Soil Physics PP-3.pptx for soil science students 1212 it is used to explore physical property of soil

For soil science students

1. 1 Hawassa University Wondo Genet College of Forestry and Natural Resources Soil Resource and Watershed Management Department Soil Physics (SRWM 2053) By: Berhanu S. Wondo Genet, Ethiopia

2. 2 CHAPTER ONE 1. INTRODUCTION 1.1 Dimensions and Units Science depends on measured numbers, most of which have units. The following sections should help you understand: what units are, how they are used, and how they are converted from one system to another. Dimensions v s Units It is fairly easy to confuse the physical dimensions of a quantity with the units used to measure the dimension. We usually consider quantities like mass , length , time , and perhaps charge and temperature , as fundamental dimensions .

3. Cont… Table 1. Basic Units of the SI System 3 Quantity Name of SI Units Abbreviation Length meter m Mass Kilogram kg Time Second s Electric current Ampere A Temperature Kelvin K Luminous Intensity Candela cd mol Quantity of substance mole

4. A measurement unit is one that is defined by a physical standard of measurements. All other units of measurements are derived from the fundamental units. The international system makes use of a series of prefixes to indicate decimal fractions or multiples of various units by some power of 10. Table 2. common prefixes used in the SI system 4 Prefix Factor Symbol Prefix Factor Symbol deci 10 -1 d deca 10 da centi 10-2 c hecto 10 2 h milli 10 -3 m kilo 10 3 k micro 10 -6 µ mega 10 6 M nano 10 -9 n giga 10 9 G pico 10 -12 p tera 10 12 T femto 10 -15 f peta 10 15 P atto 10 -18 a exa 10 18 E

5. 1.2 Conversion of Units of Measurement Information = Information x Conversion factor Conversion = Amount in units required/1 unit given Example 1: To convert micro meters to meters, the conversion factor is 1x10 -6 meter/micrometer Example 2: A meter monitoring the supply of oxygen gas recorded 0.454 m 3 . what is the volume of oxygen in dm 3 ? Solution : since 1 m 3 = 1000 dm 3 , then conversion factor = unit required/unit given = 1000 dm 3 /m 3, therefore, volume (dm 3 ) = volume (m 3 ) x conversion factor=0.454 m 3 x 1000 dm 3 /m 3 = 454 dm 3 Soil physics is one of the major subdivision of soil science that seeks to define, measure and predict physical properties and behavior of soils (natural condition in field and human activities) It is concerned with the study of state and transport of matter and energy in the soil. It consists of both fundamental (basic) study and applied study 5

6. Cont…. 1.3 Three Phases Dispersed Soil System Soil consists of three phases or physical states, such as soil solid, soil liquid and soil gas phases. Soil solid phase mineral matter (inorganic ) and organic matter Soil liquid and gas phases are occupied pore-space (void) The three phases of soils are important in management scenarios For instance, water logging condition ➜ anaerobic condition (lack of air) which is not balanced due to: Increasing water content and Decreasing air content So, the management scenarios should balance the proportions of two constituents 6

7. 7 CHAPTER TWO 2. SOIL PHYSICAL PROPERTIES AND INTER- RELATIONSHIPS The physical properties of soils determine their adaptability to cultivation and the level of biological activity that can be supported by the soil. Soil physical properties also largely determine the soil's water and air supplying capacity to plants. Many soil physical properties change with changes in land use system and its management such as intensity of cultivation, the instrument used and the nature of the land under cultivation, rendering the soil less permeable and more susceptible to runoff and erosion losses

8. 8 2.1 Soil Texture 2.1.1. Description and Importance It concerned with the size of mineral particles It is the relative proportion of particles of various sizes in a given soil. The proportion of each size group in a given soil (the texture) can not be altered and thus is considered a basic property of a soil. The analytical procedure by which the particles are separated is called particle size analysis. Soil texture is critical to understand soil behavior and management. It determines the characteristics or properties of various horizons as: A, B, C and D

9. 9 Gravel (coarse fragments >2mm in diameter ) Sand (0.05mm – 2.00mm in diameter) Voids between sands are large and promote free drainage and entry of air into the soil. Because of large size, sand particles have lower specific surface area. Thus hold little water and are prone to drought. Silt (0.05mm - 0.002mm in diameter) Are micro sand particles. Pores between silt particles are much smaller and numerous. Clay (< 0.002mm in diameter) Have large specific surface area ( as much as 10000 times) High water and nutrient absorbing capacity. When wet sticky and plastic; when dry hard. Pores between clay particles are very small and hence water and air movement is slow.

10. 10 2.1.2. Soil textural classes There are two methods: Taking of a dry/moist soil and rubbed between the thumbs and forefingers this method is called feel method. The textural triangle is used for the determination of particle size distribution based on soil separates (sand, silt and clay).

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12. 2.1.3. Stoke’s Law and Particle size analysis Mechanical Analysis : It is the procedure for determining the particle size distribution of a soil sample The first step in this procedure is adding the soil sample in an aqueous suspension The primary soil particles, often naturally aggregated, must be separated and made discrete by removal of cementing agents (such as organic matter, calcium carbonate, or iron oxides) and deflocculating the clay. Organic matter can be removed by oxidation with hydrogen peroxide; calcium carbonate can be dissolved by addition of hydrochloric acid. For the deflocculating, dispersing agents such as sodium hexametaphosphate is used. 12

13. Cont.. This method is generally called sedimentation method. After complete dispersion, either of the following methods can be used for measuring the proportion of the separates: 1. The settling velocity of the particles 2. The density of the suspension from which the particles are settling measured based on the particles of Stoke’s law Stoke’s Law and its Derivation It states that the terminal velocity of a spherical settling under the influence of gravity in a fluid of given density and viscosity is proportional to the square of the particle’s radius. A particle falling in a vacuum will encounter no resistance as it is accelerated by gravity and thus its velocity will increase as it falls. 13

14. Cont.. A particle falling in a fluid, on the other hand, will encounter a frictional resistance proportional to the product of its radius and velocity and to the viscosity of the fluid. The resisting force due to friction, F r , was shown by Stoke in 1851 to be: F r = 6 πη rV , where η = viscosity of the fluid (g/cm sec), r = radius of the particle (cm) and V = velocity of the particle (cm/sec) Initially, as the particle begins its fall, its velocity increases. Eventually, a point is reached at which the increasing resistance force equals the constant downward force and the particle then continues to fall without acceleration at a constant velocity known as the terminal velocity, V t . 14

15. Cont… The downward force due to gravity, F g , is: F g = π r 3 ( ρ s – ρ f )g 4/3 π r 3 = volume of the particle (cm 3 ) ρ s = density of the particle (g cm -3 ), ρ f = density of fluid (g cm -3 ) g = acceleration of gravity (cm sec -2 ) Setting the two forces equal, we obtain Stoke’s law: Where d or r is the diameter/radius of the particle, ρ s = 2.65 x10 3 kg m -3 , ρ f = 1x10 3 kg m -3 , η = 1x10 -3 Ns/m 2 at 20℃, and g = 9.81 N/kg are constant at specific temperature, the above equation simplifies to: V t = Kr 2 , where K is constant. 15

16. Cont… Assuming that the terminal velocity is attained almost instantly, we can obtain the time, t, needed for the particle to fall through a height, h: Rearranging and solving for the particle diameter gives: The use of Stoke’s law for measurement of particle sizes is dependent on certain simplifying assumptions that may not be in accord with reality. 16

17. Cont… Among these: The particles are sufficiently large to be unaffected by the thermal (Brownian) motion of the fluid molecules. The particles are rigid, spherical and smooth. All particles have the same density The suspension is sufficiently dilute that particles do not interfere with one another and each settles independently. The flow of the fluid around the particles is laminar, i.e., no particle exceeds the critical velocity for the onset of turbulence. Draw Backs In reality, soil particles are neither spherical nor smooth, it affects the calculated ‘d’ equivalent settling diameter Soil particles are not all of the same density, this is the only true if they have the same mineralogy- but can not hold. 17

18. 18 2.2. Soil particle and bulk densities A good soil for growing plants should generally contain about 50% pore space and 50% solids. In the soil, air and water are found within the pore spaces between the particles.

19. 19 2.2.1. Particle density ( Dp ) If this ideal soil were dried out, and all the pore space were compressed so that there was no more air and no more space, all that would remain would be the solid portion of the soil. It is usually defined as the mass or weight of a unit volume of soil solids and is called particle density. Its unit is g/cm 3 .

20. 20 Particle density depends on chemical composition, Crystal structure and Organic matter For general calculations, the average arable mineral surface soil (3 – 5% organic matter) may be considered to have a particle density of about 2.65g/cc. For most mineral soils particle density varies between 2.6g/cm 3 and 2.75g/cm 3 . This occurs because quartz , feldspars , and the colloidal silicates with densities within this range usually make up the major portion of mineral soils. When unusual amount of minerals with high particle density such as magnetite, granite, epidote , zircon, tourmaline, and hornblende are present, the particle density may exceed 2.75g/cm 3 .

21. 21 2.2.2. Bulk density (Db) It is defined as the mass or weight of a unit volume of dry soil. This volume includes both solids and pores.

