Fire Condition - Assignment Example

ASSESSMENT BRIEF Student’s Name Name of the Class Professor (Tutor) Name of School (University) The State and City Date 1. Estimate the heat release rate in free-standing pool fire above 1.6m diameter acetone spillage. Combustion efficiency is 0.75. = HC.FfA Where HC.F is heat of combustion (1789kJ/mol) A is the area f mass burning rate (35.9g/m2s) 58.08 g/mol 35.9g= 0.618113 mol 35.9g/m2s=0.618113 mol/ m2s Area= 0.82 x = 2.01088 m2 = 1789kj/mol * 2.01088 m2 * 0.618113 mol/ m2s = 2223.639kJ/s 2. Compare toxic potency of wool and rigid polyurethane in non-flaming and post-flashover burning regimes. A significant difference in fire conditions comes from non- flaming and flaming combustions. In combustions flaming conditions, air fuel ratio is very important. When the temperatures in a building increases, oxygen concentration reduces. A lot of carbon monoxide is produced when the concentration of oxygen is low. There are chemical equations predicting the production of CO in a combustion. Smouldering is important in some occasions. However, with the polyurethane, toxic materials are produced. Wool may not be that toxic 3. Calculate the heat release rate necessary to cause flashover, using three different approaches. Critically review the answers obtained from each of the different approaches used. Consider the room 3.2m x 3.2m in floor area and 2.20m in height with a door opening 1.8m height and 0.65m width. The boundary materials are gypsum plaster which is 0.016m thick. Thomas` flashover correlation equation Qfo=7.8Aroom +378(AventHvent1/2) Aroom =3.2^2 =10.24m2 Avent=(1.8*0.65) = 1.17 Hvent1/2=1.8^0.5 =1.34164 Qfo=7.8*10.24 +378*(1.17*1.34164) = 673.226 Babrauskas method Qfo=(750Avent)(Hvent1/2) = (750*1.17)( 1.34164) = 1177.29 Method three Qfo=(600Avent)(Hvent1/2) = (600*1.17)( 1.34164) =941.831 4. Compare the chemical reaction rates at two temperatures 400 and 800K. The activation energy is 180Kj/mole. For most chemical reactions, the rate of chemical reaction depends on temperature. When temperatures increases, the rate of reaction also increases. From kinetic molecular theory, increase in temperature results in kinetic energy. This makes the particles to move faster. Therefore, when reaction is taking pace at 800K, the reaction will be faster than when it is taking place at 400K. 5. Estimate the extinction coefficient of the smoke produced by flaming combustion of 0.2kg of polystyrene-foam in a square room of 2.5m height. P= t*C Where p is the extinction coefficient T is the extinction coefficient per unit mass concentration (m2/kg) C mass concentration of the smoke ( kg/m3) Volume of the room= 2.5^3 = 15.625 m3 Area of the room = 2.5^2 = 6.25 m2 t= 6.25m2/0.2kg C= 0.2kg/15.625 m3 P= 6.25m2/0.2kg * 0.2kg/15.625 m3 = 0.4/m 6. Consider a 2.5m diameter pan fire of methyl alcohol with a heat release rate intensity of of surface area. Calculate the mean flame height under normal atmospheric conditions. Under the normal atmospheric conditions, the equation for estimating the mean height of the flame is; L= -1.02D + 0.235Q2/5 Where L is the mean height of the flame D is the diameter of fire pan Q is the heat release rate L= (-1.02 * 2.5) + 0.235* 500(kW/m2)2/5 = 0.27264 m 7. Calculate the total fractional lethal dose for the mixture inhalation: CO (1000ppm), C02(9000ppm). HCN (20ppm), HCL (30ppm) for 10 minutes and 30 minutes. Compare with unity. = = 1000/1020 + 9000/10100 +20/2 = 11.87 This is higher than unity unity is 10% 8. Critically review zone and field models used for compartment fire modelling. Provide examples and critically analyse the main assumptions, limitations, advantages and disadvantages of these models. Zone models This model is always applied to the buildings which are made up of several compartments which are joined together. Each compartment is divided into control zones or volumes. There are uniform quantity of interest; energy, moment, momentum and conservation of mass. There is applied (quantitues) to every zone using ordinary differential equations or algebraic representations. One advantage of this model is the low speed, memory requirement, prior assumptions and structured output. The one major disadavatage is that there could be need for prior assumptions. Also, there are inaccuracies. An example of zone model is one commonly used is the CFAST which was developed by the national institute (HOSMER, C. 2004, 112). Field model There are several field models which are available. To predict the movement of smoke and hot gases, smoke concentration and temperature of air cabin`s seating area, a finite –difference (two- dimensional) model of air craft fire was created. Several computer models have been developed. These are low cost models that have been made to predict how the fire would spread as well as smoke in areas which are enclosed. The advantage is that these models are very efficient and accurate. The disadvatage is that they require large and faster computers. These computers must have more memory than those required by the models used for zones. 9. A new type of insulating board has been developed by that esteemed construction organisation Kaput Co. They warn that at extremely high heat fluxes it could be ignited, but they don’t think it’s very likely and it would take hours, so there’s no real risk! As an expert on the ignitability of materials you are asked to perform a thick/thin calaculation given the following data on the material: Density 2300 kg m-3 Thermal conductivity 0.85 W m-1 K-1 Specific heat capacity 847 J kg-1 K-1 Thickness of board 8mm Initial/ambient laboratory temperature 18°C Ignition temperature 415°C It is considered that if 20 kW m-2 would be enough to ignite most materials (i.e. indicative of flashover fires). Perhaps something easy to ignite would only require 10 kW m-2. Would this material ignite within ten minutes if exposed to 10 kW m-2? Useful formulae: = 0.85/(2300*0.847) = 0.00043632Wm2/K = *0.85 *2300*847() =51631s = 860 minutes = 2300* 0.847*0.008*() = 618.7 s = 10 minutes Yes for a thin material, it would ignite but a thick material would not ignite = 0.85/(2300*0.847) = 0.00043632Wm2/K thick=2 = 9.5mm thin= 2 = 1.04mm 10. The underside of a smoky layer 3m x 9m is radiating like a flat, isotropic plate at 610°C to the floor of a compartment 1.5m below. The mean emissivity is 0.41 and the floor is homogenous/flat plate at 41°C. What is the rate of heat transfer from the smoky layer to the floor? = 28.94 X=2 Y=6 A= 3*9 = 27m2 = 5.67 x 10-8 WM-2K-4 = 28.94*0.41*5.67 x 10-8*27(8834-3144) = 10866034.73W = 10.866MW Useful figures & formulae: Where and Figure 1. Parallel plates [15 marks] _________________________END OF ASSIGNMENT_________________________ Reference HOSMER, C. A. (2004). Flashover. New York, iUniverse. Read More