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SITE INFORMATION AND CORRESPONDENCE
Environmental Health - Public
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2900 - Site Mitigation Program
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PR0505525
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SITE INFORMATION AND CORRESPONDENCE
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Last modified
7/11/2019 10:54:12 PM
Creation date
7/11/2019 4:50:05 PM
Metadata
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Template:
EHD - Public
ProgramCode
2900 - Site Mitigation Program
File Section
SITE INFORMATION AND CORRESPONDENCE
RECORD_ID
PR0505525
PE
2953
FACILITY_ID
FA0002387
FACILITY_NAME
KEYSTONE AUTOMOTIVE INDUSTRIES INC
STREET_NUMBER
632
Direction
S
STREET_NAME
EL DORADO
STREET_TYPE
ST
City
STOCKTON
Zip
95203
APN
14907033
CURRENT_STATUS
02
SITE_LOCATION
632 S EL DORADO ST
P_LOCATION
01
P_DISTRICT
001
QC Status
Approved
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FORMER STOCKTON PLATING FACILITY <br /> 632 South El Dorado Street, Stockton, California <br /> Assumptions: <br /> • The distribution of dissolved hydrocarbon concentrations on the site can:be approximated by <br /> an ellipse (See Figure 1); and <br /> • The effective porosity of soil at the site is estimated to be 30%. <br /> • Data collected from the June 2002 ground water sampling event are representative of <br /> existing site conditions. <br /> The area of an ellipse can be calculated by the formula: <br /> Ae = (long radius, a)(short radius, b)(.7854) <br /> For the estimated ellipse, a= 75 ft, b=45 ft, and thickness c —45 ft <br /> The average TPH--g concentration for the ellipse is estimated to be 66,785 micrograms per liter, this <br /> concentration is equivalent to 0.00066785 grams per ml or 6.6 x 109. A ml is very nearly equivalent <br /> to a cubic centimeter of water, which by definition equals one gram, this concentration is nearly a <br /> unitless number. <br /> The area of the ellipse is given by:. <br /> Ae = 75(45)(0.7854) = 2650 ft2 <br /> The volume of the ellipse is given by: - <br /> Ve =A. x c = 2650 ft2 x 45 ft = 119,282 ft3 <br /> Water occupies the porosity in the soil,which is estimated to be 30%of the soil volume, so the total <br /> volume of water in the saturated portion of the ellipse is approximated by: <br /> Ve = (.30)(119,282ft') = 35,784.78 ft' <br /> One cubic foot of water contains 7.48 gallons,which weighs 8.337 lbs/gal,so the.'volume of the water <br /> in the'ellipse is given by: <br /> Ve = (38,784.78 ft)(7.48 gal/ft') =267,670 gallons <br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by: <br /> Me = (267,670 gallons)(8.337 lb/gal) = 2,231,566.5 lbs <br /> Multiplying Me by the hydrocarbon concentration yields the approximate mass of dissolved <br /> hydrocarbons in the saturated portion of the ellipse: <br /> Meg (2,231,566.5 ibs)(0.00066785) 1,4.901bs gasoline <br />
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