22. Cont… It is the density for a volume of soil as it exists naturally, includes any air space and organic materials in the soil clays. Moisture is not included in the sample weight. The bulk soil volume is assumed not to have changed by drying; only the water has been removed leaving empty pores. This is not true for soils with considerable amounts of swelling clays. Factors affecting bulk density soils that are loose and porous will have low bulk density Those that are more compact will have high values. Example: Particles of sandy soils generally lie in close contact, such soils have high bulk densities. 22

23. Cont… The low organic matter content of sandy soils further encourages the high bulk density value. On the other hand, particles of the fine textured surface soils, such as silt loams, clay loams and clays, ordinarily do not rest so close together. This occurs because these surface soils are comparatively well granulated, a condition encouraged by their relatively high content of organic matter. Granulation encourages a fluffy, porous condition, which results in low bulk density values. Consequently, the bulk density of a well-granulated silt loam surface soil is sure to be lower than that of a representative sandy loam. 23

24. Cont… Very compact sub-soils, regardless of texture, may have bulk densities as high as 2 g cm -3 . There is a tendency for the bulk density to rise with depth. This apparently results from: Lower content of organic matter Less aggregation and root penetration Compaction caused by the weight of the overlying layers Crop and soil management practices are also likely to influence bulk density, especially of the surface layers. For instance, addition of farm manure in large amounts tends to lower bulk density whereas intensive cultivation operates in the opposite direction. 24

25. Cont… Interpretation of Bulk Density Data Bulk densities above 1.75 g cm -3 for sands or 1.45 to 1.65 g cm -3 for silt and clays are quoted as causing hindrance to root penetration. For good plant growth, bulk densities should be below about 1.4 g cm -3 for clays and 1.6 g cm -3 for sands. Increases in soil bulk density impose the following stresses on a plant root system: The mechanical resistance to root penetration increases, so reducing the plant’s root to exploit its environment. The air-filled porosity of the soil decreases, thus restricting the air supply to plant roots and facilitating the build-up of toxic products such as carbon dioxide and ethylene. As well as decreasing total porosity, compaction of soil decreases the volume of coarse pores relative to the volume of fine ones, and hence also increases the proportion of total porosity occupied by water at any given potential 25

26. Cont… In general, hydraulic conductivity or permeability decreases with increasing bulk density, making field crops more susceptible to the adverse effects of water-logging. Example on calculation of bulk density A sample of moist soil having a wet mass of 1000 g and a volume 640 cm 3 was dried in the oven and found to have a dry mass of 800 g. calculate the dry bulk density. solution: Db = mass of dry soil/total volume of soil = 800 g/640 cm 3 = 1.25 g cm -3 Uses of Bulk Density Data Bulk density data can be used to: Calculate the total water storage capacity per soil volume Evaluate soil layers to determine if they are too compacted to allow root penetration or adequate aeration. Calculate the average weight of soil for a hectare area per unit depth. 26

27. Cont… Mass of Dry Soil = Soil Volume x Bulk Density Volume = Area x Depth Mass of Dry Soil = Area x Depth x Bulk Density Example: A given 1ha area and 15cm depth has a bulk density of 1.3g cm -3 .What is the average mass of soil in this 1ha in Kg? Solution: mass of dry soil = Area x Depth x Bulk density = 1ha x 15cm x 1.3 g cm -3 = 10,000m2 x 0.15m x 1300 Kg m -3 = 1,950,000 Kg ≈ 2,000,000Kg. Calculation of volumetric water content and storage capacity per unit volume of soil: θ v = θ m x ρ b / ρ w , where θ v = volumetric water content (L 3 /L 3 ), θ m = gravimetric water content (g/g or kg/Kg), ρ b and ρ w = soil bulk density and water density, respectively. Gravimetric water content, θ m = mass of water/mass of oven dry soil 27

28. Cont… Example: The weight of wet soil was 100 g. After drying the sample in an oven for 24 hours at 105 ℃, the sample weight was 80 g. calculate: 1. The gravimetric water content 2. The volumetric water content assuming that the bulk density is 1.2 g cm -3 . 3. The height of water corresponding to 1cm 3 volume of that soil. Solution: θ m = (100-80)/80 = 0.25 or 25% θ v = (0.25 x 1.2)/1 = 0.3 or 30% Assuming the dimension of the soil sample to be 1 cm x 1 cm x 1 cm, the volume = 1 cm 3 and area = 1cm 2 28

29. Cont… θ v = 0.30 means 0.30 cm 3 of water in 1 cm 3 of soil. And the height of water corresponding with that volume will be divided by the area, which will be: 0.30 cm 3 / 1 cm 2 = 0.30 cm of water per 1 cm 3 of soil. In general, the soil bulk density can be affected by: Pore space, Texture of soils, Organic matter, Soil depth Arrangement of particles and Compaction. 29

30. Cont… Specific gravity (SG) is the ratio of the density of a substance to the density of a reference substance; equivalently, it is the ratio of the mass of a substance to the mass of a reference substance for the same given volume. Apparent specific gravity is the ratio of the weight of a volume of the substance to the weight of an equal volume of the reference substance. The reference substance for liquids is nearly always water at its densest (at 4 °C or 39.2 °F); for gases it is air at room temperature (20 °C or 68 °F). Nonetheless, the temperature and pressure must be specified for both the sample and the reference. Pressure is nearly always 1  atm (101.325  kPa ). Being a ratio of densities, specific gravity is a dimensionless quantity . Specific gravity varies with temperature and pressure; reference and sample must be compared at the same temperature and pressure or be corrected to a standard reference temperature and pressure. Substances with a specific gravity of 1 are neutrally buoyant in water. Those with SG greater than 1 are denser than water and will, disregarding surface tension effects, sink in it. Those with an SG less than 1 are less dense than water and will float on it. 30

31. Cont.. In scientific work, the relationship of mass to volume is usually expressed directly in terms of the density (mass per unit volume) of the substance under study. It is in industry where specific gravity finds wide application, often for historical reasons. True specific gravity can be expressed mathematically as: Where ρ sample is the density of the sample and ρ H2O is the density of water. The apparent specific gravity is simply the ratio of the weights of equal volumes of sample and water in air: Where W A,sample represents the weight of the sample measured in air and W A,H2O the weight of water measured in air. It can be shown that true specific gravity can be computed from different properties: where g is the local acceleration due to gravity, V is the volume of the sample and of water (the same for both), ρ sample is the density of the sample, ρ H2O is the density of water and W V represents a weight obtained in vacuum. 31

32. 32 2.3 Soil Total Porosity ( ƒ ) and Pore-size Distribution The pore space of a soil is that portion of the soil volume occupied by air and water which  contain about 50% porosity unless compacted Voids consist of that portion of the soil volume not occupied by soil solids, either mineral or organic. Pores occur in a range of sizes, each having different effects on soil properties (‘Tortuous pathways’ best describe soil pores) Under field conditions, pores are occupied at all time by air and water. Soil particles have irregular shapes and thus leave the spaces of pores between them very irregular in shape, size and direction. Sands, for example, have large and continuous pores whereas clays have more total pore space. The amount of this pore space is determined largely by the arrangement of the solid particles. In particles that lie close together as in sands or compact subsoil, the total porosity is low whereas in those particles arranged in porous aggregates, the total pore space per unit volume will be high.

33. 33 a) Size of Pores: the following size groups can be identified Coarse (Macro) pores:   the size of these pores is >0.1 mm in diameter, and their main function is aeration and drainage by gravity They are visible to the naked eye. Measures of the coarse porosity are of some value as indirect estimates of likely water-logging and consequent anaerobium within profiles. Coarse porosity (% by volume) can be calculated as the difference between the total porosity (% by volume) and volumetric water content at field capacity (FC). allow for ready movement of air and water in soil Accommodate plant roots and small soil animals 2. Medium ( Meso ) pores : their size ranges between 0.03-0.1mm in diameter and their main function is conduction of water by rapid capillary flow 3.Fine (Micro) pores: <0.03 mm in diameter and their function is water retention and slow capillary flow.

34. Cont… Air does not move through these pores easily Almost always water filled, but little is available to plants Estimating Porosity from Densities Bulk density is easy to measure, and particle density can usually be assumed to be 2.65g/cm 3 for a mineral soil. % solid space = (Volume of solids /Total volume) x100 % % solid space = (Vs/ Vt ) x 100 solid space=(Bulk density/Particle density)x100 = (Db/ Dp ) x100 Since % pore space + % solid space = 100 → % pore space = 100 - % solid space = 100- (Db/Dpx100) ⸪ % porosity ( ƒ )  =  34

35. Cont… Percent porosity is useful, in combination with bulk density, for evaluating:    soil compaction,  maximum water holding capacity and suitability for agriculture. The amount of pore space and the proportion of pore space is very important. Therefore, heterogeneous, continuous (up to the rooting depth) and permanent pore spaces are critical for normal plant growth and development. Air-filled Porosity ( fa ): It is a measure of the relative air content of the soil. An estimate of their volume is an indication of the aeration and drainage status of a soil. fa = Va / Vt = ( Vt - Vw -Vs)/ Vt Void Ratio (e): the void ratio is also an index of the fractional volume of soil pores, and it relates that volume to the volume of solids rather than to the total volume of a soil. e = ( Va + Vw )/Vs = Vv/( Vt -Vv) e = f/(1-f) and f = e/(1+e), where Va is volume of air, Vw is volume of water, Vs is volume of solid, f is total porosity, e is void ratio, Vv is volume of void and Vt is the total volume of soil 35

36. 36 Factors influencing total pore space a) Texture : sandy surface soils have total pore space 35-50% and for medium to fine-textured soils 40-60% or even more in cases of high organic matter and marked granulation

37. Cont… b) Depth : for some compact subsoils , it may drop as low as 25 to 30 percent c) Soil Structure : cropping tends to lower total pore space below that of the virgin or uncropped soils, which usually is associated with a decrease in organic matter content and a consequent lowering of granulation. d) Bulk Density : as soil bulk density increases mainly due to compaction, total pore spaces decreases or vice versa e) Organic Matter : as the contents of soil organic matter increases, total pore spaces also increases. Soil particles are generally bound or cemented together by various substances including organic matter, colloidal clay, iron, aluminum and other hydrous oxides, and by flocculating action of certain salts. 2.4.1. Classification of Soil Structure Soil may be either:1. Structureless or 2. Structured 37

38. 38 2.4. Soil structure It is the arrangement of soil primary particles into groupings or Peds .

39. Cont… Structureless soils have no recognizable natural aggregates . Structureless material may be: a) Single grained Single grained material results when individual soil particles show little or no tendency to adhereto other particles, an example being coarse-textured sandy soils. b) Massive Massive materials result when the soil is composed of finer particles such as silts and especially clays. The natural cohesiveness of these size separates binds the individual particles together. The massive materials show little or no tendency to break apart under light pressure into smaller structural units. Structured soils (soil structure) can be classified based on: 1.Type (form) of Structure : this is determined by the shape and arrangement of peds . 39

40. Cont… 2 . The Size (class) of Structure : as differentiated by the size of the peds . 3. The Grade of Structure : as determined by the distinctness (differential between cohesion within peds and adhesion between peds ) and durability of the peds . .It relates to the degree of aggregation or the development of soil structure. In the field, a classification of grade is based on a finger test (durability of peds ) or a crushing of a soil sample. 1. Type of Soil Structure The resulting aggregates are the types of ped (natural aggregates) defined by their shape as granular, platy, blocky, prismatic, columnar, and structure less. However the four principal types of soil structure are: Spheroidal (granular and crumb), Prism-like (columnar and prismatic), plate, (platy and laminar)and block-like (angular blocky and sub-angular blocky). 40

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45. 45 Ped Development • The shape of  peds reflect the relative stresses placed on the material – Platy: Vertical stress greater than horizontal – Blocky: Equal stress – Prismatic: Horizontal greater than vertical Stresses come from weight of material above and lateral pressure from swelling of clays, inter- ped gaps form when soil dries (Clays and organic matter fall into gaps and coating)  2. Class of the Structure The various sizes of peds are designed by class of soil structure as follows:

