Page 188 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
0.5
v = C (2gh)
v
9.67 = C (2 × 9.81 × 5.4864) .
v
C = 0.93.
v
C = C ∕C
(5.48a)
v
d
c
C = 0.60∕0.93 = 0.64.
c
EXAMPLE 5.56 DETERMINATION OF DISCHARGE COEFFICIENT 0.5 (5.47b)
Determine the coefficient of discharge for an orifice 2 in. (50.8 mm) in diameter, which discharges from a tank with a head of 16 ft
3
3
(4.876 m). The discharge rate Q is measured at 0.55 ft /s (0.0156 m /s). The actual velocity at the vena contracta, v.c., is 29.0 ft/s
(8.839 m/s).
Solution 1 (US Customary System):
v = (2gh) 0.5 (5.47d)
i
2
v = (2 × 32.2ft∕s × 16 ft) 0.5 = 32.1ft∕s.
i
C = v∕v i (5.47a)
v
C = (29 ft∕s)∕(32.1ft∕s) = 0.90.
v
2 2
Area of orifice: A = ( ∕4)(2∕12) = 0.0218 ft .
o
2
3
Area at v.c.: A jet = Q∕v = (0.55 ft ∕s)∕ (29 ft∕s) = 0.019 ft .
2
2
C = A ∕A = (0.019 ft )∕(0.0218 ft ) = 0.87 (5.48a)
c jet o
C = C C = 0.90 × 0.87 = 0.78 (5.48b)
d v c
Solution 2 (SI System):
v = (2gh) 0.5 (5.47d)
i
2
v = (2 × 9.81 m∕s × 4.876 m) 0.5 = 9.78 m∕s.
i
C = v∕v i (5.47a)
v
C = (8.839 m∕s)∕(9.78 m∕s) = 0.90.
v
2 2
Area of orifice: A = ( ∕4)(0.0508) = 0.002 m .
o
2
3
Area at v.c.: A jet = Q∕v = (0.0156 m ∕s)∕ (8.839 m∕s) = 0.00176 m .
2
2
C = A ∕A = (0.00176 m )∕(0.002 m ) = 0.88 (5.48a)
c jet o
C = C C = 0.90 × 0.88 = 0.79. (5.48b)
d v c
5.10.2 Forces Developed by Moving Fluids where
t = round-trip travel time, s
Water hammer is caused by the sudden decrease in liquid
L = length of pipe, ft (m)
motion. In a water pipeline, the time of pressure wave to
c = celerity (speed of propagation) of pressure wave, ft/s
travel upstream and back (round-trip) is given by
(m/s)
t = 2L∕c (5.52)