{"id":876,"date":"2020-12-09T04:43:05","date_gmt":"2020-12-09T04:43:05","guid":{"rendered":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/?post_type=chapter&p=876"},"modified":"2020-12-29T18:26:53","modified_gmt":"2020-12-29T18:26:53","slug":"solution-to-exercise-1","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/chapter\/solution-to-exercise-1\/","title":{"raw":"Solution to Exercise 1","rendered":"Solution to Exercise 1"},"content":{"raw":"

1) A 100 cubic centimeter (cm3<\/sup>) sample of soil has an initial weight of 227.1 grams. It is oven dried at 105\u00b0C to a constant weight of 222.0 grams. The sample is then saturated with water and has a weight of 236.6 grams. Next, the sample is then allowed to drain by gravity in an environment of 100% humidity and is reweighted at 224.4 grams. Assuming that 1 cm3<\/sup> of water = 1 gram at 15.5\u00b0C:<\/p>\r\n1a) Calculate the porosity using Equation 6 of this book.<\/em>\r\n\r\n\r\n\r\n
[latex]\\displaystyle n_e=\\frac{V_I}{V_T}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n
n<\/em>e<\/em><\/sub><\/td>\r\n=<\/td>\r\neffective porosity (dimensionless)<\/td>\r\n<\/tr>\r\n
V<\/em>I<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of interconnected pore space (L3<\/sup>)<\/td>\r\n<\/tr>\r\n
V<\/em>T<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of sample (L3<\/sup>)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRecognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water in the sample is the saturated weight in grams minus the oven dried weight in grams:\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle n_{e}=\\frac{\\left(\\textup{saturated\\ weight}-\\textup{dry\\ weight}\\right)\\ as\\ volume}{100\\ {\\textup{cm}}^3}[\/latex] = [latex]\\displaystyle \\frac{236.6\\ {\\textup{cm}}^3-222.0\\ {\\textup{cm}}^3}{100\\ {\\textup{cm}}^3}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\displaystyle n_{e}=\\frac{14.6\\ \\textup{cm}^{3}}{100\\ \\textup{cm}^{3}}=0.146[\/latex] or 14.6%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n1b) Calculate the specific yield using Equation 11 of this book.<\/em>\r\n\r\n\r\n\r\n
[latex]\\displaystyle S_y=\\frac{V_D}{V_T}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n
S<\/em>y<\/em><\/sub><\/td>\r\n=<\/td>\r\nspecific yield (dimensionless)<\/td>\r\n<\/tr>\r\n
V<\/em>D<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of water that drains by gravity (L3<\/sup>)<\/td>\r\n<\/tr>\r\n
V<\/em>T<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of sample (L3<\/sup>)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRecognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water drained from the sample is the weight of drained water in grams:\r\n\r\n\r\n\r\n
[latex]\\displaystyle S_{y}=\\frac{12.2\\ \\textup{g}}{100\\ \\textup{cm}^{3}}=0.122[\/latex] or 12.2%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n 1c) Calculate the specific retention using Equation 12 of this book.<\/em>\r\n\r\n\r\n\r\n
[latex]\\displaystyle S_r=\\frac{V_R}{V_T}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n
S<\/em>r<\/em><\/sub><\/td>\r\n=<\/td>\r\nspecific retention (dimensionless)<\/td>\r\n<\/tr>\r\n
V<\/em>R<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of water retained against gravity after drainage ceases (L3<\/sup>)<\/td>\r\n<\/tr>\r\n
V<\/em>T<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of sample (L3<\/sup>)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRecognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water retained in the sample is the total water in the sample (i.e., saturated weight in grams minus the dry weight in grams) minus the weight of water drained:\r\n\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle S_{r}=\\frac{(saturated\\ weight-dry\\ weight-water\\ drained)\\ as\\ volume}{V_{T}}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\displaystyle S_{r}=\\frac{236.6\\ \\textup{g}-222.0\\ \\textup{g}-12.2\\ \\textup{g}}{100\\ \\textup{cm}^{3}}[\/latex] = [latex]\\displaystyle \\frac{2.4\\ \\textup{g}}{100\\ \\textup{cm}^{3}}[\/latex] = 0.024 or 2.4%<\/td>\r\n<\/tr>\r\n
