122
4 Mine Ventilation Networks
23 = R 23 Q
2
234 = 3.12
N s
2
m 8
7.16
m
3
s
2
= 159.83 Pa
4 = R 4 Q
2
234 = 3
N s
2
m 8
7.16
m
3
s
2
= 153.64 Pa
If R 23 is broken down we see that 23 = 2 = 3 because we have a parallel
arrangement of resistances. Thus, it is possible to calculate the airflow rates traversing
R 2 and R 3 :
R1
R2
R3
R4
R56
R7
Q 2 =
P 2
R 2
=
159.83 Pa
12
N s 2
m 8
= 3.65
m
3
s
Q 3 =
P 3
R 3
=
159.83 Pa
13
N s 2
m 8
= 3.5
m
3
s
Finally, the resistance R 56 can be broken down. As R 5 and R 6 are in a series
arrangement, the airflow Q 56 is the same through both and pressure losses 5 and
6 can be calculated, thus:
R1
R2
R3
R4
R5
R6
R7
5 = R 5 Q
2
56 = 6.5
N s
2
m 8
5
m
3
s
2
= 162.5 Pa
6 = R 6 Q
2
56 = 6
N s
2
m 8
5
m
3
s
2
= 150 Pa