Lecture: 01
Fire and Explosion
Dharmsinh Desai University, Nadiad
Gujarat - India
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Three most common chemical plant accidents are
Fires Explosion Toxic release
Yearly losses due to fires and explosions are substantial.
To prevent accidents due to fires and explosions, engineer
must be familiar with ....
The fires and explosion properties of materials.
The nature of the fire and explosion process.
The procedures to reduce fire and explosion hazard.
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Fire: (According to ISO):
- Fire is a process of combustion, characterized by the emission of heat
accompanied by smoke or f lame or both.
- Fire is a process of oxidation.
Fire Triangle:
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When fuel, oxidizer and ignition source are present at the necessary
levels/proportion, burning will occurs.
Fire will not occur if – 1. Fuel is not present or is not present in sufficient
quantities . 2. Oxidizer is not present or is not present in sufficient
quantities. 3. The ignition source is not energetic enough to initiate the
fire.
Various fuels, oxidizer and ignition sources:
Fuels: Oxidizers: Ignition sources:
Liquids: Gasoline, Acetone, Liquids: Hydrogen peroxide, - Sparks,
Ether, Pentane etc. Nitric acid, Perchloric acid - Flames,
- Static electricity,
Solids: Plastics, Wood dust, Solids: Metal peroxides, - Heat
Fibers, Metal particles etc. Ammonium nitrite
Gases: Acetylene, Propane, Gases: Oxygen, Fluorine,
carbon monooxide, Chlorine.
Hydrogen etc. 4
Classification of Fires:
Class (A) fire: Class (A) fires are those which involves cellulosic
materials like wood, paper, cloth etc.
Class (B) fire: Class (B) fires are those which involves f lammable
hydrocarbonic liquids like Kerosene, Petrol etc.
Class (C) fire: Class (C) fires are those which involves electrical
equipments like electric moter, server, computer etc.
Class (D) fire: Class (D) fires are those which involves solids such as
Sodium, potassium and zirconium etc.
Class (K/F) fire: Class (K) fires are those which involves animals and
vegetable fats such as cooking oils and greases.
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Some of the commonly used definitions related to fires:
Combustion or Fire: Combustion or Fire is a chemical reaction in which a
substance combines with an oxidant and releases energy. Part of the energy
released is used to sustain the reaction.
Ignition: Ignition of a f lammable mixture may be caused by a f lammable
mixture coming in contact with a source of ignition with sufficient energy
or the gas reaching a temperature high enough to cause the gas to autoignite.
Auto ignition temperature: A fixed temperature above which adequate
energy is available in the environment to provide an ignition source.
Flash point: The f lash point of a liquid is the lowest temperature at which
it gives off enough vapor to form an ignitable mixture with air and resulted in
a momentary f lash.
Fire point: The fire point is the lowest temperature at which a vapor above
a liquid will continue to burn once ignited; the fire point temperature is
higher than the f lash point.
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Flammability limits (LFL and UFL):
Flammability limits (e.g. LFL & UFL) provides us with the range of fuel
concentration (normally in vol %) within which a gaseous mixture can
ignited and burned.
Means a mixture (fuel vapor + air) is f lammable only when the
composition is between the LFL and UFL. As shown in figure below.
UFL
Flammable mixture
LFL
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LFL: Below the LFL, there is not enough fuel to cause ignition. Example:
Mixture of 1% Methane in air at atmospheric pressure and ambient temp., is
placed in contact with a naked f lame. Result: The Methane will be gradually
consumed as the molecules cross the high temperature regions close to the
f lame, but the combustion will not propagate to the gaseous mixture, as
would occur if the mixture were within the f lammability limit.
UFL: If the fuel concentration greater than the UFL, there is insufficient
oxygen for the reaction to propagate away from the source of ignition.
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Fig. 1: Relationships between various f lammability properties
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Flammability characteristics of liquids and Gases/vapors:
Liquids:
The f lash point is one of the major physical properties used to
determine the fire and explosion hazards of liquids.
The f lash points for pure components are easily determined
experimentally (e.g. Open cup methods or Closed cup methods)
The Flash points can be estimated for multicomponent mixtures if
only one component is f lammable and if the f lash point of the
f lammable component is known.
