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

2007

1 All correspondences should be addressed to likosw@missouri.edu or ninglu@mines.edu

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

i

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 courseat the Colorado School of Mines and their fine homeworks served as an initial starting points forchapters 5, 6, 7, 10, and 11.

ii

Chapter 1

State of Usaturated Soil

1.1 Where are the regions in the U.S. where unsaturated soils are likely encountered to significantdepth below the ground surface?

Unsaturated soils are likely encountered to significant depth below the ground surface in largeportions 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 depthand extent of the unsaturated zone. A thick unsaturated zone generally occurs in regionswhere potential evaporation outweighs annual precipitation by a factor ranging between 2and 20 within 40 degree north of equator.

1.3 What is the fundamental difference between unsaturated soils and saturated soils in terms ofpore water pressure?

Pore water pressure in a saturated soil is generally compressive and isotropic, in contrastto 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 statesof stresses. It is described in terms of (c) and internal friction angle (f). It is described interms of states of cohesion (c′) and internal friction angle(φ′) to dictate a failure shear stressat a given normal effective stress (σ′n) in τf = c′ + σ′n tan φ′. Graphically, the Mohr-Coulombfailure criterion plots as a straight line on the effective normal and shear stress graph asshown in S1.1. It is also important to note that the Mohr-Coulomb failure criterion is a linearapproximation and therefore valid only near the range of values from which it was derived.

1

Figure S1.1: Mohr-Coulomb Failure Criterion

2

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 settlementof unsaturated soils, bearing capacity for shallow foundations under moisture loading, slopestability, and land sliding.

1.7 For a given unsaturated soil under either a dry or wet condition, which one has a highersuction?

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 thegiven phenomenon. Material variables generally vary with state variables and describe thephysical characteristics of the material. These variables are intrinsic material properties. Con-stitutive laws describe the governing physical principles which demonstrate interrelationshipsbetween or among state variables and material variables. Constitutive laws are used as themathematical connection between state and material variables for the purpose of predictionand explanation of phenomena (p. 26-28).

1.9 What are the principal differences between saturated and unsaturated soil profiles of pore waterpressure, total stress, and effective stress?

Pore water Pressure Profiles- Generally vary linearly with depth, increasing hydrostaticallybelow 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 self-weight when the material is dewatered. Therefore, the total stress profile for a saturated soilextends 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 isalso important to note that effective stresses for an unsaturated soil are greater than for asaturated soil due to the tensile pore water pressure in the unsaturated soil (p. 22-23). Figures1.12 and 1.13 graphically demonstrate these concepts. For both saturated and unsaturatedconditions, 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 ahigher effective stress? Why?

According to Bishop’s effective stress concept, unsaturated conditions have higher effectivestress since pore water pressures are negative, leading to a greater effective stress. This isdemonstrated by examining the equation: σ′ = (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 saturatedand unsaturated states, respectively?

3

The shape of the pore pressure profile under the hydrostatic condition for saturated andunsaturated 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 suctionvalue? If the soil at the air dry condition has a matric suction of 100 MPa, what is the soilwater potential in joules per kilogram?

For −1000 J/kg potential the equivalent soil suction is −1000 kPa, since 100 J/kg = 100 kPa.Soil with a matric suction of 100 MPa has a soil water potential of 100, 000 J/kg since0.1 MPa = 100 J/kg (p. 40).

1.13 Three soils- clay, silt, and sand- are all equilibrated at the same matric suction, which soilhas the highest water content and why?

If clay, silt, and sand are all equilibrated at the same matric suction, then clay has the highestwater content as shown on Figure 1.20 (p. 42). Clay has the highest water content at a givenmatric suction due its charged surfaces and very high specific surface area as (p. 42) Sandand 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 po-tential of the pore water relative to a reference potential of free water. These mechanismsinclude capillary effects, short-range adsorption effects composed of particle-pore water inter-action, 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 forcefield interactions at the solid-liquid interface. Osmotic effects are the result of dissolved solutein 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 ofsoil particles could occur. Matric suction is generally used to group the aggregate of capillaryand short-range adsorption effects. Osmotic suction refers to the aggregate of osmotic effects(p. 34-35).

5

Chapter 2

Material Variables

2.1 What are the state variables that control the density of air? What is the average air densityat your location?

Temperature, pressure, and relative humidity are state variables that control the density of airas evidenced by the equation below. The average air density for Golden, CO during a typicalwinter, assuming a 10% relative humidity, an average temperature of 0� and an averagepressure of 85 kPa is as follows:

Moist air density is estimated using Tables 2.8 and 2.9.

