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Fire Modeling Problems in Building - Assignment Example

Summary
The assignment "Fire Modeling Problems in Building" focuses on the critical analysis of the main fire modeling problems while constructing sustainable buildings. The Arklow warehouse fire is considered to have been a major fire inferno that started with a loud explosion (O'Toole, 2016)…
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Extract of sample "Fire Modeling Problems in Building"

PART 2

Case Study: Arklow Warehouse Fire in Wicklow (2016)

The Arklow warehouse fire is considered to have been a major fire inferno that started with a loud explosion (O'Toole, 2016). The warehouse was made of asbestos mainly, which increased the blaze of the fire the firefighters were battling while threatening the lives of the people since asbestos is considered a significantly harmful product. The ire is reported to have started around 10 am and was stopped around 4pm (O'Toole, 2016). This shows how the fire was a major inferno and challenging to stop. Hours after the fire was stopped, the people could still encounter bits of asbestos falling on the ground. It depicted that even though the fire had been stopped, it remained a threat to the lives of the people around (O'Toole, 2016).

Lessons From The Fire

One of the main lesson learnt from this fire was the fact that buildings such as this warehouse should not be developed using materials that may cause health hazards to the people. More importantly, the asbestos materials did not only create challenges in stopping the fire but was also a key factor in threatening the lives of those around. That is; the asbestos is a key fire resistance product, but having been used in the entire large warehouse was a major problem. The main lesson learnt from the fire is that it is important to use materials that will not lead to health hazards for the employees working in the warehouse, or the people around the area in case a fire occurs that leads to the people been exposed to the fire.

The inferno was challenging to stop because the warehouse did not have any alarms or equipment has to handle such a fire. The lack of equipment’s and other systems to help avert the fire were a key factor into why the fire prolonged for so long. Therefore, it is important that such buildings install fire resistance systems including equipment’s that can help control the spread of fire. The fire in the warehouse started and 10am, and was stopped at 4pm showing that the materials in the warehouse promoted the spread of the fire. Therefore, in case of such fires, the fires are challenging to stop. The result of the fire is damaging.

Recommendations

In case of such a fire currently, the design of the building and other fire protection systems are the key factors that can be relied upon for complete control and management of the fire. The design of the building can be developed in a manner that limits the spread of both smoke and fire, which is beneficial in stopping fires.

Additionally, using automatic sprinklers among other natural ventilations as described in part one can be a major part in ensuring that the fires are controlled. There is no specific way of controlling the fires, but such strategies can be effective in controlling the spread and damage of the fire. More importantly, it is advisable for such warehouses to not use materials such as asbestos which are a threat to the employees working in the warehouse and the environment.

PART 3: Fire Modelling Problems

  • Freon systems includes the CFSs, and have been used in manufacturing aerosol sprays, packing materials and blowing agents among others. The CFCs are named through archaic numbers of three integers. That is I = carbon atoms minus one; J = hydrogen atoms minus one; and K = fluorine atoms. Halons are fluorocarbons containing some bromine without any hydrogen. Halons nomenclature involves I, j, k, l. That is; I represents the carbon atoms in the halon. J = fluorine atoms. K = chlorine atoms; l = bromine. For instance, Halon 2201 = C2F2Br1.

Halons affect the environment through releasing ozone that damages the local scale materials and plants. They have long lifespans, which stipulates they end up in the stratosphere, thus destroying the ozone layer. The stratosphere ozone layer protects the earth from the UV rays of the sun. Halons occur as powerful ozone depleters, while contributing highly to global warming.

Under the Montreal Protocol, the Halon was replaced as a fire suppressant since it was harmful to the environment through depleting the ozone layer (Snyder, 2008, 1).

According to the latest update on the Montreal Protocol, Halons have significantly decreased though Halon 1301 remains high (noaa.gov, 2010, 1). On an international level, halon replacement was the right decision based on the characteristics of the product and how it affects the ozone layer. Had it continued to be used in the past, today some of the world problems such as global warming would be very adverse.

