{"id":1889,"date":"2026-02-14T04:29:15","date_gmt":"2026-02-14T04:29:15","guid":{"rendered":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/chapter\/ch106-xhtml\/"},"modified":"2026-08-01T13:59:38","modified_gmt":"2026-08-01T13:59:38","slug":"solution-exercise-5","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/chapter\/solution-exercise-5\/","title":{"raw":"Solution Exercise 5","rendered":"Solution Exercise 5"},"content":{"raw":"
\r\n
\r\n
\r\n\r\nA production well was designed to yield 2,000 m3<\/sup>\/d from a 40 m thick confined gravel\u2011rich aquifer. The well was 40 cm in diameter and screened over 35 m. Once the well was completed, a step test was conducted by pumping the well at 1,400 m3<\/sup>\/d, 1,790 m3<\/sup>\/d and then 2,520 m3<\/sup>\/d for a total of 90 minutes with each step lasting 30 min. The time\u2011drawdown data and a semi\u2011log plot of the time\u2011drawdown data are presented here.\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\n<\/div>\r\n
\r\n\r\nStep\u2011drawdown test data for a production well. a) Time\u2011drawdown data for three steps. b) Plot of the log of time versus drawdown.\r\n\r\n<\/div>\r\n
\r\n\r\nAn Excel\u00ae data base of the time\u2011drawdown data is available on the web page for this book<\/a>\u2197.\r\n\r\n<\/div>\r\n
    \r\n \t
  1. Calculate the value of C<\/em> and B<\/em> for this system. Compute the well loss expected when pumping the well at 2,000 m3<\/sup>\/d.<\/li>\r\n<\/ol>\r\n
    \r\n\r\n\"\"\r\n\r\n<\/div>\r\n
    \r\n\r\nPlot of drawdown versus the log of time showing the calculation of \u2206s values. \u2206sstep1<\/sub><\/em> for the first 30\u2011minute step is computed as the difference from the drawdown at zero time and the drawdown at 30 min (double arrow). \u2206 sstep2<\/sub><\/em> is computed by subtracting the drawdown at 60 min from the extrapolated drawdown value from the first step (double arrow). \u2206sstep3<\/sub> is computed in a similar fashion from the drawdown at 90 minutes.\r\n\r\n<\/div>\r\n
    \r\n
    A graph of s<\/em>\/QT<\/sub><\/em> versus total Q<\/em> is prepared by computing the ratios of \u2206s<\/em>\/Q<\/em> and Q<\/em>.<\/blockquote>\r\n<\/div>\r\n
    \r\n
    \r\n\r\n[latex]\\frac{\\Delta s_{step1}}{Q_{1}} = \\frac{4.1\\ m}{1400\\frac{m^{3}}{d}} = 0.0029\\frac{d}{m^{2}}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n\r\n[latex]\\frac{\\Delta s_{step1} + \\Delta s_{step2}}{Q_{2}} = \\frac{4.1\\ m\\ + \\ 1.42\\ m}{1790\\frac{m^{3}}{d}} = 0.003\\frac{d}{m^{2}}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n\r\n[latex]\\frac{\\Delta s_{step1} + \\Delta s_{step2} + \\Delta s_{step3}}{Q_{3}} = \\frac{4.1\\ m\\ + \\ 1.42\\ m + 2.78\\ m}{2520\\frac{m^{3}}{d}} = 0.0033\\frac{d}{m^{2}}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n\r\n\"\"\r\n\r\n<\/div>\r\n
    \r\n\r\nPlot of \u2206s<\/em>\/Q<\/em> versus the log of the corresponding total QT<\/sub><\/em> for each step. The slope of the line (C<\/em>) is computed as the difference in the \u2206s<\/em>\/Q<\/em> values (e.g., horizontal red arrows) divided by the corresponding difference in pumping rate (vertical red arrows).\r\n\r\n<\/div>\r\n
    \r\n
    Reading values to define the slope from the graph, C<\/em> is determined as follows.<\/blockquote>\r\n<\/div>\r\n
    \r\n
