A number of equations have been derived for predicting the saturated hydraulic con- ductivity from the pore size distribution. One begins with Poiseuille’s equation, which describes the flux of water in a single capillary. Capillaries are interconnected in various ways to represent soil pores. A tortuosity correction is often applied to account for the increased path that must be covered by water moving through the porous material, and the equation is integrated over all pore sizes present in the soil. The result of one such model is (Scheidegger, 1960)
Ks=ρl
φfr2
ηl (6.5)
130 Steady-State Water Flow and Hydraulic Conductivity
whereφf is the total porosity,lis a tortuosity factor,ris a ‘mean hydraulic radius’ repre- sentative of the pore size distribution of the porous material andηis viscosity. A different approach was used by Childs and Collis-George (1950) and Marshall (1958). Since this approach is useful for finding unsaturated as well as saturated conductivity, it will be examined in detail.
Poiseuille’s law describes the flow of water in a cylinder of radiusR [m] and length L[m]. A hydrostatic pressure difference [N m–2] between pointsP1andP2 determines a pressure gradientP. Let us now consider a cylinder of radiusrsmaller than the total cylinder radiusR. The hydrostatic forceFh[N] is given by the pressure differenceP multiplied by the cross-sectional areaπr2:
Fh=πr2P (6.6)
The force that acts against the hydrostatic pressure is the viscous drag forceFd, acting along the interior area of the cylinder:
Fd = 2τπrL (6.7)
where τ is the shear stress [N m–2] and 2πrL is the total interior area of the cylinder.
When the flow is steady-state, the two forces are balancing each other and the shear stress can be expressed as
τ =rP
2L (6.8)
The law of viscosity for momentum transport states that the shear stress is determined by the velocity gradient and the dynamic fluid viscosityη [kg m–1 s–1]. Therefore the shear stress of eqn (6.8) can be written as
τ =rP
2L = –ηdv
dr (6.9)
The first step in the derivation of Poiseuille’s law is to compute the velocity as a function of the internal radiusr, therefore integrating eqn (6.9). Rearranging eqn (6.9) to bring the radius on one side and the velocity on the other leads to
r dr= –2Lη
P dv (6.10)
The integration is performed fromr to the total radius of the cylinderR and from the corresponding velocitiesv(r) tov(R) (note that on the cylinder wall, the velocity is zero and thereforev(R) = 0):
Saturated Hydraulic Conductivity 131 R
r
r dr= –2Lη P
0 v(r)
dv (6.11)
r2 2
R
r
= –2Lη P
0
v(r)
(6.12) R2
2 –r2
2 = –2Lη
Pv(r) (6.13)
v(r) = P
4Lη R2–r2!
(6.14) The velocity distribution shows that the velocity is zero where the radiusris equal to the radius of the cylinderR and it is maximum at the centre of the cylinder, wherer = 0.
This equation describes the well-known parabolic distribution of velocities in a cylinder.
The following code PSP_Poisseulle.py is written to compute and plot the velocity vectors in a cylinder. The cylinder and then the velocity vectors are drawn.
#PSP_Poisseulle from visual import * import numpy as np scene.background=(1,1,1) scene.title = ""
l = 5.
R = 1.
D_P = 40.
eta = 1.
# draw cylinder
angles=arange(1.1*pi,2.1*pi,pi/20.) n = 500
for i in range(n):
spring=curve(color=(0,1,1), radius=0.06) for phi in angles:
spring.append(pos=(l*(float(i)/float(n)-0.55), R*cos(phi), R*sin(phi)))
for i in range(11):
x = 0
y = (float(i)/11.-0.5)*2.*R r = np.sqrt(x*x+y*y)
if (r < R):
arrow(pos=(0,y,x),axis = (D_P * (R*R-r*r)/(eta*l*4.),0,0), shaftwidth = 0.03, color=(0,0,0))
Figure 6.1 shows the output of the program. Note that eqn (6.14) is implemented in the last line, to generate the arrows representing the velocity vectors as functions of
132 Steady-State Water Flow and Hydraulic Conductivity
Figure 6.1 Section of a capillary tube and parabolic distribution of velocities.
the radius (r). The instructionsshaftwidth = 0.03, color=(0,0,0)are Visual Python instructions to control the width of the arrow and its colour (0,0,0) for black.
