 Welcome to lecture number 14 of advanced geotechnical engineering course. In the previous lecture we have introduced ourselves two methods for measuring the permeability, we said that there are two types of methods in the laboratory, one is constant head test and falling head test and we also discussed about the chief differences between these two test methods. In this lecture which is permeability and seepage part 3, we will try to discuss about the factors affecting permeability and then we will introduce ourselves two different types of the flows and the mathematics which is connected with the seepage phenomenon. So this part of the lecture is permeability and seepage part 3. So as shown in this slide the permeability and drainage characteristics of soils are shown, the coefficient of permeability k which is actually mentioned in meter per second, if you look into this the one which is actually there in the yellow color wherein it actually has in a good drainage characteristics when it comes to this pink color, this particular range has the poor drainage characteristics and the color which is in orange here that is beyond 10 to the power of minus 8 meter per second which is actually has the practically impervious drainage characteristics. So this particular these permeabilities are possible for soils which are a clean gravel or clean sands or clean sand with gravel mixers. The poor drainage characteristics are the soils which are actually having permeability in the range of 10 to the power of minus 6 to 10 to the power of minus 8 meter per second. This is possible for very fine sands organic and inorganic sills, mixers of sands, silt and clay, glacial till, stratified clay deposits. So for this type of soils it is possible that the permeability can be in the range of 10 to the power of minus 6 to 10 to the power of minus 8 meter per second. For certain type of soils like homogeneous clay below the zone of weathering these soils can actually possess the permeability in the range of 10 to the power of minus 8 to 10 to the power of minus 11 meter per second. So we have different ranges of permeabilities and there is a unique property for the soil and the soils which can have the granular soils mostly have good permeability or very high permeability, fine grained soils have low permeability. The factors affecting the permeability if you wanted to look into it, the coefficient of permeability is a measure of the ease with which water flows to the permeability materials. So this is put forwarded by Kozny-Karman and he has proposed an equation, they have proposed an equation which is V is equal to 1 by CS and S suffixes T square and multiplied by gamma W by mu into E cube by 1 plus E into I. So this is nothing but V is equal to Ki the component K that is coefficient of permeability is indicated here by 1 by CS SS T square gamma W by mu into E cube by 1 plus E. So this is according to Kozny-Karman, this is basically valid for coarse grained soils and Taylor 1948 he has also proposed the equation reflecting the influence of the permeate and soil characteristics on the K and this is deduced by using the Poiseuille's law which is given like this V is equal to C into DE square the DE is nothing but the particle size gamma W by mu, gamma W is nothing but the unit weight of the permeate mu is nothing but the dynamic viscosity of the permeate E cube by 1 plus E into I. So both this equation assume that interconnect voids are visualized as a number of capillary cubes through which the water can flow. So we have two sets of equations one is proposed by Kozny-Karman basically is valid for coarse grained soils the other one is Taylor 1948 which is deduced based on the Poiseuille's law which is given by V is equal to C DE square gamma W by mu into U cube by 1 plus E into I. Now if you look into the Kozny-Karman equation the V which is nothing but defined as discharge velocity and the CS is defined as a shape factor for granular soils typically CS will be equal to 2.5 SS is nothing but the surface area per unit volume of solids. So the surface area of the unit volume of the soils suppose if you see here it is in the denominator and the factor T which is nothing but the tortuosity factor which is defined as ratio of the tortuous length that is nothing but a path taken by the water flowing through the soil along the voids that means that this particular length which is indicated here the wavy pattern is nothing but the tortuous path the length L is nothing but the length of the sample through which the flow is occurring. So the tortuosity is nothing but ratio of the tortuous length that is L1 to L. So for granular soils the tortuosity factor is 1.414. So here one parameter which is defined which is called as absolute permeability or intrinsic permeability which is found to be constant for typical soil skeleton. So same value will be there for a particular soil and particular state. So the permeability coefficient of permeability is now connected with capital K and gamma W by mu. So capital K is equal to capital K is nothing but now 1 by CS and SS T square into E cube by 1 plus E. So the units for absolute permeability or intrinsic permeability are generally given in Darcy's or centimeter square or meter square. So the units of absolute permeability which is also called as intrinsic permeability and which is found to be in a function of the soil skeleton and it possesses the same value for a particular soil and one Darcy is equal to 