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[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was conducted on the soil by allowing the sample to drain for 24 hours. After drainage the sample mass was 295 g. The soil was then dried and weighed 284 g. What are the specific yield [8 points], specific retention [8 points], and porosity [8 points] of the sample

Sagot :

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]

solid soil volume =[tex]205 \ cc[/tex]

saturated mass soil = [tex]361 \ g[/tex]

The weight of the soil after drainage is =[tex]295 \ g[/tex]

Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]

Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]

Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]

[tex]= 361- 295 \\\\ = 66 \ cc[/tex]

Soil water retained volume = (draining field weight - dry soil weight)

                                             [tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]

[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]

                    [tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]

(Its saturated water volume is equal to the volume of voids)

[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]

                              [tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]

[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]

                            [tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]

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