S<\/em>r<\/em><\/sub> = 0.024 or 2.4%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCheck that, as in Equation 13 of this book:\r\n\r\n\r\n\r\n\r\n
n<\/em>e<\/em><\/sub> = S<\/em>y<\/em><\/sub> + S<\/em>r<\/em><\/sub><\/td>\r\n<\/tr>\r\n
0.146 = 0.122 + 0.024 = yes 0.146<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n1d) Evaluate whether the resulting particle density is reasonable using Equation 3 of this book.<\/em>\r\n\r\n\r\n\r\n
\u03c1<\/em>b<\/em><\/sub> = (1 - n<\/em>) \u03c1<\/em>p<\/em><\/sub> + n \u03c1<\/em>f<\/em><\/sub><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n\r\n
\u03c1<\/em>b<\/em><\/sub><\/td>\r\n=<\/td>\r\nbulk density (M\/L3<\/sup>)<\/td>\r\n<\/tr>\r\n
n<\/em><\/td>\r\n=<\/td>\r\ntotal porosity (dimensionless)<\/td>\r\n<\/tr>\r\n
\u03c1<\/em>p<\/em><\/sub><\/td>\r\n=<\/td>\r\nparticle density (M\/L3<\/sup>)<\/td>\r\n<\/tr>\r\n
\u03c1<\/em>f<\/em><\/sub><\/td>\r\n=<\/td>\r\nfluid density (M\/L3<\/sup>)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRearrange to solve for particle density:\r\n\r\n\r\n\r\n
[latex]\\displaystyle \\rho _{p}=\\frac{\\rho _{b}-n\\rho _{f}}{(1-n)}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFrom the laboratory data:\r\n\r\n\r\n\r\n\r\n\r\n
\u03c1<\/em>b<\/em><\/sub><\/td>\r\n=<\/td>\r\nsaturated weight over total volume (236.6 g \/ 100 cm3<\/sup>) = 2.37 g\/cm3<\/sup><\/td>\r\n<\/tr>\r\n
n<\/em><\/td>\r\n=<\/td>\r\nassume effective porosity is total porosity = 0.146<\/td>\r\n<\/tr>\r\n
\u03c1<\/em>f<\/em><\/sub><\/td>\r\n=<\/td>\r\n1 g\/cm3<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstitute and solve for particle density:\r\n\r\n\r\n\r\n
[latex]\\displaystyle \\rho _{p}=\\frac{\\frac{2.37\\ \\textup{g}}{\\textup{cm}^{3}}-0.146\\ \\frac{1\\ \\textup{g}}{\\textup{cm}^{3}}}{(1-0.146)}=2.6\\ \\frac{\\textup{g}}{{\\textup{cm}}^3}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nYes, this is a reasonable particle density for soil. Soil is generally a mix of quartz, feldspars and clays and this particle density fall within the range of those materials, being less than quartz (2.67 g\/cm3<\/sup>), greater than feldspar (2.56 g\/cm3<\/sup>), and typical of clay densities (2.6 g\/cm3<\/sup>).\r\n\r\n1e) Calculate the void ratio using Equation 8 of this book, where e is the void ratio.<\/em>\r\n\r\n\r\n\r\n
[latex]\\displaystyle n=\\frac{e}{1+e}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRearranging to solve for e<\/em>:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
n<\/em>(1 + e<\/em>) = e<\/em><\/td>\r\n<\/tr>\r\n
n<\/em> + ne<\/em> = e<\/em><\/td>\r\n<\/tr>\r\n
n<\/em> = e<\/em> - ne<\/em> = (1 - n<\/em>)e<\/em><\/td>\r\n<\/tr>\r\n
[latex]\\displaystyle e=\\frac{n}{1-n}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\displaystyle e=\\frac{n}{1-n}=\\frac{0.146}{1-0.146}=0.17[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n1f) Calculate the initial moisture content of the sample before it was dried using Equation 9 of this book.<\/em>\r\n\r\n\r\n\r\n
[latex]\\displaystyle \\theta=\\frac{V_W}{V_T}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n
\u03b8<\/em><\/td>\r\n=<\/td>\r\nmoisture content (dimensionless)<\/td>\r\n<\/tr>\r\n
V<\/em>W<\/em><\/small><\/sub><\/td>\r\n=<\/td>\r\nvolume of water in the pore space (L3<\/sup>)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRecognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the initial volume of water in the sample is saturated sample weight minus the dry weight:\r\n\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle \\theta =\\frac{V_{W}}{V_{T}}=\\frac{(\\textup{initial\\ sample\\ weight}-\\textup{dry\\ sample\\ weight})\\ as\\ volume}{100\\ \\textup{cm}^{3}}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\displaystyle \\theta =\\frac{V_{W}}{V_{T}}=\\frac{227.1\\ \\textup{cm}^{3}-222.0\\ \\textup{cm}^{3}}{100\\ \\textup{cm}^{3}}=\\frac{5.1\\ \\textup{cm}^{3}}{100\\ \\textup{cm}^{3}}[\/latex]<\/td>\r\n<\/tr>\r\n
