Page 187 - Materials Chemistry, Second Edition
P. 187

170       Practical Design Calculations for Groundwater and Soil Remediation



              Discussion:
                1.  The answers from parts (a) and (b) should be the same. The
                   apparent difference is from truncation errors.
                2.  The  Darcy  velocity  at  20  feet  away  from  the  well  is  relatively
                   slow, at 1.4 m/day.


           Example 5.12:   Estimate the Radius of Influence of a Soil-Venting
                         Well by Using the Extracted Vapor Flow Rate
           Determine the radius of influence of a soil-venting well using the following
           information:

              •  Pressure at the venting well = 0.85 atm
              •  Flow rate measured at the extraction pump discharge = 0.21 m /min
                                                                       3
              •  Well screen length = 4 m
              •  Diameter of the venting well = 0.1 m
              •  Permeability of the formation = 1.0 darcy
              •  Viscosity of air = 1.8 ×10  poise
                                     −4
              •  Temperature of the formation = 20°C

              Strategy:
              This problem can be viewed as the reverse of Example 5.10, in which
                the radius of influence was given for estimation of the vapor extrac-
                tion flow rate. In this problem, the flow rate was given to estimate
                the radius of influence. As in the previous example, a few unit con-
                versions need to be performed first:

                 •  1 atm = 1.013 × 10  N/m 2
                                   5
                 •  1 darcy = 10  cm  = 10  m 2
                                   2
                              −8
                                       −12
                 •  1 poise = 100 centipoise = 0.1 N/s/m 2
                 •  (So, 0.018 centipoise = 1.8 × 10  poise = 1.8 ×10  N/s/m )
                                                            −5
                                             −4
                                                                    2
              Solution:
               (a)  The vapor flow rate entering the extraction well can be found by
                   using Equation (5.7):

                                    P well          0.85 
                            Q atm =    P atm   Q well =  0.21 =    1    Q well

                            Q well = 0.24 m/min= 0.004 m/s
                                        3
                                                      3
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