Permeability 1A. ‘TRODUCTION Permeability is defined as the property of a porous material which permits the passage or seepage of water (or other fluids) through its interconnecting voids. A material having continuous voids is called permeable. Gravels are highly permeable while stiff clay is the least permeable, and hence such a clay may be termed impermeable for all practical purposes. The flow of water through soils may either be a laminar flow or a turbulent flow In laminar flow, each fluid particle travels along a definite path which never crosses the path of any other particle. In turbulent flow, the paths are irregular and twisting, crossing and recrossing at random (Taylor, 1948). In most of the practical flow problems in soil mechanics, the flow is laminar. ‘The study of seepage of water through soil is important for the following enginecring problems 1. Determination of rate of settlement of a saturated compressible soil layer. 2. Calculation of seepage through the body of earth dams, and stability of slopes 3. Calculation of uplift pressure under hydraulic structures and their safety against piping. 4. Ground water flow towards wells and drainage of soil. In this chapter, we shall discuss the methods of determining the permeability of soils, and the study of various factors affecting it. The ground water flow towards the wells has been discussed in chapter 8, while the unconfined and confined flow problems of earth structures have been discussed in chapters 9 and 10 respectively. 7.2. DARCY'S LAW ‘The law of flow of water through soil was first studied by Darcy (1856) who demonstrated experimentally that for laminer flow conditions in a saturated soil, the rate of flow or the discharge per unit time is proportional to the hydraulic gradient. q=kia (71) am178 SOIL. MECHANICS AND FOUNDATIONS. FIG. 7.1, FLOW OF WATER THROUGH SOIL, or (7.2) where g=discharge per unit time A= total cross-sectional area of soil mass, perpendicular to the direction of flow i= hydraulic gradient k= Darcy's coefficient of permeability v= velocity of flow, or average discharge velocity. If a soil sample of length £ and cross-sectional area A, is subjected to differential hy hy head of water, h;-/; , the hydraulic gradient i will be equal to and, we have (7.1 a) From Eq. 7.2, when hydraulic gradient is unity, k is equal to v. Thus, the coefficient of permeability, or simply permeability, is defined as the average velocity of flow that will occur through the total cross-sectional area of soil under unit hydraulic gradient. The dimensions of the coefficient of permeability x are the same as those of velocity. It is usually expressed as cm/sec or m/day or feet/day. Table 7.1 gives some typical values of co-efficient of permeability of various soils. q= 17.1 TYPICAL VALUES OF & Soil Type T Co-fficien of permeability “Chan gravel 1.0 and greater ‘Clean sand (coarse) 10-1 x 107 ‘Sand (mixture) 1x 107-5 x 107 Fine sand Sx 107-1107 Sity sand | 2010 Silt | $x 1041 10% chy [te 10°* and _smatter |PERMEABILITY 19 7.3. DISCHARGE VELOCITY AND SEEPAGE VELOCITY ‘The Darcy's law (Eqs. 7.1, 7.2) in no way describes the state of flow within individual pores. Darcy’s law represents the statistical macroscopic equivalent of the Navier-Stokes equations of motion for the viscous flow of ground water. The velocity of flow v is the rate of discharge of water per unit of total cross-sectional area 4 of ‘soil, ‘This total area of cross-section is composed of the area of solids 4, and area of voids A,. Since the flow takes through the voids, the actual or true velocity of flow will be more than the discharge velocity. This actual velocity is called the seepage velocity v,, and is defined as the rate of discharge of percolating water per unit cross-sectional area of voids perpendicular to the direction of flow. From the definitions of the discharge velocity and seepage velocity, we have qQ=VA=VsAy w= we (7.3) An ‘The seepage velocity vis also proportional to the hydraulic gradient : ii (where ky = coefficient of percolation) wlT.4) From Darcy's law, or £ 247.5) n 7.4. VALIDITY OF DARCY'S LAW Darcy's law of linear dependency between velocity of flow v and hydraulic gradient i is valid only for laminer flow conditions in the soil. From the experiments on flow through pipes, Reynolds found that the flow is laminar so long as the velocity of flow is less than a lower critical velocity ve expressed in terms of Reynolds number as follows ved Pe 2000 (7.6) n where ve = lower critical velocity in the pipe (cm/sec) ; d= diameter of pipe (cm) Pw=density of water (g/ml) ; n= viscosity of water (g sec/cm’) and g=acceleration due to gravity (em/sec’). Based on this analogy, the flow through soils may be assumed to depend upon the dimensions of the pore spaces. In coarse grained soils, where the pore dimensions are larger, there will be greater possibility of flow becoming turbulent. Francher, Lewis and Barnes (1933) demonstrated experimentally that flow through sands remain laminer and the Darcy's law valid so long as the Reynolds number, expressed in the form below, is equal to or less than unity : ¥Da dw < ns 1 (7.7)180 SOIL MECHANICS AND FOUNDATIONS. wher v= velocity of flow (cm/sec) ; De =diameter of average particle (cm) Substituting the numerical values of physical constants and other parameters as follows: n=1x 10% g - sec/em* ; g = 981 cm/sec? Da=Dw=D (uniform soil) ; ist 00 Diy we get D ~=0.05 em =0.5 mm The above criterion gives conservative results. Scheidegger’s (1957) collected data show that critical Reynold’s number may vary from 0.1 to 75, for Darcy's law to be valid. Such a wide variation is partly due to the different interpretations to the characteristic diameter used in the equation for Reynold’s number. Allen Hazen (1892) concluded that linear dependency of the velocity and hydraulic gradient existed if the effective size of soil did not exceed 3 mm. Based on Piefke’s experimental results, Prinz concluded that the law of proportionality between velocity and hydraulic gradient deviates faster in very fine sand than in coaser soils. The critical velocities of flow for fine sands and fine gravels were found to be 0.9710 cm/sec and 5.56x10 cm/sec respectively. For the ground water flow occurring in nature and normally encountered in soil engineering problems, the Darcy’s law is generally within its validity limits. 