapour and the second to the violent vaporization of the liquid. The relative magnitude of both peak overpressures cannot be deduced from
169
experimental data, as those shown in Fig. 5-10 [25], as the overpressure from the liquid flashing is significantly directional and, thus, the registered data will depend on the position of the measurement device. Furthermore, it will depend on the mass fractions of liquid and vapour in the vessel just before the explosion. Although the blast waves from the two phenomena are often separated, the conservative assumption that both are combined is sometimes met. A third peak can be observed, corresponding to the combustion process, if the substance contained in the vessel is a fuel.
Fig. 5-10. Overpressure as a function of time for a BLEVE of propane (vessel of 5.7 m3 filled up to 80%); data recorded at 15 m from the vessel [25].
The side-on peak value of the pressure wave generated by the explosion can be estimated, once the energy involved in the explosion is known, from the equivalent TNT mass. This method implies a certain inaccuracy, as in the explosion of a vessel the energy is released at a lower velocity than in a TNT explosion and also because the volume of the vessel is much larger than that which would contain the equivalent amount of a conventional explosive. Nevertheless, the method is simple and allows useful estimations to be made. Other methods have been explained in Chapter 4. Due to the fact that the volume initially occupied by the energy released in the explosion is much larger than that which would be occupied by the equivalent mass of TNT, a correction must be made in the distance from the centre of the explosion to the point at which the pressure wave must be estimated. This correction is performed using the ‘scaled’ distance, dn, based on the similitude principle proposed by Hopkinson (see Chapter 4). As overpressure is a function of the distance and two different explosions do not cause the same overpressure at the same distance from the centre of the explosion, the scaled distance is defined as that at which the overpressure has the same value for both explosions. The scaled distance is related to the real distance and to the equivalent TNT mass by the cubic root law,
170
dn
d E WTNT 1 3
(5-26) -1/3
where dn is the scaled distance (m·kg ), and d is the real distance (m) (from the centre of the explosion) at which the overpressure must be estimated. E is the fraction of the released energy converted into a pressure wave, usually ranging between 0.4 and 0.5 for ductile breaking [27]; a conservative approach is to assume E = 0.5. Neglecting the coefficient E implies a 30 % overestimation of overpressure. From the value of dn it is possible to estimate the overpressure using the 'P vs. dn plot (Chapter 4). In the explosion of vessels (especially in the case of cylindrical vessels) the pressure wave is not equal in all directions but significantly directional; however, this cannot actually be quantified. Tanks can also fail and burst with cold liquid if the tank is sufficiently weakened [28]. In this case, the blast is weak and the resulting fireball will include a considerable ground poolfire. ______________________________________ Example 5-4 This example applies the two methods (isentropic expansion and irreversible expansion respectively) to an estimation of the effects of the explosion of a tank with a volume of 250 m3, filled to 80% capacity with propane (stored as a pressurized liquid at room temperature), which is heated by fire to 55 ºC (19 bar) and bursts; the pressure wave must be estimated at a distance of 180 m from the vessel. Solution The calculations can be solved using real data for propane [29]. Thus, the hypothesis of ideal gas must not be applied and it is not necessary to use average values of propane properties. Table 5-6 shows all the values that define the initial state of propane (first column): temperature, pressure, specific volume of saturated liquid and vapour, internal energy and specific entropy corresponding to the saturation states, mass and volume of vapour and liquid, and internal energy of each phase and of the whole mass. It has been assumed that the volume of the tank does not vary and that no mass is lost (the safety valve is closed). Therefore, the vessel evolves at constant volume until it reaches the explosion temperature of 55 ºC. All data corresponding to this state (explosion state) can be seen in the second column of Table 5-6. The states and processes considered have been plotted schematically in the T-S diagram for propane (Fig. 5-11). The third column of Table 5-6 shows the results obtained assuming an isentropic process: the final state is established by the pressure (P0 = 0.1013 MPa) and the entropy, which has the same value as the entropy just before the explosion. Finally, the fourth column shows the results obtained assuming an adiabatic and irreversible expansion. The variation in the internal energy was obtained from the intersection of the two straight lines of Fig. 5-9. Table 5-7 shows the variation in internal energy, heat and expansion work for the two models analysed. It can be observed that in the case of an adiabatic irreversible expansion the energy released by the explosion is 2.5 times less than the energy obtained when an isentropic process is assumed.
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Table 5-6 Propane properties used in the example Initial statea
Pressure, kPa Temperature, ºC Total mass, kg Mass of liquid, kg Mass of vapour, kg Vapour specific volume, m3·kg-1 Liquid specific volume, m3·kg-1 System specific volume, m3·kg-1 Total vapour volume, m3 Total liquid volume, m3 Total volume, m3 Vapour fraction Vapour specific internal energy, kJ·kg-1 Liquid specific internal energy, kJ·kg-1 System specific internal energy, kJ·kg-1 Total vapour internal energy, MJ Total liquid internal energy, MJ Total internal energy, MJ Vapour specific entropy, kJ·kg-1·K-1 Liquid specific entropy, kJ·kg-1·K-1 System specific entropy, kJ·kg-1·K-1
834.4 20 100956 100054 902 0.05539 0.001999 0.002476 50 200 250 0.008941 549.7 250.3 253.0 500 25040 25540 2.355 1.181 1.192
Explosion stateb 1901 55 100956 100007 949.1 0.02293 0.002282 0.002476 21.8 228.2 250 0.009401 582 349.2 351.4 560 34920 35480 2.33 1.501 1.508
Hypothetical final state (isentropic process)c 101.3 -42.02 100956 51508 49448 0.4136 0.001721 0.2035 20454 89 20543 0.4898 483.7 100.1 288 23920 5150 29070 2.448 0.6068 1.508
Hypothetical final state (adiabatic process and W=-P0'V)d 101.3 -42.02 100956 41288 59668 0.4136 0.001721 0.2452 24681 71 24752 0.591 483.7 100.1 326.8 28860 4133 32990 2.448 0.6068 1.695
a
This and all the other states are assumed to be in thermodynamic equilibrium. There are no temperature or pressure gradients. This state is assumed to be reached because the tank does not change its volume and no mass is lost. c This hypothetical state is specified by P0=101.3 KPa and a specific entropy equal to the tank specific entropy just before the explosion. d This hypothetical state is specified by P0=101.3 kPa and the condition 'U=-P0·'V. b
Table 5-7 Variation in the internal energy and work for the two models From explosion conditions to From explosion conditions to final final conditions (isentropic) conditions (irreversible) -6410 -2490 'U (MJ) Work (MJ) -6410 -2490
Using the equivalence of 4680 J per g of TNT, an approximate equivalent mass of TNT of 1096 kg is obtained for the assumption of isentropic expansion and 532 kg for adiabatic irreversible expansion. Assuming ductile failure of the vessel (approximately 50% of the released energy devoted to pressure wave), for the distance of 180 m, the plot of overpressure vs. scaled distance for TNT (see Chapter 4, Fig. 4-4) gives overpressure values of 5.8 kPa (isentropic expansion) and 4 kPa (irreversible expansion). These values would cause minor damage to house structures (see Chapter 7, Table 7-14). ______________________________________
172
Fig. 5-11. Temperature-entropy diagram for propane.1-2, heating of the tank; 2-3, isentropic (adiabatic and reversible) process; 2-4, adiabatic irreversible process. Taken from [24], by permission.
4.4 Using liquid superheating energy for a quick estimation of 'P A recent work [30] has proposed a new methodology, based on the use of the “liquid superheating energy”, for a quick calculation of the approximate value of 'P. In an adiabatic vaporization process, the fraction of liquid that is vaporized can only obtain the required energy from the remaining liquid mass that is cooled. As commented in section 2.3 of this chapter, if qv is the required vaporization energy per unit mass (kJ/kg), it can be expressed as a function of the enthalpy according to the following expression: qv
hgo hl
(5-6)
where hgo is the enthalpy of saturated vapour at To and hl is the enthalpy of liquid at T. If ql is the heat (also per unit mass, kJ/kg) that can be released by the remaining liquid fraction when it is cooled from the initial temperature to the boiling temperature at atmospheric pressure (To), it can be expressed as: ql
hl hlo
(5-7)
where hl and hlo are the enthalpies of the liquid at temperatures T and To respectively. In all cases, the superheating energy (SE) contained in the superheated liquid with respect to its final state immediately after the explosion (i.e. in equilibrium with its vapour at atmospheric pressure) will be the energy that will be partly converted to work to build the overpressure. Therefore, it seems quite logical to consider this superheating energy as an indicator of the severity of a given explosion. In this analysis, for simplicity, the authors did not take into account the contribution of the expansion of the vapour already existing inside the vessel just before the explosion, which
173
was considered to be negligible compared to the contribution of the liquid vaporization. The error thus introduced is relatively low and depends on the relative volumes of liquid and vapour in the vessel. As an example, for a tank with a volume V containing propane at 55 ºC and 19 bar, the error is 4% if Vliquid = 0.8V, 19% if Vliquid = 0.4V, and 40% if Vliquid = 0.2 V. 1600
Superheating energy, kJ/kg
1400 1200 1000 800 600 400 200 0 370
390
410
430
450
470
490
510
530
550
570
590
610
630
650
Temperature, K
Fig. 5-12. Variation of superheating energy as a function of temperature for water. Taken from [30], by permission.
Taking into account that the variation of enthalpy between two liquid states is very similar to the variation of energy, the difference of enthalpy values on the right-hand side of Equation (5-7) can be assumed to be the superheating energy of a liquid superheated at a temperature T (or, under very specific conditions, at the superheat temperature limit) compared to the energy that it would have if it was in equilibrium at the temperature To, i. e. SE
hl hlo
(5-27)
Superheating energy increases, of course, with liquid temperature. In Figure 5-12, SE is plotted as a function of T for water, showing the high energy content at temperatures even below Tsl-E (606.4 K). The same authors used Tsl-E (the superheat limit temperature from energy balance) and PTsl-E (the equilibrium pressure at Tsl-E) as explosion severity indicators. In Figure 5-13 (PTsl-E – Po) is plotted as a function of (Tsl-E – To) for a set of substances. Each substance is differently located to the others. From a practical point of view, it can be assumed that the closer a substance is to the origin of the coordinates, the more likely it will be to reach its uppermost limit (superheating limit), but less energy will be released when an explosion takes place. This is the case, for example, of nitrogen, n-octane or n-heptane. Instead, water is in the opposite situation: a major increase in temperature (and pressure) is required to reach the superheating limit, but the final explosion will release a very large amount of energy even at temperatures below Tsl-E. This is the reason why the explosions of water boilers are so severe.
174
14000 Water
12000
PTsl-E-Po, kPa
10000 8000 Ammonia
6000 Methanol
4000 Nitrogen
Ethylene
Methane
2000
Propylene
Ethanol
Ethane
n-Pentane Propane n-Heptane n-Butane n-Octane n-Hexane
0 0
20
40
60
80
100
120
140
160
180
200
220
240
Tsl-E-To, K
Fig. 5-13. Diverse substances in a PTsl-E – Po vs. Tsl-E – To plot. As the distance of origin of coordinates increases the released energy converted into overpressure increases. Taken from [30], by permission.
The ratio between the energy converted in the pressure wave and SE was calculated [30] for both the isentropic and the irreversible process, for a set of substances (those in Fig. 5-13). To do this, the “useful” energy of the explosion was multiplied by 0.5 (it is usually accepted that in the ductile breaking of a vessel, approximately 50% of the released energy is converted into overpressure), divided by the SE and finally multiplied by 100 to express it as a percentage. To take into account the ground effect (in a practical case, the explosion will take place at the surface of the earth or slightly above it) these percentages should be multiplied by 2 to account for reflection of overpressure wave on ground. However, if this effect has already corrected in the TNT curve used to determine 'P this correction is not required. By analyzing these data, it was observed that for an isentropic process, the energy devoted to overpressure ranges between 7 and 14% of SE, while for an irreversible process it ranges between 3.6 and 5%. The comparison of these values with experimental data from the literature showed a fairly good agreement. Therefore, this method allows a quick estimation of the 'P for a given vessel to be made, if its content and its temperature just before the explosion are known. ______________________________________ Example 5-5 Suppose the explosion of a vessel containing 2,000 kg of liquid water at T = 553 K and P = 64.24 bar (hl = 1236.5 kJ kg-1). Estimate 'P at a distance of 50 m. Additional data: hlo = 418.9 kJ kg-1; amount of TNT required to release 1 MJ = 0.214 kg. Solution Calculation of the superheating energy: SE = 1236.5 – 418.9 = 817.6 kJ kg-1
175
And, for the whole mass of water: 2000 kg · 817.6 kJ kg-1 = 1635205 kJ a). If an isentropic process is assumed, the maximum energy converted into overpressure will be the 14% of SE: 1635205 kJ · 0.14 = 228930 kJ and the equivalent mass of TNT is: WTNT = 228930 · (0.214 · 10-3) = 49 kg The scaled distance will be: dn
50 3
49
13.7 m kg-1/3
Which gives an approximate peak overpressure (obtained from the 'P vs. dn plot, Fig. 4-4) of 0.1 bar. b). If an irreversible process is assumed, the maximum energy converted into overpressure will be the 5% of SE, and then the following values are obtained: Maximum energy converted into overpressure: 1635205 kJ · 0.05 = 81760 kJ WTNT = 81760 · (0.214 · 10-3) = 17.5 kg dn = 19.2 m kg-1/3
'P = 0.068 bar ______________________________________ 4.5 Estimation of 'P from characteristic curves A new method for establishing the relationship between overpressure, impulse and distance for vessel burst has been proposed recently by González Ferradás et al. [31]. These authors obtained this relationship from the curves given by the Baker method [32]. The equations obtained, referred to as characteristic curves, were fitted by potential equations which depend on explosion energy and geometry of the vessel. A set of characteristic curves were then represented in an overpressure-impulse diagram, including the distances to the explosion centre. Thus, once the explosion energy is known, these diagrams allow the determination of overpressure and impulse as a function of distance for a given vessel geometry. In Figs. 5-14 and 5-15 the characteristic curves are plotted for spherical and cylindrical vessel bursts, respectively.
176
Explosion energy (Eexp, J)
100000
1012
3 1011
10000
1011 200 Distances (m)
3 1010
300
1010
Impulse (Pa·s)
500
1000
1000
3 109
1500 2500
109
4000
3 108
6000
100
10000 108
3 107
10 7 10
25 50 Distances (m ) 100
1 100
1000
10000
100000
1000000
10000000
Over pressure (Pa)
Fig. 5-14. Characteristic curves for a spherical vessel burst. Reprinted from [31] with
permission of John Wiley & Sons, Inc.
______________________________________ Example 5-6 Use the characteristic curves of Fig. 5-15 to estimate the peak overpressure for the explosion of the cylindrical vessel containing liquefied propane mentioned in Example 5-4. Assume adiabatic and irreversible expansion. Solution The energy released in the explosion is 2490 MJ. Interpolation in Fig. 5-15 for a distance of 180 m gives an approximate value of: 'P = 4,000 Pa. ______________________________________
177
Explosion energy (Eexp, J)
100000
1012 3 1011
1011
10000 200
3 1010 300 1010
Impulse (Pa·s)
Distances (m)
500
1000 3 109
1000 1500
109
2500 4000
3 108
6000
100
10000 108
3 107
10 7 10
25 50 100
Distances (m)
1 100
1000
10000
100000
1000000
10000000
Overpressure (Pa)
Fig. 5-15. Characteristic curves for a cylindrical vessel burst. Reprinted from [31] with
permission of John Wiley & Sons, Inc. 4.6 Missiles Projectiles from vessel explosions are one of the most difficult hazards to quantify, because of their random behaviour. The fragments thrown by the explosion have a restricted and directional action, but often with a larger radius of destructive effects than the pressure wave and the acute thermal effects of the fireball. These fragments can cause a domino effect if they destroy other tanks or equipment. The velocity required by a fragment to penetrate another similar tank ranges from 4 to 12 m·s-1, and the maximum velocity that can be reached by the fragments in a vessel explosion —a function of the conditions at which the explosion occurs, the volume of vapour initially contained in the vessel, and the shape of the vessel— ranges from 100 to 200 m·s-1.
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There are basically two kinds of projectiles from a BLEVE, as in the case of conventional container explosions: í primary projectiles, which are major pieces of the container í secondary projectiles, which are generated by the acceleration of nearby objects (pipes, bars, bricks, etc.). Table 5-8 Typical number of fragments Type of vessel Number of fragments Cylinder 2 or 3 Sphere 2 to 15 (usually less than 5)
The number of primary projectiles (i.e., major pieces of the tank) will depend on the type of failure, the heating process, the shape of the vessel, and the severity of the explosion (Table 5-8). Typically, a vessel explosion will involve a ductile failure; the cracks will propagate at lower velocity without branching. The number of fragments will be less than if it were a fragile failure.
Fig. 5-16. Common failure trends in cylindrical vessels.
In the case of cylindrical tanks, the initial crack usually follows an axial direction and then changes and follows a circumferential one (e.g., following a weld). Thus, the vessel is usually broken into two pieces, the bottom of the tank and the remainder of the vessel (Fig. 5-16). However, in cylindrical tanks there can also be 3 projectiles. If there are three fragments, there can be two types of failure. The vessel can be divided into two bottom parts and the central body, or can be divided into two fragments, one bottom and the rest; this second fragment can be divided through the imaginary line that would separate the liquid and the vapour (Fig. 5-16). The bottom usually breaks at the weld or near it; if there is no welding, it can be assumed that it will break at a distance from the end equal to 10% of the total length of the vessel. The projectiles will probably follow the direction of the cylinder axis. This can be seen in Fig. 5-17, which corresponds to the explosion of a road tanker containing liquefied natural gas [33]. The tank was broken into two major pieces. There were one longitudinal and two circular cracks, which did not overlap totally with the welding or with its transition zone. The rear part of the tank (together with part of the truck’s mechanical structure) was ejected to a distance of 80 m. The front part, travelled 125 m, where it struck a house. The motor and cabin covered a distance of 257 m. All the major pieces (both ends of the tank, some parts of the structure, the central section of the tank and the motor) and the location of the road tanker at the moment of the explosion were practically on a straight line (this accident is analized in Example 5-7).
179
R=257 m
Motor
Baffles House Front piece of the tank
Melted aluminium fragments
Remains of axles
Central part of the tank and valves
Baffles
Initial position of the tank Rear piece of the tank
0
50
100
150
200
250 m
Fig. 5-17. Distribution of fragments in the explosion of a cylindrical vessel. Taken from [33], by permission.
However, usually there is a certain scattering. Analysis of 15 accidents [34] provided the data shown in Table 5-9 (Fig. 5-18), taking into account the 45º sectors at each side of the cylinder. Data obtained experimentally with 13 BLEVEs of 400 l vessels [35] gave somewhat different results; in some cases, this author observed that the tank remained flattened on the ground with both ends attached. Table 5-9 Probability of projectile launching in cylindrical vessels Sector Probability 1 2
0.62 0.38
Fig. 5-18. Distribution of projectiles from a cylindrical vessel.
In the case of spherical vessels, it is much more difficult to predict the number of fragments or the direction followed by them. The number of projectiles will be in the range of 2 to 15, although Baum [36, 37] has reported that there will typically be less than five
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projectiles. The analysis of several cases shows that the distribution is not symmetrical; this can probably be attributed to the special position of the contact flame/vessel in each case, although other aspects (e.g., construction details) may also exert an influence. 4.6.1 Range The distance reached by projectiles from cylindrical tanks is usually greater than that reached by fragments from spherical vessels. To calculate it, the initial velocity and the mass of the fragment are required. Nevertheless, the following simple expressions have been suggested [14, 36] for the prediction of the range of cylindrical tank projectiles (tube fragments): for tanks < 5 m3 in capacity: l 90 M 0.33 for tanks > 5 m3 in capacity: l 465 M 0.1
(5-28-a) (5-28-b)
where M is the mass of substance contained in the vessel (kg) and l the range (m). The difference between the two expressions is due to the reduced relative effect of drag (ratio of drag force to tank weight) as tank size increases (when tank capacities exceed 5 m3, the tank size is increased by increasing the tank length, not the tank diameter) [14]. These expressions were obtained assuming the tank is 80% full of liquid LPG at the time of failure, and for fragments launched at an optimal angle (45º to the horizontal). As most fragments will not be launched at this angle, the real ranges will typically be less (approximately 80%) than those predicted by Eqs. (5-28). In the accident in Mexico City, a projectile from a large cylindrical vessel travelled 1100 m. In case of evacuation, the distance should therefore be at least 1 km for large cylindrical tanks. Concerning the range of projectiles from spherical vessels, analysis of 58 fragments from 7 accidents showed that 70% of the fragments reached distances of less than 200 m. However, fragments from spherical vessels have reached 600 m (Mexico City) and even 700 m. The distance reached is usually smaller in the case of fragments from spherical vessels, as they are less aerodynamic than those from cylindrical tanks. Various theoretical models have been suggested for the prediction of these maximum distances, but they are not very practical, as to apply them the mass and shape of the fragment must be known (see [38, 39] for more detailed information). Birk [35] suggested the following approximate guide for estimating projectile range: í 80 to 90% of fragments fall within 4 times the fireball radius í extreme rocketing fragments may travel up to 15 times the fireball radius í in very severe, rare cases, rocketing fragments may travel up to 30 times the fireball radius. This author suggests that personnel should be evacuated to beyond 15 to 30 times the fireball radius. Finally, Stawczyk [40], who performed a series of BLEVEs with domestic cylindrical vessels (5 and 11 kg capacity) containing propane and propane-butane mixtures, concluded that the greatest danger to life was caused by projectiles. Usually 3-5 main projectiles and several single, smaller fragments were formed. In open space, the largest fragments reached a distance of 70 m and smaller compact elements (e. g., the head) reached 200-300 m. He recommended an exclusion zone with a radius of 300 m for people and an exclusion zone of 50 m to protect the rescue services from the blast wave. In closed rooms, measured blast waves could cause serious damage to people and also disturb the building structure.