    \r\n\r\n[latex]\\begin{aligned} C &= \\ \\frac{0.0031\\frac{d}{m^{2}}\\ - \\ 0.0028\\frac{d}{m^{2}}}{2000\\frac{m^{3}}{d}\\ - 1000\\frac{m^{3}}{d}} \\\\ &= 0.0000003\\ or\\ 3x10^{- 7}\\frac{d^{2}}{m^{5}} \\end{aligned}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    B<\/em> is read from the graph as the y\u2011intercept at QT<\/sub><\/em> = 0 as B<\/em> = 0.00248 d\/m2<\/sup>.<\/blockquote>\r\n<\/div>\r\n
    \r\n
    The well loss (CQ<\/em>2<\/sup>) when pumping at 2,000 m3\/<\/sup>d would be computed as follows.<\/blockquote>\r\n<\/div>\r\n
    \r\n
    \r\n\r\n[latex]well\\ loss = \\ 0.0000003\\ \\ d^{2}\/{m\\ }^{5}\\ \\ (2000\\ m^{3}\/d)^{2}\\ \\ = \\ 1.2\\ m[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n \t
    1. Estimate the total drawdown after pumping the well for 30 min at 2,000 m3<\/sup>\/d.<\/li>\r\n<\/ol>\r\n
      \r\n
      The total drawdown (formational drawdown + well loss) when pumping at 2,000 m3<\/sup>\/d for 30 minutes is calculated as follows.<\/blockquote>\r\n<\/div>\r\n
      \r\n
      \r\n\r\n[latex]s_{T} = BQ + CQ^{2}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n
      \r\n\r\n[latex]s_{T} = 0.0028\\frac{d}{m^{2}}\\left( 2000\\frac{m^{3}}{d} \\right) + 3x10^{- 7}\\frac{d^{2}}{m^{5}}\\left( 2000\\frac{m^{3}}{d} \\right)^{2}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n
      \r\n\r\n[latex]s_{T} = 5.6\\ m + 1.2\\ m = 6.8\\ m[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n
      To compute total drawdowns at later times new values of B<\/em> would need to be determined as B<\/em> is time dependent. This would require inputting a value of T<\/em> and S in the appropriate equation (e.g., Cooper\u2011Jacob) and recomputing B for a prescribed period of pumping (t<\/em>).<\/blockquote>\r\n<\/div>\r\n
        \r\n \t
      1. Estimate transmissivity using the first 30 minutes of time\u2011drawdown data (step 1) (use the Cooper\u2011Jacob method).<\/li>\r\n<\/ol>\r\n
        \r\n\r\n\"\"\r\n\r\n<\/div>\r\n
        \r\n\r\nSemi\u2011log plot of the time drawdown data for step one. A straight\u2011line fit is performed and a value of \u2206slog-t<\/sub><\/em> determined. This value is then used in the Cooper\u2011Jacob analysis to estimate a value of T<\/em>. As the time drawdown data contains some well loss the storage coefficient is not computed.\r\n\r\n<\/div>\r\n
        \r\n
        The Cooper\u2011Jacob solution found in Section 8.2.3.<\/blockquote>\r\n<\/div>\r\n
        \r\n
        \r\n\r\n[latex]T = 2.3\\frac{Q}{4\\pi\\Delta s{\\Delta s}_{log - t}} = 2.3\\frac{1400\\ \\frac{m^{3}}{d}}{4\\ (3.14)\\ 1.2\\ m} = 214\\frac{m^{2}}{d}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
        \r\n
        The time drawdown data in step one has some well loss so T<\/em> can be determined but S<\/em> is not estimated as the projected t0<\/sub><\/em> value would be too low (Section 12.1.4). If all the step 1 data were corrected for well loss (sT<\/sub><\/em> \u2011 CQ<\/em>2<\/sup>) S could be computed from a semi\u2011log plot of the corrected data. For example, at 10 min the recorded drawdown is 3.52\u00a0m. The corrected drawdown (s<\/em>) would be as follows.<\/blockquote>\r\n<\/div>\r\n
        \r\n
        \r\n\r\n[latex]s = 3.52\\ m - 3x10^{- 7}\\frac{d^{2}}{m^{5}}\\left( 1400\\frac{m^{3}}{d} \\right)^{2} = 2.93\\ m[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
        \r\n
        Return to Exercise\u00a05<\/a><\/blockquote>\r\n<\/div>\r\n
        \r\n
        Return to where text linked to Exercise\u00a05<\/a><\/blockquote>\r\n<\/div>\r\n<\/div>\r\n
        <\/div>\r\n<\/div>","rendered":"
        \n
        \n
        \n