0.987 into 10 to the power of minus 8 centimeter square. So here we have said that coefficient of permeability is a function of the number of parameters like one is specific surface area and tortuosity factor and void ratio and the shape factor for basically for granular soils. So the factors affecting the permeability can be summarized like this. So based on the previous discussions by equations proposed by Kozny-Karman or Taylor 1948 shape and size of soil particles that is shape of the soil particle whether it is a angular or whether it is having a plate shaped particle and the size of the soil particles that means that larger the soil particle or smaller the soil particle and void ratio. So K increases with increase in the void ratio and degree of saturation also like K increases with increase in the degree of saturation. For partially saturated soils the permeability will be less because of the partial saturation. The composition of soil particles it also depends upon the mineral type of the mineral present in the soil. So soil structure and viscosity of the permeability and density and concentration of the permeability. So list of the factors affecting the permeability are the shape and size of the soil particles, void ratio, degree of saturation, composition of soil particles, soil structure and viscosity of the permeant and density and concentration of the permeant and also like the compact effort you know with the different compact efforts and different molding water contents there will be change in the permeability. So here the factors affecting the permeability what we discussed is that effect of void ratio and basically here on the left hand side the permeability which is permeability that is permeability factors like different e cube by 1 plus e, e square by 1 plus e and e square are given in the y axis and the permeability in mm per second which is actually given on the x axis. So for granular soils k is proportional to approximately e cube by 1 plus e into degree of saturation cube. So this is approximately valid for this relationship is approximate for s that is degree of saturation less than 100%. On the right hand side there is a plot which is actually shown for void ratio on the y axis and logarithmic of the clay logarithmic of k on the x axis. So it can be seen here this is for a saturated natural clay soils, saturated natural clay soils. So this factor Ck which is nothing but the permeability change factor and k0 is the permeability at void ratio e0, k0 is the permeability at initial void ratio e0. So once the pressure is applied or when the load is applied the void ratio decreases in the process what will happen the permeability changes. So the permeability change factor is defined as dE by d log k that is the difference between k2 and k1 permeability at different void ratios. So this is approximately 1 by 3 to 1 by 2 e0 for a which is approximated as Ck as 1 by 2 to 1 by 3 times the initial void ratio. So k by knowing this permeability change factor we can determine permeability in a different stages and this is k is equal to k0 into 10 raise to e – e0 by Ck where e is the void ratio at a particular time and e0 is the initial void ratio and Ck is the permeability change factor which is approximated as 1 by 3 to 1 by 2 times the e0 and k0 is the initial permeability at initial void ratio. So in this plot the relationship between the permeability change index or permeability change factor Ck and e0 for all clays tested are shown and this is after Tevan's et al 1983. So here it can be seen that the permeability change index Ck is approximated as 1 by 2 to 1 by 2 to 1 by 3 e0 and there are upper bound and lower bound values which are actually shown this is based on the clays for all the types of clays tested and reported by Tevan's et al 1983. Now the next factor is that effect of grain size the permeability of the grain size depend mainly on the cross sectional area of the pore channels. So we knew that when you have what very the pore size which is nothing but the d is proportional to the effective particle size let us say. So in that case we can approximate d is equal to ed 10 the pore size is approximated as 20% of the d 10 that means that the smaller the particle size the finer is the pore channel. Since the average diameter of the pores in a soil at a given porosity increase in proportion to the average grain size the permeability of the granular soils might be expected to increase as a square of the some characteristic grain size. So the permeability of granular soils might be expected to increase as a square of the some characteristic grain size generally it is considered as d 10 but in the recent studies indicate that d 5 that is a soil which actually has got 5% particles passing. So the traditional the empirical formula for estimating the permeability was given by Hazen and the Hazen empirical formula for predicting k basically valid for clean sands is actually given here and which is actually valid for soil which is having less than 5% fines. So k is equal to cd 10 square so what has been done is that number of sandy type of soils having less than 5% fines were taken and the constant head permeability test were conducted and the correlation actually has been plotted and which actually indicates that the k in meter per second can be obtained with constant c which is having a value of 10 to the power of minus 2 and d 10 that is effective particle size in millimeter. Once we have this c is