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4.6.2 Velocity Recently, Baum [37] proposed the following expressions, which agreed very well with his experimental data —obtained with small vessels containing water— to estimate the upper band of velocities for fragments from horizontal cylindrical pressure vessels containing a high temperature liquid. end-cap missiles: u
1.25 K 0.375 § Psat ¨¨ © P0
· ¸¸ ¹
0.085
us
(5-29)
This expression can be used for initial liquid fractions > 60 %. In it K
P S R 3f
(5-30)
Wm u s2
where P is the pressure in the vessel just before the explosion (bar) Rf is the fragment radius (m) Wm is the missile mass (kg) and us is the sound velocity in the vapour (m s-1) us
J RT
(5-31)
Mw
‘Rocket’ missiles: u
I Wm
(5-32)
where I is the impulse (integrated pressure history of the closed end), which can be calculated as follows: if P > Psat if P = Psat
I I
A P t w Psat t o t w Psat t e Psat A t o t e
(5-33) (5-34)
where A is rocket cross-section tw is the time taken for the rarefaction wave to propagate from the break to the closed end of the rocket to is the time required to achieve a fully open breach, and te is the period required to expel the vessel contents from the open end of the rocket: tw
L
(5-35)
aw
182
te
L am
to
§ W · ¨¨ ¸¸ © Psat S r ¹
(5-36) 12
(5-36)
L is the length of the rocket. aw is the propagation velocity of the depressurization wave inside the vessel; it corresponds approximately to the velocity of sound in the vapour. am is the velocity of flow at the exit cross-section; it corresponds to the velocity of sound in a bubbly flow in thermal equilibrium. ______________________________________ Example 5-7 A cylindrical vessel with a volume of 22 m3 is filled to 80% with propane at room temperature. Upon heating by an external fire, it bursts and is divided into two fragments: a cap end and the remainder of the vessel. Estimate the maximum distance reached by the fragments (Uliquid, 20 ºC = 500 kg·m-3; Uvapour, 20 ºC = 40 kg·m-3). Solution The maximum distance will be that reached by the tube fragment. The mass of butane contained in the vessel is: M
22 0.8 500 22 0.2 40
8975 kg
By applying Eq. (5-28-b):
l
465 8975 0.1
1150 m
Taking into account the overestimation due to the non-optimal launching angle, l | 920 m . ______________________________________ 5 PREVENTIVE MEASURES
In the case of an emergency that can lead to an accident of the BLEVE-fireball type, it is very difficult to improvise adequate actions to control the situation. Any plan requiring the presence of people will be very dangerous, as it is impossible to foresee when the explosion will occur. As mentioned previously, the time elapsed from the instant in which the fire starts until the moment at which the explosion occurs can vary significantly depending on the circumstances. A typical value could be 20 min, although is some cases the time has been much shorter: in the accident of San Juan Ixhuatepec (1984) it was of only 69 seconds. The actions should therefore be preventive and taken beforehand. The risk of a BLEVE can be reduced to tolerable levels if several of these measures can be taken at the same time. These are briefly discussed here.
Sloping ground. The installation must be designed in such a way that any leak of a liquid (for example, liquefied petroleum gas) could be immediately removed from the area in which there is the tank that must be protected. The ground should be smooth and with a slope of
183
2.5% (1.5% minimum); a draining system must lead to a trench or a tank far enough away to avoid contact between the flames and the tank. It must be taken into account that, in case of wind, the flames can have an inclination of 45º as well as a significant drag, and that they can reach approximately twice the diameter of the trench [41].
Thermal insulation. If the walls of the tank are blanketed with a fireproof material (with a low thermal conductivity) the heating of the vessel —and, therefore, its pressure increase— by an eventual fire is significantly delayed. Furthermore, in long emergencies, thermal insulation reduces the heat flow to the system and makes it possible for the safety valve to prevent the explosion (fireproofing relies significantly on the correct operation of the safety valve). It must be taken into account that these valves are not designed to solve these types of emergencies on their own, as their cross section should be excessive. Fireproofing ensures protection for a limited time (usually 4 to 5 hours). It is the most suitable device for road or railway tanks. In any case, thermal insulation should be a complement, and other protective systems (for example, cooling of the vessel) should be installed. Another interesting point is that the structural elements (vessel legs) should also be insulated, to avoid the falling of the vessel under excessive heating (this is what happened with two of the spherical tanks in Mexico City; even after falling, however, surprisingly they did not explode). The thermal insulation should be installed in such a way that it could be effective in the event of a fire and, also, allow the tank surface and structural elements to be inspected periodically. Cooling with water. The usefulness of water spray in protecting vessels exposed to the direct action of fire has been proven over many years. It is important to use the water from the first moments, with a layer of a certain thickness totally covering the wall to be cooled, especially those areas directly in contact with the flame. The required flowrate of water should be kept constant —in some cases, the action of firefighters and the consequent increase in water consumption have considerably decreased the pressure in the network and, thus, the water flowrate to the vessels— with a minimum value that will depend on the circumstances. To protect a fire-engulfed tank, the water flowrate will depend on the circumstances. If the safety valve is correctly designed and works normally, the water flowrate should not be reduced below 10 1·m-2·min-1 if there is direct contact with the flame [42]; a flowrate of 15 l·m-2·min-1 has also been recommended [4, 8]. As a general criterion, a flowrate of 12.5 l·m-2 ·min-1 is a good engineering specification to avoid the explosion in equipment engulfed in fire. To have an efficient cooling effect, water should be applied before the temperature of the wall reaches 80ºC. If there is no flame impingement —only thermal radiation— smaller flowrates can be used. If there is flame impingement on the wall, the thermal flux will depend on the type of flame (for a pool fire it can be approximately 100 kW·m-2, while for a highly turbulent flame it can reach 350 kW·m-2). In this case, for the zone of the wall located above the liquid surface, flowrates even larger than 25 l·min-1·m –2 may be required. Another aspect to be taken into account is that all safety elements —valves, pipes, etc.— should be designed to resist the action of fire and the high temperatures that will be reached during the emergency; otherwise, they will collapse in the first moments, especially if there is direct contact with the flames.
184
Pressure reduction. If pressure is reduced, the walls of the vessel will be exposed to less stress and the risk of explosion if the temperature increases will be lower. As a general criterion, API recommends the installation of devices able to reduce the pressure to approximately 8 bar (absolute) or to half of the design pressure in 15 minutes. If the ground is sloped and the vessel is thermally insulated, this time can be longer. The depressurization can require a remote control valve besides the safety valve. The released material should be eliminated in safe conditions [43], for example with a torch. It should also be taken into account that in some cases a strong depressurization can cause extremely low temperatures, leading to fragile conditions in the steel. Mounding or burying. The possibility of either totally or partially burying the vessel has been suggested, and is used sometimes. This provides good protection against thermal radiation for a very long time period, as well as against missiles and impacts from moving vehicles. However, this measure has some disadvantages, primarily the eventual corrosion in the tank walls, although actually this problem can be prevented. Water barriers. This is a relatively new system in which a set of sprayers generates curtains of fine water spray. The barriers retain the vapour released from the leak —thus reducing the possibility of ignition— and disperse them into the atmosphere. The action of water sprays consists mainly in reducing the density difference between the ambient air and the dense gas. Firstly, the mechanical dilution of the gas cloud by entrainment of the gas and clean air in the spray consists in momentum exchange between the droplets and the gas phase [44]. Secondly, the forced dispersion of the released gas is enhanced by cloud heating. The water barrier effectiveness depends on its own characteristics (droplets distribution, types of nozzles, water pressure, etc.). Protection from mechanical impacts. Tanks containing materials stored at temperatures higher than their boiling temperatures at atmospheric pressure must be protected from impacts from cranes or other equipment or moving vehicles. A special case —not treated here— is the protection of tank cars. Overflow. This is an incident that has caused a number of BLEVEs. Nowadays it is much less common, however, and adequate devices are installed to avoid it (level controls, safety valves). Minimum separation distances. The minimum distances between vessels are usually established by regulations and will not be discussed here. They are important from the point of view of thermal radiation and, particularly, to avoid direct contact between the flames from the fire in one piece of equipment and the wall of another vessel (thermal flux increases significantly if there is flame impingement). They do not guarantee protection, however, in the case of an explosion (blast, projectiles). Actuation on the initiating mechanisms. Diverse systems have been proposed to avoid homogeneous nucleation. These include installing aluminium mesh inside the tank, and adding nuclei which initiate boiling. These systems have proved to be useful for very specific applications (for example, small fuel containers). However, for use in large storage tanks they are still being investigated.
185
6 EXAMPLE CASES
______________________________________ Example 5-8 3 A tank with a volume of 250 m , 80% filled with propane (stored as a pressurized liquid at room temperature), is heated by a fire to 55ºC (~19 bar) and bursts. The pressure wave, as well as the consequences on people, must be estimated at a distance of 180 m (thermal effects corresponding to this case are studied in Chapter 3). Data: Room temperature = 20°C; HR = 50% (partial pressure of water vapour, 1155 Pa); J =1.14; 5
-1
-1
'hv = 4.3·10 J·kg ; 'Hc = 46,000 kJ·kg ; Tc = 369.8 K; Tboil. atm. pres. = 231.1 K; Uliquid, 20 °C = -3 -3 3 -1 -1 500 kg·m , Uliquid, 55 °C = 444 kg·m-3; Uvapour, 55 °C = 37 kg·m ; cp liquid = 2.4·10 J·kg ·K ; Po = 1.9·106 N·m-2.
Solution: First of all, the mass of propane involved is calculated: M
Vl U l , 20º C
(0.8 250 m 3 ) 500 kg m -3
100,000 kg
Estimation of pressure wave. Assuming an adiabatic-irreversible expansion, the overpressure is estimated from Eqs. (5-18), (5-24), and (5-25) (see Example 5-3): Energy released = 2490·103 kJ As the energy released per gram of TNT is approximately 4,680 J,
WTNT = 532 kg Assuming that 45% of the released mechanical energy is transformed into pressure wave (ductile breaking of the vessel):
WTNT overpressure E WTNT dn
d
E WTNT
13
180 2401 3
0.45 532
240 kg
29 m kg -1 3
With the TNT equivalence diagram (overpressure vs. dn, see Chapter 4), an overpressure of 0.041 bar is found. If isentropic expansion is assumed, a more conservative value is obtained: WTNT = 1096 kg, (WTNT)overpressure = 493 kg, dn = 23 m kg-1/3, 'P = 0.055 bar. Consequences on people. The consequences of overpressure on people (vulnerability models, probit equations) are treated in Chapter 7.
186
For 'P = 0.055 bar (the most conservative value) there would not be any direct consequences (lung damage, ear-drum rupture) on people. ______________________________________ ______________________________________ Example 5-9 On June 2002, a road tanker transporting LNG suffered an accident and exploded (one person killed, two injured) in Tivissa (Catalonia, Spain) [33]. After losing control, it turned over, tipping onto its left side and finally stopping. Flames appeared immediately and moments later the tyres started to burn. The flames then increased in size (approximately 13 m high). 20 minutes after the road accident, the tank exploded. Witnesses heard a small explosion, then a strong hiss and then the large explosion. Just after the explosion a white cloud appeared which immediately ignited. The driver died and of two people located at 200 m one suffered first degree burns and the other suffered second degree burns (both were lightly dressed). The tanker (AISI-304 SS) was cylindrical, with a diameter of 2.33 m and a length of 13.5 m. It was protected by an expanded polyurethane external insulation (130 mm thick, selfextinguishing, and covered by a 2 mm aluminium plate). It was designed for a working pressure of 7 bar, the hydraulic test being performed at 9.1 bar. With a volume of 56 m3, 85% of it was filled with liquid. The temperature of the LNG was slightly below -160 ºC and the pressure slightly below 1 bar. There were three safety valves: two 1" valves set to 7 bar and one ¾" valve set to 9 bar, located at the top of the vessel (in the vapour zone). The rear of the tank, with a length of 5 m, was ejected to a distance of 80 m. The front part, with a length of 4 m, travelled 125 m, where it struck a house. The glass windows of the house remained intact. The motor and cabin covered a distance of 257 m. All major pieces and the location of the road-tanker at the moment of the explosion were on a straight line. Calculate the effects of this explosion. As there are no BLEVEs registered with LNG, study whether this explosion could have been a BLEVE.
Solution A survey carried out using the MHIDAS database [33], containing 12,179 accidents, showed that 43% occurred in transportation and 8.6% in road transport. Only 9 involved LNG, and of these only in one was the LNG involved in a fire. Among the 60 BLEVE accidents contained in the database, none was found involving LNG. Thermal effects (see Chapter 3) Assuming that all the mass initially contained in the tank was involved in the fireball, the mass of fuel in the fireball (see table) was 19,050 kg. Liquid Vapour
Volume (m3) 45 10
Density (kg·m-3) 423 1.6
Temperature (K) 111.5 111.5
From this mass the diameter, duration and height can be obtained; for an approximate estimation, constant values of D, H and Ep will be assumed:
D
5.8 M 1 / 3
154 m
t 0.9 M 0, 25 10.6 s H 0.75 D 115 m
187
The position of the radiating surface with respect to the people can now be established (see Fig. 5-19). The two people burnt were at a distance of 200 m from the tanker. The thermal radiation at this distance will now be calculated. Assuming an atmospheric relative humidity of 50% (partial pressure of water vapour, 1155 Nm-2), the atmospheric transmissivity is:
W
2.85 (1155 154) 0.12
0.67
and the view factor: D2 Fmax 0.11 2 ªD º 4 « d» ¬2 ¼
Fig. 5-19. Schematic plot of the fireball.
The fraction of energy radiated (assuming a pressure in the vessel of 10 bar at the moment of the explosion):
K
0.00325 (10 10 5 ) 0.32
0.27
And the emissive power: Ep
K M 'H c S D 2t
0.27 19,050 46,458 S 154 2 10.6
300 kWm 2
The maximum radiation to which a person located at a distance of 200 m is exposed is thus: I 0.67 0.11 300 22 kW m 2
188
As the witnesses were standing, the receiving surface must be considered to be vertical. For a vertical surface the radiation received would be:
Iv
I cos D
22 0.86 19 kWm 2
Taking into account the duration of the fireball (10.6 s), the dose received by those people would be:
dose
t I v
43
5.3 10 6 s (W m -2 ) 4/3
According to information on vulnerability to thermal radiation (Chapter 7), 95% of the exposed individuals should receive first degree burns and 7% second degree burns. This prediction agrees with the observed consequences. Limit superheat temperature If Eq. (5-1) is applied:
Tsl Tc
0.895 Tc 0.895 191 171 K
And if the energy balance approach is applied:
Tsl-E = 176.8 K (see Table 5-3). Thus, according to the criterion exposed in section 2.2, for the explosion to be a BLEVE (if the restrictive criterion based on Tsl is applied) the LNG should have reached the temperature of 171 K or, alternatively, 176.8 K, depending on the procedure followed for calculating the superheat limit temperature from the initial value (113 K). Pressure wave According to the equilibrium data, 171 K corresponds to a pressure in the vessel of 24.5 bar and 176.8 K corresponds to 30.2 bar (Clausius-Clapeyron constants: A = 1029.5, B = 9.23). Therefore, to have a BLEVE (according to the restricted criterion of the superheat temperature limit criterion) the pressure inside the vessel before the explosion should have been at least 24.7 bar. However, taking into account the set values of the safety valves, this seems rather improbable. Furthermore, for this pressure, the explosion would have generated a pressure wave which would have broken the windows (glass and frames) of the house. As this did not happen, the pressure inside the tank at the moment of the explosion must have been lower than that required for a BLEVE according to the aforementioned criterion. Since we know that the glass windows of a house located at 125 m remained intact, this distance will be accepted as the maximum distance for breaking of glass. This corresponds (see Chapter 4) to a pressure wave of approximately 0.03 bar. From the 'P vs. dn plot a scaled distance of 43 m kg-1/3 is found. This implies an equivalent mass of TNT of:
43
125 ; WTNT = 25 kg (WTNT )1 / 3
189
Assuming E = 0.45, WTNT overpressure
25 0.45
55.5 kg
From this value, the pressure inside the tank just before the explosion can be obtained. Through a process of trial and error, a pressure of 8 bar is finally found. This would therefore be the maximum pressure at which the vessel could have exploded. However, taking into account the lack of homogeneity in the pressure wave in the different directions arising from the vessel explosions, this result cannot be considered definitive. In fact, the existence of a first explosion, then a strong hiss before the large explosion seems to confirm the two-step mode for the failure of the vessel typical of some BLEVE explosions. This accident has been recently analyzed by Pitblado [45], who considered that this tanker truck event had characteristics of a BLEVE. NOMENCLATURE
A
constant in the Clausius-Clapeyron equation (-); in Eqs. (5-33) and (5-34), vessel cross section area (m2) a constant (-) am velocity of flow at exit plane (m·s-1) aw propagation velocity of the initial rarefaction wave (m·s-1) B constant in the Clausius-Clapeyron equation (-) b constant (-) c constant (-) cp specific heat at constant pressure (J·kg-1·K-1) D diameter of fireball (m) d distance from the centre of the vessel to the point at which the overpressure must be calculated (m); distance between the flames and the target (m) dn scaled distance (m·kg-1/3) Ep emissive power (kW·m-2) Ev energy released in the vapour expansion (kJ) e constant (-) F view factor (-) f vaporization fraction (-) H height at which the centre of fireball is located (m) h height at which the top of fireball is located (m) 'Hc net heat of combustion (kJ·kg-1) hg enthalpy of the gas at the temperature T (kJ·kg-1) hgo enthalpy of the gas at To (kJ·kg-1) hl enthalpy of the liquid at the temperature T (kJ·kg-1) hlo enthalpy of the liquid at the temperature T0 (kJ·kg-1) 'hv liquid vaporization heat (kJ·kg-1) HR relative humidity (%) I radiation intensity (kW·m-2); in Eqs. (5-32), (5-33) and (5-34), impulse applied to closed end (bar·m2·s) radiation intensity on a vertical surface (kW·m-2) Iv
190
l range of cylindrical tank projectiles (m) L length of “rocket” (m) M mass of substance (kg) Mw molecular weight (kg·kmol-1) m mass of vapour existing initially (kg) N number of fragments (-) P vapour pressure; pressure in the vessel just before the explosion (bar or N·m-2) P0 atmospheric pressure (bar) Pc critical pressure (atm) Psat liquid saturation pressure (bar) Pw partial pressure of water (N·m-2) ql heat released by the remaining liquid (kJ·kg-1) qv heat required to vaporize liquid (kJ·kg-1) R ideal gas constant (8.314 J·mol-1·K-1) r vessel bore radius (m) Rf fragment radius (m) S entropy (kJ kg-1 K-1) SM tensile strength (Pa) T temperature; temperature of the substance at the moment of the explosion (K) Tc critical temperature (K) T0 boiling temperature at atmospheric pressure (K) Tsl superheat temperature limit (K) Tsl-E superheat temperature limit according to the energy balance (K) Tsl-RK superheat temperature limit from the Redlich-Kwong spinodal curve (K) Tsl-Tc superheat temperature limit at atmospheric pressure from Eq. (5-1) (K) Tsl-t superheat temperature limit according to the tangent line to the vapour pressuretemperature curve at the critical point (K) t time (s) te time required to expel the vessel content from the open end of the “rocket” (s) to time required to achieve a fully open breach (s) tw time taken for the rarefaction wave to propagate from the break to the closed end of the “rocket” (s) u missile velocity (m·s-1) Ul internal energy of the liquid (kJ·kg-1) Ug internal energy of the vapour (kJ·kg-1); missile velocity (m·s-1) us sound velocity in the vapour (m·s-1) V volume of vapour in the vessel (m3) Vi initial volume of vapour in the fireball (m3) Vr volume of the spherical vessel (m3) Vl volume of liquid in the vessel just before the explosion (m3) x vapour fraction at the final state of the irreversible process (-) W expansion work (kJ) Wm missile mass (kg) WTNT equivalent mass of TNT (kg) z wall thickness (m) D angle formed by the abscissa axis and the tangent to the saturation curve at the critical point (º); also, angle formed by the line joining the centre of the fireball and the target and the horizontal (º)
191
E J K U Ul Ug W
fraction of the energy released converted in pressure wave (-) ratio of specific heats (-) radiation coefficient (-) density (kg·m-3) liquid density (kg·m-3) vapour density (kg·m-3) atmospheric transmissivity (-) Subscripts l liquid g vapour REFERENCES [1] W. L. Walls, Fire Command, May (1979) 22. [2] R. C. Reid, Science, 203 (1979) 1263. [3] CCPS, Center for Chemical Process Safety,
“Guidelines for Evaluating the Characteristics of Vapour Cloud Explosions, Flash Fires and BLEVEs”, AIChE, New York, 1994. [4] H. Londiche, H. and R. Guillemet, Loss Prevention and Safety Promotion in the Process Industry, 1, 551. J. J. Mewis, H. J. Pasman and E. E. De Rademaeker, eds. Elsevier Science, Amsterdam, 1991. [5] R. W. Prugh, Chem. Eng. Progr., Febr. (1991) 66. [6] C. A. Mcdevitt, C. K. Chan, F. R. Steward and K. N. Tennankore, J. Hazard. Mater., 25, (1990) 169. [7] C. M. Pietersen and S. Cendejas, Analysis of the LPG Accident in San Juan Ixhuatepec, Mexico City, TNO, Report 85-0222, The Hague, 1985. [8] F. N. Nazario, Chem. Eng., Aug. (1988) 102-109. [9] ASTM STP 825, A Guide to the Safe Handling of Hazardous Materials Accidents, American Society of Testing and Materials, Philadelphia, 1983. [10] J. M. Salla, M. Demichela, J. Casal. J. Loss Prevent. Proc. Ind, 19 (2006) 690. [11] R. C. Reid, Am. Sci., 64 (1976) 146. [12] W. Towsend, C. Anderson, J. Zook, and G. Cowgill, Comparison of Thermally Coated and Uninsulated Rail Tank Cars Filled with LPG Subjected to a Fire Environment, U.S. Department of Transport, Report n. FRA-OR8D, 75-32, Washington DC, 1974. [13] R. D. Appleyard, Testing and Evaluation of the Explosafe System as a Method of Controlling the BLEVE, Report TP2740, Transportation Development Centre, Montreal, Canada, 1980. [14] A. M. Birk, J. Loss Prevent. Proc., 8, (1995) 275. [15] A. M. Birk and M. H. Cunningham, J. Hazard. Mater., 48 (1996) 219. [16] Y. W. Gong. W. S. Lin, A. Z. Gu and X. S. Lu, J. Hazard. Mater., A108 (2004) 35. [17] A. M. Birk, Z. Ye, J. Maillette, and M. H. Cunningham, AIChE Symp. Ser. 52 (1993) 122. [18] C. Mans, Ingeniería Química, Nov. (1985) 349. [19] J. E. S. Venart, IChemE Hazards XV, Symp. Ser. N. 148 (2000) 121. [20] T. Roberts and H. Beckett, Hazard Consequences of Jet Fire Interactions with Vessels Containing Pressurized Liquids: Project Final Report. HSL Report R04.029, PS/96/03, Buxton, 1996.
