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
<
\
pressure (Pa) safety valve absolute backpressure (bar) choked or critical pressure (Pa) pressure inside the vessel or the pipe (Pa) vessel design pressure (Pa) maximum pressure reached during venting (Pa) atmospheric pressure (Pa) pressure inside the pipe at a given distance from the source (Pa) pressure downstream from the hole (Pa) set pressure (Pa) vapour pressure at the storage temperature (Pa) pressure drop (Pa) heat flux (W) ideal gas constant (8.314 kJ kmole-1 K-1) Reynolds number (-) effective pool radius (m) temperature (K) temperature of the gas at choked conditions (K) temperature in the vessel or the pipe (K) temperature of the gas in the jet (K) temperature of the gas in the pipe, at a given distance from the source (K) velocity of liquid in the pipe (m s-1) velocity of the jet at the gas outlet (m s-1) speed of sound in a gas (m s-1) wind speed (m s-1) change in specific volume, liquid to vapour (m3 kg-1) initial mass of gas in the vessel (kg) initial mass of liquid (gk) mass of liquid immediately after flash (kg) mass of vapour immediately after the flash (kg) gas compressibility factor (-) safety valve outflow coefficient (-) pipe roughness (m) dimensionless factor (-) ratio of heat capacities, cp/cv (-) density of the gas (kg m-3) density of the liquid (kg m-3) safety valve outflow function (-) dimensionless factor (-)
REFERENCES [1] O. Levenspiel, Engineering Flow and Heat Exchange. Plenum Press, New York, 1984. nd [2] D. A. Crowl, J. F. Louvar. Chemical Process Safety. Fundamentals with Applications, 2
ed. Prentice Hall PTR, Upper Saddle River, 2002. [3] Committee for the Prevention of Disasters. Methods for the Calculation of Physical
Effects (the “Yellow Book”), 3rd ed. The Hague, SDU, 1997.
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D. A. Crowl. J. Loss Prev. Process Ind. 5 (1992) 73. T. C. Foster. Chem. Eng. May 4 (1981) 105. H. Montiel, J. A. Vílchez, J. Casal, J. Arnaldos. J. Hazard. Mater., 59 (1998) 211. M. Chase. Nist-Janaf Thermochemical Tables. 4th edition. Journal of Physical and Chemical Reference Data. Monograph No. 9, 1998. [8] R. C. Reid, J. M. Prausnitz, B. E. Poling. The Properties of Gases & Liquids. McGrawHill, New York, 1987. [9] Committee for the Prevention of Disasters. Guidelines for Quantitative Risk Analysis (the “Purple Book”). The Hague, SDU, 1999. [10] M. Andreassen, B. Bakken, U. Danielsen, H. Haanes, K. D. Oldshausen, G. Solum, J. P. Stensass, H. Thon, R. Wighus. Handbook for Fire Calculations and Fire Risk Assessment in the Process Industry. Scandpower A/S. Sintef-Nbl, Lillestrom, 1992. [11] H. G. Fisher, H. S. Forrest, S. S. Grossel, J. E. Huff, A. R. Muller, J. A. Noronha, D. A. Shaw, B. J. Tilley. Emergency Relief System Design Using DIERS Technology. DIERSAIChE, New York, 1992. [12] H. K. Fauske. Plant/Operations Progr. 4 (1985) 132. [13] CCPS. Guidelines for Chemical Process Quantitative Risk Analysis. AIChE, New York, 2000. [14] H. K. Fauske, M. Epstein. Source Term Considerations in Connection with Chemical Accidents and Vapor Cloud Modeling. Proc. of the International Conference on Vapor Cloud Modeling. Cambridge, MA. AIChE, New York, 1987. [15] Spirax Sarco. Steam Engineering Tutorials. Learning Modules, Module 9.1, Introduction to Safety Valves. URL: http://www.spiraxsarco.com/learn/, consulted February 2007. [16] NFPA. Flammable and Combustible Liquids Code. NFPA 30. National Fire Protection Association. Quincy, MA. 1987. [17] NFPA. Standard for the Storage and Use of Liquefied Petroleum Gases. NFPA 58. National Fire Protection Association. Quincy, MA. 1987. [18] J. Bond. Loss Prev. Bulletin, no. 65 (1985) 20. [19] J. C. Leung. AIChE Journal, 32 (1986) 1622. [20] J. E. Huff. Plant/Operations Prog., 1 (1982) 211. [21] J. M. Santamaría, P. A. Braña. Análisis y reducción de riesgos en la industria química. Editorial MAPFRE. Madrid, 1994. [22] Committee for the Prevention of Disasters. Methods for the Calculation of Physical Effects (the “Yellow Book”), 1st ed. The Hague, SDU, 1979. [4] [5] [6] [7]
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Chapter 3
Fire accidents 1 INTRODUCTION Of the various accidents that can occur in the process industry, fire is, generally speaking, the type whose effects are felt over the shortest distances: toxic gas clouds and explosions usually cover much larger areas. However, the effects of a fire can be severe as the thermal flux may affect other equipment (domino effect), thus giving rise to other events (explosions, releases) that can dramatically increase the scale of the accident. In fact, in many major accidents occurring in process plants or in the transportation of hazardous materials, fire is an initial stage, followed by a release or an explosion. Thus, different combinations may be observed: fire + larger fire, fire + explosion, fire + gas cloud, fire + BLEVE/fireball. Diverse historical analyses have demonstrated that fires are the most frequent type of accident, followed by explosions and gas clouds. Darbra et al. [1] found that, for accidents occurring in sea ports, 51% corresponded to the general case of “loss of containment”, 29% were fires, 17% were explosions and 3% were gas clouds. If only accidents leading to fire, explosions or gas clouds are considered, these values become 59.5% for fire, 34.5% for explosions and 6% for gas clouds. Another survey [2] obtained similar results for accidents occurring during the transport of hazardous substances by road and rail: 65% were fires, 24% explosions and 11% gas clouds. Both for establishing safety distances and for estimating the cooling flow rates required to protect equipment affected by thermal radiation, the usefulness of being able to foresee the effects of a fire as a function of distance is evident. The mathematical modelling of a fire allows predictions to be made regarding the possible damage to people and equipment, and the establishment of the necessary safety measures to reduce or prevent this damage. In this chapter, the main features of common fuels and of the various types of fire are analyzed and the most common mathematical models are described. 2 COMBUSTION Combustion is a chemical reaction in which a fuel reacts with an oxidant, yielding various products and releasing energy. Combustion always takes place in the gas phase: liquid fuels become vaporized due to the heat from the flames and then react with the oxygen in the air; solids are decomposed due to the high temperature and the gases released react with the oxidizer, usually oxygen from the air. The flames are these reacting gases at high temperature. Combustion products are released as smoke, which also contains unburnt fuel; in accidental
61
fires combustion is usually fairly bad, due to the poor mixing of fuel gas or vapour with oxygen, and large amounts of black smoke are generated.
Fig. 3-1. The fire triangle. For combustion to proceed, the three sides must be connected. A chemical chain reaction is also required: the fire triangle evolves towards the fire tetrahedron.
For combustion to occur, three elements are required: fuel, an oxidant and a source of ignition. These three elements can be represented by the fire triangle (Fig. 3-1): if one side is missing, combustion is impossible; if the three sides are connected, combustion is possible. Strictly speaking, another element is also required: the occurrence of a chemical chain reaction. Without this chemical reaction —for example, because of the presence of a Halon extinguisher— fire is not possible. Thus, the fire triangle evolves towards a fire tetrahedron. However, in practice some additional conditions must be fulfilled for combustion to take place: the oxidizer and the fuel must be present in sufficient quantities, the fuel must be ready to ignite (for example, its temperature must be above a minimum value) and the source of ignition must have a minimum intensity. 2.1 Combustion reaction and combustion heat In a fire, thermal energy is released because a fuel is burnt. Combustion is an exothermal chemical reaction in which a substance combines with an oxidant, commonly oxygen from the air. The fuel may be solid, but most fires in the process industry involve liquid or gas fuels. Some of the energy released is used to sustain the reaction. Thus: m
QF QL qv
(3-1)
where m is the combustion rate (kg m-2 s-1) QF is the heat flux from the flames into the fuel (kW m-2) QL is the heat lost from the fuel surface (kW m-2) and qv is the heat required to produce the gas (the heat needed to increase its temperature to boiling point plus the latent heat of evaporation in the case of a liquid) (kJ kg-1). The heat released from the combustion, 'Hc, will have two different values depending on whether the water formed with the reaction products is considered to be in the liquid or
62
vapour state. The difference will be the vaporization heat of water (44 kJ mol-1 or 2444 kJ kg-1 at 25 ºC). For example, for methane, CH4 + 2O2 o CO2 + 2H2O
'H c' = -888 kJ mol-1 (-5.55·104 kJ kg-1) if water is considered to be a liquid 'H c = -800 kJ mol-1 (-5·104 kJ kg-1) if water is considered to be a vapour. As in a fire the water leaves the system in the smoke as a vapour, commonly the 'Hc value is used and is called the net combustion heat rate. The gases undergoing the diverse chemical reactions associated with combustion are extremely hot and therefore luminous: this luminous zone is the flame. From a practical point of view, we should consider two types of combustion: one associated with premixed flames and the other one associated with diffusion flames. 2.2 Premixed flames and diffusion flames Premixed flames exist when the vapour or gas fuel and the oxidant have been well mixed before burning as, for example, in a Bunsen burner. The premixed flame is short, very hot, has a blue colour and gives off a relatively small amount of very hot gas: combustion is good. This is the type of combustion which occurs in burners and other combustion tools. If the air inlet of the burner is closed, then the fuel exiting from the burner nozzle will undergo fairly poor mixing with air, mainly by entrainment and diffusion. The flame will now be long, undulating, yellow, very luminous and sooty: this is a diffusion flame. Its temperature is significantly lower than that of the premixed flame, larger amounts of black smoke are produced, some unburnt fuel is lost and the overall efficiency of the combustion is lower. Diffusion flames are typical of most accidental fires. 3 TYPES OF FIRE The fires that can occur in industrial installations or in the transportation of hazardous substances can be classified according to the state of the fuel involved and the conditions in which ignition takes place. A simplified diagram indicating the diverse types of fire that can occur as a consequence of the loss of containment of a flammable material can be seen in Fig. 3-1. Although the combustion of solid materials may also cause large fires, the most common fuels in industrial accidents are usually liquids and gases. Therefore, the accident starts with the loss of containment of a flammable fluid. If it is a liquid, the release can create a pool on the ground —covering a limited area if a dike is present— or on water; the outcome is similar in the case of a tank that has lost its roof due to an explosion. In all of these cases, the ignition will start a pool fire (a tank fire may be considered to be a particular case of a pool fire). If the liquid is flowing, which is not common, the ensuing running liquid fire will have different features and can sometimes be very difficult to extinguish. If the liquid undergoes flash vaporization due to a sudden depressurization, a fireball will probably be created. When the material released is a gas or a vapour, if ignition takes place immediately there will be a jet fire. If ignition is not immediate, a cloud containing a flammable mixture may build up under
63
certain meteorological conditions; the ignition of this cloud will cause a flash fire. A flash fire can also occur if a pool is not ignited, due to the vaporization of the fuel. Pool and tank fires are the most frequent types of fire, followed by jet fires, flash fires and fireballs. While pool fires and jet fires may burn for long periods, a fireball will usually last less than one minute (often only a few seconds) and a flash fire is a very short phenomenon, lasting only a few tens of seconds.
Fig. 3-2. Types of fire that may occur in an industrial installation.
3.1 Pool fires Pool fires occurring in industrial accidents are characterized by turbulent diffusion flames on a horizontal pool of fuel that is vaporized. The liquid receives heat from the flames by convection and radiation and may lose or gain heat by conduction towards/from the solid or liquid substrate under the liquid layer. Once the fire has reached the steady state, there is a feedback mechanism that controls the feeding of fuel vapour to the flames. The amount of heat transferred between the fuel and the underlying interface will depend on the fuel and the substrate conditions. If the substrate is cool water and the fuel is a liquid that is initially at ambient temperature, heat losses can be substantial and the fuel vaporization can decrease to such an extent that the flame cannot be maintained. By contrast, if the fuel is cryogenic, heat will be transferred from the substrate to the fuel —at least in the first period— and combustion can be improved. The burning rate is a function of the pool diameter. Blinov and Khudiakov [3, 4] carried out an experimental study of the behaviour of pool fires with different fuels and diameters. They observed the same behaviour for all fuels. The highest burning rates corresponded to the smaller pool diameters, in the laminar regime. Burning rates then decreased as the diameter increased, reaching a minimum at approximately D = 0.1 m (Re | 20). Over the range 0.1 m < D < 1 m (20 < Re < 200) burning velocity increased with pool diameter. For larger diameters (D > 1 m, Re > 200) the flames were fully turbulent and the burning velocity was essentially constant and did not depend on the diameter. This correlation between the burning rate and the pool diameter can be explained in terms of the relative contribution of the various mechanisms to the heat transfer from the flame to the liquid fuel [5].
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In large pool fires —more than 1 m in diameter— large amounts of soot are produced due to poor combustion. This soot gives the flames their yellow colour; it then cools and gives off black smoke which absorbs radiation from the flame. This phenomenon, known as smoke blockage, has a strong effect on the overall thermal radiation emitted from the fire. Pool fires on the ground occur when a fuel release creates a pool in a dike or on the ground. The release can be instantaneous, for example due to the collapse of a tank, or continuous or semi-continuous, for example due to leakage through a hole. If there is no dike, the diameter of the pool will depend on the type of release, the combustion rate and the ground slope. In an instantaneous release, the liquid will spread out until it reaches a barrier or until all of the fuel has been burnt. In a continuous release, the size of the pool will increase until the burning rate is equal to the release flow rate, thus reaching an equilibrium diameter. The fuel (usually a hydrocarbon) can burn over a layer of water, usually created by the intervention of firemen. If the fire burns for a certain time, the water may boil, giving rise to the phenomenon known as thin-layer boilover. In large tank fires, the existence of a layer of water can lead to boilover, which is a dangerous phenomenon. Pool fires can also take place on the sea surface. In this case, the heat transfer from the fuel to the water can be significant and a situation can be reached in which the flame cannot be maintained. Furthermore, if ignition is delayed the diameter of the pool will increase and its thickness may reach a minimum value below which ignition is no longer possible. 3.2 Jet fires Jet fires are turbulent diffusion flames caused by the combustion of a flammable gas or vapour released at a certain velocity though a hole, a flange, etc. The release is not always accidental: flares are widely used in process plants to safely dispose of flammable gases. Jet fires entrain large amounts of air into the flame, due to the turbulence of the flow. Typically, the volume of entrained air can be up to five times that required for stoichiometric combustion [6]. Due to the efficiency of the mixing and the better combustion rate, temperatures in jet fire flames are often higher than those reached in natural diffusion flames in a pool fire. As a consequence, jet fires can cause serious damage to equipment, through both thermal radiation and flame impingement. 3.3 Flash fires If a flammable gas or vapour is released in certain meteorological conditions —calm or low wind speed— a cloud will be formed. This can also be caused by the release of a pressurized liquid undergoing a flash vaporization or by evaporation from a pool. This is the case of liquefied natural gas or liquefied propane: if a pool is formed, it will undergo intense vaporization. The release may be instantaneous or continuous. The flammable cloud will disperse, increasing in size, and will move according to wind direction. If it meets a source of ignition —an electrical device, a flame, an electrostatic spark— the mass between the flammability limits will burn very quickly, as the flames are propagated through the cloud. The flame propagation velocity for LPG-air mixtures has been reported to be in the range of 5-10 m s-1 [7] and to increase with wind speed. The duration of the phenomenon is very short —a few tens of seconds. The area covered by the flammable mixture will be subjected to a very strong heat flux, while outside this area the effects of the thermal radiation will be greatly reduced and in practice are often considered to be negligible. If the mass of fuel in the cloud is large, a significant blast can also occur; in this case, the accident is considered to be an explosion rather than a flash fire.
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3.4 Fireballs If a tank containing a pressurized liquid is heated, the pressure inside the tank will increase. If the walls are not able to withstand the high stress, they will eventually collapse. Upon failure, due to the instantaneous depressurization, a sudden flash of a fraction of the liquid will take place and a biphasic liquid/vapour mixture will then be released. If the substance is a fuel, as is often the case in the process industry (for example, propane), this mixture will probably ignite, creating a fireball, initially at ground level. Subsequently, the whole turbulent mass will increase in volume and will rise, producing a wake. The thermal radiation can be very strong. Due to the difficulty in foreseeing the moment at which the fireball may occur (it can happen at any moment from the beginning of the emergency, without warning) fireballs have killed many people, most of them firemen. The mechanical aspects of explosions usually associated to fireballs are described in detail in Chapter 5. 4 FLAMMABILITY Not all of the substances are equally hazardous in terms of causing fire accidents. It is well known that with some flammable liquids, for example acetone or gasoline, it is relatively easy for an accident to occur when the substance is handled without the required care, while with others, for example diesel oil, the likelihood of such a problem is reduced —although a risk still exists. This depends on the different properties of the substance, the most important of which (for practical purposes) are its flammability limits, its flash point temperature and its autoignition temperature. 4.1 Flammability limits Mixtures of a flammable gas or vapour with air are flammable only if the concentration of the vapour ranges between two values called the lower flammability limit (LFL) and the upper flammability limit (UFL). If the concentration is lower than the LFL, there is not enough fuel for combustion to occur; if the concentration is higher than the UFL, the mixture is too rich and there is not enough oxygen. The flammability limits can be obtained experimentally, using standard methods, with devices that determine whether the flame propagates through a given mixture of fuel/air. According to the experimental procedure followed, rather different values can be obtained; this explains the data scattering that is sometimes found in the literature. The flammability limits are commonly expressed as the volume percentage of fuel at 25 ºC. For example, the limits for propane are 2.1% and 9.5%. The range covered by these limits is important from the point of view of the fire hazards associated with a given substance. For example, hydrogen is a very dangerous gas: its flammability limits are 4% and 75%. This means that the probability of a flammable atmosphere in the event of a hydrogen release is very high. For gasoline, the LFL is approximately 1.4%: if there is a release or if gasoline is being handled, a flammable gasoline/air mixture will quickly appear nearby. From the point of view of safety, the LFL is probably the most important limit, as it is related to the formation of a flammable atmosphere; the UFL can be significant when flammable substances are handled in closed volumes (rooms or tanks). Flammability limits for several common substances can be seen in Table 3-1. A more complete list has been included in the annexes. The best study published in this field is probably that by Zabetakis [8], which gives a set of values obtained experimentally with the apparatus developed at the US Bureau of Mines.
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Table 3-1. Flammability limits, flash point temperature and autoignition substances in air at atmospheric pressure Substance LFL, % volume UFL, % volume Acetone 2.6 12.8 Acetylene 2.5 80.0 Aniline 1.2 11.0 Ammonia 15.0 28.0 Benzene 1.4 8.0 n-Butane 1.8 8.4 Crude oil 1 6.0 Cyclohexane 1.3 8.3 Methane 5.3 15.0 Ethylene 2.8 34.0 Ethane 3.0 12.4 Ethanol 3.5 19.0 n-Hexane 1.2 6.9 Hydrogen 4.0 74.2 Kerosene 0.7 6.0 Motor gasoline 1.4 7.6 Methane 5.3 15.0 Octane 1.0 6.7 n-Pentane 1.3 7.6 Propane 2.1 9.5 Propylene 2 11.7 Toluene 1.3 7.0
temperature for several common Tf, ºC -17.8 -17.8 14.4 ---11 -60 -18 -17 -222.5 ---135 12.8 -26 --49 -46 -222.4 13.3 -40 -104.4 --4.4
Tautoign, ºC 700 305 615 630 562 405 230-250 260 632 450 515 558 234 400 229 280 632 458 287 493 443 536
4.1.1 Estimation of flammability limits Several methods have been proposed for calculating the values of flammability limits. All of these give only approximate results, so experimental determination is always more accurate and reliable. A very simple procedure [9], proposed essentially for hydrocarbons, gives the values of the limits as a function of the stoichiometric concentration cst: LFL
0.55 c st
(3-2)
UFL
3.5 c st
(3-3)
cst is the stoichiometric concentration of fuel in air, expressed as volume percentage (fuel in fuel plus air), that allows complete combustion, consuming all the oxygen available. This method was improved by Hilado and Li [10], who proposed the following expressions: LFL
a c st
(3-4)
UFL
b c st
(3-5)
where a and b are constants that depend on the chemical structure of the substance. Their values have been included in Table 3-2.
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Table 3-2 Values of constants a and b in Eqs. (3-4) and (3-5) [10] Substance a b Linear saturated hydrocarbons 0.555 3.10 Cycloalkanes 0.567 3.34 Alkenes 0.475 3.41 Aromatic hydrocarbons 0.531 3.16 Alcohols-glycols 0.476 3.12 Ethers-oxides 0.537 7.03 Epoxides 0.537 10.19 Esters 0.552 2.88 Other compounds C, H, O 0.537 3.09 Monochlorinated compounds 0.609 2.61 Dichlorinated compounds 0.716 2.61 Brominated compounds 1.147 1.50 Amines 0.692 3.58 Compounds containing S 0.577 3.95
_____________________________ Example 3-1 Estimation of the flammability limits for ethanol using both methods.
Solution The combustion reaction is: C 2 H 5OH 3O2 o 2CO2 3H 2 O Therefore, the stoichiometric concentration is:
moles fuel 1 100 100 6.54% 100 moles fuel moles air 1 3 21 Applying Eqs. (3-2) and (3-3), c st
LFL 0.55 6.54 3.60% volume UFL 3.5 6.54 22.89% volume.
Using the values from Table 3-2 for alcohols, LFL 0.476 6.54 3.11% volume UFL 3.12 6.54 20.40% volume.
Comparing these results with the values in Table 3-1, it can be seen that Eq. (3-2) gives a fairly accurate prediction for the value of the LFL, while for the UFL Eq. (3-5) seems to be more accurate. ______________________________________
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4.1.2 Flammability limits of gas mixtures If the fuel is a mixture of different components, the flammability limits can be estimated —again with a certain margin of error, which can be significant in some cases— as a function of their respective concentrations, using the empirical expressions proposed by Le Chatelier [11]: LFLmixt
1 ci ¦ LFL i 1 i
(3-6)
UFLmixt
1 ci ¦ UFL i 1 i
(3-7)
n
n
where LFLi is the value of the LFL for component i (% volume) ci is the concentration (% vol.) of component i on a fuel basis (6ci = 100) and n is the number of combustible components in the mixture. ______________________________________ Example 3-2 Estimation of the flammability limits of a mixture containing acetone (3% by volume), benzene (2%) and ethanol (7%) in air. Solution The concentrations of the three components on a fuel basis are:
ci
% vol i
% vol acetone % vol benzene % vol ethanol mixture
Therefore: cacetone = 0.25 cbenzene = 0.17 cethanol = 0.58 Taking into account the values of the LFL and UFL for the three components (Table 3-1), the flammability limits of the mixture are: LFLmixt
1 0.25 0.17 0.58 3 1.4 3.5
UFLmixt
1 0.25 0.17 0.58 13 8 19
2.70% volume
14.1% volume
The mixture is flammable, as 3% + 2% + 7% = 12%. ______________________________________
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4.1.3 Flammability limits as a function of pressure Although most situations involving fuel/air flammable mixtures usually occur at what is essentially atmospheric pressure, storage or transportation are sometimes carried out under different conditions, and this can cause variations in the flammability limits. Flammability limits depend on pressure. Below atmospheric pressure, as pressure decreases (at constant temperature) the values of the two limits converge and the gap between them is narrowed until a certain pressure is reached at which they have the same value; at lower pressures, the flame cannot propagate through the mixture. This is due [12] to the fact that the concentration of gas is too low to sustain combustion. A method for predicting flammability limits at reduced pressures has been proposed by Arnaldos et al. [13]. However, if pressure is reduced in a tank containing a fuel at constant temperature, the increase in partial pressure of the fuel vapour can change the atmosphere above the liquid surface from a non-flammable condition to a concentration in the flammable range. This situation can occur in the fuel tanks of aircraft following take-off [5], when the aircraft climbs above a certain altitude. As the pressure rises above atmospheric pressure, flammability limits become greater. In fact, the LFL decreases only very slightly as pressure increases, but the UFL increases substantially. The variation of the UFL as pressure increases can be estimated using the following empirical expression [8]:
UFLP
UFL 20.6(log Pi 10 3 1)
(3-8)
where Pi is the absolute pressure (N m-2) and UFL is the upper flammability limit at atmospheric pressure (% volume). 4.1.4 Flammability limits as a function of temperature Flammability limits also change with temperature: as temperature increases, the range of concentrations between the two limits widens. Of the various correlations that can be found in the literature to determine this variation, the most well known are probably the following [14]: LFLT
LFL298
3.136 T 298 'H c
(3-9)
UFLT
UFL298
3.136 T 298 'H c
(3-10)
where T is the temperature (K) and 'Hc is the net heat of combustion (kJ mole-1). Another empirical expression has been proposed [15] for estimating the LFL at a temperature T as a function of the same parameter at another temperature T1: LFLT
§ T T1 · ¸¸ LFLT1 ¨¨1 © 873 T1 ¹
(3-11)
where the temperatures are expressed in K.
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______________________________________ Example 3-3 Estimation of the LFL of octane at 100 ºC. Solution The LFL of octane at 25 ºC is (Table 3-1) 1.0% volume. Therefore, § 373 298 · 1.0 ¨1 ¸ 0.87% volume. © 873 298 ¹ ______________________________________ LFL100
4.1.5 Inerting and flammability diagrams In some industrial operations, flammable mixtures of a flammable gas and air can be generated. A typical example is the extraction of a liquid fuel from a vessel. As the liquid level descends, air must enter the vessel to avoid the creation of a vacuum, which could cause the tank walls to collapse and impede liquid flow. If air is allowed to enter, it will mix with the vapour already existing in the volume above the liquid surface, creating a mixture that will probably reach a concentration within the flammable range. To avoid this occurrence, it is a common practice to introduce an inert gas (nitrogen, carbon dioxide) to reduce the concentration of oxygen, thus preventing the generation of a flammable atmosphere. In the latter case, a more complex gas system will exist – for example hydrocarbon, oxygen and nitrogen – and the flammable conditions, which must be determined experimentally, must be represented in a two-dimensional diagram. An interesting practical example is that of filling or emptying a tank, a relatively dangerous operation: an historical analysis [16] has shown that 8% of all accidents occurring in process plants and during the transportation of hazardous materials are associated with these operations. A major source of accidents when filling a tank with a flammable liquid or emptying a tank that contains a flammable liquid is the fact that when air is present in the tank or enters during the operation, a flammable mixture can build up in the vapour space above the liquid. There are a number of possible ignition sources, the most frequent being sparks created by the electrostatic charge built up between the flowing fluid and the equipment. To avoid this dangerous situation, appropriate procedures can be applied by using an inert gas. These procedures may be better explained with the help of an x vs. y diagram, where
x
inert volume 100 flammable volume
y
flammable volume inert volume 100 total volume (flammable inert air)
When an empty tank has to be filled with a flammable liquid, the tank is often initially full of air; therefore, before starting the operation x = 0 and y = 0 (see Fig. 3-3-a). In this case, as the liquid is introduced, its vapour will diffuse into the headspace, and its concentration in the air-vapour mixture will increase with time. At a certain point, this concentration will reach the LFL; from that point on, and for a certain period, a flammable mixture will exist inside the tank, with the associated risk of explosion.
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To avoid this risk, an inert gas must be introduced at the beginning of the operation. Thus, x = f and y will progressively increase. When y reaches a certain value higher than any other within the flammable region (value A in Fig. 3-3-a) the flammable liquid may be introduced. The value of x will now gradually decrease, but will always remain outside the flammability range.
Fig. 3-3. a) Filling a tank with a flammable liquid using an inert gas. b) Emptying a tank which contains a flammable liquid using an inert gas.
The opposite case is emptying a tank that contains a flammable liquid. In this case, an increasing gas headspace is built up as the liquid level descends. To avoid reduced pressure in the volume (which could impede liquid flow), gas must be allowed to enter the tank. If this gas is an oxidant (for example, air) a mixture within the flammability limits may occur, with the consequent risk of explosion. One way to remove this risk is again to use an inert gas. If the tank is being emptied, the headspace will initially be full of fuel vapour (it will contain neither air nor inert gas). Therefore, initially x = 0% and y = 100% (Fig. 3-3-b). If inert gas is added, y will remain constant (100%) and x will progressively increase until it reaches point B. As of this moment, air can be introduced while the liquid flows without any risk of creating a flammable mixture: the concentration in the head space will always be outside the flammability region. 4.2 Flash point temperature The flash point is the lowest temperature at which a flammable liquid gives off enough vapour to form a mixture with air that is flammable if an ignition source is present. It gives an idea of the ease with which a liquid can be ignited: gasoline (Tf | -42 ºC) is practically always ready to ignite; by contrast, diesel oil (Tf | 66 ºC) is fairly difficult to ignite under standard room conditions. Substances with higher flash points are less flammable or hazardous than those with lower flash points. Thus, the flash point is an essential parameter in establishing how hazardous a substance is in terms of fire risk. The flash point temperature is obtained experimentally. Two procedures are used: opencup and closed-cup. Closed-cup methods give lower values. The flash point measured may also change according to the apparatus used. This may create some confusion and an amount
72
of scattering in the published data. The flash point temperature for various substances is shown in Table 3-1. A more extensive survey can be found in the annexes. It is well known that there is a close relationship between the boiling point of a liquid and its flash point, and many correlations (commonly parabolic and hyperbolic equations) have been published, although most of them show large deviations when their predictions are compared to experimental data. Satyanarayana and Rao [17] proposed a new expression, which seems to be much more accurate, by correlating the data (closed-cup flash point temperature) for 1200 organic compounds with a high degree of accuracy (average absolute error less than 1%) with the following expression:
Tf
§ c b¨¨ T a © 0
2
c
· T0 ¸¸ e ¹
c § · ¨1 e T0 ¸ ¨ ¸ © ¹
(3-12)
2
where T0 is the boiling temperature of the substance (K) at atmospheric pressure, and a, b and c are constants (K) (see Table 3-3). Table 3-3 Constants in Eq. (3-12) [17] Chemical group a Hydrocarbons 225.1 Alcohols 230.8 Amines 222.4 Acids 323.2 Ethers 275.9 Sulphur 238.0 Esters 260.8 Ketones 260.5 Halogens 262.1 Aldehydes 264.5 Phosphorus 201.7 Nitrogens 185.7 Petroleum fractions 237.9
b 537.6 390.5 416.6 600.1 700.0 577.9 449.2 296.0 414.0 293.0 416.1 432.0 334.4
c 2217 1780 1900 2970 2879 2297 2217 1908 2154 1970 1666 1645 1807
4.3 Autoignition temperature The autoignition temperature, also called spontaneous ignition, is the temperature at which a substance is ignited in the absence of any ignition source. It has also been defined as the lowest temperature at which a substance undergoes self-heating at a sufficient rate to cause its ignition. This second definition establishes a delay prior to ignition; this is again the explanation for a certain amount of scattering often found in the experimental data. The values of the autoignition temperature for several substances have been included in Table 3-1; more data can be found in the annexes. The theoretical estimation of this parameter can be performed by taking into account the chemical structure of the substance through a rather complicated methodology [15]. Nevertheless, the most reliable values will always be those obtained experimentally under the same conditions (pressure, fuel concentration, oxygen concentration) that would exist in a real situation.
73
5 ESTIMATION OF THERMAL RADIATION FROM FIRES
The ability to predict the effects of a fire is highly useful, both from the point of view of applying preventive measures and for emergency management. If, for example, the thermal radiation intensity that will affect a tank is known, it is possible to design a deluge system to protect the tank and to avoid the domino effect. In order to establish safety distances, it is also necessary to know the effects of the different kinds of accidents; if a foam monitor is installed in a location that, in the event of a fire, will be subjected to a strong heat flux, firemen will not be able to use it. Therefore, mathematical modelling is required to predict the thermal radiation that will reach a given target located at a certain distance from the flames. A number of mathematical models have been proposed by several authors. Some of them are too simple to provide reliable information, whilst others are too sophisticated and require data that in most cases is unknown. Here, one of the most well known and commonly used models, the solid flame model, has been selected. A description of the point source model is also included because of its applicability in some situations. 5.1 Point source model The point source model assumes that the fire can be represented by a point that irradiates thermal energy in all directions (Figure 3-4). The point source is usually located in the geometrical centre of the fire. The radiated energy is a fraction of the total energy released by the combustion. It is generally assumed that this energy is radiated in all directions. Therefore, the thermal radiation intensity reaching a given target, which is assumed to be proportional to the inverse square of the distance from the source, is given by
I
Qr 4 S l p2
(3-13)
where I is the thermal radiation intensity (kW m-2) Qr is the heat released as thermal radiation per unit time (kW) and lp is the distance between the point source and the target (m). For jet fires and pool fires, Qr is a function of the burning rate m’ (kg s-1), the heat of combustion 'Hc and the radiative fractionor radiant heat fractionKrad, i.e. the fraction of the combustion energy that is transferred as thermal radiation: Qr
K rad m ' 'H c
(3-14)
The radiant fraction is a significant parameter that affects the whole fire, which in turn depends on several factors: the type of fuel, the flame temperature, the type of flame, the amount of smoke formed during combustion, etc. Values ranging between 0.1 and 0.4 have been obtained experimentally for hydrocarbons. It is rather difficult to estimate the value theoretically, and this is in fact one of the limitations of the point source model. The following expression has been proposed for pool fires with a diameter D (m) [18]:
K rad
0.35 e 0.05 D
(3-15)
74
Fig. 3-4. The point source model.
