Using Equations 3.15 through 3.22
n = 0.0417 D.50 7 = 0.0417 (.0031)
.167
16
= 0.016
(3.15)
n = 0.0482 D.50 7 = 0.0482 (.0031)
.167
16
= 0.018
(3.16)
n = 0.0473 D.75 7 = 0.0473 (0.004)
.167
16
= 0.019
(3.17)
n = 0.046 D.90 7 = 0.046 (0.0049)
.167
16
= 0.019
(3.18)
(Assume yo = 5.0 ft = 1.52 m and R = yo)
1/ 6
0.113 y0
0.113 (1.52)0.167
n=
=
= 0.019
(3.19)
y0
1.52
1.16 + 2.0 log
1.16 +2.0 log
D
0.0044
84
0.25
0.25
R
1.52
1
= 1.9
= 1.9
= 8.19
(3.20)
1/ 2
f
0.0044
D84
f1/2 = 0.122
n = 0.113 R1/ 6 f 1/ 2 = 0.113 (1.52)1/ 6 (0.122) = 0.015
(3.22)
R
1.52
+ 1.1 = 2.0 log
1
= 2.0 log
+ 1.1 = 6.18
(3.21)
1/ 2
0.0044
D 84
f
f1/2 = 0.162
n = 0.113 R1/ 6 f 1/ 2 = 0.113 (1.52)1/ 6 (0.162) = 0.020
(3.22)
Based on the range of n-values indicated above a n-value of 0.018 is selected. Alternatively,
the maximum and minimum n value could be used and a decision made on the basis of the
velocity and discharge.
Manning's n in Cobble Bed Streams
What are the Manning's n values for a cobble bed stream with the following size distribution:
=
0.19 m (190 mm)
D50
=
0.12 m (120 mm)
D16
D90
=
0.28 m (280 mm)
=
0.25 m (250 mm)
D84
D75
=
0.23 m (230 mm)
3.58
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