This appendix provides methods for estimating, as required in § 11.25 of this part, the areas where exposure of ground water or surface water resources may have occurred or are likely to occur. These methods may be used in the absence of more complete information on the ground water or surface water resources.
Ground Water
The longitudinal path length (LPL) factors in table 1 are to be applied in estimating the area potentially exposed downgradient of the known limit of exposure or of the boundary of the site. Estimates of lateral path width (LPW) are to be used when the LPW exceeds the width of the plume as determined from available data, or when the width of the plume at the boundary of the site is estimated as less than the LPW. In the absence of data to the contrary, the largest values of LPL and LPW consistent with the geohydrologic data available shall be used to make the estimates required in the preassessment screen. An example computation using the LPL and LPW factors follows table 1.
Table 1—Factors for Estimation of Areas Potentially Exposed Via the Ground Water Pathway
| Aquifer type
| Hyd. conductiv- ity/porosity factor (miles/year)
| Hydraulic gradient estimate (feet/mile)
| Time since release began (in years)
|
| Longitudinal path length (in feet)
| Lateral path width (in feet)
|
|---|
| Sand | 50 | × | × | = | | LPW = 0.2LPL
|
| Sand + silt | 0.5 | × | × | = | | LPW = 0.3LPL
|
| Gravel | 6000 | × | × | = | | LPW = 0.2LPL
|
| Sandstone | 0.01 | × | × | = | | LPW = 0.4LPL
|
| Shale | 3 × 10−6 | × | × | = | | LPW = 0.8LPL
|
| Karst Limestone or Dolomite | 10 | × | × | = | | LPW = 0.2LPL
|
| Limestone or Dolomite | 0.01 | × | × | = | | LPW = 0.4LPL
|
| Fractured Crystalline Rocks | 0.3 | × | × | = | | LPW = 0.3LPL
|
| Dense Crystalline Rocks | 1 × 10−5 | × | × | = | | LPW = 0.8LPL |
A release of hazardous substances occurs from a facility located in a glacial valley. Available data indicate the release may have occurred intermittently over a period of almost 1 year, although only one well about 300 feet downgradient of the facility boundary had detectable quantities of contaminants. The contaminated well is screened in the water table aquifer composed of gravelly sands. The facility boundary nearest the contaminated well is almost 3,000 feet in length, but a review of available data determined the release is probably localized along a 500-foot section of the boundary where a stream leaves the facility. Available water table data indicate hydraulic gradients in the valley range from 0.005 feet/mile up to 0.25 feet/mile near pumping wells. No pumping wells are known to be located near the release, and a mean hydraulic gradient of 0.1 feet/mile is estimated in the vicinity of the release site. Using the gravel factor from table 1, the LPL and LPW are estimated:
6000 × 0.1 × 1 = 600 feet (LPL)
and
600 × 0.2 = 120 feet (LPW).
Since the estimated LPW (120 feet) is less than the plume width (500 feet) determined from other available data, the greater number is used to compute the area potentially exposed:
(1) 600 feet × 500 feet = 300,000 square feet (about 6.9 acres). The available information allows an initial determination of area potentially exposed via the ground water pathway to be estimated:
(2) 300 feet × 500 feet = 150,000 square feet (about 3.5 acres).
The total area potentially exposed is the sum of (1) and (2):
6.9 + 3.5 = 10.4 acres.
Surface Water
The area of surface water resources potentially exposed should be estimated by applying the principles included in the examples provided below.
Example 1:A release occurs and most of the oil or hazardous substance enters a creek, stream, or river instantaneously or over a short time interval (pulse input is assumed). The maximum concentration at any downstream location, past the initial mixing distance, is estimated by:
Cp = 25(Wi)/(T
0.7 Q)
where Cp is the peak concentration, in milligrams/liter (mg/L),
Wi is the total reported (or estimated) weight of the undiluted substance released, in pounds,
Q is the discharge of the creek, stream, or river, in cubic feet/second, and
T is the time, in hours, when the peak concentration is estimated to reach a downstream location L, in miles from the entry point.
The time T may be estimated from:
T = 1.5(L)/Vs
where T and L are defined as above and
Vs is the mean stream velocity, in feet per second.
The mean stream velocity may be estimated from available discharge measurements or from estimates of slope of the water surface S (foot drop per foot distance downstream) and estimates of discharge Q (defined above) using the following equations:
for pool and riffle reaches Vs = 0.38(Q
0.40)(S
0.20), or
for channel-controlled reaches Vs = 2.69(Q
0.26)(S
0.28).
Estimates of S may be made from the slope of the channel, if necessary.
As the peak concentrations become attenuated by downstream transport, the plume containing the released substance becomes elongated. The time the plume might take to pass a particular point downstream may be estimated using the following equation:
Tp = 9.25 × 10
6 Wi/(QCp)
where
Tp is the time estimate, in hours, and Wi, Cp, and Q are defined above.
Example 2:A release occurs and most of the oil or hazardous substance enters a creek, stream, or river very slowly or over a long time period (sustained input assumed). The maximum concentration at any downstream location, past the initial mixing distance, is estimated by:
Cp = C(q)/(Q +
where Cp and Q are defined above,
C is the average concentration of the released substance during the period of release, in mg/L, and
q is the discharge rate of the release into the streamflow, in cubic feet/second.
For the above computations, the initial mixing distance may be estimated by:
Lm = (1.7 × 10−5)Vs B
2/(D
1.5 S
0.5)
where
Lm is the initial mixing distance, in miles,
Vs is defined above,
B is the average stream surface width, in ft,
D is the mean depth of the stream, in ft, and
S is the estimated water-surface slope, in ft/ft.
Example 3:A release occurs and the oil or hazardous substance enters a pond, lake, reservoir, or coastal body of water. The concentration of soluble released substance in the surface water body may be estimated by:
Cp = CVc/(Vw + Vc)
where
Cp and C are defined above,
Vc is the estimated total volume of substance released, in volumetric units, and
Vw is the estimated volume of the surface water body, in the same volumetric units used for Vc.
[51 FR 27725, Aug. 1, 1986, as amended at 52 FR 9100, Mar. 20, 1987]