The point source model is so simple that it does not take into account the absorption of the thermal radiation by the atmosphere or the position of the surface that receives the radiation: it is assumed that this surface faces toward the radiation source so that it receives the maximum thermal flux [19]. Eq. (3-14) can be modified to include these two factors, giving the expression: I
K rad m ' 'H cW cos M 4 S l p2
(3-16)
where Mis the angle between the plane perpendicular to the receiving surface and the line joining the source point and the target (º) and W is the atmospheric transmissivity (-). The point source model overestimates the intensity of the thermal radiation near the fire, due to the fact that it does not take into account the flame geometry. It should not therefore be used, for example, to establish separation distances between adjacent equipment. However, it predicts with reasonable accuracy the radiation intensity in the far field, at distances greater than 5 pool diameters from the centre of the fire; thus, it is sometimes used to perform conservative calculations of danger to personnel.
5.1.1 Atmospheric transmissivity The atmospheric transmissivity accounts for the absorption of the thermal radiation by the atmosphere, essentially by carbon dioxide and water vapour. This attenuates the radiation that finally reaches the target surface. The atmospheric transmissivity depends on the distance between the flames and the target. While the carbon dioxide content in the atmosphere is essentially constant, the water vapour content depends on the temperature and the atmospheric humidity. It can be estimated from the following equations:
W 1.53 ( Pw d ) 0.06 for Pw·d < 104 N·m-1
(3-17-a)
75
W 2.02 Pw d 0.09 for 104 d Pw·d d 105 N·m-1 W
2.85 ( Pw d ) 0.12 for Pw·d > 105 N·m-1
(3-17-b) (3-17-c)
where Pw is the partial pressure of water in the atmosphere (N m-2) and d is the distance between the surface of the flame and the target. Pw can be estimated by the following expression: Pw
Pwa
HR 100
(3-18)
where Pwa is the saturated water vapour pressure at the atmospheric temperature (N m-2) and HR is the relative humidity of the atmosphere (%). Pwa can be obtained from the prevailing temperature of the atmosphere [19]: ln Pwa
23.18986
3816.42 T 46.13
(3-19)
where Pwa is expressed in N m-2 and T in K. ______________________________________ Example 3-4. There is a pool fire of diesel oil with a diameter of 6 m and an average flame height of 11.5 m. Calculate the thermal radiation that reaches the vertical surface of a tank, at a height of 1.6 m above the ground; the tank wall is at a distance of 15 m from the diesel oil pool perimeter (Fig. 3-4). 'Hc diesel = 41900 kJ kg-1. Ambient temperature is 16 ºC. Atmospheric relative humidity is 79%. m = 0.05 kg m-2 s-1.
Solution The distance between the point source and the target is: 2
§ 11.5 · 2 1.6 ¸ 15 3 ¨ © 2 ¹
lp
18.5 m
and the distance between the surface of the flames and the target is:
cos D d
18 18.5
15 cos D
0.97 15 = 15.5 m 0.97
Estimation of the atmospheric transmissivity: ln Pwa
23.18986
3816.42
289.16 46.13
7.486
76
Pwa = 1783 N m-2 Pw = 1783 Pw d
79 100
1408 N m-2
1408 15.5
21824; therefore, by applying Eq. (3-17-b):
2.02 1408 15.5
W K rad
0.35 e 0.056
0.09
0.82
0.26
The overall burning rate is: m'
0.05 S
62 4
1.414 kg s-1
Therefore I
0.26 1.414 41900 0.82 0.97 4 S 18.5 2
2.8 kW m-2
(The experimental thermal radiation intensity for this situation, measured with a radiometer, was 2.3 kW m-2 [20]). ______________________________________ 5.2
Solid flame model This is the most common model used to estimate the thermal radiation from fires. It is more accurate than the point source model, even at short distances from the flame. The solid flame model assumes that the fire is a still, grey body encompassing the entire visible volume of the flames, which emits thermal radiation from its surface (Figure 3-5). The irradiance of the smoke (non visible flame) plume above the fire is partly taken into account. In fact, most models use the maximum length of the flame rather than the average one, and this includes some of the smoke volume above the flame. The shape of the flames will depend on the features of the fire. In the case of a pool fire, the pool shape will be essential: if the pool is circular, the fire will approximate to a cylinder; if there is any wind, the cylinder will be tilted. If a rectangular dike retains the liquid fuel, the fire will be assumed to be parallelepipedic. In more general cases, it is assumed that the body radiates energy uniformly from its whole surface. The thermal radiation intensity reaching a given target is
I
WFE
(3-20)
where W is the atmospheric transmissivity (-) F is the view factor (-) and E is the average emissive power of the flames (kW m-2).
77
In the following paragraphs the estimation of the view factor and the emissive power is discussed; the atmospheric transmissivity has already been discussed in Section 5.1.1.
Fig. 3-5. The solid flame model.
5.2.1 View factor The view factor, a parameter which appears in practically all thermal radiation calculations, is the ratio between the amount of thermal radiation emitted by a flame and the amount of thermal radiation received by an object not in contact with the flame. This ratio depends on the shape and size of the fire, the distance between the flame and the receiving element and the relative position of the flame and target surfaces. It can be represented by a general equation: FdA2 o A1
cos M1 cos M 2 dA1 Sd2 A1
³
(3-21)
where M1 and M2 are the angles made by the normals and dA1 on the flame and by dA2 on the receiving element, and d is the distance between the flame surface and the receiving element. View factors can be calculated for specific situations, although the corresponding mathematical expressions are usually complex and it is relatively easy to make errors when using them. This is why the most common types of fire are shown in tabulated form or in graphical plots. Tables 3-4 to 3-7 show the values of the view factor for vertical and horizontal target surfaces, for pool fires with a cylindrical and parallelepipedic shape (Fig. 3-5), respectively. Expressions for calculating view factors for these fire shapes, as well as for tilted cylinders, can be found in [21] and [22].
78
Table 3-4 Vertical view factor (Fv) for a cylindrical fire l/(D/2) 1.10 1.20 1.30 1.40 1.50 2.00 3.00 4.00 5.00 10.00 20.00 50.00
0.1 0.330 0.196 0.130 0.096 0.071 0.028 0.009 0.005 0.003 0.000 0.000 0.000
0.2 0.415 0.308 0.227 0.173 0.135 0.056 0.019 0.010 0.006 0.001 0.000 0.000
0.5 0.449 0.397 0.344 0.296 0.253 0.126 0.047 0.024 0.015 0.003 0.000 0.000
1.0 0.453 0.413 0.376 0.342 0.312 0.194 0.086 0.047 0.029 0.006 0.001 0.000
H/(D/2) 2.0 3.0 0.454 0.454 0.416 0.416 0.383 0.384 0.354 0.356 0.329 0.330 0.236 0.245 0.132 0.150 0.080 0.100 0.053 0.069 0.013 0.019 0.003 0.004 0.000 0.000
5.0 0.454 0.416 0.384 0.356 0.333 0.248 0.161 0.115 0.086 0.029 0.007 0.001
6.0 0.454 0.416 0.384 0.357 0.333 0.249 0.163 0.119 0.091 0.032 0.009 0.001
10.0 0.454 0.416 0.384 0.357 0.333 0.249 0.165 0.123 0.097 0.042 0.014 0.002
20.0 0.454 0.416 0.384 0.357 0.333 0.249 0.166 0.124 0.099 0.048 0.020 0.004
5.0 0.362 0.312 0.277 0.250 0.228 0.158 0.091 0.057 0.037 0.007 0.001 0.000
6.0 0.362 0.312 0.277 0.251 0.229 0.160 0.095 0.062 0.043 0.009 0.001 0.000
10.0 0.363 0.313 0.278 0.252 0.231 0.164 0.103 0.073 0.054 0.017 0.003 0.000
20.0 0.363 0.313 0.279 0.253 0.232 0.166 0.106 0.078 0.061 0.026 0.003 0.000
0.25 0.0606 0.0604 0.0598 0.0581 0.0494 0.0431 0.0331 0.0184 0.0149 0.0076 0.0038 0.0015
0.2 0.0490 0.0489 0.0483 0.0470 0.0400 0.0349 0.0268 0.0149 0.0121 0.0062 0.0031 0.0012
0.1 0.0249 0.0248 0.0245 0.0239 0.0203 0.0178 0.0137 0.0076 0.0062 0.0031 0.0016 0.0006
0.05 0.0125 0.0124 0.0123 0.0120 0.0102 0.0089 0.0069 0.0038 0.0031 0.0016 0.0008 0.0003
Table 3-5 Horizontal view factor (Fh) for a cylindrical fire l/(D/2) 1.10 1.20 1.30 1.40 1.50 2.00 3.00 4.00 5.00 10.00 20.00 50.00
0.1 0.132 0.044 0.020 0.011 0.005 0.001 0.000 0.000 0.000 0.000 0.000 0.000
0.2 0.242 0.120 0.065 0.038 0.024 0.005 0.000 0.000 0.000 0.000 0.000 0.000
0.5 0.332 0.243 0.178 0.130 0.097 0.027 0.005 0.001 0.000 0.000 0.000 0.000
1.0 0.354 0.291 0.242 0.203 0.170 0.073 0.019 0.007 0.003 0.000 0.000 0.000
H/(D/2) 2.0 3.0 0.360 0.362 0.307 0.310 0.268 0.274 0.238 0.246 0.212 0.222 0.126 0.145 0.050 0.071 0.022 0.038 0.011 0.021 0.001 0.003 0.000 0.000 0.000 0.000
Table 3-6 Vertical view factor (Fv) for a parallelepipedic fire H/x 10 5 3 2 1 0.75 0.50 0.25 0.20 0.10 0.05 0.02
10 0.2480 0.2447 0.2369 0.2234 0.1767 0.1499 0.1118 0.0606 0.0490 0.0249 0.0124 0.0050
5 0.2447 0.2421 0.2350 0.2221 0.1760 0.1494 0.1114 0.0604 0.0489 0.0248 0.0123 0.0050
3 0.2369 0.2350 0.2292 0.2176 0.1734 0.1475 0.1101 0.0598 0.0483 0.0245 0.0122 0.0049
2 0.2234 0.2221 0.2176 0.2078 0.1674 0.1427 0.1068 0.0581 0.0470 0.0239 0.0120 0.0048
1 0.1767 0.1750 0.1734 0.1674 0.1385 0.1193 0.0902 0.0494 0.0400 0.0203 0.0102 0.0041
w/x 0.75 0.5 0.1499 0.1118 0.1491 0.1114 0.1478 0.1101 0.1427 0.1068 0.1193 0.0902 0.1032 0.0784 0.0784 0.0599 0.0431 0.0331 0.0349 0.0268 0.0178 0.0137 0.0089 0.0069 0.0036 0.0027
0.02 0.0050 0.0050 0.0049 0.0048 0.0041 0.0036 0.0027 0.0015 0.0012 0.0006 0.0003 0.0001
The value of the maximum view factor, corresponding to a surface located perpendicularly to the direction of the radiation, can be calculated using the following expression: Fmax
Fv2 Fh2
(3-22)
79
This is a simplified expression, which can not be used if the flames are inclined crosswind with respect to the target. Table 3-7 Horizontal view factor (Fh) for a parallelepipedic fire x/w 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.5 2.0 3.0 4.0 5.0
H/w
0.1 0.0732 0.0263 0.0127 0.0073 0.0047 0.0032 0.0023 0.0017 0.0013 0.0010 0.0007 0.0004 0.0002 0.0001 0.0000 0.0000
0.2 0.1380 0.0728 0.0414 0.0257 0.0171 0.0120 0.0087 0.0065 0.0050 0.0040 0.0026 0.0015 0.0007 0.0002 0.0001 0.0000
0.3 0.1705 0.1105 0.0720 0.0485 0.0339 0.0245 0.0182 0.0139 0.0108 0.0086 0.0056 0.0032 0.0015 0.0005 0.0002 0.0001
0.5 0.1998 0.1549 0.1182 0.0899 0.0687 0.0530 0.0414 0.0327 0.0261 0.0211 0.0142 0.0084 0.0041 0.0013 0.0006 0.0003
0.7 0.2126 0.1774 0.1459 0.1190 0.0966 0.0784 0.0638 0.0522 0.0429 0.0355 0.0249 0.0152 0.0076 0.0026 0.0011 0.0006
1.0 0.2217 0.1944 0.1687 0.1452 0.1243 0.1059 0.0903 0.0767 0.0653 0.0557 0.0409 0.0265 0.0139 0.0050 0.0023 0.0012
1.5 0.2279 0.2063 0.1855 0.1660 0.1478 0.1312 0.1162 0.1028 0.0908 0.0803 0.0629 0.0440 0.0253 0.0100 0.0047 0.0026
2.0 0.2305 0.2113 0.1928 0.1752 0.1588 0.1436 0.1296 0.1169 0.1054 0.0951 0.0774 0.0572 0.0355 0.0154 0.0077 0.0043
5.2.2 Emissive power The emissive power is the radiant heat emitted per unit surface of the flame and per unit time (kW m-2); it represents the radiative characteristics of the fire. In fact, the thermal radiation emitted from the flame is really generated by the whole volume of the fire (from the hot fuel gas, the combustion products, the soot particles) and not only on its surface. Thus, the emissive power is a useful two-dimensional simplification of a complex, three-dimensional complex heat transfer problem. Two types of emissive power can be distinguished: a) point emissive power, associated with the value measured over a small area of the flame; b) average emissive power, corresponding to the emissive power of the whole flame surface. Emissive power can be expressed as a function of emissivity and flame temperature: E
V H T fl4 Ta4
(3-23)
where V is the Stefan-Boltzmann constant (W m-2 K-4) H is the emissivity (-) Tfl is the radiation temperature of the flame (K) and Ta is the ambient temperature (K). However, it is very difficult to calculate both Tfl (significantly lower than the adiabatic flame temperature) and H: as the temperature is not uniform over the flame and varies with time, and the emissivity depends on the substances present in the flame. Therefore, empirical procedures are commonly applied to estimate the value of E for the different types of fire; alternatively, experimental values from similar fires are assumed. The emissive power changes with the position in the fire, with higher values near the bottom and decreasing values as the height increases. Fig. 3-6 shows two typical mean emissive power contour plots [20, 23] produced by a 3 m gasoline pool fire (left) and a 6 m diesel pool fire (right). The vertical and horizontal dimensions were converted to a dimensionless form by dividing their values by the pool diameter (D). A high-radiance zone
80
appears near the base, approximately between H/D = 0.1 and H/D = 0.6; the values of E for this zone varied from 80-100 kW m-2 for pools of 1.5 m in diameter to 120-160 kW m-2 for pools with larger diameters. Significantly lower values of E can be observed in the top part of the fire, where luminous flame has been almost completely substituted by black smoke. 2.663
Gasoline D=3 m
kW/m
2
2.030
140
2.414 2.165
2
kW/m
140
1.839 1.647
120
1.916
120
1.456 100
1.419
80
100
1.264 H/D
1.668 H/D
Diesel D=6 m
1.170
1.073
80
0.881 60
0.921
60
0.690
0.672
0.498
40 0.423
40
0.306 20
0.174 -0.50 -0.25 0.00 x/D
20
0.115
0.25 0.50
-0.50 -0.25 0.00 0.25
0.50
x/D
Fig. 3-6. Two mean E contour plots. Left: 3 m gasoline pool fire. Right: 6 m diesel pool fire. Taken from [23], by permission.
Although an average value of E is often assumed for the entire fire surface, there are in fact two zones commonly found for many fuels: a luminous zone and a non-luminous zone, both of which have different values of emissive power. Fig. 3-7 shows the luminous and nonluminous parts of a fire (diesel oil, 6 m diameter), obtained by superimposing the IR and visible images [20, 23]. Flame shape
Luminous part
Non-luminous part
Fig. 3-7. Luminous and non-luminous parts of a pool fire of diesel oil (6 m diameter). Taken from [23], by permission.
81
The contribution of the two parts changes according to the type of fire and the properties of the fuel. An average value of the emissive power for the whole fire can be estimated by taking the two contributions into account: E av
xlum Elum 1 xlum E soot
(3-24)
where xlum is the fraction of the fire surface covered by the luminous flame and Elum and Esoot are the values of E for the luminous and non-luminous zones of the fire, respectively (kW m-2). For gasoline and diesel oil pool fires, the experimental data obtained for a circular pool with a diameter of 1.5 m d D d 6 m [23] indicate that Esoot is independent of the diameter and of the type of fuel: Esoot = 40 kW m-2. However, the average value of Elum ranged between 80 kW m-2 and 120 kW m-2, as it is a function of the diameter and the type of fuel. Elum increased with the pool diameter up to 5 m; thus, for D < 5 m: Elumgasoline Elumdiesel
53.64 D 0.474
(3-25)
28.03D 0.877
while for D t 5 m Elum = 115 kW m-2. xlum is constant for D < 5 m (xlum gasoline = 0.45, xlum diesel oil = 0.30) while for D > 5 m it decreases. There are not enough data to establish the exact decrease and the minimum value of xlum; some authors state that beyond D > 20 m xlum = 0. 0.5 0.4
xlum
0.3 0.2 0.1
2
E (kW/m )
0.0
Elum E
100 90 80 70 60 50 40 30 1
2
3
4
5
6
7 8 9 10
20
30
Diameter (m)
Fig. 3-8. Evolution of xlum, Elum and E as a function of pool diameter for gasoline. Taken from [23], by permission.
Consequently, the emissive power of fires involving gasoline, diesel oil and similar fuels can be estimated as follows:
82
E
xlum Elum 1 xlum E soot for D 20 m
(3-26)
E | E soot for D t 20 m
There is experimental evidence that E increases with pool diameter, which is essentially due to the increase of Elum (while the ratio between the surfaces of luminous and nonluminous flame remains constant); once it reaches a maximum value, E then starts to decrease as a result of the decrease in the luminous flame surface. The variation of xlum and E for gasoline is plotted in Figure 3-8. Finally, the following expression was also suggested for estimating the value of E: E
K rad m 'H c
(3-27)
A
where A is the area of the solid flame from which radiation is released (m2). The suggested conservative value for Krad is 0.35. 6 FLAME SIZE
Knowledge of the size and shape of the flames is required to estimate the effects of the fire – i.e. the radiation that will reach a given target – using the solid flame model. It is also required, when considering short distances, in order to discern whether there will be any flame impingement on nearby equipment. This can be comparatively difficult to predict for a number of reasons. First of all, the exact shape of the flames is not known, as it is always fairly irregular; the shape is usually compared to a given geometric body (a cylinder, a parallelepiped, a sphere). Secondly, the size of the flames varies with time due to the turbulence of the phenomenon, particularly for large fires. This is why average and maximum values have been defined.
1.0
Intermittency
0.8
0.6
(L/D)av 0.4
0.2
(L/D)max 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4
L/D
Fig. 3-9. Intermittency corresponding to a 3 m gasoline pool fire [20].
83
A helpful concept in defining these values is the intermittence criterion, developed by Zukoski et al. [24]. The intermittency i(L) is defined as the fraction of time during which the length of the flame is at least greater than L; this concept is represented in Fig. 3-9 for a gasoline pool fire. Thus, the average length of a flame is defined as the length at which the intermittency reaches a value of 0.5. The maximum flame length is commonly defined as the length at which the intermittency is 0.05. The size of the flames depends on the burning rate, which, in turn, depends on several factors (type of fuel, type of fire, etc.). Empirical or semiempirical equations are commonly used to estimate these values. In the followings paragraphs a selection of these equations is presented for pool fires and jet fires. 6.1 Pool fire size
6.1.1 Pool diameter When a liquid is spilled and a pool is formed, there are two possibilities: a pool on the ground or a pool on the water surface. The following paragraphs briefly discuss the difference between the two. Pools on ground If there is a dike or a barrier that contains the liquid spill, the pool diameter is fixed. If the dike is right-angled, the equivalent diameter must be used:
D
4
surface area of the pool
(3-28)
S
The size of a pool caused by a liquid spill on the ground depends on the duration and flow rate of the spill. Thus, liquid spills can be divided into two categories: - Instantaneous spills - Continuous spills If the spill is instantaneous, the pool will grow until it finds a physical barrier (for example, a dike) or until it is burnt if there is a pool fire. In the case of a continuous spill, the pool will grow until it finds a physical barrier or until the vaporization velocity or the burning rate equals the spill flow rate. Although many spills will in fact be semi-continuous (i.e. a certain amount of liquid is spilled during a given time), it is useful to have a criterion for distinguishing between instantaneous and continuous spills. The following has been proposed [25]: t cr
t spill y
(3-29)
Vl1 / 3
where tcr is a dimensionless critical time (-) tspill is the duration of the spill (s) y is the fuel burning rate (m s-1), and Vl is the total volume of the spilled liquid (m3).
84
According to this criterion, spills are instantaneous if tcr < 2 · 10-3. Otherwise, they are considered continuous. For an instantaneous spill, without any containing barriers, the pool diameter can be expressed as a function of time by the following expression [21, 31]:
D pool
Dmax
ª 3§ t « ¨ « 2 ¨© t max ¬
·ª § 2 ·§ t ¸¸ « 1 ¨¨ 1¸¸¨¨ ¹ «¬ © 3 ¹© t max
· ¸¸ ¹
2
ºº »» »¼ »¼
1/ 2
(3-30)
where Dmax is the maximum diameter (m) that the pool can reach: Dmax
§V 3 g · 2 ¨¨ l 2 ¸¸ © y ¹
1/ 8
(3-31)
and tmax is the time (s) required to reach this maximum diameter:
§ V t max 0.6743 ¨¨ l 2 ©gy
· ¸¸ ¹
1/ 4
(3-32)
However, a pool with the maximum diameter Dmax exists only for a short time. Using it to calculate the thermal radiation from a pool fire would lead to an overestimation of fire hazards [7]. Therefore, in practice it is better to use an average pool diameter value expressed as: Dav 0.683 Dmax
(3-33)
In the case of a continuous release without any physical barriers, the equilibrium diameter for a burning pool (for a situation in which the release rate equals the burning rate) and the time required to reach it can be estimated with the following expressions [21]: Deq
§ v · 2 ¨¨ l ¸¸ ©S y ¹
t eq 0.564
1/ 2
(3-34)
Deq
(3-35)
g y D
1/ 3
eq
where Deq is the equilibrium diameter (m) vl is the leak rate (m3 s-1), and teq is the time required to reach Deq (s). Pools on water In the case of a spill on water (usually seawater), if there is immediate ignition, the equations proposed for spills on smooth ground can be applied by replacing g with an “effective” value g’ defined as follows [6]:
85
§ U g ' g ¨¨1 l © U water
· ¸¸ ¹
(3-36)
where Ul is the density of the spilled liquid (kg m-3), and Uwater is the density of water (or seawater) (kg m-3). If, in the case of a continuous spill, there is delayed ignition and the pool diameter has become larger than the equilibrium diameter, the equilibrium diameter will soon be reached after ignition and this value will be maintained while the spill flow rate is constant. If the diameter has reached such a value that the thickness of the spilled layer has become very small (approximately 1.25 mm), ignition will not be possible even though an ignition source exists. If the spill is instantaneous and the ignition is delayed, the diameter will evolve as a function of time, passing through the following phases [26]. In phase 1, gravitational and inertial forces prevail: ª U Ul º D1 2.28 « w g Vl t 2 » ¬ Uw ¼
1/ 4
(3-37)
where Uw is the density of water (kg m-3) Ul is the density of the spilled liquid (kg m-3) Vl is the volume of liquid fuel released instantaneously (m3), and t is the time elapsed from the start of the release (s). In phase 2, viscous forces prevail:
D2
ª U U l g Vl 2 t 3 / 2 º 1.96 « w » Q w1 / 2 ¼ ¬ Ul
1/ 6
(3-38)
where Qw is the seawater viscosity (m2 s-1). Finally, in phase 3, the surface tension is the dominant spreading mechanism:
D3
§ f 2 t3 · ¸¸ 3.2 ¨¨ r2 © UwQ w ¹
1/ 4
(3-39)
where fr is the interface tension (N m-1) (ranging between 0.005 and 0.02 N m-1). 6.1.2 Burning rate The burning rate is usually estimated using the following expression: m mf 1 e kD
(3-40)
where mf is the burning velocity for an infinite diameter pool (kg m-2 s-1) and k is a constant (m-1). Various authors have proposed values for mf and k (Table 3-8). For large scale fires, m | mf .
86
Table 3-8 Expressions for the estimation of burning velocity Diesel oil -2 -1 Authors k, m-1 mf , kg m s Babrauskas [27] 0.034 2.80 Rew et al. [28] 0.054 1.30 Muñoz et al. [23] 0.054 0.88
Gasoline mf , kg m s-1 0.055 0.067 0.082 -2
k, m-1 2.10 1.50 1.31
An alternative way for estimating the burning rate is to apply the expression proposed by Burgess [29]: m
0.001
'H c 'hv c p T0 Ta
(3-41)
where 'Hc is the net combustion heat (kJ kg-1) 'hv is the vaporization heat of the fuel at its boiling temperature (kJ kg-1) and T0 is the boiling temperature at atmospheric pressure (K). The burning rate of a pool fire can also be expressed in m s-1. Obviously, the relationship between this burning rate and the mass burning rate is: y
m
(3-42)
Ul
6.1.3 Height and length of the flames The height of the visible flame is a function of pool diameter and burning velocity. Of the various expressions that have been proposed for the estimation of this value, the most commonly used is the one developed by Thomas [30]: H D
ª m º 42 « » ¬« U a gD ¼»
0.61
(3-43)
This expression gives the average flame height (or length, if the flame is tilted). The maximum length can be estimated as a function of the average value as follows [22]:
§L· ¨ ¸ © D ¹ max
§L· 1.52¨ ¸ © D ¹ average
(3-44)
6.1.4 Influence of wind Wind can have an influence on flame length. A recent study [31] found that the influence of wind on burning velocity is almost negligible at uw < 2 m s-1. At higher velocities, the following equation is [30] is often used: H D
ª m º 55« » «¬ U a g D »¼
0.67
u*
0.21
(3-45)
87
u* being a dimensionless wind velocity: u*
uw
§ gmD· ¸¸ ¨¨ © Ua ¹
1/ 3
; (if u* < 1 then it is assumed u* = 1)
(3-46)
where uw is the wind velocity (m s-1). Wind can also tilt the flames and significantly move their bottom part (Fig. 3-10), thus causing the flames to spill over the edge of the pool and elongating the flame base. This can be highly significant if there is equipment nearby, as the level of thermal radiation will increase.
Fig. 3-10. Flame tilt and drag in pool and tank fires. Taken from [32], by permission.
If there are short separation distances between equipment, for example adjacent tanks, the flames may engulf part of this equipment. If there is flame impingement, the thermal flux can increase dramatically, leading to the structural failure of the equipment. Therefore, both flame tilt and drag must be evaluated in the presence of wind. The flame tilt can be estimated using the following expression [25, 32]: for u* d 1
cosD= 1 cos D= 1/ u
*
for u* > 1
and the flame drag can be estimated for any fuel [32, 33] by means of:
88
0.069
§ u2 · D' (3-47) 1.5 ¨¨ w ¸¸ D © gD¹ ______________________________________ Example 3-5 Imagine the instantaneous rupture of a tank containing 3500 m3 of gasoline. The tank is located in a dike with a diameter of 60 m. There is a slow wind (uw = 1.5 m s-1). a) Estimate the maximum thermal radiation on the wall of a tank located 25 m from the dike wall. b) If for a short period the wind speed increases to 6 m s-1, will a set of valves located 14 m downwind of the dike wall be engulfed by the fire? Gasoline liquid density = 870 kg m-3. Ambient temperature = 18 ºC. Relative humidity = 70%. Air density = 1.2 kg m-3.
Solution a) Calculation of the maximum pool diameter (Eq. (3-31)), using the approximate value for the burning rate of 0.082 kg m-2 s-1 (Table 3-8): Dmax
§ 3500 3 9.81 870 2 2 ¨¨ 0.082 2 ©
· ¸¸ ¹
1/ 8
576 m.
As this value is higher than the dike diameter, the diameter of the pool will be 60 m. The approximate burning rate can now be verified for this pool diameter (Eq. (3-40)): m
0.082 1 e 1.31 60
0.082 kg m-2 s-1.
Estimation of flame height (Eq. 3-43)): H
ª º 0.082 60 42 « » ¬1.2 9.81 60 ¼
0.61
70 m
The influence of the wind can be ruled out (u* = 0.44). Emissive power: as D > 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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The SFPE Handbook of Fire Protection Engineering. NFPA-SFPE. Boston, 1988. [22] Siegel, R., Howell, J. R. Thermal Radiation Heat Transfer. Taylor & Francis, New York,
1992. [23] Muñoz, M., Arnaldos, J., Casal, J., Planas, E. Comb. and Flame 139 (2004) 263. [24] Zukoski, E. E., Cetegen, B. M., Kubota, T. Visible Structure of Buoyant Diffusion
Flames. 20th Symposium on Combustion. The Combustion Institute. Pittsburgh, 1985. [25] Mudan, K. Prog. Energy Combust. Sci. 10 (1984) 59. [26] Casal, J., Montiel, H., Planas, E., Vílchez, J. A. Análisis del riesgo en instalaciones
industriales. Alfaomega. Bogotá, 2001. [27] Babrauskas, V. Fire Technol. 19 (1983) 251. [28] Rew, P. J., Hulbert, W. G., Deanes, D. M. Trans. IChemE. 75 (1997) 81. [29] Burgess, D. S., Hertzberg, M. Radiation from Pool Flames. In Heat Transfer in Flames.
John Wiley and Sons. New York, 1974.
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[30] Thomas, P. H. The Size of Flames from Natural Fires. 9 Symposium on Combustion.
The Combustion Institute. Academic Press. New York, 1963. [31] Chatris, J. M., Quintela, J., Folch, J., Planas, E., Arnaldos, J., Casal, J. Comb. and Flame
126 (2001) 1373. [32] Lautkaski, R. J. Loss Prev. Process Ind. 5 (1992) 175. [33] Moorhouse, J. IChemE, Symp. Series n. 71 (1982) 165. [34] Hawthorne, W. R, Weddell, D., Hottel, H. Mixing and Combustion in Turbulent Gas Jets.
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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[43] Broeckmann, B., Schecker, H. G. Boilover effects in burning oil tanks. 7 International
Symposium on Loss Prevention and Safety promotion in the Process Industries, 1992. [44] INERIS. Boilover. Ministère de l’Ecologie et du Développement Durable. 2003. [45] CCPS. Guidelines for Consequence Analysis of Chemical Releases. AIChE, New York,
1999. [46] Satyanarayana, K., Borah, M., Rao, P. G. J. Loss. Prev. Proc. Ind., 4 (1991) 344-349. [47] W. E. Martinsen and J. D. Marx, Int. Conference and Workshop on Modelling and
Consequences of Accidental Releases of Hazardous Materials. CCPS, AIChE, 605, 1999. [48] M. W. Roberts, Analysis of Boiling Liquid Expanding Vapor Explosion (BLEVE) Events
at DOE Sites, Safety Analysis Workshop 2000 (http://www.efcog.org/ publication/WG%20Minutes/sawg/2000%20Conference/papers_pdf/roberts.pdf). [49] M. Demichela, N. Piccinini and A. Poggio, Process Saf. Environ., 82 (B2) (2004) 128. [50] HID, Safety Report Assessment Guide: LPG, 2001. [51] A. F. Roberts, Fire Safety J., 4 (1982) [52] EPA.CEPPO, Risk Management Program Guidance Offsite Consequence Analysis, 1999. [53] D. F. Bagster and R. M. Pitblado, Chem. Eng. Prog., July (1989) 69. [54] I. A. Papazoglou and O. N. Anezeris, J. Hazard. Mater., 67 (1999) 217.
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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
124
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
125
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.
126
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.
129
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.
130
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.
131
______________________________________ 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%.
133
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
134
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. ______________________________________
135
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
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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.
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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.
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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.
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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:
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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.:
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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
Po hlo
hl
hgo
Specific enthalpy
Fig. 5-5. The situation in which cooling 50% of the liquid vaporizes the rest of the liquid. Taken from [10], by permission.
It can be demonstrated that the condition defined by Eqs. (5-8) and (5-9) corresponds to the transfer of the maximum amount of energy between the liquid mass which cools down and the liquid mass which is vaporised, leading to a minimum amount of energy in the remaining liquid. In other words, liquid fractions that are higher or lower than 50 % at temperatures other than Tsl-E can lead to the partial or total vaporisation of the rest of the liquid, but only the 50 % fraction minimises the energy of the remaining liquid. As an example, Table 5-4 shows the data obtained when the appropriate calculations were carried out for a given substance, hexane [10]. It can be seen that the energy transferred reaches its maximum value at f = 0.5 and Tsl-E = 407.6 K. Tsl-E can be obtained from Eq. (5-9) by trial and error calculation or graphically by plotting both the left hand side and the right hand side of Eq. (5-9) —the required vaporization energy and the energy released by the cooling liquid— for each substance as a function of temperature; the intersect of these two lines will give Tsl-E.
