5 Particle density (rs)DefinitionsMass of soil particle divided by volume of soil particleSpecific gravity, SG = ratio of mass of soil particle tomass of equal volume of water of water at 4°CParticle density in cgs or tonnes m-3numerically equal to SGmean particle density depends on:ratio of OM to mineral matterconstitution of soil mineralsconstitution of OM
6 DeterminationSG bottleboiled water to remove dissolved airde-aerate for several hours with vacuum pump to removeair trapped between particlesproblem of floating OMTypical valuesorganic matter = 1.3 g cm-3quartz = 2.66 g cm-3average for clay = 2.65 g cm-3orthoclase = 2.5 to 2.6 g cm-3mica = 2.8 to 3.2 g cm-3limonite = 3.4 to 4.0 cm-3Fe (OH)3 = 3.75 cm-3normally taken as 2.65 cm-3
7 Bulk density (rb) and related parameters rb = mass of solidstotal volumeValue is effected by particle density, degree of compaction,organic matter content
8 Typical values:0.9 for organic soil (peaty) to 1.8 for compacted sandSand generally has a higher density than clay - why?What do we mean by heavy & light soils?Determination:soil coring devicesproblems of compactionoven drying at 105°Cgamma ray transmission
9 Gamma ray transmission measures density –2 probes - transmitter & detector
11 Coefficient of linear extensibility (COLE) Bulk density changes in swelling - shrinking soils.COLE is a measure of thisCompares dry with saturated soil after it comes to equilibrium.Cracks complicate the problem of determining BD ofswelling soils.Even allowing for cracks the overall density may be higheron shrinking as the surface becomes lower.
13 Total pore space (T)= volume of (air + water)vol. of (air + soil + water)
14 Volume of (air + water)= total volume (air + soil + water) - volume of soilwhere Vt and Vs are the volumes of the total sample andthe soil particles respectivelyVs = Ms /rs and Vt = Ms /rbwhere Ms is the mass of oven dry soil and rs and rb are theparticle density and bulk density respectively. So:
15 rb = (1-T)/rs and so: and : Used in agricultural (soils) research especially forcompaction studies.Typical values 0.3 to 0.6. Often expressed as a %.
16 Packing densitymeasure of compaction of particular texture classVoid ratio (e)Used mainly in engineering applicationse = volume of (air + water)volume of soile = T/(1-T) [void ratio]Typically 0.3 to 2.0Air filled porosity= volume of airvolume of total
17 Moisture content and related parameters (a) Volumetric basis:volume of watervolume of totalqv = Vw/Vt(b) Gravimetric basis:mass of watermass of soilqm = Mw/Ms
18 As Vw = Mw/rw and Vt = Ms/rb then and so qv = qmrb/rw= qmrb/1 (not dimensionally correct)in metric measurements - density of water is 1Often expressed as depth/depth for example mm/m
19 Degree of saturation (s) degree of saturation = volume of watervolume of (water + air)s = qV/TLiquid ratioLiquid ratio = volume of watervolume of solid
20 An example to tryA hole 30 cm X 30 cm x 30 cm is dug in a field. The wetsoil weighs kg. The soil is taken back to thelaboratory and oven dried. The final weight is kg.(a) What is the bulk density(b) What was the moisture content in the field(i) by volume(ii) by weight(c) If the mean particle density is 2.64, what is the totalpore space
21 Graphical representation .... Q. Why is the moisture content less at depth?)
22 Measurement of soil moisture Laboratorydefinitiveweigh, oven dry at 105°C for 24 hours, reweighif volume of hole from which sample was taken is known, bulk density can be calculated and hence volumentric moisture contentField methodsInclude:neutron scatteringgamma ray transmissiontime domain reflectometryall need calibration against laboratory method
24 H scatters and slows neutrons very effectively - elastic collisions with atomic nuclei called “thermalisation” of fast neutrons - come to same thermal (vibrational) energy as atoms at ambient temperaturehydrogen, has nucleus of about same size & mass as neutron and so has much greater thermalising effect on fast neutrons than any other elementmethod detects mostly H atoms not water per sesingle probe containing radioactive source of high-energy neutrons such as radium-beryllium or americium-beryllium or caesium-137thermal neutron density easily measuredthermal neutron density may be calibrated against water concentration on volume basis of other sources of H are constant
25 Time domain reflectometry measures dielectric constant - ability of soil totransmit electromagnetic (radar) waves -mostly but not entirely dependent on water
27 Simple parameters to characterise H2O & O2 availability Soil water potentialmatric potentialgravitational potentialpressure potential
28 Note on unitsSoil water potential is the energy density -usually per unit volumeSince dimensions of energy is ML2T-2 (force x distance)dimensions of soil water potential has dimensions ofML-1T-2Pressure is force per unit area so has units ofMLT-2/L2 = ML-1T-2Soil water potential thus has same units as pressure.It can this be expressed as bars, cm H2O, cm Hg,atmospheresSI unit of Pressure, and so energy density, is the Pascal1 kPa = 10 mb, 1 bar = 100 kPa
29 Capillarity andadsorbed watercombine to producematric potential
30 Permanent wilting point Usually taken as 15 (1500 kPa) bars, but may be more,e.g. 20 bars (2000 kPa).Water held between 1500 and 2000 kPa negligiblein virtually all soils.PWP strongly correlated with clay.In reality, a dynamic property which depends on:potential evapotranspiration,unsaturated hydraulic conductivity of the soil,type of plant.
