Introduction to Combustion and Fire - Assignment Example

FV1001 INTRODUCTION TO COMBUSTION AND FIRE Matter & International Systems of units. Theoretical questions What are the fundamental units of measurement and derived units? In the SI international system of units, the units of mass (grams), distance or length (metres), time (seconds), temperature (Kelvin), and amount of substance (moles) are classified as fundamental or base units. These units cannot be divided further. Units of quantities that are derived from these fundamental units are called derived units. These derived units can be expressed in terms of fundamental units. For example the unit of volume, litre, is derived from metres (1L = 0.001m3), while the unit for force, Newton, is derived from mass, length, and time (1N=1kg m s-2). (Lewis & Evans 2006) 2. Explain numerical prefixes for SI units. In the SI system of units, when the base units are extremely large or small for use, numerical prefixes are used to express larger or small units (Lewis & Evans 2006). The prefixes kilo (103), mega (106), and giga (109) are added to produce larger units, while deci (10-1), centi (10-2), milli (10-3), micro (10-6), nano (10-9), and pico (10-12) are added to produce small units. Examples: For small masses, the unit gram (g) may be too big, so milligram (mg, which equals 10-3g) is used. For long distances, the unit metre (m) may be too small, so kilometre (km, which equals 103m is used) (Lewis & Evans 2006). Exercises 1. Reduce the following dimensions to their simplest form: a) Pressure/power Pressure has the unit Pascal (Pa). 1 Pa = 1Nm-2 = 1kgms-2m-2 = 1kgm-1s-2 Power has the unit Watt. 1 Watt = 1Js-1 1 Nms-1 = 1 kgms-2ms-1=1 kgm2s-3 In terms of the dimensional formulae of fundamentals units (M, L, and T or mass, length and time) (SFPE Handbook 2002): Pressure = ML-1T-2 Power = ML2T-3 Pressure/Power = (ML-1T-2)/( ML2T-3) = L-3T = m-3s Therefore, Pressure/Power equals Time/Volume or (volume flow rate)-1. b) Work/Velocity Work has the unit Joule. 1J=1Nm = 1kgms-2m = 1kgm2s-2 Velocity is expressed as m/s In terms of the dimensional formulae of fundamentals units (M, L, and T or mass, length and time): Work = ML2T-2 Velocity = LT-1 Work/Velocity = (ML2T-2)/( LT-1) = MLT-1 = Momentum. Therefore, Work/Velocity is simplified to Momentum. 2. Convert in the SI units the following values a) 250 cm/microsecond 1cm/microsecond = 1×10-2m/1×10-6s = 1×104ms-1 250 cm/microsecond = 250×104ms-1 b) 100000 gigagram cm/minutes 1 gigagram cm/minutes = (1×109g )(1×10-2m)/60s=1.66×105gms-1 100000 gigagram cm/minutes = 1.66×1010gms-1 c) 150 miles/minutes 1mile/minute = 1609.344m/60s = 26.8224ms-1 150 miles/minutes = 150×26.8224ms-1 = 4023.36ms-1 2. Chemical elements and compounds Theoretical questions 1. What are chemical bonds and valence? Give examples. Chemical bonds are forces that hold atoms in a compound together. They are formed by sharing of electrons between atoms (Lewis & Evans 2006). For example, hydrogen atoms are connected to the oxygen atom in water through chemical bonds. Figure 1 Chemical bonds in water. The connecting lines between atoms represent chemical bonds. Valence is the combining power of an atom (Lewis & Evans 2006). Valence refers to the number of electrons an atom can give away, accept, or share to attain the electronic configuration of the nearest noble gas (Lewis & Evans 2006). Examples: Carbon whose electronic configuration is 1s22s2sp2 requires 4 electrons to attain neons electronic configuration of 1s22s22p6. Valence of carbon is 4. Lithium whose electronic configuration is 1s22s1 can give away one electron to attain heliums electronic configuration. Valence of lithium is 1. 