However, for computation of water flow, in capillary tubes, the total volume of water flowing in the cylinder is of interest; therefore, the velocityv(r) must be integrated over the flowing area:
Q=
v(r)dA (6.15)
The area is the cross-sectional area of the cylinder, which is a circle, and the integration can then be performed in cylindrical coordinates, where the incremental area is given by the change in radius multiplied by the change in angle:dA=dr d, whereis the angle, ranging from 0 to 2π. A double integration is then performed over the radius and the angle:
Q= R
0
2π 0
v(r)r dr d (6.16)
Since the integral over the angle is 2π, Q= 2π
R 0
v(r)r dr (6.17)
Now the integral can be solved by substituting the definition of velocity from eqn (6.14):
Q= P 4Lη2π
R 0
R2–r2!
r dr (6.18)
Q= Pπ 2Lη
R 0
R2r dr– R
0
r2r dr
(6.19) SinceRis constant, it can be moved out of the integral:
Q= Pπ 2Lη
R2
R 0
r dr– R
0
r3dr
(6.20)
Saturated Hydraulic Conductivity 133 We now solve the integral:
Q=Pπ 2Lη
"
R2 1 2r2
R
0
–1 4r4
R
0
#
(6.21)
Q= Pπ 2Lη
1 2R4– 1
4R4
= Pπ 2Lη
1
4R4 (6.22)
Equation (6.22) shows that the flow in a cylinder is proportional to the pressure gradient and the fourth power of the radius and inversely proportional to the dynamic viscosity of the fluid. This equation can be written for an infinitesimal variation of pressure with respect to the cylinder length (dP/dx):
Q= πR4 8η
dP
dx (6.23)
whereRis now the total radius of the capillary andxis the axis of flow direction. Con- sidering the discharge over a specific cross-cylinder area, it is possible to define a specific discharge
q= Q A = πR4
8η 1 πR2
dP dx = R2
8η dP
dx (6.24)
The termR2/8ηis the hydraulic conductivity, which is the product of an intrinsic con- ductivity given by the pore radius and an extrinsic conductivity given by the liquid properties (viscosity). Equation (6.24) shows that the conductivity is proportional to the square of the pore radius. If the soil is seen as a large number of capillary tubes of varying sizes, then the flux density in soil is the flux density for each capillary size present in the soil multiplied by the area of that size of capillary per unit cross-sectional area of soil. Contributions from each pore class are summed to get the total flux. Since pores are not continuous, some provision must be made for the way pores can fit together. The Childs and Collis-George (1950) model assumes that only pores in a direct sequence contribute to the flux and that flux is always controlled by the smaller of two pores in a sequence.
If a soil column were broken at some arbitrary point as shown in Fig. 6.2, the exposed face could be examined to determine the area of pores having radii betweenrandr+ dr.
This area can be expressed as
dA
A =F(r)dr (6.25)
134 Steady-State Water Flow and Hydraulic Conductivity
dA/A F(r) dr
dA/A F(r) dr
1 2
Figure 6.2 Schematic of the statistical approach for estimation of hydraulic conductivity.
whereF(r) is a pore size distribution function defined such that the total porosity of the soil is
φf = rmax
0
F(r)dr (6.26)
where rmax is the largest capillary tube (or pore). In eqn (6.25),dA/A represents the probability of finding a pore of a given size on one face of the broken soil column.
