Indicators

Storyboard

The presence of pores is one of the key aspects of soil behavior. On one hand, it enables the movement of water and/or moisture within the soil, but on the other hand, it affects the soil's mechanical properties.

Therefore, it is crucial to have indicators that describe the presence of porosity and the proportion of water they contain, as these indicators play a vital role in characterizing the hydraulic, thermodynamic, and mechanical properties of the soil.

>Model

ID:(365, 0)



Mechanisms

Iframe

>Top



Code
Concept

Mechanisms

ID:(15200, 0)



Model of solid volume, water and gas

Concept

>Top


In the soil model, the total volume ($V_t$) of the sample consists of three main components:

• the solid volume ($V_s$): This component includes the volume of all the grains present in the sample.

• the water Volume ($V_w$): Represents the volume of water contained in both the micropores and macropores of the soil.

• the gas Volume ($V_g$): Comprises the volume of gas or air contained in the sample.

The following diagram summarizes this description:

None

ID:(1642, 0)



Representation of effective depth

Image

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The effective Depth ($D_e$) refers to the depth that the water contained in a volume of soil would reach if all the solid volume were "removed," as illustrated in the following image:

None

This provides an intuitive measure of the water content in the soil.

ID:(1641, 0)



Mass model of solid, water and gas

Concept

>Top


In the soil model, the total Mass ($M_t$) of the sample consists of three main components:

• the total Dry Mass of Sample ($M_s$): This component includes the masses of all the grains present in the sample.

• the mass of water in the soil ($M_w$): Represents the mass of water contained in both the micropores and macropores of the soil.

• the mass of gas in the soil ($M_g$): Comprises the mass of gas or air contained in the sample (which can be comparatively considered as nearly zero, i.e., $M_g\sim 0$).

ID:(2084, 0)



Internal surface

Concept

>Top


One of the distinguishing properties of particulate matter, such as soil, is its internal surface area. By internal surface area, we mean the sum of all the surfaces of each of the grains. This surface area is one of the key factors for studying moisture behavior and the presence of nutrients in the soil.

When we multiply the surface area of each grain by its quantity, we obtain the total surface area. To determine the surface area of each grain, it is essential to consider its shape. It's important to remember that both sand and silt are modeled as spheres, while clay is represented as a straight parallelepiped.

ID:(1540, 0)



Model

Top

>Top



Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$w_c$
w_c
Height of a clay plate
m
$l_c$
l_c
Length and width of a clay plate
m
$\rho_p$
rho_p
Particle density
kg/m^3
$\pi$
pi
Pi
rad
$r_a$
r_a
Sand grain radius
m
$a_i$
a_i
Silt Grain Side
m
$s_c$
s_c
Surface of a grain of clay
m^2
$s_a$
s_a
Surface of a grain of sand
m^2
$\rho_w$
rho_w
Water density
kg/m^3

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$f_g$
f_g
Air Porosity
-
$s_i$
s_i
Capillary section factor in silt
m
$S_c$
S_c
Clay grain surface
m^2
$z$
z
Depth
m
$\rho_b$
rho_b
Dry bulk density
kg/m^3
$D_e$
D_e
Effective Depth
m
$V_g$
V_g
Gas Volume
m^3
$\gamma_M$
gamma_M
Inner Surface by Mass
m^2/kg
$\gamma_V$
gamma_V
Inner Surface by Volume
1/m
$M_g$
M_g
Mass of gas in the soil
kg
$M_w$
M_w
Mass of water in the soil
kg
$\Phi$
Phi
Mass porosity
-
$N_c$
N_c
Number of clay grains in the sample
-
$N_a$
N_a
Number of sand grains in the sample
-
$N_i$
N_i
Number of silt grains in the sample
-
$V_p$
V_p
Pore volume
m^3
$\theta_w$
theta_w
Relationship gravimetric water solido
-
$\theta_V$
theta_V
Relationship volumetric water land
-
$\theta_r$
theta_r
Relationship volumetric water solido
-
$\theta_s$
theta_s
Relative Saturation
-
$S$
S
Section Flow
m^2
$S_i$
S_i
Silt grain surface
m^2
$S_t$
S_t
Soil Inner Surface
m^2
$V_s$
V_s
Solid volume
m^3
$V_s$
V_s
Solid volume of a component
m^3
$S_a$
S_a
Surface of sand grains
m^2
$M_t$
M_t
Total Mass
kg
$V_t$
V_t
Total volume
m^3
$e$
e
Void ratio
-
$V_w$
V_w
Water Volume
m^3

