Evaluation of The Effects and Consequences of Major Accidents in Industrial Plants

 

Evaluation of the Effects and Consequences of Major Accidents in Industrial Plants

 

Industrial Safety Series

Vol. 1. Safety of Reactive Reactive Chemicals Chemicals (T. (T. Yoshida) Yoshida) Vol. 2. Indiv Individua iduall Behaviour Behaviour in in the Control Control of of Danger Danger (A.R. Hale Hale and A.I. Glendon) Vol. 3. Fluid Mechanic Mechanics s for Industrial Industrial Safety Safety and Environme Environmental ntal Protection Protection (T.K. (T .K. Fannelöp) V Ha zards oChemicals f Chemical Chemi cal and Rea ctions s (T. (T. Grewer) Gre Vol. ol. 4. 5. Thermal Safety Saf ety of Hazards Reactiv Re active eof Chem icals aReaction nd Pyrotechn Pyro technics ics wer) (T. Yoshida, Y. Wada and N. Foster) Vol. 6. Risk Assess Assessment ment and and Managemen Managementt in the Conte Context xt of the the Seveso Sev eso II Directive (C. Kirchsteiger, Editor and M. Christou and G. Papadakis Papadakis,, Co-editors) Vol. 7. Critica Criticall Temper Temperature atures s for the Thermal Thermal Explosion Explosion of Chemical Chemicals s (T. Kotoyori) Vol. 8. Eva Evaluati luation on of the Effects Effects and Conseque Consequences nces of Major Major Accidents Accidents in Industrial Plants (J. Casal)

 

Industr ial Safety Safety Se Series, ries, 8

Evaluation of the Effects and Consequences of Major Accidents in Industrial Plants Joaquim Joaqui m Casal Casal Centre for Studies on Technological Risk Department of Chemical Engineering Universitat Politècnica de Catalunya Barcelona, Spain

Elsevier   Am st erd erdam am – B os to n – Heidel Hei del ber berg g – L on do n – New Yor k – Ox fo ford rd Paris – San Diego – San Francisco – Singapore – Sydney – Tokyo

 

Elsevier Radarweg 29, PO Box 211, 1000 AE Amsterdam, The Netherlands Linacre House, Jordan Hill, Oxford OX2 8DP, UK

First edition 2008 Copyright © 2008 Elsevier B.V. All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [email protected]. [email protected]. Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent verification of diagnoses and drug dosages should be made Library of Congress Cataloging-in-Pu Cataloging-in-Publication blication Data  A catalog record for this book is available from the Library of Congress British Library Cataloguing in Publication Data  A catalogue record for this book is available from the British Library

ISBN: ISB N: 978 978-0-0-444 444-53 -53081 081-3 -3

For information on all Elsevier publications visit our website at books.elsevier.com

Printed and bound in Hungary 07 08 09 10 11

 

10 9 8 7 6 5 4 3 2 1

 

For Miriam. And for the swallows that build their nests under our roof every year.

 

 

V

This page intentionally left blank 

 

 

 

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 Piccinini, at the Polytechnic University of thank Turin, my whofriend taught Norberto a pioneering course aonprofessor 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.

 

VII

 

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

 

VIII

 

 

Contents

Preface v 1. I ntr oduc oductio tion n 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 12   4.2 An example case 12 12   5. Mathematical16 modelling of accidents 14  Nomenclature References 16

2. So Sour ur ce 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

 

IX

 

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 quantitative risk analysis 55 8.1 Loss-of-containment events in pressurized tanks and vessels 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. F ir e accidents 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 and flammability diagrams 71 4.2 4.1.5 Flash Inerting 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

  X

 

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 size of the the jet 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. V apour cloud cloud explosions 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. BL E V E s and ves vessel explosions explosions 147 1. Introduction 147 2. Mechanism of BLEVE 149

 

XI

 

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. A tm os ospher pher ic disper disperssion of toxic or flam m able clouds 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 Dispersion models for neutral gases (Gaussian (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

 

XII

 

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. V ulne ulnerr ability 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

 

XIII

 

8. Q uantitat ive r is iskk analysis 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 loss-of-containment events events 306 6.2 Failure repression 6.3 Human of error 306 systems 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 initiating events events 311 7.2 Event trees trees of the diverse 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

A nnex nnex 1 Constants in the Antoine equation 333  Annex 2 Flammability levels, flash temperature and heat of combustion (higher value) for different substances 335  A nnex nnex 3 Acute Exposure Guideline Levels (AEGLs) 337 A nnex nnex 4 Immediately Dangerous to Life and Health Health concentrations (IDLH) 345 Annex 5   Determining the damage to humans from explosions using characteristic curves 347

Index   353

 

XIV

 

Chapter 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, environmen environmental tal 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 magnitudee of its probable consequences: magnitud Risk = frequency · magnitude of consequences Therefore, if an accident occurs once every 50 years and its consequences are estimated to  be one hundred fatalities, 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  (“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 methodol ogies 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 meteorolog meteorological ical 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

1

 

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: Cate Ca tego gory ry A ri risk sks: s: tho those se th that at are are una unavo void idab able le and and ac acce cept pted ed wi with thou outt any any compensation compensatio n (for example, the risk of death caused by lightning). Catego Cat egory ry B risk risks: s: tho those se tha thatt are, are, str strict ictly ly spe speaki aking ng,, avoi avoidab dable le bu butt whic which h must must be

considered unavoidable in everyday life (for example, the risk of dying in a traffic accident). Catego Cat egory ry C risk risks: s: tho those se tha thatt are are cle clearl arly y avo avoida idable ble bu butt to to whic which h peop people le exp expose ose 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: Conven Con ventio tional nal ris risks: ks: tho those se rela related ted to acti activit vities ies and equ equipm ipment ent typ typica ically lly fou found nd in in most most industries (for example, electrocution). Spec Sp ecif ific ic ris risks ks:: thos thosee asso associ ciat ated ed wit with h hand handli ling ng or or usin using g subs substa tanc nces es tha thatt are are considered hazardous due to their properties and nature (for example, toxic or radioactive substances). Majo Ma jorr risk risks: s: tho those se rel relat ated ed to to exce except ptio iona nall acci accide dent ntss and and situ situat atio ions ns who whose se consequences can be especially severe as large amounts of energy or hazardous substances may be released during short periods of time. Conventional Convention al 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 implement implementing ing 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

R I SK A N A L Y SI S

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 methodolo gies can provide reasonably accurate estimates of potential accidents, the frequency of these accidents and the magnitud magnitudee 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 specific methodology methodology for this analysis. 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

2

 

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 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

3

 

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 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. approach. i.e. it establishes the potential accidents and estimates their effects and consequences 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 event nt t r ee ees  s .  plotting and solving the appropriate appropriate fault tr ees  and  and eve 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 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.

4

 

3

M A JO JO R A C C I D E N T S

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 mech me chan anic ical al:: bla blast st (p (pre ress ssur uree wav wave) e) an and d ej ejec ecti tion on of fr frag agme ment ntss

chem ch emic ical al:: re rele leas asee of to tox xic mat ater eria ials ls.. 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 T y pe pes 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:  phenom ena: 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 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. Jett fi r es . Steady state turbulent diffusion flames with a large length/diameter ratio, caused Je  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 casePool of high-speed effects are observed the fluxes, jet fire tip. fires, tank jets) fires and andbuoyancy jet fires can produce very highatheat 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. Fl as ashh fir es . 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

5

 

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 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

6

 

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.

7

 

Fig. 1-4. Simplified schematic representation of the accidents that can occur following a loss of containment, their effects and the potential associated damage.

8

 

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 (buildings and equipment), while the release of a hazardous material onto onto water, into the soil or into the atmosphere can represent a danger to people and the environmen environment. t. 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 Damag 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 consequences are caused by explosions, followed by fires and then gas clouds [8]. 1

 EXPLOSIVE  FLAMMABLE  TOXIC 0.1      y       t       i       l       i       b      a       b      o      r       P

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

breakdown costs

air 

storage (liquid & liquefied gases)

water 

warehouses

soil

land vehicles

indirect costs (loss of image, etc.)

deaths

biosphere

injured people evacuation costs

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. 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: 2

Risk = k · capital 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

D O M I N O E F FE FE C T

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 C las lasssifi ification cation of domi domino no effe effects cts 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 secondary physical 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 Small conditioners

12 --

-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 secondary accidents accidents and their relative frequencies frequencies are shown shown in Table 1-3 [11]. [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 A n example example cas casee 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 equipmentt were significant in a radius equipmen r adius 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    jet fire: flame impingement on a large pressurized pipe  failure of the pipe   fireball plus large jet fire: flame impingement on pressurized LPG tank    BLEVE plus fireball from one tank: blast, radiation effects plus mechanical impact  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

M A T H E M A T I C A L M O D E L L I N G O F AC A C C I D E NT NT S

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 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 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 damage to property (buildings (buildings and equipment). 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 mass in theofvapour cloud contributes to the blast.effects However, these solutions lead all to the overestimates the distances at which significant are observed or of can 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      10  M  cyclohexan e   0  .1  10  10,000  10,000  kg This corresponds to a scaled distance d n n   = 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 cyclohexane and a yield of 3%, 3%,

M TNT      10  M  cyclohexan e   0  .03  10  3,000  900  kg This corresponds to a scaled distance d n n  =  = 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 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 -2 m . 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 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 -1/3

n  d  E    F   I   M   M TN T    

scaled distance power (m kg (kW ) m-2) flame emissive 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 Affidabilitá e Sicurezza nella Industria Industria Chimica. IEC. 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.

