24
1 Fundamental Concepts of Fluid Mechanics for Mine Ventilation
Simplifying:
P
ρg
= −z 1 + f L
16Q
2
2π 2 g D 5
Since:
sin 15
◦
=
z 1
L
Therefore:
P =
f
z 1
sin 15 ◦
16Q
2
2π 2 g D 5 − z 1
ρg = 265.54 Pa
Note: Note that if the fluid was water instead of air, the difference in pressure due
to height would become more important due to its greater specific weight.
Exercise 1.8 The attached table provides the values corresponding to the flow of an
air stream through a duct. Calculate the value of the coefficient of friction using the
expressions of von Karman–Prandtl, Nikuradse, Colebrook, USBM and Haaland.
Variable
Symbol
Units
Value
Diameter
D
m
0.5
Flow
Q
m 3 s −1
0.9
Absolute roughness
E
m
0.0015
Air density
ρ
kg m −3
1.2
Dynamic viscosity
μ
N s m −2
0.0000186
Mean velocity
v
m s −1
4.58366
Reynolds number
Re
–
147,860
Solution
Solutions search by iterative calculations are simplified by making
1
√
f
= A and
searching for the value of that makes f (x) = A − f (A, parameters) = 0. Therefore:
Using von Karman–Prandtl’s formula (Eq. 1.25) (turbulent flow and smooth duct)
we have:
1
√
f
= −2.0 log
2.51
Re
√
f
A = 7.76057; f (x) = −9.765 × 10
−7
f = 0.01660