Guidelines on remediation of contaminated sites
Appendix
5.4
Example of calculation of evaporation of volatile substances from soil
1 Background
The following example, cf. Figure 1, features a calculation of the contributions of trichloroethylene contamination to indoor air and outdoor air concentrations. The equations are numbered in accordance with Appendix 5.3.
Figure 1
Calculation example, contributions to outdoor air and indoor air.
The following two tables show the standard values, imaginary investigation values and chemical constants used in connection with calculation.
Table 1
Standard values and investigation values
|
Standard value |
Investigation value |
Relative volumetric proportion of air, VL |
0.30 |
|
Relative volumetric proportion of water, VV |
0.15 |
|
Relative volumetric proportion of soil, VJ |
0.55 |
|
Temperature, T |
298 K = 25o C |
|
Soil particle density, d |
2.65 kg/l |
|
Soil trichloroethylene concentration, CT |
|
0.1 mg/kg |
Soil density, r |
1.7 kg/l |
|
Soil content of organic substances, foc |
0.002 |
|
Sand strata thickness, outdoors, X |
|
2.1 m |
Sand strata thickness, under floor, X1 |
|
2.0 m |
Concrete deck thickness, X2 |
|
0.08 m |
Material constant for concrete, N2 |
0.002 |
|
Building ceiling height, Lh |
|
2.3 m |
Building air renewal, Ls |
8.3 × 10-5 s-1 |
|
Pressure differential over concrete deck, DP |
5 Pa |
|
Concrete deck thickness, hb |
|
80 mm |
Reinforcement bar spacing, Db |
50 mm |
|
Reinforcement diameter, da |
3 mm |
|
Relative humidity, RF |
60 % |
|
Cement content CM |
220 kg/m3 |
|
Water/cement ratio, V/c |
0.67 |
|
Shrinkage time, ts |
7,300 days |
|
Dynamic viscosity of air, m |
1.8 × 10-5 (kg/m) × s |
|
Floor length |
|
10 m |
Floor width |
|
10 m |
The chemical constants can be found in tabular form in Appendix 5.5
Table 2
Chemical constants.
Trichloroethylene partial pressure, p |
9,900 N/m2 |
Trichloroethylene molecular weight, m |
131.39 g/mol |
Gas constant, R |
8.314 J/mol·K |
Trichloroethylene solubility, S |
1,400,000 mg/m3 |
Trichloroethylene diffusion coefficient, DL |
8.8 ·10-6 m2/s |
Trichloroethylene octanol-water ratio, Kow |
102,53 l/kg |
2 Calculations
2.1 Phase distribution in soil
The total volume of soil can be seen as the sum of the volumes of the soil phases, see Equation 1.
Equation 1
VL + VV + VJ = 1
where:
VL =relative volumetric proportion of air in soil
VV =relative volumetric proportion of water in soil
VJ =relative volumetric proportion of soil particles in soil
Maximum trichloroethylene in one cubic metre (1 m3) of soil distributed on the three phases of soil can be calculated as follows, see Equations 2-10:
In the air phase of soil (soil gas):
Equation 2
ML, max = VL · CL, max = 0.30 · 525,000 mg/m3 = 158,000 mg/m3
| where: | ML,max | = | maximum amount of trichloroethylene in soil gas (mg/m3 soil volume) |
| CL,max | = | saturated vapour concentration of contaminant (mg/m3 soil gas). |
CL,max can be calculated on the basis on the partial pressure of
trichloroethylene partial pressure by means of the law of ideal gases:
Equation 3
where:
p =trichloroethylene partial pressure (9,900 N/m2)
m =trichloroethylene molecular weight (131.39 g/mol)
R =gas constant (8.314 J/mol · K)
T =temperature (298 K = 25o C)
In the water phase of soil (soil water):
Equation 4
MV,max = VV · S = 0.15 · 1,400,000 mg/m3 = 210,000 mg/m3
| where: | MV,max | = | maximum amount of trichloroethylene in soil water (mg/m3 soil volume) |
| S | = | trichloroethylene solubility in water (1,400,000 mg/m3 soil water). |
In the partial phase of soil:
Equation 9
MJ,max = VJ · d · Koc · foc · S (mg/m3)
= 0.55 · 2.65 kg/l ·101,79 · 0.002 · 1,400,000 mg/m3 = 252,000
mg/m3
| where: | MJ,max | = | maximum amount of trichloroethylene, which has adsorbed to the organic fraction of the soil particles (mg/m3 soil volume) |
| d | = | soil particle density (2.65 kg/l) | |
| Koc | = | trichloroethylene ratio between organic carbon and water (1/kg) | |
| foc | = | soil content of organic carbon (0.002) |
The trichloroethylene ratio between carbonate and water can be estimated on the basis of
the octanol/water ratio Kow, (equation 10a).
log Koc = 1.04 · log Kow – 0.84 = 1.04 · 2.53 – 0.84 = 1.79
Equation 10a
The soil’s maximum capacity for trichloroethylene (immediately preceding NAPL) will then be:
ML,max + MV,max + MJ,max
= 158,000 mg/m3 + 210,000 mg/m3 + 252,000 mg/m3
= 620,000 mg/m3
Based on the above-mentioned assumption that the relative distribution among the three soil phases is independent of total soil concentration, the distribution of trichloroethylene in the three phases of soil can be calculated.
