 Welcome to lecture series in advanced geotechnical engineering course, we are in module 7 geotechnical physical modeling lecture number 6, module 7 lecture number 6 on geotechnical physical modeling. So we have been discussing about the scaling loss in centrifuge based on physical modeling and that we have discussed about the how we can scale down the force, work and energy and while scaling down energy we said that it can be in the form of a potential energy or in the form of a kinetic energy. So in the form of a potential energy for that what we said is that considering the definition of potential energy P e normally expressed as the energy lost by falling a mass m through the height h. So this is the P e is equal to mg h. So for getting the scale factor for centrifuge based physical modeling when we take for compare for model and prototype we get you know P e in model and P e in prototype as 1 by n cube. So that means that in the centrifuge modeling basically can offer a very effective way of investigating the effect of explosions on buildings, earthen dams and dams retaining structures etc without the need to conduct these studies in the full scale. If you wanted to conduct these tests you know in the full scale first of all they are expensive and then also can cause damage to the environment. So let us see some examples practical examples with some typical problems. So we said that the potential energy model and prototype is 1 by n cube. So in example 1 consider a requirement to model an explosion event in the field that has an energy release of 1 Giga joule. 1 Giga joule is nothing but 10 to the power of 9 joules at 10 to the power of 9 joules and then we need to model this event at 100 gravities. Now you know the energy which is required to which is planned for you know the release is about 1 Giga joules that is 1 into 10 to the power of 9 joules. So the E model is equal to E suffix p divided by n cube using the scaling considerations we have deduced in the previous slide. Then 10 to the power of 9 divided by 100 cube we get 1000 joules that means that 1 kilo joule. So this is an approximately equivalent to having an explosion from 0.239 grams of TNT. The TNT is nothing but Trinotrotoline, Trinitrotoline. So 1 gram of TNT releases an energy equal to 4.184 into 10 to the power of 3 joules of energy. So 1 gram of TNT is equivalent to 4.184 kilo joules. So by using this relationship we can calculate 1 into 10 cube divided by 4.184 into 10 to the power of 3.184 kilo joules. So with this what we get is that 0.239 grams of TNT which is required and if this is ignited at 100 gravities the equivalent energy released will be equivalent to that of 1 Giga joules that is equal to 10 to the power of 9 joules in the field. The another example let us say that we consider 3.5 grams of TNT including detonator used for modeling an explosion event at 50 gravities. So energy released in the centrifuge model is nothing but 3.5 into 4.184 into 10 raise to 3 joules. So wherein we have 14.644 into 10 raise to 3 joules that means that 14.644 kilo joules is the energy released in the centrifuge model. But equivalent energy in the field is nothing but 14.644 into 10 raise to 3 into 50 raise to 3 that is that scale factor 50 cube when you multiply you will get 1.83 into 10 to power of 9 joules which is nothing but 1.83 Giga joules. So you know this implies that the effect of explosions can be modeled on geotechnical structures using quite small charges in a centrifuge model to simulate the blast with extensive energy released in the reality. So before that you know let us also consider any typical example when you have got some 2.5 grams of TNT Trinotnitrotoline ignited at 200 gravities that means that this may not be really possible but as far as for example is concerned if you look into it then let us look what will be the equivalent energy and equivalent mass in the field. So consider let us say 2.5 grams of TNT and this is the example and then we said that we will conduct this experiment at 200 gravities that means that 200 gravities energy released in the centrifuge model which is nothing but 2.5 into 4.184 into 10 raise to 3 joules. So which is nothing but 10.46 kilo joules. Now equivalent energy in the field is 10.46 into 10 raise to 3 into 200 raise to 3 so because the gravity level is very high we have considered here 200 times the gravity with that what will what we get is that 83.68 into 10 to the power of 9 joules which is nothing but 83.68 giga joules 1 giga joule is equal to 10 to the power of 9 joules. Now equivalent mass of TNT if it is whatever is required if you wanted to estimate in the field that comes out to be 83.68 into 10 raise to 9 divided by 1 gram of TNT is equal to 4.184 into 10 raise to 3 joules. So by using that equivalency what we get is that 20,000 kg or 20 tons that is 20 tons of mass is required to in the field of 20 tons of TNT is required to generate energy which is equivalent to energy release of 83.68 giga joules. So this explains that you know these what are the three examples we have considered these exams that the effect of explosions can be modeled on geotechnical structures using quite small charges in a centrifuge model to simulate blasts with an extensive energy release in the reality. But however while you know considering these things in the centrifuge base physical modeling the safety of the equipment and other issues need to be addressed. Now after having discussed about the force work energy scale factors. Now let us look into you know different aspects of scaling loss you know what we have said is that with gm by gp is equal to n that is when gamma m it becomes n gamma p that gm by gp the