Burns and Fire Modelling - Report Example

1. Thiazole (C3H3NS) occurs natrually as part of vitamin B1 (thiamin) such as in pasta and bread. Thiazoles have anti-tumor and anti-viral properties but most thiazole compounds are flavourings. If it is combusted perfectly in air the products are carbon dioxide, water, hydrogen sulphide (H2S) and nitrogen. What is the balanced reaction? [8 marks] 2(C3H3NS)+13/2O2 COMBUSTION 6CO2 +H2O+2H2S+N2 2{2(C3H3NS)+13/2O2 COMBUSTION 6CO2 +H2O+2H2S+N2} 4(C3H3NS)+13O2 COMBUSTION 12CO2 +2H2O+4H2S+2N2 What is the RMM of thiazole? [2 marks] Relative molecular mass of C3H3NS From the periodic table the following are masses Oxygen=16g Carbon(C)=12g Hydrogen(H)=1g Nitrogen (N)=14g Sulphide(S)=32g Total RMM=(3*12)+(3*1)+(1*14)+(1*32)=85g What is the fuel-air mass ratio? [5 marks] Fuel mass=mass of thiazole=85g From periodic table, oxygen=16g Oxygen used=13O2=(26*16)=416g=21% 100%=Air Total Air=416*100/21=1981g Ratio=85/1981 =1:23.3 Approximatly 1:23 What is the oxygen depletion? [4 marks] Oxygen mass =16g 13O2=(26*16)=416g What is the yield of CO2? [2 marks] With masses from periodic table =(12*12)+(16*24)=528g What is the yield of H2O [2 marks] 2H2O=(4*1)+(2*16)=36g What is the yield of N2? [1 mark] 2N2 =4*14=56g For ideal combustion, what is the yield of carbon-monoxide? [1 mark] 12CO 12*16=192g You may assume the composition of air is 21% O2 and 79 % N2. A periodic table is provided overleaf, should you need one. [25 marks for this question] 2. Critically review and distinguish superficial, partial-thickness and full thickness burns and explain the significance of ‘eschar’. How would eschar appear on a Lund-Browder chart? [15 marks for this question] Partial thickness It is burn injury which result in entire epidermal layer being destroyed with varying thickness of the dermis, hence can be either superficial or deep .Hair follicles, Substructures-sweat remain undamaged. It is characterised by creamy in appearance and usually get healed itself by regeneration of the epithelial layer. The depth of the burn usually take up to 7 - 10 days to declare itself as superficial or deep and the resulting scarring (Enoch et al 2009).Acticoat moisten with clean water, intrasite, cling film and bandage are used as dressing Full thickness burns Injury affects the entire thickness of epidermis, dermal appendages and epithelial element, Spontaneous healing is not possible. If the burned are is left, it will heal by contraction and is characterised by its whitish leather appearance or it can change to brown cherry red or charred black. Its firm and leathery in texture. Nerve sensation is reduces or lost completely and the burned areas cannot blanch under pressure and. The eschar of resulting from burn is very inelastic and leads to shrinking of the tissues as oedema occurs. Dressing of the burned area can be done as in partial burn above (Enoch et al 2009) The escher will be show by shading the number 13 in above chart 3.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. [15 marks for this part] Zone Model. Zone models use a number of algebraic equations to solve and determine flow properties. It takes into consideration two or three zones, namely; upper and lower layer and sometimes third zone which represent the plume in the compartment with the fire. Mass transfer, momentum, energy and species are monitored from zone to another using tailored equation which takes assumptions into consideration. Two-layer approach is based on phenomena that hot gases will collect at ceiling while cooler layer will be collected below hot gases. This model is normally used to track properties associated with movement of the smoke and hot gases from source to other parts. The output depends largely on user and characteristic of the design fire. Other zone models in-cooperate sub-models to simulate specific phenomena. Advantage It is possible to run zone models for multi-compartment fire scenarios in few seconds using personal computers. Disadvantages Sometimes assumption of uniform zones are not accurate hence difficulty in modelling complex shapes Assumption of movement of hot gases to the ceiling may not be true therefore affects the final