The Solution for the Fire Modelling Problems - Assignment Example

Solve the following fire modelling problems. 1.  Find the flame height of the fire involved plastic materials of 0.55 m2 circle area with burning rate approximately 14.5 g/s at a heat of combustion of 35 kJ/g. What should be a critical diameter of the fire area that the fire reaches ceiling on height of 3 m? Heat release rate Q = Burning rate x Heat of combustion of the material = 14.5 x35 =507.5Kj/s The height of flame L = -1.02D+0.23Q2/5 (Tewarson,1995) The area of plastic is =0.55 A =   D =  =  = 0.84m L = -1.02x0.84 +0.23Q2/5 = 1.92m The height of the flame will be 1.92m To find the critical diameter for the flame to reach a height of 3m L =3m 3 = -1.02xD +0.23x507.52/5 D =  = 0.21m 2. Mixed fuel is composed by methane (volume percent is 0.25), carbon monoxide (0.35) and hydrogen (0.40). Calculate the lower flammable limit concentration for the mixture and the concentration of each component in the mixture with air. Combustion reactions are as follows For Methane case CH4 + 2O2 + 7.52N2 → CO2 + 2H2O +7.52N2 as the volume of nitrogen will be 2×79÷21=7.52. For CO 2CO + 02 +3.76N2 → 2CO2 +3.76N2 or CO + 0.5O2+ 1.88N2 → CO2+1.88N2 For H2 2H2 +O2 +3.76N2 → 2H2O +3.76N2 or H2 +0.5O2+ 1.88N2 → H2O+1.88N2 Now putting into consideration volume of the mixture The volume of air required for methane 0.25(2+7.52) = 2.38 volumes of air The volume of air required for CO 0.35(0.5+1.88) = 0.833 volumes of air Volume of air required for H2 0.4 (0.5+1.88) = 0.714 volumes of air Total volumes of air required for one volume of mixture = 3.927 The flammable concentration of the mixture = x100 = 20.29% of air mixture Methane concentration of in the mixture = 0.25x20.29= 5.07% of the air fuel mixture CO concentration in mixture = 0.35 x 20.29 = 7.1% of the air fuel mixture H2 concentration in mixture = 0.4x20.29 = 8.11% of the air fuel mixture. 3. Consider a 1.7 m diameter pan fire of petrol with heat release intensity of about 400 kW/m2 of surface area. Calculate the flame height under the normal atmospheric conditions. Using L = -1.02xD +0.23Q2/5 The flame height L = -1.02x1.7 + 0.23Q2/5 = 0.79m 4. Calculate the wavelength for infrared thermal radiation with frequency 1.5x1015 Hz and compare with the wavelength for BBC Radio 4, 92.8 MHz. The expression linking wavelength (λ ) speed of wave (c) and frequency for electromagnetic wave (ν) is given by λ = c / ν For the case of electromagnetic wave the speed is given by co = 2.99 x 108 m/s For the case of infrared thermal radiation with frequency = 1.5x1015 Hz λ =  = 1.99x10-7 m For BBC radio 4 frequency =92.8 MHz λ =  = 3.221 m the thermal radiation has a much short wavelength in comparison to the BBC radio 4 wavelength Figure 1 5.  A person with initial speed of 1.25 m/s is moving to fire exit as described on the Fig. 1. His travel consists of two parts (AB and BC). In the first part (AB) he is moving with constant speed of 1.25 m/s. When he has achieved the point B, he will start to move with constant deceleration of 0.01 m/s2 due to the crowd in the second part of his trip. What time is needed for the person to achieve the fire exit? Assume that AD is 6 m, BC is 6 m and α = 30o. Figure 1 Distance from A to B = 6/COS30 = 12m Time required to move to move from A to B = 12/1.2 = 10 S Distance from B to C is given by the linear motion equation S = ut +1/2at2 Also by use of the linear motion equation V2 = u2 +2as Where v is the final velocity (velocity at C) Uis the initial velocity = 1.2 m/s A is the acceleration = -0.01m/s2 S is the distance from B to C = 5m therefore v2= 1.22+2x -0.01x5 = 1.34 v=1.16m/s The average velocity = (v+u)/2 = (1.2+1.16)/2 = 1.18m/s Time taken in moving from B to C = distance/average velocity = 5/1.18 = 4.2 s The total time required to achieve the exit = 10 +4.2 = 14.2 s 6.  A person is moving to a fire exit through a corridor (Fig. 2). His speed is 1.25 m/s and constant during his travel. In the corridor there is a strong air movement. Speed of air movement is 0.3 m/s and width of the corridor is 10 m. Find the minimum time needed to reach the fire exit. Explain your answer and indicate the right direction for his evacuation? Figure 2 If the person in question is to move from the starting point of the evacuation it is necessary to put into consideration the effect of the air movement in the corridor. It has also been noted that if the original direction of the person of α =300 is used in addition of the air velocity component there will be a drift which will make the final point to be away from exit point. For the person to have his end point coincide with the exit point it will be necessary to change the initial direction as illustrated in the diagram such that the drift from the wind will make the person to end up at the exit C . If the diagram is assumed the diagram to be proportional to the air components CC is the air velocity component (0.3). AC is the is the initial velocity of the person (1.2m/s) BAC = α =300 CC = 0.3m/s AC = 1.2m/s Also BCA = 180-(90+30) = 600 The velocity vector c in triangle ACC can be calculated by use of the sine rule  =  Now, if a is taken as side AC ; A as angle BCA; b as side CC and B as angle CAC The through Substitution  =  SinB = Sin60 x0.3/1.2 = 0.217 CAC= B = 12.50 The resultant velocity AC (a) can be Calculated as ACC = 180-(12.5+60) = 180-72.5 = 107.50  =  a = 1.2xsin107.5/sin60 = 1.32 m/s distance AC = a  =  a = 