The following applies to the air phase of soil
Equation 11
| where: | ||
| fL | = | relative amount of trichloroethylene in soil gas in relation to total soil content (calculated pr. m3 soil). |
| ML, MV, MJ | = | actual amount of trichloroethylene in each of the three phases (mg/m3 soil). |
With a total soil concentration CT (0.1 mg trichloroethylene/kg soil volume)
the amount of trichloroethylene in air ML can be established:
Equation 13
ML = fL · CT · r · 103 = 0.25 · 0.1 mg/kg · 1.7 kg/l · 103 = 43.3 mg/m3 soil volume
where:
Ct = trichloroethylene concentration in soil (0.1 mg/kg)
r = soil density (1.7 kg/l)
The trichloroethylene concentration in soil gas, CL, is now calculated on the basis of trichloroethylene concentration in soil, CT, see equation 14.
Equation 14
CL does not exceed CL,max, which means that there is no NAPL. If free CL exceeds CL,max , there is NAPL and CL,max is used in subsequent calculations.
As the calculated soil gas concentration under building is more than 100 times greater the evaporation criteria (0.001 mg/m3) and is more than 10 times greater than the evaporation criteria in the open-air area, soil gas measuring can be carried out. If a sufficient number of soil gas measurements show that the soil gas concentration is less than 0.1 mg/m3 under the building or less than 0.01 mg/m3 in the open-air area, the site can be cleared.
If the soil gas concentrations measured are more than 10 and 100 times greater, respectively, than the evaporation criterion, calculations are made of the diffusive contribution to outdoor air, cf. Section 2.2 and the diffusive and convective contributions to indoor air, cf. Sections 2.3 and 2.4.
2.2 Diffusive contribution to the contamination concentrations in outdoor air
Equation 15
J = 5.4 × 10-5 mg/(m× s)
| where: | J | = | flux (evaporation) (mg/(m2 · s)) |
| N | = | material constant (unitless) | |
| DL | = | diffusion coefficient of trichloroethylene in air (8.8 × 10-6 m2/s) | |
| X | = | depth corresponding to concentration CL (2.1 m) | |
| Co | = | background concentration at the site (mg/m3) is set at 0, as it is much smaller than CL. |
The material constant N for sand is calculated as:
Equation 18
N = VL3,33 / (VL + Vv)2 = 0.303,33/(0.30+0.15)2 = 0.09
Equation 21
where:
Cu = diffusive trichloroethylene contribution to outdoor air (mg/m3)
v = wind velocity (1 m/s).
The diffusive contribution to outdoor air is then 0.00068 mg/m3, which is less than the evaporation criterion of 0.001 mg/m3.
2.3 Diffusive contribution to indoor air contamination concentrations
Equation 17
| where: | J | = | flux (evaporation) (mg/(m2 · s)) |
| N1 | = | material constant for sand (0.09) | |
| N2 | = | material constant for concrete (0.002) | |
| DL | = | trichloroethylene diffusion coefficient in air (8.8·10-6 m2/s) | |
| X1 | = | thickness of sand strata (2.0 m) | |
| X2 | = | thickness of concrete (0.08 m) | |
| Co | = | the background concentration at the site (mg/m3) is set at 0, as it is a lot smaller than CL. |
Equation 18
N1 = VL3,33 / (VL + Vv)2 = 0.33,33/(0.3+0.15)2 = 0.09,
N2 = 0.002 corresponding to environmentally neutral concrete.
Equation 24
| where: | Ci | = | diffusive trichloroethylene contribution to indoor air (mg/m3) |
| Lh | = | building ceiling height (2.3 m) | |
| Ls | = | air renewal in building (8.3·10-5 s-1) |
This establishes the diffusive contribution to the indoor air as 0.10 mg/m3.
2.4 Convective contribution through concrete deck to indoor air contamination concentration
2.4.1 Calculating crack lengths and widths
The floor made of 8 centimetres of reinforced environmentally neutral concrete with 20 reinforcement bars, each with 3 mm tentor steel pr. 1000 mm; this corresponds to a concrete deck in accordance with Radonvejledningen (‘The Radon Guidelines’) /3/.
