Enclosure Fire Dynamics - Lab Report Example

THIS IS A DRAFT VERSION THIS IS A DRAFT VERSION THIS IS A DRAFT VERSION ENCLOSURE FIRE DYNAMICS N1. Using Arrhenius’ law compare the chemical reaction rates at two temperatures, 450K and 900K, where the activation energy is 180kJ/mole. Comment on the results (2 marks). Solution: The Arrhenius equation for reaction rates is well-known as: R(T) = k exp(- E / ( RT ) ) For temperatures T1 and T2, the equation can be written as follows: R(T1) = k exp(- E / ( RT1 ) ) R(T2) = k exp(- E / ( RT2 ) ) Note: the pre-exponential factor (k) is not given but can be assumed to be constant (i.e. the same for each temperature). These two equations can be combined to give the ratio of reaction rates resulting in the following equation: R(T2) k exp( - E / ( RT2 )) --------- = --------------------- R(T1) k exp( - E / ( RT1 )) which simplifies to: R(T2) exp( - E / ( RT2 )) --------- = --------------------- R(T1) exp( - E / ( RT1 )) Where: T1 = Temperature 1, 450 K T2 = Temperature 2, 900 K E = Activation energy, 180 k J/mole R = Universal gas constant, 8.314 J/(mole K) Substituting these values into the equation gives: R(T2) exp( - 180 000 / ( 8.314 x 900 )) --------- = ---------------------------------------- R(T1) exp( - 180 000 / ( 8.314 x 450 )) R(T2) --------- = 2.8 x 10 ¹⁰ R(T1) Increasing the temperature from 450 K to 900 K increases the reaction rate by a factor of 2.8 x 10 ¹⁰. N2. Calculate the maximum pre-explosion heating and induction period for polyvinyl nitrite for the storage temperatures of 20C, 100C and 200C. Comment on the results (2 marks). Solution: The maximum pre-explosion temperature can be calculated using the following formula: where: T0 = Initial temperature, [ K] R = Universal gas constant, 8.314 J / (mole K) E = Activation energy for polyvinyl nitrite, 107 k J / mole. Therefore the maximun pre-explosion temperature Tpre-expl is: for 293 K, Tpre-expl = 293 + 8.134 x 293² / ( 107x10³ ) [ K ] Tpre-expl = 299.52 K for 373 K, Tpre-expl = 373 + 8.134 x 373² / ( 107x10³ ) [ K ] Tpre-expl = 383.57 K for 473 K, Tpre-expl = 473 + 8.134 x 473² / 107 [ K ] Tpre-expl = 490.00 K Induction Period. The induction period for adiabatic thermal explosion of polyvinyl nitrite can be calculated using the following formula : Where: Cp = Specific heat for polyvinyl nitrite, X J / ( mole K) R = Universal gas constant, 8.314 J / (mole K) T0 = Initial temperature, 293 K, 373 K 473 K, E = Activation energy for polyvinyl nitrite, 107 K J / mole QK0 = Heat Release for polyvinyl nitrite, X J / (kg s) Substituting these values (other than temperature) into the equation gives: Cp x 8.314 T0² tad = -------------------------------------- E QK0 exp( - E / (8.314 T0) ) Substituting in T0 = 293 K, 373 K and 473 K gives: For 293 K, tad = For 373 K, tad = For 473 K, tad = N3. Calculate the maximum diameter of a cylindrical tank, which provides secure storage of the explosive substance plumbum azide upon the temperatures 20C and 150C providing heat transfer coefficient at the tank surface of 25 W/(m² K) (2 marks). Solution: Critical condition for thermal explosion is given by the equation: where: Therefore: Now placing this into the equation for critical condition for thermal explosion and rearranging to give d: Where: h = Heat transfer co-efficient, 25 W/ (m² K) R = Universal gas constant, 8.314 J/ (mole K) T0 = Initial temperatures, 293 K and 423 K E = Activation energy for plumbum azide, X [J/mole] QK0 = Heat release for plumbum azide, X [J/(kg.s)] ρ = Density for plumbum azide, X [kg/m3] Substituting these values (other than temperature) into the formula gives: Substituting in the values of T0 = 293 K and 423 K in to the above equation gives the maximum diameter of a cylindrical tank that will provide secure storage for plumbum azide at the temperatures 293 K and 423 K: For T0 = 293, d = X m (i.e. approximately X km) For T0 = 423, d = X m (i.e. approximately X mm) N4. Given the parameters of the powder, E = 85 k J/mole, k = 1.0 W/(m K), ρ=800 kg/m³, Qk0 = 1.95x10¹¹ J/(kg s) determine the critical plane layer thicknesses (above which presents an explosion hazard) of the dried powder in an oven at 200 C and 500 C (3 marks). Solution: First