Page 142 - Electrical Installation in Hazardous Area

108 Electrical installations in hazardous areas

Thus:

[h + (v sin @)’/2g] = 0.5 x g x ?
t2 = zh/g = (vsin@)’/$ S
t2 = = (v sin @)*/$10.5 S

Adding this to tl and rationalizing this, gives a total time t as follows:

t = [(Zgh + V2 sin’ @)O.’ + v sin @]/g S
As the horizontal element of velocity is vcos where @ is the angle of
release above horizontal, the distance travelled by the liquid jet neglecting
the friction of the air through which it moves will be;

= v cos @(2gh + V2 sin2 @)0.5 + v sin @]/g m (Equation 4.14)

where the jet is horizontal, sin @ = 0 and cos @ = 1 which simplifies the
equation to:
x = (2~2h/~)0.5 m (Equation 4.15)
It is recommended that this simplified equation is used wherever the jet
release direction is not specified, as the equations are idealized and it is
unlikely that the release will in fact be ideal. Therefore all that is produced
is an estimate and, because of the turbulence in a jet which is not an ideal
orifice, the horizontal figure will in most cases give a distance which is
larger than that which will practically occur. Where this is not so, however,
then Equation 4.14 should be differentiated for a maximum and the angle
giving the maximum introduced into Equation 4.14 to obtain the necessary
distance.
Where the jet is directed at an angle below the horizontal, the equation
changes to the following:

-
x = cos sin @ - ~~hy3.5 sin @]/g m (Equation 4.16)
From Equation 4.13:

Velocity = 1.13[(P - 105)/q]0.5 m/s (Equation 4.17)

The distance travelled by the jet before it reaches the ground can now be
calculated but it is difficult to define the angle which gives the maximum
range for a leak which is not directionally defined (the normal case), and
also it may not be possible on occasion to confirm that the jet will not be
ideal. To overcome this the differentiation approach described earlier will
be necessary. At the point where the jet lands a pool will form and, if not
physically contained, will increase in size until the evaporation rate from
its surface is equal to the release rate from the leak orifice.