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Fire Protection: An Overview - Assignment Example

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"Fire Protection: An Overview" paper examines building criteria, external wall construction, and roof coverings. The paper also describes the components of a standard fire test and the drawbacks of the standard fire test and the thermal conductivity of steel…
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Name……………………………………………………………xxxxx Student number …………………………………………………xxxxx Tutor……………………………………………………………xxxxx Title………………………………......................................................... Fire Protection Course……………………………………………………………xxxxx @2010 Fire Protection Question 1(a) Equations on Burning Rate 1. dMwood ⁄dt = 0.092A√h 2. dM/dt = 0.09(Qpw2)1/3h The tow equations use differentiation in determining the burning rate. The burining rate in equation one is obtained through the differentiation of a constant 0.092, the area of an opeing A(m2) which is multiplied with the square root of its height (h). This is mainly used to determine burning rate of heat from a single opening. The burning rate in equation two is calculated through the differentiation of a constant 0.09, a cube root of the product of the convective heat release rate (Qp) and the squared width (w2) which multiplied by its height (h). Is is mainly used where there is a mass flux. From the formula, it is clear that the two equations determine the same thing but their formulas are distinct. The first equation uses the area of an opening while the second equation takes account of the width – window. 1(b) (i) Solution Specific heat of the air: Cp = 1.03kJ/(kg.K) Mass flux: dM/dt = 0.09(Qpw2)1/3h = 0.09(Qw2/1.5)1/3h Temperature: T = Ta + Qp/ CpdM/dt = Ta + Q/1.5Cp dM/dt = Ta + (Q/1.5)2/3/0.09w2/3hCp = 20 + (1053/1.5) 2/3/0.09 × 22/3 × 1 × 1.03 = 556.770C (ii) Medium clas t square fire model Q = at2 Where a is the constant of the model and Q is the heat release rate 1053 = 0.0117t2 Thus, t2 = (1053/0.00117) = 90,000 t square root of 90, 000 = 300 Heat realesed in the growth time period: E1 = ʃot Qdt = ʃot at2dt = at3/3 = 0.0117/3 × 3003 = 105300 (kJ) = 105.3 (MJ) Curve 1053 Heat Release Rate 300 Time Question 2 (a) and (b) Fire Spreads In the high-rise student bulding, fire would easily spread through the door since it is made of wood. This is because doors made of wood do not resist full load of fire for more than one hour. The fire will spread through the glazed windows which are framed of combustible uPVC plastic material it is not a good fire resistance material. The central air conditioning ducts connectint each room in the accomodation building will be the center stage of the fire spread. The conditiong ducts will transfer heat and fire form one room to another enhancing its spread and this will lead to high smoke eminating from the building. The smoke will lead to fire spreads around the building. The heat conduction (direct contact) in a fire is the transfer of thermal energy produced by the fire through matter, from a regionof higher temperature to a region of lower temperature, and acts to equalize temperature differences. Thermal radiation is the electromagnetic radiation emitted from the surface of an object which is due to the objects temperature. The radiation is generated when heat form the movement of charged particles within atoms is converted to electromagnetic radiation. (c) Use of Fourier’s Law Q = -kΔT Where q is the local heat flux (W/2) ΔT is thetemperature gradient (K/m) K is the thermal conductivity (W/(m.K)) On integration on the materials surface Q = -kʃs ΔTd, where q is the amount or heat transferred per unit minute Integrating betwen two endpoints at constant temperature Q = -kAΔT/Δx Where A is the surface area, ΔT is the temperature difference between ends: Δx is the distance between the ends. Solution for part (c) of question 2 ΔT = TA – TB = (585 + 273)K – (250+273)K = 235K Q = -kA ΔT/ Δx Q/A = -k ΔT/ Δx -1250W/m2 = - 0.76W/(m.K) × 235K/ Δx = -178.6W/m/ Δx -1250W/m2 = -178.6W/m/ Δx Therefore Δx = 178.6W/m/1250W/m2 = 0.143m 0.143m is the minimum thickness of the separated wall. Question 3 Building Criteria External Wall Construction External walls are elements of structure and the relevant period of fire resistance and depends on the use, height and size of the building in question. For instance, if the wall is 1000mm or more, a reduced standard of the fire resistance is accepted in most cases and the all only needs fire resistance from inside. The external wall of a building should not provide a medium for fire spread if it is likely to be a risk to health or safety. The use of combustible materials in the cladding system and extensive cavities may present suc a risk in tall buidings. The external surfaces of walls should meet all the provisions of Diagram 40 and 40c in Approved Document B page 95. If the building is of 18m high or more above the ground level, any insulation product, filler material used in the external wall construction should be of limited combustibility. The Approved Dovument B also gives provisions to restrict the combustibility of external walls of buildings that are less than 1000mm from the relevant boundary and, irrespective of boundary distance, the external walss of high buildings. Portal frames of reinforced concrete can support external walls requiring a similar degree of fire resistance without specific provision at the base to resist overturning. Roof Coverings The Approved Document B (102) gives provisions limiting the ue, near a