7.5. POISEUILLE’S LAW OF FLOW THROUGH CAPILLARY TUBE* The relationship governing the laminer flow of water through capillary tube is known as Poiseuille’s law. In the laminar flow, each particle of fluid flows in a direct line. At the surface of the tube, which will be wet with water, the molecular attraction holds a very thin layer of water immobile, so that the velocity at the surface of the tube is zero. The velocity increases to a maximum at the centre of the tube. The variations in velocity from point to the point are accompanied by small friction losses. FIG. 7.2. FLOW THROUGH CAPILLARY TUBE.PERMEABILITY st Fig. 7.2 (a) shows a capillary tube, of length L and radius R. The velocity distribution is shown in Fig. 7.2 (©). At any radial distance r from the centre, the velocity is v and the velocity gradient (i.e., space rate of change of velocity) is-% ‘The unit shear at the top and bottom of the cylinder of water (Fig. 7.2 d) of radius r is given by dv "ar where 1 is the co-efficient of viscosity, the approximate value of which may be taken about 10° gram sec per sq. cm. If the tube is subjected to a head of water h, at one end, and f, (h,>h,) at the other end, flow will take place and various forces acting on the cylinder of water at any radius r will be as shown in Fig. 7.2 d. Since the tube is in equilibrium, the sum of all forces acting on the cylinder must be zero. Hence hy Yw (R) = Ia Yw (HP) = tm rL)=0 or t (2a rL)=(h- ha) xP Yw (7.8) 2arl{ -n$)=(h-h) a? ae or dve Be Replacing f)-/; by A (net head causing flow) and intergrating, we get ay P+e At r=R, v=0 ge cn fe Hence is ®-?) (7.9) This is the law of variation of velocity. ‘The quantity of water flowing in the thin cylindrical sheet dr thick, is given by - tte ope dq = Qnrdr) v= FR =P) 2nrdr *. Total quantity of water flowing in the capillary tube, per unit time, is _ te ee _ le Rt angen fe Py rar = eR Replacing /' =i = hydraulic gradient enh, 7.19) an If a is the area of the tube, average velocity is given by G_ 4 _wR Vay iit i (TAL a) ane Bn ‘ wi Vay = TA or ae (7.11) Eq. 7.11 is the Poiseuille’s law in which the velocity of flow during laminar flow varies as the first power of the hydraulic gradient.182 SOIL. MECHANICS AND FOUNDATIONS Effect of shape of the capillary tube. Eqs. 7.10 and 7.11 are valid for circular capillary tube only. In engineering hydraulics, the velocity is generally designated in terms of the hydraulic radius Ry , which is defined as the ratio of area to the wetted perimeter. . aR OR _ For circular tbe Ru=7—R=5 or R=2Ry Substituting this in Eg. 7.10, we get adie Re gene 5 ia (7.12) Similarly, the rate of flow between two flat, closely spaced, parallel plates, can be found to be given by : a= wie ia (7.13) Hence, we can conclude that for the capillary tube of any other given geometrical shape of cross-section, the flow may be expressed as wRe =6 @ a ia (7.14) where C,= shape constant. Fig. 7.3 shows the shape of a irregular capillary voids. Let the total cross-sectional area of the soil mass be A, and n be the porosity FIG. 7.3. TUBE OF IRREGULAR SHAPE. a= area of flow passage = nA 2 (« eRe al e715) Hydraulic mean radius in soil pores. Kozeny’s (1927) concept of hydraulic mean radius in soils leads to the following expression A_AL Ru= If V, is the volume of solids in a soil mass having voids ratio e, the volume of the flow channel (=AL) will be e¥;. The total surface area of flow channel (= PL) is equal to the total surface area A, of the soil grains. +4716) Let D, = diameter of the spherical grain which has the same ratio of volume to surface area as holds collectively for the grains in a given soil ve A: xD? 6 Substituting this in Eq. 7.15, Ru=Se ef a)PERMEABILITY 183 (7.17) tm where C is a new shape constant, 7.6. FACTORS AFFECTING PERMEABILITY Fig. 7.17 is the Poiseuille’s law adapted for the flow through the soil pores. Comparing it with the Darcy’s law : q=kiA, we get w _é noite ‘Thus, the factors affecting permeability are : 1. Grain size ; 2. Properties of the pore fluid 3. Voids ratio of the soil ; 4. Structural arrangement of the soil particles 5. Entrapped air and foreign-matter ; and 6. Adsorbed water in clayey soils. 1. Effect of size and shape of particles. Permeability varies approximately as the square of the grain size. Since soils consist of many different-sized grains, some specific grain size has to be used for comparison. Allen Hazen (1892), based on his experiments on filter sands of particle size between 0.1 and 3 mm, found that the permeability can be expressed as k=D (7-18) k= CDS (7.19) coefficient of permeability (cm/sec); Dio = effective diameter (cm) constant, approximately equal to 100 when Dio is expressed in centimetre. Auempt have been made to correlate the permeability with specific surface of the soil particles. One such relationship is given by Kozeny (1907) where xt +-(7.20) Ken Se 1-w where k= coefficient of permeability (cm/sec per unit hydraulic gradient) porosity S,= specific surface of particles (em’/cm’) 1 = viscosity (g-sec / cm’) Kx = constant, equal to 5 for spherical particles On the basis of his experiments, Loudon (1952-53) developed the following empirical formula: logo (k $7) =a + bn (7.21) where @, 5 are constants, the values of which are 1.365 and 5.15 respectively for permeability at 10° C. As demonstrated by Loudon’s experiments, the permeability of coarse grained soils is inversely proportional to the specific surface, at a given porosity 2. Effect of properties of pore fluid. Eq. 7.18 indicates that the permeability is directly proportional to the unit weight of water and inversely proportional to its viscosity, Though the unit weight of water does not change much with the change in temperature,184 SOIL. MECHANICS AND FOUNDATIONS there is great variation in viscosity with temperature. Hence, when other factors remain constant, the effect of the property of water on the values of permeability can be expressed ky bom It is usual to convert the permeability results to a standard temperature (27°C) for comparison purposes, by expression : n ak AT.22 ko =k ¢ ) However, if change in the unit weight of water due to temperature is also taken imto account, we have the more general equation. fm Ym m2 Pe kT Yn” Pr where kay = permeability at 27° C ; tn = viscosity at 27° C k= permeability at test temperature ; 1 = viscosity at test temperature. Muskat (1937) pointed out that a more general coefficient of permeability, called the physical permeability ky is related to the Darcy's coefficient of permeability k as follows: k, z (7.22 0) In any soil, k, has the same value for all fluids and all temperatures as long as the voids ratio and the structure of the soil skeleton are not changed. 3. Effect of voids ratio. Eq. 7.18 indicates that the effect of voids ratio on the values of permeability can be expressed as k [ Ge] | Ge kl tte | lite, Laboratory experiments have shown that the factor C changes very litle with the change in the voids ratio of un-stratified sand samples. However, for clays, it varies appreciably. ‘Thus, for coarse graind soils, Eq. 7.24 reduces to > > gk 1K a a ee +723 a) Ite, Ter Hey ep Based on another con- cept of mean hydraulic radius |.) ¥/0/ 3 for the soils, the following |W % relationship is obtained : as (7.22, a) (7.22) +(7.23) (7.23 b) —rk Void Ratio (0) —> It has been found that a semi-logarithmic plot of | voids ratio versus permeabil- + Void Fao Function SS Permeabiity ity is approximately a straight (dog Seale) (log Scale) line for both coarse grained FIG. 7.4. VARIATION OF & WITH .PERMEABILITY 18s as well as fine grained soils (Fig. 7.4) 4. Effect of structural arrangement of particles and stratification. The structural arrangement of the particle may vary, at the same voids ratio, depending upon the method of deposition or compacting the soil mass. The structure may be entirely different for a disturbed sample as compared to an undisturbed sample which may possess stratification. ‘The effect of structural disturbance on permeability is much pronounced in fine-grained soils. Stratified soil masses have marked variations in their permeabilities in direction parallel and perpendicular to stratification, the permeability parallel to the stratification being always greater ( $7.13). When flow through natural soil deposits is under consideration, permeability should be determined on undisturbed soil as its natural structural arrangement. 