results Assumptions It assumes that given space can be disintegrated in small number of zones and algebraic equation used to determine properties of the flow. Velocity and temperature are assumed to be uniform Field model Fields model provide formula for fluid flow through a volume by utilizing Navier-stokes equation which is a partial differential equations for conservation of momentum, mass and energy are approximated as finite difference over a number of controlled volume. It utilizes CFD software packages. Field models, provide a method for modelling the fluid flow through a volume using numerical solutions of the Navier-Stokes equations. The solutions of partial differential equations for conservation of mass, momentum and energy are approximated as finite differences over a number of control volumes. It use thousand of control volume as compared to Zone model hence requires highly trained personnel It is also used in simulation of fire driven flow like spread of hot gases, heat radiation and monitoring smoke and water particles. It has been successful in modelling smoke motion and has been used to model fires. It has been successfully used in modelling of smoke movement and modelling of fires. It has capability to model pre-flash and localised fires in complicated bodies with smoke moving in multi-compartments. The model for covered fires area appropriate for low-speed, Heat-driven flow with an emphasis on smoke and heat transport from fires (Olenick, et al 2003) Advantage It achieves more refined results as compared to zone model It is very quick method Disadvantage It needs greater computational resources and longer simulation time ie weeks or sometimes months. It is very complex and needs more training of people using it It requires great computation and longer timeline in between project and completing all the simulation required 4. The underside of a smoky layer 7m x 10m is radiating like a flat, isotropic plate at 440°C to the floor of a compartment 1.85m below. The mean emissivity is 0.45 and the floor is homogenous/flat plate at 40°C. What is the rate of heat transfer from the smoky layer to the floor? Useful figures & formulae: (1) (2) Where and Figure 1. Parallel plates [20 marks for this part] Solution Assumming the isotropic body is black, q = ε σ (Th4 - Tc4) Ac         where σ = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant for black body Q = heat transfer per unit time (W) or Rate of Heat transfer TH= hot body absolute temperature (K) Tc = cold surroundings absolute temperature (K) Ac = area of the object  (m2) Temperatore conversion 10C=274.15K TH4=713.15K TH4=313.15K A=7m*10m=70m2 Q=F12*0.45 *70*5.6703 10-8 (W/m2K4) (713.154-313.154) = F120.624 X=a/z=7/1.85=3.78378=3.8 Y=b/z=10/1.85=5.4054=5.4 F12= 1+X2=1*3.8=15.44 1+Y2=1*5.4=30.16 F12=2/π*3.8*5.4[(ln√{(15.44*30.16)/(1+14.44+29.16)}+{(3.8√30.16)arctan(5.4/√30.16)}+{(5.4 √14.44)arctan(5.4/√14.44)}-(3.8acrtan3.8)-(5.4acrtan5.4)] =25.76 Q= F120.624 Q=15.6W 5. A new type of insulating board has been developed by that esteemed construction company Kaput Ltd. 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= 2400 kg m-3 Thermal conductivity 0.85 W m-1 K-1 Specific heat capacity 827 J kg-1 K-1 Thickness of board 7mm Initial/ambient laboratory temperature 18°C Ignition temperature 410°C It is considered that if 20 kW m-2 would be enough to ignite most materials (i.e. indicative of flashover fires). Perhaps sSomething 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: [25 marks] where ρ- density c -specific heat k- Thermal conductivity =(0.85W/mk)/(2400Kg/m3 *827J/kg)=4.2*10-7wm/JK =0.85W/mk*2400Kg/m3 *827J/kg) =1687080WJ/M3 =√(4.2*10-7wm/JK*0.007m)= 2.9*10-9wm2/JK Q’=10 kW m-2=10000W m-2 =1687080WJ/M3*π/4{(410-18)/1000}2 =1325562.86*0.00253664 =3362.48 WJ/M3 =2400Kg/m3 *827J/kg *0.007M*{(410-18)/10000} =77804.16J/m2 Work Cited Herndon D. Total burn care. 2nd ed. London: WB Saunders, 2002.print Enoch S, Roshan A, Shah M.Emergency and early management of burnsand scalds. Br Med , 2009.Print Olenick, S.M. and Carpenter, D.J. (2003). An Updated International Survey of Computer Models for Fire and Smoke, Journal of Fire Protection Engineering, 2003.Print Read More