2. What are free atoms, radicals, and ions? Free atom is an atom that becomes stable in diatomic form, but is unattached currently because it either is formed by a reaction or is preserved at extremely low pressure or in an inert frozen matrix (Friedman 1998). Example: Cl2, O2, N2, F2 and H2 are diatomic, but free atoms Cl, O, N, F, and H are reaction intermediaries. Free radicals are atoms or groups with unpaired electrons (Friedman 1998). Example OH, CH, CH2, CH3 free radicals are formed as reaction intermediaries. Ions are atoms or groups with excess or deficiency of electrons, carrying negative or positive charge (Friedman 1998). Example, Na+, OH-, Cl-, etc. Exercises 1. Explain chemical bonds in the molecule of carbon monoxide. Carbon has an electronic structure of 1s22s22p2, while oxygen has an electronic structure of 1s22s22p4. Carbon has a valence of 4 and oxygen 2. The 2p electrons in valence shell of carbon combine with unpaired 2p electrons in the valence shell of oxygen. A lone pair of electrons in oxygen forms a coordinate bond between the 2 atoms. Figure 2 Bonding in CO. Lewis structure and structural formula. 2. Give examples of single and double bonds. Single bonds: Here, a single pair of electrons is shared between atoms. Examples: Figure 3 Single bonds in chlorine and methane. Double bonds: Two pairs of electrons are shared between atoms. Examples Figure 4 Double bonds in oxygen and ethene 3. States of matter: Fluids, Solids, and Gases 1. Explain fluid characteristics: temperature, velocity, and density. The fluid characteristics of temperature, velocity, and density have different values at every point in space. So, these properties are considered continuous functions of time and position (Fox, McDonald, and Pitchard 2004). 2. What is pressure? Pressure of a gas is the force exerted by the gas per unit area. The unit for pressure is Pascal (Pa). 3. Equation of state for an ideal gas. The equation of state for an ideal gas is given as: PV = nRT Where P is the pressure of the gas, V the volume, n the number of moles of gas, T the temperature, and R the universal gas constant. Exercises 1. Demonstrate Boyles law using (P,V) and (V,T) graphs. For an ideal gas, assume n and T are constant. Then, a (P,V) graph looks like Figure 5a. When P & n are constant, the (V,T) graph looks like Figure 5b. Figure 5 a. (P,V) graph at constant n & T. b. (V,T) graph at constant P & T. From Figure 5a, at any point 1 and 2, P1V1 = P2V2 or V(1/P) This is the basis for Boyles law: For a fixed amount of gas at a constant temperature, volume is inversely proportional to pressure (Atkins & Jones 2007). From Figure 5b, it is clear that VT This is Charles law: For a fixed amount of gas at constant pressure, volume varies linearly with temperature (Atkins & Jones 2007). 2. Density of air at 0°C and 2 atm is 2.58kg/m3. Determine the molecular weight. T = 0°C = 273K P = 2atm n=No. of moles M=mass Density  =M/V= 2.58kg/m3 = 2.58g/L R = 0.082Latm/K/mol Molecular weight Mmol = M/n n=M/Mmol Substituting in the ideal gas equation PV=nRT or V/n=RT/P or MmolV/M=RT/P or Mmol=RT/P Mmol = RT/P = (2.58g/L)( 0.082Latm/K/mol)(273K)/2atm = 28.88g/mol Molecular weight of the gas is 28.88g/mol. 3. Molecular weight of carbon monoxide is 0.028kg/mol. Determine the density at T=900K and P=3atm T=900K P=3atm R = 0.082Latm/K/mol Mmol = 0.028kg/mol=28g/mol n=M/Mmol The ideal gas equation PV=nRT is written as MmolV/M=RT/P or Mmol=RT/P or =PMmol/RT  = PMmol/RT = (3atm)(28g/mol)/(0.082 Latm/K/mol)(900K) = 1.14g/L 4. Express 30°C in Fahrenheit and Kelvin scale. Tc = 30°C Temperature in Fahrenheit Tf = Tc(9/5) + 32 = (30°C)(9/5) + 32 = 86°F Temperature in Kelvin Tk = Tc + 273.15 = 303.15K 4. Chemical reactions and their rates Theoretical questions 1. Explain balancing of same atoms in stoichiometric chemical equations with example. In order to fulfil the Law of Conservation of Mass, the number of same atoms on both sides of an equation should be the same. A stoichiometric chemical is said to be balanced only when number of same atoms in reactants and products are equal. Consider the equation: CH4 + O2  CO2 + H2O (incorrect) This is not a balanced equation because of differences in number of atoms in reactants and products. Atom No. in reactants No. in products C 1 1 H 4 2 O 2 3 For both H and O, the number of atoms in reactants and products is not the same. To balance the equation, first consider the number of H atoms. It is 4 in reactants and 2 in products. To balance this, change the number of water molecules to 2. CH4 + O2  CO2 + 2H2O (incorrect) Now, the number of H atoms in reactants and products is the same. However, for O atom, the number is 2 in reactants and 4 in products. Changing the number of O2 molecules to 2 balances the equation. CH4 + 2O2  CO2 + 2H2O (correct) The number of atoms in reactants and products is: Atom No. in reactants No. in products C 1 1 H 4 4 O 4 4 2. What are stoichiometric mixtures? Give an example. A stoichiometric mixture is a mixture where all reactants are converted to products completely with no excess of reactants left. For example, a mixture of methane and oxygen where the number of moles of methane and oxygen are in the ratio 1:2 (nmethane/noxygen = 1/2) (Quintiere 2006). 