 Welcome to module 2 lecture 13. The previous class we introduced ourselves to permeability and seepage and we discussed about the quick sand condition and then we said that the quick sand condition is not a type of sand it is a condition which actually happens when the water flows vertically upward in soil in some certain situations. So this lecture is titled as permeability and seepage 2. So what we discussed in the previous lecture is about the quick sand condition in the final leg of the lecture we discussed about quick sand condition and we said that the quick sand is not a type of sand but a flow condition occurring within a cohesionless soil where its effective stress is reduced to zero due to upward flow of water. Quick sand condition is not a type of sand but a flow condition occurring within equation of soil when its effective stress is reduced to zero due to upward flow of water. So quick sand condition occurs in nature when water is being forced upward under pressurized conditions. So why does quick condition or boiling occurs mostly in fine sands or sills because of their low binding tendency. Some practical examples also we discussed they are excavations in granular materials behind the coffer dams along the alongside rivers. Every place where artesian pressures exist that is where head of water is greater than the hydrostatic water pressure or when a previous underground structure is continuous and a connectivity of place where head is higher or behind the river embankments that is levees which are actually constructed to protect floods. In reality contrary to the popular belief it is not possible to drown in quick sand. The reason is the density of the quick sand which is nothing but a mixture of water and sand particles is much greater than that of water consequently it is literally impossible for a person to be suckered into quick sand and disappear. So a person walking into quick sand would sink to about waist depth than the then float. Let us based on the discussion whatever we have made let us consider this example on the boiling condition. In this problem the sand layer of the soil profile is shown in the figure here which is under artesian conditions that means that we have got a stiff clay layer here and a sand layer and a rock strata here. And this particular sand layer is subjected to some artesian pressure that means that this particular sand layer has actually has got a continuous source of water where it can actually maintain a head of water which is equivalent to 5 meters. That means that when you measure the pressure here the head of the water is actually measured as 5 meters. Now in that situation if you would like to dig a trench here a trench is to be excavated in the clay up to a depth of 4 meters then we need to find out what will be the depth of water hitch to avoid boiling. That means the sand layer of the soil profile is shown and basically it is a stiff clay which is actually having a unit weight of 18 kilo Newton per meter cube and the depth of the trench which actually has been made as 4 meters. Suppose here determine the depth of the water hitch to avoid boiling. So this can be solved like this. We are interested in this point P so at point P the stress at this point is nothing but gamma hw that is whatever the resulting war burden which we are going to put into 6 minus 4 that is 2 into gamma clay that is the silty clay which is nothing but h gamma w plus 2 clay and at P the water pressure is 5 into gamma w. So at P the effective stress is nothing but gamma w h plus 2 gamma clay minus 5 gamma w for sigma dash is equal to 0 for boiling condition to occur here if you apply this one what we get is that h is equal to 1.33 meters. So if you are able to maintain the height of water which is in the fill say at least 1.3 meters minimum then we can actually avoid boiling condition. That means that for example say by mistake the excavation has actually happened and there is a anticipation of some boiling condition. So the one of the immediate remedy is to fill the trench with water to prevent the caving of the excavation. Now we actually have discussed that when the water flows from higher energy higher head to the lower head it exerts an energy on soil particles in the process of flowing from higher head to the lower head. So that these particular forces which are actually exerted on soil particles are called as seepage forces. So water flowing past a soil particle exerts a drag force on the particle in the direction of the flow. For example here if you consider a stem which is actually having a head hitch and this is the water level and if this is the length of the sample then water flows vertically upwards in this direction. If you consider an equilibrium of a soil particle here the weight of the particle is acting downwards, binant force acting upwards the drag forces act in the direction of the flow. The drag forces surrounding the particle they act in the direction of the flow. So this is the direction of the flow which is upward in this case. So the drag force is caused by the pressure gradient and by viscous drag which is actually occurred because of the water solid interaction and also the head pressure gradient which is actually maintained here. Now further if you look into the equilibrium of an element having cross section area A in the direction of the flow the FBD of the soil is that the weight of the soil mass is W and the pressure which is actually exerted by the water is actually acting at the base here. So this is the free body of the soil particles or a particle or a grain which is actually shown here. At critical condition H tends to become HC, at