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Re: Example 3.5 from Domenico and Schwartz

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  • Ron St. John
    Sean: First let me say that my printing of the Domenico & Schwartz text identifies this example as Example 3.3 and is on page 82, but it appears to be the same
    Message 1 of 5 , May 31, 2011
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      Sean:

      First let me say that my printing of the Domenico & Schwartz text identifies this example as Example 3.3 and is on page 82, but it appears to be the same problem. Next, let me say that I'm not defending the example problem, I'm simply trying to explain its workings.
      I have attempted to address each of your concerns is addressed below:


      1. In my experience a hydraulic gradient in a low hydraulic conductivity unit (10 ft/yr) unit like the one in the example is not that unusual. Because of the low hydraulic conductivity of the unit, the hydraulic gradient must increase to provide sufficient volumetric flow through the unit.

      2. The hydraulic head in the sand unit below the unoxidized till is not hydraulically connected to the stream/creek on the right side of the diagram. There is no discharge from the sand to the creek. They are hydraulically independent.

      3. In step #2 of the example problem, it would have been better to characterize the flow as the "Volumetric flow within the trench" not "into the trench". The area through which flow occurs is the 25 feet of saturated thickness within the oxidized till and the length over which occurs (600 ft). 25' x 600' = 15,000 ft2

      4. The vertical hydraulic gradient through the unoxidized till is 1 because the hydraulic head in the oxidized till is 25 feet above the oxidized/unoxidized till contact and the hydraulic head in the sand is 15 feet below the oxidized/unoxidized till contact. This is a total head difference between the two units of 40 feet. The illustration also identifies the thickness of the unoxidized till as 40 feet. Therefore, dh/dv equals 40'/40' or 1.

      Hope this helps,

      Ronald B. St. John, PHG, CPG
      Principal Hydrogeologist
      St. John-Mittelhauser & Associates
      1401 Branding Avenue, Suite 315
      Downers Grove, IL 60515
      Phone: (630) 427-8111
      Mobile: (630) 803-0254
      Fax: (630) 427-8129
      www.st-ma.com
    • Sean Shekarforoush
      Dear Ron, Thank you very much for your email. I really appreciate your time and effort. Your explanations are very clear and all acceptable, however, it
      Message 2 of 5 , Jun 1, 2011
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        Dear Ron,

        Thank you very much for your email. I really appreciate your time and effort.

        Your explanations are very clear and all acceptable, however, it appears that the copy of the book that you have is different to mine. A copy of that example is attached. Please take a look and let me know if it is different to yours.

        1. I agree that a hydraulic gradient of 0.072 is not unusual for units with low K, but my point is whether this gradient is practical for the conditions presented in this example or not. With this gradient, the water table should drop 7.2 metres for every 100 metre or 20 metres along the length of the site (278m x 0.072 = 20m). Try to draw the water table on the cross section for yourself.

        2. I understand that the sand unit is not discharging into the creek, however my question is whether the groundwater elevation shown for the sand unit fits the hydrog settings? Try to draw the equipotential lines and flow lines for the Example model and you will see that it won't work.

        3. the area through which the flow occurs is 15000 ft2, if we assume the saturated thickness near the creek is 25 feet which means no change in water table across the site. This assumption doesn't explain the gradient of 0.072. Also, the area presented in the example is shown to be 4572 m2 or 49,200 ft2 according to my copy of the book.

        4. the hydraulic head shown for the sand unit in y copy is 1 metre below the oxidized/unoxidized contact, therefore, dh/dv equals approximately 9/13.

        Again I appreciate your time and effort and looking forward hearing from you.

