Page 147 - Water Engineering Hydraulics, Distribution and Treatment

Darcy–Weisbach formula has the form
2
(5.10a)
h = f(L∕d)(v ∕2g)
f
b
in which f = a/R ,and a and b are numerical constants. Darcy–Weisbach formula with f = a/R is mainly used for determining the
head loss of turbulent flows, R > 4,000. The condition of the pipe wall causes complex degrees of turbulence and different values of
f for the same R. Equations (5.13) and (5.14) show the complexity of f values.
In the critical zone between R = 2,000 and R = 4,000, Eq. (5.15) represents the complexity of f values.
A Moody diagram shown in Fig. 5.11 can be used for determination of f values, in turn, for calculation of h using Eq. (5.10a),
f
when the pipe roughness, and Reynolds number, are known.
EXAMPLE 5.18 HEAD LOSS DETERMINATION USING MOODY DIAGRAM b 5.2 Fluid Transport 125
A 3,200-ft (975.4-m)-long cast iron pipeline with 12 in. (304.8 mm) diameter carries a water flow at a velocity of 15 ft/s (4.572 m/s)
◦
◦
and temperature of 68 F(20 C). Determine the Reynolds number, the resistance coefficient f, and the head loss using Moody diagram.
Solution 1 (US Customary System):
◦
2
= kinematic viscosity of water at 68 F = 1.059 × 10 −5 ft /s.
2
6
R = vd/ = (15 ft/s)(12∕12 ft)/(1.059 × 10 −5 ft /s) = 1.4 × 10 > 2,000 turbulent flow.
K = 8.5 × 10 −4 ft for cast iron pipe (from Fig. 5.11)
s
K /d = 8.5 × 10 −4 ft/(12∕12) ft = 0.0008.
s
From Moody diagram (Fig. 5.11), f = 0.02.
2
2
2
h = f(L/d)(v ∕2g) = 0.02(3,200 ft∕1 ft)(15 ft/s) /[2(32.2 ft/s )] = 223.6 ft.
f
Solution 2 (SI System):
◦
2
= kinematic viscosity of water at 20 C = 1.007 × 10 −6 m /s.
6
2
R = vd/ = (4.572 m/s)(0.3048 m)/(1.007 × 10 −6 m /s) = 1.4 × 10 > 2,000 turbulent flow.
K = 0.26 mm = 0.00026 m for cast iron pipe. (from Fig. 5.11)
s
K /d = 0.00026 m∕0.3048 m = 0.00085.
s
From Moody diagram (Fig. 5.11), f = 0.02.
2
2
2
h = f(L/d)(v ∕2g) = 0.02(975.4 m∕0.3048 m)(4.572 m/s) /[2(9.81 m/s )] = 68.18 m.
f
EXAMPLE 5.19 LAMINAR OR TURBULENT FLOW
◦
◦
Determine the type of flow occurring in a 16 in. (40.64 cm) pipe when (a) water at 60 F (15.55 C) flows at a velocity of 3.8 ft/s
◦
◦
(1.1582 m/s); and (b) heavy fuel oil at 60 F (15.55 C) flows at the same velocity. At this temperature, kinematic viscosity is 1.217 ×
2
2
2
2
10 −5 ft /s (1.1306 × 10 −6 m /s) for water, and 221 × 10 −5 ft /s (20.53 × 10 −5 m /s) for heavy oil.
Solution 1 (US Customary System):
R = vd/ = (velocity)(diameter)/(kinematic viscosity).
2
R = (3.8 ft/s)(16∕12) ft/(1.217 × 10 −5 ft /s) = 416,324 > 2,000 for water.
So the water flow is under turbulent flow condition.
2
R = (3.8 ft/s)(16∕12) ft/(221 × 10 −5 ft /s) = 2,292 > 2,000 for heavy oil.
So the heavy oil is under turbulent flow condition, but very close to laminar flow condition.
Solution 2 (SI System):
R = vd/ = (velocity)(diameter)/(kinematic viscosity).