 Let us talk about a bit on in the details on transient method. So, basically the governing equation for a line heat source which is nothing but a probe in an infinite medium is used to understand the transient method. So, this is the device where the probe is there thermocouples are inserted as I said you need two boundary conditions minimum to solve a differential equation. And when we say boundary condition basically these are temperatures and the at a given distance. So, you can fix the thermocouples in such a way that they are placed at different radial distances. So, r is known what you are measuring is theta when the entire setup is heated with the help of a line source which happens to be a thermal probe. So, if you take the axis and if you define a radial distance r this is the form the equation the famous consolidation equation where you just keep on changing the parameters theta to u to you know c and whatever. So, this is rate of change of temperature with respect to time which is equal to alpha which is equivalent of c v term or d i term which is nothing but thermal diffusivity, second derivative of rate of change of temperature with respect to distance and 1 upon r del theta by del r. Where else we have seen this equation in consolidation equation. So, three dimensional consolidation equation which you use in your PVD designs is it not three dimensional consolidation. So, the initial boundary conditions are theta equal to theta 0 at t equal to 0 and r equal to infinity ambience temperature. So, r equal to infinity is nothing but the boundary of the soil mass and the environment where I can assume at r equal to infinity at initial time t equal to 0 theta is ambient temperature. So, this is one of the boundary conditions if you put this boundary condition what you get is limit r tending to 0 2 pi k thermal resistivity term r is the radius del theta by del r equal to minus q where q is i square r i is the current r is the resistance of the nichrome wire. If you further solve this equation you will be getting this elongated form of the equation 2 minus theta 1 is nothing but the delta y and log of t 2 by t 1 is nothing but log of x. So, the slope is equal to r t as q upon 4 pi inverse. So, if you know if you conduct this experiment and plot temperature versus time on a log scale if you know the slope of the graph you just use this slope q is known pi is known r t is known. So, the first job is done that is you wanted to know the value of thermal resistivity or the inverse of resistivity is conductivity is this ok. Now, the biggest challenge is how to calibrate the probe how would you ascertain whether the measurements which are being done by the probe are correct or not. So, when you use soils and at mixtures where the contact is not a big problem you use glycerol. Glycerol is nothing but the glycerin. So, you take a certain amount of glycerol in a container and put the probe inside do the same test at different voltages and measure the values of r t. So, wherever the resistivity values of the standard fluid likely soil matches with the experimental value this is the voltage at which you have to conduct the test all right. So, once you have done this you have no problem you can go ahead and find out the thermal properties of soils, but the real challenges when you work on rocks or in the concrete what is the real challenge? The challenge is how to insert the insert the probe in the porous media. So, this is where this was the strategy devised by Krishna. What he did is he at the time of casting of let us say concrete bricks he sacrificed few thermal probes in each brick. So, this is what is defined as a brick A series that means you integrate the brick with the thermal probe. So, you cast the probe along with the concrete. The second series is you take the concrete and drill a hole place the probe and whatever the contact area is in between where the air is you fill that area with the help of a heat sink fluid. Heat sink fluid is a fluid of very high conductivity and very less resistivity. So, this is the fluid which is normally used for maintaining the proper contact between the boundaries. So, in this case you think of a hole in which the probe is sitting and because of drilling of the hole you have created lot of fractures in the rocks are the concrete all right which were not there in the undisturbed form and then to replace this air you use this sink fluid and then measure the resistivities. So, this is a calibration which is done normally for rocks where you can have corresponding to different days of curing you can find out the resistivities for brick A series and brick B series and what you notice is that the brick B series always gives you higher resistivity. Why it is so? Because as I said when you are drilling the hole you are creating lot of fractures in the rock mass or the concrete. So, this gets reflected very easily in the form of this relationship where the slope is 1.38 which shows that the brick B series is predicting resistivity is higher than the brick A series where you have made the probe as a part of the concrete or the system. So, this was a very good technique which was developed by Krishnaya we have published in International Journal of Rock Mechanics this paper I think 2003. Now, these are the results of the thermodate which I was talking about the heating and trying heating and wetting cycle. Again you know what you are doing here is you start from a very elevated temperature