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[21] D. A. Crowl, J. F. Louvar. Chemical Process Safety. Fundamentals with Applications.
2nd. Ed. Prentice Hall PTR. New Jersey, 2002. th [22] R. H. Perry, D. W. Green, J. O. Maloney, eds. Chemical Engineer´s Handbook, 7 ed. McGraw-Hill. New York, 1997. [23] J. M. Smith and H. C. van Ness, Introduction to Chemical Engineering Thermodynamics, 4th ed. McGraw-Hill, New York, 1987. [24] E. Planas, J. M. Salla and J. Casal, J. Loss Prevent. Proc. Ind.17 (2004) 431-437. [25] D. M. Johnson, J. M. Pritchard and M. J. Wickens, Large Catastrophic Releases of Flammable Liquids. Comission of the European Communities Report, contract n. EV4T.0014.UK, 1990. [26] A. Ludwig and W. Balke, Untersuchung der versagensgrenzen eines mit flüssigas gefüllen eisenbahnkesselwagens bei unterfeuerung, Rapport B. A. M. 3215, Berlin, 1999. [27] J. Casal, J. Arnaldos, H. Montiel, E. Planas-Cuchi and J. A. Vílchez, Modelling and understanding BLEVE. The Handbook of Hazardous Materials Spills Technology, 22.122.27. M. Fingas, ed. McGraw-Hill, New York, 2001. [28] A. M. Birk. In Fire and Explosion Hazards, 23, ed. by D. Bradley, D. Drysdale and G. Makhviladze. CRFES, Preston, 2001. [29] R. Tillner-Roth, Fundamental Equations of State. Shaker Verlag, Aachen, 1989 [30] J. Casal, J. M. Salla. J. Hazard. Mater. A137 (2006) 1321. [31] González Ferradás, E., Díaz Alonso, F., Sánchez Pérez, J. F., Miñana Aznar, A., Ruiz Gimeno, J. and Martínez Alonso, J. Process Safety Progr. 25 (2006) 250. [32] Baker, W. E., Kulesz, J. J., Ricker, R. E., Bessey, R. L., Westine, P. S., Parr, V. B., and Oldham, G. A. Workbook for predicting pressure wave and fragment effects of explosing propellant tanks and gas storage vessels. NASA CR-134906, NASA Scientific and Technical Information Office, Washington DC, 1977. [33] E. Planas-Cuchi, N. Gasulla, A. Ventosa and J. Casal, J. Loss Prevent. Proc., 17 (2004) 315. [34] P. L. Holden and A. B. Reeves, Chem. Eng. Symp. Ser., n. 93 (1985) 205. [35] A. M. Birk, J. Loss. Prevent. Proc., 9 (1996) 173. [36] M. R. Baum, ASME J. Pressure Vessel Technol., 110 (1988) 168. [37] M. R. Baum, J. Loss Prevent. Proc., 12 (1999) 137. [38] W. E. Baker, P. A. Cox, P. S. Westine, J. J. Kulesz, and R. A. Strehlow. Explosion Hazards and Evaluation. Elsevier, Amsterdam, 1983. [39] CPD. Methods for the Calculation of Physical Effects , part 2. C. J. H. van der Bosch and R. A. P. M. Weterings, eds. Committee for the Prevention of Disasters, The Hague, 1997. [40] J. Stawczyk, J. Hazard. Mater. B96 (2003)189. [41] T. Kletz, Hydrocarbon Proc., August (1977) 98. [42] T. E. Maddison, The Fire Protection of LPG Storage Vessels. The Design of Water Spray Systems. LPGITA Seminar, U. K. 1989. [43] Y. N. Shebeko, A. P. Shevchuck. and I. M. Smolin, J. Hazard. Mater., 50 (1996) 227. [44] A. Dandrieux, J. P. Dimbour, G. Dusserre. Loss Prevent. Proc., 18 (2005) 245. [45] R. J. Pitblado Hazard. Mater. (in press).
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Chapter 6
Atmospheric dispersion of toxic or flammable clouds 1
INTRODUCTION
In industrial installations most accidents occur due to the loss of containment in pipes and units that are used to transport or store gas or liquid materials, as they do in the transportation by road or rail of these materials. Most of these substances are a threat to health and the environment. Clearly, if the substance released can give rise to a gas or vapour cloud —as occurred in 1984 in the Bhopal accident, the worst accident in the history of the chemical industry— the prediction of its evolution is an important issue. Foreseeing the behaviour of a toxic or flammable release will give the variation in the concentration of the substance involved at different points as a function of time. This allows a reasonable estimation of the accident’s effects on people, both outdoors and indoors, on equipment and on the environment, and provides information that is crucial in designing safety measures and emergency plans. This chapter contains fundamental information for modelling the evolution of gas or vapour clouds resulting from accidental releases. 2
ATMOSPHERIC VARIABLES
The term dispersion is used in accident modelling to describe the evolution of a cloud of toxic or flammable gas or vapour in the atmosphere. The dispersion of such a cloud takes place by diffusion and, essentially, transported by the wind: the cloud moves in the wind’s direction but also perpendicular to the wind, both vertically and horizontally. In the case of gases that are heavier than air, the dispersion may proceed even against the wind direction. It is therefore a complex phenomenon, which, for mathematical modelling, requires various simplifying assumptions. Different meteorological variables influence the atmospheric dispersion of pollutants. The velocity and direction of the wind and atmospheric turbulence significantly affect the dispersion of gas clouds. Humidity and temperature have less of an effect and thermal inversion has a decisive influence, although only in specific cases. The meteorological variables are not constant but change with time, especially on a daily basis and with the seasons. To model the atmospheric dispersion in a representative way, average values of these variables are usually taken for a given zone. Averages of meteorological parameters are usually considered over one-hour periods. The main meteorological variables that affect atmospheric dispersion are described in subsequent paragraphs.
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2.1 Wind Wind is air in motion, which essentially results from atmospheric pressure and the distribution of temperature on the Earth’s surface. It is significantly influenced by the topographical features of the area. Urban heat islands, sea-land breezes, mountain valley winds and wind channelling in river valleys are examples of this influence [1]. The wind has an entrainment effect that leads to the dispersion of a gas cloud. The concentrations in a plume are inversely proportional to wind speed.
Fig. 6-1. Wind rose (Port of Barcelona, 2004). Table 6-1 Information on the wind at a given location (Port of Barcelona, 2004) Wind direction Total 1-2 m s-1 2-3 m s-1 3-5 m s-1 5-7 m s-1 N 1.299 0.446 0.372 0.355 0.109 NNE 3.810 0.440 0.938 1.550 0.7550 NE 8.340 0.898 1.321 3.060 1.402 ENE 7.791 0.875 1.241 2.626 1.510 E 4.468 0.801 0.927 1.630 0.658 ESE 3.335 0.870 1.121 0.984 0.309 SE 2.832 0.778 0.801 0.910 0.315 SSE 3.484 0.652 0.875 1.350 0.498 S 3.970 0.566 0.887 1.659 0.709 SSW 7.694 0.583 0.807 2.128 2.998 SW 7.94 0.898 1.093 2.408 2.059 WSW 8.060 1.985 2.088 2.488 1.098 W 8.998 4.067 3.123 1.442 0.292 WNW 14.759 3.930 5.372 3.850 1.201 NW 6.590 1.710 1.619 2.380 0.767 NNW 1.716 0.698 0.469 0.383 0.154
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7-9 m s-1 0.017 0.097 0.767 0.892 0.355 0.023 0.029 0.086 0.149 1.041 1.327 0.332 0.063 0.349 0.109 0.011
>9 m s-1 0 0.029 0.892 0.646 0.097 0.029 0 0.023 0 0.137 0.154 0.069 0.011 0.057 0.006 0
The information on the wind in a given geographical area is usually provided by a wind rose. A wind rose is a graphical representation of the frequency of the winds according to their direction and velocity (Fig. 6-1). For example, the wind rose plotted in Figure 6-1 indicates that in that location WNW winds were predominant, with velocities reaching 9 m/s. In a wind rose the directions are usually given in 8 or 16 sectors of 45º or 22.5º respectively, which are labelled N, NE, E, SE, S, etc. or N, NNE, NE, ENE, E, etc. North is 0 (or 360) degrees and west is 270 degrees. The wind direction is the direction from which the wind blows: a north wind (N), for example, blows from north to south. The information on the wind can also be given as a table, as shown in Table 6-1; here the frequencies of the different wind velocities are given over twelve months. Wind velocity changes with height in the Earth’s friction layer (the zone up to between 700 m and 1,000 m above sea level. Due to the presence of the ground surface, there is a drag force, parallel to the ground and opposite to the air motion, which hinders this motion; the force is transferred vertically to the entire friction layer and creates a velocity gradient. As a result, wind velocity increases with height and slows at heights close to the ground due to the frictional effect of the ground’s surface. The wind speed profile is a function of the Earth’s roughness (Fig. 6-2) and its variation as a function of height can be expressed using the following expression: u1
§z · u 2 ¨¨ 1 ¸¸ © z2 ¹
D
(6-1)
where u1 = wind velocity at a height z1 u2 = wind velocity at a height z2, and D = a coefficient that depends on atmospheric stability (atmospheric stability classes are defined in Section 2.3) and the surface roughness (see Table 6-2). Table 6-2 Value of coefficient D as a function of the atmospheric stability class Stability class Durban zone Drural zone A 0.15 0.07 B 0.15 0.07 C 0.20 0.10 D 0.25 0.15 E 0.40 0.35 F 0.60 0.55
The surface roughness length, z0, is used to define the influence of the ground roughness (which depends on the existence of trees, buildings, etc.) on the velocity profile and the air’s mechanical turbulence; however, z0 does not account for the effects of large obstacles. Table 6-3 shows the values of z0 for various types of ground surface. Usually, wind velocity is measured at a height of 10 m and this is the value generally used in dispersion calculations. Most accidental releases that can give rise to toxic or flammable clouds occur at a very low height, so this value is quite convenient. However, the wind field has significant turbulence, and this sometimes makes the determination of exact values for wind velocity and direction difficult.
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Table 6-3 Surface roughness length for different situations [2] Ground surface Open water (at least 5 km) Mud flats, snow (no vegetation, no obstacles) Open flat terrain: grass, a few isolated objects Low crops: occasional large obstacles (x/h>20)1 High crops: large scattered obstacles (15<x/h 0.3 bar 2.5% 0% 0.3 >'P > 0.1 bar 0% 0% 'P < 0.1 bar
Table 7-13 summarizes the criteria that are usually applied in risk analysis to estimate the effects of explosions on human beings. 5.2 Consequences of an explosion for buildings and structures The damage caused by an overpressure wave on a building or an industrial facility depends on the peak overpressure and on the impulse, and also on a variety of circumstances, such as the following: eventual reflection due to partial confinement, strength of the installation, etc. Data obtained from real cases are gathered as criteria for predicting the effects of overpressure on buildings and structures, which are usually expressed as a function of peak overpressure. Table 7-14 (modified from [23]) gives detailed information. Probit functions have also been obtained for the effects of explosions on buildings; see [11] for a complete analysis. As an example, two of the expressions in [12] which apply to apartment buildings are included here. Major structural damage: Y
5 0.26 ln V
(7-27)
with V
§ 17500 · ¨ ¸ © 'P ¹
8.4
§ 290 · ¨ ¸ © i ¹
9.3
(7-28)
Collapse: Y
5 0.22 ln V
(7-29)
with V
§ 40000 · ¨ ¸ © 'P ¹
7.4
§ 460 · ¨ ¸ © i ¹
11.3
(7-30)
where 'P is the peak overpressure (N m-2) and i is the impulse (N m-2 s). Equations (7-29) and (7-30) can be applied to buildings of up to four stories; for higher buildings, see [12].
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Table 7-14 Damage to buildings and structures (blast) Peak overpressure, Damage Peak overpressure, kPa bar 0.15 Annoying noise 0.0015 0.2 No structural damage; occasional breaking of 0.002 large window panes already under strain 0.3 Loud noise similar to sonic boom; occasional 0.003 glass failure 0.7 Breakage of small windows under strain 0.007 1 Typical threshold for glass breakage 0.01 2 Probability of 0.95 of no serious damage beyond 0.02 this value; some damage to house ceilings; 50% of window glass broken 3 Limited minor structural damage 0.03 3.5-7 Windows usually shattered; occasional damage to 0.035-0.07 window frames 5 Minor damage to house structures 0.05 7 Collapse of roof of a tank 0.07 8 Partial demolition of houses, made uninhabitable 0.08 7-15 Corrugated asbestos shattered. Corrugated steel or 0.07-0.15 aluminum panels fastenings fail, followed by buckling; wood panel fastenings fail, panels blown in 10 Steel frame of clad buildings slightly distorted 0.1 15 Partial collapse of walls and roofs of houses 0.15 15-20 Unreinforced concrete or cinderblock walls 0.15-0.2 shattered 18 Lower limit of serious structural damage; 50% 0.18 destruction of brickwork of houses 0.2 20 Heavy machines in industrial buildings suffer little damage; steel frame building distorted and pulled away from foundations 20-28 Frameless, self-framing steel panel building 0.2-0.28 demolished; rupture of oil storage tanks 20-40 Large trees fall down 0.2-0.4 30 Cladding of light industrial buildings ruptured. 0.3 Panelling torn-off 35 Breakage of wooden telephone poles; most 0.35 buildings destroyed, except for concrete reinforced shear wall buildings; “platting” of cars and trucks pressed inwards 35-40 Displacement of pipe bridge, failure of piping 0.35-0.4 35-50 Near-complete destruction of houses 0.35-0.5 40-55 Collapse of pipe bridge 0.4-0.55 50 Loaded tank cars/train wagons overturned; brick 0.5 walls, 20-30 cm thick, collapse 50-55 Unreinforced brick panels, 25-35 cm thick, fail by 0-5-0.55 shearing or flexure 60 Loaded train boxcars completely demolished 0.6 70 Probable total destruction of buildings; heavy 0.7 machine tools moved and badly damaged
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6 VULNERABILITY TO TOXIC SUBSTANCES
Toxic substances can enter biological organisms by various routes: ingestion (mouth, stomach), inhalation (nose or mouth, lungs), injection (cuts in the skin) and dermal absorption (skin membrane). Generally speaking, from the point of view of risk analysis and major accidents, the most usual way for human beings to be affected by toxic substances is by inhaling them. The following section is essentially devoted to this subject. Toxicants entering the body by inhalation follow the respiratory system: nose, mouth, pharynx, larynx, trachea and lungs. The first part of the system (from nose to trachea) is essentially affected by water-soluble toxicants, which form acids or bases in the mucus. The lungs are affected by substances that block the transfer of gases or react with the walls of the alveoli. The vulnerability of human beings to inhaled toxic substances is closely related to the nature and properties of the substance and the dose, i.e. it depends on the concentration in the atmosphere and how long it is inhaled for. These variables are used to establish threshold values for different situations (e.g. working environments and emergencies). Definitions for several of these values are provided in Tables 7-15 and 7-16, and a more exhaustive survey is included in Annex V. Table 7-15 Definitions for threshold values for accidents at work Threshold value Definition TLV-TWA Threshold Limit Value-Time Weighted Average: time-weighted average concentration for a normal 8 h workday and a 40 h workweek to which nearly all workers may be repeatedly exposed, day after day, without adverse effects. Excursions above the limit are allowed if compensated by excursions below it. TLV-STEL Threshold Limit Value-Short Term Exposure Limit: the maximum concentration to which workers can be continuously exposed for a period of up to 15 min without suffering a) intolerable irritation, b) chronic or irreversible tissue change, c) narcosis to a degree at which accident proneness increases, self-rescue is impaired or worker efficiency is materially reduced, provided that no more than 4 excursions per day are permitted, with at least 60 min between exposure periods and provided that the daily TLV-TWA is not exceeded. TLV-C Threshold Limit Value-Ceiling: the concentration that should not be exceeded, even instantaneously.