46. Classification of soil structure considering size and type of peds Size Block-Like (angular and sub-angular blocky structure) [mm] diameter Spheriodal (Granular and crumb structure) [mm] diameter Plate-like (Platy structure) [mm] width Prism-Like (prismatic and columnar structure) dia. Very fine < 5 < 1 < 1(very thin <10 Fine 5-10 1-2 1-2 (thin) 10-20 Medium 10-20 2-5 2-5 20-50 Coarse 20-50 5-10 5-10 (thick) 50-100 Very Coarse > 50 > 10 >10 (very thick) > 100 46

47. 3. Grade of Soil Structure The degree of aggregation of soil structure is the distinctness or durability with which the peds are observed, namely weak , moderate and strong . 47 Grade Abbreviation Description Structure less No observable aggregation or no orderly arrangement of material lines or weakness Weak 1 Poorly formed indistinct peds Moderate 2 Well formed distinct peds , moderately durable and evident, but not distinct in undisturbed soil. Strong 3 Durable peds that are quite evident in undisplaced soil, adhere weakly to one another, withstand displacement, and become separated when soil is disturbed

48. Cont… The three characteristics of soil structure are conventionally written in the order grade, class/size and type/shape or form . For example, weak fine sub-angular blocky structure , which refers to weak-grade , fine-class/size and sub-angular blocky- type/shape . 2.4.2. Effects of soil structure on soil conditions and Properties Soil structure affects the pattern of peds . This in turn affects the pore size distribution, which greatly influences water movement, heat transfer, and aeration. Moreover, structure modifies the influence of soil texture with regard to water and air relationships and the ease of root penetration. Activities such as timber, grazing, tillage, trafficking, drainage, liming and manuring impacts soils largely through their effect on soil structure, especially in the surface horizons 48

49. 49 2.4.3. Factors affecting the development of soil structure 1. Climatic influences : Alternating wetting and drying of soils have been observed to be responsible for the aggregation of soil materials into clods of varying sizes. The soil moisture condition is also important in soil aggregation. Generally speaking, soil peds are more well formed and stable in sub humid and semi-arid soils, (example chernozems and prairie soils) than in wet ( pedozols ) or desert soils ( sierozems ). In general, the best structure is that which increases the aeration and the water holding capacity of the soil, and facilitates the activities of micro-organisms. 2. Vegetation During growth, plant roots pushing through the soil tend to compress soil particle into small aggregates and break the larger aggregates present in the soil. Root exudates help to bind soil particles together into aggregates

50. Cont… The dehydration of the soil in the vicinity of the root system as water is absorbed by the plants causes local shrinkage and formation of surface fracture. 3. Soil Fauna and Microorganisms The beneficial effects on soil aggregation and its stability originate from the integrated activity of soil fauna, microorganisms and vegetation. Fungi and actinomycetes cause mechanical binding of the aggregates by the mycelia they produce The metabolic processes of the microorganisms synthesize complex organic molecules, which have cementation effect. Thus, the microbial products like bacterial gum and microbial polysaccharides play a major role in soil aggregate stabilization 50

51. Cont… Products of microbial decomposition of organic materials added to soil such as humic acids, colloidal proteins and cellulose materials also produce stabilizing effect. 4. Cation Effects The cations adsorbed by the soil colloids greatly affect aggregate formation. If sodium is the dominant cation adsorbed , the soil particles do not come together to form structural aggregates but they repel each other causing dispersed conditions. The divalent cations like Ca 2+ and Mg 2+ , on the other hand form electropositive links between the electronegative soil particles causing the soil colloids to come together and form aggregates called floccules. 51

52. 5. Organic Matter Organic matter stimulates the formation and stabilization of granular and crumb type aggregates. Organic compounds, such as polysaccharides, chemically interact with the clays and form bridges between the individual soil particles and bind them together in water-stable aggregates. 6. Clay particle and Clay-Organic Interaction Three distinct groups of colloidal materials have cementation and aggregating effect-clay particles themselves, inorganic colloids of iron and aluminum oxides and organic colloids. Clay-to-clay interaction takes place due to electrostatic and Van Dar waal’s forces leading to the formation of secondary particles. 52

53. Cont… The linkage established by polyvalent exchangeable ctions on the clay particles influence the formation of aggregates. Also polyvalent cations serve as bridges to form clay-organic complexes leading to the formation of stable aggregates. 2.4.4. Management of Soil Structure Practical management of soil structure is generally restricted to the topsoil or plow layer in regard to use of soils for plant growth. A stable structure at the soil surface is known to encourage more rapid infiltration of water during storms and results in reduced water runoff and soil erosion. This allows the soils to store more water within the profile. 53

54. Cont… The stability of peds or aggregates depends on their ability to retain their shape after being subjected to rain drop impact and tillage. In addition, for the peds to be permanent, the particles in the peds must hold together when dry peds are wetted or resist the pressure from entrapped air. Soil management practices that result in frequent additions of organic matter will tend to produce more microbial gum and increase ped formation and stability. Practices that result in keeping a vegetative cover to the management of soil structure are: Restricting tillage activities to optimum soil moisture conditions to ensure least destruction in the soil structure Adoption of suitable tillage or minimum tillage to reduce the loss of aggregates stabilizing organic matter 54

55. Cont… Covering the soil surface with crop residues or plant litter to protect the aggregates from the beating action of rain and to add the organic matter; Incorporation of organic materials (plant residues, animal manures, etc) that supply decomposition products that stabilize soil aggregates (polysaccharides, colloidal proteins, cellulose) Adoption of suitable cropping systems that increase and maintain the organic matter level of the soil ( eg . Application of fertilizers, inclusion of grasses, sods in the rotation, etc); Remember that green manuring and cover crops are good sources of organic matter. Since, each soil has distinct problems , it requires a specific soil and crop management practices for soil structure management. 55

56. 56 2.5. Soil consistency 2.5.1. Description and Significance It is a term used to describe the resistance of the soil to deformation or rupture. It is determined by the cohesive and adhesive properties of the entire soil mass. The following consistency terms are loose, friable, firm & extremely firm. Loose

57. 57 friable firm extremely firm

58. 58 Consistency is described for three moisture levels as: A. Wet soil consistence: when moisture content is at or slightly more than field capacity . B. Moist soil consistence : when moisture content is between field capacity and permanent wilting point. C. Dry soil consistence : when moisture content is below permanent wilting point. Table: Consistency of a soil is commonly associated with the textural class as follows:

59. Cont… The stickiness describes the quality of adhesion to other objects. The plasticity is the capability of being moulded by hands (malleability) When recording consistence, it is important to record the moisture status as well (wet, moist and dry soil consistence) Classification of consistence and description 59 Moisture status Consistence Description Wet Non-sticky Almost no natural adhesion of soil material to finger Slightly sticky Soil material adheres to only one finger sticky Soil material adheres to both fingers Very sticky Soil material strongly adheres to both fingers

60. wet Non-plastic No wire is formable by rolling the soil between the hands Slightly plastic Only short (<1cm) wires formed by rolling b/n hands Plastic Long wires (>1cm) formed and moderate pressure needed to deform a block of a moulded material Very plastic Much pressure needed to deform a block of mouleded Moist Loose Soil material is non-coherent Very friable Aggregates crush easily between thumb and finger Friable Gentle pressure is required to crush aggregates Firm Moderate pressure is required to crush aggregates Very firm Strong pressure is required to crush aggregates Extremely firm Aggregates can not be broken by pressure Dry Loose Non-coherent Soft Breaks under slight pressure b/n thumb and forefinger Slightly hard Breaks under moderate pressure Hard Breaks with difficulty Very hard Very resistant to pressure; cannot be broken b/n thumb and forefinger Extremely hard Extreme resistance to pressure; cannot be broken in the hand 60

61. 61 2.6. Soil color Color is the most obvious and easily determined soil property. It is mostly used in differentiating soil horizons and soil classification and can be used to determine other soil chemical properties. It can be described or represented by hue, value and chroma .

62. 62 The color is NOTED as: hue, value / chroma

63. 63 Factors influence soil color 2.6.1. Parent material from which the soil was derived Most soil minerals, such as quartz and feldspars are light in color.

64. 64 2.6.2. Organic matter content.

65. 65 2.6.3. Water content

66. 66 2.6.4. Presence of oxidation of (Fe and Mn) example red/brown colors suggest the presence of oxidized iron oxides, whereas blues chemically reduced conditions, carbonates whitish color.

67. 67 Assignment One 1. Explain the difference on water holding capacity of soils in terms of their textural composition. 2. The volume of solid and pore spaces of a certain amount of soil in the field is 1 cm 3 that weighs 1.33 g. Calculate: a) The bulk density of the soil b) The particle density of the soil if 50% compaction occurred to the soil. 3. A bulk and particle densities of a sand soil are 1.5 and 2.65 g cm -3 respectively. Calculate the total percentage of pore space. 4.Two different timber-harvest methods are being tested on adjacent forest plots with clay loam surface soil. Initially, the bulk density of the surface soil in both plots was 1.1 Mgm -3 . One year after the harvest operations, plot A soil had a bulk density of 1.48Mgm -3 , while that in plot B was 1.29Mgm -3 .Interpret these values with regard to the relative merits of systems A and B, and the likely effects on the soil’s function in the forest ecosystem.

68. 68 5. For the forest plot B in question number 4, what was the change in percent pore space of the surface soil caused by timber harvest? Would you expect that most of this change was in the micropores or in the macropores ? Explain. 6. What are the textural classes of two soils, the first with 15% clay and 45% silt, and the second with 80% sand and 10% clay? 7. You are considering the purchase of some farmland region with variable soil textures. The soils on one farm are mostly sandy loams and loamy sands, while those on a second farm are mostly clay loams and clays. List the potential advantages and disadvantages of each farm as suggested by the texture of its soils. 8. Let the soil has 160g weight after it is dried in an oven, in a core sampler with 95cm 3. Determine Db and % ƒ of the soil. 9. Voids between sands promote free drainage and entry, because of their large size and low specific surface area.

69. 69 10. Explain the effect of textural composition on the consistency of the soil? 11. What are the factors influencing total pore space of a soil? 12. A soil is sampled by a core measuring 7.6cm in diameter and 7.6cm deep. The core weight is 292g and tin weight of 8g. The total core plus wet soil plus tin is 1000g. After oven dry at 105 c the total dry soil weighed 860g. Calculate the gravimetric moisture content and bulk density. 13. A soil in the green house container has a dry bulk density of 1.4g/cm 3 and pore space of 0.35 by volume. Calculate the particle density of the soil sample. 14. Determine the soil textural class of a soil sample with a composition of 18g clay, 38g sand and 41g silt using the textural triangle. 15. Increased in volume of the soil suspension of a soil sample was 20cm 3 when 50g of dry soil was mixed in a known volume of water. calculate bulk density of the soil.

