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Hello everyone. We have been discussing the volume change behavior of soils especially clays. let us understand with simple experiments how the volume change of different clays can be understood. Let me demonstrate the variation and volume of thesoil in water right kerosene, which is non-polar. Here we have 3 different soils one is inert sand which does not react to water or kerosene and we have bentonite, which is rich in montmorillonite, which reacts with water and kerosene again because it is non-polar itdoes not react. So, the clay water interaction because of the mechanisms involved in clay water interaction, the change in the volume of the soils can be understood with simple experiments which are called index or equilibrium cement volume. And, we have kaolin which is rich in mineral. The volume of this minerals in water is smaller than in the presence of kerosene.So, let us understand why the volume of these soils of clays change with different interactions with water and kerosene. I have my research team here they would help me in doing this test that is understood now.We have 2 cylinders; in one cylinder we have sand dry sand have to and in the other cylinder will have sand kept in water. Here you can see the volume of the sand in boththe cylinders is nearly the same. So, that is because of sand has not reacted with water. So, because of that, the volume is merely the same, another hand if we have bentonite let us see what happens. Here we have taken 2 grams of bentonite clay; we have a glass jar glass beaker, the soil will be mixed on with water. But; however, for mixing, we require a lab shaker. So, that thorough mixing is possiblethe dry sample is added to the glass beaker, and water is added.So, now this is normal mixing is not possible. So, therefore, this will be placed in a shaker and will be continuously stirred or mixed in the orbital Here, we can specify the speed and we can set the time. So, normally, this is how this sample is placed and we can set the speed. And, as slow speed is increased it will start shaking and considerable time is required for bentonites, for mixing. And, once the sample isthoroughly mixed this will be transferred into a glass jar to make 100 ml solution and sedimentation will be allowed. Now, you have 2 grams of clay dry bentonite powder which is available in a 100 mlglass cylinder. Finally, be thoroughly mix it again, and then we let it settle for equilibrium, the particles slowly settle under gravity. This takes enormous time in the case of bentonites, we have a clay water cylinder which is prepared earlier, much earlier which is shown nowhere. So, here you can see that there is clear water that is formed. And, you can see the interface between clay and then water. If you look at the volume of the clay now this is significantly high. This is because of the formation of diffused layers. Let me explain onthe board now. When we have a cylinder, where we placed clay in it when it is mixed with 100 ml of water, which forms a clear clay water solution at equilibrium. This is the equilibrium volume. If sand of that amount is taken and then put it in water, it does not exhibit thatmuch of volume expansion or volume change, volume essentially remains the same as we have just seen. In the case of clay, the volume expands and the equilibrium volume is significantly higher. This is because when you consider individual particle or clay platelet, which is a thin cylinder particle or platelet. The thickness is around 10 Angstrom units nearly one nanometer for smectite particles that is bentonite. And, this is a slender particle platelet; because of the negative charge on the surface, it has a formation of diffused layersaround individual platelet. So, the diffused layer around individual particle or platelet is significantly high the thickness could extend perhaps 50 to 100 orders of magnitude higher than the thickness of the platelet. It can be an extension to the higher value depending on the surface properties of individual clay platelet or individual clay or a given clay, which defines the plasticity of the clay. So, now electrostatic potential around individual platelet varies in this manner and this is a thickness of the diffuse double layer, which is represented with small d around individual clay platelet. If, I have 2 clay platelets which are just started interacting witheach other. And, they have diffuse double layers formed, now if this is the particlethickness or platelet thickness. And this is the thickness of the diffuse double layer from 2 different particles; if one particle diffuse double layer thickness called d, then it is total value is 2 d. So, we can relate this thickness of the diffuse double layer with void ratio and from void ratio, the equilibrium volume can be related. So, it is very clear to understand how the equilibrium volume is such a significantly high value. So, here the void ratio we define void ratio as the volume of voids by volume of solids. In this case, if you consider the volume of voids as this is the volume of void space, this is a particle, this is another particle. So, the volume of voids can be obtained cause this is the space that represents the volume of voids. So, the volume of voids can be obtained as 2 d times the surface area. So, which is A divided by the volume of solids that is the thickness of the plate times the area. So, the area gets canceled from the numerator and denominator. So, essentially the void ratio is simply 2 times the thickness of the diffuse double layer divided by the thickness of the plate. If diffuse double layer thickness and thickness of the plate are known we can estimate the void ratio. However, we can avoid estimation of the thickness of the plate and plate by considering specific surface area which can be determined in the