5. Effect of degree of saturation and other foreign matter. The permeability is greatly reduced if air is entrapped in the voids thus reducing its degree of saturation. The dissolved air in the pore fluid (water) may get liberated, thus changing the permeability. Ideal condition of test are when air-free distilled water is used and the soil is completely saturated by vacuum saturation, for measuring the permeability. However, since the percolating water in the field may have some gas content, it may appear more realistic to use the actual field water for testing in the laboratory. Organic foreign matter also has the tendency to move towards crictical flow channels and choke them up, thus decreasing the permeability 6. Effect of adsorbed water. The adsorbed water surrounding the fine-soil particles is not free to move, and reduces the effective pore space available for the passage of water. According to a crude approximation after Casagrande, 0.1 may be taken as the voids ratio occupied by adsorbed water, and the permeability may be roughly assumed to be proportional to the square of the net voids ratio of (e-0.1) . 7.7 COEFFICENT OF ABSOLUTE PERMEABILITY In the previous article, we have discussed various factors on which the coefficient ‘of permeability (k) depends. Thus, the coefficient of permeability (k) depends not only on the properties of the soil mass (such as size, shape, specific surface, structural arrangement, stratification, voids ratio etc.) but also on the properties of the permeant (i.e. water) which flows through it. Let us now introduce a coefficient which does not depend upon the properties of permeant. Such a coefficient, known as coefficient of absolute permeability (K) is definied by the expression x=i(2) (7.24 a) Yw Substituting the value of nn from Eq. 7.18, we get rec( 2, ]o (7.24) lt+e The above equation indicates that the coefficient of absolute permeability is independent of the properties of permeant (i.e. water) and it depends solely on the properties of soil mass, Dimensions of K : From Eq. 7.24 (a)PERMEABILITY 18s as well as fine grained soils (Fig. 7.4) 4. Effect of structural arrangement of particles and stratification. The structural arrangement of the particle may vary, at the same voids ratio, depending upon the method of deposition or compacting the soil mass. The structure may be entirely different for a disturbed sample as compared to an undisturbed sample which may possess stratification. ‘The effect of structural disturbance on permeability is much pronounced in fine-grained soils. Stratified soil masses have marked variations in their permeabilities in direction parallel and perpendicular to stratification, the permeability parallel to the stratification being always greater ( $7.13). When flow through natural soil deposits is under consideration, permeability should be determined on undisturbed soil as its natural structural arrangement. 5. Effect of degree of saturation and other foreign matter. The permeability is greatly reduced if air is entrapped in the voids thus reducing its degree of saturation. The dissolved air in the pore fluid (water) may get liberated, thus changing the permeability. Ideal condition of test are when air-free distilled water is used and the soil is completely saturated by vacuum saturation, for measuring the permeability. However, since the percolating water in the field may have some gas content, it may appear more realistic to use the actual field water for testing in the laboratory. Organic foreign matter also has the tendency to move towards crictical flow channels and choke them up, thus decreasing the permeability 6. Effect of adsorbed water. The adsorbed water surrounding the fine-soil particles is not free to move, and reduces the effective pore space available for the passage of water. According to a crude approximation after Casagrande, 0.1 may be taken as the voids ratio occupied by adsorbed water, and the permeability may be roughly assumed to be proportional to the square of the net voids ratio of (e-0.1) . 7.7 COEFFICENT OF ABSOLUTE PERMEABILITY In the previous article, we have discussed various factors on which the coefficient ‘of permeability (k) depends. Thus, the coefficient of permeability (k) depends not only on the properties of the soil mass (such as size, shape, specific surface, structural arrangement, stratification, voids ratio etc.) but also on the properties of the permeant (i.e. water) which flows through it. Let us now introduce a coefficient which does not depend upon the properties of permeant. Such a coefficient, known as coefficient of absolute permeability (K) is definied by the expression x=i(2) (7.24 a) Yw Substituting the value of nn from Eq. 7.18, we get rec( 2, ]o (7.24) lt+e The above equation indicates that the coefficient of absolute permeability is independent of the properties of permeant (i.e. water) and it depends solely on the properties of soil mass, Dimensions of K : From Eq. 7.24 (a)186 ‘SOI, MECHANICS AND FOUNDATIONS n-(e] 318) Hence K has the dimensions of area. The units of K are : mm?,cm?,m* or darcy. 1 darcy = 0.987 x 10° em’, It is imeresting to note that’ for a given voids ratio and structural arrangement of particles, the coefficient of absolute permeability (K) is constant, irrespective of type/properties of fluid. 7.8 DETERMINATION OF COEFFICIENT OF PERMEABILITY The coefficient of permeability can be determined. by the following methods (@) Laboratory methods (1) Constant head permeability test. (2) Falling head permeability test. (b) Field methods (1) Pumping-out tests. (2) Pumping-in tests. (c) Indirect methods (1) Computation from grain size or specific surface. (2) Horizontal capillarity test (3) Consolidation test data. Permeability can be determined in the laboratory by direct measurement with the help of permeameters, by allowing the water to flow through soil sample either under constant head ot under variable head. Permeability can also be determined directly by field test, described in chapter 8 (well hydraulics). The indirect method of computing the permeability from consolidation test data has been explained in chapter 15. Empirical formulae for determination of coefficient of permeability (k) Permeability can also be computed from several empirical formulae given below. 1. Jaky’s formula Jaky (1944) found that a fair estimate of the order of magnitude of can be obtained for ail soils from the formula k= 100 Dy -(7.25 a) where D,, denotes grain size (in cm) that occurs with the greatest frequency. 2. Allen Hazen’s formula k=CDip «(7.25 b) where C is a constant, which is taken approximately equal to 100 when Dy is expressed in cm. 3. Terzaghi’s formula ‘Terzaghi (1955) developed the following formula for fairly uniform sands, which reflects the effect of grain size and void ratioPERMEABILITY 187 =200 Die +(1.25. 