scale factor is n that means that which is nothing but if we call length scale factor n suffix l as lm by lp is equal to 1 by n then that is nothing but gm by gp is equal to n is equal to 1 by nl. So using you know the fundamental principles of acceleration we can write gm by gp as lm by tm square divided by lp by tp square we equal to n. So with that what we get is that tm by tp is equal to 1 by n. Because you know if you are actually looking at physical modeling small scale physical modeling at normal gravity and if gm by gp is equal to 1 and if you substitute in lm by tm square divided by lp by tp square is equal to 1 we will get tm by tp the scale factor for time is equal to root n that is the square root of n that is 1 by square root of n that tm by tp you will get it as 1 by n to power of half. Now in case of centrifuge based physical modeling with gm by gp is equal to n is equal to 1 by nl then with that you know by equivalency what we get is that lm by tm square by lp by tp square is equal to n with that we can get tm by tp is equal to 1 by n and by using velocity definition fundamental definition where lm by tm what we have written is that v suffix m we have written as lm by tm and vp lp by tp and if you substitute for tm by tp is equal to 1 by n lm by lp is equal to 1 by n so here what we get is that 1 that means that the velocities are same as that in the field vm is equal to vp and the time is you know it takes 1 by n times less time and this is for you know external loading is concerned that means that a load or a force when it is exerted and this scale factor it may be observed that for the weight forces or external loading the scale factor for centrifuge based physical modeling for this phenomenon it works out to be tm by tp is equal to 1 by n. Now the scale factor for the strain rate for example when we wanted to apply you know certain loading at a certain you know time then you know we can define the strain rate as change in length by original length in time t so that means that we can write that is nothing but delta l by delta l by l into time t. So by for writing the scale factor for strain rate we can write as n epsilon suffix t as delta lm divided by lm tm divided by delta lp by lp tp with that you know by simplifying this we get delta lm by delta p into lp by lm into tp by tm. Now substituting for tp by tm is equal to n, lp by lm is equal to n and delta lm by delta p is equal to as when the length is reduced by 1 by n times the small change in length also is reduced by 1 by n times. So if that 1 by n into n into n you will get n so that means that the strain rate in model is n times the strain rate in the prototype so the strain rate in the model is n times the strain rate in the prototype. So the scale factor for kinetic energy so in the previous lecture we have introduced just a parameter kinetic energy. Suppose this is an example like you know if a ship impact is actually planned in a centrifuge based on physical modeling at a certain gravity then you know we have to subject the block to the certain you know impact is created by some moving mass. So for that the scale factor for kinetic energy is given as n e k is equal to half mv square in model to ratio to half mv square to prototype. So which is equal to 1 by n cube because mass is actually m by mp is equal to 1 by n cube and vm by vp is equal to 1 what we have got here. So with that the kinetic energy also is scaled as 1 by n cube and so as far as you know the weight force and external loading is concerned what we have noted is that velocities in model of prototype were 1 and the time scale factor is tm by tp is equal to 1 by n. And the scale factor for the strain rate is working out to be epsilon suffix t in model is equal to n times epsilon suffix t in prototype. Now another example for you know the energy scale factor so we are actually looking into this you know we know that the pounding or a dropping of a damping of a weight on to the loose sandy soil or sandy strata is one of the methods for improving the ground one of the methods for improving the ground. So in this slide what we have shown is that you know the application of centrifuge modeling technique for simulation of dynamic compaction. So dynamic compaction is a technique which creates a depression at each drop location and also produces an area settlement. So when we do this in a grid fashion as it is shown in this you know photograph where in a known weight is dropped from a known height. So where in these weights are in the range of 10 tons to 30 tons and the dropping heights are in the range of 10 meters to 30 meters because it depends upon the you know the availability of the crane. So in this if you wanted to model this let us look how the you know the modeling can be done and how the impact velocity is compare. So in this example the dynamic compaction let us consider in the field a 20 tons of weight is actually dropped on to the soil mass which is shown here a 20 tons of the weight dropped on to a soil a loose sandy soil and the dropping height is 20 meters. So the potential energy in the prototype which is nothing but 20,000 that is kg into 9.81 into a height is 20 mgh. So with that what we get is that 3.924 into 10 to the power of 6 joules and potential energy in the model which is nothing but PEM in model is equal to PEP by n cube so 3.92 into 10 to the power of 6 divided by 50 cube with that what we get is that 31.39 into joules. So this is the potential energy which is actually modeled in the centrifuge. So that means that we consider a small model which is reduced by 1 by n times. So what we have done is that we have produced the same soil as that in the prototype