determine the limiting value of the Frank-Kamenetskii parameter δcr and the characteristic size L for a plane layer: δcr is 0.878 for a plane layer. L is equal to half the layer thickness for a plane layer. The critical condition for thermal explosion is given by the formula: Where: ρ = Density, 800 [kg/m3] Qk0 = Heat release, 1.95x10¹¹ J/(kg s) E = Activating energy, 85 x 10³ J K /mole R = Universal gas constant, 8.314 J/mole T0 = Initial temperature, 473 K (i.e. 200+273 = 473K) 773 K (i.e. 500+273 = 773K) k = Heat conductivity = 1.0 [W/(m.K)] L = characteristic sizes of the systems, i.e. half the layer thickness. Substituting these values (except L) in to the equation gives: For temperature initial 473 K Substituting in the values of L For temperature initial 773 K N5. The initial temperature of a flammable mixture is 293 K and the initial pressure is 1 atm. The mixture is burned in a closed vessel and the adiabatic flame temperature 2300K is achieved. Calculate the pressure in the burnt mixture. Calculate the laminar burning velocity assuming a propane-air mixture under this pressure (3 marks). Solution: For pressure in the burnt mixture: Using the ideal gas equation: PV = nRT Where: P = Pressure [atm] V = Volume [m3] n = Amount of molecules R = Universal gas constant, 8.314 [J/mole K] T = Temperature [K] Creating equations for both the initial temperature and the final temperature gives: Dividing one equation by the other gives: The values for V, R and n are constant. Therefore V0 = Vf, R0 = Rf, and n0=nf and the equation can be simplified to: Making the final pressure Pf the subject gives: Where: P0 = Intitial pressure, 1.0 [atm] T0 = Intial temperature, 293 [K] Tf = Final temperature, 2300 [K] Substituting these values into the equation give: 2300 Pf = 1.0 -------- 293 Pf = 7.8498 [atm] Solution: Calculate the laminar burning velocity assuming a propane-air mixture under this pressure. Variation of burning velocity with pressure is given by the equation: Where: Vb0 = Burning velocity for propane –air, X [m/s] P = Pressure, 7. 8498 [atm] P0 = Initial ambient pressure, 1.0 [atm] n = 0.25 to 0.33 (see notes) Substituting these values into the equation gives: N6. A mixed fuel is composed of methane (volume fraction 0.3), propane (0.2), carbon monoxide (0.25) and hydrogen (0.25). Calculate the lower flammability limit for the mixture (oxidiser is air) (2 marks). Solution: The lower flammability of a mixture is given by the equation: Where: Cfi = the volume concentration for each component (volume fraction) LFLi = the lower flammability limit of each component in air. Therefore the lower flammability limit of the mixture is given by: Fuel component Volume fraction Cf Lower Flammability Limit Methane 0.3 5.0% or 0.05 propane 0.2 Carbon Monoxide 0.25 12.5% or 0.125 Hydrogen 0.25 4.0% or 0.04 Using the values in the above table gives: 1 LFL = ------------------------------------------ 0.3 0.2 0.25 0.25 ------- + ------- + ---------- + ------- 0.5 0.125 0.04 LFL = The lower flammability for the whole of the mixture in air is 5.5%. By multiplying the composition of each component by the LFL of the mixture the concentration of each component in the mixture can be calculated. N7. Consider a 2.9 m diameter pan fire of methyl alcohol with a heat release intensity of 500 kW/m² of surface area. Calculate the mean flame height under normal atmospheric conditions (2 marks). Solution: The average flame height is calculated using Heskestad’s equation: Where: Lf = Average height of flame [m] D = Diameter of pool, 2.9 [m] = Energy (Heat) release [kW]. The heat release is calculated by multiplying the surface area of the pool by the heat release rate: = 3302.59 kW Therefore: Lf = 0.235 x 3302.59 ²/⁵ – 1.02 x 2.9 Lf = 3.04 m The mean flame height is therefore 3.04 m. N8. A stack of wood pallets (1.5m x 1.5m) burns with a total heat release rate of 2300 kW under normal atmospheric conditions. Calculate the mean flame height above the top of the pallet stack (3 marks). Solution: The equation to calculate flame height is based on a circular pool (i.e. equation uses the diameter of the circle). Therefore, to use the equation we should calculate the equivalent diameter of a circle