Porosity of a specific soil

Storyboard

The microporosity of the soil depends on its composition, so it is important to be able to model it based on the proportion of different components. To do this, the volumetric factor of the various textures is first studied, and then porosity is estimated, taking into account that there is a basic component provided by clay. Additionally, the presence of sand and silt is considered, but it's important to note that clay can penetrate the spaces between the grains, which reduces the total porosity.

>Model

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Mechanisms

Concept

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Code
Concept

Mechanisms

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Porosity in different soils

Concept

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In the case of soils in general, we can study the soil texture triangle. If we describe the typical range of porosity observed for each type of soil, we can see what was discussed earlier. In the corner where sand prevails, we have a porosity that can reach up to 25%, which is optimal for a model of spheres:

Texture triangle that includes the porosity range obtained from [1] and [2].



Type $g_a$ [%] $g_i$ [%] $g_c$ [%] f [%]
Clay 0-45 0-40 55-100 40-50 [1]
Loam 23-52 28-50 8-27 40-50 [1]
Sand 85-100 0-15 0-10 25-35 [1]
Silt 0-20 80-100 0-13 35-45 [2]
Silty Clay 0-20 40-60 40-60 40-50 [1]
Sandy Clay 45-65 0-20 35-55 35-45 [1]
Clay Loam 20-45 15-53 28-40 40-50 [1]
Silty Clay Loam 0-20 40-73 28-40 40-50 [1]
Sandy Clay Loam 45-80 0-33 20-35 35-45 [1]
Silt Loam 0-50 50-88 0-28 35-45 [2]
Sandy Loam 43-85 0-50 0-20 30-40 [2]
Loamy Sand 70-90 0-30 0-15 25-35 [2]

Now, if we look at the silt corner, we can see that a porosity of 35% can be achieved, which corresponds to the level of space that cannot be filled by cubes. This means that the material is not capable of arranging itself in a way that takes advantage of the cubic structure. This is likely a consequence of attractive forces at the micron scale that result in disordered stacking.

In the last case, we can see the limit of clay, where porosity reaches a value around 40%, which again must be a consequence of the interaction between the plates that can organize groups of them but not the entire system.

In summary, we observe that in the lower left corner, where the soil is primarily composed of sand, the porosity can reach 25%. These 25% represent precisely the porosity achieved in the best-case scenario for a model of spheres.

In other words, there is inherent porosity specific to soil types, and in those with a significant presence of clay, clay dominates. The effect of sand and silt only prevails in extreme cases where the material has very little clay.

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

[2] Principles of Geotechnical Engineering, Braja M. Das, (2010), Cengage Learning

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Distribution of sand, silt and clay

Concept

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If we assume that the densities of the different components are similar:

$\rho_s\sim\rho_a\sim\rho_i\sim\rho_c$



the volume factors the volume fraction of sand in the sample ($f_a$), the volume fraction of silt in the sample ($f_i$), the volume fraction of clay in the sample ($f_c$) in terms of 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 ($g_c$) can be expressed as follows:

$f_a = (1-f)\displaystyle\frac{\rho_s}{\rho_a}g_a \sim (1-f)g_a$



$f_i = (1-f)\displaystyle\frac{\rho_s}{\rho_i}g_i \sim (1-f)g_i$



$f_c = (1-f)\displaystyle\frac{\rho_s}{\rho_c}g_c \sim (1-f)g_c$



This allows us to estimate the range of volumetric factors for different types of soils, including when the porosity ($f$) and the volume fraction of macropores in the sample ($f_m$) are null:

Type $f_a$ [%] $f_i$ [%] $f_c$ [%] $f$ [%]
Clay 0-25 0-22 30-55 40-50
Loam 13-29 15-28 4-15 40-50
Sand 60-70 0-11 0-7 25-35
Silt 0-11 44-55 0-7 35-45
Silty clay 0-9 18-27 18-27 40-50
Sandy clay 27-39 0-12 21-33 35-45
Clay loam 11-25 8-29 15-22 40-50
Silty clay loam 0-11 22-40 15-22 40-50
Sandy clay loam 27-48 0-20 12-21 35-45
Silt loam 0-30 30-53 0-17 35-45
Sandy loam 28-55 0-33 0-13 30-40
Loamy sand 49-63 0-21 0-11 25-35

