4.3 Concept of Resistance of an Airway
97
Question 4.1 Given the von Karman equation for rough conduit:
1
√
f x
= 1.76 + 2.0 log
r
ε
(a) Demonstrate that the relation between the friction factor for any section (f x ) and
the friction factor a 10 m
2 gallery (f 10 ) is given by the expression:
f x =
f 10
(0.75 + 0.25 log A)
2
Assume circular cross section for the gallery and absolute roughness (ε) of 0.15 m.
(b) Determine how many times larger is the coefficient of friction of a gallery of
4 m
2 than that of a gallery of 12 m
2 .
Answer
(a) At the turbulence level of a mine gallery, the friction factor is independent of
Re, and therefore we can employ:
1
√
f x
= 1.76 + 2.0 log
r
ε
Particularizing for (a) f 10 and A 10 and (b) f x and A, and calculating the quotient,
for a circular cross section we have that:
1
√
f 10
1
√
f x
=
1.76 + 2.0 log
√
A 10
ε
√ π
1.76 + 2.0 log
√
A
ε
√ π
Thus, for A 10 = 10 m
2 and ε = 0.15 m:
f x
f 10
=
⎛
⎝
1.76 + 2.0 log
√
10
0.15
√ π
1.76 + 2.0 log
√
A
0.15
√ π
⎞
⎠
2
Operating we get:
f x =
f 10
(0.75 + 0.25 log A)
2
(b) Using the above expression, we have:
f 12 =
f 10
(0.75 + 0.25 log 12)
2