with the same area as the wood pallets. First calculate the area of the wooden pallets: Area = 1.5 x 1.5 Area = 2.25 m² The area of a circle is given by the formula: Transpose to make diameter the subject: D = 1.6925 m Now substitute this into the average flame height formula. Where: Lf = Average height of flame [m] D = Diameter of pool, 1.6925 [m] = Energy (Heat) release, 2300 [kW]. Therefore: Lf = 0.235 x 2300 ²/⁵ – 1.02 x 1.6925 Lf = 3.47 m The mean flame height above the stack of pallets is therefore 3.47 m. N9. Estimate the heat release rate above a free-standing steady pool fire with 2.1m diameter. Assume a net heat of combustion of 33.4 MJ/kg fuel, a mass burning rate of 40 g/(m² s) and a combustion efficiency of 0.75 (2 marks). Solution: In order to estimate the heat release, first transpose the formula for calculating the burning rate to make heat flux the subject: Where: = Burning rate, 0.04 [ kg /(m² s) ] = Heat flux [ kW/m² ] Lv = Latent heat of evaporation or gasification, 33400 [kJ/kg] Therefore: = 1336 efficiency kW/m² = 1336 x 0.75 kW/m² = 1002 kW/m² Therefore: The estimate heat release is given by: Net = 1002 ( ¶ D² )/ 4 Net = 1002 ( ¶ 2.1² )/ 4 Net = 3470.53 kW The estimate heat release from the surface is 3470.53 kW. N10. Estimate the average heat flux received by the surface of a free standing Heptane pool fire (3 marks). Solution: N11. Calculate the diameter of the kerosene pool fire, which provides a 200 kW sustained fire. Assume the combustion efficiency is 0.75 (3 marks). Solution: N12. Estimate the wavelength corresponding to maximum emissive power of a hot body of dull red light. Calculate the corresponding frequency (3 marks). Solution: N13. Calculate the increase in emitted power if the temperature is increased by a factor of 2 (3 marks). Solution: The emitted power is given by: P = sigma A T⁴ T = Temperature. [K] A = is the radiating surface area. Sigma = Stefan-Boltzmann constant 5.6704 x 10⁻⁸ W / (m² K⁴). T1 = T0 T2 = 2T0 P1 = sigma A T1⁴ P1 = sigma A T0⁴ P2 = sigma A T2⁴ P2 = sigma A (2 T0)⁴ P2 = sigma A (2 T0)⁴ = 16 sigma A T0⁴ P2 = 16 P1 Increase of emitted power is about 16 times. N14. Two parallel surfaces of square shape are located opposite each other at the distance equal to the side of the square. Estimate the view factor (3 marks). N15. A 5x3.5x2.8m (height) room is fully involved in fire. The hot gas-soot media inside the room is homogenous and contains combustible products of hydrocarbon fuel. The estimated mole fraction for carbon dioxide 0.14 and for water vapour is 0.10. Soot volume fraction is 0.22x10-6. The flame inside the room is of cherry red colour. Calculate the total emissivity of the media and the rate of energy emission by thermal radiation from the broken window 2.0m x 1.5m (6 marks). N16. Estimate the rate of C0 production in the free burning of polystyrene C8H8 with an area of 4.0 m² (3 marks). N17. Calculate the mass concentration of combustion products, generated by burning flexible polyurethane, that corresponds to a 50% lethal probability after 8 and 30 minutes exposure (3 marks). N18. Estimate the extinction coefficient of the smoke produced by the flaming combustion of 0.2 kg of polystyrene-foam C8H8 in a 5m x 5m square room of 2.4m height (3 marks). N19. Consider a 2.0 m diameter pan fire of methyl alcohol. Calculate the flame height under normal atmospheric conditions, the maximum plume velocity, and the temperature at a height of 5m. Assume the radiative heat fraction is 25% (3 marks). Solution: For methyl alcohol, Mass Burning Rate, m* = 0.017 (kg/m² s) Heat of Combustiion, H = 20000 (kJ/kg) Empirical Constant, kbeta = 100 m⁻¹ The pool fire heat release rate (kW) is given by: = m H ( 1 – exp(-kbeta D) ) A Where = pool fire heat release rate (kW) m = mass burning rate of fuel per unit surface area (kg/m² s) H = effective heat of combustion of fuel (kJ/kg) A = Surface area of pool fire (m²) kbeta = empirical constant (m⁻¹) D = diameter of pool fire (m) = 0.017 x 20000 x ( 1 – exp( - 100.0 x 2.0 ) ) ( ¶ 2.0² / 4) = 1068.14 kW D = 2.0 m = Energy (Heat) release, 1068.14 kW. Lf = 0.235 (1068.14 kW)²/⁵ – 1.02 x 2 Lf = 1.78 m Maximum plume velocity is given by: u0max = 1.97 Qc ¹/⁵ u0max = 1.97 ( 0.25 x 1068.14 )¹/⁵ u0max = 6.02 m/s Temperature at height of 5 m: Delta T = 25.0 Qc ²/³ z⁻⁵/³ [ K ] Delta T = 25.0 ( 0.25 x 1068.14 )²/³ 5⁻⁵/³ K Delta T = 70.90 K N20. Calculate the plume mass flow rate as a function of height for the fire in example N19. Give the numerical values of the plume mass flow rate at the heights which area equal to the flame height and double the flame height (3 marks). Solution: Mass flow rate above the flame is given by: Height 1 = 1.78 m Height2 = 2 x 1.78 m For 1.78 m: Mass Flow rate = 0.71 ( 0.25 x 1068.14 )¹/³ ( 1.78 )⁵/³ [ 1 + 0.027 ( 0.25 x 1068.14 )²/³ ( 1. 