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Table 5-4 Flash vaporization temperature, liquid fraction and energy involved (hexane, overall mass = 1 kg) Temperature at which flash Cooled liquid Energy transferred from vaporisation starts (K) mass fraction cooled liquid to evaporated liquid (kJ) 453.2 0.10 29.6 433.2 0.28 66.5 423.2 0.37 76.4 413.2 0.45 81.3 407.6 0.50 82.1 403.2 0.54 81.7 393.2 0.62 77.7 383.2 0.69 69.8 373.2 0.77 58.0
While the values of Tsl obtained from the stability criteria (see Fig. 5-1) depend on the features of the equation of state used, calculating the superheat limit temperature from the enthalpies of Eq. (5-9), Tsl-E, only depends on the accuracy of the thermophysical data used. The temperatures Tsl-E obtained in this way for diverse substances have been included in the last column of Table 5.3. Column 8 contains the values of the superheat limit temperature obtained from the Redlich-Kwong spinodal curve. As it can be observed, there is a certain scattering among the diverse values of the superheat temperature limit. For substances with a low acentric factor, (which comply relatively well with the law of corresponding states) the values of Tsl-RK and Tsl-E are relatively similar [10]. As the acentric factor increases, the difference also increases which could be attributed to the fact that the RK equation is not suitable for describing a substance when its molecular complexity increases. 2.4 When is an explosion a BLEVE? The aforementioned theory, although accepted by many, fails to explain some of the BLEVEs that have occurred (for example, Mexico City, 1984; Feyzin, 1966) and various authors have now questioned its validity. As mentioned at the beginning of this chapter, a definition of BLEVE has been proposed that is much wider and more flexible [3]. Therefore, explosions that do not strictly meet the Reid criterion have been considered recently to be a BLEVE. Furthermore, some authors doubt the applicability of a strict criterion to a real case. In the heating of a large vessel by a fire, there are likely to be some aspects that can make it difficult to predict the BLEVE occurrence. Amongst these, we can cite the liquid temperature stratification, which can significantly affect the thermodynamic conditions inside the vessel. The contents of a tank are not heated uniformly during a fire. In various fire tests conducted on LPG tanks [12-16], the temperature of the liquid varied between the bottom of the tank (where the liquid was cooler) and the top (where it was warmer). Due to buoyancy, the temperature at the top of the tank increases faster than at the bottom. This temperature stratification plays an important role in the event of an accident. The energy contained within the vessel for a given pressure is lowered by the stratification, as the average temperature of the liquid is lower. Thus, stratification decreases the time to the explosion, as the pressure in the vessel is determined by the vapour pressure, which is a function of the temperature of the uppermost liquid layer. The stratification also decreases the severity of the accident, as there is less energy contained in the system.
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In fact, the methodologies available to estimate the overpressure from a BLEVE do not take into account whether the temperature inside the tank just before the explosion was or not higher than Tsl, i. e., these methodologies give a value of 'P according to the initial temperature and pressure and without considering whether the phenomenon of homogeneous nucleation took place. Birk et al. [17] made a distinction between ‘hot’ and ‘cold’ BLEVEs. According to this author, hot BLEVEs are those in which the liquid is at a temperature higher than the superheat temperature limit at the atmospheric pressure, whilst in cold BLEVEs the average temperature of the liquid is lower than the superheat temperature limit. According to this distinction, the effects of hot BLEVEs would be more severe than those of cold BLEVEs (what is quite logical taking into account that, in this case, temperature, pressure and energy content are higher). ______________________________________ Example 5-2 In some cases, classifying an explosion as a BLEVE will depend on the accepted definition of the term. Here is an example of such a case [18]. On 11 July 1978, at 14.30 h, a road tanker transporting propylene exploded near a campsite at Els Alfacs (Catalonia, Spain), resulting in the death of 216 people. The volume of the vessel was 44.4 m3 and it had been loaded with 23,619 kg of propylene. It was overloaded, as the maximum amount that it could legally transport was 19,350 kg. There was no safety valve. When loading finished, at 12 o’clock, the temperature of the propylene was 1 ºC. Analyze the case.
Solution Let’s consider the vessel containing a mass M of propylene (Fig. 5-6). The following relationships apply: M V
Uv
Ul
Ml M g Vl V g Mg Vg Ml Vl
Fig. 5-6. Liquid and vapour phases in the vessel.
By combining them, the following expression can be derived:
160
Vg
V Ul M Ul U g
Note that when the temperature increases, Ul and Ug decrease and therefore, as Ul >> Ug, the volume of vapour inside the vessel will decrease. Thus, it is possible to reach a temperature at which V Ul
M
This situation would be reached at a liquid density of: 44.4 m 3 U l kg m -3 23,619 kg
Ul
532 kg m -3
(Although the vessel also contained 0.03% nitrogen and 6.5% propane, pure propylene has been considered here for the sake of simplicity; furthermore, the increase in volume of the tank is negligible over the temperature range considered).
Fig. 5-7. The process on the pressure-enthalpy diagram: the line from points 4 to 5 represents the
heating of the vessel full of liquid [18].
161
This Ul value corresponds to a temperature of 8 ºC. A heat transfer calculation shows that this was the temperature of the vessel on that sunny day after 2.5 h travelling. Therefore, the increase in temperature and pressure led to the condensation of all the vapour, and the vessel became full of liquid at a pressure of approximately 744 kPa. Later on, due to heating by solar radiation, the liquid temperature continued to increase. However, the liquid could not expand, as the volume of the vessel was practically constant. Consequently, according to the liquid compressibility coefficient, the pressure in the vessel increased dramatically. The tensile strength of the steel from which the vessel was constructed was approximately 665 MPa; this value was reached after some further heating. The whole process can be seen in the pressure-enthalpy diagram (Fig. 5-7). Point 1 corresponds to the situation just after the tank was loaded (point 2, saturated liquid; point 3, saturated vapour). The line from points 1 to 4 represents the process during which all the vapour condensed, whilst the line from points 4 to 5 represents the heating of the vessel full of liquid. It can be seen from the diagram how quickly the pressure increases. Fig. 5-8 shows again the influence of overloading the tank. With a load of 23,619 kg of propylene, once the whole tank became full of liquid (at a pressure of 744 kPa and a temperature of 8.3 ºC), further heating implied an abrupt increase of pressure which led to the explosion. If the tank had been correctly loaded (19,350 kg), the vapour volume would had been maintained and the increase of pressure as temperature increased would have been much lower; under the prevailing environmental conditions, the explosion would not have occurred (a temperature of 58.8 ºC (at 2460 kPa) would have been required to reach the condition of tank full of liquid). 30000 28000
Correct load Overload
26000 24000 22000
Pressure (kPa)
20000 18000 16000
Hydraulic test pressure (900 kPa)
14000 12000
Design pressure (700 kPa)
10000 8000 6000
Tank (correct load) full of liquid (2460 kPa)
Tank (overloaded) full of liquid (744 kPa)
4000 2000 0 0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 Temperature (ºC)
Fig. 5-8. Evolution of pressure as a function of temperature inside the tank. For the overloaded tank, pressure increases abruptly after the tank has become full of liquid. For the correctly loaded tank, this situation would have been reached at 59 ºC (the explosion would not have occurred).
The high pressure caused a crack, and then the depressurisation gave rise to an explosive flash of the superheated liquid. The vessel was split into two fragments, one of which was propelled over a distance of 200 m. The propylene formed a cloud that moved towards the campsite and ignited originating a fireball.
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Although this accident was not a BLEVE according to the superheat limit temperature criterion (see Table 5-3), it can be considered as such according to the most recently accepted criteria. ______________________________________ 3 VESSEL FAILURE 3.1 Mechanism By analysing the catastrophic failure of four 4.5 m3 LPG vessels filled to different levels and subjected to jet fire attack, Venart [19] suggested a two-step process for the mechanism of vessel failure, which is described in the following paragraphs. In response to fire, the pressure inside the vessel rises to the set value of the pressure relief valve, which opens. This valve is theoretically designed to discharge enough vapour to maintain the vessel at a safe pressure irrespective of its thermal exposure (API 520 1990, API 521 1993, ASME 1992, NFPA 58 1998). As the fill level decreases, the time to first vent increases. If the fill is low, the evaporation rate will be also low and will not exceed the capacity of the valve: the pressure can remain constant with only partial valve lift or cycling (relief valves must be carefully sized for fire, because oversizing can lead to chattering during the relief and to the subsequent failure of the valve). As fill increases, the evaporation rate increases concomitantly —the wetted surface is larger— and the exiting vapour may be strongly superheated due to vapour stratification; thus, the valve capacity may be exceeded. Furthermore, the opening of the valve will depressurise the vessel, giving rise to increased boiling and the formation of a two-phase swell within the liquid. According to the fill, the valve intake can be two-phase flow; in this condition, the choke velocity will be significantly lower than that corresponding to vapour flow, the release flow rate will decrease and the pressure in the vessel will increase. In parallel, due to the heating of those parts of the vessel walls not wetted by the liquid, they can weaken and undergo some plastic deformation at the hottest points, eventually leading to the formation of a crack. This initiating fracture provides an additional relief, decreasing the flow through the pressure relief valve or even, at low fills, allowing it to close. In the vapour zone, metal wall temperatures are extremely variable under the action of fire; they can drop locally by up to 150ºC less than 500 mm away from the initial rupture site [20]. This affects metal strength and the initial crack can stop progressing in the cooler, thicker and stronger wall. The size of this initial fissure will be a function of the metal temperature, the fill level and the pressure. Subsequently, the crack can be catastrophically restarted by thermal stresses at the crack tip due to quenching by direct liquid contact with the wall, thus initiating the failure of the vessel; a possible influence of the impact of a low-void fraction swell on the superheated head of the vessel, which would function as a water-hammer pressure impulse, has also been suggested [19]. Thus, crack development and propagation during vessel failure often occurs in two steps: í the formation of an initial crack that becomes stable, forming an opening for a discharge in addition to that of the PRV í a final failure stage in which the initial crack propagates rapidly into the metal as a plain stress shear failure. In the aforementioned experimental explosions of four 4.5 m3 vessels, the time elapsed between initial crack formation and failure was a function of the vessel fill, ranging between 2 s (85% fill) and 40 s (20% fill), and the length of the initial cracks ranged from 290 mm to 580 mm.
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3.2 Pressure required for vessel failure For pressure vessels, the bursting pressure P is usually of the order of four or five times the design pressure. The pressure at which a vessel or a pipe has failed can be estimated from its dimensions, geometry and material of construction. The following expressions can be applied [21]. For cylindrical vessels with the pressure (P-P0) not exceeding 0.385 times the mechanical strength of the material (SM): P
P0
SM z r 0 .6 z
(5-10)
where P is the internal absolute pressure (Pa) P0 is the atmospheric pressure (Pa) SM is the tensile strength of the material (Pa) r is the inside radius of the vessel (m), and z is the wall thickness of the vessel (m). For cylindrical vessels with pressure (P-P0) exceeding 0.385 SM: 2
P
§z · S M ¨ 1¸ S M ¹ ©r P0 2 §z · 1 ¸ 1 ¨ ©r ¹
(5-11)
For spherical vessels with pressure (P-P0) not exceeding 0.665 SM: P
P0
2 SM z r 0 .2 z
(5-12)
For spherical vessels with pressure (P-P0) exceeding 0.665 SM: 2
P
§z · 2 S M ¨ 1¸ 2 S M ¹ ©r P0 2 §z · 1 ¸ 2 ¨ ©r ¹
(5-13)
Eqs. (5-10) and (5-11) can be applied also to estimate the maximum internal pressure in pipes. The values of the tensile strength of the material can be found in the literature [22]. Tensile strengths for some common materials are shown in Table 5-5. Of course, these expressions can be used if the vessel is at the temperature at which the values of SM have been determined. If the vessel has been locally heated by a fire, the value of the tensile strength in that zone will change. The value of pressure at failure obtained by this procedure can be used to estimate the mechanical effects (blast) of the explosion.
164
Table 5-5 Strength of materials at ambient temperature [22] Material Tensile strength (MPa) Stainless steel 304 565 Stainless steel 316 565 Stainless steel 430 517 Carbon steel C-Mn 415 Cast iron 173 Aluminium 70 Borosilicate glass 70 Nickel 452 Hastelloy C 500 Titanium 417
______________________________________ Example 5-3 Estimate the pressure at which the vessel described in Example 5-2 failed. The internal diameter of the vessel was 2.3 m and the wall thickness 8.1 mm. It was constructed with stainless steel AISI-304. At the time of the explosion, the vessel wall exposed to sun radiation was lightly lower than 20 ºC, i. e. data from Table 5-5 can be applied. Solution Assuming that P < 0.385 SM: P
P0
SM z r 0.6 z
101,325
565 10 6 (8.1 10 3 ) = 4.7 · 106 Pa 1.15 0.6 (8.1 10 3 )
As this value is lower than 0.385 SM, 4.7 MPa (is the approximate pressure at which the vessel failed. ______________________________________ 4 ESTIMATION OF EXPLOSION EFFECTS 4.1 Thermal radiation When a vessel explosion involves a flammable substance, it is usually followed by a fireball, which releases intense thermal radiation. 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). The estimation from thermal effects (thermal radiation) from a fireball are treated in Chapter 3. Although BLEVEs are often associated to a fireball, if the substance involved is not flammable —for example, water— the explosion will not be followed by a fireball and the effects will be only mechanical: pressure wave and missiles. 4.2 Mechanical energy released by the explosion When a vessel bursts, the mechanical energy contained inside is released (note that the units of pressure are energy per unit volume). The substance contained in the vessel
165
instantaneously increases in volume due to the expansion of the vapour already existing in the vessel at the moment of the explosion and to the superheated liquid, which undergoes a practically instantaneous partial vaporization (flash). The mechanical energy released in a vessel explosion is distributed between the following: í the energy of the pressure wave í the kinetic energy of the projectiles í the energy required to break the vessel. The relative distribution of the energy will change in relation to the particular conditions of the explosion. It is very difficult to accurately establish the amount of energy that will contribute to the pressure wave. A fraction of the energy released in the explosion will become kinetic energy of the propelled fragments (and a small amount of energy will also have been required to break the vessel into pieces). It has been suggested that in a fragile vessel failure 80% of the energy released contributes to the creation of the pressure wave. In the case of ductile break —in which large fragments of the vessel are propelled— the energy in the pressure wave is only 40-50%. In both cases, the remainder of the energy becomes kinetic energy of the fragments, as other contributions (heating of the environment) are negligible. Most vessels or tanks are constructed from materials that are ductile under operating conditions. A fragile failure is only found under very specific conditions, when the stress reached by the material is much higher than its plastic limit. This only occurs with tempered steel and glass. Therefore, vessel explosions usually consist of ductile breaking, producing a limited number of fragments. 4.2.1. Ideal gas behaviour and isentropic expansion Due to the values of temperature and pressure usually associated to this type of explosions, ideal gas behaviour should not be assumed. However, a number of authors assume it and apply the method explained in the following paragraphs to estimate the mechanical energy released by the explosion. Concerning the vapour initially in the vessel, the energy released by its adiabatic expansion (from the breaking pressure in the vessel up to the atmospheric pressure) is: W
m g U 2 U 1
(5-14)
where W is the expansion work of the vapour (kJ), mg is the mass of vapour already existing in the vessel at the moment of the failure (kg), U1 is the internal energy of the vapour under -1
the conditions at which the vessel bursts (kJ·kg ) and U2 is the internal energy of the vapour after the expansion up to atmospheric pressure (kJ·kg-1). Supposing that the expansion is isentropic and reversible —due to the velocity at which it takes place— and that the vapour behaves as an ideal gas [5, 23] this energy can be calculated as:
W
³ PdV
(5-15)
Introducing the relationship P·VJ = constant (J being the ratio of specific heats) and integrating:
166
W
J 1 § · § P V · ¨ § P0 · J ¸ ¸¸ ¨1 ¨ ¸ ¸ 10 ¨¨ © J 1 ¹ ¨ © P ¹ ¸ © ¹ 2
(5-16)
where W is expressed in kJ, P0 is the atmospheric pressure (bar), V is the initial volume of vapour (m3), and P is the pressure (bar) in the vessel just before the explosion. This energy can be expressed as TNT equivalent mass by using the adequate energy conversion factor (approximately 4680 J per gram of TNT),
WTNT
J 1 § · § P V · ¨ § P0 · J ¸ ¸¸ ¨1 ¨ ¸ ¸ 0.021 ¨¨ © J 1 ¹ ¨ © P ¹ ¸ © ¹
(5-17)
where WTNT is the equivalent mass of TNT (kg). Furthermore, if the vessel contains superheated liquid —as in the case of a BLEVE explosion— the released energy can be estimated by using the same method. In this case, it must be taken into account that the mass of liquid will partially vaporize suddenly when it reaches atmospheric pressure. The volume of this vapour at the pressure in the vessel just before the explosion must then be calculated; adding this fictitious volume to the real one, the equivalent mass of TNT will be:
WTNT
J 1 § · § P V * · ¨ § P0 · J ¸ ¸¸ ¨1 ¨ ¸ ¸ 0.021 ¨¨ © J 1 ¹ ¨ © P ¹ ¸ © ¹
(5-18)
where V* is the volume of vapour in the vessel plus the volume (at the pressure inside the vessel and the corresponding saturation temperature) of the vapour generated in the explosion, in m3: V
§U V Vl f ¨ 1 ¨U © g
· ¸ ¸ ¹
(5-19)
V is the volume of vapour inside the vessel before the explosion, Vl is the volume of liquid in the vessel before the explosion (m3), and f is the vaporization fraction (flash), i.e., the fraction of liquid which vaporizes in the depressurisation; its value can be calculated with the following expression obtained from a heat balance: f
c pl (T T0 )
(5-20-a)
'hvTo
where T is the temperature of the substance at the moment of the explosion T0 is the boiling temperature at atmospheric pressure (K) and 'hvT0 is the liquid vaporization heat at T0 (kJ kg-1). Or with the following equation, more accurate:
167
f
1 e
ª § c « 2.63 plT0 Tc T0 ¨ 1 §¨ Tc T ¨¨ ¨ T T « H vT0 © © c 0 ¬
· ¸ ¸ ¹
0.38
·º ¸» ¸¸ » ¹¼
(5-20-b)
where T0 is the boiling temperature of the substance at atmospheric pressure (K), Tc is the critical temperature (K), cplTo is the specific heat of the liquid (kJ·kg-1·K-1) at T0 and 'hvTo is the enthalpy of vaporization of the substance at T0 (kJ·kg-1). 4.2.2. Real gas behaviour and adiabatic irreversible expansion The methodology explained above assumes ideal gas behaviour and a reversible adiabatic (isentropic) process, and involves the use of average properties for J, cp and hg to calculate the energy released. However, these assumptions are far removed from the real phenomenon. Although in the sudden expansion associated with an explosion the condition of an adiabatic process can be assumed due to the velocity of the phenomenon, it will never be a reversible process, but rather a highly irreversible one. In fact, as the work associated with an isentropic process is the largest of all those that can be obtained from an adiabatic process, a considerable overestimation is introduced, the energy calculated in fact being the maximum possible value. A new methodology has been proposed [24] in which the expansion work is calculated assuming that the expansion is adiabatic but irreversible, so that the only work performed is that associated with the variation in volume ('V) that occurs when the contents of the vessel changes from the conditions at explosion to the point of reaching atmospheric pressure (P0 = 0.1013 MPa). As in the isentropic approach, the hypothesis that immediately after the explosion there is liquid-vapour equilibrium at atmospheric pressure and the corresponding saturation temperature is also assumed; however, the state (entropy) will be different. The real expansion work is –P0·'V, 'V being the variation in volume of the whole content of the vessel when it changes from the explosion state to the hypothetical final state. On the other hand, for an adiabatic process this work must be equal to the variation in internal energy of the vessel content, 'U: P0 'V
'U
(5-21)
This equation can be solved graphically. Fig. 5-9 shows the variation in U (for the whole content of the vessel) from the explosion situation to that of different conditions, defined by a common pressure of 0.1013 MPa and different liquid-vapour equilibrium conditions corresponding to different vapour fraction values. For both situations (just before the explosion and the final state) the whole mass of liquid plus vapour contained in the vessel was considered. In the same figure, the variation in –P0·'V was also plotted against the vapour fraction corresponding to the theoretical final condition. The intersection of both straight lines corresponds to the condition imposed by Eq. (5-21). The ordinate at this point represents both the variation in U and the energy released by the explosion. Eq. (5-21) can also be solved analytically. Taking into account the mass and energy balances, the equations of the two straight lines plotted in Fig. 5-9 can be expressed as: 'U P0 'V
U
l
U g M x M U l U i
>
P0 V g Vl M x M Vl Vi
(5-22)
@
(5-23)
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Fig. 5-9. Variation in 'U and P0·'V as a function of the vapour fraction of the theoretical final condition (see Example 5-4). Taken from [24], by permission.
where Ul is the internal energy of the liquid (MJ kg-1) Ug is the internal energy of the vapour (MJ kg-1) M is the overall mass of the substance in the vessel (kg) x is the vapour fraction at the final state of the process (-) P0 is the atmospheric pressure (MPa) 'V is the volume variation of the whole vessel content due to the explosion (m3) Vg is the specific volume of vapour at the final state of the process (m3 kg-1) Vl is the specific volume of liquid at the final state of the process (m3 kg-1). From these two equations, the intersection point can be found:
x
M P0 Vl Vi P0 M U l U i U l U g Vg Vl P0 M
>
(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.
171
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). ______________________________________
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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.
178
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
180
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<20)1 Parkland, bushes: numerous obstacles (x/h<15)1 Regular large obstacle coverage (suburb, forest) City centre with high- and low-rise buildings
z0, m 0.0002 0.005 0.03 0.10 0.25 0.5 1.0 3.0
1 x is a typical upwind obstacle distance and h the height of the corresponding major obstacles
The turbulence is constituted by circular eddies in all directions. It is important because it improves dispersion. There are two types of turbulence: a) mechanical turbulence, which is generated by the wind moving past the elements of roughness on the ground surface (buildings, vegetation) or by the shearing action between adjacent air layers moving at different speeds, and b) buoyant turbulence, which is generated by the heating or cooling of air near the ground surface; in sunny conditions, for example, the ground is heated and transfers heat to the air above it, which creates large upward-rising eddies called thermals.
Fig. 6-2. Variation of wind velocity with height for different situations.
The existence of thermals implies unstable atmospheric conditions and enhances the dispersion of pollutants in the atmosphere. Instead, at night, with a clear sky and a light breeze, the existence of an inversion layer will prevent turbulences from developing and dispersion will be hindered; this is why, during the night, odours (from a rubbish dump or a farm, for example) go farther.
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2.2 Lapse rates In the atmosphere, or more exactly, in the troposphere, the temperature of the air usually decreases with altitude. If a parcel of air rises through the atmosphere (due to mechanical turbulences, for example) it will undergo an expansion process, as the pressure decreases with altitude. If this process is assumed to be adiabatic (i.e. involving no exchange of heat with the surrounding air), the temperature of the air inside the parcel will decrease (adiabatic cooling), as it will expend energy in the expansion against its surroundings. The decrease of temperature as a function of height can be established theoretically for dry air: dry air expanding adiabatically cools at 9.8 ºC per kilometre; this is called the adiabatic lapse rate (Fig. 6-3). Usually it is assumed that the adiabatic lapse rate is approximately 1 ºC per 100 m. If the air in the parcel is saturated, latent heat is released as water vapour condenses due to the cooling process; the wet adiabatic lapse rate (approximately 0.6 ºC per 100 m) is therefore lower than the dry one. However, due to diverse circumstances, the lapse rate can be different. When the atmospheric lapse rate exceeds the dry adiabatic lapse rate, it is said to be superadiabatic. When it is less than the dry adiabatic lapse rate, it is said to be subadiabatic. If the temperature of the air is constant with height, the lapse rate is zero and the atmosphere is isothermal. The lapse rate is very important because it establishes the ability of the atmosphere to disperse any release (see next section).
Fig. 6-3. The different lapse rates and the stability of the atmosphere.
Finally, in certain circumstances, the temperature of the atmosphere increases with height, the lapse rate being therefore negative; i.e. at a certain height the temperature is higher than that of the air below. This is called thermal inversion. Although there are two types of inversion, from the point of view of accidents radiation inversion is most important. Radiation inversion is due to the different cooling rates of the earth and the air above it: the ground is cooled and it cools the adjacent layer of air; thus a layer of cool air above the
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ground is built, which is covered by hotter air. This is a common phenomenon at night (Fig. 6-4), although it can also occur during the day for geographical reasons. Another type of inversion is subsidence inversion, which is caused by a high-pressure system: the air circulating in a high pressure zone descends and undergoes a compression process that heats it and forms a layer of warm air that covers the cooler air. From the point of view of the atmospheric dispersion of gases the inversion is a very important phenomenon, as it prevents vertical dispersion and keeps any release in the lower layers of the atmosphere.
Fig. 6-4. Typical distribution of air temperature as a function of altitude during the day and at night. During the day the lapse rate is adiabatic; during the night there is an inversion layer.
2.3 Atmospheric stability Atmospheric stability is the tendency of the atmosphere to increase (instability) or to decrease (stability) the vertical displacement of air originated by mechanical causes (such as the wind). It is a function of the temperature profile and is closely related to the ability of the atmosphere to disperse pollutants. Let us assume that an atmosphere has a temperature gradient of -1.5 ºC/100 m, i.e. the temperature of the air decreases by 1.5 ºC when the height increases by 100 m (this is called a superadiabatic gradient). If a parcel of air at 20 ºC is transported 100 m upwards by mechanical turbulence, its temperature will decrease by 1 ºC due to expansion (the pressure decreases), but in the new position the temperature of the surrounding air will be 20-1.5 = 18.5 ºC; therefore, the parcel will remain warmer than the surrounding atmosphere and thus its displacement upwards will continue to accelerate (Fig. 6-5) [3]. If the parcel is transported from its initial position (at 20 ºC) to a new one 100 m below, its temperature will become 21 ºC and the temperature of the surrounding air will be 20+1.5 = 21.5 ºC; therefore, the downwards displacement of the parcel will continue to accelerate. Thus, mechanical turbulence increases and the atmosphere is said to be unstable.
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Fig. 6-5. The motion of air in unstable and stable atmospheres.
If the temperature gradient in the atmosphere is -0.5 ºC/100 m, i.e. the temperature only decreases by 0.5 ºC when the height increases by 100 m (this is called a subadiabatic gradient), then when a parcel of air whose temperature is initially 20 ºC is displaced 100 m upwards it will be surrounded by air at 19.5 ºC. Therefore, the parcel, which is at 19 ºC, heavier than the surrounding air, will be subject to a downward buoyancy force that will push it downwards in a movement opposite to the initial one. If the parcel is transported from its initial position (at 20 ºC) to a new one 100 m below, its temperature will be 21 ºC and that of the surrounding air will be 20.5 ºC; the buoyancy force will push the parcel upwards. The atmosphere—stable in this case—, tends to eliminate any movement in a vertical direction. Atmospheric stability is therefore an estimation of the state of the atmosphere that cannot be measured directly as can temperature or pressure. Stability is usually estimated as a function of the wind velocity and sun radiation. An unstable atmosphere is characterized by the existence of a significant vertical displacement of air, a negative vertical temperature gradient (the temperature decreases with height), frequent fluctuations in wind direction and strong sun radiation. Instead, a stable atmosphere implies a laminar flow of air layers (low turbulence), positive vertical temperature gradient (thermal inversion), limited fluctuations in wind direction and little sun radiation. An example of an unstable atmosphere would be a sunny summer day in which the sun radiation heats the surface of the earth, and then this surface heats the lower layers of the atmosphere, giving rise to vertical streams of air; i.e. there is considerable turbulence near the ground. An example of a stable atmosphere would be a clear summer night in which the ground cools rapidly due to the radiation to the atmosphere, causing the lower layers of the atmosphere to cool. This creates a light horizontal breeze with little turbulence. From the point of view of stability, the atmospheric conditions are classified according to six (or sometimes seven) different stability classes (Table 6-4). There are several ways to estimate the stability class from the main meteorological variables (this is valid for an estimation of the stability on the ground but not on the surface of the sea): according to the vertical thermal gradient (Table 6-5), according to the fluctuation in the wind direction (Table 6-6) or according to the sun radiation and the velocity of the wind (Table 6-7).
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Table 6-4 Definition of the atmospheric stability classes Stability class Definition A Extremely unstable B Moderately unstable C Slightly unstable D Neutral E Slightly stable F Moderately stable G* Very stable * Stability class G is rarely used
Table 6-5 Stability class according to the vertical temperature gradient Stability class Vertical thermal gradient (ºC/100 m) A < -1.9 ºC B from -1.9 ºC to -1.7 ºC C from -1.7 ºC to -1.5 ºC D from -1.5 ºC to -0.5 ºC E from -0.5 ºC to 1.5 ºC F < 1.5 ºC Table 6-6 Stability class according to hourly fluctuations in wind direction Stability class Horizontal variation in wind direction (º) A 25 B 20 C 15 D 10 E 5 F 2.5 Table 6-7 Determination of the atmospheric stability class according to the wind direction and sun radiation Surface wind velocity, m/s <2 2–3 3-5 5–6 >6
Nighttime1 overcast
Daytime insolation Strong
Moderate
Slight
More than 50%
Less than 50%
A A-B B
A-B B B-C C-D C
B C C D2
E E D2 D2
F F E D2
D2 D2 D2 Night refers to the period ranging between 1 h before sunset and 1 h after sunrise. 2 The stability class D (neutral) must be used, regardless of wind speed, for overcast conditions during the day and at night. 1
The stability of the atmosphere is closely related to the lapse rates (Fig. 6-3). Thus, a superadiabatic lapse rate implies an unstable atmosphere, with vertical displacement of air that will tend to disperse any pollutant. With a dry adiabatic lapse rate the atmosphere is said to be neutral, which corresponds to an intermediate situation between instability and stability.
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If the lapse rate is subadiabatic or isothermal, the atmosphere is stable: the vertical displacement of air is prevented and any pollutant that is released will not be dispersed vertically. A similar situation is found in the case of thermal inversion. ______________________________________ Example 6-1 In 1953, members of the German expedition to Nanga Parbat measured the following temperatures: at 6,700 m, -9.7ºC and at 6100 m, -0.6 ºC. What were the atmospheric conditions that day? Solution The lapse rate is: 9.7 (0.6) 6,700 6,100
0.015 º C m -1 - 1.5 º C per 100 m
This lapse rate implies an unstable atmosphere. ______________________________________
Fig. 6-6. The ambient lapse rate and the behaviour of continuous plumes (adapted from[1]): a) looping, b) neutral, c) coning, d) fanning, e) lofting, f) fumigating and g) trapping.
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The stability of the atmosphere can be deduced from the appearance of continuous emissions (for example, the smoke from a smokestack (Fig. 6-6) [1]. With a superadiabatic lapse rate and an unstable atmosphere, there will be a “looping” plume caused by the passing of thermal streams surrounding descending air past the point of release. The stream of pollutants released undergoes significant vertical and horizontal dispersion, although the resulting large loops may carry the plume down to the ground, thus causing transiently high concentrations. If there is an approximately dry adiabatic lapse rate, the atmosphere is neutral and the plume tends to rise vertically through the atmosphere. However, when the wind velocity is greater than 10 ms-1 or there is significant cloud cover, the plume tends to “cone”. With a subadiabatic lapse rate the atmosphere is slightly stable; vertical motion is inhibited and the plume tends again to “cone”. If there is inversion (negative lapse rate), there is practically no vertical motion and the plume spreads horizontally. If there is an inversion layer, the behaviour of the plume will depend on the relative position of the release source. If the emission takes place above the inversion layer, it is said to be “lofting”: it has practically no downward mixing and the emission is dispersed without any significant concentrations at ground level. If the source is located in the inversion layer, the plume is said to be “fumigating”; the emission will be trapped above ground and high concentrations may be reached at ground level. 2.4 Relative humidity The humidity of air has a significant influence on the dispersion of a pollutant only if there is a reaction between this pollutant and the water vapour contained into the atmosphere (as in the case of hydrogen fluoride). Also humidity has another effect if the gas that is being dispersed is at a lower temperature than the atmosphere: in this case, the humidity modifies the heat balance of the masses of air and gas and may give rise to the condensation of water. The cool clouds of liquefied petroleum gas, for example, are white because of the condensation and entrainment of water droplets. 2.5 Units of measurement Concerning the concentration of gas or vapour in air, two types of units are usually used: ppm (parts per million by volume or ml m-3) and mg m-3. To convert ppm into mg m-3 (at a temperature T) the following expression can be used:
22.4 T 1 c 3 (6-2) M w 273 P mg/m ______________________________________ Example 6-2 Due to an accidental release of SO2, two workers are exposed to an atmosphere with an average concentration of 180 mg m-3. The temperature of the air is 22 ºC and the atmospheric pressure is 1 atm. Compare this concentration with the IDLH (immediately dangerous to life and health) for this substance (100 ppm). c ppm
Solution The molecular weight of SO2 is 64 kg kmol-1. Substituting in Eq. (6-2):
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c ppm
22.4 273 22 1 180 64 273 1
68.1 ppm
The concentration to which the two workers are exposed is lower than the IDLH ______________________________________ 3
DISPERSION MODELS
Dispersion models are sets of mathematical equations that allow, for a given scenario, the prediction of the evolution of a cloud of pollutant as a function of its position and the time. An important aspect is the duration of the release. According to the time during which the release takes place, emissions can be classified into two categories: instantaneous and continuous. 3.1 Continuous and instantaneous releases Several different criteria are used to establish whether a release is instantaneous or continuous. For example, - an instantaneous emission: when the time required for the cloud to reach a given location is longer than the duration of the release. An example would be the explosion of a tank containing pressurized gas; - continuous emission: when the duration of the release is longer than the time required by the cloud to reach a given location. An example would be the plume from a stack. Or - instantaneous emission: when the substance escapes into the atmosphere in less than a minute; - continuous emission: when the substance continues to escape for more than a minute. In practice, the division between continuous and instantaneous emissions is relatively arbitrary and subjective. In fact, most emissions would show an intermediate behaviour: the duration of the release (tr) is finite, a stationary state is reached during a certain period of time and finally the cloud disperses (Fig. 6-7). The time required by the cloud to reach an observer located at a distance x can be estimated as tx = x·ux-1, i.e. it is assumed that the cloud travels at the same speed than the wind. The time required by the cloud to disperse completely from the end of the release is td = x·uw-1. Therefore, the moment at which the exposure of the observer to the cloud will end can be estimated as texp = tr + x·uw-1. Furthermore, while for an observer located near the source the emission may be continuous, for another observer located farther away it may be considered instantaneous; the concept of continuous source depends on the time during which a steady concentration exists at a given location. The following quantitative criteria have been suggested [4, 5]:
a release is considered continuous at a downwind location x, when u tr t 2 .5 x
(6-3)
and a release is considered instantaneous when
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u tr d 0.6 x
(6-4)
A continuous emission originates an elongated cloud called a plume (Fig. 6-7) which reaches a stationary state when the amount of gas fed into the cloud equals the mass of pollutant dispersed towards the atmosphere.
