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WELLS 4-37

                                          − 3
                            (225 )( .818 ×10 m/s )(4 2 . min )(60 s/min )
                              .
                                  2
                         S                                      =  1775 10   5

                                                                  .
                                          ( (300 m ) 2

                 where  r      300 m, the distance from the pumping well to the observation well.
                  (3)  Estimate drawdown
                  Calculate the drawdown with Equation 4-1.
                                        2
                                       3
                                                            5
                  With  T      2.818     10    m  /s,  S      1.775     10   , and a well diameter of 15 cm selected
                 in step (c) above ( r      15 cm/2     7.5 cm or 0.075 m), calculate  u  as
                                          2
                                  (0075m  ) (1775   10  5 )
                                             .
                                    .
                        u                                        1025   10   12
                                                                  .
                                          2
                                       3
                                                      ,
                             ( 4 2 818   10 m /s )(1100 d)( 86 400s/d)
                              .

                 and                                                         12 2             12 3
                                       .
                 Wu()   0 577216.   ln  1 025   10  12   1 025.   10  12   (1.025    10  )     (  . 1025  10  )
                                                                        22!             33!
                                                                                         ?
                                                                         ?
                         27 .03
                            3
                                  ⎛    1   ⎞
                               3
                          ( 32m/h ) ⎜ ⎝    ⎟ ⎠
                     s             3 600 s/h,  ( 27 03  ).       6 785 or.  68 m.

                                        3
                                           2
                         4 ()( )( 2 818 .  10 m/s)
                                      0
                           p
                  The 6.8 m is measured from the static piezometric surface that is 5.2 m below grade. The
                 drawdown piezometric surface will be a total of 5.2     6.8     12.0 m below the ground
                 surface. The top of the artesian aquifer is 150.8 m below grade. Thus, there is no poten-
                 tial for failure of the well due to dewatering the aquifer.
               e.  Well depth, casing length, and grout requirements
                    (1)  From the well log, select the water-bearing aquifer
                       The well penetrates an impervious layer of shale, extends through a sand layer, and
                     terminates in shale. From the well log and the static water level in the well, the sand
                     layer is determined to be an artesian aquifer.
                  (2) A fully penetrating well is selected. Therefore, the well depth is 189.5 m.
                  (3) The casing length and grout requirements are based on MSDWA rules, that is:
                       For artesian aquifers, the casing is sealed to the upper confining layer from within
                     1.5 m of the top of the aquifer to the ground surface.
               f.  Slot size
                   Plot the grain size analysis as “Cumulative % Retained” versus “Grain Size, mm.” The
                 points in  Figure 4-12  were plotted using the data provided for this example. Note that the
                 data are presented as “Cumulative % passing” and that they are plotted as “Cumulative
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