17

 

This page intentionally left blank 

 

Chapter 2 Chapter

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, r eleased, 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 which liquid, whereas if it will is located the top, releasetemperature, will probablyetc. be aHere, gas), the time during the release take at place, thethe ground 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: - Fl Flow ow of li liqui quid d thr throu ough gh a hol holee in in a ta tank. nk. - Fl Flow ow of li liqu quid id th thro roug ugh h a pi pipe pe..

20

 

-

Flow of gas Flow gas or vap vapou ourr thr throu ough gh a ho hole le.. Flow Fl ow of ga gass thr throu ough gh a pip pipe. e. Two-phase flow.

- Ev Evap apor orat atio ion n of a liq liqui uid d fro from m a poo pool. l. Some illustrative examples have also been included. For a more extensive treatment, additional literature is recommended [1, 2, 3, 4]. 2

L I Q U I D R E L E A SE SE

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]:

 u 2   dP  g  z          W    F f   0   s   2        where   g  is  is the acceleration of gravity (m s -2) z  is  is the height above an arbitrary level (m) -1 u  is  is the velocity of the fluid (m s ) P  is  is the pressure (Pa)

(2-1)

-1 W s s is    is the shaft work (kJ kg ) F f f  is the friction loss (kJ kg -1), and -3   is the density of the fluid (kg m ).

For incompressible fluids, the density is constant and

dP  P     l    l 

 

(2-2)

Furthermore, if there is no pump or turbine in the line (see Fig. 2-2), W s s  = 0. Then, the mechanical energy can be simplified to:

 u 2   P  g  z        F f   0 2        l 

(2-3)

2.1 Flow of liquid liquid thr oug ugh h a hole hole in a tank Consider a tank containing a liquid up to a certain level, and with an absolute pressure liquid.toThis pressure can be kept constant (as, for example, thevalue tank or P cont  above  padded withthe nitrogen avoid the formation of a flammable atmosphere) at awhen certain oisr 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 C D  D , in order to take into account the frictional losses in the hole:

m   Aor   l  C D 

 P   P 0

2     cont      l 

   g h l    

(2-4)

where   m  is  is the liquid mass flow rate (kg s -1) A or  is the cross-sectional area of the orifice (m 2) C D  D  is a discharge coefficient (-) P cont  c ont  is the pressure above the liquid (Pa) P 0 0  is the outside pressure (usually the atmospheric pressure) (Pa), and h l l  is  is the height of liquid above the leak (m). The discharge coefficient C D  D   is a complicated function of the leak geometry, the ratio d or    and the Reynolds number in the hole. For turbulent flow, the following values can be or  /d pipe  p ipe  and used: sharp-edged orifices, C D  D  = 0.62; straight orifices, C D  D  = 0.82; and rounded orifices, C D  D  = 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 m . As the of 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  can be calculated with the following expression [2]:

22

 

m   Aor   l  C D 

 P cont   P 0      l  g  C D 2 Aor 2 2     g h l    t     A l  t     

(2-5)

initial 

where A t  is  is the cross-sectional area of the tank (m 2) and t  is  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:

  At       t e   C D  g   Aor     1

 P   P 0   2 cont   g  h l        l    initial 

2 P cont   P 0  

  l 

 

(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 ( d or  or  = 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. -3 (     toluene  t oluene  = 867 kg m ).

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  

  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, h l l  = 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

 

h l   7.5 

2.66 2

  d 



7.365 m

4 The following table summarizes the calculation results for the various time segments. m, kg s-1 

h l l , m

0-180 180-360 360-540 540-720 720-900 900-1,080 1,080-1,260 1,260-1,440

12.80 12.68 12.57 12.45 12.34 12.23 12.11 11.99

7.500 7.365 7.231 7.099 6.968 6.832 6.704 6.577

1,440-1,620 1,620-1,800 1,800

11.88 11.76

6.451 6.324 6.201

t , s

Accumulated spilled mass, kg 2,304 4,586 6,850 9,092 11,313 13,514 15,694 17,853

The amount spilled is 22,107 kg. c) The time is calculated with Eq. (2-6):

    5 2     1  4  t e   0.62  9.81    0.05 2 

2  9.81  7.5

 19,944 s   5.54 h.

19,991 22,107

  



4    ______________________________________   _________________________ _____________ 

2.2 Flow of of liquid through 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 2.2 .1 Li quid 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   l  L u 2

(2-7-a)

d p 

24

 

or 

u  

P  d p 

(2-7-b)

2 f F    l  L where   P  is  is the pressure drop over the pipe (Pa) L  is the pipe length (m)  is the fluid velocity (m s -1) u  is d p p  is the diameter of the pipe (m), and f F  F  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   A p    l 

P  d p  2 f F    l  L

where   A p  is  is the cross-sectional area of the pipe (m 2).

(2-8)

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. 2. 3. 4.

Guess a val Guess value ue for for the the Reynol Reynolds ds numb number. er. Calcul Cal culate ate the the value value of the Fann Fanning ing frict friction ion facto factor. r. Calcul Cal culate ate the the liqu liquid id veloc velocity ity in in the pip pipe. e. Calcul Cal culate ate a new new Rey Reynol nolds ds num number ber..

25

 

Compare the two two Reynolds Reynolds numbers. numbers. If they they are not not equal, equal, correct correct the value value of Re  and  and 5. Compare 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:

  pipe    L equivalent  L  L straight 

(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

L equivalent  /d p p 

  300   170   7   40   200  900

30 20 15 90 60 20 75 16 30 Negligible

Sudden enlargement from d p p  to D  2

  d p 2  

Re    1  32   D 2  

Laminar flow in d p p 

1

Turbulent flow in d p p 

4

Sudden contraction from D  to d p p ; all conditions except high-speed gas flow where P 1/P 2   2

f F ,in d p 

  2     d p 2   1   2    D   

  d p 2   1.25   2  160   D         d p 2   f F ,in d  1.25   2   D        Re

Laminar flow in d p p 

1

Turbulent flow in d p p 

10

p 

*Taken from O. Levenspiel, Enginee Engineerr ing F low and Heat Exchange, Exchange, p. 25, Plenum Press,  New York (1984), (1984), with permission permission of Springer Springer Science and Business Business Media. Media.

26

 

2.2.2 .2 F r iction facto factorr 2.2 The Fanning friction factor (f F  F ) can be calculated as follows [1]. For laminar flow:

f F  

16

(2-10)

Re In the transition regime, the value of f F  F   is uncertain. For turbulent flow, the Colebrook equation can be used: 1.255       1     4 log    3 .7 d p  Re f F   f F     

1

where   is  is the roughness of the pipe (-) (Table 2-2). Table 2-2 Roughness of clean pipes* Pipe material , mm Riveted steel 1 - 10 Concrete 0.3 - 3 Wood stave 0.2 - 1 Cast iron 0.25 - 0.26 Galvanized iron 0.15 Asphalted cast iron 0.12 Commercial steel or wrought iron 0.043 - 0.046 Drawn tubing 0.0015 Glass 0 Plastic (PVC, ABS, polyethylene) 0 *Taken from O. Levenspiel, Enginee Engineerr ing Fl ow and Heat Exchange  Exchange . Plenum Press, New York, 1984.

(2-11)

For fully developed turbulent flow in rough pipes, f F  F  can be calculated with:

  d     4 log  3.7 p   f F       

1

(2-12)

For smooth pipes and Re  <  < 100,000, f F  F  can be calculated with the Blasius equation:

f F   0.079  Re 0.25

(2-13)

And for smooth pipes and Re  >  > 100,000, the following expression can be applied [6]:

f F   0.0232  Re 0.1507

(2-14)

The value of the Fanning factor factor can also be obtained from the classical plot of f F  F   as a function of    and  Re   (see Fig. 2-4).  /d  and   Re  (see

27

 

Fig. 2-4. The Fanning factor as a function of the Reynolds number and an d pipe roughness. Taken from O. Levenspiel, Enginee  Engineeri ri ng Fl ow and Heat Exchange  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. -4 -1 -1 -3 Data:  toluene, t oluene, 20 ºC  = 5.9 · 10  kg m  s ;  toluene, t oluene, 20 ºC  = 867 kg m ;  commercial c ommercial ste steel  el  = 0.046 mm.

28

 

Solution  For the commercial steel pipe:

 

d p 



0.046

 0.00046

100

The pressure drop through the piping system will be the difference between the pressure at the  pipe inlet and the atmospheric pressure:

P     l  g    h l   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  Re  = 4 u     Re  5.9  10  25,000  0.17  m s-1 0.1  867 d p    l 

From Fig. 2-4, f F  F  = 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   must be corrected. The following table summarizes the results of the trial-anderror procedure. Re 

 

u , m s-1

f F 

 

 (Eq. (2-7b)), m s-1 u  (Eq.