70. 70 CHAPTER THREE 3. SOIL WATER It is the water that may be evaporated from a soil by heating to between 100 and 110 ℃ until there is no further weight loss. Characterization of soil water can be done in two ways: 1. The soil moisture content, refers to soil wetness 2. The physicochemical condition or state of soil water in terms of its free energy per unit mass called potential . 3.1. The Soil-Water Content (Mass and Volume Ratios) Described as per mass or per volume fraction of soil wetness 1. Mass Wetness : θ m = Mass of water/Mass of oven dry soil θ m = M w /Ms 2. Volume wetness: θ v = V w / V t = V w /(V s + V w + V a ) 3. Relationship b/n the Two: θ v = θ m ρ b / ρ w = θ m Г b , where, Г b = bulk specific gravity of the soil(a dimensionless ratio which usually lies in the range between 1.1 and 1.7), ρ b = density of soil, ρ w = density of water

71. 71 Limitations: Works relatively better for non-swelling soils in which the bulk density, and hence bulk specific gravity, are constant regardless of wetness. For swelling soils, the bulk density must be known as a function of mass wetness. The wettest possible condition of a soil is called saturation-all the soil pores are filled with water. Saturation is easy to define for coarse-textured soil but difficult for fine-textured (especially swelling) soils, for these soils may continue to imbibe water and swell even after all pores have been filled with water. The lowest wetness in nature (field condition) is air dryness, and in the laboratory is known as oven-dry condition. 3.2. Soil-Water Content Measurement Direct methods : involves removal of soil water by : Evaporation, leaching, or a chemical process and subsequent determination of the amount of water removed.

72. Examples: Thermogravimetric /gravimetric method Chemical method-calcium carbide (Ca 2 C) method Feel or appearance method Bouyoucos’s alcohol-burning method Removal by distillation/absorption in a desiccant and measuring of the quantity removed Indirect methods: Depending on monitoring of some soil property that is function of water content. Examples: The neutron probe method /neutron scattering Gamma-ray attenuation Capacitance TDR methods 72

73. 73 1. Gravimetric/ Thermogravimetric method (Direct method) Involves the following procedures: Removing moist sample by auguring Determining the moist weight Dry in an oven set at a temperature of 100-110 o C Removing the samples and allowing them to cool in a desiccator Reweighing the sample until the sample attains a constant weight A constant weight can be assumed to have been achieved when the changes between repeated weighing, are less than 0.1% of the original sample weight. The interval between weighing should be at least half an hour. In some instance a period of 4 hrs should elapse between successive weighing.

74. Advantages of the thermogravimetric method include: Simplicity Reliability Low cost in terms of equipment requirements Disadvantage of this method include: Destructive sampling The time that the procedure takes. Errors come from: Oxidation of some organic matter Presence of appreciable amount of adsorbed water at 105 o C Examples on calculation of soil-water content The following data were obtained before and after irrigation. From these data, calculate the mass and volume wetness values of each layer and determine the amount of water (in mm) added to each layer and to the profile as a whole. 74

75. 75 Sampling Time Sample Number Depth (mm) Bulk density (g/cm3) Wet mass (g) Dry mass (g) Before Irrigation 1 0-400 1.2 110 100 2 600-1000 1.5 96 80 After Irrigation 1 0-400 1.2 180 150 2 600-1000 1.5 156 120

76. 76 solution: 1. Mass wetness θ m = wet mass – dry mass/dry mass Before irrigation sample 1. θ m = 110-100/100 = 10/100 = 0.10 sample 2. θ m = 96-80/80 = 16/80 =0.20 After irrigation sample 1. θ m = 180-150/150 = 30/150 = 0.20 sample 2. θ m = 156-120/120 = 36/120 = 0.30 2. Volume wetness θ v = θ m ρ b / ρ w , where ρ w = 1 Before irrigation sample 1. θ v = 0.1x1.2 = 0.12 sample 2. θ v = 0.2x1.5 = 0.30 After irrigation sample 1. θ v = 0.2x1.2 = 0.24 sample 2. θ v = 0.30x1.5 = 0.45

77. 77 Depth of water = Volume wetness x equivalent depth of soil d w = θ vi x Z i 1. Before irrigation sample 1. d w = 0.12x400 mm = 48 mm sample 2. d w = 0.30x400 mm = 120 mm Total water = 48 + 120 = 168 mm, ( from 0-1000mm depth) 2. After irrigation sample 1. d w = 0.24x400 mm = 96 mm sample 2. d w = 0.45x400 mm = 180 mm Total depth of water after irrigation = 96 mm + 180 mm = 276 mm Water added to each layer and the profile as a whole 1. water added to each layer a) Layer 1. Added water = 96 mm – 48 mm = 48 mm b) Layer 2. Added water = 180 mm – 120 mm = 60 mm c) Total water added = 48 mm + 60 mm = 108 mm

78. 2. Indirect methods Neutron Method ( Neutron Scattering Method) Makes use of the ability of hydrogen to slow down fast neutrons, emitted from a radioactive source, much more efficiently than other substances. Source of the fast neutrons and a slow neutron detector are housed in one unit called probe. The probe is lowered down a tube known as an access tube inserted into the soil. Fast neutrons are moderated (i.e., they are slowed down to “thermal” energies) upon collision with atomic nuclei, in particular hydrogen nuclei. 78

79. Benefits of neutron probe: Quick Once the tube is installed, the same location can be used repeatedly with time, there by providing a “standard” location. Fairly repeatable Drawbacks: Expensive Probe must be calibrated for each soil /access tube type Top 15 cm are difficult to measure due to escapage of neutrons into the atmosphere Slight radiation exposure danger Difficulty with maintenance, repair or spare parts in developing countries 79

80. 80 3.3 Energy State of Soil-Water The concept of energy is the most important concept in nature. Energy is a fundamental entity common to all forms of matter in the physical world. For instance, water retention and movement in soil is a consequence of energy effects (differences in energy cause these processes to happen) Energy is defined as the capacity of doing work-closely associated with the concept of work: Work done = Force x Distance ⇒ W = F x h According to Newton’s 2 nd law of motion, when a body is acted upon by a constant force (F), its resultant acceleration (a) is in the direction of the applied force and , hence, is given by: a = F/m ⇒ F = a x m Therefore, by Newton’s 2 nd law motion, the force required to lift the mass, m, is equal to its own weight : F = mxg , where g is the acceleration of gravity. Substituting the weight mg for F gives the following equation: w = mg x h

81. a------------ Soil Column h Reference plane m Example 1: If a 5 g mass of water is raised vertically to a distance of 1 cm, the work done will be: solution: W = 5gx1cmx980cm/sec 2 = 4,900 dynes-cm or ergs. 2. Calculate the work done in moving a mass of 2 kg of water to a distance of 5 m above the reference level. Solution: W = 2kgx5mx9.8m/sec 2 = 98 kgm 2 sec -2 = 98 Nm = 98 J 81 mm

82. Potential Energy A body is said to have potential energy, Ep , if by virtue of its position or state it is able to do work. By definition: Ep = F x h and Ep = m x g x h Example: A mass of a 5 kg of water is raised to a height of 2 m above the reference level. Calculate its potential energy. Solution: Ep = 5kgx9.8m/sec 2 x2m = 98Nm = 98 Joule Potential energy can be Positive, Zero or Negative Consider the figure below, above the reference level at position a, the element of water has positive potential energy while at point b, below the reference level, it has negative potential energy. At the reference level, it has zero potential energy. 82

83. When the mass m of water is raised from the reference level to a , work is done on the mass, m, and the mass acquires potential energy equivalent to mgh 1 In returning from a to reference level the mass of water looses the potential energy in performing the work equal to mgh 1 . Similarly, in going from the reference level to b, the mass of water looses the energy and ends up at b with mgh 2 less energy than it had at the reference level. To raise it again at the reference level an equivalent amount of work mgh 2 will have to be done. a Ep = +mgh 1 Reference Plane h 1 b h 2 Ep = -mgh 2 Figure: variation of potential energy above and below reference 83

84. Soil-water potential: Because the velocity of flow of water in soil is often very low, thus kinetic energy, E k , is negligible and its potential energy, E p , is primarily responsible for determining the state and movement of water in soil. E p varies considerably from point to point. This difference gives rise to the tendency for water to flow. The water flows from the point where the potential energy is higher to where it is lower in the direction of greatest decrease in potential energy. The potential energy is acquired by the water whether the work is done against gravitational, elastic, osmotic, electric or magnetic force The term soil-water potential denotes the specific potential energy of soil-water to that of water under standard reference state. 84

85. Total Soil-Water Potential and its Components The amount of work that must be done per unit quantity of pure free water in order to transport reversibly and isothermally an infinitesimal quantity of water from a pool of pure water at a specified elevation at atmospheric pressure to the soil water (at the point under consideration). It is impractical to make measurements specified in this formal definition. The definition implies that the soil-water potential is the difference between the energy state of soil-water and that of pure free water. The standard state generally used is that of a hypothetical reservoir of pure free water at: Atmospheric pressure Same temperature at that of soil water At a given and constant elevation 85

86. The selection of reference level is purely arbitrary. In any soil-water system the zero energy level is known as reference state. The conventional characteristics of reference state of water are the following: Pure water : the water at reference state has no dissolved gasses or solids and free of any other impurities Atmospheric pressure : standard pressure or standard atmosphere is used. 1 atmosphere = 1.013x10 6 dynes cm -2 = pressure exerted by a column of water of height, h = 1035 cm Zero hydrostatic pressure : the reference level is considered to be free flat surface with zero hydrostatic pressure. Zero elevation : all points which are measured in the soil water system, should be from this reference level. 86

87. The soil water potential is due to several forces, each of which is a component of the total soil water potential,  t . These components are due to differences in energy levels resulting from gravitational potential ( z ) , matric potential ( m ), submerged hydrostatic/pressure ( s ) and osmotic potential ( o ) All of these acts simultaneously to influence water behavior in soils.  t =  z +  m +  o +  s + … (…) indicates the possible contribution of additional potentials not yet mentioned (less significant potentials). 87

88. 88

89. 89 1.Gravitational Potential The force of gravity acts on soil-water the same ways as it does on any other body, the attraction being toward the earth’s center. The gravitational potential, Ψ z, of water at any point in a gravitational field may be defined as the work done, w, or potential energy stored Ep , per unit mass in carrying any mass m, from infinity up to that point Ψ z = work/mass = Ep /m, where Ψ z is gravitational potential in erg/g The gravitational potential or work done per unit mass of water to raise it from the reference level to a point above the reference will be: Ψ z = mgz /m = gz This shows that if the gravitational potential is expressed on per unit mass basis, it is linearly related to elevation, z, of the point above the reference plane. The slope of the line is simply g. At a height z above a reference, the gravitational potential energy Ψ z as a mass m of water occupying a volume v is:

90. 90 Ψ z = work/v = mgz /v = ρ w gz where, g = acceleration of gravity z = the height of the soil water above the reference elevation ρ w = density of water In terms of potential energy per unit of weight: Ψ z = mgz /mg = z In practice , Ψ z, is the difference in elevation of the point in question and the reference point. Therefore, Ψ z is POSITIVE if the point in question is ABOVE the reference and is NEGATIVE if the point in question is BELOW the reference Thus, the Ψ z is independent of soil properties (chemical and pressure conditions of soil-water). Ψ z depends only on the vertical distance (relative elevation) between the reference and the point in question.