IDLH levels were designed by NIOSH [24] for healthy workers in an exposure situation that is likely to cause death, immediate or delayed permanent damage to health, or prevent escape from such an environment. IDLH values have been successfully used in working with emergency situations for many years and are still being used today. Acute Exposure Guideline Levels, or AEGLs, are being developed by the National Research Council’s Committee on Toxicology [27]. The Committee developed detailed guidelines for devising uniform, meaningful emergency response standards for the general public. The criteria in the guidelines take into account sensitive individuals and are meant to protect nearly all individuals in a population. Each of the three levels of AEGL is developed for each of five exposure periods: 10 minutes, 30 minutes, 1 hour, 4 hours and 8 hours. AEGLs are intended to describe the risk to humans resulting from once-in-a-lifetime, or rare, exposure to airborne chemicals. They are being developed [27] to help both national and
271
local authorities, as well as private companies, deal with emergencies involving spills or other catastrophic exposures (acute exposures are single, non-repetitive exposures for not more than 8 hours). Table 7-16 Definitions of threshold values for emergencies Threshold value Definition IDLH Immediately Dangerous to Life and Health: maximum airborne concentration of a substance to which a healthy male worker can be exposed for up to 30 min and still be able to escape without loss of life or irreversible organ system damage. AEGL Acute Exposure Guidelines Levels. AEGL-1 is the airborne concentration of a substance above which it is predicted that the general population, including susceptible individuals, could experience significant discomfort, irritation, or certain asymptomatic nonsensory effects. However, the effects are not disabling and are transient and reversible upon cessation of exposure. AEGL-2 is the airborne concentration of a substance above which it is predicted that the general population, including susceptible individuals, could experience irreversible or other serious, long-lasting adverse health effects or an impaired ability to escape. AEGL-3 is the airborne concentration of a substance above which it is predicted that the general population, including susceptible individuals, could experience life-threatening health effects or death. ERPG Emergency Response Planning Guidelines. ERPG-1 is the maximum airborne concentration below which it is believed that nearly all individuals could be exposed for up to 1 hour without experiencing other than mild transient adverse health effects or perceiving a clearly defined objectionable odour. ERPG-2 is the maximum airborne concentration below which it is believed that nearly all individuals could be exposed for up to 1 hour without experiencing or developing irreversible or other serious health effects or symptoms which could impair an individual’s ability to take protective action. ERPG-3 is the maximum airborne concentration below which it is believed that nearly all individuals could be exposed for up to 1 h without experiencing or developing life-threatening health effects. TEEL Temporary Emergency Exposure Limits. TEEL-0 is the threshold concentration below which most individuals will experience no appreciable risk to their health. TEEL-1 is the same as ERPG-1 TEEL-2 is the same as ERPG-2 TEEL-3 is the same as ERPG-3 A list of values for these thresholds is included in Annex V.
ERPGs were defined by the American Industrial Hygiene Association [25] as concentration ranges at which adverse health effects could be observed. ERPG guidelines do not protect everyone. Hypersensitive individuals suffer adverse reactions to concentrations far below those suggested in the guidelines. In addition, ERPGs, like other exposure guidelines, are based mostly on tests on animals, thus raising the question of their applicability to humans. The guidelines are focused on one period of time: 1 hour. The ERPG Committee strongly advises against trying to extrapolate ERPG values to longer periods of time.
272
TEELs are temporary LOCs, defined by the U.S. Department of Energy [26] to use when ERPGs are not available. Like ERPGs, they do not incorporate safety factors; rather, they are designed to represent the predicted response of members of the general public to different concentrations of a chemical during an incident. 6.1 Dose and probit equations Inhaled doses of toxicants are usually defined in terms of concentration per unit time of exposure raised to a power multiplied by duration exposure (cnt). In most accidents in which toxic substances are released into the atmosphere, the concentration at a given point varies as a function of time. Furthermore, the position of a person can also change, especially if he or she is outdoors (i.e. can escape). Therefore, the dose of the toxicant must be expressed as follows: t
n
³ >c(t )@ dt
D
(7-31)
0
The probit equation for lethality in the case of inhalation of a toxic substance has the following general expression: Y
a b ln
n
t
³ >c(t )@ dt
(7-32)
0
For practical purposes, the following expression is usually substituted for the one above: Y
a b ln 6 i cin 't i
(7-33)
where ci is the average concentration during 'ti, usually expressed in ppm = ml·m-3, and t is the exposure time in minutes. For every substance, a, b and n are constant values, and n is a chemical-specific parameter greater than zero. There can be multiple n values for a single chemical that are applicable to different response endpoints; for example, n for irritation by ammonia is 4.6 while n for lethality is 2. Table 7-17 provides a, b and n values for a set of substances [23]. For the most common substances, diverse probit equations have been proposed by different authors. Note that atmospheric dispersion models usually calculate the toxic concentrations in mg·m-3. To transform these units to ppm (or vice versa), the following expression can be applied: c ppm
22.4 T 1.013 c mg/m 3 M 273 P
(7-34)
where M is the molecular weight (kg kmol-1) T is the ambient temperature (k) and P is the atmospheric pressure (bar). The values of a, b and n are available only for a limited set of substances. If they are not known for a given substance, they can be obtained approximately from experimental data obtained for animals; see [11] for a more detailed explanation.
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Eq. (7-33) can be used to calculate the value of the concentration that will originate a specific percentage of lethality for a given exposure time. For example, the value of the probit variable corresponding to 1% of lethality is Y = 2.67. Therefore, 2.67
a b ln c n t
(7-35)
where c = LC01, lethal concentration affecting 1% of the exposed population (ml m-3), and t is the exposure time (min). Then
LC01
ª 2.67b a «e « t ¬«
º » » ¼»
1/ n
(7-36)
In emergency situations, a criterion widely applied is to assume an exposure time to toxic atmospheres of 10 min. Table 7-17 Constants for lethal toxicity probit equation Substance a b n Acrolein -9.931 2.049 1 Acrylonitrile -29.42 3.008 1.43 Ammonia -35.9 1.85 2 Benzene -109.78 5.3 2 Bromine -9.04 0.92 2 Carbon monoxide -37.98 3.7 1 Carbon tetrachloride -6.29 0.408 2.5 Chlorine -8.29 0.92 2 Formaldehyde -12.24 1.3 2 Hydrogen chloride -16.85 2.0 1 Hydrogen cyanide -29.42 3.008 1.43 Hydrogen fluoridea -25.87 3.354 1 Hydrogen sulfide -31.42 3.008 1.43 Methyl bromide -56.81 5.27 1 Methyl isocyanate -5.642 1.637 0.653 Nitrogen dioxide -13.79 1.4 2 Phosgene -19.27 3.686 1 Propylene oxide -7.415 0.509 2 Sulfur dioxide -15.67 2.1 1 Toluene -6.794 0.408 2.5 a The constants are to be used in the probit equation using ppm as the concentration term and minutes as the time term. For hydrogen fluoride, enter concentration in mg/m3 instead of ppm.
Studies on the consequences of the exposure of humans or laboratory animals to toxic concentrations of elements are generally conducted for time periods different from those which are of interest in acute exposure (accidental) scenarios. In order to adjust exposure durations to a given value, the time extrapolation based on Haber’s Law is usually used.
274
Haber’s Law states that the product of the concentration and exposure time required to produce a given physiologic effect is equal to a constant level or severity of response. Therefore, when the duration of the experimental exposure differs from the duration of an exposure for which an acute exposure level must be calculated, the following relationship can be applied to establish the concentration which will give rise to the same consequences: Dose
n
c1 t1
n
c2 t 2
(7-37)
When the n value is not available, a default value can be used. Most of the published values for n range between 0.8 and 4.6. The mean value of this range is 2.2, while the interquartile range (25%-75%), where most of the n values are found, is from 1 to 2.2. When extrapolating from an exposure duration that is greater than 1 h to a 1 h exposure, the value of n = 2 is suggested by OEHHA [28], while when extrapolating from an experimental exposure duration of less than 1 h to 1 h, the value of n = 1 should be used. ______________________________________ Example 7-5 Calculate the value of LC01 for formaldehyde, for an exposure time of 10 min. Solution For formaldehyde, a = -12.24, b = 1.3 and n = 2. By applying Eq. (7-36): 1/ 2
ª 2.671.312.24 º e » LC01 « 98 ppm. « 10 » ¼» ¬« ______________________________________ As analyzed in the chapter devoted to atmospheric dispersion, a closed building with a given ventilation rate will give rise to different indoor and outdoor concentrations of a pollutant. The indoor concentration will be a function of the evolution of the outdoor concentration and of the ventilation rate (which, if unknown, can be taken to be 1 h-1). Therefore, at least for the first steps of the event, the indoor concentration will be clearly lower than outdoor concentration. That the percentage of fatalities indoors is 10% of outdoor fatalities is usually taken as a general criterion in risk analysis studies. 6.2 Substances released from a fire Toxic substances can be released from a fire due to pyrolysis, combustion, etc. For example, if wool burns, highly toxic HCN is released. Smoke can also lead to serious intoxication and even death. Table 7-18 [29] summarizes the origin of the main toxic substances which can be found in smoke. Table 7-19 (modified from [26]) shows the effects of carbon monoxide at different concentrations. The effects of carbon dioxide are summarized in Table 7-20 [31, 32]. At low concentrations, CO2 increases velocity of respiration, which thus increases the possible inhalation of other toxic gases. Most of the combustible materials require a minimum concentration of 15% oxygen in the air for the combustion to proceed. In a closed space, therefore, it could be possible for a
275
decrease in the concentration of oxygen to cause a fire to become extinguished, which could thus imply a chance for surviving. It must be taken into account, however, that the combined effect of different products of combustion (e.g. CO and CO2) may have fatal consequences. Table 7-18 The sources of the main toxicants appearing in gases released by combustion Toxicant Sources CO, CO2 All materials containing carbon HCN Wool, silk, polyacrylonitrile, nylon, polyurethane, etc. NOx Produced in small quantities from fabrics and in larger quantities from cellulose nitrate, celluloid, etc. NH3 Wool, silk, nylon and melamine; concentrations generally low in ordinary building fires HCl Materials containing chlorine, e.g. PVC, and some fire retardant treated materials SO2 Materials containing sulfur (e.g. rubber) HF Fluorinated resins HBr Fire retardant materials containing bromine Acrolein Pyrolysis of polyolefins and cellulose at low temperatures ( | 400 ºC) (fats and oils) Table 7-19 Effects of carbon monoxide on people Concentration of CO, ppm Consequences 50 Allowable exposure for 8 hours 100 Allowable exposure for several hours 400-500 No appreciable effect after 1 hour 600-700 Just appreciable after 1 hour 1,000-1,200 Unpleasant after 1 hour 1,500-2,000 Dangerous when inhaled for 1 hour 4,000 Fatal when inhaled for less than 1 hour 10,000 Fatal when inhaled for 1 minute
______________________________________ Example 7-6 An accidental release of chlorine originates a toxic cloud. A group of 12 people are exposed to the concentrations shown in the table. a) Estimate the consequences. b) Calculate the average concentration that, for an exposure time of 10 min, would imply the same lethality. Concentration, ppm 200 500 900 1,100 500 200
Time, min. 1 2 3 2 1 1
Solution a) The probit equation for chlorine (lethality) is as follows (see Table 7-17):
probit
8.29 0.92 ln 6 c 2 't
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The dose received by the group of individuals is
6 c 2 't = 2002 · 1 + 5002 · 2 + 9002 · 3 + 1,1002 · 2 + 5002 · 1 + 2002 · 1 = 5,680,000 ppm2 min. The value of the probit function is probit
8.29 0.92 ln 5,680,000 = 6.02
This implies a percentage (see Table 7-1) of 85%; this means that approximately 10 individuals will die. b) Y = 6.02. By applying Eq. (7-36): 1/ n
1/ 2
ª 6.020.928.29 º ª Y b a º e » «e » « LC01 755 ppm « 10 » « t » ¼» ¬« ¼» ¬« ______________________________________ Table 7-20 Effects of carbon dioxide on people Concentration of CO2, % Consequences 2 Increases breathing velocity by 50% 3 Doubles the breathing velocity 5 Breathing can be difficult for some individuals 10 Unconsciousness threshold reached in 30 minutes 12 Unconsciousness threshold reached in 5 minutes 15 Exposure limit: 1 minute 20 Unconsciousness occurs in less than 1 minute
7 INERT GASES
In the case of toxic gases, the fact that people are generally aware that a toxic gas is dangerous aids prevention and protection in certain dangerous situations. However, this awareness does not exist in the case of inert gases, which are often not associated with danger. Nevertheless, the release of inert gases such as nitrogen, and atmospheres that are rich in these gases can be extremely dangerous, as the gases themselves will possibly not be detected when they are inhaled (they are odorless and they are not irritants). Inert gases are dangerous because their presence leads to a reduction in the concentration of oxygen. When a human being is breathing in an atmosphere that is poor in oxygen, i.e. that has a concentration of oxygen that is less than normal, there is a risk of suffocation, which increases when the concentration of oxygen decreases and when the duration of the exposure increases. When breathing pure nitrogen or other gases such as methane and carbon dioxide which contain no oxygen, the oxygen in the lungs is washed out and replaced by the new gas. Blood from the lungs does not receive enough oxygen and the oxygen in the brain quickly becomes deficient. A few seconds of inhaling an oxygen-free gas can lead to mental failure. Usually,
277
there are no symptoms or warnings. At work, the oxygen concentration should be at least 19.5%; less than 19.5% should be considered dangerous. Other gases, such as carbon monoxide and hydrogen sulfide, are chemical asphyxiants and have an additional blocking action. Table 7-21 Consequences of exposure to poor oxygen atmospheres Concentration of O2,% Consequences 16-19 Decreased ability to work; may induce early symptoms in persons with coronary, pulmonary or circulatory problems 16 Dangerous situation 11-14 Physical and intellectual capacity decrease 8-11 Possible unconsciousness in a relatively short time 8-10 8 min: 100% fatal; 6 min: 50% fatal; 4-5 min: recovery with treatment, brain damage and death are possible 6-8 Unconsciousness occurs in a few seconds, death in 8 minutes 16 none 16 1-a (toxic cloud to S) 6 1-b (toxic cloud to W) 4 2-a (explosion)
f, year-1 0 8.4 · 10-7 3.6 · 10-7 2.5 · 10-6
Cumulative frequency, year-1 0 8.4 · 10-7 1.2 · 10-6 3.7 · 10-6
We can now plot these data on a logarithmic scale to obtain the f-N curve (Fig. 8-8).
Fig. 8-8. Societal risk f-N curve.
______________________________________ ______________________________________ Example 8-3 A potential explosion hazard caused by the accumulation of propane in the flammability range has been identified by applying HAZOP to a processing plant furnace. The dimensions of the furnace chamber are 10 m x 20 m x 5 m [10]. a). Calculate the maximum amount of propane which could accumulate in flammable conditions at ambient temperature (cool furnace) and estimate the intervention area and the alert area to be considered in the event of an explosion. b). The frequency of explosion in the furnace has been estimated as 3 · 10-5 year-1. Assuming continuous operation during the whole year, calculate the distance at which the direct effects of the explosion yield an individual risk of 10-6 year-1. Assume that the intervention distance (di) is the distance at which 'P = 0.125 bar and that the alert distance (da) is the distance at which 'P = 0.05 bar. Additional data: flammability limits for propane (Table 3-1): 2.1 – 9.5% volume; Mw propane = 44 kg kmole-1; ambient conditions: temperature = 25 ºC, pressure = 101.325 kPa.
304
Solution a). Maximum amount of propane in flammable conditions: V
0.095 10 20 5 95 m3
VU
V
PMw RT
101.325 10 44 8.314 10 273 25 3
95
3
171 kg of propane
The relationship between 'P and the distance is established by applying the equivalent TNT method (Chapter 4): di
3
WTNT d n
da
3
K
M 'H c dn 'H TNT
3
0.03 17110 d n
dn can be obtained from Fig. 4-4 for the two values of 'P:
'P = 0.125 bar dn = 12 m kg-1/3 'P = 0.050 bar dn = 24 m kg-1/3 Therefore, d
3
0.03 17110 12
45 m
d
3
0.03 17110 24
89 m
b). By applying Eq. (8-2): 10 6
3 10 5 PFi
PFi = 0.033 a probability of 33 fatalities per 1,000 exposed people.
The corresponding probit variable is (Table 7-1): Y = 3.16 By applying the probit expression for death due to pulmonary haemorrhage (Eq. (7-16)):
3.16
77.1 6.91ln 'P
'P = 110,750 Pa (or 1.1 bar) The scaled distance is obtained from Fig. 4-4:
305
dn = 3.2 m kg-1/3
Therefore, the distance at which IR = 10-6 year-1 is: d
3
0.03 17110 3.2 12 m
This is very close to the furnace. At this distance, the risk posed by the indirect effects of the explosion would be much higher than the risk due to direct effects. ______________________________________ 6 FREQUENCIES AND PROBABILITIES
To estimate the probable frequencies of the diverse accidental scenarios in an event tree, it is necessary to assign values to the frequency of the initiating event (for example, rupture of a pipe) and to the probabilities which appear in the accident sequence (for example, ignition of a flammable cloud or blast formation). For the performance of a consistent quantitative risk analysis, it is essential to define event frequencies and probabilities as realistically as possible. Under or over-estimation of these values can lead to errors of more than one order of magnitude in individual and societal risk. These data are obtained from research projects, from the historical analysis of accidents and from expert judgement. In this section the frequencies corresponding to the most common loss of containment events and the probabilities of ignition and explosion have been included. 6.1 Frequencies of most common loss-of-containment events The frequencies usually attributed to the loss of containment events included in Chapter 2, section 8, are summarized in Table 8-4. They have been take from [4], a well known reference book. 6.2 Failure of repression systems Once a loss of containment occurs, a blocking or repression system can stop or limit the outflow. A general value for the failure of a repression system is 0.05 per demand. 6.3 Human error Different values have been proposed for human error probability in recovery from an abnormal event: - Perform a critical action under conditions of moderately high stress: P = 0.05 [5] - Perform a critical action under conditions of extremely high stress: P = 0.2 – 0.25. 6.4 Probabilities for ignition and explosion of flammable spills Since no obvious conclusion can be drawn as to whether a flammable spill can encounter an ignition source, whether ignition takes place and how far (in space and time) ignition occurs from the spill location, historical data are used to standardise ignition probabilities in quantitative risk analysis. The same applies to the formation of a blast wave, given the ignition of a flammable cloud. Fig. 8-8 is a generalized event tree for the spill of a flammable material. The tree only contains three major bifurcations, to which the following check questions can be assigned: 1)
306
is the spill immediately ignited? 2) If not, is the subsequent vapour/gas cloud ignited (i.e does delayed ignition occur)? 3) If so, does the ignition cause a blast? Table 8-4 Frequencies of the most common loss of containment events [4] System Instantaneous release of Release (at constant the complete inventory rate) of the complete inventory in 10 min. Stationary vessels: Pressure vessel 5 · 10-7 year-1 5 · 10-7 year-1 -6 -1 Process vessel 5 · 10 year 5 · 10-6 year-1 -6 -1 Reactor vessel 5 · 10 year 5 · 10-6 year-1 Atmospheric tanks: Single containm. tank 5 · 10-6 year-1 5 · 10-6 year-1 Mounded tank 1 · 10-8 year-1 Heat exchangers: Haz. mat. outside pipes 5 · 10-5 year-1 5 · 10-5 year-1 -5 -1 Haz. mat. inside pipes 1 · 10 year 1 · 10-3 year-1 Road tankers and tank wagons in an establishment Pressurized tank Atmospheric tank
Instantaneous release of the complete inventory 5 · 10-7 year-1 1 · 10-5 year-1
Pipes d < 75 mm 75 mm d d d 150 mm d > 150 mm
Continuous release from a hole (largest connection size) 5 · 10-7 year-1 5 · 10-7 year-1
Continuous release from a hole with ds = 10 mm
Full bore rupture, loading/unloading hose
Full bore rupture 1 · 10-6 m-1 year-1 3 · 10-7 m-1 year-1 1 · 10-7 m-1 year-1
Discharge (maximum) from a safety relief device
4 · 10-6 h-1 4 · 10-6 h-1
1 · 10-5 year-1 1 · 10-4 year-1 1 · 10-4 year-1 1 · 10-4 year-1
1 · 10-3 year-1 1 · 10-2 year-1 Leak of loading/unloading hose (ds = 0.1·d, max. 50 mm) 4 · 10-5 h-1 4 · 10-5 h-1
Leak (d = 0.1 ds, max 50 mm) 5 · 10-6 m-1 year-1 2 · 10-6 m-1 year-1 5 · 10-7 m-1 year-1 2 · 10-5 year-1
Ignition probability is a function of: - amount released: the greater the release, the larger the area covered by the ignitable cloud and the higher the probability of it finding an ignition source; - the substance released: the more volatile and flammable the material, the more likely the ignition; - the characteristics of the surroundings. Recently, the following expressions have been obtained from an exhaustive historical analysis [6, 11] to estimate the ignition probability: P1 P1 P2
a Q b for land transportation spills
(8-10)
P1 P1 P2
c for maritime transportation spills 1 d Q f
(8-11)
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Fig. 8-8. General event tree for flammable material leaks. Taken from [11], by permission.