equal to 10 to the power of minus 2 and d 10 in millimeters the permeability can be obtained in meter per second. So by knowing the effective particle size at the first hand in order to estimate the permeability this particular relationship can be used. So c basically is a constant in this case it is equal to 10 to the power of minus 2 which includes the effect of the shape of the pore channels in the direction of the flow and the total volume of pores. So c is a constant which includes the shape of the pore channels in the direction of the flow and total volume of pores. So d 10 is selected because the smaller particles control the size of the pore channels. So in this case the Hazen actually has considered the d 10 because the smaller particles control the size of the pore channels. This particular correlation is presented by Kenny et al 1984 and this is with d 5 that is on the x axis which is represented on the log scale and the hydraulic conductivity k which is actually represented on the y axis and all sands which are actually having relative density 80% placed at 80% relative density more or less greater than 80% relative density was considered and particle size of the sands which are actually used the soil which is used in the various specimens it ranges from 0.04 to 25.4 mm and the coefficient of uniformity is in the range of 1.04 to 12 and it is said that the k the absolute permeability in millimeter square is given as 0.05 to 1 into d 5 square where d 5 is in millimeters and this particular relationship was proposed by Kenny et al 1984 and the effective grain size d 5 would be better choice compared to d 10 according to the data correlated and presented by Kenny et al 1984 the effective grain size was d 5 was reported as a better choice compared to d 10 and the factors affecting the permeability further once you discuss the effect of the degree of saturation. So we have said that with increase in degree of saturation the permeability increases. So k is actually proportional to degree of saturation at low saturation there will be reduction in the flow channels available for the flow because the part of the voids is actually occupied by the air so with increase in saturation degree of saturation the coefficient of permeability of the soil increases. So here a measured data which is presented by of thus 1987 where degree of saturation is plotted on the x axis and permeability is plotted on the y axis for a typical sand where it can show that with increase in the degree of saturation there is an increase in the permeability. Further there is important aspect which is required is that the soil fabric or soil structure or the arrangement of the soil particles within the given soil mass. The permeability of the soil deposit is significantly affected by its in place soil structure is loose granular soil would have higher void ratio than a dense soil and therefore would permit greater flow. So loose granular soil would have higher void ratio than a denser soil so and then there would be a so it would permit greater flow. Similarly when you have what a fine grained soil with flocculated structure will have higher permeability than the dispersed structure. So if you look into the two types of extreme to soil one is coarse grained soil where it can have a loose granular structure or same coarse grained soil can have a dense granular structure. A loose granular structure can have higher permeability than the denser granular structure. Similarly a fine grained soil with a flocculent structure or flocculent arrangement will have higher permeability than the dispersed structure. So here a fine grained soil with a flocculent structure will have higher permeability than the soil with a dispersed structure that is what we said. Even at similar void ratios a clay with undistributed flocculated structure will possess larger void openings than the same clay having a dispersed structure. So the path which is actually this is with a dispersed structure where if you can see and the permeability in this direction is found to be less and when you have got flow which is actually taking place in this direction because of certain available higher hydraulic gradient the permeability will be on the and higher side in this direction along the you know the flow which is actually taking place along the platelet particles. In the flow through clusters of the particles in clay soils so flow mainly controlled by the voids between the flocs and the floc size is a function of the particle size shape and environment in which these flocs have been formed and the for marine elliptic place the permeability which is actually kh and kv will be equivalent to 1 to 1.5 and the question of permeability of a soil with flocculated structure will be isotropic in nature in the sense that the flow the number of flow channels available to flow in any direction will be equal or identical for a flocculent structure whereas in case of a dispersed structure the flow along the shape of the dispersed the parallel laid particles will be higher compared to the perpendicular direction because of the increased tortuosity for the flow. So for the marine elliptic place the kh that is coefficient of permeability in the horizontal direction and the coefficient of permeability in the vertical direction the ratio can be equal to 1 to 1.5 depending upon the type of the normant in which they got deposited. Similarly when you have got the compact effort with an increasing compact