91. 91 Example: Two points in a soil, each point is located a specified vertical distance from a reference elevation (i.e., a water table). Find the difference in gravitational potential between the two points. _____________ Solution: Ψ zA = +15 cm A Ψ zB = -10 cm 15 cm △ Ψ z = 15 – (-10) _______________ Reference △ Ψ z = +25 cm 10 cm B _______________

92. 92 The reference elevation can be arbitrarily chosen. This makes the absolute magnitude of the gravitational potential almost meaningless. We are, however, usually interested in the difference in potential between two points, in which case it makes no difference where the reference is chosen. Example: The same two points in the previous example but with the reference elevation located as shown in the figure (rising water table). Find the difference in Ψ z between the two points. ۰ A ____________ Solution : Ψ zA = +5 cm 5 cm Ψ zB = -20 cm _______reference △ Ψ z = Ψ zA - Ψ zB △ Ψ z = 5-(-20) = +25 cm 20 cm ۰ B

93. 93 Example 2. Assume that a piezometer (an open-ended cylindrical tube) has been inserted into a soil profile and that this tube contains water at 0.90 m depth. On a mass, volume, and weight basis, compute the gravitational potential, assuming that the reference level is: 1. located at the soil surface 2. At the depth of the water table 3. At a drain that is located at a depth of 1.2 m Computational formula: 1. Ψ z,m = mgz /m = gz 2. Ψ z,v = mgz /v = ρ w gz 3. Ψ z,w = mgz /mg = z

94. Piezometer ___________ _________________Reference 1 (Soil surface) ______________________________ reference 2 (water table) Water ___ z = 1.2 m (reference 3) Drain Figure. Schematic of a piezometer inserted into a soil profile containing a drain at 1.2 m depth 94

95. position Basis of Expression Mass (J/kg) Volume (Pa) Weight (m) 1. Reference level at the soil surface Soil surface 0.00 0.00 0.00 Water table -8.82 -8.82 -0.90 Drain -11.76 -11.76 -1.2 2. Reference level at water table Soil surface 8.82 8.82 0.90 Water table 0.00 0.00 0.00 Drain -2.94 -2.94 -0.30 3. Reference level at the drain Soil surface 11.76 11.76 1.2 Water table 2.94 2.94 0.30 Drain 0.00 0.00 0.00 95

96. 96 2. Matric Potential Results from the affinity of the water to the whole matrix of the soil, including its pores and particle surfaces together. Consider the following conditions: 1. If hydrostatic pressure is greater than atmospheric pressure-positive pressure potential will be resulted 2. If soil-water pressure is lower than atmospheric pressure-the pressure potential is negative (tension or suction) water under a free water surface- positive pressure potential Water at free water surface-zero pressure potential Water risen in a capillary tube above the free water surface-negative pressure potential The negative pressure potential has been called matric potential resulted from adsorption of water molecule to soil solids and capillary (retention of water in the capillary pores)

97. 97 If the unit water is expressed as a weight, the Ψ m at a point is the vertical between the point in the soil and the water surface of a manometer filled with water and connected to the soil via a ceramic cup. _____ _________________ _________ Ψ m = 15 cm ____________________ Figure. An unglazed ceramic cup embedded in soil is connected to a water manometer to form a tensiometer .

98. 98 The weight matric potential of the soil water at the cup is the vertical distance from the center of the cup to the water level in the manometer, Ψ m = mgh /mg = h For the situation illustrated Ψ m = -15 cm. Alternatively, matric potential can be measured from relative vapor pressure as: where: e = vapor pressure of the soil water e o = vapor pressure of pure water at the same temperature R = universal gas constant (8.314 J/K.mol) T = Kelvin temperature M = mass in kg of a mole of water (0.018015)

99. 99 Example: Assume that the relative vapor pressure of water in soil at a temperature of 20 ℃ is equal to 0.85. Calculate the matric potential of the soil water. Solution: Ψ m = (8.314 x 293/0.018) x ln (0.85/1) Ψ m = -21,989 J/kg = -21,989 kPa = -2,249.5 m, Note: 1kpa = 0.1023m heads of water 3. Submerged Hydrostatic Pressure Potential, Ψ s Is due to the weight of water at a point under consideration Under field condition, it implies mostly to saturated soils The reference for it is atmospheric pressure Water under a free water surface has a positive pressure potential At the free water surface, the pressure potential is zero It is the vertical distance from the point in question in the soil profile to the open water surface of a piezometer connected to the point in question. The submerged pressure potential per unit volume of soil water is: Ψ s = ρ w gh , where h is the submergence depth below the free water surface (water Table)

100. 100 Similarly, pressure potential per unit weight is: Ψ p = h ______ _____ ______ Reference (water table, free Water Surf.) Ψ s = 10 cm _______ The pressure potential results from a net pressure difference of Hydraulic pressure in a saturated soil Pneumatic pressure in a pressure membrane apparatus Turgid pressure in plant cells NB:It does not include any effects that result from a curved air-water interface or adsorption to a solid

101. 101 Reason: where the system is open to the atmosphere, there is no overall pressure difference at the interface Example: Calculate the pressure potential distribution in the profile of a soil with a water table at the 60 cm depth. Solution: Soil depth (m) Submerged Pressure potential (m) 0.2 0.4 0.6 0.8 0.2 1.0 0.4

102. 102 4 . Osmotic Potential The osmotic potential is attributable to the presence of solutes in the soil-in other words to the soil solution. The solutes may be ionic or nonionic, but their net effect is to reduce the free energy of water, primarily because the solute ions or molecules attract the water molecules The magnitude of the osmotic potential can be calculated by means of the following equation: ∏ = - iCRT ∏ = osmotic pressure ( atm ) R = universal gas constant (0.082L.atm.mol -1 K -1 ) T = Kelvin temperature (273 + ℃) C =concentration of salt solution (mol/L) i = total number of + and - charges

103. 103 Example: Calculate the osmotic potential of a solution containing 0.01N NaCl at 25 ℃. Solution: ∏ Na+ = (-0.082x298x0.01)x1 = -0.25 atm ∏ Cl - = (-0.082x298x0.01) x1 = -0.25 atm ∏ Na + + ∏ Cl - = -0.50 atm Since the osmotic potential of pure free water is zero, the osmotic potential of a soil solution at the same temperature of free water is negative. Or ∏ NaCl = -2(0.082x298x0.01) = -0.50 atm = -517 cm water ionization of the various salts varies with concentration and also in the presence of other salts, it is difficult to determine accurately the concentration of each solute species in soil solution. Osmotic potential of soils cannot be accurately determined by taking the product of the osmotic pressure and the number of species and summing overall species.

104. 104 V an’t Hoff’s formula shows that the osmotic potential varies with the chemical composition of the solution and the variability is based on the number of ions produced per formula weight of salt and not the weight of dissolved solutes alone. Therefore, two other approaches have been suggested. Use of the linear regression between osmotic potential and the total dissolved solids (TDS) Ψ o = -0.056 x TDS, where TDS is in mg/liter and Ψ o is in kPa 2. Relate EC to Ψ o as: Ψ o = -0.36 x EC , where EC is in dS /m the magnitude of the Ψ o is dependent only on the concentration of the solution and on the properties of the solvent When the unit quantity of water is expressed in terms of mass, Ψ o, becomes: Ψ o = (RT/M) ln (e/ e o )

105. 105 Where, R = gas constant (8.314 J/K mol), T = Kelvin temperature, M = partial molal volume (kg/mol), e = equilibrium vapor pressure for the solution and e o = equilibrium vapor pressure for the pure solvent. When Ψ o is expressed per unit volume of water Ψ o = (RT/V) ln (e/ e o ), where V is the molar volume (the volume of 1mol of water is 18 cm 3 ). 3.4. Soil Water Retention Curve A particular soil water potential is related to the soil water content The relationship between water content and matric potential when plotted is called water retention curve (also called soil-water characteristic curve) If we apply a progressively increasing suction to a saturated soil, more water is drawn out of the soil and more of the relatively large pores that cannot retain water against the suction applied, will empty out.

106. 106 Recalling the capillary equation (h = 0.15/r), a gradual increase in suction will result in emptying of progressively smaller pore, until at high suction values, only the very narrow pores retain water. An increase in soil water suction is also associated with a decreasing thickness of the hydration envelopes covering the soil particles surfaces Increasing suction is therefore associated with decreasing soil wetness with the amount of water remaining at equilibrium being a function of the sizes and volumes of the water-filled pores and hence it is a function of the matric potential Characteristic points on the soil-water retention curve: Two points are important on SWRC 1. Field Capacity (FC) and 2. permanent wilting point (PWP)

107. 107 1.Field capacity: is the term used to described the maximum water content that the soil will hold following free drainage. It, therefore, doesn’t correspond to a fixed matric potential, but instead represents the condition of each individual soil after the larger pores have drained freely under gravity. Factors affecting FC Transmission characteristics of the profile Soil stratification Swelling and shrinking Presence of pans Occurrence of groundwater table For medium-texture soils, the moisture content corresponding to -0.33 bar is taken as FC if it cannot be done in the field.

108. 108 2.Permanent wilting point: is arbitrarily defined as the soil water content at which the leaves of sunflower plants wilt permanently, i.e., when they do not recover their turgor if subsequently placed in a saturated atmosphere. The moisture content corresponding to a matric potential of -15 bar is considered as PWP. PWP depends on: 1. climate conditions (evaporative demand, wind, cloud cover) 2. soil conditions (hydraulic conductivity) 3. plant species 3. Maximum Saturation is the wettest possible condition of a soil. This is when all the soil pores are filled with water. This corresponds to the soil water content at zero matric potential.