Coefficients c, d and f are listed in Table 8-5. Eqs. (8-10) and (8-11) are valid for any amount spilled, of course provided the probability valued obtained does not exceed 1 (very high amounts spilled using Eq. (8-10)); in this case, it can be assumed that P1 P1 P2 = 1. Table 8-5 Ignition probability. Values of the parameters a and b of Eq. (8-10) and c, d and f of Eq. (8-11)* Substance Land transportation Maritime transportation a b c d f LPG 0.022 0.32 Light fractions 0.00027 0.72 0.039 6.49 0.76 Crude oil, kerosene/jet fuel, diesel 0.00055 0.53 0.013 40.75 1.00 oil/gas oil No. 4-6 fuel oil 0.00 0.00 *Fixed plants: use a value intermediate between those for maritime and land transportation
The probability of ignition as a function of the mass released or the source flow rate and the type of substance can also be taken from Table 8-6 [4]. Table 8-6 Ignition probability, fixed plants [4] Source Continuous Instantaneous < 10 kg s-1 < 1,000 kg 10-100 kg s-1 1,000-10,000 kg >100 kg s-1 > 10,000 kg
Liquid 0.065 0.065 0.065
Substance Gas, low reactivity Gas, average/high reactivity 0.02 0.2 0.04 0.5 0.09 0.7
Explosion probability refers to the likelihood of a flammable cloud forming a blast wave once ignition has taken place; this parameter is identified with the variable P3 in Fig. 8-8. Explosion probability is affected by the properties of the substance (reactivity) and by the size of the flammable gas cloud. The amount spilled is important to the explosion probability
308
because a certain lapse of time and relatively high flammable concentrations are needed for the flame front to reach a high speed, and thus cause a significant overpressure. Explosion probability data are shown in Table 8-7 [6, 11]. Table 8-7 Explosion probability as a function of the substance, the amount spilled and the activity Activity Amount Generic explosion Specific explosion probability spilled, kg probability LPG Light Crude oil Diesel oil fractions kerosene gas oil jet fuel Fixed plants; land 1-100 0.06 0.043 0.067 0.088 0.044 transportation 100-10,000 0.30 0.22 0.34 0.44 0.22 > 10,000 0.40 0.29 0.45 0.58 0.29 Maritime transp. 100-10,000 0.25 0.33 0.38 0.18 > 10,000 0.37 0.48 0.57 0.27
In an event tree scheme such as that of Fig. 8-8, one must split values into a probability of immediate ignition (P1) and one of delayed ignition ( P1 P ). The following values can be assumed [11]: - for petrol and light fractions, a ratio of delayed to immediate ignition probabilities of 1:1 can be used; - for LPG, a ratio of 1:1 can also be used, given the great variety of data available for this material class; - for diesel/kerosene/crude oil, a value of 1:10 has been suggested to take into account the possibility that the spill may happen above the ambient temperature. 6.5 Meteorological data To limit the calculation time for the quantitative risk analysis, usually the data concerning stability class and wind velocity are grouped in a reduced number of representative sets. In [4] six representative sets are suggested. Often, two sets are used: (D, 5 m s-1 or 4 m s-1), and (F, 2 m s-1). 7 EXAMPLE CASE
______________________________________ Example 8-4 A cylindrical tank containing liquefied propane is located close to an inhabited zone (Fig. 89). The tank has a volume of 82 m3 and is connected to the loading point by a pipe with a diameter of 2”and a length of 40 m; this pipe is fitted with an automatic excess flow rate valve which interrupts the flow when the flow rate reaches 3.5 kg s-1. The pipe that connects the tank with the distribution system is also fitted with a similar valve. While loading the tank another 2” pipe (Lpipe = 40 m) transports the propane gas back to the tank car. The road tanker has a volume of 20 m3 and is connected to the loading pipe through a flexible duct with a diameter of 2.5” and l = 1.6 m. Loading operations are performed during 87 h year-1 and a highly-trained operator is always present. The propane tank is protected from flame
309
impingement by a water spray system. In winter conditions (90 days per year) an evaporating unit is used to evaporate the liquefied propane. The pipe connecting the tank with this unit is 40 m long and has a diameter of 2”.
Fig. 8-9. Schematic representation of the LPG storage unit.
Average atmospheric data: P0 = 1.013 bar; ambient temperature = 25 ºC; relative humidity = 60%; two wind velocities must be considered: 2 m s-1 and 5 m s-1; wind direction: see wind rose (Table 8-8); atmospheric stability class: D (uw = 5 m s-1) and F (uw = 2 m s-1); surface roughness length = 0.1 m. Table 8-8 Wind rose Wind direction j 1 N 2 3 4 E 5 6 7 S 8 9 10 W 11 12
Angle Probability 0 5.7 30 8.0 60 8.8 90 8.5 120 8.5 150 8.6 180 8.5 210 11.1 240 11.6 270 9.9 300 5.5 330 5.2
N 12
8
4 W
0
E
S
Identify the diverse initiating events and determine their frequencies. Determine the diverse accident sequences, calculate the frequency of the different accident scenarios, estimate the effects of the most significant accidents and establish the corresponding individual risk contours.
310
Solution 7.1 Estimation of the frequencies of initiating events Table 8-9 gives descriptions of the initiating events considered. The generic frequencies were taken from a reference book [4]. The loading time per year was taken into account for Events 1-5. Events 7 and 8 can occur only over 90 days per year. The frequency of Events 3, 4, 5, 7, 8 and 9 was estimated taking into account the length of the corresponding pipes. Table 8-9 Initiating events and their respective estimated frequencies* Event Description Frequency estimation 1 LPG release due to full-bore rupture of road tanker f = 4·10-6 h-1 · 87 h year-1 flexible pipe. 2 LPG release due to partial rupture of road tanker f = 4·10-5 h-1 · 87 h year-1 flexible pipe. 3 LPG release due to full-bore f = rupture of the pipe § 87 h year -1 · ¸ 1 10 -3 km -1 year -1 ¨¨ connecting the loading point -1 ¸ © 8760 h year ¹ with the tank.
4
5
6
7
8
9
LPG release due to partial rupture of the pipe connecting the loading point with the tank. Release of propane (gas) due to full-bore rupture of the pipe connecting the top of the tank with road tanker. Release of propane (gas) through the safety relief valve. LPG release due to full-bore rupture of the pipe connecting the tank with the evaporators. LPG release due to partial rupture of the pipe connecting the tank with the evaporators. Release of propane (gas) due to full-bore rupture of the pipe connecting the storage tank with the distribution system.
f=
Frequency f = 3.48 ·10-4 year-1 f = 3.48 ·10-3 year-1 f = 9.93·10-6 km-1yr-1 f = 3.97 ·10-7 year-1
§ 87 h year -1 · ¸ 5 10 -3 km -1 year -1 ¨¨ -1 ¸ © 8760 h year ¹ f= § · 87 h ¸ 6 10 -3 km -1 year -1 ¨¨ -1 ¸ © 8760 h year ¹
f = 4.97·10-5 km-1yr-1 f = 2 ·10-6 year-1
f = 2·10-5 year-1
f = 2 ·10-5 year-1
f=
§ 90 24 h year -1 · ¸ 1 10 -3 km -1 year -1 ¨¨ -1 ¸ © 8760 h year ¹ f= § 90 24 h year -1 · ¸ 5 10 -3 km -1 year -1 ¨¨ -1 ¸ © 8760 h year ¹ f = 6 ·10-3 km-1 year-1
f =5.96·10-5 km-1yr-1 f = 2.38 ·10-6 year-1
f =2.47·10-4 km-1yr-1 f = 9.88 ·10-6 year-1 f =1.23·10-3 km-1 yr-1 f = 4.93 ·10-5 year-1
f = 2.4 ·10-4 year-1
*Frequencies taken from Table 8-4.
We now develop the corresponding event trees to establish the different sequences from the different initiating events to the final accident scenarios.
311
7.2 Event trees of the diverse initiating events Initiating event 1. Full-bore rupture of the flexible pipe. The first safety measure is the excess flow rate valve, which would stop the release. The second safety measure if the valve fails is for the operator to close the propane exit valve manually to stop the release. If we take into account the position of the flexible pipe and its distance from the tank and the tank car, we obtain a probability of 8% (which corresponds to an angle of 30º on a horizontal plane, required for the jet fire that affects the tank) for the storage tank (which is also protected by a water spray system) and 50% for the tank car due to its proximity to the pipe. The different sequences can be seen in the event tree (Fig. 8-10): Initiating event
Failure of excess flow rate
Immediate ignition
Operator failure
Flames Flames Delayed Failure of water impinge on impinge on ignition spray system tank road tanker Yes P5=0.05 No P =0.95
Yes P4=0.083 Yes P3=0.2 Yes P2=0.5
Flame front acceleration
Accidental scenario
Tank BLEVE Jet fire
5
Yes P6=0.5
Road tanker BLEVE
No
P4 =0.917
No
Jet fire
P6 =0.5 No
No outcome
P3 =0.8
Yes P1=0.05
Yes P8=0.5
Yes P7=0.2
Full-bore rupture f = 3.48 10-4 y-1
Yes P9=0.01 No P =0.99
Flash fire + VCE Flash fire
9
No
No
P2 =0.5
P8 =0.5 No
No outcome No outcome
P7 =0.8 No
No outcome
P1 =0.95
Fig. 8-10. Event tree and accident scenarios, initiating event 1 (full-bore rupture of the flexible pipe). Table 8-10 Frequencies of the accident scenarios associated with initiating event 1 Accident scenario Frequency estimation Jet fire f = (3.48·10-4 · 0.05 · 0.5 · 0.2 · 0.08 · 0.95) + + (3.48 ·10-4 · 0.05 · 0.5 · 0.2 · 0.92 · 0.5) Flash fire
VCE Road tanker BLEVE Tank BLEVE
f = (3.48 ·10-4 · 0.05 · 0.5 · 0.2 · 0.5 · 0.01) + + (3.48 ·10-4 · 0.05 · 0.5 · 0.2 · 0.5 · 0.99)
Frequency f = 9.15 · 10-7 year-1 f = 8.7 · 10-7 year-1
This frequency has not been estimated as significant peak pressures would not be reached. f = 3.48 ·10-4 · 0.05 · 0.5 · 0.2 · 0.92 · 0.5 f = 3.48 ·10-4 · 0.05 · 0.5 · 0.2 · 0.08 · 0.05
312
f = 8 · 10-7 year-1 f = 6 · 10-9 year-1
The frequency of each accident scenario can be calculated according to the different sequences. Table 8-10 shows the corresponding values.
Initiating event 2. Partial rupture of the flexible pipe. The release flow rate would be much lower – close to the unloading flow rate in normal operation – and the excess flow valve would not be closed. The resulting jet fire would be smaller and would not reach the storage tank. The probability that the tank car will be affected by the jet fire is kept as 0.5. Fig. 8-11 shows the corresponding event tree. The possibility of a blast wave due to the explosion of a flammable vapour cloud has not been considered, as the minimum amount of flammable mixture required to produce blast would not be reached. Initiating event
Immediate ignition
Operator failure
Flames impinge Failure of water Flames impinge on tank spray system on road tanker
Yes P4=0 Yes P3=0.2 Yes P2=0.2
Delayed ignition
Flame front acceleration
Yes P5=0.05 No P =0.95
Accidental scenario No outcome Jet fire
5
Yes P6=0.5
Road tanker BLEVE
No
P4 =1
No
Jet fire
P6 =0.5 No
No outcome
P3 =0.8
Partial rupture f = 3.48 10-3 y-1
Yes P8=0.2
Yes P7=0.2
Yes P9=0 No P =1
No outcome Flash fire
9
No
No
P2 =0.8
P8 =0.8 No
No outcome No outcome
P7 =0.8
Fig. 8-11. Event tree and accident scenarios for initiating event 2 (road tanker: partial rupture of the flexible pipe).
Table 8-11 shows the estimated frequencies for the different accident scenarios. The probability of blast wave for VCE has been considered negligible. Table 8-11 Frequencies of the accident scenarios associated with initiating event 2 Accident scenario Frequency estimation Jet fire f = 3.48 ·10-3 · 0.2 · 0.2 · 1 · 0.5 Flash fire
f = 3.48 ·10-3 · 0.8 · 0.2 · 0.2 · 1
VCE
This frequency has not been estimated as it is considered that the minimum amount of flammable mixture required to produce blast would not be reached.
Road tanker BLEVE
f = 3.48 ·10-3 · 0.2 · 0.2 · 1 · 0.5
313
Frequency f = 6.96 · 10-5 year-1 f = 1.11 · 10-4 year-1
f = 6.96 · 10-5 year-1
Initiating event 3. Full-bore rupture of the pipe connecting the loading point with the storage tank. The excess flow rate valve would stop the release. In order for a BLEVE to occur it would be necessary for the jet fire to impinge on the tank or the road tanker. A probability of 0.15 is assumed for the storage tank, taking into account the length of the connecting pipe and the angle of the jet that would cause it to affect the tank (35º). In the case of the road tanker, a length of 13 m and an angle of 23º give a probability of 0.02. Fig. 8-12 shows the corresponding event tree. The estimated frequencies can be seen in Table 8-12. Failure of Failure of excess Immediate Operator Flames impinge water spray flow valve ignition failure on tank system
Initiating event
Yes P4=0.15 Yes P3=0.2 Yes P2=0.2
Flames impinge on road tanker
Delayed ignition
Flame front Accidental scenario acceleration
Yes P5=0.05 No P =0.95
Tank BLEVE Jet fire
5
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98 No
No outcome
P3 =0.8
Yes P1=0.05
Yes P8=0.2
Yes P7=0.2
Full-bore rupture f = 3.97 10-7 y-1
Yes P9=0 No P =1
No outcome Flash fire
9
No
No
P2 =0.8
P8 =0.8 No
No outcome No outcome
P7 =0.8 No
No outcome
P1 =0.95
Fig. 8-04. Event tree and accident scenarios for initiating event 3 (full-bore rupture of the pipe connecting the loading point with the storage tank). Table 8-12 Frequencies of the accident scenarios associated with initiating event 3 Accident scenario Frequency estimation Jet fire f = (3.97·10-7 · 0.05 · 0.2 · 0.2 · 0.15 · 0.95) + + (3.97 ·10-7 · 0.05 · 0.2 · 0.2 · 0.85 · 0.98)
Frequency f = 7.75 ·10-10 year-1
Flash fire
f = 3.97 ·10-7 · 0.05 · 0.8 · 0.2 · 0.2 · 1
VCE
This frequency has not been estimated as it is considered that the minimum amount of flammable mixture required to produce blast would not be reached.
Road tanker BLEVE
f = 3.97 ·10-7 · 0.05 · 0.2 · 0.2 · 0.85 · 0.02
f = 1.35 ·10-11 year-1
Tank BLEVE
f = 3.97 ·10-7 · 0.05 · 0.2 · 0.2 · 0.15 · 0.05
f = 6 · 10-12 year-1
314
f = 6.35 ·10-10 year-1
Initiating event 4. Partial rupture of the pipe connecting the loading point with the storage tank. In this case the release flow rate would be similar to the normal unloading release from the road tanker. Therefore, only the operator would be able to stop the flow. The rest of the sequences are similar to those for the previous event trees (see Fig. 8-13). Table 8-13 shows the estimated frequencies. Initiating event
Immediate ignition
Operator failure
Flames impinge Failure of water on tank spray system
Yes P4=0.15 Yes P3=0.2 Yes P2=0.2
Flames impinge on road tanker
Delayed ignition
Flame front acceleration
Yes P5=0.05 No P =0.95
Accidental scenario Tank BLEVE Jet fire
5
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98 No
No outcome
P3 =0.8
Partial rupture f = 2 10-6 y-1
Yes P8=0.2
Yes P7=0.2
Yes P9=0 No P =1
No outcome Flash fire
9
No
No
P2 =0.8
P8 =0.8 No
No outcome No outcome
P7 =0.8
Fig. 8-13. Event tree and accident scenarios for initiating event 4 (partial rupture of the pipe connecting the loading point with the storage tank). Table 8-13 Frequencies of the accident scenarios associated with initiating event 4 Accident scenario Frequency estimation Jet fire f = (2 ·10-6 · 0.2 · 0.2 · 0.15 · 0.95) + + (2·10-6 · 0.2 · 0.2 · 0.85 · 0.98)
Frequency f = 7.8 ·10-8 year-1
Flash fire
f = 2 ·10-6 · 0.8 · 0.2 · 0.1 · 1
f = 3.2 ·10-8 year-1
VCE
This frequency has not been estimated as it is considered that the minimum amount of flammable mixture required to produce blast is not reached.
Road tanker BLEVE
f = 2 ·10-6 · 0.2 · 0.2 · 0.85 · 0.02
f = 1.36 ·10-9 year-1
Tank BLEVE
f = 2 ·10-6 · 0.2 · 0.2 · 0.15 · 0.05
f = 6 ·10-10 year-1
Initiating event 5. Full-bore rupture of the pipe connecting the top of the tank with the road tanker. The rupture originates a release of propane (gas). In this case the operator should close the valve. The lay-out of the pipe is the same as in the previous cases, so the same probabilities and assumptions apply (Fig. 8-14 and Table 8-14): jet fire flames can impinge on the storage tank or on the road tanker.
315
Initiating event
Flames impinge on tank
Immediate ignition Operator failure
Flames impinge on Accidental scenario road tanker
Yes P5=0.05 No P =0.95
Yes P4=0.15 Yes P3=0.2 Yes P2=0.2
Failure of water spray system
Tank BLEVE Jet fire
5
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98
Full-bore rupture f = 2.38 10-6 y-1
No
No outcome
P3 =0.8 No
No outcome
P2 =0.8
Fig. 8-14. Event tree and accident scenarios for initiating event 5 (full-bore rupture of the pipe connecting the top of the tank with the road tanker). Table 8-14 Frequencies of the accident scenarios associated with initiating event 5 Accident scenario Frequency estimation Jet fire f = (2.38 ·10-6 · 0.2 · 0.2 · 0.15 ·0.95) + + (2.28 ·10-6 · 0.2 · 0.2 · 0.85 · 0.98)
Frequency f = 9.29 ·10-8 year-1
Road tanker BLEVE
f = 2.38 ·10-6 · 0.2 · 0.2 · 0.85 · 0.02
f = 1.62 ·10-9 year-1
Tank BLEVE
f = 2.38 ·10-6 · 0.2 · 0.2 · 0.15 · 0.05
f = 7.14 ·10-10 year-1
Initiating event 6. Release of propane gas through the safety relief valve. The frequency of the initiating event is taken from Table 8-4. As the relief valve discharges vertically and is located 1 m above the tank top, it is assumed that there would be no significant effects on either the tank or the tank car in the event of ignition (Fig. 8-15 and Table 8-15): no significant outcomes would occur. Initiating event
Flames impinge on tank
Immediate ignition
Flames impinge on road tanker
Yes P4=0
No outcome
Yes P2=0.2 Opening of PSV f = 2 10 y -5
Accidental scenario
Yes P6 = 0
No outcome
No
P4 =1
No
Jet fire
P6 = 1
-1
No
No outcome
P2 =0.8
Fig. 8-15. Event tree and accident scenarios for initiating event 6 (release of propane gas through the safety relief valve).
316
Table 8-15 Frequencies of the accident scenarios associated with initiating event 6 Accident scenario Frequency estimation Jet fire f = 2 ·10-5 · 0.2 · 1 · 1
Frequency f = 4 ·10-6 year-1
Initiating event 7. Full-bore rupture of the pipe connecting the tank with the evaporators. In this case the excess flow-rate valve should close automatically. This event may occur while the operator is away from the plant. The equipment is fitted with a set of automatic valves that stops the flow of LPG from the storage tank in the event of a fire: the sequences leading to the different accident scenarios therefore depend on the failure of this automatic shut-off system. The assumptions and probabilities are essentially the same as those applied in the previous cases (Fig. 8-16 and Table 8-16). Initiating event
Failure of excess Immediate flow valve ignition
Failure of automatic valve
Flames Flames Failure of water Delayed Flame front impinge on impinge on spray system ignition acceleration tank road tanker
Yes P4=0.15 Yes P3=0.05 Yes P2=0.2
Yes P5=0.05 No P =0.95
Accidental scenario
Tank BLEVE Jet fire
5
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98 No
No output
P3 =0.95
Yes P1=0.05
Yes P9=0 Yes P8=0.2 No
Yes P7=0.05
Full-bore rupture f = 9.88 10-6 y-1
No output Flash fire
P9 =1
No
No
P2 =0.8
P8 =0.8 No
No output No output
P7 =0.95 No
No output
P1 =0.95
Fig. 8-16. Event tree and accident scenarios for initiating event 7 (full-bore rupture of the pipe connecting the tank with the evaporators). Table 8-16 Frequencies of the accident scenarios associated with initiating event 7 Accident scenario Frequency estimation Jet fire f = (9.88·10-6 · 0.05 · 0.2 · 0.05 · 0.15 · 0.95)+ + (9.88 ·10-6 · 0.05 · 0.2 · 0.05 · 0.85 · 0.98) Flash fire
f = 9.88·10-6 · 0.05 · 0.8 · 0.05 · 0.2 · 1
VCE
This frequency has not been estimated as it is considered that the minimum amount of flammable mixture required to produce blast is not reached.
