Superficial, Partial-Thickness and Full Thickness Burns and Explanation of the Significance of Eschar - Assignment Example

Assignment Details 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? 4C3H3NS + 13O2  12 CO2 +4H2O +4 H2S +2N2 What is the RMM of thiazole? Element mass numner Total mass of elements C-12 12x12=144 H-1 1x12=12 N-14 4x14=46 S-32 4x32=128 RMM 144+12+46+128=330 What is the fuel-air mass ratio? Putting into consideration the composition of air the equation changes to 4C3H3NS + 13O2  12 CO2 +4H2O +4 H2S +2N2+48.9N2 The calculations indicates clearly that complete combustion of one volume of thiazole would requirem3 +48.9= 51.9 volumes of air is required thus fuel-to-air ratio required is 1:51.9 This puts the fuel-to-air ratio requirement at 1:14.28 to achieve complete combustion and for one volume of C2H6N2O there is a maximum of two volumes of 2CO2. What is the oxygen depletion? For 4 moles of thiazole there is 13 volumes of oygen depletion What is the yield of CO2? For 4 moles of thazole there is 12 volumes CO2 yield What is the yield of H2O? For 4 moles of thiazole there is 4 volumes H2O yield What is the yield of N2? For 4 moles of thiazole there is 2 volumes N2 yield For ideal combustion, what is the yield of carbon-monoxide? The equation where carbon monoxide is yielded is 4C3H3NS + 7O2  12 CO +2H2O +4 H2S +2N2 This shows that for 4 moles of thiazole there are 12 volumes of carbon monoxide 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? Assessment of depth of burn is vital in the determination of dressing needed and also of even more importance is that the depth of burn is a determinant factor if the wound can heal naturally without surgery being required (Andronicus M. et al ,1998). It also determines the severity of the residual scar. Burns may be classified as being epidermal (superficial), dermal (partial thickness wounds) and full thickness wounds depending on how much the skin tissues is affected and also how the underlying tissues have been affected (British Burn Association, 2008). Dermal burns are classified further into superficial partial thickness and deep partial thickness. Lund Browder Chart is of great use as it is highly accurate where by age and body proportions are put into consideration during its use. As a result of this is use is considered as the best alternative where different age groups are involved with emphasis being in the cases involving children (Manchester et al, 2009). But all said it is important to note that specific charts are to be used during the calculation process. The conditions of burns can be changed when Escher is used and making it to be important. 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. Looking closely on field models has led to the revelation that it is a great challenge to make acute predictions of thermal conditions as well as chemical compositions as far as ventilated compartment fires are concerned. Other findings made are that in formal ventilation progress is case of accessibility to compartment that is well ventilated, field models have indicated to exhibit better performance in the prediction of temperatures and chemical compositions provided there is specification of the uncertainties in the experiment (Merci , B & Vandevelde, 2007; Bishop. et al. 1992 . Types of Combustion modelling Lamimar flamelet model: in this model there is incorporation of finite rate chemistry effects which involves turbulent combustion. The assumption made is that occurrence of combustion reaction may be at microscopic level in the flames that are turbulent in nature. In this process the concentration of the dominant species and temperature mixture fraction are given out. In the flowfield prediction of Favre-averaged mixture fraction mean, the variance ξ and g, it is possible for the probability density function to be computed in the standard form. In the case involving two PDF components, the beta function which is one of the two is applicable in the turbulent jet while the second is the delta function. Chemical-kinetic computations that go together with turbulence modelling can be achieved when the flamelet model is applied (Norwegian Fire Research Laboratory, 1996). Constrained equilibrium method. In this model there is calculation of temperature, level of concentration of species by correlating the variables and the mixture fractions where evaluation can be effected by rigorously computing the laminar diffusion flame. However, by taking this route it has been found to be difficult when it comes to incorporation of radiation in the model, the being attributed to high level of energy loss which results from the diffusion flames. The solution of the problem is through the approach of constrained equilibrium where equilibrium calculations and the enthalpy value are imposed on the problem. This approach is similar to the system being cooled. In making the equilibrium calculations, it is very important to have knowledge of thermodynamic properties of the system while information concerning the reaction is disregarded and thus relaxing to equilibrium where it exist (Kerrison, 1998). This implies that when a reaction mixture is composed to equilibrium, it will in no way be dependent on the mechanism of reaction. Eddy break-up model This model involves giving an explicit equation for fuel mass fraction The equation involved is +) -  = -. In this model it is assumed that thre is very fast chemistry which means turbulent mixing is in charge of the of the process. For turbulence timescale ,  can be expressed as  = min, Where , represents fuel time-averaged mass fractions,  represents the oxidant while  is the product. The stoichiometric oxygen-to-fuel mass ratio is given by “s”,  is density of the fluid.  