Sm = tanφ/tanθ = tan 40/tan 18.5 = 2.51
Equation 6.12
Sm - S.F.2
(2.5)2 - 1.5 2
2
cos θ =
η=
cos 18.5 = 0.40
1.5 x 2.5 2
S.F. S 2
m
Riprap size is obtained from Equation 6.16
0.3 Vt2
0.3 x (2.88)2
=
=
= 0.38 m
Dm
(S s - 1) g η (2.65 - 1) x 9.81x 0.4
This is the size recommended.
6.11 SOLVED PROBLEMS FOR FILTER DESIGN (SI)
The requirements for a gravel filter are given in Section 6.7. The gradation of a filter should be
such that
D50 (Filter )
< 40
(6.24)
D50 (Base)
D15 (Filter )
5<
< 40
(6.25)
D15 (Base)
D15 (Filter )
<5
(6.26)
D85 (Base)
6.11.1 PROBLEM 1 Filter Design
The properties of the riprap and base material are given in Table 6.4. Determine if filter is
needed between the riprap and the base material.
Table 6.4. Sizes of Materials.
Base Material (Sand)
Riprap (Gravel)
D85 = 1.50 mm
D85 = 24 mm
D50 = 0.75 mm
D50 = 12 mm
D15 = 0.38 mm
D15 = 6 mm
In accordance with the recommended sizes for filters:
D50 (Riprap)
12
=
= 16
D50 (Base)
0.75
6.61
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