by putting the entire thermodate device in the oven and then after attaining certain temperature you take it out and put it in oven. So, what you saw earlier was a heating cycle that means you are heating the geomaterial the temperatures are going up. This is in combination with the previous graph where by putting the system in water you are cooling it. So, this is a cooling response of the material on a log scale and on a normal scale. So, as time increases the temperature drops down. So, this becomes a cooling curve again you can find out the stresses between the materials in the cooling and wetting cycle and so on. Well, another challenge is you have obtained the value of RT or K by doing this test how you are going to obtain the value of alpha that is the diffusion coefficient or coefficient of diffusivity. Now, this is defined as D square capital T upon T 50. Do you find this type of a term in geomechanics somewhere identical to this? Exactly. So, because the norms remain same almost you can just play with the parameters and the concepts remain same. Now, the question here is how would you get T 50 and what is T 50? So, T 50 is a time corresponding to 50 percent change in a temperature. So, mu happens to be the percentage drop in temperature. So, initial temperature minus final temperature divided by initial temperature is the percentage. So, if you plot it over here with respect to time factor T, this is the same time factor which you have used earlier. This is nothing, but C v into T upon H square if you remember alright. So, this is nothing, but your T factor. So, if depending upon the geometry of the mold that is if height is too much longer than the diameter we say H is infinity. However, if you are following the same concept where H is equal to 2 times the diameter you have the bottom curve. From here at 50 percent you can find out what is the value of capital T. If you know capital T, if you know the time at which the 50 percent drop in temperature takes place, if you know the value of the diameter of the mold in which you are doing the test you can get the value of alpha alright. So, again this is a very simple method of obtaining thermal diffusivity. So, two terms have been obtained now that is R T, K and alpha alright. Now, these are the results where R T as a function of gamma D has been plotted for different type of soils and in general what you will notice is that if gamma D is more the dry density is more the degree of compaction is more the thermal resistivity is less alright because of more contact between the grains and always remember that the resistivity of the minerals is lesser than resistivity of air clear. So, more contact between the two grains the resistivity of the composite system will be less. Now, look at the second variation of course, I asked you these questions in the beginning if you plot resistivity with respect to moisture content more the moisture content with resistivity of water is quite less as compared to air. So, pores are getting filled up with water and hence the resistivity will drop down. So, the general trend is as moisture content increases the thermal resistivity decreases alright. Now, these are the trends which show you the variation of diffusivity thermal diffusivity with respect to dry unit weight. So, truly speaking diffusivity does not depend upon density of the material do you agree with this or not? Why? The same logic is valid for D i that is diffusion coefficient not being dependent on the unit weight. Do you talk about consolidation coefficient as a function of unit weight of the material? Yes or no? C v is the property related to unit weight or not? Tell me yes or no? No. What is the significance of this? This is the fundamental response of the porous media. All these coefficients they have been so designed that they do not depend upon the density they are they are free from the matrix structure of the material. These are synthetic parameters they are basically the parameters used for non-dimensionalizing the equation. So, del u by del t equal to C v into del square u by del z square. So, C v into this term gets cancelled out alright and what you are left with is rate of change of temperature with respect to time. So, truly speaking there is no influence of density of the porous media on the thermal diffusivity. However, if you talk about the thermal diffusivity with respect to moisture content, yes there would be some change. So, as moisture content increases thermal diffusivity increases and beyond a certain point you will notice that it will become almost constant or same. So, these type of characteristics can be utilized in an algorithm mathematical algorithm. If you know the properties of the soils you can just compute their restivities or diffusivities or specific heat. Now, this is the response of specific heat with respect to dry density. So, as expected specific heat should also not depend upon the dry density. So, it almost remains practically constant, but look at the specific heat with respect to moisture content. Do you agree with this trend? More water present in the system, the specific heat of water is more. So, the specific heat of the composite system will be more. So, CP increases with increase in moisture content. Now, this is an interesting relationship where thermal conductivity has been plotted with respect to porosity. And particularly if you look at the numbers of porosities 0.1 to 0.7, what is your feeling? This