Lab 1

Principles of Soil Science Practical 1 Part 1: Sample 1 Sample 2 Total weight of oven-dried soil (in grams) 72.4 g 55.1 g 40 second hydrometer readings: Reading #1 4 g 21 g Reading #2 3 g 20.5 g Reading #3 3.5 g 19.5 g Reading #4 3 g 20 g Reading #5 3.5 g 20 g Average of 40 second hydrometer readings 3.4 g 20.2 g Temperature of suspension at 40 second readings (  C) 24  C 24.5  C Corrected 40 second hydrometer reading (grams) 24 – 18 = 6 6 * 0.25 = 1.5 3.4 + 1.5 = 4.9 g 24.5 – 18 = 6.5 6.5 * 0.25 = 1.625 1.625 + 20.2 = 21.83 g 2-hour hydrometer readings 1.5 g 3 g Temperature of suspension at 2 hour reading (  C) 24  C 24.5  C Corrected 2-hour hydrometer reading (grams) 24 – 18 = 6 6 * 0.25 = 1.5 1.5 + 1.5 = 3 g 24.5 – 18 = 6.5 6.5 * 0.25 = 1.625 1.625 + 3 = 4.63 g Amount of sand (grams) 72.4g – 4.9g = 67.5 g 55.1g – 21.83g = 33.27 g Percentage of sand (67.5g/72.4g) * 100 = 93.23 % (33.27g/55.1g) * 100 = 60.38 % Percentage of clay (3g/72.4g) * 100 = 4.14 % (4.63g/55.1g) * 100 = 8.40 % Percentage of silt 100 % - (93.23% + 4.14%) = 2.63 % 100% - (60.38% + 8.40%) = 31.22 % Question 1: The calculations for the percentage of sand, silt and clay are shown above in the table. The textural class of soil sample 1 in conjunction with a soil textural triangle is primarily sand with 93.23 % of sand. There was 4.14 percent of clay and 2.63 % of silt. The calculations for the percentage of sand, silt and clay are shown above in the table. The textural class of soil sample 2 in conjunction with a soil textural triangle is a sandy loam

with 60.38 % of sand. There was 8.40 % of clay and 31.22 % of silt. The silt component in this soil was a lot more than sample 1. Question 2: Horizon A is a more weathered layer and contains a lot of organic matter usually so this would be classified to have more organic matter overall. This is primarily because it’s the top horizon and has all the dead plants and small animals decompose on it over time. This gives horizon A a darker color than other soils. Sample 2 was a bit darker so I would assume that it was horizon A as it was also sticking to the gloves when touched. Sample 2 has a lot more silt and clay than sample 1 which is why its classified as a sandy loam. This means that the pore spaces in sample 2 are much closer to one another than in sample 1 and so it has led water holding capacity which is why it sank to the bottom of the cylinder during the experiment a lot quicker too. Sample 1 on the other hand, would not have as much or almost no organic matter as it’s a lot lighter in color. Things like the organic matter present in the soils, stays in the soil during the experiment as it’s a part of the soil and so it would definitely interfere with the analysis and influence the hydrometer readings. Question 3: Sample 1: Sand Aeration Very good Water holding capacity Low Rate of infiltration High Sample 2: Sandy Loam Aeration Good Water holding capacity Moderate Rate of infiltration Moderate Both samples have a good amount of sand present, especially sample 1 so there is a lot of space for air particles to pass through. This causes both samples to have high aeration. Sample 1 does have higher aeration than sample 2. Sample 1 has mostly sand and so bigger pore spaces which allows water to move right through. This makes it so that it has a very poor water holding capacity. Sample 2 however has a lot more silt present in it which causes it to have smaller pores which can actually hold a bit of water hence why a moderate water holding capacity. This directly relates to infiltration rate. For sands, it’s really high whereas for silts and clays with smaller pores, the infiltration rate tends to be lower.

Weight of empty container (grams) 12.69 g 12.69 g Weight of container filled with water (grams) 293.1 g 293.1 g Weight of water (grams) 280.41 g 280.41 g Volume of container (cm 3 ) = mass of water (g) / density of water (g/cm 3 ) 293.1g/ 1g/ cm 3 = 293.1 cm 3 293.1g/ 1g/ cm 3 = 293.1 cm 3 Question 1: Mass of water (g) = mass of soil sample (g) – mass of oven dry soil sample (g) Sample 1 Sample 2 455.51g - 434.51g = 21 g 420.11g – 373.9g = 46.21 g Gravimetric soil water content: Sample 1 Sample 2 Mass of water present in the soil/mass of oven dry soil 21g/434.51g = 0.048 g water / grams of dry soil If we wanted to see it in percentage: 0.048 * 100 = 4.83 % gravimetric soil moisture content Mass of water/mass of oven dry soil 46.21g/373.9g = 0.124 g water / grams of dry soil If we wanted to see it in percentage: 0.124 * 100 = 12.36 % gravimetric soil moisture content Question 2: Bulk density = weight of oven dry soil/volume of the soil (includes the volume of soil particles and the volume of pores among soil particles) which would be the volume of the container Sample 1 Sample 2 Volume of container = 293.1 cm 3 (calculated in the table above) Bulk density: 434.51g/293.1 cm 3 = 1.48 g/cm 3 Volume of container = 293.1 cm 3 (calculated in the table above) Bulk density: 373.9g/293.1 cm 3 = 1.28 g/cm 3 Question 3: Bulk density is used as an indicator for how compact a sample of soil is. The higher the bulk density, the lower the soil compaction therefore the lower the pore spaces in the soil as well. It’s calculated as the mass of the oven dry soil which is primarily the solid soil particles in a given volume of a soil sample or a field. Bulk density increases as depth increases, so the lower in the ground you go, the higher the bulk density. This is because the soil underneath the surface soil

is a lot more compacted than the soil above it (due to more mass above it, etc.) and has lower organic matter content. These factors contribute to the soils deeper to have a higher bulk density. As seen in the calculation above, soil sample 1 has a higher bulk density (1.48 g/cm 3 ) than soil sample 2 (1.28 g/cm 3 ) and this is because sample 1 primarily consists of sand whereas sample 2 is a sandy loam. Since sample 1 is mostly sand, it has larger but a lot less pore spaces as its loosely packed than sample 2 which has a higher silt and clay presence. In sample 2, there are a lot more smaller silt and clay particles and therefore have a lot of smaller pore spaces between them. Sample 1 has less pore spaces like mentioned, so it has a higher bulk density. Question 4: The gravimetric moisture content for soil sample 1 was calculated above: Mass of water/mass of oven dry soil (21g/434.51g) * 100 = 4.83 % = 0.048 g water / grams of dry soil At max saturation, 58% of the soil is occupied by water. From lecture notes: Storage capacity of a soil = field capacity – moisture content = 58% - 4.83% = 53.17% Therefore, the storage capacity of sample 1 is 53.17% or 0.5317 Volumetric storage capacity = storage capacity * bulk density sample 1 = 0.5317 * 1.48 g/cm 3 = 0.79 g/cm 3 Therefore, the volumetric storage capacity for sample 1 is 0.79 g/cm 3 Depth of wetting = amount of rain/volumetric storage capacity The surface area doesn’t matter as 1 km 2 would be the same as 1 m 2 so we only account for the depth (height) of rain which is given as 1.8 cm in the question. Depth of wetting = 1.8cm / 0.79 = 2.28 cm Therefore, the depth of wetting for 1.8 cm of rain in sample 1 soil would be 2.28 cm.