78 )⁻⁵/³] Mass Flow rate = 17.072 kg/s For 3.56 m: Mass Flow rate = 0.71 ( 0.25 x 1068.14 )¹/³ ( 3.56 )⁵/³ [ 1 + 0.027 ( 0.25 x 1068.14 )²/³ ( 3.56 )⁻⁵/³] Mass Flow rate = 43.067 kg/s N21. A fire with a constant heat release of 100 kW is developed in a compartment with overall dimensions 8.0 x 5.0m and 3.0m height. Calculate the time required for the smoke layer interface to descend to a height of 1.7m (3 marks). Solution: The time formula is given by: A = area of the room [m] Q = heat release [kW] H = Height [m] t HGL = 4.1 ( 2 x 8 x 5 + 2 x 8 x 3 + 2 x 5 x 3 ) / ( 100¹/³ 1.7²/³ ) t HGL = 45.47 s N22. A fire in a vented shopping mall of rectangular cross-section 10 m width generates smoke with the production rate of 33 kg/sec and a temperature rise of 167 K above ambient. Calculate the depth of screens required (3 marks). Solution: Given the formula, we solve for , which is : Delta T0 = 167 K z = 10 m = 167³/² / ( 25³/² x 10⁻⁵/² ) = 5459.64 kW N23. Consider a 1.8 m diameter pan fire of methyl alcohol. Calculate the maximum temperature rise under the ceiling (height = 12m) directly above the fire and 3 m away (3 marks). Solution: For methyl alcohol, Mass Burning Rate, m* = 0.017 (kg/m² s) Heat of Combustiion, H = 20000 (kJ/kg) Empirical Constant, kbeta = 100 m⁻¹ The pool fire heat release rate (kW) is given by: = m H ( 1 – exp(-kbeta D) ) A Where = pool fire heat release rate (kW) m = mass burning rate of fuel per unit surface area (kg/m² s) H = effective heat of combustion of fuel (kJ/kg) A = Surface area of pool fire (m²) kbeta = empirical constant (m⁻¹) D = diameter of pool fire (m) = 0.017 x 20000 x ( 1 – exp( - 100.0 x 2.0 ) ) ( ¶ 2.0² / 4) = 1068.14 kW The maximum temperature rise under the ceiling (H = 12 m )is given by: Delta T = 16.9 (1068.14 kW )²/³ / 12⁵/³ Delta T = 28.07 K The maximum temperature rise at H = 15 m, is given by: Delta T = 16.9 (1068.14 kW )²/³ / 15⁵/³ Delta T = 19.35 K N24. Assuming the following room geometry, 3.5 m x 3.5 m floor area, 2.3 m height with a door opening of 1.8 m height and 0.65 m width, calculate the heat release rate necessary to cause flashover using different approaches (4 marks). Solution: Tomass Flashover Correlation Equation. = 7.8 Aroom + 378 ( Avent Hvent¹/²) Where: = Heat release rate necessary for flashover (kW) Aroom = Area of all surfaces within the room, exclusive of the vent area (m²) Avent = Area of the total of all vents (m²) Hvent = The difference between the elevation the highest point of all the vents and the lowest point of all the vents (m). Aroom = 2 x 3.5 x 3.5 + 4 x 3.5 x 2.3 – 1.8 x 0.65 = 55.53 m² Avent = 1.8 x 0.65 = 1.17 m² Hvent = 1.8 m = 7.8 x 55.53 + 378 x 1.17 x 1.8¹/² = 1026.48 kW Babrauskas Models is ginve by: = Heat release rate necessary for flashover (kW) Avent = Area of vent opening (m²) Hvent = Height of vent opening (m) = 750 Avent Hvent¹/² Avent = 1.8 x 0.65 = 1.17 m² Hvent = 1.8 m = 750 x 1.17 x 1.8¹/² = 1177.28 kW CONTENTS 1. High-Level description of the combustion process……………………… 2 2. Enclosure fire development…………………………………………………………. 2 2.1 High-Level description of fire development within an enclosure 2 2.2 Cases in fire development within an enclosure………………………… 4 2.3 Detailed description of enclosure fire development…………………. 4 3. Factors affecting enclosure fire dynamics…………………………………… 6 Summary…………………………………………………………………………………………. 