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Bioremediation Fundamentals Need to know more? Send us an email or feel free to search our online series with our give away search engine: |
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Trichloroethylene (TCE) has been disposed of in open, unlined evaporation ponds for a number of years. The TCE was used to clean engine parts and was contaminated with grease and water. Soil beneath the ponds is of moderate clay (10%) and organic content (3%). The groundwater table is 50 feet or about 15.3m below the bottom of the ponds. Sketch a possible contaminant distribution in the soil matrix. Drinking water wells using the aquifer below the ponds are located a distance of about one mile. How long will it take to reach them?
Before diving into much detail let's take a look at what presumably makes up an idealized 1 m3 of contaminated soil. Assuming a volumetric water content of 0.2 the amount of matter in liquid phase Vw is given by
Vw = water content = 1 m3 * 0.2 = 200 L
Porosity given by
Porosity = (water content Vw + volume of air Va) / total volume V
Picking a value of 0.35 for porosity (0.3 to 0.6 range as per Eweis et al) we can solve for Va ( volume of matter in gas phase), as follows
Va = total volume V * porosity - Vw = 1 m3 * 0.35 - . 2 m3 = 15 m3
The total amount of contaminant is given by
s * Gb * V + Cw * Vw + Ca * Va (I)
being
s = mass of contaminant [solute] per unit dry mass of soil (unitless)
Gb = dry bulk density
Cw = liquid phase contaminant concentration, e.g. mg/L
Ca = gas phase contaminant concentration
Before proceeding let's go over three fundamental concepts, as follow
1. from soil literature the distribution of contaminant between the solid and liquid phase is given by Ksd the so-called soil distribution coefficient defined as
Ksd = s / Cw
2. The thin film model is applicable so that at the interface the ratio of the chemical in the gas to amount of said chemical in the liquid is assumed to equal the Henry's law coefficient H, id est
H = Ca / Cw
3. In the absence of other data, the soil distribution coefficient Ksd can be estimated by the following expression, id est
Ksd = c. 6.3 * 10^-7 * foc * Kow, being
foc = fractional organic carbon content (given)
Kow = octanol water partition coefficient, i.e. the ratio of the concentration of constituent in octanol to the concentration in water.
Just of the sake of number crunching let's pick some token dry bulk density say Gb, e.g. 2070 kg/m3.
Let's plug the expressions presented above into mass balance equation (I):
s * Gb * V + Cw * Vw + Ca * Va (I)
s * Gb * V + Cw * Vw + H * Cw * Va (II)
Related TCE data is as follows:
H = 0.3 (Henry's )
solubility = 1100 mg/L (water has soaked up as much TCE as it may)
Kow = 240 (partition coefficient)
Soil distribution coefficient Ksd follows from straight forward evaluation::
Ksd = c. 6.3 * 10^-7 * foc * Kow = 6.3 * 10^-7 * 0.03 * 240 = 4.536 * 10^-6 L/mg = 4.536 * 10^-3 m3/kg
Liquid-phase contaminant concentration Cw can be assumed to be equated to solubility,i .e. the maximum amount of said constituent in the liquid phase (saturation), thus we have
Cw = liquid phase constituent concentration = solubility = 1100 mg/L
Solving for s (mass of contaminant [solute] per unit dry mass of soil)
s = Cw * Ksd = 1100 mg/L * 4.536 * 10^-3 L/mg = 4.9896 10^-3
We can now calculate each of the three terms which make up mass balance (II), id est
4.9896 10^-3 * 2070 kg/m3 * 1 m3 + 1100 mg/L * 200 L + 0.3 * 1100 mg/L * 150 L =
10.328 kg (sorbed onto soil) + 0.22 kg (liquid phase portion) + 0.066 kg (gas phase fraction)
The following percents can be produced:
sorbed to soil: 10.328*100/10.614 = c. 97.4%
liquid phase: 0.22 * 100 / 10.614 = c. 2.1%
gas phase: 0.066 * 100/10.614 = c. 0.5%
A staggering amount of the constituent ends up "glued" to the soil matrix, a typical bioremediation situation. As the vast majority of contaminant mass is sorbed, the calculation of the time required to get rid of the problem can translate into having to account for both required time to desorb plus time to degrade, a sort of two-stage process. .
Contaminant Distribution in Soil Matrix |
|
| chemically or physically sorbed to soil | 97.4% |
| liquid-phase | 2.1% |
| gas-phase | 0.5% |
Now for time to get to the wells - we certainly need an accurate/reliable as possible figure for hydraulic conductivity Kc to apply Darcy's law for porous media:
Vs = - Kc* deltaH / deltaL
being
deltaH = vertical distance to groundwater (50" or 15.3m)
deltaL = 1 mile = 1609 m
As per Eweis et al, values of Kc range from approximately 10^-3 m/s for coarse sands to 10^-8 m/s for clays. In the absence of better data we simply speculate for a number of data sets as follows
| Kc (m/s) | Vs (m/d) | years to reach |
| 0.001 | 8.162 * 10^-1 | 5.5 |
| 0.0001 | 8.162 * 10^-2 | 54.8 |
| 0.00001 | 8.162 * 10^-3 | 547.6 |
| 0.000001 | 8.162 * 10^-4 | 5475.9 |
| 0.0000001 | 8.162 * 10^-5 | 54758.6 |
| 0.00000001 | 8.162 * 10^-6 | 547585.9 |
One just hopes everything would happen as per our simplified number crunching!!
A leaking underground storage tank has contaminated an aquifer Estimate the velocity at which the plume will move. Contaminant and aquifer characteristics are as follows:
Kc = 10^-3 m/s
DeltaH/deltaL = - 0.06
Gb = 1900 kg/m3
Foc = 0.01
Kow = 250
porosity = 0.25
Superficial velocity Vs can be calculated, as per Darcy, as follows::
Vs = - Kc * dh/dx = 10^-3 m/s * 0.06 * 60 * 60 * 24 = 5.184 m/day
Soil distribution coefficient Ksd is as easily computed:
Ksd = 6.3 * 10^-7 * foc * Kow = 6.3 * 10^-7 * 0.01 * 250 = 1.575 * 10^-6 L/mg = 1.575 * 10^-3 m3/kg
Retardation coefficient
R = 1 + Gb * Ksd / theta = 1 + 1900 kg/m3 * 1.575 * 10^-3 m3/kg / 0.25 = 12.97
Pore velocity V = Vs / porosity = 5.184 m/day / 0.25 = 20.736 m/day = c. 21 m/day
Contaminant velocity is then given by
Vc = pore velocity V / R = 20.736 m/day / 12.97 = c. 1.6 m/day
One hopes real world calcs were as unencumbered .. still the primer purpose is met. Good luck with your bioremediation endeavors!