317
Frequency f = 4.9 ·10-10 year-1 f = 3.9 ·10-9 year-1
Road tanker BLEVE
f = 9.88 ·10-6 · 0.05 · 0.2 · 0.05 · 0.85 · 0.02
f = 8.4 ·10-11 year-1
Tank BLEVE
f = 9.88 ·10-6 · 0.05 · 0.2 · 0.05 · 0.15 · 0.05
f = 3.7 ·10-11 year-1
Initiating event 8. Partial rupture of the pipe connecting the tank with the evaporators. In this case the set of automatic valves installed in the storage tank exit pipe may be activated (depending on the release flow rate), which stops the flow. The rest of the sequences and probabilities are similar to those for the previous case (Fig. 8-17 and Table 8-17). The possibility of a storage tank BLEVE depends on the failure of the water spray system (probability of failure of this system: see section 6.2). Initiating event
Immediate ignition
Flames Failure of Flames impinge Failure of water impinge on automatic valves on tank spray system road tanker
Yes P4=0.15 Yes P3=0.05 Yes P2=0.2
Delayed ignition
Flame front acceleration
Yes P5=0.05 No P =0.95
Accidental scenario
Tank BLEVE Jet fire
5
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98 No
No outcome
P3 =0.95
Partial rupture f =4.93 10-5 y-1
Yes P8=0.2
Yes P7=0.05
Yes P9=0 No P =1
No outcome Flash fire
9
No
No
P2 =0.8
P8 =0.8 No
No outcome No outcome
Fig. 8-17. Event tree and accident scenarios for initiating event 8 (partial rupture of the pipe connecting the tank with the evaporators). Table 8-17 Frequencies of the accident scenarios associated with initiating event 8 Accident scenario Frequency estimation Jet fire f = (4.93 ·10-5 · 0.2 · 0.05 · 0.15 · 0.95) + + (4.93 ·10-5 · 0.2 · 0.05 · 0.85 · 0.98)
Frequency f = 4.8 ·10-7 year-1
Flash fire
f = 4.93 ·10-5 · 0.8 · 0.05 · 0.2 · 1
VCE
This frequency has not been estimated as it is considered that the minimum amount of flammable mixture required to produce blast is not reached.
Road tanker BLEVE
f = 4.93 ·10-5 · 0.2 · 0.05 · 0.85 · 0.02
f = 8.38 ·10-9 year-1
Tank BLEVE
f = 4.93 ·10-5 · 0.2 · 0.05 · 0.15 · 0.05
f = 3.7 ·10-9 year-1
318
f = 3.9 ·10-7 year-1
Initiating event 9. Full-bore rupture of the pipe connecting the tank with the distribution system (release of gas phase). In the event of a full-bore rupture of the pipe connecting the storage tank with the distribution piping system, the different sequences depend on the action or failure of the automatic valves installed to stop the propane flow in the event of a fire. The final sequences are similar to those in the previous event trees. Initiating event
Immediate ignition
Failure of automatic valves
Flames impinge on tank
Failure of water spray system
Flames impinge on road tanker
Yes P5=0.05 No P =0.95
Yes P4=0.15 Yes P3=0.05
Accidental scenario Tank BLEVE Jet fire
5
Yes P2=0.2
Yes P6=0.02
Road tanker BLEVE
No
P4 =0.85
No
Jet fire
P6 =0.98 Full-bore rupture
No
f = 2.4 10-4 y-1
P3 =0.95
No outcome
No
No outcome
P2 =0.8
Fig. 8-18. Event tree and accident scenarios for initiating event 9 (full-bore rupture of the pipe connecting the tank with the distribution system). Table 8-18 Frequencies of the accident scenarios associated with initiating event 9 Accident scenario Frequency estimation Jet fire f = (2.4 ·10-4 · 0.2 · 0.05 · 0.15 · 0.95) + + (2.4 ·10-4 · 0.2 · 0.05 · 0.85 · 0.98)
Frequency f = 2.3 ·10-6 year-1
Road tanker BLEVE
f = 2.4 ·10-4 · 0.2 · 0.05 · 0.85 · 0.02
f = 4.08 ·10-8 year-1
Tank BLEVE
f = 2.4 ·10-4 · 0.2 · 0.05 · 0.15 · 0.05
f = 1.8 ·10-8 year-1
7.3 Effects of the different accident scenarios We will now calculate the effects of the most severe accident scenarios considered: storage tank BLEVE/fireball, road tanker BLEVE/fireball, release of propane creating a flammable cloud (flash fire), and jet fire. Nomenclature: see the respective chapters.
7.3.1 Initiating event 1. Release of propane (liquid) due to the full-bore rupture of the flexible pipe connecting the road tanker with the storage tank Road tanker filling degree: 10%; road tanker volume: 20 m3; dpipe = 0.0635 m; ambient temperature = 25 ºC; wind speed = 5 m s-1; stability = class D; ground roughness = 10 cm; relative humidity = 60%; LFLpropane = 2.1%; Upropane, liquid = 553 kg m-3; Upropane, gas, Tb = 2.32 kg m-3; propane boiling temperature at atmospheric pressure = -42 ºC. The maximum distance reached by c = LFL will be calculated by applying the BritterMcQuaid model. Due to the storage conditions, liquid propane undergoes a flash vaporization and is released as a two-phase flow. If it was released only as a gas the mass flow rate would be approximately
319
7.6 kg s-1. The flow rate in increased by 45% to allow for the two-phase flow; therefore, we obtain an approximate value of 11 kg s-1.
v0
11 2.32
g0
2.32 1.2 9.16 m s-2 1.2
D
§ 4.74 · ¸ ¨ © 5 ¹
4.74 m3 s-1
1/ 2
9.16 4.74 5 3 0.974
0.974 m
0.357 > 0.15 dense gas.
5 1.5 60 t 2.5 x Therefore, the release can be considered continuous up to x = 180 m. c
0.021 § 298 · 0.021 1 0.021 ¨ ¸ © 231 ¹
ª 9.16 2 4.74 º « » 55 ¬ ¼ x § 4.74 · ¨ ¸ © 5 ¹ c c0
1/ 2
0.01635 m3 m-3
1/ 5
0.425
x 0.97
0.01635
From Fig. 6-19, x = 200, x = 194 m 0.97 At this distance the release cannot be considered continuous. If the calculation is repeated for an instantaneous release: V0 = 4.74 m3
320
ª 9.16 4.741 / 3 º « » 52 ¬ ¼
0.784
x x ; 1.68 1.68
x 3
1/ 2
4.74
105 ; x = 176 m
In both cases, this distance seems too conservative. The simulation with standard computer codes gives x | 50 m. Therefore, we assume an intermediate value of x = 100 m for illustrative purposes. Furthermore, the blast would be negligible due to the small amount released. 7.3.2 Initiating event 9. Release of propane (gas) due to the full-bore rupture of the pipe connecting the tank with the distribution system Tank filling degree: 70%; tank volume = 115 m3; Ttank = 25 ºC; Ptank = 9.5 bar; Lpipe = 20 m; dpipe = 0.0508 m; H = 45·10-6 m. Estimation of the initial release flow rate: Estimated pressure at a point just in front of the pipe opening: 9.3 bar. A value of fF = 0.0048 is assumed. Applying Eq. (2-34): § § 1.15 1 ·· 2 ¨ 2 ¨1 Ma cont ¸¸ 1.15 1 ¨ © 2 ¹ ¸ §¨ 1 1·¸ 1.15 § 4 0.0048 20 · ln ¨ ¸ 2 2 ¸ ¨ 2 1.15 1 Macont ¸ ¨© Macont © 0.0508 ¹ ¹ ¨ ¸ © ¹
0
By trial and error, Macont = 0.28. Therefore, by applying Eq. (2-29): Ycont
1
1.15 1 0.28 2 2
1.00588
From Eq. (2-33): Tp
298
2 1.00588 1.15 1
Tp = 279 K
Mass flow rate through the opening (Eq. (2-19)): 1.151
mhole
S 0.0508 2 4
§ 2 · 1.151 44.1 ¸¸ 1 9.3 10 1 1.15 ¨¨ 1 . 15 1 279 8 .314 10 3 © ¹
5
321
5.25 kg s -1
Mass flow rate through the pipe (Eq. (2-24)) (Ucont = 17.1 kg m-3): 11.15 § § ·· § 1.15 · ¨ § 9.3 · 1.15 ¨ ¸¸ 5 ¸¸ ¨ ¨ 2 ¨ 9.5 10 17.1 ¨¨ 1¸ ¸ ¸ 1 1 . 15 9 . 5 © ¹ ¸¸ ¨ ¹¨ © © ¹¹ © 20 · § 4 0.0048 ¨ ¸ © 0.0508 ¹
m pipe
S 0.0508
2
4
0.54 kg s -1
A new trial is required. The following table shows the results of the calculation procedure: mhole, kg s-1 5.25 4.51 3.95 2.82 2.63
Pp, bar 9.3 8 7 5 4.7
Therefore, m = 2.63 kg s-1. Calculation of the jet fire size: Tj
§ 1.15 1 · ¨ ¸ 1.15 ¹
© 324 §¨ 1.013 25 ·¸ 9.5 © ¹
222 K
1.15
Por
Mj
uj
§ 2 · 1.151 9.5 ¨ ¸ © 1.15 1 ¹
1.15 1 §¨ 5.46 ·¸ © 1.013 ¹ 1.15 1 2.126
5.46 bar 1.15 1 1.15
1.15 8314 222 44.1
2 2.126
467 m s-1
For choked flow:
Uj dj
1.969
273 222.5
4 2.63 S 467 2.4
2.4 kg m-3
0.05466
322
mpipe, kg s-1 0.54 1.61 2.03 2.57 2.63
§ 2.4 · 0.05466 ¨ ¸ © 1.2 ¹
ds
1/ 2
0.0789 m
cst-mass = 0.06 0.024 3
9.81 0.0789 5 / 3 § 2.85 · Y 0 .2 Y 2 / 3 ¨ ¸ 2 467 © 0.06 ¹
2/3
0
By trial and error, Y = 290 Lbo
290 0.0789
22.9 m
An angle of 45º is assumed between the hole axis and the wind vector.
Lb
22.9 0.51 e 0.4 5 0.49 1 6.07 10 3 45 90
Rw
5 467
16.3 m
0.0107
s 16.3 0.185 e 20 0.0107 0.015 2.6 m C ' 1000 e 100 0.0107 0.8
343
§ 9.81 0.0789 ¨¨ 2 0 . 0789 467 2 ©
Rids
· ¸¸ ¹
1/ 3
0.0152
W1
1/ 2 ª ª 1 º 70 0.0151 343 0.0107 º § 1.2 · 0.0789 13.5 e 6 0.0162 1.5 « 1 « 1 ¨ » 1.07 m ¸ »e © 2.4 ¹ 15 ¼» ¬« ¬« ¼»
W2
16.3 0.18 e 1.5 0.0107 0.31 1 0.47 e 25 0.0107 5.05 m
Ri Lb 0
D L
§ 9.81 22.9 ¨¨ 2 467 2 0 . 0789 ©
· ¸¸ ¹
1/ 3
4.43
45 90 1 e 25.6 0.0107 8000 0.0107
9º
4.43
16.3 2 2.6 2 sin 2 9 2.6 cos 9 13.7 m
The point source model can be applied to establish the safe distance (corresponding to I = 1 kW m-2):
323
0.21 e
1
0.00323 467
0.11 2.63 46000 0.81 4S d2
Therefore, the intensity of 1 kW m-2 is found at a horizontal distance of 49 m from the release point. Estimation of the thermal effects of the road tanker BLEVE/fireball Vessel volume = 20 m3; filling degree: 50%; Upropane, 25 ºC = 500 kg m-3; relative humidity = 60 %. The propane mass is:
20 · 0.5 · 500 = 5,000 kg Fireball diameter (Eq. (3.83)):
D
5.8 5,0001 / 3
99 m
Duration (Eq. (3-81)):
t
0.9 5,000 0.25
7.5 s
Height at which the fireball centre is located: H = 0.75 · 99 = 74 m
Radiant heat fraction (Eq. (3-88)):
K rad
0.00325 9.5 10 5
0.32
0.266
Emissive power (Eq. (3-88)):
E
0.266 5,000 46,000 S 99 2 7.5
260 kW m-2
The partial pressure of water in the atmosphere is calculated by using Eqs. (3-18) and (3-19): Pw = 1857 Pa. The intensity of the thermal radiation at different distances is calculated by applying the solid flame model, e.g. for a distance x = 150 m: Distance between the flame and the target: d = 118 m. Atmospheric transmissivity:
W
2.85 1857 118
0.12
0.65
View factor:
324
F
I
99 2 § 99 · 4 ¨ 118 ¸ 2 © ¹
2
0.087
0.65 0.087 260 14.8 kW m-2
And on a vertical surface:
Iv
14.8
150 13.2 kW m-2. 167
The following table shows the values of Iv as a function of the distance from the initial position of the tank car. x (m) 50 70 80 100 120 150
Iv (kW m-2) 33.0 30.2 27.8 22.8 18.3 13.3
x (m) 200 250 300 400 500 585
Iv (kW m-2) 8.2 5.4 3.8 2.1 1.3 1.0
Estimation of the peak overpressure of the road tanker BLEVE The value of 'P is calculated by applying the concept of superheating energy (Eq. (5-27)). The gas expansion is assumed to be an irreversible process, in which approximately 5% of SE is devoted to creating overpressure. Thus, at a distance of 50 m from the initial tank car location:
Enthalpies of liquid propane: at 9.5 bar, 25 ºC, hl = 265.3 kJ kg-1; at 1.013 bar, 231 ºC, hlo = 100 kJ kg-1. SE = 265.3 – 100 = 165.3 kJ kg-1
For the total mass of propane: 165.3 kJ kg-1 · 5000 kg = 82,650 kJ Energy converted into overpressure (5%): 82,650 kJ · 0.05 = 4,135 kJ WTNT = 4,135 · (0.214 · 10-3) = 0.88 kg
At a distance of 50 m, dn = 52 m kg-1/3; from Fig. 4-4, 'P | 0.017 bar. This overpressure does not pose a danger to people. The typical threshold value of 'P for glass breakage (0.01 bar) is found at 80 m.
325
Estimation of the thermal effects of the storage tank BLEVE/fireball Vessel volume = 82 m3; filling degree: 70%; Upropane, 25 ºC = 500 kg m-3; relative humidity = 60 %. In the worst-case scenario the BLEVE occurs very quickly after flame impingement. Therefore, the propane mass is:
82 · 0.7 · 500 = 28,700 kg Fireball diameter:
D
5.8 28,7001 / 3
178 m
Duration:
t
0.9 28,7000.25
11.7 s
Height at which the fireball centre is located: H = 0.75 · 178 = 133 m
Radiant heat fraction:
K rad
0.266
Emissive power:
E
0.266 28,700 46,000 S 178 2 11.7
300 kW m-2
The partial pressure of water in the atmosphere is Pw = 1857 Pa. The intensity of the thermal radiation at different distances is calculated by applying the solid flame model for an example distance x = 250 m. Distance between the flame and the target: d = 194 m. Atmospheric transmissivity:
W
2.85 1857 194
0.12
0.614
View factor:
F
Iv
1782 § 178 · 194 ¸ 4¨ 2 © ¹
2
0.099
0.614 0.099 300
250 16.1 kW m-2 283
326
The following table shows the values of Iv as a function of the distance from the initial position of the tank car. x (m) 100 150 200 250 300
Iv (kW m-2) 35.4 23.4 21.6 16.1 12.1
x (m) 400 500 600 700 950
Iv (kW m-2) 7.3 4.8 3.1 2.4 1.0
Estimation of the peak overpressure for the storage tank BLEVE Enthalpies of liquid propane: at 9.5 bar, 25 ºC, hl = 265.3 kJ kg-1; at 1.013 bar, 231 ºC, hlo = 100 kJ kg-1. SE = 265.3 – 100 = 165.3 kJ kg-1
For the total mass of propane: 165.3 kJ kg-1 · 28,700 kg = 4.752375 · 106 kJ Energy converted into overpressure (5%): 4.752375 · 106 kJ · 0.05 = 237,620 kJ WTNT = 237,620 · (0.214 · 10-3) = 50.9 kg
At a distance of 50 m, dn = 13.5 m kg-1/3; from Fig. 4-4, 'P | 0.1 bar. This overpressure does not pose a danger to people. The typical threshold value of 'P for glass breakage (0.01 bar) is found at 300 m. 7.4 Calculation of the individual risk The individual risk will be calculated for a point located 100 m from the centre of the plant in the direction j = 3 (wind blowing from 240º; see wind rose, Table 8-8). For practical purposes, all initiating events will be assumed to occur in the centre of the installation. Meteorological data, probabilities: stability class D, PM = 0.8 stability class F, PM = 0.2 wind blowing from 240º, Pw = 0.116. The point is affected by the following accident scenarios: - flash fire - tank car BLEVE/fireball - storage tank BLEVE/fireball We will calculate the contribution of these accident scenarios to individual risk.
Flash fire For the sake of simplicity, we consider that the flammable cloud reaches the point for both stability classes (although they would actually give two different values).
327
Furthermore, we consider that the probability of death in the area covered by the flash fire is Pd = 1 (see Chapter 7, section 4.1.3). Overall frequency taking into account the different initiating events (from Tables 8-10 to 818): fflash fire = 8.7 · 10-7 + 1.11 · 10-4 + 6.35 · 10-10 + 3.2 · 10-8 + 3.9 · 10-9 + 3.9 · 10-7 = =1.12 · 10-4 year-1 P = 0.116 · (0.8 + 0.2) · 1 = 0.116
'IRflash fire = (1.12 · 10-4) · 0.116 = 1.3 · 10-5 fatalities year-1 Road tanker fireball Froad tanker fireball = 8 · 10-7 + 6.96 · 10-5 + 1.35 · 10-11 + 1.36 · 10-9 + 1.62 · 10-9 + 8.4 · 10-11 + 8.38 · 10-9 + 4.08 · 10-8 = 7.05 · 10-5 year-1
At x = 100 m, I = 22.8 kW m-2 during 7.5 s Y = -36.38 + 2.56 ln (7.5 · 22,8004/3) = 3.03; 2.5% deaths; Pd = 0.025 'IRroad tanker fireball = (7.05 · 10-5) · 0.025 = 1.76 · 10-6 fatalities year-1 Storage tank fireball f storage tank fireball = 6 · 10-9 + 6 · 10-12 + 6 · 10-10 + 7.14 · 10-10 + 3.7 · 10-11 + 3.7 · 10-9 + 1.8 ·10-8 = 2.9 · 10-8 year-1
At x = 100 m, I = 35.4 kW m-2 during 11.7 s Y = -36.38 + 2.56 ln (11.7 · 35,4004/3) = 5.67; 75% deaths; Pd = 0.75 'IRstorage tank fireball = (2.9 · 10-8) · 0.75 = 2.18 · 10-8 fatalities year-1 Therefore, the individual risk at the selected point due to all the accident scenarios considered is: IR = 1.3 · 10-5 + 1.76 · 10-6 + 2.18 · 10-8 = 1.48 · 10-5 fatalities year-1 In order to determine the influence of the meteorological conditions we calculate the individual risk at a point located at the same distance from the initiation point but in the opposite direction (j = 9): fflash fire = 1.12 · 10-4 year-1 Pw = 0.088; P = 0.088 · 1 · 1 = 0.088
'IRflash fire = (1.12 · 10-4) · 0.088 = 9.86 · 10-6 fatalities year-1
328
The individual risk caused by the two fireballs does not depend on the direction. Therefore, IR = 9.86 · 10-6 + 1.76 · 10-6 + 2.18 · 10-8 = 1.16 · 10-5 fatalities year-1 Fig. 8-19 shows the iso-risk curves for this scenario. The influence of the flash fire is clear for short distances, while for larger distances the individual risk depends essentially on the fireballs effects, which are uniform in all directions.
Fig. 8-19. Individual risk curves.