        Best Regards

        Sean Shekarforoush
      • Trevor Elliot
        I suspect that this is all down to identifying A in Example 3.5, and given two people quoting different units unfotunately issues may have arisen Re: simply
        Message 3 of 5 , Jun 2, 2011
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          I suspect that this is all down to identifying A in Example 3.5, and given two people quoting different units unfotunately issues may have arisen Re: simply from a problem of conversion between units from the First to Second Edition of Domenico & Schwartz. Looking at the 1st Edition (Example 3.5 is on p.82). The units are in ft and velocities in ft/yr. In the 2nd Edition (example 3.5 now on p.48) the units are m and m/s. I suspect the original example with dimensions in ft was converted more or less into m. HOWEVER, this doesn’t seem to have been done consistently, eg:

          Length: 834ft <----> 278m ( a factor of 3)
          Width : 600ft <----> 200m ( a factor of 3)

          BUT

          Depth: 40ft <----> 13m ( a factor of 3.077)
          Water level (oxidized till): 25ft <----> 8m ( a factor of 3.125)

          [and just for fun going vertically the given depth below top of the unoxidized till is 15ft in the original but only 1m in the 1998 version].

          For hydraulic conductivity 10ft/yr is taken to equate simply to 10^(-7)m/s, and the same hydraulic gradient of 0.072 again is used.

          The Calculation #2 in original 1990 version (1st Edition) for horizontal flow then uses an Area term quoted as being 15000 ft (sic). It should really have been given as 600ftx25ft = 15000ft^2. This IS the value then used for A in the 1990 Edition, and propagated through the calculation Unfortunately if you simply change 15000ft to m it comes to around the value apparently now being quoted as m^2 in the 1998 version (15000/4572 gives a conversion factor of 3.281). Of course 15000ft^2 should in fact translate to circa 1667m^2 (if simply using a simple conversion factor of 3).

          I believe the correct Area to use here for Calculation #2 in the 1998 (2nd Edition, metric) version therefore as given in the Example should be 8mx200m = 1600m^2. The volumetric flow into the trench then should be 1.152 m^3/s if you propagate that through.

          trevor elliot

          -------------------------------------------------
          Dr. Trevor Elliot,
          Reader in Environmental Engineering,
          Academic Coordinator, Environmental Tracers Laboratory (ETL),
          Deputy Director of Research (Environmental),
          Environmental Engineering Research Centre (EERC),
          School of Planning, Architecture & Civil Engineering (SPACE),
          Queen's University Belfast,
          David Keir Building, Stranmillis Road,
          Belfast. BT9 5AG. Northern Ireland.
          t.elliot@...
          Tel. +44 (0)28 9097 4736
          Fax. +44 (0)28 9097 4278
          http://www.qub.ac.uk/space
          http://www.qub.ac.uk/research-centres/eerc/
          http://space.qub.ac.uk:8077/EERC/elliot/default.aspx
          ________________________________________
          From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Sean Shekarforoush [sshekarforoush@...]
          Sent: 01 June 2011 21:34
          To: gwmodel@yahoogroups.com
          Subject: [gwmodel] Re: Example 3.5 from Domenico and Schwartz

          Dear Ron,

          Thank you very much for your email. I really appreciate your time and effort.

          Your explanations are very clear and all acceptable, however, it appears that the copy of the book that you have is different to mine. A copy of that example is attached. Please take a look and let me know if it is different to yours.

          1. I agree that a hydraulic gradient of 0.072 is not unusual for units with low K, but my point is whether this gradient is practical for the conditions presented in this example or not. With this gradient, the water table should drop 7.2 metres for every 100 metre or 20 metres along the length of the site (278m x 0.072 = 20m). Try to draw the water table on the cross section for yourself.

          2. I understand that the sand unit is not discharging into the creek, however my question is whether the groundwater elevation shown for the sand unit fits the hydrog settings? Try to draw the equipotential lines and flow lines for the Example model and you will see that it won't work.

          3. the area through which the flow occurs is 15000 ft2, if we assume the saturated thickness near the creek is 25 feet which means no change in water table across the site. This assumption doesn't explain the gradient of 0.072. Also, the area presented in the example is shown to be 4572 m2 or 49,200 ft2 according to my copy of the book.