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Regression for own porosity

Description

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Given the information we have for the temperature in degrees Celsius in state 2 ($t_2$), the own volume factor of sand ($p_a$), the silt own volume factor ($p_i$), the clay own volume factor ($p_c$), 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 ($g_c$), which satisfies the equation:

$ p_p = p_a g_a + p_i g_i + p_c g_c $



and we know the average values for different soil textures with their respective the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), the mass fraction of clay in the sample ($g_c$), and the microporosity own volume factor ($p_p$) as:

$ p_p = \displaystyle\frac{ f }{1- f }$



Type $g_a$ [-] $g_i$ [-] $g_c$ [-] $p_p$ [-]
Clay 0.225 0.200 0.775 0.818
Loam 0.375 0.390 0.175 0.818
Sand 0.925 0.075 0.050 0.429
Silt 0.100 0.900 0.065 0.818
Silty Clay 0.100 0.500 0.500 1.222
Sandy Clay 0.550 0.100 0.450 0.667
Clay Loam 0.325 0.340 0.340 0.818
Silty Clay Loam 0.100 0.565 0.340 0.818
Sandy Clay Loam 0.625 0.165 0.275 0.667
Silt Loam 0.250 0.690 0.140 0.667
Sandy Loam 0.640 0.250 0.100 0.538
Clayey Sand 0.800 0.150 0.075 0.429



We can perform a regression to determine the values of the temperature in degrees Celsius in state 2 ($t_2$), the own volume factor of sand ($p_a$), and the silt own volume factor ($p_i$). The result is a fit with an R-squared of 0.974 and the parameters are as follows:

Type $p$ [%] $q$ [%] p-test
Sand (a) 33.9 25.3 0.007029
Silt (i) 87.6 46.7 0.000041
Clay (c) 96.8 49.2 0.000158



In general, the level of compaction for sand with a the sand's own porosity ($q_a$) of approximately 25% corresponds to maximum compaction. However, with a the silt own porosity ($q_i$) of around 47%, it is higher than the optimum, as are the 49% for the clay own porosity ($q_c$). In any case, the factors are a good estimation given the high R-squared and low p-test values for each factor, which are significantly lower than the traditional threshold of 0.05. Attempts to consider other powers in the regression show that the linear approximation is the only one yielding coefficients below 0.05, suggesting that the samples must have distributions that do not exhibit significant mixing effects and are simply sums of components, such as aggregates.

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Model

Concept

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Variables

Symbol
Text
Variables
Units
$q_c$
q_c
Clay own porosity
-
$p_c$
p_c
Clay own volume factor
-
$V_m$
V_m
Macropore volume
m^3
$g_c$
g_c
Mass fraction of clay in the sample
-
$g_a$
g_a
Mass fraction of sand in the sample
-
$g_i$
g_i
Mass fraction of silt in the sample
-
$p_p$
p_p
Microporosity own volume factor
-
$p_a$
p_a
Own volume factor of sand
-
$f$
f
Porosity
$q_a$
q_a
Sand's own porosity
-
$q_i$
q_i
Silt own porosity
-
$p_i$
p_i
Silt own volume factor
-
$V_c$
V_c
Solid volume of clay
m^3
$V_a$
V_a
Solid volume of sand
m^3
$V_i$
V_i
Solid volume of silt
m^3
$V_t$
V_t
Total volume
m^3
$f_c$
f_c
Volume fraction of clay in the sample
-
$f_m$
f_m
Volume fraction of macropores in the sample
-
$f_a$
f_a
Volume fraction of sand in the sample
-
$f_i$
f_i
Volume fraction of silt in the sample
-

Parameters

Symbol
Text
Variables
Units
$\rho_c$
rho_c
Density of a grain of clay
kg/m^3
$\rho_a$
rho_a
Density of a grain of sand
kg/m^3
$\rho_i$
rho_i
Density of a grain of silt
kg/m^3
$\rho_s$
rho_s
Solid Density
kg/m^3


Selected parameter

Symbol
Variables
Value
Units
MKS Value
MKS Units

Calculations

Symbol
Equation
Solved
Translated

Equation

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Equation

$\displaystyle\frac{ f }{1- f }= \displaystyle\frac{ q_a }{1- q_a } g_a +\displaystyle\frac{ q_i }{1- q_i } g_i + \displaystyle\frac{ q_c }{1- q_c } g_c$

f /(1- f ) = q_a * g_a /(1- q_a ) + q_i * g_i /(1- q_i ) + q_c * g_c /(1- q_c )