25,000 70,000 300,000 400.000 500,000 520,000 530,000

0.170 0.544 2.041 2.722 3.402 3.539 3.631

0.00640 0.00520 0.00450 0.00440 0.00435 0.00433 0.00430

2.920 3.240 3.480 3.520 3.540 3.544 3.560

Finally, u  =  = 3.547 m s-1. Thus, the mass flow rate is:

 

  0.12

-1

m   3.547  867 

4  24.1 kg s .  ______________________________________   ____________________________ __________ 

29

 

3

G A S/ S / V A P O U R R E L E A SE SE

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 (  1 /   2  2   < 2,  P 1 /P 2  2   < 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 F low of of gas gas/vapour /vapour thr oug ough h a hole hole For liquids and gases with low pressure changes and low velocities ( P 1 /P 2    < 0.3) 2  < 2, M a  < (M a   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 3. 1.1 Cr iti ca call ve velocity locity 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 ( P cont  c ont ) through a hole in the wall (Fig. 2-5). If the pressure downstream from the hole ( P out  out ) decreases, the velocity of the gas through the hole increases. This velocity will increase until, at a certain value of  P   P out , it reaches the velocity of sound in that gas (at that temperature). Further decrease of P out  out  will not cause any increase in fluid velocity: the velocity of sound at P choked  choked ,   T choked  choked , 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 P choked  choked , even though P out  out   decreases further. P choked  choked   is called choked   or critical  pressure , and the velocity at the hole in these conditions is called choked  or  or cri tical velocity  velocity . Assuming isentropic expansion, the relationship between the choked pressure and the  pressure inside the tank can be expressed as:  

P choked   2     1   P cont      1

(2-15)

where   P cont   is the pressure inside the container or the pipe (Pa), and c ont  is  (-).  is the ratio of heat capacities, c p  p /c    v  v  (-). The choked velocity is the maximum possible velocity in an accidental release. It is found in most accidental gas releases. Since P out  out   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 P cont    will not produce any further increase c ont  will in the gas exit velocity. Therefore, critical velocity will be reached if the following condition is fulfilled:  

P cont 

      1   1

 P 0 

2



(2-16)

In fact,    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,    is the ratio of heat capacities.   is always than unity. For gases, it ranges from 1.1 to 1.4; therefore,    is sonic velocity will begreater reached when P cont   /P 0 0most c ont         1.9. For air (  , for example, sonic velocity is reached when P cont   /P 0 0    1.893, i.e. when the downstream absolute pressure is c ont 

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:

u s  

 T  R 10 3

(2-17)

M  v 

where   u s s is    is expressed in m s -1 R  is  is the ideal gas constant (8.314 kJ kmole -1 K -1) M v  is  is the molecular weight of the gas (kg kmole -1)   v  v , for various gases. Table 2-3 shows the ratio of heat capacities,   =  = c p p /c  The density of a gas increases with pressure. Therefore, once the critical velocity has been reached, if P cont  c ont  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 P cont  c ont . It is therefore clear that what becomes critical or choked is the velocity (m/s) of the gas rather than the flow. Thus, cri tical velocity  velocity  is  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 sonic is reached. therefore the the mass flowvelocity rate is also choked. In this case, the inlet gas density is constant and The temperature of the gas in the jet at the orifice is:

 P choked        P cont   

T choked   T cont  

        1         

  2     T cont      1    

(2-18)

where   T cont   is the temperature in the container or pipe (K). c ont  is 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   u s s,  m s-1   =  = c p p /c    v  v  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 Helium Hexane

44.0 4.0 86.2

1.215 1.660 1.062

261 1014 175

Hydrogen chloride Hydrogen Hydrogen sulphide Methane  Natural gasa  Nitrogen Oxygen Propane Propylene Sulphur dioxide

36.5 2.0 34.1 16.0 18.1 28.0 32.0 44.1 42.1 64.1

1.399 1.405 1.326 1.304 1.270 1.406 1.395 1.146 1.148 1.264

308 1314 310 449 419 352 329 253 260 221

Toluene

92.1

1.087

171

a

  86.15% CH4, 12.68% C2H6, 0.09% C4H10, 0.68% N2

32

 

3.1.2 asss flow r ate 3.1.2 M as 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:  1

M  v  2    1     m hole   Aor C D  P cont     3     1  Z T cont R 10  

(2-19)

where  m is the mass flow rate (kg s -1) C d d  is  is a dimensionless discharge coefficient (-) A or  is the cross-sectional area of the orifice (m 2), and  is the gas compressibility factor at P cont   = 1). Z  is c ont , T cont  c ont  (-) (for ideal gas behaviour, Z  =   is a dimensionless factor that depends on the velocity of the gas. For sonic gas velocity:

   1

(2-20)

and for subsonic gas velocity:  1         1   P           P  2 1           2     0  1   0      1   2    P cont      P cont         1

2

(2-21)

The value of  has been plotted as a function of P 0  0 /P    cont  c ont , 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:

L  j 

6 u  j  d or  

(2-22)

u w  -1

where u  j  is the velocity of the jet at the source (m s ) d or  or  is the diameter of the source, and -1 -1 u w  w  is the average ambient wind speed (m s ) (default value: 5 m s ).

3.1. 3. 1.33 Di scha charr ge coe coefficien fficientt C D  D   is a coefficient that takes into account the fact that the process is not isentropic. Its value is C D  D   = 1.0 for a full-bore rupture in a pipe. For sharp-edged orifices in accidental releases (high Reynolds number), some authors recommend C D  D  = 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,   =  = 1.15. Therefore:

     1  2 

 

   1

  

   1 2 

1 .15

1.15 1.15 1

  1.74

Since

P cont  10 bar   = 9.87  P 0 1.013 bar  the propane velocity at the orifice is critical. Therefore, by applying Eq. (2-19):  1

  2    1

 m hole   Aor C D  P cont         1 

M  v  Z T cont R 10

3



1.151



  0.02 4

2

  2   1.151 0.525 kg/s if  C D   0.62 44.1   C D  10 10  1 1.15   3 1 . 15  1   1 273  25 8 . 314  10       0.847 kg/s if  C D   1.0 5

The velocity at the hole can now be calculated. The conditions at the choked cho ked jet are: 1.15

     1 . 15  1       2

P choked   10 

1.15 1



5.744  bar

1.151

 5.74  1.15  

T choked   298 

 

10

 

5.744 10  44.1  11  kg m-3  8.314 10  277.2 5

  choked 

277.2  K

3

The velocities at the hole corresponding to the two mass flow rates are:

u 

u 

0.525

  0.02 2       11 4     0.847



152  m s-1

 245  m s-1

     0.02 2   11   4  

34

 

 

The speed of sound in propane at choked conditions is:

u s  



1.15  277.2  8.314 10 3

  245  m s-1

44.1  ______________________________________   ___________________________ ___________  1,00

0,95

0,90

             

=  1,10 1,15 1,20 1,25 1,30 1,35 1,40 1,45

1,10

0,85

0,80

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 P0 /Pcont

0,70

0,65

0,60

0,55

0,50

Fig. 2-6.  2-6.  as  as a function of P 0 0 /P   and  .   cont  and

3.2 Flow of gas gas/vapo /vapour ur thr ou oug gh 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] [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 cont   P 0  P cont   P p     P p   P 0  P pipe   P hole 

(2-23)

where P p   is the pressure inside the pipe just in front of the opening (Pa), and p  is P 0 0  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 (m pipe  p ipe ) must be equal to the mass flow through the hole ( m hole  h ole ):

m pipe      m hole 

(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]: P p 

 

2    P  dP 

m pipe   A p 

P cont 

(2-25)

4 f F 

  L      d p       2

where A p  is  is the cross-sectional area of the pipe (m )    (P)  (P) is the density of the gas (kg m-3) f F   is the Fanning friction factor (-) F  is L  is the length of the pipe (m), and  is the diameter of the pipe p ipe (m). d p p  is

The following relationships apply:

36

 

 P       constant      Z  

1  

(2-26)

Z   R 10 3     1  c v   M  v 

(2-27)

where Z  is the compressibility factor. For ideal gas behaviour:

Z   1 and        The integral in Eq. (2-25) may be solved analytically, assuming a constant compressibility factor and constant specific heat at a constant volume, c v v ; for Z  =  = 1: 1                P p           P   dP    P   1         cont  cont       P   1           cont   P      

P p 

(2-28)

cont 

2

-4

2

(the units are kg  m  s ). 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,  P p  p . The following procedure must be followed: 1. Guess the pressure pressure insid insidee the pipe just just in front of the pipe opening opening (P 0   < P p p  <  < P cont  0  < c ont ). 2. Calcul Calculate ate the mass flow rate rate through through the pipe opening opening with with Eq. Eq. (2-19) (C D  D  = 0.62 for a hole in the pipe wall; C D  D  = 1 for full-bore rupture). 3. Calcul Calculate ate the the mass flow rate rate through through the pipe pipe with with Eqs. (2-25) (2-25) and (2-28). 4. Compar Comparee the two two mass flow flow rates. rates. If they they are not not equal, equal, correct correct the value value of P p  p   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]:

     1 2 Y i   1   M a i  2

T p 



Y cont 

(2-29)

(2-30)

T cont  P p  P cont 

Y p 



M a cont 

Y cont 

M a p 

Y p 

(2-31)

 M a 2p  Y cont     1  4 f F  L  1   2 2  M a 2p        d p     0 2 ln  M a cont  Y p     M a cont   

   1

(2-32)

37

 

For sonic flow at the exit, M a p   = 1 and Eqs. (2-29) and (2-31) become: p  =

T p    2 Y cont  T cont 



   1

 4 f F  L    2 Y cont      1   2 2 1   0 2 ln     1 M a cont      M a cont         d p    

(2-33)