          4. the hydraulic head shown for the sand unit in y copy is 1 metre below the oxidized/unoxidized contact, therefore, dh/dv equals approximately 9/13.

          Again I appreciate your time and effort and looking forward hearing from you.

          Best Regards

          Sean Shekarforoush


          ------------------------------------

          From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Ron St. John [rons@...]
          Sent: 31 May 2011 16:42
          To: gwmodel@yahoogroups.com
          Subject: [gwmodel] Re: Example 3.5 from Domenico and Schwartz
          Sean:
          First let me say that my printing of the Domenico & Schwartz text identifies this example as Example 3.3 and is on page 82, but it appears to be the same problem. Next, let me say that I'm not defending the example problem, I'm simply trying to explain its workings.
          I have attempted to address each of your concerns is addressed below:

          1. In my experience a hydraulic gradient in a low hydraulic conductivity unit (10 ft/yr) unit like the one in the example is not that unusual. Because of the low hydraulic conductivity of the unit, the hydraulic gradient must increase to provide sufficient volumetric flow through the unit.
          2. The hydraulic head in the sand unit below the unoxidized till is not hydraulically connected to the stream/creek on the right side of the diagram. There is no discharge from the sand to the creek. They are hydraulically independent.
          3. In step #2 of the example problem, it would have been better to characterize the flow as the "Volumetric flow within the trench" not "into the trench". The area through which flow occurs is the 25 feet of saturated thickness within the oxidized till and the length over which occurs (600 ft). 25' x 600' = 15,000 ft2
          4. The vertical hydraulic gradient through the unoxidized till is 1 because the hydraulic head in the oxidized till is 25 feet above the oxidized/unoxidized till contact and the hydraulic head in the sand is 15 feet below the oxidized/unoxidized till contact. This is a total head difference between the two units of 40 feet. The illustration also identifies the thickness of the unoxidized till as 40 feet. Therefore, dh/dv equals 40'/40' or 1.
          Hope this helps,
          Ronald B. St. John, PHG, CPG
          Principal Hydrogeologist
          St. John-Mittelhauser & Associates
          1401 Branding Avenue, Suite 315
          Downers Grove, IL 60515
          Phone: (630) 427-8111
          Mobile: (630) 803-0254
          Fax: (630) 427-8129
          www.st-ma.com


          From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Sean Shekarforoush [sshekarforoush@...]
          Sent: 28 May 2011 22:14
          To: gwmodel@yahoogroups.com
          Subject: [gwmodel] Example 3.5 from Domenico and Schwartz
          Hi everyone,
          As we know, "Physical and Chemical Hydrogeology by Domenico and Schwartz" is one of the best hydrogeology reference books.
          A friend of mine asked me to explain to him "Example 3.5" from that book and he expected a clear and easy to understand explanation. However, I found it hard to explain the hydrogeological settings that fit the model presented for the example. The water table gradient (0.072) looks too high as it causes the water table to drop more than 20 meters over the 278 meters. Also, the water table shown for the sand layer can't be explained with the presence of the stream on the right hand side.
          It is also not clear how the volumetric flow to the trench is calculated (A=4572 m2 ???).
          Lastly, I can't understand why the vertical gradient for the unoxidized till is assigned to be 1? unless it is assumed that the groundwater is drained to the bottom of the oxidized till by the drain system which is hard to happen within the till layer except places near the drain system.
          Cheers
          Sean Shekarforoush

          ------------------------------------
          =================================================
        • Trevor Elliot
          Sorry that should of course read 1.152x10^(-5) m^3/s as the volumetric flow then in metric units (Example 3.5, Calculation #2, Domenico & Schwartz, 2nd
          Message 4 of 5 , Jun 2, 2011
          • 0 Attachment
            Sorry that should of course read 1.152x10^(-5) m^3/s as the volumetric flow then in metric units (Example 3.5, Calculation #2, Domenico & Schwartz, 2nd Edition, 1998).