Calculations


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

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ D_e = \theta_V z $

D_e = theta_V * z


$ e =\displaystyle\frac{ V_g + V_w }{ V_s }$

e =( V_g + V_w )/ V_s


$ f_g =\displaystyle\frac{ V_g }{ V_t }$

f_g = V_g / V_t


$ \gamma_M =\displaystyle\frac{ S_t }{ M_s }$

g_M = S_t / M_s


$ M_t = M_s + M_w $

M_t = M_s + M_w


$ \Phi = 1 - \displaystyle\frac{ \rho_b }{ \rho_p }$

Phi = 1 - rho_b / rho_p


$ \rho_b =\displaystyle\frac{ M_s }{ V_t }$

rho_b = M_s / V_t


$ \rho_w =\displaystyle\frac{ M_w }{ V_w }$

rho_w = M_w / V_w


$ S = \displaystyle\frac{ \theta_V }{ \Phi }$

S = theta_V / Phi


$ S_a = N_a s_a $

S_a = N_a * s_a


$ s_c = 2 l_c ^2 + 4 w_c l_c $

s_c = 2* l_c ^2 + 4* w_c * l_c


$ S_c = N_c s_c $

S_c = N_c * s_c


$ s_i = 6 a_i ^2 $

s_i = 6* a_i ^2


$ S_i = N_i s_i $

S_i = N_i * s_i


$ S_t = S_a + S_l + S_c $

S_t = S_a + S_l + S_c


$ \theta_r =\displaystyle\frac{ V_w }{ V_s }$

theta_r = V_w / V_s


$ \theta_s =\displaystyle\frac{ V_w }{ V_g + V_w }$

theta_s = V_w /( V_g + V_w )


$ \theta_V =\displaystyle\frac{ V_w }{ V_t }$

theta_V = V_w / V_t


$ \theta_w =\displaystyle\frac{ M_w }{ M_s }$

theta_w = M_w / M_s


$ \gamma_V =\displaystyle\frac{ S_t }{ V_t }$

V_g = S_t / V_t


$ V_p = V_w + V_g $

V_p = V_w + V_g


$ V_t = V_s + V_w + V_g $

V_t = V_s + V_w + V_g


$ s_a = 4 \pi r_a ^2 $

s_k = 4* pi * r_a ^2

ID:(15219, 0)



Total volume with water

Equation

>Top, >Model


The total volume ($V_t$) is obtained by adding the solid part of the grains, which corresponds to the solid volume ($V_s$), to the water included in the water Volume ($V_w$), and the air or, in general, the gas contained in the gas Volume ($V_g$):

$ V_t = V_s + V_w + V_g $

$V_g$
Gas Volume
$m^3$
5997
$V_s$
Solid volume
$m^3$
5995
$V_t$
Total volume
$m^3$
4946
$V_w$
Water Volume
$m^3$
5996

ID:(15089, 0)



Pore volume

Equation

>Top, >Model


The pore volume ($V_p$) is not necessarily empty; it can contain water, in particular, so we introduce the variable the water Volume ($V_w$). On the other hand, the remaining volume is considered as the gas Volume ($V_g$).

In this way, the pore volume ($V_p$) is calculated as the sum of both types of volumes:

$ V_p = V_w + V_g $

$V_g$
Gas Volume
$m^3$
5997
$V_p$
Pore volume
$m^3$
5806
$V_w$
Water Volume
$m^3$
5996

ID:(4723, 0)



Soil water volumetric ratio

Equation

>Top, >Model


An indicator that indicates the proportion of water within the total volume of the sample is the relationship volumetric water land ($\theta_V$). It is calculated by estimating the ratio between the water Volume ($V_w$) and the total volume ($V_t$):

$ \theta_V =\displaystyle\frac{ V_w }{ V_t }$

$\theta_V$
Relationship volumetric water land
$-$
5810
$V_t$
Total volume
$m^3$
4946
$V_w$
Water Volume
$m^3$
5996

ID:(4721, 0)



Solid water volumetric ratio

Equation

>Top, >Model


An indicator that signifies the proportion of water within the solid volume of the sample is the relationship volumetric water solido ($\theta_r$). It is calculated by estimating the ratio between the water Volume ($V_w$) and the solid volume ($V_s$):