and the model is rotating about a vertical axis in a horizontal plane what you looking is the top view. So this is the 50 gravities is subjected to this model. So this is the small mass so in order to calculate what will be the weight of the mass or weight of the tamper which is required to be released at 50 gravities. So mass of the tamper is nothing but MP by n cube with that 20 by 50 cube what it works out to be 160 grams of tamper if it is released from a height of say 20 by 50 is above 40 centimeters. So when you release from height of 40 centimeter on to this thing it creates the so called energy which is nothing but 0.16 that is the weight in kgs 160 grams divided by 1000 into the gravity level is not one gravity it is 50 into 9.81. So 50 into 9.81 into 0.4 is the dropping height. So with that what we get is that 31.39 into joules from the scaling loss also we got the same figure. Now if you look into the impact velocity the impact velocity is nothing but how do we obtain is that by equating potential energy with kinetic energy of MV square is equal to mgh we get V is equal to root to gh. So in the prototype the impact velocity in the field when we drop 20 tons mass at a height of 20 meters that the impact velocity is nothing but root over 2gh with that 2 into 9.81 into 20 where you get 19.81 meter per second. In the model which is at 50 gravities in the small scale model at 50 gravities the impact velocity is nothing but root over 2 into n into g into h with that what we get is that 19.81 meter per second. So here in this particular example what we have noticed is that how the enhanced gravities can be used for modeling dynamic compaction. And we also have seen the application of the energy scale factor and the merits of using this technique for the understanding the phenomena of the structure subjected to this type of energies is discussed. Now after having seen the energy scaling loss and the salient applications let us try to look into reducing the scaling loss when the seepage actually happening through a slope or when the consolidation is happening that means this when the seepage and these consolidation phenomena basically they are the diffusion events that means that when a flow occurs in the field in a slope when there is a head difference between let us say that the flow can actually happen from upstream side to downstream side if there is a head drop takes place over a length here. So consider a typical slope in the field situation wherein we have got certain difference of the head where the water is flowing from these are the flow lines which are actually shown here and these are the equipotential lines which are actually shown here a curvilinear squares and at a point h below the top surface of the slope the sigma p is nothing but rho into g into h because rho is nothing but the mass density of the soil in the slope and g is normal gravity that is 9.81 into meter per second square into h is the height. Now similarly at a given point on a typical flow line we can see that the up is equal to rho w g h the pore water pressure in the prototype is equal to of small height say h if you put a stand pipe assume that there is a h which is the head which is actually measured then rho w g h that is the mass density of the water and times the gravity into h. So if the similar situation is modeled then there is a possibility that we have to when you retain the same shape and then this is created by inducing the seepage by putting a constant source of water on the upstream side and maintaining that head of the water then it is possible for us to reduce these steady state seepage conditions. These when we have this steady state seepage conditions which are actually simulated in the model let us assume that this model is subjected to a rotation about a vertical axis in a vertical axis in a horizontal plane such that the n gravities are imposed to a model which is reduced by 1 by n times of the field situation which was actually shown in the previous slide. Then what we can say is that this is the slope with this configuration these flow net which is represented here but because of the radial acceleration field there can be minor influence the influence on the flow net with curvilinear lines because of the acceleration gravity field that is because of the radial acceleration field. So except from that the pressures which are actually there in model and prototype shall be identical. So let us look again at the point here in the total stress which is nothing but sigma m is equal to rho into n g into h by n so wherein this is equivalent to the prototype stress similarly the water pressure which is actually measured at the same point homologous point which is actually shown in the field situation u m is equal to rho w into n g into h by n which is also nothing but u p that is the u suffix p. So in fact now we have to understand we can see that the physical distances are small physical distance are small the pressures are identical and pressures are identical. So let us see how the time is actually factored or how the time scale factor is arrived in the centrifugal based physical modelling and also see how we can actually model the see phase force or see phase pressure and also how the time scale factor for the time is actually modded. So in order to do that first we let us look into how what is see phase force and what is see phase pressure and then by applying the fundamental definitions we can deduce the scaling loss. So in this particular slide what we are actually trying interested in that deducing the scale factor for the see phase force. So for see phase force so we also said that you know when we do similitude