\u03b8<\/em> = 0.051 or 5.1%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n1g) Calculate the initial degree of saturation using Equation 10 of this book.<\/em>\r\n\r\n\r\n\r\n\r\n
Degree of Saturation = [latex]\\displaystyle \\frac{\\theta}{n_e}[\/latex]<\/td>\r\n<\/tr>\r\n
Degree of Saturation = [latex]\\displaystyle \\frac{0.051}{0.146}[\/latex] = 0.35 or 35%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Return to Exercise 1<\/a><\/p>","rendered":"

1) A 100 cubic centimeter (cm3<\/sup>) sample of soil has an initial weight of 227.1 grams. It is oven dried at 105\u00b0C to a constant weight of 222.0 grams. The sample is then saturated with water and has a weight of 236.6 grams. Next, the sample is then allowed to drain by gravity in an environment of 100% humidity and is reweighted at 224.4 grams. Assuming that 1 cm3<\/sup> of water = 1 gram at 15.5\u00b0C:<\/p>\n

1a) Calculate the porosity using Equation 6 of this book.<\/em><\/p>\n\n\n\n
\"\displaystyle n_e=\frac{V_I}{V_T}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n
n<\/em>e<\/em><\/sub><\/td>\n=<\/td>\neffective porosity (dimensionless)<\/td>\n<\/tr>\n
V<\/em>I<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of interconnected pore space (L3<\/sup>)<\/td>\n<\/tr>\n
V<\/em>T<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of sample (L3<\/sup>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Recognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water in the sample is the saturated weight in grams minus the oven dried weight in grams:<\/p>\n\n\n\n\n
\"\displaystyle n_{e}=\frac{\left(\textup{saturated\ weight}-\textup{dry\ weight}\right)\ as\ volume}{100\ {\textup{cm}}^3}\" = \"\displaystyle \frac{236.6\ {\textup{cm}}^3-222.0\ {\textup{cm}}^3}{100\ {\textup{cm}}^3}\"<\/td>\n<\/tr>\n
\"\displaystyle n_{e}=\frac{14.6\ \textup{cm}^{3}}{100\ \textup{cm}^{3}}=0.146\" or 14.6%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1b) Calculate the specific yield using Equation 11 of this book.<\/em><\/p>\n\n\n\n
\"\displaystyle S_y=\frac{V_D}{V_T}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n
S<\/em>y<\/em><\/sub><\/td>\n=<\/td>\nspecific yield (dimensionless)<\/td>\n<\/tr>\n
V<\/em>D<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of water that drains by gravity (L3<\/sup>)<\/td>\n<\/tr>\n
V<\/em>T<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of sample (L3<\/sup>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Recognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water drained from the sample is the weight of drained water in grams:<\/p>\n\n\n\n
\"\displaystyle S_{y}=\frac{12.2\ \textup{g}}{100\ \textup{cm}^{3}}=0.122\" or 12.2%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1c) Calculate the specific retention using Equation 12 of this book.<\/em><\/p>\n\n\n\n
\"\displaystyle S_r=\frac{V_R}{V_T}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n
S<\/em>r<\/em><\/sub><\/td>\n=<\/td>\nspecific retention (dimensionless)<\/td>\n<\/tr>\n
V<\/em>R<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of water retained against gravity after drainage ceases (L3<\/sup>)<\/td>\n<\/tr>\n
V<\/em>T<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of sample (L3<\/sup>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Recognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the volume of water retained in the sample is the total water in the sample (i.e., saturated weight in grams minus the dry weight in grams) minus the weight of water drained:<\/p>\n\n\n\n\n\n
\"\displaystyle S_{r}=\frac{(saturated\ weight-dry\ weight-water\ drained)\ as\ volume}{V_{T}}\"<\/td>\n<\/tr>\n
\"\displaystyle S_{r}=\frac{236.6\ \textup{g}-222.0\ \textup{g}-12.2\ \textup{g}}{100\ \textup{cm}^{3}}\" = \"\displaystyle \frac{2.4\ \textup{g}}{100\ \textup{cm}^{3}}\" = 0.024 or 2.4%<\/td>\n<\/tr>\n
S<\/em>r<\/em><\/sub> = 0.024 or 2.4%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Check that, as in Equation 13 of this book:<\/p>\n\n\n\n\n