109. 109 - 100000 - 10000 Hygroscopic water - 3100 ( kPa , log) -1500 Permanent Wilting Point Soil -100 moisture Potential -10 Field Capacity -1 Total available water Saturation 10 20 30 40 50 Moisture content (% vol ) Gravitational water

110. 110 3.6. Determination of Available Water Capacity (AWC) AWC is defined as the volume of water retained between FC and PWP. Values of AWC are most commonly used for determination of the depth and frequency of irrigation required. AWC = θ FC – θ PWP Fraction available water depleted ( ƒ d ) ƒ d = ( θ FC - θ v ) = soil water deficit (SWD) θ v = current soil volumetric water content Fraction available water remaining ( ƒ r ) ƒ r =

111. 111 θ v - θ wp = soil-water balance (SWB) Total available water capacity(TAWC) within the plant root zone can be computed as follows: TAWC = AWC X RD , where RD = depth of the plant root zone (mm, cm, m) If different soil layers are there: n TAWC = ∑ AWCi x Zi i =1

112. 112 CHAPTER FOUR 4. WATER MOVEMENT IN SOILS Water movement in soils occurs under both: 1. Saturated 2. Unsaturated conditions Saturated conditions occur below the water table where water movement is predominantly horizontal, with lesser components of flow in the vertical direction. Saturated soils occur when soil pores are entirely filled with water. In this case, the water content ( θ ) is equal to the total porosity ( ƒ) and the air filled porosity (ƒ) is zero. Unsaturated conditions generally predominate above the water table ( vadose zone), localized zones of saturation can exist especially following precipitation or irrigation.

113. 113 The flow of water in soil is important for: Calculation of water balances Redistribution of water, solutes and energy within the soil-plant-atmosphere systems. 4.1. Water Flow in Capillary Tubes 4.1.1. Poiseuille’s Law: the flow of water in saturated soils At the microscopic scale of an individual pore, approximated as a water-filled cylinder of given radius, r, the volumtric flow rate (Q) is described by Poiseuille’s equation: Q = Where, ƞ = the water viscosity, ρ = water density, g = acceleration due to gravity, △H = the difference in total between two points along the cylinder and L = the distance between the two points

114. 114 4.1.2. Darcy’s Law: Measurement of saturated hydraulic conductivity and Infiltration Because it is usually not feasible to determine the size distribution and inter-connectively of pores a macroscopic approach is used to describe flow through soils. This approach was first taken by Henri Darcy, an engineer working on sand filters used to purify the water in the city of Dijon in France. The result of his experiment indicated that the quantity (volume) of water filtering through the sand bed per unit time depended on The cross-sectional area of the bed The thickness of the bed(length of the bed) The height of water on the filter bed Proportionality constant, K The famous experimental Darcy formula is: Q = A( h+L )t/L

115. 115 Where: Q = volume of water passing through the sand filter (L 3 ) L = length of the sand filter bed (L) , A = cross-sectional area of the sand filter bed (L 2 ), h = height of water on the filter bed(L), t = time (T), K = proportionality constant (LT -1 ) h A L Screen Fig. The Darcy sand filter Q

116. 4.2 . Flow of water in unsaturated soil Applies to a flow condition in which the effectively conducting pore space is not the total pore space but a partially saturated one. In unsaturated soil, the fluid may form a continuous phase or be separated by soil particles . In unsaturated soil, water flows under the influence of negative hydrostatic pressure head or suction head difference. Darcy’s law was originally conceived for saturated flow only. Richards (1931) extended it to unsaturated flow, with the provision that the conductivity is a function of the matric suction head (i.e., K= k( Ψ m ). 116

117. CHAPTER FIVE 5. INFILTRATION AND REDISTRIBUTION Infiltration is the term applied to the process of water entry into the soil generally by downward flow through all or parts of the soil surface and is the first stage of water movement in soil. The rate of infiltration relative to the rate of supply (by flooding the surface, by sprinkling or as rain) determines: How much water will enter the root zone How much will runoff Therefore, the rate of infiltration affects: The water economy of plant communities The amount of surface runoff and its attendant danger of soil erosion. Knowledge of the infiltration process as it is affected by the soil’s properties and transient conditions, and by the mode of water 117

118. supply, is, therefore, a prerequisite for efficient soil and water management. 5.1. Downward Infiltration into a Uniform Soil After irrigation or rain has continued for some time, the following zones can usually be distinguished in the water content profile: Rainfall ---------------------------- (cm/h) k sat Rate of ponding /runoff Rate ---------------------------------------------- Infiltration, I II ( i ) t ponding Time Figure 1: Infiltration rate vs time 118

119. Figure 1: Dependence of the infiltration rate upon time, under an irrigation of constant intensity lower than the initial value but higher than the final value of the soil infiltrability. Infiltration terminologies a) Infiltration rate “ i ” It is the volume flux of water flowing into the profile per unit of soil surface area under any set of circumstances. b) Infiltration capacity/ Infiltrability This is the infiltration flux resulting when water at atmospheric pressure is made freely available at the soil surface. This designates the limiting rate which the soil can absorb through its surface. As long as the rate of water delivery to the surface is smaller than the soil’s infiltrability, water infiltrates as fast as it arrives and the supply rate determines the infiltration rate and hence the process supply (or flux) controlled (I in figure 1). 119

120. If the delivery rate exceeds the soil’s infiltrability, it is the soil’s infiltrability that determines the actual infiltration rate, and thus the process becomes surface controlled or profile controlled (II in figure 1) c) Final infiltration capacity/steady state infiltration This is the asymptotically a constant rate which the soil attains after some time and from which it appears to decrease no more. The steady-state infiltrability is practically equal to the saturated hydraulic conductivity (figure 2). Decreasing infiltrability infiltrability k sat ------ ------------------------------ Steady infiltrability Time Figure 2: Time dependence of infiltrability under shallow ponding 120

121. d) Cumulative infiltration “I” This is the total amount of water infiltrating a unit cross-sectional area since the beginning of the water application (Figure 3). It is the time integral of the infiltration rate, and has a curvilinear dependence with a gradually decreasing slope. cumulative infiltration Time Figure 3: Time dependence of cumulative infiltration under shallow ponding .. 121

122. 5.4. Importance of Redistribution Process It determines the amount of water retained at various times and depth zones, It affects the water economy of plants The rate and duration of downward flow during redistribution determine: The effective soil water storage a property that is vitally important, particularly in relatively dry regions, where water supply is infrequent and plants must rely for long periods on the unreplenished reservoir of water within the rooting zone. How much water will flow through the root zone (rather than be detained, or temporarily stored within it) and hence how much leaching of solutes will take place. 122

123. 123 CHAPTER SIX 6. SOIL MECHANICS 6.1. Soil Compactibility in Relation to Soil Wetness Soil compaction is defined as the method of mechanically increasing the density of soil. Soil compaction occurs when soil particles are pressed together, reducing pore space between them. Heavily compacted soils contain few large pores and have a reduced rate of both water infiltration and drainage from the compacted layer. This occurs because large pores are the most effective in moving water through the soil when it is saturated. In addition, the exchange of gases slows down in compacted soils, causing an increase in the likelihood of aeration-related problem. While soil compaction increases soil strength, roots must exert greater force to penetrate the compacted layer.

124. 124 Soil compaction changes pore space size, distribution, and soil strength. One way to quantify compaction is to measure the change by measuring the bulk density. As the pore space is decreased within a soil, the bulk density is increased. For a given amount of compactive force, the resulting bulk density is a function of soil moisture. If one starts with a dry soil, the attainable bulk density at first increases with increasing soil wetness, beyond which the density decreases. A dry soil resists compaction because of it is Stiff matrix and High degree of particle-to-particle bonding, interlocking, and/or frictional resistance to deformation.

125. As soil wetness increases, the moisture films: Weaken the inter-particle bonds Cause swelling Reducing internal friction by “lubricating” the particles, Thus making the soil more workable and compactible . As soil wetness nears saturation, however, the fractional volume of expellable air is reduced and the soil can no longer be compacted by a given compactive effort to the same degree as before. Henceforth, any further increase in soil moisture reduces, rather than increases, soil compactibility . At saturation, no amount of kneading can cause any increase in soil bulk density(unless water is expelled). At high wetness values, water is seen to prevent closer packing of the soil matrix. 125

126. Rather,the water which hydrates the grains pushes them apart and causes swelling, thus reducing the attainable bulk density. Maximum density (g/cm3) density Bulk Optimum moisture mass wetness Figure. Typical moisture-density curve for a medium-textured soil, indicating the maximum density obtainable with a particular compactive effort. 126

127. 6.2. Causes of Soil Compaction There are several forces, natural and man-induced, that compact a soil. Listed below are several causes of soil compaction: Raindrop impact- this is certainly a natural cause of compaction, it can be seen as a soil crust (usually less than ½ inch thick at the soil surface) that may prevent seedling emergence. Rotary hoeing can often alleviate this problem. Tillage operation -continuous moldboard plowing or disking at the same depth will cause serious tillage pans (compacted layers) just below the depth of tillage in some soils. This tillage pan is generally relatively thin (1-2 inches thick), may not have a varying depth of tillage over time or by special tillage operations 127

128. Wheel traffic -This is without a doubt the major cause of soil compaction. The weight of tractors has increased from less than 3 tons in the 1940’s to approximately 20 tons today for the big four-wheel-drive units. This is of special concern because spring planting is often done before the soil is dry enough to support the heavy planting equipment. Minimal crop rotation -The trend towards a limited crop rotation has two effects: Limiting different rooting systems and their beneficial effects on breaking subsoil compaction Increased potential for compaction early in the cropping season, due to more activity and field traffic. 128

129. 6.3. Control of Soil Compaction There are four strategies commonly used in dealing with compaction: 1. Avoidance 2. Alleviation 3. Controlled traffic 4. Acceptance 1. Avoidance or Prevention Avoidance is the most desirable where it is physically and economically possible. The old age of “stay off the field until it is fit to work” still applies. Avoid all but truly essential pressure-inducing operations. However, the possible sever economic repercussions of delaying planting, harvesting, or other operations may outweigh compaction damage or loss. 129

130. 2. Alleviation At times potentially damaging compaction is unavoidable. There are two ways of alleviating and lessening the damage caused by compaction: 1. attempt to remove the compaction or 2. attempt to reduce the adverse effects of the compaction Moldboard tillage of the compacted depth has been effective in removing surface compaction, sub soiling, chiseling and others can be used to temporarily alleviate the problem. One way to reduce the adverse effects of compaction is to apply fertilizer in a way that increases the availability. Such measures may include row/band application of phosphorus or potassium. Split application of nitrogen or other practices that minimize the loss of nitrogen by denitrification may also alleviate compaction problems. 130

131. 3. Controlled traffic In a normal year, as much as 90% of the field may be tracked by equipment. The philosophy behind controlled traffic is to restrict the amount of soil traveled on by using the same wheel tracks. 70 to 90% of the total plow layer compaction occurs on the first trip across the field. By controlling traffic, the tracked area will not be compacted. 4. Acceptance Acceptance is waiting for the detrimental effects to be removed by natural forces. However, this may not be practical if there is compaction below the plow layer. The deeper the compaction and higher the clay content, the longer it will persist. 131

132. 132 6.4. Soil Consolidation The process by which a body of soil, either initially saturated or compacted to the point of saturation, is compressed in a manner that results in reduction of pore volume by expulsion of water is called consolidation. Compaction is the compression of unsaturated soil due to reduction of its air-filled pore space without change in mass wetness. Because of the slower nature of consolidation relative to compaction (water being 50-100 times more viscous than air at ordinarily encountered temperatures), consolidation is not an immediate response of soil to transient pressures such as those caused by episodic traffic or tillage. Consolidation is manifested in the gradual settlement, or subsidence, of soil under long term loading such as that due to a permanent structure (e.g., single building).