3. How does chemical reaction rate depend on concentration of reactants? What are the concentration and a mole? The rate of a chemical reaction is the change in concentration of a reactant with time. Increasing the concentration of reactants causes an increase in the frequency of collisions between reactant molecules (in gas or liquid state), causing an increase in reactant rates (Lewis & Evans 2006). Mole is the unit for amount of substance. Concentration is a measure of packing of particles in a unit volume and has the unit mol/L (Lewis & Evans 2006). 4. Explain temperature dependence of the chemical reaction rate and Arrhenius equation. For a reaction to take place, the colliding reactant molecules must possess a minimum kinetic energy that than can overcome repulsive forces between molecules. This energy is referred to as the activation energy. Increasing the temperature increases energy of molecules, so the number of collisions between molecules with higher energies increases (Lewis & Evans 2006). This variation of reaction rate with temperature is given by the Arrhenius equation: k = Ae-E/RT or ln(k)=ln(A)-(E/RT) Here k is the rate constant at temperature T, R the universal gas constant, A is the pre-exponential factor or frequency factor, and E the Arrhenius activation energy (Lewis & Evans 2006). Exercises 1. Draw qualitatively the Arrhenius temperature dependence in the coordinates (normalised reaction rate - temperature) for different values of energy E1 and E2 (E1>E2). Figure 6 Arrhenius temperature dependence for energy values E1 and E2 (E1>E2) 2. Compare the chemical reaction rates for two temperatures T1 = 300K and T2 = 600 K for activation energy E = 180 kJ/mol. R is 8.314J/(mol K). T1 = 300K T2 = 600 K E = 180 kJ/mol = 180×103J/mol R = 8.314J/(mol K) If k1 and k2 are the rate constants at temperatures T1 and T2 respectively, the combined Arrhenius equation for both temperatures is (Lewis & Evans 2006): = 36.08 The high value of E is responsible for the dramatic result. ­­­­­­­­­­­­­­­ 5. Thermal Explosion Theoretical question 1. Explain the mechanism of self-acceleration of chemical reaction for thermal explosion in adiabatic conditions. During a chemical reaction, an Arrhenius type equation gives the heat release in a small volume: Where is the heat release, the heat of combustion, A the pre-exponential factor, n the reaction order, V the volume, and Ci the concentration (Drysdale 1998). If the system is non-adiabatic, heat released goes to the surroundings: Where is the heat loss rate, h the heat transfer coefficient, S the surface area of reaction volume, and the change in temperature. System cools as = (Drysdale 1998). However, in an adiabatic system, heat released remains in the system. Therefore, >. As this continues, temperature rises rapidly and the system self-accelerates to cause thermal explosion (Drysdale 1998). 2. Consider heat balance (Semenov diagram) in a vessel with cold walls. What are critical conditions of thermal explosion? Semenov assumed that temperature within the reaction mixture is constant and that heat loss is where T is variation of temperature between reaction mixture and walls (Drysdale 1998). For three values of temperature, Semenov diagrams are as in Figure 7. Figure 7Semenov model spontaneous ignition. Source: Drysdale (1996). At critical temperature T1, heat release curve and heat loss curves intersect tangentially, so=or system cools. At T2, no intersection occurs and temperature increases rapidly, leading to thermal explosion. 