# Solution to Exercise 1

1) A 100 cubic centimeter (cm^{3}) sample of soil has an initial weight of 227.1 grams. It is oven dried at 105°C 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 cm^{3} of water = 1 gram at 15.5°C:

*1a) Calculate the porosity using Equation 6 of this book.*

where:

n_{e} |
= | effective porosity (dimensionless) |

V_{I} |
= | volume of interconnected pore space (L^{3}) |

V_{T} |
= | volume of sample (L^{3}) |

Recognizing that water weight in grams is equivalent to volume in cm^{3} because 1g of water is 1 cm^{3} 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:

= |

or 14.6% |

*1b) Calculate the specific yield using Equation 11 of this book.*

where:

S_{y} |
= | specific yield (dimensionless) |

V_{D} |
= | volume of water that drains by gravity (L^{3}) |

V_{T} |
= | volume of sample (L^{3}) |

Recognizing that water weight in grams is equivalent to volume in cm^{3} because 1g of water is 1 cm^{3} 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:

or 12.2% |

* 1c) Calculate the specific retention using Equation 12 of this book.*

where:

S_{r} |
= | specific retention (dimensionless) |

V_{R} |
= | volume of water retained against gravity after drainage ceases (L^{3}) |

V_{T} |
= | volume of sample (L^{3}) |

Recognizing that water weight in grams is equivalent to volume in cm^{3} because 1g of water is 1 cm^{3} 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:

= = 0.024 or 2.4% |

S_{r} = 0.024 or 2.4% |

Check that, as in Equation 13 of this book:

n_{e} = S_{y} + S_{r} |

0.146 = 0.122 + 0.024 = yes 0.146 |

*1d) Evaluate whether the resulting particle density is reasonable using Equation 3 of this book.*

ρ_{b} = (1 – n) ρ_{p} + n ρ_{f} |

where:

ρ_{b} |
= | bulk density (M/L^{3}) |

n |
= | total porosity (dimensionless) |

ρ_{p} |
= | particle density (M/L^{3}) |

ρ_{f} |
= | fluid density (M/L^{3}) |

Rearrange to solve for particle density:

From the laboratory data:

ρ_{b} |
= | saturated weight over total volume (236.6 g / 100 cm^{3}) = 2.37 g/cm^{3} |

n |
= | assume effective porosity is total porosity = 0.146 |

ρ_{f} |
= | 1 g/cm^{3} |

Substitute and solve for particle density:

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/cm^{3}), greater than feldspar (2.56 g/cm^{3}), and typical of clay densities (2.6 g/cm^{3}).

*1e) Calculate the void ratio using Equation 8 of this book, where e is the void ratio.*

Rearranging to solve for *e*:

n(1 + e) = e |

n + ne = e |

n = e – ne = (1 – n)e |

*1f) Calculate the initial moisture content of the sample before it was dried using Equation 9 of this book.*

where:

θ |
= | moisture content (dimensionless) |

V_{W} |
= | volume of water in the pore space (L^{3}) |

Recognizing that water weight in grams is equivalent to volume in cm^{3} because 1g of water is 1 cm^{3} 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:

θ = 0.051 or 5.1% |

*1g) Calculate the initial degree of saturation using Equation 10 of this book.*

Degree of Saturation = |

Degree of Saturation = = 0.35 or 35% |