Page 149 - Water Engineering Hydraulics, Distribution and Treatment

Solution 2 (SI System):
Velocity, v = Q∕A
3
= (0.0445 m ∕s)∕[(0.300 m) × 0.785]
= 0.62 m∕s.
Reynolds number, R = vd ∕
= (0.62 m∕s)(0.300 m)(0.851 × 1,000 kg∕m )∕(0.1N-s∕m )
= 1, 580 < 2,000, which indicates a laminar flow.
Friction factor, f = 64∕R under laminar flow conditions
= 64∕1,580
= 0.041. 2 3 2 5.2 Fluid Transport 127
2
Head loss, h = f(L∕d)(v ∕2g)
f
2
2
= (0.041)(3,048 m∕0.300 m)(0.62 m∕s) ∕(2 × 9.81 m∕s )
= 8.2m.
EXAMPLE 5.22 RATIO OF HEAD LOSS IN A PIPE TO THAT IN A PERFORATED PIPE
Show that the head loss h in a pipe is equal to three times the head loss h in a perforated pipe, having the same length, diameter, and
f
f
friction factor.
Take the flow in the imperforated pipe as Q ; assume a straight-line variation of the flow Q with distance in the perforated pipe,
0
with Q = Q at the inlet of the pipe and Q = 0 at the end of the line. Consider only losses to pipe friction, and assume no variation
0
in the value of f with a changing Q.
Solution:
2
h = f(L∕d)(v ∕2g) (5.10a)
f
Since
v = Q∕A (5.4)
then
2
2
h = f(L∕D)(Q ∕2gA ) (5.16)
f
2
2
h = (f∕D)(1∕2gA )Q L (5.17)
f
2
h = KQ L
f
where
2
K = (f∕D)(1∕2gA ).
dx
Q 0
Q x
x
I

Taking a small length along the perforated pipe (see sketch) = dx,then
L
2
h = K ∫ Q dx.
f
x
0