GGR Problem Set 1 (2)

GGR Problem Set 1 1. Compare and contrast the properties of sand, silt, and clay in soils. In your answer, explain how the different particle sizes influence soil properties and behavior. ( 9 marks ) Sand, silt, and clay particles are three common mineral components that make up soil. The particle sizes of these soils give them distinct properties and influence their behaviors, aeration, water retention, and nutrient availability. Firstly, sand particles have the largest particle size. The diameter of the sand particle sizes ranges from 2.0 to 0.05 millimeters . Sandy soils generally have large pore spaces and a grainy and loose texture. The particles are additionally rounded or angular. The large particle size and pore spaces means that sandy soils have good aeration and are generally not very compact. Combined with the low attraction to other sand particles and water, sandy soils have difficulty retaining water, and are excellent at water drainage. Sandy soils also have challenges retaining nutrients, since the large pore space in the soil allows for leaching. The particle size of sandy soils affects its texture, but also affects the soil’s behaviors. Due to large particle size and large pore space, sandy soils are unable to retain water or to hold onto usable ingredients, but they have good aeration. Secondly, the size of silt particles fall between sand and clay particle sizes, ranging from 0.05 to 0.002 millimeters . Silty soils have smaller pore spaces than sandy soils. While they have a similar shape to sandy soils, rounded or angular, they have a smooth and powdery texture when they are moist. They have less aeration than sandy soils since they have less pore space between particles. Additionally, their particles have medium attraction to other silt particles and after, which means that they can hold more water than sandy soils, but less water than clay soils. They have some ability for nutrient retention. Since silt particles are smaller than sandy particles, and have more pore space than sandy particles, they are able to retain water and nutrients more than sandy soils. Lastly, clay particles have the smallest particles, with a diameter of less than 0.002 millimeters . Clay soils have a flat platelet shape. The small particle size means that it has a sticky and malleable texture. Clay soils do not have food aeration because of the small particles’ strong attraction to other clay particles, water, and other substances. Additionally, this, along with the small pore space between particles, means that clay particles have excellent water but have a high risk of waterlogging. Cations in clay particles are attracted to the high surface area of clay particles, and ensures nutrient retention.

The proportions of sand, silt, and clay in soils determine soils’ textures, properties, and behaviors. Particle size and pore space affects water retention, drainage, aeration, and nutrient retention. 2. (a) In your own words, define loam. ( 1 mark ) A loam, also called a loamy soil, is a type of mineral soil with the proportions of 50% soil solids (minerals and organic matter) and 50% pore space (water and air). Loamy soil is special because of its mineral composition – around 40% silt, 40% sand, and 20% clay. Loam is a good soil for many garden plants because of its ability to hold moisture and drain well so that air reaches the roots of the plant. 3. (a) Using the soil texture classes triangle below (Figure 2), determine the textural classes of the following soils: ( 3 marks ) ● Soil 1: 40% sand, 10% silt, 50% clay – Sandy Clay ● Soil 2: 10% sand, 45% silt – Silty Clay ● Soil 3: 45% silt, 15% clay – Loam Figure 2- A soil texture triangle classifies soils based on the percentages of sand, silt, and clay. (Source: Weil and Brady, 2019, Figure 4.7)

of soil solids, and is given by the formula: D p = . ??𝑖?ℎ? ?? ??𝑖? ???𝑖?? ?????? ?? ??𝑖? ???𝑖?? Bulk density is found by dividing the weight of oven dry soil with the total volume of soil, including solids and pore space. It is given in the formula: D b = . ??𝑖?ℎ? ?? ??𝑖? (???? ??𝑦 ??𝑖?) ?????? ?? ??𝑖? (???𝑖?+????) (b) Calculate the particle density and the bulk density of the following soil given. ● weight of soil solids = 1 Mg (Note: 1 megagram (Mg) = 1 million grams) ● volume of soil solids = 0.55 m 3 ● volume of pores = 0.2 m 3 ( 4 marks ) Particle Density D p = ??𝑖?ℎ? ?? ??𝑖? ???𝑖?? ?????? ?? ??𝑖? ???𝑖?? D p = 1 𝑀? 0.55 ? 3 D p = 1.818 Mg/m 3 Bulk Density D b = ??𝑖?ℎ? ?? ??𝑖? (???? ??𝑦 ??𝑖?) ?????? ?? ??𝑖? (???𝑖?+????) D b = 1 𝑀? (0.55+0.2) ? 3 D b = 1.333 Mg/m 3 THEREFORE, the particle density of the soil is 1.818 Mg/m 3 and the bulk density is 1.33 Mg/m 3 , rounded to the third decimal.

