Fire Protection Physics - Assignment Example

Solutions to Fire Protection Questions: Name: Course: Instructor: Date: 1. (a) Comparison of two burning rate equations The equations are: dMwood = 0.092 A√h (Kg/s) …………………………. (Equation 1) dt dM = 0.09 (Qp w2) ⅓ h (Kg/s) ………………………… (Equation 2) dt Where; A (m) is the area of opening, w (m) is the width and h (m) is the height of opening, while Qp (kW) is convective heat release rate. Equation 1 illustrates the model for combustible of wood fuel with constant heat release rate in a room with a single opening. From the equation the duration of burning period t(s) can be calculated by; t = Mwood ÷ dMwood dt where; Mwood (Kg) is the mass of the wood The corresponding heat release rate Q (MW) is Q = dMwood ∆h dt Where; dMwood is constant, hence Q is constant too dt Equation 2 is applied to a mass flow from an opening, for instance a room fire with a window. Knowing the mass flux from; dM = 0.71Qp ⅓ (Z - Zo) 5/3 dt Hence the temperature can be calculated as T = [Ta + (Qp ÷ Cp dM)] = [Ta + (Qp ÷ 1.5Cp dM)] = Ta + [(Q/1.5) ⅔ ÷ 0.09 w⅔h Cp dt dt Therefore from above it is evident that, equation 1 can be applied to calculate the mass, duration of burning period of combustible matters, the area of opening and height of opening in a room with a single opening. On the other hand equation 2 can be applied to compute the convective heat release rate, height and width of the opening, and further more the average temperature of the room. Equation 2 engages mass flux while equation 1 does not. (b) Solution: The specific heat of the air: Cp = 1.03kJ/ (kg. K) The mass flux: dM = 0.09 (Qp w2) ⅓ h = 0.09 (Q w2) ⅓ h dt 1.5 The temperature: T = [Ta + (Qp ÷ Cp dM)] = [Ta + (Qp ÷ 1.5Cp dM)] = Ta + [(Q/1.5) ⅔ ÷ 0.09 w⅔h Cp dt dt = 20+ [(1053/1.5) ⅔ ÷ 0.09 x 2⅔ x 1 x 1.03 = 556.77(oC) Hence, average temperature of room after flashover is 556.77(oC) The flashover time instant: Q = at2 1053 = 0.0117 t12 Therefore, t1 = √ (1053 ÷ 0.0117) = √90000 = 300 (s) Heat released in the growth time period: E1 = ∫0 Q dt = ∫0 at2 dt = (at13 ÷ 3) = (0.0117 ÷ 3) x 3003 = 105300(KJ) = 105.3(MJ) The duration of the growth time period: tg = Eg ÷ Q = 105.3 MJ ÷ 1.053MW = 100(s) Thus temperature of room during the growth time period is; T = Ti + 345log10 (1 + 8t) Where; Ti (oC) is the initial temperature, T (oC) is the temperature and t (minutes) is the time T = Ti + 345log10 (1 + 8t) = 556.77 + 345log10 (1 + (8 x 100)) = 1558.52(oC) The heat released during the burning time period: E2 = E – E1 = (315.9 -105.3) = 210.6(MJ) Where; E is the fire load in the room The duration of the burning: tb = E2 ÷ Q = 210.6 MJ ÷ 1.053MW = 200(s) The temperature during the burning period is; T = Ti + 345log10 (1 + 8t) = 556.77 + 345log10 (1 + (8 x 200)) = 1662.29(oC) Figure 1 below represents the temperature versus time curve. Figure 1: The temperature versus time curve 2 (a) The three fire spreads are; Heat from the fire, as the source of ignition; Air movement through the central air conditioning; and the free convection air currents driven by the fire itself (b). The unburnt combustible uPVC plastic material obtains radiation from the flames of the fire and furthermore from the smoke layer and adjacent structures. The burning rate amplifies with incident radiation, so producing more heat and smoke. Besides, the unburnt wood fuel is heated more rapidly to channel spontaneous ignition point by the augmented radiant energy. Thus the fire grows and spreads more quickly. This radiation of heat energy is called “positive feedback” in view of the fact that the heat comes initially from the fuel by burning its vapour and is convected to the smoke layer, then is fed back to the fuel by radiation. Air movement through the central air conditioning can cause the flames to rise at an angle so that they extend beyond the adjacent unburnt wood fuel, increasing the ‘view factor” and hence incident radiation to the combustible fuel and other structures. In addition to the free convection air currents driven by the fire itself circulate the air as well as smoke, effectively hence increasing the convected heat transfer within the building. (c). Solution: ∆T = TA - TB = (585 + 273) K – (250 + 273) K = (858 – 523) = 335K Q = -kA ∆T => Q = -k ∆T ∆x ∆x -1250W/m2 = 0.76W/(m.K) x 