n = 0.342 D.50 7 = (0.0342) (0.0102).167 = 0.016
16
(3.15)
n = 0.395 D.50 7 = (0.0395) (0.0102).167 = 0.018
16
(3.16)
n = 0.388 D.75 7 = (0.0388) (0.0131).167 = 0.019
16
(3.17)
n = 0.038 D.90 7 = 0.038 (0.0161).167 = 0.019
16
(3.18)
(Assume yo = 5.0 ft and R = yo)
0.0927 y1/ 6
0.0927 (5.0)0.167
0
n=
=
= 0.019
(3.19)
5.0
y
1.16 + 2.0 log
1.16 +2.0 log 0
D
0.0144
84
0.25
R
1
5.00
= 1.9
= 1.9
= 8.20
(3.20)
D
1/ 2
0.0144
f
84
f1/2 = 0.122
n = 0.0927 R1/ 6 f 1/ 2 = 0.0927 (5.0)1/ 6 (0.122) = 0.015
(3.22)
R
5.0
1
= 2.0 log
+ 1.1 = 2.0 log
+ 1.1 = 6.18
(3.21)
D
1/ 2
0.0144
f
84
f1/2 = 0.162
n = 0.0927 R1/ 6 f 1/ 2 = 0.0927 (5.0)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 based 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:
=
0.62 ft
(190 mm)
D50
=
0.39 ft
(120 mm)
D16
=
0.92 ft
(280 mm)
D90
D84
=
0.82 ft
(250 mm)
D75
=
0.75 ft
(230 mm)
n = 0.342 D1/06 with D 50 in ft. n = 0.0342 (0.62)1/ 6 = 0.032
(3.15)
5
n = 0.395 D1/06 with D 50 in ft. n = 0.0395 (0.62 ft)1/ 6 = 0.037
(3.16)
5
3.63
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