effort the permeability decreases. So for example here on the y axis there is a gamma D which is plotted here which is also shown here and the water content on the x axis. So as we have seen for a typical clay initially this particular portion is the optimum moisture content and this side is the wet side of optimum and this side is the dry side of optimum. So at this point the density is actually maximum so lower void ratio will be there and here the density is less and the higher void ratio is possible and so as the water content is increased you can notice that the soil fabric changes from more or less from the flocculent structure to a dispersed structure. So the particles undergo in the process of the compaction the particles undergo rotation by about 90 degrees. In the sense that what will happen is that the particles finally once they reach the wet side of optimum they start getting arranged parallel to each other. So in the process what we can say is that the wet side of optimum the predominant soil structure in case of when you compact the soils is dispersed in nature the same soil with higher lower water content but same density can actually have a flocculent structure. At the maximum dry unit weight and optimum water content the soil structure is neither flocculent nor dispersed it actually has got the blend of both flocculent and dispersed structures. So here with the increase in the compact effort this is the lower compact effort and this is the standard proctor compact effort say and this is the modified proctor compact effect and with increase in compact effort there will be a decrease in the permeability because with increase in compact effort there is an increase in the density and then decrease in the void ratio that means that the permeability decreases. So here variation of the K with water content and gamma D is actually discussed here the importance of the fabric is brought out here so in this particular slide which is actually shown here this is the compaction curve and this is the 100% saturation line or zero air voids line and this is the line of optimums that is with increase in compact effort the compaction curves the maximum peaks will be occurring here and on the plot below what you see is the logarithmic K versus water content. So what we notice that initially the permeability will be high and once it reaches to the optimum optimum water content the permeability takes a dip and that is decreases and further there is an increase in the permeability but towards the wet side of optimum you can see here up to certain extent here there is a possibility that the permeability is actually decreasing towards the wet side of the optimum. So the dry side of optimum the dense aggregates with larger voids will be there because of this also we discussed that the flocculent fabric or flocculated structure will be there because of that the high permeability is resulted into when we consider the wet side of optimum the more uniform distribution of particles with small voids hence the low permeability can result and especially this is attributed to the dispersed fabric which is prevalent on the wet side of optimum. At same gamma D that is the dry unit weight you can notice that the permeability of the wet clay is actually less than the permeability of the dry clay. So that is the reason why for certain type of applications it is advised to compact the clay especially for constructing clay barriers it is advised to compact the clay towards the wet side of optimum because the permeability will be less because there it is not the strength of the soil which is important the soil which is actually having the target permeability is important. So here which is actually given like again minimum k for given compact effort for example here the same plot which is actually shown here but would like to draw your attention to the two curves which is actually shown here this is the curve A and this is the curve B. So curve A if you notice here k minimum is actually occurring at Wm is equal to at the Wopt that is the optimum water content that is at maximum gamma D minimum void ratio. So the k minimum is actually occurring at the Wm is equal to Wopt in case of curve B that is here which is actually shown here curve B which is here where k minimum is actually occurring at Wm greater than Wopt. So the fabric is actually more important than decreasing. So this indicates that the particular arrangement is actually more important than the decrease in the gamma D. So in the field always use Wm greater than or equal to Wopt to get low permeability especially for clay barriers that is what we actually have discussed in the previous slide also. So in this particular slide variation of k with water content gamma D a real test data is actually reported by Daniel and Benson 1980 is presented here on the vertical axis what you see is the permeability is given in k in centimeter per second and the molding water content is actually given on the x axis and this is basically a silty clay with liquid limit 37% and plastic limit 23% hence the plasticity index is about 14%. So these are the three types of compactive efforts are actually represented here are considered one is the low compaction low proctor compaction that means that in this case the energy compactive energy is less compared to the standard proctor medium is nothing but the standard proctor and high effort is nothing but the modified proctor. So you can see that the effect