corresponds to soils or this corresponds to different material. Actually this corresponds to rocks and gravels. So, you will not find so, high porosities in case of soils which are compacted. But when people are working in mines and the mineralogy and particularly those who are working are sciences, they require these type of grass much more. So, this training of the graph was done with the help of the field thermal probe which I showed you 1 meter long. And with the help of this you can find out the thermal resistivities of the crushed rocks and the gravels. And what you will notice is that the conductivity will decrease as the porosity increases. So, I have drawn a line over here. Now, this is what is known as the critical porosity, beyond which there is no significant change in thermal conductivity of the material. So, what would you prefer as a designer? You would like to work in the range of porosity less than this or more than this. Less than this, less than this the material is very unstable. Any unit drop in conductivity, restivity is going to cost you enough amount of money remember. So, the best is whenever there is no change in the conductivity. So, this is the safest limit. The logic says you compact the soil mass so that you are somewhere near in this range. The practical problem is you cannot create this state of the material where porosity is a very very high. You agree or no? Very high porosity means what? It is a good compaction state or less compaction state less compaction state. So, most of the time you work in the field in this range only though it is not a very preferable state of the material. But then these type of nomograms will help you in understanding and designing a cable system. Next time whenever you pass through a place where cables are being laid, we just please stop for some time, request the person to show you a cable. The cable engineering is one of the biggest engineering electrical engineering. I do not know whether you have ever seen an electrical cable or not. Electrical cables are not so simple devices. So, first you should get a chance to see somewhere and then I am sure you will appreciate that what it takes to design a big electrical cable. There will be small small units of the cables floating in some fluid which acts as a coolant and these cables are laid in kilometers long areas. So, in power system particularly those who are working in electrical engineering they deal with this type of designs and we help them as far as geotechnical engineering aspects are concerned. So, based on these tests you can get some generalized relationships. Let me quickly take you through. These generalized relationships were developed by my student David in 2000 known as Didi Tham. What he did is he tried to correlate the dry single phase soils. I say the single phase you can understand that dry soil has no water and hence the connotation is that this is a single phase system. So, 1 upon RT is some constant, some empirical values multiplied by unit weight gamma D. You please do not remember these things just try to understand that how gamma D is related to resistivity and so on. For moist single phase soils depending upon the type of the soil you have parameter B coming in the picture, parameter C coming to the picture depending upon the moisture content. So, if the soil is totally dry moisture content does not come to the picture if moisture content is there this equation becomes important. RT is in degree centigrade centimeter per watt, W is in percentage, gamma D is in gram per centimeter cube. ABC depend on the percentage fraction of the soil and its moisture content and determining these parameters is a big challenge. So, these are the tables which are given by him depending upon the fraction of the soil you have a parameter. So, if it is a clay soil you can use 0.219, soil soil 0.219 based on hundreds of experiments I think more than 5600 experiments you might have done. The B parameter depends upon the moisture content and the fraction of the soil you have some values and the parameters C which again depends upon the moisture content. Now, this is how the mathematical algorithm was developed. Once you have A, B and C the biggest question is A, B and C are also a function of moisture content and fraction of the material. So, when you are working in composite soils what A, B, C values should be taken. So, this algorithm comes very handy for clay and silt phase the weightage function associated with each parameter would be phase of the soil phase is nothing, but the percentage fraction of the soil when moisture content falls in the range of 5 and 2. The minimum of the absolute value of phase percentage when moisture content is more than 5 and then silty sand, fine sand, coarse sand and gravel you can use these functions. So, this is what the ultimately the software was which we developed for finding out the thermal properties. This work is still remaining because whatever has been done till now is only for resistivity or the conductivity. For diffusivity and heat capacity such exercise has to be done, but the logic says that they are not much susceptible to change in gamma D. So, we are not very curious to obtain them and CP also is not much required in most of the situations. So, even if you know alpha value, you know RT value, you know the density of the material you can always compute CP value, but those of you who want to become an entrepreneur you can work in this area and you can