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Phases of Unsaturated Soils – Overview

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Video 1
Physical laws like the law of conservation of mass, energy, etc govern the behavior of all materials including soil. Prediction of a phenomenon requires the understanding of physical laws related to it. From these physical laws governing equations for the phenomenon can be derived while the constitutive relations related purely to the material are useful in solving these equations.Constitutive relationships are interrelations between different state variables that are associated with a phenomenon. With a qualitative understanding of the state variables related to a system,these interrelations can be used in the prediction of a physical phenomenon. The Coulomb’s law of friction, compressibility relation in consolidation, soil-water characteristic curve, etc are constitutive relationships.  State Variables The state variables represent, the state of a system. For example, void ratio, which defines the state of the soil such as effective stress and soil water content, is a statevariable in consolidation. Similarly, water content, pore water pressure, excess pore water pressure are state variables in consolidation behavior. Pore-air pressure and suction, which is the difference between pore-air pressure and pore water pressure, arealso state variables related to unsaturated soils,  Material ConstantsMaterial constants are the proportionality constants of constitutive relationships. These may vary from one state to another state, and also one material to another. For example, the angle of internal friction and cohesion, are the material constants in shear strength behavior of soils. The coefficient of volume compressibility(mv), which is the inverse of bulk modulus of the soil, is also a material constant. Other examples of material constants are the viscosity of water, hydraulic conductivity, diffusion coefficient, air permeability, etc. Material constants help in understanding the behavior of material and in the prediction of physical phenomena associated with it. PHASES OF UNSATURATED SOIL The unsaturated soil is a multi-phase system with pore-air, pore water, and solids as itsthree phases. The air phase is generally bounded by the pore space, and not occupied by water or any liquid. The matter within this pore space may be air, vapor, or a combination of both. Similarly, the liquid phase is generally bounded by the pore space not occupied by the air. The matter within this pore space may be miscible or immiscible solutions likewater, oil, non-aqueous phase liquids etcetera, or combination of two or more. The solid phase consist of the soil grains are particles and may range from relatively fine-grained materials such as silts and clays to organic material to relatively coarse-grained materials such as sand and gravel sand or gravel. However, each of these phrases by no means pure. Because when you take a soil sample in its partly saturated condition when you apply pressure in the triaxial setup; in triaxial setup, if you have taken an unsaturated sample and without saturation, if you compress the sample, the pore-air which is available in the soil mass may get compressed, and the pore-air may get dissolved in pore water. Or, if due to the temperature changes fluctuations due to temperature fluctuations, even the pore water may get evaporated and pore water may exist as vapors in air. So, therefore, water vapor dissolves in pore-air; and gases and solids dissolve in pore waters. So, for example, dissolved air, dissolved salts. Due to extreme variations in individual phase behaviors, because the air phase if youconsider, which is highly compressible; water phase, which is highly incompressible because, for the geotechnical load applications, we consider the water pressure the compressibility of for geotechnical load applications, we consider water is incompressible. So, water is incompressible, the air is highly compressible for the applied loads.So, the behavior of each pore phase and equilibrium between these phases is studied separately. Solids, soil solids are also relatively incompressible, but elastic compression can be considered when a load is applied on the soil. The state variables and materialconstants of the pore-air phase and pore water phase separately are discussed here. Air density, the density of air acts as a driving force for the pore-air floor, and vapor phase transport in the soils. It also plays an important role in the equilibrium between pore-air and free water or pore water, and the atmospheric air. Air density, density isnothing but mass by volume, but then air density we cannot measure in this way. So, we use ideal gas law. And the density of air is expressed as au a the atmospheric pressure times molar mass divided by R T. R is a gas constant and T is the temperature in Kelvin So, the molar mass is defined as the molar mass. The molar mass of a substance is its mass divided by its amount of substance, which is a constant for any given substance and expressed in gram per mol or kg per kilo mol. To compute the density of air, we need to understand the composition of air. When you consider the air, nitrogen is dominant, which is about 78 percent by volume and oxygen is 21 percent by volume, and trace gases there with less than 1 percent. So, individually if you see the oxygen the volume ratio compared to dry air is about 0.21 and nitrogen 0.781, carbon dioxide percentage is very very less 0.03, (Refer Time: 09:09), hydrogen very very less insignificant amount, argon, neon, helium, krypton, xenon, all these different components available in dry air. If you consider the molar mass of individual components of gas components, oxygen has 32, because it is O 2 16 times to there is 32 kg per kilo mol or gram per mol. Nitrogen 28.02, carbon dioxide 44.01, likewise and if you consider the molar mass in air, this volume ratio is multiplied with molar mass, then you get molar mass in air, this is a volume fraction and this is a molar mass. So, this is a molar mass