of the compactive effort on the optimum where you can see that different with the increase in the compactive effort there is a decrease in the optimum water content and the second issue is that the typical distinct variation of the permeability with molding water content. So with an increasing molding water content there is a decrease in the permeability and we see that beyond optimum for all the different all the types of compactive efforts irrespect of the compactive efforts you can see that it increase which actually happens beyond the optimum. So this is for high effort we can see that the permeability decreases and here you can see that so beyond the optimum content but when we come to the wet side of optimum as we discussed in the previous slide we have to note that the type of the arrangement the particular arrangement plays a key role than the density which is achieved. Another connecting to our discussion in affecting the permeabilities effect of soil type the volume of the water that can flow through a soil mass is related more to the size of the void openings than the number of the total number of voids. So if you note down the k coefficient of permeability of the coarser end soil is always greater than k of the fine grain soils even though if you look into the void ratios are frequently greater for the fine grain soils see fine grain soils can actually have very high void ratios. So if you say that k increases with the increase in void ratio which this argument is not really true when it comes to this particular factor. So the k of the coarse grain soils is greater than k of fine grain soils in fact the k of the coarser end of permeability of the sandy soil is about 1 million times than that of the 1 million times of the clay soil. So the k of coarse grain soils is a function of the particle size, gradation and particle shape roughness and void ratio of the medium. If you consider the coefficient of permeability of the coarse grain soils when you distort the factors it is function of the particle size, gradation, particle shape, roughness and the void ratio of the soil medium. And k of the fine grain soils which is a function of the type of the clay mineral and adsorbed ions and where the particle surface forces actually predominate. So when you have this one so if you look into this and if you consider the application of fluid mechanics to that then we will be able to understand why the k of coarse grain soils is greater than k of fine grain soils. So in this particular slide what is actually shown is a typical flow which is actually happens through a coarse grain soil having you know let us assume that if you have got a grain here and if you have got a grain here and because of the presence of roughness the velocity with which the water is actually flowing through the voids is actually decreases in the sense that along the boundary walls the because of the frictional drag the velocity drops to 0 and but at the mid distance from the that is d by 2 if d is the diameter of the pore at d by 2 from the edge of the wall it can be seen that the velocity is actually maximum here. So the typical velocity distribution if you assume by using the flow water through two parallel plates and two parallel plates are actually considered as the edges of the you know the soil particles and the flow through the pore channel in a sandy soil is represented like this. Then we actually consider you know clay soil we actually have the adsorbed water that is the adsorbed water which is actually there and then there is a possibility that because of the flow which is actually taking place this adsorbed water layer and then because of the frictional effect the velocity here also drops down to 0 but at the center there will be maximum but when we consider the magnitude of this and magnitude of this particular V max in the sandy soil V max in the clay soil there is a marginal difference will be there similarly when you have got say depressed double layer with decreased adsorbed layer then there can be possible that you know more water flow can take place but however if you look into this velocity distribution though it is analogous but this is actually several magnitudes less than the flow to the pore channel in sandy soil. So the phenomena of the higher permeability of the coarse-grained soil can be explained using the concept of the water flow through the a conduit so because of this the particular coarse-grained soil will actually have higher permeability but when it comes to fine-grained soil we actually have said that one is that adsorbed you know when the water is actually flowing through the adsorbed water layer there is a decrease in the velocity distribution moreover it is nothing but the type of the mineral which is actually present for example a for a fine-grained soils when void spaces are very small all lines of flow are physically close to the wall of the conduit and therefore only low velocity occurs in place basically the flow is already occurs in small channels as further amped because of the some of the water voids is held or adsorbed to the clay particles reducing the flow area further and restricting the flow. So because of this particular explanation with the whatever we have discussed so far we can say that the clay of the coefficient of the permeability of the clay soil is much less than the coefficient of permeability of sandy soil. Now further one of the other factors which we have discussed is that effect