335K = -254.6W/m ∆x ∆x Therefore ∆x = 254.6W/m = 0.204(m) 1250W/m2 Hence the minimum thickness of the separated wall in order to avoid fire spread via heat conduction through the wall is 0.204(m) 3 (a) In relation to the reduction of fire spread, external walls have three major functions that include; prevention of fire from low storey to upper storey; confining the fire within the building until it’s controlled or burns itself out; and inhibiting spread of fire across the relevant boundaries of buildings. Thus the building criteria should observe that the external wall construction have a higher fire resistance. These building criteria’s are as discussed in the following paragraphs. (b) In relation to the reduction of fire spread, roof covering contributes to reducing spread of fire in similar ways like the external wall. The only variation is that, the external wall prevents fire spread by radiation to adjoining buildings while the roof covering does so by transmitting the fire from burning building to others. For this case the building criteria should observe that the roof covering construction has a higher fire resistance. These building criteria’s are as discussed in the following paragraphs. 4 (a) The three components that comprise a standard fire test are; the fire furnace, the test specimen, and the measurement equipment. The fire furnace refers to the place where the test specimen is heated for a given period of time in order to analyze its resistance to fire, as well as to investigate the properties of the materials that form it. There are two types of fire furnaces which include; the vertical and horizontal fire furnaces. The test specimen implies to the sample to be tested for fire resistance, which for instance could be a wall which is made of materials such as concrete, fibre composite, glass, wood, or steel reinforced. The measurement equipment monitors and records; the changes in the test specimen performance when subjected to a fire, incident heat flux, dimensions and boundary end conditions of the specimen, and effect of temperature rise within the specimen structure members on the relevant properties of the specimen. (b). The standard fire test to analyze the fire resistance of a building structure, comprises of; the fire furnace, the building structure and the measurement equipment. The building structure is loaded and then heated within the measured temperature regime in the fire furnace following a prescribed temperature-time relationship until failure of the building structure occurs. If the building structure is a beam or a slab it will be heated from beneath, columns on all sides, whereas walls are heated from the backside of wall surface. The temperature-time variation is regulated by fire test standard, for instance, ISO 834 specifies the temperature-time curve as follows; T = Ti + 345log10 (1 + 8t) Where; Ti (oC) is the initial temperature, T (oC) is the temperature and t (minutes) is the time Through comparison of the standard fire curve ISO 834 to the one obtained from the standard fire test, and through analysis of presence of any cracks or deformation the building structure can be passed to be either satisfactory to use in the construction of the building or unsatisfactory. The three criteria to terminate the test are; insulation, integrity and load-bearing capacity. Insulation refers to the ability of the element to provide insulation from high temperatures, in other words it is the ability of the element to resist transfer of excessive heat. Failure in this case would occur when the mean temperature cannot be attained by the unexposed surface. Integrity, refers to the ability of the test specimen to withstand cracks or openings from occurring in the separate element such that ignition may not occur on the unexposed side, or the capacity to resist fire penetration. Failure in this case would take place when cracks or openings occur on the separating element such that ignition takes place on the unexposed side. Load-bearing capacity refers to the capability of the test specimen to maintain load-bearing capacity which is applied to load-bearing construction; in otherwise it is the ability of the element to resist collapse. Failure occurs when the element loses the load-bearing capacity when it cannot