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In the case of soil, it can be modeled by assuming that it contains multiple pores that form small capillaries running through it. Based on this, equations for laminar flow through tubes can be applied, and the hydraulic resistances of capillary networks can be calculated. These resistances depend on the porosity and, thus, on the proportion of the different components.

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Concept

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The individual pores come together to form chains that create capillaries through which water flows.

To model this phenomenon, it is necessary to estimate both the radius of these capillaries and their length, considering that they are typically not straight.

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Concept

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The section Flow ($S$) includes the pore section ($S_p$) generated by the number of capillaries ($N_p$):

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Concept

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If we observe a cross-section of the soil, we'll notice that capillaries pass through the spaces between the grains. By doing so, their number is similar to that of the grains themselves, so we can assume that the number of capillaries ($N_p$) is similar to the number of grains present in the cross-section:

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The total flow is calculated as the sum of the individual flows through the different pores:

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If we assume that all the pores are identical, we can obtain the total flow ($J_{Vt}$) by multiplying the volume flow ($J_V$) individually by the number of capillaries ($N_p$).

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Equation

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The flow of liquid in a porous medium such as soil is measured using the variable the flux density ($j_s$), which represents the average velocity at which the liquid moves through it. When modeling the soil and how the liquid passes through it, it is found that this process is influenced by factors such as the porosity ($f$) and the radius of a generic grain ($r_0$), which, when greater, facilitate the flow, whereas the viscosity ($\eta$) hinders passage through capillaries, reducing the flow velocity.

The modeling eventually incorporates what we will call the hydraulic conductivity ($K_s$), a variable that depends on the interactions between the radius of a generic grain ($r_0$), the porosity ($f$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the viscosity ($\eta$), and the generic own porosity ($q_0$):

$K_s \equiv \displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }$

Conditions

Variables

Relation

Development

Since the flux density ($j_s$) is related to the radius of a generic grain ($r_0$), the porosity ($f$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the viscosity ($\eta$), the generic own porosity ($q_0$), the height difference ($\Delta h$), and the sample length ($\Delta L$) through the equation:

$j_s =-\displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }\displaystyle\frac{ \Delta h }{ \Delta L }$

We can define a factor that we'll call the hydraulic conductivity ($K_s$) as follows:

$K_s \equiv \displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }$

This factor encompasses all the elements related to the properties of both the soil and the liquid that flows through it.

the hydraulic conductivity ($K_s$) expresses how easily the liquid is conducted through the porous medium. In fact, the hydraulic conductivity ($K_s$) increases with the porosity ($f$) and the radius of a generic grain ($r_0$), and decreases with the generic own porosity ($q_0$) and the viscosity ($\eta$).

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If we examine the literature, we can find estimates of hydraulic conductivity for different soil textures, which are shown here as a function of their exponent (i.e., -7 is indicated for a hydraulic conductivity of 1E-7 m/s):

The results are summarized in the following table:

These data were obtained from the following literature, which is referenced in the hydraulic conductivity column:

[1] "Geotechnical Engineering Principles and Practices" by Donald P. Coduto et al., Prentice Hall (1999)

[2] "Soil Mechanics and Foundations" by Muni Budhu, John Wiley & Sons. (2011)

[3] "Introduction to Environmental Engineering" by Mackenzie Davis and David Cornwell, McGraw Hill (2022)

[4] "Principles of Geotechnical Engineering" by Braja M. Das, CL-Engineering (2009)

[5] "Soil Mechanics in Engineering Practice" by Karl Terzaghi and Ralph B. Peck, John Wiley & Sons. (1996)

[6] "Soil Mechanics: Concepts and Applications" by William Powrie, CRC Press (2013)

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Parameters

$\rho_w$

rho_w

Liquid density

kg/m^3

$\pi$

pi

Pi

rad

$\eta_w$

eta_w

Water Viscosity

Pa s

Variables

$L$

L

Capillary length

m

$J_V$

J_V

Hagen-Poiseuille Law

m^3/s

$J_{Vt}$

J_Vt

Hagen-Poiseuille law for soil

m^3/s

$\Delta h$

Dh

Height of liquid column

m

$N_p$

N_p

Number of capillaries

-

$S_p$

S_p

Pore section

m^2

$f$

f

Porosity

-

$R$

R

Porosity Radio

m

$\Delta p_2$

Dp_2

Pressure Difference 2

Pa

$J_{Vt}$

J_Vt

Sample Flow

m^3/s

$S_1$

S_1

Section in point 1

m^2

$S_2$

S_2

Section in point 2

m^2

$S$

S

Section or Area

m^2

$S_t$

S_t

Section soil

m^2

$J_{Vt}$

J_Vt

Total flow

m^3/s

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

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Calculations

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Solved

Translated

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Variable Given Calculate Target : Equation To be used

