106
4 Mine Ventilation Networks
ρ=1.15 kg m
-3
Q=80 m
3 s
-1
Solution
(a) Resistance due to friction:
R =
K O L
A 3
⎛
⎝ ρ
1.2
kg
m 3
⎞
⎠ =
0.012
Ns
2
m 4 · 16 m · 500 m
15 3 m 6
·
1.15
1.2
= 0.0273
N s
2
m 8
(b) Resistance due to shock in the curve union:
X = 0.86
R shock =
Xρ
2 A 2 =
0.86 · 1.15
kg
m 3
2 · (15 m 2 ) 2 = 0.00219
N s
2
m 8
(c) The hydraulic diameter (D h ) is determined as:
D h =
4 A
O
=
4 · (5 · 3) m
2
(2 · 5 + 2 · 3)m
= 3.75 m
The equivalent length is, therefore:
L eq =
1.2
kg
m 3
X
8K
D h =
1.2
kg
m 3 · 0.86
8 · 0.012
kg
m 3
· 3.75 m = 40.31 m
(d) The total aerodynamic resistance (R T ) is obtained as:
R T =
K O
L + L eq
A 3
⎛
⎝
ρ
1.2
kg
m 3
⎞
⎠