ρa,moist =udωd

RT− 0.611

(ωd

ωv− 1

)exp

(17.27

T − 273.2T − 36

)ωvRH

RT

The first part of the expression is estimated as 1.084 kgm3 (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 = 1.084kg

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 tothe absolute humidity in equilibrium with free water at the same temperature. RH is alsoequivalent to the ratio of vapor pressure in equilibrium with a given solution and the saturatedvapor 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 viscosityof air? The viscosity of which phase, air or water, is more sensitive to temperature changesbetween 0 and 100� ?

νw/νa = 8.77×10−4kg/m·s1.845×10−5kg/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

6

humidity variation? If the vapor pressure remains unchanged, at what temperature will dewformation occur?

RH = uv

uv,sat. With uv constant, RH depends on the saturated vapor pressure which is related

to change in temperature according to the following equation: uv,sat = 0.611 exp(17.27 T−273.2

T−36

).

At 15�, uv,sat = 1.70 kPa. At 30�, uv,sat = 4.24 kPa. With uv = 1.6 kPa, RH ranges from93.8% to 37.7%. If the vapor pressure remains unchanged, dew formation will occur whenuv,sat = uv = 1.6 kPa. This condition is met when T = 287.2 K, or 14.0�. Dew formationwill 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 + 1Gsw

=1

1 + 12.7(3)

= 0.89 = 89%

2.6 A closed room is filled with humid air. If the temperature rises significantly, does the relativehumidity increase or decrease?

In a closed room filled with humid air, a significant temperature rise will result in a decreaseof relative humidity (RH) since RH is inversely proportional to temperature (by Equations2.11 and 2.12).

2.7 Can the vapor pressure of soil gas be greater than the saturation pressure at the same temper-ature and pressure? Why or why not?

No, the vapor pressure of soil gas cannot be greater than saturation pressure at the same tem-perature and pressure because the saturation pressure is the maximum pressure correspondingto 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 mathematicallyevidenced by the following definition:

θ =Vw

Vt=

Vw

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.

7

2.10 When the temperature of unsaturated soil increases, does the surface tension at the air-waterinterface increase or decrease?

When the temperature of unsaturated soil increases, the surface tension at the air-waterinterface 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 95kPa, respectively? What is the relative change in dry-air density if the tempreature rises to40� and the air pressure remains unchanged? If the temperature is kept at a constant valueof 25� , how much pressure change is required to cause the dry-air density to decrease by15% compared to 95 kPa?

The density of dry air at T = 25� and P = 95 kPa is

ρd =udωd

RT=

(95 kPa)(28.966 × 10−3 kg/mol)(8.314 N ·m/mol ·K)(298.2 K)

= 1.110 kg/m3

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)(8.314 N ·m/mol ·K)(313.2 K)

= 1.057 kg/m3

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-airdensity to decrease by 15 % compared to 95 kPa (ρd = 0.85 · 1.110 kg/m3 = 0.944 kg/m3) iscomputed as follows:

ud =(ρd)(RT )

ωd=

(0.944 kg/m3)((8.314 N ·m/mol ·K)(298.2 K))28.966 × 10−3 kg/mol

= 80.76 kPa

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 fora sandy soil as 10−3 m, and a specific discharge for both air and water as 10−2 m/s, identifythe 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 meanpore size for a sandy soil as 10−3 m, and a specific discharge, q, for both air and water as10−2 m/s, the Reyonds number for air and water are:

Rea =ρvd

µ=

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

8

Rew =ρvd

µ=

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

These Reynolds numbers indicate that the flow regime for air is linear and laminar since thevalue is less than one. The Reynolds number between 1 and 100 for water indicates thatalthough 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 isthe dew-point temperature if the vapor density is maintained constant but the temperature dropsduring the night? (d) What is the absolute humidity if temperature in the soil is maintainedconstant but vaporization is allowed to occur? (e) What is the free energy per unit mass ofthe 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 abso-lute 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

ωwln(RH) = − (8.314 N ·m/mol ·K)(295.2 K)

18 kg/kmolln(0.80)

E = 30.43 kJ/kg

2.14 At a prevailing temperature of 25� and a pressure of 95 kPa, how much does the density ofair change from a completely dry state to a 100% relative humidity state?

From Table 2.8 and Table 2.9:

RH(%) ρa(kg/m3)0 1.110

100 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 pressureinside the water meniscus for a capillary tube with diameter of 0.001 mm? If the temperatureincreases to 50� , what is the pressure inside the meniscus?

ua − uw =2Ts

R

9

where ua = air pressure, uw = pressure inside the water meniscus, Ts = the surface tensionbetween 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

R= 101.3 kPa− 2 · 72.75 mN/m

0.0005 mm= −189.7 kPa

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

uw = ua − 2Ts

R= 101.3 kPa− 2 · 67.91 mN/m

0.0005 mm= −170.3 kPa

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 withincreasing temperature

2.16 If a tensiometer were used to measure matric suction of unsaturated soil at an elevationof 500 m above sea level, what would be the approximate maximum possible reading of thetensiometer?