In this case the FP temperature is estimated by determining the
temperature at which the partial pressure in vapour of the f lammable
component in the mixture is equal to the pure component vapour
pressure at its FP.
Raoult's law ( )is used to compute the VP pure component,
based on the partial pressure required to f lash. 10
Numerical Example: 01
Methanol has a f lash point of 54 °F, and its vapor pressure at this
temperature is 62 mm Hg. What is the f lash point of a solution
containing 75% methanol and 25% water by weight?
Solution:
The mole fractions of each component are needed to apply Raoult's law.
Assuming a basis of 100 lb of solution, we can construct the following:
Raoult's law:
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Using a graph of the VP versus temperature, shown in Fig. Below, the f lash
point of the solution is 20.5°C, or 68.9°F.
Fig. 2: Saturation vapor pressure for methanol
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Gases and Vapour:
The Flammability limits for vapours are determined experimentally in a
specially designed closed vessel apparatus as shown in Fig. below;
Step-1: Vapor-air mixtures of known
concentration are added and then
ignited.
Step-2: The maximum explosion
pressure is measured.
Step-3: This test is repeated with
different concentrations to establish the
range of f lammability for the specific
gas.
Fig. 3: Test apparatus for acquiring vapor explosion data
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Example: Methane –Air mixture
Fig. 4: Maximum pressure for methane combustion in a 20-L sphere.
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Gases/Vapour mixture:
LFLs and UFLs for mixtures are computed using the Le Chatelier
equation as given below:
... (1) ... (2)
LFLi = Lower f lammable limite for component (i) in volume % of component (i)
in fuel and air.
UFLi = Upper f lammable limit for component (i) in volume % of component (i) in
fuel and air.
yi = Mole fraction of component (i) on a combustible basis.
n = Number of combustible species.
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Le Chatelier's equation is empirically derived and is not universally
applicable.
Mashuga and Crowl derived Le Chatelier's equation using
thermodynamics. The derivation shows that the following assumptions
are inherent in this equation and these assumptions were found to be
reasonably valid at the LFL and less so at the UFL .
The product heat capacities are constant.
The number of moles of gas is constant.
The combustion kinetics of the pure species is independent and
unchanged by the presence of other combustible species.
The adiabatic temperature rise at the f lammability limit is the
same for all species
Proper usage of Le Chatelier's rule requires f lammability limit data at
the same temperature and pressure.
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Numerical Example: 02
What are the LFL and UFL of a gas mixture composed of 0.8% Hexane,
2.0% Methane, and 0.5% Ethylene by volume?
Solution:
The mole fractions on a fuel-only basis are calculated in the table. The LFL and UFL
data are given.
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Eq. (1) is used to determine the LFL of the mixture:
Eq. (2) is used to determine the LFL of the mixture:
12.9%
Flammable = 3.3%
2.75%
Because the mixture contains 3.3% total combustibles, it is f lammable. 18
Flammability Limit Dependence on Temperature:
In general, the f lammability range increases with temperature. The
following empirically derived equations are available for vapors:
... (3)
... (4)
ΔH = Net heat of combustion (kcal/mole)
T = Temperature (oC)
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Flammability Limit Dependence on Pressure:
Pressure has little effect on the LFL except at very low pressures (40 mm
Hg absolute), where f lames do not propagate.
The UFL increases significantly as the pressure is increased, broadening
the f lammability range.
An empirical expression for the UFL for vapors as a function of pressure
is given below;
... (5)
P = The pressure (mega pascals absolute) and
UFL = The upper f lammable limit at 1 atm.
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Numerical Example: 03
If the UFLfor a substance is 11.0% by volume at 0.0 MPa gauge, what is the
UFL at 6.2 MPa gauge?
Solution:
The absolute pressure is P = 6.2 + 0.101 = 6.301 MPa.
Now, the UFL is determined using Eq. (5).
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BOOKS AND REFERENCES:
1. Crowl D.A. and Louvar J.F., Chemical Process Safety: Fundamentals With
Applications.
2. Lees F.P. Lee’s Loss Prevention in Process industries: Hazard Identification,
Assessment and control
3. Kletz T, What Went Wrong? Case Histories of Process Plant Disasters: How
They Could Have Been Avoided
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