0) where D, = effective grain size (ive. the diameter of the sphere for which the ratio of its volume to its surface area is the same as the similar ratio for a given assemblage of soil particles. 4. Kozney’s formula d) Ken Si 1-0 specific surface of particles ( cm’ cm’) where Ss 1 = viscosity ( g - sec /em*) Kx= constant, equal to 5 for spherical particles 5. Louden’s formula Logi, (k S where @, b are constants, the value of which are 1.365 and 5.15 respectively for permeability at 10°C, Soil type, numerical values of k and methods of its determination Table 7.2 gives the soil types, numerical values of k and methods of its determination, atbn TABLE 7.2 0” 0 T Practically Sightly pervious impervious Fino anc, sandy si Clean gravel and sand —" oo | erminaon by constant head permearneter Determinati Determination by consolidation test Fat clay 1 8610 10 t0* Cotfciont of permeability k, cvs188 SOIL MECHANICS AND FOUNDATIONS 7.9. CONSTANT HEAD PERMEABILITY TEST Fig. 7.5 shows the diagrammatical representation of constant head test. Water flows from the overhead tank consisting of three tubes: the inlet tube, the overflow tube and the outlet tube. The constant hydraulic gradient i causing the flow is the head h (i.e. difference in the water levels of the overhead and bottom tanks) divided by the length L of the sample. If the length of the sample is large, the head lost over a length of specimen is measured by inserting piezometric tubes, as shown in Fig. 7.5 (b). | L3—+ overtiow [parvave ‘Measuring Jar @) CO) FIG. 7.5. CONSTANT HEAD TEST. If Q is the total quantity of flow in a time interval t, we have from Darcy's law, q= Seki 21 Ob 1 ret Obt A726) where A = total cross-sectional area of sample. When steady state of flow is reached, the total quantity of water Q in time f collected in a measuring jar. The observations are recorded as shown in Table 7.3 [See Experiement 14].PERMEABILITY 189 7.10. FALLING HEAD PERMEABILITY TEST ‘The constant head permeability test is used for coarse-grained soil only where a reasonable discharge can be collected in a given time. However, the falling head test is used for relatively less permeable soils where the discharge is small. Fig. 7.6 shows the diagrammatical representation of a falling head test arrangement A stand pipe of known cross-sectional area a is fitted over the permeameter, and water is allowed to run down, The water level in the stand pipe constantly falls as water flows. Observations are started after steady state of flow has reached. The head at any time instant ¢ is equal to the differece in the water level in the stand pipe and the bottom tank, Let fy and fh, be heads at time intervals ¢, and f, (42> 4) respectively. Let A be the head at any intermediate time interval 1, and—dh be the change in the head in a smaller time imerval dt (minus sign has been used since h decreases as 1 increases). Hence, from Darcy’s law, the rate of flow q is given by ‘Y Funnel dha) _ q = hiA where i= hydraulic gradient at time txt Ak a at Integrating between two time limits, we get AK "dh aL h hn Ak (1, — t) = loge or aL (ty = ti) = loge la Denoting % - 1 + We get B= 2.3 & tops Mt 27.27) FIG. 7.6. FALLING HEAD TEST The laboratory observations consist of measurement of the heads A, and A, at two chosen time intervals 1, and f . The observations are recorded as shown in Table 7.4. 7.11. THE JODHPUR PERMEAMETER The Jodhpur Permeameter was designed and developed by Dr. Alam Singh (1958) at the Soil Engineering Laboratory of M.B.M.Engineering College, Jodhpur. The apparatus is meant for studying the permeability characteristics of all types of soil samples under different conditions of laboratory as well as in the field. Both falling head and constant head test can be performed on remoulded as well as undisturbed specimens. Remoulded specimens can be preparded either by static or by dynamic compaction method. The Jodhpur permeameter (Model III) comprises the following :190 SOIL MECHANICS AND FOUNDATIONS i. Permeameter mould 2. Top cap fitted with water inlet nozzle and air release valve 3. Dynamic compaction base plate 4, Perforated base plate 5. Perforated top plate 6 1 8 Static compaction flanged end-plugs, 2 Nos., 3 cm and 2.5 cm high Compaction collars, 2 Nos. 3 em and 2.5 cm high Split collar 9. 2.5 kg Dynamic Ramming Tool (DRT) 10. Rod temper I Bottom tank 12, Constant head tank fitted with air intake tube 13, Spare brass tube and rubber stopper for vacuum saturation 14, Set of three stand pipes, fixed to the back of constant head tank 15. Core cutter 16. Dolly for core cutter 17, Centering ring for cutter 18. Wire gauge, pad of filter paper, copper, wool pad, bolts, nuts, spanner. funnel, pinch cocks and flexible tubing. The permeameter mould which is a cylinder of internal capacity 300 ml, 50 cm? cross-sectional area (79.8 mm diameter) and 6 cm effective height, has two studs fixed to the side lugs which aid in assembling the mould, the top cap and the perforated base plate (or the collars and the compacting base plate as the case may be). A rubber gasket (washer) under the top cap ensures water tightness. The permeameter assembly is placed in the bottom tank having a water outlet which permits accurate contol for water level for falling heads tests. The bolt head under the perforated base plate keeps the permeameter mould assembly raised a little in the bottom tank thus allowing free flow of water through the base perforations, To test undisturbed specimen, the 0.3 litre core cutter (50 cm’ in cross-section * 6 cm high) with dolly attached to top is pushed into the undisturbed block of soil. The soil surrounding the outside of the cutter dolly is cut, and the cutter and dolly full of undisturbed soil is removed. The soil is cut flush with top and bottom ends of the cutter, afier removing the dolly. The centering ring for the cutter is placed over the perforated base plate and the core with the undisturbed specimen is placed centrally over the perforated ase plate with the cutting edge downwards. The top is then tightened over the cutter The remoulded specimen can be prepared either by static compaction, or by. dynamic compaction, at any desired density. The weight of the wet soil, to compact it at a givey density and water content, is first calculated. To compact it by static compaction, the 3 cm collar is attached to the bottom end of the 0.3 litre mould and 2.5 cm collar to the top end. The split collar is placed around the 2.5 cm flanged end plug. The: mould assembly is supported over the 2.5 cm end plug with the 2.5 cm collar resting jon the split collar. The calculated weight of the wet soil is put into the mould and the top plug is inserted, The entire assembly is kept in a press and the sample is compacted. AfterPERMEABILITY wi ‘compaction, the 3 cm plug and 3 cm collar are removed and, after putting the fine mesh gauge etc., perforated base plate is fixed over it. The mould is turned upside down, the plug and the collar are removed, and the top perforated plate and top cap are fixed To compact the specimen dynamically, using the rod temper, the mould is fixed upside down on the dynamic compaction plate, and the collar is fixed to its other end. The wet soil of pre-calculated quantity is then compacted into the mould by means of the rod temper, in two or three layers. After