133. 133 Hence the practical importance of consolidation lies not so much in agriculture as in engineering, where the amount and uniformity of foundation settlement are of vital concern to those interested in the stability and safety of structures. In general, granular soils such as dense sands are the least compressible soils, and such consolidation as does take place usually occurs relatively rapidly due to the soil’s high permeability Clays and other fine-textured soils, with higher porosity, generally have a much greater total compressibility, but the rate of their consolidation is likely to be slow, sometimes exceeding over a period of mouths or even years.

134. 134 CHAPTER SEVEN SPATIAL AND TERMPORAL VARIABILITY OF SOIL PHYSICAL PROPERTIES 7.1 Spatial Variability of Soil Physical Properties Characterizing the spatial variability and distribution of soil properties is important in predicting the rates of ecosystem processes with respect to natural and anthropogenic factors and in understanding how ecosystems and their services work. In agriculture, studies of the effects of land management on soil properties have shown that cultivation generally increases the potential for soil degradation due to the breakdown of soil aggregates and the reduction of soil cohesion, water content and nutrient holding capacity. Cultivation, especially when accompanied by tillage, has been reported to have significant effects on topsoil structure and thus the ability of soil to fulfill essential soil functions and services in relation to root growth, gas and water transport and organic matter turnover.

135. Soil properties vary considerably under different crops, tillage type and intensity, fertilizer types and application rates. Consequently, the physical properties of the soil are also affected by many factors that change vertically with depth, laterally across fields and temporally in response to climate and human activity. Since this variability affects plant growth, nutrient dynamics, and other soil processes, knowledge of the spatial variability of soil physical properties is therefore necessary. Spatial and temporal dependency and distribution of soil characteristics will be investigated and assessed using several geo-statistical tools, GIS mapping, time series analysis and traditional statistics. Geo-statistics provides the basis for the interpolation and interpretation of the spatial variability of soil properties. 135

136. Information on the spatial variability of soil properties leads to better management, decisions aimed at correcting problems and at least maintaining productivity and sustainability of the soils and thus increasing the precision of farming practices. A better understanding of the spatial variability of soil properties would enable refining agricultural management practices by identifying sites where remediation and management are needed. This promotes sustainable soil and land use and also provides a valuable base against which subsequent future measurements can be proposed. Despite the importance of this topic in agriculture, the literature is not abundant on the variability of soil physical properties in most of countries Furthermore, existing studies on the spatial variability of soil properties have focused on the top soil (0–20 cm) with less or no studies at deeper soil depths (30–100 cm). 136

137. 137 Water commonly has a leading role among the factors responsible for spatial and temporal yield variability and is a major input resource for precision management. When the application of water or water quality (salinity) is non-uniform in the field, the resulting soil moisture properties may be an important factor in causing spatial variations in crop yield. Yield variability within surface-irrigated fields has been related to the spatial variability of available soil water due to non-uniform irrigation. Spatial dependence can occur at scales varying from a few meters to several kilometers. For example, a short-range (3-4 m) and a long-range (16-35 km) spatial dependence for sand and clay, respectively. Variations in soil properties can also be expressed by dividing the coefficient of variation (CV) into different ranges, for example, least (<15%), moderate (15-35%) and most (>35%)

138. Site-specific farming assumes that fields used for agricultural production are not uniform but rather exhibit large spatial variability in their properties. Knowledge of the spatial variability of soil physical and hydraulic properties is important for precision farming and also for the evaluation of agricultural management practices.  It is well known that soil physical and hydraulic properties vary over space and time from field to field as well as within fields. These variations are affected and controlled by many factors such as vegetation, agricultural management practices, previous farming practices and weather conditions. Point samples are important contributors to our understanding of the dynamics of a field.  However, collecting soil and other samples is time consuming to acquire and expensive to analyze. 138

139. In addition, the accuracy of maps developed from point data depends on mathematical models to emulate the spatial variation of the variables of interest. Spatial modeling requires intensive sampling efforts that increase in complexity as within field variation increases. These types of data must be integrated with remote sensing technologies to optimize benefits and reduce costs.  Soil data is collected on grid and transect sampling schemes under various soil conditions, tillage systems, crop rotations and irrigation systems.   139

140. CHAPTER EIGHT: SOIL TILLAGE AND IT EFFECTS 140

141. 141 8.1 Soil Tillage Tillage is the agricultural preparation of soil by mechanical agitation of various types, such as digging, stirring, and overturning. Examples of human-powered tilling methods using hand tools include shoveling, picking, mattock work, hoeing, and raking. Examples of draft-animal-powered or mechanized work include ploughing (overturning with moldboards or chiseling with chisel shanks), rot tilling, rolling with cult packers or other rollers, harrowing, and cultivating with cultivator shanks (teeth). Small-scale gardening and farming, for household food production or small business production, tends to use the smaller-scale methods above, whereas medium- to large-scale farming tends to use the larger-scale methods. Any type of gardening or farming, but especially larger-scale commercial types, may also use low-till or no-till methods as well.

142. Tillage is often classified into two types, primary and secondary. There is no strict boundary between them so much as a loose distinction between tillage that is deeper and more thorough (primary) and tillage that is shallower and sometimes more selective of location (secondary). Primary tillage such as ploughing tends to produce a rough surface finish, whereas secondary tillage tends to produce a smoother surface finish, such as that required to make a good seedbed for many crops. Harrowing and roto-tilling often combine primary and secondary tillage into one operation. "Tillage" can also mean the land that is tilled . The word " cultivation " has several senses that overlap substantially with those of "tillage“. In a general context, both can refer to agriculture. 142

143. Within agriculture, both can refer to any of the kinds of soil agitation described above. Additionally, "cultivation" or "cultivating" may refer to an even narrower sense of shallow, selective secondary tillage of row crop fields that kills weeds while sparing the crop plants. 8.1.1 Tillage Systems 8.1.1.1 Reduced tillage Reduced tillage leaves between 15 and 30% residue cover on the soil or 500 to 1000 pounds per acre (560 to 1100 kg/ha) of small grain residue during the critical erosion period. This may involve the use of a chisel plow, field cultivators, or other implements. See the general comments below to see how they can affect the amount of residue. 143

144. 8.1.1.2 Intensive tillage Intensive tillage leaves less than 15% crop residue cover or less than 500 pounds per acre (560 kg/ha) of small grain residue. This type of tillage is often referred to as conventional tillage but as conservational tillage is now more widely used than intensive tillage (in the United States), it is often not appropriate to refer to this type of tillage as conventional. Intensive tillage often involves multiple operations with implements such as a mold board, disk, and/or chisel plow. Then a finisher with a harrow, rolling basket, and cutter can be used to prepare the seed bed. 8.1.1.2 Conservation tillage Conservation tillage leaves at least 30% of crop residue on the soil surface, or at least 1,000 lb/ac (1,100 kg/ha) of small grain residue on the surface during the critical soil erosion period. 144

145. This slows water movement, which reduces the amount of soil erosion. Conservation tillage also benefits farmers by reducing fuel consumption and soil compaction. By reducing the number of times the farmer travels over the field, farmers realize significant savings in fuel and labor. In most years since 1997, conservation tillage was used in US cropland more than intensive or reduced tillage. However, conservation tillage delays warming of the soil due to the reduction of dark earth exposure to the warmth of the spring sun, thus delaying the planting of the next year's spring crop of corn. a) No-till : Never use a plow, disk, etc., ever again and aims for 100% ground cover. 145

146. b) Rotational Tillage: Tilling the soil every two years or less often (every other year, or every third year, etc.). c) Zone tillage/Strip-Till: is a form of modified deep tillage in which only narrow strips are tilled, leaving soil in between the rows untilled. This type of tillage agitates the soil to help reduce soil compaction problems and to improve internal soil drainage The Purpose of Zone tillage is designed to only disrupt the soil in a narrow strip directly below the crop row. In comparison to no-till, zone tillage creates approximately a 5-inch-wide strip that simultaneously breaks up plow pans, assists in warming the soil and helps to prepare a seedbed. When combined with cover crops, zone tillage helps replace lost organic matter, slows the deterioration of the soil, improves soil drainage, increases soil water and nutrient holding capacity, and allows necessary soil organisms to survive. 146

147. 8.2 Effects of Tillage Practices 8.2.1 Positive Effects of Plowing Tillage Loosens and aerates the top layer of soil, which facilitates planting the crop. Helps mix harvest residue, organic matter (humus), and nutrients evenly into the soil Mechanically destroys weeds Dries the soil before seeding (in wetter climates tillage aids in keeping the soil drier) When done in Autumn, helps exposed soil crumble over winter through frosting and defrosting, which helps prepare a smooth surface for spring planting 8.2.2 Negative Effects of Plowing Tillage Dries the soil before seeding . Soil loses a lot of nutrients, like nitrogen fertilizer, and its ability to store water. 147

148. Decreases the water infiltration rate of soil (Results in more runoff and erosion since the soil absorbs water slower than before) Tilling the soil results in dislodging the cohesiveness of the soil particles thereby inducing erosion. Increases chemical runoff ( Eutrophication:nutrient runoff into a body of water) Reduces organic matter in the soil . Reduces microbes, earthworms, ants, etc. Destroys soil aggregates . Compaction of the soil, also known as a tillage pan . Can attract slugs, cut worms, army worms, and other harmful insects to the left over residues. Crop diseases can be harbored in surface residues . 148

149. 149 CHAPTER NINE: SOIL AERATION 9.1 Definition soil aeration is the process by which air in the soil is replaced by air from the atmosphere. Soil aeration is one of the most important determinants of soil productivity. In the process of respiration-plants absorb oxygen and release carbon dioxide. Adequate root respiration requires that the soil itself be aerated. This is to say the gaseous exchange between soil air and the atmosphere should be at a rate that prevents deficiency of oxygen and excess of carbon dioxide from developing in the root zone. Under conditions of restricted aeration, soil microorganisms might compete with the roots of higher plants.

