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ENV-2E1Y - Fluvial Geomorphology 2004 - 2005 Slope Stability and Related Topics Flownet of seepage of water through soil around an obstruction Section 2 Seepage, Flow of Water, Pore Water Pressures N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 Slope Stability and Related Topics 2. Seepage of Water in Soils/Permeability 2.1 Introduction Both the first an second are clearly important (the first arises NOTE: FIRST A HEALTH WARNING!!!! There are directly from the depth below the water table, whereas there is some sections of this handout which are in shaded boxes. ample evidence (in the form of springs to indicate water flow These are not essential parts, but complement the main through soils). course. Thus for some of you who are doing ENV-2B31 (Mathematics I) you may object the simplistic approach What about the velocity head? sometimes used in the handout. There is a more rigorous approach for you in these boxes. In other cases, the boxes Typical velocities even in coarse sands will rarely exceed 10 show additional information which may be derived which may mm s-1, and the magnitude of the last term in this extreme be of use in other courses (e.g. Hydrogeology). Please consult case will be (remember to convert to metres!!!!):- the note before each box. ( 0. 01)2 0. 000005 m in terms of total head or 0.00005 2 x 10 kPa in pressure terms. For another set of notes for this section see the University of West of England WEB Site on the topic. This is exceedingly small, and in most soils, the velocity will http://fbe.uwe.ac.uk/public/geocal/SoilMech/water/water.htm be many orders of magnitude less than this so in future we can A knowledge of the factors affecting the flow of water through conveniently neglect the velocity head term. soils is important as the permeability of a soil affects the way in which a soil consolidates which in turn affects its 2.2 Hydraulic Gradient mechanical properties. Equally water pressures may build up within the soil and as seen the demonstrations can greatly Associated with water pressure is the hydraulic gradient affect the ability of a soil to resist shearing. There are three which is a measure how fast the water pressure is changing. component parts to the water pressure:- This in turn affects how fast water will flow and what the immediate pressure within the soil will be. i) that pressure arising from a static head Now consider flow between two points A and B (see Fig. 2.1). [ at any point at a height Z above the measuring datum, this pressure will be w Z where w is The hydraulic gradient is defined as the rate of change of head the unit weight of water.] of water with distance (in the direction of measurement) ii) excess pore water pressure (i.e. a pressure head differential which actually causes water flow [ this has the symbol u] w v2 iii) a velocity head and equals 2g The total pore water pressure (often abbreviated to pwp) w v2 wZ u = 2g .......2.1 For those who have done the Hydrology or Oceanography options you may already be familiar with Bernoulli's equation Fig. 2.1 Flow of water in a channel of simple horizontal cross of fluid flow i.e. section u v2 H Z w 2g ........2.2 Since the flow is horizontal, there is no effect from the static total = position + pressure + velocity head of water. head head head head The head of water at A is h1 i.e. equation 2.1 is Bernoulli's equation expressed in terms of and at B h2 pressure rather than head of water. [To convert from The excess pressure of the water at A (as the velocity is small) equation 2.1 to 2.2 all we need do is divide by w. is w h1 How significant are these three terms in water flow in soils? and at B w h2 14 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 The gradient of the head drop (or pressure drop) is known as Note the -ve sign as h decreases as we go in direction of the hydraulic gradient (i). In this simple situation, the flow of water, hydraulic gradient is:- i.e. from B to A. h1 h2 i the total pwp at A is = uw = w (Z + h) NOTE: The hydraulic gradient as defined above is while the total pwp at B is = uw + duw dimensionless (i.e. has no units). In some other disciplines, it is defined in terms of pressure (rather = w (Z + dZ + h + dh) than head) and thus is no longer dimensionless. In this case, i.e. the difference in head is duw = w (dZ + w (h1 h2 ) dh) i In this case, there are units associated with the w dZ is the pressure arising from the head difference hydraulic gradient (i.e. kN m-3). To keep things simple we shall use the first definition in this course. wdh is the pressure arising from the excess pressure in the standpipes which will cause flow of water The above is a simple description of what