compaction, the collar is removed and, after placing the fine mesh gauge, the perforated base plate is fixed, The mould assembly is then turned upside down, the compaction base plate is detached, and the top cap is fixed. Alternatively, if permeability at Proctor's maximum dry density and at a moulding water content equal to the optimum value is required, first the maximum dry density and optimum water content is determined. (This can be done by Jodhpur Mini compactor test, Proctor test or by compaction in permeameter mould itself). The soil is then compacted at the optimum water content in two layers in the 0.3 litre permeameter mould (mould assembly as described in the above para) with 15 blows of 2.5 kg DRT given to cach layer. After the compaction, the compaction collar is removed, the excess soil is trimmed off, and the perforated base plate is fixed, as described in the above para. Example 7.1. Calculate the co-efficient of permeability of a soil sample, 6 cm in height and 50 cm* in cross-sectional area, if a quantity of water equal 10 430 mil passed down in 10 minues, under an effective constant head of 40 cm. On oven-drying, the test specimen has mass of 498 g. Taking the specific gravity of soil solids as 2.65, calculate the seepage velocity of water during the test Solution : Given : Q= 430 ml ; 1=10 x 60 =600 seconds 4=30cm? ; L=6 cm ; h=40 cm 1 430 6 «2l 430 6 2.15% 107 From Eq. 7.26, kee oo 3p 7 215% 10" cm/sec = 2.15 x 10° x 864 = 1.86 m/day (Since 1 cm/sec = 864 m/day) 430 2 Now Boersy = 1435 «107 emisee Alternatively, veki =2.15 x 103% 2 = 1.435 x 107 cm/sec Ma _ Gow, _ 2.65 x1 Now a= 506 Pa 1.66 aes +e vy _ 1.435 x 107 0.373 Example 7.2. In a falling head permeameter test, the initial head (1 = 0) is 40 cm The head drops by 5 cm in 10 minutes. Calculate the time required to run the test for192 SOIL. MECHANICS AND FOUNDATIONS the final head 10 be at 20 cm, If the sample is 6 cm is height and 50 cm’ in cross-sectional area, calculate the coefficient of permeability, taking area of stand pipe =0.5 cm’. Solution: In a time interval ¢= 10 minutes, the head drops from initial value of hy =40 to hy =40-5=35 cm aL joy, ts From Eq, 7.26, we have k= 2.3 $ logo 5* 23ab j,i H or 1-25 hopn t= mlogw Ft where m - 2h at = constant for the set up 40 10 10 a 10=mlogo 5 OF m=—y =p pee = 172.5 units loge 33 = hy hy 1 = m login 52 = 172.5 logie = Now, let the time interval required for the head to drop from initial value of h; = 40 cm to a final value of h»=20 cm, be ¢ minutes. t= 172.5 log 2 = 172.5 x 0.301= 51.9 minutes 23 aL ; 23 aL Again, m= O20 = 172.5. units Faas “mminute (Since 1 used 10 compute m was in minutes) EE XOS 6 crujsec = 1.335 x 10* cm/sec. 50 x 172.5 x 60 Alternatively, k= 2.3% opie = Boe Jogn 2 = Example 7.3. Due 10 a rise of temperature, the viscosity and unit weight of the percolating fluid are reduced to 75% and 97% respect-ively. Other things being constant, calculate the percentage change in co-efficient of permeability. Solution : Let k, , Yu, and n, represent the coefficient of permeability, weight and Viscosity at the increased temperature. Dropping the suffix 1 to represent these quantities at the standard (or original) temperature, we have 1.335 « 10° cm/sec beat and k)=4™ where A = constant Aim or heen T km tw. 7 Now Ye, = 0.97 Ww andy) =0.75 _ 0.97) _ by =k 55D = 1.295 k Increase in k= 29.5.%.(To Face Page 1!PERMEABILITY 193 Example 7.4. A constant head permeability test was run on a sand sample 16 cm in length and 60 cm? in cross-sectional area. Porosity was n,=40%. Under a constant head of 30 cm, the discharge was found to be 45 cm’ in 18 seconds. Calculate the coefficient of permeability. Also, determine the discharge velocity and seepage velocity during the test. Estimate the permeability of the sand for a porosity of ny = 35% or Solution 161 2929 10° em/s =2.L From Eq. 7.26, aeOe. oo Discharge velocity, Seepage velocity, Again, from Eq. 7.23 (a) nm “=n Sa 0.35° aoo3t 22 x 10°? Cer 1.26 x 107? cm/s my Ceny a-04y Example 7.5. Permeability tests were performed on a soil sample, under different voids ratio and different temperatures and the following results were obtained. Test No. Voids ratio (e) Temperature °C k(cm/s) 1 0.65 25° 04x10 ¢ 2 1.02 40° 1.9.x 10°* Estimate the coefficient of permeability at a temperature of 20° C for a voids ratio of 0.80. Given the following physical properties of water: At 20° C, n = 10.09 «10°* and p, = 0.998 g/cm’ At 25° Gm = 8.95 «10° g sec/em? and py = 0.997 g/cm* At 40° C, 9 = 6.54 x 10°‘ g sec/em’ and py =0.992 g/cm* Solution Step 1: Let us first convert both the test results to a temperature of 20° C kM Pw From Eq. 7.22, ht Pm ome: a kT” Pwr For first test. (kim = 0.4 x 10-4 x 895% 10 = 0.355 x 10°‘ cms 10.09 x 10-* * 0.997194 SOIl. MECHANICS AND FOUNDATIONS 19-1 8:54 x10" 0.998 10.09 x 10°* 0.992 Step 2: Now convert these values for a void ratio of 0.8. Using Eq. 7.23 (a): (_e_) _ os _ i{ e_) _ 065) _ ( 2) — 102 ($2) Soom: Tre), 10.8 "0166: T+e J, 1+ 102” 5 For second test ado = 1.9 x = 1.239 x 10°' cm/s 4 0.284 0.166 -4 0.284 0.535 Taking the average, the probable value is, k = 6.4 x 10° cm/sec 7.12, THE CAPILLARITY-PERMEABILITY TEST The capillarity-permeability test or the horizontal capillarity test is used to determine the coefficient of permeability k ‘as well as the capillary height h. of the soil sample. Fig. 7.7 shows the set up for the test. For the first test, (k,)os = 0.355 x 10 = 0.607 x 10°* cmisec. For the second test (K)os = 1.239 « 10 = 0.670 x 10-* emisec. FIG. 7.7. CAPILLARITY-PERMEABILITY TEST. Dry soil sample is placed in a transparent lucite or glass tube, about 4 cm in diameter and 35 cm long, at a desired density. Water is allowed to flow from one end, under constant head /t and the other end is kept open to atmosphere through air vent tube. At any time interval 1, after the commencement of the test, let the capillary water travel through a distance x, from point A to B. At point A, there is a pressure head h,, while at the point B, there is a pressure deficiency (i.c., a negative head) equal to h, of water. *. Hydraulic head lost in causing the flow from A to B= hp ~(—h,)=hy+ he hy + hie Hydraulic gradient = From Darcy’s law, v=4i or nvs=ki (assuming 100% saturation)PERMEABILITY 19s where vs= seepage velocity, parallel to the direction of rE If the coefficient of permeability is designated as k, at a partial saturation S the above expression may be rewritten as Ath x a & Snve= kui or sn ky x de = 5 (lo + he) dt Integrating between the limits x, and x; for x, with the corresponding values of 1 and 1, we get JP ade = i+ ng fr ae ” Sn =x? _ ky ona Fa ot he) (7.28 a) If S=100 %, the above expression reduces to Pa Tan = Fo + he) (7.28) In the above equation, there are two unknowns : k and , . The first set of observations (upto about first half length of the tube) are taken under a head (/), . As the capillary saturation proceeds, the values of x are recorded at various time intervals 1. A plot of x° with =, =i set of observations (for the next half of the tube) are taken under an increased head and the plotting of observed values of x against ¢ gives the value of the quantity x, 1 gives a straight line, the slope of which gives the value of ~ m, say). The second (=m, say). Knowing the two slopes, the values of k and h, can be found by the simultaneous solution of the following two equation: 2 x) ny The degree of saturation can be found by taking the wet mass of the soil sample at the end of the test. The porosity is computed from the known dry mass, volume and specific gravity. 