   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 P p p . If there is, furthermore, a certain flow rate through the pipe, it must be taken into account in order to estimate the value of P p  p .  ______________________________________   _________________________ _____________ 

Example 2-4 A constant pressure source (5 bar, 288 K) of natural gas ( M v  = 17.4;   =   = 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, C D  D  = 1.0. For a smooth polyethylene pipe (Fig. 2-4), a value of the Fanning factor f F  F   = 0.002 will be assumed. Guessed pressure: P p  p  = 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     2 1  M a cont      1.27  1    1 2  4  0 .002  330   0         ln  1  1 . 27   2 2     M a cont  2 0 . 164 1.27  1 M a cont             

i.e.:

2  2   0.27 M a cont    1  1.135 ln   2  M a 2  20.44  0 2 . 27 M a  cont    cont   

By trial and error, M a cont   = 0.2. c ont  = Therefore, by applying Eq. (2-29):

Y cont   1 

1.27  1 2

0.2 2

 1.005

From Eq. (2-33):

38

 

T p  288



2  1.005 1  .27  1

T p p  =  = 255 K Mass flow rate through the opening (Eq. (2-19)): 1.27 1

m hole  

  0.164

2

4

  2   1.271 17.4  1.0 4 10  1 1.27   16 kg s -1 3 255  8.314 10  1.27  1  5

 

-3 Mass flow rate through the pipe (Eq. (2-25)) (     cont  c ont  = 3.68 kg m ):

2

m pipe  

  0.164 2 4

    111.27.27          1.27   4   5  1    5 10  3.68  1  1.27    5                  4.33 kg s -1     330   4  0.002    0.164 

Since two different results have been obtained, a new trial is required. The results of the calculation procedure are summarized in the following table: P p p , 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

-1 m pipe  p ipe , kg s 

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, C D  D  = 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, f F   = 0.003, is now assumed F  = (it will be checked later). Guessed pressure: P p  p  = 4.5 bar. To estimate the temperature of the gas, by trial and error the value of M a cont  c ont  is found: M a cont  c ont  =

0.17. The temperature of the gas just in front of the opening is calculated with Eqs. (2 29) and (2-33): T p  p  = 254.7 K. The mass flow rates are now: 1.27 1

  0.05

m hole  

4

2

  2   1.271 17.4 -1 0.62 4.5 10  1 1.27   254.7  8.314 10 3  1.04 kg s  1.27  1   

5

39

 

2

m pipe  

  0.164 2 4

11.27         1.27      1 . 27 4 . 5       5      5 10  3 . 68 1      1  1.27     5                   2.56 kg s -1   330   4  0.003    0.164 

The two values are different. By trial and error, the following values are ultimately obtained:  = 1.13 kg s-1. P p p  = 4.9 bar, m  = 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 f F 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: P p, bar

m, kg s-1

Re

f F*

4,90

1,13

891,53

0,01795

4,50

1,04

819,78

0,01952

4,47

1,03

813,17

0,01968

4,46

1,03

812,52

0,01969

*f F  calculated with Eq. (2-10) because there is laminar laminar flow.

Therefore, P p  =4.46 bar and m  =  = 1.03 kg s-1. p =4.46  ______________________________________   ______________________________ ________  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 T im e-depe -depende ndent nt gas r ele leas asee 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 initial  t       W     

m t   m initial   exp  

(2-35) -1

where   m (t(t))  is the mass flow rate at the th e time t after the onset of the leak (kg s )

40

 

-1 m initial  i nitial  is the mass flow rate at the onset of the leak (kg s ) W  is  is the initial mass of gas in the vessel (kg), and  is the time after the onset of the leak (s). t  is

 ______________________________________   ___________________________ ___________ 

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): 1.411

m hole  

  0.0254

2

4

 

  2   1.411 28   1.015 kg s -1 3 298  8.314 10  1.41  1 

0.62 14 10 5  1 1.41

(A value of C D  D  = 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   0.82  kg s-1.    142.2  

m 30  1.015 exp  

Fig. 2-8 shows the variation of mass flow rate as a function of time.

Fig. 2-8. Flow of a gas g as through a hole as a function of time.

 ______________________________________   ____________________________ __________ 

41

 

4

T W O -P - P H A SE SE F L O W

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 Flas Flashing hing 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 dw l l   implies a decrease in the temperature of the remaining liquid dT :

dw l  

w l  c pl 

H v 

dT 

(2-36)

This expression can be integrated between the initial temperature of the liquid before depressurization, T cont  c ont , and the final temperature of the mixture (the atmospheric boiling point if the mixture is released into the atmosphere), T b  orderr to cal calcul culate ate the mas masss of vapo vapour: ur: b , in orde c pl  T cont  T b  

w v   w il     1   e 

  

H v 

   

(2-37)

where   w v v  is  is the mass of vapour (kg)  is the initial mass of liquid (kg) w i l  is H v  is the mean latent heat of vaporization between T cont   T b b  (kJ kg-1), and c ont  and  T  -1 -1  and T b   (kJ kg  K  ). c pl  pl  is the mean heat capacity of the liquid between T cont  c ont  and b  (kJ The ratio between the mass of vapour formed and the initial mass of liquid is usually called the vaporization fraction:

 w v    1  e  f   w il 

c p T cont  T b  

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 significantly preferential complicated. vaporization of the more volatile components and the calculation of  f  f  becomes

4.2 T wo-phas wo-phasee dis discchar ge 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]:

G ne  

 

H v 

(2-39)

v lv  N T cont  c p l 

-2 -1 where   G ne  ne  is the discharge rate (kg m  s )  is the latent heat of vaporization at boiling temperature of liquid (kJ kg-1) H v  is T cont  c ont  is the storage temperature (K)  is the change in specific volume brought about by the change from liquid to vapour v l v  is (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  

H v 2 2 P cont   P 0   l  C D 2 v lv 2 T  cont  c p l 



L L e 

(2-40)

where   P cont  c ont  is the pressure inside the vessel (Pa) P 0 0  is the atmospheric pressure (Pa) L  is the pipe length to opening (ranging from zero to 0.1 m) (m)  is the liquid density (kg m-3)   l l  is C D  D  is the discharge coefficient (-), and L e  = 0.1 m. Eq. (2-40) shows that the degree of non-equilibrium varies directly with the pipe length L . For   L /d    = 0, there is no flashing and Eq. (2-39) reduces to the orifice equation for

p 

incompressible liquid flow. For L     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 /d p    ratio p  ratio and Eq. (2-39) becomes:

43

 

G e  

 

H v 

v lv  T cont  c p l 

(2-41)

If the physical properties ( H v , v l v ) are unknown, Eq. (2-41) can be expressed as:

G e   

dP 

T 

(2-42)

dT  c p  The effect of subcooling on the discharge rate is:

G sub   2  P    cont   P v       l 

(2-43)

where   P v v  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]:

G e 2 G 2 p   C D   G sub   N   

2

(2-44)

where   G 2p   is the discharge mass flow rate (kg m-2 s-1), and 2p  is C D  D  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

SA F E T Y R E L I E F V A L V E S

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) d isc) 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 fo llowed and on each specific case. For compressible fluids, it normally no rmally 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 D is iscchar ge fr om a safe safety ty r elief valve The discharge flow from a safety relief valve depends on the pressure inside the equipment the cross-sectional flow area of the valve. The relationship between three variables isand established by the existing methods for sizing safety valves. The these 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:

A sv  

0.1791 m sv 

  w  P cont 

T  Z  M  v 

and for liquid applications: 0.6211 m 

(2-45)

Asv  

sv 

  w 

   P  l 

cont 

(2-46)

 P   back 

45

 

2

where  A sv  is  is the minimum cross-sectional flow area of the safety valve (mm ) -1 m sv  sv  is the discharge mass flow rate (kg h )  is the absolute relieving pressure (bar) P cont  c ont  is  is the absolute backpressure (bar) P back  b ack  is  is the inlet temperature (K) T  is -3   l l  is the liquid density (kg m )  is the molecular weight (kg kmol -1) M v  is  is the com compre pressi ssibil bility ity fac factor tor (-) Z  is  is an outflow coefficient specified by the manufacturer (-), and  w  w  is   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 (P cont  c ont   = 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 c p  p   of the liquid between -1 -1 -3 these two temperatures = 2.54 kJ kg  K  ;  l l  =  = 440 kg m ; k  =  = 1.15; molecular weight = 44.1; 2 A v  = 3.25 cm .

