92
4 Mine Ventilation Networks
(b) For the bare rock:
P = 0.015
N s
2
m 4
4.71 m · (85) ·
50
m
3
s
2
1.76 m 2
3
1.1
1.2
= 2494.43 Pa
Note that the relationship between the two pressure losses is the same as the one
between their Atkinson friction factors.
Exercise 4.3 A circular duct is traversed by a flow of 180 m
3 s
−1 of air under a
pressure difference between its extremes of 2000 Pa. Determine the flow through a
circular duct of the same material as the previous one, half radius and double length,
when the pressure difference between the ends of the duct becomes 7000 Pa. Solve
this using the Atkinson equation.
Solution
Particularizing the Atkinson expression for a circular duct, one has:
P = K
O L Q
2
A 3 =
2πr L Q
2
πr 2
3 =
2πr L Q
2
π 3 r 6 =
2L Q
2
π 2 r 5
Applying it to the two proposed cases 1 and 2:
P 1 =
2L 1 Q
2
1
π 2 r
5
1
P 2 =
2L 2 Q
2
2
π 2 r
5
2
Dividing both expressions and simplifying:
P 1
P 2
=
2L 1 Q
2
1
π 2 r
5
1
2L 2 Q
2
2
π 2 r
5
2
=
L 1
r
5
1
Q
2
1
L 2
r
5
2
Q
2
2
You have:
P 2 =
L 2
r
5
2
Q
2
2
L 1
r
5
1
Q
2
1
P 1
Particularizing:
P 2 =
2
6 Q
2
2
Q
2
1
P 1