in modeling we have linear similitude and dynamic similitude and kinematic similitude. So the dynamic similitude is nothing but you know the force in model and prototype has to have a certain scale factor. So for you know in this case of say for the governing loss in centrifuge based physical modelling where sigma m by sigma p is equal to 1 sigma dash m by sigma dash p is equal to and U m by U p is equal to 1 then in that case the force square factor that is nf n suffix f has to be you know nl square is equal to 1 by n square. So that means that whether it is a see phase force whether it is a weight force or whether it is an external load or whether it is due to certain you know type of activity the scale factor in order to fulfill the dynamic similitude similarity the scale factor the force nf has to be nl square. So with that understanding let us try to look into how for in order to maintain that nf is equal to nl square for dynamic similitude then other factors will be subjected to variation. So considered here a small element of soil where the flow is actually occurring within the flow net and this is the one flow line this is the another flow line. So here perpendicular to this there are the this is one equipotential line and this is another equipotential line assume that this length is delta l and this length is also delta l that means that perpendicular to this plane where the flow is occurring the area is delta l into 1 and the volume of the element is nothing but delta l into delta l into 1 that is that we have considered perimeter length perpendicular to plane of this figure. This is point A and this is point B at point A the head of water is h1 at point B head of water is h2 and h1 greater than h2 so the flow actually occurs from point A to B and h1 minus h2 is equal to delta h. Now what we do is that we calculate the force applied to sand particles and that is obtained as gamma w into h1 into delta l into 1 that means that gamma w into h1 is the pressure acting on the area through which the flow is occurring is that delta l into 1 that is gamma w into h1 into delta l into 1. And similarly at point B gamma w into h2 gamma w into h2 into delta l into 1. Now by simplifying this what we do is that gamma w into h1 minus h2 that is delta h into delta l. Now what we do is that we write gamma w into delta h by delta l into delta l square into 1 so the adjustment of the terms we have done. So writing i is equal to delta h by delta l and v is equal to delta l square into 1. So what we have got the expression for seepage force is that j is equal to gamma w into i into v. So gamma w into v is called as the weight of the fluid phase that means that then the seepage force is equal to i hydraulic gradient into W f that is the weight of the fluid phase. So the seepage pressure is nothing but defined as seepage force per unit volume in that case seepage pressure p suffix s is equal to i gamma w. So what we have derived is that i gamma w. So this occurs in the direction of the flow. Now what we do is that by adopting considering the Darcy's law, the Darcy's law states that v is equal to Ki and the another condition is that for the ranges of the hydraulic gradient whatever you apply the flow has to be in the laminar regime. So we know that the flow can be in the laminar regime or it can when the velocities are very high it can go into transient and then into turbulent zone even in the normal practice when the flow is taking place through coarse grained soils or gravel the laminar regime cannot be guaranteed. So using the Darcy's law i is equal to v by K you know what we can do is that we substituted fact that is seepage force is equal to i gamma w v. So by writing v by K into gamma w into v and by using this for sigma m by sigma p is equal to sigma dash m by sigma dash p is equal to u m by a p is equal to 1 and with gamma w into v is equal to W f that is weight force then we can write the scale factor for seepage force as seepage force in model and prototype is equal to v m by K v p K p by K m because it is at this point and then W f in model W f by f in prototype. So this is what actually we have got so by using the previous discussions the force weight force can be scaled on by 1 by n square but we have two terms which is actually not known to us and that is nothing but one is perimbalt of the soil. So perimbalt of the soil is actually defined as the property of the soil and which is actually is with which the water can flow through the soil. So but we found that the K which is nothing but quotient of perimbalty or Darcy's quotient of perimbalty is found to be function of K that is called absolute perimbalty or intensive perimbalty rho W that is the mass density of the fluid or permeant and G is nothing but the gravity at which this flow is occurring and mu that is the mu W that is the flow of that is the dynamic viscosity of the permeant or the water. So now K absolute perimbalty is found to be you know according to Kozeny-Karman equation we can say that K is nothing but a function of soil skeleton and wherein K is nothing but 1 by CD. So CD is nothing but the shear factor and SS is nothing but specific surface area and T square is nothing but the tortuosity. So this tortuosity is nothing but when water is flowing through the solids this is nothing but the length of the actual path taken by the water to the imaginary length that is length of the sample. So that is nothing but the tortuous path is about 1.414 times you know the length what we assume L for granular soils. So what we have seen is that K which is actually function of these shape factor whether it is angular or sub angular or specific surface area that means