$ f = f_m + \displaystyle\frac{ q_a }{1- q_a } f_a + \displaystyle\frac{ q_i }{1- q_i } f_i + \displaystyle\frac{ q_c }{1- q_c } f_c $

f = f_m + q_a * f_a /(1- q_a ) + q_i * f_i /(1- q_i ) + q_c * f_c /(1- q_c )


$\displaystyle\frac{1}{1- q_a } f_a + \displaystyle\frac{1}{1- q_i } f_i + \displaystyle\frac{1}{1- q_c } f_c + f_m = 1$

f_a /(1- q_a ) + f_i /(1- q_i ) + f_c /(1- q_c ) + f_m = 1


$ f_a = (1- f )\displaystyle\frac{ \rho_s }{ \rho_a } g_a $

f_a = rho_s *(1 - f )* g_a /( rho_a )


$ f_a = \displaystyle\frac{ V_a }{ V_t }$

f_a = V_a / V_t


$ f_c = (1- f )\displaystyle\frac{ \rho_s }{ \rho_c } g_c $

f_c = (1 - f )* rho_s * g_c / rho_c


$ f_c = \displaystyle\frac{ V_c }{ V_t }$

f_c = V_c / V_t


$ f_i = (1- f )\displaystyle\frac{ \rho_s }{ \rho_i } g_i $

f_i = (1 + f )* rho_s * g_i / rho_i


$ f_i = \displaystyle\frac{ V_i }{ V_t }$

f_i = V_i / V_t


$ f_m = \displaystyle\frac{ V_m }{ V_t }$

f_m = V_m / V_t


$ p_a = \displaystyle\frac{ q_a }{1- q_a }$

p_a = q_a /(1- q_a )


$ p_c = \displaystyle\frac{ q_c }{1- q_c }$

p_c = q_c /(1- q_c )


$ p_i = \displaystyle\frac{ q_i }{1- q_i }$

p_i = q_i /(1- q_i )


$ p_p = \displaystyle\frac{ f }{1- f }$

p_p = f /(1- f )


$ p_p = p_a g_a + p_i g_i + p_c g_c $

p_p = p_a * g_a + p_i * g_i + p_c * g_c

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Sand volume factor

Equation

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Similar to how proportions between the masses of each component and the total mass are defined, we can establish an analogous system using volumes. With this in mind, we can define the volume fraction of sand in the sample ($f_a$) in relation to the total volume ($V_t$). This will allow us to calculate the quantity of the solid volume of sand ($V_a$) in the context of the total volume ($V_t$).

$ f_a = \displaystyle\frac{ V_a }{ V_t }$

$V_a$
Solid volume of sand
$m^3$
$V_t$
Total volume
$m^3$
$f_a$
Volume fraction of sand in the sample
$-$

ID:(10369, 0)



Silt volume factor

Equation

>Top, >Model


Similar to how proportions between the masses of each component and the total mass are defined, we can establish an analogous system using volumes. With this in mind, we can define the volume fraction of silt in the sample ($f_i$) in relation to the total volume ($V_t$). This will allow us to calculate the quantity of the solid volume of silt ($V_i$) in the context of the total volume ($V_t$).

$ f_i = \displaystyle\frac{ V_i }{ V_t }$

$V_i$
Solid volume of silt
$m^3$
$V_t$
Total volume
$m^3$
$f_i$
Volume fraction of silt in the sample
$-$

ID:(10367, 0)



Clay volume factor

Equation

>Top, >Model


Similar to the way proportions between the masses of each component and the total mass are defined, we can establish an analogous system using volumes. With this in mind, we can define the volume fraction of clay in the sample ($f_c$) in relation to the total volume ($V_t$). This will allow us to calculate the quantity of the total volume ($V_t$) in the context of the total volume.

