Page 145 - Water Engineering Hydraulics, Distribution and Treatment

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5.2 Fluid Transport
4. In the critical zone between R = 2,000 and R =
4,000, both R and ∕r make their appearance in
sion lines and pipe networks. Among the reasons are the
the semiempirical equation of Colebrook and White
following:
(1937–1938):
1. Because the relative roughness ∕r is a key to f,it
is not possible to find r (or d) directly when h , v
√
√
1∕ f = 1.74 − 2 log ( ∕r + 18.7∕R f)
f
or Q, , and water temperature (or ) are given. A
trial-and-error solution is required.
The magnitudes of absolute roughness in the f:R and
2. Because transmission lines may include noncircular
f:( ∕r) relationships depend on the angularity, height, and
as well as circular conduits, additional f:R diagrams
other geometrical properties of the roughness element and its
are needed.
distribution. Common magnitudes of ∕r have been evalu-
ated for large pipes by Bradley and Thompson (1951). (5.15) of fluid transport problems encountered in water transmis-
3. Because entry 2 also often applies to partially filled
Despite the logic and inherent conceptual simplicity of sections, additional diagrams are also necessary for
the combination of friction-factor diagram and Weisbach them. For such sections, moreover, trial-and-error
formulation, there are important reasons why water engi- solutions must be performed whenever the depth of
neers do not make use of them for the routine solution flow is unknown.
EXAMPLE 5.15 HEAD LOSS FOR LAMINAR FLOW
This example introduces Poiseuille’s equation for laminar flow when Reynolds number R is smaller than or equal to 2,000.
2
2
h = (32 Lv)∕( d ) = (32 Lv)∕(gd ) (5.12b)
f
where
h = head loss, ft, m
f
= absolute viscosity
d = pipe diameter, ft, m
3
3
= specific weight, 62.4 lb/ft , 9.8 kN/m for water
L = pipe length, ft, m
v = average velocity, ft/s, m/s
= / = kinematic viscosity
2
g = acceleration due to gravity, 32.2 ft/s , 9.81 m/s 2
Develop Eqs. (5.10a) and (5.12a) using Poiseuille’s equation (Eq. 5.12b) shown above.
Solution:
2
2
2
h = (32 Lv)/( d ) = (32 Lv)/(gd ) = (32 Lv)/(gd )(2v∕2v).
f
2
2
2
h = (64 /vd)(L/d)(v ∕2g) = (64∕R)(L/d)(v ∕2g) = f(L/d)(v ∕2g).
f
2
h = f(L/d)(v ∕2g), which is Eq. (5.10a).
f
f = 64∕R, which is Eq. (5.12a).
f and R are the resistance coefficient and the Reynolds number, respectively.
EXAMPLE 5.16 VELOCITY, REYNOLDS NUMBER, AND HEAD LOSS UNDER LAMINAR FLOW CONDITIONS
The elevation of reservoir A’s water surface is 210 m (688.98 ft), while the elevation of reservoir B’s water surface is 200 m
(656.17 ft). The two reservoirs are connected by a pipe which is 610 m (2,001.31 ft) long and 5 mm (0.0164 ft) in diameter.
3
3
Assume (a) water flows from reservoir A to reservoir B under laminar flow conditions; (b) = 9.8 kN/m = 62.4 lb/ft ;and (c) =
◦
◦
2
2
1.003 × 10 −3 N-s/m = 2.05 × 10 −5 lb-s/ft at 20 C(68 F). Determine the velocity, the Reynolds number, and head loss per unit
length.