118
4 Mine Ventilation Networks
Solution
Let us start by calculating the equivalent resistance (R 3 ) for parallel branches A
and B:
1
R 3
=
1
R A
+
1
R B
Substituting values we have that:
1
R 3
=
1
0.7
N s 2
m 8
+
1
3.1
N s 2
m 8
Therefore:
R 3 = 0.3216
N s
2
m 8
Then, the total resistance (R T ) is:
R T = R 1 + R 3 + R 2
R T = (0.25 + 0.3216 + 0.25)
N s
2
m 8 = 0.8217
N s
2
m 8
Thus, the total airflow is (Q T ):
Q T =
P T
R T
=
600 Pa
0.8217
N s 2
m 8
= 27.022
m
3
s
According to the mass conservation principle:
Q T = Q A + Q B = 27.022
m
3
s
The pressure loss in the parallel branch (with equivalent resistance R 3 ) can be
calculated as:
P 3 = R 3 Q 2
T = 0.3216
N s 2
m 8
27.022
m 3
s
2
= 234.83 Pa
Given that pressure loss is the same for all parallel branches:
ΔP 3 = ΔP A = ΔP B
4 Mine Ventilation Networks
Solution
Let us start by calculating the equivalent resistance (R 3 ) for parallel branches A
and B:
1
R 3
=
1
R A
+
1
R B
Substituting values we have that:
1
R 3
=
1
0.7
N s 2
m 8
+
1
3.1
N s 2
m 8
Therefore:
R 3 = 0.3216
N s
2
m 8
Then, the total resistance (R T ) is:
R T = R 1 + R 3 + R 2
R T = (0.25 + 0.3216 + 0.25)
N s
2
m 8 = 0.8217
N s
2
m 8
Thus, the total airflow is (Q T ):
Q T =
P T
R T
=
600 Pa
0.8217
N s 2
m 8
= 27.022
m
3
s
According to the mass conservation principle:
Q T = Q A + Q B = 27.022
m
3
s
The pressure loss in the parallel branch (with equivalent resistance R 3 ) can be
calculated as:
P 3 = R 3 Q 2
T = 0.3216
N s 2
m 8
27.022
m 3
s
2
= 234.83 Pa
Given that pressure loss is the same for all parallel branches:
ΔP 3 = ΔP A = ΔP B