2. Explain temperature-time curve for adiabatic thermal explosion. Figure 8 Temperature-Time curve for adiabatic thermal explosion. Source: Drysdale (1996). During adiabatic thermal explosion, the auto-ignition temperature is lowered with increase in induction period. This is because the there is no heat loss from the system and the system continues to heat up with time. 2. Consider qualitatively the Semenov diagram for thermal explosion in a vessel with cold walls. Explain the effect of initial temperature and size of vessel on critical conditions for thermal explosion to take place. Figure 7 depicts the Semenov diagram for thermal explosion in a vessel with cold walls. At T = T0, heat loss is greater than heat produced. At T = T1, heat loss equals to the heat produced. This implies system cools rapidly in spite of heat energy generated by the reaction. At T = T2, heat loss is less than the heat produced. So, temperature rises continuously until thermal explosion. 3. Calculate the induction period for adiabatic thermal explosion of flammable pyrotechnic mixture PVN for initial temperature T0 = 400K. Induction period is inversely proportional to the temperature (Figure 8). Induction period  1/T0= 1/400 = 0.0025s. 6. Forms of Heat Transfer Theoretical question 1. Explain the physical mechanisms of conduction, convection, and radiation heat transfer. Thermal conduction. Thermal conduction is the flow of heat through a temperature gradient. Electrons in the conducting material gain kinetic energy and move from higher temperature region to lower temperature region, increasing the temperature. Rate of thermal conduction is defined as: Where, Where is heat transfer rate, k is the thermal conductivity, A is the surface area of cross-section, T is change in temperature, and x is thickness of conducting surface (Quintiere 2006). Convection. Convection is the heat transfer caused by moving fluid currents of the medium. The heat flux in convection is defined as: q˝ = h(Tf-Ts) Where q˝ is the heat flux, h is the heat transfer coefficient, Tf the gas temperature and Ts the solid surface temperature (Stec 2008). Radiation. Radiation is the heat transfer through electromagnetic waves to an object located at a distance from the heat source. Radiation flux is defined as: q˝ = εσTs4 Where q˝ is the heat flux, ε the emissivity, σ the Stefan-Boltzmann constant, and T the temperature (Quintiere 2006). Exercises 1. Calculate the rate of heat transfer through a 0.4 m2 of a plaster wall 3 cm thick. One side of wall is 625°C and other 50°C. Thermal conductivity of plaster is 0.50.5 W/m x°C. k = 0.5 Wm-1(°C)-1 A = 0.4m2 T = 625°C-25°C = 600°C x = 3 cm = 3×10-2 m (Q/T) = (0.5 Wm-1(°C)-1×0.4m2) 600°C/(3×10-2 m) = 4000W Rate of heat transfer is 4000W. 2. A concrete wall (emissivity 0.55) is at 800°C. Calculate the radiative heat flux emitted by sq. meter of the wall ε = 0.55 Ts = 800°C = 1073K σ = 5.67×10-8Wm-2K-4 q˝ = εσTs4 = (0.55)( 5.67×10-8Wm-2K-4)(1073K)4 = 41337.5 Wm-2 Radiational heat flux is 41,337.5 Wm-2. 7. Ignition Theoretical questions 1. How to access hazards associated with runaway reactions? In hazards associated with thermal runaway or a runaway reaction, the ignition takes place at the centre of a body where the temperature is highest. So, the body is likely to appear unharmed on the outside, while it is completely charred inside. This could be hazardous to fire fighters. The Frank-Kamenetskii thermal explosion theory that considers temperature gradients in a thermal explosion, is used in such fire investigations, as it helps in predicting the critical ambient temperature for the body (Quintiere 2006). 2. Describe the process of ignition of a solid combustible material by a hotplate. Explain evolution of temperature field in the solid material. Consider a solid combustible material (sides are insulated) with heat of combustion Δhc, thickness ro, cross-sectional area S, density ρ, heat capacity c, conductivity k, and mass loss rate of Ae-E/RT. This is placed on a hot plate whose edges are insulated, whose temperature is Tb, and whose power output is P. We consider a small coordinate x measured from the materials surface. Let ξ =x/ ro and φ = T/Tb. The differential equation for such a process is (Quintiere 2006) : Heat is transferred through convection at the boundary of the material whose convective coefficient is hc and ambient temperature is T∞. We assume φ-1 is very small