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IMASS CALCULATIONS OF EXTRACTED HYDROCARBONS (Gasoline) LODITHEATER 107 North School Street, Lodi, California • The hydrocarbon mass removed during the operating period can be calculated using the following equation M = C•Q•t where M = cumulative mass recovered(kg) C = vapor concentration(kg/m3) Q= extraction flow rate (m3/hr) t= operational period(hrs) 12-10-01 to 01-29-02 using C = (9,171 +2,156 +4,257)/ 3 µg/1= 0 0052 kg/m3 (TPH-g and BTEX) Q = 35 cfm(average) =60 m3/hr I t = 1008 (sum of known operation) 0 0052 • 60 • 1008 = 314 kg of gasohne 314 kg • 2 2046 lbs/kg= 692 6 lbs of gasoline ' to convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs 692 6 lbs of gal • 0 16 gal/lbs = 110 gallons of gasoline I01-29-02 to 02-11-02 using C = (4,257 +3,870) /2 gg/l= 0 004 kg/rn' (TPH-g and BTEX) Q= 35 cfm(average) = 60 m3/hr t = 288 (sum of known operation) 0 004 • 60 • 288 = 69 kg of gasoline 69 kg • 2 2046 lbs/kg= 152 4 I ofgasoline to convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs 152 4 lbs of gal • 0 16 galllbs= 24 4 gallons of gasoline 02-11-02 to 03-11-02 using C= (3,870+3480) /2 µg/l= 0 0036 kg/m3 (TPH-g and BTEX) Q= 30 cfm(average)= 50 m3/hr t = 718 (sum of known operation) 0 0036 • 50 • 718 = 129 24 kg of gasoline 129 24 kg • 2 2046 lbs/kg= 285 lbs of gasoline to convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs 285 Is of gal • 0 16 gal/lbs =45 6 gallons of gasoline 03-11-02 to 04-15-02 I using C = (3480+3,333)/ 2 go= 0 0034 kg/rn? (TPH-g and BTEX) Q= 30 cfm(average)= 50 m3/hr t = 840(sum of known operation) 0 0034 • 50 • 840= 143 kg of gasoline 143 kg • 2 2046 lbs/kg= 315 lbs of gasoline to convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs • 315 lbs of gal • 0 16 gal/lbs = 50 gallons of gasoline A total of 230 gallons or 656 kg of gasolme (TPH-g and BTEX) were removed over St three months of re-mttlatzon of the SVE system I Advanced GeoEnvfronmental,Inc