Problem Solution Manual for Unsaturated Soil Mechanics

Problem Solution Manual for Unsaturated Soil Mechanics by Ning Lu and William J. Likos, Wiley, 2004 This manual is prepared by the following individuals: Phillip J. Wolfram, Alexandra Wayllace, William J. Likos1 , and Ning Lu1

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2007 All correspondences should be addressed to [email protected] or [email protected]

Contents 1 State of Usaturated Soil

1

2 Material Variables

6

3 Interfacial Equilibrium

11

4 Capillarity

20

5 State of Stress

28

6 Shear Strength

36

7 Suction and Earth Pressure Profiles

52

8 Steady Flows

66

9 Transient Flows

71

10 Suction Measurement

80

11 Hydraulic Conductivity Measurement

87

12 Suction and Hydraulic Conductivity Models

90

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Acknowledgments This Solution Manual was not possible without the generous help of Silvia Simoni, Adam Prochaska, and Tom Bonnie. These individuals were participants in the Unsaturated Soil Mechanics course at the Colorado School of Mines and their fine homeworks served as an initial starting points for chapters 5, 6, 7, 10, and 11.

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Chapter 1

State of Usaturated Soil 1.1 Where are the regions in the U.S. where unsaturated soils are likely encountered to significant depth below the ground surface? Unsaturated soils are likely encountered to significant depth below the ground surface in large portions of the United States in the arid or semiarid regions (Figure 1.8) 1.2 What kind of climatic conditions tent to lead to the formation of a thick unsaturated zone? Precipitation, evaporation, and evapotranspiration are factors that contribute to the depth and extent of the unsaturated zone. A thick unsaturated zone generally occurs in regions where potential evaporation outweighs annual precipitation by a factor ranging between 2 and 20 within 40 degree north of equator. 1.3 What is the fundamental difference between unsaturated soils and saturated soils in terms of pore water pressure? Pore water pressure in a saturated soil is generally compressive and isotropic, in contrast to unsaturated soils where pore water pressure is generally, but not necessarily, tensile (p. 20-21). 1.4 Describe and illustrate the Mohr-Coulomb failure criterion. The Mohr Coulomb criterion delineates a failure envelope for a material defining critical states of stresses. It is described in terms of (c) and internal friction angle (f ). It is described in terms of states of cohesion (c0 ) and internal friction angle(φ0 ) to dictate a failure shear stress at a given normal effective stress (σn0 ) in τf = c0 + σn0 tan φ0 . Graphically, the Mohr-Coulomb failure criterion plots as a straight line on the effective normal and shear stress graph as shown in S1.1. It is also important to note that the Mohr-Coulomb failure criterion is a linear approximation and therefore valid only near the range of values from which it was derived.

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Figure S1.1: Mohr-Coulomb Failure Criterion

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1.5 When the state of stress (i.e., Mohr circle) in a soil reaches the Mohr-Coulomb criterion, what is the state of stress called? Failure state. 1.6 Give three examples of unsaturated soil mechanics problems in geotechnical engineering. Transient and steady seepage in unsaturated embankment dams, consolidation and settlement of unsaturated soils, bearing capacity for shallow foundations under moisture loading, slope stability, and land sliding. 1.7 For a given unsaturated soil under either a dry or wet condition, which one has a higher suction? The dry soil has a higher suction (p. 39, 42-43). 1.8 What are state variables, material variables, and constitutive laws? State variables are those variables that completely describe the state of the system for the given phenomenon. Material variables generally vary with state variables and describe the physical characteristics of the material. These variables are intrinsic material properties. Constitutive laws describe the governing physical principles which demonstrate interrelationships between or among state variables and material variables. Constitutive laws are used as the mathematical connection between state and material variables for the purpose of prediction and explanation of phenomena (p. 26-28). 1.9 What are the principal differences between saturated and unsaturated soil profiles of pore water pressure, total stress, and effective stress? Pore water Pressure Profiles- Generally vary linearly with depth, increasing hydrostatically below the water table (saturated soils), and decreasing hydrostatically above the water table (unsaturated soils). Total Stress Profiles- For unsaturated soils, total stress decreases due to change in the selfweight when the material is dewatered. Therefore, the total stress profile for a saturated soil extends to a greater magnitude than the profile for the same soil under unsaturated conditions. Effective Stress Profiles- The effective stress for a saturated soil at ground surface is 0. It is also important to note that effective stresses for an unsaturated soil are greater than for a saturated soil due to the tensile pore water pressure in the unsaturated soil (p. 22-23). Figures 1.12 and 1.13 graphically demonstrate these concepts. For both saturated and unsaturated conditions, horizontal stresses are dependent upon vertical stresses according to formula 1.5a (p. 23). 1.10 According to Bishop’s effective stress concept, which state, saturated or unsaturated, has a higher effective stress? Why? According to Bishop’s effective stress concept, unsaturated conditions have higher effective stress since pore water pressures are negative, leading to a greater effective stress. This is demonstrated by examining the equation: σ 0 = (σ − ua ) + χ(ua − uw ) where matric suction (ua − uw ) is positive. 1.11 What is the shape of the pore water pressure profile under hydrostatic conditions in saturated and unsaturated states, respectively? 3

