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ARCHIVED REPORTS XR0000757
Environmental Health - Public
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EHD Program Facility Records by Street Name
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BEVERLY
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2900 - Site Mitigation Program
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PR0540667
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ARCHIVED REPORTS XR0000757
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Entry Properties
Last modified
2/8/2019 9:49:27 AM
Creation date
2/8/2019 9:25:10 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
2900 - Site Mitigation Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0000757
RECORD_ID
PR0540667
PE
2960
FACILITY_ID
FA0023252
FACILITY_NAME
WELDON CHURCH PROPERTY
STREET_NUMBER
104
Direction
W
STREET_NAME
BEVERLY
STREET_TYPE
PL
City
TRACY
Zip
95376
CURRENT_STATUS
01
SITE_LOCATION
104 W BEVERLY PL
QC Status
Approved
Scanner
WNg
Tags
EHD - Public
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' HYDROCARBON-IMPACTED GROUND WATER <br /> VOLUME ASSUMPTIONS AND CALCULATIONS <br /> r <br />' Based on the average hydrocarbon concentration of TPH-g and benzene in impacted ground <br /> water samples collected at the site, the following calculation of mass of dissolve petroleum <br />' hydrocarbons has been made. <br /> Hydrocarbon-impacted ground water samples have been recovered from monitoring wells MW 1 <br />' and MW4. Over the last 5 ground water monitoring events, the average TPH-g and benzene <br /> concentrations between the two wells has been 1,187µg/1 and 164.1 µg/l, respectively. <br />' The plume of impacted ground water is assumed to extend laterally just to the margins of <br /> unimpacted wells in a regular, symmetrical distribution, as shown on Figure 5. Such a plume <br /> would be 180 feet long by 110 feet wide, in an elliptical pattern. The plume thickness is an <br />' assumed 15 feet for the purposes of this calculation. <br /> The area of an ellipse is given by: A = nab, where a and b are the length of the axes <br />' So... A = -n(90 ft 55 ft = 15,550 ft' <br />' Multiplying by the plume thickness yields the plume volume <br /> V =Ah= (15,550 ft2)(15 ft)=233,263 ft3 <br /> The impacted ground water occupies the pore spaces in the soil, so the volume of impacted <br /> water, assuming 25% effective porosity is <br /> V = (233,263 ft3 0.25 = 58,318 ft' <br />' Converting cubic feet to gallons and then to liters, we have <br />' VW =(58,318 ft)(7.48 gal/ft3)(3.785 llgal) = 1651030.9 1 <br /> Multiplying the volume of impacted water by the contaminant concentrations, divided by 109 <br />' yield the mass of each contaminant in kilograms; <br /> MTPH_g= (1651030.91)(1187 ,ug/1) ; 10' = 1.95977 kg gasoline, or about 2 kilograms. <br /> Mbenzene (— 1651030.91)(164.1 y 1) - 109 = 0.271 kg benzene. <br /> 1 <br /> 1 <br />' Advanced GeoEnvironmental,Inc. <br />
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