carry the applied loading. (c). Three drawbacks of the standard fire test approach consist of; Expense, Reproducibility, and fire scenario limitations. The standard fire test approach is expensive and time-consuming weighed against numerical methods in addition to analytic approach. For the case of reproducibility; not only the different furnace tests but also the same furnace and specimen cannot produce similar results. So for one case, it is required to carry out more similar tests than one, which also makes the test more expensive, time-consuming and laborious. On the side of fire scenario limitations, is that, the temperature in the test furnace does not represent the fire exposure to the elements in reality and also the development of temperature in the furnace is not matched to the fire history in practice. 5 (a). The strength of the structural steel reduces with increase in temperature, because the stress-strain relationship is lost, which a well defined yield between the elastic and plastic portions. The relationship instead becomes highly non linear, with both strength and stiffness decreasing, hence the steel-constructed building collapsing or its strength reducing. Thermal expansion (a measure of materials ability to expand or contract in response to temperature changes) of steel increases linearly up to 700oC, when there is temporary sudden shrinkage with any further increase in temperature; which is caused by phase transformation from pearlite to austenite and rearrangement of the crystal structure. This indeed makes steel to lose its strength. (b) Three fire protection measures for the steel structure can be achieved through; insulating the element with spray material or board type protection; shielding; and filling hollow sections with concrete or liquid to form a heat sink. Passive fire protection materials insulate the steel structure from high temperatures, since they have lower thermal conductivity compared to steel. For instance boards are fixed dry usually to columns; beams are more commonly protected with spray materials. The insulation can also be done by mastics which form a thick protective coating that is impervious at ambient and high temperatures. The hollow columns of steel structure can gain enhanced fire resistance by filling them with concrete. The mechanism here is that, during a fire, heat will flow through the steel to the low conductivity concrete. As the steel loses its yield strength with increasing temperature the load is transferred to the concrete. Adding fibre or bar reinforcement to the concrete can attain enhanced periods of fire resistance. A two hour fire resistance can be achieved using this technique. Shielding can be achieved by use of shelf angle floor beams, which are beams with angles bolted to the web to support the floor slab thus shielding the upper part of the beam from the fire leaving only the bottom of the beam exposed. A one hour fire resistance can be achieved using this technique. (c). Solution: Given that the building will collapse if the structure reduces to 0.60 times of the original strength, the percentage of strength reduction (r value) is 40% Using; 100r = -10.0 + 0.064Tc (Equation 3) Where; Tc is fire temperature at the time of collapse and r is the percentage of strength reduction (r value) 100 x 40/100 = -10.0 + 0.064 Tc 40 = -10.0 + 0.064 Tc 40 + 10.0 = 0.064 Tc Tc = (40 + 10.0) ÷ 0.064 = 781.25 oC Therefore, fire temperature at the time of collapse is 781.25 (oC) The time instant when the building collapses can be calculated from the ISO 834 equation given as; Tc = Ti + 345log10 (1 + 8t) (Equation 4) Where; Ti (oC) is the initial temperature, Tc (oC) is the temperature at the time of collapse and tc (minutes) is the time at the instant of collapse But, Ti is calculated from 100r = -10.0 + 0.064 Ti 100 x 100/100 = -10.0 + 0.064 Ti 1 = -10.0 + 0.064 Ti Ti = (1 + 10.0) ÷ 0.064 = 171.88 oC Therefore, fire temperature at the time of collapse is 171.88(oC) Hence, replacing values in equation 4 above; 781.25 = 171.88 + 345log10 (1 + 8 tc) log10 (1 + 8 tc) = (781.25 - 171.88) ÷ 345 log10 (1 + 8 tc) = 1.7663 1 + 8 tc = 101.7663 1 + 8 tc = 58.38 8 tc = 58.38 + 1 tc = 59.38 ÷ 8 = 7.42 (minutes) Therefore, the time instant when building will collapse is 7.42 (minutes) Read More