The shape of the pore pressure profile under the hydrostatic condition for saturated and unsaturated states is linear, as shown in Figure S1.2 where z = 0 is the ground surface:

Figure S1.2: Comparison of saturated and unsaturated hydrostatic soil profiles 1.12 If an unsaturated soil has a water potential of −1000 J/kg, what is the equivalent soil suction value? If the soil at the air dry condition has a matric suction of 100 M P a, what is the soil water potential in joules per kilogram? For −1000 J/kg potential the equivalent soil suction is −1000 kP a, since 100 J/kg = 100 kP a. Soil with a matric suction of 100 M P a has a soil water potential of 100, 000 J/kg since 0.1 M P a = 100 J/kg (p. 40). 1.13 Three soils- clay, silt, and sand- are all equilibrated at the same matric suction, which soil has the highest water content and why? If clay, silt, and sand are all equilibrated at the same matric suction, then clay has the highest water content as shown on Figure 1.20 (p. 42). Clay has the highest water content at a given matric suction due its charged surfaces and very high specific surface area as (p. 42) Sand and silt have lower specific surface areas than clay. 1.14 Describe the major physical and physicochemical mechanisms responsible for soil suction. Soil suction is caused by the physical and physicochemical mechanisms that decrease the potential of the pore water relative to a reference potential of free water. These mechanisms include capillary effects, short-range adsorption effects composed of particle-pore water interaction, and osmotic effects. Capillary effects are caused by curvature of the air-water interface. Short-range adsorption effects are composed of electrical double layer and van der Waals force field interactions at the solid-liquid interface. Osmotic effects are the result of dissolved solute in the pore water. With the same chemical concentration, osmotic pressure of pure solution 4

could be different with that of pore water as interaction between solute and solid surface of soil particles could occur. Matric suction is generally used to group the aggregate of capillary and short-range adsorption effects. Osmotic suction refers to the aggregate of osmotic effects (p. 34-35).

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Chapter 2

Material Variables 2.1 What are the state variables that control the density of air? What is the average air density at your location? Temperature, pressure, and relative humidity are state variables that control the density of air as evidenced by the equation below. The average air density for Golden, CO during a typical winter, assuming a 10% relative humidity, an average temperature of 0 ‰ and an average pressure of 85 kPa is as follows: Moist air density is estimated using Tables 2.8 and 2.9. µ ¶ µ ¶ ud ωd ωd T − 273.2 ωv RH ρa,moist = − 0.611 − 1 exp 17.27 RT ωv T − 36 RT kg The first part of the expression is estimated as 1.084 m 3 (from Table 2.8), and the second part of the expression can be estimated as 0.000 (from Table 2.9), resulting in

ρa,moist = 1.084

kg m3

2.2 What is the physical meaning of relative humidity? Relative humidity (RH) is the ratio of absolute humidity in equilibrium with any solution to the absolute humidity in equilibrium with free water at the same temperature. RH is also equivalent to the ratio of vapor pressure in equilibrium with a given solution and the saturated vapor pressure in equilibrium with free water. 2.3 At 25 ‰ and 101.3 kPa (1 atm), what is the ratio of the viscosity of water to the viscosity of air? The viscosity of which phase, air or water, is more sensitive to temperature changes between 0 and 100 ‰ ? −4

8.77×10 kg/m·s νw /νa = 1.845×10 −5 kg/m·s = 47.53 ≈ 50. Water is more sensitive to viscocity changes than air, changing by a magnitude of about 10, between 0 ‰ and 100 ‰.