7.13. PERMEABILITY OF STRATIFIED SOIL DEPOSITS In nature, soil mass may consisit of several layers deposited one above the other ‘Their bedding planes ‘may be horizontal, inclined or vertical. Each layer, assumed to be homogeneous and isotropic, has its own value of coefficient of permeability. The average permeability of the whole deposit will depend upon the direction of flow with relation to the direction of the bedding planes. We shall consider both the cases of flow : (i) parallel to the bedding planes and (ii) perpendicular to the bedding planes. 2 (ig + he) and ( =m = (hg + he)196 SOIL. MECHANICS AND FOUNDATIONS 1, Average permeability parallel to the bedding planes. t Let Z,. ZjsZ_= thickness of + layers and k, ,k, permeabilities of the layers. For flow to be parallel — to the bedding planes, the hydraulic gra- 4 1a dient i will be the same for all the ¥ layers. However, since v= ki and since t a % k is different, the velocity of flow will 2 [we be different in different layers. rs Let k,= average permeability of the eit E soil deposit parallel to the bedding plane. FIG. 7.8. FLOW PARALLEL TO BEDDING PLANE. Total disharge through the soil deposit = Sum of discharge throuh the individual layers “ GQ t gat onda or = kei Z= by iZy+ he iZy + wash i Ze or b= BB tht thm (where Z=Z,+ Zr Zs) ..(7.29) 2, Average permeability perpendicular to the bedding planes. In this case, the velocity of flow, and hence the unit discharge, will be the same through each layer. However, the hydraulic gradient, and hence the head loss through each layer will be different. Denoting the head loss through the layers by hy, fy.......f and the total head loss as h, we have hah +h. But hy =i) Zi she h=i tht 2) Now, if k; = average permeability perpendicualr to the bedding plane, we have Also Substituting these values in (i), we get WZ_ vi, va Wn +B Zz keg (7.30 or iB, (7.30) Bk It can be shown that for any stratified soil mass ke is always greater than k:. For example, consider FIG. 7.9. FLOW PERPENDICULAR a three-layer system, having k, = 2, ks = 1, ks =4 units, ‘TO BEDDING PLANEPERMEABILITY 197 Let 2=4, Z=1 and Z=2 units, Z=4+1+2=7 units 2x4) +(x DF Qx4) 7 ke > ke Example 7.6. A strafed soil deposit consists of four layers of equal thickness. The co-efficient of permeability of the second, third and fourth layers are respectively yrd,4 and twice of the coefficient of permeability of the top layer. Compute the average permeabilities of the deposit, parallel and perpendicular to the direction of the stratification in terms of the permeability of the top layer. Sloution : Let the thickness of the top layer be Z and its permeability be &. Total thickness of deposit =4Z Now Example 7.7. Fig. 7.10 shows an aquifer inclined at 12° 10 the horizontal. Two observation wells, dug upto the aquifer, at a horizontal distance of 80 m show a difference of 6 m in the water levels. Taking coefficient of permeability of aquifer soil as 1.2 mm/sec, determine the discharge through the : ‘aquifer, per unit width. The thickness of aquifer normal to the direction of flow is 3.2 m. Solution : Length of travel (L) be- tween wells A and B A Impervious soil > STH HHA 32m Inctined aquifer SEE a a Now, from Darey law, flow + 0 m —______> per unit width of aquifer is given by FIG. 7.10198 ‘SOIL MECHANICS AND FOUNDATIONS Example 7.8. A capillarity permeability test was performed in two stages. In the first stage, the wetted surface advanced from its initial position of 3 cm to 10 cm in 8 minutes, under a head of 60 cm at the entry of water. In the second stage, the wetted surface advanced from 10 cm to 23 cm in 24 minutes under a head of 230 cm. At the end of test, the degree of saturation was found 10 be 92% and the porosity was 32%. Determine the capillarity head and the coefficient of permeability. Solution For the first stage of test, we have, from Eq. 7.28 (a) 10-3 2ky 8x 60 To Oth or ey (60 + he) = 0.0279 =) Modifying Eq. 7.28 (a) for the second stage of test. We get 237 - 10° 2 kw r 24x 60 ~ 092% 032 230 + hd or ++(2) From (1) and (2), we have a sn From which, we get he=237.2 em _ 0.0279 ite4 Hence from (1), y= GA = 0.939 x 10-* cm/sec 7.14. EXAMPLES FROM COMPETETIVE EXAMINATIONS Example 7.9. Calculate the coefficient of permeability of a soil sample 6 cm is height and SOce? in cross-sectional area, if a quantity of water equal to 450 mi passed down in 10 minutes under an effective constant head of 40 cm. On oven drying, the test specimen weighs 495 g. Taking the. specific gravity of soil solids as 2.65, calculate the seepage velocity of water during the test (Givil Services Exams. 1989) Solution: Given : L=6 cm ; A= $0 cm? ; Q= 450 ml , ¢= 10 minutes ; h = 40 cm This question was set from example 7.1 of this book. All the data are exactly the same except that Q has been changed from 430 ml to 450 mi and mass has been changed from 498 ¢ 10 495 & Eq. 7.26 : = 2.25 x 10°? em/sec= 1.944 m/day Ma__ 495 eT x6 0.606 ; n = —&— = —2 Te 1+0.622 = 3.975 x 10°* em/sec Example 7.10. Jn a capillary permeability test conducted in two stages under a head of 50 cm and 200 cm, in the first stage the wetted surface rose from 20 mm to 80 mm in 6 minutes. In the second stage it rose from 80 mm to 200 mm in 20 minutes. If the degree of saturation is 90% and porosity is 30%, determine the capillary head and the coefficient of permeability. (Civil Services Exam. 1991)PERMEABILITY 19 Solution: s aoe (50+ or @ For second stage, SS = yh (200 + he) or 200 + he) = 2.268 alii) Solving (i) and (i), we get he=170.59em and k=6.12% 107% cm/min = 1.02 10°* emisee. Example 7.11. To determine the capillary head and the permeability of soil, a tube containing @ soit with a void ratio of 0.6 was kept horicomally in a trough filled with water with its centre at @ depth of 75 cm from the water level. Water was found 10 advance from 1.5 cm to 12 cm in 8.5 ‘minutes. In another test with the same soil kept at a depth of 22.5 cm below the water level, water was found to advance from 13 cm to 21 cm in 10 minutes, Determine the capillary head and the coefficient of permeability of the soil. (Civil Services Exam. 1993) Solution ; ¢=0.6 ; hence n=—£—=—98_ 9.375 Ite 1+06 Assume the soil wo be fully saturated. yg For first stage, FHS the) oF KTS + helm 3.1268 “o For the secon stage, TPES + He) oF S$ HRSA “i From () and (i), he =16.27 em and k=0.1315 cavmin =2.19 « 10°? cm/sec Example 7.12. A horizomal stratified soil deposit consists of three uniform layers of thickness 6,4 ‘and 12 m respectively. The permeabilities of these layers are 8 x 10‘ cm/s, 52x 10°“ cm/s and 6 = 10~‘ cmys, find the effective average permeability of the deposit in the horizomal and vestical direction. (Civil Services Exam, 2001) Solution : Refer Example 7.4. From Fa, 729, 6x 8x10 +4 x52 10-*H12x6x 104 rom Fa, 7.29, = 644412 = OES Oa 792 10-tem/s 68+ 472+ 1276 Example 7.13. A capillary permeability test was conducted