Page 161 - Water Engineering Hydraulics, Distribution and Treatment

0.010 Smooth cement lining; best planed timber
0.012 New wood flumes; lined cast iron
0.013 Vitrified sewer pipe; concrete pipe; metal flumes
0.015 Average clay pipe; average cast iron pipe; average cement-lined pipe
0.023 Well-maintained earth canals
0.027 Dredged earth canals
0.040 Canals cut in rocks
0.030 Rivers in good condition
Therefore, selection of n = 0.015 for an average cement-lined open channel appears to be appropriate. 5.2 Fluid Transport 139
EXAMPLE 5.32 OPEN CHANNEL FLOW
Determine the expected flow in a 4.2-ft (1.28-m)-wide rectangular cement-lined open channel. The channel is laid on a slope of 4.8
ft in 10,000 ft (4.8 m in 10,000 m) and water flows 2.2 ft (0.67 m) deep. Use the Kutter’s C assuming the roughness coefficient n
equals 0.015.
Solution 1 (US Customary System)
√
v = C rs (5.18a)
r = A/P = (4.2 ft × 2.2 ft)/(2.2 ft × 2 + 4.2 ft) = 1.074 ft.
w
s = 4.8 ft∕10,000 ft = 0.00048.
0.5
C = (41.65 + 0.00281∕s + 1.811∕n)∕[1 + (n∕r )(41.65 + 0.00281∕s)] (5.24)
0.5
C = (41.65 +0.00281∕0.00048 + 1.811∕0.015)/[1 + (0.015∕1.074 )(41.65 + 0.00281∕0.00048)].
C = 100.
√
Q = Av = AC rs
√
= (4.2 × 2.2)(100) 1.074 × 0.00048
3
= 21.0ft ∕s.
Solution 2 (SI System):
√
v = 0.552C rs (5.18b)
r = A/P = (1.28 m × 0.67 m)/(0.67 m × 2 + 1.28 m) = 0.327 m.
w
s = 4.8 m∕10,000 m = 0.00048.
From Solution 1,C = 100.
√
Q = Av = 0.552 AC rs
√
= 0.552(1.28 × 0.67)(100) 0.327 × 0.00048
3
= 0.59 m ∕s.

EXAMPLE 5.33 OPEN CHANNEL FIELD INVESTIGATION—KUTTER METHOD
3
3
A water flow of 21.0 ft /s (0.59 m /s) was measured in a rectangular open channel 4.2 ft (1.28 m) wide with water flowing 2.2 ft
(0.67 m) deep. Determine the roughness factor n for the channel lining, if the channel slope is 0.00048. Use the Kutter’s method.
Solution 1 (US Customary System):
√
v = C rs (5.18a)