        A production well was designed to yield 2,000 m3<\/sup>\/d from a 40 m thick confined gravel\u2011rich aquifer. The well was 40 cm in diameter and screened over 35 m. Once the well was completed, a step test was conducted by pumping the well at 1,400 m3<\/sup>\/d, 1,790 m3<\/sup>\/d and then 2,520 m3<\/sup>\/d for a total of 90 minutes with each step lasting 30 min. The time\u2011drawdown data and a semi\u2011log plot of the time\u2011drawdown data are presented here.<\/p>\n<\/div>\n

        \n

        \"\"<\/p>\n<\/div>\n

        \n

        Step\u2011drawdown test data for a production well. a) Time\u2011drawdown data for three steps. b) Plot of the log of time versus drawdown.<\/p>\n<\/div>\n

        \n

        An Excel\u00ae data base of the time\u2011drawdown data is available on the web page for this book<\/a>\u2197.<\/p>\n<\/div>\n

          \n
        1. Calculate the value of C<\/em> and B<\/em> for this system. Compute the well loss expected when pumping the well at 2,000 m3<\/sup>\/d.<\/li>\n<\/ol>\n
          \n

          \"\"<\/p>\n<\/div>\n

          \n

          Plot of drawdown versus the log of time showing the calculation of \u2206s values. \u2206sstep1<\/sub><\/em> for the first 30\u2011minute step is computed as the difference from the drawdown at zero time and the drawdown at 30 min (double arrow). \u2206 sstep2<\/sub><\/em> is computed by subtracting the drawdown at 60 min from the extrapolated drawdown value from the first step (double arrow). \u2206sstep3<\/sub> is computed in a similar fashion from the drawdown at 90 minutes.<\/p>\n<\/div>\n

          \n

          A graph of s<\/em>\/QT<\/sub><\/em> versus total Q<\/em> is prepared by computing the ratios of \u2206s<\/em>\/Q<\/em> and Q<\/em>.<\/p><\/blockquote>\n<\/div>\n

          \n
          \n

          [latex]\\frac{\\Delta s_{step1}}{Q_{1}} = \\frac{4.1\\ m}{1400\\frac{m^{3}}{d}} = 0.0029\\frac{d}{m^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          [latex]\\frac{\\Delta s_{step1} + \\Delta s_{step2}}{Q_{2}} = \\frac{4.1\\ m\\ + \\ 1.42\\ m}{1790\\frac{m^{3}}{d}} = 0.003\\frac{d}{m^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          [latex]\\frac{\\Delta s_{step1} + \\Delta s_{step2} + \\Delta s_{step3}}{Q_{3}} = \\frac{4.1\\ m\\ + \\ 1.42\\ m + 2.78\\ m}{2520\\frac{m^{3}}{d}} = 0.0033\\frac{d}{m^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n

          \"\"<\/p>\n<\/div>\n

          \n

          Plot of \u2206s<\/em>\/Q<\/em> versus the log of the corresponding total QT<\/sub><\/em> for each step. The slope of the line (C<\/em>) is computed as the difference in the \u2206s<\/em>\/Q<\/em> values (e.g., horizontal red arrows) divided by the corresponding difference in pumping rate (vertical red arrows).<\/p>\n<\/div>\n

          \n

          Reading values to define the slope from the graph, C<\/em> is determined as follows.<\/p><\/blockquote>\n<\/div>\n

          \n
          \n

          [latex]\\begin{aligned} C &= \\ \\frac{0.0031\\frac{d}{m^{2}}\\ - \\ 0.0028\\frac{d}{m^{2}}}{2000\\frac{m^{3}}{d}\\ - 1000\\frac{m^{3}}{d}} \\\\ &= 0.0000003\\ or\\ 3x10^{- 7}\\frac{d^{2}}{m^{5}} \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n

          B<\/em> is read from the graph as the y\u2011intercept at QT<\/sub><\/em> = 0 as B<\/em> = 0.00248 d\/m2<\/sup>.<\/p><\/blockquote>\n<\/div>\n

          \n

          The well loss (CQ<\/em>2<\/sup>) when pumping at 2,000 m3\/<\/sup>d would be computed as follows.<\/p><\/blockquote>\n<\/div>\n

          \n
          \n

          [latex]well\\ loss = \\ 0.0000003\\ \\ d^{2}\/{m\\ }^{5}\\ \\ (2000\\ m^{3}\/d)^{2}\\ \\ = \\ 1.2\\ m[\/latex]<\/p>\n<\/div>\n<\/div>\n