critical condition H tends to become HC. So it is nothing but Gs plus E by 1 plus E gamma W into Al, Gs plus E by 1 plus E into gamma W into Al. So this is nothing but the saturated unit weight acting over area A and length of the sample L that is the weight and based on the pressure here HC plus L into A into gamma W which is nothing but the upward pressure is actually acting at this point. So by simplification what we get is that IC is equal to HC by L is equal to Gs minus 1 by 1 plus E. So Gs minus 1 means Gs is nothing but the specific gravity of the solid particles and E is the void ratio of the soil matrix. So the critical gradient HC is equal to Gs minus 1 by 1 plus E and considering the free body diagram of the grain in the direction of the flow at I is equal to IC. So with this once we if you simplify this we will be able to calculate what is the seepage pressure, seepage force acting in a given volume where the flow is actually occurring and if you are able to take the seepage force per unit volume of the fluid phase where the flow is occurring you will be able to get the seepage pressure. The seepage pressure which is nothing but the seepage force per unit volume the units of the seepage pressure are nothing but kilo Newton per meter cube. So now the free body diagram of the grain once we consider the weight of the dry soil particle is equal to frictional drag force acting on the soil grain in the direction of the flow plus the bionic force. So frictional drag on the acting on the soil particle can be obtained by Wd that weight of the soil particle minus B. Weight of the soil particle can be obtained by dry unit weight of the soil particle that is Gs gamma W by 1 plus E into A into L minus B is nothing but the volume of the solid into gamma W. So this we can write it as Fsc as Gs gamma W by 1 plus E into Al minus V minus Vv by V into V gamma W. So what we have done is that Vs what we have replaced by V minus Vv by V into V gamma W. So factor this frictional force that is nothing but the applied on the soil particles is nothing but Gs gamma W by 1 plus E into Al minus 1 minus N into Al into gamma W. So this is upon simplification what we get is that Fsc that is the seepage force acting or applied on the soil particles is equal to Gs minus 1 by 1 plus E into gamma W into V. So this once we write it as at I is equal to Ice it can be written as Fsc that is the frictional force which is actually applied in the direction of the flow as Ic gamma W where Ic is equal to Hc by L which is nothing but Gs minus 1 by 1 plus E into gamma W into V. V is the volume or which the fluid flow is actually occurring with a higher head to the lower head. So if H is actually say less than Hc where the critical head is not yet reached then the seepage force Js is nothing but I gamma Wv. So when H tends to become Hc that means that when the H tends to become the critical head which actually can make the effective stress equal to 0 in that situation the seepage pressure is given as Ic gamma W. In case the when H is less than Hc when I is less than Ic the seepage pressure is given as Psc is equal to I gamma W. Both the seepage force and seepage pressure act in the direction of the flow. The seepage force can also be applied by this particular deduction for this to happen consider a rectangular element let us assume that this element is actually having dimensions in two dimensions like delta L in this direction and delta L in this direction and assume that this is the direction of the flow because here the pressure head maintained is H1 and here the pressure head is maintained H2 then the assuming that the H1 is greater than H2 the total head the difference is nothing but delta H it is actually happening over L and L that means that the head available here is H1 – H2 that is delta H by the time actually the water flows are erupts out of this particular point it is having a head available that is the head loss which actually happens is almost complete 100 percent that is delta H is equal to 0 here and delta H is equal to full head will be here. So area per unit width of the section is nothing but delta L into 1 if you consider if we consider the unit width perpendicular to the flow direction means that 1 meter here 1 units here then the area over which the flow is actually happening is delta L into 1 and volume affected by the seepage force is nothing but delta L into delta L into 1. So delta L square into 1 is the volume of the fluid phase. Now by writing the expression for force applied to sand particles as a difference of force applied on the left hand side here and the right hand side here from here to here there is a drop which is actually occurring and this is the direction of the flow and this is called as this distance between these two points is called as the flow channel. So the force applied here is nothing but gamma W H1 into delta L into 1 force applied here is gamma W H2 into delta into 1 as H1 is greater than H2 gamma H1 minus gamma H2 into delta L into delta 1. So here there will be a bracket which is here. So with this what we can write is that the force applied to sand particles is nothing but gamma W into delta H into delta L into 1. Now by writing gamma W into delta H by delta L into delta L square 1 we can write delta L square 1 as the volume and delta H by delta L is nothing but the hydraulic gradient. So with that I can write I gamma W V. So this is nothing but the seepage force which is I gamma W V, I is the hydraulic gradient which is nothing but here the definition is that delta H by delta L because delta H is the drop between the points left hand section and right hand section which is shown