2.4 Temperature varies between 15 ‰ in the night and 30 ‰ in the afternoon at a certain location. If the ambient vapor pressure remains constant at 1.6 kPa, what is the range of the relative

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humidity variation? If the vapor pressure remains unchanged, at what temperature will dew formation occur? RH =

uv uv,sat .

With uv constant, RH depends on the saturated vapor pressure which is related ³ ´ to change in temperature according to the following equation: uv,sat = 0.611 exp 17.27 T T−273.2 . −36 At 15 ‰, uv,sat = 1.70 kPa. At 30 ‰, uv,sat = 4.24 kPa. With uv = 1.6 kPa, RH ranges from 93.8 % to 37.7 %. If the vapor pressure remains unchanged, dew formation will occur when uv,sat = uv = 1.6 kPa. This condition is met when T = 287.2 K, or 14.0 ‰. Dew formation will therefore occur at 14.0 ‰. 2.5 If a saturated swelling soil has a specific gravity of 2.7 and gravimetric water content of 300 %, what is the volumetric water content? θ=

1 1+

1 Gs w

=

1 1+

1 2.7(3)

= 0.89 = 89%

2.6 A closed room is filled with humid air. If the temperature rises significantly, does the relative humidity increase or decrease? In a closed room filled with humid air, a significant temperature rise will result in a decrease of relative humidity (RH) since RH is inversely proportional to temperature (by Equations 2.11 and 2.12). 2.7 Can the vapor pressure of soil gas be greater than the saturation pressure at the same temperature and pressure? Why or why not? No, the vapor pressure of soil gas cannot be greater than saturation pressure at the same temperature and pressure because the saturation pressure is the maximum pressure corresponding to a given equilibrium state (state of same temperature and pressure). 2.8 Can volumetric water content be greater than 100% in unsaturated soil? Volumetric water content cannot be greater than 100 % in unsaturated soil as mathematically evidenced by the following definition: θ=

Vw Vw = Vt Vw + Vs + Va

where θ = volumetric water content, Vw = volume of water, Vs = volume of solids, and Va = volume of air. 2.9 Is degree of saturation a mass-based or volume-based quantity? Degree of saturation (S) is a volume-based quantity since it is defined as: S=

Vw Vv

where Vw = volume of water, and Vv = volume of voids.

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2.10 When the temperature of unsaturated soil increases, does the surface tension at the air-water interface increase or decrease? When the temperature of unsaturated soil increases, the surface tension at the air-water interface decreases, as shown in Figure 2.12 and Table 2.10 (Lu, 2004). 2.11 What is the density of dry air if the prevailing temperature and pressure are 25 ‰ and 95 kPa, respectively? What is the relative change in dry-air density if the tempreature rises to 40 ‰ and the air pressure remains unchanged? If the temperature is kept at a constant value of 25 ‰ , how much pressure change is required to cause the dry-air density to decrease by 15% compared to 95 kPa? The density of dry air at T = 25 ‰ and P = 95 kPa is ρd =

ud ωd (95 kPa)(28.966 × 10−3 kg/mol) = = 1.110 kg/m3 RT (8.314 N · m/mol · K)(298.2 K)

If the temperature rises to 40 ‰ and the air pressure remains unchanged, the air density is ρd =

(95 kPa)(28.966 × 10−3 kg/mol) = 1.057 kg/m3 (8.314 N · m/mol · K)(313.2 K)

Relative change of 0.053 kg/m3 or a 4.8 % decrease. If temperature is kept constant at 25 ‰, the pressure change required to cause the dry-air density to decrease by 15 % compared to 95 kPa (ρd = 0.85 · 1.110 kg/m3 = 0.944 kg/m3 ) is computed as follows: ud =