and use the dimensionless temperature variable θ = E/(R T∞)( φ-1). The equation then changes to (Quintiere 2006) Where, δ is the Damkohler number. The boundary conditions explain evolution of temperature field in the material. They would be (Quintiere 2006): At the hotplate surface (x=0): -kdT/dx = P/S At ambient, -kdT/dx =hc(T- T∞) Exercise 1. Calculate minimum external heat flux to ignite TNT if temperature of external convective flow is 200°C. T∞ = 200°C 8. Premixed Flame Theoretical questions 1. Why do flames propagate through a combustible mixture? Flame propagation takes place due to heterogeneous combustion of particles. When a combustible mixture is ignited, flame propagates when the concentrations are adequate and there are no external forces causing extensive cooling (Quintiere 2006). 2. Determine the flame front and flame propagation velocity. The combustion zone of a flame is called flame front. This region is perpendicular to the flow of combustion mixture. Flame propagation velocity is the velocity with which the premixed flame moves in a direction perpendicular to its surface through adjacent unburned mixture (Quintiere 2006). Flame velocity is calculated as ratio of chemical energy release rate to the energy required to raise temperature to the ignition temperature. 3. Explain fluid dynamics of premixed flame on the Bunsen burner. The premixed flame in a Bunsen burner is in the shape of a cone sitting at the open end of the vertical burner tube. Figure 9 Flame in a Bunsen burner. Source: Quintiere (2006) Here, the flame-front is in a state of quasi-equilibrium with the combustion mixture, giving a configuration where flame propagation speed and local flowrate (normal to the flame front and parallel to the tube) are equal. For a flame speed of Su, vtube the flowrate parallel to the tube, 2φ the cone angle: Su = vtubesin φ 4. Describe three regions in a plane premixed flame. What is adiabatic flame temperature? A plane premixed flame has the following regions or zones (Drysdale 1998, Quintiere 2006): 1. Pre-heating zone: Here, temperature increases by conduction of heat from hotter parts of the flame and combustion does not take place until a certain temperature is reached. Here, fuel diffuses forward, while combustion products diffuse backward. 2. Reaction zone: Most of the combustion process takes place here. Reaction here is induced by diffusion of free radicals from hotter parts of the flame. This region constitutes the visible flame. 3. Post-flame region: This is a high temperature region where free radicals formed are recombined. Cooling takes place subsequently. Figure 10 Zones in a plane flame. Source Drysdale (1998) Adiabatic flame temperature is the maximum possible temperature gained by a reaction at constant pressure. This is given by the balance between heat released in a combustion reaction (ΔHrθ) at the initial temperature (Ti) of reactants and the heat required to raise temperature of products of the final frame temperature (Tf): Where Cp is the heat capacity (SFPE Handbook 2002). Exercise 1. Estimate the ratio of gas velocity in fresh mixture and gas velocity of combustion products in a premixed flame. 9. Detonation Theoretical Questions 1. Describe the internal structure of a detonation wave. Detonation is a type of explosion where the combustion wave or flame propagates at supersonic speed through unburned fuel (SFPE Handbook 2002). Here, pressure rise is non-uniform and instantaneous during propagation of the shock wave. Figure 11 Structure of detonation wave. Source: Kubota Figure 11 depicts the detonation wave of a reactive gas at a one-dimensional steady state flow condition. At the front of the detonation wave, pressure (p1), temperature (T1) and density (ρ1) rise as the shock wave passes. This increase in temperature causes an exothermic reaction behind the shock wave, building up the temperature further. Behind the shockwave, pressure, temperature, and density decrease until they reach steady state condition (p2, T2, ρ2). This point where they reach steady state is known as Chapman-Jouguet (CJ) point. 2. Discuss the methods to prevent the initiation of explosion and to reduce possible damage. 