Question

The lower flammability limit (LFL) and upper flammability limit (UFL) are important safety limits which describe the minimum and maximum volume fraction in air, respectively, that a flammable substance can burn if ignited. If the range of concentration of benzene in air in which ignition could take place is 1.4 - 8.0 vol%, what would be the corresponding temperatures for air saturated with benzene in the vapour space of a storage tank assuming a total pressure in the vapour space of 100 kPa?

Answer 1

For air saturated with benzene, partial pressure of benzene would be equiavlent to vapor pressure of benzene in air.

a) Now for LFL, vol% of benezne = 1.4 mol%, partial pressure of benzene = total pressure * mol fraction of benzene

since for vapor phase, mole fraction of a component is same as volume fraction of the component.

partial pressure of benzene = 100*1.4/100 = 1.4 kPa or 1.4*7.5 mmHg = 10.5 mmHg

Now using antonie equation for benzene, T (⁰C)= B/(A-Log₁₀p) -C where p is in mmHG, for benzene we know from literature that A = 6.88, B = 1196.8 & C = 219.16

On solving the above equation for p = 10.5 mm Hg, T = -14.88 DegC

a) Now for HFL, vol% of benezne = 8 mol%, partial pressure of benzene = total pressure * mol fraction of benzene

since for vapor phase, mole fraction of a component is same as volume fraction of the component.

partial pressure of benzene = 100*8/100 = 8 kPa or 8*7.5 mmHg = 60.0 mmHg

Now using antonie equation for benzene, T (⁰C)= B/(A-Log₁₀p) -C where p is in mmHG, for benzene we know from literature that A = 6.88, B = 1196.8 & C = 219.16

On solving the above equation for p = 60.0 mm Hg, T = -15.4 DegC