(ρd )(RT ) = ωd

(0.944 kg/m3 )((8.314 N · m/mol · K)(298.2 K)) = 80.76 kPa 28.966 × 10−3 kg/mol This value corresponds to a decrease in pressure of 14.24 kPa. 2.12 Estimate the viscosity of air and water at a temperature of 50 ‰ . Given a mean pore size for a sandy soil as 10−3 m, and a specific discharge for both air and water as 10−2 m/s, identify the flow regimes for the air and water, respectively. Estimates for the viscosity of air and water at a temperature of 50 ‰, are as follows: µa = 1.96 × 10−5 kg/(m · s) µw = 5.32 × 10−4 kg/(m · s) Assume that density of water is 1000 kg/m3 and the density of air is 1 kg/m3 . Given a mean pore size for a sandy soil as 10−3 m, and a specific discharge, q, for both air and water as 10−2 m/s, the Reyonds number for air and water are: Rea =

ρvd (1 kg/m3 )(10−2 m/s)(10−3 m) = = 0.51 < 1 µ 1.96 × 10−5 kg/(m · s)

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Rew =

ρvd (1000 kg/m3 )(10−2 m/s)(10−3 m) = = 18.8 < 100 µ 5.32 × 10−4 kg/(m · s)

These Reynolds numbers indicate that the flow regime for air is linear and laminar since the value is less than one. The Reynolds number between 1 and 100 for water indicates that although the flow regime for water is laminar, it is not linear. 2.13 The relative humidity at equilibrium in an unsaturated soil is measured to be 80% at 22 ‰ . (a) What is the vapor pressure in the soil? (b) What is the vapor density of the soil? (c) What is the dew-point temperature if the vapor density is maintained constant but the temperature drops during the night? (d) What is the absolute humidity if temperature in the soil is maintained constant but vaporization is allowed to occur? (e) What is the free energy per unit mass of the pore water? Relative humidity at equilibrium in an unsaturated soil is measured to be 80 % at 22 ‰. Assume ambient pressure is 101.3 kPa so that Table 2.6 can be used. (a) The vapor pressure in the soil, uv , is 2.114 kPa (80 % of the value from Table 2.6). (b) The vapor density of the soil, ρv , is 15.507 g/m3 (80 % of the value from Table 2.6). (c) If the vapor density is maintained constant but the temperature drops during the night, the dew-point temperature is approximately 18.3 ‰ (as determined by finding the absolute humidity for ρv = 15.507 g/m3 , or alternatively by equation 2.14). (d) If temperature in the soil is maintained constant but vaporization is allowed to occur, the absolute humidity is 19.384 g/m3 (from Table 2.6). (e) The free energy per unit mass of the pore water is: E=−

RT (8.314 N · m/mol · K)(295.2 K) ln(RH) = − ln(0.80) ωw 18 kg/kmol E = 30.43 kJ/kg

2.14 At a prevailing temperature of 25 ‰ and a pressure of 95 kPa, how much does the density of air change from a completely dry state to a 100% relative humidity state? From Table 2.8 and Table 2.9: RH( %) 0 100

ρa (kg/m3 ) 1.110 1.096

1.110 kg/m3 − 0.014 kg/m3 = 1.096 kg/m3 . 2.15 If the ambient air pressure is 101.3 kPa and the temperature is 20 ‰ , what is the pressure inside the water meniscus for a capillary tube with diameter of 0.001 mm? If the temperature increases to 50 ‰ , what is the pressure inside the meniscus? ua − uw =

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2Ts R

where ua = air pressure, uw = pressure inside the water meniscus, Ts = the surface tension between the water-air interface, and R = radius of curvature of the capillary meniscus. From Table 2.10, Ts = 72.75 mN/m for 20 ‰. uw = ua −

2Ts 2 · 72.75 mN/m = 101.3 kPa − = −189.7 kPa R 0.0005 mm

The pressure inside the water meniscus is -189.7 kPa. If temperature increases to 50 ‰, Ts = 67.91 mN/m. uw = ua −

2Ts 2 · 67.91 mN/m = 101.3 kPa − = −170.3 kPa R 0.0005 mm

At the increased temperature of 50 ‰, the pressure inside the water meniscus is -170.3 kPa. It is important to note that pressure within the water meniscus decreases in magnitude with increasing temperature 2.16 If a tensiometer were used to measure matric suction of unsaturated soil at an elevation of 500 m above sea level, what would be the approximate maximum possible reading of the tensiometer? If a tensiometer were used to measure matric suction of unsaturated soil at an elevation of 500 m above sea level, the approximate maximum possible reading of the tensiometer would be approximately 91 kPa (from Figure 2.17).