1. Explosion venting helps to limit damage due to explosion. In this process, combustion gasses during a deflagration are discharged to preserve pressure below the enclosure damage threshold (SFPE Handbook 2002). Effective explosion systems kick off early in the deflagration, have large vent area and vent combustion gases without restrictions (SFPE Handbook 2002). 2. Explosion suppression systems prevent initiation of explosion. These systems can detect and suppress an incipient explosion before pressure rises to the enclosure damage threshold. A sensor, which is either a pressure or a flame radiation detector, senses embryonic explosion in the early stages of deflagration and triggers discharge of suppression agents into the system (SFPE Handbook 2002). Exercises 1. Calculate the velocity of steady-state freely propagating strong detonation if the ratio of specific heat capacities is 1.3 and heat of combustion 775 kJ/kg. Velocity of a detonation, c, is given as: Where is the ratio of specific heat capacities and is the heat of combustion. We have, = 1.3 =775kJ/kg=775×103 J kg-1 Therefore, c=(1.3×775×103 J kg-1)1/2 =1003.74ms-1 The velocity of steady-state freely propagating strong detonation is 972ms-1. 10. Diffusion combustion. Fire plume. Theoretical questions 1. Explain how a candle burns. A liquid with high firepoint (paraffin or candle wax) ignites easily when absorbed on to a wick, which is a porous medium with low thermal conductivity. When the wax-soaked wick is lit, pyrolysis of wax causes a rapid increase in local temperature (600-800°C) as thinness of the wax coating prevents heat loss due to convection and the wick also acts as an insulator. This leads to ignition and flame spread over the wick surface (Drysdale 1998). At the flame centre, the amount of oxygen is insufficient for complete oxidisation and formation of conjugated double bonds, cyclization, and aromatization occurs, which glow, giving a bright yellow flame (Stec 2008). 2. Compare a jet fire and a buoyancy dominated flame. Buoyancy dominated flames occur when condensed fuels are burnt. These flames are laminar and their height increases steadily. As the diameter of fuel bed increases, degree of turbulence increases (Drysdale 1998). Jet fires on the other hand occur on burning premixed fuels. At high flowrates, turbulence of flame increases as flame stability near the burner rim is lost due to entrainment of surplus air at the base of the flame (Drysdale 1998). They are dependent on momentum of the fuel vapour. 3. Explain the low value of Froude number for natural fires. Natural fires are driven by buoyancy, so their flame heights are low. As higher Froude numbers correspond to increasing flame height, natural fires have lower Froude numbers (Drysdale 1998). 4. Explain qualitatively the burning process of liquid and solid materials. Rate of burning of gaseous materials is equal to rate of supply of fuel. However, for liquid and solid materials, the rate of supply of volatile compounds from the fuel surface depends on heat transfer rate from flame to fuel. So rate of burning m" is expressed as: m" = (QF" - QL")/Lv Where QF" - QL" gives the heat transfer rate and Lv is the latent heat of vaporization. 11. Combustible Liquids and Solids. Theoretical Questions 1. Describe the flash point, fire point and auto-ignition temperature for combustible liquids. How these characteristic are measured in laboratories? Flashpoint. Flashpoint of a liquid fuel is the lowest temperature where a combustible vapour-air mixture exists on the fuel surface. For a liquid fuel, flashpoint is determined by the lowest temperature where the vapour-air mixture ignites, causing a flash (Drysdale 1998). Fire point: To achieve burning of a liquid at flashpoint, the temperature of the liquid should be increased to a value called fire point. Auto-ignition temperature is the lowest temperature at which a combustible mixture can self-ignite (Quintiere 2006). 2. Discuss the main factors influencing flame spread over solid materials. The following factors affect flame spread over solids: 1. Ambient wind: Flame can spread upwind (opposed flame spread) or downwind (wind-aided flame spread) (SFPE Handbook 2002). This wind could be meteorological, externally caused, or fire-induced. 