of the permeate like if you have got the permeate which is actually given as you know k is proportional to unit weight of the permeate and the viscosity of the permeate variation of the gamma w that is the unit weight of the water the permeate with temperature is negligible but variation of mu with the temperature is not negligible so higher the you know dynamic viscosity of the permeate the lower will be the permeability. So with increase in you know viscosity of the pore fluid the permeability of the soil can be decreased so variation of the mu with the temperature is not negligible but if you are able to increase let us say that the pore fluid is actually replaced with another pore fluid having higher mu the coefficient of the permeability can be brought down. And further we also discussed that from the Kozeny-Karman equation effect of the specific surface area. So here if indicates that higher the specific surface area lower will be the permeability that means that higher will be the specific surface area means for example when you take kaolinite, illite and montamelite the illite mineral the montamelite minerals have actually very high specific surface area compared to the kaolinite mineral particles so that means that with an increase in the specific surface area the permeability of the soil decreases and also exhibits the this is attributed to the more adsorption. The classification of the soils according to their coefficient of permeability if you look into it it can be given as degree of so the soil can be classified based on the different values of the permeability when you say that permeability value in meter per second if it is greater than 10 to the power of minus 3 we say that the soil actually has got high permeability and when the permeability is in the range of 10 to the power of minus 3 to 10 to the power of minus 5 meter per second we can say that the soil is actually having medium permeability and in low which is in between 10 to the power of minus 5 to 10 to the power of minus 7 meter per second and very low that is between 10 to the power of minus 7 to 10 to the power of minus 10 meter per second and the soil is said to be classified based on the permeability as practically impervious if the permeability is less than 10 to the power of minus 9 meter per second. So as of now we discussed it for the homogeneous soils but we may not actually get the homogeneous soil deposits frequently. So the effect of the you know coefficient of permeability of the saturated soils or the stratified soils, so in this particular case a layer 1, layer 2, layer 3, layer 4 the water can actually flow through parallel to the layers or water can actually flow up to downwards that is in the vertical direction. That means that in a given soil when you have got a status, so the water can flow in horizontal direction as well as vertical direction or upward direction because of some artesian conditions. So in that case how to determine the equivalent permeability which we require to understand. In general the natural soil deposits are stratified and if the stratification is continues the effective coefficient of permeabilities in the horizontal vertical directions can be readily calculated. So in this particular discussion if you simplify by using our for determining the this particular condition of flow occurring parallel to the layers. That means that if you have got say h1, h2, h3 to hn, n number of layers in a bedded horizontally the flow in the horizontal direction that is parallel to the layers when it is actually happening. Let us assume that when we have the left hand side limb and the difference in head between these two is say hl which is nothing but the head loss between this point and this point and so the input is nothing but the water which is q. So these soils can actually have permeabilities k1, k2, k3, k4 to so on to kn. So the equivalent permeability in the horizontal direction is that k equivalent in the horizontal direction are kh and the total thickness of the soil layer is nothing but h1 plus h2 plus h3 so on to hn. So here the condition is that qn is equal to q out with that what will happen is that the flow gets divided into depending upon the permeability of the soil which is apportioned as q is equal to q1 plus q2 plus q3 so on to qn. So the condition here is that the head loss is actually occurs over you know length of the sample l and the discharge q is equal to q1 plus q2 plus q3 so on to qn and then q which is actually comes out. So with the base on that discussion for the flow in the horizontal direction parallel to the layers for horizontal flow the head drop hl or the same flow path length l will be the same for each layer. So because as the head loss which actually occurs over a length of the sample l though it is actually having the different type of the soil layers and the hydraulic gradient which actually gets dissipated in layer 1, layer 2, layer 3, I1 is equal to I2 is equal to I3 is equal to In. So the flow rate through a layered block of soil of breadth B, B is the unit perpendicular to the plane of the figure which we considered. So with that we can say the KIA which is nothing but KH that is the equivalent permeability in the horizontal direction, I which is nothing but the hydraulic gradient and H is the thickness of the soil strata and B is the breadth perpendicular to the plane of the figure which we considered. So KIA which is nothing but KHIBH similarly for layer 1, layer 