in the previous slide and gamma W into V, V is nothing but the volume of the fluid phase which is nothing but delta L into delta L into 1. Seepage pressure is can be given by seepage force per unit volume which is nothing but I gamma W. So the critical hydraulic gradient and quick condition, the quick condition occurs at a critical upward gradient Ic when the seepage force just balances the bionic weight of the soil and shear stresses on the sides of the element are neglected. So here the shear stresses on the sides of the elements are neglected. So the critical hydraulic gradient is typically around 1 for many soils fluidized beds in chemical engineering systems rely on the deliberate generation of quick condition to ensure that the chemical process can occur efficiently. So in some cases where fluidized beds in order to generate the chemical processes the quick conditions are deliberately generated so that this actually serves for the purpose. So the critical hydraulic gradient is typically around 1 and in considering these deliberations what we have discussed, what we have neglected is that shear stresses on the sides of the element are neglected. So let us consider another example problem. The problem is described like this here. We need to determine and plot the total stress pore water pressure and effective stress diagram or we need to plot if in case 1 H is equal to 1 meter, case 2 H is equal to 4 meters and H is equal to 2 meters. So this is the H which is actually shown here the difference in water level between the this portion of the stem to the this level at a small a. So this particular portion of the stem actually has got a cross section area a and having soil thickness of 2 meters and this is the C point is at the midpoint mid distance above d and this level is actually considered as a datum. The saturated unit weight of the sand particles is given as 20 kilo Newton per meter cube. So this level is C, D, B and A. So in case 1 the hydraulic gradient is nothing but the length of the soil sample is 2 meters. So I is nothing but H by 2. So here when the H is equal to 1 meter it is 0.5 and H is equal to 4 meters it is 4 by 2 that is 2 and case 3 I is equal to 1. So in one case I is equal to IC another case is I greater than IC another case I less than IC. So in the three conditions based on the three conditions now in the case 1 I is equal to 0.5 in all the three cases which is actually shown here because this H is actually greater than this the water actually flows in vertically upward direction. So the flow occurs in the vertically upward direction. So this is the datum and this D is at the small d is at the this at the datum level and C is at the midpoint and B is at the top surface top of this soil surface and A is the top of the water surface. So at point A that is at the top of the water surface on the right hand side stem the pressure head is 0 and the elevation head is 4 meters because it is 4 meters above the assumed datum. So total head is 4 meters at point B the pressure head is 2 meters elevation head is 2 meters total head is nothing but pressure head plus elevation head which is nothing but the 4 meters. At point D pressure head is 5 meters and elevation head is 0 because it is on the atom. So the total head is 5 meters at point C here what actually happened is that 50% of the head which is actually available as actually already dissipated. So the pressure head is now is 3.5 meters elevation head is nothing but 1 meter because it is 1 meter above the assumed datum so total head available is 4.5 meters. So with this when we plot this diagram with the D, C, B at these points and A at this point the total stress can be given like this the total stress diagram in units of kilo Newton per meter square can be plotted as 20, 40 and 60. So at this point actually it is 20, 40 and 60 and pore water pressure with the whatever the total heads we have derived we can actually say it is 20, 35 and 50 kilo Pascal's so or kilo Pascal's or kilo Newton per meter square. So in this case here it is nothing but the effective stress which is nothing but sigma minus u we can say that which still it is actually having a 10 kilo Pascal's effective stress at the base and 5 kilo Pascal's or 5 kilo Newton per meter square at the mid height and here it is 0. So this is for a case of i is equal to 0.5. In the second case what we discussed is that head is equal to 4 meters which becomes i is equal to 2 with that we can actually say the point A now is again pressure head is 0, elevation head is 4 meters, total head is 4 meters and point B pressure head is 2 meters, elevation head is 2 meters, total head is 4 meters and point D pressure head is 8 meters, elevation head is here it is 0 because it is on the datum and total head is 8 meters and point C where pressure head is 5 meters, elevation head is 1 meter, total head is 6 meters. So here when you plot the diagram based on the head which is actually available we can write the total stress as like this which is same and then here in this case the because of the change in the head conditions here because it is 4 meters it actually gives 20, 50 and 80 kilo Pascal's or kilo Newton meter square. So this indicates that when you take sigma minus u this indicates that all the pressures are actually negative that means that quick sand condition would have already occurred because the head is so high the quick sand, quick sand condition would have already occurred. In the third case where i is equal to 1 is maintained with that case you know the point at A the pressure head is 0, elevation head is 4 meters and total head is 4 meters and point B pressure head is 2 