laboratory.So, the specific surface area is nothing, but the surface area per unit mass which has units of meter square per gram. So, here if I take the surface area of individual particle or platelet, the surface area consists of this area and area of the edges. If I ignore the area ofedges you have 2 times the area divided by the mass of solids, if I multiply with the volume of solids and divide with the volume of solids, I can write this as 2 A divided by the density of solids. And, this volume of solids again I can write it as this is density and this volume of solids again I can write it as t p times area again area gets canceled. And so, therefore, the specific surface area is nothing, but 2 divided by the density of solids times thickness of platelet. I can replace this thickness of the platelet from the previous equation. Now, I can write the void ratio as d times rho s times SSA. So, we got rid of t p now, the thickness of the platelet now or this can be written as d times specific gravity times rho w times surface area-specific surface area. Generally, thespecific surface area of clays, like bentonites is very high, which can be as high as 800 meter square per gram. So, therefore, this is a significantly higher value maybe 800 meter square per gram. So, this is the density of water 1 gram per centimeter cube and this is the constant value, it may be around 2.7 this is the case of diffuse double layer, which can be several orders ofmagnitude higher than the thickness of the plate. This may be around 50 times if we angstrom then 500 Angstrom units or 1000 angstrom units we can use. And, if you calculate the void ratios will be significantly high we can be as high as more than 100. So, similarly, this void ratio can be related to the equilibrium sediment volume, because this is the total volume, and knowing the total volume and volume of solids, we can relate this to void ratio. So, therefore, the void ratio is related to the thickness of the diffuse double layer and specific surface area. Because once there is the formation of a diffuse double layer around the clay platelet. So, the interaction of these 2 diffuse double layers would govern the total volume of the soil. So, as this diffuse double layer thickness is significantly high, the volume is significantly high. So, this can be verified by conducting an experiment in kerosene which is nonpolar. So, therefore, there is no formation of diffuse double layers. So, when you conduct the experiment the equilibrium volume is significantly lower. So, this our equilibrium volume in water and equilibrium volume in kerosene or any other non-polar liquids. So, the difference between these 2 volumes would infer our the swell potential of the soil or how much it can significantly swell due to the clay water interaction or due to the formation of diffuse double layers. So, this is the difference between the equilibrium sediment volume of bentonite in water and kerosene. In the case of water, the equilibrium sediment volume is significantly high somewherehere, still, the soil is about to settle and still it has not come to equilibrium and it may take further some more days, but anyways the equilibrium sediment volume will be somewhere here. So, this is the volume around 60 55 ml, around 55 58 ml is the volume,when 2 grams of dry bentonite powder is mixed with water, the equilibrium sediment volume is nearly 58 ml. So, in the same powder, the same 2 grams of bentonite powder when it is mixed with kerosene. So, the volume is significantly low. So, this is the volume of clay when there is no formation of diffuse double layers. So, the volume is non-measurable by this even, because this is much lower than even 5 ml.So, this shows the potential of bentonite in the presence of water.
We have been discussing the volume change period of kaolinite. Kaolinite in a peculiar manner it behaves it does not swell. However, it collapses or it exhibits collapse in the presence of water or water which has pH higher than the isoelectric point of the edge. So, generally, if you take distilled water, the isolated point of the edges is about 3 to 4 and the distilled water pH is around 7 sometimes it may be 6.5 and 7. So, therefore, due to the changes in the charge in the particles, the particle interaction is different, it exhibits different behavior. So, if you if I consider kaolinite particle or platelet. So, the particle thickness or the platelet thickness would be around 10 nanometers nearly 100 angstroms units, which is nearly 10 times larger than the smectite particle, which is seen from the scanning electromicroscope. So, now particle thickness is larger therefore, the specific surface area is small, the specific surface area is smaller compared to the smectite and the specific surface area varies in the range of 10 to 12 for kaolinite and the is in between the specific surface area is about 18 meter square per gram. So, because the thickness of the particle would govern the surface area would influence the specific surface area. So, when the particles are smaller surface area and the cations change capacity is also small because there is a strong bond between individual basic units of kaolinite of the 1 is to 1 mineral. So, therefore, the charge is also compensated except that there is a small charge which is due to the isomorphism of substitution, which is called a permanent charge, but which isa smaller value. So, the ccs the cation exchange capacities would 5 meq per 100 gram. Apart from the permanent charge the charges on the kaolinite platelet us would change depending on the interaction with different (Refer Time: 18:08), edges generally have a positive charge and the face has a negative charge.So, therefore, when the particles interact in dry environment particles would form or arrange themselves into this structure is called edge to face or structure. So, this