$ f_c = \displaystyle\frac{ V_c }{ V_t }$

$V_c$
Solid volume of clay
$m^3$
$V_t$
Total volume
$m^3$

ID:(10368, 0)



Macropore volume factor

Equation

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Similar to how proportions between the masses of each component and the total mass are defined, we can establish an analogous system using volumes. With this in mind, we can define the volume fraction of macropores in the sample ($f_m$) in relation to the total volume ($V_t$). This will allow us to calculate the quantity of the macropore volume ($V_m$) in the context of the total volume ($V_t$).

$ f_m = \displaystyle\frac{ V_m }{ V_t }$

$V_m$
Macropore volume
$m^3$
$V_t$
Total volume
$m^3$
$f_m$
Volume fraction of macropores in the sample
$-$

ID:(15084, 0)



Soil condition

Equation

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The condition for clayey soils based on the solid volume of sand ($V_a$), the solid volume of silt ($V_i$), the solid volume of clay ($V_c$), the macropore volume ($V_m$), the own volume ($V_z$), and the clay own porosity ($q_c$) is expressed as:

$ V_z = \displaystyle\frac{1}{1- q_a } V_a + \displaystyle\frac{1}{1- q_i } V_i + \displaystyle\frac{1}{1- q_c } V_c $



When we use the equation for the total volume ($V_t$) in terms of the own volume ($V_z$) and divide it by the total volume ($V_t$), we can rewrite it in terms of the volume fraction of sand in the sample ($f_a$), the volume fraction of silt in the sample ($f_i$), the volume fraction of clay in the sample ($f_c$), and the volume fraction of macropores in the sample ($f_m$) as follows:

$\displaystyle\frac{1}{1- q_a } f_a + \displaystyle\frac{1}{1- q_i } f_i + \displaystyle\frac{1}{1- q_c } f_c + f_m = 1$

$q_c$
Clay own porosity
$-$
$q_a$
Sand's own porosity
$-$
$q_i$
Silt own porosity
$-$
$f_c$
Volume fraction of clay in the sample
$-$
$f_m$
Volume fraction of macropores in the sample
$-$
$f_a$
Volume fraction of sand in the sample
$-$
$f_i$
Volume fraction of silt in the sample
$-$

With the equation of the total volume ($V_t$) in terms of the own volume ($V_z$) and the macropore volume ($V_m$):

$ V_t = V_m + V_z $



replacing the own volume ($V_z$) in terms of the solid volume of sand ($V_a$), the solid volume of silt ($V_i$), the solid volume of clay ($V_c$), the macropore volume ($V_m$), and the clay own porosity ($q_c$) with:

$ V_z = V_a + V_i + \displaystyle\frac{1}{1- q_c } V_c$



we obtain:

$V_t = V_a + V_i + \displaystyle\frac{1}{1-q_c}V_c+V_m$



If we divide this equation by the total volume ($V_t$) and use the definitions of the volume fraction of sand in the sample ($f_a$)

$ f_a = \displaystyle\frac{ V_a }{ V_t }$



for the volume fraction of silt in the sample ($f_i$)

$ f_i = \displaystyle\frac{ V_i }{ V_t }$



for the volume fraction of clay in the sample ($f_c$)

$ f_c = \displaystyle\frac{ V_c }{ V_t }$



and for the volume fraction of macropores in the sample ($f_m$)

$ f_m = \displaystyle\frac{ V_m }{ V_t }$



we obtain the following relationship:

$\displaystyle\frac{1}{1- q_a } f_a + \displaystyle\frac{1}{1- q_i } f_i + \displaystyle\frac{1}{1- q_c } f_c + f_m = 1$

ID:(15086, 0)



Soil porosity

Equation

>Top, >Model


The pore volume ($V_p$) in a clayey material, which is a function of the volume of the macropore volume ($V_m$), the sand's own porosity ($q_a$), the silt own porosity ($q_i$), the clay own porosity ($q_c$), the solid volume of sand ($V_a$), the solid volume of silt ($V_i$) and the solid volume of clay ($V_c$):

$ V_p = V_m + \displaystyle\frac{ q_a }{1- q_a } V_a + \displaystyle\frac{ q_i }{1- q_i } V_i + \displaystyle\frac{ q_c }{1- q_c } V_c $



can be rewritten by dividing the equation by the total volume ($V_t$) and expressing the equation in terms of the porosity ($f$), the volume fraction of macropores in the sample ($f_m$), the volume fraction of sand in the sample ($f_a$), the volume fraction of silt in the sample ($f_i$) and the volume fraction of clay in the sample ($f_c$), resulting in:

$ f = f_m + \displaystyle\frac{ q_a }{1- q_a } f_a + \displaystyle\frac{ q_i }{1- q_i } f_i + \displaystyle\frac{ q_c }{1- q_c } f_c $

$q_c$
Clay own porosity
$-$
$f$
Porosity
$-$
$q_a$
Sand's own porosity
$-$
$q_i$
Silt own porosity
$-$
$f_c$
Volume fraction of clay in the sample
$-$
$f_a$
Volume fraction of sand in the sample
$-$
$f_i$
Volume fraction of silt in the sample
$-$

The calculation of the pore volume ($V_p$) can be performed using the volumes of the macropore volume ($V_m$), the solid volume of clay ($V_c$), and the clay own porosity ($q_c$) with the following equation:

$ V_p = V_m + \displaystyle\frac{ q_a }{1- q_a } V_a + \displaystyle\frac{ q_i }{1- q_i } V_i + \displaystyle\frac{ q_c }{1- q_c } V_c $



By dividing this equation by the total volume ($V_t$), we can use the porosity ($f$)

$ f =\displaystyle\frac{ V_p }{ V_t }$



along with the volume fraction of macropores in the sample ($f_m$)

$ f_m = \displaystyle\frac{ V_m }{ V_t }$



and the volume fraction of clay in the sample ($f_c$)

$ f_c = \displaystyle\frac{ V_c }{ V_t }$



which simplifies to

$ f = f_m + \displaystyle\frac{ q_a }{1- q_a } f_a + \displaystyle\frac{ q_i }{1- q_i } f_i + \displaystyle\frac{ q_c }{1- q_c } f_c $

.

ID:(2074, 0)



Calculation of sand volume factor

Equation

>Top, >Model


How the volume fraction of sand in the sample ($f_a$) was defined in terms of the solid volume of sand ($V_a$) and the total volume ($V_t$):

$ f_a = \displaystyle\frac{ V_a }{ V_t }$



Therefore, with the density of a grain of sand ($\rho_a$), the solid Density ($\rho_s$), the porosity ($f$) and the mass fraction of sand in the sample ($g_a$), you can calculate the factor using:

$ f_a = (1- f )\displaystyle\frac{ \rho_s }{ \rho_a } g_a $

$\rho_a$
Density of a grain of sand
2.63e+3
$kg/m^3$
$g_a$
Mass fraction of sand in the sample
$-$
$f$
Porosity
$-$
$\rho_s$
Solid Density
$kg/m^3$
$f_a$
Volume fraction of sand in the sample
$-$

To calculate the volume fraction of sand in the sample ($f_a$), you can employ the definition with the solid volume of sand ($V_a$) and the total volume ($V_t$) as follows:

$ f_a = \displaystyle\frac{ V_a }{ V_t }$



the solid volume of sand ($V_a$) can be expressed with the density of a grain of sand ($\rho_a$) and the dry mass of sand in the sample ($M_a$) using the equation:

$ V_a =\displaystyle\frac{ M_a }{ \rho_a }$



For the total volume ($V_t$), you can work with the solid volume ($V_s$) and the pore volume ($V_p$) using the equation:

$ V_t = V_s + V_p $



utilizing the expression for the porosity ($f$):

$ f =\displaystyle\frac{ V_p }{ V_t }$



With both equations, you obtain the expression:

$V_t = \displaystyle\frac{1}{1-f} V_s$



Using the definition of the solid Density ($\rho_s$) with the total Dry Mass of Sample ($M_s$) and the solid volume ($V_s$):

$ \rho_s = \displaystyle\frac{ M_s }{ V_s }$



you can express the total volume ($V_t$) as:

$V_t = \displaystyle\frac{M_s}{(1-f)\rho_s}$



This way, you obtain the expression for the volume fraction of sand in the sample ($f_a$) as:

$f_a= \displaystyle\frac{V_a}{V_t}= \displaystyle\frac{M_a}{M_s} \displaystyle\frac{(1-f)\rho_s}{\rho_a}$



which, with the equation for the mass fraction of sand in the sample ($g_a$):

$ g_a =\displaystyle\frac{ M_a }{ M_s }$



reduces to:

$ f_a = (1- f )\displaystyle\frac{ \rho_s }{ \rho_a } g_a $

ID:(15093, 0)



Silt volume factor calculation

Equation

>Top, >Model


How the volume fraction of sand in the sample ($f_a$) was defined in terms of the solid volume of silt ($V_i$) and the total volume ($V_t$):

$ f_i = \displaystyle\frac{ V_i }{ V_t }$



Therefore, with the density of a grain of silt ($\rho_i$), the solid Density ($\rho_s$), the porosity ($f$) and the mass fraction of silt in the sample ($g_i$) you can calculate the factor using:

$ f_i = (1- f )\displaystyle\frac{ \rho_s }{ \rho_i } g_i $

$\rho_i$
Density of a grain of silt
2.65e+3
$kg/m^3$
$g_i$
Mass fraction of silt in the sample
$-$
$f$
Porosity
$-$
$\rho_s$
Solid Density
$kg/m^3$
$f_i$
Volume fraction of silt in the sample
$-$

ID:(15094, 0)



Clay Volume Factor Calculation

Equation

>Top, >Model


How the volume fraction of sand in the sample ($f_a$) was defined in terms of the solid volume of clay ($V_c$) and the total volume ($V_t$):

$ f_c = \displaystyle\frac{ V_c }{ V_t }$



Therefore, with the density of a grain of clay ($\rho_c$), the solid Density ($\rho_s$), the porosity ($f$) and the mass fraction of clay in the sample ($g_c$), you can calculate the factor using:

$ f_c = (1- f )\displaystyle\frac{ \rho_s }{ \rho_c } g_c $

$\rho_c$
Density of a grain of clay
2.8e+3
$kg/m^3$
$g_c$
Mass fraction of clay in the sample
$-$
$f$
Porosity
$-$
$\rho_s$
Solid Density
$kg/m^3$
$f_c$
Volume fraction of clay in the sample
$-$

ID:(15095, 0)



Porosity equation calculation

Equation

>Top, >Model


The porosity ($f$) is a function of the volume fraction of macropores in the sample ($f_m$), the volume fraction of sand in the sample ($f_a$), the volume fraction of silt in the sample ($f_i$), the volume fraction of clay in the sample ($f_c$), the sand's own porosity ($q_a$), the silt own porosity ($q_i$), and the clay own porosity ($q_c$):

$ f = f_m + \displaystyle\frac{ q_a }{1- q_a } f_a + \displaystyle\frac{ q_i }{1- q_i } f_i + \displaystyle\frac{ q_c }{1- q_c } f_c $



Using the relationships between volumetric factors and mass factors 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 ($g_c$), assuming the absence of macropores and equal densities for the three components, we obtain:

$\displaystyle\frac{ f }{1- f }= \displaystyle\frac{ q_a }{1- q_a } g_a +\displaystyle\frac{ q_i }{1- q_i } g_i + \displaystyle\frac{ q_c }{1- q_c } g_c$

$q_c$
Clay own porosity
$-$
$g_c$
Mass fraction of clay in the sample
$-$
$g_a$
Mass fraction of sand in the sample
$-$
$g_i$
Mass fraction of silt in the sample
$-$
$f$
Porosity
$-$
$q_a$
Sand's own porosity
$-$
$q_i$
Silt own porosity
$-$

The porosity ($f$) is a function of the number of silt grains in the sample ($N_i$), the volume fraction of sand in the sample ($f_a$), the volume fraction of silt in the sample ($f_i$), the volume fraction of clay in the sample ($f_c$), the sand's own porosity ($q_a$), the silt own porosity ($q_i$), and the clay own porosity ($q_c$):

$ f = f_m + \displaystyle\frac{ q_a }{1- q_a } f_a + \displaystyle\frac{ q_i }{1- q_i } f_i + \displaystyle\frac{ q_c }{1- q_c } f_c $



Given that with the solid Density ($\rho_s$), the density of a grain of sand ($\rho_a$), and the mass fraction of sand in the sample ($g_a$) we have:

$ f_c = (1- f )\displaystyle\frac{ \rho_s }{ \rho_c } g_c $



and with the density of a grain of silt ($\rho_i$) and the mass fraction of silt in the sample ($g_i$) we have:

