Page 148 - Water Engineering Hydraulics, Distribution and Treatment

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Water Hydraulics, Transmission, and Appurtenances
Chapter 5
2
−6
m /s) = 416,324 > 2,000 for water.
R = (1.1582 m/s)(0.4064 m)/(1.1306 × 10
So the water flow is under turbulent flow condition.
2
−5
R = (1.1582 m/s)(0.4064 m)/(20.53 × 10
m /s) = 2,292 > 2,000 for heavy oil.
So the heavy oil is under turbulent flow condition, but very close to laminar flow condition.
EXAMPLE 5.20 PIPE SIZING UNDER LAMINAR FLOW CONDITIONS
◦
◦
Determine the pipe diameter for delivering 96 gpm (363.3 L/min) of medium fuel oil at 41 F(5 C), assuming that the kinematic
2
2
viscosity = 6.55 × 10 −5 ft /s = 6.085 × 10 −6 m /s.
Solution 1 (US Customary System):
3
2
2
Q = Av = (area)(velocity) = (d × 0.785 ft )(v ft/s) = 96 gpm = 0.2139 ft /s.
3
2
2
2
v = Q/A = (0.2139 ft /s)/(0.785d ft ) = 0.272∕d ft/s.
R = vd/ = (velocity)(diameter)/(kinematic viscosity).
2
2
R = (0.272∕d ft/s)(d ft)/(6.55 × 10 −5 ft /s) = 2,000.
Pipe diameter = 2.07 ft.
Solution 2 (SI System):
2
2
3
Q = Av = (area)(velocity) = (d × 0.785 m )(v m/s) = 363.3 L/min = 0.3633∕60 m /s.
2
2
3
2
v = Q/A = (0.0.3633∕60) m /s/(0.785 d m ) = 0.0077∕d m/s.
R = vd/ = (velocity)(diameter)/(kinematic viscosity).
2
2
R = (0.0077∕d m/s)(d m)/(6.085 × 10 −6 m /s) = 2,000.
Pipe diameter = 0.63 m.
EXAMPLE 5.21 HEAD LOSS USING DARCY–WEISBACH FORMULA
2
2
Oil of absolute viscosity 0.0021 lb-s/ft (0.1 N-s/m ) and specific gravity 0.851 flows through 10,000 ft (3,048 m) of 12 in. (300 mm)
3
3
stainless steel pipe at a flow rate of 1.57 ft /s (0.0445 m /s). Determine the Reynolds number, the friction factor, and the head loss in
the pipe.
Solution 1 (US Customary System):
Velocity, v = Q∕A
2
3
= (1.57ft ∕s)∕[(1 ft) × 0.785]
= 2ft∕s.
Reynolds number, R = vd ∕ = vd ∕ g
3
2
2
= (2 ft∕s)(1 ft)(0.851 × 62.4lb∕ft )∕[(0.0021 lb-s∕ft ) × (32.2ft∕s )]
= 1, 570 < 2,000, which indicates a laminar flow.
Friction factor, f = 64∕R under laminar flow conditions
= 6∕1,570
= 0.041.
2
Head loss, h = f(L∕d)(v ∕2g)
f
2
2
= (0.041)(10,000 ft∕1 ft)(2 ft∕s) ∕(2 × 32.2ft∕s )
= 25.5ft.