how to measure the from B to A. hydraulic gradient. Strictly speaking we should be talking in terms of the differential coefficient:- In the alternative definition of hydraulic gradient used in some disciplines, dh du i i ds ds and the units are kN.m-3 where s refers to a general direction of measurement. For most of you it is sufficient to accept the above definition, 2.3 The Permeameter but if you want the full derivation it is given in the following box (see WARNING given in introduction about these boxes). The permeability of a soil dictates how quickly water will flow within the soil, and more importantly how quickly excess water pressure will dissipate. The permeability of a soil may be measured with a Permeameter. For silts and sands it is common to use a constant head Permeameter (Fig. 2.3). For clays the permeability is so low and a falling head Permeameter is used. 2.3.1 Constant Head Permeameter (Fig. 2.3) The apparatus consists of a vertical cylindrical tube in which is placed the sample. Below and above the sample are porous stones. Water is fed from a supply to a constant head reservoir and to the base of the sample. After passing through the sample the water flows into a measuring cylinder which is used to estimate the flow rate. In the more accurate permeameters there are two pressure at a fixed distance apart Fig. 2.2 Illustration of two types of water pressure. in the cylinder wall. Fine bore capillary tubes are inserted into sample and connected to manometers to measure water Let A and B be two points separated by a small distance pressure at the two points, and thus the hydraulic gradient ds, and let the excess p.w.p. be du = w dh may be determined. then the hydraulic gradient (denoted by i) is given by:- The experiment starts with the constant head reservoir at a given level. Water is allowed to pass through the sample for a h i few minutes (depending on the nature of the material under s test) until a steady state is reached. The level of water in the In the limit as ds ---> 0 pressure tappings is measured and water is allowed to flow into the measuring cylinder for a known period of time. dh or ……….(2.3) i i 1 du ds w ds The flow rate Q may be estimated (volume of water collected in cylinder divided by time). We also measure the total internal cross-sectional area of the cylinder (At) . 15 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 of water, the density of water, and a shape factor (i.e. relating to the shape of the voids). Since this is an introductory course, and the use of non-dimensional parameters for hydraulic gradient leads to a simpler set of units, this will be used in this course and is consistent with those used in the textbooks on the reading list.. 2.5 Experimental Results from Permeameter The results of a replication of Darcy's experiment are shown as line A. For many soils, the linearity of the line is good, confirming Darcy's Law, but in peats, the data is often very non-linear. Fig. 2.3 Constant Head Permeameter The velocity va of the water as it flows in the part of the cylinder above the sample may be obtained from:- Q va At ........................................(2.4) va is also known as the apparent velocity (i.e. the velocity the water would have when passing through the soil if the solids occupied zero volume. It is less than the actual velocity. Fig. 2.4 Experimental Permeability Results on Leighton Buzzard Sand (after Schofield and Wroth, 1968). 2.4 Darcy's Law As the head is increased, the upward pressure gradient across In 1856 Darcy, using equipment similar to that described the sample increases, and eventually at point C the upward above and continued the experiment by raising the level of the force equals the downwards force from self-weight and constant head reservoir while keeping the height of the sample LIQUEFACTION occurs (i.e. we have created a quicksand). constant. Two points were noted. First, the flow rate The soil "boils" at this point. If we now reduce the pressure increased, and secondly the hydraulic gradient also increased. gradient by lowering the constant head reservoir, we will find More readings were taken with further increases in the height that the sample will settle again, but will occupy a greater of the constant head reservoir. volume than previously. Darcy found that the apparent velocity (v) was proportional We may now repeat the experiment, but this time, although still displaying a linear trend, the points lie on a line with a to the hydraulic gradient (i) higher gradient. As might be expected, the rate of flow of dh water in the loose sample is greater than for the dense sample i.e. va ki k . .........