in two stages under a head of 60 cm and 180 cm respectively at the entry end. In the first stage, the wetted surface moved from 1.5 em 10 7 cm in 7 minutes. In the second stage, it advanced from 7 cm 10 18.5 cm in 24 minutes. The degree of saturation at the end of the test was 85% and the porosity was 35%. Determine the 4.909 « 10-4 cm/s From Eq. 7.30, ke capillary head and the coefficient of permeability. (Engg. Services Exam. 1990) Solution: Refer Eq. 7.28 (a) Pu1S hy First. stage Tass thd200 ‘SOIL MECHANICS AND FOUNDATIONS or ky (60 + ie) = 0.9934 “ 1S-T lke M085 «035 or fy (180 + he) = 1.8175 =) 180 + he 1.8175 OF. 0.9934 Hence from (), ky = 6.87 « 10°? cm/min = 1.148 x 10°* em/see Example 7.14, What will be the ratio of average permeability in horizontal direction 10 that in the vertical direction for a soil deposit consisting of three horizontal layers, if the thickness and permeability of the second layer are twice of those of the first and those of the third layer twice those of second? (Engg. Services Exam. 2002) Solution : Refer Example 7.4. Let ky and 2) stand for first layer. Second stage (180 + he) From (i) and (if) = 1.8296, from which he = 84.65em For second layer = ky=2k and a=2z For third layer : ba2h-4h and g-2a=4q Total zeat2atdaqTa kath) Qa) +4h) 4a) From Fg. 7.29, fy BBP ON COT ON GS) 21, From Eq. 7.30, fy=- 4 eT, ay 2a yd a Rk A 21,308 hora Example 7.15. In a falling head permeameter test on a silty clay sample, the following results were obtained » sample length 12 mm; sample diameter 80 mm; initial head 1200 mm; final head 400 ‘mm: time for fall in head 6 minutes ; stand pipe diameter 4 mm. Find the coefficient of permeability Of the soil in. mmisec. (Gate Exam, 1998) Solution : Refer Example 7.2 Given | La 12 om; A=% = 50.265 en?: a =F (0.47 =0.1257 en? 1=6 min.=360 sec. y= 120 cm; y= 40 cm 0.1257 x 12 = 2.303 % togy 1257 x 12 From 64. 7.27, & = 2.303 logy f= 2.303 SFT logio “20. 9.159 x 10°F em/see Example 7.16. Estimate the flow quantity (in titres per second) through the soil in the pipe shown below. The pressure heads at two lo- cations are shown in Fig.7.11. The internal diameter of the pipe is 1 m ‘and the efficient of permeability of soil is 1x 10™* m/sec. (Gate Exam. 20001) Sotion g=kidwkat ant x 10 215 Ey = 0.03935 10-5 m/see FIG. 7.11PERMEABILITY 201 7.15. LABORATORY EXPERIMENTS EXPERIMENT 14 : DETERMINATION OF PERMEABILITY BY CONSTANT HEAD TEST Object and scope. The object of the experiment is to determine the coefficient of permeability of sei in the laboratory by constant head test using Jodhpur Permeameter, Materials and equipment. () Jodhpur permeameter complete with all accessories, (ii) De-aired water (iy Balance 10 weigh 1 1 g, (iv) 4.75 mim and 2 mm IS sieves, (») Mixing pan or basin, (vi) Stop watch, (vi) Graduated measuring cylinder, (viii) Metre scale, (ix) Beaker, (x) Thermometer, (x) Containers, for water content determination, (xii) Straight edge or trimming knife. Test procedure (a) Preparation of statically compacted remoulded specimen 1. Take 800 10 1000 g of representative specimen of soil and mix water t0 it so that its water content is raised to the optimum water content for the soil determined by Proctor's test. If permeability is to be determined at any other water content, raise the water content of the soil wo the desired value Leave the soil mix for some time in ait-tight container 2. For the given volume (V) of the mould, calculate the mass (M) of the soil mix so as 10 give the desired dry. density (p,), using the following expression M=pg(ltw)V Take the mass of the above soil accuate to | g 3. Assemble permeameter for static compaction, For this, attach the 3 cm collar to the bottom end of the 0.3 litre mould and 2.5 cm collar to its top end. Support the mould assembly over the 2.5 cm end plug with the 2.5 cm collar resting on the split collar Kept around the 2.5 cm end plug. ‘The 0.3 litre mould should be lightly greased from inside. 4. Put the weighed quantity of soil (step 2) into the mould assembly. Insert the top 3 cm end plug into the top collar. The soil may be tamped with hand while being poured into the mould. Keep the entire assembly into a compression machine and remove the split collar. Apply compressive force fon the assembly ll the flanges of both the end plugs touch the corresponding. collars. 5. Maintain the load for about 1 minute and then release it, Remove the tp 3-cm plug and collar. Place a filter paper or fine wire gauge on the top of the specimen and fix the perforated base plate on it. 6. Turn the mould assembly upside down and remove the 2.5 cm end plug and collar. Place the top perforated plate on the top of the soil specimen and fix the top can on to it, after inserting the sealing gasket The specimen is now ready for the permeability test () Preparation of dynamically compacted remoulded specimen 1. Take 800 to 1000 g of representative specimen of soil and raise its water content to the optimum water content. Leave the soil mix in an airtight container for some time, 2. Assemble the permeameter for dynamic compaction. For this, grease the mould lightly from inside and place it upside down on the dynamic compaction base. Find the mass of the assembly accurate to 1 g. Put the 3 cm collar to the other end. 3. Compact the wet soil mix in two layers, with 15 blows of the 2.5 kg dynamic ramming tool, given to each layer. Remove the collar and trim off the excess soil. Find the mass of mould assembly with soil. The difference of the two masses taken in steps (2) and (3) would give the mass (M) of the soil compacted. 4. Place filter paper or fine wire mesh on the top of soil specimen and fix the perforated base plate on to it202 SOIL MECHANICS AND FOUNDATIONS 5. Turn the assembly upside down and remove the compaction plate. Place the top perforated plate fon the top of the soil specimen and fix the top cap on to it, after inserting the sealing gasket. ‘The specimen is now ready for the permeability test. (©) Saturation of compacted specimen To saturate the compacted specimen, place the permeameter mould in the vacuum desiccator and open air release valve. Fill the desiccator with de-aired water till the water level reaches well above the top cap and the water inlet nozzle is submerged. Apply vaccum of about $ to 10 cm of mercury and maintain it for some time. Increase this vacuum slowly in steps, to about 70 cm of mercury. In every increment, sufficient time should be given so that the air bubbles come out without vibrating the specimen. Take out specimen when the saturation is complet. (@) Constant head test 1. Place the mould assembly in the bottom tank and fill the bottom tank with water upto its outlet. 