{"id":100,"date":"2022-07-14T00:09:37","date_gmt":"2022-07-14T00:09:37","guid":{"rendered":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/chapter\/solution-exercise-6\/"},"modified":"2023-03-11T17:27:09","modified_gmt":"2023-03-11T17:27:09","slug":"solution-exercise-6","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/chapter\/solution-exercise-6\/","title":{"raw":"Solution Exercise 6","rendered":"Solution Exercise 6"},"content":{"raw":"
\r\n

Current injected [I<\/em>]:\u00a02\u00a0mA<\/p>\r\n

Apparent resistivity [\u03c1<\/em>]:\u00a0250\u00a0ohm-m<\/p>\r\n

Noise level:\u00a015\u00a0mV<\/p>\r\n

[latex]\\displaystyle K_{g}=\\frac{2\\pi }{\\frac{1}{\\overline{AM}}-\\frac{1}{\\overline{AN}}-\\frac{1}{\\overline{BM}}+\\frac{1}{\\overline{BN}}}[\/latex]<\/p>\r\n

[latex]\\displaystyle K_{g,Wenner}=\\frac{2\\pi }{\\frac{1}{5\\mathrm{m}}-\\frac{1}{10\\mathrm{m}}-\\frac{1}{10\\mathrm{m}}+\\frac{1}{5\\mathrm{m}}}[\/latex]<\/p>\r\n

=\u00a0<\/em>31.4\u00a0m<\/p>\r\n

For a positive geometric factor with a dipole-dipole array, you need to have the electrodes configured as A B N M<\/em> (this also gives a positive resistance). If you configure the electrodes as A B M N<\/em> (as often written) the geometric factor is negative (so is the resistance). Using A B N M<\/em> the geometric factor is:<\/p>\r\n

[latex]\\displaystyle K_{g,dipole-dipole}=\\frac{2\\pi }{\\frac{1}{\\overline{AM}}-\\frac{1}{\\overline{AN}}-\\frac{1}{\\overline{BM}}+\\frac{1}{\\overline{BN}}}[\/latex] = [latex]\\displaystyle \\frac{2\\pi }{\\frac{1}{9\\mathrm{m}}-\\frac{1}{6\\mathrm{m}}-\\frac{1}{6\\mathrm{m}}+\\frac{1}{3\\mathrm{m}}}[\/latex] = 56.5 m<\/p>\r\n

If you do the calculation as A B M N<\/em>, you get a geometric factor of -56.5.<\/p>\r\n

We then need to calculate the apparent resistivity:<\/p>\r\n

\u03c1<\/em>a<\/em><\/em><\/sub>\u00a0=\u00a0<\/em>K<\/em>g<\/em><\/em><\/sub>\u0394V\/I<\/em><\/p>\r\n

\u03c1<\/em>a<\/em><\/em><\/sub>_Wenner<\/em><\/em><\/sub>\u00a0=\u00a0250\u00a0ohm-m\u00a0=\u00a0[latex]\\frac{31.4\\ \\textrm{m}\\ V\\ \\textrm{mV}}{2\\ \\textrm{mA}}[\/latex]<\/p>\r\n

solving for V<\/em> yields V<\/em>\u00a0=\u00a015.9\u00a0mV<\/p>\r\n

\u03c1<\/em>a<\/em><\/em><\/sub>_dipole<\/em><\/em><\/sub>-<\/em><\/em><\/sub>dipole<\/em><\/em><\/sub>\u00a0=\u00a0250\u00a0ohm-m\u00a0=\u00a0[latex]\\frac{56.5\\ \\textrm{m}\\ V\\ \\textrm{mV}}{2\\ \\textrm{mA}}[\/latex]<\/p>\r\n

solving for V<\/em> yields V<\/em>\u00a0=\u00a08.8\u00a0mV<\/p>\r\n

A negative voltage would be correct for A B M N<\/em> configuration, as the geometric factor is negative.<\/p>\r\n

In summary:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Array<\/strong><\/td>\r\na(m)<\/strong><\/td>\r\nn<\/strong><\/td>\r\nK<\/em>g<\/em><\/sub><\/strong><\/td>\r\nV(mV)<\/strong><\/td>\r\nSNR<\/strong><\/td>\r\n<\/tr>\r\n
Wenner<\/td>\r\n5<\/td>\r\n-<\/td>\r\n31.4<\/span><\/td>\r\n15.9<\/span><\/td>\r\n1.06<\/span><\/td>\r\n<\/tr>\r\n
Dipole-dipole<\/td>\r\n3<\/td>\r\n3<\/td>\r\n56.5<\/span><\/td>\r\n8.8<\/span><\/td>\r\n0.59<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Return to Exercise 6<\/a><\/p>\r\n\r\n<\/div>","rendered":"