(2.5) and hence the coefficient of permeability (k) has increased. ds The results shown in Fig. 2.4 were obtained for the same k is known as the coefficient of permeability Leighton Buzzard Sand as used in most of the demonstrations described in section 1. NOTE: k has the units of velocity (i.e. m.s-1) because the hydraulic gradient in non-dimensional. In some branches of The cylinder containing the sample has a uniform cross- Science, the hydraulic gradient is measured in terms of the section and thus there is a uniform hydraulic gradient within pressure (rather than head). In such cases, permeability is the soil sample and equals up the cylinder and it is equal to measured in units of m3.s-1.kN-1. The two sets of readings h differ only by a factor of the unit weight of water. There is i s further confusion in that the term hydraulic conductivity is sometimes used instead of permeability. where dh is the difference in the level of the pressure tappings, This latter term, used in Hydrogeology, differs from and ds is the distance between the tappings. permeability in that it also attempts to allow for the viscosity 16 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 The volume of the reduction in water in the capillary tube in a The initial voids ratio for the (initially medium dense) sand time t will equal the flow of water through the sample in the may be calculated as follows:- same period. Let m be the total mass of sand dh khA its length a and dt L then volume occupied = A h t dh A 1 1 m i .e. a k dt h L0 and volume of sand grains = Gs w h 2 aL h k 2.3 log 10 0 ...............(2.7 ) where Gs is the specific gravity of the solid particles. Hence At 1 h1 A m G AGs w Thus to measure the permeability using a falling head e s w 1 Permeameter, the sample is enclosed in the sample tube and m m Gs w ...........(2.6) the capillary tube is filled so that the water level is nearly at the top. A valve remains closed preventing the water flowing through the sample until the experiment is ready to begin. The height of the water in the capillary tube is measured and a stop 2.6 Falling Head Permeameter (Fig. 2.5). clock is started as the valve is opened. The time for the level in the capillary tube to fall to a second level is noted. The When the permeability of fine grained silts/clays is to be measurements needed are thus:- two heights on the capillary determined it is found that the flow rate is so low that it cannot tube, the diameter of the capillary tube, the length and be measured accurately. A falling head Permeameter is then diameter of the sample. employed. 2.7 Formation of a Quicksand - Piping Let a be cross-section of capillary tube A be cross-section of sample tube Referring back to graph in section 2.5. When point C was L be length of sample tube reached the sand appeared to boil as in a quicksand and piping ho be initial height of water at t = 0 occurred. h be height at time t and h1 be height at time t1 The seepage force is that force exerted by the water seeping through the soil. In the situation above, at point C, the seepage force equals the submerged weight of the sand. Let critical value of dh at which the quicksand occurs be dhcrit. the corresponding critical hydraulic gradient (icrit) will be given by:- Q va At the upwards force = Atw dhcrit = At du whilst the downwards force = At'dz thus the critical hydraulic gradient occurs when h crit A ' Gs 1 i crit t z At w 1e .........(2.8) Fig. 2.5 Falling Head Permeameter During piping the volume occupied by the sand increases and From Darcy consequently the voids ratio. The second line (loose in Fig. Q = kiA 2.4) was obtained by repeating the experiment at the new voids ratio. 17 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 At a condition of piping, the sand becomes completely 1) estimations of pump capacities in water resources, and buoyant and it is as though the effect of gravity had been 2) stability calculations on slopes. reduced to zero. The flow of water through soils is directly analogous to two If flow of water had been downwards, the effect of gravity other processes namely HEAT FLOW and the flow of would be increased. This effect can be used in model analysis electricity. Those of you who have done the Energy to study slope stability using models of a slope.. Conservation (ENV-2D02) course will already be familiar with the equations defining heat flow through a component of 2.8 Typical Values of k a building (such as the walls). Those of you who did Physics at school or who have done Geophysics will be familiar with gravels k > 10 mms -1 electrical conductivity. sands 10 mms-1 > k > 10-2 mms-1 These analogies are helpful, as we can model both heat flow and water flow using an electrical analogue model which is silts 10-2 mms-1 > k > 10-5 mms-1 easier to change than is either a brick wall or a layer of sediment. clays k < 10 -5 mms-1 In HEAT FLOW kA 2.9 Actual Seepage Velocity Q ( 1 2 ) where Q is the heat flow rate The actual velocity of seepage through the pores must be 1 is the internal temperature greater than the apparent velocity as calculated by equation 2.4. 