Page 247 - Materials Chemistry, Second Edition
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230       Practical Design Calculations for Groundwater and Soil Remediation



                               60  Y



                               30
                      Stagnation         Groundwater  ow
                        point
                                                                    X
                                0
                    –100         0          100         200         300

                              –30      Extraction
                                         well

                              –60

           FIGURE 6.4
           Capture zone of two extraction wells (Example 6.7).


               (b)  Determine the optimum distance between these two wells,
                   which is 0.32Q/Bu (from Table 6.1):

                                                   3
                       0.32Q  =  (0.32)(50 gal/min)(1 ft /7.48 gal)  = 19.2 ft
                         Bu       (80ft)(1.39  × 10 −3  ft/min)
                   The distance of each well to the origin is half of this value
                       = 0.16Q/Bu = 9.6 ft.
               (c)  Determine the sidestream distance from the well to the envelope
                   of the capture zone at the line of the pumping well, Q/2Bu:


                                               3
                           Q  =  (50gal/min)(1 ft /7.48 gal)  = 30.0 ft
                                                −3
                          2Bu   (2)(80ft)(1.39  × 10 ft/min)

               (d)  Determine the downstream distance from the well to stagnation
                   point of the envelope, Q/2πBu:

                                                3
                          Q   =   (50gal/min)(1 ft /7.48 gal)  = 9.6 ft
                                                  −3
                         2 πBu  (2)( π)(80ft)(1.39  × 10 ft/min)

               (e)  Determine the sidestream distance of the envelope far upstream
                   of the pumping wells, Q/Bu:

                                               3
                            Q  =  (50gal/min)(1 ft /7.48 gal)  = 60.0 ft
                                               −3
                           Bu   (80ft)(1.39  × 10 ft/min)
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