2, layer 3 if you write layer 1 we can write it as K1 I1 BH1 similarly for layer 2 K2 IBH2. So when I computing the flow in the horizontal direction as Q is equal to Q1 plus Q2 plus Q3 so on to Qn I can write now by simplification KH is equal to K1 H1 plus K2 H2 plus K3 H2 so on to KN Hn divided by H1 plus H2 plus H3 so on to Hn. So this is summation which is given as KH is equal to M is equal to 1 to N KHM into HM divided by HM M is equal to 1 to N. So KH is nothing but the equivalent coefficient of permeability in the horizontal direction. So for determining the equivalent permeability when you have got stratified layers when the flow is actually occurring through the parallel to the layers we can determine the permeability, equivalent permeability has KH is equal to K1 H1 plus K2 H2 plus K3 H3 so on to KN Hn divided by H1 plus H2 plus H3 so on to Hn. Similarly when you consider the stratified soils and permeability particularly the flow in the vertical direction that means that when the flow actually happens perpendicular to the layers. So in this case here because of the higher head here the water actually takes water flows upwards like this but we have got different layers of thicknesses like K1 having thickness of H1, layer having H2 having permeability K2, layer having thickness H3 and having permeability K3 so on to layer having thickness Hn to having permeability with KN. So but here what is actually happening is that the it is the head which is you know gets apportioned is your I is equal to you know I1 plus I2 plus I3 but what actually happens is that the Q which is actually entering in the soil strata stratified soil and coming out to be equal that is Q is equal to Q1 is equal to Q3 is equal to Qn. With this condition now we can write down for a flow in the vertical direction that is perpendicular to the layers for a vertical flow the flow rate Q through area A of each layer is same. So the head drop across a series of layers that is we can say that the head drop which is nothing but the head loss in layer 1 and then the total head loss is equal to head loss in layer 1 plus head loss in layer 2 so on to head loss in the layer n. So we can write now with I is equal to H by that is delta H by L where L is nothing but the thickness of the layer when you put into that I can write as HL as IH that is in terms of H is the total thickness of the stratified layers and I1 is the hydraulic gradient occurred in the layer 1 and H1 is the thickness of the layer 1 I2 is the hydraulic gradient occurred in the layer 2 and H2 is the thickness of the layer 2 similarly I3H3 and so on to INHR. By substituting V is equal to Ki that is nothing but I is equal to V by K so in the case on the left hand side we can write as V by KV into H is equal to V by K1 into H1 plus V by K2 into H2 plus V by K3 into H3 so on to V by KN into Hn. So this particular expression when you further simplify for the flow in vertical direction that is perpendicular to layers we can write it as KV is equal to H1 plus H2 plus H3 so on to Hn divided by H1 by K1 plus H2 by K2 plus H3 by K3 so on to Hn by KN this is Hn by KN. So the K vertical permeability is nothing but M is equal to 1 to N Hm divided by M is equal to 1 summation to N to Hm by K vertical permeability of the particular layers. So the equivalent permeability of equivalent coefficient of permeability in the vertical direction so this can be used for both vertical flows flow occurring perpendicular to the soil status in the vertical direction the KV can be given as H1 plus H2 plus H3 so on to Hn to H1 by K1 plus H2 by K2 plus H3 by K3 so on to Hn by KN. So the main points about the stratified soils we should understand is that in general for stratified soils what we have seen is that KH is not equivalent to KV. So when we say that the horizontal permeability is not equivalent to vertical permeability then we say that the soil is anaestropic in nature. In case where a soil deposits permeabilities are not the same in all direction then we say that the properties are anaestropic if the properties of the same in all the directions then it is called isotropic for example when you are actually constructing an embankment with a material obtained from a borough area and when you are achieving the identical permeability because of the compaction then we can say that the permeability is isotropic in nature but particularly when we are actually constructing earthen dams with different types of soils or when we are considering flow occurring in soil status then they are generally anaestropic in nature. So for stratified soils we always we say that KH is always greater than KV the reason which is actually attributed to if you look into it if you consider from the you know coefficient of earth pressure at rest if you look into if you recall that one and which is nothing but K is equal to KH by K is equal to sigma H by sigma B when K is equal to say 0.5 sigma H is equal to 0.5 times sigma V that means that for some stratified soils KH is actually less than KV sigma H is less than sigma V and for that KH will be greater than KV. So more voids or more spaces are available in the horizontal plane under the consideration that means that is with which the flow can takes place in along the horizontal direction is relatively higher compared to the vertical direction and because of this particular the number of voids which are actually available for the water to flow through in the horizontal direction is there higher