Page 139 - Electrical Installation in Hazardous Area

Calculation of release rates and extents 105

vertical extents of explosive atmosphere should be doubled wherever the
density is more than 1.5 times the density of air, or less than 0.7 times the
density of air. This will mean, for instance, doubling the upward extent for
such gases as hydrogen in such circumstances and doubling the downward
direction for such gases as pentane - indeed this latter will apply for most
of the hydrocarbon gases

4.1.1 Examples of gas and vapour release

Example 1

A pipe flange gasket fails producing an orifice of 4 x lo-’ m2. The contained
gas is ethylene (M = 28) and the pressure inside the pipe is 4 x 10’ N/m2.
Ambient temperature is 22°C. The contained gas is at ambient tempera-
ture. Assuming the jet direction cannot be defined with confidence, what
will be the hazardous area surrounding the pipejoint? The lower explosive
limit of ethylene in air at 22°C temperature and atmospheric pressure is
2.7 per cent.
The release pressure is in excess of critical pressure and thus:
G = 0.006aP (M/T)0.5 kg/s (Equation 4.1 )
G = 0.006 x 4 x lo-’ x 3 x 105 (28/295)O.’ kg/s

G = 0.03 kg/s
Dispersal is by jet energy and so distance to LEL is:
X = 2.1 x 103(G/E2 x M’.’ x T0.5)0.55 m (Equation 4.6)

X = 2.1 x 103(0.03/2.7* x 28l.’ x 295°.5)0.55 m
X = 2.7 m

The hazardous area will be a sphere extending 2.7m from the source of
release in all directions.

Example 2

An obstruction is present at a horizontal distance of lm from the leak
location described in example 1. What will the extent of the hazardous are
be? The mass release remains at 0.03 kg/s as in the first example and it is
necessary to calculate the amount of gas in the mixture at the obstruction
as follows:
B = 100 - (100 - LEL) L/X %(Equation 4.8)

B = 100 - (100 - 2.7)1/27 Yo
B = 64%