sell their software it will be in great demand. Now, based on the studies these type of relationships are also developed for rocks. So, what you notice here is that RT, alpha and CP they have been plotted with respect to porosity. So, if porosity is more the thermal resistivity will be higher. Is this correct? If thermal diffusivity is high porosity will be less. So, if porosity is more diffusivity will be almost less and specific heat of the rocks again will depend upon if CP is more porosity is more because more water is there in the system. So, the question now is is this part clear to you that how material characterization can be done porous media characterization can be done based on thermal properties. When you talk about porous media characterization the most important parameter is to obtain porosity. All your latest techniques of TDR, FDR you know sand replacement method or whatever even your nuclear density gauges which are used are working on the principle of determination of porosity or the moisture content. So, if you have this type of a nomogram alright and if you know the resistivities you can obtain the porosity and vice versa. If you know the diffusivity you can obtain the porosity and vice versa and if you know this specific heat you can find out the porosity vice versa. So, this is the beauty of this type of skin. Any questions? Yes. Yes. So, more porosity means more water and if you have more water in the system no. See more porosity means you see this could be a state of dry material also. So, when we talk about RT it is a composite effect of the voids and the water. So, here you should interpret it like more porosity means less compacted state. This is how you will you will read this graph that means here the porosity is more because the system is less compacted. Less dense. So, a less dense system will always show you more resistivity, but when you come to this graph because as I said water plays a very important role on CP. So, here CP is a function of moisture content and hence moisture content is a function of porosity. Look at this graph CP is a function of moisture content. So, moisture content is a function of porosity. So, this type of a graph can be utilized very easily to obtain moisture content of the soil mass just by measuring CP value, but the interpretation here is less porosity good compacted material dense material and hence resistivity is less alright. When it is RT then you have to take in terms of density only more than the amount of water. Yes, because what is the reason because RT includes again in it the type of the material and the moisture content and the density you must have seen the response of RT with respect to gamma D is tremendous. The drop is much more rapid between RT and gamma D. Similarly, the drop is much more between RT and moisture content. So, gamma D and moisture content can be put together to define porosity. So, in other words what you are doing here is you are defining this porosity as a function of gamma D and moisture content yes clear. So, here both the terms are coming into the picture. So, truly speaking this RT includes in it the state of compaction and the moisture content associated with it. However, there is no way to do it for CP because CP is blind of gamma D and CP is only you know it understands only the moisture content. So, in other words this is nothing, but truly speaking on y axis you have moisture content on x axis you have porosity term. Now, this is the state of confusion truly speaking because alpha remains again not too much dependent upon gamma D, but then alpha depends a little bit on moisture content. So, this also is a moisture content versus porosity response. So, that means you have two graphs where you can iterate the values of the unknowns depending upon how many unknowns you have. And this type of situation is used for defining the compactness of the geomaterial. The best example of these type of nomograms which are being used in practice right now is you must have heard of the name geogauge and nuclear density gauges which railways are using quite a lot for checking the compaction quality of the embankments on which the railway tracks are being laid. It is a very big project which is taken up by Indian railways. So, if I ask you to find out the elastic modulus of the compacted soil mass what parameters you require you require gamma D and moisture content together simultaneously is it not. So, if you know the two parameters from the same location you can quantify its elastic modulus which is required for designing the railway tracks for dynamic loading. So, all this should go in the algorithm when you talk about the material properties and that is where I was saying that porosity does not understand whether this is mechanical loading or a thermal loading is it not it is a fundamental behavior of the material. So, this material property happens to be a pivotal material property in linking the material mechanical response and the thermal response. In other words if any one of the responses are not required the system is liable to give you the response to a certain effect and one of the effects would be which we are talking about here is how heat is going to migrate into the system. This is part clear Sneha any doubts these trends are basically very very generic in nature they are not very very particular trends yes see saturated saturated material will always show you a certain constant value of RT is it not because