$ \theta_r =\displaystyle\frac{ V_w }{ V_s }$

$\theta_r$
Relationship volumetric water solido
$-$
5811
$V_s$
Solid volume of a component
$m^3$
6038
$V_w$
Water Volume
$m^3$
5996

ID:(4722, 0)



Void ratio

Equation

>Top, >Model


The relationship between the volume of water and the solid volume compares the amount of water to the amount of solids in the soil. However, since the volume of water can vary, it is interesting to compare the pore volume ($V_p$), or alternatively the sum of the gas Volume ($V_g$) and the water Volume ($V_w$), with the solid volume ($V_s$) to define the void ratio ($e$) as follows:

$ e =\displaystyle\frac{ V_g + V_w }{ V_s }$

$V_g$
Gas Volume
$m^3$
5997
$V_s$
Solid volume of a component
$m^3$
6038
$e$
Void ratio
$-$
5813
$V_w$
Water Volume
$m^3$
5996

ID:(4728, 0)



Air porosity

Equation

>Top, >Model


The porosity ($f$) is defined as the relationship between the pore volume ($V_p$) and the total volume ($V_t$). Similarly, the air Porosity ($f_g$) is defined based on the volume not occupied by water, which is the relationship between the gas Volume ($V_g$) and the total volume ($V_t$):

$ f_g =\displaystyle\frac{ V_g }{ V_t }$

$f_a$
Air Porosity
$-$
5808
$V_g$
Gas Volume
$m^3$
5997
$V_t$
Total volume
$m^3$
4946

ID:(4724, 0)



Relative saturation

Equation

>Top, >Model


The relative Saturation ($\theta_s$) is calculated as the proportion of porosity occupied by water, defined by the water Volume ($V_w$), divided by the sum of the water Volume ($V_w$) and the gas Volume ($V_g$), expressed as:

$ \theta_s =\displaystyle\frac{ V_w }{ V_g + V_w }$

$V_g$
Gas Volume
$m^3$
5997
$\theta_s$
Relative Saturation
$-$
5812
$V_w$
Water Volume
$m^3$
5996

ID:(4727, 0)



Effective Depth

Equation

>Top, >Model


The relationship volumetric water land ($\theta_V$) allows us to estimate the effective Depth ($D_e$) that the water would reach if the soil were removed to a depth of the depth ($z$), which is calculated using the following equation:

$ D_e = \theta_V z $

$z$
Depth
$m$
4945
$D_e$
Effective Depth
$m$
4966
$\theta_V$
Relationship volumetric water land
$-$
5810

If you have a volume of soil with width and length $L$ and the depth ($z$), its volume is represented by the equation:

$V_t = L^2z$



With the effective Depth ($D_e$) representing an important variable, the volume of water can be calculated as follows:

$V_w = L^2D_e$



Furthermore, with the equation

$ \theta_V =\displaystyle\frac{ V_w }{ V_t }$



we can relate these variables as follows:

$\theta_V = \displaystyle\frac{V_w}{V_t} = \displaystyle\frac{D_e}{z}$



Therefore, the variable the effective Depth ($D_e$) can be calculated using the following expression:

$ D_e = \theta_V z $

ID:(3231, 0)



Total Mass

Equation

>Top, >Model


The total Mass ($M_t$) is calculated by adding the total Dry Mass of Sample ($M_s$) and the mass of water in the soil ($M_w$) together, as follows:

$ M_t = M_s + M_w $

$M_s$
Mass of gas in the soil
$kg$
5994
$M_w$
Mass of water in the soil
$kg$
5999
$M_t$
Total Mass
$kg$
5807

ID:(4247, 0)



Apparent density

Equation

>Top, >Model


In general, density is defined as the ratio of mass to volume of a material. In the case of soil, which contains porosity, when using the total volume ($V_t$), it includes the solid volume ($V_s$), the water Volume ($V_w$), and the pore volume ($V_p$). Typically, apparent density is calculated for dry material, i.e., without water ($M_w \sim 0$), so that the total Mass ($M_t$) is equal to the total Dry Mass of Sample ($M_s$):

$M_t\sim M_s$



It's important to note that this is an approximation, as when soil is dried, a small amount of water always remains, making it very difficult to accurately measure the solid mass without water.