in the air that is 6.704 contribution from oxygen, and 21.88 from nitrogen. And summation if 28.97 kg per kilo mol, this is the total molecular mass of air. So, if you know the total mass of the total molecular mass of air, you can compute the density of air using this particular relationship, where you have if you consider the atmospheric air pressure as 101.325 kPa, and molecular mass we just obtain that is 28.97 kg/k-mol, and gas constant is 8.314 joule per mole Kelvin and temperature is 298 Kelvin. So, here 101.325 kilo Pascal kilo Newton per meter square times 28.97 kg per kilo mol divided by 8.314 joules can be expressed as Newton meter per mol Kelvin into 298 Kelvin. So, Kelvin gets canceled. The kilo, kilo gets canceled; mol gets canceled. And you have kilo here and Newton should be cancelled here. And you have kilogram kg per meter cube that is a density of water, units of the density of air. So, the density of air is justthis, 101.325 times 28.97 divide by 8.314 into 298 which is 1.185 kg per meter cube. So, here the conversion for temperature from degree Celsius to Kelvin, this is the way one can express. If you see the sensitivity of air density, the air density changes with temperature and pressure. So, the density of air, which is expressed as u a times molecular mass by R T change in density is equal to it depends on both air pressure and temperature. Therefore, ∆u a divided into M divided by R T minus ua M by R T square into ∆T. So, using u/v principle, then ∆ρa/ρ is nothing but ∆ua/ua minus ∆T/T. So, this is the expression for understanding the sensitivity of air density with pressure and temperature. This is the sensitivity of air density with pressure. When the pressure changes from thestandard value of 1.0, 1.0, 1.0, 101.325 kilo Pascal to different values, it may decrease or it may increase depending on the elevation from the mean sea level; when it decreases, the density of air decreases. So, the decrease can be directly estimated the delta rho a by rho can directly be estimated by from this particular expression delta u by u a, where you do not need to estimate the rho a. Here using this particle expression, directly the change can be understood. For example, this is a change to 80 then 80 minus 101.325 divided by 101.325 times 100 should give  you minus 21.1 percent. This is the relative change of density with the change in the pressure. As all as we all know that the atmospheric pressure changes with altitude when the altitude increases when you go up when you go when the altitude increases from the mean sea level the atmospheric pressure decreases. The decrease in the atmospheric pressure causes changes in the density of air as we observed from the earlier expression.So, for example, if an empty water bottle, which is sealed well at an altitude of nearly 4500 meters above the mean sea level, where you have the pressure about 58 kilo Pascal, and brought down to different elevations, altitudes, an altitude of nearly about 2500meters near the mean sea level, then the bottles get crushed, the empty bottle get crushed. Because the ambient pressure got increased relative to the pressure, which is inside the bottle, the bottle get crushed. Similarly, as the atmospheric pressure is smaller compared to the atmospheric pressure, atmospheric pressure at higher altitudes is smaller compared to the atmospheric pressureat mean sea level close to the sea levels. When you for cooking of lentils or any other grams, you require certain pressure to be maintained as the atmospheric pressure is lower, you require more gas consumption, you require more energy to cook the dal orlentils. Because the required pressure for lentils to be cooked is fixed. And therefore you need to give more energy to get the lentils cooked. So, the gas consumption at higher elevations is more compared to the lower elevations lower altitudes are close to the mean sea levels. So, therefore as the elevations change, as the atmospheric pressure change there is a change in the density of air, which can be seen. When you apply pressures on soils when there is a gradient of density that is experienced then there is a flow of air that takes place. Similarly, this is about the sensitivity of air density with temperature. As we have seen the density of air changes with temperature in this particular manner when air pressure is constant and the pressure is constant. So, the change in the temperature from at 20 degrees a standard temperature, the temperature is 293.15 Kelvin and the density of air is 1.204. And relative to this, when the temperature changes, how much relative change in density absorbed is seen in this particular column. Here, the temperature should be substituted in Kelvin. So, if there is a change in temperature of temperature from 20 to 30 when there is a change in temperature from 293.15 to 303.15 standard temperature is this so which is plus 3.3 in percentage, this needs to be multiplied with 100 to get in percentage. So, this is the change in density of air is a percentage change in density out of the density of air due to temperature. This is the implication of the sea breeze, which we often see on the seashore as the ocean is cooler compared to land. Because, water is a heat sink, therefore water is cooler compared to land. So, therefore there is a breeze when the air comes on to the land, which gets heated up quickly, and the density of air decreases which is lighter. So, therefore it raises and raising air helps in forming clouds, and this cooler air as this produces cooler air, and this cooler air is heavier, and then again it deposits. So, this is the cycle that follows, and you get a cooler air when you are near the seashore or water body in summer. So, this is, in summary, the density of air density with temperature and pressure. Here the pressure is represented as gauge pressure. So, this is not absolute pressure this is a gaugepressure, so for a given pressure with change in the temperature, this is how the density changes. And for very high pressures also it is mentioned here that may not be important in geotechnical practice.