water has a certain constant value of thermal resistivity. So, in that case the density will take over. So, I do not know whether I will next lecture I will show you a graph where let me answer your question here. Yes, if you take this relationship where gamma D is changing and W is on the x axis. In fact, this cannot be answered with respect to this graph you have to plot RT with respect to gamma D. See if you plot RT with respect to moisture content this is how the curve would be if you plot RT with respect to gamma D this is how the curve would be all right. Now, here this is the water content here this is the gamma D effect. So, truly speaking these relationships are three dimensional nature. When I draw a line somewhere here what this corresponds to this corresponds to water that means this much component of the resistance is coming because of the water and whatever is above this line is because of the compacted state which is gamma D. So, truly speaking you write RT is a function of moisture content plus gamma D where this is the base line for water and this is the role of water and this is the role of the fabric structure. So, there is a sort of add on similarly in this case also if I draw a line over here this is the resistivity of the water approximately 110 degree centigrade centimeter per watt and whatever is getting added up over here this is because of the compacted state. So, this is how your dry density that is a contact between grain to grain and the presence of water in the pores is playing a composite role is this ok or not. See this is a base line. So, the resistivity for water is minimum that means you think of a slurry a slurry will always show you a resistivity of this which is equal to water. You keep on adding different soils they all will come and merge over here. That means this is where the matrix and the mineralogy and the compacted state is contributing to the overall resistivity of the composite material. Look at the second situation you have the moisture content you keep on adding moisture content suppose moisture content is 0 resistivity is going to be very high. So, this is moisture content equal to 0 by the time the moisture content becomes too much what happens this curve is going like this and becomes like this got it. So, this is W equal to 100 percent. So, there is a rotation of the graph in the anticlockwise direction. So, this is W 0 certain value certain value you keep on increasing this is the direction in which the moisture content is increasing. Now look at this graph in this graph this is the direction in which gamma D is increasing. Now if you go much more into the philosophy you keep on increasing gamma D the maximum gamma D will be a state where all the pores are going to be filled up with either air that is one situation maximum contact that is maximum gamma D, but all the pores are filled up with air the second situation would be maximum density, but all the pores are filled up with water. So, again there are two extremes for the maximum possible gamma D even it is it is becoming confusing because you are you are treating them as individual parameters. So, please remember always that when you say RT versus W there are some other parameters and mechanisms which are controlling the whole mechanism which cannot be ignored exactly that is what. So, it is a composite function you may have this function you know maybe subsiding this effect or this function subsiding this effect or there could be a 50-50 you know drop between the two or there could be 70, 30, 30, 70. So, the whole idea of showing you this was that this is the mechanism based on which you can define the thermal response of the material. For that matter if you allow me to show you another graph the classical curve how would you interpret this at this point what is happening? At this point what is happening? If you ignore third axis your interpretation would be completely wrong look at this what is different here the saturation is different though you have the same gamma D. Look at the second situation you may have a you may have a same saturation line ok. So, one parameter is same different moisture contents and different gamma D's. The third interpretation is for the same moisture content I may have how many saturation states not 3 gen that is what the interesting thing is you will report it ok. Now, this is the graph now this is the saturation line draw a line over here I can have another compaction state I have another compaction state I have another compaction state and so on. You are forgetting this state of compaction this state of compaction this state of compaction this state of compaction you are not taking into account meaning thereby for the same moisture content I can vary the saturation also depending upon the compaction state. This is a very interesting situation you know the three phase system itself is difficult to follow. But anyway I hope you can follow now the best way to understand this is that there is a certain background value over bit another mechanism takes over. So, this mechanism is slurry this mechanism is soil matrix and so on. So, I hope you have got the answer to your question ok. Any other question? Yes this is Vinayal sorry see quickly I will show you if you plot R T versus gamma D this will be for clay I will show you in the next class this will be for all silt this will be for sands. So, this is how the material will sit together I will show it to you in the next lecture.