that whether it is actually having large specific surface area or you know for example when you have got clays clays actually have got higher specific area so they have the low permeability and T that is the tortuosity and also the void ratio E cube by 1 plus E and then gamma W that is the mass density the unit weight of the fluid and mu W that is the dynamic viscosity fluid. Now when we have K which is actually you know intrinsic permeability or absolute permeability which is a function of soil skeleton when you are actually representing the same soil skeleton as that in the prototype in the centrifuge model then with Km is equal to Kb we can say that Km is equal to Kp and mass density of the water in model and prototype identical and dynamic viscosity of the water and model and prototype are identical. So with that when you say that when you substitute what you get is that K is proportional to G that is when G increased by N times in a small scale model in 1 by N times model then you know K is equal to Km by Kp is equal to Gm by Gp is equal to Ngp by Gp is equal to N so this indicates that the permeabilities are you know N times that of the prototype but it actually leads to you know a question that when you say that K is proportional to G that indicates that at 0 gravity does it mean that the soil is impervious that is that means that K tend to 0. So this is a you know some you know the conflict or what you can say the fallacy which is you know comes out with this you know particular you know understanding is that when you take by this expression K is equal to K rho WG by mu W it implies that when other factors are same that K proportional to G so that the permeability scale factor is works out to be Km is equal to Nkp that means that the soil which is having a permeability of 1 into 10 to power of minus 9 meter per second and it appears to have a permeability of N times you know 10 to power of minus 9 meter per second if N is equal to 100 it is something like 10 to power of minus 7 meter per second that means that a clay soil will tend to have permeability as that of silty soil. So this you know need to be understood and how this conflict can be addressed we shall look into it. So in order to do this you know let us look once again you know the correct the definition of hydraulic gradient itself hydraulic gradient itself. So if you are actually saying you know this V is equal to Ki where I is equal to H by L so with that what we can write is that K is nothing but K rho WG by mu W into H by L. So here by rearranging these terms what we do is that this rho WG H you when you put it within the parenthesis as shown here and K rho WGH divided by L mu W. So by writing K by mu W and rho WGH is nothing but gamma W into H and is nothing but you know H is the pressure the head drop. So we can write like a K by mu W into delta P by L delta P by L is that pressure drop or difference pressure drop over a length L that is nothing but K by mu W into there is nothing but a pressure gradient pressure drop over a length L where L is the length of the sample through which the flow is actually taking place. Now so with this you know it actually comes out that this also you know lead to one you know interesting you know deduction if you look into it we can write K is equal to K rho WG by mu W. Now by readjusting the terms K is nothing but the coefficient of permeability of the soil in the laboratory divided by G when we take K by rho WG as gamma W here and this K capital K that is nothing but the absolute permeability divided by mu W. So K by gamma W divided by capital K by mu W with that by using this and by you know because absolute permeability is known it is measured in Darcy's and it is the function of soil skeleton and which actually has got a units of meter square but you know in order to you know operate you know determine the permeability or velocity or you know seepage velocity in the velocities of the models what we can do is that we can actually use this replace this with K by gamma W. So when we substitute for K by gamma W for K by capital K by mu W with K by gamma W we can write V Darcy's you know discharge velocity is equal to small K that is the coefficient of permeability divided by gamma W into delta P by N. So in a given centrifuge model if you are able to measure a pressure difference between two points and over a length L then we can actually calculate what is the you know discharge velocity by knowing the coefficient of permeability of the soil in the laboratory. So this K is nothing but coefficient of permeability of the soil determined in normal gravity in the laboratory, gamma W is nothing but unit weight of the soil that is unit weight of the water K is nothing but the coefficient of permeability in the laboratory which is determined either by falling head test or constant head test and gamma W is nothing but the unit weight of water. So by knowing delta P by L we can actually calculate discharge velocity and we know that the discharge velocity and seepage velocity is you know seepage velocity is the actual velocity which is actually passing through the grains. So that Vs is equal to V by n where n is nothing but the porosity. So for a given soil skeleton when the porosities are identical then we can say that the seepage velocity also can be obtained by using this expression. So if we are using the same soil in the centrifuge model and prototype then we should expect that same permeability k for soils in both. So by enabling the modifications and by defining the hydraulic gradient as delta P by L and this you know the conflict which was you know which was introduced k is proportional to g. Does it mean that