8 REFERENCES…………………………………………………………………………………… 9 1. High-Level description of the combustion process. The analysis of the combustion process involves two main areas: the flow of energy and the mass transfer. As an example, the case of a candle shall be used. The means via which energy flows include conduction, radiation and convention. Of those, radiation is the most significant method, as the energy release field (flame) radiates thermal energy and directly heats the surrounding air. Convection then transfers heat up and away from the combustion area (1). The mass transfer starts with the wax being liquefied. As molten wax is vaporized and burned around the wick, its mass has to be replaced. This causes more liquid wax to be drawn upwards, via the wick. Around the wick, the high temperature causes the chemical bonds of the wax to break down and it is turned into gas. At that point, oxygen supply is limited and the vaporized wax is partly burned. The combustion products, together with the unburned gaseous wax then move upwards, where with ample oxygen supply the burning of wax is completed. 2. Enclosure fire development. 2.1 High-Level description of fire development within an enclosure. Following ignition, and as the fire spreads, it generates increasing amounts of energy. At this early stage, there is an abundance of oxygen and the fire is said to be fuel-controlled. For the present, the properties of the enclosure have negligible effect on the fire. Following the fire generation, the air within and surrounding the flame is heated, and therefore becomes lighter than the ambient air. The light air, together with air which is drafted by the resultant current, moves upwards towards the ceiling. This is called the fire plume. Next, air surrounding the flame area moves laterally towards the fire. Gradually, a mass flow is established, via which the fire is supplied with more oxygen and the enclosure air is heated. Following the generation of the plume and its movement upwards, it will eventually meet the ceiling. At that point the plume will spread away from the plume center and cover the area bellow the ceiling. This is known as the ceiling jet. For a plethora of enclosure-type fires, it has been observed that the enclosure will eventually be divided into two distinct layers. Namely, lower layer containing cold air which has yet to move towards the fire and an upper layer of hot plume gases, that is a mixture of combustion gases and fresh air that was involved in the plume creation. Following this phase, the hot layer is expanded. The fire causes most of the enclosure air to move laterally towards it (as described earlier), enclose it in the plume and eventually move upwards. So, more and more air is heated and the volume of cold air is reduced. This hot air, however, will transfer heat to the cold air zone via radiation. Furthermore it will transfer thermal energy via radiation and convention to the higher parts of the enclosure, as the ceiling and the walls. Most importantly however, it will provide additional heat to the fire area, causing a positive-feedback mechanism, and eventually increasing the burning rate of the fire. Another very important phase in the development of the fire is the flashover. According to the ISO organization (1), this is: “The rapid transition to a state of total surface involvement in a fire of combustible material within an enclosure”. In practical terms, the flashover may be thought as the transition from the developing to the fully developed stages of the fire. One mechanism which may cause the flashover is the top layer of the enclosure continuously increasing its temperature. Then, heat radiation from the boundary of this hot-air layer may become so intense that other flammable material may self-ignite. This causes an instant and very intense increase in the energy release rate, greatly and abruptly increasing the strength of the fire. At the next step the fire is known as “fully-developed”. Now all flammable mater in the enclosure is actively involved in the fire development. 