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ARCHIVED REPORTS_XR0008355
EnvironmentalHealth
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3500 - Local Oversight Program
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PR0545864
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ARCHIVED REPORTS_XR0008355
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Last modified
7/21/2020 9:58:42 AM
Creation date
7/21/2020 9:35:35 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
3500 - Local Oversight Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0008355
RECORD_ID
PR0545864
PE
3528
FACILITY_ID
FA0004530
FACILITY_NAME
MARLOWE PROPERTY
STREET_NUMBER
4648
STREET_NAME
WATERLOO
STREET_TYPE
RD
City
STOCKTON
Zip
95215
CURRENT_STATUS
02
SITE_LOCATION
4648 WATERLOO RD
P_LOCATION
99
P_DISTRICT
002
QC Status
Approved
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EHD - Public
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1 <br />' MASS-VOLUME CALCULATIONS OF EXTRACTED GASOLINE HYDROCARBONS <br />' Assumptions <br /> • Utilizing analytical data collected from air samples(Table 3) and from field measuiemerts collected during <br />' the SVE pilot test(Table 4),the volume of extracted gasoline hydrocarbons at the site can be approximated <br /> The hydrocarbon mass removed during the operating period can be calculated using the following equation <br />' M =C • Q • t <br /> where M =cumulative mass recovered(kg) <br />' C= vapor concentration (kg/m') <br /> Q= extraction flow rate(m'/hr) <br /> t=operational period <br />' Mass-volume calculations were separated into the following four intervals <br /> 1) 23 July to 06 August 1999 <br /> I 2) 06 August to 16 August 1999 <br /> 3) 16 August to 20 August 1999 <br /> 4) 20 August to August 27 1999 <br /> I <br /> The average TPH-g concentration and flow rates during each time interval was utilized for calculating the approximate <br /> 10 <br /> mass/volume of extracted gasoline hydrocarbons <br /> 28 July to 06 August 1999 <br /> 1) The average TPH-g concentration for this time interval was determined to be 440�gll The average flow rate <br /> was calculated at 1 45 inches of water or approximately 62 cubic feet per minute, the operational period was <br /> equal to 174 hours of running time <br /> C = 440,ug/l =0 000440 kg/m' <br /> Q ; 1 45 inches water = 62 ft'/min <br /> I= 174 hours <br /> Converting Q to m'/hour yields <br /> Q= (62 ft'/mm)-(60 min hour)•(0 0283168m'/ft')= 105 34 m'/hour <br /> M w C • Q - t=(0 000440 kg/m') •(105 34 m'/hour) - (174 hours)= 8 06 kg of gasoline <br /> Converting kg of gasoline to pounds of gasoline yields <br /> I8 06 kg of gas • 2 2046 lbs of gaslk-of gas = 17 77 lbs of gasoline <br /> Converting pounds of gasoline to gallons of gasoline yields <br /> 17 77 lbs of gas • 0 16 gal of gas/lbs of gas =2 84 gallons of gasoline <br /> I <br />
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