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You should bear in mind that this calculation is based on an initial temperature of 300 K and the chemical reaction has been simplifi ed. In some tables the data is based on another temperature, but particularly in experiments where the correct reaction formula is used indirectly. Sample calculation: Upper fl ammability limit Methane releases 800 KJ/mol and C p for the incoming products is taken from Table 4. The equation DH c = S (C p × DT) is used, where DT is the difference between the adiabatic fl ame temperature and the initial temperature. The initial temperature is set to 300 K and the adiabatic fl ame temperature is set to 1973 K, which is somewhat higher than normal. 11 The reaction formula is as follows: XCH4 + CH4 + 2O2 + 2 × 79 N2 P CO2 + 2H2O + 2 × 79 N2 + XCH4 21 21 800000 = 54.3 + 2 × 41.2 + X × 34.9 + × 2.33 + 2 × 79 32.7 (1973 – 300) 21 The equation gives X = 1.17 and using this, we can calculate the proportion of methane in the mixture. 1.17 + 1 = 18% volume 1.17 + 1 + 2 + (2)79 21 According to the result of the equation, 18% of the reactants are methane. 18% is converted to g/m 3 using the density for methane gas. The equation will then be: 16/29 × 1.2 = 0.65 kg/m 3 This will then give the mass as 0.65 × 0.18 =117 g/m 3 . 179
You should bear in mind that this calculation is based on an initial temperature of 300 K and the reaction formula has been simplifi ed. In some tables the data is based on another temperature, but particularly in experiments where the correct reaction formula is used indirectly. Sample calculation: Temperature impact on lower fl ammability limit We will start with a sample calculation where the lower fl ammability limit for methane has been calculated. The initial temperature, T 0, is changed in the calculations and replaced by 500 K. Calculations are very approximate, but provide a rough estimate of the temperature’s effect on the lower fl ammability limit. The equation DH c = S (C p × DT) is used, where DT is the difference between the adiabatic fl ame temperature and the initial temperature. The initial temperature is set to 500 K. The reaction formula is the one used earlier (see Sample calculation: lower fl ammability limit). 800000 = 54.3 + 2 × 41.2 + X × 34.9 + X × 79 32.7 + 2 × 79 × 32.7 (1600 – 500) 21 21 The equation gives X = 2.18 and using this, we can calculate the proportion of methane in the mixture. 1 = 4.8% volume 2.18 + 1 + 2 + (2.18 + 2) 79 21 According to the result of the equation, 4.8% of the reactants are methane. This can be compared with the result for the lower fl ammability limit calculation from the earlier example, 5.7%, where the initial temperature was assumed to be 300 K. If the temperature continues to rise, in theory, even small amounts of fuel will be combustible. The upper fl ammability limit’s variation with temperature can be calculated using the same method. 180
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Lars-Göran Bengtsson Enclosure fi
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The content of this book may not be
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4.4 Risk assessment 99 Smoke gases
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Foreword The aim of this book is to
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Overview This book assumes that its
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Figure 1. Fire growth curve featuri
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Fire extinguished/burnt out Just sm
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chapter 2 How a fi re starts People
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smoke gases in the compartment, but
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Pyrolysis gases material is subject
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Ignition time Ignition time can als
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Smouldering fi res can therefore re
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2.4.1 Thermal inertia kρc The fl a
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grow to 50 cm, then a 1 m high fl a
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Flammability in solid materials is
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6. The fl ame spread rate varies wi
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Figure 22. Smoke gases start to acc
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Figure 25. The different sections i
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The production of unburnt gases is
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Heat release rate from a fuel surfa
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Carbon monoxide (CO) is the next mo
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ment to accelerate, but this is obv
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is far too high for a premixed fl a
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is fi lled with fuel molecules, con
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Figure 33. Flames on the under side
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Flammability limits Vent For a prem
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CH4 + 2 + 2 + + 2O2 = CO2 + 2H2O +
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Temperature impact on fl ammability
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energy is required, which can be ex
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ustion takes place in under-ventila
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If turbulence affects the system th
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Neutral plane The differences in pr
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Where the differences in temperatur
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Positive pressure Negative pressure
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Dp = 353(1/Ta - 1/Tg)gh Let us take
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Figure 51 (top). The pressure condi
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Vent just during the early stage of
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In Figure 57 calculations have not
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3.4 Summary In many situations the
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oom and in a fairly closed room (on
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Figure 59. The fi re has spread to
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Figure 60. Flashover occurrence ove
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90 precisely the exact moment when
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The amount of energy released when
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Figure 63. Thermal balance on a fue
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Figure 64. Flame spread under the c
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Access to fuel Figure 67. Fire’s
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The signs indicating that a fl asho
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Figure 69. Different signs indicati
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Smoke gas temperature 700 600 500 4
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106 clear that we need to reach the
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This example illustrates how crucia
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110
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112
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114
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Figure 78. Fire pulsating. Figure 7
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Figure 81 (above). The fi re is abo
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Figure 85. The fi re resumes its de
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122 is suffi ciently big. A fi re
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Figure 89. A backdraught entails a
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Figure 90. The picture shows a vent
Page 129 and 130: 128 The current which results in sm
Page 131 and 132: 130 fans should not be used. You sh
Page 133 and 134: Figure 97. A larger part is premixe
Page 135 and 136: Figure 101. There are still combust
Page 137 and 138: Figure 103. Sauna. A common scenari
Page 139 and 140: Fires in enclosed spaced with minim
Page 141 and 142: Figure 107. The smoke gases are str
Page 143 and 144: Figure 110. The fi xed lance versio
Page 145 and 146: Four fi re scenarios: 1. The fi re
Page 147 and 148: 146 to occur. If a backdraught occu
Page 149 and 150: 148 Fire at 62 Watts Street The New
Page 152 and 153: chapter 7 Smoke gas explosion Throu
Page 154 and 155: ignition source required to ignite
Page 156 and 157: ity is also affected by turbulence
Page 158 and 159: 7.3.2 Course of action If smoke gas
Page 160 and 161: Test your knowledge! 1. Why do smok
Page 162: Corridor Space in between Sauna app
Page 165 and 166: 164 gases the fl ames would come ou
Page 167 and 168: Figure 119. Plan drawing of the bas
Page 169 and 170: 168 Glossary Adiabatic fl ame If al
Page 171 and 172: Heat of combustion, This measures t
Page 173 and 174: 172 Suggested solutions to test que
Page 175 and 176: 174 13. In a closed room there will
Page 177 and 178: 176 6. By cooling down the smoke ga
Page 179: Sample calculations Flammability li
Page 183 and 184: Backdraught Sample calculation: The
Page 185 and 186: E = (T f/T i)( N b/N u) Tf Ti Nb Nu
Page 187 and 188: v* = v/(g × h × b) 0.5 , which gi
Page 189 and 190: 188 gram), Institutionen för brand
Page 191 and 192: Index 62 Watts street 148 Adiabatic
Page 193 and 194: 192 List of fi gures Drawn illustra
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Enclosure fires

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