2 is the external temperature A is the cross section area Q is the thicknessof wall va At ........................................(2.4 and k is the thermal conductivity repeated) In the FLOW of ELECTRICITY If vs is the actual velocity, and Av is the actual cross kA I ( E1 E2 ) section of the voids where I is the heat flow then Q = vsAv = vaAt - i.e. the water flowing through E1 is the inletpotential the soil pores must equal the apparent flow mentioned earlier. This is the continuity equation.. E2 is the outletpotential A is the cross section area A V v 1e vs va t va t a va is the thicknessof wall i.e. Av Vv n e and k is the electrical conductivity ..........(2.9) In the FLOW of WATER in SOILS where Vt and Vv are the volume of the total sample and kA that of the voids respectively. Q ( h1 h2 ) for dense Leighton Buzzard Sand vs is approximately 2.6va where Q is the flow rate h1 is the intlet head NOTE: In the symbols, upper case V is used for volumes, h2 is the outlethead but lower case v is used for velocities. A is the cross section area is the thicknessof wall 2.10 Flow of Water in Soils and k is the permeability In the permeameter the shape of the sample is cylindrical and analysis of the flow of water through the soil is simple. In The solution of the problem through a cylinder is nature, however, the boundary conditions are never as simple straightforward and can become quite complex in more and ways must be found to enable estimates to be made of general situations. The flow of water (or flow of heat, or flow seepage. of electricity) is governed by Laplace's equations and suitable solutions to these equations. There are five ways in which This is important for two reasons: solutions may be achieved:- 18 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 This is the two-dimensional version of Laplace's equation, 1) Mathematical solutions and solution to seepage problems from which we can a) exact solutions for certain simple obtain the water pressure at any point require solution of situations this fundamental equation within appropriate boundary b) solutions by successive approximate - conditions. In this course we shall be using graphical e.g. relaxation methods solutions, but if you are doing ENV-2B31 2) Graphical solutions (Mathematics), you may like to attempt a mathematical 3) Solutions using the electrical analogue solution of the problem to simple boundary value 4) Solutions using models conditions. The example shown in section 2.14 may also be solved provided that you transform the co-ordinates Mathematical solutions are beyond the scope of this course using conformal transformation beforehand. (although anyone doing the Maths options may wish to follow section 2.11, and then attempt some solutions as applications There must be continuity: i.e. the water flowing in across for the Maths Course). the bottom and left hand border must equal the water flowing out across the other two borders. The Relaxation Method requires the development of computer software. An example is already in use in dynamic heat flow v x v z computations in the Energy Conservation Course as a self v x z v z x ( v x . x )z ( v z . z)x contained computer package, and it is hope to adapt this for x z use in water flow in soils for this course in the next few years. v x v z i. e. 0 ..........( 2. 10) Graphical Solutions are the method we shall adopt in this x z course for convenience we will make the substitution = ki Electrical Analogue method are nice to use in practicals, and were in fact used in the past in this course when more time i i i. e. k and k was devoted to the topic. x x z z Solutions using models. These are expensive to construct, but nevertheless can give useful information particularly in thus vx and v z x z complex situations when it becomes difficult to define the conditions easily in mathematical terms. A model of water Substituting for vx in Equation (2.10) we get flow around an obstruction was constructed many years ago and is now used in the Hydrogeology option. 2 2 0 ............(2.11) 2.11 Flow Equations - 2-D case x 2 z 2 This section contains the Mathematical derivation of the basic Note the analogy with flow of electricity: equations governing the flow of water in two-dimensions. This is strictly for those who are mathematically inclined - for 2V 2V those who are not - skip the following highlighted box. 0 x 2 z 2 where V is the electrical potential or voltage or heat flow 2 2 0 x 2 z 2 where = temperature By analogy is termed the potential in the flow of water. Fig. 2.5 Schematic of two-dimensional flow of water As solutions to the above equations already exist for heat/electricty flow, we may use these soltuions to solve flow of water problems. 