31 Field capacitythe upper limit of available water;traditionally defined as the moisture content of a soil48 hours after saturation and subsequently being allowedto drain;a high proportion of irrigation water added abovefield capacity is “wasted”;FC has also been considered to be:0.33 bars [33 kPa] in USA or0.1 bars [10 kPa] in the UKFC also sometimes considered as the mean soil moisturecontent in winter (cold climates) when the potentialevapotranspiration is small (and so drainage is main factorgoverning equilibrium moisture content.
32 The tension equivalent to FC will be at least equal to the air entry potential - see below.FC, PWP and AWC are strongly dependent on texture, OM and BD
33 Air capacityDefined as the air content (%) at field capacity.Used in poaching studies.Low air capacity usually means poor aeration.Available water capacityDifference between FC and PWP (%)often x soil depth to give mm
34 Exercise The moisture content of a soil at field capacity was found to be 27.3% by weight. At wilting point, themoisture content was 19.7%. After oven drying of avolumetric sample, it was found that the bulkdensity was 1.42 g cm-3.What is the available water capactiy as a percentageof the volume?A crop has a rooting depth of 1.5 m. How much wateris potentially available to the crop in mm equivalent.If irrigation is to take place when the AWC is depletedby 40%, how much water would need to be added?
35 Effect of bulk density on air capacity, wilting point & field capacity
40 Dynamic nature of FC, PWP, AWC It is important to realise that FC, PWP and AWC arecommonly conceived as static soil properties but thatin reality, the are used as proxies for characteristics of dynamic system.They do not take into account:field conditions such as underlying horizons;rainfall and or irrigation frequency and amount;hydraulic conductivity of the soil;run-off characteristics;roots extension;water infiltration and redistribution;drainage from soil profile;some water may drain at the same time asevapotranspiration takes place;ground cover changes;
41 crop height changesclimate, especially evapotranspiration rate effect thevaluesBeware of too simplistic a view.Even so, FC, PWP and AWC are very useful concepts.
42 Measurement of soil potential TensiometersAfter Richards, 1965
43 Electrical resistance methods Gypsum blocksGranular Matrix Sensorse.g. WATERMARK sensor from Irrometer Co, USA
44 Relationship of soil water potential to soil vapour pressure If vapour between soil particles is in equilibrium with heldwater, the vapour pressure is influenced by the “pull” ofthe soil water ...where :Yt is the sum of matric and osmotic potential is the density of the water at the prevailing temperature,R is the Universal Gas ConstantM is the molecular weight of waterT is the Temperature (°K)e is the vapour pressure in the soil porese0 is the saturated vapour pressure of free water at theparticular temperature
45 The phenomenon is used as the basis of: (a) the determination of the potential of a soil in thelaboratory (often in order to determine the moisturerelease characteristics) by allowing a filter paper ofknown pore size / moisture release characteristics tocome into equilibrium with the moist air over the soilwhich is also in equilibrium with the soil waterpotential.(b) to determine the soil water potential in the field bydetermining the humidity of the soil air using athermocouple psychrometer
47 Moisture release characteristics Determinationpressure plate apparatussand; sand/kaolin bath apparatusfilter paper - allow to come into equilibrium and weighpapersolution - mixture so that vapour pressure is known andthis can be equated to soil potential, allow soil to comeinto equilibrium with solutionuse of pF scale
48 Filter paper methodtop filter paper not incontact - measuressum of matric andosmoticpotential of soilbottom filter paper isin pore contact someasures matricpotential
52 Air entry potential (fe) Also known as air entry value or bubbling pressure= pressure at which largest pores begins to emptyRelated to structure and field capacity.fe corresponds to the largest pore sizewhereand
53 f es is the air entry potential when the bulk density is f is in J/kg 1.3 g cm-3f is in J/kgdg the geometric mean particle diameter, is in mm, andsg is the geometric standard deviation of the particle sizesin mm (ranges from 1 to 30).
