Environmental Engineering Reference
In-Depth Information
TABLE 5.13. Residual Saturation of Petroleum Fuels
water. The long time scales required for flowing ground-
water to remove residual nAPLs is illustrated in the
following example.
middle
Fuel
Soil
Gasolines
Distillates
Oils
Coarse gravel
0.0063
0.013
0.025
EXAMPLE 5.14
Coarse sand
0.019
0.038
0.075
Fine sand/silts
0.05
0.10
0.20
A cubic meter of aquifer has a porosity of 0.3 and con-
tains TCE at a residual saturation of 20%. If the density
of TCE is 1470 kg/m
3
, the solubility of TCE in water
is 1100 mg/L, and the mean seepage velocity of the
groundwater is 0.02 m/day, estimate the time it would
take for the TCE to be removed by dissolution.
Source of data
: American Petroleum Institute (1989).
where
ρ
f
is the density of the nAPL (mL
−3
) (see Appen-
dix B.2),
n
is the porosity of the soil (dimensionless), and
ρ
s
is the density of the soil (mL
−3
). In applying the
residual saturations shown in Table 5.13, values of
ρ
f
are
typically 750 kg/m
3
for gasoline, 800 kg/m
3
for middle
distillates, and 900 kg/m
3
for fuel oils.
Solution
From the data given,
n
= 0.3, and the residual saturation,
S
r
, is 0.20, hence the residual volume of TCE in 1 m
3
of
aquifer is given by
EXAMPLE 5.13
3
Volume of TCE
=
( .
0 20 0 3 1
)( . )( )
=
0 06
.
m
Estimate the residual mass fraction in mg/kg when spills
of gasoline in medium sand are cleaned up by pumping
free product from the surface of the water table. Assume
that the porosity of the aquifer is 0.23 and the density
of the sand is 2600 kg/m
3
.
Since the density of TCE is 1470 kg/m
3
, 0.06 m
3
corresponds to Equation (0.06)(1470) = 88.2 kg of
TCE. With a solubility of 1100 mg/L = 1.1 kg/m
3
, the
volume of water required to dissolve the 88.2 kg of
TCE is given by
Solution
88 2
1 1
.
.
Dissolution water required
=
=
80 2
.
m
3
From the data given,
n
= 0.23,
ρ
s
= 2600 kg/m
3
, and the
density of gasoline can be taken as
ρ
f
= 750 kg/m
3
. Inter-
polating in Table 5.13 between coarse sand and fine sand
gives
S
r
= 0.035 for medium sand. Substituting the data
given into Equation (5.79) gives the mass fraction,
M
f
,
at residual saturation as
Since the seepage velocity of the groundwater is
0.02 m/day, assuming that the contaminated volume is a
1 × 1 × 1 m block of aquifer, the time required for
80.2 m
3
of water to flow through the 1 m
3
of contami-
nated aquifer is given by
ρ
nS
n
f
r
M
=
80 2
0 02 1 1
.
80 2
0 02 0 3 1
.
f
ρ
(
1
−
)
+
ρ
nS
Time
=
=
=
13 367
,
days
s
f
r
.
n
(
×
)
.
( . )( )
750 0 23 0 035
2600 1 0 23
( .
)( .
)
=
=
36 6
.
years
(
−
.
)
+
750 0 23 0 035
( .
)( .
)
=
0 0030
.
kg/kg
=
3000
mg/kg
Hence the residual nAPL will generate a contami-
nant plume at saturation level (1100 mg/L) for 36.6
years! This result should be considered as somewhat
approximate, since dissolution rates are highly depen-
dent on the range and size distribution of nAPL blobs
(Schnoor, 1996).
Therefore, when all the free product gasoline is
removed from the contaminated soil, approximately
3000 mg/kg will remain trapped in the pores of the solid
matrix. This trapped gasoline will eventually be removed
by such processes as evaporation, dissolution, and bio-
logical and chemical degradation.
5.6.2 Raoult's Law
Even at residual saturation levels, nAPLs are capable
of contaminating large volumes of water and cannot be
removed easily except by dissolution in flowing ground-
In cases where nAPLs consist of a mixture of sub-
stances, then the solubilities of the individual substances
in the surrounding water are less than the solubilities of
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