Equation 27
dw = k · da = 1 · 3 mm = 3 mm
| where: | dw | = | crack-determining diameters of reinforcement bars |
| da | = | nominal reinforcement diameter (3 mm) | |
| k | = | 1, due to the fact that the reinforcement is tentor steel |
Equation 29
| where: | aw | = | crack parameter (mm) |
| hb | = | thickness of concrete deck (80 mm) | |
| Db | = | reinforcement bar spacing (50 mm) |
The free shrinkage strain es can be
calculated as
Equation 30
es = ec · kb · kd · kt = 0.0333% · 1.035 · 0.866 · 0.989 = 0.0295 %
| where | es | = | shrinkage strain (%) |
| ec | = | base shrinkage (%), see Equation 30 | |
| kb | = | coefficient taking into account influences from the composition of concrete (unitless), see Equation 31 | |
| kd | = | coefficient taking into account influences from geometry (unitless), see Equation 33 | |
| kt | = | coefficient taking into account influences from shrinkage time (unitless), see Equation 34 |
Base shrinkage can be calculated as
Equation 31
where RF = relative humidity (60 %)
kb is calculated on the basis of the composition of concrete
Equation 32
kb = 0.007 · CM · (v/c + 0.333) · v/c
kb = 0.007 · 220 · (0.67 + 0.333) · 0.67 = 1.035
where
CM = cement content (220 kg/m3)
v/c = water/cement ratio (0.67)
The equivalent radius r and kd are calculated by means of the following formulae:
Equation 33
where r = equivalent construction radius (mm)
Equation 34
Influence of time:
Equation 35
where ts = shrinkage time (7,300 days)
to, a and b are auxiliary parameters (unitless)
Equation 36
2.4.2 Calculating reinforcement tension
The calculation is carried out in accordance with Beton-Bogen (‘The Concrete Book’) /1/.
The reinforcement ratio j is:
Equation 37
| where: | j | = | reinforcement ratio (unitless) |
| As | = | cross-section area of reinforcement (28.27 mm2) | |
| Ab | = | cross-section area of concrete (16,000 mm2). |
Elastic strain n is
Equation 38
| where: | n | = | elastic strain (unitless) |
| Es | = | coefficients of elasticity for steel (210,000 MPa) | |
| Eb | = | coefficients of elasticity for concrete (20,000 MPa). |
Compression stress of reinforcement
Equation 39
According to the standard specifications of concrete DS 411 /2/ the crack width can be calculated by means of the formula:
Equation 40
| where: | w | = | crack width (mm) |
| ss | = | reinforcement tension (60.8 MPa) | |
| aw | = | crack parameter (1,667 mm) |
The formula then applies to aw < 2,000.
2.4.3 Calculating crack spacing
According to Beton-Bogen (‘The Concrete Book’) /1/ the smallest crack spacing can be calculated as
Equation 41
where lm = smallest crack spacing (mm).
The average crack spacing is calculated as:
Equation 42
lw = 1.5 × lm = 1.5 · 424 mm = 636 mm
where lw = average crack spacing (mm).
Total crack length is calculated as:
Equation 43
| where: | ltot | = | total crack length (m) |
| ll | = | floor length (10 m) | |
| lb | = | floor width (10 m) |
Volume flow q pr. m2 floor area will be:
2.4.2 Calculating air transport through cracks
Equation 45
| where | q | = | volume flow pr. m2 floor area ((m3/s)/m2) |
| D P | = | pressure differential over concrete deck (5 Pa) | |
| m | = | dynamic viscosity of gas (1.8·10-5 (kg/m) · s) | |
| Ag | = | floor area (100 m2), i.e. ll · lb | |
| w | = | crack width (0.111 mm) | |
| hb | = | concrete-deck thickness (80 mm) | |
| ltot | = | total crack length (294 m) |
The concentration over floor Ck can be calculated as follows:
Equation 48
| N1 | = | material constant for sand = 0.09 |
| DL | = | trichloroethylene diffusion coefficient in air = 8.8 ·10-6 m2/s |
| CL | = | concentration of contaminants in soil gas at a contamination = 140 mg/m3 soil gas |
| x1 | = | thickness of sand stratum under floor = 2.0 m |
| Nb | = | material constant for concrete = 0.002 |
| xb | = | thickness of concrete deck = 0.08 m |
| q | = | 1.16 × 10-6 (m3/s)/m2 |
| LS | = | air renewal in building = 8.3 × 10-5 s-1 |
| Lh | = | building ceiling height 2.3 m |
When these values are substituted into the equation the result is:
Equation 48
The calculated diffusive contribution (Ci = 0.10 mg/m3) as well as the total contribution (Ck = 0.23 mg/m3) are greater than the evaporation criterion for trichloroethylene of 0.001 mg/m3.
As the evaporation criterion for indoor air has been exceeded, the indoor air must be investigated; alternatively remedial measures to ensure an acceptable indoor air must be carried out.
If soil gas measuring has been carried out, the relevant values are to be used in calculating equation 15 or 17 in Section 2.2 and 2.3, being substituted as CL while taking into account the distance to the measuring point Xn.
References
| /1/ | Herholdt, A.D., Justesen, C.F.P., Nepper Christensen, P.
& Nielsen, A. 1985: Beton-Bogen (‘The Concrete Book’). [Tilbage] |
| /2/ | Dansk Ingeniørforenings norm for betonkonstruktioner
(‘Danish standard specifications for concrete constructions’), 1984. Dansk
Standard DS411. [Tilbage] |
| /3/ | The National Housing and Building Agency, 1993: Radon og
Nybyggeri (‘Radon and New building’) [Tilbage] |