Video 2 
The next one is the density of water. Water density can directly influence the physical and mechanical behavior of unsaturated soils because many unsaturated soil parameters depend on the water density. The density of water, in turn, is a function of temperature.Most materials like mercury, gold, etcetera decrease their volume consistently on the temperature decreases. When water is cool down steadily from its boiling point, the volume decreases up to 4 degrees 4 degree Celsius and then water expands until it freezes. The freezing point temperature of the water is 0 degrees, where water undergoes a phase transformation from liquid to solid. So, if you plot the density versus temperature, the density of water versus temperature. As the temperature decreases the density increases, because volume decreases up to 4 degrees, but beyond that the density drops. So, here at 4 degrees, you see that the densityis maximum for water. So, because of which here it is maximum, and of beyond that it decreases again. So, therefore when it when water changes its phase from when phase transformation occurs for water from water to solids that is ice, ice has lower density lesser density compared to the water. So, ice can float in water. As the ice starts to melt upon heating; a reverse phenomena, if you are going from hereto here; As the ice starts to melt upon heating some of the hydrogen bonds break, and water molecules can slide closer close to closer together, when compared to its solidstate it is ice. The ice is about 10 percent less dense than cold water at 4 degrees Celsius. So, the volume expansion of water below 4 degrees temperature is called the anomalous expansion of water. We all know this. Such expansion causes bursting of frozen pipes and frostbites on human skin in the cold countries. So, the ice, which is less than 4 degree Celsius, canfloat on water due to these consequences. The ice can displace more water volume than its weight. Moreover, the conduction of heat is dependent on the density of the substance a lighter material, which is less dense, conducts less heat compared to the heavier material or substance that is dense substance. Because of which the aquatic life survives, because in cold countries when cold places in cold places when the waters freeze. It acts as a barrier thermal barrier, and it conducts thermal. It is a poor conductor it is a poor thermal conductor, because of which the waters inside maintain warmer temperatures, because of which the aquatic life survives. This also has implications on geotechnical practice in cold countries, where thermal expansion of water etcetera cause bursting of pipes, and changes in the pore structure,freezing of the volume expansion of pore water due to dropping of temperatures may affect the soil behavior. Here the effect of pressure on water density, if we consider as we have already discussed that water is relatively incompressible for the geotechnical loads we encounter. However, due to the electrochemical properties of clays the density of water changes withsaturation. As we have discussed earlier clay particles carry negative charge on its surface due to isomorphous substitution, because of which positive ions are exchangeable positive ions or exchangeable cations are held on the surface strongly exchangeable cations are held on the surface. When water is available due to the concentration gradient the ions this positive ions should diffuse away from the surface. However, due to the strong electrical forces or due to the strong affinity towards the positive ions the clay surface tries to hold ions at the surface. So, at equilibrium, you have a distribution of these ions distribution of ions and forming a diffuse double layer around a clay particle. This we have seen earlier. So, when the therefore if you consider expansive soils like bentonites, and black cotton soils, where you have a high percentage of mod morganite mineral content, they have a very high surface area. And they essentially have very high surface charge density that can be estimated from its surface area and cation exchange capacity. So, pure mod morganite may have a surface area of about 800 meter square per gram. And it can have a CEC of 110 milliequivalent per 100 gram due to such a high value of the surface area, and cation exchange capacity the surface change density has a surface charge density is very high. It can have very high values of specific surface area and cation exchange capacity, because of which in air-dry state self soil sample is left outside. It can absorb water from the atmosphere and can hold it around the particle surface. And forms a thin film around the particle surface, so which is the diffuse double layer atmospheric conditions the hygroscopic moisture content can be as high as 15 percent in highly expensive mod morganite rich clays and clay soils. So, it also depends on the ambient conditions, where at high humid conditions, it can absorb more water comparedto dry conditions, and temperature also plays a role. Therefore, in clays because of the strong bond due to because in clays, if you observe in clays, the thick layer here; thick line here; around the clay particles is a diffuse double layer, which is very strong strongly held around