7xsin90/sin60 = 8.1m Thus the time required for the person to reach the exit = 8.1/1.32 = 6.14s For the person moving at 1.2m/s to end up at the exit point he will need to reduce her starting angle to (30-12.5) = 17.50. Due to the effect of the strong wind the person will be drifted so that the angle will increase to the desired 300 and the resultant velocity will be 1.32 m/s. 7. How different is the result for the previous question, if air movement changes its direction on opposite (U = - 0.3 m/s). If there is a change in the direction of wind the velocity vector will change as shown in the diagram There will be no change in the distance to be travelled as the start and exit point have not been changed AC = 8.1m Angle BAC = 300 ; BCA = 600 ; ACC = 180-60 = 1200 In terms of velocity AC = 1.2m/s; CC = 0.3m/s Using the sine rule angle CAC can be established  =  SinA = 0.3sin120/1.2 = 0.22 A = angle CAC = 12.70 Angle ACC = 180-(120+12.7) = 47.30 By application of the sine rule velocity AC can be established  =  a = 1.2sin47.3/sin120 = 1.02 m/s Velocity vector AC = 1.02m/s Time taken to reach the exit point = 8.1/1.02 = 7.94s It can be seen that the time taken to reach the exit has increased compared to the previous case. The person will also be required to start at an angle of 30 +12.7 = 42.70C so that after the drift due the wind the angle will reduce to 300C. 8. A compartment is fully involved in fire. The flame inside the room is dull red. Calculate thermal radiation emission [W/m2] from the compartment considering the gray body model with ε = 0.75. Thus emissivity of the media = =0.75 Where = Energy emitted per unit area A = Area of opening = 1m2  = 0.75 σ =Stefan Boltzmann Constant, 5.67x10-8 W/m2/K4 T= 7000 C (973.15) which is the temperature of dell red flame  = 1 x 0.75 x 5.67x10-8 x973.154 = 38.138kW. 9. Explain nomenclature of halon and freon systems (use examples of Halon 1211 and Freon CFC 11). Make a definition of environmental damage potential of halons and freons. Compare environmental damage potentials and atmospheric lifetimes of Halon 1211 and CFC 11. Freon are compounds that are composed of the CFCs and HCFCs and they are normally made through halogen exchange with the most basic being the chlorinated methanes and ethanes. On the other hand halons are usually generated through the free-radical reactions involving chloroflourocarbons where the C-H bonds are replaced by C-Br bonds. A good example of Freon formation is where chlorodifluoromethane is formed from chloroform HCCl3 + 2 HF → HCF2Cl + 2 HCl While halon example is the formation of 2-bromo-2-chroro-1,1,1-triflouroethate commonly known as halothane. CF3CH2Cl + Br2 → CF3CHBrCl + HBr In naming Freons there is a numbering system that is used with a prefix of Freon-, R- , CFC- or HCFC being used. In the nomenclature the number on the extreme right will give the number of fluorine atoms present, next to it the number hydrogen plus 1 atoms is given, followed by the number of carbon atoms minus 1 with zeroes being omitted and then the remaining atoms being Chlorine. Thus Freon CFC 11 shows a methane derivative with one Flourine atoms with no hydrogen while Halons have four numbers (Rodney,2004). By 1960s both Freons and halons were highly available and their effectiveness in firefighting could not go unnoticed (Fine 2011). By the early 1980s halons were commonly used in aircrafts, large vehicles, aircrafts in addition to their application in computer facilities and galeries. At this point concern had started being raised of the effect of these products on the ozone layer. In the Vienna Convention did not cover halons as at this point it was believed that their use in the extinguishing systems was of very small volume and was not expected to have any significant effect in addition of it being of very high importance to human safety for considering restriction. 10. Compare the chemical reaction rates at three temperatures – 100, 500 and 1500 K. The activation energy is 150 kJ/mole. Make your conclusion how temperature affects chemical reaction rate. According to Fire Tests (1993) Taking the universal gas constant R = 8.314 [J/mole.K] A is taken as 1011 Activation energy (E) = 150kJ/K For 100K Rate = 1011exp) = 192482423918 molecules per second For 500K Rate = 1011exp) = 278874557807 molecules per second For 1500K Rate = 1011exp) 402053796103 molecules per second References Rodney C. (2004). Scientific American Inventions and Discoveries, p.351. John Wiley & Songs, Inc., New Jersey. ISBN 0-471-24410-4 Fine R. A. (2011). Observations of CFCs and SF6 as ocean tracers. Ann Rev Mar Sci.; 3:173-95. Babrauskas, V. (1995). “Burning Rates,” SFPE Handbook of Fire Protection Engineering, 2nd ed., National Fire Protection Association, Quincy, MA,. Särdqvist, S., “Initial Fires: RHR, Smoke Production and CO Generation from Single Items and Room Fire Tests (1993). ISRN LUTVDG/TVBB--3070--SE, Department of Fire Safety Engineering, Lund University, Lund, Sweden. Babrauskas, V. & Greyson, S. (1992). Heat Release in Fires, E. & F. N. Spon, London. Tewarson, A., “Generation of Heat and Chemical Compounds in Fires,” SFPE Handbook of Fire Protection Engineering, 2nd ed., National Fire Protection Association, Quincy, MA, 1995. Tewarson, A. (1996). “Flammability,” Physical Properties of Polymers Handbook, Ed. Mark, J.A., NFPA, (1985). Guide for Smoke and Heat Venting, NFPA 204M, National Fire Protection Association, Quincy,MA,. Read More