            \n
          1. Estimate the total drawdown after pumping the well for 30 min at 2,000 m3<\/sup>\/d.<\/li>\n<\/ol>\n
            \n

            The total drawdown (formational drawdown + well loss) when pumping at 2,000 m3<\/sup>\/d for 30 minutes is calculated as follows.<\/p><\/blockquote>\n<\/div>\n

            \n
            \n

            [latex]s_{T} = BQ + CQ^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

            \n
            \n

            [latex]s_{T} = 0.0028\\frac{d}{m^{2}}\\left( 2000\\frac{m^{3}}{d} \\right) + 3x10^{- 7}\\frac{d^{2}}{m^{5}}\\left( 2000\\frac{m^{3}}{d} \\right)^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

            \n
            \n

            [latex]s_{T} = 5.6\\ m + 1.2\\ m = 6.8\\ m[\/latex]<\/p>\n<\/div>\n<\/div>\n

            \n

            To compute total drawdowns at later times new values of B<\/em> would need to be determined as B<\/em> is time dependent. This would require inputting a value of T<\/em> and S in the appropriate equation (e.g., Cooper\u2011Jacob) and recomputing B for a prescribed period of pumping (t<\/em>).<\/p><\/blockquote>\n<\/div>\n

              \n
            1. Estimate transmissivity using the first 30 minutes of time\u2011drawdown data (step 1) (use the Cooper\u2011Jacob method).<\/li>\n<\/ol>\n
              \n

              \"\"<\/p>\n<\/div>\n

              \n

              Semi\u2011log plot of the time drawdown data for step one. A straight\u2011line fit is performed and a value of \u2206slog-t<\/sub><\/em> determined. This value is then used in the Cooper\u2011Jacob analysis to estimate a value of T<\/em>. As the time drawdown data contains some well loss the storage coefficient is not computed.<\/p>\n<\/div>\n

              \n

              The Cooper\u2011Jacob solution found in Section 8.2.3.<\/p><\/blockquote>\n<\/div>\n

              \n
              \n

              [latex]T = 2.3\\frac{Q}{4\\pi\\Delta s{\\Delta s}_{log - t}} = 2.3\\frac{1400\\ \\frac{m^{3}}{d}}{4\\ (3.14)\\ 1.2\\ m} = 214\\frac{m^{2}}{d}[\/latex]<\/p>\n<\/div>\n<\/div>\n

              \n

              The time drawdown data in step one has some well loss so T<\/em> can be determined but S<\/em> is not estimated as the projected t0<\/sub><\/em> value would be too low (Section 12.1.4). If all the step 1 data were corrected for well loss (sT<\/sub><\/em> \u2011 CQ<\/em>2<\/sup>) S could be computed from a semi\u2011log plot of the corrected data. For example, at 10 min the recorded drawdown is 3.52\u00a0m. The corrected drawdown (s<\/em>) would be as follows.<\/p><\/blockquote>\n<\/div>\n

              \n
              \n

              [latex]s = 3.52\\ m - 3x10^{- 7}\\frac{d^{2}}{m^{5}}\\left( 1400\\frac{m^{3}}{d} \\right)^{2} = 2.93\\ m[\/latex]<\/p>\n<\/div>\n<\/div>\n

              \n

              Return to Exercise\u00a05<\/a><\/p><\/blockquote>\n<\/div>\n

              \n

              Return to where text linked to Exercise\u00a05<\/a><\/p><\/blockquote>\n<\/div>\n<\/div>\n

              <\/div>\n<\/div>\n","protected":false},"author":8,"menu_order":98,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1889","chapter","type-chapter","status-publish","hentry"],"part":605,"_links":{"self":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/1889","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/wp\/v2\/users\/8"}],"version-history":[{"count":7,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/1889\/revisions"}],"predecessor-version":[{"id":4147,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/1889\/revisions\/4147"}],"part":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/parts\/605"}],"metadata":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/1889\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/wp\/v2\/media?parent=1889"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/pressbooks\/v2\/chapter-type?post=1889"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/wp\/v2\/contributor?post=1889"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/books.gw-project.org\/an-introduction-to-hydraulic-testing-in-hydrogeology\/wp-json\/wp\/v2\/license?post=1889"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}