150. Gases can move in either of the following two ways: 1. air phase-in the pores which are drained of water, provided the pores are interconnected and open to the atmosphere. 2. in dissolved form through the water phase. The rate of diffusion of gases in the air phase is generally greater than in the water phase, hence soil aeration is dependent largely upon the volume fraction of air-filled pores. 9.2 Volume Fraction of Soil Air Since water and air compete for the same pore space, their volume fractions are so related that an increase of the one generally entails a decrease of the other. The volume fraction of air can be calculated from: ƒa = ƒ- θ where ƒa = volume fraction of air, ƒ = the total porosity, θ = the volume fraction of water 150

151. 151 Volume fraction of air is a transient property. To use the volume fraction of air as an index of soil aeration, it is necessary to specify some characteristic and reproducible wetness value generally used is field capacity. The analogous soil aeration index is field air-capacity. Field air-capacity is defined as a fractional volume of air in a soil at the field capacity water content. Factors affecting air capacity are: 1. Soil texture Sandy soils is 25% or more Loamy soils is between 15% and 20% Clayey soils is likely to fall below 10% of the total soil volume.

152. 152 2. Soil structure Strongly aggregate soils: consists of 20-30% soil air (considerable volume of macroscopic (inter-aggregate) pores) Compacted soil: 5% soil air Limitations of using air capacity as an index of soil aeration: The value is difficult to determine with a satisfactory degree of accuracy. Rate of exchange of soil air rather than simply the content of soil air constitutes the decisive factor (entrapped air) 9.3 Composition of Soil Air In well aerated soil, the composition of soil air is close to that of the external atmosphere, as the oxygen consumed in the soil is readily replaced from the atmosphere. This is not the case in poorly aerated soil.

153. 153 Composition of soil air can differ greatly from the composition of the external atmosphere because of several factors: Time of the year Temperature Soil moisture content Depth below the soil surface Root growth Microbial activity pH Rate of exchange of gases through the soil surface. For example, there is high concentration of CO 2 in the soil than in the atmosphere due to: Production of CO 2 from aerobic respiration by the roots of higher plants and numerous macro and microorganisms Oxidation of carbonaceous matter products CO 2 CO 2 is 0.03% in the air to 0.3% in the soil, whereas O 2 remains about 20%.

154. 154 9.4 Mechanisms of Gaseous Exchange between Soil and Atmosphere Two distinct mechanisms have been identified for the interchange of gases between the soil and the atmosphere: Convection/mass flow and Diffusion 1. Convection In convection also called mass flow , the moving force consists of a gradient of total gas pressure and it results in the entire mass of air streaming from a zone of higher pressure to one of lower pressure. The convective flow of air in the soil can be formulated using an equation analogous to Darcy’s law for water flow (q = - k ∇H ) , as follows: (1)

155. 155 where qv = the volume convective flux of air (volume flowing through a unit cross-sectional area per unit time) k = permeability of the air-filled pore space, ƞ = viscosity of soil air ∇P = the three-dimensional gradient of soil air pressure In one dimension, equation (1) takes the from: q v = (2) If the flux is expressed in terms of mass (rather than volume) per unit area and per unit time, then the equation is: q m = (3) Where: q m = the mass convective flux and ρ = the density of soil air

156. 156 Recalling that the density of gas depends on its pressure and temperature, and assuming that soil air is an ideal gas, in which the relations of mass, volume, and temperature are given by the equation: PV = nRT (4) where: P = pressure, V = volume, n = number of moles of gas, R = the universal gas constant per mole, T = temperature (K) Since the density ρ = M/V, and the mass M is equal to the number of moles n times the molecular weight m, we have: ρ = (m/RT)P (5) Substituting (5) in (4), we obtain: q m =

157. 157 2. Diffusion of Soil Air The moving force in case of diffusion is a gradient of partial pressure (or concentration) of any constituent member of the variable gas mixture which we call air, and it causes the molecules of the unevenly distributed constituent to migrate from a zone of higher to lower concentration. The diffusion transport of gases such as CO 2 and O 2 in the soil occurs partly in the gaseous phase and partly in the liquid phase Diffusion through the air-filled pores maintains the exchange of between the atmosphere and the soil, whereas diffusion through water films of various thickness maintains the supply of oxygen to, and disposal of CO 2 from, live tissues. For both pathways, the diffusion process can be described by Fick’s law: q d = - D o dc / dx (1)

158. 158 Where: q d = the diffusion flux (mass diffusing across a unit area per unit time), D o = the diffusion coefficient (having dimension of area per time), C = concentration (mass of diffusing substance per volume), x = distance and dc/ dx = concentration gradient If partial pressure P is used in stead of concentration of the diffusing component, we get: q d = -(Do/ β )( dP / dx ) (2) where β = the ratio of the partial pressure to concentration. The diffusion coefficient in the soil, (Ds) must be smaller than that in bulk air, owing to: Limited fraction of the total volume occupied by continuous air-filled pores, and The tortuous nature of the air-filled pores Hence, we expect Ds to be some function of the air-filled porosity, ƒa .

159. 159 Diffusion workers have found different relations between Ds and ƒa for various soils.Some examples: = ĸƒ a 2 ➱ Ds = D o ĸƒ a 2 ( Buckingham,1904) (3) = 0.66 ƒ a ➱ Ds = 0.66D o ƒ a ( Penman, 1940) (4) 0.66 is a tortuosity coefficient, suggesting that the apparent path is about two-thirds the length of the real average path of diffusion in the soil. 9.5 Soil Respiration and Aeration Requirements The overall rate of respiration due to all biological activity in the soil, i.e., the amount of oxygen consumed and the amount of CO 2 produced by the entire profile determines the aeration requirement of the soil. Plant roots require a constant and adequate supply of oxygen for their respiration. Generally, it is possible to grow most of the crop plants when the oxygen concentration exceeds 10%.

160. 160 9.6 Characterization of Soil Air The aeration status of a soil can be characterized in many ways: the most convenient ways are: 1. the gaseous oxygen content in soil air 2. air capacity or aeration porosity 3. the oxygen diffusion rate (ODR) 4. the oxidation-reduction ( redox ) potential 1.The gaseous oxygen content in soil air The oxygen content of soil air is regulated by three factors: Amount of macropores as influenced by texture and structure Soil water content which affects the proportion of air-filled pores Consumption of oxygen by plant roots and microorganisms In well aerated surface soils the oxygen content of soil air is about 20.73% which is very close to that of atmospheric air (20.95%). In such soils, oxygen consumed by plants and microorganisms is quickly replaced.

161. 161 In poorly aerated soils (fine-texture with poor structure): Only few macropores Most of the pores are micropores filled with water Leave very little space for air storage Water blocks the pathways through which air could exchange with atmospheric air. The overall effect of these is to reduce the oxygen content of the soil air and create an anaerobic condition. 2. Air capacity or aeration porosity Determination of aeration porosity gives an idea about how much of the soil volume can be expected to be filled with air a few hours after a heavy rainfall or irrigation. A soil is generally well-aerated for crop plants if it has an aeration porosity of 10% or more of the total volume unless there is a high groundwater table. NB. Factors affecting aeration porosity include texture, structure, organic matter, soil compaction, groundwater table.

162. 162 3. Oxygen diffusion rate (ODR) It is the rate at which oxygen of soil air exchanges with the oxygen of the atmospheric air by the process of diffusion. It is expressed as gram per square centimeter per minute. ODR largely determines the rate at which oxygen can be replenished when it is consumed for respiration of plant roots and micro-organisms or when it is driven out of the soil pore spaces by water. For continued supply of oxygen to plant roots and for satisfactory growth of most plants and microorganisms the ODR should be at least 40x10 -8 g/cm 2 -min, because when ODR value drops below this value, the growth of most plants appear to suffer. Factors affecting ODR are: 1. Soil water content: ODR increases with decreasing soil water content. Diffusion of oxygen through air-filled pores is about 10,000 times faster than it is through water.

163. 163 The decrease of ODR with increase in soil water content is due to reduction in pore volume for soil air, blockage of pores through which oxygen diffusion takes place. 2. Soil texture: Affects ODR through its effect on pore size distribution which affects water retention. ODR decreases with increase in fineness of soils. 3. Soil depth: generally, ODR decreases with increase in soil depth. As a result, oxygen concentration decreases with increase in depth. 4. Soil compaction: Compaction affects ODR by reducing the amount of macropores as well as total porosity. ODR decreases with increase in bulk density of soil. 4. Oxidation- reduction potential In flooded or partially saturated soils, the molecular oxygen may not be present and the oxygen diffusion rate may be too low or may approach to zero. The aeration status of such soils may be better characterized with the help of oxidation-reduction potential which is a measure of the intensity of reduction in soil in the absence of oxygen.

164. 164 CHAPTER TEN: SOIL TEMPERATURE AND HEAT FLOW 10.1 Soil Temperature Temperature is the measure of hotness of a body whereas heat is the physical cause due to which the hotness or coldness is felt. Soil temperature determines: Rates and directions of soil physical processes Rates of energy and mass exchange with the atmosphere-including evaporation and aeration Types and rates of chemical reactions in the soil Strongly influences biological processes: a) seed germination b) seedling emergence and growth c) root development d) microbial activity

165. 165 Soil temperature is dynamic. It varies in response to: Changes in the radiant (how much heat reaches the soil surface) Changes in thermal and latent heat exchange processes 10.2 Modes of H eat Energy Transfer in Soils There are three principle modes of energy transfer from one place to another: Radiation is the process in which heat can transmit from a hotter body to a colder body without intervention of any material medium a) Radiation: The radiant energy travels in straight line with the speed of light. By this process, heat reaches on the surface of the earth from the sun, although the vast space between the sun and the atmosphere of the earth is vacuum. By radiation, energy is emitted in the form of electromagnetic waves from all bodies above 0K.

166. 166 The radiation received on earth surface supply the energy for evaporation, transpiration and used to heat up the soil. b) Convection It is a processes in which heat is transferred from one point to another by actual movement of the heated material particles from a place of higher temperature to one of lower temperature. Liquids and gasses are generally heated by the process of convection. c) Conduction Conduction is a process in which heat is transmitted from the hotter to the colder part of a body or from a hotter body to a colder body in contact with each other by internal molecular motion without any transference of material particles.

167. 167 10.3 Effect of Soil Temperature on Plant Growth Soil temperature affects the following plant processes: Germination of seeds Root growth and activity Crop growth and yield Effect on Germination Seed germination is highly affected by soil temperature. The effect varies with crops. Root growth and activity Elongation of roots and activity of roots such as uptake to water and nutrients are highly affected by soil temperature. These processes become sluggish as the temperature drops below the optimum. 10.4 Management of Soil Temperature The temperature of the soil can be modified by: 1. covering the soil surface (mulch) 2. manipulating the surface conditions (tillage) 3. drainage, irrigation, evaporation (soil wetness control) Mulching Muches make a barrier to the transfer of heat or water vapor and affect the soil thermal regime by interception of the incoming radiation