k is equal to 0 at you know the you know at 0 gravities is can be addressed like this. So by defining the hydraulic gradient you know as you know some change in pressure head over a given distance and this is different in the model and prototype. If we say that the hydraulic gradient is different in model and prototype because here the length is reduced by one by n times that is that physical distance is reduced by one by n times but the pressure change in pressure head between model and prototype are identical. So the hydraulic gradient however is actually defined now if you define this hydraulic gradient as you know as a change in pressure over a given distance and this difference is actually and if it is treated as different in model and prototype this can be addressed. So let us look how this can be looked into that means that IM by IP which is what we say that delta P by L in model and prototype. Now if you look into is if pressure head delta P in this centrifuge model and prototype will be the same then although they occur over a much smaller distance in the centrifuge model and with LM by LP is equal to one by n. So what we can write is that IM is equal to NIP when we say that IM is equal to NIP when we say that V is equal to Ki that is the Ki that means that I when we define as a change in pressure over a length L so when we that we can say that IM is equal to NIP. Then we can actually have the scaling law of K which is you know maintained as identical as that in the prototype that means that Km is equal to Kp that means that the only the pressure over a length L you know is n times that in the prototype. So that means that the scaling law for the hydraulic gradient actually suggests that the centrifuge models will have much higher hydraulic gradients than the prototype. So if you are having a hydraulic gradient of 1 then it can be you know 10 at you know 10 times gravity. So with this definition if you look into that for a the gravity gradients are high they say this scaling law whatever we have actually discussed for hydraulic gradient with the discussion what we had in the previous slide this suggests that the scaling law for hydraulic gradients suggests that the centrifuge models will have much higher hydraulic gradients than in the prototypes. Now with the definitions whatever we have discussed we can write that K by mu w in model is equal to K by mu w in prototype and delta BAL model is equal to n times delta P by L prototype that means that delta P by L is nothing but IM is equal to NIP and Km is equal to Kp. So here we are actually maintaining the Km is equal to Kp and with that you know what we get is that the velocities discharge velocity is Vm is equal to Nvp that is that n times in the model. So as the physical distance are small and the velocities n times you know Vp and similarly the seepage velocity as Vs is equal to V by n and with the porosities are identical for a given soil skeleton in model prototype. So Vs in model is equal to n Vsp so the scaling law for seepage velocity is n which is indicate that seepage velocity in centrifuge model will be relatively high. So the seepage velocity in the centrifuge models will be relatively high n times that of in the prototype. This result is actually quite important you know we need to see that you know we shall ensure that the laminar regime is prevalent in centrifuge models. So for that you know when we say that the Reynolds number when it is you know when we define the Reynolds number the Reynolds number can be defined as rho Vd by mu where rho is the mass density of the permeant or fluid V is the velocity and d is nothing but you know the effective particle size let us say td10 by mu w is nothing but the dynamic viscosity. So when we set this you know when we calculate this threshold values and except when we actually have got you know the very coarse grained particles then what it says is that for if you say that re less than 1 according to we are 1979 you know for flow of water through soils it says that you know for flow to be laminar the Reynolds number of in the in soil Reynolds number have to be less than equal to 1. So when we take you know RE is equal to 1 then we can calculate the threshold you know the G levels then we found that the G levels are actually for when you have got some coarser particles only it says that it should not have you know you know the G level more than 45 or so but when you have got the range of the particle model soil which are actually used in centrifugal based medical molding that G levels which are actually required for going in from laminar for crossing the laminar regime is you know very high that means that the G levels are very high they are actually beyond the scopes of the scope of the centrifuge equipments which are actually available in the world now. So that means that by all means it is possible for us to ensure the laminar regime for the type of elastic what we actually occur because of the virtue of the IM is equal to NIP with that Vm is equal to NVP is you know occurs so because of that you know this so called issue is important but you know when you look into the scale factor for you know Reynolds number for same soil in model prototype when we have this the Reynolds number scale factor works out to be N times Reynolds number in the prototype Reynolds number in the model is equal to N times Reynolds number in the prototype. So this Reynolds number with what we have done is that because Vm is equal to NVP so because of that what will happen is that you know the soil D10 model is equal to D10 in prototype and when we have got identical pore fluid in model and prototype we can say that Reynolds number