the particles. And this is free water light blue color is light blue colour is free water, and thick blue is absorbed water; and whichis due to the formation of the diffused double layer. This white colour is air pockets. Since, an unsaturated soil where you have a three-phase system, it is observed that from physical observations using X-ray diffraction techniques etcetera that as a gravimetricwater content decreases. The density of water decreases slightly from one gram per centimeter cube, but then when the gravimetric water content is decreased beyond 10 percent. It is observed that the density increases even 7 to 1.4-gram percent meter cube, so that means the density of adsorbed water is very high, which can be as high as 1.4 gram per centimeter cube. This is from a review paper by Martin, the full reference is given here Adsorbed Water and Clay review in Clays and Clay Minerals. This has some strong implications on soil behavior, because when you estimate the degree of saturation for at different gravimetric water contents are in unsaturated states, you often see that the degree of saturation crosses 100 percent; So, this anomaly because the density of water is not considered into account. For example, if you take a column, compacted with soil, maybe at air-dry straight, and then after that, you connect to a reservoir, a water reservoir. And allow the water to flow into you should have air went, so that air can escape when water goes in. So, then water, when it diffuses into the soil after sometime you do not allow the water to go out, and let it completely get saturated. You can measure its weight with time, and then when the sample gets completely saturated, you can extrude the sample from the column, and measure its water content. So, you know its mass, total mass, and total volume, because the volume of the column is known. So, inside diameter and then the length of the sample is known, so that you canestimate the volume, mass is known, and water content is known. From mass and volume, you can estimate the density. Knowing the density and water content you canestimate the dry density from rho by 1 plus w. Knowing the dry density, you can estimate the void ratio using this expression, you can estimate the void ratio. And knowing the void ratio, and water content and G s specific gravity, you can estimate the degree of saturation. In this process when you estimate the degree of saturation, when you plot degree of saturation versus gravimetric water content, nearly is a saturated state you see that the degree of saturation estimated is more than 100 percent in highly expensive clays line bentonites. So, this is because unless you adjust the ρw to higher values, you will not get the degree of saturation to 100 percent. So, this can be clearly seen when you experimentally estimate a degree of when you experimentally measure the water content and estimate the degree of saturation. So, after discussing the state variable such as the density of water and density of air, letus discuss some material constant such as viscosity of air and viscosity of water. Viscosity is defined as the ability of fluid to deform under shear stresses; Viscosity of soil pore-air and pore water has a direct bearing on the hydraulic conductivity. And important viscosity is an important material constant for analyzing the flow problems. Viscosity definitions in fluid mechanics, you see that there is one dynamic viscosity and kinematic. So, dynamic or absolute viscosity, which has units of Newton’s second per meter square or centipoise expresses resistance to shearing flows of the fluid. Kinematic viscosity is expressed as the ratio of dynamic viscosity to the density of the fluid, which is used in analyzing the Reynolds number. So, if you plot the dynamic viscosity of both air and water with temperature, an interesting observation is that the water, viscosity decreases with an increase intemperature. But, air viscosity increases with an increase in temperature. At standard temperature 20 degrees, the air viscosity is 0.018 centipoise or 1.8 into 10 power minus 5 Newton second per meter square. Because 1 poise is equal to 0.1 Newton second per meter square there is a conversion. So, at the same temperature, the viscosity of water dynamic viscosity of water is 1 nearly 1 centipoise are 1 into 10 power minus 3 Newtonsecond per meter square. It is interesting to note that the dynamic viscosity of air increases with temperature contrary to the water where the viscosity decreases. An increase in air temperatureincreases the air viscosity because the gas molecules get additional energy and colloid more often. This increases the momentum transfer between stationary and moving molecules which is what causes the air viscosity. Water viscosity, in contrast, decreaseswith temperature. Warm water is useful, therefore washing clothes has the lowest viscous water will have more ability to penetrate into the small pores of the cloth and remove the dirt. The use of detergents would enhance the dirt removal process. The flow characteristics of water through the soils will also be improved with the low viscosity as the conductivity of the water increases. Therefore, the viscosity has a direct bearing on the flow behavior.Thank you.