6.5 Precise Calculation of the Regulator Area
243
it is 60 Pa. Estimate the area of the regulator: (a) by the expression of the equivalent
orifice, (b) by the precise method and (c) determine the deviation between both
expressions.
Solution
(a) Directly applying Eq. 6.24:
A =
1.2
√ P
Q =
1.2
√
60 Pa
· 4.1
m
3
s
= 0.63 m
2
(b) The velocity is calculated as:
V =
Q
S
=
4.1
m
3
s
4.3 m 2 = 0.95
m
s
The dynamic pressure is, therefore:
P v =
1
2
ρV
2
=
1
2
1.2
kg
m 3
0.95
m
s
2 = 0.54 Pa
The shock loss factor can then be calculated as:
X =
P x
P v
=
60 Pa
0.54 Pa
= 111.11
Then, the coefficient N is:
N =
Z
X + Z + 2
√
X
=
2.5
111.11 + 2.5 + 2
√
111.11
= 0.136 (< 0.2)
Therefore, the area of the regulator is:
A 0 = N A a = 0.136 · 4.3 m
2
= 0.585 m
2
(c) The difference between the two approximations, expressed as a relative error
(E r ), is:
E r =
(0.63 − 0.585)
0.585
100 = 7.7%
Question 6.5 (a) Indicate the main consequences of an air velocity outside the
normal range for a gallery. (b) Specify the measures you would take if the mine you
are working in has a gallery with an excessive airflow speed.