DHARM
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 155
We know,
k =
qrr
zz
log ( / )e
()
21
2
2
1
π−^2
=
2 303 2 70 15 100
60 31 3 29
10
22
.log(/)
(. )
××
×−π
cm/s
= 1.18 × 10–1 mm/s.
Example 5.13. Determine the order of magnitude of the composite shape factor in the Poiseulle’s
equation adapted for flow of water through uniform sands that have spherical grains and a
void ratio of 0.9, basing this determination on Hazen’s approximate expression for permeabil-
ity.
Poiseulle’s equation adapted for the flow of water through soil is:
k = D e
e
s^2 C
3
1
..
()
γ.
μ+
with the usual notation, C being the composite shape factor.
By Allen Hazen’s relationship,
k = 100 D 102
D 10 is the same as the diameter of grains, Ds, for uniform sands.
∴ k = 100 Ds^2. Here Ds is in cm while k is in cm/s.
Substituting –
100 Ds^2 = D
e
e
s^2 C
3
1
..
()
.
γ
μ+
C =^100
1
.. 3
μ ()
γ
+e
e
; here μ is in N-Sec/cm^2 and γ is in N/cm^3.
C =
100 10 10 1 9
981 10 09
76
33
×××
××
−.
.(.)
since μ = 10–3 N-sec/cm^2 (at 20°C) and γ = 9.81 kN/m^3
= 0.002657.
Example 5.14. A cohesionless soil has a permeability of 0.036 cm per second at a void ratio of
0.36. Make predictions of the permeability of this soil when at a void ratio of 0.45 according to
the two functions of void ratio that are proposed.
k 1 : k 2 =
e
e
e
e
1
3
1
2
3
() (^112) |
---|
++() |
0.036 : k 2 = (. ) |
. |
: |
(. ) |
. |
036 |
136 |
045 |
145 |
33 |
= 0.546 : 1 |
∴ k 2 = |
1 |
0546. |
× 0.36 mm/s = 6.60 × 10–1 mm/s |
Also, k 1 : k 2 = e 12 : e 22 |
0.036 : k 2 = (0.36)^2 : (0.45)^2 |
= 0.1296 : 0.2025 |