Therefore, we define the dry bulk density ($\rho_b$) as the ratio of the mass of water in the soil ($M_w$) to the total volume ($V_t$):

$ \rho_b =\displaystyle\frac{ M_s }{ V_t }$

$\rho_b$
Dry bulk density
$kg/m^3$
5804
$M_t$
Total Mass
$kg$
5807
$V_t$
Total volume
$m^3$
4946

ID:(4719, 0)



Water density

Equation

>Top, >Model


When working with water, it's also crucial to consider the variable the water density ($\rho_w$), which is calculated using the mass of water in the soil ($M_w$) and the water Volume ($V_w$) with the following equation:

$ \rho_w =\displaystyle\frac{ M_w }{ V_w }$

$M_w$
Mass of water in the soil
$kg$
5999
$\rho_w$
Water density
$kg/m^3$
6000
$V_w$
Water Volume
$m^3$
5996

ID:(4730, 0)



Gravimetric water-solid ratio

Equation

>Top, >Model


If we wish to indicate the extent to which the soil contains water, we can introduce an indicator called the relationship gravimetric water solido ($\theta_w$), which is calculated as the ratio of the mass of water in the soil ($M_w$) to the total Dry Mass of Sample ($M_s$), using the following equation:

$ \theta_w =\displaystyle\frac{ M_w }{ M_s }$

$M_s$
Mass of gas in the soil
$kg$
5994
$M_w$
Mass of water in the soil
$kg$
5999
$\theta_w$
Relationship gravimetric water solido
$-$
5809

ID:(4720, 0)



Surface of a grain of sand

Equation

>Top, >Model


Since we model a grain of sand as a sphere, its the surface of a grain of sand ($s_a$) can be calculated based on the sand grain radius ($r_a$) as follows:

$ s_a = 4 \pi r_a ^2 $

$\pi$
Pi
3.1415927
$rad$
5057
$r_a$
Sand grain radius
$m$
10096
$s_a$
Surface of a grain of sand
$m^2$
5800

ID:(3167, 0)



Surface of a silt grain

Equation

>Top, >Model


Since we model a grain of silt as a cube, its the surface of a silt grain ($s_i$) can be calculated based on the silt Grain Side ($a_i$) as follows:

$ s_i = 6 a_i ^2 $

$s_i$
Capillary section factor in silt
$m$
6563
$a_i$
Silt Grain Side
$m$
10097

ID:(3169, 0)



Surface of a grain of clay

Equation

>Top, >Model


Since we model a clay grain as a rectangular parallelepiped, its the surface of a grain of clay ($s_c$) can be calculated based on the the length and width of a clay plate ($l_c$) and the height of a clay plate ($w_c$) of the clay grain as follows:

$ s_c = 2 l_c ^2 + 4 w_c l_c $

$w_c$
Height of a clay plate
$m$
5989
$l_c$
Length and width of a clay plate
$m$
5991
$s_c$
Surface of a grain of clay
$m^2$
5993

ID:(4361, 0)



Surface of sand grains

Equation

>Top, >Model


The surface of sand grains ($S_a$) can be calculated from the number of sand grains in the sample ($N_a$) and the surface of a grain of sand ($s_a$) as follows:

$ S_a = N_a s_a $

$N_a$
Number of sand grains in the sample
$-$
4941
$s_a$
Surface of a grain of sand
$m^2$
5800
$S_a$
Surface of sand grains
$m^2$
6036

ID:(929, 0)



Surface of silt grains

Equation

>Top, >Model


The silt grain surface ($S_i$) can be calculated from the number of silt grains in the sample ($N_i$) and the surface of a silt grain ($s_i$) as follows:

$ S_i = N_i s_i $

$s_i$
Capillary section factor in silt
$m$
6563
$N_i$
Number of silt grains in the sample
$-$
10100
$S_i$
Silt grain surface
$m^2$
6037

ID:(33, 0)



Surface of clay grains

Equation

>Top, >Model


The clay grain surface ($S_c$), which can be calculated from the number of clay grains in the sample ($N_c$) and the surface of a silt grain ($s_i$) as follows:

$ S_c = N_c s_c $

$S_c$
Clay grain surface
$m^2$
5801
$N_c$
Number of clay grains in the sample
$-$
10101
$s_c$
Surface of a grain of clay
$m^2$
5993

ID:(35, 0)



Inner Surface of the Soil

Equation

>Top, >Model


Since grains only have smaller sections in contact, we can assume, in a first approximation, that their entire surface is available for absorbing water and supporting life. Therefore, we introduce the concept of the "interior soil surface" and describe it as the sum of all the grain surfaces. In this way, if the soil Inner Surface ($S_t$) is obtained as the sum of the surface of sand grains ($S_a$), the silt grain surface ($S_i$), and :

$ S_t = S_a + S_l + S_c $

$S_c$
Clay grain surface
$m^2$
5801
$S_l$
Silt grain surface
$m^2$
6037
$S_t$
Soil Inner Surface
$m^2$
4939
$S_a$
Surface of sand grains
$m^2$
6036

ID:(3166, 0)



Internal surface area per mass

Equation

>Top, >Model


The issue with the soil Inner Surface ($S_t$) is that it depends on the sample size and, therefore, does not provide an indicator of the soil's surface capacity.

