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ARCHIVED REPORTS_XR0008453
Environmental Health - Public
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3500 - Local Oversight Program
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PR0545864
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ARCHIVED REPORTS_XR0008453
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Last modified
7/21/2020 10:00:02 AM
Creation date
7/21/2020 9:53:31 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
3500 - Local Oversight Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0008453
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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i <br /> MASS-VOLUME CALCULATIONS OF EXTRACTED GASOLINE HYDROCARBONS <br />' Assumptions <br /> • Utilizing analytical data collected from air samples (Table 3)and from field measurements collected during <br /> I 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(m3/hr) <br /> t=operational period <br /> Mass-volume calculations were separated into the following four intervals <br /> I) 27 August to 27 September 1999 <br /> 2) 27 September to 18 October 1999 <br /> 3) 18 October to 19 November 1999 <br /> 4) 19 November to December 14 1999 <br /> IThe average TPH-g concentration and flow rates during each time interval was utilized for calculating the approximate <br /> mass/volume of extracted gasoline hydrocarbons <br /> 27 August to 27 September 1999 <br /> 1) The average TPH-g concentration for this time interval was determined to be 1,193 ugll The average flow rate <br /> was calculated at 0 70 inches of water or approximately 45 cubic feet per minute, the operational period was <br /> equal to 696 hours of running time <br /> C = 1193/2g/1=0 001 193 kg/m' <br /> Q = 0 70 inches water = 45 ft' <br /> /min <br /> t=696 hours <br /> Converting Q to m'/hour yields <br /> Q =(45 ft3/min) •(60 min/liour) •(0 0283168m3/ft3) =76 45 m3/hour <br /> M=C• Q • t=(0 001193 kg/m') •(76 45 m3/hour) • (696 hours) =63 47 kg of gasoline <br /> Converting kg of gasoline to pounds of gasoline yields <br /> 63 47 kg of gas - 2 2046 lbs of gas/kg of gas= 139 94 lbs of gasoline <br /> IConverting pounds of gasoline to gallons of gasoline yields <br /> I139 94 lbs of gas • 0 16 gal of gas/lbs of gas = 22 39 gallons of gasoline <br /> I <br /> I <br />
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