If a tensiometer were used to measure matric suction of unsaturated soil at an elevation of500 m above sea level, the approximate maximum possible reading of the tensiometer wouldbe approximately 91 kPa (from Figure 2.17).

10

Chapter 3

Interfacial Equilibrium

3.1 A liter of water at 25� can dissolve 0.0283 L of oxygen when the pressure of oxygen inequilibrium with the solution is 1 atm. Derive the Henry’s law constant for oxygen in waterfrom this information.

Mi/ωi

Vl= KHiuai

Since there are 22.4 L per each mole of an ideal gas like oxygen:

1 mol

22.4 L(0.0283 L) = 1.2634 × 10−3 mol O2.

Notice that Mi/ωi = moles of gas. Therefore, Mi/ωi = 1.2634 × 10−3 mol. Solving equa-tion 3.1 for KHi:

Mi/ωi

Vlui= KHiKHi =

1.2634 × 10−3 mol O2

(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 airpressure 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 isρair = 1.2 kg/m3 and the density of water is ρw = 998 kg/m3.

Hai =ρair

ρwhair =

1.2 kg/m3

998 kg/m30.01708 = 2.05 × 10−5

3.3 If the air pressure changes to 10 bars in the previous problem, what is the mass coefficient ofsolubility?

If the air pressure changes to 10 bars (1 MPa) in the previous problem, the density of air is

ρair =uaωa

RT=

(106 N/m2)(29× 10−3 kg/mol)(8.314 N ·m/mol ×K)(298 K)

= 11.71 kg/m3

11

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

Hai =ρair

ρwhair =

11.71 kg/m3

998 kg/m30.01708 = 2.00 × 10−4

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 containerand the pore water. Assume the total mass of water vapor in each side is much less than theamount of water in the capillary tubes. Also assume that the initial water levels in the tubesare 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 thetank has a lower vapor pressure than the right. After opening the valve, the vapor pressureson each side of the tank equalize, increasing the relative humidity on the left side of the tankwith reference to the initial state. Similarly, the relative humidity of the right side of the tankis decreased with respect to the initial equilibrium state.Since the relative humidity on the right side of tank has decreased, the water content of thecapillary tube decreases due to capillary evaporation resulting in a decreasing water level inthese tubes. Conversely, the tubes in the left side of the tank experience an increased relativehumidity, which results in capillary condensation, increasing the water level in the tubes. Thewater level of the tubes in the left side of the container will increase and the water level in thetubes in the right side of the container will decrease. Assuming that the contact angle doesnot change as water content changes, the tubes on the right side of the tank will completelyevaporate and the tubes on the left side of the tank will fill up to reflect the change in watervolume on the right.

3.5 Calculate the hydrostatic pressure of water at 28� in spherical raindrops with (a) 5 mmdiameter and (b) 0.2 mm diameter.

The surface tension between air and water is 71.4 mN/m, as determined from interpolation oftable 2.10 (Lu, 76). Assuming an ambient air pressure of 101.3 kPa, the hydrostatic pressureof spherical raindrops at 28 � is as follows for diameters of 5 mm and 0.2 mm.

ua − uw =2Ts

R=⇒ uw = ua − 2Ts

R

For a diameter of 5 mm:

uw = 101.3 kPa− 2(71.4 mN/m)−2.5 mm

= 101.3 kPa + .057 kPa = 101.36 kPa

For a diameter of 0.2 mm:

uw = 101.3 kPa− 2(71.4 mN/m)−0.1 mm

= 101.3kPa + 1.428kPa = 102.73 kPa

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 rangeof matric suction? (b) What is the range of pore water pressure? (c) What is the range ofvapor 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 cosα

r

For r = 10−4 m:

ua − uw =2(71.79 mN/m) cos(0)

10−4 m= 1.436 kPa

For r = 10−7 m:

ua − uw =2(71.79 mN/m) cos(0)

10−7 m= 1.436 MPa

Therefore, matric suction ranges from 1.44 kPa to 1.44 MPa

(b) Assume an air pressure of 101.3 kPa.uw = ua − (ua − uw).Therefore, the range for pore water pressure is 99.86 kPa to −1.33 MPa.