represents laminar viscosity while  represents turbulent viscosity.  on the other hand gives laminar Prandtl number,  represents turbulent Prundtl number t is the time transient cases. Also in the equation CR and CA are constants which are determined empirically depending on the mixing model and the reaction rate chemistry. 4. The underside of a smoky layer 14m x 9m is radiating like a flat, isotropic plate at 510°C to the floor of a compartment 2.15m below. The mean emissivity is 0.41 and the floor is homogenous/flat plate at 35°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 The formulae to be used (1) Where  is the view factor between the two surfaces where heat is being transferred  Stefan Boltzmann Constant, 5.67x10-8 W/m2/K4  is the temperature of the hot surface (K) in this case 273+510=783K  is the temperature of the cold surface (K) in our case 273+35=308K A is the area of the radiating surface in the case 8x11 In order to calculate the view factor the formula used is (2) Where =9/2.15= 4.19 and = 14/2.15=6.51 Now applying the formula by first calculating the components in the brackets independently the results will be 1- = 1.29 2- 4.19 3- 6.51 4- 4.16*tan-1 (4.16) = 320.86 5- 6.51*tan-1 (6.51)= 529.05 Now substituting for the components in the bracket F12=  Then applying Q= (38.61)(0.41)(5.67*10-8)(14x9)((783)4-(308)4) Q= 41.49 Mw 1 2 3 4 3 2 1 0 1  atomic number Table 3. Periodic table of elements 2 H  symbol He Hydrogen  name Helium 1  relative atomic mass 4 3 4 5 6 7 8 9 10 Li Be B C N O F Ne Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon 7 9 11 12 14 16 19 20 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar Sodium Magnesium Transition Elements Aluminum Silicon Phosphorus Sulphur Chlorine Argon 23 24  27 28 31 32 35.5 40 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton 39 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 85 88 89 91 93 96 (98) 101 103 106 108 112 115 119 122 128 127 131 55 56 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba 57-71 Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Cesium Barium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon 133 137 178 181 184 186 190 192 195 197 201 204 207 209 (209) (210) (222) 87 88 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra 89-103 Rf Db Sg Bh Hs Mt Uun Uuu Uub Uut Uuq Uup Uuh |Uuh Uuo Francium Radium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Ununnilium Unununium Ununbium Ununtritium Ununquadium Ununpentium Ununhexium Ununhexium Ununoctium (223) (226) (261) (262) (263) (262) (265) (266) (269) (272) (277) - (289) - (289) - (293)  57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium 139 140 141 144 (147) 150 152 157 159 163 165 167 169 173 175 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium (227) 232 231 238 (237) (244) (243) (247) (247) (249) (254) (253) (256) (254) (257) 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 2300 kg m-3 Thermal conductivity 0.82 W m-1 K-1 Specific heat capacity 824 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 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: Assuming material to be thin and rearranging We have Substituting = 8.667kw The heat flux is less than 10kw and this means that the heat flux exposure of 10kw on the material for 10 minutes is high enough to ignite thematerial. References Andronicus M. et al (1998) Non-accidental burns in children. Burns 24: 552–8 Bishop , S & Drysdale, D (1995) Experimental Comparison With A Compartment Fire Model. International Communications in Heat and Mass Transfer. British Burn Association (2008) Emergency management of severe burns course manual, UK version. Wythenshawe Hospital. Hudspith J and Rayatt S (2004) First aid and treatment of minor burns. Br Med J 328:1487–9 La Hei et al(2006).Laser Doppler imaging of paediatric burns: burn wound outcome can be predicted independent of clinical examination. Burns 32: 550–3 Manchester Enoch S, Roshan A, Shah M (2009). Emergency and early management of burns and scalds. Br Med J 338: 937–41 Merci , B & Vandevelde, P (2007) Experimental study of natural roof ventilation in full-scale enclosure fire tests in a small compartment . Fire Safety Journal. 42 () p523-535 Novozhilov, V (2001) Computational fluid dynamics modeling of compartment fires. Progress in energy and combustion science. 27 p611-666 Read More