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  

2.54   320  231  1  exp       0.468 w il  358    

w v 

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 



A g    w  P cont   

M v 

0.1791

T  Z 



A g   0.45  0.7  16.3

44.1

0.1791

320

Liquid flow: 0.532 m sv  

 P back   Al   w    l  P cont    0.6211



Al  0.7   440 16.3  1.013 0.6211

Furthermore,

A g  + A l  = 325 By solving these three equations, the discharge d ischarge mass flow rate is obtained:  = 6,535 kg h-1 m sv  sv  =  ______________________________________   ____________________________ __________  6

R E L I E F D I SC H A R G E S

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 R elie elieff flow r ate for for ves vesssels subj ec ectt to exter external nal fir fi r e 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 expo sed 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: - sph pheerical tank: 55 55% % - horizont ntaal ta tank: 75 75% - rec rectan tangula gularr tank tank : 100% (ex (exclu cludin ding g the the top sur surfac face) e) - ve vert rtic ical al ta tank nk:: 100% 100% (up to a hei height ght of 9 m). 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  Q     63,080   Aexp 2

2

18.6 m  < Aexp < 92.9 m 2

2

92.9 m  < Aexp < 260 m

0.566    Q  =  = 224,130 Aexp

0.338    Q  =  = 630,240 Aexp

(2-45-a) (2-45-b) (2-45-c)

Aexp  260  m2 and P design  design  < 0.07 barg  Q  4,130,000

(2-45-d)

0.82  > 0.07 barg    Q  =  = 43,185 Aexp Aexp > 260 m2 and P design  design  >

(2-45-e)

where   A exp  is  is the exposed vessel surface (m2), and  is the heat flux (W). Q  is

48

 

For tanks containing LPG (pressurized tanks), NFPA58 [17] recommends the following expression: 0.82  = 70,945 Aexp Q  =

(2-45-f) 2

where A exp  is the entire surface area of the vessel (m ). 3. The gas discharge rate is calculated using the following expression:

m  

F  Q  (2-46)

H v 

where   m  is  is the discharge rate (kg s -1)  is a reduction factor (-), and F  is -1  is the latent heat of vaporization at the boiling temperature of the liquid (kJ kg ). H v  is The reduction factor F  depends  depends on the protective measures applied: -   A exp   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. - ade adequa quatel tely y desig designed ned wat water er spr spray ay sys system tem and dra draina inage: ge: F  =  = 0.3. - ad adeq equa uate te th ther erma mall in insu sula lati tion on:: F  =  = 0.3. - wat water er spra spray y syste system m plus plus the therma rmall insu insulat lation ion plus drai drainage nage:: F  =  = 0.15. - no none ne of the the afor aforem emen enti tione oned d prote protect ctiv ivee meas measur ures es:: 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. -1 Data: H v n-hexane = 334 kJ kg .

Solution  The area exposed to fire is.







2

Aexp  0.75    4  7     4 2 / 2  103.7  m

Taking into account the drainage system, F  =  = 0.5. The heat flux received by the tank is: 0.338  = 630,240 Aexp Q  =  630,240 103.7 0.338   3,025,790 W

And the required discharge mass flow rate is:

m  

F  Q  0.5  3,025,790   4,530  kg h-1 = 1.26 kg s-1 H v  334

 ______________________________________   ______________________________ ________ 

6.2 R elief flow r ate for for ve vesssels undergo undergoing ing a r unaway r eac action tion 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: - ene energy rgy rel releas eased ed in the poly polymer meriza izatio tion n proc process ess:: 900, 900,000 000 MJ. - ene energy rgy rel releas eased ed in the com combus bustio tion n of of the the vapo vapour ur pha phase se 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 Take n 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 p ressure of the device. The relief mass flow rate can be estimated from the mass and heat balances by applying some simplifying assumptions: - the dis discha charge rge mas masss flow flow rate rate is ess essent ential ially ly cons constan tant. t. - th thee hea heatt of of rea react ctio ion n per per un unit it ma mass ss (q ) is practically constant. - the phy physic sical al and the thermo rmodyna dynamic mic pro propert perties ies are con consta stant. nt. 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   G  A 

W  q 

 V  H v    c   T    v 

2

(2-47)

W  v 

lv 





where   m  is  is the discharge rate (kg s -1) G  is  is the discharge rate per unit cross-sectional area of venting (kg m -2 s-1)  is the initial mass contained in the vessel (kg) W  is q  is  is the heat flow released by the chemical reaction (kW kg-1) V  is  is the volume of the vessel (m 3) c v v  is  is the specific heat at constant volume (kJ kg -1K -1) H v  is  is the latent heat of vaporization of the liquid (kJ kg-1) v l v  is  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 2

 dT    dT          dt   set    dt   max 

c v  

(2-48)

The ratio (H v  /v l v ) can be obtained from P- T   data by applying the following relationship obtained from the Clapeyron relation:

H v  v lv 

 T set 

dP  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     1  G  0.9 v lv   c p  T  

0.5

(2-50)

-1 -1 where H v  must be expressed in J kg -1 and c p  p  in J kg  K  .

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 m 3; W  =  = 9,500 kg; P s s =    = -1 4.5 bar; T s s =    = 482.5 K; (dT/dt) set  =   = 0.493 K s  (sealed system); P max    = 5.4 bar; T m  max  = m  = 492.7 K; -1 (dT/dt) max  = 0.662 K s . Material properties: P s s  = 4.5 bar  3

-1

Specific volume, liquid, m  kg   Specific volume, gas, m3 kg-1  -1 -1 C pl  pl , kJ kg  K   

 

0.001388 0.08553 2.470 310.6

H v , kJ kg-1

P m  m  = 5.4 bar 0.001414 0.07278 2.514 302.3

Solution 

From Eq. (2-48), and assuming c v v    c p  p  for an incompressible fluid:

q  

1 2

2.47 0.662  0.493 1.426  kJ kg-1 s-1

The mass discharge rate and the mass flux can be calculated from Eqs . (2-47) and (2-50): 9,500 1.426

m  

2

 

  255  kg s-1

13.2 310.6  2  .47  492.7  482.5 9,500 0.08553 - 0.001388   0.5

    1    G  0.9   0.08553  0.001388  2,470  482.5  310,600

And the corresponding vent area is: 255 A   0.084 m2 3040  ______________________________________   _________________________ _____________ 

52

-2 -1

3040 kg m s

 

7

E V A PO PO R A T I O N O F A L I Q U I D F R O M A P O O L

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 Po Pool ol size size 7.2.11 Pool on ground 7.2. 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)

 

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).

7.2.22 Pool on water 7.2. 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 E vapo vaporr atio ation n of boiling boiling liquids 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. surrou ndings. 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  T s   T pool 

(2-52)

  s  t 

-2 where   Q pool  p ool  is the heat flux reaching the pool from the ground (W m ) k s s is    is the thermal conductivity of the soil (see Table 2-4) (W m -1 K -1)

   is the temperature of the soil (K) T s s is T pool   is the temperature of the liquid pool (K) p ool  is the thermal diffusivity of the soil (m 2 s-1) (see Table 2-4), and  s s is   t  is  is the time after the spill (s). Table 2-4 Values of thermal conductivity and thermal diffusivity of various solid media [18, 19] Substance   k s s,  W m-1 K -1  s s,  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 -7

Gravel Carbon steel Concrete

2.5 45.0 1.1

11 · 10 127 · 10-7 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   m pool   is the mass boiling rate (kg s -1) p ool  is A  is the area of the pool (m2), and  is the latent heat of vaporization (expressed in J kg -1). H v  is 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 E vapo vaporr atio ation n of non-boiling 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 (P v v   – P amb  ). The mass transfer process will be significantly amb ). 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 w 0.78 r 0.11

M  v  P 0   P v   P amb    ln1   R T  P  P  v      0

(2-54)

where   G pool   is the evaporation rate (kg m-2 s-1) p ool  is -1 u w   is the wind speed (m s ) w  is r  is the radius of the circular pool (m) M v  is the molecular weight of the liquid (kg kmol-1) -1 -1 R  is  is the ideal gas constant (expressed in J kmol  K  )  is the temperature (K) T  is P v v  is  is the vapour pressure of o f the liquid at its temperature (Pa) P amb   is the partial pressure of the liquid in the atmosphere (Pa), and amb  is P 0 0  is  is the atmospheric pressure (Pa). For  P   P v v  < 2·104 Pa, Eq. (2-54) may be simplified to:

G pool   2 10 3  u w 0.78   r   0.11

M v  R T 

P v   P amb  

(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: M v  = 86; P v 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): 3   0.78

G pool   2 10  3

0.11

11





  -2 -1   0 . 00224 kg m  s 8.314 10 3 273  20    1.0132 10 5  16,130  86  1.0132 10 5

 

ln1 

16,130  0

And, for the whole pool:

m  =  = 0.00224 kg m-2 s-1 · 380 m2 = 0.851 kg s-1  ______________________________________   ___________________________ ___________  8

G E N E R A L O U T F L O W G U I D E L I N E S F O R Q U A N T I T A T I V E R I SK A N A L Y SI S

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 L os osss-of-co -of-contain ntain ment ev events ents in pr essur ized tank tan k s and vess 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 L os osss-of-co -of-contain ntain ment events events in atm os ospher pher ic tank s In atmospheric storage tanks, pressure is atmospheric or slightly higher. The following LOCs are usually considered: Instantaneous release of the complete inventory: - Di Dire rect ctly ly in into to th thee atm atmos osph pher ere. e. - Fro From m the the tan tank k into into an unim unimpai paired red sec seconda ondary ry cont contain ainer. er. Continuous release of the complete inventory in 10 min at a constant release r elease rate: - Di Dire rect ctly ly in into to th thee atm atmos osph pher ere. e. - Fro From m the the tan tank k into into an unim unimpai paired red sec seconda ondary ry cont contain ainer. er. Continuous release from a hole with a diameter of 10 mm: - Di Dire rect ctly ly in into to th thee atm atmos osph pher ere. e. - Fro From m the the tan tank k into into an unim unimpai paired red sec seconda ondary ry cont contain ainer. er.

8.3 L os osss-of-co -of-contain ntain ment events events in pip es 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, C D  D  = 1.0. In other cases, C D  D  = 0.62. Assume that the pipe has no bends and a wall roughness of approximately 45 m.