19 2.12 Graphical Solutions - Flow Nets (Fig. N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 2.12 Graphical Solutions - Flow Nets (Fig. 2.6). In this course we shall only concern ourselves with the graphical solution to Laplace's Equation. While this may at first sight seem rather crude, it is nonetheless very effective and can be used for all shapes of flow channel provided that a little care is used when drawing. We do this by drawing flownets - i.e. we draw lines parallel to the line of flow of water (flowlines), and lines at right angles which are lines of equal pressure or equipotentials. Experience certainly helps, and you may wish to practice by drawing arbitrary shapes to define boundaries and attempt to fill in the appropriate flow net. In the permeameter we had a situation where flowlines are vertical and lines of equal pressure (equi-potentials) are horizontal. Fig. 2.7 Asymmetric Flownet Note: equipotentials are still orthogonal to flow lines and it can be shown that all regions bounded by two adjacent flowlines and two adjacent equipotentials are curvilinear rectangles of similar proportion (or in most cases squares - See figure 2.8). The following is merely a proof of the this, you may skip the following box unless you are particularly interested. Consider a single flow channel initially of width a1 and a spacing between equipotentials of b1. The potential drop between all equipotentials is Dh. At a second point in the flow channel where its width is now a2 the distance between the equipotentials is b2. By continuity we must have the same amount of water flowing in both parts of the flow channel. Fig. 2.6 Flowlines and Equipotentials in a simple situation where all flow lines are parallel Two points to note:- 1) flowlines and equipotentials are at right angles to one another. 2) the cylinder walls are also flowlines. The distances between the equipotentials are equal and thus the head drops between the equipotentials are also equal. Fig. 2.8 Each flow channel is defined by two adjacent flow lines define Let a1 be the width of channel when velocity is v1 the region in which water moves, i.e. water will not move from and a2 be the width of channel when velocity is v2 one channel to another (apart from a minor effect from diffusion). then q = v1a1 = v2a2 = v3a3 What happens if cylinder is not of constant cross-section? (see Fig. 2.7) kh kh but v1 = k i1 = and v2 = k i2 = b1 b2 By symmetry, flow lines must diverge, but since the package of water A must still remain within the flowlines c - c, d - d a1 a its velocity will change and become less. If the dashed lines thus q kh kh 2 represent equipotentials, then they must become progressively b1 b2 further apart as one proceeds towards the right. 20 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 a1 a nf k H Hence for equal pressure drops we must have 2 nf q f b1 b2 the total seepage = nd .............. (2.12) i.e. the ratio is constant. ============= Thus we must draw rectangles of constant proportion. In other words if we require to determine the total volume of However this is difficult when we have curvilinear figures water seeping we need only draw a flow net and count the unless a = b (i.e. we have squares - a special case of the above number of pressure drops and flowlines. requirement). To work out the pressure at any point, which is what we really General Flownets solution - (See figure 2.9) want, the procedure is equally simple and will be illustrated with reference to a particular example (Section 2.13). Solutions are relatively straightforward. We need:- 1) draw the appropriate flownet 2) count the number of pressure drops in the flow net (over 2.13 Seepage around an obstruction - (See figure 2.10) the relevant distance) 3) count the number of flow lines 4) do a simple calculation as given by equation 2.12 below. The example shown in Fig. 2.10 has a vertical obstruction which is preventing the normal flow of water. It also represents the model rig in the Soil Mechanics Lab which was Let total pressure drop between AB and CD be H and let formerly used as a practical in the fore-runner to this course, there be nd pressure drops and nf flow lines. but recently has been used in the Hydrology courses. Assuming uniform permeability k and h to be pressure drop On one side of the obstruction, the water level is maintained across one square of side. at a high level while the water level on the other side is at h H