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Chapter 3

Interfacial Equilibrium 3.1 A liter of water at 25 ‰ can dissolve 0.0283 L of oxygen when the pressure of oxygen in equilibrium with the solution is 1 atm. Derive the Henry’s law constant for oxygen in water from this information. Mi /ωi = KHi uai Vl Since there are 22.4 L per each mole of an ideal gas like oxygen: 1 mol (0.0283 L) = 1.2634 × 10−3 mol O2 . 22.4 L Notice that Mi /ωi = moles of gas. Therefore, Mi /ωi = 1.2634 × 10−3 mol. Solving equation 3.1 for KHi : Mi /ωi 1.2634 × 10−3 mol O2 = KHi KHi = Vl ui (1 L)(1 atm)(1.01325 bar / atm) KHi = 1.247 × 10−3 M/bar 3.2 What is the mass coefficient of solubility of air at a temperature of 20 ‰ and a total air pressure of 1 atm if the volumetric coefficient of solubility of air, hair , is 0.01708? At a temperature of 20‰ and a total air pressure of 1 atm (101.3 kPa), the density of air is ρair = 1.2 kg/m3 and the density of water is ρw = 998 kg/m3 . Hai =

ρair 1.2 kg/m3 hair = 0.01708 = 2.05 × 10−5 ρw 998 kg/m3

3.3 If the air pressure changes to 10 bars in the previous problem, what is the mass coefficient of solubility? If the air pressure changes to 10 bars (1 MPa) in the previous problem, the density of air is ρair =

ua ωa (106 N/m2 )(29 × 10−3 kg/mol) = = 11.71 kg/m3 RT (8.314 N · m/mol × K)(298 K) 11

The density of water is approximately the same as before, ρwater = 998 kg/m3 . Hai =

11.71 kg/m3 ρair hair = 0.01708 = 2.00 × 10−4 ρw 998 kg/m3

3.4 Two different sizes of capillary tubes are in the divided container shown in Fig. 3.25 (r1 = 10−6 m and r2 = 10−4 m). Each side has reached equilibrium between the air in the container and the pore water. Assume the total mass of water vapor in each side is much less than the amount of water in the capillary tubes. Also assume that the initial water levels in the tubes are very low compared to the overall lengths of the tubes. Describe the equilibrium position(s) of the water levels in the tubes when the valve is opened. Since r1 < r2 , and vapor pressure is directly proportional to r, uv1 < uv2 , the left side of the tank has a lower vapor pressure than the right. After opening the valve, the vapor pressures on each side of the tank equalize, increasing the relative humidity on the left side of the tank with reference to the initial state. Similarly, the relative humidity of the right side of the tank is decreased with respect to the initial equilibrium state. Since the relative humidity on the right side of tank has decreased, the water content of the capillary tube decreases due to capillary evaporation resulting in a decreasing water level in these tubes. Conversely, the tubes in the left side of the tank experience an increased relative humidity, which results in capillary condensation, increasing the water level in the tubes. The water level of the tubes in the left side of the container will increase and the water level in the tubes in the right side of the container will decrease. Assuming that the contact angle does not change as water content changes, the tubes on the right side of the tank will completely evaporate and the tubes on the left side of the tank will fill up to reflect the change in water volume on the right. 3.5 Calculate the hydrostatic pressure of water at 28 ‰ in spherical raindrops with (a) 5 mm diameter and (b) 0.2 mm diameter. The surface tension between air and water is 71.4 mN/m, as determined from interpolation of table 2.10 (Lu, 76). Assuming an ambient air pressure of 101.3 kPa, the hydrostatic pressure of spherical raindrops at 28 ‰ is as follows for diameters of 5 mm and 0.2 mm. ua − uw =