(c) The vapor pressure can be estimated as uv = exp(− 2Tsνw cos α

rRT

)uv0. At T = 25 �,

uv0 = 3.167 kPa.For r = 10−4m:

uv = exp

(− 2(71.79 mN/m)(0.018 m3/kmol) cos 0

(10−4 m)(8.31432 J/mol · K)(298.15 K

)(3.167 kPa)

uv = 3.16697 kPa

For r = 10−7m:

uv = exp

(− 2(71.79mN/m)(0.018 m3/kmol) cos 0

(10−7m)(8.31432 J/mol · K)(298.15 K

)(3.167 kPa)

uv = 3.13415 kPa

Therefore, the range of vapor pressure is from 3.167 kPa to 3.134 kPa.

(d) Relative humidity is uv

uv0.

For r = 10−4m:RH =

3.167 kPa

3.167 kPa≈ 100%

For r = 10−7m:RH =

3.13415 kPa

3.167 kPa= 0.9896 = 98.96%

3.7 Table 3.5 (p. 125) shows data comprising the soil-water characteristic curve during a dryingprocess for an unsaturated soil. Assuming the drying process has a contact angle of zero, andthe wetting process has a contact angle of 30◦, calculate and plot the soil-water characteristiccurve for the wetting process.

The matric suction for the drying process can be computed from equation 3.34. The matricsuction for the wetting process is the matric suction for the drying process multiplied bycos 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 relativehumidity (%) and capillary tube radius (m) for a temperature of 20� and contact angle of0◦, 30◦, and 60◦. (c) Discus the general characteristics of each plot.

Matric suction is given by

ua − uw = −RT

νwln(RH)

where νw = 0.0018m3/kmol, and R = 8.314 J/mol ·K.

(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 largescale.

(b) A plot of relative humidity and capillary tube radius for a temperature of 20 � andcontact 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 agiven 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 agiven relative humidity.

15

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 isapproximately 35 % at 1 Afor small contact angles. Qualitatively, this indicates that forsmall 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-watermenisci varies between 10−6 and 10−5 m, the contact angle is zero, and the air pressure iszero, answer the following:(a) What is the range of vapor pressure in the soil pores? (b) Whatis the range of relative humidity in the soil pores? (c) What is the range of pore water pressurein the soil pores?

At a temperature of 20� , Ts = 72.75mN/m. Saturated vapor pressure is uv0 = 2.337 kPa.

(a) For r = 10−6m:

uv0 = exp

(− 2(72.75 mN/m)(0.018 m3/kmol)

(10−6m)(8.314 J/mol ·K)(293.15 K)

)(2.337 kPa)

uv0 = 2.3345 kPa

16

For r = 10−5m:

uv0 = exp

(− 2(72.75 mN/m)(0.018 m3/kmol)

(10−5m)(8.314 J/mol ·K)(293.15 K)

)(2.337 kPa)

uv0 = 2.3368 kPa

Therefore, vapor pressure ranges from 2.335 to 2.337 kPa in the soil pores.

(b) For r = 10−6m:

RH =2.3345 kPa

2.337 kPa= 0.9989 = 99.9%

For r = 10−5m:RH =

2.3368 kPa

2.337 kPa= 0.9999 = 100.0%

Therefore, relative humidity ranges from 99.9 % to 100.0 % in the soil.

(c) Since air pressure is zero, uw = RTνw

ln(RH).

For r = 10−6m:

uw =(8.314 J/mol ·K)(298.15 K)

0.018 m3/kmolln(0.9989)

uw = −145.5 kPa

For r = 10−5m:

uw =(8.314 J/mol ·K)(298.15 K)

0.018 m3/kmolln(0.9999)

uw = −14.55 kPa

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 porewater above the water table in the field, what is the corresponding range of the height abovethe water table?

If the negative pore water from the previous example were to draw soil pore water above thewater table in the field, the corresponding height is calculated by recognizing that the porewater pressure corresponds to a head, which is the height of the water above the water table.Assume γw = 9.81 kN/m3. For r = 10−6m:

h =uw

γw=

145.5 kN/m2

9.81 kN/m3= 14.83m

For r = 10−5m

h =uw

γw=

14.55 kN/m2

9.81 kN/m3= 1.483m

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

17

3.11 Complete or answer the following for Figure 3.26 (p. 127). (a) Label the wetting and dryingbranches of the characteristic curve for the soil on Fig. 3.26a. (b) What is the saturated watercontent during drying for the soil on FIg 3.26a? (c) Estimate the air-entry pressure duringdrying 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 volumetricwater contents.

(e) The residual water content for soil B, as shown on Figure S3.4, is 0.02, or 2%.

18

Figure S3.4: Soil Water Characteristic Curves

19