Fig. 6-7. The evolution of a gas cloud originated by a continuous release.
The instantaneous emission of material forms a cloud (puff) that moves with the wind while it disperses into the atmosphere. Taking the surface corresponding to a given concentration, the volume surrounded by it decreases as the puff moves from the release point (Fig. 6-8). Correspondingly, the volume of the cloud (including the outer lower concentrations) increases as the puff moves with the wind. In the region that lies between the criteria expressed in Eqs. (6-3) and (6-4), the release is a transient one. Diverse criteria have been proposed to estimate the concentration in this region. For example, Britter and McQuaid suggest calculating the concentration using both continuous and instantaneous models and, because there is an overestimation when they are applied to a transient release, to take the smaller of the two values as the upper bound of concentration.
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Fig. 6-8. The evolution of a puff of material released instantaneously (the concentration is the same in the three cloud envelopes).
3.2 Effective height of emission Following the release, there can be some plume rise due to the momentum of the release and to the buoyancy effects originated by the density differences between the fluid released and the ambient air (Fig. 6-9). For hot plume effluents, such as smoke from a smokestack, buoyancy effects will be most important; as a rule of thumb, if the exit temperature is about 10 to 15 K higher than the air temperature, buoyant rise will be greater than that due to the momentum [6]. The influence of buoyancy will persist until the temperature of the plume falls to that of ambient air, essentially due to the entrance of entrained air; momentum will also be dissipated after a short time (typically 30-40 s in a smokestack).
Fig. 6-9. Plume rise due to momentum and buoyancy.
This plume rise ('H) should be added, even in the case of a ground level release, to the height of the source to obtain the effective height (H) of the release. It is this effective height that must be introduced into the appropriate equations to predict atmospheric dispersion. A number of equations, both empirical and theoretical, have been proposed to estimate the plume rise. Among them, the set of expressions proposed by Briggs [6] includes a specific
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calculation for diverse conditions. The Holland equation [7], which is easier to apply and is also widely used, is also a good way of obtaining the value of 'H: 'H
us d u
ª § Ts Ta ·º ¸¸» «1.5 .68Pd ¨¨ © Ts ¹¼» ¬«
(6-5)
where d is the diameter inside the stack (m) P is the atmospheric pressure (bar) Ts is the temperature of the gas inside the stack (K) Ta is the air temperature (K) us is the exit velocity of the gas inside the stack (m s-1) and u is the wind velocity (m s-1). The effective source height is then H
H ph 'H
(6-6)
where Hph is the physical height above ground of the source (m). If the velocity of the gas emitted from the source is low, the plume may be entrained into the wake originated by the stack. To avoid this, the following condition must be fulfilled: us t 1.5u. 4
DISPERSION MODELS FOR NEUTRAL GASES (GAUSSIAN MODELS)
Gaussian dispersion equations are used to describe the atmospheric dispersion of gases whose density is similar to that of air. Therefore, they can be applied to pollutants whose molecular weight is similar to that of air or to that of heavier-than-air gases that are sufficiently diluted at the source; this is often the case of the pollutants contained in combustion gases (SO2, NO2, CO2) or vapours from spilled liquids. Most of these equations are based on the general equation suggested by Pasquill [8] and Gifford [9]: dc dt
wx § wx · w § wx · w § wx · ¨ K x ¸ ¨¨ K y ¸¸ ¨ K z ¸ wx © wx ¹ wy © wy ¹ wz © wz ¹
(6-7)
This equation gives the variation of the concentration over time in the x direction (downwind) as a function of the variables in all directions of a three-dimensional space; Kx, Ky and Kz are the values of the diffusivity in the three directions. The Gaussian models are based on the following simplifying assumptions: - The mass flow rate of the emission is essentially continuous over time. - No material is removed from the plume by chemical reaction: all the mass emitted from the source remains in the atmosphere. - There are no gravitational effects on the material emitted. - The meteorological conditions are essentially constant over time during the period of transport from source to receptor. - The ground roughness is uniform in the dispersion area. There are no obstacles, such as mountains or buildings, and the ground is horizontal.
208
-
The cloud is transported by the wind. The time-averaged concentration profiles in the crosswind direction, both horizontal and vertical (perpendicular to the transport direction), can be represented by a Gaussian or normal distribution (Fig. 6-10).
4.1 Continuous emission A coordinate system with three orthogonal axes, x, y and z, is usually used to specify the concentration. The origin of the system is located at ground level at the position of the source (i.e. at the emission point for ground level emissions and under the point of emission for sources above ground). The x axis is orientated in the direction of transport of the centreline of the pollutant plume, the y axis is crosswind and the z axis is vertical.
Fig. 6-10. Horizontal and vertical Gaussian distributions of pollutant concentration.
The Gaussian model takes into account the following facts [6]: - The concentrations at a given point are directly proportional to the emission rate. - Parallel to the x axis, the concentrations are inversely proportional to wind speed. - Parallel to the y axis (crosswind) the concentrations are inversely proportional to the plume’s crosswind spreading, Vy. As Vy increases with the downwind distance from the source, the dispersion increases and the concentration correspondingly decreases. - Parallel to the z axis (crosswind, vertical) the concentrations are inversely proportional to the plume’s vertical spreading, Vz. As Vz increases with the downwind distance from the source, the concentration correspondingly decreases. All these relationships can be expressed by the following equation, which gives the concentration at a point x downwind, y crosswind and at a height z above the ground, resulting from an emission that takes place at an effective height H above the ground:
209
c( x, y , z , H )
2 ·º § y 2 · ª § z H 2 · § ¸ D exp¨ z H ¸» exp¨ 2 ¸ «exp¨ ¨ 2V 2 ¸» ¨ 2V ¸ « ¨ 2V 2 ¸ 2SuV yV z z z ¹ © ¹¼ © y¹ ¬ ©
m
(6-8)
where m is the mass flow rate of the emission (kg s-1). D is the ground reflection coefficient. It takes into account the eventual adsorption of the pollutant on the soil. For a complete adsorption D = 0 and for a complete reflection D = 1. Usually the conservative value D = 1 is assumed. This reflection contribution is considered to be emitted by a virtual source located at a height z = -H at the same horizontal distance x from the receptor.
Fig. 6-11. Dispersion coefficient Vy for a continuous release (rural area).
210
Fig. 6-12. Dispersion coefficient Vz for a continuous release (rural area).
Fig. 6-13. Dispersion coefficient Vy for a continuous release (urban area).
211
Fig. 6-14. Dispersion coefficient Vz for a continuous release (urban area).
To apply Equation (6-8), the values of the standard deviations or dispersion coefficients (also called Pasquill-Gifford dispersion coefficients) Vy and Vz are required. They are functions of the distance downwind from the emission point and the atmospheric conditions. They can be obtained from Figures 6-11 to 6-14 or can be calculated (for 100 m < x < 10,000 m) using the expressions shown in Table 6-8 [10, 11]. Table 6-8 Equations for the calculation of the dispersion coefficients Vy and Vz for a continuous emission (the distance x is in metres) [10, 11] Stability class Vy (m) Vz (m) Rural conditions A 0.22x(1+0.0001x)-0.5 0.20x B 0.16x(1+0.0001x)-0.5 0.12x C 0.11x(1+0.0001x)-0.5 0.08x(1+0.0002x)-0.5 -0.5 D 0.08x(1+0.0001x) 0.06x(1+0.0015x)-0.5 E 0.06x(1+0.0001x)-0.5 0.03x(1+0.0003x)-1 -0.5 F 0.04x(1+0.0001x) 0.016x(1+0.0003x)-1 Urban conditions A-B 0.32x(1+0.0004x)-0.5 0.24x(1+0.0001x)+0.5 -0.5 C 0.22x(1+0.0004x) 0.20x D 0.16x(1+0.0004x)-0.5 0.14x(1+0.0003x)-0.5 E-F 0.11x(1+0.0004x)-0.5 0.08x(1+0.0015x)-0.5
212
A distinction is made between urban and rural areas. In urban areas there is more turbulence due to the presence of buildings and to the somewhat warmer temperature (heat island effect); this is why in urban or industrial areas the real atmospheric stability is lower than that which would correspond to the prevailing meteorological conditions. In an urban area, a plume is dispersed more rapidly than in a rural area. Thus, different dispersion coefficients are used for urban and rural areas. Equation (6-8) can be modified for specific situations. Thus, for receptors at ground level (z = 0), it can be simplified to c ( x , y , 0, H )
2 § y2 · ¸ exp§¨ H ·¸ exp¨ ¨ 2V 2 ¸ ¨ 2V 2 ¸ SuV yV z y ¹ z ¹ © ©
m
(6-9)
For a ground level release (H = 0), it can be simplified to c( x , y , z ,0)
2 § y2 · ¸ exp§¨ z ·¸ exp¨ ¨ 2V 2 ¸ ¨ 2V 2 ¸ SuV yV z y ¹ © z ¹ ©
m
(6-10)
Equation (6-8) can be simplified further if the concentration at ground level is calculated on the plume’s central line (i.e. y = 0): c ( x , 0 , 0, H )
§H2 · ¸ exp¨ ¨ 2V 2 ¸ SuV yV z z ¹ © m
(6-11)
Equation (6-11) can be simplified again if the concentration at ground level is calculated for the plume’s central line, for a ground level release: c( x , 0,0,0 )
m
(6-12)
S uV y V z
If several plumes —from different sources— overlap, the resulting concentration of a pollutant at a given location will be the sum of the concentrations of this pollutant due to each plume. ______________________________________ Example 6-3 Near a village there are three factories, each with a smokestack that is releasing sulphur dioxide. You wish to know the concentration of this pollutant at a given point M (see the figure). Factories A and B authorize you to carry out measurements at their respective stacks, while Factory C simply declares that they have a very good gas treatment system so they are not polluting the environment. You measure, simultaneously, the SO2 emission from Stacks A (85 g s-1) and B (70 g s-1) and the concentration of SO2 at ground level at point M (1·10-4 g m-3). The effective heights of these stacks are HA = 60 m and HB = 72 m. You also manage to get hold of the following information concerning the stack at Factory C: height, 53 m; internal diameter, 0.8 m; smoke exit velocity, 9 m s-1; smoke temperature, 300 ºC. The meteorological conditions are as
213
follows: the day is overcast; u = 6 m s-1; atmospheric temperature = 20 ºC and atmospheric pressure = 1.013 bar. Assume it is a rural area. Determine the rate at which SO2 is being released from Factory C’s smokestack.
Fig. 6-15. Overlapping of several plumes.
Solution From Table 6-6, the stability class is D. The dispersion coefficients can be obtained from Table 6-8 or from Figures 6-10 and 6-11. Factory A:
Vy
0.08 500 1 0.0001 500
Vz
0.06 500 1 0.0015 500
0.5
39 m
0 .5
22.7 m
0.5
23.6 m
Factory B:
Vy
0.08 300 1 0.0001 300
Vz
0.06 300 1 0.0015 300
0.5
14.9 m
0.5
54 m
Factory C:
Vy
0.08 700 1 0.0001 700
Vz
0.06 700 1 0.0015 700
0. 5
29.3 m
The concentration of sulphur dioxide at point M (ground level) due to factory A can be calculated with Eq. (6-9):
c(500,50,0, 60)
§ § 50 2 · 85 60 2 · ¨ ¸ exp¨¨ exp 2 ¸ ¨ 2 22.7 2 ¸¸ S 6 39 22.7 © ¹ © 2 39 ¹
Concentration at point M (ground level) due to factory B:
214
6.86 10 5 g m -3
c(300,50,0,72)
§ § 70 50 2 · 72 2 · ¸ exp¨¨ ¸ exp¨¨ 2 ¸ 2 ¸ S 6 23.6 14.9 © 2 23.6 ¹ © 2 14.9 ¹
9.5 10 9 g m -3
Therefore, factory C should be the origin of:
1 10 4 6.86 10 5 9.5 10 9
3.14 10 5 g m -3
Calculation of the effective height of the stack of factory C (Eq. (6-5)): 'H
9 0.8 ª § 573 293 ·º 1.5 2.68 1.013 0.8¨ ¸» « 6 ¬ © 573 ¹¼
HC
53 3 56 m
3m
Therefore, the SO2 release rate of stack C is:
§ 130 2 · § 56 2 · m ¸ exp¨¨ ¸ ; m 104.5 g s -1 exp¨¨ 2 ¸ 2 ¸ S 6 54.1 29.3 © 2 54.1 ¹ 2 29 . 3 © ¹ ______________________________________ c( 700,130,0,56)
3.14 10 5
4.2 Instantaneous emission In the case of an instantaneous release, the concentration at a given location will depend on the position (x, y, z), on the effective height of the source (H) and on the time (t). The concentration can be estimated using the following expression: 2 ·º · ª § z H 2 · § ¸ «exp¨ ¸ D exp¨ z H ¸» ¸» ¨ 2V 2 ¸ ¸ « ¨ 2V 2 z z ¹¼ © ¹ ¹ ¬ © (6-13) where m is now the mass of pollutant released instantaneously. Again, the conservative value D = 1 is usually assumed. The dispersion coefficients Vx, Vy and Vz for the case of instantaneous release can be obtained from Figures 6-16 and 6-17 or can be calculated using the expressions shown in Table 6-9 [10, 11].
c( x, y , z , H ,t )
§ x ut 2 y2 ¨ exp 2V z2 2S 3 / 2 V xV yV z ¨© 2V x2 m
Table 6-9 Equations for the calculation of the dispersion coefficients Vx, Vy and Vz for an instantaneous release (distance x is in metres) [10, 11] Stability class Vx or Vy (m) Vz (m) A 0.18x0.92 0.60x0.75 B 0.14x0.92 0.53x0.73 C 0.10x0.92 0.34x0.71 0.92 D 0.06x 0.15x0.70 0.92 E 0.04x 0.10x0.65 F 0.02x0.89 0.05x0.61
215
Fig. 6-16. Dispersion coefficient Vy (=Vx) for an instantaneous release.
Fig. 6-17. Dispersion coefficient Vz for an instantaneous release.
216
The ground-level (z = 0) concentration can be obtained from:
c( x, y ,0, H ,t )
§ x ut 2 §H2 · y 2 ·¸ ¨ ¸ ¨ exp exp 2 2¸ ¨ 2V 2 ¸ ¨ 2S 3 / 2V xV yV z 2 V 2 V x z ¹ z ¹ © © m
(6-14)
And the ground-level concentration on the x axis (y = 0) is given by: c( x,0,0, H ,t )
2· § x ut 2 · § ¨ ¸ exp¨ H ¸ exp ¨ 2V 2 ¸ ¨ 2S 3 / 2V xV yV z 2V x2 ¸¹ z ¹ © ©
m
(6-15)
To establish the maximum concentration to which a person located at a given position can be exposed, Equation (6-13) can be applied to calculate the concentration at a distance x downwind (y = 0) from the source at ground level (z = 0) and at t = x u-1, i.e. when the centre of the cloud reaches that person:
c ( x , 0, 0 , x / u )
m 2S 3 / 2V xV yV z
exp
H2
(6-16)
2V z2
The time during which the cloud is passing over a given location can be estimated [2] by the following expression: tO
2V x u
§ c 2 ln¨¨ © clim
· ¸¸ ¹
(6-17)
where c is the maximum concentration, calculated with Equation (6-16), and clim is the limit concentration defining the beginning and the end of the cloud (for example, 1% of c). ______________________________________ Example 6-4 Due to a road accident, there is a leak of chlorine from a tank. Although the leak is quickly stopped, 4 kg of chlorine are released; the release can be considered instantaneous. Downwind, on the road, several cars have stopped at a distance of 200 m. Calculate the time required for the centre of the cloud to reach the cars, the maximum concentration at which the people in the cars will be exposed, and the time during which the cloud (defined by the isoplete corresponding to the ERPG-2) will be passing over those people. Meteorological conditions: u = 2 m s-1, T = 20 ºC, overcast conditions. Solution The time required for the cloud to reach the cars depends on the distance from the point of release and on the wind speed: 200 t 100 s 2
217
The meteorological conditions indicate that stability class D must be assumed (Table 6-7). The values of the dispersion coefficients can be obtained from Figs. 6-16 and 6-17 or from the expressions in Table 6-9:
0.06 200 0.92
Vx
Vy
Vz
0.15 200 0.7
7.9 m
6.1 m
Then, the maximum concentration at the location where the cars are can be estimated with Eq. (6-16): C ( 200,0,0,0,100)
4 2S
3/ 2
2
7.9 6.1
1.33 10 3 kg m -3
1330 mg m -3
457 ppm
For chlorine, ERPG-2 = 3 ppm. Then, the time during which the cloud will be passing over the cars is (Eq. (5-17)): 2 7.9 § 457 · 2 ln¨ ¸ 25 s 2 © 3 ¹ ______________________________________ tO
4.3 Short-term releases An emission into the atmosphere can neither be continuous nor instantaneous; for example, the release from a safety valve may last 10 or 15 min until corrective actions stop the emergency venting. Specific equations for this case were obtained [12, 13] from Equation (6-13) by considering a short-term continuous release as being an infinite number of overlapping elementary puffs. After considerable manipulation, the following expressions were obtained for the calculation of the maximum concentration at a given distance from the source:
ª § x «erf ¨¨ «¬ © 2V x
c max
cc 2
c max
ª § ut r cc «erf ¨ ¨ ¬« © 2 2V x
· § ¸ erf ¨ x ut r ¸ ¨ 2V x ¹ ©
·º ¸» ¸ ¹»¼
·º ¸» ¸ ¹¼»
ut r 2
for
x!
for
xd
ut r 2
(6-18)
(6-19)
where cc is the concentration obtained from Equation (6-8) for a continuous emission (ppm or mg m-3) x is the downwind distance to the receptor (m) Vx is the dispersion coefficient ( | Vy for short duration releases) (m) tr is the duration of the emission (s) u is the wind velocity (m s-1) erf is the error function (see Annex at the end of the chapter) (-).
218
5
DISPERSION MODELS FOR HEAVIER-THAN-AIR GASES
The dispersion of a material released into the atmosphere can be greatly influenced, especially in the first moments, by its density. Many hazardous materials that can be emitted in the case of an accidental loss of containment are heavier than air, for various reasons [2]: - Molecular weight. If the molecular weight of a material is greater than that of air, as that of chlorine or propane, the gas or vapour will be heavier than air. The perfect gas law shows that the density of a gas or vapour is proportional to its molecular weight, the relative density with respect to air being
U rel
Mw M wair
(6-20)
Temperature. If the material released is a cold gas, or it is cooled because of the evaporation from a pool of liquid, it may again be heavier than air. An example would be a release of liquefied natural gas. - Formation of an aerosol. If a pressurized liquid undergoes strong depressurization, flashing can occur and a mixture of aerosol and vapour will be released, with an average density higher than that of the vapour. Furthermore, the evaporation of these droplets will have a cooling action and will maintain the cloud at a lower temperature for longer. - Chemical reactions. Some reactive materials can undergo chemical reactions that modify their molecular weight during their dispersion into the atmosphere (polymerization, hydration, etc.) Common dense gases are chlorine, hydrogen fluoride, ammonia, propane and LNG. Chlorine behaves like a dense gas both in liquefied and gaseous releases. Its normal boiling temperature is -34 ºC, so in the release of liquefied chlorine its density is even higher; chlorine gas releases are heavier than air because of its molecular weight, 71. Hydrogen fluoride has a molecular weight of 20 and a normal boiling point of 20 ºC, but nevertheless it often behaves like a dense gas [14]. This is due to the fact that under certain circumstances it undergoes an oligomerization reaction, which has a formula of (HF)n; thus, saturated vapour of HF has a molecular weight of 78.2 at its normal boiling point [15]. Furthermore, in moist air an HF-H2O liquid phase is usually formed, and the density of the cloud depends on the atmospheric humidity, although generally it is higher than that of air until the dispersion reduces it. Ammonia has a molecular weight of 17 and a normal boiling point of -33 ºC. A release of liquid ammonia originates a dense gas cloud. If there is no liquid spray, the density of the cloud will increase when it mixes with air and it is likely to become a neutral gas; generally, if there is a fraction of liquid spray, the cloud will be dense. Propane has a molecular weight of 44 and a normal boiling temperature of -42 ºC, so releases, whether of propane gas or of liquefied propane, are always dense. Finally, LNG originates dense releases, not because of its molecular weight (which is approximately 16), but due to temperature effects (its normal boiling point is -161 ºC). If the material released is heavier than air, it will first descend to the ground and then spread radially. The steps of the dispersion process are very clear in the case of an instantaneous release (Fig. 6-18.): - Buoyancy step: the initial puff, which is usually assumed to be a cylinder, is pushed down by gravity and spreads radially as its height is reduced. This process lasts a few -
219
-
-
seconds and is relatively complex due to the influence of gravity, wind, heat transfer from the environment and its mixing with air. Stable and stratified dispersion: the cloud is dispersed by the wind. Due to the high density, a stratified flow exists on the ground, which reduces the entrance of air and the dilution of the cloud. Neutral dispersion: the cloud reaches a density similar to that of the surrounding air and Gaussian models can be applied. It is generally accepted that the transition from dense gas models to Gaussian models can be carried out at Ucloud-Uair < 0.1 kg m-3.
Fig. 6-18. Initial evolution of a cloud of heavier-than-air material.
The mathematical modelling of the dispersion of dense gases is much more complex than that of lighter-than-air or neutral gases, and a number of authors have proposed procedures to solve this problem. The existing models can be classified into the following categories: - Modified conventional models, which use models like that by Pasquill-Gifford and involve the empirical adjustment of several coefficients and parameters. The predictions of these models are not satisfactory. - Box models, in which the dispersion is simulated by assuming a source with a simple initial geometry (for example, a cylinder). The evolution of the cloud is calculated by solving the set of equations that describes the effect of gravity on height and diameter, the entrance of air into the cloud, the momentum, heat and mass balances, etc. These equations are integrated as a function of time by numerical procedures. These models require a great deal of information on the phenomenon that is difficult to obtain and are usually limited to flat, unobstructed terrain, without any slopes or buildings. - K-theory models, which are based on the solution of a set of equations that describe the conservation of mass, energy momentum, etc., for certain boundary conditions. Non-flat terrain and obstacles can be taken into account. The numerical solution is rather complex, as a set of partial differential equations in transient regime must be solved, and the system must be subjected to an appropriate discretization.
220
5.1 Britter and McQuaid model Generally, solving a problem of dense gas dispersion is complicated and laborious, and computer codes are often applied. However, a relatively simple model was proposed by Britter and McQuaid [4, 15], which will be briefly described here. The model consists of a set of empirical correlations based on dimensional analysis and experimental data obtained both from field tests and wind tunnel tests. It considers the concentration in a cloud to be a function of a set of dimensionless groups; atmospheric stability is not taken into account, as its influence was shown to be negligible; ground roughness is not considered either. Continuous and instantaneous ground-level releases are considered for flat terrain. To establish whether buoyancy effects have an influence and whether the dense gas model should be applied, the following criteria are proposed: For continuous releases 1/ 3
§ g o vo · ¨ 3 ¸ ©u D¹
t 0.15
(6-21)
For instantaneous releases
g oVo 1/ 2
(6-22)
t 0.20
uD
where u is the average wind speed at a height of 10 m, go is the initial corrected gravity, which influences the initial negative buoyancy of the cloud: go
g U U a
(6-23)
Ua
(U is the density of the released gas), uw is the wind velocity taken at z = 10 m and D is a characteristic dimension of the source. For continuous releases 1/ 2
D
§ vo · ¨ ¸ ©u ¹
(6-24)
and for instantaneous releases D Vo1/ 3
(6-25)
The authors propose the criteria expressed in Equations [6-3] and [6-4] to determine whether a release is continuous or instantaneous. 5.1.1 Continuous release By applying several approximations to simplify the problem, the following relationship is found:
221
x
vo /u 1/ 2
§ g 2v f ¨ o5 o ¨ u ©
1/ 5
· ¸ ¸ ¹
(6-26)
where x is the downwind distance at which the average concentration on the plume axis is c. By using experimental data, the authors obtained the plot in Figure 6-19, which represents Equation (6-26). In this figure, the area bounded by a dotted line indicates the region in which the method was validated with full-scale experimental data. The vertical dotted line on the left indicates the limit of passive dispersion (Gaussian model), established at ( g o2 vo u w5 )1/5 = 0.2. The concentration values obtained from this figure are time-averaged (approximately 10 min) values; due to the fluctuations, higher values can be reached over short times. According to Britter and McQuaid, short-duration concentrations can be 1.6 times higher than c.
Fig. 6-19. Britter-McQuaid correlation for dense gas continuous release. Taken from [4], by permission.
222
The area bounded by a given concentration (time-averaged) can be estimated [4] from the following expressions (see Fig. 6-20), obtained from full-scale results: Lu
0.5 D 2 lb
(6-27)
Lh
Lho 2.5 lb1/ 3 x 2 / 3
(6-28)
Lho
D 8 lb
(6-29)
where lb = vogou-3. However, this is a conservative estimate of the area. A more realistic estimate, which would be consistent with experimental observations, could be achieved by connecting the cloud boundaries at 2xn/3 (xn being the downwind distance to the relevant concentration) with xn. Finally, the height of the plume, for a given concentration, can be estimated, from the conservation of mass, as Lv
vo uLh
(6-30)
Fig. 6-20. Area bounded by a given concentration. Taken from [4], by permission.
5.1.2 Instantaneous release From a dimensionless analysis, the following relationship is found: x Vo1/ 3
1/ 2
§ g V 1/ 3 · f ¨ o 2o ¸ ¸ ¨ u ¹ ©
(6-31)
Figure 6-21 shows the plot of the correlation based on Equation (6-31); again, the fullscale data region and the limit of passive dispersion are indicated. The curves correspond to the values indicated for c/co, c being the maximum concentration along the downwind path of the cloud centre.
223
Fig. 6-21. Britter-McQuaid correlation for dense gas instantaneous release. Taken from [4], by permission.
To estimate the arrival time of a cloud (for a given concentration contour) at a point located at a distance x from the source, an advection velocity of 0.4u is assumed (from fullscale experimental data). The cloud radius at that point is
R
R
2 o
1.2 g oVo 1/ 2 t
1/ 2
(6-32)
and the arrival time can be obtained from x
0.4ut Ro2 1.2 g oVo 1 / 2 t
1/ 2
(6-33)
Finally, the departure time of the cloud can be estimated from the following expression: x
0.4ut d Ro2 1.2 g oVo 1 / 2 t d
1/ 2
(6-34)
224
5.1.3 Finite duration release As stated in Section 3.1, in the in-between region of the criteria corresponding to continuous and instantaneous releases, the concentration should be calculated for both cases, taking as the upper bound of concentration the smaller of the two values. Non-isothermal releases Heat transfer from the environment to a cold release will reduce the negative buoyancy and the horizontal cloud spreading. In the case of non-isothermal releases, the value of the concentration in the plume must be corrected [4] to apply the Britter-McQuaid correlations. The initial volume in a nonisothermal release is vo and, after mixing with a volume of air va, the related concentration is vo/(vo + va). However, the corresponding initial volume in an isothermal correlation must be vo(Ta/To). Therefore, the effective concentration that must be used in the calculations is c
c
§T · c 1 c ¨¨ a ¸¸ © To ¹ ______________________________________ Example 6-5.
(6-35)
Full-scale experiments were carried out [14] with liquefied propane releases. The data below correspond to one of the experiments [4]. Liquid spill rate: 0.065 m3 s-1 Duration: 270 s Wind speed: 5.2 m Density of liquefied propane: 553 kg m-3 Density of propane gas at boiling point (-42ºC): 2.32 kg m-3 Ambient air temperature: 20 ºC Ambient air density: 1.21 kg m-3 Calculate the maximum distance downwind at which the lower flammability limit (LFL = 2.1% in volume) was reached. Solution The release flow rate was vo
0.065 m 3 s -1 553 kg m -3 2.32 kg m
-3
15.5 m 3 s -1
The buoyancy of the initial release was go
§ 2.32 kg m -3 1.21 kg m -3 · ¸ 9.81 m s -2 ¨ -3 ¨ ¸ 1.21 kg m © ¹
9 m s -2
Characteristic dimension of the source (Equation (6-24)):
225
D
§ 15.5 m 3 s -1 · ¸ ¨¨ -1 ¸ © 5.2 m s ¹
1/ 2
1.73 m
Equation (6-21) must be applied to see whether the behaviour is that of a dense gas, as follows: § 9 m s -2 15.5 m 3 s 1 · ¨ ¸ ¨ 5.2 m s -1 3 1.73 m ¸ © ¹
1/ 3
0.83 ! 0.15 dense gas.
Now Equation (6-3) will indicate the distance from the source up to which the release may be considered to be continuous: 5.2 m s -1 270 s t 2.5 x Therefore, the release can be considered to be continuous up to x d 562 m
The concentration of propane in the cloud must be corrected for the non-isothermal condition (Equation (6-35)): c
0.021 § 293 · 0.021 1 0.021 ¨ ¸ © 231 ¹
0.017 m 3 /m 3
Now the dimensionless groups corresponding to Figure 6-19 must be calculated: § 9 215.5 · ¸ ¨¨ 5 ¸ © 5.2 ¹
1/ 5
0.80
and x 15.5/5.2 1 / 2
x 1.73
As co = 1.0, c/co = 0.017. Therefore, from Fig. 6-19: x 197 x 340 m. 1.73 ______________________________________
226
6
CALCULATING CONCENTRATION CONTOUR COORDINATES
The prediction of the contour, for a given concentration, of a plume caused by a release into the atmosphere of substances either heavier or lighter than air is very useful in the prediction of plume behaviour. A set of analytical equations has been proposed [16] to achieve this. The procedure is based on the Ooms equations. 6.1 The Ooms integral plume model According to Ooms [17] and Ooms et al. [18], the profiles of the different parameters (velocity, density and concentration) that define a plume on a plane perpendicular to its axis at a given point can be established, if cylindrical geometry and a Gaussian shape are assumed, by the following expressions:
ª r2 º u s (r ,T ) ucosT u *s exp « 2 » «¬ b »¼
(6-36)
U a U s*exp«
r2 º » «¬ O2 b 2 »¼
(6-37)
ª r2 º c s r ,T c s*exp « 2 2 » «¬ O b »¼
(6-38)
U s (r , T )
ª
This model, as well as others derived from it, allows u s* , U s* , cs* and bs to be calculated from the momentum, heat and mass balances. However, these models do not allow the contours corresponding to constant concentration (isoplete) values to be determined. In the following paragraphs, we describe a method for carrying out this calculation. 6.2 Determining concentration contour coordinates A point located on the plume axis at a distance s (measured on the axis) from the origin of the plume is defined by a pair of coordinates (x, z) (see Fig. 6-22). After a series of geometric considerations [17], the points (xt, zt) and (xb, zb) located at a distance r on a line perpendicular to the plume axis at the point (x, z) can be determined.
Upper isoplete The value of zt can be calculated from the following expression: zt
z
O 2 b 2 x , z c * x, z ln cloc 1 tg 2T
(6-39)
and, afterwards, the value of xt can be obtained from
227
zt z xt x
cotgT
(6-40)
All the parameters appearing in these expressions are known: the values of c*(x, z), b(x, z), T, x and z are determined when the Ooms model is solved numerically to calculate the plume axis; O is a known constant ( 1.35 ) and cloc is the concentration corresponding to the isoplete, i.e. a known and fixed value.
Fig. 6-22. Plume isopletes in the dispersion of a gas into the atmosphere.
Lower isoplete Following the same procedure, first the value of the coordinate zb can be calculated from zt
z
O 2 b 2 x, z c * x, z ln cloc 1 tg 2T
(6-41)
and then the coordinate xb can be calculated using the following expression:
zb z xb x
cotgT
(6-42)
To carry out these calculations, the Ooms model must previously be solved to obtain the values of b and c* along the path of the plume and those of the coordinates (x, z) of the plume axis. The simplest option is to use a given calculation model with proven reliability. The example included in the following paragraphs was solved using the Ooms model incorporated into the DEGADIS calculation code [19].
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______________________________________ Example 6-6 The case analyzed is the release of propylene from a pressurized storage tank. Two different wind speeds are considered, as well as three release rates. The data concerning the emission conditions and the meteorological situation are included in the following table. Characteristics of propylene relief device discharge Parameters Values used Meteorological conditions Wind velocity (m s-1) at z = 10 m 1; 5 Surface roughness (m) 0.01 Stability class E Ambient temperature (K) 293 Atmospheric pressure (atm) 1 Relative humidity (%) 60 Emission conditions Molecular weight 42 LFL (%) 2 Temperature (K) 323 Specific heat (J kg-1 K-1) 1620 Release rate (kg s-1) 1.88; 10.2; 18.22 Release height (m) 30 Release diameter (m) 0.25
Solution The results obtained are plotted in Figures. 6-23 and 6-24. Figure 6-23 shows the plume shape as a function of the discharged mass flow rate. Two isopletes were calculated, which correspond to the lower flammability limit (LFL) and half this value (0.5 LFL). It can be observed how, as the release rate decreases, the influence of the momentum decreases and the plume inclines more significantly in the direction of the wind. This allows us to study whether danger may arise due to the nearness of other facilities.
Fig. 6-23. Dispersion of a propylene release for three different mass flow rates.
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Figure 6-24 shows the influence of the wind speed for a given release rate. A low wind speed allows the LFL isoplete to reach greater horizontal distances, while a higher wind speed disperses the propylene cloud faster.
Fig. 6-24. Dispersion of a propylene release for two different wind velocities.