is equal to N times Reynolds number in prototype. But the question is that if you want the Reynolds number identical as that in the prototype then you know two ways one we can think of one is to you know scale down the particles that means that D10 in the prototype in the model is D10 in the prototype by N that means that when we have actually scaled down the particles that means the scaling down of the gradation is one option. The second option is that to increase the you know the dynamic viscosity of the fluid that means that if you are able to increase the dynamic viscosity of the fluid then by say let us say in a 1 by N times reduced model we can say that the dynamic viscosity is increased by N times. So that means that we have a if you have a higher viscous fluid there is a possibility that you know Reynolds number in model and prototype can be maintained identical. So but however the interaction of the fluids particularly with high viscous fluids with saturation with all types of fluids all types of soils like you know silty sand clay soils is very difficult. So and also from the scaling discussion and with the requirement of threshold gravity levels it is you know not required or not mandatory to change the pore fluid as far as the see page phenomenon is concerned particularly when we are actually investigating see page of water see page of water through embankment dams or you know certain geotechnical structure. So what though the velocities are N times this result is actually important and we also have to see that. So what we are saying is that when we say a Reynolds number is equal to 1 less than 1 or so we say that the laminar regime is you know to the more or less is you know satisfied. Now after having scaled down the seepage force and seepage you know pressure and then let us look into you know reduce what is the time scale factor. Now we said that from the definition whatever we have that velocity in model that is Darcy's discharge velocity in model is N times discharge velocity in prototype. So from the fundamental definition of velocity we can write Tm by Tp is equal to Lm by Vm is equal to Lp by Vp. So Lm by Vm is equal to Lp by Vp this Vm and Vp V is nothing but the Darcy's velocity and with that we can write Lm by Lp is equal to Vp by Vm is equal to 1 by N square. So that means that here if you look into it time which is actually velocity is N times as that in prototype the hydraulic gradient is N times that of the prototype. So if you are having a soil with Km is equal to Kp and Vm by Vp is equal to N. So for Km not with it says that Tm by Tp is 1 by N square. So Tm by Tp is equal to 1 by N square that indicates that the time required for flow to takes place is 1 by N square times that of the prototype. That means let us consider 50 gravities and the model is actually occurring subjected to 50 gravities and let us say that if you wanted to have a certain seepage time wanted to see let us say 50 Tm by Tp is equal to 1 by 50 square. So let us say that if it takes you know 1 year so 1 year time 52 days time in the prototype divided by 1 by 50 square. So what we get is that you know if you look into the after simplification you will get an off and over that means that the 52 days of prototype time is equivalent to off and over at 50 gravities. So this indicates that you know the so called seepage phenomenon is actually is very rapid and there is a possibility that you know this can be done and it can be used for number of applications like wherever the like contaminant you know contaminant transport. Let us say that in example of contaminant transport where it can actually have or occur in 1 by N square times that of the time it takes in the prototype. So by some chance actually the Km is not equal to Kp that means that the prototype soil and model soil they have a difference in permeability then we can actually also account for that you know in the expression whatever the logic we have discussed with Im is equal to Nip. So we can write that Vm by Vp is equal to N into Km by Kp. So by using again the same definition and substituting here what we get is that Tm by Tp is equal to Kp by Km into 1 by N square. So this Kp by Km is you know for example if they are 1 then they get cancelled then what we get is that Tm by Tp is 1 by N square. So you know this Tm by Tp when there is Km is not equal to Kp we can actually account for this where in the ratio into 1 by N square. So the above equation is valid but if some reason the soils in model prototype have different permeabilities. If the soils in model prototype have different permeabilities then the time scale factor for Cp is Tm by Tp is equal to Kp by Km into 1 by N square. So now after having discussed you know the Cp scale factors then we also have you know phenomenon like consolidation a consolidation of a soil. Say for example we know that the consolidation of a soil takes place over a long period time depending upon the type of the clay and its permeability. So the consolidation of soil is a diffusion process and basically that occurs when excess pore water pressures are generated in soil due to application of rapid loading. So when you have you know rapid loading then there is a possibility that the dissipation of pore water pressure takes place excess pore water pressures that is excess pore water pressures nothing but the pore water pressures above the hydrostatic pressures. And this takes over a period of time. So with the elapses of the time these excess pore water pressures decrease and excess effective stress in the soil increases. So the void ratio of the soil changes allowing for the settlement to take place. So when we have this you know