Equations

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Equation

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The porosity ($f$) can be calculated from the pore section ($S_p$) and the section Flow ($S$) using the following formula:

$f = \displaystyle\frac{ S_p }{ S }$

Conditions

Variables

Relation

Development

With the height the infinitesimal distance ($ds$), the volume of the section Flow ($S$) is

$S ds$

and the volume of the pores with the pore section ($S_p$) is

$S_p ds$

Therefore, the porosity ($f$) is calculated as

$f = \displaystyle\frac{S_p ds}{S ds} = \displaystyle\frac{S_p}{S}$

resulting in the following equation:

$f = \displaystyle\frac{ S_p }{ S }$

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Equation

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Since the cross-sectional area of a pore of tube radius ($R$) is $\pi R^2$ and the number of capillaries ($N_p$) is related to that, we can calculate the pore section ($S_p$) as follows:

$S_p = N_p \pi R ^2$

Conditions

Variables

$N_p$

Number of capillaries

$-$

6039

$\pi$

Pi

3.1415927

$rad$

5057

$S_p$

Pore section

$m^2$

6040

$R$

Porosity Radio

$m$

6021

Relation

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Equation

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With the porosity ($f$) and the section Flow ($S$), we obtain the pore section ($S_p$), which when divided by the calculated capillary section of the tube radius ($R$), yields the number of capillaries ($N_p$) as follows:

$N_p =\displaystyle\frac{ f S }{ \pi R ^2}$

Conditions

Variables

$N_p$

Number of capillaries

$-$

6039

$\pi$

Pi

3.1415927

$rad$

5057

$f$

Porosity

$-$

5805

$R$

Porosity Radio

$m$

6021

$S_t$

Section soil

$m^2$

6041

Relation

Development

As the porosity ($f$), calculated with the pore section ($S_p$) and the section Flow ($S$) using

$f = \displaystyle\frac{ S_p }{ S }$

together with the equation for the calculation of the pore section ($S_p$) based on the number of capillaries ($N_p$) and the tube radius ($R$) through

$S_p = N_p \pi R ^2$

results in

$f = \displaystyle\frac{N_p\pi R^2}{S}$

can be solved for the number of capillaries ($N_p$), yielding

$N_p =\displaystyle\frac{ f S }{ \pi R ^2}$

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Equation

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If we assume that the number of capillaries is equal to the number of visible grains in a section, we can demonstrate that for a grain radius of the radius of a generic grain ($r_0$) and a porosity ($f$), the tube radius ($R$) will be equal to:

$R = \sqrt{\displaystyle\frac{ f }{1- f }} r_0$

Conditions

Variables

Relation

Development

If we consider the area in the section that does not contain pores, subtracting the pore section ($S_p$) from the section Flow ($S$) and dividing it by the area of a generic grain with a radius of radius of a generic grain ($r_0$), we obtain the number of visible grains in the cross-section:

$\displaystyle\frac{S-S_p}{\pi r_0^2}=\displaystyle\frac{(1-f)S}{\pi r_0^2}$

where we use the relation for the porosity ($f$):

$f = \displaystyle\frac{ S_p }{ S }$

If the number of grains is equal to number of capillaries ($N_p$) with the expression:

$N_p =\displaystyle\frac{ f S }{ \pi R ^2}$

where the radius is the tube radius ($R$). With this, we obtain the relation:

$\displaystyle\frac{(1-f)S}{\pi r_0^2}=\displaystyle\frac{fS}{\pi R^2}$

resulting in:

$R = \sqrt{\displaystyle\frac{ f }{1- f }} r_0$

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Equation

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If we equate the volume of a capillary to the volume of a chain of grains multiplied by the sample length ($\Delta L$), we obtain a relationship between the capillary length ($l$) and the porosity ($f$) given by:

$l = q_0 \displaystyle\frac{(1- f )}{ f } \Delta L$

Conditions

Variables

Relation

Development

The volume of the capillary can be calculated from the tube radius ($R$) and the capillary length ($l$), which is equal to the volume of a chain of grains of the radius of a generic grain ($r_0$) and the tube length ($\Delta L$) multiplied by the generic own porosity ($q_0$):

$\pi R^2 l = q_0 \pi r_0^2 \Delta L$

This, in conjunction with the porosity ($f$) in the relationship

$R = \sqrt{\displaystyle\frac{ f }{1- f }} r_0$

results in the following relationship:

$l = q_0 \displaystyle\frac{(1- f )}{ f } \Delta L$

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Equation

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The total flow ($J_{Vt}$) is calculated by multiplying ($$) by the value of the volume flow ($J_V$) in each capillary, as follows:

$J_{Vt} = N_p J_V$

Conditions

Variables

$J_V$

Hagen-Poiseuille Law

$m^3/s$

6023

$N_p$

Number of capillaries

$-$

6039

$J_{Vt}$

Total flow

$m^3/s$

6042

Relation

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Equation

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If we apply the Hagen-Poiseuille equation to the volume flow ($J_V$) for the case of capillaries with the tube radius ($R$) expressed in terms of the porosity ($f$) and the capillary length ($l$) as a function of the sample length ($\Delta L$), we can calculate the total flow ($J_{Vt}$) using

The result can be expressed in terms of the section Flow ($S$), the viscosity ($\eta$), the generic own porosity ($q_0$), the radius of a generic grain ($r_0$), and the pressure difference ($\Delta p$):

$J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p$

Conditions

Variables

$L$

Capillary length

$m$

6022

$J_{Vt}$

Hagen-Poiseuille law for soil

$m^3/s$

6043

$f$

Porosity

$-$

5805

$R$

Porosity Radio

$m$

6021

$\Delta p_2$

Pressure Difference 2

$Pa$

5820

$S_t$

Section soil

$m^2$

6041

$\eta_w$

Water Viscosity

$Pa s$

6020

Relation

Development

To calculate the total flow ($J_{Vt}$) using the number of capillaries ($N_p$) and the volume flow ($J_V$) for each capillary through

$J_{Vt} = N_p J_V$

we obtain the number of capillaries ($N_p$) with the porosity ($f$), the section Flow ($S$), and the tube radius ($R$) through

$N_p =\displaystyle\frac{ f S }{ \pi R ^2}$

and the Hagen-Poiseuille law using the viscosity ($\eta$), the pressure difference ($\Delta p$), and the capillary length ($l$) is calculated with

$J_v = - \displaystyle\frac{\pi R^4}{8\eta}\displaystyle\frac{\Delta p}{l}$

Using the relationship for the tube radius ($R$) in terms of the radius of a generic grain ($r_0$)

$R = \sqrt{\displaystyle\frac{ f }{1- f }} r_0$

and for the capillary length ($l$), the generic own porosity ($q_0$), and the sample length ($\Delta L$)

$l = q_0 \displaystyle\frac{(1- f )}{ f } \Delta L$

we obtain

$J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p$

This equation corresponds to the Kozeny-Carman equation, which was developed by Kozeny and Carman to model the flow of a liquid through a porous medium and was published in:

• In regards to the capillary conduction of water in soil., ("Ueber kapillare Leitung des Wassers im Boden."), J. Kozeny, Sitzungsber Akad. Wiss., Wien, 136(2a): 271-306 (1927)

• Fluid flow through granular beds., P.C. Carman, Transactions, Institution of Chemical Engineers, London, 15: 150-166 (1937)

• Flow of gases through porous media., P.C. Carman, Butterworths, London (1956)

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Equation

>Top, >Model

$R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$

Conditions

Variables

$S_1$

Section in point 1

$m^2$

5257

$S_2$

Section in point 2

$m^2$

6017

$S$

Section or Area

$m^2$

5405

Relation

Development

With Darcy's law, where the pressure difference ($\Delta p$) equals the hydraulic resistance ($R_h$) and the total flow ($J_{Vt}$):

$\Delta p = R_h J_V$

Thus, with the equation for the soil with the section Flow ($S$), the radius of a generic grain ($r_0$), the viscosity ($\eta$), the generic own porosity ($q_0$), the porosity ($f$), the pressure difference ($\Delta p$), and the sample length ($\Delta L$):

$J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p$

Therefore, the hydraulic resistance ($R_h$) is:

$R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$

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Equation

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In the case of a tube through which a liquid with the water density ($\rho_w$) flows due to the pressure difference ($\Delta p$) generated by a ($$) under the influence of gravity represented by the gravitational Acceleration ($g$) and calculated with the equation:

this can be used in the Hagen-Poiseuille equation, along with the definition of the flux density ($j_s$) in terms of the total flow ($J_{Vt}$), which in turn depends on the radius of a generic grain ($r_0$), the generic own porosity ($q_0$), the porosity ($f$), the viscosity ($\eta$), and the sample length ($\Delta L$):

$j_s =-\displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }\displaystyle\frac{ \Delta h }{ \Delta L }$

Conditions

Variables

$L$

Capillary length

$m$

6022

$\Delta h$

Height of liquid column

$m$

5819

$\rho_w$

Liquid density

$kg/m^3$

5407

$f$

Porosity

$-$

5805

$R$

Porosity Radio

$m$

6021

$J_{Vt}$

Sample Flow

$m^3/s$

6044

$S_t$

Section soil

$m^2$

6041

$\eta_w$

Water Viscosity

$Pa s$

6020

Relation

Development

In the case of capillaries through which a liquid with the liquid density ($\rho_w$) flows due to the pressure difference ($\Delta p$) generated by ($$) under the influence of gravity represented by the gravitational Acceleration ($g$) and calculated with the equation:

$\Delta p = \rho_w g \Delta h$

this can be applied in the Hagen-Poiseuille equation, in terms of the total flow ($J_{Vt}$), which in turn depends on the radius of a generic grain ($r_0$), the generic own porosity ($q_0$), the porosity ($f$), the viscosity ($\eta$), the section or Area ($S$), and the sample length ($\Delta L$) as described in the equation:

$J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p$

Together with the definition of the flux density ($j_s$):

$j_s = \displaystyle\frac{J_{Vt}}{S}$

We have:

$j_s=\displaystyle\frac{J_{Vt}}{S}=\displaystyle\frac{ r_0 ^2 }{8 q_g }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }\displaystyle\frac{ \Delta h }{ L }$

resulting in:

$j_s =-\displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta }\displaystyle\frac{ \Delta h }{ \Delta L }$

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Equation

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To establish a relationship between hydraulic conductivity and mass factors, we introduce the capillary scaling factor ($\gamma$) with the radius of a generic grain ($r_0$), the sand grain radius ($r_a$), the generic own porosity ($q_0$), and the sand's own porosity ($q_a$) as

$\gamma = \displaystyle\frac{ r_0 ^2}{ r_a ^2}\displaystyle\frac{ q_a }{ q_0 }$

Conditions

Variables

Relation

.

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Equation

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From the definition of the hydraulic conductivity ($K_s$) and the capillary scaling factor ($\gamma$), we can express the hydraulic conductivity ($K_s$) in terms of the sand grain radius ($r_a$), the sand's own porosity ($q_a$), the porosity ($f$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the viscosity ($\eta$) as:

$K_s = \displaystyle\frac{ r_a ^2}{8 q_a }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ \rho_w g }{ \eta } \gamma$

Conditions

Variables

Relation

.

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Equation

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The calculation of the capillary scaling factor ($\gamma$) is derived from the porosity ($f$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the viscosity ($\eta$), excluding the sand grain radius ($r_a$) and the sand's own porosity ($q_a$), through the following equation:

$\gamma=\displaystyle\frac{8K_sq_a}{r_a^2}\displaystyle\frac{(1-f)^2}{f^3}\displaystyle\frac{\eta}{\rho_wg}$

If we want to relate the capillary scaling factor ($\gamma$) to the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), and the mass fraction of clay in the sample ($g_c$), we observe that while the former varies by 6 orders of magnitude, the latter vary by only 2 orders of magnitude. Therefore, it makes sense to establish a relationship with the natural logarithm of $\gamma$. Thus, we perform a regression using the equation:

$\ln \gamma = s_a g_a + s_i g_i + s_c g_c$

Conditions

Variables

Relation

with the capillary section factor in sand ($s_a$), the capillary section factor in silt ($s_i$), and the clay capillary section factor ($s_c$).

The mean data for each range are as follows:

The regression results in a linear relationship with an R-squared value of 0.9975 and the following coefficients, along with the values to assess the quality of the coefficients:

Given the p-test values, we can assume that all coefficients are highly relevant.

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