Page 49 - Advanced Mine Ventilation

32 Advanced Mine Ventilation

1

2
2 H 3
H 3 ¼ R 3 Q 3 or Q ¼
3
R 3
1 1 1
2 2 2
H 1 H 2 H 3
Hence Q ¼ Q þ Q þ Q ¼ þ þ
2
1
3
R 1 R 2 R 3
Hence H 1 ¼ H 2 ¼ H 3 ¼ H.
If R eqv is the equivalent resistance for network of three airways,
H ⎡ 1 1 1 ⎤
Q= = H ⎢ + + 0.5 ⎥
R eqv ⎣ R 1 0.5 R 2 0.5 R 3 ⎦
Hence
1 1 1 1
¼ þ þ (2.21)
0.5 0.5 0.5 0.5
R eqv R 1 R 2 R 3
An example:
Three airways in (Fig. 2.3B) are in parallel with a total air quantity of 90,000 CFM.
Their resistances are

in. min 10
2 10
R 1 ¼ 10
ft 6
2 10
in. min 10
R 2 ¼ 15
ft 6
2 10
in. min 10
R 3 ¼ 20 .
ft 6
Calculate the quantity of air in each split and the head loss.

⎡ ⎤ 2
⎢ 1 ⎥
R eqv = ⎢ 1 1 1 ⎥ × 10 -10
⎢ 10 + ⎢ 15 + 20 ⎥ ⎣ ⎥ ⎦

2
= 1.574 × 10 -10 inch-min /ft 6

Hence

2
H ¼ R eqv Q ¼ 1:574 10 10 ð90; 000Þ 2
2
¼ 1:574 81 10 ¼ 1:275 in.