meters, elevation head is 2 meters, total head is 4 meters. At point D now because of 2 meters head which is actually maintained h is equal to 2 meters with that i is equal to 1 is maintained for that with elevation head is equal to 0 the total head is 6 meters and point C which is pressure head is 4 meters, elevation head is 1 meter and total head is 5 meters. With this we can actually write for the case 3 as at this point sigma total stress is same but pore water pressure u which is nothing but 20, 40, 60 when we take a subtraction of sigma minus u at all points you can say that it indicates 0 that means that this is when i when head of 2 meters is it is subjected when you neglect all the frictional forces it can be said that this just subjected to a quick sand condition, just subjected to a quick sand condition. Another example problem which is relevant to the field conditions here a large open excavation was made in a stratum of clay with a saturated unit weight of 17.6 kilonewton per meter cube. When the depth of the excavation reached its 1.5 meter the bottom rose up that means the heaving of the bottom is taken place and gradually cracked and flooded from water by mixture of sand and water. So subsequent boring showed that the clay was underlined by a bed of sand with its surface at a depth of 11 meters. So compute the elevation for which the water would have resin from the sand surface into the drill hole before the excavation was started. So we need to calculate elevation for which the water would have resin from the sand surface into a drill hole before the excavation started. So the problem is that 11 meter thick soil underlined by a previous layer and in which the excavation which is actually planned to a depth of 7.5 meters and H is the head which actually may cause might have caused the heaving. So for that to be determined what we need to do is that what we have saw the way we have done in the previous problem we can calculate here at point A the effective stress and this can actually can lead to the failure when the effective stress at point A is equal to 0. So we can actually by knowing the saturated unit weight we can actually calculate 17.6 into 11 minus 7.5 this is which is available and this particular portion of the soil has been excavated minus 10 into H that is nothing but the 10 is the unit weight of the water which is actually taken as 10 kilo Newton per meter cube gamma into H. So with that we can actually get the answer as H is equal to 6 meters that means that when the head actually suppose the sand layer here at this point when it is subjected to head of artesian head of 6 meters then there is a possibility of so that might this particular head might have actually caused the so called heaving of the excavation in that particular problem. Another example problem in this case we have the problem statement is like this determine and plot the total vertical stress pore water and effective vertical stress distribution at levels A B and C. So the soil strata which is actually given like this it is 6 meter is silty clay and below that there is a core sand. So A is top of the surface and water table is actually assumed to be at this surface so the hydrostatic water surface is actually at this surface and the C which is nothing but the joint interface between silty clay and core sand and 3 meter is the thickness of the clay layer of the 3 meter is thickness of the core sand layer core sand layer thickness is which is actually having a 3 meters and the specific gravity of the solids is gs is equal to 2.65 and void ratio is 0.833 and the silty clay which is actually having gs is equal to 2.7 and natural moisture content of 45.2 percent and the head of the water which is actually measured here is assumed that it is 4 meter above this ground water table. So the given problem an artesian pressure exists in the lower sand layer so that is what actually what we have understood the artesian pressure prevalence in the lower problem lower sand layer at A B and C and B is C is located 1 meter above the datum so in this given problem and this particular surface is assumed as datum this is actually assumed as datum and here this particular point is 1 meter above the datum that is C and A is actually 3 meter above the datum that is at this point and then we have here there is a saturated sand layer and here it is a dry soil layer and so here we need to determine sigma u and sigma dash at point A B and C in the previous problem this solution need to be worked out based on the given data here however this problem solution we can actually look into it this problem is different from the problem statement which is given in the previous slide in this this is also subjected to artesian pressure but what exactly happens is that the head actually loss occurs from point when the flow actually takes place from point B to point A. So the total head at A which is nothing but total head at A is nothing but 2 meters of water that is elevation head is 3 meters and pressure head is 2 meters that is 2 meters of that hydrostatic pressure is there and because of that the total head at point A is 3 5 meters and total head at B which is because here it is subjected to this 3 plus 2 5 plus 1 6 plus 1 7 meters. So 7 plus this being the datum what we get is the total head at B is equal to 7 meters. So because of this the delta H which is actually available the head loss here is nothing but 2 meters occurring over the loss the flow is occurring over a length of 3 meters at C is a point which is actually 1 meter from the datum. So at point B that is sigma can be obtained as 18.51 to 1 that is the gamma D into 1 meter