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ARCHIVED REPORTS_XR0011932
EnvironmentalHealth
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EHD Program Facility Records by Street Name
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W
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WILSON
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101
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2900 - Site Mitigation Program
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PR0541653
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ARCHIVED REPORTS_XR0011932
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Last modified
10/10/2020 10:49:58 PM
Creation date
7/9/2020 8:22:05 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
2900 - Site Mitigation Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0011932
RECORD_ID
PR0541653
PE
2965
FACILITY_ID
FA0023871
FACILITY_NAME
TOP FILLING STATION
STREET_NUMBER
101
Direction
S
STREET_NAME
WILSON
STREET_TYPE
WAY
City
STOCKTON
Zip
95205
APN
15125307
CURRENT_STATUS
01
SITE_LOCATION
101 S WILSON WAY
P_LOCATION
01
QC Status
Approved
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LSauers
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EHD - Public
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. APPENDIX G <br /> SOIL VAPOR EXTRACTED VOLUME-MASS CALCULATIONS <br />' TOP FILLING STATION <br /> 101 South Wilson Way, Stockton, California <br /> Assumptions <br />' Utilizing field measurements and analytical data collected from air samples between 22 June <br /> 2005 and 07 September 2005, the volume and mass of extracted gasoline hydrocarbons at <br /> the site can be approximated <br /> The hydrocarbon volume-mass removed dune the o <br />' following equation g P eratin <br /> g period can be calculated using the <br /> M=C • Q • t <br />' where M=cumulative mass recovered (kg) <br /> C =vapor concentration (kg/m') <br /> rQ=extraction flow rate (m'/hr) <br /> t=operational period(hrs) <br /> 1 <br /> 22 June 2005 - 15 July 2005 <br />' The average analytical TPH-g concentration for this time interval was determined to be 1,440 1 <br /> (06-22-05= 480 µg/1 and 07-15-05 = 2,400 µlb <br /> µg/1, avg = 1,440 µg/1) The average flow rate was <br /> calculated at 145 inches of water or approximately 59 cubic feet per minute (nomograph), the <br /> operational period was equal to 552 hours of Winning time (Table 6) <br /> TPH-g(total petroleum hydrocarbons quantified as gasoline) <br /> C = 1,440 µg/1= 0 001440 kg/m' <br /> Q = 145 inches water = 59 ft'/min <br /> t=552 hours (23 days) <br /> Converting Q to m'/hour yields <br /> Q = (59 fl/min) • (60 min/hour) • (0 0283168m'/f}') = 100 24 m3/hour <br /> M= C • Q • t=(0001440k m3 1002 ' <br /> kg/m') ( 4 m /hour) • (552 hours) =79.7 kg of TPH-g <br /> Converting kg of TPH-g to pounds of TPH-g yields <br /> Advanced GeoEnvironmeiiw,Inc <br />
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