2.2 Cases in fire development within an enclosure. Considering a non-perfectly sealed enclosure we herein examine three important scenarios according to which the dynamics of an enclosure fire may develop. Firstly, when combustion gases which contain sufficient amount of unburned fuel flow via openings to adjacent rooms, they will then mix with fresh, oxygen-rich air. An ignition cause (such as a spark) may then be sufficient to ignite this flammable mixture and therefore transfer the fire to the neighboring room. Second, as the air at the top layer becomes hotter and therefore (via expansion) its volume increases, it lowers its border with the cold air zone as well. When the hot air lowers enough to come in contact with windows, heated window glass may break. This will cause the fire to possibly expand outside the enclosure according to the mechanism described earlier, but also to supply the fire with fresh air. This may cause the destructive phenomenon described next. The flow of cold air inwards to the enclosure will generate turbulence to the air inside the compartment. This will result in a perfect mixture of fresh air with fire gases that contain unburned amounts of fuel. A point of intense heat concentration, or some ignition source, is then sufficient to cause a violent and explosive combustion of this mix. Even worse, this abrupt increase in temperature will result in the enclosure air expanding out of any openings and thus expanding the explosion front outside the enclosure. 2.3. Detailed description of enclosure fire development. Two main dimensions of enclosure fire development are studied in bibliography. Firstly, the development in terms of flow through openings and second the development in terms of temperature. The development in terms of flow via openings is divided in four phases. Those are: Phase One. Provided that the opening is lower than the ceiling, the expanding hot gases at the top part of the enclosure will displace the cold air at the lower part. This displaced cold air will then flow outwards via the available openings. Phase Two. The hot gases increase in volume and gradually lower the boundary with the cold air layer. At some point, the hot gases are low enough to meet the upper part of the opening and therefore start flowing outwards. Now, there is flow of both hot and cold air outwards from the opening. Phase Three. Following Phase two where hot gases also started flowing outwards from the window, to maintain mass equilibrium this hot air has to be substituted. It is therefore replaced with cold air which now flows inwards the enclosure via the window. Phase Four. This phase effectively involves the fully-developed fire described earlier. Here, the enclosure is said to contain smoke which is “well-mixed”, i.e. smoke of uniform composure and temperature. Describing the development of fire in terms of temperature, Walton and Thomas (2) identify five phases. Those are: Ignition. This is the process that can cause an exothermic chemical reaction. Apart from intended ignition (e.g. the spark plug of an internal combustion engine), ignition may also be caused spontaneously. As an example, the air-fuel mixture within an engine cylinder which experiences pressure and temperature conditions that exceed certain values. This will cause the mixture to self-ignite, in a process which is actually more efficient than spark plug-caused ignition. Growth. This is the rate at which the fire develops. It is dependent on the geometrical properties of the fuel mater and the fire field, the chemical properties of the fuel, the abundance or otherwise of oxygen and a number of other factors. The growth phase may be quite prolonged, as is the case of a smothering fire, or very short and abrupt as is the case of flaming combustion. Pre-flashover fire. This is mostly determined by the growth phase, that is, the development of the energy release rate. This rate is very important, as it effectively determines the time available to humans to evacuate the fire theatre. Flashover. The formal definition of flashover has already been provided in previous sections. Yet, the term “flashover” cannot be readily applied in all conditions. This is due to the fact that every fire has different characteristics as well as factors which affect its behavior and development. Other technical definitions of flashover include: 1) When the enclosure temperature reaches 600°C and 2) The thermal energy radiation towards the floor is between 15 and 20 KW/m². Fully-developed fire. Here the fire consumes oxygen at the maximum rate. As a consequence, the energy release