2Ts 2Ts =⇒ uw = ua − R R

For a diameter of 5 mm: uw = 101.3 kP a −

2(71.4 mN/m) = 101.3 kP a + .057 kP a = 101.36 kP a −2.5 mm

For a diameter of 0.2 mm: uw = 101.3 kP a −

2(71.4 mN/m) = 101.3kP a + 1.428kP a = 102.73 kP a −0.1 mm

3.6 For a bundle of capillary tubes of various sizes ranging between 10−7 and 10−4 m in radii, assume the contact angle is zero, T= 25 ‰, and answer the following: (a) What is the range of matric suction? (b) What is the range of pore water pressure? (c) What is the range of vapor pressure? (d)What is the range of relative humidity? The air-water surface tension, Ts , is 71.79 mN/m. 12

(a) The matric suction, ua − uw for a given radius is given as ua − uw =

2Ts cos α r

For r = 10−4 m: ua − uw =

2(71.79 mN/m) cos(0) = 1.436 kP a 10−4 m

ua − uw =

2(71.79 mN/m) cos(0) = 1.436 M P a 10−7 m

For r = 10−7 m:

Therefore, matric suction ranges from 1.44 kP a to 1.44 M P a (b) Assume an air pressure of 101.3 kP a. uw = ua − (ua − uw ). Therefore, the range for pore water pressure is 99.86 kP a to −1.33 M P a. ¢ ¡ νw cos α (c) The vapor pressure can be estimated as uv = exp − 2Ts rRT uv0 . At T = 25 ‰, uv0 = 3.167 kP a. For r = 10−4 m: µ ¶ 2(71.79 mN/m)(0.018 m3 /kmol) cos 0 uv = exp − (3.167 kP a) (10−4 m)(8.31432 J/mol · K)(298.15 K uv = 3.16697 kP a For r = 10

−7

m:

µ ¶ 2(71.79mN/m)(0.018 m3 /kmol) cos 0 uv = exp − (3.167 kP a) (10−7 m)(8.31432 J/mol · K)(298.15 K uv = 3.13415 kP a Therefore, the range of vapor pressure is from 3.167 kP a to 3.134 kP a. (d) Relative humidity is For r = 10−4 m:

uv uv0 .

RH = For r = 10−7 m: RH =

3.167 kP a ≈ 100% 3.167 kP a

3.13415 kP a = 0.9896 = 98.96% 3.167 kP a

3.7 Table 3.5 (p. 125) shows data comprising the soil-water characteristic curve during a drying process for an unsaturated soil. Assuming the drying process has a contact angle of zero, and the wetting process has a contact angle of 30◦ , calculate and plot the soil-water characteristic curve for the wetting process. The matric suction for the drying process can be computed from equation 3.34. The matric suction for the wetting process is the matric suction for the drying process multiplied by cos αwet cos αdry . A plot of the SWCC for the wetting process is included below. 13

Figure S3.1: Soil-Water Characteristic Curve for Wetting Process 3.8 (a) Plot the relationship between matric suction (kPa, log scale) and relative humidity (%, linear scale) for temperatures of 20, 40, and 60 ‰. (b) Plot the relationship between relative humidity (%) and capillary tube radius (m) for a temperature of 20 ‰ and contact angle of 0◦ , 30◦ , and 60◦ . (c) Discus the general characteristics of each plot. Matric suction is given by

RT ln(RH) νw where νw = 0.0018m3 /kmol, and R = 8.314 J/mol · K. ua − uw = −

(a) For temperatures of 20, 40, and 60 ‰ : As shown by Figure S3.2, the SWCC for the range of temperatures is almost similar, demonstrating that the temperature dependence on SWCC is not significant at a large scale. (b) A plot of relative humidity and capillary tube radius for a temperature of 20 ‰ and contact angle of 0◦ , 30◦ , and 60◦ is shown in Figure S3.3. At 20 ‰, Ts = 72.75mN/m. νw = 0.0018m3 /kmol, and assume T = 298.15 K. It is important to note that as the contact angle increases, the relative humidity at a given capillary tube radius increases. 14

Figure S3.2: Matric Suction vs. Relative Humidity for T = 20, 40, and 60 ‰ (c) In Figure S3.2, the temperature does not significantly change the matric suction at a given relative humidity.

