Environmental Engineering Reference
In-Depth Information
the load on the piston is increased, the free air above the porous
stone will instantaneously be compressed in accordance with
Boyle's law. Then some air will move into the porous stone
in accordance with Henry's law. Henry's law states that the
mass of gas dissolved in a fixed quantity of liquid at a constant
temperature is directly proportional to the absolute pressure
of the gas above the solution (Sisler et al., 1953). The process
of air flow into the porous stone will continue over time until
the free air pressure is equal to the pressure of the air in the
porous stone. Each time the piston load is increased, there is
an immediate volume change in the free air followed by a slow
process where air flows into the porous stone. Eventually, all
the free air will have moved into the porous stone, and any
additional applied load will be carried by the porous stone
(i.e., water).
The above analogy cannot totally simulate the processes
that occur in an unsaturated soil. In the presence of a solid,
such as soil particles, the air and water pressures can have
different magnitudes. The air and water pressures in a soil
can also change at differing rates during the loading process.
In the analogy, the free air and the water (i.e., porous stone)
have the same pressure. Differences between the air and water
pressures are later discussed and shown to be of significance
to the compressibility formulation of air-water mixtures.
The mass of air going into or coming out of water is
time dependent. This time-dependent process can either be
ignored or taken into consideration, depending upon the
engineering problem under consideration. The amount of
air that can be dissolved in water is referred to as solubil-
ity, and the rate of solution is referred to as diffusivity. The
volume of dissolved air in water is essentially independent
of air or water pressures. This can be demonstrated using
the ideal gas law and Henry's law. The ideal gas law can
be rearranged and applied to a gas dissolving in water at a
particular temperature and pressure:
where:
V
d
=
volume of dissolved air in water,
M
d
=
mass of dissolved air in water, and
u
a
¯
=
absolute pressure of the dissolved air.
The absolute pressure of dissolved air is equal to the abso-
lute pressure of the free air under equilibrium conditions.
Referring to the piston and porous stone analogy, an increase
in the piston load will instantaneously increase the pressure
in the free air, and therefore more free air will commence
entering the porous stone (i.e., water). After some time, an
equilibrium condition will be reached where the pressures
in the free air and the dissolved air are equal. If the piston
load is then increased, the process will be repeated.
The mass of dissolved air under equilibrium conditions is
dependent upon the corresponding absolute air pressure as
stated by Henry's law. If the temperature remains constant
throughout the process, the ratio between the mass of air
and the absolute pressure in the dissolved air is constant:
M
d
1
¯
M
d
2
¯
u
a
1
=
u
a
2
=
const
(2.35)
where:
M
d
1
,
u
a
1
=
¯
mass and absolute pressure of the dissolved
air, respectively, at condition 1, and
M
d
2
,
u
a
2
=
¯
mass and absolute pressure of the dissolved
air, respectively, at condition 2.
The volume of dissolved air in water,
V
d
, is computed
from the gas law. At a constant temperature, the volume of
dissolved air in water is a constant for different pressures.
The ratio between the mass of each gas that can be dissolved
in a liquid and the mass of the liquid is called the coeffi-
cient of solubility,
H
. Table 2.12 presents the coefficients
M
d
¯
RT
ω
a
V
d
=
(2.34)
u
a
Table 2.12 Solubility of Gases in Water (Under a Pressure of 101.3 kPa)
Coefficient of Solubility,
H
a
Volumetric Coefficient of Solubility,
h
b
Temperature (
◦
C)
Air
c
Air
d
Oxygen
Nitrogen, Argon, etc.
10
−
6
10
−
6
10
−
6
14
.
56
×
23
.
87
×
38
.
43
×
0
0.02918
10
−
6
10
−
6
10
−
6
4
13
.
06
×
21
.
59
×
34
.
65
×
0.02632
10
−
6
10
−
6
10
−
6
10
11
.
25
×
18
.
82
×
30
.
07
×
0.02284
10
−
6
10
−
6
10
−
6
20
9
.
11
×
15
.
51
×
24
.
62
×
0.01868
10
−
6
10
−
6
10
−
6
30
7
.
55
×
13
.
10
×
20
.
65
×
0.01564
Source:
From Dorsey, 1940.
a
At standard atmospheric pressure.
b
h
=
ρ
w
/ρ
a
H.
c
Units of grams of air per gram of water.
d
Units of volume of air per volume of water.
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