Again, these results are very useful in predicting the meteorological conditions in which a controlled discharge can be carried out without the risk of a flash fire occurring. ______________________________________ 7 DISPERSION OF DUST
A release may contain dust or aerosol particles entrained by a gas or a vapour. Once these particles have been emitted into the atmosphere, they will move as a plume or a puff. If the particles are heavy, i.e. if they are large or have a high density, they will settle relatively quickly and atmospheric dispersion will be limited. However, if they are small and light, atmospheric turbulence will hinder any deposition and the plume will be dispersed as if it was a gas. There are a number of examples of airborne dust or aerosol particles; amongst them we can cite several cases of atmospheric dispersion of soy dust (giving rise to allergy outbreaks) and of airborne transmission of pathogenic agents. The deposition of particulate material from a gas cloud is a field of great uncertainty that has been studied by Clancey [20]. The terminal velocity of a spherical particle (viscous flow) may be calculated from the force balance by applying Stokes’s law: ut
d p2 g U p U a
(6-43)
18P
This expression is valid for Reynolds numbers Re<1. This equation really applies to spherical particles; although for non-spherical particles a correction coefficient should strictly be applied, the error is negligible for our purposes.
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For particles with a diameter of less than 20 Pm or with a terminal velocity of less than 1 cm s-1, the effects of atmospheric turbulence are still more significant than those due to gravity [20], although other authors [2] limit this condition to particles whose diameter is less than 10 Pm. Therefore, the effect of dry deposition on the concentration in the plume can be neglected, in contrast to the effect due to dispersion in the atmosphere. Besides dry deposition, airborne particles may be separated from the atmosphere by wet deposition, i.e. by the action of falling raindrops. However, it has been proved [21] that only for submicron particles (which can be entrained by turbulence into the rain-forming layers) is wet deposition much greater than dry deposition. 8
ATMOSPHERIC DISPERSION OF INFECTIOUS AGENTS
Few papers have been published on the atmospheric dispersion of pathogenic agents, although the potential consequences of these phenomena mean that they may be extremely important both socially and economically. Clouds of viruses, and also, in some cases, clouds of bacteria, may be formed as a result of different events, such as a release from an industrial plant; this was the case of an accidental emission of foot-and-mouth disease in Denmark in 1966, which was probably from a plant in which a vaccine had been produced without an adequate air-filtration system. In 1979, an explosion led to the emission of an aerosol of anthrax (a very dangerous bacterium, which has been considered a possible agent of biological warfare) at a military biological research institute in Sverdlovsk (Ekaterinburg), Russia; due to the outbreak, some 70-80 people died (see Example 6-9 at the end of this chapter). However, the spreading of a virus may simply be due to an emission caused by infected persons or animals. For obvious reasons, this is relatively frequent in farms, with severe consequences for neighbouring farms. Research of a number of outbreaks occurring in different countries has shown that the atmospheric transport of viruses is a mechanism by which epidemics of some diseases can propagate, not only to premises that are relatively near, but also over large distances over the sea and even from one country to another. There are two significant differences between airborne virus dispersion and gas dispersion. The first is that, in the case of a virus, the dose may be enough to infect an animal or a person even over very large distances if the atmospheric conditions are favourable. On the other hand, viruses may lose their virulence over time, depending on the environmental conditions. Generally speaking, two different scenarios may be considered: firstly, the airborne transport of virus over distances of up to about 10 km, which is relatively common, and secondly, transport, sometimes over sea, over much greater distances (up to 100 km), which is less frequent. The first scenario is considered here. 8.1 Emission source Concerning the emission source, in the case of persons or animals a virus can be excreted for a period of a few days, at a rate that depends on the species of animal, the period of infection and the type of virus. Emission can therefore be substituted by a certain flow rate of air corresponding to the respiratory rate of the infected subject, which contains a given concentration of the virus. The initial temperature of this air is body temperature (approximately 38 ºC) but it can be assumed that it rapidly cools to room temperature. The duration of the emission is usually a few days, much longer than in the case of industrial accidental releases. In the case of a release from an industrial facility, a given mass of emitted
231
air and a given concentration of pathogenic agent must be considered; depending on the type of event, a gas flow-rate during a certain time, or even an instantaneous emission, should also be considered. 8.2 Dispersion of airborne pathogenic agents Pathogenic agents are transported as an aerosol, the granulometric distribution of which depends on the source. For excretions from persons or animals, this is approximately as follows: 60% of particles with a diameter of approximately 6 Pm, 25% in the range from 3 to 6 Pm and 15% with a diameter of less than 3 Pm. In the latter size range, dispersion is indistinguishable from a gas, but the eventual deposition of these particles on the ground must be taken into account. It may be generally accepted that an airborne virus is transported in particles whose average diameter is approximately 6 Pm. For this diameter, Equation (6-43) gives a terminal velocity of 0.001 m s-1, which corresponds to a Reynolds number of approximately 3.5·10-4. Therefore, taking into account the low value for terminal velocity, it may be assumed that due to the effect of atmospheric turbulence the deposition or transfer of the virus to the ground is negligible. Furthermore, as stated in Section 6, for 6 Pm-diameter particles wet deposition is negligible, especially for distances of up to 10 km from the source. Although several authors have concluded that there is a significant relationship between rain and the spreading of disease, it seems that this is essentially due to the fact that during precipitation very favourable conditions for airborne dispersion of viruses exist: relative humidity is very high (and, therefore, virus activity and potential infectivity are high), the lower layers of the atmosphere are relatively stable and wind speed is adequate for transporting virus particles. Of course, all these considerations concern only the aforementioned 6 Pm particles; virus particles surrounded by mucus or attached to dust may be deposited when the air is calm or may be washed out by rain. Atmospheric humidity is an important parameter, as it has a significant influence on virus survival after emission. In order for a virus to remain active, the relative humidity must be higher than 60%; under these conditions, a virus may be active for many hours, but at lower humidities it rapidly becomes inactive. In dry climates, therefore, humidity constitutes a significant limitation to the atmospheric propagation of epidemics. Furthermore, bright days imply the action of ultraviolet light on the virus, which may be killed as a result. As with a gas, the emission of an infectious agent will give rise to a plume in the direction of the wind, with a given horizontal and vertical dispersion. The shape and concentration of the plume will be a function of the wind velocity and meteorological conditions, especially atmospheric stability. Taking into account the usual low virus concentration (in mass per unit volume) in the air emitted from the source, it may be accepted without introducing any significant error that the emission is a neutral gas (with the same density as air). Therefore, it may be assumed that the plume follows a Gaussian distribution and that the equations included in previous sections apply. Eventually, a decay term should be added to take into account the killing action of the ultraviolet light. 8.3 Epidemics: dispersion of airborne viruses A specific aspect of the airborne transmission of pathogenic agents is the atmospheric dispersion of viruses, which can be important in the transmission of certain epidemics between farms.
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The probability of the airborne transmission of a disease is related to the total amount of a virus inhaled by each animal in the target farm and to the minimum infective dose required to produce infection. The amount of virus depends on the quantity released at source, its survival in the atmosphere and the amount inhaled in the period during which the farm is inside the plume. The total dose of virus inhaled by an animal recipient in TCID50 units can be obtained from the following expression:
Dose
c Vresp t exp
(6-44)
where c is the virus concentration in ID50 m-3 (ID50 is defined as the dose that will infect 50% of test animals), Vresp is the respiratory rate of each animal in m3 min-1 and texp is the duration of exposure in min. In the case of epidemics, the periods during which the source is emitting are much longer than in the case of toxic gas releases. Although wind direction tends to vary during the day, if a predominant direction is maintained the plume may cover a given location for many hours and allow indoor concentrations to reach levels that are much closer to those found outdoors; in some cases, the only result of bringing in cattle immediately was to delay the infection for one or two days. ______________________________________ Example 6-7 Between 6 January and 6 February 1967, an outbreak of foot-and mouth disease (FMD) was confirmed in 29 farms in Hampshire [21]: a slaughter policy was applied to 2,774 cattle, 414 sheep, 4,708 pigs and 6 goats. The disease initially spread from the local abattoir, where pigs had fed on infected meat. In an analysis of this case carried out by Sellers and Forman, four farms were selected (numbers 2, 5, 11 and 12 respectively) in which the airborne virus had probably been the only mechanism of infection. By taking into account the date of detection of the disease, the incubation period and the direction of the wind, they were able to select the days on which virus transport could have taken place. By considering the information on meteorological conditions and the source, the following values were established [21] concerning two representative situations: a) Atmospheric stability class D Wind speed: 5 m s-1 Relative humidity: 70% Cloud cover: 70% Air temperature: 4 ºC Ground roughness: 3 cm (open country) Release flow rate: 0.1 m3 min-1 Release concentration: 0.16 ID50 m-3 Release temperature: 15 ºC b) The same values as for (a) but with the following changes: Atmospheric stability class: E Wind speed: 2.5 m s-1 Cloud cover: 50%. More information on the scenario has been included in the table below. An excretion rate of 5.8x106 ID50 per day from an infected pig and 1.22x105 ID50 per day from an infected cow were proposed; 4x108 ID50 per day per infected pig has also been proposed (both values should be used in the calculations) [22, 23]. For FMD, the minimum
233
dose required to infect an animal has been reported to be 10 ID50 for cows and sheep and 400 ID50 for pigs. The respiratory rates for an adult pig and cow are 0.025 and 0.1 m3 min-1 respectively. Information on the infected farms Origin Infected farm Distance, km Abattoir Abattoir Abattoir Abattoir
2 5 11 12
6 2.5 3 10
Hours of wind > 2.5 m s-1 48 48 48 48
Hours of wind > 5 m s-1 36 36 36 36
Analyze the atmospheric dispersion to confirm whether farms 2, 5, 11 and 12 could have been infected by the airborne virus.
Solution The virus that causes FMD, one of the most important of animal diseases, has been shown to be stable in aerosols from most natural fluids at relative humidities above 55%; if one considers that the delay between the excretion of the virus and its propagation to the farms was less than one hour, any loss of activity can be discounted in the calculations. The ALOHA code was used to perform the simulation [24, 25]. The distance at which a given concentration is reached is —for a given set of meteorological conditions— a function of the release flow rate (which is, of course, a function of the number of animals infected at the source). All the remaining variables are constant.
Fig. 6-25. Dependence of plume concentration at a given point (on the plume centreline downwind, at 2.5 km from the source) on the release flow rate (atmospheric stability class D; wind velocity 5 m s-1). Taken from [26], by permission.
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Figure 6-25 shows the variation in concentration at a given point, on the plume centreline downwind, as a function of the release flow rate; the variation follows a linear trend, as could be inferred from the form of Equation (6-12). Figure 6-26 shows the variation of the concentration at a given distance from the source as a function of the wind velocity. As the velocity increases, the turbulence in the atmosphere also increases and the dispersion is greater; this means that the plume is diluted and the concentration decreases. The variation has been plotted for three atmospheric stability classes (D, E and F) in the range of wind velocities in which they can exist.
Fig. 6-26. Influence of wind velocity on virus concentration at a given distance downwind (2 km) for different atmospheric stability classes. Taken from [26], by permission.
The doses estimated from the simulation for the different farms, for the two meteorological conditions considered, are given in the Table. Estimated mean doses* (ID50) of FMD per cow for atmospheric stability classes D and E Farm Atmospheric stability class D Atmospheric stability class E 2 5 11 12
0.34/12.3 1.20/82.8 0.92/62.8 0.14/9.4
0.80/55.3 2.50/172.5 2.00/135.6 0.36/25.1
* Doses predicted by assuming virus excretion rates of 5.8·106/4·108 per infected pig.
If we consider the highest excretion criterion, all the farms could have received an ID50 that would have been enough to infect animals. ______________________________________
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9
ESCAPING
Although the size of the cloud, its duration and its concentration are essential in establishing the dose received by people exposed to it, in a real case two other aspects must be taken into account: escape and possible protection by sheltering. The influence of escaping can be significant in certain cases. It will affect people outdoors and people inhabiting in poorly constructed houses with a very high air renewal rate (this was the case at Bhopal, where people lived in huts). Its influence on the concentration and exposure time can be estimated without any difficulties; an escape velocity of 4 m·s-1 is usually assumed. It must be mentioned that in such a situation people should escape following the crosswind direction (in practice, this is not that easy to do), which is the shortest way to leave the cloud. If people escape in the wind direction, they will move with the cloud, i.e. inside the cloud, thus increasing the exposure time and the dose. It must be taken into account that people who are running away breathe at a higher speed and thus inhale a greater amount of the material in the plume. In the case of Bhopal, people escaped towards two large hospitals following the same direction as the wind and keeping inside the plume. This fact certainly increased the consequences of the accident. 10 SHELTERING
The concentration of a pollutant in a closed house with a given ventilation rate will be different to that outdoors. If the pollutant is only present in the outdoor atmosphere for a certain period of time, as is the case in accidental releases, the concentration inside a building will obviously be a function of the concentration outdoors and of the velocity at which the air in the building is renewed (influx of fresh air by ventilation). The concentration indoors, at least during a certain time, will be lower than it is outdoors (especially if ventilation is reduced). A closed building is, therefore, a protection in the event of a toxic release. If, furthermore, people inside the building act to reduce ventilation (closing doors and windows, stopping air renewal systems, plugging any gaps, etc.), the protection by sheltering can be decisive in the case of relatively short emergencies —as accidental releases usually are— and therefore it is a measure often applied. Instead, evacuation implies that people leaves the buildings to flee, thus being exposed to the toxic cloud. 10.1 Concentration indoors The evolution of the concentration indoors can be calculated as a function of concentration outdoors by applying relatively simple mathematical models. Three cases should be analyzed: continuous emission, temporary release and instantaneous release.
10.1.1 Continuous release If the outdoor concentration co is constant, the indoor concentration will increase continuously up to a certain value due to the air that enters the building by ventilation. Inside a building with a volume V and a ventilation rate v, the concentration of pollutant can be found by solving the mass balance as expressed in Fig. 6-27. v is the ventilation rate, the number which indicates how many times per hour the entire contents of a room will be provided with fresh air. It is a function of the building type, and it is constituted by the fresh air that enters through joints and seams due to the wind and the temperature difference between the outside and inside. Usually, it is assumed that doors and windows are closed; if they are open, the
236
ventilation rate will be much higher; it is usually assumed also that there is no mechanical ventilation. If the ventilation rate is unknown, a defect value of 1 h-1 can be assumed (for a building with the windows completely open it can range between 10 h-1 and 15 h-1).
Fig. 6-27. Concentration indoors and outdoors.
The general mass balance can be expressed by the following equation [27]:
v co
v ci
dM poll dt b
v ads ci
(6-45)
where Mpoll is the mass of pollutant inside the building. By integrating Eq. (6-45) with the limiting condition ci = 0 as tb = 0, the following expression is obtained that relates the values of indoor and outdoor concentrations:
ci t b co
v v v ads
ª « 1 «1 « § v v ads t b « exp¨ V © ¬
º » » ·» ¸» ¹¼
(6-46)
In this equation, vads has been introduced to account for the rate at which the substance is eventually eliminated inside the building by various mechanisms: adsorption on the walls, respiration, etc. For reactive gases (SO2, NO2) the adsorption rate can be assumed to have an approximate value of 0.5 h-1, and for inert gases it can be assumed to be zero. If this rate is unknown or is taken to be negligible, Eq. (6-46) can be simplified slightly and, for sufficiently long periods, indoor concentrations can be equal to outdoor concentrations (Fig. 6-28); this is not usually the case in real emergency situations. Furthermore, often the concentration outdoors will not be constant due to fluctuations of the wind direction (which depend on the atmospheric stability conditions) which will modify the position of the plume, but this effect usually is not corrected in this type of calculations.
10.1.2 Temporary release A temporary or transient release is a constant emission lasting a short period of time which is longer than the time required for the gas to travel to that building. For a temporary release, the outdoor concentration changes as follows: co = 0 before the arrival of the cloud at the building (tb < 0,);
237
co = co(x) from the arrival of the cloud until the moment at which it starts to disappear (0 < tb < tO), co = 0 from the moment at which the cloud starts to disappear (tb > tO) where tO is the time during which the plume moves over the building.
Fig. 6-28. Continuous source. Evolution ci/c0 ratio as a function of time for two different situations.
Therefore, two different situations must be taken into account: 0 d t b d t O and tb > tO . a) 0 d t b d t O . In this case, the solution of the pollutant mass balance is the same as for a continuous external source; the relationship between indoor and outdoor concentrations is given by Eq. (6-41). b) tb > tO . From the instant tb = tO , the cloud has already passed and the indoor concentration will start to decrease due to the ventilation that will start to introduce pollutantfree air into the building. The integration of Eq. (6-45) with the limiting condition ci(t) = ci max as tb = tO gives the following expression, which relates indoor to outdoor concentrations: ci t b co
ª § v v ads t O · º v ª v v ads t b º «exp¨ ¸ 1» exp « » v v ads ¬ © V V ¬ ¼ ¹ ¼
(6-47)
The variation of co and ci as a function of time according to this equation can be seen in Fig. 6-29 for a given case. The outdoor concentration can be considered to be approximately constant during a time tO , and when the passage of the cloud finishes it decreases quickly. The indoor concentration will increase continuously from zero to a maximum value (at t = tO ) while the cloud passes over the building, and afterwards it decrease gradually to zero due to the entrance of fresh air by ventilation. The longer the duration of the release, the higher will be the concentration reached indoors. In this case, measures to reduce the ventilation rate should be applied (taping seams and joints, shutting-off mechanical ventilation) in order to keep a lower concentration indoors, this implying a lower dose for the people inside the building.
238
Fig. 6-29. Temporary source. Evolution of co and ci as a function of time.
10.1.3 Instantaneous release This case is more complicated than the previous ones, because the value of co changes continuously as a function of time, first increasing up to a maximum value and then decreasing, while the “puff” of released pollutant passes over the building. The following relationship applies for the concentration at a distance x from the instantaneous source located at a height H: co x, y, z , t mFx x, t b Fy y, t b Fz z , t b
(6-48)
where [3]
Fx x, t b
§ x ut b 2 · ¨ ¸ exp ¨ 2V 2 ut ¸ 2S V x2 (ut ) x b ¹ ©
(6-49-a)
Fy y, t b
§ · y2 ¨ ¸ exp 2 2 ¨ ¸ 2 V ut 2S V y (ut ) y b ¹ ©
(6-49-b)
Fz z , t b
ª § z H 2 «exp¨ 2 2S V (ut ) ¬« ¨© 2V z ut b
1
1
1
2 z
· § z H 2 ¸ exp¨ ¸ ¨ 2V 2 ut z b ¹ ©
·º ¸» ¸ ¹¼»
(6-49-c)
In these expressions, the values of the dispersion coefficient V are a function of the distance (utb) and H is the source height. The differential equation that represents the pollutant balance is also in this case the same as Eq. (6-45), which must fulfill the condition ci = 0 at tb = 0. This equation can be integrated with considerable manipulation, using the Laplace transform, and the following expression obtained:
239
ci t b
v § mFz Fy ¨ Q p 1 ¨ V L v v ads ¨ 2u ¨ p V ©
· ¸ ¸ ¸ ¸ ¹
(6-50)
where Q p
§V2 İxp¨ x2 ¨ 2u ©
§ xu · ¨¨ p 2 ¸¸ Vx ¹ ©
2
· ¸erf §¨ V x §¨ p xu ·¸ ·¸ ¸ c ¨ u 2 ¨© V x2 ¸¹ ¸¹ © ¹
(6-51)
The final equation for ci(tb) that results from the resolution of the inverse Laplace transform is
ci (t b )
v mFy Fz §K2 x 2 ·¸ § v v ads V exp¨ exp¨ 2 ¸ ¨ V 2u 2V x ¹ © © 2
§ ut b K ·¸º ·ª § K · ¸ erf ¨¨ ¸ «erf ¨ » 2 ¸¹¼» ¹ ¬« © 2 ¹ ©V x 2
(6-52)
where
K
§ v v ads · 2 ¨ ¸V x ux © V ¹ uV x
(6-53)
A typical variation of the values of co and ci for the case of instantaneous source can be seen in Figure 6-30.
Fig. 6-30. Instantaneous source. Evolution of c0 and ci as a function of time.
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Usually, if windows and doors are closed, the natural ventilation rate is relatively small and the value of the concentration indoors is much lower than that of the concentration outdoors until most of the “puff” has passed over the building. It has been suggested that in rooms located at the downstream side of a building the ventilation rate is 0.6 times that corresponding to the rooms at the wind side; this means that in upstream rooms the concentration is always smaller and that they offer a better protection. Both in temporary and instantaneous releases, from a certain moment the concentration indoors will be higher than the concentration outdoors (see Fig. 6-30): the passage of the cloud has practically finished, but due to the low ventilation rate the concentration indoor decreases slowly. From this moment, people indoors should leave the building (the end of the emergency should be communicated to the population); the windows and doors should be open to allow the entrance of fresh air. After this, some pollutant could still remain in closed rooms or cellars. 10.1.4 A simplified approach Davies and Purdy [28] have proposed a simple method to describe the evolution of the concentration indoors, assuming a top hat concentration distribution of the cloud outdoors. These authors propose the following expression to calculate the concentration indoors as a function of time: ci (t )
co >1 exp nt b @
(6-54)
where n is the number of air changes per hour and tb is also expressed in hour. ______________________________________ Example 6.8 There is a release from a hole in a pipe, at ground level: 4 kg s-1 of ammonia gas at 20 ºC during 30 minutes. At 350 m downwind there is a house, with a ventilation rate of 1.2 h-1. The adsorption rate can be assumed to be approximately 0.2 h-1. Ambient temperature is 20 ºC, the sky is completely overcast, wind velocity is 4.5 m s-1, and the surroundings are rural. Calculate whether the maximum concentration indoors will reach the value of IDLH for ammonia (300 ppm). Solution First of all the concentration of ammonia outdoors must be estimated. Taking into account the release time and the distance, for a ground level release and a point on the plume’s central line, Eq. (6-12) can be applied. From Figs. 6-11 and 6-12 the approximate values of the dispersion coefficients are obtained (stability class: D): Vy = 27 m and Vz = 17 m. Therefore: co ( 350,0, 2,0 )
4 kg s 1 S 4.5 ms 1 27 m17 m
0.00062 kg m -3
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The maximum value of the concentration indoors will correspond to the instant t = tOSo, both Eqs. (6-46) or (6-47) can be applied:
1.2 h V m h 1.2 0.2 h -1 V m 3 h 1 -1
ci
0.00062 kg m -3
3
1
ª º « » 1 » «1 « § 1.2 0.2 h -1 V m 3 h 1 0.5 · » ¸¸ » « exp¨¨ V m 3 h 1 © ¹ ¼» ¬«
0.00026 kg m -3 380 ppm. Therefore, the IDLH value will be exceeded indoors. If Eq. (6.54) is applied: ci ( 0.5 h )
0.00062 kg m -3 >1 exp (1.2 0.2 0.5@ 0.0003 kg m-3 440 ppm
(the adsorption rate has been included in n). ______________________________________ 11 EXAMPLE CASE
______________________________________ Example 6-9 On 2 April 1979, an accidental release of anthrax occurred at Compound 19, a military microbiological facility at Sverdlovsk (Ekaterinburg), Russia. Officially 68 people died due to the anthrax outbreak, although it has been suggested that the number of fatalities could have been up to 80; 300 to 400 people were affected. Meselson et al. [29] and Meselson [30] have studied this event in detail and the following information is taken from those papers. During the event, there was NNW wind with a speed of approximately 5 m s-1. Due to the lack of information, a nominal release height of 10 m is assumed. Atmospheric stability class: D The dispersion took place in open country. Negligible infectivity decay rate and deposition velocity can be assumed (d < 5 Pm, as produced by a laboratory aerosol generator). Release flow rate (unknown): Q spores/s (continuous release) or Q’ spores (instantaneous release). Release time: unknown. As for the source strength (unknown), the researchers [29, 30] obtained an attack rate at a ceramics factory (Fig. 6-31) of approximately 2%. Two other independent teams of researchers had obtained a dose-response relation for the inhalation of anthrax by primates and although it implies a significant uncertainty when it is applied to the actual population, used this, in conjunction with their own dispersion estimates, to estimate that at the ceramics factory the dose causing 2% fatalities ranged between 9 spores and 1,300 spores. If these two estimates bracket the actual value, the weight of spores released as an aerosol would have ranged approximately between a few milligrams and one gram (the number of spores per milligram is taken to be 109). Estimate the concentration profile of the plume.
242
Fig. 6-31. Probable location of 62 patients when they suffered the exposure [29] and the position of the plume (modified from [29]).
Solution Figure 6-31 shows the location of 62 patients (most of them dead) whose position (at work or at home) could be identified by the researchers. Clearly this distribution matches the area covered by the release from Compound 19 with a NNW wind. Due to the lack of information on the source (Soviet officials described it as an “explosion”), two situations can be considered: a continuous (albeit short) emission and an instantaneous emission. a) Continuous emission From Figure 6-31 it can be seen that all of the 62 people who were affected were within a distance of 4,500 m from the source (whose exact location inside Compound 19 is unknown). At such a distance, strictly speaking, a release could be considered continuous, according to the criterion established in Equation (6-3), if the release time were to be approximately 37 min. However, taking into account the description of the event as an “explosion”, it seems that in this case a rather shorter release time must be considered; therefore, a hypothesis of tr = 8 min is assumed. Equation (6-8) can be applied and Vy and Vz can be obtained from the corresponding expressions in Table 6-8 (rural conditions, stability class D). Here, the ALOHA [23] code is used.
243
If a release of 10 mg of anthrax is assumed (i.e. 1.25 mg min-1), the plume can be plotted for a given isoplete (c = 40 spores m-3) (Fig. 6-22). The maximum concentration on the plume’s central line, that is, at the ceramics factory, is 90 spores m-3. The maximum dose at the ceramics factory (located at x = 2.8 km, y = 0) is 721 spores min m-3. If we assume a breathing rate of 0.03 m3 min-1, such as that of a person engaged in light work, the number of spores inhaled by a person outdoors would be 721 · 0.03 = 22 spores. For a person located indoors, the number of spores inhaled would be 10. If a release of 1 g of anthrax is assumed, the plume (for the same concentration isoline of 40 spores m-3) covers a much larger area. The maximum dose at the ceramics factory is now 7.21·104 spores min m-3 and the number of spores inhaled by a person outdoors is 2,160 spores (indoors: 1,146).
Fig. 6-32. Variation of the concentration of anthrax (spores m-3) as a function of time at the ceramics factory (instantaneous release of 1 g).
b) Instantaneous emission The calculation was again performed with the ALOHA code [22]. ALOHA considers that an instantaneous release lasts 1 min. Thus, for a release of 1 g, the maximum concentration at the ceramics factory is 24,300 spores m-3. One person at this position would inhale 2,160 spores (the same amount as for a continuous emission lasting 8 min). The variation of the concentration at the ceramics factory as a function of time, both outdoors and indoors, has been plotted in Figure 6-32. The values obtained are of the same order of magnitude as those proposed by Meselson et al. [29, 30]. However, we should emphasize again that these are just estimates that follow from a series of assumptions on the source strength and on the dose-response relations for the infectious aerosol and the human population exposed to it. ______________________________________ NOMENCLATURE
b
local characteristic width of the plume (m)
244
c c* cc ci clim cloc co cs cs * D d dp erf erfc g go H Hph 'H h Ki lb Lh Lho Lu m Mpoll Mw n P r Re T t Ta tb td texp To Ts tr tO u us
concentration (ppm or mg m-3); in Figs. 6.19 and 6.21, ground level concentration on the plume axis (vol/vol) effective concentration corrected for the non-isothermal release (ppm) concentration for a continuous emission (ppm or mg m-3) indoor concentration (ppm or mg m-3) limit concentration defining the beginning and the end of the cloud (ppm or mg m-3) concentration of pollutant at the isoplete (kg m-3) outdoor concentration (ppm or mg m-3); in Figs. 6.19 and 6.21, initial concentration of the release (vol/vol) concentration of pollutant at a given point of the plume (kg m-3) concentration of pollutant at the plume’s axis (kg m-3) characteristic dimension of the source (m) inside stack diameter (m) particle diameter (m) error function complementary error function gravitational acceleration (m s-2) initial corrected gravity (m s-2) effective release height (m) height above ground of the source (m) plume rise (m) height of major obstacles (m) eddy diffusivity in x, y or z direction (area time-1) buoyancy length scale (m) half-width of the plume (m) half-width of the plume at the source (m) upwind extent of the plume (m) emission mass flow rate (kg s-1 or g s-1); amount of pollutant released instantaneously (kg) mass of pollutant inside the building (kg) molecular weight (kg kmol-1) number of air changes per hour (h-1) atmospheric pressure (bar or atm) distance from the isoplete to the plume’s axis on a plane perpendicular to the axis (m) Reynolds number (-) temperature (K) time elapsed from the start of the emission (s) air temperature (K) time elapsed from the moment at which the plume reaches the building (s or h) departure time of the cloud with a given concentration contour (s) duration of exposure (s); in Eq. (6-44) (min) source temperature for non-isothermal releases (K) stack gas temperature (K) duration of the emission (s) time during which the plume moves over the building (s) wind velocity (m s-1) stack gas exit velocity (m s-1); in Eq. (6-36), plume velocity relative to the coordinate system at a given point of the plume (m s-1)
245
us* ut V v va vads Vo vo Vresp x xb xt y z zb z0 zt
D T O P U Ua Us Us * Ur V
plume velocity relative to the surrounding atmosphere at a point on plume axis (m s-1) terminal velocity of a particle (m s-1) volume of the building (m3) ventilation rate (m3 s-1) volume of air mixed with the released material in the plume (m3) rate at which the pollutant is consumed inside the building (m3 s-1) initial volume of the dense gas instantaneously released (m3) initial plume flow rate (m3 s-1) respiratory rate of a person or animal (m3 min-1) Cartesian coordinate (m); downwind distance from the source (m) horizontal coordinate of the lower isolplete (m) horizontal coordinate of the upper isoplete (m) Cartesian coordinate (m) height above ground surface (m); Cartesian coordinate (m) vertical coordinate of the lower isoplete (m) surface roughness length (m) vertical coordinate of the upper isoplete (m) coefficient in Eq. (6-1) (-); reflection coefficient in Eqs. (6-8) and (6-13) (-) angle between the plume axis and the horizontal (rad) turbulent Schmidt number ( | 1.35 ) viscosity of the gas (kg m-1 s-1) density of the gas (kg m-3) density of air (kg m-3) density of the particle (kg m-3) density of the gas at the plume axis minus density of the atmosphere surrounding the plume (kg m-3) relative density with respect to air (-) dispersion coefficient in x, y or z direction (m)
ANNEX 6-1
The error function is defined as the integral erf (x) =
2
S
x
³e
t 2
dt for x t 0
0
To evaluate the erf (x) numerically it can be approximated by [11]: erf x 1 be x
2
for x t 0
where b = 0.254829592a – 0.284496736 a2 + 1.421413741 a3 – 1.453152027 a4 + 1.061405429 a5 a = (1 + 0.3275911x)-1
246
For negative values of x: erf (-x) = -erf (x) All values of erf (x) range between 0 and 1, and all the values of erf (-x) range between 0 and -1. REFERENCES [1] H. S. Peavy, D. R. Rowe and G. Tchobanoglous, Environmental Engineering. McGraw-
Hill International Editions. New York, 1985. [2] TNO. Committee for the Prevention of Disasters. Methods for the calculation of physical
effects (Yellow Book). Directorate General of Labour. The Hague, 1997. [3] M. L. Davies and D. A. Cornwell, Introduction to Environmental Engineering. McGraw-
Hill, Inc. New York, 1991. [4] R. E. Britter and J. McQuaid. Workbook on the dispersion of dense gases. Health and
Safety Executive. Sheffield, 1988. [5] Committee for the Prevention of Disasters. Methods for the determination of possible
damage (Green Book). The Hague, 1992. [6] S. B. Turner, Workbook of Atmospheric Dispersion Estimates. Lewis Publishers. Boca
Raton, 1994. [7] J. Z. Holland. A meteorological survey of the Oak Ridge area: final report covering the
period 1858-1952. Weather Bureau, 554-559. Atomic Energy Comm., Report ORO-99. Washington DC, 1953. [8] F. Pasquill. Meteorol. Mag., 90 (1961) 33 [9] F. A. Gifford. Nuclear Safety, 2 (1961) 47. [10] R. F. Griffiths. Atmosph. Env., 28 (1994) 2861. [11] G. A. Briggs. Diffusion Estimation for Small Emissions. Report ATDL-106. Environmental Research Laboratories. Washington, DC, 1974. [12] E. Palazzi, M. Defaveri and G. Fumarola. Atmosph. Env., 16 (1982) 2785. [13] M. R. Beychok. Fundamentals of Stack Gas Dispersion. M. R. Beychok, Irvine, CA, 1994. [14] J. S. Puttock, K. Macfarlane, A. Prothero, F. J. Rees, P. T. Roberts, H. W. Witlox and D. N. Blewitt. J. Loss Prev. Proc. Ind., 4 (1991) 16. [15] F. P. Lees. Loss Prevention in the Process Industries. Butterworth Heinemann, Oxford, 1996. [16] J. A. Vílchez, E. Planas-Cuchi, J. Casal and J. Arnaldos. J. Loss Prev. Proc. Ind., 15 (2002) 507. [17] G. Ooms. Atmosph. Env., 6 (1972) 899. [18] G. Ooms, A. P. Mahieu, F. Zelis. In First International Symposium on Loss Prevention and Safety Promotion in the Process Industries (211-219). Elsevier. Amsterdam, 1974. [19] Environmental Protection Agency. Complete DEGADIS, 1990. Executable and source FORTRAN Code (933KB, ZIP). Available online at http://www.epa.gov/ttn/scram/. [20] V. J. Clancey. Chem. Proc. Haz., 6 (1977) 121. [21] R. F. Sellers and A. J. Forman. J. Hygiene, Cambr., 71 (1973) 15. [22] A. I. Donaldson, N. P. Ferris, J. Gloster. Res. Veter. Sci. 33 (1982) 384. [23] A. I. Donaldson, C. F. Gibson, R. Oliver, C. Hamblin and R. P. Kitching. Res. Veter. Sci., 43 (1987) 339.