NOMENCLATURE
D d dn dor ds E F f fF fi
fireball diameter (m) distance between the surface of the flames and the target (m) scaled distance (m kg-1/3) orifice or outlet diameter (m) effective orifice diameter (m) flame emissive power (kW m-2) view factor (-) frequency (year-1) Fanning friction factor (-) frequency of the accident scenario i (year-1)
329
fN Fv H hl hlo HR I IRav IRx,y IRx,y, i L LBv Lpipe M m Ma Mw Ni Rw P P PFi px,y Po Pw Re s T t Ta Tcont T0 Tj Tp usound uw x
D DE H Krad Ua Uf-a Uj W
frequency of all accident scenarios with N or more fatalities (year-1) view factor, vertical surface (-) height at which the fireball centre is located (m) enthalpy of the liquid at temperature T (kJ·kg-1) enthalpy of the liquid at temperature T0 (kJ·kg-1) relative humidity of the atmosphere (%) intensity of the thermal radiation reaching a given target (kW m-2) average individual risk (exposed population) (year-1) total individual risk of fatality at the geographical location x, y (year-1) individual risk of fatality at the geographical location x, y from the accidental scenario i (year-1) length of the visible flame (m) vertical distance between the gas outlet and the flame tip (m) pipe length (m) mass of substance (kg) mass flow rate (kg s-1) Mach number (= u us-1) (-) molecular weight (kg kmole-1) number of fatalities from each accident scenario (-) ratio between wind velocity and jet velocity at the gas outlet (-) pressure (N m-2 or bar) probability (-) probability that the accidental scenario i results in a fatality at location x, y (-) number of people at location x, y (-) atmospheric pressure (N m-2) partial pressure of water in the atmosphere (N m-2) Reynolds number (-) lift-off distance (m) temperature (K) fireball duration (s) ambient temperature (K) temperature inside the container (K) boiling temperature at atmospheric pressure (K) jet temperature at the gas outlet (K) temperature of the gas at a given point of the pipe (K) speed of sound in a given gas (m s-1) wind speed (m s-1) horizontal distance between the centre of the fireball and the target (Fig. 3-14) (m) tilt angle of a jet fire (º) angle between the axis of the orifice and the line joining the centre of the orifice and the tip of the flame (º) pipe roughness (m) radiant heat fraction (-) air density (kg m-3) density of the fuel-air mixture (kg m-3) density of gas in the outlet (kg m-3) atmospheric transmissivity (-)
330
REFERENCES [1] Health and Safety Executive. Canvey: An Investigation of Potential Hazards from
operations in the Canvey Island/Thurrock Area. HM Stationery Office. London, 1978. [2] Health and Safety Executive. Canvey: A Second Report. A Review of the Potential
Hazard from Operations in the Canvey Island/Thurrock Area Three Years after Publication of the Canvey Report. HM Stationery Office. London, 1981. [3] Rijnmond Public Authority. A Risk Analysis of 6 Potentially hazardous Industrial Objects in the Rijnmond Area-A Pilot Study. D. Reidel, Dordrecht, 1982. [4] Committee for the Prevention of Disasters. Guidelines for Quantitative Risk Analysis (the “Purple Book”). The Hague, SDU, 1999. [5] Center for Chemical Process Safety. Guidelines for Chemical Process Quantitative Risk Analysis, 2nd ed. AIChE. New York, 2000. [6] A. Ronza. PhD thesis. UPC. Barcelona, 2007. [7] M. Considine. The Assessment of Individual and Societal Risks. SRD Report R-310, Safety and Reliability Directorate, UK Atomic Energy Authority. Arrington, 1984. [8] C. M. Pietersen, B. F. P. Van het Veld. J. Loss Prev. Process Ind., 5, 60, 1992. [9] D. C. Hendershot. A Simple problem to Explain and Clarify the principles of Risk Calculation. 1997. http://home.att.net/-d.c.hendershot/papers/pdfs/riskland.pdf (consulted 19/V/2007). [10] J. A. Vílchez. Personal communication. [11] A. Ronza, J. A. Vílchez, J. Casal. J. Hazard. Mater., doi:10.1016/j.jhazmat 2005.11.057, 2007. [12] TNO. The RISKCURVES Software. Apeldoorn, 2000.
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Annex 1
Constants in the Antoine equation log10 P
A
B T C
P: saturation vapor pressure (bar); T: temperature (K). Selected from National Institute of Standards and Technology Chemistry WebBook. http://webbook.nist.gov/chemistry/ (consulted on 9 May 2007)
Substance Acetaldehyde Acetic acid Acetone Acetylene Acrolein Acrylonitrile Allene Allyl chloride Ammonia Aniline Benzene 1,3 Butadiene Butanal Butane Butanol 1-Butene n-Butyl acetate Butyl chloride Carbon disulfide Cyclohexane Cyclopropane Diethyl ether Diethylamine Diisopropyl ether
Temperature range, K 293 – 377 290 – 391 259 – 507 214 – 308 208 – 326 222 – 351 152 – 238 286 – 317 239– 371 304 – 457 288 – 354. 198 – 272 304 – 347 273 - 425 296 – 391 196 – 269 333 – 399 256 – 352 277 – 353 293 – 355 183 – 241 250 – 329 305 – 334 297 – 340
333
A 3.68639 4.68206 4.42448 4.66141 4.11586 4.06661 3.79809 2.24083 4.86886 4.34541 4.01814 3.99798 3.59112 4.35576 4.54607 4.24696 4.26803 3.99588 4.06683 3.96988 4.05015 4.02200 2.86193 3.96649
B 822.894 1642.54 1312.253 909.079 1167.888 1255.939 755.286 365.121 1113.928 1661.858 1203.835 941.662 952.851 1175.581 1351.555 1099.207 1440.231 1182.903 1168.62 1203.526 870.393 1062.64 559.071 1135.034
C -69.899 -39.764 -32.445 7.947 -41.56 -41.853 -39.159 -154.919 -10.409 -74.048 -53.226 -32.753 -82.569 -2.071 -93.34 -8.256 -61.362 -54.885 -31.616 -50.287 -25.063 -44.93 -132.974 -54.92
Dimethyl ether Dimethylamine Ethanol Ethyl acetate Ethyl chloride Ethylene Ethylene oxide Formaldehyde Furan Heptane Hexane Hydrocyanic acid Hydrogen Isobutane Isopentane Isopropyl alcohol Isopropylamine Methane Methyl methacrylate Methyl acetate Methyl acrylate Methyl alcohol Methyl chloride Methyl ethyl ketone Methyl formate Monomethyl amine Octane Pentane Propane Propyl amine Propylene Propylene oxide Tetrahydrothiophene Toluene Triethyl amine Trimethyl amine Vinyl acetate Vinyl chloride m-Xylene p-Xylene o-Xylene
195 – 248 201 – 280 273 – 352 289 – 349 217 – 286 149 – 188 273 – 305 164 – 251 275 – 334 299 - 372 286 - 343 256 - 319 21 - 32 261 - 408 289 - 302 330 - 362 277 - 334 110 - 190 312 - 362 275 - 329 229 - 353 288 - 357 303 - 416 314 - 370 294 - 305 190 - 267 217 - 297 269 - 341 277 - 361 296 - 351 165 - 226 292 - 344 333 - 373 273 - 323 323 - 368 193 - 277 295 - 345 165 - 259 273 - 333 286 - 452 273 - 323
4.11475 4.29371 5.37229 4.22809 4.16181 3.87261 5.84696 4.28176 4.10003 4.02832 4.00266 4.67417 3.54314 4.3281 3.91457 4.861 4.01507 4.22061 5.37785 4.20364 4.32327 5.20409 4.91858 3.9894 0.25097 4.5199 5.2012 3.9892 4.53678 4.05136 3.97488 3.55046 5.00861 4.14157 2.98368 4.01613 4.34032 3.98598 5.09199 4.14553 4.93755
334
894.669 995.445 1670.409 1245.702 1052.821 584.146 2022.83 959.43 1060.801 1268.636 1171.53 1340.791 99.395 1132.108 1020.012 1357.427 985.65 516.689 1945.56 1164.426 1338.663 1581.341 1427.529 1150.207 6.524 1034.977 1936.281 1070.617 1149.36 1044.028 795.819 802.487 1979.981 1377.578 695.814 970.297 1299.069 892.757 1996.545 1474.403 1901.373
-30.604 -47.869 -40.191 -55.189 -32.078 -18.307 62.656 -29.758 -45.416 -56.199 -48.784 -11.592 7.726 0.918 -40.053 -75.814 -59.079 11.223 -7.569 -52.69 -43.516 -33.5 45.137 -63.904 -278.54 -37.574 -20.143 -40.454 24.906 -62.314 -24.884 -81.348 2.346 -50.507 -128.271 -34.06 -46.183 -35.051 -14.772 -55.377 -26.268
Annex 2
Flammability limits, flash temperature and combustion (higher value) for different substances
heat
of
Selected from: * Sax, N.I., Lewis, R.J. Dangerous Properties of Industrial Materials. Seventh Edition. Van Norstrand Reinhold. New York, 1989. ** Suzuki, T., Koide, K. Correlation between Upper Flammability Limits and Thermochemical Properties of Organic Compounds. Fire and Materials 18 (1994) 393397.
Substance Acetaldehyde Acetic acid Acetone Acetylene Acrolein Acrylonitrile Allyl bromide Allyl chloride Aniline Benzene 1,3 Butadiene Butanal Butane Butanol 1-Butene n-Butyl acetate Butyl chloride Carbone monoxide Cyclohexane Cyclopropane Diethylamine Diisopropyl ether
LFL*, % vol.
UFL*, % vol.
4 5.4 2.6 2.5 2.8 3.1 4.4 2.9 1.3 1.4 2 2.5 1.9 1.4 1.6 1.3 1.9 12.5 1.3 2.4 1.8 1.4
57 16 12.8 82 31 17 7.3 11.2 8 11.5 12.5 8.5 11.2 9.3 7.5 10.1 74.2 8.4 10.4 10.1 7.9
335
Flash temp.*, K 'H**, kJ kg-1 235.4 315.9 255.4 255.4 < 255.4 272.0 272.0 241.5 343.1 262.0 197.0 266.5 213.1 308.1 210.9 265.4 263.7 256.1 255.1 245.4
17344 15422 31360 49907 33718 37027 42266 46966 34841 49510 36809 48419 30887 46970 49698 42033 39571
Dimethyl ether Dimethylamine Ethanol Ethyl acetate Ethylene Ethylene oxide Formaldehyde Furan Gasoline Heptane Hexane Hydrocyanic acid Hydrogen Isoamyl alcohol Isobutane Isopentane Isopropyl alcohol Isopropylamine Kerosene Methane Methyl acetate Methyl acrylate Methyl alcohol Methyl chloride Methyl cyclohexane Methyl ethyl ketone Methyl formate Methyl methacrylate Monomethyl amine Octane Pentane Propane Propyl amine Propylene Propylene oxide Toluene Triethyl amine Trimethyl amine Vinyl acetate Vinyl chloride m-Xylene o-Xylene p-Xylene
3.4 2.8 3.3 2.2 2.7 3 7 2.3 1.3 1.05 1.2 5.6 4.1 1.2 1.9 1.4 2.5 2.3 0.7 5.3 3.1 2.8 6 8.1 1.2 1.8 5.9 2.1 4.95 1 1.5 2.3 2 2.4 2.8 1.27 1.2 2 2.6 4 1.1 1 1.1
27 14.4 19 11 36 100 73 14.3 7.1 6.7 7.5 40 74.2 9 8.5 7.6 12 10.4 5 15 16 25 36.5 17 6.7 11.5 20 12.5 20.75 4.7 7.8 9.5 10.4 10.1 37 7 8 11.6 13.4 22 7 6 7
336
232.0 255.4 286.3 268.7 253.1 358.1 237.6 227.6 269.3 250.1 255.4 315.9 < 222.0 284.8 235.9 338.7 - 358.1 50.6 263.1 270.4 285.4 < 273.1 269.3 267.6 254.1 283.1 273.1 286.5 < 233.1 168.7 235.9 165.4 235.9 277.6 266.5 266.5 265.4 265.1 298.1 290.1 298.1
31702 39239 30588 25804 50310 29687 19008 31006 47711 48671 49363 48893 34126 40315 55505 21978 24377 23845 46855 34374 16703 34936 48250 49017 50338 40535 48902 33459 42847 40859 41330 24566 43276 43292 43282
Annex 3
Acute Exposure Guideline Levels (AEGLs) Source: http://www.epa.gov/oppt/aegl/pubs/chemlist.htm (consulted on 8 May 2007) Notes: all values are in ppm. Changes to proposed values may occur. * t 10% LEL; ** t 50% LEL For values denoted as * safety considerations against the hazard(s) of explosion(s) must be taken into account. For values denoted as ** extreme safety considerations against the hazard(s) of explosion(s) must be taken into account. NR: not recommended due to insufficient data Substance Acetaldehyde (interim) AEGL1 AEGL2 AEGL3 Acetone (interim) AEGL1 AEGL2 AEGL3 Acetone cyanohydrin (interim) AEGL1 AEGL2 AEGL3 Acetonitrile (interim) AEGL1 AEGL2 AEGL3
10 min
30 min
60 min
4h
8h
45 340 1100
45 340 1100
45 270 840
45 170 530
45 110 260
200 9300* **
200 4900 8600*
200 3200 5700*
200 1400 2500
200 950 1700
2.5 17 27
2.5 10 21
2.0 7.1 15
1.3 3.5 8.6
1.0 2.5 6.6
13 490 1000
13 490 1000
13 320 670
13 130 280
13 86 180
0.030 0.44 6.2
0.030 0.18 2.5
0.030 0.10 1.4
0.030 0.10 0.48
0.030 0.10 0.27
1.5 68 480
1.5 68 480
1.5 46 180
1.5 21 85
1.5 14 58
Acrolein (interim) AEGL1 AEGL2 AEGL3
Acrylic acid (interim) AEGL1 AEGL2 AEGL3
337
Substance Acrylonitrile (proposed) AEGL1 AEGL2 AEGL3
10 min
30 min
60 min
4h
8h
4.6 290 480
4.6 110 180
4.6 57 100
4.6 16 35
4.6 8.6 19
2.1 4.2 36
2.1 4.2 25
2.1 4.2 20
2.1 4.2 10
2.1 4.2 10
0.42 0.33 150
0.42 0.33 40
0.42 0.33 18
0.42 1.8 3.5
0.42 1.2 2.3
30 220 2700
30 220 2600
30 160 1100
30 110 550
30 110 390
48 72 120
16 24 40
8.0 12 20
2.0 3.0 5.0
1.0 1.5 2.5
NR 0.30 0.91
NR 0.21 0.63
NR 0.17 0.50
NR 0.040 0.13
NR 0.020 0.060
130 2000* **
73 1100 5600*
52 800 4000*
18 400 2000*
9.0 200 990
NR 39 100
NR 27 71
NR 22 56
NR 11 23
NR 5.6 11
0.120.12 8.1 84
0.12 3.5 36
0.12 2.0 21
0.12 0.70 7.3
0.12 0.41 7.3
8.3 160 820
8.3 160 820
8.3 130 480
8.3 81 170
8.3 53 97
0.50 2.8 50
0.50 2.8 28
0.50 2.0 20
0.50 1.0 10
0.50 0.71 7.1
Allyl alcohol (interim) AEGL1 AEGL2 AEGL3
Allyl Amine (interim) AEGL1 AEGL2 AEGL3
Ammonia (interim) AEGL1 AEGL2 AEGL3
Aniline (final) AEGL1 AEGL2 AEGL3
Arsine (final) AEGL1 AEGL2 AEGL3
Benzene (interim) AEGL1 AEGL2 AEGL3
Benzonitrile (interim) AEGL1 AEGL2 AEGL3
Bromine trifluoride (interim) AEGL1 AEGL2 AEGL3
n-Butyl acrylate (interim) AEGL1 AEGL2 AEGL3
Chlorine (final) AEGL1 AEGL2 AEGL3
338
Substance
10 min
30 min
60 min
4h
8h
0.15 1.4 3.0
0.15 1.4 3.0
0.15 1.1 2.4
0.15 0.69 1.5
0.15 0.45 0.98
NR 120 4000
NR 80 4000
NR 64 3200
NR 40 2000
NR 29 1600
10 130 480
10 85 320
10 66 250
10 40 150
10 32 120
0.035 0.17 4.0
0.035 0.17 2.3
0.024 0.12 1.6
0.012 0.061 0.82
0.0087 0.043 0.58
5.7 53 570
5.7 53 160
5.7 24 72
5.7 14 43
5.7 10 30
8.3 66 950
8.3 45 410
8.3 36 240
8.3 19 71
8.3 9.4 41
NR 12 25
NR 12 25
NR 9.7 20
NR 6.1 13
NR 4.8 10
NR 80 360
NR 80 360
NR 45 200
NR 14 63
NR 7.9 35
0.90 14 100
0.90 14 70
0.90 14 56
0.90 14 35
0.90 14 35
0.1 23 64
0.1 16 45
0.1 13 35
0.1 3.1 8.9
0.1 1.6 4.4
1 100 740
1 43 250
1 22 120
1 11 31
1 11 31
Chlorine dioxide (final) AEGL1 AEGL2 AEGL3
Chloroform (interim) AEGL1 AEGL2 AEGL3
Dimethylamine (proposed) AEGL1 AEGL2 AEGL3
Dimethyl sulfate (interim) AEGL1 AEGL2 AEGL3
Epichlorohydrin (interim) AEGL1 AEGL2 AEGL3
Ethyl acrylate (interim) AEGL1 AEGL2 AEGL3
Ethylene diamine (final) AEGL1 AEGL2 AEGL3
Ethylene oxide (interim) AEGL1 AEGL2 AEGL3
Formaldehyde (interim) AEGL1 AEGL2 AEGL3
Hydrazine (interim) AEGL1 AEGL2 AEGL3
Hydrogen bromide (interim) AEGL1 AEGL2 AEGL3
339
Substance
10 min
30 min
60 min
4h
8h
1.8 100 620
1.8 43 210
1.8 22 100
1.8 11 26
1.8 11 26
2.5 17 27
2.5 10 21
2.0 7.1 15
1.3 3.5 8.6
1.0 2.5 6.6
1.0 95 170
1.0 34 62
1.0 24 44
1.0 12 22
1.0 12 22
NR 1.8 5.4
NR 1.0 3.1
NR 0.73 2.2
NR 0.37 1.1
NR 0.26 0.78
0.75 41 76
0.60 32 59
0.51 27 50
0.36 20 37
0.33 17 31
NR 33 120
NR 23 85
NR 18 68
NR 11 17
NR 7.5 8.5
NR 7.5 15
NR 7.5 15
NR 4.9 10
NR 2.0 4.3
NR 1.3 2.8
6.7 76 280
6.7 76 280
6.7 61 220
6.7 38 140
6.7 25 71
2.0 16 32
2.0 16 32
1.0 13 25
1.0 6.5 13
1.0 6.5 13
670 11000* **
670 4000* 14000*
530 2100 7200*
340 730 2400
270 520 1600
NR 5.3 16
NR 1.8 5.5
NR 0.90 2.7
NR 0.23 0.68
NR 0.11 0.34
Hydrogen chloride (final) AEGL1 AEGL2 AEGL3
Hydrogen cyanide (final) AEGL1 AEGL2 AEGL3
Hydrogen fluoride (final) AEGL1 AEGL2 AEGL3
Hydrogen selenide (interim) AEGL1 AEGL2 AEGL3
Hydrogen sulphide (interim) AEGL1 AEGL2 AEGL3
Isobutyronitrile (interim) AEGL1 AEGL2 AEGL3
Malononitrile (interim) AEGL1 AEGL2 AEGL3
Methacrylic acid (interim) AEGL1 AEGL2 AEGL3
Methacrylonitrile (interim) AEGL1 AEGL2 AEGL3
Methanol (interim) AEGL1 AEGL2 AEGL3
Methyl hydrazine (final) AEGL1 AEGL2 AEGL3
340
Substance
10 min
30 min
60 min
4h
8h
NR 0.40 1.2
NR 0.13 0.40
NR 0.067 0.20
NR 0.017 0.05
NR 0.008 0.025
NR 59 120
NR 59 86
NR 47 68
NR 30 43
NR 19 22
17 150 720
17 150 720
17 120 570
17 76 360
17 50 180
0.53 43 170
0.53 30 120
0.53 24 92
0.53 6.0 23
0.53 3.0 11
0.50 20 34
0.50 15 25
0.50 12 20
0.50 8.2 14
0.50 6.7 11
NR 160 320
NR 110 220
NR 90 180
NR 55 110
NR 38 76
NR 4.3 13
NR 1.6 4.7
NR 0.83 2.5
NR 0.24 0.71
NR 0.13 0.38
0.17 0.50 19
0.17 0.50 9.6
0.17 0.50 4.8
0.17 0.50 2.6
0.17 0.50 1.9
0.013 0.53 1.6
0.013 0.37 1.1
0.013 0.30 0.90
0.013 0.077 0.23
0.013 0.037 0.11
19 29 NR
19 29 NR
15 23 NR
9.5 15 NR
6.3 12 NR
NR 1.0 3.0
NR 0.70 2.1
NR 0.53 1.6
NR 0.33 1.0
NR 0.17 0.52
Methyl isocyanate (final) AEGL1 AEGL2 AEGL3
Methyl mercaptan (interim) AEGL1 AEGL2 AEGL3
Methyl methacrylate (interim) AEGL1 AEGL2 AEGL3
Nitric acid (interim) AEGL1 AEGL2 AEGL3
Nitrogen dioxide (interim) AEGL1 AEGL2 AEGL3
N,N-Dimethylformamide (inter.) AEGL1 AEGL2 AEGL3
Oxygen difluoride (proposed) AEGL1 AEGL2 AEGL3
Peracetic acid (interim) AEGL1 AEGL2 AEGL3
Perchloromethyl mercaptan (int.) AEGL1 AEGL2 AEGL3
Phenol (interim) AEGL1 AEGL2 AEGL3
Phenyl mercaptan (proposed) AEGL1 AEGL2 AEGL3
341
Substance
10 min
30 min
60 min
4h
8h
NR 0.60 3.6
NR 0.60 1.5
NR 0.30 0.75
NR 0.080 0.20
NR 0.040 0.090
NR 4.0 7.2
NR 4.0 7.2
NR 2.0 3.6
NR 0.50 0.90
NR 0.25 0.45
0.34 2.5 7.0
0.34 2.5 7.0
0.34 2.0 5.6
