Generating the Thermal Radiation Factor - Assignment Example

Part B ANSWER: Let us find X and Y; And Therefore, And This implies that X2 and Y2 By inserting the values in the below formulae we will get the thermal radiation factor: Which reduces to; Hence, F12[1.342+16.320+36.540-5.634-11.408] F12 Therefore, F12 The Rate of Heat Transfer: Q = F2€∂A (TH4 – TC4) F12 ═ 0.706 € ═ 0.32 ∂ ═ 5.67 X 10−8Wm−2K−4 A ═ 120m2 TH4 ═ 713K TC4═ 326K Therefore, Q ═ (0.706) (0.32) (5.67 X 10−8) (120) (7134) (3264) This means that; Q ═ 385930 W ═385.93 kW PART C ANSWER a) FSE standard room comprising of sheet metal stud framework is lined in three layers (Babrauskas, 18). Gypsum wallboard and one layer of calcium silicate board that is thick. Doubletrees of naked chromel alumel thermocouples are used to keep track of the temperature of the doorway. Two platinum-rhodium thermocouples are added upon noticing an increase in temperature above that of chrome Lalu Mel upper range. Both uncooled and cooled extraction probes are positioned to allow gas layers be in a position of sampling the higher combustion layers, a lower layer, and finally exterior of the doorway. Temperature data is used uniform with inside measurements of the thermocouple. This is used in calculating the doorway mass flow. The area of the doorway is determined by Ah1/2 ventilation parameter (Fire, 1982). A represents entirely geometrically scaled summing area of the opening of the ventilation and h the opening height. Doorway mass flow is calculated by applying the same formula as the FSE. Scaling of the doorway of RSE by use of ventilation parameter Ah1/2 is so accurate. The scaling technique includes the geometric factor of scaling for RSE which is 0.4. Scaling up of the mass flow rate from RSE to FSE, increase by an element of 6.25 is expected. Ventilation technique thereof helps in the scaling of mass flows both in and out of the enclosures. b) Smoke is a product of combustion that is condensed (Mulholland, 1995, p. 210). Smoke aerosol has got different appearances ranging from light coloured ones to black which is soot produced during combustion. Smoke is produced under two combustion conditions. The conditions namely are pyrolysis and shouldering. It has got an overall effect of the amount and colour of the smoke. Pyrolysis takes place at the surface of fuel due to an elevation in temperature. It is as a result of radiant flux hitting the surface. Being that pyrolyzing samples temperature is less than that of gas phase flame, the vapour that is evolving from the surface rises. Constituents of low vapour pressure in turn condense. Smoke droplets that look like light coloured smoke are formed. Shouldering combustion too leads to smoke production. It is just that in this case, combustion is usually self-sustaining whereas pyrolysis needs external heat. Smoke has various properties such as light extinction, detection, and visibility, sedimentation velocity, aerodynamic diameter, coefficient of its scattering, condensation rate etcetera. Engineers can identify smoke spread in buildings in the following ways, use of active smoke control systems such as sprinkler heads that are activated by any detection of smoke. PART D a) Combustion is a high-temperature reaction that occurs between fuels and oxidants. It is mostly oxygen in the atmosphere that generates gaseous products often referred to as smoke. For there to be fire, there are there necessary conditions that must be in place. The conditions include the presence of oxygen, heat and finally fuel. This is often referred to as the fire triangle (Stott, 2000, 336). Oxygen being a high oxidizer, meaning it accepts electrons, it helps in combustion. It is very electronegative. Heat produced during fire outbreak makes combustion self-sustaining. Without the heat, combustion cannot move on. Fuel stand to be the third. It acts a catalyst that stimulates burning. There has to be burning fuel for combustion to be witnessed. For burning process to end or extinguishes, one of the three elements is removed. b) Several methods of fighting fire are known. One of them is cooling. In this case, heat is transferred. This process is more efficient by using water as the primary cooling agent. The second method is smothering. It is the removal of oxygen during the combustion process. A natural gas such as carbon dioxide is used. This neutralizes the oxygen necessary for burning to continue. Lastly, there is chain breaking mechanism. The combustion chain is broken by adding an element to the fire. An example is by adding foam to fire. Cooling of flames is done by use of large amounts of water. It is when the fire is involving the ordinary combustible material. The fires that often occur in vapour and mixture of air mostly on the surface of liquids that are combustible i.e. gasoline, the most appropriate method of controlling it is by isolating the source of fuel. It is because most of these class B materials do burn with high intensity, unlike class (A) materials. In the case of hydrocarbon fires, the most suitable solution is, and highly active agent for handling the condition is the use of foam that is of little expansion. Foam is applied and let to flow evenly on the surface of the hot material. T creates a strong and not easily broken layer of smothering blanket. A fire that is caused by volatile liquid products is often extinguished by use of chemical agents that are of dry nature. The chemical interferes with the reactants in the combustion process. Fire gets to stop since no combustion is taking place. PART E ANSWER The effect that occurs in the body of the human to time The equation that was proposed to be the mastermind of explaining the effect of the temperature on the human skin can be simplified up to: Ts ═ T0 + √ (t/ᴫ) TS ═ 317K, Which is the temperature that is taken to be 44 as we are after the specific temperature that inflicts pain and burn sensation to the skin. T0 ═ 310K, Which is the normal temperature of the body at room conditions. Q ═ 100w M-2S-1, Which is assumed to be constant all through the period of the time stated. ß ═ 1.7 kWs1/2m-2K-1 Therefore, the only parameter to be deduced is the time t; º C Response 37 Normal body temperatures 44 The human skin starts to feel paining 48 The human skin gets a first degree burning injuries 55 The human skin succumbs a second degree burns and injury 62 A phase in which burning human tissue becomes numbing 72 Human skin is instantly destroyed The time (t) is the duration that a human will feel the burn and pain sensation with the heat flux maintained as constant at 100w m-2s-1. Evidently, rearranging the equation above, we can come up with the following equation: T ═ {ß [(TS – To)]/2Q} 2 T ═ 3.14 [{1700(317-310)/200}] 2 T ═ 11122 s T ═ 11122/3600 T ═ 3hrs When we have the below parameters; The copper wire volume: It finally gives, The mass of the copper wire is therefore given by; The Energy dissipated, E per hour is given by; (Beaver& Powers, 2010) But the system is such 0.1% of the above energy gets lost as a result of the wire heating up. It therefore means that 99.9% of the original energy shall still be within the system. Assuming that the wire was initially at room temperature; Therefore; Final temperature at the end of an hour was therefore; Note: The assumption that I made, i.e. the initial temperature was taken as room the temperature due to the fact, copper is a very good conductor and it would easily assume the present temperature. In this case it is 298K Reference Babrauskas, V., Engineering Variables to Replace the Concept of ‘No combustibility.' Fire Technology, pp.1-21. Mulholland, G.W., 1995. Smoke production and properties. SFPE handbook of fire protection engineering, 3, pp.2-258. Stott, P., 2000. Combustion in tropical biomass fires: a critical review. Progress in Physical Geography, 24(3), pp.355-377. Fire, J.C., 1982. Effect of Ventilation on the Rates of Heat, Smoke, and Carbon Monoxide Production in a Typical Jail Cell Fire. Read More