247
[24] National Oceanic and Atmospheric Administration. ALOHA, Version 5.5. NOAA,
Seattle, WA 98115, 1992. [25] J. Casal, M. Moreso, E. Planas-Cuchí and J. Casal. Veter. Rec., June (1997) 672. [26] J. Casal, E. Planas-Cuchi, J. M. Moreso and J. Casal. J. Hazard. Mater., 43 (1995) 229. [27] J. Casal, E. Planas-Cuchi and J. Casal. J. Hazard. Mater., A68 (1999) 179. [28] P. C. Davies and G. Purdy. Toxic risk assessment: the effect of being indoors. IChE
North Western Branch Papers, 1986, 1, 1.1. [29] M. Meselson, J. Guillemin, M. Hugh-Jones, A. Langmuir, I. Popova, A. Shelokov and O.
Yampolskaya. Science, 266 (1994) 1202. [30] M. Meselson. The ASA Newsletter, 01-6, No. 87 (2001) (http://www.asanltr.com/).
248
Chapter 7
Vulnerability 1 INTRODUCTION Once the effects of an accident —thermal radiation, overpressure, evolution of the concentration of a toxic material— have been established, its consequences must be determined; i.e. its effects on individuals, buildings, equipment, and the environment must be estimated. This estimate can be made using tabulated data or graphs, or by applying analytical models. The results will be, once more, an approximation. The validity of the estimate will depend on the correct application of the models and —as in other engineering fields— on the analyst’s criteria and experience. This chapter provides an overview of the different methods which can be used to estimate the consequences of an accident, as well as several examples of application. 2 POPULATION RESPONSE TO AN ACCIDENT Individuals in a population will not all respond in exactly the same way when exposed to the effects of an accident (such as thermal radiation, dose of a toxicant, overpressure from an explosion). For a given intensity or dose, some will only be slightly affected, while others will be severely affected; most will show an intermediate response. The overall response of a population to such a situation can usually be represented using a normal distribution (Fig. 7-1), according to the following expression: f ( x)
ª 1 § x P ·2 º exp « ¨ ¸ » V 2S ¬« 2 © V ¹ ¼» 1
(7-1)
where f(x) is the percentage of individuals experiencing a given response to the effects of the accident x is the response V is the standard deviation and P is the mean. If the data plotted in Fig. 7-1 are available for different dosage values, the mean response can be plotted as a function of the dose. For convenience, the logarithm of the dose is plotted in the graph in Fig. 7-2 (for the case of second degree burns), in such a way as to obtain an approximately straight line in the middle of the curve, thus making the graph easier to use.
249
Fig. 7-1. Normal distribution: the response of a population to a given major accident.
These curves can be obtained for exposure to thermal flux, overpressure, etc. However, in practice, it is better to work with analytical expressions, which are more practical from the point of view of calculations. This is why probit equations are usually used.
Fig. 7-2. Response (% of individuals suffering second-degree burns) as a function of the logarithm of the dose (t in s; I in W m-2) for a given case (exposure to thermal radiation).
3 PROBIT ANALYSIS A function that relates the magnitude of an action —for example, thermal radiation from a fire— to the degree of damage it causes, i.e. a relationship between the dose and the response,
250
is required to estimate the consequences of an accident. The most frequently applied method is the probit analysis, which relates the probit (from probability unit) variable to the probability. The probit variable Y is a measure of the percentage of a population subjected to an effect at a given intensity (V) which will suffer a given damage. This variable follows a normal distribution and has an average value of 5 and a normal deviation of 1. The relationship between the probit variable (Y) and the probability (P) can be expressed as follows [1]: P
1 2S
Y 5
ª V2º »·dV 2 ¼
³ exp«¬
f
(7-2)
In practice, the probability (which varies from 0 to 1) is substituted by a percentage (from 0 to 100), which is much more practical for the purposes of risk analysis. Figure 7-3 shows a plot of the relationship expressed in Equation (2) between probability (which is expressed as a percentage) and the probit variable, and the corresponding data are tabulated in Table 7-1. Equation (7-2) conveniently transforms the sigmoidal shape corresponding to the doseresponse relationship (for example, overpressure-percentage of fatalities) into a straight line (Fig.7-4, for a given case) when plotted using a linear scale for the probit function.
Fig. 7-3. Relationship between the probit variable and the percentage [1].
If the percentage of the population experiencing a certain response is not plotted versus the intensity of the action, but rather versus its logarithm, the resulting curve is much closer to a normal distribution. Then, the following relationship can be established between the probit variable and V as follows:
Y
5
ln V P
(7-3)
V
251
where P and V are the mean and the standard deviation, respectively, of the normal distribution.
Fig. 7-4. The dose-response relationship applied to the case of death from lung hemorrhage (explosion) ('P, N m-2).
Equation (7-3) can be modified to give the following expression, which is normally used to calculate the value of the probit variable Y: Y
a b·ln V
(7-4)
where a and b are constants that are experimentally determined from information on the accidents or, in some cases, from experimentation on animals. V is a measure of the dose of the damaging effect, which may be just one parameter (for example, overpressure in the case of an explosion) or a combination of various parameters (for example, a combination of concentration and time in a toxic gas release, or of thermal radiation and time in the case of a fire). Once the value of Y is determined, the probit variable must be converted into the percentage of individuals affected, in order to estimate the real consequences of the accident for the population (i.e. the number of individuals injured or dead). This can be done using the data in Table 7-1. Another expression that relates the percentage of individuals affected by a given impact to the probit variable, which may be suitable for certain calculation procedures, is as follows:
252
Percentage
ª Y 5 § Y 5 ·º ¸» 50 «1 erf ¨¨ ¸ © 2 ¹»¼ ¬« Y 5
(7-5)
Table 7-1 Relationship between the probit variable and the percentage [1] % 0 1 2 3 4 5 6 0 2.67 2.95 3.12 3.25 3.36 3.45 10 3.72 3.77 3.82 3.87 3.92 3.96 4.01 20 4.16 4.19 4.23 4.26 4.29 4.33 4.36 30 4.48 4.50 4.53 4.56 4.59 4.61 4.64 40 4.75 4.77 4.80 4.82 4.85 4.87 4.90 50 5.00 5.03 5.05 5.08 5.10 5.13 5.15 60 5.25 5.28 5.31 5.33 5.36 5.39 5.41 70 5.52 5.55 5.58 5.61 5.64 5.67 5.71 80 5.84 5.88 5.92 5.95 5.99 6.04 6.08 90 6.28 6.34 6.41 6.48 6.55 6.64 6.75 0.0 0.1 0.2 0.3 0.4 0.5 0.6 99 7.33 7.37 7.41 7.46 7.51 7.58 7.65
7 3.52 4.05 4.39 4.67 4.92 5.18 5.44 5.74 6.13 6.88 0.7 7.75
8 3.59 4.08 4.42 4.69 4.95 5.20 5.47 5.77 6.18 7.05 0.8 7.88
9 3.66 4.12 4.45 4.72 4.97 5.23 5.50 5.81 6.23 7.33 0.9 8.09
Vílchez et al. [2] proposed a set of analytical expressions for converting probit variables into percentages and vice versa, which are very useful for performing the calculations by means of a computer code or a worksheet; the results thus obtained show excellent agreement with the values taken from the figures and tables commonly used to calculate the percentage of people injured in a given accident, even in the asymptotic regions of the probit function (see Annex 7-1). Currently, the probit method is widely used to estimate the consequences of certain major accidents on individuals. ______________________________________ Example 7-1 By gathering real information on several accidental explosions, the data in Table 7-2 on the percentage of individuals that died due to pulmonary hemorrhage can be obtained. Using these data, the probit equation for death due to pulmonary hemorrhage must be obtained. The peak overpressure 'P should be taken as the intensity of the effect. Table 7-2 Peak overpressure, N m-2 1.03 · 105 1.20 · 105 1.39 · 105 1.50 · 105 1.74 · 105 2.02 · 105
% of fatalities 1 10 40 60 90 99
Solution To find a correlation of the type in Equation (7-4), the ln of the peak overpressure and the values of the probit function that correspond to the percentages in Table 7-1 are required.
253
These data can be calculated and obtained from Table 7-1 and are shown in Table 7-3. Table 7-3 Peak overpressure, N·m-2 1.03 · 105 1.20 · 105 1.39 · 105 1.50 · 105 1.74 · 105 2.02 · 105
% of fatalities 1 10 40 60 90 99
ln 'P 11.544 11.696 11.845 11.917 12.067 12.218
Y 2.67 3.72 4.75 5.25 6.28 7.33
The value of Y has been plotted vs. ln 'P in Fig. 7-4. A straight line is obtained. The following equation corresponds to this line:
Y
77.1 6.91·ln 'P
This is the probit equation for predicting the percentage of individuals who will die due to pulmonary hemorrhage (see Section 5). ______________________________________ It should be taken into account that when different probit equations are used to estimate diverse consequences (for example, first-degree burns, second-degree burns, or lethality) on a given population, different categories will overlap. Thus, all those individuals suffering second-degree burns will appear to have also suffered first-degree burns, and all those individuals who die due to thermal radiation will also have suffered second-degree burns (see Example 7-2). 4 VULNERABILITY TO THERMAL RADIATION 4.1 Damage to people The main effects of thermal flux (usually radiation) on individuals are burns to the skin, the severity of which depends on the intensity of the radiation (kW/m2) and on the dose received. Burns are classified into the following three categories: í First-degree burns are superficial injuries; the skin becomes red and painful; í Second-degree burns are deeper injuries (0.1 mm); the skin becomes red and blisters form; í Third-degree burns are deep injuries (1-2 mm); victims lose all sensation in the burned area, and the skin will have been destroyed and be yellow or black in color. For second- and third-degree burns there is a given probability of mortality. The existence of these burns on a large surface area of the body, and the consequent destruction or degradation of the skin —the envelope which protects the body from the environment— causes the loss of fluid and greatly increases the probability of infections and death. The probability of surviving in such a situation is a function of the percentage of body surface area burned and of the age of the person (see Table 7-4) [3]. Furthermore, in the case of burns affecting 30% or more of the body surface area, the person may go into a state of shock and die [4]. The “tolerable limit” for people —a concept which is not particularly accurate— is considered to be on the order of 5 kW·m-2.
254
As a reference, we can remember that the sun’s radiation on a sunny day is approximately 1 kW·m-2. Table 7-4 Relationship between percentage of body area burned, age and mortality [3] Age
% area burned
t 93 88-92 83-87 78-82 73-77 68-72 63-67 58-62 53-67 48-52 43-47 38-42 33-37 28-32 23-27 18-22 13-17 8-12 3-7 0-2
year 0 4 1 .9 .9 .8 .7 .6 .5 .4 .3 .2 .2 .1 .1 0 0 0 0 0 0 0
5 9 1 .9 .9 .8 .7 .6 .5 .4 .3 .2 .2 .1 .1 0 0 0 0 0 0 0
1014
1519
2024
2529
3034
3539
4044
4549
5054
5559
6064
6569
7074
7579
80
1 .9 .9 .8 .8 .7 .6 .4 .3 .3 .2 .1 .1 0 0 0 0 0 0 0
1 .9 .9 .8 .8 .7 .6 .5 .4 .3 .2 .1 .1 0 0 0 0 0 0 0
1 1 .9 .9 .8 .7 .6 .5 .4 .3 .2 .2 .1 .1 0 0 0 0 0 0
1 1 .9 .9 .8 .8 .7 .6 .5 .3 .3 .2 .1 .1 0 0 0 0 0 0
1 1 1 .9 .9 .8 .7 .6 .5 .4 .3 .2 .2 .1 .1 0 0 0 0 0
1 1 1 .9 .9 .8 .8 .7 .6 .5 .4 .3 .2 .1 .1 .1 0 0 0 0
1 1 1 1 .9 .9 .8 .7 .7 .6 .4 .3 .3 .2 .1 .1 0 0 0 0
1 1 1 1 1 .9 .9 .8 .7 .6 .5 .4 .3 .2 .2 .1 .1 0 0 0
1 1 1 1 1 .9 .9 .9 .8 .7 .6 .5 .4 .3 .2 .1 .1 .1 0 0
1 1 1 1 1 1 .9 .9 .9 .8 .7 .6 .5 .4 .3 .2 .1 .1 0 0
1 1 1 1 1 1 1 1 .9 .9 .8 .8 .7 .6 .4 .3 .2 .1 .1 0
1 1 1 1 1 1 1 1 1 1 1 .9 .8 .7 .6 .4 .3 .2 .1 .1
1 1 1 1 1 1 1 1 1 1 1 1 .9 .9 .7 .6 .5 .3 .2 .1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 .9 .8 .6 .5 .3 .2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 .9 .7 .5 .4 .2
Fig. 7-5. Time before one feels pain as a function of thermal radiation (modified from [7]).
A number of authors have studied the time elapsed before one feels pain as a function of the heat flux [6]. Data obtained experimentally from volunteers showed that a person feels
255
pain when the skin reaches a temperature of 45 ºC at a depth of 0.1 mm, and this may be expressed as follows:
td
(35 / I ) 4 / 3
(7-6)
where td is the time elapsed before one feels pain, in seconds, and I is the thermal flux, in kW·m-2. Figure 7-5 shows the relationship between radiation intensity and the time before individuals feel pain when their unprotected skin is exposed to thermal radiation. The data are those obtained by Buettner [7], and the line corresponds to Equation (7-6).
Fig. 7-6. Consequences of thermal radiation as a function of intensity and exposure time [8].
At 5 kW·m-2, the time before one feels pain (assuming the skin to be unprotected) is approximately 13 s, and in 20 s this thermal flux can cause second-degree burns. Generally, it is assumed that thermal fluxes lower than 1.5 kW·m-2 cause no pain, independently of the exposure time (although pain and burns can appear later for long exposure times, which are not usually associated with accident scenarios). When the temperature of the skin reaches 55 ºC, blistering appears. Table 7-5 gives, in an approximate way, the levels of damage for different thermal fluxes. Figure 7-6 shows the variation of consequences as a function of time and thermal radiation intensity.
256
Table 5 Approximate levels of damage for different thermal fluxes Thermal flux, Effect kW m-2 1.4 Harmless for individuals not wearing special protection 1.6 Will cause no discomfort at long exposures 1.7 Minimum required to feel pain 2.1 Minimum required to feel pain after 1 min 4.0 Enough to cause pain after an exposure of 20 s; blistering of the skin is likely; 0% lethality 4.7 Causes pain in 15-20 s, burns after 30 s 7.0 Maximum tolerable for firefighters who are totally protected (classical protective clothing) 11.7 Thin, partially insulated steel may lose its mechanical integrity 12.5 Plastic insulation of electrical wires melts; melting of plastic tubing; 100% lethality 15.0 Critical radiation intensity* for wood (flame ignition without contact with the surface) 25.0 Thin, insulated steel may lose its mechanical integrity 35.0 Critical radiation intensity for wood and textiles (without flame ignition) Threshold value for the ignition of buildings 37.5 Damage to process equipment, collapse of structures *Critical radiation intensity: minimum radiation at which ignition can occur
4.1.1 Probit equations There are various probit equations for estimating the effects of thermal flux on human beings. The dose, in this case, is usually taken to be (t·I4/3). Some of the most widely used of these equations [9], which are obtained by adapting equations previously proposed by Eisenberg et al. for nuclear explosions to the case of hydrocarbon fires, are shown below: First-degree burns Y
39.83 3.0186 ln t I 4 / 3
(7-7)
Second-degree burns Y
43.14 3.0186 ln t I 4 / 3
(7-8)
Lethality Y
36.38 2.56 ln t I 4 / 3
(7-9)
where t is the (effective) exposure time in s and I is the radiation intensity in Wm-2. The duration of the exposure to thermal radiation, for short fires, such as, for example, fireballs, can be measured. For fires of longer duration, 30-40 s has been suggested for urban areas [10]. In other cases, if no specific information is available, the exposure time can be assumed to be the time required to cover a distance at which the radiation has decreased to 1 kW m-2.
257
4.1.2 Clothing These expressions do not take into account any protection provided by clothing. Clothing has a protective effect unless it is ignited; clothing becoming ignited, however, causes severe burns, which entails a high probability of death. The threshold value for the ignition of clothing (t = 20 s) is approximately 35 kW·m-2. If clothing is not ignited, at least 80% of the skin is thought to be protected; face and neck (7%), armpits (8%) and hands (5%) remain unprotected. However, even with this protective effect, and given a typical age distribution in the population and their diverse lethality versus burnt area, with severe burns on 20% of the body surface approximately 14% of the population would die [11]. Considering the protective effect of clothing, lethality can be estimated using the following expression [12]: Y
37.23 2.56 ln t I 4 / 3
(7-10)
Equations (7-7), (7-8), (7-9) and (7-10) are plotted in Figure 7-7. 4.1.3 Escape It is usually assumed that individuals take 5 s to react [12], and can escape at a velocity ranging from 4 to 6 m/s; 4 m/s has been suggested as a general (albeit rather conservative) value [11]. The exposure time, as a general criterion, will then be the time taken to react plus the time taken to reach shelter or to reach a distance where I d 1 kW·m-2. Thus, the dose received by an individual who escapes, for an open area, is [11]:
Dose
I
4/3 0
t eff
I
t
4/3 0
t r ³ I t tr
4/3
dt
I
4/3 0
ª 3 d0 «t r 5 u « ¬
5 / 3 ° § · ½°º u t exp t r ¸¸ ¾» ®1 ¨¨1 °¯ © d 0 ¹ °¿»¼
(7.11)
where I0 is the value of I at the initial position for the individual, in W m-2; d0 is the distance between the flame surface and the initial position of the individual, in m; teff is the effective exposure duration (equivalent time for an average radiation intensity I0); texp is the total exposure time in s (if the duration of the fire t is shorter than texp, then texp = t); tr is the reaction time in s; and u is the escape velocity in m s-1. In an urban area (high building density), the distance to the nearest shelter is assumed to be approximately 20 m; this implies texp = tr + tesc = 10 s. The radiation intensity for this case may be considered to be approximately constant and equal to I0. In an intermediate situation (a so-called built-up area) that has a lower building density, the distance to the nearest shelter can be postulated to be 50 m; Equation (7-11) should be applied. The threshold ignition for a building is considered to be approximately 35 kW·m-2, although this value does depend on the type of building. In risk studies, if I t 35 kW·m-2 it is generally assumed that all the people inside the building will die, and that they will survive if I<35 kW·m-2. A special case is that of a flash fire. There are two possibilities: í People outside the zone covered by the fire are subjected to thermal radiation, but because the duration is very short, it is usually assumed that there is little damage. However, this criterion should be taken with caution, as in several cases people who
258
were outside the cloud received serious burns (for example, in the flash fire occurred at Lynchburg in 1972). í People inside the fire come into direct contact with the flames, undergo severe burns and their clothing probably ignites. It is usually assumed that all the people die. If a building ignites, meaning that there are secondary fires (which is usually the case), it is assumed that the people located inside the building die; if the building does not ignite, they are protected.
Fig. 7-7. Probit functions for thermal radiation (t in s and I in W·m-2).
______________________________________ Example 7-2 Take a fireball that involves 19,050 kg of LNG (the case described in Example 6-7 in Chapter 6). Estimate the effects of thermal radiation on a group of 40 individuals, who are located —at the beginning of the fireball— at a distance of 180 m.
259
Solution By performing the calculations described in Chapter 3, the thermal radiation intensity on a vertical surface located at a distance x of 180 m can be found: I vertical = 24 kW m-2 The distance between the flames’ surface and the target is d = 137 m. The distance at which I vertical = 1 kW m-2 can also be calculated: x1kWm 2
900 m
If people escape at a velocity of 4 m/s, they will need to reach that distance in the following time: t escape
900 180 4
180 s
Then texp = 5 + 180 = 185 s tfireball = 10.6 s (Example 6-7, Chapter 6). As tfireball < texp it will be assumed that texp = tfireball. Therefore, the effective exposure duration is t eff
3 d0 tr 5 u
5 / 3 ° § · ½° u t exp t r ¸¸ ¾ ®1 ¨¨1 °¯ © d 0 ¹ °¿
5
3 137 ° § 4 10.6 5 ·¸ ®1 ¨1 5 4 °¯ © 137 ¹
5 / 3
½° ¾ °¿
9 .6 s
and the corresponding dose is then dose
24,000 4 / 3 9.6
6,645,885 W m -2
4/3
s
By applying the probit equations for thermal radiation, the percentage of the population affected is found to be the following: First-degree burns
probit
39.83 3.0186 ln 6645885 7.59 99.5 %
Second-degree burns probit
43.14 3.0186 ln 6645885
4.28 24 %
Lethality
probit
36.38 2.56 ln 6645885 3.84 12 %
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Therefore, approximately 99% of the people exposed will suffer first-degree burns, 24% will suffer second-degree burns and 12% will die. Because the effects overlap, this means that five people (approximately) will die, five will suffer second-degree burns and 30 will suffer firstdegree burns. ______________________________________ 4.1.4 Effect of hot air Thermal radiation can significantly heat the surrounding air, and this consequently has an effect on human beings. Table 7-6 gives information on the physiological response to these conditions [14]. Table 7-6 Consequences of the increase of air temperature on people T, ºC Physiological response 125 Significant difficulties for breathing 140 Tolerable for 5 min 150 Limit temperature for escape 160 Quick and unbearable pain (if skin is dry) 180 Irreversible injuries in 30 s 205 Tolerance time for the breathing system: less than 4 min (if skin is wet)
4.2 Material damages Heat flux may cause a variety of damage to materials: surfaces or paint becoming discolored or peeling off, deformation of structural elements, ignition of combustible materials, breakage (of glass, for example), failure of structural elements, etc. The damage depends on the heat flux, on the properties of the material and on the physical features of the element (shape, thickness, etc.). It is not possible — except in a few cases — to give general rules for predicting this damage, and specific analysis should take into account the transient heat transfer phenomenon. Nevertheless, several simplified criteria can be applied to estimate certain values. The ignition of solids depends on the surface temperature. The velocity at which the temperature of a solid increases depends on the heat flux and on its capacity to store energy. When the solid surface reaches the ignition temperature (Tig), ignition takes place. A minimum heat flux (usually called critical heat flux) is required to reach Tig. If the heat flux is equal to or higher than the critical heat flux, then ignition will occur after a time tig. Two situations are usually considered: thin and thick materials. In thermal terms, thin materials are those in which temperature is uniform throughout its thickness during the heating process; their physical thickness is approximately 1-2 mm. The ignition time can be estimated using the following expression:
t ig
U cp l
Tig T0
(7-12)
I
where tig is the ignition time (s) cp is the specific heat at constant pressure (kJ kg-1 K-1) l is the thickness of the material (m) T0 is the initial temperature of the material (K)
261
Tig is the temperature of ignition (K) I is the heat flux (kW m-2), and U is the density of the material (kg m-3). Thick materials are those for which l > 2 mm. For these materials,
t ig
ª Tig T0 º 2 k U cp « » 3 ¬ I ¼
2
(7-13)
where k is the thermal conductivity at an average temperature in kW m-1 K-1. Table 7- 6 Typical ignition times for thick solids [15] Heat flux, kW·m-2 Time, s Material 10 300 Plexiglas, polyurethane foam, acrylate carpet 20 70 Wool carpet 150 Paper on gypsum board 250 Wood particleboard 30 5 Polyisocyanurate foam 70 Wool/nylon carpet 150 Hardboard
Equations [7-12] and [7-13] give only approximate results, and significant errors may occur when they are applied to different situations. More accurate expressions for the different cases which occur in practice can be found in [16]. Table 7-6 gives typical ignition times for thick solids and Table 7-7 gives the values of Tig and the critical heat flux for various thick materials. Table 7-7 Ignition properties of construction materials [17] Material kUcp, (kWm-2K)2s Plywood, plain (0.63 cm) 0.46 Particleboard (1.27 cm) 0.93 Hardboard (6.35 mm) 1.87 Fiber insulation board 0.46 Polyisocyanurate (5.08 cm) 0.02 Foam, rigid (2.54 cm) 0.03 Foam, flexible (2.54 cm) 0.32 Polystyrene (5 cm) 0.38 Polycarbonate (1.52 mm) 1.16 PMMA, Type G (1.27 cm) 1.02 PMMA, polycast (1.59 mm) 0.73 Carpet (wool, untreated) 0.25 Gypsum board, FR (1.27 cm) 0.40 Asphalt shingle 0.70 Fiberglass shingle 0.50 Glass-reinforced polyester (2.24 mm) 0.32 Glass-reinforced polyester (1.14 mm) 0.72 Aircraft panel, epoxy fiberite 0.24
262
qcritical, kW·m-2 16 18 10 14 21 20 16 46 30 15 9 20 28 15 21 16 17 28
Tig, ºC 390 412 298 355 445 435 390 630 528 378 278 435 510 378 445 390 400 505
___________________________________ Example 7-3 Calculate the ignition time for a 1.27 cm-thick particleboard slab subjected to a radiant heat flux of 25 kW·m-2. Solution The value of thermal inertia (kUCp) for the particleboard is (Table 7-7) 0.93 (kWm-2k-1)2s. By applying Equation (7-13): 2
t ig
2 ª Tig T0 º 2 2 ª 412 20 º k U cp « = 152 s » = 0.93 « 3 3 ¬ 25 »¼ ¬ I ¼
This result agrees quite closely with the experimental data obtained by Quintiere [15]. ______________________________________ 5 VULNERABILITY TO EXPLOSIONS 5.1 Damage to human beings The damage caused by explosions to human beings can be classified into the following two groups: - Direct consequences due to the blast. - Indirect consequences due to fragments (primary and secondary), due to body displacement, or caused by the collapse of buildings. Approximate planning guidelines [18] for scenarios in which individuals are injured due to explosions are presented in Table 7-8. Table 7-8 Consequences of overpressure on people Overpressure, bar Consequences <1 Individuals should be reasonably safe inside a reinforced structure away from windows or lying on the ground if outdoors 0.07-0.014 Casualties or fatalities are to be expected as a result of missiles or selfimpact against objects >0.21 Dynamic pressure loading is likely to hurl a human being to the ground 0.34 Eardrums rupture 1 Lung damage
5.1.1 Direct consequences The main direct effect of an explosion on human beings is the sudden increase in pressure that occurs as a blast wave passes. The human body stands up to pressure relatively well, as it is made up in large part by water, a non-compressible fluid. Therefore, the direct damage essentially occurs in those parts of the body that contain air pockets and can therefore be crushed, i.e. ears and lungs. The overpressure depends on the position of the human body and on the eventual reflection of the pressure wave. Fig. 7-8 [19] summarizes the main possible situations. If the body does not obstruct the incident wave, P equals the side-on peak overpressure of the blast wave 'P (‘a’ in Fig. 7-8). If the body is in such a position that it is engulfed by the blast wave
263
(‘b’ in Fig. 7-8) and thus disturbs it (standing is the most hazardous body exposure position), the resulting overpressure on the chest wall is the peak overpressure plus the pressure Pw caused by the action of the explosion wind on the body.
Fig. 7-8. Different positions of the human body with respect to the pressure wave. Taken from [19], by permission
Therefore,
P
'P Pw
'P
5'P 2 2'P 14 10 5
(7-14)
If the blast wave reflects against a wall near the body (‘c’ and ‘d’ in Fig. 7-8), then the reflected pressure Pr acts on the body: P
Pr
8'P 2 14'P 10 5 'P 7 10 5
(7-15)
(in the case of a shockwave, another expression should be applied [11]). Lethality due to pulmonary hemorrhage can be estimated using the probit expression proposed by Eisenberg et al. [20]: Y
77.1 6.91 ln P
(7-16)
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To determine what percentage of the eardrum is damaged, the following expression can be used [12]: Y
12.6 1.524 ln P
(7-17)
In Equations (7-14) through (7-17), P is the overpressure in N·m-2.
5.1.2 Indirect consequences The indirect effects include secondary effects (impact by fragments from the exploding device itself or from other objects that are also subjected to the blast wave) and tertiary effects (essentially, displacement of the human body due to the blast wave and impact against the ground or an obstacle). 5.1.2.1 Body displacement Due to the explosion wind which follows the pressure wave, the body can be displaced over a certain distance. Besides the injuries which can occur when the body tumbles onto the ground, there is a high probability of death if it collides with fixed obstacles. This will depend on the velocity of the body, the shape of the obstacle and the part of the body involved in the impact. The probability of fatal consequences is very high if the skull undergoes the impact; the probability of fracture to the base of the skull is related to the impact velocity. Table 7-9 gives several threshold values [21]. Table 7-9 Criteria for skull-base fracture due to impact Impact velocity, m·s-1 Criterion 3.0 Safe 4.0 Threshold 5.5 50% 7.0 | 100%
Concerning the situation in which the whole body collides with an obstacle, the following values have been proposed (Table 7-10) [21]: Table 7-10 Lethality due to impact on the whole body Impact velocity, m·s-1 Criterion 3.0 Safe 6.5 Threshold 16.5 50% | 100% 42.0
The body velocity can be obtained from Figures 7-9 and 7-10 [22]. These figures were obtained for an individual who has a mass of 70 kg. The following probit function was obtained [11] for the percentage of individuals that survive an impact to the head:
y
5.0 8.49 ln S
(7-18)
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Fig. 7-9. Relationship between body velocity, peak overpressure and impulse for skull-base fracture.
Fig. 7-10. Relationship between body velocity, peak overpressure and impulse for impact on the whole body.
where
S
2.43 10 3 4 10 8 'P i 'P
(7-19)
i being the impulse of the overpressure wave (N·m-2·s) and 'P the side-on peak overpressure (N·m-2).
266
The percentage of individuals that survive an impact to the whole body can be estimated using the following expression: Y
5.0 2.44 ln S
(7-20)
where 7.38 10 3 1.3 10 9 'P i 'P
S
(7-21)
5.1.2.2 Fragments Two types of fragments are essentially considered: cutting and non-cutting fragments. Cutting fragments —typically, glass fragments— penetrate the skin. Non-cutting fragments or debris —for example, a brick— cause high compressive stresses in the body. Data on the effects of fragments are relatively scarce. Table 7-11 [19] and Table 7-12 show several criteria for non-penetrating fragments that have masses of 4.5 kg and 10 g, respectively. Table 7-11 Critical velocities for the impact against the skull that has a mass of 4.5 kg Event Impact velocity, m·s-1 Cerebral concussion 3.0 4.6 Skull fracture 3.0 4.6 7.0
of a piece of debris criteria Safe Threshold Safe Threshold Near 100%
Table 7-12 Critical velocities for the impact of a fragment of glass that has a mass of 10 g Impact velocity, m·s-1 Criteria 15 Threshold for skin injuries 30 Threshold for severe injuries 55 Severe injuries (50%) 90 Severe injuries (100%)
The probability of survival can be estimated using the following expressions [11]. For fragments whose mass is greater than 4.5 kg: Y
13.19 10.54 ln v
(7-22)
where v is the velocity of the fragment (m·s-1). For fragments whose mass is between 0.1 and 4.5 kg: Y
17.56 5.30 ln k
(7-23)
with
267
1 Wm v 2 2
k
(7-24)
where Wm is the mass of the fragment (kg). For fragments whose mass is between 0.001 and 0.1 kg: Y
29.15 2.10 ln S
(7-25)
Wm v 5.115
(7-26)
with
S
______________________________________ Example 7-4 The explosion of a vapor cloud causes, at a distance of 200 m, a pressure wave of 40 kPa. At the same distance, and located in front of a wall facing the pressure wave, there are 22 human beings. Estimate the consequences of the explosion on these individuals.
Solution Blast wave is reflected against the wall; the reflected pressure is (Equation 7-15):
P
8 'P 2 14 'P 10 5 'P 7 10 5
Pr
8 40,000 2 14 40,000 10 5 40,000 7 10 5
92,970 N m -2
Eardrum damage: probit
12.6 1.524 ln 92,970
4.83 43 %
Lethality: probit
77.1 6.91ln 92,970 1.95 0 %
Therefore, approximately 7 individuals (43%) will suffer eardrum damage; there will be no fatalities. ______________________________________ 5.1.3 Collapse of buildings When a building is destroyed due to an earthquake or an explosion —both phenomena can be considered similar, from the point of view of the consequences for human beings— some of the individuals inside will die. If a building collapses, children and the elderly generally have a lower probability of survival. From the information available, it also seems that for practically all age categories more women than men are seriously or fatally injured. Even if practically the whole building collapses, human beings can survive in the remaining volumes inside the collapsed structure. Although the consequences depend on the size, age and type of a building (more serious or fatal injures occur in older and in larger
268
buildings), it can generally be assumed that the collapse of a building will produce a casualty rate of 100%: 20%-50% will die and 80%-50% will be injured. Table 7-13 Several criteria which are usually applied in risk analysis to estimate the effects of explosions on human beings Peak overpressure Fatalities indoors Fatalities outdoors 100% 100% 'P > 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.