thickness plus 19 into 2 that is saturated unit weight of the sand layer below the hydrostatic water table plus 17 into 3. So with that I have got 107.5 kilo Newton per meter square and at point B U is equal to 10 into 7 so 70 kilo Newton per meter square. So sigma dash is nothing but 107.5 minus 70 which is nothing but 37.5 kilo Newton per meter square and at point A we can actually obtain the above the that is at point A is at this point sigma is at this point sigma is nothing but 18.5 into 1 plus 19 into 2 which is equal to 56.5 kilo Newton per meter square and U is equal to 10 into 2 which is 20 kilo Newton per meter square. So sigma dash is equal to 56.5 minus 20 is nothing but 36.5 kilo Newton per meter square. So at point A the effective stress is this much at point B the effective stress is this much almost of the same order and at point C which is in between point A and B where total stress is given as 90.5 and pore water pressure is given as 10 into pressure head at C. So here based on the discussion here the head which is actually available here the full head drop is actually available which can take place is 2 meters and this is the length of the so hydraulic gradient here is nothing but 2 by 3 meters that is 2 by 3.67 is the hydraulic slope of this line. So here at 50% of the clay thickness that is 1.5 meters we will have the head of only 1 meters but here because of this head which is actually available is 1.33 meters. So we can actually write the pore water pressure is nothing but 10 into 5.33 which is nothing but 53.33 kilo Newton per meter square. The effective stress is nothing but total stress minus pore water pressure which is nothing but 38.2 kilo Newton per meter square. So the total head at C is nothing but total head at B minus i into z which we can write it as 7 minus 2 by 3 is nothing but the hydraulic gradient that is nothing but a 2 meter head loss over a length of the soil layer as 3 meters that is 2 by 3 into z is equal to 1. So with that 6.33 meters the pressure head at C can be obtained like this total head at C is nothing but 6.33 minus elevation head is 1 meter. So because of that the pressure head at C what we consider is 5.33 is explained here. So with this for the given problem what we discussed in the previous slide the total stress diagram which is actually shown like 56.5 the units are in kilo Newton per meter square the units are in kilo Newton per meter square 56.5 kilo Newton per meter square 90.5 and not 7.5 and the water pressure is nothing but 20 and at this point it is 53.3 and 70 and here the effective stress is nothing but 36.5 38.2 and 37.5 is having a constant effective stress of about 37.5 average effective stress of about 38 kilo Pascal's. So the measurement of soil permeability is so we have said that permeability is a property of the soil where different types of soils exhibit different values and this is actually used for different conditions different applications in civil engineering construction and it is very it will be interesting to know how this permeability is actually measured. The permeability can be measured from either from the indirect methods or it can be measured from the laboratory test or through the field test. So the rate of flow of water q that is volume over time t through a cross section area is found to be proportional to the hydraulic gradient I according to Darcy's law. So which is nothing but v is equal to q by a is equal to ki or q is nothing but the ki a. So discharge over a cross section area which is nothing but q is equal to ki a where i is nothing but the hydraulic gradient over a length l that is nothing I is nothing but head loss over a length l hydraulic gradient is equal to head loss over a length l where v is the flow velocity and k is the coefficient of permeability with the dimensions velocity over length over time. See the coefficient of permeability of a soil is a measure of the conductance that is the reciprocal of the resistance that is provided that provides the flow of water through its pores. So it is a property of the soil which actually determines which actually tells the ease with which the water can flow through the soils. As that has been mentioned in the in the previous discussion the value of the value of the coefficient of permeability k depends upon the average size of the pores and is actually related to the particle sizes and their packing. So we actually have said that the pore diameter can be approximated as void ratio times the effective particle size. So if you take effective particle size of a sand and effective particle size of a clay and clay is actually having very low effective particle size with that we can say that the pore diameters of the clay is very fine compared to a sand. So it is approximated that 20% of effective particle size is regarded as pore diameter. So the value of the coefficient of permeability in a prima facie depends on average size of the pores and is related to the particle sizes and their packing and particle shape particularly whether it is angular or whether it is rounded or sub rounded the shape and the soil structure and that is nothing but the arrangement of the soil particles whether it is a flocculated or whether it is a dispersed structure or whether it is bulk structure. So it depends upon the whether it is a loose bulk packing or dense packing in case of a bulk particles. The ratio of the permeabilities of typical sands and gravels to that of the clays is to those of the clays is of the order of 1 million times. That means that the ratio of the permeability of the typical sands and gravels to those of typical clays is generally of the order of 10 