A Semenov Diagram of Flashover Fire Development - Assignment Example

This critical condition is the flashover.

While considering the flashover period and.

In the above equation, AF will represent the projected surface area of external fuel that experiences the compartment's direct heating. The equation accounts for the approximate effect of temperature and oxygen, and the effects of distribution in the compartment.

The parameter represents a view factor of the radiation. For the scenarios possible from fire consuming an object fully to spreading, there is a small effect of the burning rate of oxygen as its concentration does not significantly decrease to flashover (Rockett, 1976). The loss rate is given by the equation below;

With T being a function of time, the independent variable t is taken to be dependent as inverse function t = t (T) is available in principle.

The diagram below (Semenov diagram) shows the energy release behavior and energy loss behavior as a function of pool fires.

The diagram depicts the behavior in “low flame absorbing surface fuels”, in this case, walls. The loss curve decreases with increasing time and increasing ventilation factor (), causing the flashover, which is a critical condition to take place (Karlson and Quintere, 2000).

The critical temperatures take place at 3000C to 6000C. This means that liquid fuels require low temperatures and solids that are charging use high temperatures. The common criterion used for flashover is 5000C. In both cases explained here, the flashover results from a single burning item, and a ventilation limit is attained after the flashover.

When a second item is ignited by the first, the resulting Semenov diagram is as follows;

For the contact of the flames, the ignition temperature applies with a small ignition time of approximately 10 to 100 seconds.

For the concurrent spread of the fire, the growth rates tend to be faster and thus a critical condition is reached at low temperatures of the compartment. The dependence of the flame spread area on Q and also on the surface temperatures make the spread to be more sensitive to feedback (Epstein, 1988).

The Arrhenius equation

This equation studies the relationship between the relationship rate increase and the increase in temperature. This increase in reaction rate is not linear with temperature and is given by:

K = rate constant,

Ea = activation energy

R = gas constant (8.314 J/mol K)

T = temperature (K)

A = is the frequency factor that is particular for some certain reaction

k2/k1 = exp (-45.9 kJ/mol/8.314 J/mol K(1/873-1/278))

= exp (-5520.81*(0.0011-0.0036))

= exp13.802025

Question 3: (LO1 chemistry)

Consider a small sample of putrescene C4H12N2

1. 
   a)    Assuming that putrescence burns completely; calculate the
   stoichiometric fuel-to-air ratio.
   b)    Calculate the maximum yields of.
   c)    Calculate the maximum yields of and.
   d)    calculate the maximum depletion of.
   
   

SOLUTION

Putrescene (C4H12N2)

Burning of Putrescene

C4H12N2 (s) + 7O2 (g)                                                                                      4CO2 (g) + 6H2O (g) + N2 (g)

The RMM of Putrescene is

4(12.011) + 12(1.0079) + 2(14.0067) = 48.044 + 12.0948 + 28.0134 = 88.1522

= 88.1 amu (3 s.f.)

Taking a sample of 0.88 grams of Putrescene

With complete combustion of Putrescene, 1 mole of putrescene reacts with 7 moles of oxygen to yield 4 moles of Carbon dioxide, 6 moles of water, and 1 mole of nitrogen gas.

Thus, the ratio of C4H12N2: O2: H2O: CO2: N2 is 1:7:6:4:1

The molar ratio is:

C4H12N2: O2: CO2: N2 = 1:7:4:1

Taking into consideration the composition of air, 21% O2 and 79 % N2

To calculate the amount of oxygen required for complete combustion of C4H12N2 completely;

Get the number of C4H12N2 in moles,

From the sample of 0.88g of C4H12N2

Moles of C4H12N2 in 0.88g = 0.88g÷88.1a.m.u. = 0.0099886493

                        = 0.01moles of C4H12N2

Therefore; Number of moles of O2 consumed in the combustion = 0.01 × 7 = 0.07 moles of O2 gas

                        Mass of O2 gas = 0.07 mol × 2 × 15.9994g/mole = 2.24grams

• Thus, the stoichiometric fuel-to-air ratio is given by

AFR = mass of air ÷ Mass of fuel =2.24 ÷ 0.88 = 2.5:1 or 5:1

• A number of moles of CO2 yield = 01 × 4 = 0.04 moles of CO2 gas.

                                       Mass of CO2 gas = 0.04 mol × (31.9988 + 12.011) g/mole

0.04 mol × 44.0099g/mole = 1.76grams

• Number of moles of N2 yield = 0.01 × 1 = 0.01moles of N2

                        Mass of N2 gas = 0.01 mol × 2 × 14.0067g/mole = 0.28grams          

• Maximum depletion of air is;

Ratio of air to fuel × mass of fuel

2.5 × 0.88 = 2.2 grams

Question 4

Compartment dimensions = 15m by 9m by 2.8m

Doorway = 0.8m by 2.1m

Walls = 4cm thin

Density = 400kg/m3

Specific heat capacity = 913J/Kg/K

Heat transfer coefficient =?

Total area = wall area+ floor area + ceiling area – openings

= (15m*9m*2) + (4*2.8m*9m) – (0.8m*2.1m) = 370.8m2-1.68 m2

= 369.12m2

Thermal penetration time is given by: