Page 163 - Water Engineering Hydraulics, Distribution and Treatment

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Solution 2 (SI System):
A = 1.28 × 0.67 = 0.858.
2
r = A∕P = (0.858 m )∕(0.67 m × 2 + 1.28 m) = 0.327 m.
w
0.67
(s)
v = (1∕n)(r)
Q = Av.
0.67
0.59 = (0.858)[(1∕n)(0.327)
(0.00048) ].
0.5
0.67
n = (0.858)[(0.327) 0.5 (0.00048) ]∕0.59. 0.5 v = 1 n (r) 0.67 (s) 0.5 5.2 Fluid Transport (5.26)
n = 0.015.
EXAMPLE 5.35 APPLICATION OF MANNING FORMULA FOR A PIPE FLOWING FULL
Determine the velocity and flow of a circular pipe 24 in. (061 m) in diameter, which is flowing full. The pipe roughness coefficient
is 0.013 and the slope of its energy line is s = 0.0004.
Solution 1 (US Customary System):
2
For a pipe flowing full r = A/P = ( D ∕4)/( D) = D∕4.
w
v = (0.59∕n)(D) 0.67 (s) 0.5 (5.28)
0.5
v = (0.59∕0.013)(24∕12) 0.67 (0.0004) .
v = 1.44 ft∕s.

Q = (0.46∕n)(D) 2.67 (s) 0.5 (5.29)
0.5
Q = (0.46∕0.013)(24∕12) 2.67 (0.0004) .
3
Q = 4.5ft ∕s = 2,020 gpm.
Using the nomogram (Appendix 15), for n = 0.013:
v = 1.45 ft∕s.
Q = 2,000 gpm.

Solution 2 (SI System):
v = (0.40∕n)(D) 0.67 (s) 0.5 (5.30)
0.5
v = (0.40∕0.013)(0.61) 0.67 (0.0004) .
v = 0.44 m∕s.

Q = (0.31∕n)(D) 2.67 (s) 0.5 (5.31)
0.5
Q = (0.31∕0.013)(0.61) 2.57 (0.0004) .
3
Q = 0.127 m ∕s.
Using the nomogram (Appendix 15), for n = 0.013:
v = 0.44 m∕s.
3
Q = 0.126 m ∕s.