boundary, of roof coverings which will not give adequate protection against the spread of fire over them. Roof covering refers to constructions which may consist of one or more layers of material, but does not refer to the roof structure as a whole. The separation distance is the minimum distance from the roof to the relevant boundary, which may be a notional boundary. Plastic rooflights, when used in rooflights, a rigid thermoplastic sheet product made from unplasticised PVC, which achieves a Class 1 rating for surface spread of flame when tested. Uwired glass, when used in rooflights, at least 4mm thick can be regarded as having an AA designation (National Class). Thatch and wood shingles should be regarded as having an AD designation when used in rooflights. Question 4 (a) Components of a standard fire test a). Fire furnance b). Test specimen and c). Measurement equipment In fire furnance, a material or building is tested whether it is fire resistance or for how long can it withstand give amount of fire. A test specimen is used to measure the fire resistance in respect to temperature while a measurement equipment is used to determine if there cracks/damages on the wall or on the test specimen. Generally, in the standard test the tested element is loaded and then heated with measured temperature regime in the furnance following a prescribed temperature-time relationship until failure of the element occurs. Traditionally beams and slabs are heated from beneath, whereas columns are heated on all sides. The temperature-time relationship variation is regulated by the fire test standard. (b) In testing the fire restistance of a building structure, the test construction is assessed against three criteria of failure. The criterias include; Insulation – this indicates the average temperature on an unexposed surface achieves a temperature. Integrity – this criteria is used to evaluate on cracksor openings in a separate element such as an ignition can occur on the esposed side. Load-bearing capacity – in this criteria, the element in test loses load bearing capacity when it is nolonger able to carry the applied loading. (c) Drawbacks of the standard fire test 1. It is expensive Carrying out a standard fire test is an expensive affair and time consuming if compared with numerical methods and analytic approach 2. Specimen Limitations In carrying out a standard test, there is limitation on the size and type of the specimens. The test is only capable of studying the failure mode of a single structural element and this means that it cannot be used to examine failure mode of complex structures. 3. Fire scenario limitations The standard test does not reflect the actual fire exposure on elements and the development of the furnance in standard test is not in harmony with the fire history across the globe. Question 5 (a) Thermal conductivity of steel and increasing temperatures reduce its strength. Steel usually loses its form on increasing temperature. The behaviour of the structure becomes highly non-linear with increasing temperature, with both strength and stiffness decreasing. At higher temperatures the concept of 1% proof stree is typically adopted. The stress is measured at a strain equal to 1% permanent deformation of its original length and strength. Thermal expansion is a measure of a materials ability to expand in response to temperature changes. The coefficient of thermal expansion, is defined as the expansion of a unit length of material when it is raised by 10C. (b) There are three ways in which fire protection of steel is realised. Through shielding, insulating the element with spray material and filling hollow sections with concrete. The use of concrete was the most common mode of fire protection for steel but the methods has several disadvantages. It is expensive, it increases the weight of the structure and it is time consuming when applying on site. Insulating helps in protecting the material from high temperatures. Steel is commonly proctected with spray materials. The advantage of using spray is that it is cheap and are easy to apply on the material. The application is always wet and might cause delays on site to do other works. Shielding is another method of fire protecting steel. Paints are usually stable at low temperatures but swell at around 200 C to provide a charred layer of low conductivity material to insulate steel. (c) Structure strength reduction 0.60 times of the original stregth 100r = -10.0 + 0.064T Where r is the stregnth reduction and T temperature T (C) ISO 834 specifies the temperature time curve as T = Ti + 345log10(1+8t) Where, Ti is the intial temperature; t – temperature; t(min) time Getting T 100(0.60) = - 10.0 + 0.064T 70 = 0.064T T = 70/0.064 = 1093.75 0.064Ti = 100(0.9) + 10 Ti = 100/0.064 = 1562. 5 Time 1093.75 = 1562.5 + 345log10(1 + 8t) -468.75 = 345log10(1 + 8t) Log10(1 + 8t) = -468.75/345 = -1.4 (1 + 8t) = 10(-1.4) 1 + 8t = 0.04 8t = 0.96 t = 0.12 References British Standard 476, British Standard Institution, 2007 CIBSE Guide E – Fire engineering, The Chartered Institution of Building Services Engineerings, London, 1997 Lamont, S, The behaviour of multi-storey composite steel formed structures in response to compartment fire, PHD dissertation, University of Edinburgh, 2001 Purkiss, J A, Fire safety engineering, design of structures, Elsevier, second edition, 2007 Seward, D, Understanding structures, Macmillan, 2004 The Buiding Regulations 2006 – Approved Document B – Fire Safety, Her Majesty’s Stationary Office, 2006 Read More
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