90
4 Mine Ventilation Networks
We finally obtain (Eq. 4.8):
P = K 1.2
⎛
⎝
ρ
1.2
kg
m 3
⎞
⎠
O
L + L eq
Q
2
A 3
(4.8)
Exercise 4.1 Consider there to be a 6.5 m × 4.5 m rectangular airway of 800 mm
of absolute roughness traversed by an airflow of 30 m
3 s
−1 . You are asked to:
(a) Determine the Atkinson friction factor of the airway.
(b) Comment on the extent to which the results obtained are close to those expected
in a mining gallery.
Data:
Nikuradse expression:
1
√
f
= −2.0 log
ε
D h
3.71
where
• E: Absolute roughness,
• D h : Hydraulic diameter (m),
• Air density: 1.16
kg
m 3 , and
• Air dynamic viscosity: 1.86 · 10
−5 N ·
s
m
2 .
Solution
(a) The hydraulic diameter is calculated as:
D h =
4 · 6.5 · 4.5
2 · 6.5 + 2 · 4.5
m = 5.318 m
The average air speed can be calculated as:
v =
Q
A
=
30
m
3
s
6.5 · 4.5 m 2 = 1.025
m
s
With this value and the dynamic viscosity of the air, we obtain:
Re =
ρ D h v
μ
=
1.16
kg
m 3 · 5.318 m · 1.025
m
s
1.86 × 10 −5 N
s
m 2
= 33,9951.72
4 Mine Ventilation Networks
We finally obtain (Eq. 4.8):
P = K 1.2
⎛
⎝
ρ
1.2
kg
m 3
⎞
⎠
O
L + L eq
Q
2
A 3
(4.8)
Exercise 4.1 Consider there to be a 6.5 m × 4.5 m rectangular airway of 800 mm
of absolute roughness traversed by an airflow of 30 m
3 s
−1 . You are asked to:
(a) Determine the Atkinson friction factor of the airway.
(b) Comment on the extent to which the results obtained are close to those expected
in a mining gallery.
Data:
Nikuradse expression:
1
√
f
= −2.0 log
ε
D h
3.71
where
• E: Absolute roughness,
• D h : Hydraulic diameter (m),
• Air density: 1.16
kg
m 3 , and
• Air dynamic viscosity: 1.86 · 10
−5 N ·
s
m
2 .
Solution
(a) The hydraulic diameter is calculated as:
D h =
4 · 6.5 · 4.5
2 · 6.5 + 2 · 4.5
m = 5.318 m
The average air speed can be calculated as:
v =
Q
A
=
30
m
3
s
6.5 · 4.5 m 2 = 1.025
m
s
With this value and the dynamic viscosity of the air, we obtain:
Re =
ρ D h v
μ
=
1.16
kg
m 3 · 5.318 m · 1.025
m
s
1.86 × 10 −5 N
s
m 2
= 33,9951.72