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I <br /> Table 2. Total Mass of Dissolved Benzene <br /> Dissolved Benzene plume calculation summary: <br /> For the calculated volumes the Benzene mass range is 90-135 grams. <br /> V, contains a total plume mass of 90 grams of Benzene. <br /> V2 contains a total plume mass of 1.34 grams of Benzene. <br /> A)Benzene area of plume with concentrations greater than I Oug/L (ppb) <br /> MW-21 =814ug/L ��Y � 1 = AN vn If :- <br /> RI <br /> RI =40' _ 000 cod IdIQ t6 f t <br />' <br /> R2=20' <br /> A=n(10)(20) =628 ft2 <br /> Vi =(628 ft)(10 ft)(0 58) =3,642 f ' (28.317 L/f3)= 103,000 L groundwater <br />' V2 =(628 ft)(15 ft)(0.58) =5,463 ft (28 317 L/ft) = 155,000 L groundwater <br /> 103,000 L x 814 ug/L=8 396E+07 ug benzene=84 g Benzene ,67ce3 g <br /> 155,000 L x 814 u = 1 259E+08u benzene= 126 S Benzene <br />' B ) Benzene area of plume with concentrations greater than 1 ug/L (ppb) <br /> MW-3R= 9 1 ugIL, MW-2=3 ug/L, AVG=6 ug/L <br /> Ri = 100' <br />' R2= 80' <br /> A=7c(30)(100)=9420 f 2 <br /> V1 =(9420 ft2)(10 ft)(0 58)= 36,424 f ' (28.317 L/fi)= 1.031E+06 L <br />' V2 =(9420 ft)(15 ft)(0 58)= 54,636 ft(28.317 L/fl)= 1.547E+06 L <br /> Note: Volumes for the outer plume must have the previously calculated volume deducted. <br />' Vi = 1 031E+06 L— 103,000 L=9.284E+051., groundwater <br /> V2 = I.547E+06 L— 155,000 L= 1.392E+06 L groundwater <br /> 9 284E+05 L x 6 ug/L=5 571E+06 ug benzene=5.6 g Benzene <br />' 1 392E+06 L x 6 u = 8 353E+06 ug benzene=8..4&B <br /> I <br /> I <br /> • <br />' 3 <br />