2. Geometrical orientation. Flame spread in solids depends on the orientation: vertical or horizontal, facing upwards or downwards (SFPE Handbook 2002). 3. Other factors. Other factors that affect flame spread include heat transfer for laminar and turbulent flows, flame radiation, and chemical kinetic factors (SFPE Handbook 2002). Exercise 1. Calculate the average flame height for a pool gasoline fire. Diameter of pool is 4m. Average flame height Lf is: Lf=0.232/5-1.02D Where D is pool diameter and is the heat release: , where is the mass burning rate and is heat of combustion Mass burning rate per unit area for gasoline = 0.011kgm-2s-1. D = 4m Radius r = 4m/2 = 2m For the entire circular pool, =0.011 kgm-2s-1 ×(π(2m)2) = 0.138kg s-1 for gasoline = 44.1kJg-1. = (0.138kg s-1×44.1 ×103 kJ/kg) = 6092.5 kJs-1 Lf = 0.23 (6092.5)2/5-1.02 (4m) = 3.4m So average flame height of gasoline pool with diameter 4m is 3.4m. 12. Fire as a combustion system. Theoretical questions 1. Define fuel load and mass burning rate. 2. Define heat of combustion, heat release rate, and heat efficiency. Heat of combustion is the energy release rate per unit loss of fuel mass. Heat release rate is the energy released for loss of entire fuel mass. Combustion efficiency is the ratio of total heat release rate to theoretical heat release rate. 3. Describe three zones in turbulent diffusion flame (fire plume) A fire plume has three zones or regimes (Drysdale 1998): 1. The flame zone, or the near field, which is characterised by a persistent flame and an accelerating flow of burning gases. 2. The intermittent zone, which is characterised by intermittent flaming and near constant flow velocity. 3. The buoyant plume characterised by decreasing velocity and temperature with height. 4. Describe smoke production in fires. What is the range of the smoke conversion factor? For a fire environment, smoke production is one of the basic characteristics. Smoke production takes place in the combustion conditions of flaming, pyrolysis, and smouldering, and all these conditions affect quantity and quality of smoke. Smoke conversion factor is the ratio of mass of smoke produced to mass of fuel burned. This can range from 0 to 0.20. 5. Why is thermal radiation important in fire? Importance of thermal radiation in fire: 1. Radiation is the most frequent mode for energy feedback from flames to burning material. Hence, rate of burning depends on thermal radiation. 2. Radiation transfer spreads flame to nearby combustibles. The speed of fire spread depends directly on radiation intensity. Exercise: 3. Calculate the heat release rate for PMMA (m=0.035 kg/m2 s, heat of combustion 25.2 MJ/kg, combustion efficiency 0.55. Mass burning rate m"= 0.035kgm-2s-1 Theoretical heat of combustion = 25.2 ×106Jkg-1 Combustion efficiency = 0.55 Combustion efficiency = (Effective heat of combustion)/(Total heat of combustion). Total heat of combustion = Theoretical Heat of combustion × combustion efficiency = 25.2 ×106J /kg ×0.55 = 13.86 ×106Jkg-1 Heat release rate = Total heat of combustion× Mass burning rate = 13.86×106Jkg-1×0.035 kgm-2s-1 = 485.1 kJ m-2s-1 13. Fire in enclosures Theoretical question 1. What is a positive thermal feedback for fires in enclosures leading to flashover? During burning of a material in an enclosure, the heat released increases temperature of hot layer of gasses, enclosure walls, and ceiling. Some of this heat hits the fire surface, increasing mass loss rate. This increase in mass loss rate is positive thermal feedback. Radiation of heat to gases, enclosure walls, and ceiling causes fire to spread to distant areas in the enclosure, leading to flashpoint. 2. Explain temperature-time dependence for fire development in an enclosure. 3. Describe the conditions necessary for flashover to occur in terms of radiant heat flux at floor level, temperature of a hot upper layer. Conditions necessary for flashover Heat fluxes should be low to cause piloted ignition. A heat flux of 20 kWm-2 to the floor may result in flashover. This flow of heat flux is associated with upper layer temperatures of 500-600°C. In burning enclosures, the conditions at lower heat release rates are identical to open burning. However, at higher heat release rates of 1.5-1.75MW, flashover occurs. In fires reaching flashover, the interface between upper and lower hot layers is near the floor. Airflow is maximum for a given upper gas temperature. 