Page 47 - Advanced Mine Ventilation

30 Advanced Mine Ventilation

Thus the EO for the mine in the previous problem:

-4
3.9 × 10 × 250,000
EO = = 48.75 ft 2
4
Eq. (2.19) with slight modification can also be used to calculate the size of a regu-
lator in a mine airway. The regulator is needed to restrict the airflow in a given split to
deliver a fixed quantity of air.
The EO of the regulator that can dissipate a head of H is first determined by
Eq. (2.19), and then a correction is made for contraction factor, K 2 . An example
will illustrate it better.
2
Given Q ¼ 100,000 CFM, A ¼ 80 ft , find the size of a square regulator to drop
two inches of W.G.
Hence
-4
3.9 × 10 (100,000)
EO =
2K
2
=
EO 17.5 ft 2

where K 2 is equal to 2.5. Contraction factors, K 2 , for a number of edges of a regulator
are listed below [8].

Type of Edge K 2
Rounded 1.5
Smooth 2.0
Square 2.5
Sharp 3.8

2.8 Ventilation Airways in Series/Parallel

In a ventilation system, two basic combinations of airways often arise: in series or in
parallel. These can be analyzed mathematically, but the entire network analysis will
need a computer. Computer simulation of a mine ventilation network will be discussed
later in the text.
Fig. 2.3A shows three airways in series, and Fig. 2.3B shows three airways in
parallel.

2.8.1 Airways in Series
Fig. 2.3A shows that the same air quantity flows through all three roadways. The total
head loss in the three airways is the sum of head losses in each roadway. If the