Page 148 - Electrical Installation in Hazardous Area

1 14 Electrical installations in hazardous areas

It is now necessary to know how far beyond the edge of (and above)
the pool the explosive atmosphere will extend and for this Equation 4.26 is
used giving:

X = {[5.8 x lo-’ x 17.3 x 0.22 x 295]/[2.1 x 19]}’.14
X = 1.75 (say 2) m

Thus there will be a hazardous area 2m high to a distance in all directions
of 12m from the point at which a vertical line from the leak touches the
ground. It is now necessary to consider the second assumption which is
that the entire release forms a mist but in this case, as the driving pressure
is so low, it can be ignored. Thus the following solution holds for this
circumstance: There will be no sigruficant hazardous area created around
the pipe where the leak occurs but from a point on the ground vertically
below the leak a 2 m high hazardous area will exist for a radius of 22 m (10
+ 10 + 2).

Example 4

With the same liquid transmission system and leak orifice as that described
in example 3, isoprene is transmitted at a gauge pressure of 2 x lo5 N/m2.
Temperature is 22°C as before.
Relevant parameters for isoprene are:

Liquid density = 680kg/m3
Vapour density = 3kg/m3
Boiling point = 34°C
Partial pressure = 7.1 x lo4 N/m’ (0.71 Atm.)
Molecular weight = 68
Lower explosive limit = l%inair.

The initial approach to this problem is exactly the same as that for
Example 3 in that the first action is to determine the mass rate of liquid
release using Equation 4.13
G = 1.13 x 4 x x [680 x 2 x 105]0.5 kg/s

G = 0.53 kg/s
It is now necessary to determine the distance travelled by the liquid jet
before it strikes the ground using Equation 4.15:

Velocity of Jet = 1.13(105/680)0.5 = 13.7 m/s
Distance = (2 x 13.7’ x 3/9.81)0.5 m
Distance = 10.7 (say 11) m
The pool will now form with its centre at a horizontal distance of 11 m
from a point on the ground vertically below the leak orifice. It is now