273
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
276
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 <6 Immediate unconsciousness 4 Coma in 40 s, convulsions, respiration ceases, death
The consequences of underoxigenation may vary significantly from one individual to another; Table 7-21 provides some information on the consequences of exposure to rare atmospheres. It should be noted that inert gases offer no warning: most of them are colorless, odorless and tasteless. Accidents typically occur in closed or semi-closed spaces that are poorly ventilated and have a null or defective entrance of fresh air, such as underground galleys, tanks and wells. The build-up of atmospheres that have a low oxygen concentration in these places can be due to a variety of reasons: inert gas may be produced by an inertization process, gases may be released as a result of the fermentation of organic material, oxygen may become depleted due to corrosion, etc. A special case is the danger associated with the cryogenic use of liquid nitrogen: upon evaporation, its volume increases 700 times and the gas, which is very cold, tends to accumulate in the lower cavities. ______________________________________ Example 7-7 In a chemical plant, a buried tank (5 m in diameter, 5 m in height) is made inert by introducing 157 m3 of nitrogen (at room temperature and atmospheric pressure). Due to a series of changes (the enterprise is sold to a new owner), this section of the plant is closed. Three years later, a new process is started. One worker enters the tank (which has remained closed all this time) to clean it. A few moments later, his companion sees him lying down, unconscious. He enters the tank to help him and loses consciousness. Later on, firefighters recover the workers, who are both dead. Explain this event. Use the simplifying assumption that the entrance of the nitrogen implied the purging of the same volume of air as was initially in the tank. Solution It is evident that the entrance of nitrogen originated a decrease in the concentration of oxygen in the tank. Concentration of nitrogen in the tank (with a volume of 314.2 m3):
278
157 (314.2 157) 0.78
c N2
314.2 c N 2
0.89 (89% volume)
Concentration of oxygen:
314.2 157 0.21 314.2
0.105 (10.5% volume)
The data from Table 7-21 indicate that at this oxygen concentration a person will probably fall unconscious in a relatively short time. In this case, the person remains in that atmosphere and, after a certain time, dies. ______________________________________ 8 INFLUENCE OF SHELTERING
The consequences of an accident on population are not exactly the same if people are outdoors than if they are indoors, as a dwelling can offer a significant protection in some accidental scenarios (in fact, even a car may imply some protection in case of thermal radiation). Therefore, the distribution of people indoors/outdoors —which varies with time and with the type of activity— should be taken into account when estimating the consequences of an accident. The fraction of the population present during daytime is usually taken as 1.0 for industrial areas and 0.7 for residential areas. During night-time, the fraction of the population present is 1.0 for residential areas and 0 for industrial areas (or 0.2 if work is done in night shifts). As for the fraction of the population present indoors, 0.93 has been proposed for daytime and 0.99 for night-time; these values can be lower for warm zones such us, for example, the Mediterranean countries, as well as for certain zones (as, for example, tourist resorts). 8.1 Thermal radiation In the case of a fire, inside the flame envelope (the LFL contour for a flash fire), due to the high thermal flux and the ignition of clothing, the mortality rate among people outdoors is assumed to be 100%. In flash fires, people located inside a building will not undergo direct flame contact; nevertheless, the flash fire will give rise to secondary fires and the survival of these people will depend on the circumstances: with a conservative approach, a mortality of 100% is often assumed as well. Outside the flame envelope, the consequences depend on the type of fire. For a flash fire —a very fast phenomenon— the mortality rate is often assumed to be zero; however, in some accidents of this type (for example, the one occurred in Lynchburg in 1972 [33]) people who were outside the flammable cloud received serious burns. For pool fire, jet fire and fireball the consequences on people outdoors are a function of thermal radiation intensity and on the exposure time (maximum exposure time: 20 s). People inside a building are safe except if the building is ignited; in this case, all people inside will probably die. The threshold for the ignition of buildings is assumed to be 35 kW m-2 (however, this varies with the materials).
279
8.2 Blast In the case of an explosion, being outdoors is better than being inside a building, because if the building collapses or is severely damaged people inside will be killed or injured (see Table 7-13 and section 5.1.3). It is accepted that 10% of the houses inside the 0.1 barg contour are severely damaged and 1 out of eight persons inside them is killed. 8.3 Toxic exposure Staying indoors reduces significantly the toxic dose, as the concentration indoors is lower than the concentration outdoors during the passage of the cloud. Once the cloud has passed and there is no pollutant outdoors, there is still some pollutant indoors, so people must leave the building, which must be ventilated. The relationship between both concentrations —indoors and outdoors— is a function of the building ventilation rate; a mathematical model can be developed to establish this relationship (see Chapter 6). The ventilation rate varies with the type of building; if it is unknown, a default value of 1 h-1 can be assumed. If the concentration indoors is not calculated, a general criterion usually assumed is that the indoor mortality rate in the case of toxic exposure is 10% of the corresponding outdoor rate. 9 RELATIONSHIP BETWEEN THE NUMBER OF PEOPLE KILLED AND THE NUMBER OF PEOPLE INJURED IN MAJOR ACCIDENTS
As mentioned before, in quantitative risk assessment of accident scenarios rough hypothesis are often used to evaluate the magnitude of the consequences. These hypotheses are sometimes essential in order to draw iso-risk curves, as they complement the physical effect and vulnerability calculations. As an example, it has been mentioned before in this chapter that for blast the data shown in Table 7-13 are widely accepted as a general criterion; similar values have been proposed for indoor/outdoor mortality in the case of toxic exposure, etc. These hypothesis, which are mostly rules of thumb based on experience, do not necessarily agree with real data from actual cases, although they are normally of the same order of magnitude. Moreover, they save a lot of effort and reduce the time needed to carry out quantitative risk assessment. Risk analysis, especially when major accident scenarios are concerned, seek to estimate/predict the overall number of people affected, i. e., fatalities, injuries and possibly evacuees. Standard quantitative risk assessment focuses mainly on calculating the number of fatal victims as well as calculating the distances that define the areas to be evacuated. The number of injured people is seldom evaluated, as it would involve significant additional effort and in most cases little or no information is available. These estimations are based on calculating the effects of an accident, and they are independent from each other. Each estimation refers to a specific vulnerability criterion, and may include some of the shortcuts mentioned above. A statistical relationship has been proposed [34] to link the number of fatalities (NK) in an accident involving hazardous materials to the number of injured people (NI). Such a relationship can be especially useful in certain cases. For example, if NK is calculated for the area surrounding a plant where a major accident may hypothetically happen, a shortcut for estimating the expected number of injured people would be very useful for emergency planning purposes. An injured person is someone who needs to be hospitalized quickly. From
280
the point of view of emergency management, estimating the number of injured people quickly in the case of hazardous materials accidents can be therefore of vital importance. Ronza et al [34] performed a historical analysis using a sample of 975 accidents, with at least one fatality, occurred between 1975 and 2005; several measures were applied to reduce data bias: accidents occurred before 1975 were excluded, since they happened in a technological setting very different from the present one. In addition, the following sets of accidents were excluded: accidents due to handling/manufacturing conventional explosives, accidents due to sabotage, and the only accident (Bhopal, 1984) with more than 2,000 fatalities, because of their exceptional and atypical nature. As the data were scattered, a complex statistical treatment was applied using two multivariate analysis techniques, principal component analysis and clustering, to identify the data subsets that could undergo a selective, specific statistical treatment. Further treatment of these subsets led to a set of mathematical expressions that can be used to estimate the probable number of injured people as a function of the number of fatalities for all accidents, as well as for gas cloud, fire and explosion events respectively: for all accidents:
NI
6 N K0.77
for gas clouds:
NI
34 N K0.54
for fires:
NI
3N K0.86
for explosions:
NI
7 N K0.76
where NI is the number of injured people and NK is the number of fatalities. Due to the particular form of the data distribution, these equations return NI values which are not expected to be exceeded (with a probability of 75%). As it can be seen, the accidents that involve gas clouds result in the greatest number of people injured, followed by explosion and fire accidents, respectively. 10 ZONING ACCORDING TO VULNERABILITY
The area potentially affected by an accident can be divided into different zones according to the type of hazard (chemical, mechanical or thermal), the intensity of the effects and the vulnerability of people and property. The following types of zone are often considered: Intervention Zone (IZ): accidents involve a degree of damage which warrants the immediate application of protective measures. Alert Zone (AZ): accidents involve effects that, although perceptible by the population, do not justify intervention, except in order to protect critical groups of people. Domino Effect Zone (DZ): zones in which potential accident propagation must be considered. Various countries have established official threshold values for the physical and chemical effects (overpressure, thermal radiation, concentration or dose of a toxicant) of the different types of major accident in order to delimit these zones. As an example, Table 7-22 shows a set of these values (applied in Spain), classified as a function of the type of accident.
281
Table 7-22 Definition of Intervention Zone, Alert Zone and Domino Effect Zone Threshold value Dangerous Accident Description of phenomenon type planning zones Intervention Zone Alert Zone Dose of toxicant: Dose of toxicant: dose = cnmax·texp
Chemical type
Toxic dispersion
Unconfined vapour cloud explosion
Mechanical type
Confined vapour cloud explosion
Physical explosion
Flash fire
Thermal type
Pool fire
Jet fire
dose = Cnmax·texp
cmax: maximum concentration of Zone covered toxicant in air by the t : exposure time dispersion of exp n: depends on the toxic clouds substance
cmax: maximum concentration of toxicant in air texp: exposure time n: depends on the substance
cmax calculated from AEGL-2, ERPG-2 and TEEL-2 Zone affected ('P) by the ignition of a 'P = 125 mbar cloud of flammable vapour Zone affected ('P) by the ignition of a 'P = 125 mbar confined cloud of flammable vapour Zone affected ('P) by the expansion of 'P = 125 mbar a pressurized gas Clouds of Lower flammable flammability limit gas/vapour within flamm. (LFL) limits Thermal radiation Zone dose: surrounding 250 (kW·m-2)4/3 the pool Equivalent to 5 kW m-2 for 30 s Thermal radiation dose: Zone surrounding 250 (kW·m-2)4/3 the jet Equivalent to 5 kW m-2 for 30 s
cmax calculated from AEGL-1, ERPG-1 and TEEL-1 'P = 50 mbar
282
DZ
--
'P = 160 mbar
'P = 50 mbar
'P = 160 mbar
'P = 50 mbar
'P = 160 mbar
50% LFL
--
Thermal radiation dose: 115 (kW·m-2)4/3 Equivalent to 3 kW m-2 for 30 s Thermal radiation dose: 115 (kW·m-2)4/3 Equivalent to 3 kW m-2 for 30 s
Thermal radiation intensity: 8 kW m-2 Thermal radiation intensity: 8 kW m-2
11 EXAMPLE CASE
______________________________________ Example 7-8 On the 30 January 1996, the ship Vento di Scirocco entered the harbor of Barcelona. There was a small fire on board in a container containing sodium hydrosulfite [35]. Later on, the fire grew considerably, and engulfed another container containing hexane drums and threading through several others containing pesticides, monomers and toluene diisocyanate (note that separation criteria for keeping flammable products separate from those that can produce toxic gases had not been observed, thus infringing the corresponding IMDG rule). After several hours, the fire brigade removed the containers from the vicinity of the fire, and finally, after 8 hours, they extinguished the fire. A series of explosions (of hexane drums) had occurred, followed by fireballs. The fire gave rise to a large smoke plume, which was present from 11:30 a.m. until 7:30 p.m., when it moved out to sea. The smoke behaviour was that of a neutral gas: neither buoyant rise nor gravity lowering were observed. a). Analyze what would have happened had the wind been blowing (with the same speed) in the direction of the city. b). Calculate the consequences (lethality) if a group of 6 workers had been exposed to the smoke plume (in the worst case) at a distance of 100 m from the ship, during 15 min. Further data: the container in which the fire started contained 21,500 kg of sodium hydrosulfite (Na2S2O4). The container burned for 5 h. Two atmospheric situations must be considered, corresponding to the two extreme values of the registered wind velocities (u1 = 2.5 m/s and stability class = B, and u2 = 8.5 m/s and stability class = C). Solution a). Sodium hydrosulfite can undergo an exothermal decomposition reaction when exposed to heat or flame: 2Na2S2O4 o SO2 + Na2SO3 + Na2S2O3
(a)
Sodium hydrosulfite can also react exothermally with water, according to the following steps: 2Na2S2O4 o 4Na+ + 2S2O422S2O42- + H2O o SO32- + 2HSO32HSO3- + 2H+ o 2SO2 + 2H2O
(b) (c) (d)
The smoke produced by the accident was a mixture of gases, produced by the decomposition of sodium hydrosulfite and the combustion of hexane, of which the main toxic component was SO2. An estimate of the toxic dispersion must be therefore based on this gas, as the composition of the remaining gases is unknown. SO2 could be produced by the mechanisms described above (Equation (a) and Equations (b), (c) and (d). The overall amount of SO2 released by these two mechanisms would be 3,676 kg (Equation (a)) and 7,352 kg (Equations (a), (b) and (c)) respectively. It is impossible, with the information available, to establish the real contribution of each mechanism. However, the most interesting estimation is probably that which will enable us to establish the limiting values for the emission.
283
Fig. 7-11. Simulation of the atmospheric dispersion.
Therefore, the following limiting situations must be simulated: 1. all the SO2 generated, according to Equation (a); 2. all the SO2 generated, according to Equations (b), (c) and (d). This assumption gives the following average emission rates for the period (5 h) during which the sodium hydrosulfite container was burning: 1. 200 g/s 2. 400 g/s. A simulation —using the ALOHA code— can be performed for these two situations and for the two sets of meteorological conditions. Taking into account the smoke behaviour, Gaussian models can be applied; for these models only information concerning source strength and meteorological conditions is required. The results of two simulations (u1 = 2.5 m/s and stability class = B and the two average emission rates) can be seen in Figs. 7-11 and 7-12. In figure 7-12, IDHL, TLV-STEL and TLV-TWA values were plotted for SO2 (IDHL = 285 mg/m3, TLV-STEL = 14.2 mg/m3, TLV-TWA = 5.7 mg/m3 and AEGL 2 = 0.75 ppm). The concentration decreases with distance downwind and soon becomes lower than what could be called dangerous: it becomes less than IDHL at approximately 100 m, less than TLV-STEL at approximately 300 m and less than AEGL 2 at approximately 1,300 m. However, it is obvious that exposure to concentrations lower than IDHL for periods higher than 30 min can result in equivalent doses. Therefore, the distance corresponding to equivalent toxic effects for the period during which the emission took place (approximately 5 h) must be calculated. To do this, Equation (7-35) (Haber’s Law) can be applied using n = 1 for SO2 (see Table 7-17): Dose
285 30
c 2 300; c2 = 28.5 mg·m-3
Then, for a period of 5 h, the concentration that would produce the same effect as IDHL for a period of 30 min would be 28.5 mg·m3.
284
Fig. 7-12. Simulation of SO2 concentration as a function of distance. Threshold values have also been plotted.
Fig. 7-13. Intervention zones for the largest emission flow rate. Taken from [35], by permission.
Two distances could therefore be considered from the point of view of the toxic effects on the population: 1) a distance downwind where after a short exposure (30 min) toxic effects could require medical assistance (but without endangering people’s lives);
285
2) a distance downwind at which the aforementioned effects would appear only after a long exposure (approximately 5 h). The corresponding maximum radial distances are plotted in Fig. 7-13 for the worst case (emission of 400 g/s of SO2). The maximum distances covered are rather small. Nowadays, the “Intervention Zone” and the “Alarm Zone” are usually established taking as threshold values AEGL 2 and AEGL 1, respectively, which are much lower than IDHL. For SO2, AEGL 2 = 0.75 ppm (2 mg m-3) and AEGL 1 = 0.2 ppm (0.55 mg m-3). The zone covered by AEGL 2 has a radius of approximately 1,200 m (and is located inside the port area), and the AEGL 1 reaches a distance of 1,600 m; these values are much more conservative than those corresponding to the estimation based on the IDHL. In fact, even if the wind had blown towards the city (and all other meteorological conditions had remained constant) the emission of gases would not have posed any danger to the population. b). At a distance d = 100 m, c | 100 ppm (Fig. 7-12). Therefore,
Y
15.67 2.1 ln 100 15
-0.31
According to Table 7-1, nobody would have died. ______________________________________ NOMENCLATURE
a b c Cp d d0 I i I0 k l M NI NK n P Pr Pw 'P Q t T Tig
constant in the probit equation (-) constant in the probit equation (-) concentration (ppm = ml m-3 or mg m-3) specific heat at constant pressure (kJ kg-1 K-1) distance between the flame surface and the target (m) distance between the flame surface and the initial position of the person (m) radiation intensity or heat flux (kW m-2) impulse of the overpressure wave (N m-2 s) value of I at the initial position of the target (kW m-2) thermal conductivity (kW m-1 K-1) thickness (m) molecular weight (kg kmol-1) number ok killed people (-) number of injured people (-) parameter in the probit equation (-) probability (-); overpressure (N m-2); pressure (N m-2). reflected pressure (N m-2) pressure caused by the explosion wind (N m-2) side-on peak overpressure (N m-2) percentage (-) exposure time (effective) (s) temperature, K temperature of ignition (ºC or K)
286
T0 td teff texp tig tr u V v vb w Wm x Y
V P U
initial temperature (ºC or K) time required to feel pain, s effective exposure duration (s) total exposure time (s) ignition time (s) reaction time (s) escape velocity (m s-1); intermediate variable (Annex 1) dose or intensity of a given effect (variable units) velocity of the fragment (m s-1) body velocity (m s-1) intermediate variable (Annex 1) mass of the fragment (kg) response to a given action; horizontal distance (m) probit variable (-) standard deviation (-) mean value (-) density (kg m-3)
ANNEX 7-1
Analytical expressions were proposed [2] for converting both probit variables to percentages of affected people and percentages of affected people to probit variables, based on correlations found in the literature for the normal Gaussian distribution function. Expressions to find the percentage from the probit variable The percentage Q can be calculated using an approximation to the normal integral, as follows:
s
1 1 r Y 5
(A-1)
(note that Y 5 is an absolute value), ª 1 ª Y 5 2 º º m (b1 s b2 s 2 b3 s 3 b4 s 4 b5 s 5 ) « exp « »» 2 ¼ ¼» ¬ ¬« 2S
(A-2)
with: r = 0.2316419 b1 = 0.319381530 b2 = -0.356563782 b3 = 1.781477937 b4 = -1.821255978 b5 = 1.330274429 using s and m as intermediate variables. Then the percentage Q of people affected by a given accident can be obtained as follows:
287
Q 100 (1 m) if Y ! 5
(A-3)
Q 100 m if Y d 5
(A-4)
The estimated error for Q in this approximation is H Y < 7.5·10-6. Expressions to find the probit variable from percentage The value of the probit variable Y can be calculated using an approximation to the inverse normal integral: P
Q / 100
(A-5)
u
§ 1 · ln¨ 2 ¸ if 0 P d 0.5 ©P ¹
(A-6)
u
§ 1 · ¸ if 0.5 P 1 ln¨¨ 2 ¸ © 1 P ¹
(A-7)
c0 c1u c2 u 2 1 d1u d 2 u 2 d 3u 3
(A-8)
w u
with: c0 = 2.515517 c1 = 0.802853 c2 = 0.010328 d1 = 1.432788 d2 = 0.189269 d3 = 0.001308 using u and w as intermediate variables. Then , the probit variable Y can be obtained as follows:
Y
5 w if 0 P d 0.5
(A-9)
Y
5 w if 0.5 P 1
(A-10)
The estimated error for Y in this approximation is H Y < 4.5·10-4. The accuracy of these conversions (probit/percentage, percentage/probit) is very good even in the asymptotic regions of the probit function, where values must be obtained if very low individual risk levels must be estimated. REFERENCES [1] D. J. Finney, Probit Análisis. Cambridge University Press, London, 1971.
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[2] J. A. Vílchez, H. Montiel, J. Casal and J. Arnaldos, J. of Loss Prevent. Proc., 14 (2001)
193. [3] J. P. Bull, The Lancet, Nov. (1971) 1133. [4] J. M. Martin, IEC-Monografies de la Secció de Ciències, 6 (1991) 139. [5] T. T. Lie, Eigenschappen van warmtebeschermende Kleding en Keuringsmethoden.
Brand, 1963. [6] K. S. Mudan, Prog. Energ. Combust., 10 (1984) 59. [7] K. Buettner, J. Appl. Physiol., 3 (1951) 703. [8] N. A. Eisenberg, C. J. Lynch and R. J. Breding, Vulnerability Model: A Simulation
System for Assessing Damage Resulting from Marine Spills, USCG, AD/A-015 245, NTIS Report n. CG-D-137-75, 1975 [9] C. K. Tsao and W. W. Perry, Modifications to the vulnerability model: a simulation system for assessing damage resulting from marine spills (VM4). US Coast Guard, AD/A-075231, NTIS report n. CG-D-38-79, 1979. [10] A. H. Rausch, N. A. Eisenberg and C. J. Lynch, Continuing Development of the Vulnerability model, Department of Transportation, USCG, Washington D. C., Report n. CG-D-53-77, 1977 [11] DGLMSAE. Methods for the determination of possible damage (Green book). The Hague, 1992. [12] C. M. Pietersen, J. Loss Prevent. Proc., 10 (1990) 136. [13] I. Hymes, The Physiological and Pathological Effects of Thermal Radiation. U. K. Atomic Energy Authority, SRD, Warrington, SRD R 275, 1983. [14] A. J. Pryor, Full-Scale Evaluation of the Fire Hazard of Interior Wall Finishes. SRI. San Antonio, 1968. [15] J. G. Quintiere, Principles of Fire Behaviour. Delmar Publishers, Albany, 1998. [16] V. Babrauskas, Ignition Handbook. Fire Science Publishers, Issaquah, 2003. [17] J. G. Quintiere and M. Harkleroad. New Concepts for Measuring Flame Spread Properties. NBSIR 84-2943, National Bureau of Standards, 1984, Gaithersburg, MD. [18] W. C. Brasie and D. Simpson. Guidelines for estimating damage from chemical explosions. CEP Technical Manual, Loss Prevention, vol. 2, 1967. [19] W. E. Baker, P. A. Cox, P. S. Westine, J. J. Kulesz, and R. A Strehlow, Explosion hazards and evaluation. Elsevier Scientific Publishers. Amsterdam, 1983. [20] N. A. Eisenberg et al., Vulnerability model. A simulation system for assessing damage resulting from marine spills. NTIS AD-A015-245.Springfield, VA, 1975. [21] C. S. White, The nature of the problems involved in estimating the immediate casualties from nuclear explosions. CEX-71.1, Civil Effects Study. U. A. Atomic Energy Commission, DR-1886. 1971. [22] J. G. Bowen, E. R. Flechter and D. R. Richmond. Estimate of man’s tolerance to the direct effects of air blast. Lovelace Foundation for Medical Education and Research. Alburquerque, 1968. [23] CCPS, Guidelines for chemical process quantitative risk analysis. AIChE, New York, 1989. [24] National Institute for Occupational Safety and Health. Pocket guide to chemical hazards. U. S. Department of Health and Human Services, Washington (D. C.), 1994. [25] American Industrial Hygiene Association. Emergency response planning guidelines. Akron (Oh), 1991. [26] U. S. Department of Energy. ERPGs and TEELs for chemicals of concern. Rev. 16. Report WSMS-SAE-00-0001. http://tis-hq.eh.doe.gov/web/Chem_Safety/teel.html.
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[27] NRCCT. Acute exposure guideline levels for selected airborne chemicals, vol. 1.
Washington, D. C. National Academy of Press, 2000. [28] OEHHA. Office of Environmental Health Hazard Assessment (EPA), Technical support
document for the determination of acute reference exposure levels for airborne toxicants, California, 1998. [29] Scandpower, Handbook for fire calculations and fire risk assessment in the process industry. Scandpower, Kjeller, 1992. [30] R. E. Reinke and C. F. Reinhardt, Fires, toxicity and plastics, modern plastics, 1973. [31] J. L. Bryan, Fire Safety J. 14 (1986) 11. [32] H. L. Kaplan, A. F. Grand and G. E. Hartzell, Fire Safety J., 7 (1984) 22. [33] CCPS, Center for Chemical Process Safety, “Guidelines for Evaluating the Characteristics of Vapour Cloud Explosions, Flash Fires and BLEVEs”, AIChE, New York, 1994. [34] A. Ronza, M. Muñoz, S. Carol and J. Casal, J. Hazard. Mater., A133 (2006) 46-52. [35] E. Planas.Cuchi, J. A Vílchez, F. X. Pérez-Alavedra and J. Casal, J. Loss Prevent. Proc. 11 (1998) 323.
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Chapter 8
Quantitative risk analysis 1 INTRODUCTION The tools required for estimating the consequences of accidents are mathematical models for predicting accident effects (thermal radiation, overpressure, dose) and vulnerability models. The final risk is determined by multiplying these consequences by the frequency of the accidents over time. The ultimate aim of risk assessment for a given case, such as a process plant located close to an urban area, is to calculate the distribution of risk over the affected area by estimating the existing hazards and the consequences of the different accident scenarios and by applying the appropriate frequencies or probabilities. Quantitative risk analysis (QRA) consists of a set of methodologies for estimating the risk posed by a given system in terms of human loss or, in some cases, economic loss. Major accidents with severe consequences continue to occur despite the systematic application of classic safety measures such as engineering codes and checklists. Quantitative risk analysis provides an advanced and complementary approach that can be considered a further step towards safer industry. The first milestones in the quantitative risk analysis of handling, storing and processing hazardous materials are the two Canvey Reports developed by the UK Health and Safety Executive [1, 2] and the Rijnmond Report, issued by the Rijnmond Public Authority [3]. These studies were commissioned by public authorities in response to industry and public concerns about the risks posed by large concentrations of hazardous materials in handling and storage facilities. The Canvey Reports made a significant contribution, particularly in the field of accident frequencies, which were obtained statistically from historical data or by expert judgement. The Rijnmond Report gathered a large amount of failure frequency data and showed a high level of efficiency in presenting results, which were shown as f-N curves and risk contours. Various publications provide information on QRA methodologies. Two of the most important examples are the “Purple Book” published by TNO [4] and the Guidelines published by CCPS [5]. In this chapter the essential concepts associated with individual and societal risks and risk mapping are presented. Basic data on the frequency of loss-of-containment events are provided and the most common accident sequences that lead to the different accident scenarios are considered. Two simple examples are used to illustrate the essential aspects of quantitative risk analysis procedures, which apply the concepts of individual and societal risk and make only a brief reference to the calculation of effects and consequences. Then, the QRA methodologies and the mathematical models discussed in the previous chapters are applied to a real case.
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2 QUANTITATIVE RISK ANALYSIS STEPS Quantitative risk analysis is based on the application of mathematical models to determine the consequences of previously identified accident scenarios and on the use of the corresponding frequencies to estimate the resulting risk over a given area. QRA is performed through a series of steps or activities which are schematized in Fig. 8-1. Collection of relevant information Geographical situation
Climate data
Physical and chemical data of products
Technical data about the process or system analysed
Hazard scenario identification
Frequency estimation
Event trees (with probability data)
Effects and consequences analysis
Estimation of individual risk
Estimation of global risk for the population
Fig. 8-1. Main steps in a quantitative risk analysis [6].
The first step is to gather all of the relevant information. This may include geographical information about the location of the studied area: whether it is level or mountainous, the presence of rivers, its latitude and longitude, etc; climate data: humidity, solar radiation, wind
292
direction and velocity (wind rose), atmospheric stability, etc; the physical and chemical properties of the substances involved; and, finally, information about the process and the plant or system analyzed. Once this information has been obtained it is possible to identify the different hazard scenarios that must be considered: a hole in a pipe, a full pipe rupture, the explosion of a vessel, etc. Simplifying guidelines have been proposed for generic risk analysis (see Chapter 2, section 8). Once the possible types of release have been identified it is necessary to estimate their respective frequencies. This can be done by using generic data obtained from various sources. Table 8-4 to 8-6 show a selection of the most significant values. Each release type leads to diverse sequences which, in turn, cause different types of accident. Therefore, it is necessary to develop the corresponding event trees. By introducing the corresponding probability at each branch in the sequence it is possible to estimate the frequency of the different accident scenarios. The following nomenclature is used. The initiating event, for example the rupture of a pipe, is an incident. The physical result of an incident, for example the generation of a toxic cloud, is an accident; the accident can evolve according to an accident sequence, which can lead to several potential accident scenarios. Finally, the accident scenario is one of several specific situations at the end of an accident sequence, such as the development of a toxic cloud in a given direction.
Fig. 8-2. Example of an incident, accident sequence and accident scenarios with corresponding frequencies and probabilities.
The event tree plotted in Fig. 8-2 illustrates these definitions and the use of frequencies and probabilities. An impact on a tank that contains a flammable gas creates a hole (estimated frequency: f year-1); this is the incident or initiating event.
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The accident sequence at this point depends on the occurrence or non-occurrence of immediate ignition. If there is immediate ignition (probability P1) the resulting accident scenario is a jet fire, the frequency of which is obtained by combining the frequency of the initiating event and the probability of immediate ignition: (f · P1) year-1. If immediate ignition does not occur (probability: 1-P1) there are two subsequent possibilities: “delayed ignition” and “non-ignition”. In the case of delayed ignition (probability: P2) there are two further possibilities: either the amount of flammable product in the cloud (within the flammability limits) and the congestion/confinement are conducive to a vapour cloud explosion (probability: P3), or a flash fire occurs (probability: 1-P3) with no mechanical effects. In the first case, the final accident scenario is a vapour cloud explosion (plus a flash fire) the frequency of which is estimated as [f (1-P1) P2 P3] year-1. In the case of a flash fire the frequency is [f (1-P1) P2 (1-P3)] year-1. Finally, if there is no ignition the gas will disperse in the atmosphere (frequency: [f (1-P1) (1-P2)] year-1). Mathematical accident models and vulnerability models must be used to calculate the effects and consequences of each accident scenario. By combining the consequences with their corresponding frequencies we can obtain the individual risk at any desired distance and plot iso-risk curves. Once we know the individual risk at any point it is possible to estimate the overall risk to the population. 3 INDIVIDUAL AND SOCIETAL RISKS Risk is a function of the frequency of an accident and the magnitude of its consequences. The possible consequences are injury to the population or economic losses. Quantitative risk analysis often takes into account human loss. The possible types of harm to people are injury or fatality. However, a significant amount of additional work is required in order to analyze injuries (for example, second degree burns or harm due to a toxic gas) and a significant level of uncertainty affects the analysis in the case of multiple hazards. As a result, most risk assessments only consider fatal effects. The following definitions of individual and societal risks and the risk contour maps are based on fatal effects. 3.1 Definition of individual and societal risks The results of QRA for human loss are often expressed as individual and/or group or societal risks. Individual risk is the risk to a person in the vicinity of a hazard [7]. This definition includes the nature of injury to the individual, the probability of each type of injury and the time period during which the injury may occur. If individual risk is expressed in terms of fatalities it can be defined as a function of spatial coordinates which represent the probability that an individual at a fixed outdoor point during a one-year period will die as a result of an accident at a plant, installation or transport route. The unit used to measure individual risk is year-1. Individual risk at a given location can be expressed as [5]: i n
IR x , y
¦ IR i 1
(8-1)
x , y ,i
where IRx,y is the total individual risk of fatality at the geographical location x, y (probability of fatality per year or year-1)
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IRx,y,i is the individual risk of fatality at the geographical location x, y under accident scenario i (probability of fatality per year or year-1) and n is the total number of accident scenarios considered in the analysis. IRx,y,i can be expressed as a function of frequency and probability:
f i PFi
IR x , y ,i
(8-2)
where fi is the frequency of the accident scenario i (year-1) and PFi is the probability that the accident scenario i will result in a fatality at location x,y. The value of PFi is obtained by applying the accident effects models and vulnerability models. Finally, fi is calculated from the frequency of the incident (initiating event) and the probability that the sequence of events leading to the accident scenario i will occur: fi
f incident i Psequence i
(8-3)
where fincident i is the frequency of occurrence of the incident or initiating event (year-1) and Psequence i is the overall probability of the sequence of events that lead to the accident scenario i (-). Initiating events with a frequency of less than 1·10-9 year-1 are not usually considered in quantitative risk analysis. The same applies (when calculating the risk to the external population) to accidents where the distance at which the probability of lethality is 1% is within the limits of the plant. Major accidents can affect areas in which a large number of people are concentrated. In these cases the societal risk is used to measure the risk to a group of people. Societal risk is the expected number of casualties per year measured in casualties·year-1. It is calculated using demographic data for a given area: Societal risk =
³
(individual risk)·[population density (x, y)] dx dy
(8-4)
When estimating the societal risk in a given area it is possible to consider specific parameters such as time-of-day effects (different situations during the day or at night, for example), day-of-week effects (working days or weekends) and the protection provided by shelter (people indoors will be protected against certain accident scenarios). Finally, average individual risk can be calculated as the average of the individual risk of all people exposed to risk from a given facility. It is very important to know how this population has been selected, as the inclusion of people at low or no risk can introduce a significant bias. Average individual risk is often calculated for the exposed population, but it could also be calculated for the total population in a given area without taking into account whether all or only some of the people are exposed to risk from the facility. This can significantly lower the value of the average risk. The average individual risk for the exposed population is calculated as [5]: IR av
¦ IR x, y
x, y
p x, y
(8-5)
where IRav is the average individual risk (exposed population) (year-1) and px,y is the number of people in location x, y (-)
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The average individual risk can also be expressed (mainly for employees) as the fatal accident rate, i.e. the number of fatalities per 108 person-hours of exposure. 4 RISK MAPPING 4.1 Individual risk contours Individual risk contours or iso-risk curves represent the levels of individual risk around the installation analyzed. An iso-risk curve connects all of the geographical locations around a hazardous activity with an equal individual risk, i.e. all of the locations with the same overall probability of lethality. This is the clearest and most common way of producing a graphical representation of the risk over a given area. In order to produce an iso-risk curve it is necessary to calculate the respective contributions of different accident scenarios, each of which has its own probability and lethality: the resulting overall risk is the sum of the risks corresponding to each one of the accident scenarios. In order to establish the iso-risk curves it is necessary to perform the calculations corresponding to [8]: - the different accident scenarios (hole in a pipe, collapse of a tank, etc.), - the different reactions of the system to a particular incident (system failure, operator intervention, etc.), - the different meteorological conditions (atmospheric stability, wind direction and speed), - the different ignition locations and times. Individual risk contours can be circular if the physical effects of the accidents spread uniformly in all directions or more irregular if the intensity of the effects differs according to the direction. We can therefore distinguish between radial risk and directional risk. Radial risk produces circular iso-risk curves, where the value of risk decreases with the radius. Accident scenarios typically associated with this type of risk are fire, BLEVE, other vessel explosions and certain vapour cloud explosions: thermal radiation and overpressure essentially spread uniformly in all directions. Directional risk produces irregular iso-risk curves due to the non-homogeneous distribution of wind direction as a function of time (wind rose), the various possible grades of atmospheric stability and certain physical directional parameters. The accident scenarios associated with this type of risk are the atmospheric dispersion of flammable or toxic substances or the ejection of missiles in the explosion of cylindrical vessels. When radial and directional risks overlap, the overall iso-risk curves are obtained by adding the risks corresponding to each accident scenario and the shape of the curve is altered accordingly. 4.2 Procedure The following steps are used to calculate the individual risk curves. 1. Define a mesh of points in which to calculate the individual risk. The mesh must include the activity/plant analyzed and should cover at least the maximum distance over which any lethal consequences of the accident scenarios are observed. Mesh cells are usually 25 m x 25 m for distances of less than 300 m and 100 m x 100 m for larger distances. 2. Establish coordinates (x, y) for the point at which the accident scenario (j) originates. Establish the frequency (fji) of the initiating event for this accident scenario.