may be another (Refer Time: 18:35) particles and these are otherparticles. So, if you see there is a macropore volume that exists between individual platelet us and there is a micropore volume and this is the macropore volume. So, if kaolinite is mixed with kerosene in a cylinder, the platelets would arrange themselves in this manner and edge to face interaction they have and the particle or equilibrium volume is higher because here the particles would not interact with the fluid. So, therefore, the volume is higher. However, if the distilled water is used in place of kerosene so, distilled water which has a pH higher than the isoelectric point of the edges, the charge of the edge would change to negative. So, therefore, now edges and particle face both have negative charge, particleswould arrange themselves into the the face to face manner. So, therefore, in the same jar when water is mixed with, when kaolinite is mixed with water, the particles would arrange themselves into the the face to face manner. So, the the final sediment volume is smaller, this is the equilibrium sediment volume. Essentially the kaolinite and bentonite when there mixed with water the particle arrangement is face to face only, there is dispersed condition. However, in the case of kaolinite because diffuse double layer does not exist around the clay platelet because of low surface area and low CEC. So, the volume is smaller. In the case of bentonite due to high surfacecharge, the diffuse double layer thickness is larger though volume is very high. So, therefore, when it is compared with kerosene volume is much lower, when it appears that the. So, therefore, the volume is a lower compacted state when kaolinite is in a compacted state when the structure is built on kaolinite, it can take whatever the load that is put on the soil in a dry state. But; however, when the soil has saturated the structure of the kaolinite would change to face to face arrangement and which causes collapse on the soil. Here the kaolinite is mixed with water and the kaolinite platelet us. Now they would disperse and finally, settle with time and at equilibrium, this is the volume that exhibits that is exhibited by the kaolinite. So, the volume is about 18 ml in water. So, there is a clear solution that is formed, and but as the pH value increases. So, there is a lot of dispersion that takes place and the clear solution may not be visible. So, in the case of kerosene, the volume is higher, the volume is about 28 ml, again the volume of the equilibrium sediment volume would vary with the amount of kaolinite mineral that is present in kaolin soil and the surface properties. So, when the kaolinite is mixed with water. So, there is a dispersion that forms because particles are now arranged in the face to face arrangement, face to face manner. So, it would take enormous time to settle and come to equilibrium.So, when it comes to equilibrium there is a thickness of sediment volume that would be much higher compared to kerosene or much higher compared to the sediment volume in the presence of kerosene. So, the dispersion still after formation on this equilibriumvolume, there is some dispersion usually that is left above the clear solution, the clear solution.
So, this is the moisture sensor called EC 5 developed by decagon, which is an inexpensive sensor for the measurement of moisture content in the field and laboratory. This determines the volumetric water content directly based on the dielectric property of the soil using the capacitance principle. So, you can also obtain the dielectric properties of different fluids, if you immerse it in water based on the dielectric property it will give you the volumetric water content. And, if you put it in kerosene as the dielectric constituent is much smaller are significantly lower compared to water the volumetric content water also changes. So, apart from this, this can be directly embedded in soil in the field or in the laboratory columns to estimate the volumetric water content which changes in the moisture content. So, this is another moisture sensor called 5 TM which is also developed by decagon. So, these 2 sensors can be used in the soil for the estimation of volumetric water content. So, now, we will demonstrate how the sensors respond in the presence of water in the presence of soil with different moisture contents. The 5 TM moisture sensor is connected to data logger called EM 50. Now, this is connected to the computer to read the data. The data logger can record and store the data for a very long time. So, currently, it shows the value is nearly close to 0. So, the volumetric water content is close to 0, which is the volume of water by total volume. So, it is a meter cube per meter cube. Now, the sensor will be dipped in water. Let us observe the change in the reading when the sensor is dipped. The reading change has to close to 1. So, which is validating the sensor is working or not. Now, this sensor will be dipped in soil and soil will be flooded and the moisture content will be changed within the soil so, that we can record the readings. Now, here the sensor will be embedded in the column, where sand is poured. Here thesand is poured at very loose density then the sensor will be embedded. The sensor will be directly kept in the soil, the orientation of the sensor could be in any direction, it could behorizontal or vertical, but; however, the orientation of the sensor is sensor should beperpendicular to the soil axis. So, otherwise, when there is a movement of water, that would influence the readings of the sensor. So, whether our sensor can be embedded. This is now the sensor is embedded in the soil, the sensor is connected to the data Now, we are set to go, now we connect this column with a water reservoir, when the water enters into the sand column, the changes in the moisture readings let