Geology Reference
In-Depth Information
Mass of water
308.17
305.40
2.77 g
Mass of oven-dried soil
305.40
53.60
251.80 g
2.77
251.80
*
100
=
1.10%
Water content
=
The dry unit weight of the soil specimen can now be calculated:
112.2
1.10
Dry unit weight
100
111.0 lb
ft
3
100
>
The volume of solids, volume of voids, and void ratio can be obtained as
follows:
111.0
(
2.71
)(62.4)
=
0.6564 ft
3
Volume of solids
=
0.3436 ft
3
Volume of voids
1
0.6564
0.3436
0.6564
=
0.523
Void ratio,
e
=
(For a more complete explanation of the determination of void ratio, see
the section “Numerical Example” in Chapter 18.)
[B] Permeability Test
For trial no. 1 of the permeability test, the standpipe's cross-sectional
area,
a
, is
1.83 cm
2
; length of specimen,
L
, is
15.80 cm
; specimen's cross-
sectional area,
A
, is 81.07 cm
2
; total time,
t
, for the water in the stand-
pipe (burette) to drop from
h
1
to
h
2
is
32.3 s
; hydraulic head at the
beginning of the test,
h
1
, is
150.0 cm
; and hydraulic head at the end of
the test,
h
2
, is
20.0 cm
. With these data known, coefficient of perme-
ability
k
can be computed using Eq. (19-1):
2.3
aL
At
log
h
1
h
2
k
(19-1)
(2.3)(
1.83
)(
15.80
)
(81.07)(
32.3
)
log
150.0
20.0
10
2
cm
k
2.222
s
>
To correct for permeability at 20°C, the ratio of the viscosity of water at
22°C
to that at 20°C is determined from Table 18-2 to be 0.9531. The
permeability at 20°C is therefore
10
2
)(0.9531)
10
2
k
20°C
(2.222
2.118
cm/s
In like manner, calculation of
k
20°C
for trials no. 2 and 3 yields 2.099
10
-
2
×
and 2.158
10
-
2
cm/s, respectively. Taking an average gives
×
(2.118
+
2.099
+
2.158)
*
10
-
2
3
=
2.125
*
10
-
2
cm
Average
k
20¡ C
=
s
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