3m/s AC = 1.2m/s Also BCA = 180-(90+30) = 600 The velocity vector c in triangle ACC can be calculated by use of the sine rule  =  Now, if a is taken as side AC ; A as angle BCA; b as side CC and B as angle CAC The through Substitution  =  SinB = Sin60 x0.3/1.2 = 0.217 CAC= B = 12.50 The resultant velocity AC (a) can be Calculated as ACC = 180-(12.5+60) = 180-72.5 = 107.50  =  a = 1.2xsin107.5/sin60 = 1.32 m/s distance AC = a  =  a = 7xsin90/sin60 = 8.

1m Thus the time required for the person to reach the exit = 8.1/1.32 = 6.14s For the person moving at 1.2m/s to end up at the exit point he will need to reduce her starting angle to (30-12.5) = 17.50. Due to the effect of the strong wind the person will be drifted so that the angle will increase to the desired 300 and the resultant velocity will be 1.32 m/s. 7. How different is the result for the previous question, if air movement changes its direction on opposite (U = - 0.3 m/s). If there is a change in the direction of wind the velocity vector will change as shown in the diagram There will be no change in the distance to be travelled as the start and exit point have not been changed AC = 8.

1m Angle BAC = 300 ; BCA = 600 ; ACC = 180-60 = 1200 In terms of velocity AC = 1.2m/s; CC = 0.3m/s Using the sine rule angle CAC can be established  =  SinA = 0.3sin120/1.2 = 0.22 A = angle CAC = 12.70 Angle ACC = 180-(120+12.7) = 47.30 By application of the sine rule velocity AC can be established  =  a = 1.2sin47.3/sin120 = 1.02 m/s Velocity vector AC = 1.02m/s Time taken to reach the exit point = 8.1/1.02 = 7.94s It can be seen that the time taken to reach the exit has increased compared to the previous case.

The person will also be required to start at an angle of 30 +12.7 = 42.70C so that after the drift due the wind the angle will reduce to 300C. 8. A compartment is fully involved in fire. The flame inside the room is dull red. Calculate thermal radiation emission [W/m2] from the compartment considering the gray body model with ε = 0.75. Thus emissivity of the media = =0.75 Where = Energy emitted per unit area A = Area of opening = 1m2  = 0.75 σ =Stefan Boltzmann Constant, 5.67x10-8 W/m2/K4 T= 7000 C (973.15) which is the temperature of dell red flame  = 1 x 0.75 x 5.67x10-8 x973.154 = 38.138kW. 9. Explain nomenclature of halon and freon systems (use examples of Halon 1211 and Freon CFC 11).

Make a definition of environmental damage potential of halons and freons. Compare environmental damage potentials and atmospheric lifetimes of Halon 1211 and CFC 11. Freon are compounds that are composed of the CFCs and HCFCs and they are normally made through halogen exchange with the most basic being the chlorinated methanes and ethanes. On the other hand halons are usually generated through the free-radical reactions involving chloroflourocarbons where the C-H bonds are replaced by C-Br bonds.

A good example of Freon formation is where chlorodifluoromethane is formed from chloroform HCCl3 + 2 HF → HCF2Cl + 2 HCl While halon example is the formation of 2-bromo-2-chroro-1,1,1-triflouroethate commonly known as halothane. CF3CH2Cl + Br2 → CF3CHBrCl + HBr In naming Freons there is a numbering system that is used with a prefix of Freon-, R- , CFC- or HCFC being used. In the nomenclature the number on the extreme right will give the number of fluorine atoms present, next to it the number hydrogen plus 1 atoms is given, followed by the number of carbon atoms minus 1 with zeroes being omitted and then the remaining atoms being Chlorine.

Thus Freon CFC 11 shows a methane derivative with one Flourine atoms with no hydrogen while Halons have four numbers (Rodney,2004). By 1960s both Freons and halons were highly available and their effectiveness in firefighting could not go unnoticed (Fine 2011). By the early 1980s halons were commonly used in aircrafts, large vehicles, aircrafts in addition to their application in computer facilities and galeries.

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