            trevor
            ________________________________________
            From: Trevor Elliot
            Sent: 02 June 2011 15:26
            To: gwmodel@yahoogroups.com
            Subject: [gwmodel] Re: Example 3.5 from Domenico and Schwartz

            I suspect that this is all down to identifying A in Example 3.5, and given two people quoting different units unfotunately issues may have arisen Re: simply from a problem of conversion between units from the First to Second Edition of Domenico & Schwartz. Looking at the 1st Edition (Example 3.5 is on p.82). The units are in ft and velocities in ft/yr. In the 2nd Edition (example 3.5 now on p.48) the units are m and m/s. I suspect the original example with dimensions in ft was converted more or less into m. HOWEVER, this doesn’t seem to have been done consistently, eg:

            Length: 834ft <----> 278m ( a factor of 3)
            Width : 600ft <----> 200m ( a factor of 3)

            BUT

            Depth: 40ft <----> 13m ( a factor of 3.077)
            Water level (oxidized till): 25ft <----> 8m ( a factor of 3.125)

            [and just for fun going vertically the given depth below top of the unoxidized till is 15ft in the original but only 1m in the 1998 version].

            For hydraulic conductivity 10ft/yr is taken to equate simply to 10^(-7)m/s, and the same hydraulic gradient of 0.072 again is used.

            The Calculation #2 in original 1990 version (1st Edition) for horizontal flow then uses an Area term quoted as being 15000 ft (sic). It should really have been given as 600ftx25ft = 15000ft^2. This IS the value then used for A in the 1990 Edition, and propagated through the calculation Unfortunately if you simply change 15000ft to m it comes to around the value apparently now being quoted as m^2 in the 1998 version (15000/4572 gives a conversion factor of 3.281). Of course 15000ft^2 should in fact translate to circa 1667m^2 (if simply using a simple conversion factor of 3).

            I believe the correct Area to use here for Calculation #2 in the 1998 (2nd Edition, metric) version therefore as given in the Example should be 8mx200m = 1600m^2. The volumetric flow into the trench then should be 1.152 m^3/s if you propagate that through.

            trevor elliot

            -------------------------------------------------
            Dr. Trevor Elliot,
            Reader in Environmental Engineering,
            Academic Coordinator, Environmental Tracers Laboratory (ETL),
            Deputy Director of Research (Environmental),
            Environmental Engineering Research Centre (EERC),
            School of Planning, Architecture & Civil Engineering (SPACE),
            Queen's University Belfast,
            David Keir Building, Stranmillis Road,
            Belfast. BT9 5AG. Northern Ireland.
            t.elliot@...
            Tel. +44 (0)28 9097 4736
            Fax. +44 (0)28 9097 4278
            http://www.qub.ac.uk/space
            http://www.qub.ac.uk/research-centres/eerc/
            http://space.qub.ac.uk:8077/EERC/elliot/default.aspx
            ________________________________________
            From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Sean Shekarforoush [sshekarforoush@...]
            Sent: 01 June 2011 21:34
            To: gwmodel@yahoogroups.com
            Subject: [gwmodel] Re: Example 3.5 from Domenico and Schwartz

            Dear Ron,

            Thank you very much for your email. I really appreciate your time and effort.

            Your explanations are very clear and all acceptable, however, it appears that the copy of the book that you have is different to mine. A copy of that example is attached. Please take a look and let me know if it is different to yours.

            1. I agree that a hydraulic gradient of 0.072 is not unusual for units with low K, but my point is whether this gradient is practical for the conditions presented in this example or not. With this gradient, the water table should drop 7.2 metres for every 100 metre or 20 metres along the length of the site (278m x 0.072 = 20m). Try to draw the water table on the cross section for yourself.

            2. I understand that the sand unit is not discharging into the creek, however my question is whether the groundwater elevation shown for the sand unit fits the hydrog settings? Try to draw the equipotential lines and flow lines for the Example model and you will see that it won't work.