n<\/em>e<\/em><\/sub> = S<\/em>y<\/em><\/sub> + S<\/em>r<\/em><\/sub><\/td>\n<\/tr>\n
0.146 = 0.122 + 0.024 = yes 0.146<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1d) Evaluate whether the resulting particle density is reasonable using Equation 3 of this book.<\/em><\/p>\n\n\n\n
\u03c1<\/em>b<\/em><\/sub> = (1 – n<\/em>) \u03c1<\/em>p<\/em><\/sub> + n \u03c1<\/em>f<\/em><\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n\n
\u03c1<\/em>b<\/em><\/sub><\/td>\n=<\/td>\nbulk density (M\/L3<\/sup>)<\/td>\n<\/tr>\n
n<\/em><\/td>\n=<\/td>\ntotal porosity (dimensionless)<\/td>\n<\/tr>\n
\u03c1<\/em>p<\/em><\/sub><\/td>\n=<\/td>\nparticle density (M\/L3<\/sup>)<\/td>\n<\/tr>\n
\u03c1<\/em>f<\/em><\/sub><\/td>\n=<\/td>\nfluid density (M\/L3<\/sup>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Rearrange to solve for particle density:<\/p>\n\n\n\n
\"\displaystyle \rho _{p}=\frac{\rho _{b}-n\rho _{f}}{(1-n)}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the laboratory data:<\/p>\n\n\n\n\n\n
\u03c1<\/em>b<\/em><\/sub><\/td>\n=<\/td>\nsaturated weight over total volume (236.6 g \/ 100 cm3<\/sup>) = 2.37 g\/cm3<\/sup><\/td>\n<\/tr>\n
n<\/em><\/td>\n=<\/td>\nassume effective porosity is total porosity = 0.146<\/td>\n<\/tr>\n
\u03c1<\/em>f<\/em><\/sub><\/td>\n=<\/td>\n1 g\/cm3<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Substitute and solve for particle density:<\/p>\n\n\n\n
\"\displaystyle \rho _{p}=\frac{\frac{2.37\ \textup{g}}{\textup{cm}^{3}}-0.146\ \frac{1\ \textup{g}}{\textup{cm}^{3}}}{(1-0.146)}=2.6\ \frac{\textup{g}}{{\textup{cm}}^3}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Yes, this is a reasonable particle density for soil. Soil is generally a mix of quartz, feldspars and clays and this particle density fall within the range of those materials, being less than quartz (2.67 g\/cm3<\/sup>), greater than feldspar (2.56 g\/cm3<\/sup>), and typical of clay densities (2.6 g\/cm3<\/sup>).<\/p>\n

1e) Calculate the void ratio using Equation 8 of this book, where e is the void ratio.<\/em><\/p>\n\n\n\n
\"\displaystyle n=\frac{e}{1+e}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Rearranging to solve for e<\/em>:<\/p>\n\n\n\n\n\n\n\n
n<\/em>(1 + e<\/em>) = e<\/em><\/td>\n<\/tr>\n
n<\/em> + ne<\/em> = e<\/em><\/td>\n<\/tr>\n
n<\/em> = e<\/em> – ne<\/em> = (1 – n<\/em>)e<\/em><\/td>\n<\/tr>\n
\"\displaystyle e=\frac{n}{1-n}\"<\/td>\n<\/tr>\n
\"\displaystyle e=\frac{n}{1-n}=\frac{0.146}{1-0.146}=0.17\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1f) Calculate the initial moisture content of the sample before it was dried using Equation 9 of this book.<\/em><\/p>\n\n\n\n
\"\displaystyle \theta=\frac{V_W}{V_T}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n
\u03b8<\/em><\/td>\n=<\/td>\nmoisture content (dimensionless)<\/td>\n<\/tr>\n
V<\/em>W<\/em><\/small><\/sub><\/td>\n=<\/td>\nvolume of water in the pore space (L3<\/sup>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Recognizing that water weight in grams is equivalent to volume in cm3<\/sup> because 1g of water is 1 cm3<\/sup> use water weight where water volume is needed, so the initial volume of water in the sample is saturated sample weight minus the dry weight:<\/p>\n\n\n\n\n\n
\"\displaystyle \theta =\frac{V_{W}}{V_{T}}=\frac{(\textup{initial\ sample\ weight}-\textup{dry\ sample\ weight})\ as\ volume}{100\ \textup{cm}^{3}}\"<\/td>\n<\/tr>\n
\"\displaystyle \theta =\frac{V_{W}}{V_{T}}=\frac{227.1\ \textup{cm}^{3}-222.0\ \textup{cm}^{3}}{100\ \textup{cm}^{3}}=\frac{5.1\ \textup{cm}^{3}}{100\ \textup{cm}^{3}}\"<\/td>\n<\/tr>\n
\u03b8<\/em> = 0.051 or 5.1%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1g) Calculate the initial degree of saturation using Equation 10 of this book.<\/em><\/p>\n\n\n\n\n
Degree of Saturation = \"\displaystyle \frac{\theta}{n_e}\"<\/td>\n<\/tr>\n
Degree of Saturation = \"\displaystyle \frac{0.051}{0.146}\" = 0.35 or 35%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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