8.4 L os osss-of-co -of-contain ntain ment events events in pum ps 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 L os osss-of-co -of-contain ntain ment events events in r elie elieff devices devices Discharge at the maximum discharge rate.

56

 

8.6 L os osss-of-co -of-contain ntain ment events for stor age in war eho ehous uses es 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 L os osss-of-co -of-contai ntainm nm ent events events in tr ans anspor portt units uni ts in an establ establis ishm hm ent LOCs for road tankers and tank wagons in an establishment: - Ins Instan tantan taneou eouss rel releas easee of the com comple plete te inv invent entory ory.. - Cont Continu inuous ous relea release se from from a hole (wit (with h the size size of the the largest largest conne connecti ction). on). If If the tank tank is partly filled with liquid, a liquid phase release from the largest liquid connection is assumed. - Ful Full-b l-bore ore ruptu rupture re of the the loadi loading/ ng/unl unloadi oading ng hose hose (outfl (outflow ow from from both both sides) sides).. - Lea Leak k in the the loadin loading/u g/unlo nloadi ading ng hose hose (effec (effectiv tivee diamet diameter er of 10% 10% of the the nomina nominall diameter, with a maximum of 50 mm). - Ful Full-b l-bore ore ruptu rupture re of the the loadin loading/u g/unloa nloadin ding g arm (out (outflo flow w from bot both h sides) sides).. - Lea Leak k in the load loading/ ing/unl unloadi oading ng arm arm (effec (effectiv tivee diamet diameter er of 10% 10% of the the nomina nominall diameter, with a maximum of 50 mm). - Fir Firee under under the the tank tank (relea (release se from from the the connec connectio tions ns under under the the tank tank follo followed wed by by ignition or fire in the surroundings of the tank): instantaneous release of the complete inventory of the tank. Ships in an establishment: - Ful Full-b l-bore ore ruptu rupture re of the the loadin loading/u g/unloa nloadin ding g arm (out (outflo flow w from bot both h sides) sides).. - Lea Leak k in the load loading/ ing/unl unloadi oading ng arm arm (effec (effectiv tivee diamet diameter er of 10% 10% of the the nomina nominall -

-

diameter, with a, maximum Exte Ex tern rnal al im impa pact ct, large lar ge sp spil ill: l:of 50 mm). Gas tanker: continuous release of 180 m 3 in 1800 s. o 3 Semi-gas tanker (refrigerated): continuous release of 126 m  in 1800 s. o Single-walled liquid tanker: continuous release of 75 m 3 in 1800 s. o 3 Double-walled liquid tanker: continuous release of 75 m  in 1800 s. o Exte Ex tern rnal al im impa pact ct,, sma small ll sp spil ill: l: o Gas tanker: continuous release of 90 m 3 in 1800 s. o Semi-gas tanker (refrigerated): continuous release of 32 m3 in 1800 s. Single-walled liquid tanker: continuous release of 30 m 3 in 1800 s. o Double-walled liquid tanker: continuous release of 20 m 3 in 1800 s. o

8.8 Po Poo ol evapo evaporr atio ation n If the liquid spillwill is contained bund, theand spreading the and liquid willwill be limited: the maximum pool size be the sizeinofa the bund the poolofsize shape be defined. The effective pool radius can then be calculated from the bund area as follows:

r pool    

Abund 

(2-56)

 

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. ro ughness. A minimum value of 5 mm can be assumed.

57

 

As for the maximum surface for unconfined pools, the following values have been proposed: 2 2 1,500 m  (D  = 44 m) for pools on land and 10,000 m  (D  = 113 m) for pools on water.

8.9 O utflow and atmospheric atmospheric dis dispers persion ion 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 ( M r el ). ). - Calculate the mass released in each time segment, M segment  = M r e l/number of segments. - Calculate the duration of the first time segment, t segment   = time required to release M segment . segment  = - Calculate the release rate in the first time segment = M segment  /t segment  segment . - 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 A bund   bund area (m2) A exp   vessel surface exposed to external fire (m2) A or  cross-sectional area of the orifice (m2) A p   A sv   A t   C D  D   c p p   c p l   

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)

c v v   d p p   d or  or   F   f   F f f   f F  F   2p   G 2p  G sub  sub   h l l   H v   L  L j   M a   M v   m   m sv  sv   N  

specific heat at constant volume (kJ kg-1 K -1) pipe diameter (m) diameter of the orifice (m) reduction factor (-) vaporization fraction (-) -1 friction loss term (J kg ) 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   pressure (Pa) P back  b ack   safety valve absolute backpressure (bar) P choked  choked  choked or critical pressure (Pa) P cont  c ont   pressure inside the vessel or the pipe (Pa) P design  design   vessel design pressure (Pa) P max  maximum pressure reached during venting (Pa) max   P 0 0   atmospheric pressure (Pa) P p p   pressure inside the pipe at a given distance from the source (Pa) P out  pressure downstream from the hole (Pa) out   set pressure (Pa) P s s   vapour pressure at the storage temperature (Pa) P v v   pressure drop (Pa) P   Q   heat flux (W) R   ideal gas constant (8.314 kJ kmole-1 K -1) Re   Reynolds number (-) r pool   effective pool radius (m) T   temperature (K) T choked  conditions (K) choked   temperature of the gas at choked conditions temperature in the vessel or the pipe (K) T cont  c ont   temperature of the gas in the jet (K) T  j   T p p   temperature of the gas in the pipe, at a given distance from the source (K) -1 u   velocity of liquid in the pipe (m s ) u  j   velocity of the jet at the gas outlet (m s-1) -1 u s s   speed of sound in a gas (m s ) u w   v l v  W   w i l   w l l   w v v   Z    w  w               l l     

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, c p   (-) p /c    v  v  (-) -3 density of the gas (kg m ) -3 density of the liquid (kg m ) 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”), 3 rd  ed. The Hague, SDU, 1997.

59

 

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. th M. Chase. Nist-Janaf Thermochemical Tables. 4   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 [4] [5] [6] [7]

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”), 1 st ed. The Hague, SDU, 1979.

60

 

Chapterr 3  Chapte

F ir e acc accidents idents 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 prev ent 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

C O M B US U ST I O N

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 v aporized 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, p roceed, the three sides must be connected. co nnected. 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 tri an   (Fig. 3-1): if one side is angle  gle  (Fig. 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 C ombus ombustion tion r eac action tion and combus combustion tion heat 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:

Q    F   Q L m   q v 

(3-1)

where m  is  is the combustion rate (kg m -2 s-1) Q F   is the heat flux from the flames into the fuel (kW m-2) F  is Q L  is the heat lost from the fuel surface (kW m-2) and q v v  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, H c , will have two different values depending on whether the water formed with the reaction products is considered to be in the liquid or

62

 

-1

vapour state. The difference will be the vaporization vapo rization heat of water (44 kJ mol  or 2444 kJ kg at 25 ºC). For example, for methane,

-1

CH4 + 2O2   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 H c  value is used and is called the net  combustion  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 fl ame ames  s .

2.2 Pr emixed flames and diffusion 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 o ccurs 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

T Y PE S O F F I R E

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 poo pooll fir e  (a  (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   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.

3.11 Po 3. Poo ol fir es

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 experimental study of the behaviour of The poolhighest fires with different and diameters. Theyanobserved the same behaviour for all fuels. burning ratesfuels 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].

64

 

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.

3.22 Je 3. Jett 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.

3.33 Flas 3. Flash h fir es 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 exp losion rather than a flash fire.

65

 

3. 3.44 Fire Fireba ball llss 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.

4.11 Flammability limits 4. Mixtures of a flammable gas or vapour with air are flammable only if the concentration of the vapour ranges between two values called the lowe lowerr flam flamma mabili bili ty limit  (LFL)  (LFL) and the upper  flammability limit  (UFL).   (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.

66

 

Table 3-1. Flammability limits, flash point temperature and autoignition temperature for several common substances in air at atmospheric pressure Substance LFL, % volume UFL, % volume T f f , ºC T autoign  a utoign ,  ºC Acetone 2.6 12.8 -17.8 700 Acetylene 2.5 80.0 -17.8 305 Aniline 1.2 11.0 14.4 615

Ammonia Benzene n-Butane Crude oil Cyclohexane Methane Ethylene Ethane Ethanol n-Hexane Hydrogen Kerosene Motor gasoline Methane Octane n-Pentane Propane Propylene Toluene

15.0 1.4 1.8 1 1.3 5.3

28.0 8.0 8.4 6.0 8.3 15.0

---11 -60 -18 -17 -222.5

630 562 405 230-250 260 632

2.8 3.0 3.5 1.2 4.0 0.7 1.4 5.3 1.0 1.3 2.1 2 1.3

34.0 12.4 19.0 6.9 74.2 6.0 7.6 15.0 6.7 7.6 9.5 11.7 7.0

---135 12.8 -26 --49 -46 -222.4 13.3 -40 -104.4 --4.4

450 515 558 234 400 229 280 632 458 287 493 443 536

4.1. 4. 1.11Several Estimation Es timation of flammabili of flamm limits for calculating the values of flammability limits. All methods have abili beentyproposed 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 c st  st :

L F L  0.55 c st 

(3-2)

U F L  3.5 c st 

(3-3)

c st    is the stoichiometric concentration of fuel in air, expressed as volume percentage (fuel st  is 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:

L F L    a  c st 

(3-4)

U F L   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.