            3. the area through which the flow occurs is 15000 ft2, if we assume the saturated thickness near the creek is 25 feet which means no change in water table across the site. This assumption doesn't explain the gradient of 0.072. Also, the area presented in the example is shown to be 4572 m2 or 49,200 ft2 according to my copy of the book.

            4. the hydraulic head shown for the sand unit in y copy is 1 metre below the oxidized/unoxidized contact, therefore, dh/dv equals approximately 9/13.

            Again I appreciate your time and effort and looking forward hearing from you.

            Best Regards

            Sean Shekarforoush


            ------------------------------------

            From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Ron St. John [rons@...]
            Sent: 31 May 2011 16:42
            To: gwmodel@yahoogroups.com
            Subject: [gwmodel] Re: Example 3.5 from Domenico and Schwartz
            Sean:
            First let me say that my printing of the Domenico & Schwartz text identifies this example as Example 3.3 and is on page 82, but it appears to be the same problem. Next, let me say that I'm not defending the example problem, I'm simply trying to explain its workings.
            I have attempted to address each of your concerns is addressed below:

            1. In my experience a hydraulic gradient in a low hydraulic conductivity unit (10 ft/yr) unit like the one in the example is not that unusual. Because of the low hydraulic conductivity of the unit, the hydraulic gradient must increase to provide sufficient volumetric flow through the unit.
            2. The hydraulic head in the sand unit below the unoxidized till is not hydraulically connected to the stream/creek on the right side of the diagram. There is no discharge from the sand to the creek. They are hydraulically independent.
            3. In step #2 of the example problem, it would have been better to characterize the flow as the "Volumetric flow within the trench" not "into the trench". The area through which flow occurs is the 25 feet of saturated thickness within the oxidized till and the length over which occurs (600 ft). 25' x 600' = 15,000 ft2
            4. The vertical hydraulic gradient through the unoxidized till is 1 because the hydraulic head in the oxidized till is 25 feet above the oxidized/unoxidized till contact and the hydraulic head in the sand is 15 feet below the oxidized/unoxidized till contact. This is a total head difference between the two units of 40 feet. The illustration also identifies the thickness of the unoxidized till as 40 feet. Therefore, dh/dv equals 40'/40' or 1.
            Hope this helps,
            Ronald B. St. John, PHG, CPG
            Principal Hydrogeologist
            St. John-Mittelhauser & Associates
            1401 Branding Avenue, Suite 315
            Downers Grove, IL 60515
            Phone: (630) 427-8111
            Mobile: (630) 803-0254
            Fax: (630) 427-8129
            www.st-ma.com


            From: gwmodel@yahoogroups.com [gwmodel@yahoogroups.com] On Behalf Of Sean Shekarforoush [sshekarforoush@...]
            Sent: 28 May 2011 22:14
            To: gwmodel@yahoogroups.com
            Subject: [gwmodel] Example 3.5 from Domenico and Schwartz
            Hi everyone,
            As we know, "Physical and Chemical Hydrogeology by Domenico and Schwartz" is one of the best hydrogeology reference books.
            A friend of mine asked me to explain to him "Example 3.5" from that book and he expected a clear and easy to understand explanation. However, I found it hard to explain the hydrogeological settings that fit the model presented for the example. The water table gradient (0.072) looks too high as it causes the water table to drop more than 20 meters over the 278 meters. Also, the water table shown for the sand layer can't be explained with the presence of the stream on the right hand side.
            It is also not clear how the volumetric flow to the trench is calculated (A=4572 m2 ???).
            Lastly, I can't understand why the vertical gradient for the unoxidized till is assigned to be 1? unless it is assumed that the groundwater is drained to the bottom of the oxidized till by the drain system which is hard to happen within the till layer except places near the drain system.
            Cheers
            Sean Shekarforoush

            ------------------------------------
            =================================================
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