Page 162 - Water Engineering Hydraulics, Distribution and Treatment

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Water Hydraulics, Transmission, and Appurtenances
Chapter 5
r = A∕P = (4.2ft × 2.2ft)∕(2.2ft × 2 + 4.2ft) = 1.074 ft.
w
√
Q = Av = AC rs.
√
21.0 = (4.2 × 2.2)(C) 1.074 × 0.00048.
√
C = (21.0)∕(4.2 × 2.2) 1.074 × 0.00048
= 100.
Substitute C = 100, r = 1.074 ft, and s = 0.00048 into Eq. (5.24):
0.5
C = (41.65 + 0.00281∕s + 1.811∕n)∕[1 + (n∕r )(41.65 + 0.00281∕s)]
0.5
100 = (41.65 + 0.00281∕0.00048 + 1.811∕n)∕[1 + (n∕1.074 )(41.65 + 0.000281∕0.00048)]. (5.24)
By the trial-and-error method, assuming n = 0.011, 0.013, 0.015, 0.017, and so on, n is found to be 0.015.
Solution 2 (SI System):
√
v = 0.552AC rs (5.18b)
r = A∕P = (1.28 m × 0.67 m)∕(0.67 m × 2 + 1.28 m) = 0.327 m (1.074 ft).
w
√
Q = Av = 0.552AC rs.
√
0.59 = 0.552(1.28 × 0.67)(C) 0.327 × 0.00048.
C = 100.
Substitute C = 100, r = 1.074, and s = 0.00048 into Eq. (5.24):
0.5
C = (41.65 + 0.00281∕s + 1.811∕n)∕[1 + (n∕r )(41.65 + 0.00281∕s)] (5.24)
0.5
100 = (41.65 + 0.00281∕0.00048 + 1.811∕n)∕[1 + (n∕1.074 )(41.65 + 0.000281∕0.00048)].
By the trial-and-error method, assuming n = 0.011, 0.013, 0.015, 0.017, and so on, n is found to be 0.015.

EXAMPLE 5.34 OPEN CHANNEL FIELD INVESTIGATION—MANNING 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 Manning’s method.
Solution 1 (US Customary System):

1.486
v = (r) 0.67 (s) 0.5 (5.25)
n
2
A = 4.2 × 2.2 = 9.24 ft .
r = A∕P = (9.24)∕(2.2ft × 2 + 4.2ft) = 1.074 ft.
w
0.5
v = (1.486∕n)(1.074) 0.67 (0.00048) .
Q = Av.
0.5
21.0 = 9.24[(1.486∕n)(1.074) 0.67 (0.00048) ].
0.5
n = 9.24[(1.486)(1.074) 0.67 (0.00048) ]∕21.0.
n = 0.015.