to the power of 6. A small proportion of the fine material in coarse sand soil can lead to a significant reduction in the permeability. That means that if you are able to add a small proportion of fine grained soil to the coarse grained soil and that can influence the permeability. The number of tests can be used to measure the estimate of the permeability in the prima in primarily in the laboratory the two types of tests which are actually there called the constant head test used basically for high permeable soils and falling head test used basically for relatively impermeable soils. So the constant head test is actually basically used for high permeable soils and falling head test is used basically for relatively impermeable soils. As it was mentioned earlier there are indirect methods some correlations which are actually available based on the grain size distributions that is based on the D5, D10 and other particle sizes basically for sandy soils some correlations are actually available by knowing the effective particle size will be able to estimate the coefficient of permeability based on the grain size distribution. Also from the oidometer test or a consolidometer test when the soil is actually subjected to consolidation in a consolidometer test indirectly we can actually compute the permeability based on coefficient of consolidation and coefficient of volume compressibility and the unit weight of the water. And these are the possible laboratory methods in the field there are methods which are actually possible are called pumping test and borehole test which are popular known as the Packer's test which are actually conducted with single packer or double packer systems. So in determining the permeability of coarse grained soils particularly the large quantities of flow occurs in short periods of time and a small quantity of flow occur over long period of time for fine grained soils. So in case of fine grained soils very small quantity of the flow occurs over a long period of time. In case of a permeability of the coarse grained soils what will happen is that in a short span of time a large quantity of flow takes place. So two aspects that need to be careful attention for all types of soils are that to ensure that flow occurs only through the soil not at the interface between the soil and the mould in which the soil is contained. So in the laboratory while determining particularly the leakage from the edges required to be arrested that is that to ensure we have to ensure that the flow occurs to the soil not at the interface between the soil and the mould in which the soil is contained and the soil sample is fully saturated before recording observations. So one is that the either coarse grained soil or fine grained soil the soil sample need to be completely saturated. So for determining permeability of a soil sample by using concentrated permeability test. So in the concentrated test the setup which is actually shown here has got a sample of length L and having some head of water which is actually maintained constant and so here the head of the water which actually maintained is say H over a length L. So the hydraulic gradient is nothing but the H by L. So any additional amount of water which actually flows through this one is actually collected as a discharge here. So in a given time over the cross sectional area of the sample perpendicular to this direction of the flow. So the direction of the flow is in this direction. So that is nothing but the A. So the discharge which is actually connected in a time t is nothing but Q is equal to A into V into t and V is nothing but the we can write it as K into H by L into t where K is the coefficient of permeability which is required to be determined. So by rearrangement of the terms here by measuring the discharge in time t and over a length of the sample A and A is the cross sectional area and H is the head which is actually maintained with that we can actually determine coefficient of permeability of the soil through by using constant test. In the constant test one need to establish the steady state conditions. After establishing the ensuring the steady state conditions 3 to 4 readings need to be evolved. The average of this readings can be reported as the average permeability. And main features of the constant test include it is suitable for soils having coefficient of permeability in the range of 10 to power of minus 2 to 10 to power of minus 5 meter per second which applies to clean sand and sand gravel mixers with less than 10 percent fines. And it can be suitable for soils when usually in their completely disturbed or remolded states such as for drained materials and filters to confirm that their performance will be adequate. So it can be suitable for soils when usually in their completely disturbed remolded states such as for drained materials and filters to confirm that their performance will be adequate. In this particular slide where the setup for the falling head test or variable head test is actually shown and here because at the amount of water which is actually flowing through the soil or entering into the soil is less. So a stem which is actually having a very small cross section area is actually provided on top and like the constant head test in a given amount of time the head drop from H1 to H2 is actually measured. So here the cross section area of the stand pipe is say A and the cross section area of the sample is say A. So this is actually called as a rigid wall perimeter. At time t naught where head is H1, at time t1 the head is actually H2. So let H be the head of the water at any time t and let in time t that dt the head drop by