120
4 Mine Ventilation Networks
Solution
(a) Calculation of the equivalent resistance.
R 2 − R 3 are in parallel:
1
√
R 23
=
1
√
R 2
+
1
√
R 3
1
√
R 23
=
1
12
N s 2
m 8
+
1
13
N s 2
m 8
R 23 = 3.121
R 23 − R 4 are in series:
R 234 = R 23 + R 4 = (3.121 + 3)
N s
2
m 8 = 6.121
N s
2
m 8
R 5 − R 6 are in series:
R 56 = R 5 + R 6 = (6.5 + 6)
N s
2
m 8 = 12.5
N s
2
m 8
R 234 − R 56 are in parallel:
1
√
R 23456
=
1
√
R 234
+
1
√
R 56
=
1
6.121
N s 2
m 8
+
1
12.50
N s 2
m 8
R 23456 = 2.121
N s
2
m 8
R 1 − R 23456 − R 7 are in series:
R eq = R 1234567 = R 1 + R 23456 + R 7 = (4 + 2.121 + 2)
N s
2
m 8 = 8.121
N s
2
m 8
(b) Once the equivalent resistance of the circuit (R eq ) has been calculated and
since the total pressure loss (P T ) is known, the total airflow rate (Q T ) can
be calculated as:
Q T =
P T
R eq
=
1200 Pa
8.12
N s 2
m 8
= 12.16
m
3
s
(c) Calculation of pressure losses for the airway resistances in series.