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fav • A S S O C I A T E S I N C . During the short-term vapor extraction test on well VE2, extraction flowrates ranging from 2.2 to f . 5.5 cubic feet per minute (cfm) resulted in a vacuum of up to 0.30 inches of water in well VE3 located 46 feet away. Vapor well VE3 is screened from 30 to 50 feet bgs. During the long-term vapor extraction test on well VE3, extraction flowrates ranging from 53.5 to 78.9 cfin resulted in 1 a vacuum of up to 2.7 inches of water in well VE 1 located 16 feet away and 0.44 inches of water 1 in well MW4 located 55.5 feet away. Vapor well VE2 is screened from 50 to 70 feet bgs. Vapor well VEl is screened from 30 to 40 feet bgs. Monitoring well MW4 is screened from 70 to 90 feet bgs. is The actual radius of influence for each vapor extraction well will vary depending on subsurface f ' geologic conditions, the extent and distribution of hydrocarbon-impacted soil, and site-specific operating conditions of the vapor extraction system. The effective radius of influence should be evaluated following the installation and long-term operation of a vapor extraction system. f � During the long-term pilot test VOC vapor concentrations were measured at concentrations ranging from 180 to 2,832 parts per million by volume (ppmv). Laboratory results of the vapor f ' samples indicated that volatile hydrocarbons were detected in the influent samples. 1- Approximately 9.2 pounds of TPHg and BTEX were removed during the long vapor extraction pilot test as shown below: Average concentration of TPHg in two influent vapor samples collected from well VE3 yields: t = (6.2 ppmv+3200 ppmv) = 1,603 ppmv Y , 2 + Assuming an average flowrate of 67.4 cfm during the 5-hour vapor extraction test and using the values supplied by SunStar Laboratories of 24.5 liters of vapor per mole of TPHg and a molecular weight of 86.27 grams per mole of TPHg yields: 1 ` t; = 1.603 1 TPHq x 67.4 ft3 x 60 min x 28.32 1 air x 1 mole TPHg x 86.27g x 1 Ib 1,000,0001 air min hour T ftp— 24.5 1 vapor mole TPHg 454 g qr - - 1.4 Ib TPHg = 7.0 lbs of TPHg during the five hour test L hour The same calculation for benzene with 24.5 liters of vapor per mole of benzene and a molecular ` weight of 78 grams per mole of benzene is shown below: (� = (0 ppmv+44 ppmv) = 22 ppmv 2 r = 22 1 benzene x 67.4 W x 60 min x 28.32 1 air x 1 mole benzene x 78 .q x 1 Ib ` 1,000,000 I air min hour _Tft' 24.5 1 vapor mole benzene 454 g . = 0.0176 Ib benzene = 0.09 lbs of benzene during the five hour test hour SAEnviromentaIV5291treports4ifs.doc 4 i