2 Connect the outlet tube of the constant bead tank 1 the inlet nozzle of the permeameter, after removing the air in the flexible rubber tubing connecting the tube. Adjust the hydraulic head by either adjusting the relative heights of the permeameter mould and the constant head tank, or by raising or lowering the air intake tube within the head tank 3. Start the stop watch, and at the same time put a beaker under the outlet of the bottom tank. Run the test for some convenient time interval. Measure the quantity of water collected in the beaker during that time. 4. Repeat the test twice more, under the same head and for the same time interval. Tabulation of observations. Test observations are tabulated as illustrated in Table 7.3, TABLE 7.3 DATA AND OBSERVATION SHEET FOR CONSTANT HEAD PERMEABILITY TEST Sample No. A46 Moulding water content : 145 Specific gravity : 2.68 Dry density : 1.72 g/cm? Voids ratio : 0.56 1. Hydraulic head (cm) 2. Length of the sample (L) (cm) 3. Hydraulic gradient 1 4. Cross-sectional area of sample (om 50 ‘5. Time interval (1) (sec) 00 6. Quanity of ow (Q):() Test (mn) 860 (i) Wes (al) 855, (ii) ML tes al) 862 Average (a) 859 7. Coefficient of permeability (cm/sec) 2.86 107 8, Test temperature eo) 2 9. Permeability at 27°C (em/sec) 2.57 107 Calculations. The coefficient of permeability is calculated from Eq. 7.26. ee2 El toh APERMEABILITY 203 EXPERIMENT IS : DETERMINATION OF PERMEABILITY BY FALLING HEAD TEST Object and scope. The object of the experiment is to determine the coefficient of permeability ‘of soil in the laboratory by falling head test using Jodhpur Permeameter. Materials and equipment. Same as in Experiment 14, Test Procedure 1, Prepare the remoulded soil specimen in the permeameter and saturate it as explained in Experiment 14, 2. Keep the permeameter mould assembly in the bottom tank and fill the bottom tank with water ‘upon its. outlet. 3. Connect the water inlet nozzle of the mould to the stand pipe filled with water. Permit water to flow for some time till steady state of flow is reached 4, With the help of the stop watch, note the time interval required for the water level in the stand pipe to fall from some convenient initial value 10 some final value 5. Repeat step (4) at least twice and determine the time for the water level in the stand pipe to drop from the same initial head to the same final valve. 6. In order to determine the inside ara of cross-section of the stand pipe, collect the quantity of water contained in between two graduations of known distance apart. Find the mass of this water accurate to 0.1 g. The mass in grams divided by the distance, in em, between the two graduations will give the inside area of cross-section of the sand pipe. ‘Tabulation of observations. ‘The test observations are tabulated as illustrated in Table 7.4 TABLE 7.4 DATA AND OBSERVATION SHEET FOR FALLING HEAD PERMEABILITY TEST Sample No. A/107 Specific gravity : 2.68 Moulding water content : 12% Void ratio 0.0 Dry density : 1.67 g/cm? 1, Area of stand pipe (a) (cm) 0.785 | 2. Cross-sectional area of soil sample (A) (em?) | so 3. Length of the sample (L) | 6 | 4. Initial head (np cm | 40 | 5. Final head ta) ‘em 20 | 6. Timeimeral | (1 tet (see) 56 «i test (ses) 7 (iit) UL test (sec) | 3S 1 Average | 56 7. Coefficient of permeability ates (cm/sec) 17 «10? ‘Temperature 8. Test temperature co. 2 Coefficient of Permeability at 27°C (cm/sec) l 1.05 «10% Calculations. The coefficient of permeability is calculated from Eg.7.27 :PERMEABILITY 203 EXPERIMENT IS : DETERMINATION OF PERMEABILITY BY FALLING HEAD TEST Object and scope. The object of the experiment is to determine the coefficient of permeability ‘of soil in the laboratory by falling head test using Jodhpur Permeameter. Materials and equipment. Same as in Experiment 14, Test Procedure 1, Prepare the remoulded soil specimen in the permeameter and saturate it as explained in Experiment 14, 2. Keep the permeameter mould assembly in the bottom tank and fill the bottom tank with water ‘upon its. outlet. 3. Connect the water inlet nozzle of the mould to the stand pipe filled with water. Permit water to flow for some time till steady state of flow is reached 4, With the help of the stop watch, note the time interval required for the water level in the stand pipe to fall from some convenient initial value 10 some final value 5. Repeat step (4) at least twice and determine the time for the water level in the stand pipe to drop from the same initial head to the same final valve. 6. In order to determine the inside ara of cross-section of the stand pipe, collect the quantity of water contained in between two graduations of known distance apart. Find the mass of this water accurate to 0.1 g. The mass in grams divided by the distance, in em, between the two graduations will give the inside area of cross-section of the sand pipe. ‘Tabulation of observations. ‘The test observations are tabulated as illustrated in Table 7.4 TABLE 7.4 DATA AND OBSERVATION SHEET FOR FALLING HEAD PERMEABILITY TEST Sample No. A/107 Specific gravity : 2.68 Moulding water content : 12% Void ratio 0.0 Dry density : 1.67 g/cm? 1, Area of stand pipe (a) (cm) 0.785 | 2. Cross-sectional area of soil sample (A) (em?) | so 3. Length of the sample (L) | 6 | 4. Initial head (np cm | 40 | 5. Final head ta) ‘em 20 | 6. Timeimeral | (1 tet (see) 56 «i test (ses) 7 (iit) UL test (sec) | 3S 1 Average | 56 7. Coefficient of permeability ates (cm/sec) 17 «10? ‘Temperature 8. Test temperature co. 2 Coefficient of Permeability at 27°C (cm/sec) l 1.05 «10% Calculations. The coefficient of permeability is calculated from Eg.7.27 :208 SOIL. MECHANICS AND FOUNDATIONS PROBLEMS 1. Find the average horizontal and vertical permeabilities of a soil mass made up of horizontal layers. The first and second layers have the same thickness of 0.5 metre each. The thi layer is one metre thick. The coefficients of permeability of the first, second the the third layers are respectively 1x 10° cm/sec, 2 107 cm/sec and 5x 10 cm/sec [Ans ky = 5.5 « 10> cm/sec ; ky= 0.79 x 10° cm/sec} 2. The coefficient of permeability of a soil sample is found to be 1x 10° cm/sec at a voids ratio of 0.4. Estimate its permeability at a voids ratio of 0.6. Jans, 2.95 x 10° em/sec,, or 2.25 x 10° cm/sec} EF 3. A Soil sample of height 6 cm and area of cross-section of 100 cm? was subjected to falling head permeability test. In a time interval of five minutes, the head dropped from 60 cm to 20 cm. If the cross-sectional area of the stand pipe is 2 cm’, compute the coefficient of permeability of the soil sample. If the same sample is subjected to a constant head of 18 cm, calculate the total quantity of water that will be collected after flowing through the sample. [Ans, 4.4 x 10 emisec ; 475 ml] 4. A glass cylinder 50 cm* in inside cross-sectional area and 40 cm high is provided with a screen at the bottom and is open at the top, Saturated sand is filled in the cylinder upto a height of 10 cm above the screen. The cylinder is then filled with water upto its top. Determine the coefficient of permeability in cm/sec if the water level drops from the top of the cylinder through a distance of 20 cm in half an hour. [Ams, 3.83 x 10"cm/sec] 5. If during a permeability test on a soil sample with a falling head permeameter, equal time intervals are noted for drops of head from hy to /; and again from fz to fy . find a relationship between fy, fp and hy [Ans. iin= Vint] 6. A falling head permeameter accommodates a soil sample 6 cm high and SO cm” in cross-sectional area, The permeability of the sample is expected to be 1x 10* cmv/sec. If it is desired that the head in the stand pipe should fall from 30 cm to 10 cm in 40 minutes, determine the size of the stand pipe which should be used. [Ans. 1.53 em dia.)