0.34 1.3 3.5
0.34 0.83 1.8
10 50 370
10 50 180
6.6 33 110
2.6 13 45
1.7 8.3 28
45 330 1100
45 330 1100
45 260 840
45 170 530
45 110 260
NR 6.7 20
NR 4.7 14
NR 3.7 11
NR 0.90 2.7
NR 0.47 1.4
NR 83 170
NR 25 50
NR 12 23
NR 2.5 5.1
NR 1.2 2.4
0.067 0.11 0.33
0.067 0.11 0.33
0.053 0.087 0.26
0.033 0.057 0.17
0.017 0.028 0.083
20 230 1900*
20 160 1900*
20 130 1100*
20 130 340
20 130 340
0.20 0.75 30
0.20 0.75 30
0.20 0.75 30
0.20 0.75 19
0.20 0.75 9.6
35 230 1600
35 230 1600
35 230 1200
35 120 580
35 81 410
Phosgene (final) AEGL1 AEGL2 AEGL3
Phosphine (interim) AEGL1 AEGL2 AEGL3
Phosphorus trichloride (interim) AEGL1 AEGL2 AEGL3
Piperidine (interim) AEGL1 AEGL2 AEGL3
Propionaldehyde (interim) AEGL1 AEGL2 AEGL3
Propyl chloroformate (proposed) AEGL1 AEGL2 AEGL3
Propyleneimine (interim) AEGL1 AEGL2 AEGL3
Selenium hexafluoride (proposed) AEGL1 AEGL2 AEGL3
Styrene (interim) AEGL1 AEGL2 AEGL3
Sulfur dioxide (interim) AEGL1 AEGL2 AEGL3
Tetrachloroethylene (interim) AEGL1 AEGL2 AEGL3
342
Substance
10 min
30 min
60 min
4h
8h
NR 0.66 2.2
NR 0.66 2.2
NR 0.52 1.7
NR 0.33 1.1
NR 0.17 0.55
NR 7.6 38
NR 2.2 13
NR 1.0 5.7
NR 0.21 2.0
NR 0.094 0.91
200 3100* **
200 1600* 6100*
200 1200 4500*
200 790 3000*
200 650 2500*
0.020 0.24 0.65
0.020 0.17 0.65
0.020 0.083 0.51
0.010 0.021 0.32
0.010 0.021 0.16
0.19 27 44
0.19 8.9 27
0.19 4.4 14
0.19 1.1 2.6
0.19 0.56 1.5
260 960 6100
180 620 6100
130 450 3800
84 270 1500
77 240 970
0.60 37 170
0.60 12 56
0.60 6.2 28
0.60 3.1 7.0
0.60 3.1 7.0
1.8 190 790
1.8 64 270
1.8 32 130
1.8 16 33
1.8 16 33
450 2800 12000*
310 1600 6800*
250 1200 4800*
140 820 3400
70 820 3400
130 2500* **
130 1300* 3600*
130 920* 2500*
130 500 1300*
130 400 1000*
Tetranitromethane (final) AEGL1 AEGL2 AEGL3
Titanium tetrachloride (interim) AEGL1 AEGL2 AEGL3
Toluene (interim) AEGL1 AEGL2 AEGL3
2,4-Toluene diisocyanate (final) AEGL1 AEGL2 AEGL3
trans-Crotonaldehyde (interim) AEGL1 AEGL2 AEGL3
Trichloroethylene (interim) AEGL1 AEGL2 AEGL3
Trichloromethyl silane (interim) AEGL1 AEGL2 AEGL3
Trimethylchlorosilane (interim) AEGL1 AEGL2 AEGL3
Vinyl chloride (interim) AEGL1 AEGL2 AEGL3
Xylenes (interim) AEGL1 AEGL2 AEGL3
343
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Annex 4
Immediately Dangerous to Life and Health concentrations (IDLH) Selected from National Institute for Occupational Safety http://www.cdc.gov/niosh/idlh/intrid4.html (consulted on 9 May 2007). Substance Acetaldehyde Acetic acid Acetone Acrolein Acrylonitrile Ammonia Aniline Benzene Bromine n-Butyl acetate n-Butyl alcohol Carbon dioxide Carbon disulfide Carbon monoxide Carbon tetrachloride Chlorine Chloroacetaldehyde Chlorobenzene Chlorobromomethane Chloroform Cresol Cyclohexane Cyclohexanol Cyclohexanone Diethylamine Diisobutyl ketone Dimethylamine Dimethylformamide
IDLH, ppm 2,000 50 2,500 2 85 300 100 500 3 1,700 1,400 40,000 500 1,200 200 10 45 1,000 2,000 500 250 1,300 400 700 200 500 500 500
Substance Dimethylsulphate Dioxane Ethanolamine Ethyl acetate Ethyl acrilate Ethyl alcohol Ethylamine Ethyl benzene Ethyl bromide Ethyl butyl ketone Ethyl chloride Ethylene oxide Ethyl ether Fluorine Formaldehyde Formic acid Furfural n-Heptane Hexachloroethane n-Hexane Hydrazine Hydrogen bromide Hydrogen chloride Hydrogen cyanide Hydrogen fluoride Hydrogen peroxide Hydrogen sulfide Iodine
345
and
Health:
IDLH, ppm 7 500 30 2,000 300 3,300 600 800 2,000 1,000 3,800 800 1,900 25 20 30 100 750 300 1,100 50 30 50 50 30 75 100 2
Substance Isobutyl acetate Isobutyl alcohol Isopropyl acetate Iropropyl alcohol Isopropylamine Isopropyl ether L. P. G. Methyl acetate Methyl acetylene Methyl acrylate Methyl alcohol Methylamine Methyl bromide Methyl chloride Methyl chloroform Methylcyclohexane Methyl isocyanate Methyl mercaptan Methyl metacrylate Methyl styrene Monomethyl aniline Naphta (coal tar) Naphtalene Nitric acid Nitric oxide Nitrobenzene Nitrogen dioxide Nitromethane Nitrotoluene (o, m, p isomers) Octane Oxigen difluoride Ozone
IDLH, ppm 1,300 1,600 1,800 2,000 750 1,400 2,000 3,100 1,700 250 6,000 100 250 2,000 700 1,200 3 150 1,000 700 100 1,000 250 25 100 200 20 750 200 1,000 0.5 5
Substance n-Pentane Perchloromethyl mercaptan Petroleum distillates (naphta) Phenol Phosgene Propane n-Propyl acetate n-Propyl alcohol Propylene oxide Pyridine Selenium hexafluoride Styrene Sulfur dioxide Sulfur monochloride Sulfur pentafluoride 1,1,2,2-Tetrachloroethane Tetrachloroethylene Tetrahydrofuran Tetranithomethane Toluene Toluene 2,4-diisocyanate o-Toluidine Tributyl phosphate 1,1,2-Trichloroethane Trichloroethylene 1,2,3-Trichloropropane Triethylamine Trifluorobromomethane Turpentine Vinyl toluene Xylene (o, p, m isomers) Xylidine
346
IDLH, ppm 1,500 10 1,100 250 2 2,100 1,700 800 400 1,000 2 700 100 5 1 100 150 2,000 4 500 2.5 50 30 100 1,000 100 200 40,000 800 400 900 50
Annex 5
Determining the damage to humans from explosions using characteristic curves Damage from explosions can be taken from tables relating overpressure to the expected degree of damage, or calculated from probit equations (see Chapter 7). Probit equations, relating the magnitude of the action (overpressure and impulse) to the percentage of the exposed population that will suffer a certain degree of damage, are the most widely used methodology to determine damage to human beings. Recently, a set of diagrams have been published in which probit equations and characteristic overpressure-impulse-distance curves allow a direct assessment of damage to humans. Although using these diagrams implies a certain lack of accuracy, damage can be directly assessed in only one step. Damage to humans from explosions as a function of TNT equivalence The characteristic curves (see Chapter 4) are diagrams in which the overpressure and impulse at each distance from an explosion are plotted as a function of the TNT equivalent mass. These curves were combined -with the adequate numerical treatment- with selected probit equations for eardrum rupture, death due to displacement and head impact (skull fracture), death due to displacement and whole body impact and death due to lung hemorrhage, to obtain figures A-5.1, A-5.2 and A-5.3 [1]. From these characteristic curves the consequences on humans can be directly assessed, avoiding calculations and allowing an overview of the evolution of the damage caused by explosions. When a more accurate result is needed, fundamental equations should be used.
347
Impulse (Pa·s)
100000
3
10 kg TNT 100 m
500 10000
200 m
150 50 99
Percentage death lung haemorrhage (% )
1
5
10
25
50
75
20
90
10 5 2
50 m 25 m
1000 100000
1000000
Side-on overpressure (Pa)
Fig. A-5-1. Percentages of exposed population that would die due to lung hemorrhage (black solid lines) as a function of distance (d) (thin grey lines) and TNT equivalent mass (thick grey lines). Taken from [1], by permission.
Impulse (Pa·s)
100000
3
100 m
10 kg TNT 500
10000
1
Percentage death whole body impact (%)
1000 100000
150
200 m
5
10
25
50
75
90
95
99
50 20 10 5 2
50 m 25 m
1000000
Side-on overpressure (Pa)
Fig. A-5-2. Percentages of exposed population that would die if their body hits a rigid object (black solid lines) as a function of distance (d) (thin grey lines) and TNT equivalent mass (thick grey lines). Taken from [1], by permission.
348
100000
10 3 kg TNT
99
Percentage death skull fracture (%)
50
500
200 m
1
10000
Impulse (Pa·s)
99
600 m
400 m
75
90
95
150 50 20 10
50 25
5
1000
5
2
10
1
1 Percentage eardrum rupture (%)
0.5
0.15 100 m
25 m
100
50 m
10000
100000
1000000
Side-on overpressure (Pa)
Fig. A-5-3. Percentages of exposed population that would suffer eardrum rupture (black solid lines) or would die if their head hits a rigid object (semi-dotted lines) as a function of distance (thin grey lines) and TNT equivalent mass (thick grey lines). Taken from [1], by permission.
Damage to humans from vapour cloud explosions as a function of the Multi-energy method Characteristic overpressure-impulse-distance curves for vapour cloud explosions, obtained from the Multi-energy model (see Chapter 4) for a charge strength of 10, were combined with different probit equations to determine the damage as a function of distance [2] for the following consequences: eardrum rupture, death due to head impact, death due to whole body impact, death due to lung hemorrhage (Figs. A-5.4, A-5.5 and A-5.6).
349
12
Explosion energy (Eexp, J)
3 10
100000
12
10
11
Impulse (Pa·s)
3 10
11
10 150 m
10
3 10 99 200 m
95 90
10000
10
10
75 50 25
5
1
10
100 m
Percentage death body impact ( % )
50 m
25 m
1000 100000
1000000
Overpressure (Pa)
Fig. A-5-4. Percentages of exposed population that would die due to whole body impact (black lines) as a function of d (thin grey lines) and explosion energy (thick grey lines) for vapour cloud explosions with a multi-energy charge strength of 10. Taken from [2], by permission. 12
100000
100 m
3 10
11
3 10
11
10 Impulse (Pa·s)
150 m
10000
10
3 10
Explosion energy (Eexp, J)
12
10
200 m 99
10
10
95 90 75
50 25 1 Percentage death lung 50 m haemorrhage (%)
5
10
25 m
1000 100000
Overpressure (Pa)
1000000
Fig. A-5-5. Percentages of exposed population that would die due to lung hemorrhage (black lines) as a function of d (thin grey lines) and explosion energy (thick grey lines) for vapour cloud explosions with a multi-energy charge strength of 10. Taken from [2], by permission.
350
3 1012
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3 1011 1011
Percentage death skull fracture (%)
95 50 300 m
5
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95
200 m
3 1010
Explosion energy (E exp, J)
1012
90
400 m Impulse (Pa·s)
600 m
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1010
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Percentage eardrum rupture (%)
3 109
10 5
109 100 m 3 108 50 m 100
108 25 m
10 10000
100000
1000000
Overpressure (Pa) Fig. A-5-6. Percentages of exposed population that would suffer eardrum rupture (black solid lines) or would die due to skull fracture (semi-dotted lines) as a function of distance (thin grey lines) and explosion energy /thick grey lines) for vapour cloud explosions with a multi-energy charge strength of 10. Taken from [2], by permission.
REFERENCES [1] Díaz Alonso, F., González Ferradás, E., Sánchez Pérez, J. F., Miñana Aznar, A.,
Ruíz Gimeno, J., Martínez Alonso, J. J. Loss Prev. Process Ind. 20 (2007) 187. [2] Díaz Alonso, F., González Ferradás, E., Jiménez Sánchez, T. de J., Miñana Aznar,
A., Ruíz Gimeno, J., Martínez Alonso, J. J. Hazard. Mater. (2007), doi: 10.1016/j.jhazmat.2007.04.089.
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Index
A
Gaussian models, 208-218 heavier-than air gases, 219 lapse rate, 197 mechanical turbulence, 198 overlaping plumes, 211-212 plume rise, 207 puff, 206 radiation inversion, 201 sheltering, 234-240 short-term releases, 218 thermal inversion, 198 vulnerability to toxic substances, 271 Atmospheric stability, 200 Atmospheric stability classes, 200 Atmospheric transmissivity, 75-76 Autoignition temperature, 73 Autoignition temp. diverses substances, 67
Accident sequence, 293 Accident types, 120 Accidental scenarios, 293 frequency, 295 Acute Exposure Guideline Levels, 271-272 AEGL, diverse substances, 337-344 Aerosol atmospheric dispersion, 232 AIHA, 271 Airborne pathogenic agents, 231-235, 242244 Alert zone, 281-282 ALOHA code, 234 Anderson, C., 159 Andreassen, M., 40, 65, 70, 85, 92, 275 Anezeris, O. N., 109 Antoine equation, 94 Antoine equation, coefficients for diverse substances, 333 Antrax, atmospheric dispersion, 242-244 Appelyard, R. D., 159 Arnaldos, J., 27, 70, 80, 81, 85, 87, 171, 227, 253 ASTM, 151 Atmospheric dispersion, 195-248 aerosol dispersion, 232 buoyant turbulence, 198 concentration contour coordinates, 227 continuous/instantaneous releases, 205 continuous emission, 209 dispersion coefficients, 210-212, 215
B Babrauskas, V., 87, 262 Bagster, D. F., 107 Baker, Q. A., 133, 135, 141-142 Baker-Strehlow-Tang method, 133-135 Baker, W. E., 122, 124, 125, 176, 181,263, 264, 267 Bakken, B., 40 Balke, W., 169 Baum, H. R., 74, 180, 181, 182 Beckett, H., 163 Bennett, R. H., 135, 141-142 Bessey, R. L., 176 Beychok, M. R., 218
353
Bhopal accident, 50 Birk, A. M., 159, 160, 171, 180, 181 Blast scaling, 123 Blast wave, 119, 169 BLEVE, 147-193 characteristic curves, 178 definition, 147 ductile and fragile breaking, 171 energy released, 165-169, 173-175 irreversible expansion, 168, 173 isentropic expansion, 166, 173 liquid superheating, 151-152 mechanism, 149 México City accident, 150, 183 missiles, 178-183 most frequent causes, 149 overpressure, 166-174 preventive measures, 190 pressure wave, 169 probable number of fatalities, 138 propane, 319-323 propylene, 160 spinodal curve, 151-152 superheat limit temperature, 153-159 superheating energy, 173-176 Tivissa accident, 187-190 Blewitt, D. N., 219, 225 Blinov, V. I., 64 Bhopal accident, 10, 195 Boilover, 100-103 description, 100-101 effects, 103 factor of propensity, 102-103 heat wave, 100 heat wave speed, 102 lethality, 103 temperature, 101 thin-layer boilover, 65-101 Bond, J., 50, 54 Borah, M., 104 Bowen, J. G., 265 Braña, P. A., 54 Brasie, W. C., 263 Breding, R. J., 256 Briggs, G. A., 207, 212, 215 Britter, R. E., 205, 206, 221-225 Broeckmann, B., 100, 103 Bryan, J. L., 275
Brzustowski, T. A., 92, 94 Bull, K., 255 Burgess, D. S., 87, 120 Burning rate, 86-87 infinite diameter pool, 86 estimation, 87 Buettner, K., 255 C Cañizares, P., 137 Carbon dioxide, effects, 277 Carbon monoxide, effects, 276 Carol, S., 10, 11, 119, 120, 138, 280 Casal, J., 10, 61, 70, 71, 80, 87, 119, 138, 152, 156, 168, 179, 225, 232, 253, 307 CCPS, 1, 44, 99, 104, 122-123, 126, 138, 147, 159, 269, 273, 291, 294, 295, 306 Casal, Jordi, 234, 235, 237 Cendejas, S., 150 Cetegen, B. M., 84 Chamberlain, G. A., 94 Chan, C. K., 150 Characteristic curves, explosions, 125, 177, 347-349 Chase, M., 32 Chatris, J. M., 85, 87 Choked pressure, 31 Choked velocity, 31 Clancey, V. J., 230, 231 Clausius-Clapeyron equation, 154 Clothing, protection, 258 Collapse of buildings, 268-269 Combustion, 61 combustion heat, 63 combustion heat diverse substances, 335-336 combustion reaction, 61 Combustion energy-scaled distance, 122 Concentration in gas, units, 204 Consequences of accidents, 249-288 explosions, 263-270 hot air, 261 inert gases, 277-278 Probit analysis, 250-270 thermal radiation, 254-263 toxic substances, 271-279 Considine, M., 294 Continuous emission, 209
354
Cooling with water, 184 Cornwell, D. A., 200 Cowgill, G., 159 Cox, P. A., 122, 124-125, 181, 263, 267 CPD, 21, 25, 33, 35, 54, 56-58, 131, 181, 198, 205, 219, 291, 306-309, 311 Crescent City accident, 106 Critical velocity, 30 Croce, P. A., 78, 85 Crowl, D. A., 21, 22, 23, 54, 121, 164 Cube root scaling law, 121 Cunningham, M. H., 159, 160
puff, 206 radiation inversion, 201 sheltering, 234-240 short-term releases, 218 sun radiation, 202 thermal inversion, 198 Dispersion of dust, 9, 230 Dispersion of infectious agents, 229-233 dispersion of airborne viruses, 232 dispersion of antrax, 242-244 emission source, 231 foot-and-mouth disease, 233 Domino effect, 12-13 classification, 12 definition, 12 example, 12-13 zone, 281-282 Donaldson, A. I., 233 Dose, 108, 271 Drysdale, D., 64, 70 Dusserre, G., 185 Ductile breaking of vessels, 171 Dust explosions, 9 Dust, atmospheric dispersion, 9, 230 Dynamic pressure, 122
D Damage, types, 9-11 Dandrieux, A., 185 Danielsen, U., 40 Darbra, R. M., 61, 119 Davies, M. L., 200 Davies, P. C., 241 Deanes, D. M., 87 Defaveri, M., 218 Deflagrations, 121 Delvosalle, C., 12 Demichela, M., 105, 152, 156-159 Detonations, 122 DGLMSAE, 258, 264, 265, 269 Díaz Alonso, F., 127, 176-178, 347 Diffusion flames, 63 Dimbour, J. P., 185 Dimensionless factor