us observe. Now, the column is connected to a water reservoir. So, that can be seen here when we allow the water to flow. So, water slowly enters into the sand. We can see the movement of water into the sand column. So, now, the water rises in the sand column. So, the movement of water can be seen. Now, the water rises slowly and reaches the sensor, and as the water comes near to the sensor the readings of volumetric water content will change. The sensors have a very small radius of influence. So, when the water enters into that zone the volumetric water content will start changing. Now, you can see the changes in the water content, the volumetric water content is about 0.163 now. This will be continuously stored the data volumetric water content data willbe continuously stored, but if you want to see it we need to scan it and then see at any given point what is the volumetric water content. So, the value has changed now 2.306. So, that will keep changing and at the end of the test, we can plot the changes in the volumetric water content with time, at the place where the sensor is embedded in the soil. We are not able to see in the sharp wettingfront here because the soil is compacted in a loose state and then the densities would slightly change and therefore, there are some fringes that appear here. So, the volumetric water content is nearly 3.346 to 0.516 now, the sharp variation in the colour can be seenin the soil. This movement of water also can be predicted, if the head is constant using green method provided the saturated water content of the sand is known and the initial head of the soil is known. Now, theta value which is 2.356 slightly increased 2.36, because a head reduced now the hydraulic gradient is reduced. So, the movement is slower now. Here, we are not maintaining any constant head. So, therefore, this moisture sensor is very useful even in the field to determine the movement of water or the variation in the moisture content within the soil with time due to rainfall infiltration or any other conditions. So, that it can be used in wetting in use landslides problems. It is a T 8 tensio meter, the tensiometer generally the basic tensiometer has a higher intric porous disk. Now, the tip can be shown this tip is immersed in distilled water. And, this higher intric disk would not allow the air to enter into the system until the matric suction reaches the air entry pressure of the porous disk. So, now this is the whole system and which is connected to water reservoir inside. And, the sensor to read the negative pressure which is transmitted to the porous disk, when it is connected to a soil column or when it is connected to a soil which containsnegative water pressure. So, this is the whole setup this is the T 8 tensiometer, the distill one which is developed by the again, which can be connected to a data logger and thereadings can be read. The tensiometer it can measure in the range of nearly 1 Kilopascalto 100 Kilopascal. However, the cavitation would limit it is measurement range to go beyond 100 kilopascal. Even there is an air entry of the high air entry disk allows it to use for more than 100 kilopascal. The cavitation pressure which is nearly the atmospheric pressure minus the vapor pressure, this value is smaller than 100 kilopascal which would limit the location of this tensiometer for measuring the suction value beyond 100 Kilopascal. Due to dissolved soils etcetera, the range of this particular tensiometer will again decrease. So, this tensiometer can be readily used in the field along with moisture sensors for establishing soil water characteristic, the curve has to observe the changes in the suction with time. However, the response time of the tensiometer is different from the responsetime of the moisture sensors. Moisture sensors respond quickly compared to tensiometers. So, this tensiometer's response time varies with the initial condition of the tensiometer and the moisture content in the soil. So, the hydraulic equilibrium when it has toestablish the hydraulic equilibrium time would be equivalent to your response time. So, hydraulic equilibrium time would be higher in the case of expansive soils. So, therefore, it takes several minutes to hours to get an equilibrium value on the other hand the moisture sensors reduce immediately. So, therefore, the caution should befollowed to obtain equilibrium moisture content and equilibrium suction values in the laboratory, but in the from the field data when we read. The data will be not accurate or the data would not allow due to relating the moisture content and suction immediatelybecause the response times are different.
This is an mps 6 sensors that are developed by decagon. So, this is also to measure soil water potential or matrix suction in the field and laboratory. This can be embedded in the soil just like the other moisture sensors. And, if you look at the size compared to thetensiometer of the size is very small. And, this is inexpensive and this can be readily used in the field. The range of suctions it can be measured would vary from 9 kilo Pascal lowest suction to the maximum value of 1000 kilo Pascal. So, this is 100 MPa 100 megaPascal’s which is given by the company. So, this has fixed matrix porous ceramic disks 2 disks that are placed of these disks arethe size of 36 7 and in between there is a printed circuit board. And, outside you will have stainless steel screens. Now, when this is embedded there is a hydraulic equilibrium between the sensor and the soil that takes place. The sensors are first of all very useful for estimating the matrix suction or soil water potentials, in highlyexpansive clays, because the suction range varies significantly for the soils. However, recent studies by (Refer Time: 36:08) in 2016 show that the sensors would not respond correctly beyond 3000 kilo Pascal or 3 MPa when it