  • Fahrenheit and Rankine use the same graduation/ interval. However, they differ in that Rankine uses absolute zero as temperature reference while Fahrenheit uses a freezing point of 32 oF while the boiling point is 212 oF. In the novel, Fahrenheit 451 shows the high temperatures that the book catches fire and burns. Thus, it is a representation of how the temperatures are high. The temperatures keep rising in the same manner the book caught fire and burned.
  • (ε = 0.75)

The solution applies the equation of Stefan-Boltzmann Law:

q = σT4 j/m2s A

σ = 5.6703 * 10-8 watt/ m2K4

Q = the transfer of heat per unit

T= temperature absolute in Kelvin

A = Area of emitting body

(Barauskas, 2004)

Assuming the temperature is at 500oC of the burning compartmental room; the radiation emission can be attained as follows

0.75 (5.67 * 10 -8 W/m2 .K4) (500K)4 =

  • frequency 1014 Hz

The frequency of radio waves is about 108 Hz, whereas the frequency of gamma rays is about 1020 Hz. electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10−7 m (700 nm, or 4.3 × 1014 Hz)

Frequency for infrared = 108Hz

Wavelength = VT; V/F. Therefore; 3 * 108/ 1014

= 3 * 10-6m; V = 3* 1014m/s

FM (Frequency modulation: 88 -108 MHz

(Assumption) FM = 100.4 * 106 Hz

3 * 108/ 100.4 *106

= 2.988m

Therefore, infrared has a smaller wavelength compared to the others.

  • Reaction rate = 1/t

Activation energy = 100kj/mole

100c

Temperature

Change

-25c 1/t (reaction rate)

As temperature increases the total energy of molecules increases. Therefore, the number of the molecules having the activation energy required for the reaction also increases.

  • Element CH COH

Volume present2.550.350.10

LFLmix = 1 / ∑xi/ lfl

Le chatelier mixing ratio

Xi = molar fraction of the i-th component

LFL = lower flammability of the i-th component of the mixture

∑ 1/ (0.55/ 0.55 + 0.35/ 0.35 + 0.1/ 0.1) = 1/3

LFL = 0.3333, which is the volume percent.

  • Diameter pan = 0.80m

Heat release intensity – 450km/m2 Surface area at standard temperature and pressure.

P = 1.204 kg/m3

Cp = 1.005kj/ kg

T ͚ = 293k

g =9.81m/s2

L = 1.02D + 0.235 (Q 2/5)

To = x (Q 2/5 c / (Z –ZO)) 5/3 + T ͚

Flame Height = 1.89m

  • B C

9 = -0.01m/s2

60BC = 15m

30AD = 3M

A 3m D

SOH CAH TOA

Tan 30 = 9/3

A = 3 Tan 3 d

= 1.7321

Cos 30 = 3/H

Hyp = 3/ cos 30

Hyp = 3.464

T = D/S = 3.464/ 1.25 = 2.7713 sec to B

= (v – u/ t)

T = (v-u/9)

U2 = v2 – 295

V2 = u2 -295

V2 = 1.5625 – 0.3

V2 = 1.2625

V = 1.1236

t = 1.1236 – 1.5625/ - 0.01

For BC time = 43.89 seconds

Total time = 43.89 + 2.7713

= 46.6613 seconds

    BC

    10mVo = 1.25m/s

    U = 0.45

    30

    A

    Cos 30 = 10/ Hyp

    Hyp = 10/ cos 30 = 11.5470

    Base = 10 tan 30 = 5.7735

    Time = distance/ speed = 10/ (1.25 + 0.45)

    Time = 10/ 2.8 = 12.5 seconds (A to B)

    Time B to C = Distance/ speed = 5.7735/ (1.25 + 0.45) = 3.3962 seconds

    Total minimum time required = 12.5 + 3.3962

    = 15.8962

    Therefore, as the person is moving through the corridor, the air movement opposes his movements. Thus, it creates an obstacle but when he moves from point B to the fire exit, the air moves in his direction. Thus, it boosts his movements.

    • Time = Distance / Speed = 10/ (1.24 - 0.45)

    = 12.5 seconds

    Tb = D / S = 5.7735/ 0.8 (1.25 -0.45)

    Tb = 7.216875

    Total time = 1.25 + 7.216875

    = 19.716875 seconds

    In this case, the total time increases since at both instances the air movement acts against his directions.

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