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Figure S3.3: Relative Humidity vs Contact Angle In Figure S3.3 , relative humidity increases as the radius of the capillary tube increases, with a relative humidity of 100 % attained at approximately 1 µm. Relative humidity is approximately 35 % at 1 ˚ Afor small contact angles. Qualitatively, this indicates that for small pore sizes on the order of 1 µm and smaller, there will be large matric suction. 3.9 For unsaturated sand undergoing a drying process at 20 ‰, where the radius of the air-water menisci varies between 10−6 and 10−5 m, the contact angle is zero, and the air pressure is zero, answer the following:(a) What is the range of vapor pressure in the soil pores? (b) What is the range of relative humidity in the soil pores? (c) What is the range of pore water pressure in the soil pores? At a temperature of 20‰ , Ts = 72.75mN/m. Saturated vapor pressure is uv0 = 2.337 kP a. (a) For r = 10−6 m: µ uv0 = exp −

2(72.75 mN/m)(0.018 m3 /kmol) (10−6 m)(8.314 J/mol · K)(293.15 K) uv0 = 2.3345 kP a 16

¶ (2.337 kP a)

For r = 10−5 m: µ uv0 = exp −

2(72.75 mN/m)(0.018 m3 /kmol) (10−5 m)(8.314 J/mol · K)(293.15 K)

¶ (2.337 kP a)

uv0 = 2.3368 kP a Therefore, vapor pressure ranges from 2.335 to 2.337 kPa in the soil pores. (b) For r = 10−6 m: RH =

2.3345 kP a = 0.9989 = 99.9% 2.337 kP a

RH =

2.3368 kP a = 0.9999 = 100.0% 2.337 kP a

For r = 10−5 m:

Therefore, relative humidity ranges from 99.9 % to 100.0 % in the soil. (c) Since air pressure is zero, uw = For r = 10−6 m: uw =

RT νw

ln(RH).

(8.314 J/mol · K)(298.15 K) ln(0.9989) 0.018 m3 /kmol uw = −145.5 kP a

For r = 10−5 m: uw =

(8.314 J/mol · K)(298.15 K) ln(0.9999) 0.018 m3 /kmol uw = −14.55 kP a

Therefore, the pore water pressure ranges from -14.6 kPa to -145.5 kPa. 3.10 If the negative pore pressure in the sand from the previous example acts to draw the soil pore water above the water table in the field, what is the corresponding range of the height above the water table? If the negative pore water from the previous example were to draw soil pore water above the water table in the field, the corresponding height is calculated by recognizing that the pore water pressure corresponds to a head, which is the height of the water above the water table. Assume γw = 9.81 kN/m3 . For r = 10−6 m: h=

uw 145.5 kN/m2 = = 14.83m γw 9.81 kN/m3

h=

14.55 kN/m2 uw = = 1.483m γw 9.81 kN/m3

For r = 10−5 m

The corresponding range of the height above the water table is 1.5 m to 14.8 m.

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3.11 Complete or answer the following for Figure 3.26 (p. 127). (a) Label the wetting and drying branches of the characteristic curve for the soil on Fig. 3.26a. (b) What is the saturated water content during drying for the soil on FIg 3.26a? (c) Estimate the air-entry pressure during drying for the soil on Fig 3.26a. (d) Which soil on Fig 3.26a is the more fine-grained? (e) Estimate the residual water content for soil B. (a) Please see Figure S3.4 (b) The saturated water content during drying for the soil on Figure S3.4 is 0.3, or 30 %. (c) The air-entry pressure during drying for the soil is approximately 10 kPa. (d) Soil A is more fine-grained since matric suction occurs over a wider range of volumetric water contents. (e) The residual water content for soil B, as shown on Figure S3.4, is 0.02, or 2%.

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Figure S3.4: Soil Water Characteristic Curves

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