Operation of Fire Protection Systems - Math Problem Example

Report Assignment Name: Course: Professor: Institution: Date: Report Assignment Question One A compartment is fully involved in fire. The flame inside the room is cherry red. Considering the gray body model (ε = 0.75) calculate thermal radiation emission from the compartment and compare it with maximum radiant heat flux for indefinite skin exposure. Solution: For a compartment that is fully involved in fire, the Stefan-Boltzmann equation governing transfer of heat by radiation is used. The area is a unit (1m2). A cherry-red flame burns with a temperature range of between 800-1000oC. If the compartment is fully fired, then the temperature of the flame is 1000oC (1273.15K).Assuming the compartment is mainly made using normal concrete with a Stefan’s constant of , the value of thermal radiation emission within a square meter of area can be calculated using the equation above. q = ( = 111.7kW/ Thermal radiation emission or heat flux is the rate at which heat energy will be transferred through electromagnetic waves per unit area of the compartment. The Stefan-Boltzmann law is the relationship that governs the radiation emitted from a hot object. The maximum radiant heat flux for indefinite skin exposure is 1kW/m2. Since the value of thermal radiation obtained in this case is larger than the maximum indefinite skin exposure, a skin exposed in this environment will be destroyed (Thirumaleshwar, 2009). Question Two Mixed fuel is composed by methane (volume percent is 0.45), carbon monoxide (0.15) and hydrogen (0.40). Calculate the lower flammable limit concentration for the mixture and the concentration of each component in the mixture with air. Solution: The lower flammable limit (LFL) or lower explosive limit (LEL) refers to the minimum required percentage concentration of a substance in air for ignition. Below this lower concentration limit, the mixture cannot ignite and is described as lean. If a flammable mixture or gas gets to a concentration level beyond 10% of the lower explosive limit, one is advised to evacuate the area. The net LFL of the mixed fuel is calculated from the equation below: Net LEL =1/ ( ) Where: P1, P2 and P3 are volume fractions of methane, carbon monoxide and hydrogen respectively. LEL1, LEL2 and LEL3 are lower explosive limits for methane, carbon monoxide and hydrogen respectively. Therefore, LEL = 1/ ( ) = 1(10.23+3.33+3.75) =1/ 17.31 Net lower flammable limit = 1/17.31 = 5.78% Below this flammable limit, the mixture of fuel will be too lean to burn unless the pressure and temperature are increased. This lower flammable limit decreases as temperature increases. This means that a mixture below its LFL range may burn if sufficiently heated at a given temperature. LFL in liquids is typically near the concentration of saturated vapour at the flash point. However, as a result of different liquid properties, the LFL and the flash point have a varying relationship that is spread (Wypych, 2001). Question Three Consider a 1.2 m diameter pan fire of petrol with heat release intensity of about 490 kW/m2 of surface area. Calculate the flame height under the normal atmospheric conditions. Solution: The flame height under normal atmospheric conditions pressure, temperature, wind direction and speed can be calculated using Hesketad Equation given below: ) Where: L - Flame height (m) D – Diameter of fire Q – Heat release intensity Therefore, if the values for D and Q are substituted in the above equation. ) L = -1.224+2.8=1.58m The flame rises to a height of 1.58m when the petrol burns in a 1.2m diameter pan under normal atmospheric conditions (Cote, 2003). Question Four Compare the chemical reaction rates at three temperatures – 100, 500 and 1000 K. The activation energy is 100 kJ/mole. Make your conclusion how temperature affects chemical reaction rate. Solution: Assuming a gas constant value of 8.314 J/mol K and a pre-exponential factor of 9 M-1s-1, a comparison of reaction rates can be made by calculating the chemical reaction rate constants. The values for activation energy and temperature given are substituted in the equation below: Ine i) At -100oC M-1s-1 ii) At 500oC M-1s-1 iii) At 1000oC = -(100 X 1000) / (8.314)(1000) + ln9 M-1s-1 From the calculations above, it can be observed that the value of chemical reaction constant rate () increases as the temperature increases. This shows that chemical reactions are more rapid at higher temperatures and slow down as the temperature decreases. As the temperature increases, molecules that make matter move faster, colliding more vigorously, which causes bond cleavages and this results in increased random motion which favors the rate of reaction. Most of the chemical reactions are dependent on thermal activation which enable particles to acquire enough kinetic energy required for reaction (Wypych, 2001). Question Five Calculate the wavelength for infrared thermal radiation with frequency Hz. Compare the result with the wavelengths for Smooth Radio 100.4 and visible radiation for the human eye. Solution: The wavelength is expressed in terms of velocity and frequency using the relationship shown below: Wavelength ( = The velocity of light is m/sec. For the infra-red thermal radiation with a frequency of Hz, the wavelength is calculated form the relationship above as: Wavelength = m The radio waves lie on one end of the electromagnetic spectrum with the lowest frequency and longer wavelengths. From the relation of velocity, wavelength and frequency, increase in frequency reduces the wavelength. Radio waves have a wavelength >0.1m. Visible radiation has a wavelength between and. As it can be seen, radio waves have longest wavelength compared to the wavelength of thermal infra-red and visible light. Visible light has the shortest wavelength compared to that of infra-red radiation and radio waves. Thermal infra-red radiation have a wavelength that lies between the radio waves and visible light. Question Six 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 7 m, BC is 3 m and α = 30o. Calculating the distance AB Using the formula Cos = So, AB = AD*Cos =7 Cos 30 = 6.06m Time taken to travel over the distance AB = (Distance AB/Velocity) = (6.06m/1.2m/sec) = 5.05 sec. Calculating the time taken to travel from point B to C Using the formula for calculating acceleration a =)/t Where: a- represents negative acceleration – represents initial velocity at point B (1.25m/sec) – represents final velocity at point C (0m/sec) – represents the time taken to travel between point B and C Therefore, t =)/a = (1.25-0)/0.01 = 125 sec Total time required to reach the fire exit = 5.05+125 = 130.05 seconds. V0 = 1.2 m/s C Fire Exit BC = 3 m AD = 7 m A D Figure 1 Question Seven 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? B C Fire Exit U = 0.3 m/s AB = 10 m = 30° α V0 = 1.25 m/s A Figure 2 Solution: The resultant velocity is the sum of the person’s speed and the speed of wind. Therefore, To reach point C at the fire exit, one can move from point A to B and then from point B to C. The shortest distance that will take the minimum time is AC. The minimum time taken to reach point C from point A is calculated below. = 11.55m The minimum time taken to move to the fire exit = Question Eight How different is the result for the previous question, if air movement changes its direction on opposite (U = - 0.3 m/s). Solution If the wind changes its direction, the resultant velocity will be reduced so that, Therefore, The time taken to move to the fire exit = The wind blowing in the opposite direction reduces the speed of motion and a person will take a few more minutes to reach the fire exit. References Read More

