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2900 - Site Mitigation Program
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PR0540534
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Last modified
2/10/2020 4:13:24 PM
Creation date
2/10/2020 3:33:12 PM
Metadata
Fields
Template:
EHD - Public
ProgramCode
2900 - Site Mitigation Program
File Section
WORK PLANS
RECORD_ID
PR0540534
PE
2960
FACILITY_ID
FA0023179
FACILITY_NAME
SUPER CENTER MART
STREET_NUMBER
701
Direction
E
STREET_NAME
CHARTER
STREET_TYPE
WAY
City
STOCKTON
Zip
95206
APN
14734322
CURRENT_STATUS
01
SITE_LOCATION
701 E CHARTER WAY
P_LOCATION
01
QC Status
Approved
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EHD - Public
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APPENDIX C <br /> PRE-REMEDIATION DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS <br /> PAGE 3 OF 5 <br /> Multiplying the volume of water by the average dissolved TPH-g concentration (52,200 pg/1)yields <br /> the approximate mass of TPH-g in groundwater: <br /> V'100= (323,010/) (0.000053 kg//)= 17.11 kg TPH-g <br /> A conversion factor of 2.205 Ib/kg.can be used to convert kilograms of TPH-g to pounds of TPH-g: <br /> MTPHy = (17.11 kg TPH-g)(2.205 Ib/kg) = 37.75 lbs. TPH-g <br /> Dividing M100 by the weight of one gallon of TPH-g, 6.25 lbs/gallon, will yield the approximate <br /> volume of TPH-g dissolved into the groundwater of the middle contour interval: <br /> V10,000 = (M10,000)/6.25lb/gal) = 37.75 lbs./6.25 lbs/gal = 6.04 gallons of dissolved TPH-g <br /> 3) For the estimated 1,000 pg/l second outer contour interval, A1,000 = 4,147 ft2, h = 20 ft <br /> (average thickness) <br /> Utilizing TPH-g data from wells MW-6, (2,300 lag/1), MW-8A (4,000 pg/1) the 10,000 pg/l contour <br /> line and the 1,000 pg/l (Figure 5), the average TPH-g concentration is estimated to be 4,325 pg/l <br /> (equivalent to 0.000004325 kg/1). <br /> Area: V1,000=A1,000 x h =4,147 ft2 x 20 ft = 82,940 ft2 <br /> The volume is given by: V1,000=V1,000 410,000 = 82,940 ft3-39,700 ft3=43,240 ft3 <br /> Water occupies the porosity in the soil, which is estimated to be 30% of the soil volume, so the <br /> total volume of water in the saturated portion of the ellipse is approximated by: <br /> V1,000= (0.30)(43,240 ft3)= 12,972 ft3 <br /> One ft3 is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: <br /> V1000= (12,972 ft3)(7.48 gal/ft3)= 97,030 gallons <br /> One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by: <br /> V1,000= (97,030 gallons)(3.79 Ugal)= 367,745/water <br /> Multiplying the volume of water by the average dissolved TPH-g concentration (4,325 pg/1)yields <br /> the approximate mass of TPH-g in groundwater: <br /> V1000= (367,745/) (0.000004325 kg//)= 1.5 kg TPH-g <br /> A conversion factor of 2.205 lb/kg.can be used to convert kilograms of TPH-g to pounds of TPH-g: <br /> MTPH-g = (1.5 kg TPH-g)(2.205 Ib/kg) = 3.5 lbs. TPH-g <br />
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