\n

Current injected [I<\/em>]:\u00a02\u00a0mA<\/p>\n

Apparent resistivity [\u03c1<\/em>]:\u00a0250\u00a0ohm-m<\/p>\n

Noise level:\u00a015\u00a0mV<\/p>\n

[latex]\\displaystyle K_{g}=\\frac{2\\pi }{\\frac{1}{\\overline{AM}}-\\frac{1}{\\overline{AN}}-\\frac{1}{\\overline{BM}}+\\frac{1}{\\overline{BN}}}[\/latex]<\/p>\n

[latex]\\displaystyle K_{g,Wenner}=\\frac{2\\pi }{\\frac{1}{5\\mathrm{m}}-\\frac{1}{10\\mathrm{m}}-\\frac{1}{10\\mathrm{m}}+\\frac{1}{5\\mathrm{m}}}[\/latex]<\/p>\n

=\u00a0<\/em>31.4\u00a0m<\/p>\n

For a positive geometric factor with a dipole-dipole array, you need to have the electrodes configured as A B N M<\/em> (this also gives a positive resistance). If you configure the electrodes as A B M N<\/em> (as often written) the geometric factor is negative (so is the resistance). Using A B N M<\/em> the geometric factor is:<\/p>\n

[latex]\\displaystyle K_{g,dipole-dipole}=\\frac{2\\pi }{\\frac{1}{\\overline{AM}}-\\frac{1}{\\overline{AN}}-\\frac{1}{\\overline{BM}}+\\frac{1}{\\overline{BN}}}[\/latex] = [latex]\\displaystyle \\frac{2\\pi }{\\frac{1}{9\\mathrm{m}}-\\frac{1}{6\\mathrm{m}}-\\frac{1}{6\\mathrm{m}}+\\frac{1}{3\\mathrm{m}}}[\/latex] = 56.5 m<\/p>\n

If you do the calculation as A B M N<\/em>, you get a geometric factor of -56.5.<\/p>\n

We then need to calculate the apparent resistivity:<\/p>\n

\u03c1<\/em>a<\/em><\/em><\/sub>\u00a0=\u00a0<\/em>K<\/em>g<\/em><\/em><\/sub>\u0394V\/I<\/em><\/p>\n

\u03c1<\/em>a<\/em><\/em><\/sub>_Wenner<\/em><\/em><\/sub>\u00a0=\u00a0250\u00a0ohm-m\u00a0=\u00a0[latex]\\frac{31.4\\ \\textrm{m}\\ V\\ \\textrm{mV}}{2\\ \\textrm{mA}}[\/latex]<\/p>\n

solving for V<\/em> yields V<\/em>\u00a0=\u00a015.9\u00a0mV<\/p>\n

\u03c1<\/em>a<\/em><\/em><\/sub>_dipole<\/em><\/em><\/sub>–<\/em><\/em><\/sub>dipole<\/em><\/em><\/sub>\u00a0=\u00a0250\u00a0ohm-m\u00a0=\u00a0[latex]\\frac{56.5\\ \\textrm{m}\\ V\\ \\textrm{mV}}{2\\ \\textrm{mA}}[\/latex]<\/p>\n

solving for V<\/em> yields V<\/em>\u00a0=\u00a08.8\u00a0mV<\/p>\n

A negative voltage would be correct for A B M N<\/em> configuration, as the geometric factor is negative.<\/p>\n

In summary:<\/p>\n\n\n\n\n\n
Array<\/strong><\/td>\na(m)<\/strong><\/td>\nn<\/strong><\/td>\nK<\/em>g<\/em><\/sub><\/strong><\/td>\nV(mV)<\/strong><\/td>\nSNR<\/strong><\/td>\n<\/tr>\n
Wenner<\/td>\n5<\/td>\n–<\/td>\n31.4<\/span><\/td>\n15.9<\/span><\/td>\n1.06<\/span><\/td>\n<\/tr>\n
Dipole-dipole<\/td>\n3<\/td>\n3<\/td>\n56.5<\/span><\/td>\n8.8<\/span><\/td>\n0.59<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Return to Exercise 6<\/a><\/p>\n<\/div>\n","protected":false},"author":1,"menu_order":43,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-100","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/100","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":11,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/100\/revisions"}],"predecessor-version":[{"id":384,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/100\/revisions\/384"}],"part":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapters\/100\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/wp\/v2\/media?parent=100"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/pressbooks\/v2\/chapter-type?post=100"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/wp\/v2\/contributor?post=100"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/books.gw-project.org\/electrical-imaging-for-hydrogeology\/wp-json\/wp\/v2\/license?post=100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}