An alternative is to normalize the the soil Inner Surface ($S_t$) value with the total Mass ($M_t$), resulting in the indicator the inner Surface by Mass ($\gamma_M$):

$ \gamma_M =\displaystyle\frac{ S_t }{ M_s }$

$\gamma_M$
Inner Surface by Mass
$m^2/kg$
5803
$S_t$
Soil Inner Surface
$m^2$
4939
$M_t$
Total Mass
$kg$
5807

ID:(4718, 0)



Internal surface area by volume

Equation

>Top, >Model


The issue with the soil Inner Surface ($S_t$) is that it depends on the sample size and, therefore, does not provide an indicator of the soil's surface capacity.

An alternative is to normalize the value of the soil Inner Surface ($S_t$) using the total volume ($V_t$), resulting in the indicator the inner Surface by Volume ($\gamma_V$):

$ \gamma_V =\displaystyle\frac{ S_t }{ V_t }$

$\gamma_V$
Inner Surface by Volume
$1/m$
5802
$S_t$
Soil Inner Surface
$m^2$
4939
$V_t$
Total volume
$m^3$
4946

ID:(4717, 0)



Mass porosity

Equation

>Top, >Model


The mass porosity ($\Phi$) is initially defined in the same way as the porosity ($f$), however, it is estimated based on the dry bulk density ($\rho_b$) and the particle density ($\rho_p$) through:

$ \Phi = 1 - \displaystyle\frac{ \rho_b }{ \rho_p }$

$\rho_b$
Dry bulk density
$kg/m^3$
5804
$\Phi$
Mass porosity
$-$
10169
$\rho_p$
Particle density
$kg/m^3$
10168

The definition of the porosity ($f$) is carried out with the solid volume ($V_s$) and the total volume ($V_t$), which can be modified with the total Dry Mass of Sample ($M_s$) and the definition:

$ \rho_b =\displaystyle\frac{ M_s }{ V_t }$



resulting in:

$\Phi=1-\displaystyle\frac{V_s}{V_t}=1-\displaystyle\frac{V_s}{M_s}\displaystyle\frac{M_s}{V_t}=\displaystyle\frac{V_s}{M_s}\rho_b$



Although the relationship between the total Dry Mass of Sample ($M_s$) and the solid volume ($V_s$) corresponds to the solid Density ($\rho_s$), this density can be estimated using the particle density ($\rho_p$), leading to

$ \Phi = 1 - \displaystyle\frac{ \rho_b }{ \rho_p }$

ID:(15128, 0)



Mass saturation degree

Equation

>Top, >Model


The relative mass saturation ($\theta_S$) is initially defined in the same way as the relative Saturation ($\theta_s$), using volumes. However, instead of using the porosity ($f$), you can use the mass porosity ($\Phi$) instead, resulting in a mass-based degree of saturation:

$ S = \displaystyle\frac{ \theta_V }{ \Phi }$

$\Phi$
Mass porosity
$-$
10169
$\theta_V$
Relationship volumetric water land
$-$
5810
$S$
Section Flow
$m^2$
6011

The relative Saturation ($\theta_s$) is calculated using the water Volume ($V_w$) and the gas Volume ($V_g$) through

$ \theta_s =\displaystyle\frac{ V_w }{ V_g + V_w }$



As with the porosity ($f$) and the total volume ($V_t$),

$V_w + V_g = f V_t$



and since the relationship volumetric water land ($\theta_V$) is

$ \theta_V =\displaystyle\frac{ V_w }{ V_t }$



then

$\theta_s=\displaystyle\frac{V_w}{V_w+V_g}=\displaystyle\frac{V_w}{fV_t}=\displaystyle\frac{\theta_V}{f}$



If the porosity ($f$) is estimated using volume and replaced with the one estimated with mass the mass porosity ($\Phi$), we get

$ S = \displaystyle\frac{ \theta_V }{ \Phi }$

ID:(15129, 0)