Environmental Engineering Reference
In-Depth Information
Table D.1. Example of the design of earthen canals for two sediment diameters and two side slopes according
to the Lacey method.
SI-units
Lacey regime theory
Qd in m mf o
P
v
A
R
B o
y
k s
2.5
0.00040
2
1.00399
0.000272
7.646
0.515
4.850
0.634
3.662
0.891
42.3
3
0.00040
2
1.00399
0.000264
8.376
0.531
5.646
0.674
4.224
0.928
42.5
4
0.00040
2
1.00399
0.000252
9.672
0.557
7.176
0.742
5.223
0.995
42.9
6
0.00040
2
1.00399
0.000235
11.846
0.596
10.060
0.849
6.912
1.103
43.4
8
0.00040
2
1.00399
0.000224
13.678
0.626
12.785
0.935
8.351
1.191
43.7
10
0.00040
2
1.00399
0.000216
15.293
0.649
15.398
1.007
9.631
1.266
44.0
12
0.00040
2
1.00399
0.000210
16.752
0.669
17.925
1.070
10.797
1.332
44.2
15
0.00040
2
1.00399
0.000202
18.730
0.695
21.588
1.153
12.388
1.418
44.5
2.5
0.00015
2
0.61482
0.000120
7.646
0.438
5.712
0.747
2.006
1.261
48.5
3
0.00015
2
0.61482
0.000117
8.376
0.451
6.649
0.794
2.700
1.269
48.7
4
0.00015
2
0.61482
0.000111
9.672
0.473
8.450
0.874
3.782
1.317
49.1
6
0.00015
2
0.61482
0.000104
11.846
0.506
11.847
1.000
5.486
1.422
49.7
8
0.00015
2
0.61482
0.000099
13.678
0.531
15.056
1.101
6.897
1.516
50.1
10
0.00015
2
0.61482
0.000095
15.293
0.551
18.133
1.186
8.141
1.599
50.4
12
0.00015
2
0.61482
0.000093
16.752
0.568
21.108
1.260
9.270
1.673
50.7
15
0.00015
2
0.61482
0.000089
18.730
0.590
25.422
1.357
10.807
1.772
51.0
2.5
0.00040
1.5
1.00399
0.000272
7.646
0.515
4.850
0.634
4.693
0.819
42.3
3
0.00040
1.5
1.00399
0.000264
8.376
0.531
5.646
0.674
5.276
0.860
42.5
4
0.00040
1.5
1.00399
0.000252
9.672
0.557
7.176
0.742
6.318
0.930
42.9
6
0.00040
1.5
1.00399
0.000235
11.846
0.596
10.060
0.849
8.087
1.042
43.4
8
0.00040
1.5
1.00399
0.000224
13.678
0.626
12.785
0.935
9.597
1.132
43.7
10
0.00040
1.5
1.00399
0.000216
15.293
0.649
15.398
1.007
10.938
1.208
44.0
12
0.00040
1.5
1.00399
0.000210
16.752
0.669
17.925
1.070
12.159
1.274
44.2
15
0.00040
1.5
1.00399
0.000202
18.730
0.695
21.588
1.153
13.823
1.361
44.5
2.5
0.00015
1.5
0.61482
0.000120
7.646
0.438
5.712
0.747
3.856
1.051
48.5
3
0.00015
1.5
0.61482
0.000117
8.376
0.451
6.649
0.794
4.427
1.095
48.7
4
0.00015
1.5
0.61482
0.000111
9.672
0.473
8.450
0.874
5.441
1.173
49.1
6
0.00015
1.5
0.61482
0.000104
11.846
0.506
11.847
1.000
7.155
1.301
49.7
8
0.00015
1.5
0.61482
0.000099
13.678
0.531
15.056
1.101
8.615
1.404
50.1
10
0.00015
1.5
0.61482
0.000095
15.293
0.551
18.133
1.186
9.912
1.492
50.4
12
0.00015
1.5
0.61482
0.000093
16.752
0.568
21.108
1.260
11.093
1.570
50.7
15
0.00015
1.5
0.61482
0.000089
18.730
0.590
25.422
1.357
12.704
1.671
51.0
Example
•
10m 3 /s and sediment with size
Given is a canal with a discharge Q
=
d
=
0.001m
•
Hence, f
=
1.58745 and P
=
15.293m
•
The bottom slope S o =
0.4636m/km
13.22m 2 and R
•
After trial and error follows v
=
0.76m/s, A
=
=
0.86m
•
For a side slope m
=
2 and for the above given data follows:
B o =
10.65m and y
=
1.04m.
•
Using Manning gives n
=
0.026 and according to Strickler k s =
38.7
 
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