54 Example calculation of dg and sg (based on Campbell, p.9) It is assumed thatclay has d < mmsilt has < d < 0.05 mmsand has 0.05 < d < 2 mmThe predictor equations assumes that particle sizedistribution is log normalLogarithm of geometrical mean is given by:ln dg = S mi ln diwhere the di are the textural class sizes and mi are theamounts in each classThe di for the size classes are calculated from(lower limit + upper limit)/2
55 dclay = 0.001 mm; ln(dclay) = - 6.91 thus:dclay = mm; ln(dclay) =dsilt = mm; ln(dsilt) =dsand = mm; ln(dsand) =If a soil is 0.6 clay, 0.25 silt and 0.15 sand, thenln dg = (0.6 x ) + (0.25 x ) + (0.15 x 0.025)= == mm
56 (ln sg )2 = f1(ln d1)2 + f2 (ln d2)2 + f3 (ln d3)2 - (ln dg)2 Substituting this in the above equation,the standard air entry potential is - 12 J kg-1To make allowances for bulk density,we need first to calculate sg.The normal standard deviation is given by:In a similar way, the logarithmic standard deviation is givenby:(ln sg )2 = f1(ln d1)2 + f2 (ln d2)2 + f3 (ln d3)2 - (ln dg)2The geometric SD is the antilog of the SD of the logtransformed values.Thus: ln sg = 2.42 and so sg = e2.42 = 11.24and b = 2 x x = = 26.25
57 Thus for this soil,For bulk densities of 1.1, 1.3 and 1.5, the air entry potentialswould be:- 0.84,- 12J kg -1respectively
59 Kaolin table (100 cm to 400 cm potential) H should be added to difference between atmospheric pressure andpressure in aspirator bottle
60 Pressure plate methodfor potentials from1 bar to 15 bars
61 Prediction of matric potential Not reliable but some workers use equations of the form:where qs = saturation % (vol) and Fe, the air entry potentialis calculated as before from:
62 ClaysTreated here because flocculation in the field is an essentialpart of reclamation of sodic soils.Flocculation occurs when clay particles “stick” togetherbecause of electrical forces to form larger particles andhence improves the hydraulic properties of the clayFlocculation changes the hydraulic conductivities and themoisture holding properties of clay soils.Clay = 0.2 m -2m ; Colloidal clay: =< 0.2m ; m = 10-6 mAdsorption: concentration of one material at surfaceof anotherAbsorption: uptake of one material into another
63 Colloidal material is surrounded by thin layer of solution which is different in composition from the solution(relatively) far away from the particles.Layer moves with the particle.Micelle: colloidal particle + hydration shellIntermicellar fluid : solution between micelles
65 Micelles usually negatively charged because of: isomorphic substitution:Si++++ in the clay may be substituted by Fe+++ or Al+++which makes the clay short of + charge and sonegatively charged - smectite or illite type materialsFe++ or Mg++ may replace Fe+++ or Al+++ in alumina orgibbsite sheetsionisation at the surface:e.g. appearance of OH - at the surface and edges ofmicelleH2O adsorption and subsequent ionisation & diffusionof H+ leaving a net negative charge because of the OH –preferential adsorption of anions from solution forexample the adsorption of CO3- onto calcium carbonateleaving an associated ion in solution
66 Some mutual attraction occurs between particles because of edge effects
67 Double layerDouble layer is name given to accumulation of positively charged ions around negatively charged micelles – some attached, some in solutioncontrols flocculation and dispersiondependent oncation typecation concentrationpHthe thicker the double layer, the greater the net repulsion and the more dispersed a soil becomesimportant in structure and aggregation and reclamation of saline and sodic soils
68 Negative ion concentration in solution increases with distance & positive ion concentration decreasesnegatively chargedsoil particle+a layer of cations directlysatisfies some of thenegative charge+diffuse second layer eventuallyreaches sameconcentration as surroundingbulk solution
69 Helmholtz model - assumes the charge concentration Different models:Helmholtz model - assumes the charge concentrationdecreases linearly with distanceGouy-Chapman model - assumes charge concentrationdecreases exponentially with distanceStern model - assumes linear decrease in the Sternlayer near the surface and then an exponentialdecrease - thickness of Stern layer normallytaken as equal to the ionic radius of the adsorbedspecies.Taylor & Ashcroft, Fig. 5.7
70 The double layer is depressed by increasing the valency of the ions in the intermicellar solution and hence thepacking of the charge near the micelle surface.This is known as the depression of the double layer.
