Page 52 - Materials Chemistry, Second Edition
P. 52

Site Assessment and Remedial Investigation                        35



             It was also found that free-floating gasoline product was present in the
           two monitoring wells located within the excavated area. The apparent
           thickness of the product in each of these two wells was converted to its
           actual thickness in the formation, and they are 1 and 2 ft, respectively.
           The effective porosity and total bulk density of soil are 0.35 and 1.8 g/
           cm , respectively.
              3
             Assuming that the leakage impacted a rectangular block of soil, defined by
           the bottom of the tank pit and the water table, with length and width equal
           to those of the tank pit, estimate the following:

             (a)  Total volume of the soil stockpiles (in cubic yards)
             (b)  Mass of TPH in the stockpiles (in kilograms)
             (c)  Volume of the impacted soil left in the vadose zone (in cubic meters)
             (d)  Mass of TPH, benzene, and toluene in the vadose zone (in kilograms)
             (e)  Mass fraction and mole fraction of benzene and toluene in the leaked
                gasoline
             (f)  Volume (in gallons) and mass (in kilograms) of the free product
             (g)  Total volume of gasoline leaked (in gallons) [Note: Neglect the dis-
                solved phase in the underlying aquifer.]



              Solution:
               (a)  Assuming a fluffy factor of 1.2, total volume of the soil stockpiles
                   	   = (volume of the soil in situ)(soil fluffy factor)
                   	   =  [(volume of tank pit) − (volume of USTs)](soil fluffy factor)
                   	   = [(20′ × 30′ × 18′) − (3)(5,000 gal)(ft /7.48 gal)](1.2)
                                                     3
                   	   = (8,795 ft )(1.2) = 10,550 ft  = 391 yd 3
                               3
                                             3
               (b)  Mass of TPH in the stockpiles
                   	   = [(V)(ρ )](X) = (M )(X)
                             t
                                      s
                   where
                                 X = (10 + 200 + 400 + 800)/4 = 352.5 mg/kg
                       ρ  (in situ soil) = (1.8 g/cm ) (28,317 cm /ft )(kg/1,000 g)
                                                           3
                                                        3
                                             3
                        t
                       	  		       = 51.0 kg/ft  = 1,376 kg/yd 3
                                             3
                       ρ  (stockpiles) = ρ  (in situ soil) ÷ (soil fluffy factor)
                                      t
                        t
                       	  		       = 51.0 kg/ft  ÷ 1.2 = 42.5 kg/ft 3
                                             3
                ∴  Mass of TPH in the stockpiles
                       = [(8,795 ft )(51.0 kg/ft )](352.5 mg/kg)
                                          3
                                3
                   	   = [(10,550 ft )(42.5 kg/ft )](352.5 mg/kg)
                                           3
                                 3
                   	   = 1.58 × 10  mg = 158 kg
                                8
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