112
4 Mine Ventilation Networks
In order to double the flow, the fan power needs to be increased by 8 times.
Exercise 4.9 The main fan of a mine supplies an air flow rate of 50 m
3 s
−1 to overcome a total pressure loss of 3 kPa in the circuit (excluding outlet losses). Determine
the equivalent orifice of the mine and make some indications on its ease of ventilation.
Solution
By direct application of the expression of the equivalent orifice, we have:
A =
1.2
√
Q
A =
1.2
√
3000 Pa
· 50
m
3
s
= 1.07 m
2
As a result, it is difficult to ventilate this mine.
4.8 Resistances in Series
In a series arrangement, the circulating airflow (Q) is the same in all resistances and
the pressure loss ( is the sum of the losses in each one of them (Fig. 4.9).
So, mathematically:
= 1 + 2 + 3
Hence, applying Eq. 4.13 for each pressure differential, we have:
R eq Q
2
= R 1 Q
2
1 + R 2 Q
2
2 + R 3 Q
2
3
Given that:
Q = Q 1 = Q 2 = Q 3
Then:
R eq Q
2
= R 1 Q
2
+ R 1 Q
2
+ R 1 Q
2
Fig. 4.9 Association of
aerodynamic resistances in
series
R 1
R 2
R 3
Q
ΔP
ΔP 1
ΔP 2
ΔP 3