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3. Establish meteorological data: stability class and its corresponding probability (PM). A value of 0.8 is usually assumed for class D and 0.2 for class F. Establish the probability of the wind direction for each sector (wind rose) (Pw). P = 1 for accident scenarios that produce thermal radiation. 4. Determine the probability of the final accident scenario (Pjf). Determine the area covered by lethal effects and the probabilities of death (Pd) for each stability class and wind direction for each mesh point.
Fig. 8-3. Procedure for the calculation of the individual risk curves.
5. Calculate the contribution to individual risk ('IRji jf M,w) for the accident scenario considered, including the frequency of the initiating event, the probability of the final accident scenario, the probability of the stability class, the probability of the wind direction and the probability of death at a given point of the mesh: 'IR ji , jf , M , w
f ji Pjf PM Pw Pd
(8-6)
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6. The overall individual risk (IR) at a given point of the mesh is: IR
¦¦¦¦ 'IR ji
jf
M
w
ji , jf , M , w
(8-7)
This calculation process is shown in Fig. 8-3. Iso-risk curves are usually plotted for orders of magnitude. It is a fairly complex process which is usually performed by applying appropriate computer codes [12]. The following section provides an example of the calculation process for the individual risk at a given point (see also the Example case). 4.3 Societal risk Societal risk can be also presented as f-N curves. These are obtained by plotting the cumulative frequency (f) of accident scenarios that cause N or more fatalities per year as a function of N (usually on a log-log scale).
Fig. 8-4. f-N curve for a propane dehydrogenation plant.
In order to calculate a societal risk f-N curve it is important that the frequencies and the number of fatalities are combined correctly. The number of fatalities Ni from each accident scenario is: Ni
¦p x, y
x, y
PFi
(8-8)
where px,y is the number of people at location x, y (-) and PFi is the probability that the accident scenario i will cause a fatality at location x, y (-). The number of fatalities and the associated frequency must be estimated for each accident scenario. The data are then expressed as cumulative frequencies with
298
fN
¦f i
i
for all accidental scenarios i for which N i t N
(8-9)
where fN is the frequency of all accident scenarios with N or more fatalities (year-1) fi is the frequency of accident scenario i (year-1), and Ni is the number of fatalities in accident scenario i (-). f-N curves are mainly used in land use planning to ensure that residential areas, schools, hospitals and other public spaces fall outside specific risk contours. Fig. 8-4 shows an example of an f-N curve for a given chemical plant. Obviously, the frequency decreases as the number of fatalities increases, i.e. the accidents that cause a large number of fatalities occur far less frequently than accidents that cause few fatalities. 5 INTRODUCTORY EXAMPLES OF RISK CALCULATION
Three simple examples are used to illustrate the concepts described in the previous sections: the first is a general case involving several accident scenarios and the second deals with an explosion in a specific unit. ______________________________________ Example 8-1 A pharmaceutical substance is solved in ethanol (20% mass). 300 kg h-1 of this solution must be dried to a humidity of 0.05 kg (kg dry solid)-1. Drying must be performed in a spray drier, using 1,110 kg h-1 of dry air, previously heated. a) Calculate the concentration of ethanol in the exit air (at 70 ºC). Will the operation be correct? b) In case of an explosion, two operators would die. At the plant there are 70 workers per shift (two shifts of 8 h, 250 days per year). The plant has been working for10 years, without any previous accidents. Calculate the value of the FAR after the explosion, and the frequency (fatalities per person and year). Data: vaporization heat of ethanol = 850 kJ kg-1. Solution a) 300 · 0.2 = 60 kg h-1 of dry solid
300-60 = 240 kg h-1 ethanol 60 · 0.05 = 3 kg of ethanol in the dried solid Vaporized ethanol: 240 – 1 = 239 kg h-1 Volume: 239 kg 22.4 m 3 kmole 1 273 70 46 kg kmole -1 273
146.2 m3 h-1
Air flow rate at exit: 1,110 kg h-1 · 0.98 m3 kg-1 = 1088 m3 h-1
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c
146.2 146.2 1088
0.118 m3 m-3 i.e. 11.8% volume.
Flammability limits for ethanol/air mixtures: 3.3–19% volume: the operation would not be safe. b) Time worked at the plant over the last 10 years: 70 · (2 · 8) · 250 = 2.8 · 105 h · person FAR =
2 · 10 8 2.8 10 5
714
8 250 714 0.014 fatalities person-1 year-1. 8 10 ______________________________________ ______________________________________ Example 8-2 Consider a processing plant in which the following possible incidents have been identified: 1. The release of a toxic gas due to the full-bore rupture of a pipe (length: 4 m). 2. The release of a flammable gas due to the catastrophic failure of a vessel. The following simplifying hypotheses are assumed: - Both events are caused at the centre of the plant. - Only two typical weather conditions are present: the atmospheric stability is constant and wind blows from the north (70% of the time) or from the east (30% of the time). - The probability of death due to an accident at a given location is either 0 or 1. Additional data (Table 8-4): frequency of vessel catastrophic failure: 5·10-6 year-1; frequency of rupture of a pipe: 3·10-7 m-1 year-1; probability of ignition of a flammable cloud: 0.5. Estimate both the individual and the societal risk for the population distribution in the area shown in Figs. 8-6 and 8-7. (Note: this example is based on the case published by Hendershot [9]).
Frequency =
Solution The event trees in Fig. 8-5 show the various accident scenarios that can occur and their respective frequencies. The possibility of a flash fire in the case of a flammable gas release is not considered. For Event 1, the probability that the plume moves in a certain direction is not usually included in the event tree but instead is considered in a later step; in this case this probability has been included in order to provide a clearer picture of all possibilities considered. The frequencies of each accident scenario are calculated by multiplying the frequency of the initiating event by the probabilities of each branch: ftoxic cloud to south = (3 · 10-7) · 4 · 0.7 = 8.4 · 10-7 year-1 ftoxic cloud to west = (3 · 10-7) · 4 · 0.3 = 3.6 · 10-7 year-1 fexplosion = (5 · 10-6) · 0.5 = 2.5 · 10-6 year-1.
Now the effects and consequences must be established for each accident scenario.
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Fig. 8-5. Event trees and frequency estimations for the two incidents considered.
After applying accident and vulnerability models, assume that the following consequences are estimated for each accident scenario: - Toxic cloud (both to the south and to west): all people in a circular sector of radius 200 m and 20º width are killed (probability of death = 1) and all people outside this area are unaffected (probability of death = 0). - Explosion: all people within 130 m of the explosion centre are killed (probability of death = 1) and all people beyond this distance are unaffected (probability of death = 0).
Fig. 8-6. Individual risk contour map (the dotted line represents the facility boundary).
Fig. 8-6 shows the areas covered by each accident scenario and the facility contour. In some areas the effects of more than one scenario overlap (Areas III and IV, which are subjected to both the explosion and the toxic cloud).
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The individual risk from an accident scenario in a given area can be calculated by multiplying the probability of death (P = 1) by the frequency of the scenario. In areas affected by more than one accident scenario the individual risks corresponding to each scenario are added to obtain the total individual risk. For example, in Area IV: IRIV = 8.4 · 10-7 + 2.5 · 10-6 = 3.34 · 10-6 fatality · year-1. Table 8-1 shows the total individual risks for the various areas. Fig. 8-6 shows the risk contours corresponding to these IR values. In QRA, IR contours are usually plotted for orders of magnitude of risks (10-4, 10-5, etc.). Table 8-1 Total individual risk corresponding to each area Area Accident scenarios IR, year-1 I 2-a 2.5 · 10-6 II 2-a 2.5 · 10-6 III 2-a, 1-b (west) 2.86 · 10-6 IV 2-a, 1-a (south) 3.34 · 10-6 V 1-b (west) 3.6 · 10-7 VI 1-a (south) 8.4 · 10-7
Assume the distribution of people shown in Fig. 8.7. These people are subject to the risk covering each area. Consequently, the 14 people located to the south-east beyond the distance reached by the lethal effects of the explosion are not exposed to any risk, while the 12 people located further to the south (Area VI) are exposed to a certain degree of risk.
Fig. 8-7. Distribution of population in the accident area (* indicates employees).
The average individual risk to the population subject to risk is therefore calculated as:
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IR av
4 2.5 10 6 3 2.5 10 6 8 3.34 10 6 4 3.34 10 6 6 3.6 10 7 12 8.4 10 7 4 3 8 4 6 12
1.89 10 6 year -1
If the calculation is based on the total population in the area (including the 14 people not subject to any risk) the average individual risk is: IR av
4 2.5 10 6 3 2.5 10 6 8 3.34 10 6 4 3.34 10 6 6 3.6 10 7 12 8.4 10 7 4 3 8 4 6 12 14
1.37 10 6 year -1
However, it is more common to calculate the individual risk on the “external” population, which excludes the employees inside the plant contour. In this case the individual risk to the affected population would be calculated as: IR av
6 3.6 10 7 4 3.34 10 6 12 8.4 10 7 1.16 10 6 year-1 6 4 12
The average individual risk to on-site employees is: IR av
4 2.5 10 6 3 2.5 10 6 8 3.34 10 6 2.95 10 6 year-1 4 38
However, this is not an interesting value, as a) employees do not work 24 h per day, 365 days per year, and b) they are exposed to additional risks due to their activities. Therefore, IRav should not be used to calculate the fatal accident rate for employees. Typically, for a process plant operator FAR | 4; the risks are originated by major accidents (50%) and by conventional risks (50%). The f-N curve is used to represent the societal risk. It is necessary to know the estimated number of fatalities (external population) corresponding to each accident scenario (Table 8-2) in order to plot the curve. Table 8-2 Estimated number of fatalities for each accident scenario Number of Accident scenario f, year-1 fatalities, N 1-a (south) 8.4 · 10-7 16 1-b (west) 3.6 · 10-7 6 2-a 2.5 · 10-6 4 2-b 2.5 · 10-6 0
The data in Table 8-2 can be used to calculate the cumulative frequency (Table 8-3). This is done by rearranging the accident scenarios according to descending number of fatalities and calculating the frequency of N or more fatalities.
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Table 8-3 Cumulative frequency data Estimated number Accident scenarios of fatalities, N included > 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.
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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.
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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
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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
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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
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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.
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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
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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
100000
3 1011 1011
Percentage death skull fracture (%)
95 50 300 m
5
10000
95
200 m
3 1010
Explosion energy (E exp, J)
1012
90
400 m Impulse (Pa·s)
600 m
50
1010
25
1000
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.
351
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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 <, 35 Discharge coefficient, 22, 33 Discharge from a safety relief valve, 45 Dispersion coefficients, 207-212, 215-216 Dispersion models, 198-240 aerosol dispersion, 232 buoyant turbulence, 198 concentration contour coordinates, 227 continuous/instantaneous releases, 205 continuous emission, 209 dispersion coefficients, 210-212, 215216 Gaussian models, 208-218 heavier-than air gases, 219 infectious agents, 229-233 mechanical turbulence, 198 overlaping plumes, 211-212 plume rise, 207
E Edwards, J. D., 11 Effective height of emission, 207 Effective orifice diameter, 91 Eisenberg, N. A., 256, 257, 264 Els Alfacs accident, 160-163 Emergency Response Planning Guidelines, 272 Emissive power, 80-83, 94, 107 average, 82-83 contours, 81 diesel oil, 82 estimation, 83 gasoline, 82 luminous and non-luminous flame, 81 soot, 82 Emissivity, 80 Emmons, 99 EPA, 107, 228 Epstein, M., 44 Equivalent length of pipe fittings, 26 Error function, 244-245
355
Escaping, 234, 258 Evaporation, 53-55 from a pool on ground, 53, 84-85 from a pool on water, 53, 85-86 of boiling liquids, 53-54 of non-boiling liquids, 54-55 Event trees, 4, 293 External events, 2 Explosions, vapour cloud, 119-196 Baker-Strehlow-Tang method, 131 blast energy of TNT, 125 blast wave, 121 blast scaling, 123 characteristic curves, 127, 347 combustion energy-scaled distance, 124, 133 comparison of methods, 136-137 congestion, 133 cube root scaling law, 123 deflagrations, 123 detonations, 122 distribution, 119 dynamic pressure, 122 flame speed Mach numbers, 134 free-air and ground explosions, 124 high explosives, 122-124 Hopkinson scaling law, 123 impulse, 122 minimum amount of fuel, 120 multi-energy method, 129-132, 347 peak side-on overpressure, 122 positive phase duration, 131 Port Hudson explosion, 122 reflected pressure, 264 reflection factor, 124 probable number of fatalities, 138-140 Sachs scaled distance, 124 Scaled distance, 123-124 shock wave, 121-122 TNT equivalency method, 125, 347 types, 7 vulnerability, 263-270 wind, 264 yield factor, 125 Eyre, J. A., 65
Failure, probability, 305 Fanning friction factor, 25, 27-28 Fatal Accident Rate, 296, 300, 303 Fault trees, 4 Fauske, H. K., 43, 44 Félez, S., 119 Ferris, N. P., 231 Finney, D. J., 251-253 Fire, 61-117 body area burned, 255 fireball, 104-109 flammability, 66-72 flash fires, 65, 99 hot air, 261 jet fires, 83-97 pool fires, 84-87 substances released, 275-276 tetrahedron, 62 triangle, 62 types, 64-65 vulnerability, 254-263 Fireball, 104-109 emissive power, 107 fireball duration and diameter, 105 fireball geometry, 104 ground diameter, 104 height reached, 105-106 fireball and target position, 107 propane, 109-112, 319-323 radiant heat fraction, 106 thermal features, 106-108 thermal radiation dose, 108 variable D, H, E model, 108 view factor, 108 Fisher, H. G., 42 Filling and emptying a tank, 72 Flames, 63 diffusion, 63 drag, 88 emissive power, 80-83 height and length, 87 premixed, 63 size, 83, 92, 95, 99 speed Mach numbers, 132 tilt, 88 Flames impingement on vessel, 48 Flammability, 66 Flammability limits, 66
F FACTS database, 3
356
G Gas concentration, units, 204 Gas/vapour release, 30-41 critical velocity, 30 dimensionless factor, 35 discharge coefficient, 33 flow of gas through a pipe, 35-41 flow of gas through a hole, 30-35 speed of sound, 31 propane release, 33-35 Gasulla, N., 179, 187 Gaussian models, 206-216 Gibson, C. F., 233 Gifford, S. A., 208, 220 Gloster, J., 233 Gong, Y. W., 159 González Ferradás, E., 127, 176-178, 347 Grand, A. F., 275 Green, D. W., 164 Griffiths, R. F., 212, 215 Grossel, S. S., 42, 60 Ground explosions, 124 Gu, A. Z., 159 Guillemet, R., 147, 184 Guillemin, M., 242, 243, 244
estimation, 67-68 of gas mixtures, 69 as a function of pressure, 70 as a function of temperature, 70 of diverse substances, 67, 335-336 of hydrogen, 66 Flammability diagrams, 71-72 Flash fires, 65, 99 size of the flames, 99 heat fluxes, 99 Flashing liquids, 42 Flash point temperature, 72 estimation, 73 Flash temperature of diverse substances, 335-336 Flechter, E. R., 265 Flixborough accident, 128 Foot-and-mouth disease, 233 Flow, 21-60 incompressible, 30 of flashing liquids, 42 of liquid through a hole, 21 of liquid through a pipe, 24 of gas/vapour through a hole, 30 of gas/vapour through a pipe, 35 of propane (gas), 317-318 relief discharges, 44-52 time-dependent gas release, 40 two-phase flow, 42-44 f-N curve, 298, 303-304 Folch, J., 85, 87 Forman, A. J., 231, 233 Forrest, H. S., 42 Foster, T. C., 23 Fragile breaking of vessels, 171 Fragments, explosion, 178-183, 267 Free-air explosions, 124 Frequencies and probabilities, 306-309 failure of repression systems, 306 human error, 306 ignition and explosion, 306-309 loss of containment events, 306-307 meteorological data, 309 Friction factor, 25, 27-28 Friction loss, 21 Frothover, 100 Full-bore rupture of a pipe, 35-40 Fumarola, G., 216
H Haanes, H, 40 Haber’s law, 272, 274, 282 Hamblin, C., 233 Hamins, A. 74 Harkleroad, M., 262 Hartzell, G. E., 275 Hawthorne, W. R., 92 Hazard, definition, 1 HAZOP, 3-4 Hayhurst, C. J., 129, 130, 137 Heat wave, boilover, 100-102 Heavier-than-air gases, 219 Heat of combustion diverse substances, 335-336 Hendershot, D. C., 300 Hertzberg, M. 87 HID, 105 High explosives, 120 Hilado, G., 67 Historical analysis, 2 Hodin, A., 102, 103
357
Jet flow, 90 effective orifice diameter, 91 gas temperature, 91 gas density, 91 gas velocity, 91 Mach number, choked flow, 92 Jet temperature, 91 Jiménez Sánchez, T. J., 347 Johnson, D. M., 167, 169, 170 Jones, G. W., 67
Holden, P. L., 180 Hole in a pipe, 35-40 Hole in a tank, 21, 30 Holland, J. Z., 208 Hopkinson scaling law, 123 Hot air, consequences on people, 261 Hotel, H. C., 64, 92 Howell, J. R., 78, 87 HSE, 291 Huff, J. E., 42, 52 Hugh-Jones, M., 242, 243, 244 Hulbert, W. G., 87 Human error, 306 Hydrogen fluoride dispersion, 219
K Kalghatgi, G. T., 95 Kaplan, H. L., 275 Kern, G. R., 95 Ketchum, D. E., 133-134 Khudiakov, G. N., 64 Killed versus injured people, 280-281 Kitching, R. P., 233 Kletz, T., 184 Koen, 11 Kubota, T., 84 Kulesz, J. J., 122, 124, 125, 176, 181, 263, 264, 267
I Ignition properties of materials, 262 Immediately Dangerous to Life and Health, 272, 345-346 IDLH, diverse substances, 345-346 Ignition properties of materials, 262 Incident, 293 Individual risk, 294-296, 301, 327 INERIS, 103 Inerting, 71 Inert gases, effects, 277-278 Infectious agents, dispersion, 229-244, 240-242 Initiating event, 293 Injured people, number, 280-281 Instantaneous releases, 205 Intervention zone, 281-282
L Langmuir, A., 242, 243, 244 Lannoy, A., 129 Lapse rates, 197 Lautkaski, 88 LC01, 274-275 Le Chatelier, H., 69 Le Chatelier method, 69 Lees, F. P., 219 Leung, J. C., 50-52, 54 Levenspiel, O., 21, 26-28, 30, 37 Li, J., 67 Lift-off distance, 92, 95 Lin, W. S., 159 Linares, J. L., 137 Liquid release, 21-29 flow through a hole in a tank, 21-24 flow through a pipe, 24-29 discharge coefficient, 22 friction factor, 25, 28 roughness of clean pipes, 27 Liquid superheating, 151-152 LNG explosion and fireball, 188-190
J Jet fires, 83-97 Chamberlain model, 94-97 effective orifice diameter, 91 emissive power, 83, 94 flames size and shape, 92-96 Hawthorne et al. model, 92-93 in a calm situation, 92 influence of wind, 94-97 jet flow, 90 jet temperature, 91 jet tilt, 96 length of the flames, 95 lift-off distance, 92, 95 of propane, 317-319
358
Lobato, J., 137 Londiche, H., 147, 184 Loss of containment general guidelines, 56 pressurized tanks and vessels, 56 atmospheric tanks, 56 pipes, 56 pumps, 56 relief devices, 56 warehouses, 57 transport units in an establishment, 57 pool evaporation, 57 outflow and atmospheric dispersion, 58 Louvar, J. F., 21, 22, 54, 121, 164 Lu, X. S., 159 Ludwig, A., 169 Lynch, C. J., 256, 257 Lynchburg flash fire, 259
Merckx, W. P. M., 129, 130, 137 Meselson, M., 242, 243, 244 México City accident, 150, 183 MHIDAS database, 3 Michaelis, P., 102, 103 Miñana Aznar, A., 127, 176-178, 347 Missiles, 178-183 number of fragments, 179 common failure trends, 179 direction, 180 range, 181 velocity, 182 Mizner, G. A., 65 Montiel, H., 9, 27, 86, 119, 171, 253 Moran, K. C., 129, 130, 137 Moorhouse, R., 88 Mounding of tanks, 183 Mudan K. S., 78, 84, 85, 88, 255 Muller, A. R., 42 Multi energy method, 129-130, 347 Muñoz, M., 61, 77, 80, 81, 83, 87, 280
M Macfarlane, K., 219, 225 Maddison, T. E., 184 Mahieu, A. P., 227 Maillette, J., 160 Major accidents, 5-11 definition, 5 distribution, 9 domino effect, 12-13 effects, 8 general scheme, 8 involving fire, 5-6 involving explosions, 7 mathematical modelling, 14 types of damages, 11 Maloney, J. O., 164 Mans, C., 160 Marshall, V. C., 124, 138 Martin, J. M., 254 Martínez Alonso, J., 127, 176-178, 347 Martel, B., 70, 73 Martinsen, W. E., 105, 106, 107, 136-137 Marx, J. D., 105, 106, 107 Mathematical modelling of accidents, 14 overestimation, 15 Mavrothalassitis, G., 102, 103 Mcdevitt, C. A., 150 McGrattan, K. B., 74 McQuaid, J., 205, 206, 221-225 Mechanical energy balance, 21
N Nazario, F. N., 151, 184 NFPA, 48, 49 NIOSH, 271 Noronha, J. A., 42 NRCCT, 271 O OEHHA, 275 Oggero, A., 61 Oldham, G. A. 176 Oldshausen, K. D., 40 Oliver, R., 233 Olson, D. B., 135, 141-142 Ooms, G. 227 Outflow function for SRV, 46 Overpressure, 50, 263-270 damage to human beings, 263-269 consequences on buildings, 269-270 pressure wind, 264 reflected, 264 Overtemperature, 51 P Paese accident, 106 Palazzi, E., 218
359
Papazoglou, I. A., 109 Parr, B. V., 176 Pasquill, F., 208, 220 Pathogenic agents, dispersion, 231-235, 242-244 aerosol dispersion, 232 antrax dispersion, 242-244 dispersion of airborne viruses, 232 dose, 233 emission source, 231 foot-and-moth disease outbreak, 233 ID50, 233 influence of wind and stability, 235 respiratory rates, 234 Sverdlovsk accident, 242-244 Peavy, H. S., 196, 203, 204 Pérez-Alavedra, X., 283 Perry, R. H., 164 Perry, W. W., 257 Piccinini, N., 3, 105 Pierorazio, A. J., 133-134 Pietersen, C. M., 150, 258, 265, 269, 296 Pipe roughness, 27 Pitblado, R. M., 107, 190 Planas, E., 9, 61, 70, 71, 80, 81, 85, 86, 87, 168, 171, 179, 187, 227, 234, 237, 283 Plume rise, 207 p-N curve, 120 Poggio, A., 105 Point source model, 74-76 Poling, B. E., 32 Pool, 84-86 equivalent diameter, 84 evaporation from a pool, 53 non-boiling liquids, 54 pool diameter, 84 pools on ground, 84 pool size, 53 pools on water, 85 Pool fires, 20-23, 84-87 burning rate, 86 critical time, 84 emissive power, 20-23 equilibrium diameter, 85 equivalent diameter, 84 flame tilt and drag, 88 height and length of the flames, 87 influence of wind, 87
instantaneous spills, 84 pool diameter, 84 pools on ground, 84 pools on water, 85 size, 84 Poor oxygen atmospheres, 278 Population response to an accident, 249 Popova, I., 242, 243, 244 Prausnitz, J. M., 32, 75 Premixed flames, 63 Pressure wave, 119, 121,169 Pressure wind, 264 Priolo accident, 12-13, 106 Pritchard, D. K. J., 122, 123, 126, 169, 170 Probable number of fatalities, explosion, 138-140 Probit analysis, 250-253 Probit equations, 257-258, 264-268, 273, 347 thermal radiation, 254-263 blast, direct consequences, 263-265 blast, body displacement, 265-267 explosion, fragments267-268 Probit-percentage, table, 253 Probit-percentage, analytical, 287-288 Propylene explosion, 160 Propylene release, 229 Prothero, A., 219, 225 Prugh, R. W., 149, 166 Pryor, A. J., 261 Puff dispersion, 206 Purdy, G., 241 Puttock, J. S., 219 Q Quantitative risk analysis, 4, 291-330 accident sequence, 293 Fatal Accident Rate, 296 example case, 309 f-N curves, 296, 297 individual risk, 294-296, 301, 327 individual risk curves, 323-324, 329 iso-risk curves, 296-298 risk mapping, 296 societal risk, 295-296 steps, 292 Quintela, J., 85, 87 Quintiere, J. G., 262
360
R Radiant heat fraction, 106 Raj, 99 Rao, P. G., 73, 104 Rausch, A. H., 257 Redlich-Kwong equation of state, 152 Rees, F. J., 219, 225 Reeves, A. B., 180 Reid, R. C., 32, 75, 147, 153, 154 Reinhardt, C. F., 275 Reinke, R. E., 275 Relative humidity, 204 Relief discharges, 44-52 evolution of pressure and temp., 50 from a safety valve, 44-47 Leung method, 51 vessels subject to external fire, 48 vessels with runaway reaction, 49 outflow function, 46 Repression systems, failure, 306 Rew, P. J., 87 Richardson number, 96 Richmond, D. R., 265 Ricker, R. E., 176 Rijnmond report, 291 RPA, 291 Risk, 1 analysis, 2-3 analysis steps, 3 classification, 2 definition, 1 individual, 294-296, 301, 327 iso-risk curves, 296 mapping, 296 societal, 295-296 quantitative risk analysis, 4, 291-331 Roberts, A. F., 106 Roberts, M. W., 105 Roberts, P. T., 219, 225 Roberts, T., 163 Robertson, N. J., 129, 130, 137 Rodrigo, M. A., 137 Ronza, A., 7, 10, 119, 280, 292, 307-309 Roughness of clean pipes, 27 Rowe, D. R., 196, 203, 204 RPA, 291 Ruíz Gimeno, J., 127, 176-178, 347 Runaway reactions, 49
S Sachs, R. G., 124 Sachs scaled distance, 124 Sachs scaled impulse, 124 Sachs scaled overpressure, 124 Sáez, C., 137 Safety relief valves, 44-47 discharge from a SRV, 45-47 minimum orifice area, 45-46 outflow function, 46 Salla, J. M., 151-152, 156-159, 168, 173 Sant Carles de la Ràpita accident, 160 Sánchez Pérez, J. F., 127, 176-178, 347 Santamaría, J. M., 54 Satyanarayana, K., 73, 104 Scaled distance, 122-124, 171 Schecker, H. G., 100, 103 Sellers, R. F., 231, 233 Seveso accident, 50 Sevilla, S., 117 Shaw, D. A., 42 Shebeko, Y. N., 185 Shelokov, A., 242, 243, 244 Sheltering, 236, 279-280 Sherwood, T. K., 75 Shevchuck, A. P., 185 Ship, fire, 283 Shock wave, 120 Short-term release, 218 Siegel, R., 78, 87 Simpson, D., 263 Sloping ground, 183 Slopover, 100 Smith, J. M., 166 Smolin, Y. M., 185 Societal risk, 295-296 Solid flame model, 77-83 atmospheric transmissivity, 75-76 view factor, 78-80 emissive power, 80-83 Solum, G., 40 Sonic flow, 31 Source term, classification, 20 Speed of sound, 31-32 Spinodal curve, 151-154 Spirax-Sarco, 44, 46 Stawczyk, J., 181 Stensaas, J. P. 40
361
U U. S. Department of Energy, 273, 275
Steward, F. R., 150 Strehlow, R. A., 122, 124, 125,181, 263, 264, 267 Substances released from a fire, 275-276 Sun radiation, 202 Superheating energy, 173-176 Superheat limit temperature, 153-159 Surface roughness length, 198 Sverdlovsk accident, 242-244
V van den Berg, A. C., 129, 130, 137 van het Veld, B. F. P., 296 van Ness, H. C., 166 Vaporization fraction, 42, 167-168 Vapour cloud explosions, 119-143 Baker-Strehlow-Tang method, 131-133 blast, 121 blast scaling, 121 blast wave, 121-122 characteristic curves, 125 combustion energy-scaled distance, 124 comparison of methods, 134-135 cube root scaling law, 122 distribution, 117-121 detonations, 120 deflagrations, 121 flame speed Mach number, 132 Flixborough explosion, 126 free-air explosions, 122-123 ground explosions, 122-123 Hopkinson scaling law, 121 multi energy method, 127-130 probable number of fatalities, 136-138 propane storage area, 130 Sachs scaled overpressure, 122 Sachs scaled distance, 122 TNT equivalency method, 123-126 scaled distance, 123-124 Texas City refinery accident, 138-141 Venart, J. E. S., 163 Vento di Scirocco accident, 281-284 Ventosa, A., 179, 187 Vessel explosions characteristic curves, 178 ductile and fragile breaking, 171 energy released, 166-169, 173-175 irreversible expansion, 168, 173 isentropic expansion, 166, 173 missiles, 178-183 overpressure, 170 pressure wave, 169 pressure required for vessel failure, 164 superheating energy, 173-176
T Tchobanoglous, G., 196, 203, 204 Temporary Emergency Exposure Limits, 272 Tennankore, K. N., 150 Tensile strength of diverse materials, 165 Terminal velocity of particles, 230 Texas City refinery accident, 140-143 Thermal conductivity of solid media, 54 Thermal diffusivity of solid media, 54 Thermal insulation of vessels, 184 Thermal inversion, 199 Thermal radiation intensity, 74-83, 254 levels of damage, 257 point source model, 74-76 solid flame model, 77 tolerable limit, 254 vulnerability, 254-263 Thin-layer boilover, 65, 101 Thomas, J. K., 133-134 Thomas, P. H., 87 Thon, H., 40 Tilley, B. J., 42 Tillner-Roth, R., 171 Time-dependent gas release, 40 Tivissa accident, 187-190 TNO, 94, 198, 231, 291, 298 TNT equivalency method, 125, 347 TNT overpressure curve, 126 Top event, 4 Townsend, W., 159 Toxic substances, vulnerability, 271-277 Tsao, C. K., 257 Turner, S. B., 207, 209 Two-phase flow, 42-44 flashing liquids, 42-43 vaporization fraction, 42 two-phase discharge, 43
362
tensile strength of materials, 165 View factor, 78-80, 108 Viruses atmospheric dispersion, 232 Vílchez, J. A., 10, 11, 27, 71, 86, 119, 120, 138, 171, 227, 253, 283, 304, 307-309 Vulnerability, 249-290 dose-response relationship, 252 effects of inert gases, 277-278 influence of sheltering, 279-280 injured people, number, 280-281 poor oxygen atmospheres, 278 population response to an accident, 249 probit analysis, 249-252 probit equations, 257-258, 264, 273 probit variable-percentage, table, 253 to thermal radiation, 254-263 to overpressure, 263-270 to toxic substances, 271-277 zoning, 281-282 Vulnerability to explosions, 263-270 Haber’s law, 272, 282 eardrum damage, 265 fragments, 267-268 impact, 265-266 indirect consequences, 265 lethality, pulmonary hemorrhage, 264 probit equations, 264-268 reflected blast wave, 262 Vulnerability to thermal radiation, 254-263 body area burned-age-mortality, 255 effects of hot air, 261 escape, 258 ignition properties of materials, 262 levels of damage, thermal fluxes, 257 material damages, 261-262 probit equations, 257-258 tolerability limit, 254 Vulnerability to toxic substances, 271-277 Acute Exposure Guideline Levels, 271 Emergency Response Planning Guidelines, 272 effects of carbon dioxide, 277 effects of carbon monoxide, 276 effects of inert gases, 277-278 LC01, 274-275 poor oxygen atmospheres, 278 probit equation, 273 probit equation constants, 274
substances released from a fire, 275 Temporary Emergency Exposure Limits, 272 Threshold Limit Value-TWA, 271 Threshold Limit Value-STEL, 271 Threshold Limit Value-C, 271 W Walls, W. L., 147 Water barriers, 185 Water tank explosion, 175 Weddell, D., 92 Westine, P. S., 122, 124, 125, 176, 181, 263, 264, 267 White, C. S., 265 Wickens, M. J., 169, 170 Wighus, R., 40 Windhorst, 11 Wind direction fluctuations, 202 Wind, influence, 87, 94 Wind, pressure, 264 Wind rose, 196 Wind speed profile, 197 Witlox, H. W., 219, 225 Y Yampolskaya, O., 242, 243, 244 Ye, Z., 160 Yield factor, explosions, 123 Z Zabetakis, M. G., 66, 70, 122 Zelis, F., 227 Zoning according to vulnerability, 281 Zook, J., 159 Zukoski, E. E., 84
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