approximately the level of the soil. Water will flow in around where qf is the flow the obstruction and upwards on the downward side of it. Thus nd per unit cross-section there will be an upward flow of water immediately kH ByDarcy' s Law v ki and downstream of the obstruction, and if the upward force of the nd water exceeds the downward submerged weight of the column kHa a x 1 is the cross- of soil, then piping will occur (i.e. a quicksand will form). and q f kia nd section between flow lines. We first draw the flownet and note that piping will occur if but a seepage force at A >= buoyant weight of column of sand. Hence the total flow in a single flow channel is given by:- If the difference in the height of the water on the two sides of the obstruction is H, then the pressure drop between two kH equipotential lines will be wH / nd. qf nd and if the number of flowlines is nf this space is left blank for notes: 21 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 Fig. 2.9 Generalised Flownet 22 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 Fig. 2.10 Flow of water around an obstruction 23 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 NOTE: we must include w as we are now dealing with if we know the voids ratio, then from as this may indicate whether fine material is washed out from the soil because of pressure, and not merely head of water. high actual seepage velocities. 1e If the number of equipotential drops from the base of the vs va obstruction to the surface of the soil on the downstream side is From equation (2.9) e Nab. N ab w H At any point within our flow net we can estimate the hydraulic nd gradient by dividing the head drop between two equipotentials Then the upward seepage force = by the distance between them ' where is H and the downward force of the soil = i.e. nd the depth of penetration of the obstruction into the soil. A quicksand will occur if Once again the following is not essential to the course, but may be considered if you wish. N ab w H ' nd Alternatively we may measure the actual seepage velocity on a model determine permeability by hence rearranging equation 2.16. Nab H ' ..................(2.14) nd w i.e. v s nd e As an example in the use of a factor of safety, we may predict k . such a factor (Fs) as follows H 1e and by Darcy actual downward force of the soil Factor of safety = ------------------------------------------------ kH 1 e kH force required to just resist seepage force va i. e. v s . nd e nd …(2.16) ' nd Fs . .........................( 2. 15) 2.14 Flow nets (Summary) w H N ab Rules for drawing flow nets:- In the above example, nd = 10 and Nab 3.5 1) All impervious boundaries are flow lines. 2) All permeable boundaries are equipotentials and noting that very approximately ' = w 3) Phreatic surface - pressure is atmospheric, i.e. excess pressure is zero. 10 A further requirement is that the change in head Fs between adjacent equipotentials equals the vertical 3. 5H distance between the points on the phreatic surface. i.e. the distance must exceed 0.35 times the difference in 4) All equipotentials are at right angles to flow lines head of water. 5) All parts of the flow net must have the same geometric proportions (e.g. square or similarly shaped rectangles). The following box is not essential to the course, however it is 6) Good approximations can be obtained with 4 - 6 flow complementary to the course and to aspects of the Hydrology channels. More accurate results are possible with higher courses. numbers of flow channels, but the time taken goes up in We are sometimes interested in the actual seepage velocity vs proportion to the number of channels. The extra precision is usually not worth the extra effort. If we wished to work out the quantity of water flowing around the obstruction then from equation 2.12, and noting Note: that nd = 10 and the number of flow channels is approximately 4.25. · a difficulty in flownet analysis is the determination of the top flow line in flow down a slope or through an earth nf dam. Q n f q f kH 0. 