4.2 Friction Losses: Atkinson Expression
91
Thus indicating fully turbulent flow.
The relative roughness is:
ε
D h
=
800 × 10
−3 m
5.318 m
= 0.1504
The Re indicates that turbulent flow and since the conduit is rough, the Nikuradse
formula applies. Thus, substituting in Nikuradse expression we have:
1
√
f
= −2.0 log
0.1504
3.71
f = 0.0967
Then the Atkinson friction factor is:
K 1.2
N s
2
m 4
= 0.15 f = 0.0145
(b) In mining galleries, conditions f > 0.005 and Re > 10
6 usually occur. Under
these conditions, if we look for example at the Moody diagram, the coefficient
of friction becomes independent of Re.
Exercise 4.2 Calculate the pressure loss taking place due to friction in an 85-m long
raise of circular cross section and 1.5 m diameter when an anemometer installed in
it registers an airflow rate of 50 m
3 s
−1 . Assume that surface of the raise is:
(a) Shotcrete
K 1.2 = 0.004
N s
2
m 4
, and
(b) Bare rock
K 1.2 = 0.015
N s
2
m 4
.
The air density is 1.1 kg m
−3 .
Solution
(a) Applying the Atkinson equation with L eq = 0 (there are no accessory elements
or relevant cross-sectional changes):
P = K 1.2
O
L + L eq
Q
2
A 3
⎛
⎝ ρ
1.2
kg
m 3
⎞
⎠
For shotcrete:
P = 0.004
N s
2
m 4
4.71 m · (85) ·
50
m
3
s
2
1.76 m 2
3
1.1
1.2
= 665.18 Pa