In the case involving two PDF components, the beta function which is one of the two is applicable in the turbulent jet while the second is the delta function. Chemical-kinetic computations that go together with turbulence modelling can be achieved when the flamelet model is applied (Norwegian Fire Research Laboratory, 1996). Constrained equilibrium method. In this model there is calculation of temperature, level of concentration of species by correlating the variables and the mixture fractions where evaluation can be effected by rigorously computing the laminar diffusion flame.

However, by taking this route it has been found to be difficult when it comes to incorporation of radiation in the model, the being attributed to high level of energy loss which results from the diffusion flames. The solution of the problem is through the approach of constrained equilibrium where equilibrium calculations and the enthalpy value are imposed on the problem. This approach is similar to the system being cooled. In making the equilibrium calculations, it is very important to have knowledge of thermodynamic properties of the system while information concerning the reaction is disregarded and thus relaxing to equilibrium where it exist (Kerrison, 1998).

This implies that when a reaction mixture is composed to equilibrium, it will in no way be dependent on the mechanism of reaction. Eddy break-up model This model involves giving an explicit equation for fuel mass fraction The equation involved is +) -  = -. In this model it is assumed that thre is very fast chemistry which means turbulent mixing is in charge of the of the process. For turbulence timescale ,  can be expressed as  = min, Where , represents fuel time-averaged mass fractions,  represents the oxidant while  is the product.

The stoichiometric oxygen-to-fuel mass ratio is given by “s”,  is density of the fluid.  represents laminar viscosity while  represents turbulent viscosity.  on the other hand gives laminar Prandtl number,  represents turbulent Prundtl number t is the time transient cases. Also in the equation CR and CA are constants which are determined empirically depending on the mixing model and the reaction rate chemistry. 4. The underside of a smoky layer 14m x 9m is radiating like a flat, isotropic plate at 510°C to the floor of a compartment 2.15m below. The mean emissivity is 0.

41 and the floor is homogenous/flat plate at 35°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 The formulae to be used (1) Where  is the view factor between the two surfaces where heat is being transferred  Stefan Boltzmann Constant, 5.67x10-8 W/m2/K4  is the temperature of the hot surface (K) in this case 273+510=783K  is the temperature of the cold surface (K) in our case 273+35=308K A is the area of the radiating surface in the case 8x11 In order to calculate the view factor the formula used is (2) Where =9/2.15= 4.19 and = 14/2.15=6.51 Now applying the formula by first calculating the components in the brackets independently the results will be 1- = 1.29 2- 4.19 3- 6.51 4- 4.16*tan-1 (4.16) = 320.86 5- 6.

51*tan-1 (6.51)= 529.05 Now substituting for the components in the bracket F12=  Then applying Q= (38.61)(0.41)(5.67*10-8)(14x9)((783)4-(308)4) Q= 41.49 Mw 1 2 3 4 3 2 1 0 1  atomic number Table 3. Periodic table of elements 2 H  symbol He Hydrogen  name Helium 1  relative atomic mass 4 3 4 5 6 7 8 9 10 Li Be B C N O F Ne Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon 7 9 11 12 14 16 19 20 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar Sodium Magnesium Transition Elements Aluminum Silicon Phosphorus Sulphur Chlorine Argon 23 24  27 28 31 32 35.

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