Page 146 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
Solution 1 (US Customary System):
h = 688.98 ft − 656.17 ft = 32.81 ft.
f
d = 0.0164 ft.
L = 2,001.31 ft.
◦
2
−5
lb-s/ft at 68 F.
= 2.05 × 10
3
= 62.4 lb/ft
2
h = (32 Lv)/( d ).
f
2
3
2
32.81 ft = [32(2.05 × 10
v (velocity) = 0.41 ft/s.
Check the Reynolds number. −5 lb-s/ft )(2,001.31 ft)(v ft/s)]/[(62.4 lb/ft )(0.0164 ft) ].
R = vd/ = vd / = vd /g .
3
2
2
R = [(0.41 ft/s)(0.0164 ft)(62.4 lb/ft )]/[(32.2 ft/s )(2.05 × 10 −5 lb-s/ft )] = 630.
Flow is laminar because R is below 2,000.
3
2
Q = Av = [(0.0164 ft) × 0.785](0.41 ft/s) = 8.66 × 10 −5 ft /s.
H = h /L = 32.81 ft∕2,001.31 ft = 0.0164 ft/ft.
f
f
Solution 2 (SI System):
h = 210 m − 200 m = 10 m.
f
d = 5mm = 0.005 m.
L = 610 m.
◦
2
2
2
= 1.003 × 10 −3 N-s/m at 20 C. (Note: 1 N-s/m = 0.021 lb-s/ft )
3
3
= 9.8 kN/m = 9.8 × 1,000 N/m .
3
= 1,000 kg/m .
2
h = (32 Lv)/( d ).
f
2
3
2
10 m = [32(1.003 × 10 −3 N-s/m )(610 m)(v m/s)]/[(9.8 × 1,000 N/m )(0.005 m) ].
v (velocity) = 0.125 m/s.
Check the Reynolds number.
R = vd/ = vd / .
2
3
R = [(0.125 m/s)(0.005 m)(1,000 kg/m )]/(1.003 × 10 −3 N-s/m ) = 630.
Flow is laminar because R is below 2,000.
3
2
Q = Av = [(0.005 m) × 0.785](0.125 m/s) = 2.5 × 10 −6 m /s.
H = h /L = 10 m∕610 m = 0.0164 m/m.
f
f
EXAMPLE 5.17 HEAD LOSS UNDER TURBULENT FLOW CONDITIONS
Compare Poiseuille’s equation with Darcy–Weisbach formula, and discuss their applications for determining the head loss under
turbulent and laminar flow conditions.
Solution:
From previous example, it can be seen that Poiseuille’s equation (Eq. 5.12b) can be rearranged to have a form
2
h = f(L∕d)(v ∕2g) (5.10a)
f
in which f = 64∕R, which is Eq. (5.12a).
Poiseuille’s equation is used for hydraulic analyses under laminar flow conditions, R < 2000.