(c) Imagine the soil in 4(b) was uniform throughout a given area and that half of the area was cultivated using conventional methods for many years while the other half of the area was left uncultivated. If the above value for bulk density was for the cultivated soil, would you expect the value of bulk density for the uncultivated soil to be higher or lower than for the cultivated soil? Explain your reasoning ( 3 marks ) The bulk density of the uncultivated soil would likely be lower than the bulk density of cultivated soil. Over time, the cultivation practices that are used will destroy organic matter, compact the soil, and disrupt soil structure. Cultivation also makes the soil more susceptible to erosion, which reduces the porosity of the soil, and leads to an increased bulk density of the cultivated soil. Uncultivated soil, on the other hand, is more likely to maintain the natural soil structure, with better aeration and more pore space. Therefore, the uncultivated soil will likely have lower bulk density than the cultivated soil. 5. (a) Calculate the particle density and bulk density of a soil given: ● Weight of soil solids = 4.8 Mg ● volume of soil (solids + pores) = 3 m ● volume of pores = 0.7 m 3 ( 4 marks ) Particle Density D p = ??𝑖?ℎ? ?? ??𝑖? ???𝑖?? ?????? ?? ??𝑖? ???𝑖?? D p = 4.8 𝑀? (3−0.7) ? 3 D p = 2.087 Mg/m 3 Bulk Density D b = ??𝑖?ℎ? ?? ??𝑖? (???? ??𝑦 ??𝑖?) ?????? ?? ??𝑖? (???𝑖?+????) D b = 4.8 𝑀? 3 ? 3 D b = 1.6 Mg/m 3 THEREFORE, the particle density of the soil is 2.087 Mg/m 3 and the bulk density is 1.6 Mg/m 3 , rounded to the third decimal place.

(b) On a neighboring plot of land (plot B) with the same soil type and same initial bulk density as plot A, a different harvesting operation is performed. In year 3, the bulk density of the surface soil in plot B was 1.35 Mg/m 3 . Discuss the differences and changes in bulk density between the two plots of land. In your answer, consider the impact of different harvesting operations on soil bulk density over time. ( 3 marks ) Plot A sees an increase in bulk density over time, while Plot B sees a decrease in bulk density over time. Plot A might see an increase in bulk density because of harvesting practices that result in soil compaction. This could include the use of heavy machinery, tire pressure on soil, and even a lack of soil management. Plot B, on the other hand, likely sees a decrease in bulk density due to better harvesting practices. This could include practices like crop rotation, attempting to reduce compaction, or improving the structure of the soil. 7. (a) A given parcel of land has dimensions of 1 km x 3 km. If the required concentration of fertilizer for this parcel of land is 1125 kg / hectare, how many kilograms of fertilizer are required? Note: watch your units! ( 2 marks ) Area of land = 1 km x 3 km = 3 km 2 1 km 2 = 100 hectares 3 km 2 = 300 hectares Fertilizer required = 1125 kg/ hectare So, for 300 hectares. 1125 kg x 300 hectares = 337,500 kg. THEREFORE, 337,500 kg of fertilizer are required for the parcel of land. (b) A neighboring farmer has a parcel of land with the same dimensions and a clay loam soil. They require the same concentration of fertilizer for this parcel of land as given in 7(a). How many kilograms of fertilizer are required for their parcel of land? ( 1 mark ) The neighboring farmer requires the same concentration of fertilizer for their parcel of land that is equal in dimensions. This means that the neighboring farmer requires the same amount of kilograms of fertilizer for their parcel of land, 337,500 kg. THEREFORE, 337,500 kg of fertilizer are required for the neighbor’s parcel of land.

8. (a) Figure 3 shows two soil profiles. Based on the Soil Taxonomy classification system, soil (A) would be classified as a Mollisol and soil (B) would be an Ultisol. Draw or clearly annotate each soil profile in Figure 3 highlighting some of the important characteristics that distinguish these profiles (e.g., colours, major horizon designations, horizon depths, vegetation types, etc.). ( 4 marks ) (a) Mollisol (b) Ultisol Figure 3 - Examples of a Mollisol (A) and an Ultisol (B) soil profile. (Source: USDA, n.d. https://www.nrcs.usda.gov/resources/education-and-teaching-materials/the-twelve-orders-of-soil-taxono my/ ).