Page 153 - Electrical Installation in Hazardous Area

Calculation of release rates and extents 1 19

and this amount is given by Equation 4.27 as follows:

G, = 1.81(288 - 225)2.26/469
G, = 0.55
This gives the liquid which is assumed to reach the ground as:

GI = 1.81 - 0.5.5 = 1.26 kg/s

That vaporizing at the orifice will behave as a vapour and will be jet
dispersed in accordance with Equation 4.6 as follows:-

X = 2.1 x 103(0.56/22 x 421.5 x 288°.5)0.5 m
X = 11.5 (say 12) m
Due to the nature of the release it is necessary to assume that some mist
will occur and an allowance will be made for this by increasing the distance
by 50%. Thus:
x = 18 m
The mass of liquid reaching the ground will be assumed to vaporize
immediately and thus no pool will form. The vapour released will be wind
dispersed and Equation 4.7 is appropriate. In this case no mist will be
assumed.
X = lOB(1.29 x 288142 x 2)0.55 m
X = 24.5 m

The distance travelled by the jet before it reaches the ground is calculated
as before. There will be a hazardous area extending 24.5 m plus the distance
travelled by the jet before it reaches the ground in a horizontal plane, and
there will be a hazardous area around the source of release extending in
all above horizontal directions for 18m which will be projected vertically
down to the ground hazardous area.

4.4 Summary of use of equations

The foregoing equations are used as follows:

4.4.1 Gas and vapour releases

First determine the mass release of gas or vapour using Equation 4.1 if the
release pressure is greater than 2 x lo5 N/m2, and Equation 4.3 otherwise.
Then convert this to a volume using Equation 4.4 If the release pressure is
greater than 2 x 1 - O5 N/m2 calculate the extent of hazardous area using
Equation 4.6. If the release pressure is less than 2 x lo5 N/m2 calculate