67

 

Table 3-2 Values of constants a  and  and b  in  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 Epoxides Esters Other compounds C, H, O Monochlorinated compounds Dichlorinated compounds Brominated compounds Amines Compounds containing S

0.537 0.537 0.552 0.537 0.609 0.716 1.147 0.692 0.577

7.03 10.19 2.88 3.09 2.61 2.61 1.50 3.58 3.95

 _____________________________   _________________________ ____ 

Example 3-1 Estimation of the flammability limits for ethanol using both methods.

Solution  The combustion reaction is:

C 2 H 5O H   3O 2   2CO 2  3H 2O  Therefore, the stoichiometric concentration is:

c st  

moles fuel

1

100 

moles fuel  moles air 

1 3

100

100  6.54%

21

Applying Eqs. (3-2) and (3-3),

L F L  0.55  6   .54  3.60%  volume U F L  3.5  6 .54  22.89%  volume. Using the values from Table 3-2 for alcohols,

L F L  0.476  6.54  3.11%  volume U F L  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 L F L , while for the U F L  Eq. (3-5) seems to be more accurate.  ______________________________________   __________________________ ____________ 

68

 

4.1.2 .2 Fl amm ammabili abili ty limi ts of gas mixture mixturess 4.1 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]:

L F L mixt  

1 n 

c i 

(3-6)

 LFL i 1

U F L mixt  

i 

1 n 

c i  i 1 U F L i 

(3-7)



where  L F L i  is the value of the L F L  for component i  (% volume) c i i  is  is the concentration (% vol.) of component i  on  on a fuel basis (  c  c i i  = 100) and  is the number of combustible components in the mixture. n  is  ______________________________________   ____________________________ __________ 

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:

c i  

% vol i

% vol acetone  % vol benzene  % vol ethanolmixture 

Therefore:  = 0.25 c acetone  acetone  =  = 0.17 c benzene  benzene  =  = 0.58 c ethanol  ethanol  = Taking into account the values of the LFL and UFL for the three components (Table 3-1), the flammability limits of the mixture are:

L F L mixt  

U F L mixt  

0.25



1 0.17



0.58

3

1.4

3.5

0.25

1 0.17

0.58

13



8



 2.70%  volume

 14.1%  volume

19

The mixture is flammable, as 3% + 2% + 7% = 12%.  ______________________________________   ___________________________ ___________ 

69

 

4.1.3 ammabil bil ity li mits as a function function of pr essure 4.1.3 Fl amma 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 o f 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]:

U F L P   U F L  20.6  (logP i   10 3   1)

(3-8) -2

where P i i  is  is the absolute pressure (N m ) and UFL is the upper flammability limit at atmospheric pressure (% volume).

4.1.4 4.1 .4 Fl amm ammabil abil ity l imi ts as a function function of tempe temperr ature 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]:

L F L T   L F L 298 

3.136

U F L T   U F L 298 

3.136

  T   298 H c 

(3-9)

  T   298 H c 

(3-10)

where T  is  is the temperature (K) and

H c  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  as a function of the same parameter at another temperature T 1:

   

L F L T   L F L T  1  1

T   T 1    873  T 1 

(3-11)

where the temperatures are expressed in K.

70

 

 ___________________________  ______________________________________  ___________ 

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 

L F L100  1.0 1   

  0.87%  volume. 873  298 

 ______________________________________   ___________________________ ___________ 

4.1.5 4. 1.5 Inerting and flamm flammabili abili ty diagr diagr ams 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.  vs. y  diagram,  diagram, where

x  

inert volume flammable volume

 100

flammable volume  inert volume

y   total volume (flammable  inert  air)  100 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.

71

 

To avoid this risk, an inert gas must be introduced at the beginning of the operation. Thus, x  =  =   and y  will  will progressively increase. When y  reaches  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   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 Flas Flash h point point temperatur temperatur e 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 ( T f f      -42 ºC) is practically always ready to ignite; by contrast, diesel oil ( T f f      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: 2

c 

  c    T  b   e 

0

 T     0   T f   a 

(3-12)

c  2

1  e T            0

where T 0   is the boiling temperature of the substance (K) at atmospheric pressure, and 0  is a , b  and  and c  are  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

b 537.6 390.5 416.6 600.1

c 2217 1780 1900 2970

Ethers Sulphur Esters Ketones Halogens Aldehydes Phosphorus  Nitrogens Petroleum fractions

700.0 577.9 449.2 296.0 414.0 293.0 416.1 432.0 334.4

2879 2297 2217 1908 2154 1970 1666 1645 1807

275.9 238.0 260.8 260.5 262.1 264.5 201.7 185.7 237.9

4.33 A uto 4. utoig ignitio nition n te tempe mperr ature 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

E ST ST I M A T I O N O F T H E R M A L R A D I A T I O N F R O M F I R E S

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 , has been in selected. A description of the point source model   is flameofmodel  also included because its applicability some situations.

5.1 Point source source mode modell 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  

Q r  4  l p 2

(3-13)

where I  is  is the thermal radiation intensity (kW m -2) Q r r  is the heat released as thermal radiation per unit time (kW) and l p p  is the distance between the point source and the target (m). For jet fires and pool fires, Q r r  is a function of the burning rate m’  (kg s-1), the heat of combustion H c  and   and the radiative fraction or radiant heat fraction  r ad , i.e. the fraction of the combustion energy that is transferred as thermal radiation: rad iation:

Q r     r ad 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]:

 r ad 

 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  

 r ad  m ' H c  cos  

(3-16)

4  l p 2

where    is the angle between the plane perpendicular to the receiving surface and the line  joining the source point and the target (º) and  is the atmospheric transmissivity (-).   is 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 5.1 .1 Atmo Atmossphe pherr ic tr ans ansmis misssivi ty 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:   0.06   for P w   < 104 N·m-1    1.53  ( P  ·d  < w ·d  w   d )

(3-17-a)

75

 

 

0.09

   2.02 P w   d 

4

  for 10

5

-1

 P w ·d     10  N·m

  0.12   for P w   > 105 N·m-1    2.85  ( P  ·d  > w ·d  w   d )

(3-17-b) (3-17-c)

-2 where P w  w   is the partial pressure of water in the atmosphere (N m ) and d   is the distance  between the surface of the flame and the target. P w  w  can be estimated by the following expression:

P w    P wa 

H R 

(3-18)

100 -2

where P wa  wa  is the saturated water vapour pressure at the atmospheric temperature (N m ) and H R  is the relative humidity of the atmosphere (%).

wa  can be obtained from the prevailing temperature of the atmosphere [19]: P wa 

ln P wa 

 23.18986 

3816.42

(3-19)

T   46.13

where P wa   is expressed in N m -2 and T  in  in K. wa  is  ______________________________________   _________________________ _____________ 

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). H c   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  1.6   15  32  18.5  m l p       2   and the distance between the surface of the flames and the target is: cos   

18 18.5

  0.97

d   15  15  = 15.5 m cos   0.97 Estimation of the atmospheric transmissivity: ln P wa 

 23.18986  

3816.42

289.16  46.13

 7.486

76

 

-2 P wa  wa  = 1783 N m

79

-2 P w  w  = 1783    1408  N m

100

P w  d   1408  15.5   21824; therefore, by applying Eq. (3-17-b):  

   2.02 1408  15.5  r ad 

0.09

    0.82

 0.35  e 0.056  0.26

The overall burning rate is: 62

m   0.05  

4

  1.414 kg s 1

Therefore

I  

0.26  1.414  41900  0.82  0.97 4  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

Soli lid d fl flaa me mo mode dell

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

    F  E  I  

(3-20)

where     is  is the atmospheric transmissivity (-) F  is  is the view factor (-) and E  is  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. 5. 2.11 Vie View w factor 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:

F dA

 

2  A1

cos  1 cos  2

 

A1

 d 2

dA1

(3-21)

where 1 and 2 are the angles made by the normals and dA 1  on the flame and by dA 2  on the receiving element, and d  is  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 (F v ) for a cylindrical fire H/(D/2) 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

2.0 0.454 0.416 0.383 0.354 0.329 0.236 0.132 0.080 0.053 0.013 0.003 0.000

3.0 0.454 0.416 0.384 0.356 0.330 0.245 0.150 0.100 0.069 0.019 0.004 0.000

Table 3-5 Horizontal view factor (F h h ) for a cylindrical fire H/(D/2)

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

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

2.0 0.360 0.307 0.268 0.238 0.212 0.126 0.050 0.022 0.011 0.001 0.000 0.000

3.0 0.362 0.310 0.274 0.246 0.222 0.145 0.071 0.038 0.021 0.003 0.000 0.000

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

0.5 0.1118 0.1114 0.1101 0.1068 0.0902 0.0784 0.0599 0.0331 0.0268 0.0137 0.0069 0.0027

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 3-6 Vertical view factor (F v v ) for a parallelepipedic fire w/ x  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

0.75 0.1499 0.1491 0.1478 0.1427 0.1193 0.1032 0.0784 0.0431 0.0349 0.0178 0.0089 0.0036

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:

F max    F v 2  F h 2

(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 (F h h ) for a parallelepipedic fire H /w  x/w  0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.1 0.0732 0.0263 0.0127 0.0073 0.0047 0.0032 0.0023 0.0017 0.0013 0.0010