The probability of finding a continuous pore from one face to the other is the product of the probabilities of finding a pore of given size on either face. The contribution to conductivity for pores of radiusrandr+dris therefore
ρlr2 8η
dA A
2
= ρlr2
8η
F(r)dr F(r)dr (6.27)
The hydraulic conductivity is the integral over all pore classes, or K= ρl
8η r
0
r 0
r2F(r)dr F(r)dr (6.28)
The pore size distribution function can be inferred from a moisture retention function if the pores are assumed to be cylindrical so that the capillary rise equation (5.18) gives an estimate of the largest water-filled pore at any given matric potential. Combining eqs (5.18) and (5.27) gives
r = – 2γ
ρlψe
θ θs
b
(6.29) Since the pores are assumed to be cylindrical,F(r)dr=dθis the change in water content associated with draining all pores between radiusr andr+dr. Substituting forr2 and F(r)drin eqn (6.28) and integrating gives
Saturated Hydraulic Conductivity 135
K= γ2θs2
2ρlηψe2(2b+ 1)(2b+ 2) θ
θs
2b+2
(6.30) The saturated hydraulic conductivity is the value from eqn (6.30) whenθ =θs. Equa- tion (6.30) says that the saturated hydraulic conductivity of a soil is determined by three soil properties: θs,ψe and b. Of these, ψe is the most important. For soils with similar θsandbvalues,
Ksψe2 = constant (6.31)
This important result is the basis for scaling in studies of variability in hydraulic prop- erties of soils. It says that at a given water content (again assuming constantθs andb), variation inψmis directly related to variation inψe, and variation inKis inversely related toψe2. It is apparent that for water flow and retention calculations,ψeis an important soil parameter. We will investigate scaling properties of hydraulic properties in Chapter 7.
6.3.1 Calculating Saturated Conductivity from Soil Texture Data
The relationships between moisture retention and soil texture in Chapter 5 and the result obtained in eqn (6.31) suggest that saturated hydraulic conductivity is related to soil texture. Table 6.1 gives saturated hydraulic conductivity data from Rawlset al.(1992) Table 6.1Typical saturated hydraulic conductivities for the soil textural classes
Textural class dg ψe b Ks[g s m–3]a
[μm] [J kg–1]
Rawlset al.(1992) Eqn (6.32) Eqn (6.34)
Sand 211.11 –0.7 1.7 5.800 3.697 2.814
Loamy sand 121.77 –0.9 2.1 1.700 2.117 1.994
Sandy loam 61.67 –1.5 3.1 0.720 0.419 0.703
Loam 19.82 –1.1 4.5 0.370 0.978 0.333
Silt loam 10.53 –2.1 4.7 0.190 0.142 0.167
Sandy clay loam 25.16 –2.8 4.0 0.120 0.056 0.196
Clay loam 7.10 –2.6 5.2 0.064 0.072 0.101
Silty clay loam 3.35 –3.3 6.6 0.042 0.035 0.033
Sandy clay 11.36 –2.9 6.0 0.033 0.050 0.055
Silty clay 2.08 –3.4 7.9 0.025 0.030 0.017
Clay 1.55 –3.7 7.6 0.017 0.023 0.017
aTo convert to values in kg s m–3(or cm s–1), multiply these values by 10–3.
136 Steady-State Water Flow and Hydraulic Conductivity
for the various textural classes. There is a general trend towards lower conductivity with smaller mean particle diameter, but the variation within each texture class is large, so one must be careful in using average data like those in Table 6.1.
Correlation of saturated hydraulic conductivity with air entry potentials given in Table 6.1 gives
Ks= 0.0013|ψe|–3 (6.32)
which is similar to eqn (6.31), but with a power of 3 instead of 2 on the air-entry potential. Saturated hydraulic conductivity calculated using eqn (6.32) with represen- tative values for silt and clay contents are also shown in Table 6.1. Equation (6.32) was derived by integrating from the smallest to the largest pores, and considers all pores in the soil, even though the small pores have almost nothing to do with saturated hydraulic conductivity. Rawlset al.(1992) suggest a different approach that focuses on the largest pores. They suggest computing saturated hydraulic conductivity from the equation
Ks=Bφe4 (6.33)
where φe is the porosity between saturation and –33 J kg–1 water potential (the largest soil pores). If we assume a power-law water retention curve, this equation becomes
Ks= 0.07
θs
"
1 – ψe
–33 1/b#4
(6.34)
whereBhas been replaced with 0.07, the value that best fits the data in Table 6.1. The values predicted by eqn (6.34) are shown in Table 6.1. Even though this equation has only one free parameter (a value ofθs= 0.5 was assumed for all textures), it fits the data with only about half the error of eqn (6.32). It also has the advantage that it shows an explicit bulk density dependence (throughθs) ofKs.