amount dH. So we can actually write the quantity of water flowing to the sample in time dt from the Darcy's law can be obtained as we can write it as dQ is equal to K into I into A into dt. So which we can write it as K into H by L into A into dt. So the quantity of the discharge can also be expressed as because in a small cross section of pipe when there is a drop of water from H1 to H2 when H1 is greater than H2 what will happen is that there is the dQ is equal to minus A into dH we can write. As the time increases the head decreases. So we can actually write minus A into H1 to H2 into dH with that we can actually equate these two discharges and integrating from t is equal to t naught to t1 we can by simplifying that we can write it as K is equal to 2.303 Al by At log to the base 10 H1 by H2. So this is actually expression which is required to be used for a falling head test. The main features of the constant head and falling head test basically include in the constant head test the permeability is computed on the basis of fluid that passes through the soil sample while in the falling head test K is computed on the basis of the fluid flowing into the sample. So in the case of fine grained soil the what we in fact is measured is water entering into the specimen. In case of a constant head test is that the permeability is actually computed on the basis that flow which is actually occurring through the sample. So the main distinct difference between a constant head test and falling head test is nothing but in the constant head test the water flows through the soil in the falling head test water flows into the soil. Here in the case of a falling head test the type of the soil which is actually being discussed or being determined for permeability is fine grained soil which is actually highly influenced by the you know which is actually having the effect of the soil structure and particularly the mineralogy which actually plays a bigger role. So in the constant head test the time required to accumulate the fluid volume necessary to perform computation and basically extreme care would be required to prevent leaks in the apparatus and evaporation of the discharge water and with the falling head test the duration of the test is actually shortened and care is required to prevent evaporation of water in the inlet tube. Another problem which is actually which we have given discussed in the slide which is shown here the problem is again shown here. So this problem can be treated as an assignment problem where here the soil strata which is actually having 6 meters and the silty clay which is actually having Gs is equal to 2.7 and water content is given as 4.45.2 percent and coarse sand which is nothing but Gs is equal to 2.65 and E is equal to 0.833 and this particular layer actually has got an artesian condition. So with that it is the problem statement is actually like determine and plot the total vertical stress pore water pressure and effective vertical stress distribution at levels A and C and B is at a point at midpoint of this that is actually at this point that is 6 meter is the thickness of the clay layer. So this silty clay layer so B is at a mid distance from the this particular point. So in this particular lecture what we try to discuss is that we have introduced the you know seepage force how it can be measured as how it can be calculated one is we said that seepage force J is equal to I gamma Wv and the seepage pressure is nothing but seepage force per unit volume which is nothing but I gamma W which is the units of the seepage pressure are kilo Newton per meter cube and which is actually resulted in the this see both seepage force and seepage pressure occur in the direction of the flow then we have introduced ourselves to a couple of practical problems where when there is a soil status subjected to artesian conditions how we can actually compute these total stresses pore water pressures and effective stresses because of the resulting upward flow because of the artesian conditions and then we have actually discussed the methods for measuring permeability of the soil particularly we have actually have got two methods one is constant head permeability test and following head permeability test these are actually widely used for in the laboratory in the following head permeability test there are also one like two distinct apparatus which are actually used one is called rigid wall perimeters other one is flexible parameters in case of rigid wall perimeters the stresses cannot be applied on to the soil sample but however in case of flexible perimeters the confinement stresses can be applied and then the permeability test can be done by using the similar concepts of following head permeability test. So with this the effect of the confinement stresses on the the permeability of the soil being tested can be obtained and the main distinct difference between a constant head test and the following head test what we discussed is that in the case of a constant head test the water actually flows through the soil and in case of a following head test the water actually enters into the soil and however what is actually measured is that the small amount of change which actually occurs and we also said that the for a typical sand and clay the distinct difference is actually about you know the permeability of the sand is 10 to the power of 6 times more than the a particular typical clay that means that distinctly different to permeability is the reason and the factors affecting the permeability of soils and the some field testing methods along with the some testing data we will be discussing in the subsequent lectures.