Mechanics of Fluids

straightforward manner without iteration. Specially prepared charts, based
on these alternative dimensionless groups, are available.
From here onwards u represents the mean velocity over the cross-section,
and for simplicity the bar over the u is omitted.
Example 7.1 Determine the head lost to friction when water
flows through 300 m of 150 mm diameter galvanized steel pipe
at 50 L · s −1 .
Solution
For water at, say, 15 ◦ C, ν = 1.14 mm 2 · s −1 .
u = 50 × 10−3m3 · s−1 (π/4)(0.15) 2 = 2.83 m · s−1
m2 Re = ud
ν = 2.83 m · s−1 × 0.15 m
1.14 × 10−6 m2 = 3.72 × 105
· s−1 For galvanized steel k = 0.15 mm, say. ∴ k/d = 0.001
From Fig. 7.4 f = 0.00515 so
h f =
4 × 0.00515 × 300 m
0.15 m
(2.83 m · s−1 ) 2
= 16.81 m, say 17 m ✷
19.62 m · s−2 Example 7.2 Calculate the steady rate at which oil (ν =
10 −5 m 2 · s −1 ) will flow through a cast-iron pipe 100 mm diameter
and 120 m long under a head difference of 5 m.
Solution
As yet Re is unknown since the velocity is unknown. For cast iron (in
new condition) k = 0.25 mm, say. Hence k/d = 0.0025 and Fig. 7.4
suggests f = 0.0065 as a first trial. Then from eqn 7.1
5m=
4 × 0.0065 × 120 m
0.10 m
whence u = 1.773 m · s −1 . Therefore
Re = 1.773 m · s−1 × 0.10 m
10 −5 m 2 · s −1
u 2
19.62 m · s −2
= 1.773 × 10 4
These values of Re and k/d give f = 0.0079 from Fig. 7.4. A recalculation
of u gives 1.608 m · s −1 , hence Re = 1.608 × 10 4 . The
corresponding change of f is insignificant. The value u = 1.608 m · s −1
Variation of friction factor 255