Question

A large cloud of butane is released and eventually ignited, producing a VCE. Estimate the quantity of butane released if the blast shatters windows 2 miles from the source of the ignition.

(Make sure its easy to read)

Answer 1

Given: Distance from zero-ground point of explosion, r=2 miles=3218.69 m

From the Table Damage Estimates for Common Structures based on over pressure, obtain the over pressure value for the shattering of windows.

Over pressure range is 0.5-1 psig. Consider an average value of 0.75 psig.

Over pressure, p₀=0.75 psig

The scaled overpressure, p_s, is

where p_o is the Peak side-on over pressure

p_a is the Ambient pressure, p_a=14.7 psig

Obtain the scaled distance from the Correlation between scaled distance and explosion peak side-on overpressure for a TNT explosion occuring on a flat surface.

For p_s=0.051, scaled distance, z_e=32 m/kg^1/3

Equivalent mass of TNT, m_TNT, is obtained using the expression

Cubing both sides

where r is the Distance from zero-ground point of explosion

z_e is the Scaled distance

Mass of butane, m, is obtained using the expression

where E_TNT is the Energy of explosion of TNT, E_TNT=4686 kJ/kg

is the Empirical explosion efficiency. The efficiency for a VCE is between 1%-10%. Assume =5%=0.05

H_c is the Energy of explosion of butane

Lower heating value of combustion is used for energy of explosion.

Obtain the lower heating value of combustion from the Flammability Data of Selected Hydrocarbons.

For butane,  H_c=2657.5 kJ/mol

where 58 g/mol is the Molar mass of butane

Thus the mass of butane is

Thus the quantity of butane released is m=2.08*10⁶ kg