Lal R., Shukla M.K. Principles of Soil Physics

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where B and n are constants and can be set at 4 and 1015 respectively; and f

is the air

porosity at Φ

−33 kPa. Using regression techniques Rawls and Brakensiek (1985) have

also proposed relationships between soil texture and porosity to predict K

of soil.

12.3 LAPLACE EQUATION

For flow through soil systems where potential and gradient at every point in the flow

domain remain constant, Darcy’s law is sufficient for such a flow description. Most

processes in nature do not conform to such simple steady and stationary flow processes.

The potential gradient and flux may vary both in the direction and magnitude, such a

process can be described by the law of conservation of matter or the continuity equation.

It states that if outflow is not equal to inflow then the difference between inflow and

outflow is the change in storage, which may be positive or negative depending upon the

magnitude of inflow and outflow.

Consider a three-dimensional (3-D) space as shown in Fig. 12.7. The flow is q

, q

and q

in x-, y-, and z-directions, respectively.

If water content of this 3-D space is θ, then the rate of increase in q is equal to the rate

of change in θ. For a one-dimensional system it can be represented as follows:

(12.37)

The one-dimensional flow according to Darcy’s law can be given as

(12.38)

where H is the hydraulic head and K

is the hydraulic conductivity.

FIGURE 12.7 Flow through a 3-

dimensional space.

Combining Eqs. (12.37) and (12.38)

(12.39)

Principles of soil physics 344

The hydraulic head consists of pressure head and gravitational or positional head above a

reference or datum i.e., H=Φ

+Φ

[Eq. (12.1)]

(12.40,

For horizontal flow if Z=0 and the Eq. (12.40) reduces to

(12.41)

For vertical flow and Eq. (12.40) changes to

(12.42)

(12.43)

In saturated uncompressible soils, the rate of change of moisture content is zero.

Therefore, the left hand term of Eqs. (12.39)-(12.43) is zero. If the soil is homogeneous

then Eq. (12.43) reduces to

(12.44)

For a three dimensional system where conductivity is K

, K

, and K

in X-

Y-, and Z-

directions, the Eq. (12.44) can be written as

(12.45)

If K

, i.e., for isotropic soil, the above equation reduces to a second order partial

differential equation known as the Laplace equation.

(12.46)

The solution of a one-dimensional Laplace equation (boundary value problem) requires

two boundary conditions and is exactly similar to Darcy’s law. The solution of the

Laplace equation in a two-dimensional flow domain requires four or more boundary

conditions to be satisfied. One of boundary conditions for P is at the boundary while the

other is for the derivative of P with respect to the normal. The remaining two boundary

conditions are required in the region of application of the Laplace equation. For a more

elaborate description on the Laplace equation and its solution, readers are referred to

Kirkham and Powers (1972).

12.4 HOMOGENEITY AND HETEROGENEITY

Water flow in saturated soils 345

Under field conditions, the K

may vary from point to point, and in different directions.

When K

remains unchanged in one particular direction and at different locations in the

same hydrologic group, the soil in that area is called homogeneous. When K

is the same

in all directions, the soil is termed as isotropic. The soil is termed anisotropic when K

varies with change in direction. Anisotropy in the soils results from the structural

variability of soils, which may result in macro or micropore flows with a directional bias.

When K

varies according to the direction of flow, it is termed as asymmetrical

(Maasland and Kirkham, 1955). Consider an anisotropic soil with variable K

as K

and

in horizontal and vertical directions, respectively. The average conductivity (K

) for

this type of a soil is the geometric mean of the conductivities.

(12.47)

Non-homogeneity in soils results from layering or horizonation. The physical properties

of soils in different layers vary, resulting in nonhomogeneity. For further explanation of

anisotropy and inhomogeneity, readers can refer to Bear et al. (1968).

12.5 FLOW THROUGH A LAYERED SOIL

Consider a soil profile as given in Fig. 12.8. The soil profile consists of four isotropic

layers. The K

of these four layers is K

, K

, and K

, respectively, from soil surface

downward. The thickness of soil layers is z

, z

, and z

and the potential heads at the

interface of two layers are starting from the top layer are H

, H

, and H

, respectively.

The total thickness of the soil profile (z) is: z=z

FIGURE 12.8 Hydraulic conductivity,

thickness and head in a 4-layered soil.

The effective conductivity (K

) of the soil profile can be calculated as the weighted

conductivity as follows

K*z=K

(12.48)

Principles of soil physics 346

(12.49)

Alternatively, Darcy’s law can be used to calculate the overall K

of the soil profile. The

steady flow through layer 1 can be given as

(12.50)

Accordingly, the steady flow (v) through the entire soil profile can be given as

(12.51)

(12.52)

where K

is the effective conductivity of the soil profile. If we apply Darcy’s law on each

layer, following sets of equations result:

Layer 1:

(12.53)

Layer 2:

(12.54)

Layer 3:

(12.55)

Layer 4:

(12.56)

Note as a consequence of steady flow, the flux is the same through each layer.

Rearranging Eqs. (12.53)–(12.56) in terms of pressure heads and then adding results in

(12.57)

Equating the right-hand side of Eqs. (12.52) and (12.57) gives the weighted conductivity

of the soil profile, similar to Eq. (12.49).

Water flow in saturated soils 347

Example 12.1

Consider two soil columns given in the following figure. Horizontal flow is taking place

from one and the vertical flow is taking place in the other. The saturated hydraulic

conductivity of sand inside both columns is 5×10

−3

cm/sec. Estimate the hydraulic head,

pressure head, and gravitation head at both ends of the columns. Calculate the Darcy’s

flux through both soil columns. Use the data shown in the figures. Make necessary

assumptions.