rate peaks and the availability of oxygen controls the fire. The fire is therefore called ventilation-controlled, as the enclosure ventilation determines the oxygen supply to the fire. Post-flashover. Following flashover and the fully-developed fire phases, the fire reaches a point where the maximum temperature has been achieved. This is of great importance to the design of constructions, as this maximum temperature also defines the maximum stress which will be inflicted upon a construction by the fire. Decay. As the available fuel is reduced, its supply now controls the fire development. So the fire now is fuel-controlled. The fire now is gradually extinguished, as the concentration of fuel decreases. 3. Factors affecting enclosure fire dynamics. Two main parameters determine the dynamics of enclosure fires. Those are the enclosure properties and the fuel properties. A brief description of those is provided. We first examine the enclosure properties, by beginning with the enclosure geometry. The most important factor here is the size of the enclosure. It has been shown that the closer the ceiling and the walls are to the fire field, the faster the fire growth. This is mainly attributed to the fact that in a smaller enclosure the positive-feed back mechanism from the hot-air layer towards the fire and other flammable material is stronger. A further significant factor is the enclosure openings. Those will determine the ventilation of the enclosure, and thus the oxygen supply to the fire. So, the opening position, height and dimensions become very important, especially at the phases of the fire when it ventilation-controlled. To continue, the type of enclosure internal surfaces must be studied. Here, we are mainly interested in the heat absorption capabilities of the surfaces. Briefly, a material such as a brick wall can absorb large amounts of thermal energy from the hot gases, thereby lowering their temperature. In contrast, a layer of some thermal insulator applied on the walls of the enclosure, will cause the fire gases to retain their high temperature. Regarding the fuel properties, we study the chemical and the physical-geometrical properties of the fuel mater. Firstly, the type of fuel is most important, as this determines the energy density of the material. In home enclosures, we may have wooden furniture as the fire fuel. This has low energy density. In industrial environments, however, the fuel may be a liquid or a gas, used in some manufacturing process, which may be far richer in energy density. The chemical state of the fuel is also important. For example, a liquid fuel shall evaporate and subsequently mix with oxygen much faster than solid-state fuels. This, in turn, determines the levels of hazard to humans and to constructions. For example, a rapidly expanding fire will be of great threat to humans, but will cause small thermal stress on structures. In contrast, a slow-burning material may cause the opposite hazards. Finally, the fire dynamics are also influenced by the physical location and the geometrical shape of the fuel. As an example, if the flammable material is placed in such a way that the flow of hot gases may convey thermal energy to it, then the fire develops more quickly. This is one reason, for example, as to why a fire may spread much faster in the vertical, rather in the horizontal direction. The shape of the combustion material is also a contributing issue, as this determine the area of the fuel exposed to fresh air, and thus to oxygen. Summary. The material presented herein is a brief introduction to enclosure fire properties, characteristics and dynamics. The layout logic follows the study of (1). Concluding this essay, from a fire-safety perspective for humans the most important terms-mechanisms described here are the backdraft phenomenon and the pre-flashover and flashover phases of the enclosure fire. REFERENCES 1. Enclosure fire: Björn Karlsson and James G. Quintiere; September 28, 1999, 336pp., ISBN 9780849313004 2. Walton, W.D. and Thomas, P.H., “Estimating Temperatures in Compartment Fires”, in The SFPE Handbook of Fire Protection Engineering, 2nd ed., National Fire Protection Association, Quincy, MA, 1995 Read More