the void ratio of the soil changes and that allows the settlement to take place. So let us see how you know the scaling loss for time for consolidation can be reduced. So the governing equation for consolidation three dimensional three dimensions can be written and this is actually reduced by using the rate of change of volume to rate of change of void ratio. So we can write that dou u by dou t is equal to C v into dou square u by dou x square plus dou square u by dou y square plus dou square u by dou z square. So using C v is equal to k into 1 plus e naught by A v gamma w where A v is coefficient of compressibility which is nothing but delta e by delta sigma. So C v is equal to coefficient of consolidation, k is the coefficient of permeability, e naught is the initial void ratio, gamma w is the unit weight of the water. So let x is equal to n xm, so x and y and z are the prototype dimensions. So x is written as n times xm that means that when we take xm by x is equal to 1 by n. So it is nothing but xm by xp is equal to 1 by n. So we have for simplicity and convenience we have written x is equal to n xm and y is equal to nym and z is equal to nzm. And with that um also we write it as nu that is the scale factor for the pore water pressure and gamma w model is equal to n gamma w, gamma w the n gamma w is the scale factor for the unit weight of the you know pore fluid, tm is equal to n tt. See the interest is that how much time it will take for when the consolidation is taking place in a enhanced gravity models, then km is equal to nk key. So by substituting this in this you know governing equation of consolidation then what we get is that dou nu u by dou n tt is equal to nk k that is nothing but for k what we substituted is that nk k into 1 plus e0 and for a same soil skeleton for same degree of consolidation the void ratios are assumed to be same. So nk k plus 1 plus e0 into av that is also assumed to be same and v into gamma w. So this n this is nothing but n gamma w n this is the scale factor for the unit weight of water plus into dou square nu u by dou x by n whole square what we have done is that for x we have written x by n whole square plus because this is we are writing for model in the centrifuge model plus dou square u by dou into y by n whole square plus dou square u by dou into z by n whole square. So by simplification readjusting the terms what we get is that dou u by dou t is equal to nt by nu into nu nk n square divided by n gamma w into k into 1 plus e0 by av gamma w into dou square u by dou x square plus dou square u by dou by square plus dou square u by dou z square. So if you look into this except this terms you know this term actually obtained from here by simplifying this but except this term and this is actually appear to be that equivalent to then the prototype. So we need to see for similarity in model and prototype we have to have say identical differential equation governing differential equation in model centrifuge model n prototype that n square nk k because here nu and nu will get cancelled. So n square is actually nothing but you know which is by simplification of this terms we will get dou square u by dou x by n whole square those that n square is actually by taking common we have got here n square nk k nt n gamma w has to be equal to 1. So again using k that is absolute permeability times gamma w by kinematic viscosity and by using this we can say that nk by n gamma w is equal to 1 so that is that because these are identical and then k increases n times gamma w increases n time so nk by n gamma w is equal to 1 with that by substituting here what is for the n square nk nt n by n gamma w to be equal to 1 that nt is equal to tm by tp is equal to as to be equal to 1 by n square that means that this scaling for law for the time of consolidation basically suggest that the consolidation of the soil in centrifuge model occurs n square times faster compared to the prototype. That means that if you consider you know if 30 minutes or half an hour run of centrifuge at 50 gravities is equal to 52 days of consolidation in the field that is nothing but 52 days divided by you know the 50 square when you converted into minutes you will get as 30 minutes. So this is you know the scale factor implies that the time of the consolidation you know the suggest that the consolidation of soil in a centrifuge model occurs n square times faster compared to the prototype. So most of the scale factors like time permeability and all the scale factors several investigators over a period of time they validated this by using again the experimental evidence which is actually the centrifuge based physical modeling. So in this particular lecture what we have understood is that we are try to do some examples on the energy scale factors and then we have tried to see how a seepage phenomenon can be modeled in a centrifuge and how we can actually compute velocities in a centrifuge based physical model by knowing or by measuring the pressure over a length L and then also we discussed about how a consolidation phenomenon can be modeled and how the time scale factor. So for the time scale factor for consolidation as well as the time scale for the diffusion is works out to be 1 by n square that in the prototype that is that time in model is 1 by n square times of the that in the prototype. In the next lecture we will like to look into the capillary flow when it is actually happening what will be the rate of rise of capillarity and what will be the time of capillarity in the centrifuge based physical modeling and then subsequently we will look into how a dynamic event like earthquake can be modeled and what will be the scale factors.