71 smaller cations, like Mg2+, will decrease double layer thickness Effect of cation type++smaller cations, like Mg2+, willdecrease double layer thicknesslarger cations, like Na+, willincrease double layer thickness
72 If concentration of intermicellar solution is increased, concentration of ions in double layer reaches concentration ofintermicellar concentration nearer the micellesurface
74 pH Kaolinite edge OH O Si Si Si OH Al Al Al OH neutral pH lower pH (+1/2)OHSiAl(+1/2)(-1/2)lower pHhigher pHOHOSiAl(-1/2)(-1)
75 London - van der Vaals forces In addition to repulsive forces caused by accumulatedpositive charges, there is also an attractive force betweenclay particles caused by London - van der Vaals forces.These forces, which occur even between electrically neutralatoms, are due to the fact that, although the averageelectrical field of a neutral spherical atom is zero,the instantaneous field is not zero but fluctuates with themovements of the electrons in the atom (or ion).When 2 atoms (or ions) approach, they can synchronise theirelectronic motions so that the electrical charge in one surgestowards the other when the fluctuations in this second atomhappen to leave its nuclear field somewhat exposed in thisparticular direction.
77 Depending on relative strength of forces of attraction and forces of repulsion, attractive van der Waals forces maypredominate in which case flocculation takes place.
78 Figure . In addition to attractive van der Waals forces, there are forces of repulsion between particles in suspension. The potential energy of attraction and also that of repulsion varies with distance from the particle surface. Curve 1 is an example of repulsion, and curve 6 is an example of attraction. These two curves vary with the colloid and the kinds and amounts of electrolytes. Thus, when curves 1 and 6 are summed for different conditions, they produce curves similar to 2 to 5. Curve 2 represents a stable suspension because the energy of repulsion predominates. Addition of more electrolyte will suppress the double layer and produce a curve similar to 3, 4 or 5 depending on the amount added. With curve 3, there is still an energy barrier to flocculation, but when the particles surmount the energy barrier and approach closer than point C, the forces of attraction predominate and the particles stick together. A suspension represented by curve 5 flocculates spontaneously and cannot be redispersed unless the curve is shifted back toward curve 2. This shift may be accomplished through expanding the double layer by changing the kinds and/or amounts of the electrolyte.Fig in Taylor & Ashcroft
79 The thickness of the double layer is altered by both the concentration and the ratio of divalent to monovalent ions(and the ratio of tri-valent ions to mono-valent ions).If the constitution and concentration of the soil solutionis changed, the constitution of the ions in the double layerwill change.Replacement of monovalent ions by divalent ions in the doublelayer makes it thinner and so more easy for Van der Waalsforces to take over and make the particles stick togetherLangmuir equationRelates the amount of adsorption onto clay particles to the concentration of the solution.Look it up.
80 Specific surfaceImportant effect on:cation exchangeretention and release of various chemicals(nutrients and pollutants)swelling of claysretention of waterengineering properties(e.g. plasticity, cohesion, strength )
81 am = As/Msav = As/Vsab = As/Vtwhere a is the specific surface,As is the total surface area in the sample,Ms is the mass of solids,Vs is the volume of the solids,Vt is the total volume of the sample.Suffixes m, v and b refer to whether specific surface ison a mass basis, volume of solids basis or volume oftotal soil basis.
82 Measured from amount of gas absorbed at certain T and P. Can also be estimated from particle size distribution &distribution of mineralsNB. surface area/volume for sphere = 6/d = avTypical values
83 Texture and particle size distribution Definitions of sand, silt, clay1 = 10-6 m = 10-3 mmsand: silt: clay : < 2
86 Mechanical analysisSeparation of particlesOM removed by H2O2sometimes CaCO3 cementing agent removed by HCldeflocculation by adding Calgon(sodium hexametaphosphate)mechanical agitation (shaking, stirring, ultrasound)SievingUse sieves down to 0.05 mm (very fine sand)
87 Fd = 6phru Sedimentation Theory Falling particle in a fluid experiences a downward forceand resistance force (drag) in opposite direction.Stokes (1851) found that the drag was given by:Fd = 6phruu is terminal velocity,h is the viscosity,r is the radius of the sphere
88 6phru = 4/3 p r3 g(rs - rf) When the two forces are in equilibrium, particle reaches a “terminal velocity”. In that condition,downward force on the particle= gravity - upthrust due to fluid densityUpthrust = weight of particle - weight of fluid displacedweight of particle = 4/3 p r3 rs gwhere rs is the particle densityweight of water displaced = 4/3 p r3 rf gwhere rw is the density of fluidSo upthrust is 4/3 p r3 (rs - rf) gAt terminal velocity,6phru = 4/3 p r3 g(rs - rf)
89 which can be rearranged as: where d is the diameter of the particle.Since u = h/t whereh is height dropped andt is the time elapsedt = h/uand soand
90 Pipette methodAll particles > d(h, t) will have settled out by time tProportion of original can be determined by taking a sampleAfter 8 hours only clay is left in suspensionHydrometerMeasures density of remaining soil suspension instead oftaking a sampleX-ray transmission methodsTransmission related to density.Gives continuous distribution