Solution

Assuming a reference at the middle of soil column, the pressure head at A=20 cm and

gravitational head=0. The pressure head and gravitational head at B are both zero. The

Darcy’s flux through the soil column

Assuming a reference point at the bottom of the vertical soil column, the pressure and

gravitational head for the vertical flow are 20 cm and 50 cm, respectively, at point A′. At

point B′ both are zeros. The Darcy’s flux through this soil column

Example 12.2

Consider a two-layer soil profile as given in the figure below. The thickness of these two

layers is 40 and 30 cm, and hydraulic conductivities are 4×10−

cm/sec and 5×10

−5

cm/sec, respectively. If the head at the top of layer one is 80 cm, calculate: (a) flux

through soil profile and (b) hydraulic and pressure heads at the interface between layers.

Principles of soil physics 348

Solution

Remember the reference point is at the bottom of soil profile. For a steady flow, the

flux q can be given by q=[∆H/(R

)], where R

and R

are hydraulic resistances and

are equal to z

and z

, respectively; where z

and z

are the thickness of layer 1 and

2, respectively. Transforming these values into an equation

for q

If we apply Darcy’s law on the top layer of the soil, the hydraulic head difference will

be H

surface

−H

interface

. Remember Darcy’s equation

interface

denotes the hydraulic head at the interface, since gravitational head at the

interface is 40 cm. The pressure head at the interface will be equal to

Pressure head+Hydraulic head−gravitational head=123−40=83 cm

Example 12.3

If the saturated hydraulic conductivities of layers in Example 12.2 are 5×10

−2

and 6 ×10

−3

cm/sec. Calculate the effective conductivity and Darcy’s flux.

Solution

We can calculate the effective conductivity as

Water flow in saturated soils 349

and the flux through the entire soil column can be calculated using Darcy’s law:

PROBLEMS

1. Assume a 0.1 m long cylindrical capillary tube of uniform diameter of 0.01 m,

which is filled with saturated sand. The tube is placed horizontally and flow is taking

place from right to left. If the flow rate across the capillary is 0.2 m3/h and the pressure

difference across the capillary is 0.02 m, calculate the hydraulic conductivity of soil in

the capillary.

2. Define the conductivity obtained in the above problem. If the head difference is

raised to 0.04 m. What changes do you expect in saturated hydraulic conductivity and

flow? Explain.

3. What is a unit hydraulic gradient? What will be the pressure head difference in

problem 1 for the condition of unit gradient flow? What will be the flow rate under unit

hydraulic gradient? If the soil column in problem 1 is rotated so that the direction of flow

in the column becomes vertical, what will be the head at the inlet for a unit hydraulic

gradient flow?

4. Consider a soil core of equal diameter and length of 0.08 m. Water is allowed to

flow vertically downward under a constant head of 0.04 m. The constant rate of effluent

collected from bottom of core was 0.04 m/s. Calculate the saturated hydraulic

conductivity of soil.

5. In the above problem if head at inlet varies from 0.04 to 0.02 m in time in a time

interval of 2 min, calculate the saturated hydraulic conductivity of soil core. Making

appropriate assumptions calculate the intrinsic permeability of soil.

6. Consider a layered soil system as shown in Fig. 12.8. The thickness of layers from

top to bottom is 10, 20, 20, and 40 cm and the saturated hydraulic conductivities are 0.4,

0.8, 0.6, and 1 cm/h respectively. Calculate the overall conductivity of the soil.

7. If a constant head of 10 cm is applied on the top of the soil layer. Calculate the flux

from the bottom of soil profile. (Hint: Use K value obtained in problem 6). What will be

the Darcy flux at a depth of 30 cm from soil surface if steady flow takes place in the soil

column?

Principles of soil physics 350

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Principles of soil physics 352

Water Flow in Unsaturated Soils

Unsaturated flow of water is a more commonly prevailing condition in the field than

saturated flow. An unsaturated soil zone, or vadose zone, provides a continuum of water-

unsaturated subsurface porous media connecting the soil/atmospheric interface and

underlying saturated groundwater zone. It has several functions including (i) storage of

water and nutrients, and (ii) transmission of water and other substances. The storage of

water and nutrients is vital to the biosphere, and the water transmission is important for

replenishing the aquifers. Unsaturated flow conditions are more complex and very often

do not have direct solutions. Instead indirect methods, approximations, and numerical

methods are more commonly used in the solution of unsaturated flow problems. This

chapter describes the various laws governing the unsaturated flow through porous soil

system.

13.1 UNSATURATED ZONE OR VADOSE ZONE

A soil matrix is considered unsaturated when some of the pores are filled with water and

the remaining pores with air. The unsaturated zone of soil refers to that portion of the

subsurface above the water table, which contains both air and water in the pores. Its

thickness can vary from zero to several meters ranging from a marshy situation to an arid

environment. An unsaturated zone stores the water, nutrients, and other substances and is

of importance to the biosphere. It stores only a tiny fraction of fresh water and, therefore,

plays a minor role in the hydrologic cycle. It is the transmission zone, which redistributes

the water. Therefore, this zone controls the ground water replenishment as well as

evaporation from soil surface. The unsaturated zone experiences transport processes of

various kinds, chemical reactions, biological activity of roots, rodents, worms,

microbiota, and other organisms. It is also a zone of human activity and is used for the

cultivation and disposal of waste. This zone is also drastically disturbed by surface

mining and construction of civil structures (e.g., buildings, roads, etc). The vadose zone

is in direct contact with the atmosphere through gaseous fluxes of water vapor and

greenhouse gases (CO

, CH

, and N

O).

13.2 MECHANISMS OF UNSATURATED FLOW