4. Explain fuel-controlled and oxygen-controlled regimes of fire in an enclosure Fire in an enclosure initially develops from a small incipient fire. If this is not suppressed, the fire grows to a maximum size that is either fuel controlled or oxygen-controlled. Fuel controlled fire depends on amount of fuel present, while oxygen-controlled fire depends on the oxygen supply (Quintiere 2006). 5. What is fire resistance and its criteria? Fire resistance is the ability of an element of building construction to continue functioning as a barrier during fire. 6. Describe the main flow pattern associated with fire development in enclosures. Figure 12 Main flow patterns in fire enclosures. Source: Quintiere (2006). A hot ceiling flow that occupies one-tenth of the room height. A cold jet occurs at the floor. Between these two, a four-layer flow pattern comprising recirculating flows occupies most of the space. The cold lower layer flows and hot upper layer flows give rise to a well-defined layer interface in the middle of the four-layer pattern (Quintiere 2006). 7. Explain backdraft. A sudden supply of oxygen when allowed into a fire in enclosure combines with the fuel-rich hot gases, causing a rapid rise in combustion and pressure. This could result in failure of other windows and walls in the enclosure. This occurrence is known as back draft. 14. Fire in enclosures. Fire modelling. 1. Explain the main assumptions used in field modelling. Given an example of field modelling study. In field modelling or computational fluid dynamics (CFD) modelling of fires, a volume under consideration is divided into many sub-volumes. The laws of conservation of mass, energy, and momentum are applied to each subvolume. Field modelling is used in places where traditional building regulations may not be readily applicable such as in shopping malls, airports, or atrium hotels (SFPE Handbook 2002). 2. What are the advantages and limitations of field models? Advantages (SFPE Handbook 2002) Problems and successes demonstrated elsewhere can often be used in the fire context. Field modelling allows the study of extremely complicated problems that are only partially amenable to reduced scale physical modelling. Limitations (SFPE Handbook 2002) Field modelling does not provide a full, direct numerical simulation that solves the exact equations. Some uncertainties are introduced as a result of numerical methods used in solving continuous equations. 3. What is zone modelling of fires? Objectives of zone modelling? Zone modeling of fires uses computer programs that are designed to predict conditions resulting from a fire in an enclosure (SFPE Handbook 2002). These models solve the equations on the basis of zone assumptions that describe the fire-induced conditions within an enclosure. (SFPE Handbook 2002) 4. Why are validation and verification necessary for fire modelling? Validation is necessary to determine the models degree of accuracy in representing a real world situation from the user perspective. Verification is essential to determine if an accurately implemented model accurately represents the developers conceptual description and the solution. References Atkins, P., & Jones, L. (2007). Chemical principles: The quest for insight, 4e. Oxford: Oxford University Press. Drysdale, DD (1998). An introduction to fire dynamics, 2e. Chichester: Wiley. Fox, R., McDonald, A., Pritchard, P. (2004). Introduction to fluid mechanics, 6e. Hoboken, NJ: John Wiley & Sons. Friedman, R. (1998). Principles of fire protection chemistry and physics, 3e. NFPA, Quincy MA. Kubota, N. (2002). Propellants and explosives: Thermochemical aspects of combustion. Weinheim: Wiley-VCH. Lewis, R. and Evans, W. (2006). Chemistry, 3e. Hampshire: Palgrave Macmillan. Quintiere, J. G. (2006). Fundamentals of fire phenomena. Chichester: Wiley. SFPE Handbook of Fire Protection Engineering, 3rd ed. (2002). Quincy MA: NFPA. Stec, A. (2008). Fire and its ingredients. University of Central Lancashire. Read More