is compared the values are compared with WP 4 or hydrometer readings, because initially when the sensor is wet the moisture flows from the sensor to the soil. So, during that time the surrounding soil swells, and the moisture flow stops. So, because of that, the readings are not accurate. The other hand when the sensor is initially dry, but the soil is wet then the moisturemovement takes place from the soil to the sensor soil volume decreases. Due to which there is a discontinuity with the soil due to which there are some measurement errors. So, even though this is an inexpensive sensor that could be used for measuring the matrixsuction in the field, it has several limitations due to workability conditions in the field. Now, we have kept moisture sensor and tensiometer in the column and we are filling sandy soil sand is filled to the nearly top level. Now, this is level and now it is ready, now we will allow the water to flow through by connecting it to the reservoir. So, now, this is connected to a reservoir. Now, water is going into the column due to the provided hydraulic gradient. Now, let us see how the readings change. So, the first port is connected to the tensiometer and the second port is connected to 5 TM that can be seen here. So, the first port has the value of suction equal to minus 19 Kilopascal and the volumetric water is 0.057 meter cube per meter cube scan, as more water infiltrates, the readings change. Now, the new reading is 0.211 for the moisture sensor, and suction is nearly seen 20.89. So, the response tensiometer is higher. So, therefore, the suction value is not changing, even though the water level is changing. So, the moisture content value is changing.So, therefore, we cannot directly use moisture content and suction data to establish SWCC from this method. The moisture content now changed to 0.315 and whether the tensiometer reading is 21.57 kilo Pascal. So, it can be seen that the response times for these 2 cells are different. And, it is a vapor technique where soil sample is kept in a closed desiccator. And, at the bottom, the soil solution is placed in we can clearly see the soil sample, which is kept above the soil solution. However, the vapors from the soil solution interact with the soil, and depending on the vapor pressure of the soil solution, either moisture will be taken from the soil or moisture will be absorbed by the soil. Finally, after equilibrium, the vapor pressure of the soil pore water will be the same as the vaporpressure of the soil solution. We can place different soil solution at the soil solutions at the bottom with different concentrations. So, therefore, they exert different vapor pressures on the soil sample. So, the vapor pressures of the soil solutions can be determined using WP 4 where the solution can be kept in double P 4, and directly the total suction value can be determined. Based on the relative humidity measurement from WP 4, the total suction will be obtained. And, when this suction or the vapor pressure is in equilibrium with the soil pore water, soil also will have the same suction at equilibrium. So, use a different pore fluid at the different equilibrium territorials, whatever the water content soil achieves at equilibrium time one can establish the soil water characteristic on the drier side. So, this is a useful technique. However, it is high time taking, but this could be the only alternative in the laboratory for estimating the soil-water characteristic curve on the dry sides. This technique is very useful to estimate the, to control the suctions, to a high value, as high as 300 mega Pascal’s. The technique is simple, but regularly need to monitor the weight of the soil sample. Once the weight of the soil sample is constant; that means, there is an equilibrium that is established between the soil solution and in the sample. So, when the soil sample moisture content can be obtained at a particular level. And, soil solution vapor pressure can be obtained from the WP 4 technique, which gives thesuction value within the soil sample and suction, and moisture content with these 2 values are known we obtain 1 beta point on the SWCC on the drier side. So, this is a dew point potentiometer or WP 4 designed by decagon. This works based on the principle of measuring dew point due to condensation. Initially, the soil may have particular vapor pressure, when the sample is kept inside this WP 4 instrument or equipment. There is a closed chamber in our temperature controller, when the temperature is reduced drastically it reaches the vapor pressure becomes equal to the saturated vapor pressure at temperature point under that particular temperature, which makes the formation of dew point. So, that will be deducted.So, as soon as it is deducted, it records the dew point temperature as well as the initial temperature. Based on these 2 readings, it is possible to estimate the initial relative humidity of the soil pore system. So, from that using Kelvin’s equation it is possible to determine the total suction of the soil. So, this can be calibrated using different soil solutions when a KCl solution is used with a concentration of 0.5 at temperature 20 it gives a total suction value of 2.184. And, same KCl when it is used at a saturated solution that is 342 grams of solute should be added in one liter of water then it gives a suction value of 21.9. Then, the soil solution is taken and in the cup half full, then it is placed in this door, and when you close it the solution when you put it on. Now, the temperatures will be gradually reduced inside then this vapor pressure wherever this is equal to the saturated with pressure, corresponding temperature is a dew point temperature. So, that reading it gives here KCl 1 molar solution is used. So, the corresponding suction value is 4.6 mega Pascal. Here we can see the changes in the temperature.