compared to you know in the vertical direction and because predominantly because of you know low horizontal stresses but however in case of some over consolidated soils where the locking of the stresses takes place this particular you know deliberation is not valid. So for stratified soils basically normally consolidated in nature where sigma H is less than sigma V and the permeability is mostly that is KH is actually greater than KV. So this is an example problem based on the study flow parallel to the soil layers here in this particular problem there is an impermeable layer at the bottom most and then that top surface of the impermeable layer is actually given as the datum or considered as a datum and coarse sand which actually has got permeability of 2 into 10 to the power of minus 4 meter per second and it is having a thickness of 3 meters above that there is a 4 meter medium sand and 6 meter coarse sand K is equal to 10 to the power of minus 4 meter per second and medium sand actually has got K is equal to 0.5 10 to the power of minus 4 meter per second. Now at point A that is the height above this thing is about 10 meters and the total length is about 100 meters and the head loss is actually occurring from say A to B so we need to determine the equivalent permeability so the solution is actually as follows and we also assume that there is an impermeable layer at the top so the flow actually takes place parallel to the layers which are actually shown three layers which are the coarse sand layer and medium sand layer and coarse sand layer below. The solution for this problem works out like this total head at A which is nothing but the 13 meters plus 10 meters that means that here from depending upon the location the thickness is that 6 plus 4 this is 13 meters so the total head at A which is given as 13 plus 10 23 meters and the total head at B that is pressure head plus elevation head which actually works out to be 17.5 plus 1.5 that is at B this is above 1.5 meters so because of that so this is 3 plus 4 7 plus 3 10 plus 3 13 plus 10 so 23 is the head here and total head at B is about 19 meters so the difference of these two which is nothing but the head loss between point A and point B which is actually shown in the figure which is here point A and point B the head loss is actually is about 4 meters or a length of 100 meters between A and B so hydraulic gradient is nothing but 4 by 100 that is nothing but 0.04 using now in determining the equivalent permeability in the horizontal direction when the as the flow is occurring parallel to the layers it can be given as k1h1 plus k2h2 plus k3h3 divided by h1 plus h2 plus h3 so with that we can say that kh is equal to 1.077 into 10 to the power of minus 4 meter per second and once we get this one the total flow can be estimated as which is nothing but khi and capital H which is nothing but the summation of h1 plus h2 plus h3 or summation of the flow which is actually taking place in layer 1 plus layer 2 plus layer 3 that is q1 plus q2 plus q3 so here q is equal to q1 plus q2 plus q3. Now let us consider one more example problem in determining the permeability wherein in this particular arrangement which is actually shown calculate the volume of the water discharged in 20 minutes the cross sectional area of the soil is 400 mm square and this ordinate which is actually here is 225 mm and this horizontal distance is 150 mm and this distance above this where the inflow and the surplus flow is actually discharged takes place the war flow takes place here so this height is 375 mm and above this horizontal line this height is about 150 mm so the war flow that is discharged takes place from this here. So the permeability of the soil which is actually placed here is having 4 mm per second so the solution for this problem works out like this we need to estimate amount of flow which actually takes place in 20 minutes so by converting 20 minutes into seconds 20 into 60 to 100 seconds and area of the cross section which is perpendicular to the flow direction which is nothing but the area which is given as 4000 mm square which is can be converted into 4000 into 10 to the power of minus 6 meter square and the permeability of the soil is in meter per second it is 4 into 10 to the power of minus 3 meter per second. Now here the length of the sample is nothing but the square of the vertical ordinate and horizontal ordinate and with that we can actually get as 0.375 meters so the delta H by L the hydraulic gradient is nothing but by considering 225 plus 375 minus 150 we will be able to get this as delta H divided by 375 that is the length of the soil sample so delta H by L is equal to 1.2 by using Q is equal to K into delta H by L into T so we need to estimate the amount of water which actually flows over 20 minutes duration so that is given as A which is nothing but 4000 into 10 to the power of minus 6 meter square and the permeability which is actually given as 4 into 10 to the power of minus 3 meter per second and hydraulic gradient is 1.2 so with that it works out to be 23.04 into 10 to the power of minus 3 meter cube which is nothing but about 23.04 liters. So in this particular lecture on the permeability and the seepage part 3 we discussed about the factors influencing the permeability and we actually have solved some couple of problems. In the next lecture we will try to discuss about the flow seepage theories and then some relevant discussions pertaining to seepage theory.