0.2 0.1380 0.0728 0.0414 0.0257 0.0171 0.0120 0.0087 0.0065 0.0050 0.0040

0.3 0.1705 0.1105 0.0720 0.0485 0.0339 0.0245 0.0182 0.0139 0.0108 0.0086

0.5 0.1998 0.1549 0.1182 0.0899 0.0687 0.0530 0.0414 0.0327 0.0261 0.0211

0.7 0.2126 0.1774 0.1459 0.1190 0.0966 0.0784 0.0638 0.0522 0.0429 0.0355

1.0 0.2217 0.1944 0.1687 0.1452 0.1243 0.1059 0.0903 0.0767 0.0653 0.0557

1.5 0.2279 0.2063 0.1855 0.1660 0.1478 0.1312 0.1162 0.1028 0.0908 0.0803

2.0 0.2305 0.2113 0.1928 0.1752 0.1588 0.1436 0.1296 0.1169 0.1054 0.0951

1.2 1.5 2.0 3.0

0.0007 0.0004 0.0002 0.0001

0.0026 0.0015 0.0007 0.0002

0.0056 0.0032 0.0015 0.0005

0.0142 0.0084 0.0041 0.0013

0.0249 0.0152 0.0076 0.0026

0.0409 0.0265 0.0139 0.0050

0.0629 0.0440 0.0253 0.0100

0.0774 0.0572 0.0355 0.0154

4.0 5.0

0.0000 0.0000

0.0001 0.0000

0.0002 0.0001

0.0006 0.0003

0.0011 0.0006

0.0023 0.0012

0.0047 0.0026

0.0077 0.0043

5.2.2 5. 2.2 Emis Emisssive power power The emissive power is the radiant heat emitted per unit surface of the flame and per unit -2 time (kW m ); 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:

  T fl 4  T a 4  E        

(3-23)

where     is  is the Stefan-Boltzmann constant (W m -2 K -4)  is the emissivity (-)   is T fl fl  is the radiation temperature of the flame (K) and T a a  is the ambient temperature (K). However, it is very difficult to calculate both T fl fl  (significantly lower than the adiabatic flame temperature) and  : 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 values appliedfrom to estimate the value of E   for the different types of fire; alternatively, experimental 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   for -2 -2 this zone varied from 80-100 kW m  for pools of 1.5 m in diameter to 120-160 kW m  for  pools with larger diameters. Significantly lower values of E  can  can be observed in the top part of the fire, where luminous flame has been almost completely substituted by black smoke. 2.663 2.414 2.165

Gasoline D=3 m

kW/m

2

140

120

1.916 1.668 1.419    D    /    H 1.170

2.030 1.839 1.647

Diesel D=6 m

2

kW/m

140

120

1.456 100

1.264

100

1.073    / 80    D    H 0.881

80

0.921

60

0.690

60

0.672

40

0.498

40

0.423 0.174

0.306 20

0.115

20

-0.50 -0.25 0.00 x/D

0.25 0.50

-0.50 -0.25 0.00 0.00 0.25 0.50

x/D

Fig. 3-6. Two mean E  contour  contour plots. Left: 3 m gasoline pool p ool fire. Right: 6 m diesel pool fire. Taken from [23], by permission.

Although an average value of E  is   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   x lu m  E lum   1  x lum  E soot 

(3-24)

where x l um  is the fraction of the fire surface covered by the luminous flame and   for the luminous and non-luminous zones of the fire, E l um  and E soot  soot  are the values of E  for -2 respectively (kW m ). For gasoline and diesel oil pool fires, the experimental data obtained for a circular pool with a diameter of 1.5 m  D     6 m [23] indicate that E soot   is independent of the diameter and soot  is -2 of the type of fuel: E soot    = 40 kW m . However, the average value of E l um  ranged   ranged between 80 soot  = -2 -2 kW m  and 120 kW m , as it is a function of the diameter and the type of fuel. E l um  increased with the pool diameter up to 5 m; thus, for D  < 5 m:

E lumgasolin e   53.64 D 0.474

(3-25)

E lumdiesel   28.03D 0.877 -2

while for D     5 m E lulu m  115 kW m . x l um  is constant for   D  < 5 m ( x lum l um gasoline   = 0.45, x lum l um die diessel oil  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 x lulu m ; some authors state that beyond D  > 20 m x l um  =  = 0. 0.5 0.4 0.3   m   u 0.2    l

  x 0.1 0.0

 Elum  E

   ) 100 90   m 80    /    W 70    k 60    (    E 50

   2

40 30 1

2

3

4

5

6

7 8 9 10

20

30

Diameter (m)

Fig. 3-8. Evolution of x lulu m , E lulu m  and  and E  as  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   x lu m E lu m   1  x lu m  E soot  for D  20 m E   E soot  for D  20 m

(3-26)

There is experimental evidence that E  increases   increases with pool diameter, which is essentially due to the increase of E l um   (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 x l um   and E   for gasoline is plotted in Figure 3-8. Finally, the following expression was also suggested for estimating the value of E :

E  

  H c    r ad  m  A

(3-27)

where  A is the area of the solid flame from which radiation is released (m 2). The suggested conservative value for  r ad  is 0.35. 6

F L A M E SI SI Z E 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

0.8

    y     c     n 0.6     e      t      t      i     m     r     e 0.4      t     n      I

(L/D)av

0.2

(L/D)max

0.0

0.0 0. 0 0. 0.2 2 0. 0.4 4 0. 0.6 6 0. 0.8 8 1. 1.0 0 1. 1.2 2 1. 1.4 4 1. 1.6 6 1. 1.8 8 2. 2.0 0 2. 2.2 2 2. 2.4 4 2. 2.6 6 2. 2.8 8 3. 3.0 0 3. 3.2 2 3. 3.4 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 Po Poo ol fir e size

6.1. 6. 1.11 Poo Pooll diam diameete terr 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 equiv alent diameter must be used:

D   4

surface area of  the pool

(3-28)

 

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: - Ins nsttantaneous spi spills - Continuous sp 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 u ntil it finds a physical p hysical 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)

V l 1 / 3

where t cr  cr  is a dimensionless critical time (-)  is the duration of the spill (s) t spill  s pill  is -1  is the fuel burning rate (m s ), and y  is V l l  is the total volume of the spilled liquid (m3).

84

 

-3 According to this criterion, spills are instantaneous if   t cr  cr   < 2 · 10 . Otherwise, they are considered continuous.

For anasinstantaneous spill,by without any containing barriers, expressed a function of time the following expression [21, 31]:the pool diameter can be

 D pool   D max  

2    t      3   t       2    1   1   2  t max      3   t max      

1/ 2

(3-30)

where D max  is the maximum diameter (m) that the pool can reach:

 V l 3 g   D max  2  2    y   

1/ 8

(3-31)

and t max  is the time (s) required to reach this maximum diameter:

  V    t max  0.6743  l  2   g  y     

1/ 4

 

(3-32)

However, a pool with the maximum diameter D max  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:

D av   0.683 D max

(3-33)

the casepool of a (for continuous release without physical the burning equilibrium for In a burning a situation in which theany release ratebarriers, equals the rate)diameter and the time required to reach it can be estimated with the following expressions [21]:

  v    D eq   2  l      y   t eq    0.564

1/ 2

(3-34)

D eq 

(3-35)

g  y D  

1/ 3

eq 

where D eq  is  is the equilibrium diameter (m)  is the leak rate (m 3 s-1), and v l l  is

t eq  is the time required to reach D eq  (s). Pools on water 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

 

 

      water    

g '  g 1 

  l 

(3-36)

where    l l  is the density of the spilled liquid (kg m-3), and  is the density of water (or seawater) (kg m -3).   water  water  is 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: 1/ 4    w    l  2  D 1  2.28  g V l  t       w  

(3-37)

-3

where    w   is the density of water (kg m ) w  is -3  is the density of the spilled liquid (kg m )   l l  is   Vl is the volume of liquid fuel released instantaneously (m3), and

 is the time elapsed from the start of the release (s). t  is In phase 2, viscous forces prevail:

     w    l   g V l 2 t 3 / 2  D 2 1.96   1/ 2  w     l  

1/ 6

(3-38)

where   w is the seawater viscosity (m2 s-1). Finally, in phase 3, the surface tension is the dominant spreading mechanism:

  f r 2 t 3    D 3  3.2  2        w  w  

1/ 4

(3-39)

where f r r  is the interface tension (N m -1) (ranging between 0.005 and 0.02 N m-1).

6.1. 6. 1.22 Bu Burning rning r at atee The burning rate is usually estimated using the following expression:  

 kD 

m  m  1   e 

(3-40)

where m   is the burning velocity for an infinite diameter pool (kg m -2 s-1) and k  is  is a constant -1

(m ). Various authors have proposed values for m  and  k  (Table  (Table 3-8). For large scale fires, m 

 m  .

86

 

Table 3-8 Expressions for the estimation of burning velocity Diesel oil Authors Babrauskas [27] Rew et al. [28] Muñoz et al. [23]

m  , kg m-2 s-1 0.034 0.054 0.054

Gasoline -1

k , m 2.80 1.30 0.88

m  , kg m-2 s-1 0.055 0.067 0.082

-1

k , m 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  h v   c p  T 0  T a  

(3-41)

where   H c  is  is the net combustion heat (kJ kg-1) -1  is the vaporization heat of the fuel at its boiling temperature (kJ kg ) and h v v  is T 0 0  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    l 

(3-42)