LEL1, LEL2 and LEL3 are lower explosive limits for methane, carbon monoxide and hydrogen respectively. Therefore, LEL = 1/ ( ) = 1(10.23+3.33+3.75) =1/ 17.31 Net lower flammable limit = 1/17.31 = 5.78% Below this flammable limit, the mixture of fuel will be too lean to burn unless the pressure and temperature are increased. This lower flammable limit decreases as temperature increases. This means that a mixture below its LFL range may burn if sufficiently heated at a given temperature. LFL in liquids is typically near the concentration of saturated vapour at the flash point.

However, as a result of different liquid properties, the LFL and the flash point have a varying relationship that is spread (Wypych, 2001). Question Three Consider a 1.2 m diameter pan fire of petrol with heat release intensity of about 490 kW/m2 of surface area. Calculate the flame height under the normal atmospheric conditions. Solution: The flame height under normal atmospheric conditions pressure, temperature, wind direction and speed can be calculated using Hesketad Equation given below: ) Where: L - Flame height (m) D – Diameter of fire Q – Heat release intensity Therefore, if the values for D and Q are substituted in the above equation. ) L = -1.224+2.8=1.58m The flame rises to a height of 1.

58m when the petrol burns in a 1.2m diameter pan under normal atmospheric conditions (Cote, 2003). Question Four Compare the chemical reaction rates at three temperatures – 100, 500 and 1000 K. The activation energy is 100 kJ/mole. Make your conclusion how temperature affects chemical reaction rate. Solution: Assuming a gas constant value of 8.314 J/mol K and a pre-exponential factor of 9 M-1s-1, a comparison of reaction rates can be made by calculating the chemical reaction rate constants.

The values for activation energy and temperature given are substituted in the equation below: Ine i) At -100oC M-1s-1 ii) At 500oC M-1s-1 iii) At 1000oC = -(100 X 1000) / (8.314)(1000) + ln9 M-1s-1 From the calculations above, it can be observed that the value of chemical reaction constant rate () increases as the temperature increases. This shows that chemical reactions are more rapid at higher temperatures and slow down as the temperature decreases. As the temperature increases, molecules that make matter move faster, colliding more vigorously, which causes bond cleavages and this results in increased random motion which favors the rate of reaction.

Most of the chemical reactions are dependent on thermal activation which enable particles to acquire enough kinetic energy required for reaction (Wypych, 2001). Question Five Calculate the wavelength for infrared thermal radiation with frequency Hz. Compare the result with the wavelengths for Smooth Radio 100.4 and visible radiation for the human eye. Solution: The wavelength is expressed in terms of velocity and frequency using the relationship shown below: Wavelength ( = The velocity of light is m/sec.

For the infra-red thermal radiation with a frequency of Hz, the wavelength is calculated form the relationship above as: Wavelength = m The radio waves lie on one end of the electromagnetic spectrum with the lowest frequency and longer wavelengths. From the relation of velocity, wavelength and frequency, increase in frequency reduces the wavelength. Radio waves have a wavelength >0.1m. Visible radiation has a wavelength between and. As it can be seen, radio waves have longest wavelength compared to the wavelength of thermal infra-red and visible light.

Visible light has the shortest wavelength compared to that of infra-red radiation and radio waves. Thermal infra-red radiation have a wavelength that lies between the radio waves and visible light. Question Six 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.

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