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Unit 1 Learning
Activities
Soil Texture Answers
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| Hydrometer Analysis: A sedimentation
procedure was used to identify the soil separates. The corrected
40-second and 2-hour readings are given. (Temperature affects the
density of water, so a temperature correction is necessary when using
hydrometer analysis.) |
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| Sample |
Sample size, g | 40-second reading, g/l | 2-hour reading, g/l |
| One |
50 |
28 |
7 |
| Two |
50 |
32 |
21 |
| Three |
50 |
48 |
11 |
| Four |
100 |
12 |
2 |
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| Determining
the particle size distribution: The hydrometer measures the density of
the solution (what remains in suspension), so the 40-second reading
measures the amount of silt + clay in suspension, while the 2-hour
reading measures the amount of clay in suspension. So, the clay can be
determined directly, and the others by difference. The amount of sand
is the difference between the sample size and the 40-second reading.
The amount of silt is the difference between the 40-second and 2-hour
readings. The solutions are given below the p. 15 problems. |
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| 1.1.5.3 Particle Size Distribution: |
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| Sand, % |
Silt, % |
Clay, % |
Soil Texture Class |
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| 60 |
25 |
15 |
sandy loam |
| 35 |
45 |
20 |
loam |
| 25 |
40 |
35 |
clay loam |
| 50 |
10 |
40 |
sandy clay |
| 80 |
15 |
5 |
loamy sand |
| 20 |
30 |
50 |
clay |
| 5 |
60 |
35 |
silty clay loam |
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| Hydrometer Analysis Results |
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| Sample |
Sand, % | Silt, % | Clay, % | Soil Texture Class |
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| One |
100*(50-28)/50= 44 |
100*(28-7)/50= 42 |
100*(7/50)= 14 |
loam |
| Two |
100*(50-32)/50= 36 |
100*(32-21)/50= 22 |
100*(21)/50= 42 |
clay |
| Three |
100*(50-48)/50= 4 |
100*(48-11)/50= 74 |
100*(11)/50= 22 |
silt loam |
| Four |
100*(100-12)/100= 88 |
100*(12-2)/100= 10 |
100*(2)/100= 2 |
sand |