$ f_i = (1- f )\displaystyle\frac{ \rho_s }{ \rho_i } g_i $



Furthermore, with the density of a grain of clay ($\rho_c$) and the mass fraction of clay in the sample ($g_c$) we have:

$ f_c = (1- f )\displaystyle\frac{ \rho_s }{ \rho_c } g_c $



it is possible, in the case where the densities are equal:

$\rho_s\sim\rho_a\sim\rho_i\sim\rho_c$



and macropores do not exist:

$f_m\sim 0$



to obtain the following relationship:

$\displaystyle\frac{ f }{1- f }= \displaystyle\frac{ q_a }{1- q_a } g_a +\displaystyle\frac{ q_i }{1- q_i } g_i + \displaystyle\frac{ q_c }{1- q_c } g_c$

ID:(10370, 0)



Own volume factor of sand

Equation

>Top, >Model


Using the sand's own porosity ($q_a$), you can define the own volume factor of sand ($p_a$) as follows:

$ p_a = \displaystyle\frac{ q_a }{1- q_a }$

$p_a$
Own volume factor of sand
$-$
$q_a$
Sand's own porosity
$-$

ID:(15087, 0)



Own volume factor of silt

Equation

>Top, >Model


With the silt own porosity ($q_i$), you can define the silt own volume factor ($p_i$) as follows:

$ p_i = \displaystyle\frac{ q_i }{1- q_i }$

$q_i$
Silt own porosity
$-$
$p_i$
Silt own volume factor
$-$

ID:(15088, 0)



Own volume factor of clay

Equation

>Top, >Model


With the clay own porosity ($q_c$), you can define the clay own volume factor ($p_c$) as follows:

$ p_c = \displaystyle\frac{ q_c }{1- q_c }$

$q_c$
Clay own porosity
$-$
$p_c$
Clay own volume factor
$-$

ID:(15098, 0)



Porosity own volume factor

Equation

>Top, >Model


With the porosity ($f$), you can define the microporosity own volume factor ($p_p$) as follows:

$ p_p = \displaystyle\frac{ f }{1- f }$

$p_p$
Microporosity own volume factor
$-$
$f$
Porosity
$-$

ID:(2076, 0)



Equation for own porosity

Equation

>Top, >Model


Using the porosity ($f$), the sand's own porosity ($q_a$), the silt own porosity ($q_i$), the clay own porosity ($q_c$), 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 ($g_c$),

$\displaystyle\frac{ f }{1- f }= \displaystyle\frac{ q_a }{1- q_a } g_a +\displaystyle\frac{ q_i }{1- q_i } g_i + \displaystyle\frac{ q_c }{1- q_c } g_c$



can be rewritten with the definitions of the microporosity own volume factor ($p_p$), the own volume factor of sand ($p_a$), the silt own volume factor ($p_i$), and the clay own volume factor ($p_c$) as

$ p_p = p_a g_a + p_i g_i + p_c g_c $

$p_c$
Clay own volume factor
$-$
$g_c$
Mass fraction of clay in the sample
$-$
$g_a$
Mass fraction of sand in the sample
$-$
$g_i$
Mass fraction of silt in the sample
$-$
$p_p$
Microporosity own volume factor
$-$
$p_a$
Own volume factor of sand
$-$
$p_i$
Silt own volume factor
$-$

With the variables the porosity ($f$), the sand's own porosity ($q_a$), the silt own porosity ($q_i$), the clay own porosity ($q_c$), 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 have the following relationship:

$\displaystyle\frac{ f }{1- f }= \displaystyle\frac{ q_a }{1- q_a } g_a +\displaystyle\frac{ q_i }{1- q_i } g_i + \displaystyle\frac{ q_c }{1- q_c } g_c$



If we consider the relationship for the microporosity own volume factor ($p_p$) as:

$ p_p = \displaystyle\frac{ f }{1- f }$



The relationship for the own volume factor of sand ($p_a$) as:

$ p_a = \displaystyle\frac{ q_a }{1- q_a }$



The relationship for the silt own volume factor ($p_i$) as:

$ p_i = \displaystyle\frac{ q_i }{1- q_i }$



And the relationship for the clay own volume factor ($p_c$) as:

$ p_c = \displaystyle\frac{ q_c }{1- q_c }$



Then the overall result is:

$ p_p = p_a g_a + p_i g_i + p_c g_c $

ID:(1542, 0)