425 kH · It is difficult to give hard and fast rules - best try and see. nd However, a good approximation is a parabola. If, for example, the permeability ( k ) = 1 mms-1 and if a · In regions of sharply varying flow channels, there are field situation, H = 10 m more advanced methods which allow the subdivision of squares into quarter size squares just at the key points, Q = 0.425 x 10-3 x 10 = 4.25 litres.s-1. but these techniques are beyond the scope of this course. =============== 24 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 The next two sections (i.e. 2.15 and 2.16) of this handout are not essential for the course, but you should appreciate 2.16 Flow through soil having differing in general terms what is going on. You will not be permeabilities in x and z directions expected to do numeric problems based on the following. The above discussion has centred on the assumption that 2.15 Flow between regions of different permeability permeability is uniform in all directions. Not infrequently there is a difference, through lamination, Flowlines and equipotentials in Fig. 2.11 in Region A form between the horizontal and vertical permeabilites. squares of side a. In region b, these become rectangles of size b x . Let permeability be ka and kb then by Let these permeabilities be kx and kz respectively. continuity flow in channel must be same in both A and B. Hence:- We may drawn an equivalent flow net if we first adjust the scale of the original diagram so that we multiply all horizontal distances by a factor of kz (See figure 2.12) kx If we do this, we can then still use square (or rather curvilnear square) flow nets on the transformed diagramd diagram provided that we recognise that the effective permeability is also modified as follows:- The equivalent permeability is then k = (kxkz) Fig. 2.11 Water flow across a boundary between regions of The geometric modification is a form of conformal different permeability transformation (which those doing ENV-2B31 may have come across). Indeed the problem of the seepage around a b ka b q k a h k b h so an obstruction may also be solved by transforming the a kb actual scale into a square net using an appropriate a b a conformal transformation, although in this case it does and AB and AC require elliptical functions in the transformation. sin a sin b cos a cos b ka b tan b For details see pages 59-64 of Critical State Soil so kb tan a Mechanics by Schofield and Wroth (1968) - publisher Thus in passing between two zones flowlines are refracted McGraw Hill - It is in the Library. and I also have a copy such that the ratio of the tangents of angles of incidence are for those interested proportional to the permeabilities. Note: when drawing flow nets at boundaries, squares in one zone will become rectangles in the second zone. Fig. 2.11 Example of transformation to allow for anisotropic conditions 2.17 Uplift on Obstructions This pressure arises from both the static and excess heads. For an obstruction which has a flat base (which is horizontal, When water seeps under a large obstruction such as a boulder the static head will be constant along its length and equal to which is partly buried in the soil, the obstruction experiences the depth of the base below the water level on the down stream an uplift from the total water pressure exerted on the base. end. The excess head will vary along the underside of the obstruction. 25 N. K. Tovey ENV-2E1Y Fluvial Geomorphology 2004 – 2005 Section 2 Fig. 2.13 Uplift on a rectangular boulder by water passing underneath If the total uplift force exceeds the downward force from the Because the base of the obstruction is 2m below the surface self weight of the obstruction, then the object will be displaced the uplift force from the static head is 2w multiplied by downstream. To assess the likelihood of this happening, a width (i.e. 6w kN per metre length). From the excess head flownet is drawn for the water seeping through the soil. A (see graph), the upward force is the area under the curve graph is plotted with the x-axis as the distance along the base from the upstream face, and the water pressure as gauged multiplied by w. In this example (since the line is nearly from the equipotentials (+ the static head) as the y-axis. The linear), the upward force = 6w kN per metre length, i.e. in total force is then evaluated by working out the area under the this case it equals the static head uplift. The total uplift will curve. If this exceeds the with of the obstruction then it will then be the sum of these two components, i.e. 12w kN m-1. be displaced This uplift will considerably reduce the ability of the The pressure distribution shows the excess head an has been obstruction to resist movement through the pressure of water drawn from position of intersection of equipotentials along the (and has significance both as potential boulder blockages in a base of the obstruction. The equipotentials are nearly equi- river and also as man-made drop structure built in river distant from each other in this region and the distribution is engineering works to dissipate energy (see RDH's part of the nearly linear. By counting equipotentials, the head at the Course). There may also be erosion from the possible upstream head is 0.75 of total head, and at the down stream quicksand which may form at the down stream end of the end it is 0.25 of the total head. obstruction. space for further notes. 26