Page 239 - Materials Chemistry, Second Edition
P. 239

222       Practical Design Calculations for Groundwater and Soil Remediation



                   To determine the drawdown in the pumping well, set r at the
                       pum ping well = well radius = (2/12) ft = 0.051 m, and use
                       Equation (6.2):

                                         2
                              (200)( 2 h 2  − 35 )
                       40 =                      → h 2  = 29.2 ft
                           1,055 log[(2/12)/10]
                         or
                                                              2
                                    1.366 [(200)(0.0410)]( 2 h 2  − 10.7 )
                       [(0.15)(1440)] =                            → h 2  = 9.0 m
                                            log(0.051/3.0)
                   So the drawdown in the extraction well = 40 − 29.2 = 10.8 ft (or
                       = 12.2 − 9.0 = 3.2 m)
               (b)  To determine the radius of influence of the pumping well, set r at
                   the radius of influence (r ) at the location where the drawdown
                                         RI
                   is equal to zero. We can use the drawdown information of the
                   pumping well as:

                                        2
                          (200)(29.2 2  − 40 )
                     40  =                     → RI r  = 580 ft
                         1,055 log[(2/12)/ RI r  ]
                       or
                                                             2
                                  1.366 [(200)(0.0410)](9.0 2  − 12.2 )
                     [(0.15)(1440)] =                             → RI r  = 168 m
                                          log(0.051/ RI r  )
                   Similar results can also be derived from using the drawdown
                       information of the observation well as:

                                      2
                            (200)(35 2  − 40 )
                       40 =                 → RI r  = 598 ft
                           1,055 log[10/ RI r  ]
                         or
                                                            2
                                   1.366[(200)(0.0410)](10.7 2  − 12.2 )
                       [(0.15)(1440)] =                          → RI r  = 181 m
                                            log(3/ RI r  )
              Discussion:
                1.  As discussed in Example 6.2, the h –h  term in Equation (6.1) (for
                                                     1
                                                  2
                   confined aquifers) can be replaced by  s –s . However, no anal-
                                                          1
                                                       2
                   ogy can be made here, that is, h –h  in Equation (6.2) cannot be
                                                   2
                                                2
                                                   1
                                                2
                   replaced by s –s .
                               2
                                  2
                               1
                                  2
                2.  The  differences  in  the  calculated  r   values  in  part  (b)  come
                                                    RI
                    mainly from the unit conversions and data truncations.
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